AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 9 Classification of Elements- The Periodic Table Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 9th Lesson Structure of Atom

10th Class Chemistry 9th Lesson Structure of Atom Textbook Questions and Answers

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Question 1.
Newlands proposed the law of octaves. Mendeleeff suggested eight groups for elements in his table. How do you explain these observations in terms of modem periodic classification? (AS1)
(OR)
Correlate various tables proposed on classification of elements.
Answer:

• According to Newlands, every eighth element starting from a given element jsembles in its properties to that of the starting element, when elements are ranged in ascending order of their atomic weights.
• According to Newlands, the properties of fluorine and chlorine are similar and sodium and potassium are similar. Same aspect is given by modern periodic table.
• Mendeleeff divided it into horizontal rows and vertical columns. He called them peribds and groups respectively. Modem periodic table also gives the same.
• According to Mendeleeff, the elements of same group have similar properties. Modern periodic table also proposed the same thing.
• Mendeleeff gave the general formula for first group elements as R,0, and general formula for second group elements as RO. We can find the same thing in modern periodic table.
• The elements of particular group possess same common valency. Same was proposed by modern periodic table.

Question 2.
What are the limitations of Mendeleeff’s periodic table? How could the modern periodic table overcome the limitations of Mendeleeff’s table? (AS1)
(OR)
How can the limitations of Mendeleeffs table be overcome with the help of modern periodic table?
Answer:
Limitations of Mendeleeffs periodic table :
1) Anomalous pair of elements :
Certain elements of highest atomic weights precede those with lower atomic weights.
Eg : Tellurium (atomic weight 127.6) precedes iodine (atomic weight 126.9).

2) Dissimilar elements placed together :
a) Elements with dissimilar properties were placed in same group as sub-group A and sub-group Bt
Eg : Alkali metals like Li, Na, K, etc. of IA group have little resemblance with coinage metals like Cu, Ag, Au of IB group.

b) Cl of VII A group is a non-metal and Mn of VII B group is a metal.

Method of overcoming the limitations of Mendeleeffs periodic table by modern periodic table :
1. In modern periodic table, elements are arranged in the ascending order of their atomic numbers. So this arrangement eliminated the problem of anomalous series.
Eg : Though Tellurium (Te) has more atomic weight than Iodine (I), its atomic number is one unit less compared to Iodine.

2. The elements with similar outer shell (valence shell) electronic configurations in their atoms are in the same column called group in modern periodic table. So the elements have similar properties overcoming the Mendeleeffs second limitation.

Question 3.
Define the modern periodic law. Discuss the construction of the long form of the periodic table. (AS1)
(OR)
What are the salient features of modern periodic table?
Answer:
Modern periodic law :
‘The physical and chemical properties of elements are the periodic function of the electronic configurations of their atoms”.

Construction of the long form periodic table :

1. Based on the modern periodic law, the modern periodic table is proposed.
2. This periodic table is known as long form of the periodic table.
3. Long form periodic table is the graphical representation of Aufbau principle.
4. The modern periodic table has 18 vertical columns called groups and 7 horizontal rows known as periods.
5. There are 18 groups, represented by using Roman numerals I to VIII, with letters A and B in traditional notation, (or) 1 to 18 by Arabic numerals.
6. There are 7 periods. These periods are represented by Arabic numerals 1 to 7.
7. The number of main shells present in the atom of particular atom decides to which period it belongs.
8. First period consists 2 elements, 2^nd and 3rd periods contains 8 elements each, 4^th and 5^th periods contains 18 elements each, 6 period contains 32 elements and 7^th period is incomplete.
9. The elements are classified into s, p, d and f block elements.
10. Inert gases are placed in 18^th group.

Question 4.
Explain how the elements are classified into s, p, d and f-block elements in the periodic table and give the advantage of this kind of classification. (AS1)
(OR)
How is the periodic table classified based upon the entering of differenciating electron? Explain that classification. What is the advantage of such classification?
Answer:
1) Depending upon which sub-shell the differentiating electron enters, the elements are classified into s, p, d and f-block elements. They are

1. s – block elements,
2. p – block elements,
3. d – block elements,
4. f – block elements.

2) s – block elements :
i) If the differentiating electron enters in s-sub-shell, then the elements are called s-block elements.
ii) IA (1), IIA (2) group elements belong to this block.

3) p – block elements :
i) If the differentiating electron enters in p-sub-shell, then the elements are called p-block elements.
ii) IIIA(13), IV A (14), V A (15), VIA (16), VIIA (17) belong to p-block.

4) d – block elements :
i) If the differentiating electron enters in d-sub-shell, then the elements are called d – block elements.
ii) I B, II B, III B, IV B, V B, VI B, VII B, VIII B belong to d-block elements.
iii) They are also called transition elements.

5) f – block elements :
i) If the differentiating electrons enter in f-sub-shell, then the elements are called f-block elements.
ii) These are divided into two types
a) Lanthanides (41 elements),
b) Actinides (5f elements).
iii) These are also called as inner transition elements.

Advantage of this classification :
1) The systematic grouping of elements into groups made the study simple.
2) Each period begins with the electron entering a new shell and ends with the complete filling of s and p-sub-shells of that shell.

Question 5.
Given below is the electronic configuration of elements A, B, C, D. (AS1)

Answer:
According to electronic configuration
A = Be B = Mg C = P D = Ne
1. Which are the elements coming within the same period?
Answer:
A and D i.e. Be and Ne coming within the same period. [They have same valence shell (n = 2)]

2. Which are the ones coming within the same group?
Answer:
A and B i.e., Be and Mg coming within the same group. [They have same valence subshell with same valency (2s² and 3s²)]

3. Which are the noble gas elements?
Answer:
D, i.e. Ne is the noble gas element. [It has valency as ‘O’ and it has ‘8’ electrons in valence shell].

4. To which group and period does the element ‘C’ belong?
Answer:
Element ‘C’ i.e. ‘P’ belongs to 3^rd period and VA group.

Question 6.
Write down the characteristics of the elements having atomic number 17. (AS1)
1) Electronic configuration ___________
2) Period number _____________
3) Group number _____________
4) Element family ____________
5) No. of valence electrons ___________
6) Valency _____________
7) Metal or non-metal ____________
Answer:

1. 1s² 2s² 2p⁶ 3s² 3p⁵
2. 3
3. VII A or 17
4. Halogen family
5. 7
6. 1
7. Non-metal

Question 7.
a) State the number of valence electrons, the group number and the period number of each element given in the following table : (AS1)

Answer:

b) State whether the following elements belong to a Group (G), Period (P) or neither Group nor Period (N). (AS1)

Answer:

Question 8.
Elements in a group generally possess similar properties, but elements along a period have different properties. How do you explain this statement? (AS1)
(OR)
Elements in a group possess similar properties, but elements along a period have different properties. Explain the reason.
Answer:

• Physical and chemical properties of elements are related to their electronic configurations, particularly the outer shell configurations.
• Therefore, all the elements in a group should have similar chemical properties.
• Similarly, across the table from left to right in any period, elements get an increase in the atomic number by one unit between any two successive elements.
• Therefore, the electronic configuration of valence shell of any two elements in a period is not same. Due to this reason, elements along a period possess different chemical properties.

Question 9.
s – block and p – block elements except 18^th group elements are sometimes called as ‘Representative elements’ based on their abundant availability in the nature. Is it justified? Why? (AS1)
(OR)
Which elements are called representative elements? Why?
Answer:

• s, p – block elements are called representative elements because these are the elements which take part in chemical reactions because of incompletely filled outermost shell.
• These elements undergo chemical reactions to acquire the nearest noble gas configuration by losing or gaining or sharing of electrons.
• So they are called representative elements.

Question 10.
Complete the following table using periodic table. (AS1)

Answer:

Question 11.
Complete the following table using the periodic table. (AS1)

Answer:

Question 12.
The electronic configuration of the elements X, Y, and Z are given below.
a) X = 2
b) Y = 2, 6
c) Z = 2, 8, 2
i) Which element belongs to second period?
Answer:
Y belongs to second period.

ii) Which element belongs to second group?
Answer:
Z belongs to second group,

iii) Which element belongs to 18^th group?
Answer:
X belongs to 18^th group.

Question 13.
Identify the element that has the larger atomic radius in each pair of the following and mark it with a symbol (✓). (AS1)
(i) Mg or Ca
(ii) Li or Cs
(iii) N or P
(iv) B or Al
Answer:

Question 14.
Identify the element that has the lower ionization energy in each pair of the, following and mark it with a symbol (✓). (AS1)
(i) Mg or Na (ii) Li or O (iii) Br or F (iv) K or Br
Answer:

Question 15.
In period 2, element X is to the right of element Y. Then, find which ofitheydements have : (AS1)
i) Low nuclear charge
Answer:
Y has low nuclear charge.

ii) Low atomic size
Answer:
X has lower atomic size,

iii) High ionization energy
Answer:
X has higher ionization energy.

iv) High electronegativity
Answer:
Xhas high electronega^vity.

v) More metallic,character
Answer:
Y has more metallic character.

Question 16.
How does metallic character change when we move
i) Down a group?
ii) Across a period?
Answer:
i) Down a group :
When we move from top to bottom in a group, the metallic character increases.

ii) Across a period:
When we move left to right in a period, the metallic character decreases.

Question 17.
Why was the basis of classification of elements changed from the atomic mass to the atomic number? (AS1)
(OR)
Which atomic property is more suitable for classification of elements? Why?
Answer:

• The first attempt to classify elements was made by Dobereiner.
• Dobereiner’s attempt gave a clue that atomic masses could be correlated with properties of elements:
• Newlands’ law of octaves also followed the same basis for classification but this law is not valid for the elements that had atomic masses higher than calcium.
• Mendeleeff’s classification also based on the atomic masses of elements, but it lead to some limitations like Anomalous pair of elements and Dissimilar elements placed together.
• Moseley by analyzing the X-ray patterns of different elements was able to calculate the number of positive charges in the atoms of respective elements.
• With this analysis, Moseley realized that the atomic number is more fundamental
  characteristic of an element than its atomic weight. ,
• So, he arranged the elements in the periodic table according to the increasing order of their atomic number.
• This arrangement eliminated the problem of anomalous series and dissimilar elements placed together in Mendeleeff’s classification.

Question 18.
What is a periodic property? How do the following properties change in a group and period? Explain. (AS1)
I. a) Atomic radius
b) Ionization energy
c) Electron affinity
d) Electronegativity
II. Explain the ionization energy order in the following sets of elements: (AS1)
a) Na, Al, Cl
b) Li, Be, B
c) C, N, O
d) F, Ne, Na
e) Be, Mg, Ca
Answer:
Periodic property:
The property in which there shall be a regular gradation is called periodic property.

I. a) Atomic radius :
Period :
Atomic radius of elements decreases across a period from left to right because the nuclear charge increases due to increase in atomic number.

Group :
Atomic radius increases from top to bottom in a group due to addition of new shell.

b) Ionization energy:
Period :
When we move from left to right it does not follow a regular trend but generally increases due to increase in atomic number.

Group :
In a group from top to bottom, the ionization energy decreases due to increase in atomic size. –

c) Electron affinity:
Period :
Electron affinity values increase from left to right in a period.

Group :
Electron affinity values decrease from top to bottom in a group.

d) Electronegativity :
Period :
Electronegativity increases from left to right in a period.

Group :
Electronegativity decreases from top to bottom in a group.

II. Ionization energy order :
a) Na, Al, Cl
b) Li, Be, B
c) C, N, O
d) F, Ne, Na
e) Be, Mg, Ca
Answer:
a) In a period ionisation energy increases so the order is Na < kl < Cl.
b) Beryllium has stable configuration 1s² 2s². So it has more ionisation energy. So the order is Li < B < Be.
c) Nitrogen has half-filled p-orbitals. So it has greater ionisation energy. So the order is C < O < N.
d) Ne is inert gas right to F. Whereas Na is a metal ion in third period. So, the order is Na < F < Ne. e) In a group ionisation energy decreases. So the order is Be > Mg > Ca.

Question 19.
Name two elements that you would expect to have chemical properties similar to Mg. What is the basis for your choice? (AS2)
Answer:

• The two elements which have chemical properties similar to Magnesium are Beryllium and Calcium.
• The basis for my expectation is that they belong to same group as we know elements belonging to same group have similar properties.

Question 20.
On the basis of atomic numbers predict to which block the elements with atomic number 9, 37, 46 and 64 belong to? (AS2)
Answer:

1. The element with atomic number 9 belongs to p-block.
2. The element with atomic number 37 belongs to s-block.
3. The element with atomic number 46 belongs to d-block.
4. The element with atomic number 64 belongs to f-block.

Question 21.
Using the periodic table, predict the formula of compound formed between and element X of group 13 and another element Y of group 16. (AS2)
Answer:
The valency of 13^th group elements is 3.
The valency of 16^th group elements is 2.
The formula of compound is X₂Y₃.

Question 22.
An element X belongs to 3rd period and group 2 of the periodic table. State (AS2)
a) The no. of valence electrons
b) The valency.
c) Whether it is metal or a non-metal.
Answer:
a) The number of valence electrons are 2.
b) The valency of element is +2.
c) It is a metal.

Question 23.
An element has atomic number 19. Where would you expect this element in the periodic table and why? (AS2)
Answer:
The clement with atomic number 19 is in 4^th period and first group of the periodic table.
Reason :

1. Electronic configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s or [Ar]4s¹
2. The differentiating electron enters into 4^th shell. Hence it belongs to 4^th period.
3. The differentiating electron is in ‘s’ orbital. So it belongs to ‘s’ block.
4. The outermost orbital has only one electron. Hence it belongs to first group.

Question 24.
Aluminium does not react with water at room temperature but reacts with both dil. HCl and NaOH solutions. Verify these statements experimentally. Write your observations with chemical equations. From these observations, can we conclude that Al is a metalloid? (AS3)
Answer:

• Aluminium reacts with dil. HCl and releases hydrogen gas with formation of Aluminium chloride.
  
• Aluminium reacts with NaOH solution and releases hydrogen gas.
• 
• The above two reactions says that Aluminium is amphoteric.
• Aluminium does not react with water at room temperature.
• This concludes that the properties of Aluminium are in between a metal and non¬metal. So it behaves like a metalloid.

Question 25.
Collect the information about reactivity of VIIIA group elements (noble gases) from internet or from your school library and prepare a report on their special character when compared to other elements of periodic table. (AS4)
Answer:
Reactivity of Noble gases :

• The noble gases show extremely low chemical reactivity.
• He and Ne do not form chemical compounds.
• Xenon, krypton and argon show only minor reactivity.
• The reactivity order follows like this : Ne < He < Ar < Kr < Xe < Rn.
• Xenon can form compounds like XeF₂, XeF₄ and XeF₆, etc.

Reasons for low reactivity :

• The extremely low reactivity of noble gases is due to stable electronic configuration.
• But as we move from top to bottom the reactivity increases. So xenon can form some compounds with high electronegative elements.

Question 26.
Collect information regarding metallic character of elements of IA group and prepare report to support the idea of metallic character increases in a group as we move from top ro bottom. (AS4)
Answer:
Metallic character of IA group elements :

1. Alkali metals exhibit many of the physical properties common to metals but their densities are lower than those of other metals.
2. Alkali metals have one electron in their outer shell which is loosely bound.
3. They have largest atomic radii of the elements in their respective periods.
4. The lower ionization energies result in their metallic properties and high reactivities.
5. An alkali metal can easily lose its valence electron to form positive ion.
6. So they have greater metallic character.
7. The metallic character increases as we move from top to bottom in group due to addition of another shell, it is easy to lose electron.

Question 27.
How do you appreciate the role of electronic configuration of the atoms of elements in periodic classification? (AS6)
(OR)
How does electronic configuration help in the classification of elements in modern periodic table?
Answer:
The quantity is electronic configuration.

1. Modern periodic table is based on electronic configuration. So elements are arranged in ascending order of their atomic numbers.
2. The chemical properties of elements depend on valence electrons. The elements in same group have same number of valence electrons. So the elements belonging to same group have similar properties.
3. So the construction of modern periodic table mainly depends on electronic configuration.
4. Thus electronic configuration plays a major role in the preparation of modern periodic table. So its role is thoroughly appreciated.

Question 28.
Without knowing the electronic configurations of the atoms of elements Mendeleeff still could arrange the elements nearly close to the arrangements in the Modern periodic table. How can you appreciate this? (AS6)
Answer:

• Mendeleeff took consideration about chemical properties while arranging the elements. So the arrangement of elements is close to arrangement of elements in Modern periodic table.
• For this, he violated his periodic law.
• He left some gaps for elements, later those elements are discovered.
• So the efforts of Mendeleeff should be thoroughly appreciated.

Question 29.
Comment on the position of hydrogen in periodic table. (AS7)
Answer:

• Hydrogen is the element which has easier atomic structure than any other element.
• Electron configuration of hydrogen is Is1. It has one proton in its is nucleus and one electron in its is orbital.
• Hydrogen combines with halogens, oxygen and sulphur to form compounds having similar formulae just like alkali metals.
• Similarly, just like halogens, hydrogen also exists as diatomic molecule and combine with metals and non-metals to form covalent compounds.
• As alkali metals hydrogen can lose one electron and accept one electron as halogens.
• So in periodic table, its place may be in IA or VIIA group.
• But based on electronic configuration of hydrogen, it is placed in IA group.

Question 30.
How do the positions of elements in the periodic table help you to predict its chemical properties? Explain with an example. (As7)
Answer:
1) The physical and chemical properties of atoms of the elements depend on their electronic configuration, particularly the outer shell configurations.

2) Elements are placed in the periodic table according to the increasing order of their electronic configuration.

3) The elements in a group possess similar electronic configurations. Therefore all the elements in a group should have similar chemical properties.
Ex : Consider K

• It is the element in 4^th period 1^st group.
• Electron configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹.
• Differentiating electron enters into s-orbital. Hence it belongs to s-block.
• It is on the left side of the periodic table. Hence it is a metal.
• It is ready to lose one electron to get octet configuration. Hence its reactivity is more.
• It is Alkali metal.
• All alkali metals react with both acids and bases and releases H₂ gas.

Fill In The Blanks

1. Lithium, ……………… and potassium constitute a Dobereiner’s triad.
2. ……………… was the basis of the classifications proposed by Dobereiner, Newlands, and Mendeleeff.
3. Noble gases belong to ……………… group of periodic table.
4. The incomplete period of the modern periodic table is
5. The element at the bottom of a group would be expected to show …………….. metallic character than the element at the top

Answer:

1. Sodium
2. Atomic weight
3. VIIIA or 18 group
4. 7^th
5. higher

Multiple Choice Questions

1. Number of elements present in period – 2 of the long form of periodic table …………
A) 2
B) 8
C) 18
D) 32
Answer:
B) 8

2. Nitrogen (Z = 7) is the element of group V of the periodic table. Which of the following is the atomic number of the next element in the group?
A) 9
B) 14
C) 15
D) 17
Answer:
C) 15

3. Electronic configuration of an atom is 2, 8, 7. To which of the following elements would it be chemically similar?
A) Nitrogen (Z = 7)
B) Fluorine (Z = 9)
C) Phosphorous (Z – 15)
D) Argon (Z = 18)
Answer:
B) Fluorine (Z = 9)

4. Which of the following is the most active metal?
A) lithium
B) sodium
C) potassium
D) rubidium
Answer:
D) rubidium

10th Class Chemistry 9th Lesson Classification of Elements-The Periodic Table InText Questions and Answers

10th Class Chemistry Textbook Page No. 129

Question 1.
Can you establish the same relationship with the set of elements given in the remaining rows?
Answer:
Yes, we can establish the approximately same relationship between other elements given in the table.

Question 2.
Find average atomic weights of the first and third elements in each row and compare it with the atomic weight of the middle element. What do you observe?
Answer:
The atomic weight of middle element is arithmetic mean coverage of first and third elements.

10th Class Chemistry Textbook Page No. 135

Question 3.
What is atomic number?
Answer:
The number of positive charges (protons) in the atom of element is the atomic number of element.

10th Class Chemistry Textbook Page No. 142

Question 4.
How does the valency vary in a period on going from left to right?
Answer:
It does not follow a regular trend when we move from left to right in a period. First, it increases and then decreases and finally ‘O’ for inert gases.

Question 5.
How does the valency vary on going down a group?
Answer:
The valency is constant when we move from top to bottom in a group because the number of valence electrons are same for same group elements.

10th Class Chemistry Textbook Page No. 144

Question 6.
Do the atom of an element and its ion have same size?
Answer:
No, the positive ion has smaller size than neutral atom whereas negative ion has greater size than neutral atom.

Question 7.
Which one between Na and Na⁺ would have more size? Why?
Answer:

• The atomic number of Sodium is 1 and it has 11 protons and 11 electrons with outer electron as 3s¹ whereas Na⁺ ion has 11 protons but only 10 electrons.
• The 3s shell of Na⁺ has no electron in it.
• So the outer shell configuration is 2s²2p⁶.
• As proton number is more than electrons, the nucleus of Na⁺ ion attracts outer shell electrons with strong nuclear force.
• As a result the Na⁺ ion shrinks in size.
• Therefore, the size of Na⁺ ion is less than Na atom.

Question 8.
Which one between Cl and Cl^– would have more size? Why?
Answer:

• The electronic configuration of chlorine (Cl) atom is 1s2 2s² 2p⁶ 3s² 3p⁵ and the electronic configuration of chloride (Cl^–) ion is 1s² 2s² 2p⁶ 3s² 3p⁶.
• Both chlorine and chloride ions have 17 protons each but there are 17 electrons in chlorine atom, whereas 18 electrons in chloride ion.
• Therefore, the nuclear attraction is less in Cl^– ion when compared with chlorine atom.
• Therefore the size of the chlorine (Cl) atom is less size than chloride of Cl^– ion.

Question 9.
Which one in each of the following pairs is larger in size? Why?
a) Na, Al
b) Na, Mg⁺²
c) S^2-, Cl^–
d) Fe²⁺, Fe³⁺
e) C^4-, F^–.
Answer:
a) Na has larger size because Sodium and Aluminium are third period elements in which Na is left to Al. As we move from left to right in a period atomic size decreases.

b) Mg²⁺ has smaller size because Mg²⁺ has 10 electrons and 12 protons whereas Na has 11 electrons and 11 protons. So the distance between nucleus and outermost orbital is less in Mg²⁺ due to greater nuclear attraction.

c) S^2- has, larger size because S^2- has 18 electrons and 16 protons and Cl^– has 18 electrons and 17 protons. So nuclear attraction over outermost orbital is more in Cl^– when compared with S^2-. So S^2- has larger size.

d) In Fe²⁺ it has 26 protons and 24 electrons whereas for Fe³⁺ it has 26 protons and 23 electrons. So nuclear attraction over outermost orbital is more in Fe³⁺. So Fe³⁺ has smaller size (or) Fe²⁺ has larger size.

e) C^4- has 6 protons and 10 electrons whereas F^– has 9 protons and 10 electrons. So nuclear attraction is less in C^4-. So size of C^4- is more than F^–.

10th Class Chemistry Textbook Page No. 129

Question 10.
What relation about elements did Dobereiner want to establish?
Answer:
Dobereiner wanted to give a relationship between the properties of elements and their atomic weights.

Question 11.
The densities of calcium (Ca) and barium (Ba) are 1.55 and 3.51 gem-3 respectively. Based on Dobereiner’s law of triads can you give the approximate density of strontium (Sr)?
Answer:
Molecular weight is directly proportional to density.

So density of strontium is mean of calcium and barium according to Dobereiner.
∴ Density of strontium = \(\frac{1.55+3.51}{2}\) = 2.53.

10th Class Chemistry Textbook Page No. 130

Question 12.
Do you know why Newlands proposed the law of octaves? Explain your answer in terms of the modern structure of the atom.
Answer:

• John Newlands found that when elements were arranged in the ascending order of their atomic weights, they appeared to fall into seven groups.
• Each group contained elements with similar properties.
• If we start with hydrogen and move down, the next eighth element is fluorine, and then next eighth element is chlorine and the properties of these elements are similar.
• Similarly, if we start from Lithium their eighth element is Sodium and next eighth element is potassium. These show similar properties.

Question 13.
Do you think that Newlands’ law of octaves is correct? Justify.
Answer:
No, there are some limitations of Newlands’ model:

• There are instances of two elements fitted into the same slot. Eg : Cobalt and Nickel.
• Certain elements, totally dissimilar in their properties, were fitted into the same group.
• Law of octaves holds good only for the elements up to Calcium.
• Newlands’ periodic table was restricted to only 56 elements and did not leave any room for new elements.
• Newlands had taken consideration about active pattern sometimes without caring the similarities.

10th Class Chemistry Textbook Page No. 134

Question 14.
Why did Mendeleeff have to leave certain blank spaces in his periodic table? What is your explanation for this?
Answer:
1) Mendeleeff predicted that some elements which have similar properties with the elements in a group are missing at that time.
2) So he kept some blanks in the periodic table by writing ’eka’ to the name of the element immediately above the empty space.
3) Later these elements are discovered and they are fitted into those empty spaces.

Question 15.
What is your understanding about Ea₂O₃, EsO₂?
Answer:

• Mendeleeff predicted that after aluminium there was another element namely eka- aluminium (Ea) and after silicon, there was another element namely eka-silicon (Es).
• He also gave the formulae of those oxides as Ea203 and Es02.
• Later those elements are discovered namely gallium and germanium and Ea₂O₃ and EsO₂ as Ga₂O₃ and GeO₂.

10th Class Chemistry Textbook Page No. 135

Question 16.
All alkali metals are solids but hydrogen is a gas with diatomic molecules. Do you justify the inclusion of hydrogen in first group with alkali metals?
Answer:
No, hydrogen shows the properties of both alkali metals and halogens. Still the position of hydrogen has some questions. So it was kept just above alkali metals in first group.

10th Class Chemistry Textbook Page No. 141

Question 17.
Why are lanthanoids and actinoids placed separately at the bottom of the periodic table?
Answer:
The properties of these elements do not coincide with other elements because the valence electron enters 4f and 5f orbitals respectively. So they are placed separately at the bottom of the periodic table.

Question 18.
If lanthanoids and actinoids are inserted within the table, imagine how the table would be?
Answer:
It looks very big in size, and it is difficult to identify, as these elements have similar properties.

10th Class Chemistry Textbook Page No. 145

Question 19.
Second ionization energy of an element is higher than its first ionization energy. Why?
Answer:

• The energy required to remove an electron from unipositive ion is called second ionisation energy.
• It is difficult to remove an electron from unipositive ion when compared with neutral atom due to an increase in nuclear attraction.
• So always second ionisation energy is higher than first ionisation energy.

10th Class Chemistry Textbook Page No. 146

Question 20.
The calculated electron gain enthalpy values for alkaline earth metals and noble gases are positive. How can you explain this?
Answer:

• Generally alkaline earth metals having one, two or three valence electrons prefer to lose electrons in order to get inert gas configuration. So it is difficult to add electron to alkaline earth metals. So they have positive electron gain enthalpy values.
• Noble gases are stable. So they do not prefer to take electrons. So they have positive electron gain enthalpy.

Question 21.
The second period element, for example, ‘F’ has less electron gain enthalpy than the third period element of the same group for example ‘Cl’. Why?
Answer:

• Electron gain enthalpy values decrease in a group as we go down and increase from left to right along a period.
• But the size of Fluorine is small compared chlorine.
• So it is difficult to add electron to fluorine.
• So fluorine has less electron gain enthalpy.

10th Class Chemistry 9th Lesson Classification of Elements- The Periodic Table Activities

Activity – 1

Question 1.
Observe the following table. Establish the relationship of other elements given in the table.

Answer:

Activity – 2

Question 2.
Some main group elements of s-block and p-block have family names as given in the following table.
Observe the long form of a periodic table and complete the table with proper information.

Answer:

Activity – 3

Question 3.
Collect valencies of first 20 elements.
Answer:

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 2 Respiration.

AP State Syllabus SSC 10th Class Biology Important Questions 2nd Lesson Respiration

10th Class Biology 2nd Lesson Respiration 1 Mark Important Questions and Answers

Question 1.
What are the end products of Aerobic and Anaerobic Respirations?
Answer:
End products of aerobic respiration: Carbon dioxide, Water, Energy
End products of anaerobic respiration: Ethanol / Lactic acid, Carbon dioxide, Energy

Question 2.
In which organisms, blood does not supply the Oxygen?
Answer:
Arthropoda organisms (or) Insects (OR) Tracheal respiratory Organisms.

Question 3.
Hari said that stem also respires along with leaves. How do you support him?
Answer:
Lenticels on stem also help in gaseous exchange in some woody plants along with stomata.

Question 4.
Arrange the apparatus as above and heat the glucose. What will happen to lime water when glucose burns?
Answer:
Lime water turns milky due to carbon dioxide (CO₂).

Question 5.
What is the role of mitochondria in anaerobic respiration?
Answer:
The release of energy from glucose in the presence of oxygen occurs in mitochondria. In anaerobic respiration, as oxygen is absent, mitochondria have no role in respiration.

Question 6.
Fermented idli, dosa produce smell. Name the microorganism responsible for producing such smell.
Answer:
Yeast is responsible for producing such smell in fermented idli, dosa.

Question 7.
In what compound, the energy released during the breakdown of glucose is stored?
Answer:
“ATP” (Adenosine Triphosphate).

Question 8.
Label a and b in the given diagram.

Answer:
(a) Matrix, (b) Cristae.

Question 9.
Name chemical substance produced in human muscles during Anaerobic respiration.
Answer:
Lactic acid is produced in human muscles during Anaerobic respiration.

Question 10.
Why is Diazene Green solution added to the Glucose solution in anaerobic respiration experiment?
Answer:
Diazene Green solution is added to the Glucose solution in anaerobic respiration experiment to check the presence of oxygen in glucose solution.

Question 11.
Name the food material on which trypsin acts and name the end products.
Answer:
i) protein ii) end products – peptones.

Question 12.
“Respiration is the energy releasing process.” Write your opinion on this statement.
Answer:
The given statement is absolutely correct. We respire to use the oxygen to oxidise our food and release energy. This is similar like burning but a slower process. With the help of respiratory enzymes, energy released can be stored in the form of ATP for later use.

Question 13.
Identify the figure.

Answer:
Aerial roots in Mangrove plants.

Question 14.
Can we say that combustion and respiration are almost same actions? What evidences do you have for this?
Answer:

1. In both these processes sugar is converted to carbon dioxide and water.
2. Both these processes require oxygen.
3. Both combustion and respiration releases energy.

Question 15.
What is the role of epiglottis in respiration and swallowing food?
Answer:
The epiglottis is a flexible flap at the superior end of the pharynx in the throat. Epiglot¬tis acts as a lid over glottis and prevents food from entering into larynx. Air from pharynx enters the larynx while food enters into oesophagus.

Question 16.
What is the function of haemoglobin?
Answer:
During respiration haemoglobin carries oxygen to the cells and CO, from cells to lungs.

Question 17.
What is respiration?
Answer:
Respiration is the process by which food is broken down to release energy.

Question 18.
What does the word respiration mean in Latin?
Answer:
In Latin the word respiration means “to breathe”.

Question 19.
Who did comprehensive work on properties of gases, their exchange and respiration?
Answer:
Lavoisier and Priestly.

Question 20.
What was the gas liberated on heating powdered charcoal in a bell jar?
Answer:
It was fixed air. In those days carbon dioxide was known as fixed air.

Question 21.
What is oxygen debt?
Answer:
It is the inadequate supply of oxygen when we undertake strenuous exercise.

Question 22.
What is vitiated air?
Answer:
It is the term used then to show air from which the component needed for burning had been removed.

Question 23.
What is the total lung capacity of human being?
Answer:
The total lung capacity of human being is nearly 5800 ml.

Question 24.
Who was the renowned chemist who wrote a textbook of Human Physiology?
Answer:
John Daper was the renowned chemist who wrote a textbook of Human Physiology.

Question 25.
What happens when air passes through nasal cavities?
Answer:

1. Air is filtered in nasal cavity by mucus lining and the hairs growing from its sides, remove some of the tiny particles of dirt in the air.
2. The temperature of the air is brought close to that of the body.

Question 26.
What is the function of epiglottis?
Answer:
Epiglottis controls the movement of air and food towards their respective passages.

Question 27.
What is breathing?
Answer:

1. Breathing is the process of inhaling and exhaling.
2. The mechanism by which organisms obtain oxygen from the environment and release CO₂ is called breathing.

Question 28.
What are pleura?
Answer:
Pleura are the two membranes that protect lungs from injury.

Question 29.
What is the concentration of oxygen at a height of 13 km from the sea level?
Answer:
At a height of 13 km above sea level the concentration of oxygen is much lower about one-fifth as great as at sea level.

Question 30.
What is cellular respiration?
Answer:
Oxidation of glucose or fatty acids takes place in the cells releasing energy. Hence this process is known as cellular respiration.

Question 31.
Where does aerobic respiration occur in eukaryotic cells?
Answer:
Aerobic respiration occur in cytoplasm and mitochondria of eukaryotic cells.

Question 32.
What is Glycolysis?
Answer:
It is the first stage of respiration. In this breakdown of glucose molecule into two molecules of 3 carbon compound called pyruvic acid or pyruvate releasing energy.

Question 33.
What is the fate of pyruvate in the absence of oxygen in animals?
Answer:
In the absence of oxygen pyruvate will be converted to lactic acid and release small amount of energy in animals.

Question 34.
In which type of respiration pyruvate is converted into carbon dioxide and water?
Answer:
In aerobic respiration pyruvate is converted into carbon dioxide and water.

Question 35.
What is the main reason for feeling pain in muscles after strenuous exercise?
Answer:
Due to the anaerobic respiration in muscles large amounts of lactic acid is accumulated and this results in muscular pains.

Question 36.
What is fermentation?
Answer:
In the absence of oxygen, yeast cells convert pyruvic acid to ethanol. This process is called fermentation.

Question 37.
What is the method used to separate ethanol from the yeast glucose mixture in anaerobic respiration?
Answer:
The method used to separate ethanol from the yeast glucose mixture in anaerobic respiration is fractional distillation.

Question 38.
In which organisms does exchange of gases take place through diffusion?
Answer:
In Amoeba, hydra and planarians exchange of gases takes place through diffusion.

Question 39.
In tracheal respiratory system which carry air directly to the cells in the tissues?
Answer:
Trachioles, the fine branches of trachea carry air directly to the cells in the tissues.

Question 40.
What are the respiratory organs in fishes?
Answer:
Gills or bronchiae are the respiratory organs in fishes.

Question 41.
What is cutaneous respiration?
Answer:
If the respiration occurs through skin, it is known as cutaneous respiration, e.g : Leech, Earthworm and Frog.

Question 42.
What are the other areas on the plant body through which gaseous exchange take place?
Answer:
The areas on the plant body through which geseous exchange take place are the surface of roots, lenticels on the stem.

Question 43.
What is the full form of ATP? How is it formed?
Answer:
I) ATP stands for Adenosine triphosphate.
2) ATP is used to supply energy in the cells for the carrying all the metabolic processes.

Question 44.
What are the factors that control respiration?
Answer:
Oxygen and temperature are the two important factors that control the process of respiration.

Question 45.
What are the substances that are used for the production of energy in all living organisms?
Answer:
Glucose and fatty acids are used for the production of energy in all living organisms.

Question 46.
How many types of respiration are present? What are they?
Answer:
There are two types of respiration. They are :

1. Aerobic respiration and
2. Anaerobic respiration.

Question 47.
Where is energy stored in ATP?
Answer:
Energy is stored in the terminal phosphate bond in ATP which is having three phosphates attached to a molecule of Adenosine.

Question 48.
What are the power houses of the cell?
Answer:
Mitochondria are the power houses of the cell.

Question 49.
What is the main difference between respiration and combustion?
Answer:
In respiration several intermediates are produced and in combustion, there are no such intermediates are produced.

Question 50.
What is the equation that represents respiration?
Answer:
The equation that represents respiration is

Question 51.

.
What are the sites of cellular respiration?
Answer:
Mitochondria are the sites of cellular respiration.

Question 52.
What are cristae in mitochondria?
Answer:
The inner membrane of mitochondria is thrown into several folds called cristae.

Question 53.
What is the net gain of ATP molecules in Glycolysis?
Answer:

1. Four ATP molecules are produced when one molecule of glucose is converted to two molecules of pyruvate but two are consumed.
2. The remaining two ATP molecules are net gain in glycolysis.

Question 54.
How many ATP molecules are produced when one glucose molecule is completely oxidised?
Answer:
A net gain of 38 ATP molecules are formed from the total oxidation of one glucose molecule.

Question 55.
What are the three stages present in complete oxidation of glucose molecule?
Answer:
The three stages present in complete oxidation of glucose molecule are

1. Glycolysis
2. Kreb’s cycle and
3. Electron transport.

Question 56.
Why does oxidation of fatty acids give more energy?
Answer:
Oxidation of fatty acids give more energy due to the presence of more carbon atoms in them.

Question 57.
What are aquatic and terrestrial animals?
Answer:
Animals that live in water are called aquatic animals and that live on land are known as terrestrial animals.

Question 58.
Why is the rate of breathing in aquatic organisms much faster than terrestrial organisms?
Answer:

1. The amount of oxygen dissolved in water is low when compared to the amount of oxygen present in air.
2. Therefore the rate of breathing in aquatic animals is much faster than in terrestrial animals.

Question 59.
Which part of the roots is involved in the exchange of respiratory gases?
Answer:
The part of roots that are involved in the exchange of respiratory gases are root hairs.

Question 60.
What is the average breathing rate in an adult mem at rest?
Answer:
The average breathing rate in an adult man at rest is about 15 to 18 times per minute.

Question 61.
Why is the trachea prevented from collapsing?
Answer:
The walls of the trachea are supported by several ‘C’ shaped cartillagenous rings. They prevent the trachea from collapsing and closing.

Question 62.
Why deos the percentage of carbon dioxide increase in exhaled air?
Answer:
During oxidation of glucose carbon dioxide is produced as waste product. Hence the concentration of carbon dioxide increases in exhaled air.

Question 63.
How does breathing take place in mangrove plants?
Answer:
In mangrove plants breathing takes place through specialised structures called breath¬ing roots or pneumatophores.

Question 64.
How does respiration take place in plants where roots are present in wet places?
Answer:
The plants which have their roots in very wet places have much larger air spaces, connect the stems with the roots, making diffusion from upper parts.

Question 65.
Which form a continuous network all over the plant?
Answer:
The stomatal openings lead to a series of spaces between the cells inside the plant which form a continuous network all over the plant.

Question 66.
What are the reasons for the animals to develop different types of respiratory organs?
Answer:
Body size, availability of water, habitat in which they live and the type of circulatory system are some of the reasons for the animals to develop different types of respiratory organs.

Question 67.
Why do fishes die when taken out of water?
Answer:
Fishes do not have lungs to utilise oxygen for breathing. They have gills which can utilize only dissolved oxygen from water.

Question 68.
What would be the consequences of deficiency of haemoglobin in our bodies?
Answer:
Deficiency of haemoglobin in blood can affect the oxygen supplying capacity of blood to body cells. It can also lead to a disease called Anaemia.

Question 69.
What are the stages of respiration in man?
Answer:
Respiration in man occurs in two stages 1) Inhalation (or) Inspiration 2) Exhalation (or) Expiration.

Question 70.
Which part plays major role in respiration of man?
Answer:
Diaphragm plays a major role in respiration in man.

Question 71.
Which part plays major role in respiration of woman?
Answer:
In woman ribs play a major role in respiration.

Question 72.
How are lungs protected?
Answer:
Lungs are protected by two membranes called pleura. A fluid between these membranes protects the lungs from injury.

Question 73.
What is the composition of exhaled air?
Answer:
Exhaled air contains 16% of oxygen, 4% of carbon dioxide and 79% of nitrogen.

Question 74.
Why are red blood cells red in colour?
Answer:
Red blood cells are red in colour due to the presence of haemoglobin in their cytoplasm.

Question 75.
How is haemoglobin made up of?
Answer:
Haemoglobin is made up of a protein called globin, Iron (Hearn) and organic molecule called porphyrin.

10th Class Biology 2nd Lesson Respiration 2 Marks Important Questions and Answers

Question 1.
(a) Which gas turns lime water milky in this experiment?
Answer:
Carbondioxide (or) CO₂

(b) Which gas do you think might be present in less quantities in the air we breath out as compared to air around us?
Answer:
Oxygen (or) O₂

Question 2.
Balu said that, “Plants perform Photosynthesis during day time. They respire during night time”.
Do you agree with Bain? Why? Why not?
Answer:

1. I do not agree with Balu’s statement.
2. Photosynthesis depends on light for energy but respiration does not depend on light.
3. Hence, photosynthesis takes place during day time only whereas respiration takes place both day and night.

Question 3.
The sportsman who participated in 100 mtr race get more muscle pains. But the sportsman who participates in 5 km’s race get less muscle pains. What is the reason?
Answer:

1. Accumulation of lactic acid results in muscular pain.
2. During 100 m race a well trained athlete can hold his breath and afterwards he pants.
3. In this case, the muscles are using energy released during the anaerobic break down of glucose, lactic acid is produced.
4. The presence of lactic acid in the blood is the main cause of muscle fatigue. Whether it is 100 mtr race or 5 km race.
5. If the body is rested long enough the tiredness goes.

Question 4.
What happens if there is no epiglottis in human beings?
Answer:

1. Food may enters into the larynx.
2. Food may enters into the lungs leading to the death.
3. May not speak properly.
4. Entry of food and air may not be regulated properly.

Question 5.
Write two chemicals and two materials required to conduct the experiment “Heat and Carbon dioxide are evolved during anaerobic respiration”.
Materials required: Thermosflask, splitted corks, thermometer, wash bottle, glass tubes.
Chemicals required: Liquid paraffin, glucose solution, bicarbonate solution, Janus green B and Yeast cells.

Question 6.
Observe the below diagram.
A) To which biosystem is this picture related?
Answer:
Respiratory system.

B) Write the names of the parts of A, B.
Answer:
A – alveolus; B – blood capillary network

C) To which system are they linked with?
Answer:
Respiratory system; circulatory system.

D) Which process is happening here? What happens as a result of it?
Answer:
Gaseous exchange between alveolus of lungs and blood capillaries. Due to this the CO₂, present in blood capillaries enter alveolus and oxygen present in alveolus en¬ter blood capillaries.

Question 7.
A person reached a specific distance once on foot and once by running. In which situation his legs pain? Why?
Answer:

1. When a person runs to reach a specific distance gets pain in his legs.
2. This is due to the production of lactic acid in the muscles.
3. Due to the Anaerobic respiration glucose in muscles converts into lactic Acid.
4. Accumulation of lactic acid causes pain in leg muscles.

Question 8.
What is the advantage of the wet and warm passage of air from the nostrils to capillaries?
Answer:
When the air passes in nasal cavity and in the pharynx some changes take place.

1. The mucus layer and hair in the nasal cavity removes the dust particles in the air.
2. The temperature of the air brought to the body temperature.
3. Moistening the air.

Question 9.
In the experiment of anaerobic respiration with yeast
i) Why was liquid paraffin poured on glucose?
ii) What did you understood about anaerobic respiration?
Answer:
i) The supply of oxygen from the air can be stopped by pouring liquid paraffin on glucose.
ii) Anaerobic respiration takes place in the absence of oxygen. In this glucose molecule is incompletely oxidised. The end products of anaerobic respiration are ethyl alcohol or lactic acid and CO₂.
During anaerobic respiration small amount of energy is liberated (2ATP). Anaero¬bic respiration occurs in many anaerobic bacteria and human muscles cells. The anaero¬bic respiration can be represented as:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂+ 56 K.Cal.

Question 10.
See the below table. Write what you know from it.

Answer:

1. The inhaled air consists of 21% of oxygen whereas the exhaled air contains 16% of oxygen only. This is due to utlilisation of oxygen during cellular respiration in the body. Hence the difference occurs.
2. Inhaled air contains 0.04% of carbondioxide whereas exhale air contains 4% of carbondioxide.
   The concentration of CO₂ is increased a lot due to the release of CO₂ during cellular respiration in the body.
3. Both inhale and exhale air contains 79% of nitrogen because nitrogen has no role to play in cellular respiration.

Question 11.
What is the pathway of air from nostril to alveolus?
Answer:
Draw a flow chart of Respiratory passage of Humans.

Question 12.
What happens when a baker prepares a dough by mixing yeast in it?
Answer:

1. The yeast is commonly used for fermenting bread is saccharomyces cerevisiae.
2. Baker’s yeast has the advantage of producing uniform, quick, and reliable results because it is obtained from pure culture.
3. Water is mixed with flour, salt and the fermenting agent.
4. The mixed dough is then allowed to rise one or more times.
5. Then loaves are formed and the bread is baked in air oven.

Question 13.
How does respiration in amoeba and hydra occur through diffusion? (OR)
What are the similarities in respiration of amoeba and hydra?
Answer:

1. Amoeba and hydra are aquatic organisms.
2. Respiration in them occurs through diffusion.
3. As oxygen is used by these organisms in respiration, its concentration is reduced in cytoplasm. Hence oxygen diffuses into cytoplasm from surrounding water.
4. During respiration CO₂ is continuously produced, its concentration increases in the cytoplasm, hence it diffuses into surrounding water.

Question 14.
Write a short note on ATP. (OR) Expand ATP.
Answer:

1. From the break down of glucose the energy is released and stored up in a special compound known as ATP (Adenosine Triphosphate).
2. It is a small parcel of chemical energy. The energy currency of these cells is ATP an energy rich compound that is capable of supplying energy whenever needed within the cell.
3. Each ATP molecule gives 7200 calories of energy. This energy is stored in the form of phosphate bonds.
4. If the bond is broken, the stored energy is released.

Question 15.
How do Dolphin and Crocodile respire?
Answer:

1. The aquatic animals like dolphin and crocodile respire with the help of lungs.
2. They come out of the water for air.
3. These two animals were lived on land initially.
4. Later they lived in water and developed several adaptations to live in water.

Question 16.
Why are Mitochondria called “Power houses of cell”? (QR)
What is the energy producing organ in a cell? How does it produce energy?
Answer:

1. Cellular respiration in prokaryotic cells like that of bacteria occurs within the cytoplasm.
2. In eukaryotic cells cytoplasm and mitochondria are the sites of reaction.
3. The produced energy is stored in mitochandria in the form of ATP.
4. Hence, mitochondria are called “Power houses of cell”.

Question 17.
Write the rate of respiration in different age groups of human beings.
Answer:

1. Newborn child: 32 times per minute
2. Children of 5 years: 26 times per minute
3. Man of 25 years: 15 times per minute
4. Man of 50 years: 18 times per minute

10th Class Biology 2nd Lesson Respiration 4 Marks Important Questions and Answers

Question 1.
Write about respiration in mangroves that grow in marshy lands.
Answer:

1. Mangroves grown near the marshy places respire through aerial roots or respiratory roots.
2. The root hairs exchange the gases from their surface.
3. They obtain oxygen from the airspaces present between the soil particles.
4. The plants grown in marshy places are adapted to develop aerial roots above the soil surface which helps in gaseous exchange.

Question 2.
a) What is the aim of this experiment?
Answer:
Heat is liberated during respiration.

b) What change do you observe in thermometer readings?
Answer:
Reading increases in the thermometer.

c) In your opinion, where did this heat come from?
Answer:
The heat comes from the germinating seeds which respire and releasing heat.

d) What precaution should we take, while doing this experiment?
Answer:
The bulb of the thermometer should be dip in the germinating seeds (or) sprouts.

Question 3.
You have conducted this experiment in your classroom. Now answer the following questions.
a) What do you prove by conducting this experiment?
Answer:
To test the production of heat and carbon dioxide during anaerobic respiration.

b) Why do you heat glucose solution?
Answer:
To remove the dissolved oxygen in the glucose solution.

c) How do you confirm that glucose solution is free from oxygen after heating it?
Answer:
By adding diazine green (Janus green B) solution to glucose solution, it turns to pink.

d) What are the changes you notice in the lime water?
Answer:
Lime water turns milky white.

Question 4.

i) What change did you observe in the thermometer in the given experiment?
Answer:
Raise in the temperature

ii) Where does the heat come from?
Answer:
From the germinating seeds during respiration

iii) What result you will get, if you perform this experiment with dry seeds?
Answer:
No change of temperature in thermometre.

iv) What are the apparatus used in this experiment?
Answer:
Glass jar, germinating seeds, cork, thermometer.

Question 5.
Observe the set of apparatus and answer the following questions.
i) Which process do we know with the help of this experiment?
Answer:
Combustion.

ii) How does this process differ with respiration?
Answer:
Respiration occurs in the presence of water.
Combustion occurs in the absence of water.

iii) What are the similarities between this process and respiration?
Answer:
In both processes energy is released.

iv) Which gas turns lime – water milky?
Answer:
Carbon-di-oxide (CO₂)

Question 6.
Look at the following experiment. Answer the questions.

a) What is the aim of the experiment?
Answer:
The aim of the experiment is CO₂ is released during anaerobic respiration.

b) Which agent is used to find the presence of oxygen?
What changes do you observe when oxygen is present in Glucose solution?
Answer:
To find the presence of oxygen diazine green (Janus Green B) solution is used. The blue diazine green solution turns pink when oxygen is present in the glucose solution.

c) Why is liquid paraffin poured on glucose solution?
Answer:
By pouring liquid paraffin on glucose solution, the supply of oxygen from the air can be cut off.

d) Which gas is released during the experiment? How can you prove it?
Answer:
Carbon dioxide is released.
The released CO₂ passes into lime water it turns milky.

Question 7.
Observe the following diagram and answer the following questions.

1. What do we call the membranes that cover the lungs?
2. What is the functional unit of lungs ?
3. Which part produces the sound ?
4. What does ‘X’ denote ?

Answer:

1. Pleura
2. Alveoli
3. Larynx
4. Trachea

Question 8.
Observe the diagram and answer the following questions.

a) What does the given diagram indicate?
b) What is the part ‘X’ in the diagram?
c) What is the function of the given picture?
d) To which system the given picture belongs to?
Answer:
a) The given diagram indicates mitochondria.
b) Matrix
c) Performing cellular respiration and releasing energy in the form of ATP.
d) Respiratory system.

Question 9.
Observe the experimental setup and answer the given questions.
A) What is the aim of this experiment?
B) What are the apparatus required for this experiment?
C) What changes do you observe in thermometer during this WKm experiment?
D) What will happen, if dry seeds are taken instead of germinating seeds in this experiment?
Answer:
A) Heat is liberated during respiration.
B) Glass jar, Germinating seeds, Cork and Thermometer.
C) We can notice the raise in temperature after observing the thermometer readings.
D) There will be no change of temperature in the thermometer. We can’t prove the aim of the experiment.

Question 10.
Observe the below diagram and answer the following questions:
i) What does the above setting (diagram) indicate?
Answer:
The above setting (diagram) indicates to prove that carbon dioxide and heat are liberated during anaerobic respiration by yeast cells.

ii) Why is boiled and cooled glucose covered with paraffin?
Answer:
To prevent supply of air, boiled and cooled glucose is covered with paraffin.

iii) What is the use of adding diazine green to glucose solution? What change you notice in glucose solution?
Answer:
Diazine green is added to glucose solution to know whether oxygen is present or not in glucose solution. When the availability of oxygen is less the diazine green changes to pink colour.

iv) Why is lime water used in this experiment?
Answer:
To know whether carbon dioxide is released or not in this experiment lime water is used. Carbon dioxide changes lime water to milky white.

v) Why is bulb of thermometer dipped in the glucose water?
Answer:
To know the rise in temperature of glucose solution when heated, the bulb of thermometer is dipped in the glucose water.

Question 11.
Explain with the help of a flow chart, the path way of air in humans.
Answer:

Question 12.
Study the graph and answer the following questions :
Graph showing effects of vigorous excercise on the concentration of lactic acid in blood.
i) What was the concentration of lactic acid in blood to start with?
ii) What was the greatest concentration of lactic acid reached during the experiment?
iii) What is the concentration of lactic acid after 25 minutes of exercise?
iv) What is the relationship between lactic acid and muscle pain?
Answer:
i) 20 mg/cm³
ii) 20 minutes (Or) at “B” point,
iii) 101 mg/cm³
iv) If concentration of lactic acid increases, muscle pains also increases.

Question 13.
Observe the following :

Write the answers to the following questions:
i) How many Pyruvic acid molecules form from one Glucose?
Answer:
2 Pyruvic acid molecules.

ii) What conditions influence Pyruvic acid to participate in Aerobic and Anaerobic respiration?
Answer:
Presence of oxygen

iii) In which we get more energy in both Aerobic and Anaerobic respirations?
Answer:
Aerobic respiration

iv) The chemical that is formed in human muscles during Anaerobic respiration.
Answer:
Lactic acid

Question 14.
Why does the exchange of gases happen only in alveoli, though arteries are present in pharynx, trachea and bronchus?
Answer:

1. Alveoli are tiny air sacs in the lungs surrounded by capillaries
2. They are numerous and only single cell thickness
3. They increase the efficiency of gas exchange.
4. Due to the difference in a gradient of O₂ oxygen diffuse from alveoli to blood capillaries.

Question 15.
What are the events or steps in respiration?
Answer:
The following are the events or steps in respiration.

1. Breathing: Air moves into lungs and out of lungs.
2. Gaseous exchange in lungs: Exchange of gases between alveoli and blood.
3. Gas transport by blood: Transport of oxygen from blood capillaries of alveoli to body cells and return of carbon dioxide.
4. Gaseous exchange in cells: Exchanging oxygen from blood into the cells and carbon dioxide from cells into the blood.
5. Cellular respiration: Using oxygen in cell processes to produce carbon dioxide and water, releasing energy to be used for life processes.

Question 16.
What will happen if the respiratory tract is not moist? (OR)
Why respiratory tract should be moist?
Answer:

1. If the respiratory tract is not moist the dirt particles in the inhaled air will not be removed from air in the nasal cavities and reaches lungs and creates problems to lungs.
2. The temperature of the inhaled air is brought close to that of the body for the smooth passage in the respiratory tract. If it is dry, it is not possible.
3. If the surface dries out, gas exchange will happen at a very reduced rate since fast moving gaseous oxygen molecules do not efficiently cross the alveoli membrane.
4. The reduced gas exchange is most likely not enough to support blood oxygenation for vital functions.
5. Hence respiratory tract should be moist for smooth exchange of gases.

Question 17.
Explain the process of transportation of gases through the blood.
Answer:

1. The relative amount of gases and their combining capacity with haemoglobin and other substances in blood determine their transport via blood in the body.
2. When oxygen present in the air is within normal limits (around 21%) then almost all of it is carried in the blood by binding to haemoglobin, a protein present in the red blood cells.
3. As oxygen is diffused in the blood, it rapidly combines with the haemoglobin to form oxyhaemoglobin.
4. Not only can haemoglobin combine with oxygen, but it can easily broken into haemoglobin and oxygen.
5. Carbon dioxide is usually transported as bicarbonate, while some amount of it combines with haemoglobin and rest is dissolved in blood plasma.
   

Question 18.
Why is human life impossible at higher altitudes without a supplementary supply of oxygen? (OR)
The concentration of oxygen in air decreases as we go up from sea level. Explain briefly.
Answer:

1. If haemoglobin is exposed to air at sea level, every molecule in air combines with oxygen to form oxyhaemoglobin.
2. At a height of 13 km above sea level, the concentration of oxygen is much lower about l/5th of a sea level.
3. Under these conditions about half as many molecules of oxygen combine with haemoglobin to form oxyhaemoglobin.
4. Blood cannot carry enough oxygen to the tissues.
5. Hence human life is impossible at such a high altitude without a supplementary supply of oxygen.
6. Provision for such a supply is built into modern aircraft which have pressurized cabins that maintain an enriched air supply.

Question 19.
What are the different ways in which glucose is oxidised to provide energy in various organisms? Give one example of each.
How does oxidation of glucose occur in various organisms?
Answer:

1. Glucose is the most commonly used sugar for deriving energy in plants, animals and in microorganisms.
2. In all these organisms glucose is oxidized in two stages.
3. The first stage is known as Glycolysis. It occurs in cytoplasm.
4. During glycolysis glucose is converted to two molecules of pyruvic acid.
5. In the second stage if oxygen is available pyruvic acid is converted to C02 and water, large amount of energy is released. This is known as aerobic respiration. It occurs in most of the plant and animal cells.
6. If oxygen is inadequate or not available, pyruvic acid is converted into ethanol and carbon dioxide. This is anaerobic respiration taking place in yeast cells that is called fermentation.
7. If oxygen is not available in muscle cells, the pyruvic acid is converted into lactic acid.
   

Question 20.
Write the adaptations seen in plants living in water logged conditions.
(OR)
What are the adaptations seen in magrove plants?
Answer:

1. Most plants can aerate their roots by taking in the oxygen through lenticels or through the surface of their root hairs.
2. But plants which have their roots in very wet places, are unable to do this.
3. They are adapted to these water logged conditions by having much larger air spaces which connect the stems with the roots, making diffusion from the upper parts much more efficiently.
4. The problem of air transportion is more difficult for trees and may not survive with their roots permanently in water.
5. To overcome this problem the mangrove tree of the tropics which raise up aerial roots above the surface and takes in oxygen.

Question 21.
Describe the mechanism of branchial or gill respiration in fishes.
(OR)
Briefly explain the process of exchange of gases in fishes during respiration.
Answer:

1. Some aquatic animals like fishes have developed special organs for respiration which are known as gills or branchiae.
2. Blood is supplied to gills through capillaries which have thin walls where gases are exchanged. Gills are present in the gill pouches or branchial pouches.
   
3. Gills are provided with leaf-like folds called gill lamellae.
4. Fish keeps its mouth open and lowers the floor of the oral cavity. As a result water from outside will be drawn into the oral cavity.
5. Now the mouth is closed and the floor of the oral cavity is raised.
6. Water is pushed into the pharynx and is forced to gill pouches through internal branchia apertures.
7. When water passes through gill lamellae exchange of gases takes place, that is oxygen diffuses from water to blood and CO₂ from blood into water.
8. Then water flows through external branchia aperture.

Question 22.
Explain briefly about Pranayama- the art of breathing. (OR)
How can the capacity of lungs be improved by yoga?
Answer:

1. To improve breathing capacity the saint Patanjali developed Yogabyasa.
2. The art of breathing in Yogabyasa is called Pranayama. Prana means gas, ayama means journey.
3. In Pranayama practice air is allowed to enter three lobes of lungs in order to in¬crease the amount of oxygen to diffuse into blood.
4. More amount of oxygen available to brain and tissues the body will be more active.
5. It is very important to practise Pranayama regularly to make our life healthy and active.
6. All people irrespective of age and sex should practise Pranayama under the guidance of well trained Yoga Teacher to improve the working capacity of lungs.

Question 23.
What are the experiments carried out by Lavoisier to understand the property of gases?
Answer:

1. In his early experiments Lavoisier thought that the gas liberated on heating powdered charcoal in a bell jar kept over water in a trough was like fixed air i.e., carbon dioxide.
2. The next series of experiments deals with the combustion of phosphorous in a bell jar. From this he showed that whatever it was in the atmospheric air which combined with the phosphorous was not water vapour.
3. This was respirable air, a component of air that also helped in burning.
4. The air that we breathe out precipitated lime water while that after heating metal did not.
5. From this, he concluded that there were two processes involved in respiration.
6. Lavoisier carried out another experiment by which he showed that about one sixth of the volume of ‘vitiated air’ consists of chalky acid gas (fixed air).
7. Either eminently respirable air is changed in the lungs to chalky acid air; or an exchange takes place, the eminently respirable air being absorbed, and an almost equal volume of chalky acid air being given up to the air from the lungs.
8. Lavoisier had to admit that there were strong grounds for believing that eminently respirable air did combine with the blood to produce the red colour.

Question 24.
Explain the evolutionary changes in energy-releasing system.
(OR)
What are the different respiratory systems in animal groups?
Answer:
Exchange of gases is a common life process in all living organisms, but it is not same in all.

1. Diffusion:
   1. Single-celled organisms like amoeba or multicellular organisms like hydra and planarians obtain oxygen and expel carbon dioxide directly from the body by the process of diffusion.
   2. In multicellular animals special organs are evolved.
   3. Body size, availability of water and the type of circulatory system are some of the reasons for the animals to develop different types of respiratory organs.
2. Tracheal respiratory system : In insects tracheal respiratory system is present in which small branches of trachea called trachioles carry air directly to the cells in the tissues.
3. Bronchial respiration : In fishes gills are utilised for the exchanges of gases. Blood is supplied to gills through capillaries which have thin walls for exchange of gases. This is called bronchial respiration.
4. Cutaneous respiration: 0 Respiration through skin is called cutaneous respiration.
   Eg: i) Earth worms and leeches.
   ii) Frog, an amphibian can respire through lungs and skin.
5. Pulmonary respiration : Most of the higher animals respire with the help of lungs. This type of respiration is known as pulmonary respiration. Eg: Mammals.

Question 25.
Describe the structure of mitochondria with the help of a diagram. (OR)
Which cell organelle is called energy currency or power house of cell? What do you know about its construction?
Answer:
Mitochondria is known as energy currency or power house of cell.
Structure of mitochondria:

1. Mitochondria are sac-like structures present in the cytoplasm of the cells.
2. Mitochondria have two compartments-an inner compartment and an outer compartment. The substance in the inner compartment is called matrix.
3. The matrix is surrounded by a membrane called inner membrane of mitochondria.
4. The inner membrane is thrown into several folds called cristae. The cristae extended into the matrix.
5. The space between the folds is continuous with the outer compartment.
6. On the inner membrane, projecting into the matrix are a large number of particles called elementary particles.
7. These particles have a spherical head and a stalk. They are attached to the inner membrane by their stalk and the head portion of the particle is in the matrix.
8. The outer compartment is surrounded by another membrane – the outer membrane. The outer membrane is smooth and has no projections.
9. The inner membrane, the matrix and the elementary particles in the mitochondria have large number of enzymes and other required proteins for the respiration and energy production.

Question 26.
Draw and label mitochondria. Why should we call it cell of power ?
Answer:

Oxidation of glucose molecule occurs in the mitochondria, ot cell. This is known as cellular respiration. The energy produced during cellular respiration stored in the form of ATP molecule. Energy producing cellular respiration occurs in mitochondria hence we call it cell of power or power house of the ceil.

Question 27.
Describe how oxygen enters the blood in lungs with the help of block diagram.
(OR)
How does gaseous exchange occur in lungs?
Answer:

1. Gaseous exchange takes place within the lungs by diffusion from the alveoli to blood capillaries and vice versa. Alveoli in lungs are numerous and only one cell thick.
   
2. Alveoli are surrounded by capillaries that are also one cell thick.
3. Blood, dark red in colour flows from the heart through these capillaries and collects oxygen from the alveoli.
   At the same time, carbon dioxide passes out of the capillaries and into the alveoli.
4. When we breathe out, we get rid of carbon dioxide.
5. The bright red, oxygen rich blood is returned to the heart and pumped out to all parts of the body.

Question 28.
What is the role of diaphragm and ribs in respiration? Are both active in man and woman?
Answer:
Diaphragm:

1. Diaphragm is a muscular dome shaped tissue present at the floor of the chest cavity separating abdomen from respiratory system.
2. Diaphragm expands downwards into the abdomen thus increasing chest cavity. This allows the lungs to expand as we inhale.
3. As the diaphragm contracts upwards thus decreasing the chest cavity, it allows the air to expel from the lungs.
   Ribs:
4. The ribs protect the lungs and expand as we inhale to facilitate space for the lungs to expand. The ribs then contract expelling the air from the lungs.
5. The intercostal muscles present between the ribs help in contraction and relaxation of ribs.
6. In man, diaphragm plays a major role in the respiration, while in woman, the ribs play a major role.

Question 29.
Why are alveoli so small and uncountable in number? (OR)
How do alveoli increase the area for exchange of gases?
Answer:

1. The pouch-like air sacs at the ends of the smallest branchioles are called alveoli.
2. The walls of the alveolus are very thin and they are surrounded by very thin blood capillaries.
3. It is in the alveoli that gaseous exchange takes place.
4. There are millions of alveoli in the lungs. The presence of millions of alveoli in the lungs provides a very large area for the exchange of gases.
5. And the availability of large surface area maximises the exchanges of gases.

Question 30.
Write a brief note on respiration in plants. (OR)
Does respiration occur in plants? Explain briefly about it.
Answer:

1. In most plants exchange of gases takes place through stomata.
2. There are other areas on the plant body like surface of roots, lenticels on stem, etc. the gaseous exchange takes place.
   
3. Some plants have specialized structures like breathing roots of mangrove plants as well as the tissue in orchids.
4. Breathing roots and tissue in orchids help plants to take oxygen to produce energy and release carbon dioxide.
5. Inside the plants openings lead to a series of spaces between the cells which form a continuous network all over the plant.
6. The whole system works by diffusion.
7. As the oxygen is used up by the cells a gradient develops between the cells and the air in the spaces.
8. So oxygen passes in between the air spaces and the air outside stomata and lenticels.
9. In the same way, as more carbon dioxide is given out by the cells, a gradient occurs in the reverse direction and it passes out.

Question 31.
Write a brief note on tracheal respiration in insects.
Answer:

1. In insects blood do not contain haemoglobin, and blood is white in colour. Hence it cannot carry oxygen.
2. For respiration insects adopt a special system called tracheal system.
   
3. This system consists of a series of tubes called trachea.
4. These trachea open out through small apertures called spiracles on either side of the body.
5. All tracheal tubes of each side join and form a longitudinal tracheal trunk.
6. Trachea divide into a number of branches called tracheoles which carry air directly to the tissues.
7. As the air moves in and out of the trachea, oxygen present in the air diffuses into the cells and CO₂ diffuses into the air from the cells.

Question 32.
Write about the mechanism of respiration in human beings. (OR)
How does exchange of gases take place in human beings?
Answer:

1. Respiration in man occurs in two stages. They are inspiration and expiration.
2. During inspiration air from outside enters into the lungs by increasing the chest cavity.
   
3. Increase in the chest cavity is made by pulling the diaphragm down and pushing the ribs forward.
4. As the air pressure in the lungs is reduced, air from outside enters the lungs through external nostrils, nasal cavities, internal nares, pharynx, epiglottis, larynx, trachea, bronchi and branchioles and finally reach the alveoli where exchange of gases takes place.
5. During expiration the diaphragm and ribs come back to original positions.
   
6. This reduces the volume of chest cavity.
7. So the volume of lungs is decreased and air under pressure comes out of the lungs.

Question 33.
Study the graph given below and analyse the reasons for accumulation of lactic acid in blood after strenuous exercise.

Answer:

1. This graph shows the relation between time accumulation of lactic acid in the muscles.
2. At the beginning, the amount of lactic acid in the blood is very less.
3. Gradually it is increased by vigorous exercise.
4. Within 15 minutes it goes to maximum level which causes muscle pain.
5. Then the lactic acid is removed from muscles in an hour.
6. Muscles produce energy by anaerobic respiration.
   C₆H₁₂O₆ → lactic acid + CO₂ + energy
7. In the vigorous exercise, muscle work rapidly and produce more lactic acid.
8. That’s why lactic acid concentration is increased in muscle after strenuous exercise.

Question 34.
Observe the above graph of lactic acid accumulation in the muscles of an athlete and answer the following questions.
a) What was the concentration of lactic acid in the blood to start with?
Answer:
It is 20 mg/km³.

b) What was the greatest concentration reached during the experiment?
Answer:
101 mg/cm³.

c) If the trend between points C and D were to continue at the same rate, how long might it take for the original lactic acid level to be reached once again?
Answer:
55 minutes.

d) What does high level of lactic acid indicate about the condition of respiration?
Answer:
It indicates the accumulation of lactic acid in muscles through anaerobic respiration. The presence of lactic acid in the blood is the main cause of muscular pain and fatigue.

Question 35.
Describe the structure of human lungs with the help of a diagram.

Answer:

1. A pair of lungs is present in the chest cavity one on either side of the heart.
2. Lungs are spongy and elastic. They are enclosed by two membranes called pleura.
3. Space between the two membranes of pleura is filled with fluid. Pleura protects the lungs from injury.
4. Right lung is larger than the left lung.
5. Right lung is made of three lobes while the left lung has only two lobes.
6. Lung has several thousands of alveoli which are supplied with blood capillaries.
7. Pulmonary artery brings deoxygenated blood from heart to lungs.
8. After entering the lung, this artery divides into several arterioles and capillaries and supplies deoxygenated blood to alveoli.
9. Gas exchange occurs in the alveoli.
10. Oxygenated blood is carried from the lung to heart by the pulmonary vein.

Project work
Question 1.
Observe and analyse the questions in the table given below.

A) In which age group rate of heart beat is more?
B) In which age group rate of heart beat is less?
C) Why heart beat in Athletics is less?
D) What are reasons for more rate of heart beats differences between the newly born and children?
Answer:
A) In newly borned babies which are in 0 – 3 months of age group rate of heart beat is more i.e., 100 to 150 times.
B) In athletics the rate of heart beat is less i.e., 40 – 60 times / minute.
C) The heart of athlete pump more blood per beat due to increased cardio-vascular fitness in the structure of the heart. The muscles in the heart wall thicken and the heart pumps more blood with each beat.
D)

1. Mothers who have special medical conditions such as thyroid diseases or diabetes may give birth to new borns who are temporarily tachscardic from altered hormone and glucose levels. Tachycardia is a medical term for a very rapid heart beat.
2. Some infants are born with accessory electrical tissue in the heart causes epi¬sodes of rapid heart rate.
3. In wolf – parkinson syndrome – white syndrome there are extra cells and an ac-cessory path way, causing additional heart beats.

Question 2.
Observe the table given below and analyse the questions.

A) Why heart beat is less in animals with more weight?
B) Why heart beat is more in animals with less weight?
C) What is the relationship between weight of the body and rate of heart beat?
D) Why the weight of heart is less than body weight?
Answer:
A) The animals with more weight usually have weighted hearts. In one heart beat the large-sized hearts sends high amounts of blood to circulatory system. It takes time for the fulfilment of heart. Hence heart beat is less in animals with more body weight.
B) Usually the heart is very small in less weight animals. When the animal shrinks or contracts , its heart actually decrease the volume of blood proportionately. It can compensate for the reduced volume by increasing the rate at which it can supply blood to all body parts.
C) As the weight of the body of the animal increases the rate of heart beat per minute decreases. And also as the weight of the body decrease the rate of heart beat increases.
D) Usually the body of an organism is made by number of organs which makes the body functional. As all the body parts constitute the whole organism, the heart one of the organ is usually has less weight than body weight of an animal.

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 7 Coordination in Life Processes Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 7th Lesson Coordination in Life Processes

10th Class Biology 7th Lesson Coordination in Life Processes Textbook Questions and Answers

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Question 1.
What do you mean by hunger pangs?
Answer:

1. When glucose levels in the blood fall, hunger pangs occur in the stomach.
2. Ghrelin hormone is secreted in the stomach when it goes empty.
3. Ghrelin is secreted from certain cell walls of the stomach.
4. Hunger contractions (hunger pangs) start to occur in the stomach due to secretion of ghrelin hormone.
5. Hunger pangs continue up to 30 – 45 minutes.
6. An increase in Ghrelin levels results in sensation of hunger and motivation to consume food.

Question 2.
What are the organ systems involved in the digestion of food which we eat?
Answer:
The organ systems involved in the digestion of food are:

1. endocrine system,
2. nervous system,
3. muscular system,
4. circulatory system and
5. excretory system.

Question 3.
Rafi said smell also increases our appetite. Can you support this statement ? How?
Answer:

1. Yes. I support the statement made by Rafi that smell also increases our appetite.
2. When we smell, the air borne substances get dissolved in the watery film of nasal mucus.
3. The chemoreceptors which are otherwise called olfactory receptors present in the nose trigger signals in the form of nerve impulses to the brain where the smell is detected.
4. However interactions between the senses of taste and smell enhance our appetite.

Question 4.
Write a note on peristalsis and sphincter function in the stomach.
Answer:
A) Peristalsis functions in stomach:

1. Contraction and relaxation of the muscles present in the stomach bring in a wave-like motion called peristalsis.
2. The contraction of the stomach muscles squeeze and mix the food with the acids and juices of the stomach.
3. Mechanical mixing of food in stomach occurs by peristalsis. This allows the mass of food to further mix with the digestive enzymes.
4. Peristalsis in stomach is involuntary and under the control of the autonomous nervous system.

B) Sphincter functions in the stomach:

1. Pyloric sphincter is present at the opening of the stomach into the duodenum (small intestine).
2. The pyloric sphincter allows only small quantities of food into duodenum at a time.

Question 5.
Observe the given part of the digestive system. What is it ? What is it’s role during digestion ?
Answer:

1. The given part is the LARGE INTESTINE in the human digestive system:
   
2. If creates peristaltic waves to send waste material called stool or faeces into the rectum.
3. In the large intestine water and mineral salts are absorbed from the waste material.
4. The faecal matter containing indigested food material, bile pigments and dead bacteria is formed in the large intestine.
5. It is stored in the rectum and expelled out of the body through the Anus.

Question 6.
Give reasons.
a) If we press the tongue against the palate, we can recognize taste easily.
Answer:
Reason :

1. When the tongue is pressed against the palate, the food substance is pressed against the opening of the taste bud letting it to reach the taste cells and triggering taste signals.
2. Finally, the taste is recognized in the brain.

b) We can’t identify taste when food is very hot.
Answer:
Reason:

1. Most of the taste buds on the tongue are killed when the food is hot.
2. This prevents the person from identifying the taste.
3. The perception of taste decreases when the temperature of the food rises beyond 35°C.
4. But we don’t pay attention to it because we become worried about the burning feeling.

c) If glucose level falls in blood, we feel hungry. (OR) €EI June 2015
We feel hungry if the blood glucose level falls down. 03 June 2019
Answer:
Reason :

1. When glucose levels in the blood fall, we get hunger pangs in stomach.
2. Hungry feeling start to occur in the stomach due to hunger generating signals that the brain from the stomach due to the secretion of the hormone ‘Ghrelin’.

d) Small intestine is similar to a coiled pipe. (OR)
Why is the small intestine long and coiled?
Answer:
Reason :

1. The length of small intestine is 7 meters long.
2. It is coiled so as to fit in the human body.
3. It is coiled to increase surface area and maximum nutrient absorption when food passes through it.

e) Urination increases when we take lot of fluids.
Answer:
Reason:

1. When we take lot of fluids, the kidneys will efficiently throw the water out by forming more urine than usual.
2. When there is excess water in the body, the brain usually produces less of a hormone called vasopressin.
3. This in turn causes the kidneys to produce a lot of dilute urine, until excess water is removed.

f) The process of digestion goes on in a person whose central nervous system has been largerly affected.
Answer:
Reason :

1. The enteric nervous system embedded in the walls of the long tube of our gut or alimentary canal control gut functions often independently of the brain.
2. The mass of neural tissue of enteric nervous system filled with important neurotransmitters reveals that it does much more than merely handle digestion.

Question 7.
Write differences between the following:
a) Bolus – Chyme
b) Small intestine – Large intestine
c) Mastication – Rumination
d) Propulsion – Retropulsion
Answer:
a) Bolus – Chyme:

b) Small intestine – Large intestine:

c) Mastication – Rumination:

d) Propulsion – Retropulsion:

Question 8.
How can you say that mouth is a munching machine?
Answer:

1. Mouth plays a major role in changing the texture of food.
2. The circular muscles of the mouth enable the food to be pushed into the oral cavity and to be moved around.
3. The teeth grind, chew and shred the food.
4. The surface muscles of the jaw help in biting and chewing actions, while the deep muscles of the jaw move up, down, forward and backward during food mastication.
5. The teeth help in cutting and grinding while tongue movements evenly spread out the food and help in mixing it with saliva.
6. The muscles of the mouth enable the food to be pushed in the oral cavity and to be moved around.
7. By all this we can say that mouth is a munching machine.

Question 9.
What is mastication? Explain the role of different sets of teeth in this process.
Answer:
Mastication: Food can not be swallowed directly. So it must be grinded, chewed and shred. This process of grinding, chewing and shredding is called ’MASTICATION’.
Role of different sets of teeth in mastication:

1. In the oral cavity of human beings, 32 teeth and they are 4 different types to perform different functions.
2. They are a) Inscisors b) Canines c) Premolars d) Molars
   a) Inscisors: These are eight in number and have chisel shape. They have sharp edges and are specialized for cutting and biting the food.
   b) Canines: These are four in number and are with pointed edges- These are designed for piercing and tearing food.
   c) Premolars: These are eight in number and are also called cheek teeth. These are designed for crushing and grinding food.
   d) Molars: These are 12 in number. They are designed for crushing and grinding food.

Question 10.
During the journey of food from the mouth to stomach through oesophagus, how does muscular system coordinate in this process?
Answer:
Movement of food (bolus) from mouth to stomach through oesophagus is influenced by the coordination between muscular system and nervous system.
Muscular movement in mouth:

1. The circular muscles of the mouth enable the food to be pushed into the oral cavity and to be moved around.
2. The surface muscles of the jaw help in biting and chewing actions while the deep muscles move the jaw up and down, forward and backward during mastication.

Muscular movement in oesophagus:

1. The wall of the oesophagus is made up of two kinds of smooth muscles.
2. The inner layer consists of circular muscles and the outer layer is of longitudinal muscles.
3. Contraction of circular muscles lead to narrowing of the oesophagus behind the bolus. So the food is squeezed downwards.
4. Contraction of the longitudinal muscles in front of the bolus widens the tube, results in shortening of particular part of oesophagus.
5. In this way food moves easily and enters the stomach. ,
6. Contraction and relaxation of oesophagus muscles bring peristaltic movements.

Question 11.
Is there any reason for the intestine to be coiled with many folds ? In what way it is helpful during the process of digestion? (OR)
Small intestine is similar to a coiled pipe. Give reason.
Answer:

1. Yes, there is a reason for the intestine to be coiled with many folds.
2. The coiled and folded nature of intestine slows down the passage of food along the intestine and afford an increased surface area for absorption.
3. It also increases the surface area for the intestine to increase the absorption of nutrients through finger-like projections villi.
4. The folded and coiled intestine absorbs nutrients and water more than they breakdown.

Question 12.
What is the function of peristalsis in these parts?
a) Oesophagus b) Stomach c) Small intestine d) Large intestine
Answer:
a) Oesophagus:
Peristalsis helps in pushing the food, down the oesophagus into the stomach.

b) Stomach:

1. Peristalsis helps in storing food, breaking down food down and mixing it with juices secreted by stomach lining.
2. Peristalsis in stomach helps in partial digestion of food.

c) Small intestine:

1. Peristalsis pushes the digesting food through small intestine.
2. It helps to mix the chyme to help in the digestive process.
3. Peristalsis also helps in absorbing nutrients from the digesting food into the blood.

d) Large intestine:
Peristaltic movements help to propel feces along the large intestine through colon, to the rectum for expulsion from the body.

Question 13.
How can you justify the enteric nervous system as the second brain of the gut?
(OR)
Write about the nervous system present beneath the digestive system.
Answer:

1. The enteric nervous system, the second brain consists of sheaths of neurons embedded in the walls of the long tube of our gut, or alimentary canal.
2. The second brain measures about nine meters to end from the oesophagus to the anus.
3. The second brain contains some 100 million neurons, more than in either the spinal cord or the peripheral nervous system.
4. This multitude of neurons in the enteric nervous system enables us to “feel” the inner world of our gut and its contents.
5. Enteric nervous system contains mass of neural tissue filled with important neurotransmitters.
6. This reveals that it does much more than merely handling digestion or inflict the occasional nervous pang of hunger.
7. Enteric nervous system stimulates and coordinates the breaking down of food, absorbing nutrients and expelling of waste.
   Thus equipped with its own reflexes and senses, the second brain can control several gut functions often independently of the brain.
8. Several scientists also believe that the enteric nervous system is a way too complicated to have evolved only to make sure things move through and out of our gut smoothly.
9. Hence we can justify that the enteric nervous system as the second brain of the gut.

Question 14.
Rajesh feels hungry upon seeing food. Sheela says no more food as she is not hungry. What makes Rajesh hungry and what suppresses Sheela’s hunger?
Answer:

1. When glucose levels in the blood fall, we get hunger pangs in the stomach. Feeling hungry lies in the physiology of blood circulation.
2. The secretion of the hormone GHRELIN starts in the stomach when it goes empty.
3. Hunger contractions start to occur in the stomach due to hunger generating signals that reach the brain from the stomach due to ghrelin.
4. So Rajesh effected with secretion of ghrelin in his stomach and he felt hungry when he saw food.
5. Sheela was not affected with secretion of ghrelin in her stomach. She felt that her stomach is full.
6. When we feel the stomach is full and there is no need of food any more, another hormone LEPTIN is secreted that supresses hunger.

Question 15.
How are taste and smell related?
(OR)
What is the relationship between taste and smell?
Answer:

1. Taste and smell are closely related.
2. This close relationship is most readily seen in how we perceive the flavours of food.
3. Anyone with severe cough and cold cannot make out the difference in the taste of certain food items.
4. Actually, what is really being affected is the flavour of the food , or the combination of taste and smell.
5. That is because not only the taste but also the food odours are being defected.
6. However, interactions between the senses of taste and smell enhance our perceptions of the foods we eat.
7. Smell of the food flavour gives a similar taste to food.

Question 16.
List out the sphincter muscles of the food canal you have observed and give a brief description.
(OR)
What are the different types of sphincter muscles present in the digestive canal?
Answer:
There are different sphincter “muscles in the food canal. They are :

1. Esophageal sphincter
2. Cardiac sphincter
3. Pyloric sphincter
4. Anal sphincter and
5. Ileocecal value or sphincter.

1. Esophageal sphincter:

1. This allows entry of bolus into the oesophagus,
2. It also reduces back flow from the oesophagus to pharynx.

2. Cardiac sphincter:

1. It is present where oesophagus meets the stomach,
2. It’s location is almost directly in front of the heart,
3. It allows food from oesophagus into the stomach.

3. Pyloric sphincter:

1. It is present at the opening called pylorus, located at the end of the stomach,
2. Pyloric sphincter allows only small quantities of food into duodenum at a time.

4) Anal sphincter:

1. It is located at the anus, the end of the digestive system,
2. The release of waste is controlled partly, voluntarily by anal sphincter.

5) Ileocecal value:

1. It is situated at the junction of the small intestine (ileum) and the large intestine (colon),
2. It’s function is to limit the colonic contents into the ileum.

Question 17.
What happens if salivary ducts are closed?
(OR)
What happens if saliva is not produced by salivary glands?
Answer:
If salivary ducts are closed the following consequences may occur:

1. If salivary ducts are closed saliva will not release into the mouth, the mouth becomes dry.
2. We can’t taste the food.
3. Saliva secretes an enzyme called salivary amylase which breaks down the carbohydrates in food into dextrose and maltose sugar.
4. If saliva is not secreted in the mouth carbohydrates remains unchanged digestion will not completed.
5. Due to absence of saliva food will not get moistened, chewing and swallowing becomes difficult.
6. If salivary ducts are closed, the salivary glands become swollen and it leads to pain.

Question 18.
If size and shape of small intestine is like oesophagus what will happen?
Answer:

1. If size and shape of small intestine is like oesophagus the height of the person should be more than 22 feet as the length of the small intestine is about 22 feet. So it will not fit in the human body.
2. For the complete digestion of the food to occur, it has to stay in small intestine for 3 to 4 hours.
3. Otherwise digestion will not be completed and nutrients are not absorbed into blood . in the small intestine.
4. The tube like nature of small intestine as that of oesophagus will not provide increased surface area for complete absorption of nutrients.

Question 19.
Prepare a questionnaire to understand nervous coordination in the digestion process.
(OR)
What are the questions you are going to ask your teacher about nervous coordination in the digestion process?
(OR)
Prepare a questionnaire for an interview with a doctor to understand the nervous coordination in the digestion process.
Answer:

1. What is enteric nervous system?
2. Why the enteric nervous system is known as second brain?
3. How nervous system plays an important role in digestion of food?
4. Which nervous system controls the release of saliva from salivary glands?
5. Which cranial nerve controls the movement of muscles in the jaw?
6. The mechanism for swallowing is under the control of?

Question 20.
What experiment do you perform to understand action of saliva on flour? Explain it’s procedure and apparatus that you followed.
(Activity – 7)
(OR)
i) Explain the procedure followed in the experiment conducted to understand the action of saliva on starch.
ii) What apparatus and chemicals are used to do this experiment?
(OR)
Write the experiment you have performed in your school laboratory to know the action of Saliva on flour. How did you test the pH of Saliva?
(OR)
Explain the experiment conducted by your teacher in your class to show the action of saliva on the starch.
Answer:
Aim: To understand the action of saliva on flour.
Apparatus:

1. Two test tubes
2. Water
3. Ata flour
4. Water glass and
5. Iodine solution.

Procedure:

1. A test tube is taken and half of the test tube is filled with water.
2. A pinch of flour is added.
3. To mix flour with water, test tube as shaken.
4. Few drops of flour mixture is taken in a watch glass and test for the presence of starch by putting a drop of diluted tincture Iodine in it.
5. A blue black colour confirms the presence of starch.
6. Divide the mixture into two equal halves by transferring it to another test tube.
7. Both the test tubes have the same amount of solution.
8. Add a teaspoon of saliva to one of the test tube and mark it.
9. Do not add anything in the other test tube.
10. After 45 minutes add a drop of dilute Tincture Iodine solution to test tubes containing the solutions.

Observation:

1. Saliva containing flour mixture becomes slurry. It doesn’t show any change.
2. But in the second test tube with only flour mixture turns blue colour.

Conclusion:

1. Saliva in the test tube converts flour into sugar. So no change is observe here.
2. In the second test tube containing only flour mixture turns blue black.

(OR)
i) Answer:

1. Add a spoon of starch to the water taken in a test tube to make starch solution. Starch solution is divided into two equal parts in two test tubes.
2. Add a little saliva to one of the above test tube and the other without adding saliva and keep them a side for some time.
3. Add a drop of dilute iodine to both the test tubes. The test tube added with saliva doesn’t turns to blue black because the starch is broken down into smaller units called sugar by the saliva.
4. The other part which is not added with saliva turns into blue black, as it has starch without any change.
   So this experiment proves saliva converts complex carbohydrates into simple sugars.

ii) Answer:
Test tubes, water glass, flour, iodine, saliva.

Question 21.
Suggest a simple experiment to prove the role of palate in recognizing taste.
(OR)
How do you prove that palate has a role in recognizing taste?
(OR)
“Taste is connected to the tongue and palate”. Write an activity to prove it and write your observations.
Answer:
Experiment to prove the role of palate in recognizing taste
Aim: To prove the role of palate in recognizing taste.
Apparatus: Sugar crystals, Stop watch.
Procedure:

1. Place some sugar crystals on the tongue.
2. Keep mouth opened and see that tongue does not touch the palate.
   
3. Record the time from the moment we placed the crystals on the tongue till we get the taste by using stop watch.
4. Now repeat the test by placing the sugar crystals on the tongue pressing it against the palate.
5. Record the time from placing sugar crystals to getting the taste.

Observation: From the above activity we know that taste can be identified easily when the tongue is pressed against the palate.
Result: We can easily identify the taste of the substance we can identify the taste of it in lesser time.

Question 22.
Collect information related to feeling hunger from your school library and prepare a note on it.
Answer:

1. Generally we feel hunger when our stomach goes empty.
2. When our stomach goes empty Ghrelin hormone is released and it causes the feeling of hunger.
3. As well leptin hormone also released, but it causes the feeling of suppression of hunger.
4. Not only these, when we saw a colourful and good smelling food, automatically we feel hunger. So, the sight of seeing a good food also causes feeling of hunger.
5. If the food doesn’t have good colour, texture, good smell we can’t eat, even though it is a good food.
6. All the feelings towards food are recieved by our eyes, nose, tongue.
7. From these, information is sent to brain, from there signals are sent to stomach and causes feelings either positive or negative.
8. Hormones are also responsible for most of the feelings of hunger.

Question 23.
Draw the block diagram of showing sensation of taste from food material to brain.
Answer:

Question 24.
Draw a neatly labelled diagram showing a peristaltic movement in oesophagus Explain the importance of mucus on the walls of food pipe.
(OR)
In the food pipe, the food bolus is propelled into the stomach by peristaltic movement Represent this action with a diagram.
Answer:
Importance of mucus on the walls of food pipe :

1. The walls of the food pipe secrete a slippery substance called mucus.
2. Mucus lubricates and protects the oesophgeal walls from damage.
3. This helps the food bolus to slide down easily.
4. Besides this the saliva in the bolus also helps in easy movement.

Question 25.
Draw a schematic diagram of villus in small intestinp. Explain how digestive system coordinates with circulatory system.
(OR)
Draw the diagram of villi in small intestine and lable its parts.
Answer:
Coordination of digestive system with circulatory system :

1. After the completion of digestion in small intestine, it enters into the blood through villi.
2. In blood the digested food gets oxidised after that energy is released.
3. Releasing of energy from fot^d occur only when it enter into the blood.
4. So, there is a great coordination between digestive and

Question 26.
The mere smell or sight of food stimulates hunger . Describe the process through a neat diagram.
Answer:

1. The mere smell even the mere sight of delicious food stimulates the hunger.
2. When we smell, the air borne substances get dissolved in the watery film of nasal mucus.
3. The chemoreceptors in nose are otherwise called olfactory receptors which trigger signals in the form of nerve impulses to the brain where smell is defected.
4. The amount of the neurosecretory protein hormone ghrelin in the blood increases as a result of visual stimulation images of food.
5. Hunger contractions start to occur in the stomach due to hunger generating signals that reach the brain.
6. It is believed that the diencephalon in fore brain and vagus nerve (10th cranial nerve) plays an important role in carrying these signals to brain.
7. Hunger pangs continue up to 30-45 minutes.
8. Increase ¡n ghrelin levels results in sensation of hunger and motivation to consume food.

Question 27.
With the help of a diagram show the movement of food from mouth to the stomach. What muscles and nerves are involved in the movement of food and what is this action called as ? (OR)
Draw a diagram of peristaltic movement of food in oesophagus of elementary canal. Write how it performs.
Answer:
The muscles involved in movement of food from mouth to stomach :

1. The circular muscles of the mouth enable the food to be pushed into the oral cavity and to be moved around.
2. The surface muscles of the jaw help in biting and chewing actions and move the jaw up, down, forward and backward.
3. Contraction and relaxation of circular and longitudinal muscles in the oesophagus bring in a wave like motion that propels the bolus into stomach by the action called peristalsis.
4. As the food approaches the closed ring of pyloric sphincter the surrounding muscles relax and allow the food to pass.
5. The muscles of the upper part of the stomach have to relax and accept large volumes of swallowed material.

Nerves involved in the movement of food:

1. The fifth cranial nerve has been found to control the movement of muscles in the jaw.
2. Peristalsis in oesophagus is under the control of autonomous nervous system.

Question 28.
Prepare a cartoon on Pavlov’s experiment with a suitable caption.
Answer:

Question 29.
How do you appreciate stomach as a churning machine? How does this coordination
Answer:

1. The stomach acts like a churning machine, churning the food around to break it into even smaller pieces.
2. Mechanical mixing of food in stomach occurs by peristalsis, which is waves of muscular contractions that move along the stomach wall.
3. This allows the mass of food to further mix with the digestive enzymes.
4. Due to churning of food in stomach, a mixture that resembles thick cream called chyme is formed.
5. All the muscles in the stomach works with such a great coordination.
6. Hence we can call stomach as a churning machine.

Question 30.
There is a great variety in diversified life processes, express your feelings in the form of a poem.
Answer:

POEM
Life is a process of discomfort and pain.
Learn to handle it all with courage and sane
Remember the bliss waiting for you
Nutrition, excretion, respiration all are essential in our life.
Reproduction makes our species go on and on forever.
Autotrophs and heterophs are brothers.
Depend on each other In a systematic way.
So have courage to learn more tricks.
Search for happiness.
In life that survive.

Question 31.
Suggest any two important habitual actions to your friend while eating food, keeping in view of this chapter.
Answer:

1. Masticate the food thoroughly in the mouth.
2. Do not swallow food without chewing properly or do not eat in hurry.
3. Eat small quantities of food at regular intervals for efficient digestion and absorption of nutrients.
4. Eat the food that emit good smell and has good taste to eat.
5. Do not eat too hot and too cold items.

Fill in the blanks.

1. 3:2:1:2 is the ratio of our dentition. Here 1 represents ———–.
2. Large protein molecules are broken down in ———– of digestive tract.
3. ———– is the strong acid which is secreted during digestion.
4. Olfactory receptors present in ———– triggering signals to brain.
5. pH of saliva is ———– in nature.

Answer:

1. Canine
2. Stomach
3. HCl
4. Nose
5. Alkaline nature

Fill in the blanks with suitable words given below.

Fluctuations of hormone —–(i)—– levels results in sensation of hunger and motivation of consuming food. When you feel your stomach is full and there is no need of food any more. Another hormone —–(ii)—– that gets secreted suppresses hunger. When we take food into the mouth it has to be chewed thoroughly. For this purpose the —–(iii)—– muscles help in chewing actions, while the —–(iv)—– muscles of the jaw moves the jaw up,down, forward and backward during food mastication. The —–(v)—– nerve controls the muscles of the jaw. Under the action of —–(vi)—– nervous system Saliva is released by the salivary glands moistens the food to make chewing and swallowing easier. The salivary —–(vii)—– in the saliva breaks down the starch into sugars. As a result of chewing the food is transported into the oesophagus by the action of swallowing which is coordinated by the swallowing centre in the —–(viii)—– and the —–(ix)—–. The tongue which is gustatory recognizes the taste and —–(x)—– nerve plays an important role in sensation of taste.

Choose the right ones.
i) leptin, ghrelin, gastrin, secretin.
ii) ghrelin, leptin, secretin, gastrin.
iii) deep muscles, surface muscles, circu lar muscles, striated muscles.
iv) surface muscles, deep muscles, neck muscles, long muscles.
v) fifth cranial nerve, second cranial nerve, fifth facial nerve, spinal nerve.
vi) central nervous system, peripheral nervous system, autonomous nervous system.
vii) lipase, sucrase, galactase, amylase.
viii) medulla oblongata, cerebrum, 8th spinal nerve, cranial nerve, 7th cranial nerve.
ix) Pons varoli, brain stem, medulla oblongata, mid brain.
x) 6th cranial nerve, 5th cranial nerve, 10th cranial nerve, optic nerve.

Answer:
i) ghrelin
ii) leptin
iii) surface
iv) deep
v) fifth cranial
vi) autonomous
vii) amylase
viii) medulla oblongata
ix) brain stem
x) fifth cranial nerve

Choose the correct answer.

1. In which of the following situations you can taste quickly? [ ]
   A) Put sugar cristals on tongue
   B) Put sugar solution on tongue
   C) Press the tongue slowly against the palate
   D) Swallow directly without grinding and shredding
   Answer: C
2. Peristalsis is because of [ ]
   A) Contraction of longitudinal muscles
   B) Contraction of circular muscles
   C) Under control of autonomous nervous system
   D) Digestive secretions
   Answer: C
3. Sphincter that helps in opening of stomach into duodenum [ ]
   A) Cardiac
   B) Pyloric
   C) Anal
   D) Gastric
   Answer: B
4. Glucose and amino acids are absorbed through the following part of villus [ ]
   A) epithelial cells
   B) blood capillary
   C) lymphatic vessel
   D) all
   Answer: A
5. The region in brain portion that controls hunger signals [ ]
   A) medulla
   B) diencephalon
   C) cerebrum
   D) mid brain
   Answer: B
6. Human organism is an internal combustion machine because of …. [ ]
   (OR)
   Human being is an “Internal combustion machine” because he/she
   A) assimilation of energy from food
   B) liberate CO₂ during respiration
   C) expel waste food at the end state of digestion
   D) secrete powerful digestive juices
   Answer: A

10th Class Biology 7th Lesson Coordination in Life Processes InText Questions and Answers

10th Class Biology Textbook Page No. 145

Question 1.
How do we know that we need food?
Answer:
When we feel hungry then we know that we need food.

10th Class Biology Textbook Page No. 146

Question 2.
What plays a major role to identify stale food?
Answer:
Smell or odour plays a major role to identify stale food.

Question 3.
If you are having a tasty dish do you think the smell of it increases your appetite?
(OR)
Do the smell of tasty dish increases our appetite?
Answer:
Yes, the smell of tasty dish increases our appetite.

Question 4.
What are your observations after chewing cumin, sounf, potato and apple?
Answer:
If we chew cumin, sounf, potato and apple we observe in order to taste the food material the food should dissolve in saliva.

10th Class Biology Textbook Page No. 148

Question 5.
Are there any other sensation that affect taste?
Answer:
Temperature (hotness), coldness are the sensations that affect taste.

Question 6.
What happens to your taste sensation while sipping hot milk or tea?
Answer:
We find something more tasty while we sipping hot milk or tea.

Question 7.
What do you think could be the range of temperature for us to relish food items?
Answer:
30°C to 35°C could be the range of temperature for us to relish food items.

10th Class Biology Textbook Page No. 149

Question 8.
What do you think that would happen if the salivary glands did not function in our mouth?
Answer:

1. If the salivary glands do not function in our mouth, saliva will not release and the food do not get moistened and chewing it is difficult.
2. The taste of the food cannot be identified.
3. Carbohydrates in the food cannot be broken down and changed to dextrose and maltose molecules.

Question 9.
Suppose your taste buds were affected what would happen to your interest in haying food?
Answer:
If my taste buds were affected I cannot identify the taste of food and also loose interest in having food.

10th Class Biology Textbook Page No. 151

Question 10.
Does the level of saliva secretion change due to presence of food in the mouth?
Answer:
Yes, the level of saliva increases due to the presence of food in the mouth.

Question 11.
Can the process of chewing go on in the absence of saliva?
Answer:
Yes, the process of chewing go on in the absence of saliva. But it is very difficult to chew food and swallow it.

Question 12.
Does the saliva have any other roles to play?
Answer:
The enzyme present in saliva that is salivary amylase converts large molecules of carbohydrates into small molecules of sugar like maltose and dextrose.

Question 13.
What is the use of such an increase in surface area of food?
Answer:
It helps in more surface area for the enzyme to act.

Question 14.
What about the nature of medium for salivary amylase to act on food component?
Answer:
The nature of medium for salivary amylase to act on food component is alkaline.

Question 15.
If we swallow food material directly without mastication what will happen?
Answer:
If we swallow food material directly without mastication, the food will not get digest easily and completely.

Question 16.
Do you think the pH of our mouth changes?
Answer:
Yes, the pH of our mouth changes from acidic to alkaline by the release of saliva from salivary glands.

10th Class Biology Textbook Page No. 152

Question 17.
What are the different systems that contribute to the proper functioning of digestion in the mouth?
Answer:
Endocrine, muscular, nervous systems contribute to the proper functioning of digestion in the mouth.

Question 18.
After the digestive process in the mouth where does the food move to?
Answer:
After the digestion in the mouth the food move to oesophagus.

10th Class Biology Textbook Page No. 153

Question 19.
What are the systems that come into play for swallowing food?
Answer:
Skeletal system, nervous system and digestive systems come into play for swallowing food.

Question 20.
What does the schematic diagram tell us about the oesophagus?
Answer:
The schematic diagram of the oesophagus tells about its structure, secretion and function.

Question 21.
What kind of the tube is oesophagus?
Answer:
Oesophagus is a muscular and an elastic tube.

Question 22.
How does mucus help in passage of food?
Answer:
Mucus helps in lubricating and protecting the oesophageal wall and helps the bolus to slide down easily in the oesophagus.

10th Class Biology Textbook Page No. 154

Question 23.
What makes the movement of the food bolus in the oesophagus easy?
Answer:

1. Mucus helps the food bolus to slide down easily.
2. Peristaltic movements of the walls of oesophagus also make the movement food bolus in the oesophagus easy.

Question 24.
Why do you think the stomach is structured like a bag rather than a tube like an oesophagus?
(OR)
Why stomach is structured like a bag rather than like a tube?
Answer:

1. The food taken has to remain in the stomach for a long time for proper digestion with digestive juices and enzymes.
2. Different kinds of muscles churns the food by contraction and relaxation to form chyme.
3. If it was like a tube the food would just pass down without under going much changes and cannot remain in the stomach for long time.

Question 25.
What sets such processes into action?
Answer:

1. When the food is in the oral cavity, the nerves in the cheek and tongue are stimulated.
2. These carry messages in the form of nerve impulses to the brain.
3. They messages are transmitted from the brain, to the wall of the stomach, and stimulate the gastric glands to produce gastric juice.

10th Class Biology Textbook Page No. 155

Question 26.
What stimulates stomach muscle into action?
Answer:
The nervous system stimulates stomach muscle into action.

Question 27.
What causes the stomach to churn and mix the food?
Answer:
The contractions of the stomach muscles squeeze and mix the food with the acids and juices of the stomach.

Question 28.
Why should only a small quantity of food be passed from stomach to duodenum?
Answer:
For the complete digestion of chyme, only a small quantity of it be passed from stomach to duodenum.

Question 29.
What is involved in bringing about peristalsis?
Answer:
Contraction and relaxation of the muscles present in various parts of gut bring about peristalsis.

Question 30.
What is the direction of peristalsis? Which end of the gut does it begin?
Answer:
The direction of peristalsis is forward direction that is from mouth to anus.

Question 31.
What happens if the direction of peristalsis is reversed?
Answer:
If the direction of peristalsis is reversed the food present in the gut moves backwards.

10th Class Biology Textbook Page No. 157

Question 32.
Why do you think small intestine is long and coiled?
Answer:
The small intestine is long and coiled because the food has to stay for more time for complete digestion and absorption.

Question 33.
What process is involved in this process of absorption?
Answer:
Selective absorption of nutrients by the villi of small intestine is involved in absorption.

Question 34.
What is the relation between finger-like structures and paper folds?
Answer:

1. Finger-like structures increase the surface area.
2. The space inside the paper folds is very much high. So area is increased.
3. So increase in surface area is the relation between finger-like structures and paper folds.

Question 35.
What systems do you think are working together?
Answer:
The digestive system and circulatory system are working together.

Question 36.
Do you think those systems work together in the whole length of the digestive canal? Why / Why not?
Answer:
No, these systems are not working together in the whole length of the digestive canal. The digested food material is absorbed only in the small intestine but not elsewhere in the gut.

10th Class Biology Textbook Page No. 158

Question 37.
Often you may have experienced that if you have tension for some reason you start having loose motions. What does this show us?
Answer:
If we are tensed for some reason, the enteric nervous system or second brain loses control over the gut. Hence without our intervention, loose motions occur.

10th Class Biology Textbook Page No. 159

Question 38.
What moves out of the gut?
Answer:
The indigested food material moves out of the gut.

Question 39.
Two major pathways of waste expulsion are shown above. Which of the two do you think happens exclusively through the gut?
Answer:
Indigested food matter is expelled in the form of stool from the gut.

Question 40.
What controls the exit of stools from the body?
Answer:
The two muscular layers present In the anal sphincter control the exit of stools from the body.

Question 41.
Do you think the control is voluntary? Why / Why not?
Answer:
Yes, the control is voluntary in adults and it is involuntary in infants.

Question 42.
Did we have a sphincter in any other part of the digestive canal? Where was it?
Answer:
Yes. We have a sphincter at the opening where stomach opens into duodenum (small intestine).

10th Class Biology Textbook Page No. 160

Question 43.
What is the fate of the digested substances that move into blood from the intestine?
Answer:
The digested substances reach each cell of the body through circulatory system. There it gets oxidised and release energy.

Question 44.
Where is the energy stored?
Answer:
The energy is stored in the cells as ATP.

Question 45.
Which system do you think will remove the excess salts from our body?
Answer:
The excretory system remove the excess salts from our body.

Question 46.
What would be the path of salt removed from gut to the outside of our body?
Answer:

1. The digested food containing salts will be absorbed into the blood stream in small intestine.
2. The circulatory system supplies this to kidney through renal artery.
3. In the kidney salts are filtered and sent out of the body along with urine.
4. Some of the salts also supplied to the skin. They will be sent out of the body in the form of sweat.

10th Class Biology 7th Lesson Coordination in Life Processes Activities

Activity – 1
Observe the following table, identify and tick those options that you think makes you feel hungry.
Table

i) What stimulates hunger?
Answer:
Smell of food, sight of food and need of food stimulates hunger.

ii) What would be the result of stimulation of hunger?
Answer:
Hunger pangs occur in the stomach.

iii) Which system do you think would send the signals to make us realize that we are hungry?
Answer:
Nervous system.

iv) What kinds of controls are exercised during sensation of hunger? Are they hormonal or nervous or both?
Answer:
They are both hormonal and nervous.

v) Can you suggest any four systems involved in the process of generating hunger sensation?
Answer:
Digestive system, Endocrine system, Circulatory system and Nervous system.

Activity – 2
Observation of how our taste is affected by the sense of smell. (OK) How our taste is affected by the sense of smell?

1. First, close your nose with your fingers.
2. Pop in some zeera in your mouth and chew it for some time.
3. After that chew some sounf.
4. Could you recognise the taste?
5. How long it taken to know the taste?
6. After sometime wash your mouth and repeat the activity by chewing a piece of an apple followed by a potato (remember to close your nose)

Could you know the taste of both or did it taste the same? Why?
Answer:
No, because taste buds couldn’t send the taste signals to brain.

Observation :

1. We can taste the food that is in the form of liquid only.
2. Only after the dissolved food enters into the cup like taste buds, the sense of taste is carried to the brain for analysis. Then only we will know the taste of food material.
3. Similarly olfactory receptors which trigger signals in the form of nerve impulses to the brain where smell is detected.

i) What happens when we put a food material in our mouth?
Answer:
Our mouth salivates.

ii) Name the parts in the mouth that help us to taste food.
Answer:
Papillae (taste buds), palate.

Activity – 3

1. Take a pinch of asafoetida powder/garlic and rub it on hand kerchief/tissue paper.
2. Close your eyes and smell it.
3. Then try to identify taste of different types of food materials with the help of your friend.

i) Does garlic have a stronger scent than apple? How do you think the stronger scent affect your sensation of taste?
Answer:
Yes, garlic have a stronger scent than apple. The stronger scent motivate us to eat different types of foods.

ii) How many food materials you have identified correctly?
Answer:
Seven.

iii) Write a few lines on the relation between smell and taste.
Answer:

1. Taste and smell are intimately entwined. This close relationship is most apparent in how we perceive the flavour of food.
2. Taste itself is focussed on distinguishing chemicals that have sweet, salty, sour, bitter or umami taste.
3. However interactions between the senses of taste and smell enhance our perceptions of the foods, we eat.

iv) Have you ever felt that a particular food is tasty just by looking at it?
Answer:
I felt so many times. In general, we prefer the food material, which is attractive to our eyes and flavour to nose, then we taste it.

Activity – 4
What is the role of different parts of the mouth in helping us to taste keeping sugar crystals over the tongue?

1. Place some sugar crystals on the tongue and keep mouth opened and see that your tongue dosen’t touch the palate.
2. Record the time from the moment you placed the crystals on your tongue till you got the taste by using stop watch.
3. Now repeat the test by placing the sugar crystals on the tongue and pressing it against the palate.
4. Record the time from placing sugar crystals to getting the taste. Or put a drop of sugar solution on your tongue using a dropper.

Observation :
Based on the above activity we know that taste can be identified easily when the tongue is pressed against the palate.
i) Can we taste on dry tongue?
Answer:
No. We can’t taste on dry tongue.

ii) Which way helped you taste faster? Why?
Answer:
Taste can be identified faster when the tongue is pressed against the palate.
When the tongue is pressed against the palate the food substance is pressed against the opening of the taste bud letting it reach taste celjs triggering taste signals. Finally the taste is recognised in the brain.

Activity – 5
How do you show that the breakdown of food by using the model of chalk piece kept in vinegar?

1. Break a piece of chalk into two halves.
2. Crush one half to tiny pieces leaving the other as it is.
3. Take two small mineral wafer bottles QA ltr bottle) cut them into two equal halves and discard the upper portion.
4. Now we have two beakers from the lower cut portion.
5. Fill them half with vinegar and add the crushed chalk to one beaker and the other uncrushed half chalk to the other.
6. Observe them after half-an-hour or so.

i) Which one dissolved faster the crushed chalk or the whole one?
Answer:
Beaker with crushed chalk dissolved faster than the whole one.
This experiment tells us the need of mechanical crushing of food the mouth to increase surface area for action of substances that aid in digestion.

ii) How does this process of mechanical crushing go on in the mouth?
Answer:
Mechanical crushing of food goes in the mouth by chewing.

iii) Which parts in the mouth are involved in this?
Answer:
Teeth and tongue.

iv) What are the systems involved in this process?
Answer:
Digestive system, Nervous system, Muscular system.

Activity – 6
Observe the diagram and answer the questions and fill the table.
i) Observe the model or chart of jaw, on the basis of the figure try to guess what are the functions molars could be?
Answer:
Chewing and grinding.

ii) What do you think could be the function of inscisors?
Answer:
The function of inscisors is biting the food.

iii) Which set of teeth helps in grinding food?
Answer:
Premolars and molars.

iv) Which set helps in tearing food?
Answer:
Canines help in tearing food.

v) What is your dental formula?
Answer:
is the dental formula.

Table

Activity – 7
Testing pH of mouth at intervals of one hour.

1. Collect a strip of pH paper with a colour chart from a chemistry teacher.
2. Take a small piece of the pH paper and touch it to your tongue.
3. Match the colour with the colour chart and note the pH.
4. Take some readings after having your food at lunch break.
5. Compare your readings with that of your friend.
6. Take at least 4 readings.

Table

i) What is the usual range of pH of your mouth Acidic (or) Basic
Answer:
The usual range of pH of our mouth is more or less 6.5. At low level the saliva is acidic and at high level it is basic.

ii) Did you observe any change in pH after eating? What may have caused the change?
Answer:
Yes, 1 observed the change in pH. Bicarbonates in saliva changed that into basic.

iii) In what kind of pH do you think salivary amylase acts well?
Answer:
In the pH range 7.2 to 7.4 it acts well in alkaline medium.

iv) Does this type of food have any role to play on the pH of our mouth?
Answer:
If the food that enters is acidic it will be converted into basic stuff and then it is swallowed.
Based on the above tests we know that salival secreted causes the medium to change to alkaline as it aids in action salivary amylase.

Activity – 8
Making a model of oesophagus to observe how bolus moves forward.

1. Take a piece of waste cycle tube and insert one or two potatoes into it.
2. Lubricate the inner side of the tube with oil.
3. In the same way smear oil over potatoes.
4. Insert oil coated potatoes in the tube.
5. Now try to push the potatoes by squeezing the tube.

i) How do you squeeze the tube to make the potatoes pass through?
Answer:
By pushing the potato from behind.

ii) Do you think that the muscles in the wall of the oesophagus have to do something like this?
Answer:
Yes.

iii) How did oil help you in pushing the potatoes through the pipe?
Answer:
Oil acted as a lubricant to push the potato easily in the forward direction.

Activity – 9
How is the stomach protected from the secretions of its own acids?

1. Take two similar green leaves.
2. Grease one leaf with petroleum jelly, leave the other free.
3. Add 1 or 2 drops of some weak acids on both the leaves.
4. Observe them after half-an-hour or so and write your observations.

i) Which leaf was effected by the acid?
Answer:
The leaf to which petroleum jelly was not applied.

ii) What kind of change did you observe in the leaves?
Answer:
The colour of the leaf changes.

iii) What saved the other leaf from the effect of acid?
Answer:
Petroleum jelly.
From the above activity we can conclude that mucus secreted by the walls of stomach protects stomach from the harmful effects of hydrochloric acid.

Activity – 10
Paper tube and folded papers.

1. Provide students with a piece of paper.
2. Let them calculate the area of one side of the paper and make a roll of it.
3. Try to fill the tube by inserting few folded papers as much as possible in it.
4. Pull out the papers from the tube, unfold them and find out the whole area of the papers.

i) Compare the area of the folded papers with that of the roll. Do you find any increase in the area ? If so try to find out the reasons?
Answer:
Area is increased. The space inside the folded papers is very much high. So we can insert as many folded papers as we could.
From the above activity we can infer that the inner surface of the small intestine contains thousands of finger-like projections called villi which increase the surface area of absorption of nutrients in small intestine.

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 6 Reproduction – The Generating System Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 6th Lesson Reproduction – The Generating System

10th Class Biology 6th Lesson Reproduction – The Generating System Textbook Questions and Answers

Improve your learning

Question 1.
Why do fish and frog produce a huge number of eggs each year?
Answer:

1. External fertilisation occurs in fish and frog.
2. The female lays a vast number of eggs in water and male releases some millions of sperms on to them in water.
3. Eggs may be carried to a long-distance by water currents or they may be eaten by predators.
4. So fertilisation is a chance factor and controlled by nature.
5. Fertilisation occurs externally hence it is inevitable to give rise to vast number of eggs by fish and frog.

Question 2.
Give examples and explain what is meant by external fertilisation?
Answer:

1. Fusion of the male and female gametes is called fertilisation.
2. The fertilized egg (ovum) is called zygote.
3. If the fertilisation occurs outside the body of the female organism then it is called external fertilisation.
4. External fertilisation is seen in fish, frog and earthworm, etc.

Question 3.
Write the differences between
a) Grafting – Layering b) Stamen – Carpel.
Answer:
a) Differences between Grafting and Layering.

b) Differences between Stamen and Carpel.

Question 4.
Describe the mode of fertilization in plants with a diagram.
Answer:

1. Fertilisation is the process of fusion of male and female gametes.
2. For the fusion of male and female gametes pollen grains have to reach the surface of the stigma. This is called pollination.
3. Pollen grains received by the stigma germinate and give rise to pollen tubes. Only one pollen tube finally reaches the embryo-sac.
4. This tube will have two male nuclei, which migrate to the tip of the pollen tube at the time of fertilization.
   
5. Usually the pollen tube enters the ovule through micropyle and discharges the two male gametes into its embryo-sac.
6. One male nucleus (n) or (gamete) approaches the egg and fuses with it to form a diploid (2n) zygote. This is first fertilisation.
7. The other male nucleus (n) or (gamete) reaches the secondary nucleus (2n) and fuses with it to form the endosperm nucleus which will be triploid. This is second fertilization in the embryo-sac.
8. Thus double fertilisation occurs in embryo-sac which is unique in flowering plants.

Question 5.
What are the different modes of asexual reproduction? Cite them with examples.
Answer:
Asexual reproduction takes place by six different methods. They are:

1. Fission
2. Budding
3. Spore formation
4. Regeneration
5. Fragmentation and
6. Vegetative propagation.

1. Fission: Single-celled organisms split into two equal offsprings or more offsprings. Ex: Paramoecium, bacteria.

2. Budding: A growth on the body as a bud grows to form identical copy of parent. Ex: Yeast.

3. Spore formation : Spores are produced in the sporangium.
Ex : Rhizopus, mucor, bacteria, ferns and mosses.

4. Regeneration :
a) Ability of organisms to give rise to new individual organisms from their body parts.
b) That is if the individual is some how cut or broken up into many pieces, many of these pieces grow into separate individuals.
Ex: Hydra and Planaria.

5. Fragmentation:
a) In multicellular organisms with relatively simple body organisation breaks up into smaller pieces upon maturation.
b) These pieces of fragments grow into new individuals.
Ex: Flatworms, Spirogyra, moulds, lichens.

6. Parthenogenesis: In the process generally the female gametes or ova develop into zygote without fertilization.
Ex : Bees, ants and wasps.

7. Vegetative propagation: When a vegetative part like stem, root and leaf can produce a new organism it is called vegetative propagation. It is of two types.
a) Natural propagation:
Bryophyllum Dahlia, Carrot, Radish
b) Artificial propagation:
i) Layering: Eg: Nerium, Guava, Orange, Rose
ii) Cutting: Eg: Rose, Hibiscus, Sugarcane
iii) Grafting: Eg: Sapota, Guava, Mango, etc.

Question 6.
In what ways does sexual reproduction differ from asexual one? State at least three reasons.
(OR)
What are the differences between sexual and asexual reproduction?
Answer:

Apart from the above differences Sexual and Asexual reproduction differs in the following ways:

1. The unit of reproduction in sexual reproduction is gamete whereas in asexual repro-duction it may be whole parent body or bud or a fragment or a single somatic cell.
2. Sexual reproduction need more time to complete and it is less time in asexual reproduction.

Question 7.
How are sperm cells adapted for their function?
Answer:

1. Sperm cell is adapted to its function by carrying genetic information to an egg.
2. Its body consists of four parts :
   1. Head
   2. Neck
   3. Middle piece and
   4. Tail.
3. Sperm has a streamlined body that allows it to move quickly.
4. Nucleus is present in the head.
5. Special structure on the head called acrosome helps in penetrating the ovum during fertilisation.
6. The head of sperms also contain enzymes to breakdown and digest the zone on the egg through which it penetrates and fertilise it.
7. Large number of mitochondria are present in the mid-region, so it is able to produce a lot of energy in order to operate tail.
8. Its tail allows it to swim towards the egg for fertilisation.

Question 8.
The menstrual cycle prepares the uterus for a fertilised egg. How long is an average menstrual cycle from start to finish?
Answer:

1. The cycle of changes that occur in the female reproductive system is called menstrual cycle.
2. The average menstrual cycle from start to finish is 28 days long.

Question 9.
When the foetus is growing inside the uterus it needs nutrients? What provides these nutrients?
Answer:

1. When the foetus is growing inside the uterus it needs nutrients for its growth and development.
2. These nutrients are provided by the mother’s blood through a special structure called placenta.
3. Placenta is a tissue formed by the cells from the embryo and the mother.
4. It is formed around 12 weeks of pregnancy and becomes an important structure for nourishment of the embryo.
5. The foetus is connected to placenta in mother’s body through umbilical cord.

Question 10.
Which type of substances are absorbed by foetus from the mother?
Answer:

1. The digested food from the mother travel through the mother’s blood stream and exchange to the blood stream of the foetus through the placenta.
2. In addition to ingested food the mother’s body continuously breakes down muscles, fat and bones.
3. Releasing proteins, fat and calcium to the mother’s blood that can be absorbed through the placenta to provide nutrients to the foetus.
4. Some hormones also transfer to baby.

Question 11.
What is the job of Amniotic sac?
Answer:

1. The amniotic sac is a bag of fluid inside a women’s womb (uterus) where the embryo and foetus develops and grows.
2. The cavity within the amnion becomes filled with fluid called amniotic fluid.
3. The embryo or unborn baby floats and moves in the amniotic fluid.
4. Amniotic sac and amniotic fluid give protection against minor mechanical injury.
5. This fluid also provides a stable temperature and assists in maintaining a consistent body temperature for the unborn child.

Question 12.
What are the advantages of sexual reproduction?
(OR)
What are uses of sexual reproduction?
Answer:
Advantages of sexual reproduction:

1. Sexual reproduction promotes diversity of characters in the offsprings by providing genetic variation.
2. It plays an important role in the origin of new species have different characters.
3. This genetic variation leads to the continuous evolution of various species to form better and still better organisms.
4. Sexual reproduction influences in diversity of characters in offsprings.
5. It helps to maintain the continuty of organisms.
6. It leads to new generation to tolerate adverse or diseased conditions.
7. It helps the spreading of population to new areas.

Question 13.
How does reproduction help in providing stability to population of species?
Answer:

1. The reproduction is directly linked to the stability of the population of species because it helps in replacing the lost or aging population with the new population and thus ensures the survival of the species.
2. The process of reproduction ensure continuity of life on earth.
3. In the absence of reproduction one particular species will disappear with time.
4. Reproduction induces variations in the population which help the population to tide over adverse environmental conditions and adapt to changing environment.
5. Reproduction also helps to generate copies of individuals which are suited to a particular environment.

Question 14.
Write the differences between mitosis and meiosis.
Answer:

Question 15.
What happens to the wall of the uterus during menstruation? (OR)
What changes occur in the wall of the uterus during menstruation?
Answer:

1. During the early stage of menstruation the cells in the wall of uterus increase in number by repeated mitotic divisions.
2. The inner lining of uterus becomes thick and soft with lot of blood capillaries in it.
3. These changes in the uterus are necessary because in case the ovum released by the ovary gets fertilised by the sperm, then the uterus has to keep this fertilised ovum for further development and supply it with food and oxygen etc.
4. If fertilisation does not occur the additional growth of the uterus is detached and expelled along with some amount of blood.

Question 16.
“All unicellular organisms undergo only mitotic cell division during favourable conditions” – Do you support this statement? Why?
Answer:

1. Unicellular organisms undergo mitotic cell division not only during favourable conditions but also in unfavourable conditions.
   So I don’t support the given statement.
2. When the organism finds favourable conditions it deserves to perform division called Fission. Ex: Amoeba.
3. Sometimes scarcity of food or to save it life, some organisms undergo mitotic cell division.
4. Organisms like paramoecium undergo a type of sexual reproduction called conjugation during unfavourable conditions.

Question 17.
What would be the consequences if there is no meiosis in organisms that reproduce sexually?
(OR)
What happens if Meiosis does not take place in reproductive cells?
Answer:

1. If meiosis did not occur, a fusion of gametes would result in a doubling of the chromosomes for each successive reproduced generation.
2. For example, in case of man egg cells and sperm cells like other cells must contain 46 chromosomes.
   
3. This results in the formation of abnormalities in each generation.
4. If meiosis did not occur, there is no genetic variation in the offsprings produced by random fusion of the gametes.
5. New characters will not appear in the offsprings.
6. The process of evolution will be stopped.

Question 18.
Vicky’s father wants to grow a single plant having two desirable characters colourful flowers and big fruits. What method will you suggest to him and why?
Answer:

1. I will suggest the method of grafting to him.
2. Grafting enables us to combine the most desirable characteristics of two plants into a single plant with colourful flowers and big fruits.
3. By grafting method, a very young scion can be made to flower and produce fruits fast when it is grafted to the stock.
4. Vicky’s father can grow one of the two plants as stock and second plant can be graft to some of its branches as scions.
5. Then he can get the plant with both the desirable characters.

Question 19.
Uproot an onion plant and take a thin section of its root tip. Stain it and observe under the microscope. Draw as you see and identify the stages of the cell division.
Answer:

Question 20.
Visit a nearby village and collect information on how farmers grow sugarcane, flowering plants like chrysanthemum, primerose and vegetables like stem tubers, plump gourd (dondakaya), etc. Make a report and present in class.
Answer:
Report on growing methods of various plants :

Question 21.
Collect the information from school library or using internet what vegetative methods are followed in your district as well as in your state to propagate various plants of economic importance. Represent it with a graph.
Answer:
Vegetative methods followed in our district as well as in our state to propagate various plants of economic importance.
I) Natural vegetative propagation: In this method of vegetative propagation, a part of the plant which may be stem, root-leaf or flower gets detached from the body of the mother plant.

1. Vegetative propagation: Roots of radish, carrot, dahlia develop adventious buds which grow into leafy shoots.
2. Vegetative propagation by stems: Stolons – Vallisneria, offsets – Eichhornia, Rhizome – Banana, Ginger Bulbs – Alliumcepa (Onion); Corn – Colacasia; Tuber – Potato.
3. Vegetative propagation by leaves: Bryophyllum.
4. Vegetative propagation by modified flowers (Bulbils): Agave.

II) Artificial vegetative propagation: Certain flowering plants have the capacity to develop a part of their somatic body into a new independent plant. In artificial vegetative propagation such plants are identified and special techniques are applied to obtain new independent plant.

1. Cutting (Stems): Sugarcane, Roses, Hibiscus, Citrus plants.
2. Cutting (Root): Lemon, Tamarind.
3. Layering: Jasmine, Strawberry, Gooseberry.
4. Grafting: Rubber, Apple, Pear, Citrus, Mango, Guava.
5. Propagation by tissue culture technique: Lily, Rose, Magnolia, Fern, Banana for micropropagation, a small amount of tissue from a suitable part of the parent plant is excised and grown on a nutrient medium under aseptic conditions.

Question 22.
Make a flow chart to show the cell cycle and explain cell division describing different stages of mitosis.
Answer:

Different stages of mitotic cell division:

Question 23.
Draw neat labelled diagrams of male and female reproductive systems of plant.
Answer:

Question 24.
Observe the following part of flowering plant and prepare a note.
Answer:
The given diagram is the structure of ovule which is present in the ovary (carpel) of plant.

1. An ovule is an egg shaped structure attached by a stalk (Funicle) to the inner side of the ovary.
2. Depending upon the species of plant involved, an ovary may have one, two, several or even hundreds of ovules.
   
3. At the centre of each ovule is a microscopic embryo sac filled with food and water.
4. The embryo-sac is composed of gametophyte cells.
5. The majority of flowering plants have an embryo sac consisting of seven cells and eight nuclei.
6. They are one egg (female gamete), two synergids, one central cell (secondary nucleus) and three antipodals.
7. Central cell contains two nuclei, they are called polar nuclei.

Question 25.
Prepare a flow chart to explain the process of sexual reproduction in plants
Answer:

Question 26.
Draw a neat labelled diagram to explain plant fertilisation. Write few points on Pollen grain.
Answer:

Pollen grains:

1. Pollen grains develop in anther.
2. Anther consists of spore-forming tissue. Some of the cells in the spore-forming tissue develop as pollen mother cells.
3. Each pollen mother cell undergoes meiosis to form four daughter cells which develop into pollen grains
4. Pollen grains are haploid (‘n’) and are otherwise known as microspores or male gametes.
5. They contain only one set of chromosomes (‘n’).
6. The study of pollen grain is called palynology.
7. During pollination pollen grains are dispersed by wind and insects.
8. Pollen grains are formed in large numbers. They are light in weight and are easily carried by wind currents.

Question 27.
How will you appreciate cell division that helps in perpetuation of life?
Answer:

1. Perpetuation means continuation of life.
2. The perpetuation of life is based on the cell division.
3. Both mitotic and meiotic divisions are very essential for perpetuation of life.
4. In unicellular organisms, the mitotic cell divisions form the entire organisms.
5. Mitotic division is essential for all types of asexual reproductions.
6. In sexually reproducing organisms meiotic cell division helps in formation of gametes with haploid number of chromosomes which fuses to form diploid zygote during fertilization.
7. Zygote further divides by mitosis again and grows into an embryo and then to offspring.
8. Thus both mitotic and meiotic cell divisions play a key role in perpetuation of life. Without cell division, there is no perpetuation of life.

Question 28.
What precautions will you take to keep away from various sexually transmitted?
Answer:
Precautions to be taken to keep away from various sexually transmitted diseases:

1. Avoid sex with any one who has genital sores, a rash, discharge or other symptoms.
2. The only time unprotected sex is a safe if the partners have sex only with each other.
3. I use latex condoms every time 1 participate in sex. I use it for the entire sex act.
4. I avoid sharing towels or under clothing.
5. I wash genital organ before and after intercourse.
6. I will get a vaccination for hepatitis B. This is a series of three shots.
7. I will get tested for HIV for every six months.
8. I will not drunk or take drugs. Under these conditions, I may fail to have safe sex.
9. I consider that not having sex is the only way, sure way to prevent sexually transmitted diseases.
10. Sexual act is supposed to be an act between husband and wife. Hence I will not participate in sex before marriage.
11. When I grow up and get married, I will be upright and faithful to the life partner and will not behave immorally.
12. Hence as a student I will concentrate on studies and create activities to achieve success in life.

Question 29.
Conduct a seminar on child marriages and foeticide.
Answer:
child marriages can also be defined as “any marriage carried out under the age of 18 years and involves the girl who is physically, psychologically, mentally and physiologically immature”.

Causes of Child marriages :

1. Poverty in the families due to unemployment.
2. Illiteracy among people.
3. Ignorance of consequences of child marriages.
4. Strong superstitions among people about child marriages.
5. Religious and traditions are also responsible for occurance of child marriages.

Steps to prevent Child marriages:

1. By creating awareness among the people about the ill effects of child marriages.
2. By educating the people especially in rural areas of backward districts.
3. By providing employment opportunities to all the people.
4. Performing live plays, skits and stage shows to create enlightenment among the rural illiterate women about the pathetic future of kids who are forced to get married.

Foeticide:

1. Foeticide is the illegal practice of killing a foetus. Female foeticide is prevalent in our country as a major social evil.
2. Some kinds of complications in pregnancy can also demand surgical termination of pregnancy after 8 weeks of conception.
3. This is where the abortion is legal and doctors may have to suggest for discontinuous of pregnancy for the sake of health of the mother. However the technique of surgical termination is misused by some people by getting rid of the unborn child.
4. Abortion of foetus is an act of murder. God is the author of life and nobody should have right to take it.

Choose the correct answer.

1. The part of the female reproductive system produces the eggs [ ]
A) Ovary
B) Epididymis
C) Cervix
D) Fallopian tube
Answer: A

2. The term that we use to describe a sperm cell fusing with an egg cell [ ]
A) Fragmentation
B) Fermentation
C) Fertilisation
D) Fusion
Answer: C

3. Which part of the male reproductive system produces (human) the sperm cells ? [ ]
A) Vas deference
B) Epididymis
C) Seminiferous tubules
D) Scrotum
Answer: C

4. How does the sperm break through the egg cell membrane? Choose the option you think is right.   [ ]
A) Tears a hole in the membrane
B) Dissolves the membrane with chemicals
C) Bites through the membrane with teeth
D) Squeezes through gaps in the membrane
Answer: B

5. Why are egg cells larger than sperm cells? Choose the option you think is right. [ ]
A) Egg cells have more cells in them
B) Have food store to help growth after fertilisation
C) Have thicker cell membranes
D) Have larger nuclei
Answer: B

6. Which of these things will affect the way a foetus grows? Choose the option you think is right. [ ]
A) Chemicals in cigarette smoke
B) Alcohol
C) Drugs
D) All of the above
Answer: D

7. Which of the following is the correct sequence of steps in the human life cycle? Choose the right option. [ ]
A) Babyhood, childhood, adolescence, adulthood
B) Childhood, babyhood, adulthood, adolescence
C) Adolescence, babyhood, adulthood, childhood
D) None of the above
Answer: A

10th Class Biology 6th Lesson Reproduction – The Generating System InText Questions and Answers

10th Class Biology Textbook Page No. 117

Question 1.
How do you think bacteria were dividing to form curd?
Answer:
Curdling indicates that the increase in number of bacteria by fission.

10th Class Biology Textbook Page No. 118

Question 2.
How do you think this process (parthenogenesis) happens?
Answer:
This process occurs by the development of female gamete or ovum directly develops into zygote without fertilisation.

Question 3.
What about animals?
Answer:

1. Normally the fertilized ovum (zygote) develops into a young one. But the unfertilized ovum also develops into a young one generally the male.
2. The process of development of young ones from unfertilized ovum is called parthenogenesis.
3. In this process sperms develop by mitotic division whereas ova develop by meiotic division.
4. This strange kind of reproduction occurs in animals like bees and wasps.

Question 4.
Is regeneration can also be known as a type of fragmentation? Do you agree? Why? Why not?
Answer:

1. Yes. I agree that regeneration could be also called as a type of fragmentation. Because in both cases pieces or parts from the body of the organism can develop into a new individual.
2. Fragmentation and regeneration occur in multicellular animals.
3. Fragmentation occurs in organisms with relatively simple body organisation.
4. Whereas regeneration occurs in organisms with fully differentiated body organisation.

Question 5.
Which type of fission would produce larger colonies in less period of time. Why?
Answer:
Multiple fission would produce larger colonies in less period of time because more number of daughter cells are formed by multiple fission.

Question 6.
Which mode of asexual reproduction provides maximum scope of choice of desirable characters?
Answer:
Parthenogenesis provides maximum scope of choice of desirable characters.

10th Class Biology Textbook Page No. 120

Question 7.
What are the characters that would you like to select?
Answer:
The characters that I would like to select are plant with large number of fruits which are big in size and taste sweet with less or no seeds.

Question 8.
What mode of propagation would help you to produce the plants with selected characters?
Answer:
Mode of propagation that would help me to produce the plants with selected characters is grafting.

Question 9.
Whether they reproduce by budding or fission or fragmentation, organisms are copies of their parents. Is it true? Why ?
Answer:

1. Yes, it is true that organisms reproduced by budding or fission or fragmentation are copies of their parents.
2. Because budding or fission or fragments are not the methods of sexual reproduction.
3. No gametes were formed or fused in these methods.
4. Exchange of chromosomes or crossing over do not take place. Hence the offsprings produced are similar to their parents.

10th Class Biology Textbook Page No. 122

Question 10.
Do you find any similarities between rhizopus and fern spores and sporangia?
Answer:

1. Both rhizopus and fern reproduce asexually through spores.
2. In both the spores are microscopic, unicellular bodies produced in the sporangia.

Question 11.
What about mushrooms, how do they grow? Discuss in your class.
Answer:
Fungi grow from the fragmentation of hypae. They also form buds which are bulged from out side of cells which detaches after division of the nucleus. A special reproduc¬tive sac called sporangium produces asexual spores which are released outside. Fun¬gal sexual reproduction includes plasmogamy, Karyogamy and gametangia.

10th Class Biology Textbook Page No. 123

Question 12.
Think why testis are located outside the abdominal cavity?
Answer:
The testis are located outside the abdominal cavity because the temperature required for proper functioning of spermatogenesis is generally 2 to 3 degrees less than the body temperature,

10th Class Biology Textbook Page No. 127

Question 13.
What function do you think is served by petals and sepals?
Answer:

1. Calyx consists of sepals give protection to the flower particularly in bud condition.
2. Corolla consists of bright coloured petals and are useful in attracting insects for pollination.

Question 14.
Draw the diagram of the flower that you collect and label the parts shown and write their functions.
Answer:
Functions of flower parts:

1. Calyx: Consists of sepals – protects flower in bud conditions.
2. Corolla: Consists of coloured petals – helps in pollination.
3. Androecium: Consists of stamens – produce male gametes pollen grains,
4. Gynoecium: Female reproductive part – produce ovules inside the ovary. Stigma receives pollen grain.

10th Class Biology Textbook Page No. 128

Question 15.
How does the male reproductive cell fertilise the female reproductive cell in flowers of such plants (pea plants)?
Answer:

1. In self pollinated plants anthers are usually present above the stigma.
2. Pollen from the anthers drop on the stigma.
3. A process that takes place as the flowers close for the night,
4. And sometimes occurs before they are completely developed and ready to open.
5. At the time of pollination slightest movement of the flower’s petals stimulate the stamen to dislodge its pollen and transfer it to the near by stigma in pea plant.
6. Pollination usually occurs before the flower is fully open, .

10th Class Biology Textbook Page No. 129

Question 16.
How many nuclei are present in the pollen grain?
Answer:
Pollen grain has two nuclei. One is called a tube cell and the another is generative cell.

10th Class Biology Textbook Page No. 131

Question 17.
Which floral part may be seen in a fruit?
Answer:
Sometimes calyx may remain with fruit.

Question 18.
How cotyledons are useful for the plant?
Answer:

1. The cotyledons digest and absorb the endosperm.
2. They make the stored food available for the growth of the epicotyl and hypocotyl.
3. The cotyledons of some flowering plants, digest, absorb, and store the foods from the endosperm as the ovule is maturing into a seed. Eg: beans
4. As a consequence, the cotyledons become greatly enlarged because of stored food and the endosperm disappears more or less completely.

10th Class Biology Textbook Page No. 137

Question 19.
What differences do you find in mitosis and meiosis? Write in a tabular form.
Answer:

Question 20.
What would happen if the gametes do not have half the chromosome number as the skin parent?
Answer:

1. Daughter cells formed in meiosis are gametes.
2. These gametes have half of the chromosomes in number as the parent,
3. If the gametes do not have half of the chromosomes in number as the parent, when they fuse, they form zygote with double the number of chromosomes when compared to parent cell.
4. If it continues, cells in the offspring will have thousands of chromosomes within few generations.
5. If the chromosome number increases in a species it leads to the formation of abnormalities.

Question 21.
How would it affect the progeny formed by sexual reproduction?
Answer:
If the progeny have thousands of chromosomes in them, it results in formation of abnormalities in each generation.

Question 22.
Why did the government of India fixed the legal marriage age of boys (21 years) and girls (18 years)?
Answer:

1. Government of India fixed legal marriage age of boys as 21 years, and girls as 18 years.
2. This is because teenage mothers are not prepared mentally or physically for motherhood.
3. Early marriage and motherhood cause health problems for the mother and child.
4. It may also cause mental agony, as teenage mother is not ready for responsibilities of motherhood.

Question 23.
Do you feel that it is a social responsibility to control birth after having one or two children?
Answer:

1. Yes, it is a social responsibility of every individual to control birth after having one or two children.
2. If we don’t control birth after having one or two children, population will grow rapidly.
3. If the population increases we will not be able to provide all the facilities such as education, medicine, employment etc., to all the people.
4. It shows impact on the economic conditions of the family and the society.
5. The quality of life will decrease.

Question 24.
What do you understand by the term ‘Healthy Society’?
Answer:

1. If all the people in a society are in the state of complete physical, mental and social well being, then the society is said to be healthy society.
2. To be in a healthy society, every one in the society should take care of their own personal hygiene and cleanliness of the surroundings.
3. Avoiding child marriages, unprotected sex and creating awareness among the people regarding adverse effects of these are very essential to form a healthy society.

Question 25.
Will you encourage child marriage? Why?
Answer:

1. No. I will never encourage child marriage.
2. This is because, the sexual act always has potential to lead to pregnancy.
3. In case of child marriage, the age of girls is less than 18 years and so they are not prepared mentally or physically for motherhood.
4. If they got pregnancy the health of them and their child will be affected.

Question 26.
How does HIV is transmitted?
Answer:
Hiv is transmitted by unsafe sexual contacts, using infected devices,infected blood
transfusion, from an infected mother to child, etc.

10th Class Biology Textbook Page No. 138

Question 27.
Social discrimination against AIDS patients is also a social evil. Can you support this? Why?
Answer:

1. Yes, I will support this statement.
2. The persons suffering from HIV/AIDS are shown lot of social discrimination in the society.
3. This happens even with their own family members.
4. This is due to lack of awarness among public about spreading of disease, illiteracy misconception about AIDS.
5. If everyone knows how it will not spread they will treat HIV + ves with love and effection.
6. HIV +ves are patients. It will spread through sexual contact, blood transfusion, mother to child and not with other modes.
7. Hence they can live with us without any discrimination they need our love and family support.
8. If anybody shows discrimination, it is definitely a social evil.

10th Class Biology Textbook Page No. 140

Question 28.
Do you think you have moral right to kill a foetus?
Answer:
No, I don’t think that we have moral right to kill a foetus. It is our social responsibility to protect the foetus.

Question 29.
Why doctors are prohibited to do sex determination through ultrasound scanping for pregnant women?
Answer:
Knowing the sex of foetus inside mother’s womb is a severe crime as per the acfmade by government. Ultrasound tests are mend to know the growing condition of the foetus and also to see whether it is suffering with severe ailments. By knowing the sex of the foetus, if it is female, people are ready for aborting it. This leads to reduction in male female ratio in the country. That’s why doctors are prohibited to do sex determination through ultrasound scanning for pregnant women.

10th Class Biology 5th Lesson Reproduction – The Generating System Activities

Activity – 1

Formation of bacterial colony in milk.
Answer:

1. Take a tea spoon full of curd and mix it thoroughly with around tea spoon full of (half of the glass) luke warm milk in a bowl.
2. Take another tea spoon full of curd and mix it with 30 tea spoon full of cold milk in another bowl,
3. Cover both the bowls and note the initial time.
4. Keep observing every hour to see whether curd has formed.
5. Curdling indicates that the increase in number of bacteria.
6. Note the time taken for formation of curd in both the bowls.

• Does it take the same time to form curd in both the bowls?
  Answer: No. Formation of curd in the bowl containing luke warm milk takes nearly 5 – 6 hours. In the second bowl in which cold milk is present no curdling took place.
• What does the time taken to form nearly 30 times the size of the bacterial colony indicate?
  Answer: Time taken to form nearly 30 times the size of the bacterial colony indicates how fast bacteria are growing.

Activity – 2

Examine Rhizopus or common mould under Microscope.
(OR)
Write the materials required and the procedure followed by you to observe Rhizopus in the lab.
(OR)
Write the procedure which you follow to observe bread-mould Sporangium m your laboratory. What precautions do you take during the activity?
(OR)
Suneetha wanted to observe Rhizopus on the piece of bread,
(i) Suggest the apparatus needed. (ii) Write the procedure to be followed.
Answer:
Aim: To grow and examine rhizopus or common mould.
Materials required: Bread, plastic bag, plain glass slide, cover slip, water, eye dropper, disposable gloves, compound microscope.
Procedure to grow mould:

1. Take a soft bread and leave it in the open for about an hour so it is exposed to contaminants in the air.
2. Place the bread in a plastic bag, sprinkle water over it, so it is damp and seal the bag living some air inside.
3. Check on the piece of bread every few days and add more water if it is becoming dried out.
4. We can find whitish thread like growth with masses of black, gray and green fine dotted structures, the black dotted structure is that of bread mould.
5. A good sample of mould may take up to two weeks to form.
6. Using this mould make a slide and observe under the microscope.

Procedure to make a slide:

1. Place a drop of water in the centre of the slide, using an eye dropper if you have one, or the tip of a clean finger.
2. Using a tooth pick, scrape some of the mould off, and place it on the drop of water.
3. Take the coverslip and set it at an angle to the slide so that one edge of it touches the water drop.
4. Then carefully lower it over the drop, so that the coverslip covers the specimen without trapping air bubbles underneath.
5. Use the corner of a tissue paper or blotting paper to blot up any excess water at the edges of the coverslip.
6. View the slide with a compound microscope, starting with a low objective.

Observations:

1. The common bread mould plant consists of fine thread like projections called hyphae and thin stems having knob like structures called sporangia.
2. Each sporangium contains hundreds of minute spores.

Precautions:

1. This should not be done by those with allergies to mould or with severe asthma.
2. Avoid opening the plastic bag as much as you can.
3. If you touch the bread, be sure to thoroughly wash your hands afterwards.

Activity – 3

Observation of pollen grain.

1. Take a slide and put a few drops of water on it.
2. Take any flower like hibiscus, tridax, marigold, etc. Tap the anther over the drop of water.
3. We will see small dot like structures in water. These are pollen grains.
4. Take permanent slide of pollen grain from your lab.
5. Also see a permanent slide of pollen grain from our lab.
6. Observe under a microscope. We will make a drawing of what we observe and compare with the given diagram.

Observation: Pollen grain germinates only on stigma. Pollen grain consists of two to three cel Is surrounded by a protective wall in Angiosperms. In gymnosperms the pollen grain consists of several living cells.

Activity – 4

Seed germination.
Answer:

1. Soak a few groundnut or bengal gram (chana) seeds overnight.
2. Drain the excess water and cover the seeds with wet cloth. Leave them for a day.
3. Keep sprinkling water at regular intervals so that they do not dry up.
4. Open the seeds carefully and observe the parts.

Observation: The seed is germinated i.e., the seed embryo is developed into seedling plumule which grows into plant.

Activity – 5

Observe different stages of mitotic cell division. (OR)
Describe various stages of mitosis with the help of neat diagrams.
Answer:

1. Take permanent slides which shows different stages of mitotic cell division from your lab kit.
2. Observe carefully under microscope.
3. Draw diagrams what you observe, and compare your observations with the following chart.

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 9 Our Environment Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 9th Lesson Our Environment

10th Class Biology 9th Lesson Our Environment Textbook Questions and Answers

Improve your learning

Question 1.
What happens to the amount of energy transferred from one step to the next in a food chain?
Answer:

1. Energy is transferred along food chains from one trophic level to the next.
2. The amount of available energy decreases from one stage to the next.
3. This is because not all the food can be fully digested and assimilate.
4. Hair, feathers, insect exoskeletons, cartilage and bone in animal foods, cellulose and lignin in plant foods cannot be digested by most animals.
5. These materials are excreted or made into pellets of indigested remains.
6. Assimilated energy is available for the synthesis of new biomass through growth and reproduction.
7. Organisms also lose some biomass by death disease or annual leaf-drop.
8. Moreover at each tropic level, organisms use the most of the assimilated energy to fulfill their metabolic requirements – performance of work, growth and reproduction.
9. Most of the energy is lost in the form of heat during biological processes.
10. Only a small fraction goes to the consumer at next tropic level.

Question 2.
What do pyramids and food chain indicate in an ecosystem?
Answer:

1. The ecologists used the idea of pyramid to show relationship among organisms in an existing food chain.
2. Ecological pyramids are of three types. They are pyramid of biomass, pyramid of number, pyramid of energy.
3. Pyramid of biomass indicates the available biomass in an ecosystem; pyramid of number indicates the organisms present and pyramid of energy indicates the available energy in an ecosystem.
4. The food chain in an ecosystem indicates how energy is transferred from one organism to another.
5. The starting point of a food chain are producers and it ends with top carnivores.
6. A food chain represents a single directional transfer of energy.

Question 3.
Write a short note on pyramid of number for any food chain. What can we conclude from this pyramid of numbers?
i) tree ii) insect iii) woodpecker
(OR)
What is a pyramid of numbers? Write a brief note on the pyramid of numbers with the help of a block diagram.
Answer:
a.

1. The number of organisms in a food chain can be represented graphically in a pyramid of number.
2. Each bar represents the number of individuals at each tropic level in a food chain.
3. In the pyramid of numbers, from the first – order consumers to the large carnivores, there is normally an increase in size, but decrease in number.
4. For example, in a wood, the aphids are very small and occur in astronomical numbers.
5. The ladybirds which feed on them are distinctly larger and not so numerous.
6. The insectivorous birds which feed on the ladybirds are larger still and are only present in a small numbers, and there may only be a single pair of hawks of much larger size than the insectivorous birds on which they prey.

b.

1. In the given pyramid, the producer is a large tree, primary consumers are small
   insects which are numerous in number and secondary consumers are woodpeckers which are comparatively less in number than insects.
2. From this pyramid of number, we can conclude that sometimes the pyramid of numbers does not look like a pyramid at all.
3. This could happen if the producer is a large plant or if one of the organisms at any tropic level is very small.
4. Whatever the situation, the producer still goes at the bottom of the pyramid.

Question 4.
What is biomass? Draw a pyramid of biomass for the given food chain.
i) grass leaves ii) herbivores iii) predators iv) hawk
Answer:

1. Biomass is organic material of biological origin that has ultimately derived from the fixation of carbon dioxide and the trapping of solar energy.
2. This includes trees, shrubs, crops, grasses, algae, aquatic plants* agricultural and forest residues and all forms of human, animal and plant waste.
3. Any type of plant or animal material that can be converted into energy is called “Biomass”.
4. 

Question 5.
How is using of toxic material affecting the ecosystem? Write a short note on bioaccumulation and biomagnifications.
Answer:

1. Use of toxic materials such as pesticides, herbicides and fungicides creates new problems in the ecosystem.
2. As these toxic materials are often indiscriminate in their action and vast numbers of other animals may be destroyed.
3. Some of them may be predators which naturally feed on these pests, others may be the food for other animals.
4. Thus causing unpredictable changes in food chains and upsetting the balance within the ecosystem.
5. Some toxic substances have a cumulative effect.
6. Some of them are degradable, can be broken down into harmless substances in a comparatively short time usually a year.
7. Others are non-degradable which are potentially dangerous as they accumulate in the bodies of animals and pass right through food web.
8. This process of entering of pollutants in a food chain is known as “Bioaccumulation”.
9. The tendency of pollutants to concentrate as they move from one tropic level to the next is known as “Biomagnifications”.

Question 6.
Should we use pesticides as they prevent our crop and food from pests or we should think of alternatives? Write your view about this issue and give sound reason for your answer.
(OR)
Why should we think of alternatives to pesticides? Give reasons.
Answer:

1. We should think of alternatives. This is because these pesticides are toxic chemical whose usage leads to Bioaccumulation and Biomagnifications.
2. When we use pesticides, they prevent our crop and food from pests effectively but indiscriminately destroys a vast number of other animals.
3. This is causing unpredictable changes in food chains and upsetting the balance within the ecosystem.
4. Most of the chemical pesticides that contain mercury, arsenic or lead are non -degradable.
5. They enter into food chain, accumulate in the bodies of animals and pass right through food web.
6. Being further concentrated at each step until animals at the top of the pyramid may receive enough to do considerable harm.
7. This is one of the reasons for ever decreasing number of butterflies, bees, small and large birds.
8. Some of the pesticides are nerve poisons and might bring about changes in behaviour.
9. As the human beings are at the end of the food chain, these pesticides may get accumulated in our bodies also. This shows some adverse effects on us, when their concentration becomes sufficiently high.

Question 7.
What is a tropic level? What does it represent in an ecological pyramid?
Answer:

1. The various steps in a food chain at which the transfer of food takes place is called tropic level.
2. Tropic level means the feeding level of the organism.
3. In an ecological pyramid, the first tropic level represents the primary producers, and their number, biomass or energy.
4. Second tropic level represents the herbivores or primary consumers and their number, biomass or energy.
5. The tropic level represents the lower carnivores or secondary consumers and their number, biomass or energy.
6. The fourth tropic level represents the higher carnivores or tertiary consumers and their number, biomass or energy.

Question 8.
If you want to know more about the flow of energy in an ecosystem, what questions do you ask?
Answer:
I will ask the following questions to know more about flow of energy in an ecosystem.

1. How does the energy flow in an ecosystem from one organism to other?
2. Is the energy transformation from one level to other 100% efficient?
3. What per cent of energy transfers from one level to other?
4. What happens to the remaining energy?
5. How does the ecosystem lose its energy during energy transformation?
6. Which tropic level in an ecosystem has more energy and which has less?
7. What is the ultimate source of energy in an ecosystem?

Question 9.
What will happen if we remove predators from food web?
Answer:

1. Removal of organisms from any tropic level of a food chain or food web disturbs the ecosystem and leads to ecological imbalance.
2. If we remove predators from food web, the prey population will increase enormously as there is no natural control over them.
3. The producers population will decrease rapidly as the organisms feeding on them increase.
4. After few generations the prey population also begins to decrease as some of the preys begin to die due to starvation.
5. Some adaptations may also be developed by the organisms to bring the ecological balance.
6. But it may take some generations, till that the ecosystem will be disturbed and imbalanced.
7. For example, if we remove all the predators (carnivorous) from a forest ecosystem, the herbivorous animal population will increase as there are no carnivores to hunt them.
8. As a result plant population will decrease as the ever increasing herbivores feed more and more on plants.
9. After some generations the herbivore population begins to decrease as the decreasing number of plants are not sufficient to feed.
10. Then some herbivorous animals may adapt to feed on other herbivores to increase their survival.
11. Then scope for survival will increase for plants again which leads to ecological balance.
12. But this may take lot of time to evolve new predators and to form ecological balance.

Question 10.
Observe a plant in your kitchen garden, and write a note on producer-consumer relationship.
Answer:
When I observe a plant in kitchen garden, I came to know the following things.

1. Though it may be relatively small, a garden is a complete ecosystem.
2. It has the same components as any other large and elaborate ecosystems had.
3. The plant in a kitchen garden is a producer as it produces their own food from sunlight.
4. There are two types of consumers in this ecosystem, a) Primary consumers and b) Secondary consumers.
5. Primary consumers feed on plants. This tropic level consists of caterpillars, bees and butterflies.
6. Secondary consumers feed on primary consumers. This tropic level consists of birds, garden lizards and spiders.
7. Fungi, bacteria, insects and worms make up decomposers.
8. The producer and consumer relationship can be shown in the following food chain.
   Plant → Plant eaters such as caterpillars, bees, butterflies → Meat-eaters such as birds, garden lizards, spiders
   Producers → Primary consumers → Secondary consumers
9. The pyramid of number appears like this.
   
10. The pyramid of Biomass appears like this
    
11. The pyramid of energy appears like this
    

Question 11.
What type of information do you require to explain the pyramid of biomass?
Answer:
To explain the pyramid of biomass, we require the following information.

1. The type of ecosystem.
2. Producers in the ecosystem.
3. Primary consumers in the ecosystem.
4. Secondary consumers in the ecosystem.
5. Tertiary consumers in the ecosystem.
6. Number of organisms at each tropic level.
7. Size of organisms at each tropic level.
8. Weight of organisms at each tropic level.
9. All forms of waste produced at each tropic level and
10. In total, total amount of biomass produced at each tropic level.

Question 12.
Draw a pyramid of numbers considering yourself sis top level consumer. Pyramid of numbers
Answer:
Pyramid of numbers
Ex: 1

Ex: 2

Question 13.
Prepare slogans to promote awareness in your classmates about eco-friendly activities.
All the living things have the right to live on this earth along with us. Prepare slogans to promote awareness in public about the conservation of biodiversity.
(OR)
Which slogans do you prefer to promote awareness in your locality about eco-friendly activities?
Answer:

1. Live and let live.
2. If we protect the environment, it protects us.
3. Conserve nature – Conserve life.
4. Save mother earth.
5. Earth needs you.
6. Go ecofriendly.
7. Clean the environment, live happily.
8. Heal our planet! Turn it into a better planet.
9. Plant a tree for your environment.
10. Think ecofriendly and live ecofriendly.
11. Earth enables you to definitely stand. Allow it to stand the actual way it is.
12. You’ve only got one planet. Don’t trash it.

Question 14.
Suggest any three programmes on the prevention of soil pollution in view of avoiding pesticides.
(OR)
Suggest any four eco-friendly methods for prevention of soil pollution in view of avoiding pesticides. (OR)
In your area, soil is polluted by the enormous usage of pesticides. Suggest any two programmes for the prevention of soil pollution.
Answer:
To prevent the soil pollution caused by pesticides following programmes should be implemented.

1. Rotation of crops :
   1. Same crop should not be grown in the same field in successive seasons.
   2. Rotation of crops reduce occurance of pests and damage due to pests will be decreased.
2. Biological control: Introducing natural predator or parasite of the pest.
3. Sterility: Sterilising the males of a pest species reduces the population of pests.
4. Genetic strains: The development of genetic strains which are resistant to certain pest.
5. Studying the life histories of the pests: When this is done it is sometimes possible to sow the crops at a time when least damage will be caused.

Choose the correct answer.

1. What does a food chain always start with?
   A) The herbivore
   B) The carnivore
   C) The producer
   D) None of these
   Answer: C
2. Which of the following do plants not compete for?
   A) Water
   B) Food
   C) Space
   D) All the above
   Answer: B
3. Ban all pesticides, this means that
   A) Control on the usage of pesticides
   B) Prevention of pesticides
   C) Promote eco-friendly agricultural practices
   D) Stop biochemical factories
   Answer: C
4. According to Charles Elton
   A) Carnivores at the top of the pyramid herbivore
   B) Energy trapping is high at the top of the pyramid
   C) No producers at the top of the pyramid
   D) A and C
   Answer: D

10th Class Biology 9th Lesson Our Environment InText Questions and Answers

Question 1.
Are all terrestrial ecosystems similar?
Answer:

1. No. All the terrestrial ecosystems are not similar.
2. Basing on variations in climatic conditions such as rainfall, temperature and the availability of light, there are various kinds of ecosystems.
3. The major types of terrestrial ecosystem are
   1. Tundra,
   2. Coniferous forest,
   3. Deciduous forest,
   4. Savannah,
   5. Tropical forest and
   6. Deserts.

Question 2.
If we want to show a food chain consisting of grass, rabbit, snake and hawk then connect the given picture of organisms by putting arrows and make a food chain.
A) Name the producers and consumers in the above food chain.
B) Try to guess what does the arrows marked by you are indicate?
C) Identify at least four other food chains from your surroundings. Name the producers and different levels of consumers in those food chains.
Answer:
Grass → Rabbit → Snake → Hawk
A) In the above food chain grass is the primary producer. Rabbit is the primary consumer, snake is the secondary consumer and hawk is the tertiary consumer.
B) The arrows indicate the flow of energy from one organism to another. So these are always pointed from the food to the feeder.
C)

1. Plant → insect → frog → bird
2. Plant → insect →  frog → snake
3. Aquatic plants → insects → fish → crane
4. Plant → mice → snake → vulture
5. Plant → aphids → spiders → birds

Question 3.
Why do most of the food chains consists of four steps?
Answer:

1. Most of the food chains are quite short and mostly consists of four steps.
2. This is because only 10% of the energy present in a tropic level transfers to the other tropic level.
3. Remaining energy is dissipated as heat produced during the process of respiration and other ways.
4. Thus about three steps in a food chain very little energy is still available for use by living organisms.

Question 4.
Why do the number of organisms get decreased as we move from producer to different level of consumers?
Answer:

1. As we move from producers to different levels of consumers the energy available will decrease gradually.
2. Only ten per cent of the energy present in one tropic level transfer to another tropic level.
3. Biomass also decreases gradually as only 10 – 20% of the biomass is transferred from one tropic level to the next in a food chain.
4. As there is less energy & less biomass available at top levels, number of organisms also less, generally.
5. So, the number of organisms get decreased as we move from producer to different level of consumers.

Question 5.
Draw the pyramid of number for the following food chains.
i) Banyan → insects → woodpecker
ii) Grass → rabbit → wolf
A) Are the pyramid of number having same structure in both of the above two cases as compare to the example given in the earlier paragraph?
B) If there is a difference, then what it is?
Answer:

A) No. The pyramid of number in the above two cases doesn’t have the same structure as compared to the example given in the textbook.
B)

1. In the example (given in the textbook), number of organisms at producers level is more. This number gradually decreased in consumers level step by step. So the pyramid of number formed has typical pyramid shape with broad base and the narrow apex.
2. But in the first case given here, on a single Banyan tree, a large number of insects live and feed. These insects become food for few Woodpeckers. So producers number is less than primary and secondary consumers, and secondary consumers are less than primary consumers. So the pyramid of number does not look like a pyramid. It consists of narrow base, broad middle part and medium apex.
3. In the second case, grass which are large in number become food for few rabbits. Rabbit provides food for several wolves which are comparatively less in number than grass. So primary consumers are less in number than secondary consumers and producers. So the pyramid of number for this food chain also does not look like a pyramid. It consists of broad base, narrow middle part and medium apex. Thus it differs from case (i) also.

Question 6.
Think why the pyramids are always upright?
Answer:

1. In ecology not all the pyramids are always upright.
2. Pyramid of number may be upright, inverted or partly upright.
3. Pyramid of biomass may be upright or inverted.
4. But the pyramid of energy is always upright.
5. This is because energy will decrease when we move from producers to the high level consumers.
6. Only 10% of the energy from one tropic level transfers to the other through food chain.
7. So the energy at base is more, gradually decreases, and very less at the top.
8. As a result the energy pyramid is always upright.

Question 7.
Observe the data given in the following table.

i) In which year lake-water spread area is more? Why?
Answer:
In the year 1967. Because lake was not brought under cultivation.

ii) How do you think weeds are more in the lake?
Answer:
Excessive nutrient addition, especially from anthropogenic sources, led to explosive weed growth. Ex: Eichornia, pistia.

iii) What are the reasons for decrease in lake area?
Answer:

1. In 1996, almost entire lake was brought under cultivation.
2. Industries came along in ever growing intensity in the catchment area of the lake.

iv) How do the above reasons lead to pollution?
Answer:

1. Consequently, the drains and rivulets carry substantial quantity of various types of pollutants into the lake.
2. The major sources of pollution are agricultural runoff containing residues of several agrochemicals, fertilizers, fish tank discharges, industrial effluents containing chemical residues.

v) How was the threat to the lake due to pollution discovered?
Answer:

1. The water of the lake turned alkaline in nature, turbid, nutrient rich, low in dissolved oxygen and high in biochemical oxygen demand.
2. Water borne diseases like diarrhoea, typhoid, amoebiasis and others are said to be common among the local inhabitants who are unaware of the state of pollution in the lake water.
3. Vector borne diseases were also increased.

vi) What could be the reasons for the migration of birds to this lake?
Answer:
To avoid extreme cold weather conditions in Northern Asia and Eastern Europe birds migrate to Kolleru lake.

Question 8.
Observe the following table showing different activities in the lake and their influence.
Legend:
(+) means has influence on the mentioned problem
(-) means has no influence on the mentioned problem
i) What are the factors that affected the number of migratory birds to decrease?
Answer:
Aquaculture practices.

ii) Do you find any relationship between biological and physical problems?
Answer:
Yes. Aquaculture practices have influence on these problems.

iii) What are the reasons for chemical problems ?
Answer:
Agricultural practices, aquaculture practices, industrial activities and human activities are the reasons for chemical problems.

iv) What happens if the dissolved oxygen reduce in lake water ?
Answer:
If the dissolved oxygen reduces in lake water, sufficient amount of oxygen will not be available to organisms that live in the lake.
This leads to the death of organisms in the lake.

v) Is BOD of turbid and nutrient rich water high or low? What are its consequences?
Answer:
High. Its consequences are water borne diseases and death of organisms.

vi) People living in catchment area of Kolleru faced so many problems. Why?
Answer:
Vector borne disease increased. The lands adandoned are useless for agriculture.

Question 9.
Name any two pesticides / insecticides you have heard about.
Answer:
DDT, Aldrin, Malathian, Altrazine, Monocrotophos, Endosulphan etc.

Question 10.
How are the food grains and cereals being stored in your house and how dojyou protected them from pests and fungus?
Answer:
To protect food grains and cereals from pests and fungus, we will follow the following rules in our house.

1. First of all we will dry and clean our grain before storing.
2. We will avoid moisture in bagged grains by storing them on wooden structures, bamboo mats or polythene covers.
3. We use domestic bins or improvised storage structures such as Gaade, Kotlu, Paatara, RCC bins and flat bottom metal bins etc.
4. We fumigate the storage room with Ethylene Di-bromide (EDB) ampoules to avoid insect damage.
5. We use anticoagulant for rat control in houses.

Question 11.
Where from pollutants enter to the water sources?
Answer:
The used water from industries and run off water containing agricultural effluents bring pollutants into water sources. Municipal and domestic sewage also pollute water sources.

Question 12.
How can you say fishes living in water having heavy metals in their bodies?
Answer:
The bioaccumulation of heavy metals in tissues of fish particularly in liver, kidney and gills were analysed and found their presence.

Question 13.
Researchers found that pollution levels increase during monsoon season. Why they found so?
Answer:

1. Pollution levels increase during monsoon season in water bodies.
2. During monsoon season heavy rainfall occurs.
3. The rain water brings residues of agrochemicals, fertilizers and different types of organic substances, municipal and domestic sewage.
4. Hence pollution levels increase in monsoon season.

Question 14.
Why did people also suffer from various diseases after consuming fishes living in local water reservoir?
Answer:

1. The heavy metals could find their way into human beings through food chain.
2. This bioaccumulation cause various physiological disorders such as hypertension, sporadic fever, renal damage, nausea etc.

Question 15.
What is the food chain that has been discussed in the above case?
Answer:
The food chain discussed in the above occurrence is Crops → Locust → Sparrow → Hawk.

Question 16.
How did the campaign disturb the food chain in the fields?
Answer:

1. Crop yields after the campaign were substantially decreased.
2. Though the campaign against sparrows ended it was too late.
3. With no sparrows to eat the locust populations, the country was soon swarmed.
4. Locust coupled with bad weather led to the great Chinese famine.

Question 17.
How did these disturbances affect the environment?
Answer:

1. The number of locust increased.
2. Use of pesticides against locust population further degraded the land.

Question 18.
Is it right to eradicate a living organism in an ecosystem? How is it harmful?
Answer:

1. No, it is not right to eradicate a living organism in an ecosystem.
2. It disturbs the existing food chain.

Question 19.
Were the sparrows really responsible? What was the reason for the fall in crop production?
Answer:

1. No, the sparrows were not really responsible for the loss of food grain.
2. With no sparrows to eat the locust population crops were damaged and this led to fall in crop production.

Question 20.
What was the impact of human activities on the environment?
Answer:

1. The human activities badly affected the environment.
2. Use of pesticides against the pest degraded the land.

Question 21.
What do you suggest for such incidents not to occur?
Answer:

1. I suggest to use organic manures and organic insecticides to kill the insects.
2. Rotation of crops is the best method to protect the crops from pests.
3. We should not kill any organism on this earth because every organism has a role to play.
4. Think before you start action.

10th Class Biology 9th Lesson Our Environment Activities

Activity – 1

Observe any water ecosystem in your surroundings and identify the different food chains and food web operating in this ecosystem. Write the following details in your notebook.

WORKSHEET

1. Names of the students in a group: ——————— Date: ———–
2. Name of the ecosystem: ———————
3. Topography: ———————
4. Names / Number of plants (producers) identified: ———————
5 Names / Number of animals identified: ———————
6. Identify the different types consumers and name them & mention their number below :
Herbivores (Primary consumers): ———————
Carnivores (Secondary consumers): ———————
Top carnivores (Tertiary): ———————
7 Food relationships among them: food habits/preferences: ———————
8 Show / draw the different food chains: ———————
9. Showcase the food web: ———————
10. List out all abiotic factors existing in the ecosystem: ———————
( A check list can be given, and asked to tick)
11. Is there any threat to the ecosystem ? Yes / No ———————
If yes, what ? and how ? ———————
Suggest few remedial measures ———————
Answer:
Student’s Activity.

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 5th Lesson Refraction of Light at Plane Surfaces

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Textbook Questions and Answers

Improve Your Learning

Question 1.
Why is it difficult to shoot a fish swimming in water? (AS1)
(OR)
If the fish is swimming in water, why it is difficult to shoot?
(OR)
A shooter finds it difficult to shoot a fish swimming in water. Why?
Answer:
Due to refraction of light, it is difficult to shoot a fish swimming in water.

Reason :
The light rays coming from the fish towards shooter, bend at water-air interface. So, shooter sees only image of the fish, but not actual fish.

Question 2.
The speed of light in a diamond is 1,24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s. (AS1)
Answer:
Speed of light in diamond = 1,24,000 km/s
Speed of light in vacuum = 3,00,000 km/s

Question 3.
Refractive index of glass relative to water is 9/8. What is the refractive index of water relative to glass? (AS1)
Answer:
Refractive index of glass relative to water = \(\frac{n_{g}}{n_{w}}=\frac{9}{8}\)
∴ Refractive index of water relative to glass = \(\frac{\mathrm{n}_{\mathrm{w}}}{\mathrm{n}_{\mathrm{g}}}=\frac{8}{9} \cdot\left[\because \mathrm{n}_{12}=\frac{1}{\mathrm{n}_{21}}\right]\)

Question 4.
The absolute refractive index of water is 4/3. What is the critical angle ? (AS1)
Answer:
Absolute refractive index of water = 4/3

Critical angle of water = C = 48°5′ = 48.5°.

Question 5.
Determine the refractive index of benzene if the critical angle is 42°. (AS1)
Answer:
Critical angle of benzene = 42°.

Question 6.
Explain the formation of mirage. (AS1)
(OR)
How is the mirage formed? Explain.
(OR)
A person walking on a road observed some water being present on the road but when he went there actually he did not find water. What is that actually formed called? Explain that process.
(OR)
Sometimes during the hot summer at noon time on tar roads, it appears that there is water on the road, but there would really be no water. What do you call this phenomenon? Explain why it happens.
(OR)
Why do you see a mirage the road on a hot summer day?
Answer:

• During hot summer day, air just above the road surface is very hot and the air at higher altitudes is cool.
• We know that refractive index of air increases with density.
• So, the cooler air at the top has greater refractive index than hotter air just above the road.

• Light travels faster through the thinner hot air than the denser cool air above it.
• On hot days, the temperature decreases with height.
• Thus the refractive index of air increases with height.
• When the light from a tall object such as tree or from the sky passes, through a medium just above the road whose refractive index decreases towards ground, suffers refraction and takes a curved path because of total internal reflection.
• This refracted light reaches the observer in a direction shown as in second figure.
• This appears to the observer that the ray is reflected from ground.
• Hence we will see water on road, which is the virtual image of sky and an inverted image of tree on the road.
• Such virtual images of distant high objects cause the optical illusion called ‘mirage’.

Question 7.
How do you verify experimentally that \(\frac{\sin i}{\sin r}\) is a constant? (AS1)
(OR)
Explain the experiment that shows the relation between angle of incidence and angle of refraction through figure.
(OR)
Write an experiment to obtain the relation between angle of incidence and angle of refraction.
Answer:
Aim:
Identifying relation between angle of incidence and angle of refraction.

Materials required :
A plank, white chart, protractor, scale, small black painted plank, a semi-circular glass disc of thickness nearly 2 cm pencil and laser light.

Procedure :

1. Take a wooden plank which is covered with white chart.
2. Draw two perpendicular lines, passing through the middle of the paper as shown in the figure (a).
3. Let the intersecting point be O.
4. Mark one line as NN which is normal to the another line marked as MM.
5. Here MM represents the line drawn along the interface of two media and NN represents the normal drawn to this line at ‘O’.
6. Take a protractor and place it along NN in such a way that its centre coincides with ‘O’ as shown in figure (b).
7. Then mark the angles from 0° to 90° on both sides of the line NN.
8. Repeat the same on the other side of the line NN.
9. The angles should be represented on circular line.
10. Now place semi circular glass disc so that its diameter coincides with the interface line (MM) and its centre coincides with the point O.
11. Take the laser light and send it along NN in such a way that the laser propagates from air to glass through the interface at point O and observe the way of laser light coming from other side of disc.
12. There is no deviation.
13. Send Laser light along a line which makes 15° (angle of incidence) with NN and see that it must pass through point O.
14. Measure its corresponding angle of refraction.
15. Repeat the experiment with angle of incidences of 20°, 30°, 40°, 50° and 60° and note the corresponding angles of refraction.

Observation :

• Find sin i, sin r for every i and r note down the values in table.

• Evaluate \(\frac{\sin i}{\sin r}\) for every incident angle i.
• We get \(\frac{\sin i}{\sin r}\) as constant.
• That is the relationship between angle of incidence and angle of refraction.
• The ratio of sin i and sin r is called refractive index.

Question 8.
Explain the phenomenon of total internal reflection with one or two activities. (AS1)
Answer:
Procedure :

1. Place the semi-circular glass disc in such a way that its diameter coincides with interlace line MM and its center coincides with point O’.
2. Now send light from the curved side of the semicircular glass disc.
3. The light travels from denser medium to rarer medium.
4. Start with angle of incidence (i), equals to 0° and observe for refracted on other side of the disc.
5. It does not deviate into its path when entering rarer medium.
6. Send laser light along angles of incidence 5°, 10°, 15°, etc. and measure the angle of refraction.
7. And tabulate the results in the given table.

Observation :

• Make a table shown below and note the values ‘i’ and ‘r’.

• At particular angle of incidence, the refracted ray does not come out but grazes the interface separating the air and glass. This angle is called critical angle.
• When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e. light never enters rarer medium. This phenomenon is called total internal reflection.

Question 9.
How do you verify experimentally that the angle of refraction is more than angle of incidence when light rays travel from denser to rarer medium? (AS1)
(OR)
When the light rays travel from denser to rarer medium, how can you explain, the angle of refraction is more than angle of incidence experimentally?
Answer:
Procedure :

• Take a metal disc. Use a protractor and mark angles along its edge as shown in the figure.
• Arrange two straws at the centre of the disc, in such a way that they can be rotated freely about the centre of the disc.
• Adjust one of the straws to make an angle 10°.
• Immerse half of the disc vertically into the water, filled in a transparent vessel. While dipping, verify that the straw at 10° must be inside the water.
• From the top of the vessel, try to view the straw which is inside the water as shown in the figure.
• Then adjust the other straw which is outside of the water until both straws look like they are in a single straight line.
• Then take the disc out of the water and observe the two straws on it. You will find that they are not in a single straight line.
• Measure the angle between the normal and second straw. Note the values in the following table.

• Do the same for various angles and find the corresponding angles of refraction and note them in the table.

Observation :
We will find the angle of refraction is more than angle of incidence.
i. e., r > i.

Conclusion :
When light travels from denser (water) to rarer (air) it bends away from the normal.

Question 10.
Take a bright metal ball and make it black with soot in a candle flame. Immerse it in water. How does it appear and why? (Make hypothesis and do the above experiment) (AS2)
Answer:

• The black metallic .ball appears to be raised up in the water because the path of the ray changes its direction at the interface, separating the two media, i.e., water and air.
• This path is chosen by light ray so as to minimize time of travel between ball and eye.
• This can be possible only when the speed of light changes at interface of two media.
• In another way the speed of light is different in different media.

Hypothesis :
Speed of light changes when it travels from one medium to another medium.

Question 11.
Take a glass vessel and pour some glycerine into it and then pour water up to the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel.
1) What changes do you notice?
2) What could be the reasons for these changes? (AS2)
Answer:

1. We cannot see the glass rod in glycerine but we can see the rod in water.
2. We can also observe an apparent image of glass rod in water.
3. Reasons:
   i) Glycerine has essentially same refractive index as glass.
   ii) So, any light passing through these is bent equally.
   iii) Since both are transparent, it is not possible for our eye to distinguish the boundary by a change in the angle of reflection, and the glass seems to vanish.
   iv) But, the refractive index of glass and water are different.
   v) So the glass rod is visible to our eye in water. .

Question 12.
Do Activity-7 again. How can you find critical angle of water? Explain your steps briefly. (AS3)
Answer:

Procedure:

1. Take a cylindrical transparent vessel.
2. Place a coin at the bottom of the vessel.
3. Now pour water until you get the image of the coin on the water surface.
4. This is due to total internal reflection.

Critical angle of water :

1. Refractive index of water = 1.33
2. The sine of critical angle of water = \(\frac{1}{\text { Refractive index }}\)
3. Sin C = \(\frac{1}{\text { 1.33 }}\) ⇒ sin C = 0.7518.
   ∴ C = 8.7°
4. ∴ The critical angle of water = 48.7°.

Question 13.
Collect the values of refractive index of the following media. (AS4)

Answer:

Question 14.
Collect information on working of optical fibres. Prepare a report about various uses of optical fibres in our daily life. (AS4)
(OR)
What do you know about the working of optical fibres and make a report of various uses of optical fibres in our daily life?
(OR)
How are the optical fibres working? What are the various uses of optical fibres in our daily life?
Answer:
1) An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer (10-6 m).
2) A bunch of such thin fibres form a light pipe.

Working :
1. Optical fibre having three parts-namely core (n = 1, 7), clading (n = 1, 6) and shielding.
2. The ray of light AB gets refracted at point ‘B’ into core and incident at ‘C’ with angle of incidence i (i > c).

3. The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
4. The light is thus transmitted along the fibre.
5. The optical fibre is also based on ‘Fermat’s principle.

Uses :

1. Optical fibres are used in ‘endoscopy’ to see the internal organs like throat, stomach, intestines, etc.
2. Optical fibres are used in transmitting communication signals through light pipes.
3. Optical fibres are used in international telephone cables laid under the sea, in large computer networks, etc.
4. In optical fibre about 2000 telephone signals appropriately mixed with light waves may be simultaneously transmitted through a typical optical fibre.

Question 15.
Take a thin thermocol sheet. Cut it in circular discs of different radii like 2 cm, 3 cm, 4 cm, 4.5 cm, 5 cm etc. and mark centers with sketch pen. Now take needles of length nearly 6 cm. Pin a needle to each disc at its centre vertically. Take water in a large opaque tray and place the disc with 2 cm radius in such a way that the needle is inside the water as shown in figure. Now try to view the free end (head) of the needle from surface of the water.

1) Are you able to see the head of the needle?
Now do the same with other discs of different radii. Try to see the head of the needle, each time.
Note : The position of your eye and the position of the disc on water surface should not be changed while repeating the Activity with’other discs.
2) At what maximum radius of disc, were you not able to see the free end of the needle ?
3) Why were you not able to view the head of the nail for certain radii of the discs ?
4) Does this Activity help you to find the critical angle of the medium (water) ?
5) Draw a diagram to show the passage of light ray from the head of the nail in different situations. (AS4)
Answer:

1. Yes, we can see head of the needle.

2. Height of the pin = 6 cm
Radius of the disc = x cm
Critical angle of water = C = 48.7°
Tan C = 48.7°
\(\frac{x}{6}\) = 1.138 ⇒ x = 6.828 cm
So, at radius of 6.8 cm we cannot see the free end of the needle.

3. Because the light rays coming from object undergoing total internal reflection by touching the surface of disc.

4. Yes, we can find critical angle.
Refractive index of air (n₂) = 1.003 ; Refractive index of water (n₁) = 1.33
Sin C = \(\frac{n_{2}}{n_{1}}=\frac{1.003}{1.33}\) = 0.7541 ⇒ C = 48.7°

5.

Question 16.
Explain the refraction of light through the glass slab with a neat ray diagram. (AS5)
(OR)
Draw a glass slab diagram and explain the refraction of light through glass slab.
(OR)
Write the procedure of a lab Activity to understand lateral shift of light rays through a glass slab.
(OR)
How can you find lateral shift using glass slab?
Answer:
Aim :
A) Determination of position and nature of image formed by a glass slab.
B) Understanding lateral and vertical shift.
C) Determination of refractive index of given glass slab.

Materials required :
Plank, chart paper, clamps, scale, pencil, thin glass slab and pins.

Procedure :

1. Place a piece of chart on a plank. Clamp it. Place a glass slab in the middle of the paper.
2. Draw border line along the edges of the slab by using a pencil. Remove it. You will get a figure of a rectangle.
3. Name the vertices of the rectangle as A, B, C and D.
4. Draw a perpendicular at a point on the longer sides (AB) of the rectangle.
5. Now draw a line, from the point of intersection where side AB of rectangle and perpendicular meet, in such a way that it makes 30° angle with the normal.
6. This line represents the incident ray falling on the slab and the angle it makes with normal represents angle of incidence.
7. Now place the slab on the paper in such a way that it fits in the rectangle drawn. Fix two identical pins on the line making 30° angle with normal, such that they stand vertically with same height.
8. By looking at the two pins from the other side of the slab, fix two pins in such a way that all pins appear to be along a straight line.
9. Remove the slab and take out pins. Draw a straight line by joining the dots formed by the pins up to the edge CD of the rectangle.
10. This line represents emergent ray of the light.
11. Draw a perpendicular to the line CD where our last line drawn meets the line CD.
12. Measure the angle between emergent ray and normal.
13. This is called angle of emergence.
14. The angle of incidence and angle of emergence are equal.
15. Incident emergent rays are parallel.
16. The distance between the parallel rays is called shift.

Question 17.
Place an object on the table. Look at the object through the transparent glass slab. You will observe that it will appear closer to you. Draw a ray diagram to show the passage of light ray in this situation. (AS5)
Answer:

Question 18.
What is the reason behind the shining of diamond and how do you appreciate it? (AS6)
(OR)
For which reason is the diamond shining and how is it appreciable?
Answer:

• The critical angle of diamonds is very low, i.e., 24.4°.
• So if a light ray enters diamond, it undergoes total internal reflection.
• It makes the diamond shine brilliant.
• So total internal reflection is main cause of brilliance of diamonds.
• Majority of people are attracted towards diamonds due to this property.
• So we have to thoroughly appreciate total internal reflection for brilliance of diamonds.

Question 19.
How do you appreciate the role of Fermat’s principle in drawing ray diagrams? (AS6)
(OR)
How do you admire the role of Fermat’s principle in drawing ray diagrams? Fermat’s principle: The light ray always travels in a path which needs shortest possible time to cover distance between two points.
This principle has lot of importance on optics. This is used in

1. Laws of reflection (i.e., angle of incidence = angle of reflection)
2. Laws of refraction (Snell’s law)
3. To derive refractive index of a medium.
4. To derive refractive index of glass slab.
   So, I appreciate the Fermat’s principle.

Question 20.
A light ray is incident on air-liquid interface at 45° and is refracted at 30°. What is the refractive index of the liquid? For what angle of incidence will the angle between reflected ray and refracted ray be 90°? (AS7)
Answer:
i) Given that angle of incidence (i) = 4S°
angle of refraction (r) = 30°

ii) Given that angle between reflected and refracted ray is 90°.
We know angle of incidence = angle of reflection
∴ Angle of refraction (r) = 90 – angle of incidence
= 90 – i

Critical angle = 54.7°. This angle is also known as polarising angle.

Question 21.
Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a mirror surface from a certain viewing position. (AS7)
Answer:
When a test tube is immersed at a certain angle in a tumbler of water appears to have
a mirror surface from a certain viewing positions due to total internal reflection.

Explanation :

• The critical angle for glass is 42°.
• The glass and air in test tube works as denser and rarer mediums.
• The rays of light while travelling through water strike glass – air interface of test tube at an angle of more than 42° (i > c) they get totally internal reflected as shown figure.
• When these reflected rays reach the eye, they appear to come from the surface of test tube itself.
• Now the test tube appears like silvary.

Question 23.
In what cases does a light ray not deviate at the interface of two media? (AS7)
Answer:

1. When a light ray incident is perpendicular to the interface of surface, it does not undergo deviation.
2. When a light ray incident is more than critical angle, it does not undergo deviation (refraction) but it undergoes reflection to come back into the original medium.

Question 25.
When we sit at camp fire, objects beyond the fire seen swaying. Give the reason for it. (AS7)
(OR)
What are the reasons for the objects beyond the fire seen swaying, when we sit at camp fire?
Answer:

• The temperature of the surrounding air changes due to convection of heat by the camp fire.
• This leads to chang in density and refractive index of air, continuously.
• The continuous change in refractive index of air changes the refracted path of the light ray.
• This is the cause for swaying of an object.

Question 26.
Why do stars appear twinkling? (AS7)
(OR)
What is the reason for the appearance of stars like twinkling?
Answer:

• The twinkling of a star is due to atmospheric refraction of star light.
• The atmosphere consists of a number of layers of varying densities.
• When light rays coming from a star pass through this layers and undergo refraction for several times.
• Thats why stars appear twinkling.

Question 27.
Why does a diamond shine more than a glass piece cut to the same shape? (AS7)
(OR)
What is the reason for shining of diamond brightly as compared to glass piece cut?
Answer:

• The critical angle of a diamond is very low (i.e., 24.4°).
• So if a light ray enters a diamond it definitely undergoes total internal reflection.
• Whereas it is not possible with glass piece cut to the same shape.
• So diamond shines more than a glass piece.

Fill In The Blanks

1. At critical angle of incidence, the angle of refraction is ……………… .
2. n₁ sin i = n₂ sin r, is called ……………… .
3. Speed of light in vacuum is ……………… .
4. Total internal reflection takes place when a light ray propagates from …………. to …………… medium.
5. The refractive index of a transparent material is 3/2. The speed of the light in that medium is …………… .
6. Mirage is an example of ……………… .
Answer:

1. 90°
2. Snell’s law
3. 3 × 10⁸ m/s
4. denser, rarer
5. 2 × 10⁸ m/s
6. optical illusion / total internal reflection

Multiple Choice Questions

1. Which of the following is Snell’s law?

Answer:
B)

2. The refractive index of glass with respect to air is 2. Then the critical angle of glass air interface is ………………….
A) 0°
B) 45°
C) 30°
D) 60°
Answer:
C) 30°

3. Total internal reflection takes place when the light ray travels from …………….. .
A) rarer to denser medium
B) rarer to rarer medium
C) denser to rarer medium
D) denser to denser medium
Answer:
C) denser to rarer medium

4. The angle of deviation produced by the glass slab is …………… .
A) 0°
B) 20°
C) 90°
D) depends on the angle formed by the light ray and normal to the slab
Answer:
D) depends on the angle formed by the light ray and normal to the slab

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Additional Questions and Answers

Question 1.
Derive Snell’s law.
(OR)
Prove n₁ sin i = n₂ sin r.
(OR)
Derive the Snell’s formula from Fermat’s principle.
(OR)
Derive the formula in realtion with and of incidence and angle of refraction.
Answer:
Let X be the path and A be the point above X and B be the point below X.
Now, we have to find the way from A to B.

1. Let us try to calculate how long it would take to go from A to B by the two paths through point D and another through point C.
2. If we draw a perpendicular DE, between two paths at D, we see that the path on line is shortened by the amount EC.
3. On the other hand, in the water, by drawing corresponding perpendicular CF we find that we have to go to the extra distance DF in water. These times must be equal since we assumed there was no change in time between two paths.
4. Let the time taken by the man to travel from E to C and D to F be ∆t and v₁ and v₂ be the speeds of the running and swimming. From figure we get,
   EC = v₁ ∆t and DF = v₂ ∆t
   ⇒ \(\frac{\mathrm{EC}}{\mathrm{DF}}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\) ………….. (1)
5. Let i and r be the angles measured between the path ACB and normal NN, perpendicular to shore line X.

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces InText Questions and Answers

10th Class Physics Textbook Page No. 56

Question 1.
Why should you see a mirage as a flowing water?
Answer:

• A mirage is a naturally occuring optical phenomenon, in which light rays are bent to produce a displaced image of distant objects or the sky.
• As light passes from colder air (higher place) to warmer air (lower place), the light ray bends away from the direction of the temperature gradient.
• Once the rays reach the viewer’s eye, the visual cortex interprets it as if it traces back along a perfectly straight “line of sight”. However this line is at a tangent to the path the ray takes at the point it reaches the eye.
• The result is that an “inferior image ” of the sky above appears on the ground.
• The viewer may incorrectly interprets this sight as water that is reflecting the sky, which is to the brain, a more reasonable and common occurrence.

Question 2.
Can you take a photo of a mirage?
Answer:

• Yes, I can take a photo of a mirage.
• Our eye can catches the total internal reflected rays.
• So, camera lens also catches the same.

10th Class Physics Textbook Page No. 46

Question 3.
What difference do you notice in fig 2(a) and Fig 2(b) with the respect to refracted rays?
(OR)
Draw the ray diagram of refraction in between denser and rarer medium.

Answer:
In figure 2(a) the light ray bends towards normal whereas in 2(b) the light ray bends away from the normal.

Question 4.
Is there any relation between behaviour of refracted rays and speed of the light?
Answer:
Yes. The speed of light changes when it travels from one medium to another medium. So the light may bend towards normal or away from normal.

10th Class Physics Textbook Page No. 47

Question 5.
Why db different material media possess different values of refractive indices?
Answer:
Refractive index depends on nature of material. So different media have different values of refractive indices.

10th Class Physics Textbook Page No. 48

Question 6.
On what factors does the refractive index of a medium depend?
Answer:
Refractive index depends on (1) Nature of material and (2) Wavelength of light used.

10th Class Physics Textbook Page No. 49

Question 7.
Can we derive the relation between the angle of incidence and the angles of refraction theoretically?
Answer:
Yes, we can derive the relation between angle of incidence and angles of refraction theoretically. We can get nt sin i = n2 sin r.

10th Class Physics Textbook Page No. 53

Question 8.
Is there any chance that angle of refraction is equal to 90° ? When does this happen?
Answer:
Yes, when angle of incidence is equal to critical angle then angle of refraction is equal to 90°.

10th Class Physics Textbook Page No. 54

Question 9.
What happens to light when the angle of incidence is greater than critical angle?
Answer:
When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e., light never enters rarer medium. This phenomenon is called total internal reflection.

10th Class Physics Textbook Page No. 57

Question 10.
How does light behave when a glass slab is introduced in its path?
Answer:
The light ray undergoes refraction two times.

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Activities

Activity – 1

Question 1.
Procedure :
Take some water in a glass tumbler. Keep a pencil in it. See the pencil from one side of glass and also from the top of the glass.
Observation:

1. How does it look?
Answer:
From the side it appears to be bent. From the top it appears as it is raising.

2. Do you find any difference between two views?
Answer:
Yes, the position of pencil is different.

Activity – 2

Question 2.
Procedure :

1. Go to a long wall (of length of 30 feet) facing the Sun. Go to the one end of a wall and ask someone to bring a bright metal object near the other end of the wall.
2. When the object is few inches from the wall, it will distort and we will see a reflected image on the wall as though the wall were a mirror.

Observation:

Why is there an image of the object on the wall?
Answer:
The image is due to refraction of light.

Activity-3 Refraction

Question 3.
Procedure: –

1. Take a shallow vessel with opaque walls such as a mug, a tin or a pan.
2. Place a coin at the bottom of the vessel.
3. Move away from the vessel until we cannot see the coin (fig. 2). Ask someone to fill the vessel with water. When the vessel is filled with water the coin comes back into view (fig. 3).

1. Why are you able to see the coin when the vessel is filled with water?
Answer:
The ray of light originated from the coin does not reach your eye when the vessel is empty. Hence you are not able to see the coin. But the coin becomes visible after the vessel is filled with water.

2. How is it possible? Do you think that the ray reaches your eye when the vessel is filled with water?
Answer:
Yes, it reaches the second instance.

3. What happens to the light ray at interface between water and air?
Answer:
It bends towards the normal.

4. What could be the reason for this bending of the light ray in the second instance?
Answer:
It is based on Fermat’s principle, which states that the light ray always travels in a path which needs shortest possible time to cover the distance between the two points.

Activity – 4

Question 4.
Prove that when light ray travels from denser to rarer medium it bends away from the normal.
Answer:


Procedure :

1. Take a metal disc. Use protractor and mark angles along its edge as shown in the figure.
2. Arrange two straws from the centre of the disk.
3. Adjust one of the straws to the angle 10°.
4. Immerse half of the disc vertically into the water, filled in a transparent vessel.
5. Inside the water the angle of straw should be at 10°.
6. From the top of the vessel try to view the straw which is inside the water.
7. Then adjust the other straw which is outside the water until both straws are in a single straight line.
8. Then take the disc out of the water and observe the two straws on it.
9. We will find that they are not in a single straight line.
10. It could be seen from the side view while half of the disc is inside the water.
11. Measure the angle between the normal and second straw. Draw table for various angles and corresponding angles of refraction.

Observation :
We observe that ‘r’ is greater than ‘i’ in all cases and when light travels from
denser to rarer medium it bends away from the normal.

Activity – 6

Question 5.
Why can we not see a coin placed in water from the side of glass?
Answer:
Procedure :

1. Take a transparent glass tumbler and coin.
2. Place a coin on a table and place glass on the coin.
3. Observe the coin from the side of the glass. We can see the coin.
4. Now fill the glass with water and observe the coin from the side of the glass tumber.
5. Now we cannot see the coin because the coin rises up due to refraction.

Activity – 7

Question 7.
Why can we see the coin in water from top? What is the phenomenon behind that?
Answer:

Procedure :

1. Take a cylindrical transparent vessel. Place a coin at the bottom of the vessel.
2. Now pour water until we will get the image of the coin on the water surface.
3. This is due to total internal reflection.

Activity – 8

Question 8.
Write an Activity to find refractive index of glass slab by calculating vertical shift.
(OR)
Explain the experiment with glass slab in determination of refraction through vertical shift.
Answer:

Procedure :

1. Take a glass slab and measure the thickness of the slab.
2. Take a white chart and fix it on the table.
3. Place the slab in the middle of the chart.
4. Draw line around it.
5. Remove the slab from its place.
6. The lines form a rectangle. Name the vertices of it as A. B, C and D. ‘
7. Draw a perpendicular to the longer line AB of the rectangle at any point on it.
8. Place slab again in the rectangle ABCD.
9. Take a needle. Place at a point P in such a way that its length is parallel to the AB on the perpendicular line at a distance of 20 cm from the slab.
10. Now take another needle and by seeing at the first needle from the other side of the slab, try to keep the needle so that it forms a straight line with the first needle. 1 1)
11. Remove the slab and observe the positions of the needles.
12. They are not in same line.
13. Draw a perpendicular line from the second needle to the line on which the first needle is placed.
14. Take the intersection point as Q.
15. The distance between P and Q is vertical shift.
16. We will get the same vertical shift placing needle at different distances.
17. 

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 4 Excretion – The Wastage Disposing System Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 4th Lesson Excretion – The Wastage Disposing System

10th Class Biology 4th Lesson Excretion – The Wastage Disposing System Textbook Questions and Answers

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Question 1.
What is meant by excretion? Explain the process of formation of urine.
(OR)
What are the different stages in urine formation? Explain what happens in those stages.
(OR)
Explain various steps in the mechanism of urine formation.
Answer:
Excretion: Excretion is a biological process involved in separation and removal of toxic wastes from the body.
Formation of urine involves four stages:

1. Glomerular filtration
2. Tubular reabsorption
3. Tubular secretion and
4. Concentration of urine.

1. Glomerular filtration:

1. Blood flows from renal artery to glomerulus through afferent arteriole.
2. Filtration of blood occurs in the glomerulus.
3. Glomerular filtrate is also known as primary urine which almost equal to blood in chemical composition except the presence of blood cells.

2. Tubular reabsorption:

1. The primary urine passes into proximal convoluted tubule.
2. Useful substances like glucose, amino acids, sodium chloride, potassium ion, bicarbonate ion, water are reabsorbed into peritubular network.

3. Tubular secretion:

1. After reabsorption in proximal convoluted tubule (PCT) the urine travels through the loop of Henle into distal convoluted tubule.
2. Here some other wastes like extra salts ions of K⁺, Na⁺, Cl^– and H⁺ secrete from peritubular capillaries into distal convoluted tubule which are surrounded by peritubular network.

4. Concentration of urine:

1. 75% of water content of the nephric filtrate is reabsorbed in the region of proximal convoluted tubule and 10% of water passes out of filtrate through osmosis in the area of loop of Henle.
2. The concentration of urine takes place in the area of collecting tubes in the presence of hormone called vasopressin. The hormone is secreted only when concentrated urine is to be passed out.

Question 2.
How are waste products excreted in Amoeba?
(OR)
Write the process of excretion in amoeba.
Answer:

1. Specific excretory organs are absent in unicellular organisms like amoeba.
2. Amoeba possess osmoregulatory organell called contractile vacuole.
3. It collects water and wastes from the body, swells up, reaches the surface and bursts to release its content to outside.
4. The main excretion takes place through body surface by simple diffusion.
5. The waste material carbon dioxide is removed by diffusion through the cell membrane.

Question 3.
Name different excretory organs in human body and excretory material generated by them.
Answer:
Different excretory organs in human body and excretory material generated by them are as follows.

Question 4.
Deepak said that ‘Nephrons are functional and structural units of kidneys’. How will you support him?
(OR)
How can you say that kidney is suitable for the filtration of biological waste from blood in man?
Answer:
I support Deepak’s statement that nephrons are functional units of kidneys because

1. Nephron’s chief function is to regulate the concentration of water and soluble substances like sodium salts by filtering the blood, reabsorbing what is needed and excreting the rest as urine.
2. Nephron eliminates wastes from the body, regulates blood volume and blood pressure, controls levels of electrolytes and metabolites and regulates blood pH.
3. Its functions are vital to life and are regulated by the endocrine system.
4. Hence, nephrons are the functional units of kidneys.

Question 5.
How do plants manage the waste materials?
Answer:

1. Plants can get rid of excess water by a process like transpiration and guttation.
2. Waste products may be stored in leaves, bark and fruits.
3. When these dead leaves, bark, and ripe fruits fall off from the tree, the waste products in them are got rid of.
4. Some of the plant waste gets stored in the fruits in the form of solid bodies called Raphides. e.g: Yam.
5. Several plants prepare chemicals and store them in roots, leaves, seeds to protect against herbivores.
6. The plants excrete carbon dioxide produced as a waste during respiration only at night time.
7. The plants excrete oxygen as a waste only during the daytime.
8. The plants get rid of wastes by secreting them in the form of gums and resins.
9. Plants also excrete some waste substances into the soil around them.

Question 6.
Why do some people need to use a dialysis machine? Explain the principle involved in it.
Answer:

1. Kidneys are vital organs for filtration of nitrogenous waste material from blood and for survival of a person.
2. Factors like infections, injury, very high blood pressure, very high blood sugar restrict blood flow to kidneys.
3. This leads to accumulation of poisonous wastes in the body and leads to death.
4. When both kidneys are damaged DIALYSIS machine is used to filter the blood of a person.
5. This process is called HAEMODIALYSIS.
6. In this process blood is taken out from the main artery, mixed with an anticoagulant such as HEPARIN and then pumped into the apparatus called DIALYZER.
7. In this apparatus, blood flow through channels or tubes and are embedded in the dialyzing fluid.
8. The membrane separates the blood flowing inside the tube and dialyzing fluid by osmosis.
9. The dialyzing fluid has the same composition as that of plasma, except nitrogenous wastes.
10. Therefore, nitrogenous waste materials move out from the blood freely, there by cleaning the blood of it’s wastes.
11. This process is called DIALYSIS. Cleaned blood is pumped back to the body through a vein after adding HEPARIN.

Question 7.
What is meant by osmoregulation? How is it maintained in human body?
Answer:

1. The process of constant maintaining of the water and salt contents in the body is called osmoregulation.
2. The kidneys are the main osmoregulatory organs in human body.
3. The function of kidney is to filter blood and maintain the dissolved ion concentrations of body fluids.
4. The nephron is the functional unit of the kidney, which actively filters blood and generates urine.
5. The hormone VASOPRESSIN maintains osmotic concentration of body fluids.
6. 75% of water content of nephric filtrate is reabsorbed in the PCT (proximal convoluted tubule) and only 10% of water passes out of filtrate through osmosis in the area of loop of Henle.
7. Thus osmoregulation is maintained in human body.

Question 8.
Do you find any relationship between circulatory system and excretory system? What are they?
Answer:

1. Circulatory system delivers oxygen, nutrients, water, hormones and other essential to each cell of the body.
2. And it also transports all the carbon dioxide and other waste products of the cells to the lungs to be expired (exhaled) or to the kidneys to be excreted.
3. The excretory system is closely related to the circulatory system by virtue of the process of cleansing the blood of waste, removing excess fluids and generally keeping other fluids in balance.
4. Excretory system releases hormones to elevate blood pressure and accelerate red blood cell production.
5. Kidney stimulates the red blood cell production by erythropoetin and regulates blood pressure with the secretion of renin.

Question 9.
Give reasons.
A) Always vasopressin is not secreted.
Answer:

1. Vasopressin is secreted only when concentrated urine is to be passed out.
2. If vasopressin is not produced in sufficient quantities, very large amount of urine (more than 15 litres per day, normal persons excrete about one litre in a day) is excreted by the person which produces dilute urine.
3. Deficiency of vasopressin causes DIABETES INSIPIDUS.
4. Therefore vasopressin is always not secreted but only to concentrate urine.

B) When urine is discharged, in beginning it is acidic in nature later it becomes alkaline.
Answer:

1. Urine contains 96% of water, 2.5% of organic substances and 1.5% of inorganic solutes.
2. Urine is acidic in the beginning of it’s formation.
3. Gradually it becomes alkaline due to the decomposition of urea to form AMMONIA.

C) Diameter of afferent arteriole is bigger than efferent arteriole.
Answer:

1. Glomerulus develops from an afferent arteriole. It gives rise to an efferent arteriole.
2. The diameter of afferent arteriole is bigger than the efferent arteriole.
3. The narrower out let of efferent arteriole exerts pressure in the glomerulus and enables the blood to remain more time, thus helps in proper filtration of blood by ultrafiltration.

D) Urine is slightly thicker in summer than in winter.
Answer:

1. In summer, a large quantity of water is lost in the form of sweat.
2. To save the body from sun’s heat and to maintain water balance sweat glands secrete more sweat.
3. The remaining waste materials get concentrated.
4. Due to the high concentration of the remaining waste materials urine is slightly thicker in summer than in winter.

Question 10.
Write differences between
A) Functions of PCT and DCT
B) Kidney and Artificial kidney
C) Excretion and Secretion
D) Primary metabolites and Secondary metabolites
Answer:
A) Functions of PCT and DCT:
Functions of PCT:

1. Reabsorbs glucose, amino acids, phosphate, potassium, urea and other organic solutes from the filtrate into the peritubular network.
2. The PCT regulates pH of the filtrates by exchanging hydrogen ions in the interstitium for bicarbonate ions in the filtrate.
3. It is also responsible to secreting organic acids such as creatinine and other bases into the filtrate.
4. Proximal convoluted tubule regulates the pH of the filtrate in the kidneys.

Functions of DCT:

1. It maintains a proper concentration and pH of the urine.
2. Extra salts, ions of K⁺, Na⁺, Cl^– and H⁺ secrete from peritubular capillaries into DCT.
3. It secretes ammonium ions and hydrogen ions.
4. It is relatively impermeable to water but in the pressure of Antidiuretic hormone

B) Kidney and Artificial kidney:

(ADH) its permeability to water increases making urine concentrated.

C) Excretion and Secretion:

D) Primary metabolites and Secondary metabolites:
(OR)
What are primary and secondary metabolites in plants? Give examples.
Answer:

Question 11.
There is a pair of bean-shaped organs ‘P’ in the human body towards the back, just above the waist. A waste product ‘Q’ formed by the decomposition of unused proteins in liver is brought into organ ‘P’ through blood by an artery ‘R’. The numerous tiny filters ‘S’ present in organ ‘P’ clean the dirty blood goes into circulation through a vein ‘T’. The waste substance ‘Q’ other waste salts and excess water form a yellowish liquid ‘U’ which goes from organ ‘P’ into a bag like structure ‘V’ through two tubes ‘W’. This liquid is then thrown out of the body through a tube ‘X’.

(a) What is (i) organ P and (ii) waste substance Q?
(i) Organ P is kidney and
(ii) Waste substance Q is urea.

(b) Name (i) artery R and (ii) vein T.
Answer:
(i) Artery R is Renal artery and
(ii) Vein T is Renal vein.

(c) What are tiny filters ‘S’ known as?
Answer:
The tiny filters S are Nephrons.

(d) Name (i) Liquid (ii) Structure V (iii) Tubes W (iv) Tube X.
Answer:
(i) Liquid U is urine.
(ii) Structure V is urinary bladder.
(iii) Tube W is ureters.
(iv) Tube X is urethra.

Question 12.
The organ ‘A’ of a person has been damaged completely due to a poisonous waste material ‘B’ has started accumulation in his blood, making it dirty. In order to save this person’s life, the blood from an artery in the person’s arm is made to flow into long tubes made of substance ‘E’ which are kept in coiled form in a tank containing solution ‘F’vThis solution contains three materials ‘G’, ‘H’ and ‘I’ and similar proportions to those in normal blood. As the person’s blood passes through long tubes of substance ‘E’, most of the wastes present in it go into solution ‘F’ The clean blood is then put back into a vein in the person for circulation.
(a) What is organ A?
Answer:
Kidney.

(b) Name the waste substance B.
Answer:
Urea.

(c) What are (i) E and (ii) F?
Answer:
(i) Long tubes ‘E’ are made of cellulose.
(ii) Solution ‘F’ is dialysing fluid contains three materials like: water, glucose and salts.

(d) What are G, H and I?
Answer:
Waste molecules, nutrient molecules and water.

(e) What is the process described above known as?
Answer:
Dialysis.

Question 13.
Imagine what happens if waste materials are not sent out of the body from time to time.
(OR)
What happens when the waste products are not sent out from the body?
Answer:

1. If waste materials are not sent out of the body from time to time, our body is filled with extra water and waste products.
2. ThIs condition ¡s called UREMIA. 1f kidneys stop working completely it leads to UREMIA.
3. Our hands and feet may swell.
4. We feel tired and weak because our body needs clean blood to function properly.
5. These waste materials turn into toxic and it leads to death.

Question 14.
To keep your kidneys healthy for long period what questions will you ask a nephrologist/urologist?
Answer:

1. How can I prevent the formation of stones in kidney?
2. Does renal failure hereditary?
3. What are the dietary measures to be taken for normal functioning of kidney?
4. How diabetes harm kidneys?
5. What shall I do to keep my kidneys healthy for a long time?
6. What are the factors responsible for kidney failure?
7. How can we know that there is a problem In the kidneys?
8. Is there any relationship between blood pressure and kidney function?
9. What are ‘the signs of kidney failure?
10. Why is smoking bad for kidneys?

Question 15.
What are the gum yielding trees in your surroundings? What procedure you should follow to collect gum from trees?
Answer:
In our surroundings neem, acacia, eucalyptus, sapota are some of the gum yielding plants.
Process of collecting gum from trees:

1. Gum will flow naturally from cracks in the bark of acacia and neem trees.
2. Gum is collected about four weeks after stripping and can be repeated every few weeks thereafter for several months.
3. Mature plants are selected for taking gum.
4. Suitable stem parts are selected.
5. Grooves are made by break the bark away about three foot from the ground and 10 inches wide with a sharp sickle or knife.
6. We can fit a container at the lower end tightly, so that when gum begins to seep out, it will drip into it.
7. The plants are to be kept undisturbed for at least two or three weeks.
8. When the container is seen filled with gum they are removed.
9. It is collected and stored for supply and used as adhesives, binding agents, in the preparation of medicines etc.
10. Break some shallow notches in a ‘V’ shape, with the point of the ‘V’ diectly above the centre of the bucket.
11. Leave the bucket attached to the tree until the gun begins to seep out and drains into it.
12. Remove any nails or other metal things from the tree and after taking down the gum collection bucket.

Question 16.
Collect the information about uses of different kinds of alkaloids, take help of Library or Internet.
Answer:
Common alkaloids in plants and their uses are

Question 17.
Draw a neat labelled diagram of L.S of kidney.
(OR)
Draw a neat labelled diagram of internal structure of Kidney. Write the function of Renal artery and Renal vein.
Answer:

1. Renal artery supplies oxygenated blood with nitrogenous waste – products to kidneys.
2. Renal vein collects nitrogenous waste free and deoxygenated blood from kidneys.

Question 18.
Draw a diagram of a Nephron, and explain its structure.

Describe the structure of nephron with the help of diagram.
(OR)
Answer:
Structure of the Nephron :

Each Nephron has basically two parts:

1. 1. Malpighian body
   2. Renal tubule

1. Malpighian Body:

1. It consists of Bowman’s capsule and bunch of fine blood capillaries called the glomerulus.
2. Glomerulus develops from afferent arteriole. It gives rise to an efferent arteriole.

Renal Tubule :

1. It has three parts.
   1. Proximal Convoluted Tubule (PCT),
   2. Loop of Henle
   3. ‘U’ shaped second or Distal Convoluted Tubule (DCT)
2. Distal convoluted tubule open into a COLLECTING TUBE. Collecting tubes form PYRAMIDS and CALYCES which open into PELVIS.
3. Pelvis leads into the URETER.
4. All the parts of the renal tubule are covered by a network of PERITUBULAR CAPILLARIES fromed from EFFERENT ARTERIOLE
5. The peritubular capillaries join to form RENAL VENULE, which joins the other venules to form RENAL VEIN.

Question 19.
Draw a block diagram showing the path way of excretory system in human being.
Answer:
Block diagram showing the path way of excretory system in human beings.

Question 20.
If you want to explain the process of filtration in kidney, what diagram you need to draw ?
Answer:

Question 21.
List out the things that makes you amazing in excretory system of human being.
(OR)
How do you appreciate the functioning of excretory system of human being?
Answer:

1. In human beings during metabolic activities many waste products like water, nitrogenous compounds like ammonia, urea, uric acid, bile pigments, excess salts, etc. the toxic wastes are produced.
2. To excrete all these waste materials in human beings there is an excretory system consisting of a pair of kidneys, a pair of ureters, urinary bladder and urethra.
3. Each kidney is made up of more than one million nephron. These are the structural and functional units of excretory system.
4. Per day 170 litres of water is filtered by kidneys. Out of this 168.2 litres is reabsorbed. This is a wonderful mechanism.
5. Total amount of urine excreted per day is about 1.6 to 1.8 litres.
6. The kidneys have a great reserve power. If one kidney is removed due to disease or damage, the other kidney can take up the function of both the kidneys.
7. It is so amazing that the 10 cm in length, 5-6 cm in breadth and 4 cm thickness size of the kidney can do filtration of blood and can remove all the poisonous substances from the human body and keeping the organism healthy.
8. Skin is also an excretory organ responsible for elimination of wastes in the form of sweat along with various toxins.

Question 22.
You read about ‘Brain dead’ in this chapter. What discussions would you like to have? Why do you think so?
Answer:

1. When some one is brain dead, there is no blood flow or oxygen to their brain.
2. The brain stops functioning in any capacity.
3. Because the ventilator is breathing for the person organs such as the heart and liver continue to receive oxygen and one able to function for a few days after the brain has dead.
4. Unless damaged by injury we can transplant organs like kidney liver, heart, lungs, pancreas, skin, bone, intestine and eyes (Retina) from brain dead patients.
5. The process of transplantation of organs from brain dead patients to another is called cadaver transplantation.
6. There is very less awareness among people about organ donation.
7. Society needs much awareness in organ donation so that we can save many lives who are in need of different organs for their survival.
8. Those who are willing to donate their organs have to sign in an application form at the transplantation facility hospital.
9. Some voluntary organisations like jeevandan.org working on this aspect.

Question 23.
We the people have very less awareness about organ donation, to motivate people to write slogans about organ donation.
Answer:
Slogans about organ donation :

1. Organ donation saves lives.
2. Give a life, gift of life.
3. Donate organs today for better tomorrow.
4. The measure of a life, after all is not its duration but its donation.
5. Change your thoughts and you change your world.
6. Organ donation is icky (disagreeable) but recycle yourself is sticky.

What you have seen?

1. Have a heart, save a life.
2. The gift of life pass it on.
3. Organ donation is a gift for life.

Question 24.
After learning this chapter what habits you would like to change or follow for proper functioning of kidneys?
(OR)
What habits you would like to follow for the proper functioning of kidneys?
Answer:

1. I always eat the home cooked meals.
2. I should adapt to take low salt diet.
3. I eat the food that are lower in protein which can help to decrease the levels of protein, avoid eating cakes, pies, cookies, candies and pastries (junk foods), take fruits containing simple carbohydrates.
4. I would like to drink sufficient water nearly 5-6 litres per day.
5. I never forget drinking more water after meals.
6. I avoid taking high phosphorus foods like cheese, cola, soft drinks, butter but I replace them with butter milk, sharbat etc.
7. I eat low potassium levels fruits like apples, watermelon, cucumber, grapes, cherries, carrots, etc. because high potassium levels cause irregular heart beats.
8. I avoid high sodium contents like potato chips, bacon (cured meat from pig) cheese, meat, canned vegetables, caned soups frozen dinners and table salt, which damage kidneys.

Fill in the blanks.

1. Earthworm excretes its waste material through ———–.
2. The dark coloured outer zone of kidney is called ———–.
3. The process of control of water balance and ion concentration within organism is called ———–.
4. Reabsorption of useful product takes place in ———– part of nephron.
5. Gums and resins are the ———– products of the plants.
6. Bowman’s capsule and tubule taken together make a ———–.
7. The alkaloid used for malaria treatment is ———–.
8. The principle involved in dialysis is ———–.
9. Rubber is produced from ———– of Heavea braziliensis.
10. ———– performed first Kidney Transplantation.

Answer:

1. nephridia
2. cortex
3. osmoregulation
4. tubular
5. secondary metabolic
6. malphigian tubule
7. quinine
8. Osmosis and filtration
9. latex
10. Dr. Charles Hufnagel

Choose the correct answer.

1. The structural and functional unit of human kidney is called [ ]
   A) Neuron
   B) Nephron
   C) Nephridia
   D) Flame cell
   Answer: B
2. The excretory organ in cockroach  [ ]
   A) Malpighian tubules
   B) Raphids
   C) Ureters
   D) Nephridia
   Answer: A
3. Which of the following is the correct path taken by urine in our body? [ ]
   A) Kidney, urethra, ureters, bladder
   B) Kidney, ureters, bladder, urethra
   C) Kidney, bladder, ureters, urethra
   D) Kidney, urethra, bladder, ureters
   Answer: C
4. Malpighian tubes are excretory organs in [ ]
   A) Earthworm
   B) Housefly
   C) Flatworm
   D) Hen
   Answer: B
5. Major component of urine is [ ]
   A) Urea
   B) Sodium
   C) Water
   D) Creatine
   Answer: C
6. Special excretory organs are absent in [ ]
   A) Birds
   B) Amoeba
   C) Sponges
   D) A and B
   Answer: B
7. Which of the following hormones has direct impact on urination? [ ]
   A) Adrenal
   B) Vasopressin
   C) Testosterone
   D) Estrogen
   Answer: B
8. Amber colour to urine due to [ ]
   A) Urochrome
   B) Bilirubin
   C) Biliverdin
   D) Chlorides
   Answer: A
9. Sequence of urine formation in the nephron is [ ]
   A) Glomerular filtration → Tubular reabsorption → Tubular secretion
   B) Tubular reabsorption → Tubular secretion → Glomerular filtration
   C) Tubular secretion → Glomerular filtration → Tubular reabsorption
   D) Tubular reabsorption → Concentration of urine → Tubular secretion
   Answer: A
10. Part of the nephron that exists in outer zone of kidney [ ]
    A) Loop of the Henle
    B) PCT
    C) DCT
    D) Bowman’s capsule
    Answer: D
11. After having lunch or dinner one can feel to pass urine, because of [ ]
    A) Stomach pressures on bladder
    B) Solids become liquids
    C) Water content in food material
    D) Sphincter relaxation
    Answer: D

10th Class Biology 4th Lesson Excretion – The Wastage Disposing System InText Questions and Answers.

10th Class Biology Textbook Page No. 75

Question 1.
What products would the organism be able to take up for other activities?
Answer:
Oxygen produced in the photosynthesis process is used by all the living organisms on the earth for breathing. Without oxygen life would not be possible on the earth. No organism can survive.
Carbon dioxide produced in respiration will be taken up by plants to prepare food materials, in the process of photosynthesis. Plants are called autotrophs because they produce food for ail the animals and human beings directly or indirectly.

Question 2.
What are the products would cause harm to the body, if they are not removed?
Answer:
Nitrogenous compounds like ammonia, urea, uric acid, bile pigments, excess salts are toxic to the body. So they should be removed.

Question 3.
What happens if harmful products are not removed from our body every day?
Answer:
If harmful products are not removed from our body they get accumulated in the body and becomes toxic. This leads to the death of the person.

Question 4.
What are the substances present in blood?
Answer:
The substances present in blood are Glucose, Sodium, Potassium, Chlorides, Urea, Creatinine, Uric acid, Cholesterol, Triglycerides, Calcium Phosphorous, Bilirubin, Proteins, Albumin.

Question 5.
What are the substances present in urine?
Answer:
Protein, Creatinine, Calcium, Phosphorous, Uric acid, Sodium, Potassium are the sub-stances present in urine.

Question 6.
What are the substances present both in blood and urine?
Answer:
Sodium, Potassium, Chlorides, Urea, Uric acid, Calcium, Phosphorous are the sub¬stances present in blood and urine in common.

Question 7.
Which substances are present above the normal limits both in the blood and urine?
Answer:
Urea, Uric acid, Cholesterol, Calcium, Creatinine are present above the normal limits both in the blood and urine.

Question 8.
What do you think reading above normal limits indicates?
Answer:
If any material is above the normal limits, it causes health problem which leads to a disease and damage of the organs.

10th Class Biology Textbook Page No. 77

Question 9.
What are the materials needed to be removed from our body?
Answer:
Carbon dioxide, water and nitrogenous compounds such as ammonia, urea and uric acid are the waste materials needed to be removed from our body.

Question 10.
From where are these materials removed?
Answer:
Carbon dioxide is eliminated through lungs while small amounts of water is eliminated through body surface (sweating) and through lungs during respiration. An excretory organ system is present for excreting the nitrogenous wastes along with salts, excess water.

Question 11.
What are the organs that separate excretory materials?
Answer:
Flame cells in flat worms, Nephridia in Annelids, Malphigian tubules in Arthropods, kidneys in all vertebrates are the excretory organs that separate excretory materials and eliminate them.

Question 12.
Why do you think the body must remove waste substances?
Answer:
For the smooth functioning of the body the body should be healthy. Waste materials are the toxic substances, which are harmful to the body. So they should be eliminated.
If they remain in the body, it leads to the unhealthy conditions to the organisms.

10th Class Biology Textbook Page No. 79

Question 13.
Think why the diameter of the efferent arteriole is less than that of afferent arteriole?
Answer:
The diameter of the efferent arteriole is less than afferent arteriole so as to create pressure in the glomerulus to filter the waste materials. Due to this, blood remains in glomerulus more time.

Question 14.
Why the nephron is considered to be the structural and functional unit of the kidney?
Answer:

1. The kidney is made up of more than one million nephrons.
2. Hence it is called as the structural unit of the kidney.
3. Filtration of blood to remove nitrogenous substances occur in nephron.
4. So it is called as functional unit of the kidney.

10th Class Biology Textbook Page No. 80

Question 15.
Which arteriole has more diameter, afferent or efferent?
Answer:
Afferent arteriole has more diameter than efferent arteriole.

Question 16.
What are the substances that are filtered into the glomerular capsule?
Answer:
The substances that are filtered into the glomerular capsule are waste molecules, nutrient molecules and water.

Question 17.
If you drink more water, will you pass more urine?
Answer:
Yes. If we drink more water, we will pass more urine.

Question 18.
What are the substances reabsorbed into peritubular network from Proximal Convoluted Tubule (PCT)?
Answer:
Glucose, Amino acids, Vitamin – C, Potassium, Calcium, Sodium chlorides and 75% of water are the substances reabsorbed into peritubular network from proximal convoluted tubule.

Question 19.
What are the substances that secretes into Distal Convoluted Tubule (DCT)?
Answer:
The substances that secretes into Distal Convoluted Tubule are extra salts, ions of K\ Na+, C/“ and H+ ions.

10th Class Biology Textbook Page No. 81

Question 20.
Why more urine is produced in winter?
Answer:

1. Due to the cold condition, our body does not lose the water in the form of sweat.
2. In the same condition, blood flow to our internal organs is increased, to keep our organs warm.
3. The increase in the blood flow to the kidneys causes more blood to be filtered.
4. These two causes more urine is produced in winter.

Question 21.
What happens if reabsorption of water does not take place?
Answer:
If reabsorption of water does not take place :

1. water levels decreases in the body and leads to malfunctioning of metabolism.
2. osmoregulation of body fluids will be disturbed.
3. circulation of blood does not take place due to increase in its concentration.
4. the useful substances like nutrients, salts, hormones, vitamins are not absorbed they excrete out.

10th Class Biology Textbook Page No. 83

Question 22.
What happens if both kidneys fail completely?
Answer:

1. Complete and irreversible kidney failure is called End Stage Renal Disease (ESRD).
2. If kidneys stops working completely, our body is filled with extra water and waste products.
3. This condition is called UREMIA.
4. Our hands or feet may swell. We may feel tired and weak because our body needs clean blood to function properly.
5. The solution to this problem is either dialysis or kidney transplantation.

10th Class Biology Textbook Page No. 84

Question 23.
Is there any long term solution for kidney failure patients?
Answer:
The best long term solution for kidney failure is kidney transplantation. A functioning kidney which is a good match to the body is used in transplantation from a donor preferably a closed relative.

Question 24.
Where is the transplanted kidney fixed in the body of a kidney failure patient?
Answer:
Kidney transplantation involves placing a healthy kidney into the body where it can perform all of the functions that a failing kidney cannot. The new kidney is placed on the lower right or left side of the abdomen where it is surgically connected to nearby blood vessels.
Placing the kidney in this position allows it to be easily connected to blood vessels and the bladder. The vein and the artery of the new kidney are attached to the body’s vein and artery. The new kidney’s ureter is attached to the body’s bladder to allow urine to pass out of the body.

Question 25.
What about the failed kidneys?
Answer:

1. In most cases the old kidneys will not be removed.
2. This is because even failed kidneys release chemicals that help the body work.
3. However if those kidneys have a disease that cause on going problems such as persistent kidney infections or intestinal blockage, then the transplantion would be considered removing the old kidney.
4. The two most common medical conditions requiring “native nephrectomy” or “congenital reflux” disease and polycystic disease.

Question 26.
Can donor survive his / her life with single kidney without any complications?
Answer:

1. There are some risks.
2. There is a chance of affecting his/her life span or life style is extremely low with surgery and anaesthesia.
3. The risk of minor complications like wound infection because the kidney donar operation is a major surgical procedure.
4. Donar feels less energy and need about 4 to 6 weeks to return to their activities.

Question 27.
What are the other excretory organs of human body?
Answer:
Lungs, Skin, Liver are the other excretory organs of human body.

10th Class Biology Textbook Page No. 85

Question 28.
Collect information on sebum and prepare a news bulletin, display it on bulletin
Answer:

1. Sebum is a complex mixture of naturally produced fats, oils, waxes, cholesterol etc.
2. It is secreted by SEBACEOUS glands present in the DERMIS of skin.
3. The fat and OMS present in the sebum do not originate directly from the fats and oils consumed in the diet.
4. SEBUM is manufactured and stored by the sebaceous glands under the direction of biological process.
5. Sebaceous glands are found in great number on the face, scalp and on all parts of the skin.
6. The function of sebum : it prevent the skin from becoming dry.

Question 29.
People in cold countries get very less/no sweat. What changes occur in their skin and in other excretory organs?
Answer:

1. Regulation of body temperature is one of the functions of skin.
2. To maintain the body temperature, skin produces sweat.
3. In cold countries environment is very cool. So there is no production of sweat.
4. In these conditions there will be more pressure on other excretory organs.

10th Class Biology Textbook Page No. 86

Question 30.
Do plants excrete like animals?
Answer:

1. Plants do not have specific organs to excrete the waste materials which are formed during metabolism.
2. As in animals carbon dioxide, water, ammonia and other nitrogenous wastes are also formed in plants.
3. In plants, carbon dioxide released during respiration is sent out through stomata of leaves.
4. Plants discharge the excess water in the form of water vapour during transpiration by leaves. Due to this, the heat in the plants is reduced.
5. In germinating seeds carbon dioxide formed during respiration is released into atmosphere.
6. Plants have the capacity to utilise the by- products of one metabolic activity as the raw materials for another metabolic activity.
7. For example, oxygen released during photosynthesis is utilised for respiration; carbon dioxide released during respiration is utilised for photosynthesis. Plants convert nitrogen and ammonia into nitrates.

Question 31.
How do plants manage or send out waste products from its body?
Answer:

1. Plants can get rid of excess water by a process like transpiration and guttation. Waste products may be stored in leaves, bark and fruits.
2. When these dead leaves, bark, and ripen fruits fall off from the trees, then waste products in them are got rid off.
3. In some plants, waste gets stored in the fruits in the form of solid bodies called Raphides. Example : Yam (pendalam).
4. Several compounds are synthesized by the plants for their own use specially for defense.
5. Several plants prepare chemicals and store them in shoots, leaves, seeds for protection against herbivores.
6. Most of the chemicals are unpleasant to taste and hence herbivores do not prefer to eat such plants. Some of the chemicals are toxic and may even kill the animal that eats them.
7. Some of the plants release attractants for other organisms which will help the plants for pollination. For example, plants having root nodules secrete chemicals to attract rhizobia into the surroundings of the roots and form a symbiotic relationship with the rhizobium.

10th Class Biology Textbook Page No. 87

Question 32.
Why do plants shed their leaves and bark periodically?
Answer:

1. In summer season, transpiration rate is very high in plants.
2. To reduce the transpiration rate plants shed their leaves.
3. And some waste materials are also stored in leaves and removed by shedding of leaves.
4. Bark is the outer zone of plants, which contain phloem to transport food materials.
5. In the growing period cambium produces bark tissue. When new bark tissue is produced, the old bark is removed.
6. In this process the old and dead cells filled with waste materials are also removed.

Question 33.
Name the alkaloids which are harmful to us.
Answer:
The harmful alkaloids are :

1. Nicotine is harmful. It causes cancer to lung, throat, tongue and affects the nervous system.
2. Morphine which is used as a pain killer, may effect kidneys if they are used more.
3. Cocaine, scopolamine (Datura) etc are the alkaloids which are harmful to us.
4. Nitrogenous substances are also found in the walls of pollen grain, if they enter our body. Ex : they cause allergy.

10th Class Biology Textbook Page No. 89

Question 34.
Do roots secrete?
Answer:
Yes, roots also can secrete a portion of their peculiar secretions back into the soil.

Question 35.
Do you think is there any relation between reduction in yielding and root secretions?
Answer:

1. Yes, there is a relation between reduction in yielding and root secretions.
2. For example plants like apple where a single apple crop is yielded for 4 or 5 years continuously in the same soil, it fails to produce fruits.
3. It will not give proper yield even if you use lot of fertilizers.

Question 36.
Why do we get peculiar smell when you shift the potted plants?
Answer:

1. Some peculiar secretions are secreted and sent out from roots into soil.
2. When we shift the potted plant we get peculiar smell due to the chemical reaction of the secretion in the soil to make it fertile.

Think and Discuss

10th Class Biology Textbook Page No. 82

Question 1.
Do cells need excretion?
Answer:
Yes, the cells need excretion to keep all the body organs healthy.

Question 2.
Why are we advised to take sufficient water ?
Answer:
We are advised to take sufficient water because filtration of waste products becomes easy and there is a free flow of urine including salts, And also body temperature will be maintained.

Question 3.
Why do some children pass urine during sleep at night until 15 or 16 years of age?
Answer:
Children pass urine during sleep at night because one or more of the following reasons.

1. The child’s urinary bladder is maturing more slowly than usual. The capacity of the bladder may be reduced.
2. The risk of child passing urine in night time is due to the hereditory factor that one or both parents wet the bed as children.
3. Some children wet the bed because they do not make enough levels of vasopressin which reduces the urine production.
4. Physical or emotional problems may cause bed wetting.
5. A stressful situation can trigger bed wetting include moving to a new house, changing schools, the death of a loved one or being sexually given bad effect.

10th Class Biology Textbook Page No. 84

Question 4.
Why are weeds and wild plants not affected by insects and pests? (OR)
Ravi went to his cotton field. There he observed some of the cotton leaves were affected by insects. He also observed that weeds in between are not affected. Give reasons.
Answer:
Some plants like weeds and wild plants prepare chemicals which are unpleasant to taste and some chemicals are toxic and may even kill. So insects and pests do not touch them.

10th Class Biology 4th Lesson Excretion – The Wastage Disposing System Activities

Activity – 1

How do you study the external and internal features of kidney ? (OR)
Explain the procedure and observations of the experiment conducted to observe internal structure of the kidney.
Answer:
Aim: Studying the external and internal features of a kidney.
Materials required: Freshly collected specimen of sheep/goat’s kidney from the butcher or 3D model of a kidney sharp blade/ scalpel, tray and a jug of water.
Procedure for observation:

1. 1. Wash the kidney so that blood is completely drained from it.
      
   2. Put the kidney in the tray and observe it carefully.
   3. Note down the observation in the notebook.

1. With the help of sharp blade take the longitudinal section of the kidney and observe internal structure.
2. Draw what you have observed and compare it with the figure given.
3. After observing the L.S. of kidney answer the following questions given under.

What is the shape of kidneys?
Answer:
The shape of kidneys is bean shape.

What is the colour of kidney ?
Answer:
Colour of kidney is reddish brown.

Do you find any attachments on upper portion of kidney?
Answer:
Yes. Adrenal glands are attached to kidneys on upper portion.

Is the internal structures similar to fig – 2?
Answer:
Yes. It is similar to internal structure of fig – 2.

What is the colour of the outer part in L.S. of kidney?
Answer:
The colour of the outer part of the kidney is Dark.

In L.S. of kidney where do you find dark brown colour portion?
Answer:
Dark colour portion is found on the outer zone of kidney.

How many tubes are coming out from kidney fissure ?
Answer:
Two tubes are coming from kidney.

Andhra Pradesh SCERT AP State Board Syllabus SSC 10th Class Maths Textbook Solutions Study Material Guide Pdf free download 2024-2025 in English Medium and Telugu Medium are part of AP Board Solutions Class 10.

Students can also go through AP 10th Class Maths Notes to understand and remember the concepts easily. Students can also read AP 10th Class Maths Chapter Wise Important Questions for board exams.

AP Board Solutions Class 10 Maths – AP 10th Class Maths Textbook Solutions

AP Board Solutions Class 10 Maths Chapter 1 Real Numbers

• Chapter 1 Real Numbers Exercise 1.1
• Chapter 1 Real Numbers Exercise 1.2

AP 10th Class Maths Textbook Solutions Chapter 2 Polynomials

• Chapter 2 Polynomials Exercise 2.1
• Chapter 2 Polynomials Exercise 2.2

AP 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables

• Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1
• Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2
• Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

AP 10th Class Maths Solutions Pdf Chapter 4 Quadratic Equations

• Chapter 4 Quadratic Equations Exercise 4.1
• Chapter 4 Quadratic Equations Exercise 4.2
• Chapter 4 Quadratic Equations Exercise 4.3

AP Maths Solutions Class 10 Chapter 5 Arithmetic Progressions

• Chapter 5 Arithmetic Progressions Exercise 5.1
• Chapter 5 Arithmetic Progressions Exercise 5.2
• Chapter 5 Arithmetic Progressions Exercise 5.3
• Chapter 5 Arithmetic Progressions Exercise 5.4

AP State Board 10th Class Maths Solutions Chapter 6 Triangles

• Chapter 6 Triangles Exercise 6.1
• Chapter 6 Triangles Exercise 6.2
• Chapter 6 Triangles Exercise 6.3

AP SCERT 10th Maths Textbook Solutions Chapter 7 Coordinate Geometry

• Chapter 7 Coordinate Geometry Exercise 7.1
• Chapter 7 Coordinate Geometry Exercise 7.2

AP Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry

• Chapter 8 Introduction to Trigonometry Exercise 8.1
• Chapter 8 Introduction to Trigonometry Exercise 8.2
• Chapter 8 Introduction to Trigonometry Exercise 8.3

AP Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry

• Chapter 9 Some Applications of Trigonometry Exercise 9.1

Maths Textbook Class 10 Solutions Chapter 10 Circles

• Chapter 10 Circles Exercise 10.1
• Chapter 10 Circles Exercise 10.2

AP Board Solutions 10th Class Maths Chapter 11 Areas Related to Circles

• Chapter 11 Areas Related to Circles Exercise 11.1

SCERT Class 10 Maths Solutions Chapter 12 Surface Areas and Volumes

• Chapter 12 Surface Areas and Volumes Exercise 12.1
• Chapter 12 Surface Areas and Volumes Exercise 12.2

AP 10th Class Maths Study Material Chapter 13 Statistics

• Chapter 13 Statistics Exercise 13.1
• Chapter 13 Statistics Exercise 13.2
• Chapter 13 Statistics Exercise 13.3

Maths 10th Class Solutions AP Chapter 14 Probability

• Chapter 14 Probability Exercise 14.1

AP 10th Class Maths Textbook Solutions (Old Syllabus)

10th Class Maths Textbook Solutions Chapter 1 Real Numbers

• Chapter 1 Real Numbers Ex 1.1
• Chapter 1 Real Numbers Ex 1.2
• Chapter 1 Real Numbers Ex 1.3
• Chapter 1 Real Numbers Ex 1.4
• Chapter 1 Real Numbers Ex 1.5
• Chapter 1 Real Numbers Optional Exercise
• Chapter 1 Real Numbers InText Questions

10th Class Maths Guide Chapter 2 Sets

• Chapter 2 Sets Ex 2.1
• Chapter 2 Sets Ex 2.2
• Chapter 2 Sets Ex 2.3
• Chapter 2 Sets Ex 2.4
• Chapter 2 Sets InText Questions

Maths 10th Class Guide Chapter 3 Polynomials

• Chapter 3 Polynomials Ex 3.1
• Chapter 3 Polynomials Ex 3.2
• Chapter 3 Polynomials Ex 3.3
• Chapter 3 Polynomials Ex 3.4
• Chapter 3 Polynomials Optional Exercise
• Chapter 3 Polynomials InText Questions

AP 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables

• Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1
• Chapter 4 Pair of Linear Equations in Two Variables Ex 4.2
• Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3
• Chapter 4 Pair of Linear Equations in Two Variables Optional Exercise
• Chapter 4 Pair of Linear Equations in Two Variables InText Questions

AP 10th Class Maths Guide Pdf Free Download Chapter 5 Quadratic Equations

• Chapter 5 Quadratic Equations Ex 5.1
• Chapter 5 Quadratic Equations Ex 5.2
• Chapter 5 Quadratic Equations Ex 5.3
• Chapter 5 Quadratic Equations Ex 5.4
• Chapter 5 Quadratic Equations Optional Exercise
• Chapter 5 Quadratic Equations InText Questions

AP State 10th Class Maths Textbook Solutions Pdf Chapter 6 Progressions

• Chapter 6 Progressions Ex 6.1
• Chapter 6 Progressions Ex 6.2
• Chapter 6 Progressions Ex 6.3
• Chapter 6 Progressions Ex 6.4
• Chapter 6 Progressions Ex 6.5
• Chapter 6 Progressions Optional Exercise
• Chapter 6 Progressions InText Questions

SSC 10th Class Maths Guide Pdf Free Download Chapter 7 Coordinate Geometry

• Chapter 7 Coordinate Geometry Ex 7.1
• Chapter 7 Coordinate Geometry Ex 7.2
• Chapter 7 Coordinate Geometry Ex 7.3
• Chapter 7 Coordinate Geometry Ex 7.4
• Chapter 7 Coordinate Geometry Optional Exercise
• Chapter 7 Coordinate Geometry InText Questions

AP 10th Class Maths Solutions Pdf Chapter 8 Similar Triangles

• Chapter 8 Similar Triangles Ex 8.1
• Chapter 8 Similar Triangles Ex 8.2
• Chapter 8 Similar Triangles Ex 8.3
• Chapter 8 Similar Triangles Ex 8.4
• Chapter 8 Similar Triangles Optional Exercise
• Chapter 8 Similar Triangles InText Questions

Maths Solutions Class 10 State Board Chapter 9 Tangents and Secants to a Circle

• Chapter 9 Tangents and Secants to a Circle Ex 9.1
• Chapter 9 Tangents and Secants to a Circle Ex 9.2
• Chapter 9 Tangents and Secants to a Circle Ex 9.3
• Chapter 9 Tangents and Secants to a Circle Optional Exercise
• Chapter 9 Tangents and Secants to a Circle InText Questions

AP 10th Class Maths Study Material Pdf 2022 Chapter 10 Mensuration

• Chapter 10 Mensuration Ex 10.1
• Chapter 10 Mensuration Ex 10.2
• Chapter 10 Mensuration Ex 10.3
• Chapter 10 Mensuration Ex 10.4
• Chapter 10 Mensuration Optional Exercise
• Chapter 10 Mensuration InText Questions

Maths Material 10th Class Chapter 11 Trigonometry

• Chapter 11 Trigonometry Ex 11.1
• Chapter 11 Trigonometry Ex 11.2
• Chapter 11 Trigonometry Ex 11.3
• Chapter 11 Trigonometry Ex 11.4
• Chapter 11 Trigonometry Optional Exercise
• Chapter 11 Trigonometry InText Questions

AP State Board Class 10 Maths Solutions Chapter 12 Applications of Trigonometry

• Chapter 12 Applications of Trigonometry Ex 12.1
• Chapter 12 Applications of Trigonometry Ex 12.2
• Chapter 12 Applications of Trigonometry Optional Exercise
• Chapter 12 Applications of Trigonometry InText Questions

10th Class Maths Study Material 2021-2022 Chapter 13 Probability

• Chapter 13 Probability Ex 13.1
• Chapter 13 Probability Ex 13.2
• Chapter 13 Probability Optional Exercise
• Chapter 13 Probability InText Questions

10th Class Maths Textbook Pdf English Medium Chapter 14 Statistics

• Chapter 14 Statistics Ex 14.1
• Chapter 14 Statistics Ex 14.2
• Chapter 14 Statistics Ex 14.3
• Chapter 14 Statistics Ex 14.4
• Chapter 14 Statistics InText Questions

Old Syllabus

10th Class Maths Material 2022 Chapter 1 వాస్తవ సంఖ్యలు

• Chapter 1 వాస్తవ సంఖ్యలు Ex 1.1
• Chapter 1 వాస్తవ సంఖ్యలు Ex 1.2
• Chapter 1 వాస్తవ సంఖ్యలు Ex 1.3
• Chapter 1 వాస్తవ సంఖ్యలు Ex 1.4
• Chapter 1 వాస్తవ సంఖ్యలు Ex 1.5
• Chapter 1 వాస్తవ సంఖ్యలు Optional Exercise
• Chapter 1 వాస్తవ సంఖ్యలు InText Questions

AP SSC 10th Maths Solutions Chapter 2 సమితులు

• Chapter 2 సమితులు Ex 2.1
• Chapter 2 సమితులు Ex 2.2
• Chapter 2 సమితులు Ex 2.3
• Chapter 2 సమితులు Ex 2.4
• Chapter 2 సమితులు InText Questions

AP State Maths Textbook Solutions Chapter 3 బహుపదులు

• Chapter 3 బహుపదులు Ex 3.1
• Chapter 3 బహుపదులు Ex 3.2
• Chapter 3 బహుపదులు Ex 3.3
• Chapter 3 బహుపదులు Ex 3.4
• Chapter 3 బహుపదులు Optional Exercise
• Chapter 3 బహుపదులు InText Questions

AP 10th Maths Solutions Chapter 4 రెండు చరరాశులలో రేఖీయ సమీకరణాల జత

• Chapter 4 రెండు చరరాశులలో రేఖీయ సమీకరణాల జత Ex 4.1
• Chapter 4 రెండు చరరాశులలో రేఖీయ సమీకరణాల జత Ex 4.2
• Chapter 4 రెండు చరరాశులలో రేఖీయ సమీకరణాల జత Ex 4.3
• Chapter 4 రెండు చరరాశులలో రేఖీయ సమీకరణాల జత Optional Exercise
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10th Class Maths Guide Telugu Medium Pdf Chapter 5 వర్గ సమీకరణాలు

• Chapter 5 వర్గ సమీకరణాలు Ex 5.1
• Chapter 5 వర్గ సమీకరణాలు Ex 5.2
• Chapter 5 వర్గ సమీకరణాలు Ex 5.3
• Chapter 5 వర్గ సమీకరణాలు Ex 5.4
• Chapter 5 వర్గ సమీకరణాలు Optional Exercise
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SCERT Maths Textbook Class 10 Solutions Chapter 6 శ్రేఢులు

• Chapter 6 శ్రేఢులు Ex 6.1
• Chapter 6 శ్రేఢులు Ex 6.2
• Chapter 6 శ్రేఢులు Ex 6.3
• Chapter 6 శ్రేఢులు Ex 6.4
• Chapter 6 శ్రేఢులు Ex 6.5
• Chapter 6 శ్రేఢులు Optional Exercise
• Chapter 6 శ్రేఢులు InText Questions

AP 10th Class Maths Guide Pdf Download Chapter 7 నిరూపక రేఖాగణితం

• Chapter 7 నిరూపక రేఖాగణితం Ex 7.1
• Chapter 7 నిరూపక రేఖాగణితం Ex 7.2
• Chapter 7 నిరూపక రేఖాగణితం Ex 7.3
• Chapter 7 నిరూపక రేఖాగణితం Ex 7.4
• Chapter 7 నిరూపక రేఖాగణితం Optional Exercise
• Chapter 7 నిరూపక రేఖాగణితం InText Questions

10th Class Maths Textbook SSC Solutions 2021 Chapter 8 సరూప త్రిభుజాలు

• Chapter 8 సరూప త్రిభుజాలు Ex 8.1
• Chapter 8 సరూప త్రిభుజాలు Ex 8.2
• Chapter 8 సరూప త్రిభుజాలు Ex 8.3
• Chapter 8 సరూప త్రిభుజాలు Ex 8.4
• Chapter 8 సరూప త్రిభుజాలు Optional Exercise
• Chapter 8 సరూప త్రిభుజాలు InText Questions

AP SSC 10th Class Maths Guide Pdf Chapter 9 వృత్తాలకు స్పర్శరేఖలు మరియు ఛేదనరేఖలు

• Chapter 9 వృత్తాలకు స్పర్శరేఖలు మరియు ఛేదనరేఖలు Ex 9.1
• Chapter 9 వృత్తాలకు స్పర్శరేఖలు మరియు ఛేదనరేఖలు Ex 9.2
• Chapter 9 వృత్తాలకు స్పర్శరేఖలు మరియు ఛేదనరేఖలు Ex 9.3
• Chapter 9 వృత్తాలకు స్పర్శరేఖలు మరియు ఛేదనరేఖలు Optional Exercise
• Chapter 9 వృత్తాలకు స్పర్శరేఖలు మరియు ఛేదనరేఖలు InText Questions

10th Class Maths Textbook SSC Solutions Telangana 2021 Chapter 10 క్షేత్రమితి

• Chapter 10 క్షేత్రమితి Ex 10.1
• Chapter 10 క్షేత్రమితి Ex 10.2
• Chapter 10 క్షేత్రమితి Ex 10.3
• Chapter 10 క్షేత్రమితి Ex 10.4
• Chapter 10 క్షేత్రమితి Optional Exercise
• Chapter 10 క్షేత్రమితి InText Questions

AP SCERT 10 Class Maths Solutions Chapter 11 త్రికోణమితి

• Chapter 11 త్రికోణమితి Ex 11.1
• Chapter 11 త్రికోణమితి Ex 11.2
• Chapter 11 త్రికోణమితి Ex 11.3
• Chapter 11 త్రికోణమితి Ex 11.4
• Chapter 11 త్రికోణమితి Optional Exercise
• Chapter 11 త్రికోణమితి InText Questions

SSC 10th Class Maths Textbook Pdf Free Download Chapter 12 త్రికోణమితి అనువర్తనాలు

• Chapter 12 త్రికోణమితి అనువర్తనాలు Ex 12.1
• Chapter 12 త్రికోణమితి అనువర్తనాలు Ex 12.2
• Chapter 12 త్రికోణమితి అనువర్తనాలు Optional Exercise
• Chapter 12 త్రికోణమితి అనువర్తనాలు InText Questions

AP 10th Class Maths Textbook Pdf 2021-2022 Chapter 13 సంభావ్యత

• Chapter 13 సంభావ్యత Ex 13.1
• Chapter 13 సంభావ్యత Ex 13.2
• Chapter 13 సంభావ్యత Optional Exercise
• Chapter 13 సంభావ్యత InText Questions

10th Class Maths Textbook Chapter 14 సాంఖ్యకశాస్త్రం

• Chapter 14 సాంఖ్యకశాస్త్రం Ex 14.1
• Chapter 14 సాంఖ్యకశాస్త్రం Ex 14.2
• Chapter 14 సాంఖ్యకశాస్త్రం Ex 14.3
• Chapter 14 సాంఖ్యకశాస్త్రం Ex 14.4
• Chapter 14 సాంఖ్యకశాస్త్రం InText Questions

AP State Board Syllabus 10th Class Textbook Solutions & Study Material

• AP Board 10th Class Maths Solutions
• AP Board 10th Class Physics Solutions
• AP Board 10th Class Chemistry Solutions
• AP Board 10th Class Biology Solutions
• AP Board 10th Class Social Studies Solutions
• AP Board 10th Class English Solutions
• AP Board 10th Class Hindi Solutions
• AP Board 10th Class Telugu Solutions
• AP 10th Class Telugu Important Questions
• AP 10th Class Maths Chapter Wise Important Questions
• AP 10th Class Physical Science Important Questions
• AP 10th Class Biology Important Questions
• AP 10th Class Social Studies Important Questions
• AP 10th Class Maths Bits with Answers
• AP 10th Class Physical Science Bits with Answers
• AP 10th Class Biology Bits with Answers
• AP 10th Class Social Bits with Answers
• AP State Board 10th Class Maths Notes
• AP State Board 10th Class Physical Science Notes
• AP State Board 10th Class Biology Notes
• AP State Board 10th Class Social Studies Notes

Andhra Pradesh SCERT AP State Board Syllabus 10th Class Hindi Textbook Solutions and Study Material Pdf Sparsh Bhag 2 स्पर्श भाग 2 & Sanchayan Bhag 2 संचयन भाग 2 are part of AP Board Solutions Class 10 Hindi.

Students can also read AP 10th Class Hindi Important Questions for exam preparation.

AP Board Solutions Class 10 Hindi Sparsh & Sanchayan – AP 10th Class Hindi Study Material Guide Pdf Download

AP Board Solutions Class 10 Hindi Sparsh स्पर्श

AP Board 10th Class Hindi Textbook Solutions पद्य खंड

• 1st Lesson कबीर की साखी
• 2nd Lesson मीरा के पद
• 3rd Lesson मनुष्यता
• 4th Lesson पर्वत प्रदेश में पावस
• 5th Lesson तोप
• 6th Lesson कर चले हम फ़िदा
• 7th Lesson आत्मत्राण

AP 10th Class Hindi Study Material गद्य खंड

• 8th Lesson बड़े भाई साहब
• 9th Lesson डायरी का एक पन्ना
• 10th Lesson तताँरा-वामीरो कथा
• 11th Lesson तीसरी कसम के शिल्पकार शैलेंद्र
• 12th Lesson अब कहाँ दूसरे के दुख से दुखी होने वाले
• 13th Lesson पतझर में टूटी पत्तियाँ : (i) गिन्नी का सोना (ii) झेन की देन
• 14th Lesson कारतूस

AP Board Solutions Class 10 Hindi Sanchayan संचयन

AP 10th Class Hindi Guide Sanchayan

• 1st Lesson हरिहर काका
• 2nd Lesson सपनों के-से दिन
• 3rd Lesson टोपी शुक्ला
• पत्र लेखन
• निबंध लेखन
• शब्दकोश

AP 10th Class Hindi Study Material Pdf (Old Syllabus)

• Chapter 1 बरसते बदल (कविता)
• Chapter 2 ईदगाह (कहानी)
• यह रास्ता कहाँ जाता है? (पठन हेतु नाटक)
• Chapter 3 हम भारतवासी (कविता)
• ★ शांति की राह में (उपवाचक – निबंध)
• Chapter 4 कण-कण का अधिकारी (कविता)
• Chapter 5 लोकगीत (निबंध)
• उलझन (पठन हेतु कविताएँ)
• Chapter 6 अंतर्राष्ट्रीय स्तर पर हिंदी (पत्र)
• ★ दो कलाकार (उपवाचक – कहानी)
• Chapter 7 भक्ति पद (कविता)
• Chapter 8 स्वराज्य की नींव (एकांकी)
• माँ मुझे आने दे! (पठन हेतु कविता)
• Chapter 9 दक्षिणी गंगा गोदावरी (यात्रा-वृत्तांत)
• ★ अपने स्कूल को एक उपहार (उपवाचक – कहानी)
• Chapter 10 नीति दोहे (कविता)
• Chapter 11 जल ही जीवन है (कहानी)
• क्या आपको पता है? (पठन हेतु पी.पी.टी. निबंध)
• Chapter 12 धरती के सवाल अंतरिक्ष के जवाब (साक्षात्कार)
• ★ अनोखा उपाय (उपवाचक – अनूदित कहानी)
• पत्र लेखन
• निबंध लेखन
• शब्दकोश

Andhra Pradesh SCERT AP State Board Syllabus 10th Class English Textbook Solutions and Study Material Pdf are part of AP Board Solutions Class 10.

AP 10th Class English Guide Study Material Questions and Answers – AP Board Solutions Class 10 English

AP 10th Class English Guide First Flight (Prose)

AP 10th Class English Textbook Answers Pdf – AP 10th Class English Study Material Pdf

• 1st Lesson A Letter to God
• 2nd Lesson Nelson Mandela: Long Walk to Freedom
• 3rd Lesson Two Stories about (I) Flying His First Flight (II) Black Aeroplane
• 4th Lesson From the Diary of Anne Frank
• 5th Lesson Glimpses of India (I) A Baker from Goa (II) Coorg (III) Tea from Assam
• 6th Lesson Mijbil the Otter
• 7th Lesson Madam Rides the Bus
• 8th Lesson The Sermon at Benares
• 9th Lesson The Proposal

AP 10th Class English Study Material First Flight (Poetry)

AP Board Solutions 10th Class English – AP 10th Class English Textbook Solutions

• 1st Poem Dust of Snow, Fire and Ice
• 2nd Poem A Tiger in the Zoo
• 3rd Poem How to Tell Wild Animals, The Ball Poem
• 4th Poem Amanda
• 5th Poem The Trees
• 6th Poem Fog
• 7th Poem The Tale of Custard the Dragon
• 8th Poem For Anne Gregory

AP 10th Class English Textbook Answers Footprints without Feet

AP 10th Class English Study Material Questions And Answers Supplementary Reader Footprints without Feet

• 1st Lesson A Triumph of Surgery
• 2nd Lesson The Thief’s Story
• 3rd Lesson The Midnight Visitor
• 4th Lesson A Question of Trust
• 5th Lesson Footprints without Feet
• 6th Lesson The Making of a Scientist
• 7th Lesson The Necklace
• 8th Lesson Bholi
• 9th Lesson The Book that Saved the Earth

AP 10th Class English Workbook Answers Words and Expressions

Words and Expressions Class 10 Solutions – AP 10th Class English Workbook Words and Expressions Solutions

• Unit 1 A Letter to God
• Unit 2 Nelson Mandela: Long Walk to Freedom
• Unit 3 Two Stories about Flying
• Unit 4 From the Diary of Anne Frank
• Unit 5 Glimpses of India
• Unit 6 Mijbil the Otter
• Unit 7 Madam Rides the Bus
• Unit 8 The Sermon at Benaras
• Unit 9 The Proposal

AP 10th Class English Textbook Answers Solutions (Old Syllabus)

Unit 1 Personality Development

• A. Attitude is Altitude
• B. Every Success Story is Also a Story of Great Failures
• C. I Will Do It

Unit 2 Wit and Humour

• A. The Dear Departed Part 1
• B. The Dear Departed Part 2
• C. The Brave Potter

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Unit 3 Human Relations

• A. The Journey
• B. Once Upon a Time (Poem)
• C. What is My Name?

Unit 4 Films and Theatre

• A. Rendezvous with Ray
• B. Maya Bazaar
• C. A Tribute

Unit 5 Bio-Diversity

• A. Environment
• B. Or will the Dreamer Wake? (Poem)
• C. A Tale of Three Villages

Unit 6 Nation and Diversity

• A. My Childhood
• B. A Plea for India (Poem)
• C. Unity in Diversity in India

Andhra Pradesh SCERT AP State Board Syllabus 10th Class Telugu Important Questions and Answers are part of AP Board Solutions Class 10.

Students can also read AP Board Solutions Class 10 Telugu for board exams.

AP 10th Class Telugu Important Questions and Answers

• 1st Lesson ప్రత్యక్ష దైవాలు Important Questions
• 2nd Lesson బతుకు గంప Important Questions
• 3rd Lesson శతక మాధుర్యం Important Questions
• 4th Lesson ఉపన్యాస కళ Important Questions
• 5th Lesson జలియన్ వాలా బాగ్ Important Questions
• 6th Lesson ప్రకృతి సందేశం Important Questions
• 7th Lesson చేజారిన బాల్యం Important Questions
• 8th Lesson జీవని Important Questions
• 9th Lesson రాజధర్మం Important Questions
• 10th Lesson కన్యాశుల్కం Important Questions
• 11th Lesson యుద్ధ విజేత Important Questions
• 12th Lesson సూక్తి సుధ Important Questions
• ఉపవాచకం రామాయణం
• ఉపవాచకం Chapter 1 బాలకాండ
• ఉపవాచకం Chapter 2 అయోధ్యాకాండ
• ఉపవాచకం Chapter 3 అరణ్యకాండ
• ఉపవాచకం Chapter 4 కిష్కింధాకాండ
• ఉపవాచకం Chapter 5 సుందరకాండ
• ఉపవాచకం Chapter 6 యుద్ధకాండ

Old Syllabus

• Chapter 1 మాతృభావన Important Questions
• Chapter 2 అమరావతి Important Questions
• Chapter 3 జానపదుని జాబు Important Questions
• Chapter 4 వెన్నెల Important Questions
• Chapter 5 ధన్యుడు Important Questions
• Chapter 6 శతక మధురిమ Important Questions
• Chapter 7 మా ప్రయత్నం Important Questions
• Chapter 8 సముద్ర‌లంఘ‌నం Important Questions
• Chapter 9 మాణిక్యవీణ Important Questions
• Chapter 10 గోరంతదీపాలు Important Questions
• Chapter 11 భిక్ష Important Questions
• Chapter 12 చిత్రగ్రీవం Important Questions

Andhra Pradesh SCERT AP State Board Syllabus SSC 10th Class Social Studies Textbook Solutions Study Material Guide Pdf free download in English Medium and Telugu Medium are part of AP Board Solutions Class 10.

Students can also go through AP 10th Class Social Notes to understand and remember the concepts easily. Students can also read AP 10th Class Social Important Questions for board exams.

AP Board Solutions Class 10 Social – AP 10th Class Social Study Material Pdf Download

AP 10th Class Social Study Material English Medium

AP Board Solutions Class 10 Social: History (India and the Contemporary World – II)

• 1st Lesson The Rise of Nationalism in Europe
• 2nd Lesson Nationalism in India
• 3rd Lesson The Making of a Global World
• 4th Lesson The Age of Industrialisation
• 5th Lesson Print Culture and the Modern World

10th Class Social Study Material Pdf: Geography (Contemporary India – II)

• 1st Lesson Resources and Development
• 2nd Lesson Forest and Wildlife Resources
• 3rd Lesson Water Resources
• 4th Lesson Agriculture
• 5th Lesson Minerals and Energy Resources
• 6th Lesson Manufacturing Industries
• 7th Lesson Lifelines of National Economy

10th Class Social Guide: Political Science (Democratic Politics – II)

• 1st Lesson Power Sharing
• 2nd Lesson Federalism
• 3rd Lesson Gender, Religion and Caste
• 4th Lesson Political Parties
• 5th Lesson Outcomes of Democracy

Social Guide 10th Class: Economics (Understanding Economic Development)

• 1st Lesson Development
• 2nd Lesson Sectors of the Indian Economy
• 3rd Lesson Money and Credit
• 4th Lesson Globalisation and the Indian Economy
• 5th Lesson Consumer Rights

AP 10th Class Social Study Material Pdf Download English Medium (Old Syllabus)

• Chapter 1 India: Relief Features
• Chapter 2 Ideas of Development
• Chapter 3 Production and Employment
• Chapter 4 Climate of India
• Chapter 5 Indian Rivers and Water Resources
• Chapter 6 The People
• Chapter 7 People and Settlement
• Chapter 8 People and Migration
• Chapter 9 Rampur: A Village Economy
• Chapter 10 Globalisation
• Chapter 11 Food Security
• Chapter 12 Sustainable Development with Equity
• Chapter 13 The World Between Wars 1900-1950 Part 1
• Chapter 14 The World Between Wars 1900-1950 Part 2
• Chapter 15 National Liberation Movements in the Colonies
• Chapter 16 National Movement in India–Partition & Independence
• Chapter 17 The Making of Independent India’s Constitution
• Chapter 18 Independent India (The First 30 years – 1947-77)
• Chapter 19 Emerging Political Trends 1977 to 2000
• Chapter 20 Post – War World and India
• Chapter 21 Social Movements in Our Times
• Chapter 22 Citizens and the Governments

AP 10th Class Social Study Material Pdf Download Telugu Medium (Old Syllabus)

• Chapter 1 భారతదేశం: భౌగోళిక స్వరూపాలు
• Chapter 2 అభివృద్ధి భావనలు
• Chapter 3 ఉత్పత్తి, ఉపాధి
• Chapter 4 భారతదేశ శీతోష్ణస్థితి
• Chapter 5 భారతదేశ నదులు, నీటి వనరులు
• Chapter 6 ప్రజలు
• Chapter 7 ప్రజలు – నివాస ప్రాంతాలు
• Chapter 8 ప్రజలు – వలసలు
• Chapter 9 రాంపురం : గ్రామ ఆర్థిక వ్యవస్థ
• Chapter 10 ప్రపంచీకరణ
• Chapter 11 ఆహార భద్రత
• Chapter 12 సమానత – సుస్థిర అభివృద్ధి
• Chapter 13 ప్రపంచ యుద్ధాల మధ్య ప్రపంచం 1900-1950 : భాగం-I
• Chapter 14 ప్రపంచ యుద్ధాల మధ్య ప్రపంచం 1900-1950 : భాగం-II
• Chapter 15 వలస పాలిత ప్రాంతాలలో జాతి విముక్తి ఉద్యమాలు
• Chapter 16 భారతదేశ జాతీయోద్యమం-దేశ విభజన, స్వాతంత్య్రం: 1939-1947
• Chapter 17 స్వతంత్ర భారత రాజ్యాంగ నిర్మాణం
• Chapter 18 స్వతంత్ర భారతదేశం (మొదటి ముప్ఫై సంవత్సరాలు-1947-1977)
• Chapter 19 రాజకీయ ధోరణుల ఆవిర్భావం : 1977-2000
• Chapter 20 ప్రపంచ యుద్దాల తరువాత ప్రపంచం, భారతదేశం
• Chapter 21 సమకాలీన సామాజిక ఉద్యమాలు
• Chapter 22 పౌరులు, ప్రభుత్వాలు