{"id":9004,"date":"2023-11-04T14:09:12","date_gmt":"2023-11-04T08:39:12","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=9004"},"modified":"2023-11-10T16:47:29","modified_gmt":"2023-11-10T11:17:29","slug":"ap-board-9th-class-physical-science-important-questions-chapter-9","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-9th-class-physical-science-important-questions-chapter-9\/","title":{"rendered":"AP Board 9th Class Physical Science Important Questions Chapter 9 Floating Bodies"},"content":{"rendered":"

AP State Syllabus AP Board 9th Class Physical Science Important Questions<\/a> Chapter 9 Floating Bodies<\/p>\n

AP State Syllabus 9th Class Physical Science Important Questions 9th Lesson Floating Bodies<\/h2>\n

9th Class Physical Science 9th Lesson Floating Bodies 1 Mark Important Questions and Answers<\/h3>\n

Question 1.
\nScientifically what is density?
\nAnswer:
\nDensity is defined as mass per unit volume.
\n\"AP<\/p>\n

Question 2.
\nDefine relative density.
\nAnswer:
\nRelative density is the ratio of density of an object to the density of water.
\n\"AP<\/p>\n

Question 3.
\nWhat instrument is used to determine the purity of milk?
\nAnswer:
\nPurity of milk can be determined by using lactometer.<\/p>\n

Question 4.
\nWhat instrument is used to know the relative density of liquids?
\nAnswer:
\nRelative density of liquids can be determined by using hydrometer.<\/p>\n

\"AP<\/p>\n

Question 5.
\nWhat is atmospheric pressure?
\nAnswer:
\nAir in our surroundings exerts pressure on the surface of earth called atmospheric pressure.
\nAtmospheric pressure P0<\/sub> = \u03c1hg<\/p>\n

Question 6.
\nState Archimedes’ principle.
\nAnswer:
\nArchimedes’ principle states that when a body is immersed in a fluid it experiences an upward force of buoyancy equal to the weight of fluid displaced by the immersed portion of the body.<\/p>\n

Question 7.
\nState Pascal’s principle.
\nAnswer:
\nPascal’s principle states that external pressure applied to an enclosed body of fluid is transmitted equally in all directions throughout the fluid.<\/p>\n

Question 8.
\nWhat is the principle involved in working of hydraulic lift?
\nHydraulic lift works using Pascal’s law.<\/p>\n

\"AP<\/p>\n

Question 9.
\nIn what direction does the buoyant force on an object immersed in a liquid act?
\nAnswer:
\nBuoyant force on an object acts upward direction.<\/p>\n

9th Class Physical Science 8th Lesson Gravitation 2 Marks Important Questions and Answers<\/h3>\n

Question 1.
\nDefine density and give its units.
\nAnswer:
\nDensity is defined as the mass per unit volume.
\n\"AP
\nUnits : Unit for density is gm\/cm\u00b3 (or) kg\/m\u00b3.<\/p>\n

Question 2.
\nDefine relative density and give its units.
\nAnswer:
\nRelative density is the ratio of density of an object to the density of water.
\n\"AP
\nUnits : Relative density has no units.<\/p>\n

Question 3.
\nWhy does the wooden block float on water even though it is heavier than a marble?
\nAnswer:
\nThe relative density of wooden block is less than the relative density of water whereas relative density of marble (glass) is more than that of water. Hence wooden block floats on water and marble sinks in water.<\/p>\n

\"AP<\/p>\n

Question 4.
\nWhat do we mean by ‘heavy’, what do we mean by ‘light’?
\nAnswer:
\n‘Heavy’ and ‘light’ can be decided based on the density of the object. Among the objects of equal volume, the one which weighs more can be treated as ‘heavy’ and the other one can be treated as ‘light’.<\/p>\n

Question 5.
\nRelative density of silver is 10.8. What is density of silver in S.I. unit?
\nAnswer:
\nRelative density of silver = 10.8
\n\"AP
\nDensity of water in S.I. system = 0\u00b3 kg\/m\u00b3.
\nDensity of silver = Relative density \u00d7 Density of water = 10.8 \u00d7 10\u00b3 kg\/m\u00b3.<\/p>\n

Question 6.
\nWhy does a block of plastic released under come up to the surface of water?
\nAnswer:
\nDensity of plastic is less than water. As the density of water is more than plastic, water exerts upward force which is known as buoyant force. So, the block of plastic come up to the surface of water.<\/p>\n

Question 7.
\nThe volume of 50 g. of a substance is 20 cm\u00b3. If the density of water is 1 g\/cm\u00b3, will the substance float or sink?
\nAnswer:
\n\"AP
\nThe density of substance is more than water so the substance sinks in the water.<\/p>\n

Question 8.
\nAn iron nail sinks in water. While a large ship floats on the surface of water. Give reason.
\nAnswer:
\nThe weight of water displaced by iron needle is less than its own weight, so it sinks. On the other hand, the ship displaces water more than its own weight, so it floats on water.<\/p>\n

Question 9.
\nThe density of gold is 19 g\/cm\u00b3. Find the volume of 95 g of fold.
\nAnswer:
\n\"AP<\/p>\n

Question 10.
\nA floating boat displaces water weighing 6000 newtons.
\na) What is the buoyant force on the boat?
\nb) What is the weight of the boat?
\nAnswer:
\na) By Archimedes’ principles, the buoyant force acting on the object by it.
\nSo, the buoyant force on the boat = 6000 newtons.<\/p>\n

b) According to the principle of floating
\nWeight of object = Weight of liquid displaced by it
\n\u21d2 Weight of the boat = 6000 newtons<\/p>\n

\"AP<\/p>\n

Question 11.
\nThe volume of 50 g. of a substance is 20 cm . If the density of water is 1 g cm-3<\/sup>, will the substance float or sink?
\nAnswer:
\n\"AP
\n2. Density of water = 1 g cm-3<\/sup>.
\nAs the density of substance is greater than that of water, so, the substance will sink in water.<\/p>\n

Question 12.
\nWrite any two differences between density and relative density.
\nAnswer:
\n\"AP<\/p>\n

9th Class Physical Science 8th Lesson Gravitation 4 Marks Important Questions and Answers<\/h3>\n

Question 1.
\nDerive an expression for atmospheric pressure.
\nAnswer:
\n1) All objects on the surface of the earth are subject to constant atmospheric pressure.<\/p>\n

2) We know that the pressure is thrust (force) per unit area.
\n\"AP<\/p>\n

Question 2.
\nHow can you measure the atmospheric pressure?
\nAnswer:
\n1) Atmospheric pressure can be calculated with the help of the height of mercury column.<\/p>\n

2) The weight of the mercury column in the tube is equal to the force applied on it by the mercury in the bowl due to atmospheric pressure.
\n\u2234 Weight of mercury column (W)
\n= Mass of mercury (m) \u00d7 g
\n= Volume \u00d7 density \u00d7 g
\n= Cross sectional area of the tube \u00d7 height of the tube \u00d7 density \u00d7 g
\n= Ah\u03c1g<\/p>\n

Let P0<\/sub> be the atmospheric pressure.
\nForce on the column due to atmospheric pressure = P0<\/sub> A
\nThen Ah\u03c1g = P0<\/sub> A \u21d2 P0<\/sub> = h\u03c1g
\nHere ‘\u03c1’ and ‘g’ are constants. So the height of mercury column depends on the atmospheric pressure.
\nWe know height of mercury column h = 76 cm = 76 \u00d7 10-2<\/sup> m
\nDensity of mercury (\u03c1) = 13.6 gm\/cc = 13.6 \u00d7 10\u00b3 kg\/m\u00b33
\nAcceleration due to gravity (g) = 9.8 m\/sec2
\nP0<\/sub> = h\u03c1g = 76 \u00d7 10-2<\/sup> \u00d7 13.6 \u00d7 10\u00b3 \u00d7 9.8 = 1.01 \u00d7 105<\/sup> kg m\/m\u00b2s\u00b2
\n1 kg m\\sec\u00b2 = 1 Newton
\nP0<\/sub> = 1.01 \u00d7 105<\/sup> N\/m\u00b2
\nThis value is called atmospheric pressure.
\n1 Atmosphere = 1.01 \u00d7 105<\/sup> N\/m\u00b2 = 1.01 \u00d7 105<\/sup> Pascal<\/p>\n

\"AP<\/p>\n

Question 3.
\nCalculate the pressure at a depth ‘h’ in a liquid.
\nAnswer:
\n\"AP
\n1) Let us consider a container which contains a liquid in it of density ‘ p<\/p>\n

2) Consider cylindrical column of height ‘h’ from the surface of the liquid of cross sectional area “A”.<\/p>\n

3) The volume of the liquid column V = Ah
\nMass = Volume x density
\n\u21d2 m = Ah\u03c1
\nWeight W = mg = Ah\u03c1g
\nFrom Newton’s law the net force on the liquid column is zero, because it is at rest.<\/p>\n

4) The forces acting on the water column are
\na) Weight (W), vertically down
\nb) Force on top surface due to atmospheric pressure (P0<\/sub> A) acting vertically down.
\nc) Force on the bottom surface of the column due to static pressure of liquid (PA), acting vertically up.<\/p>\n

5) From Newton’s law we get<\/p>\n

PA = P0<\/sub>A + W \u21d2 PA = P0<\/sub>A + h\u03c1gA =\u21d2 PA = A (P0<\/sub> + h\u03c1g)
\n\u2234 P = P0<\/sub> + h\u03c1g
\nP = Pressure at depth h’ from the surface of the liquid
\nP0<\/sub> = Atmospheric pressure
\n\u2234 The pressure inside the liquid at a constant depth is constant.<\/p>\n

Question 4.
\nCalculate the pressure difference at different levels of depth in fluids.
\nAnswer:
\n\"AP
\n1) Let us consider a cylindrical column of liquid of height ‘h’ with cross sectional area
\n‘A’ and let ‘P’ be the density of the liquid.<\/p>\n

2) The pressure P1<\/sub> in the liquid at depth h, is
\nP1<\/sub> = P0<\/sub> + h1<\/sub>\u03c1g ………(1)<\/p>\n

3) Similarly pressure P2<\/sub> in the liquid at depth h2<\/sub> is
\nP2<\/sub> = P0<\/sub> + h2<\/sub>\u03c1g …………..(2)<\/p>\n

4) From (1) and (2), P2<\/sub> – P1<\/sub> = h2<\/sub>\u03c1g – h1<\/sub>pg
\nP2<\/sub> – p1<\/sub> = \u03c1g (h2<\/sub> – h1<\/sub>)<\/p>\n

5) From the figure h2<\/sub> – h1<\/sub> = h
\n\u2234 P2<\/sub> – p1<\/sub> = h\u03c1g<\/p>\n

6) The pressure difference between two levels in that liquid = h\u03c1g<\/p>\n

7) Here \u03c1 and g are constants, so the pressure difference increases with an increase in depth.<\/p>\n

Question 5.
\nState Pascal’s principle and explain it with an example.
\n(OR)
\nWrite Pascal’s law. Draw the diagram of any device works based as Pascals’ law.
\nAnswer:
\nPascal’s principle :
\nPascal’s principle states that external pressure applied to a enclosed body of fluid is transmitted equally in all directions throughout the fluid volume and the walls of the containing vessel.
\n\"AP<\/p>\n

Explanation :<\/p>\n

    \n
  1. Look at the figure.<\/li>\n
  2. Here we have an enclosed volume of fluid in a U-shaped tube.<\/li>\n
  3. The fluid is enclosed in the tube by two leak-proof pistons in each arm.<\/li>\n
  4. The ratio of cross-section areas of the right and left tubes is A2<\/sub> : A1<\/sub> and A2<\/sub> > A1<\/sub><\/li>\n
  5. When a force F1<\/sub> is applied to the left piston, the excess pressure acting on the fluid volume is F1<\/sub>\/A1<\/sub>.<\/li>\n
  6. According to Pascal’s principle, this excess pressure is transmitted equally throughout the fluid volume.<\/li>\n
  7. The excess pressure in the right-side tube (Area A2<\/sub>) is also F1<\/sub>\/A1<\/sub> and since its area is A2<\/sub>, the upward force acting on the right piston is F2<\/sub> = A2<\/sub> \u00d7 F1<\/sub>\/A1<\/sub>, which is much larger in magnitude than F1<\/sub><\/li>\n
  8. Thus the application of Pascal’s principle results in a large upward force on the right piston when a small downward force is applied on the left piston.<\/li>\n<\/ol>\n

    \"AP<\/p>\n

    Question 6.
    \nThe volume of a solid of mass 500g is 350 cm\u00b3.
    \na) What will be the density of this solid?
    \nb) What will be the relative density of the solid?
    \nc) Will it float or sink in water?
    \nAnswer:
    \n\"AP
    \nc) Relative density of solid is greater than the relative density of water. So, it will sink in water.<\/p>\n","protected":false},"excerpt":{"rendered":"

    AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 9 Floating Bodies AP State Syllabus 9th Class Physical Science Important Questions 9th Lesson Floating Bodies 9th Class Physical Science 9th Lesson Floating Bodies 1 Mark Important Questions and Answers Question 1. Scientifically what is density? Answer: Density is defined as mass per …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/9004"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=9004"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/9004\/revisions"}],"predecessor-version":[{"id":36454,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/9004\/revisions\/36454"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=9004"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=9004"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=9004"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}