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Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory

Very Short Answer Questions

Question 1.
Define mean free path. [A.P. Mar. 18, 17, 15; T.S. Mar. 17, 15]
Answer:
The average distance covered by a molecule between two successive collisions is called the mean free path.

Question 2.
How does kinetic theory justify Avogadro’s hypothesis and show that Avogadro Number in different gases is same ?
Answer:
For two different gases, we have \(\frac{P_1 V_1}{N_1 T_1}=\frac{P_2 V_2}{N_2 T_2}\) = K_B (constant)
If P, V, T are same, then N is also same for two gases.
N is called Avogadro’s number. According to Avogadro’s hypothesis, that the number of molecules per unit volume is same for all gases at a fixed temperature and pressure.
In this way kinetic theory justify Avogadro’s hypothesis.

Question 3.
When does a real gas behave like an ideal gas ? [T.S. Mar. 16; Mar. 14]
Answer:
At low pressures and high temperatures real gases behave like an ideal gas.

Question 4.
State Boyle’s Law and Charle’s Law. [A.P. Mar. 18; T.S. Mar. 15]
Answer:
Boyle’s law : At constant temperature, the pressure of a given mass of gas varies inversely with volume.
Charle’s law: At constant pressure, the volume of the gas is proportional to its absolute temperature.

Question 5.
State Dalton’s law of partial pressures. [T.S. Mar. 18; A.P. Mar. 16; Mar. 14]
Answer:
The total pressure exerted by a mixture of non-reacting gases occupying a vessel is equal to the sum of the individual pressures which each gas would exert if it is alone occupied the whole vessel.
i.e., P = P₁ + P₂ + ………….

Question 6.
Pressure of an ideal gas in container is independent of shape of the container – explain. [T.S. Mar. 17]
Answer:
The kinetic theory expression for the pressure of a given mass of an ideal gas in a container is 1 -2.
P = \(\frac{1}{3}\) nm\(\bar{v}^2\), where \(\overline{\mathrm{V}}\) indicates mean-square-speed, n is number of molecules, m is mass of molecule. From the above equation, shape of the container is immeterial. Hence pressure of an ideal gas is independent of shape of the container.

Question 7.
Explain the concept of degrees of freedom for molecules of a gas.
Answer:
The degrees of freedom of a particle indicate the number of independent motions which the particle can undergo, or the number of independent methods of exchanging energy.

A monoatomic molecule (He) has three degrees of freedom, a diatomic (H₂, O₂) one has five, while a triatomic (H₂O) one has six.

Question 8.
What is the expression between pressure and kinetic energy of a gas molecules ? [AP – Mar. ’17, ’16, ’15]
Answer:
By kinetic theory pressure, P = \(\frac{1}{3} m n \overline{V}^2\) and kinetic energy = \(\frac{1}{2} m n \overline{V}^2\) where m is the mass of the molecules, n is the number of moles per unit volume, v is the mean-square-speed.
∴ P = \(\frac{2}{3}\left(\frac{1}{2} m n \overline{V}^2\right)\) = \(\frac{2}{3}\) (kineticenergy)

Question 9.
The absolute temperature of a gas is increased 3 times. What will be the increase in rms velocity of the gas molecule ?
Answer:
Case (i) : The r.m.s velocity of gas molecule, \(\overline{V}_1=\sqrt{\frac{3 R T_1}{M}}\)
Case (ii) : The r.m.s velocity of gas molecule, \(\overline{V}_2=\sqrt{\frac{3 R T_2}{M}}\)
\(\frac{\overline{\mathrm{V}}_2}{\overline{\mathrm{V}}_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=\sqrt{\frac{3 \mathrm{~T}_1}{\mathrm{~T}_1}}\)  [∵ T₂ = 3T₁] ⇒ \(\overline{\mathrm{V}}_2=\sqrt{3} \overline{\mathrm{V}}_1=1.732 \overline{\mathrm{V}}_1\)
∴ Increase in r.m.s velocity of gas molecule = \(\overline{\mathrm{V}}_2-\overline{\mathrm{V}}_1=1.732 \overline{\mathrm{V}}_1-\overline{\mathrm{V}}_1=0.732 \overline{\mathrm{V}}_1\)

Short Answer Questions

Question 1.
Explain the kinetic interpretation of Temperature.
Answer:
Kinetic interpretation of Temperature : Since the pressure of the gas is given by
P = \(\frac{1}{3} \mathrm{mn} \overline{\mathrm{V}}^2\), where m = mass of the gas, n = \(\frac{\mathrm{N}}{\mathrm{V}}\) = Number of molecules per unit volume
\(\overline{V}\) = r.m.s velocity of gas, P = \(\frac{1}{3} m \frac{N}{V} \overline{V}^2\) ⇒ PV = \(\frac{1}{3} \mathrm{mN} \overline{\mathrm{V}}^2\)
Also for a gram molecule of the gas, PV = RT
Hence RT = \(\frac{1}{3} m N \overline{V}^2 \Rightarrow 3 \frac{R T}{N}=m \overline{V}^2 \text { or } \frac{1}{2} m \overline{V}^2=\frac{3}{2} K_B T\) [∵ \(\frac{R}{N}\) = K_B]
Here K_B is Boltzman constant. So mean kinetic energy of a molecule is \(\frac{3}{2}\) K_B T, which depends upon the temperature. As temperature increases mean kinetic energy of the molecules also increases.

Question 2.
How specific heat capacity of mono atomic, diatomic and poly atomic gases can be explained on the basis of Law of equipartition of Energy ? [Mar. 13]
Answer:
Specific heat capacity:
1) Mono atomic gases : According to law of equipartition of energy, a molecule of mono atomic gas has only 3 (translational) degrees of freedom, i.e., f = 3.
The molar specific heat of the gas at constant volume is given by C_v = \(\frac{f}{2}\) R where f is degree of freedom.
∴ C_v = \(\frac{3}{2}\)R = 3 cal/mole k [∵ R = 2 cal/mole – k]
The molar specific heat at constant pressure is given by
C_p = (\(\frac{f}{2}\) + 1)R = (\(\frac{3}{2}\) + 1) R = \(\frac{5}{2}\)R = 5 cal/ m0le – k

2) Diatomic gases : A molecule of diatomic gas has 5 degrees of freedom, 3 translational and 2 rotational i.e., f = 5
Therefore C_v = \(\frac{f}{2}\) R = \(\frac{5}{2}\)R = 5 cal/mole – k, C_p = (\(\frac{f}{2}\) + 1) R = \(\frac{7}{2}\) R = 7 cal/mole – k

3) Polyatomic gases : Polyatomic molecule has 3 translational, 3 rotational degrees of freedom, i. e., f = 6
Therefore C_v = \(\frac{f}{2}\)R = 3R = 6 cal / mole – k, C_p = (\(\frac{f}{2}\) + 1) = 4R = 8 cal / mole – k.

Question 3.
Explain the concept of absolute zero of temperature on the basis of kinetic theory.
Answer:
Concept of absolute zero on the basis of kinetic theory :
Since the pressure of the gas is given by
P = \(\frac{1}{3} \mathrm{mn} \overline{\mathrm{V}}^2\), where m = mass of the gas, n = \(\frac{\mathrm{N}}{\mathrm{V}}\) = number of molecules per unit volume.
V = r.m.s velocity of gas, P = \(\frac{1}{3} \frac{\mathrm{mN}}{\overline{\mathrm{V}}} \overline{\mathrm{V}}^2\) ⇒ PV = \(\frac{1}{3} m N \overline{V}^2\)
Also for a gram molecule of the gas, PV = RT
Hence RT = \(\frac{1}{3} M \overline{V}^2\) [∵ M = mN = molecular weight]
\(\overline{\mathrm{V}}^2=\frac{3 R T}{M}\)
∴ \(\overline{v}=\sqrt{\frac{3 R T}{M}}\)
Putting T = 0 in the above equation \(\overline{\mathrm{V}}\) = 0. Hence pressure becomes zero. Then the gas converts into liquids. Thus, this temperature is called absolute zero.

Question 4.
Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.
Answer:
Since the pressure of the gas is given by
P = \(\frac{1}{3} \mathrm{mn} \overline{\mathrm{V}}^2\), where m = mass of the gas, n = \(\frac{\mathrm{N}}{\mathrm{V}}\) = number of molecules per unit volume
\(\) = r.m.s velocity of gas, P = \(\frac{1}{3} m \frac{N}{V} \overline{V}^2\) ⇒ PV = \(\frac{1}{3} \mathrm{mNV}^2\)
Also for a gram molecule of the gas, PV = RT
Hence RT = \(\frac{1}{3}\) mNV²
3\(\frac{\mathrm{R}}{\mathrm{V}}\)T = \(\mathrm{m} \overline{\mathrm{V}}^2\) or \(\frac{1}{2}\) mV² = \(\frac{3}{2}\)K_BT [∵ \(\frac{\mathrm{R}}{\mathrm{V}}\) = K_B]
Where KB is the Boltzmann constant and T is the absolute temperature. ∴ E = \(\frac{3}{2}\)K_BT
Hence the average kinetic energy of a molecule is proportional to the absolute temperature of the gas.

Question 5.
Two thermally insulated vessels 1 and 2 of volumes V₁ and V₂ are joined with a valve and filled with air at temperatures (T₁, T₂) and pressures (P₁, P₂) respectively. If the valve joining the two vessels is opened, what will be the temperature inside the vessels at equilibrium.
Answer:
During adiabatic process, there is no loss of energy, i.e. K.E_T of molecules before mixing = K.E_T of molecules after mixing.

Question 6.
What is the ratio of r.m.s speed of Oxygen and Hydrogen molecules at the same temperature ?
Answer:
The r.m.s speed of gas is \(\overline{\mathrm{V}}=\sqrt{\frac{3 R T}{M}} \Rightarrow \frac{\overline{\mathrm{V}}_0}{\mathrm{~V}_{\mathrm{H}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}}}{\mathrm{M}_{\mathrm{O}}}}\), Given M_H = 2 and M₀ = 32
\(\frac{\overline{\mathrm{V}}_0}{\overline{\mathrm{V}}_{\mathrm{H}}}=\sqrt{\frac{2}{32}}=\frac{1}{4}\)
∴ V₀ : V_H = 1 : 4

Question 7.
Four molecules of a gas have speeds 1, 2, 3 and 4 km/s. Find the rms speed of the gas molecule.
Answer:
Given V₁ = 1 km/s; V₂= 2 km/s; V₃ = 3 km/s; V₄ = 4 km/s; = ?
V_rms = \(\sqrt{\frac{V_1^2+V_2^2+V_3^2+V_4^2}{n}}=\sqrt{\frac{1^2+2^2+3^2+4^2}{4}}\)
= \(\sqrt{\frac{1+4+9+16}{4}}=\sqrt{\frac{30}{4}}=\sqrt{7.5}\)
= 2735 kms^-1

Question 8.
If a gas has ‘f’ degrees of freedom, find the ratio of C_P and C_V.
Answer:
If a gas has ‘f’ degrees of freedom, then
C_V = \(\frac{f}{2}\)R and C_P = C_V + R = \(\frac{f}{2}\) R + R = (\(\frac{f}{2}\) + 1)R
Hence the ratio of the two specific heats of a gas is, r = \(\frac{C_p}{C_V}=\left[\frac{\frac{f}{2}+1}{\frac{f}{2} R}\right]=1+\frac{2}{f}\)

Question 9.
Calculate the molecular K.E of 1 gram of Helium (Molecular weight 4) at 127°C. Given R = 8.31 J mol^-1 K^-1.
Answer:
Given, t = 127°C, T = 273 + 127 = 400; R = 8.31 J mol^-1 K^-1
K.E. = \(\frac{3}{2}\)K_BT = \(\frac{3}{2}\) × 1.38 × 10^-23 × 400 = 8.28 × 10^-21 J.

Question 10.
When pressure increases by 2%, what is the percentage decrease in the volume of a gas, assuming Boyle’s law is obeyed ?
Answer:
The gas obeys the Boyles law PV = constant, on differentiating, PdV + VdP = 0
PdV = – V dP ⇒ \(\frac{\mathrm{dV}}{\mathrm{V}}=\frac{-\mathrm{dP}}{\mathrm{P}}\) ⇒ \(\frac{\mathrm{dV}}{\mathrm{V}}\) × 100% = –\(\frac{-\mathrm{dP}}{\mathrm{P}}\) × 100%
% change in volume = – percentage change in pressure
∴ % change in volume = – 2%
Here negative sign indicates decrease in volume.

Long Answer Questions

Question 1.
Derive an expression for the Pressure of an ideal gas in a container from Kinetic Theory and hence give Kinetic Interpretation of Temperature.
Answer:
Let us consider cubical vessel of side T with perfectly elastic wall containing gas molecules. Let the three sides of the cube be taken as co-ordinate axis. Consider a molecule moving with velocity V, in any direction at any instant. The components of V₁ along the three sides are V_x, V_y and V_z respectively. Then
\(\overline{V}^2=V_x^2+V_y^2+V_z^2\) ……………………. (1)
If ‘m is the mass of this molecule, it transfers a momentum mV when it strikes the face ABCD of the cube. Since the wall is perfectly elastic, this molecule is reflected back with a velocity – V_x and momentum – mV_x. So the change in momentum = mV_x – (- mV_x)
= 2mV_x
This molecule then travels towards the opposite face, collide with it, rebounds and travels again towards the face ABCD. The distance travelled between two successive collisions is 2l. Time taken between two successive collisions is \(\frac{2 l}{\mathrm{~V}_{\mathrm{x}}}\)

∴ Number of collisions per second = \(\frac{V_x}{2 l}\)
Change in momentum per second = (2mV_x) × \(\left(\frac{\mathrm{V}_{\mathrm{x}}}{2 l}\right)=\frac{\mathrm{mV}_{\mathrm{x}}^2}{l}\)
Thus force exerted by this molecule = \(\frac{\mathrm{mV}_{\mathrm{x}}^2}{l}\) [∵ F = \(\frac{\mathrm{dP}}{\mathrm{dt}}\)]
Hence, force exerted by N such molecule in the X-direction.
f₁ = \(\mathrm{N} \frac{\mathrm{mV}_{\mathrm{x}}^2}{l}=\frac{\mathrm{mNV}_{\mathrm{x}}^2}{l}\)
∴ Pressure exerted by the molecules in the x-direction is P_x = \(\frac{f_1}{l^2}=\frac{m N \overline{V}_x^2}{l^3}\)
Similarly, pressure exerted by the molecules in the y and z-directions are
P_y = \(\frac{\mathrm{m}}{l^3} \mathrm{~N} \overline{\mathrm{V}}_{\mathrm{y}}^2\) and P_z = \(\frac{\mathrm{m}}{l^3} \mathrm{~N} \overline{\mathrm{V}}_{\mathrm{z}}^2\)
Since the pressure exerted by a gas in all the directions is same, the average pressure
P = \(\frac{P_x+P_y+P_z}{3}=\frac{m N}{3 l^3}\left[V_x^2+v_y^2+V_z^2\right]=\frac{m N}{3 V} \overline{V}^2\) [ from (1) and V = l³]
Here V² is the mean square velocity of the molecule, V is the volume of the vessel.
If M is the mass of the gas then M = mN
∴ P = \(\frac{1}{3} \frac{M^2}{V}=\frac{1}{3} m n \overline{V}^2\) [∵ n = \(\frac{N}{V}\)]
This pressure is actually the pressure exerted by an ideal gas.

b) Kinetic interpretation of temperature :
Since the pressure of the gas is given by
P = \(\frac{1}{3} m n \overline{V}^2\), where m = mass of the gas, n = \(\frac{N}{V}\) = Number of molecules per unit volume
\(\overline{V}\) = r.m.s velocity of gas

Here K_B is Boltzmann constant. So mean kinetic energy of a molecule is \(\frac{3}{2}\) K_BT, which depends upon the temperature. As temperature increases means kinetic energy of the molecules also increases.

Additional Problems

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
Solution:
Here, diameter,
d = 3A°, r = \(\frac{\mathrm{d}}{2}=\frac{3}{2}\) A° = \(\frac{3}{2}\) × 10^-8 cm.
Molecular volume,
v = \(\frac{4}{3}\) πr³N, where N is Avagadros No.
= \(\frac{4}{3}\) × \(\frac{22}{7}\) (1.5 × 10^-8)³ × (6.023 × 10²³)
= 8.52 cc.
Actual volume occupied by 1 mole of oxygen at STP v = 22400 cc
= \(\frac{v}{v^{\prime}}=\frac{8.52}{22400}\) = 3.8 × 10^-4 = 4 × 10^-4.

Question 2.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C.) Show that it is 22.4 litres.
Solution:
For one mole of an ideal gas, ρv = RT
∴ v = \(\frac{\mathrm{RT}}{\rho}\)
Put R = 8.31 J mole^-1 k^-1, T = 273k, ρ = 1 atm = 1.013 × 10⁵ Nm^-2
v = \(\frac{8.31 \times 273}{1.013 \times 10^5}\) = o.0224m³ = 0.0224 × 10⁶
cc = 22400 cc = 22.4 lit.

Question 3.
Figure 14.8 shows plot of PV/T versus P for 1.00 × 10^-3 kg of oxygen gas at two different temperatures.

a) What does the dotted plot signify ?
b) Which is true: T₁ > T₂ or T₁ < T₂ ?
c) What is the value of PV/T where the curves meet on they y-axis ?
d) If we obtained similar plots for 1.00 × 10^-3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis ? If not, what mass of hydrogen yields the same value of PV/T (for low pressurehigh temperature region of the plot) ? (Molecular mass of H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mole^-1 K^-1.)
Solution:
a) The dotted plot shows that \(\frac{\mathrm{PV}}{\mathrm{T}}\) (= μR) is a constant quantity, independent of pressure p. This sign signifies the ideal gas behaviour.

b) The curve at temperature T₁ is closer to dotted plot than the curve at temperature T₂. As the behaviour of a real gas approaches the behaviour of a perfect gas when temperature is increased, therefore T₁ > T₂.

c) Where the two curves meet, the value of
\(\frac{\mathrm{PV}}{\mathrm{T}}\) on y-axis is equal to μR.
As mass of oxygen gas = 1.00 × 10^-3 kg = 1g
\(\frac{\mathrm{PV}}{\mathrm{T}}\) = nR (\(\frac{1}{32}\)) × 8.31 Jk^-1.

d) If we obtained similar plots for 1.00 × 10^-3 kg of hydrogen, we will not get the same value of \(\frac{\mathrm{PV}}{\mathrm{T}}\) at the point, where the curve meets on the y-axis. This is because molecular mass of hydrogen is different from that of oxygen.
For same value of \(\frac{\mathrm{PV}}{\mathrm{T}}\) mass of hydrogen required is obtained from.
\(\frac{\mathrm{PV}}{\mathrm{T}}\) = nR = \(\frac{\mathrm{M}}{2.02}\) × 8.31 = 0.26
M = \(\frac{2.02 \times 0.26}{8.31}\) gm = 6.32 × 10^-2 grams.

Question 4.
All oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol^-1 K^-1, molecular mass of O2 = 32 u).
Solution:
Initially in the oxygen cylinder,
v₁ = 30 lit = 30 × 10^-3 m³
P₁ = 15 atm = 15 × 1.01 × 10⁵ Pa;
T₁ = 27 + 273 = 300 k.
If the cylinder contains n₁ mole of oxygen gas, then
P₁v₁ = nRT₁
(or) n₁ = \(\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{RT}_1}=\frac{\left(15 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 300}\)
= 18.253 .
For oxygen, molecular weight, M = 32 g
Initial mass of cylinder,
M₁ = n₁ M = 18.253 × 32 = 584.1 g
Finally in the oxygen cylinder, let n₂ moles of O₂ be left, were,
v₂ = 30 × 10^-3 m³, P₂ = 11 × 1.01 × 10⁵ Pa, T₂ = 17 + 273 = 29K
Now
n₂ = \(\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{RT}_2}=\frac{\left(11 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 290}\)
= 13.847
∴ Final mass of oxygen gas in the cylinder,
m₂ = 13.847 × 32 = 453.1 g
∴ Mass of oxygen taken out = m₁ – m₂ = 631.0 g.

Question 5.
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grown when it reaches the surface, which is at a temperature of 35 °C ?
Solution:
v₁ = 1.0 cm³ = 1.0 × 10^-6 m³,
T₁ = 12 °C = 12 + 273 = 285 k
P₁ = 1 atm + h₁ ρg = 1.01 × 10⁵ + 40 × 10³ × 9.8
= 493000 Pa.
When the air bubble reaches at the surface of lake, then
v₂ = 2, T₂ = 35 °C = 35 + 273 = 308 K,
P₂ = 10 atm = 1.01 × 10⁵ pa.
Now \(\frac{P_1 v_1}{T_1}=\frac{P_2 v_2}{T_2}\) (or) v₂ = \(\frac{P_1 v_1 T_2}{T_1 P_2}\)
∴ v₂ = \(\frac{(493000) \times\left(1.0 \times 10^{-6}\right) \times 308}{285 \times 1.01 \times 10^5}\)
= 5.27 × 10^-6m³.

Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.
Solution:
Here, v = 25.0 m³, T = 27 + 273 = 300 k,
k = 1.38 × 10^-23 Jk^-1
Now Pv = nRT = n(NK) T = (nN) kT = N’kT
When nN = N’ = total no. of air molecules in the given gas.
N’ = \(\frac{\mathrm{Pv}}{\mathrm{kT}}=\frac{\left(1.01 \times 10^5\right) \times 25}{\left(1.38 \times 10^{-23}\right) \times 300}\) = 6.10 × 10²⁶

Question 7.
Estimate the average thermal energy of helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Solution:
i) Here, T = 27 °C = 27 + 273 = 300 k
Average thermal energy
= \(\frac{3}{2}\) kT = \(\frac{3}{2}\) × 1.38 × 10^-23 × 300
= 6.2 × 10^-21 J

ii) At T = 6000 k. Average thermal energy = \(\frac{3}{2}\)kT = \(\frac{3}{2}\) × 1.38 × 10^-23 × 6000
= 1.24 × 10^-19 J.

iii) At T = 10 million K = 10⁷ K.
Average thermal energy
= \(\frac{3}{2}\)kT = \(\frac{3}{2}\) × 1.38 × 10^-23 × 10⁷
= 2.1 × 10^-16 J.

Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (poly-atomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases ? If not, in which case is u_rms the largest ?
Solution:
All the three vessels (at the same temperature and pressure) have same volume. So in accordance with the avagadro’s law, the three vessels will contain equal number of respective molecules, being equal to
Avagadro’s number N = 6.023 × 10²³
As v_rms = \(\sqrt{\frac{3 k T}{m}}\)
i.e., v_rms ∝ \(\frac{1}{\sqrt{m}}\) at a given temp, therefore, rms speed of molecules will not be same in the three cases.
As neon has the smallest mass, therefore, rms speed will be largest in case of neon.

Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Solution:
Let c and c’ be the rms velocity of argon and a helium gas atoms at temperature TK and TK’ respectively.
Here μ = 39.9, μ = 4.0,
T = 2, T’ = -20 + 273 = 253 K.

Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂ = 28.0 u).
Solution:
Here, λ = 2, f = 2
P = 2 atm = 2 × 1.013 × 10⁵ Nm^-2
T = 17 °C = (17 + 273) K = 290 k
σ = 2 × 1 = 2A° = 2 × 10^-10 m
K = 1.38 × 10^-23 J molecule^-1 k^-1,
μ = 28 × 10^-3 kg
λ = \(\frac{\mathrm{KT}}{\sqrt{2} \pi \sigma^2 \rho}\)
= \(\frac{1.38 \times 10^{-23} \times 290}{1.414 \times 3.14\left(2 \times 10^{-10}\right)^2 \times 2.026 \times 10^5}\)
λ = \(\frac{1.38 \times 29 \times 10^{-7}}{1.44 \times 3.14 \times 2.06}\) = 1.11 × 10^-7 m.
v_rms = \(\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 \times 8.31 \times 290}{28 \times 10^{-3}}}\) = 508.24 m/s.
Collision frequency = no. of collisions per second = \(\frac{v_{r m s}}{\lambda}=\frac{508.24}{1.11 \times 10^{-7}}\) = 4.58 × 10⁹

Question 11.
A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom ?
Solution:
When the tube is held horizontally, the mercury of length 76 cm traps length of air = 15 cm. A length of 9 cm of the tube will be left at the open end. Fig. (a). The pressure Of air enclosed in tube will be atmospheric pressure. Let area of cross section of the tube be 1 sq.cm.
∴ P₁ = 76 cm and v₁ = 15 cm³.
When the tube is held vertically, 15 cm air gets another 9 cm of air (filled in the right hand side position) and let h cm of the mercury flows out to balance the atmospheric pressure. Fig. (b). Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively.
The pressure of air = 76 – (76 – h) = h cm of mercury

∴ v₂ = (24 + h) cm³ and P₂ = hcm.
If we assume that temperature remains constant, then
P₁v₁ = P₂v₁ or 76 × 15 = h × (24 + h) or h² + 24h – 1140 = 0
(or) h = \(\frac{-24 \pm \sqrt{(24)^2+4 \times 1140}}{2}\) = 23.8 cm
(or) -47.8 cm.
Since h cannot be negative, therefore h = 23.8 cm.
In vertical position it flows out (23.8) cm.

Question 12.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³s^-1. Identify the gas.
[Hint: Use Graham’s law of diffusion : R₁/R₂ = (M₂/M₁) where R₁, R₂ are diffusion rates of gases 1 and 2 and M₁ and M₂ their respective molecular masses. The law is a simple consequence of kinetic theory.]
Solution:
According to Graham’s law of diffusion
\(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\)
Where
r₁ = diffusion rate of hydrogen = 28.7 cm³s^-1
r₂ = diffusion rate of unknown gas = 7.2 cm³s^-1
M₁ = molecule wt. of hydrogen = 2u M2 = ?
M₂ = ?
∴ \(\frac{28.7}{7.2}=\sqrt{\frac{\mu_2}{\mu_1}} \text { or } M_2=\left(\frac{28.7}{7.2}\right)^2 \times 2\) = 31.78
≈ 32 µ

Question 13.
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so- called law of atmospheres.
n₂ = n₁ exp [-mg(h₂ – h₁)/K_BT] Where n₂, n₁ refer to number density at heights h₂ and h₁ respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column :
n₂ = n₁ exp [-mg N_A (ρ – P) (h₂ – h₁)/(ρRT]
Where ρ is the density of the suspended particle and ρ that of surrounding medium. [N_A is Avogadro’s number and R the universal gas constant.]
[Hint: Use Archimedes principle to f ind the apparent weight of the suspended particle.]
Solution:
According to the law of atmospheres n₂ = n₁ exp. [-\(\frac{m g}{K_B T}\) (h₂ – h₁)] ………… (1)
Where n₂ n₁ refer to number density of particles at heights h₂ and h₁ respectively. If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we will have to take effective weight of suspended particles.
Let
v = average vol. of suspended particle
ρ = density of suspended particle
ρ’ = density of liquid
m = mass of one suspended particle
m’ = mass of equal vol. of liquid displaced
According to Archimedis principle, effective weight of one suspended particle.
= actual weight – weight of liquid displaced
= mg – m’ g
= mg – v P’g = mg – (m/P) P’g = mg \(\left(\frac{1-\rho^{\prime}}{\rho}\right)\)
Also boltzman constant, K_B = \(\frac{R}{N A}\) ………….. (i)
Where R is gas constant and NA is Avagadros number.
Putting mg \(\left(\frac{1-\rho^{\prime}}{\rho}\right)\) in place of mg and value of K_B in (i).
We get
n₂ = n₁ exp \(\left[-\frac{\text { mgNA }}{R T} \cdot\left(1-\frac{\rho^{\prime}}{\rho}\right)\left(h_2-h_1\right)\right]\)
Which is the required relation.

Question 14.
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

[Hint: Assume the atoms to be tightly packed1 in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of few Å].
Solution:
If r is the radius of the atom, then volume of each atom = \(\frac{4}{3}\) πr³
Volume of all atoms in one mole of substance
= \(\frac{4}{3}\) πr³ × N = μ/P
r = \(\left[\frac{3 \mu}{4 \pi \mathrm{PN}}\right]^{\frac{1}{3}}\)
For carbon p = 12.01 × 10^-3 kg,
P = 2.22 × 10³ kg m^-3
r = \(\frac{3 \times 12.01 \times 10^{-3}}{4 \times \frac{22}{7} \times\left(2.2210^3\right) \times\left(6.02310^{23}\right)}\)
r = 1.29 × 10^-10 m = 1.29A°
Similarly for gold
r = 1.59A°
For liquid nitrogen r = 1.77A°
For lithium r = 1.73A°
For liquid fluorine r = 1.88A°

Textual Examples

Question 1.
The density of water is 1000 kg m^-3. The density of water vapour at 100° C and 1 atm pressure is 0.6 kg m^-3. The volume of a molecule multiplied by the total number gives. What is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure.
Answer:
For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is 1000/0.6 = /(6 × 10^-4) times, large. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. 6 × 10^-4.

Question 2.
Estimate the volume of a water molecule using the data in Example 1.
Answer:
In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water = 1000 kgm^-3. We know that 1 mole of water has a mass approximately equal to (2 + 16)g = 18 g = 0.018 kg.
Since 1 mole contains about 6 × 10²³ molecules (Avogadro’s number).
The mass of a molecule of water is (0.018) / (6 × 10²³) kg = 3 × 10^-26 kg.
Volume of a water molecule = (3 × 10^-26 kg) (1000 kg m^-3)
= 3 × 10^-29 m³
= (4/3)π (Radius)³
∴ Radius » 2 × 10^-10 m = 2Å

Question 3.
What is the average distance between atoms (interatomic distance) in water ? Use the data given in Examples 1 and 2.
Answer:
A given mass of water in vapour state has 1.67 × 10³ times the volume of the same mass of water inliquid state (e.g. 1). When volume increases by 10³ times the radius increases by V^1/3 or 10 times,
i.e., = 10 × 12 = 2Å.
So the average distanceis 2 × 10 40 Å.

Question 4.
A vessel contains two non-reactive gases : neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3 : 2. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of Ne = 20.2 u, molecular mass of O₂ = 32.0 u.
Answer:
Each gas (assumed ideal) obeys the gas law. Since V and T are common to the two gases, we have P₁V = μ₁ RT and P₂V = μ₂ RT. i.e., (P₁ / P₂) = (μ₁ / p₂). Here 1 and 2 refer to neon and oxygen respectively.
Since (P₁ / P₂) = (3/2) (given), (μ₁ / μ₂) = 3/2.

i) By definition μ₁ = (N₁ /N_A) and μ₂ = (N₂/N_A) where N₁ and N₂ are the number of molecules of 1 and 2, and N_A is the Avogadro’s number. Therefore,
(N₁/N₂) = (μ₁/μ₂) = 3/2.

ii) We can also write μ₁ = (m₁/ M₁) and μ₂ = (m₂ / M₂) where m₁ and m₂ are their molecular masses. (Both m₁ and M₁; as well as m₂ and M₂ should be expressed in the same units). If ρ₁ and ρ₂ are the mass densities of 1 and 2 respectively, we helve
\(\frac{\rho_1}{\rho_2}=\frac{m_1 / V}{m_2 / V}=\frac{m_1}{m_2}=\frac{\mu_1}{\mu_2} \times\left[\frac{M_1}{M_2}\right]\)
= \(\frac{3}{2} \times \frac{20.2}{30.2}\) = 0.947

Question 5.
A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixtures is 27° C. Obtain the ratio of (i) average kinetic energy per molecule and (ii) root mean square speed v_rms of the molecules of the two gases. Atomtic mass of argon = 39.9 u. Molecular mass of chlorine = 70.9 u.
Answer:
The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to (3/2) k_BT. It depends only on temperature and is independent of the nature of the gas.
i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1 : 1.
ii) Now 1/2 mv_rms² = average kinetic energy . per molecule = (3/2) K_BT where m is the mass of a molecule of the gas. Therefore,
\(\frac{\left(\mathrm{V}_{\mathrm{rms}}^2\right)_{A \mathrm{C}}}{\left(\mathrm{V}_{\mathrm{ms}}^2\right)_{\mathrm{cl}}}=\frac{(\mathrm{m})_{\mathrm{C}}}{(\mathrm{m})_{\mathrm{Ar}}}=\frac{(\mathrm{M})_{\mathrm{Cl}}}{(\mathrm{M})_{\mathrm{Ar}}}=\frac{70.9}{39.9}\) = 1.77
where M denotes the molecular mass of the gas.
\(\frac{\left(\mathrm{V}_{\text {rms }}\right)_{\mathrm{Ar}}}{\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{cl}}}\) = 1.33

Question 6.
a) When a molecule (or an elastic ball) hits a (massive) wall, it rebounds with the same speed. When a ball hits a massive bat held firmly, the same thing happens. However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower ? (ch. 6 will refresh your memory on elastic collisions).
b) When gas in a cylinder is compressed by pushing in a piston, its temperature rises. Guess at an explanation of this in terms of kinetic theory using (a) above.
c) What happens when a compressed gas pushes a piston out and expands. What would you observe ?
d) Sachin Tendulkar uses a heavy cricket bat while playing. Does it help him in anyway ?
Answer:
a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving towards the ball with a speed V relative to the wicket, then the relative speed of the ball to bat is V + u towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative to bat, is V + u moving away from the bat. So relative to the wicket the speed of the rebounding ball is V + (V + u) = 2V + u, moving away from the wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than u if the bat is not massive. For a molecule this would imply an increase in temperature.
You should be able to answer (b) (c) and (d) based on the answer to (a).

Question 7.
A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0° C ? (R = 8.31 J mol^-1 K^-1).
Answer:
Using the gas law PV = μRT, you can easily show that 1 mol of any (ideal) gas at standard temperature (273 K) and pressure (1 atm = 1.01 × 10⁵ Pa) occupies a volume of 22.4 litres. This universal volume is called molar volume. Thus the cylinder in this example contains 2 mol of helium. Further, since helium is monatomic, its predicted (and observed) molar specific heat at constant volume, C_v = (3/2) R, and molar specific heat at constant pressure, C_p = (3/2) R + R =(5/2) R. Since the volume of the cylinder is fixed, the heat required is determined by C_v.
Therefore,
Heat required = no. of moles × molar specific heat × rise in temperature
= 2 × 1.5 R × 15.0 = 45
R = 45 × 8.31 = 374 J.

Question 8.
Estimate the mean free path for water molecule in water vapour at 373 K. Use information from Exercises 1 and above. l = 2.9 × 10^-7 m = 1500d.
Ans. The d for water vapour is same as that of air. The number density is inversely proportional to absolute temperature.
So n = 2.7 × 10²⁵ × \(\frac{273}{373}\) = 2 × 10²⁵ m^-3
Hence, mean free path l = 4 × 10^-7 m.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 13th Lesson Thermodynamics Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 13th Lesson Thermodynamics

Very Short Answer Questions

Question 1.
Define Thermal equilibrium. How does it lead to the Zeroth Law of Thermodynamics?
Answer:
If the temperatures of the two systems are equal, then they are said to be in thermal equilibrium.
Zeroth law of thermodynamics: “If two systems (A, B) are in thermal equilibrium with the third system (c) separately, then two systems (A, B) thermal equilibrium with each other.”

Question 2.
Define Calorie. What is the relation between calorie and mechanical equivalent of heat?
Answer:
Calorie: It is defined to be the amount of heat required to raise the temperature of 1g of water from 14.5°C to 15.5°C.
The relation between calorie and mechanical equivalent of heat is given by 1 cal = 4186 J Kg^-1K^-1 or 1 cal = 4.186 J g^-1K^-1.

Qeustion 3.
What thermodynamic variables can be defined by
a) Zeroth Law
b) First Law ?
Answer:
a) Temperature
b) Internal energy.

Question 4.
Define specific heat capacity of the substance. On which factors does it depend ?
Answer:
Specific heat capacity :
The amount of heat required to raise the 1 gm of substance through 1°c or 1 k is called specific heat capacity (or) the heat capacity per unit mass is called specific heat.
S = \(\left(\frac{1}{m}\right) \frac{\Delta Q}{\Delta T}\)
It depends on 1) natuture of the substance 2) temperature.

Question 5.
Define molar specific heat capacity.
Answer:
Molar specific heat capacity : The amount of heat required to raise the 1gm mole of substance through 1°C or 1°K is called molar specific heat capacity.

Question 6.
For a solid, what is the total energy of an oscillator ?
Answer:
For a mole of a solid, the total energy of an oscillator, U = 3K_B T × N_A = 3RT.

Question 7.
Indicate the graph showing the variation of specific heat of water with temperature. What does it signify ?
Ans.

Significance : the specific heat of water varies slightly with temperature in the range 0 to 100°C.

Question 8.
Define state variables and equation of state.
Answer:
State variables : The state of a system is described by pressure P, temparature T, density p (intensive variables) and internal energy U, Volume V, total mass M (Extensive variables). These are called state variables.

Equation of state : The connection between the state variables is called the equation of state.

Question 9.
Why a heat engine with 100% efficiency can never be realised in practise ?
Answer:
The efficiency of heat engine is η = 1 – \(\frac{Q_2}{Q_1}\)
For Q₂ = 0, η = 1, i.e., the engine will have 100% efficiency in converting heat into work.
The first law of thermodynamics does not rule out such an engine. But experience shows that such an ideal engine with η = 1 is never possible.

Question 10.
In summer, when the valve of a bicycle tube is opened, the escaping air appears cold.
Answer:
This happens due to adiabatic expansion of the air of the tube of the bicycle.

Question 11.
Why does the brake drum of an automobile get heated up while moving down at constant speed ?
Answer:
The work done by the brake drum on wheel is converted into heat due to friction.

Question 12.
Can a room be cooled by leaving the door of an electric refrigerator open ?
Answer:
No, a room can not be cooled by leaving the door of a refrigerator open, rather it will get slightly heated.

Question 13.
Which of the’ two will increase the pressure more, an adiabatic or an isothermal process, in reducing the volume to 50% ?
Answer:
Isothermal process obeys , P₁V₁ = P₂V₂
But V₂ = \(\frac{\mathrm{v_1}}{2}\) ⇒ P₁V₁, = P₂ \(\frac{\mathrm{v_1}}{2}\)
∴ P₂ = 2P₁
Adiabatic process obeys, P₁V₁^γ = P₂V₂^γ
P₂ = \(P_1\left[\frac{v_1}{v_2}\right]^\gamma=p_1\left[\frac{v_1}{\frac{v_1}{2}}\right]\)
∴ P₂ = 2^γ P₁ Where r = \(\frac{C_p}{C_v}\)
∴ pressure is more in adiabatic than isothermal.

Question 14.
A thermoflask containing a liquid is shaken vigorously. What happens to its temperature ? .
Answer:
Work is done by the liquid on the wall of flask, since it is vigorously shaken. Hence Internal energy and temperature of the liquid increases.

Question 15.
A sound wave is sent into a gas pipe. Does its internal energy change ?
Answer:
Yes, a sound wave is sent into a gas pipe internal energy increases.

Question 16.
How much will be the internal energy change in

1. isothermal process
2. adiabatic process

Answer:

1. Change in internal energy during isothermal process is dU = 0 [∵ U is constant)]
2. a) Change in internal energy during adiabatic compression increases,
   b) Change in internal energy during adiabatic expansion decreases.

Question 17.
The coolant in a chemical or a nuclear plant should have.high specific heat. Why ?
Answer:
In chemical and nuclear plants more heat is liberated. To absorb this heat, the coolant should have a property of high specific heat for small rise in temperature.
Since S = \(\frac{1}{m} \frac{d Q}{d t}\) ⇒ S ∝ \(\frac{1}{d t}\)

Question 18.
Explain the following processes

1. Isochoric process
2. Isobaric process

Answer:

1. Isochoric process : A process takes place at constant volume is called Isochoric process. During this process, no work is done on or by the gas, its internal energy and its temperature changes.
2.  Isoboric process : A process takes place at constant pressure is called isobaric process During this process temperature, internal energy changes. The heat absorbed goes partly to increase internal energy and partly to do work.

Short Answer Questions

Question 1.
State and explain first law of thermodynamics.
Answer:
Statement: The amount of heat supplied to system is equal to the algebraic sum of the change in internal energy of the system and the amount of external work-done.

Explanation : When ∆Q is the quantity of heat is supplied to system, a part of it is used to increase its internal energy ∆U and the rest is used in doing external work ∆W. The mathematical expressions of this law is ∆Q = ∆U – ∆W. It is a special case of law of conservation of energy.

Question 2.
Define two principal specific heats of a gas. Which is greater and why ?
Answer:
Two principal specific heats of a gas are

1. Molar specific heat capacity at constant pressure
2. molar specific heat capacity at constant volume.

1. Molar Specific heat capacity at constant pressure (C_p): The amount of heat required to raise the temperature of 1 gm – mole of a gas through 1°C at constant pressure is called molar specific heat at constant pressure.
i.e., C_p = \(\frac{1}{\mu} \frac{\Delta Q}{\Delta T}\) where μ is no of moles.

2. Molar Specific heat capacity at constant volume (C_v): The amount of heat required to raise the temperature of 1gm – mole of a gas through 1°C at constant volume is called molar specific heat at constant volume i.e., C_v = \(\frac{1}{\mu} \frac{\Delta Q}{\Delta T}\)

Explanation of C_p is greater that C_v: When a gas is heated at a constant pressure, it expands. The heat supplied to it is used partly in raising its temperature and partly in doing work against the external pressure. If, on the other hand, the gas is heated at constant volume, no work is done. Therefore, the heat supplied is to be used only in raising the temperature. Hence the amount of heat required to be supplied to a gas to raise its temperature by 1 °C (say) at constant pressure will be greater than the amount required at constant volume.

Question 3.
Derive a relation between the two specific heat capacities of gas on the basis of first law of thermodynamics
Answer:
The mathematical equation of first law of thermodynamics for 1 mole of gas is given by
∆Q = ∆U + P∆V
if ∆Q is absorbed heat at constant volume, ∆V = 0
C_v = \(\left[\frac{\Delta Q}{\Delta T}\right]_V=\left[\frac{\Delta U}{\Delta T}\right]_V=\left[\frac{\Delta U}{\Delta T}\right]\) ……………. (1)
Where the subscript V is dropped in the last step, since U of an ideal gas depends only on temperature. (The subscript denotes the quantity kept fixed.) If, on the other hand, ∆Q is absorbed heat at constant pressure,
C_p = \(\left[\frac{\Delta Q}{\Delta T}\right]_P=\left[\frac{\Delta U}{\Delta T}\right]_P+P\left[\frac{\Delta V}{\Delta T}\right]_P \Rightarrow\left[\frac{\Delta Q}{\Delta T}\right]_P=\left[\frac{\Delta U}{\Delta T}\right]+P\left[\frac{\Delta V}{\Delta T}\right]_P\) ………………… (2)
The subscript p.can be dropped from the first term since U of an ideal gas depends only on T. Now, for a mole of an ideal gas.
PV = RT
Which gives P\(\left[\frac{\Delta V}{\Delta T}\right]_P\) = R
Putting the equations (1) and (3) in (2), we get
C_p = C_v + R
∴ C_p – C_v = R

Question 4.
Obtain an expression for the work done by an ideal gas during isothermal change.
Answer:
Expression for the work done by an ideal gas during isothermal change : Let a certain mass of gas expands from volume V₁ to V₂ isothermally at constanst temperature T. Let the pressure changes form P₁ to P₂.
The work done during the small change in volume dv at a pressure ‘p’ is dw = pdv
The total work done during the change of volume from V₁ to V₂ is W = \(\int d w=\int_{V_1}^{V_2} p d v\) …………. (1)
The Isothermal change is represented by
PV = constant
PV = μRT ⇒ P =\(\frac{\mu \mathrm{RT}}{\mathrm{V}}\) where μ = no. of moles
substituting this value in equation (1)
W = \(\int_{V_1}^{v_2} p d v=\int_{V_1}^{V_2} \frac{\mu R T}{V} d V=\mu R T \int_{V_1}^{V_2} \frac{d v}{V}\)
W = μRT log_e \(\frac{V_2}{V_1}\)
∴ Work done during the isothermal change W = 2.3026 RT log₁₀ \(\frac{V_2}{V_1}\) .

Question 5.
Obtain an expression for the work done by an ideal gas during adiabatic change and explain.
Answer:
Expression for the work done by an ideal gas during adiabatic change :
During an adiabatic chnage the state of an ideal gas changes from (P₁, V₁, T₁) to (P₂, V₂, T₂)
The workdone during a small change in volume dV at a pressure p is dw = pdV
Total work done by gas from volume V₁ to V₂ is ,
W = \(\int\) dw = \(\int_{v_1}^{v_2}\) pdv ………………. (1)
The adiabatic relation between pressure and volume is given by PV^r = constant K …………. (2)
P = \(\frac{k}{v^r}\) …………… (3)
and P₁V₁^r = P₂V₂^r = K
Substituting eg. (3) in (1) becomes

This is equal to the work done during the adiabatic change.

Question 6.
Compare isothermal and an adiabatic process.
Answer:
Isothermal change

1. Changes in volume and pressure of a gas taking place at constant temperature are called isothermal changes,
2. Temperature of the gas remains constant.
3. The gas remains in good thermal contact with the surroundings and heat is exchanged.
4. Internal energy remains constant. Change in internal energy ∆U = 0
5. This process takes place slowly.
6. Boyle’s law PV = Constant holds good.
7. Work done W = RT log_e\(\frac{V_2}{V_1}\) .

Adiabatic change

1. Changes in volume and pressure of a gas taking place in a thermally isolated system are called adiabatic changes.
2. Temperature of the gas changes.
3. The gas is isolated from the surroundings and heat is not exchanged ∆Q = 0.
4. Internal energy changes.
5. This process takes place quickly.
6. PV^r = Constant
7. Work done W = \(\frac{R}{(r-1)}\)(T₁ – T₂)

Question 7.
Explain the following processes
i) Cyclic process with example
ii) Non cyclic process with example
Answer:
i) Cyclic process: “A process in which the system after passing through various stages (of pressure, volume and temperature changes) returns to its initial state1′ is called as a cyclic process.
For cyclic process P.V graph is a closed curve. The area under P-V graph gives work done by the substance. In a cyclic process there will be no change in the internal energy i.e., ∆U = 0
The first law of thermodynamics states that
∆Q = ∆U + ∆W
∴ For cyclic process ∆Q = ∆W
In cyclic process, the total heat absorbed by the system equals the work done by the system. Example : Fleat engine is a device by which a system is made to undergo a cyclic process that results in conversion of heat to work.

ii) Non cyclic process : A process which is not cyclic is called Non-cyclic process. In Non-cydic process, the system does not attains its initial state. Irreversible process is a Non cyclic process. Workdone in non-cyclic process is given by the area between the curve and volume axis.

Example :

1. Diffusion of liquids or gases
2. Free expansion of a perfect gas.

Question 8.
Write a short note on Quasi – static process.
Answer:
A quasi-static process is defined as “an infinitesimally slow process in which at each and every intermediate stage the system remains in thermal and mechanical (thermodynamic) equilibrium with the surroundings through out the entire process.”

In this process the pressure and temperature of the surroundings at each and every stage will differ from those of the system only in a very small amount.

Any process taking place sufficiently slowly, not involving acceleration motions and large temperature gradients can be considered as quasi -static process.
Example : isothermal process, adiabatic process.

Question 9.
Explain qualitatively the working of a heat engine.
Answer:
“A device used to convert heat energy into work (or mechanical energy)” is called heat engine. A heat engine converts heat into work.
A heat engine consists of the following essential parts :

1. A body at a higher temperature T₁. Heat Q₁ is extracted from this body and hence it is called a “hot reservior” (or source).
2. The engine contains working substance. In a steam engine the working substance is the steam. In a diesel engine the working substance is a mixture of fuel vapour and air.
3. A body at lower temperature T₂. Heat Q₂ is rejected by the working substance to this body and hence it is called a cold reservoir (or sink).

The difference in heat absorbed (Q₁) and heat rejected (Q₂) is equal to the workdone by the system (W). This is because the engine is operated in a cyclic process. W = Q₁ – Q₂.

The heat engine operates in a cyclic process and in each cycle, the work done (W) by the system is transferred to the environment via some arrangement (shaft) to drive vehicles.

The efficiency (η) of aheat engine is defined as the ratio of the workdone (W) by the engine to the amount of heat absorbed (Q₁) by the engine.
Efficiency (η) = \(\frac{\mathrm{W}}{\mathrm{Q}_1}=\frac{\mathrm{Q}_1-\mathrm{Q}_2}{\mathrm{Q}_1}=1-\frac{\mathrm{Q}_2}{\mathrm{Q}_1}\)

Long Answer Questions

Question 1.
Explain reversible and irreversible processes. Describe the working of Carnot engine. Obtain an expression for the effieciency. [AP – Mar. ’18, ’17, ’16, 15; TS – Mar. ’17, ’15, ’14]
Answer:
Reversible process : A process that can be retraced back in the opposite direction in such away that the system passes through the same states as in the direct process, and finally the system and the surroundings return in their original states, is called a reversible process.
A reversible process is only a purely idealised process.
Examples:

1. Slow isothermal and slow adiabatic changes
2. Peltier effect and seeback effect.
3. Fusion of ice and vaporisation of water.

Irreversible process : “A process that cannot be retraced back in the opposite direction” is called as irreversible process.

In this process the system does not pass through the same intermediate states as in the reversible process.
All natural processes such as conduction, radiation, radioactivity etc, are irreversible.
Example:

1. Work done against friction.
2. Heat produced in conductors by passing a current through it. (Joule heating)
3. Diffusion of gases.

Carnot Engine : A reversible heat engine operating between two temperatures is called a carnot engine. The cycle operating it is known as carnot’s cycle. In this cycle the working substance(say an ideal gas) is taken through a cycle by means of two isothermals and two adiabatics. The four operations are shown in P-V (Indicator) diagram.

Step 1 → 2 : Isothermal expansion of the gas taking its state from (P₁, V₁, T₁) to (P₂, V₂, T₂). It is shown in curve (a).
The heat absorbed by the gas (Q₁) from the reservoir at temperature
T1 equals the workdone by the gas. i.e.
W_{1 → 2} = Q₁ = μRT₁ log_e \(\frac{V_2}{V_1}\)

Step 2 →3 : Adiabatic expansion of the gas from (P₂, V₂, T₁) to (P₃, V₃, T₂). It is shown in curve (b). The work done by the gas is gÑen by
W_{2 → 3} = \(\frac{\mu R\left(T_1-T_2\right)}{(r-1)}\)

Step 3 → 4: Isothermal compression of the gas from (P₃, V₃, T₂) to (P₄, V₄, T₂). It is shown in curve (c).
Heat released (Q₂) by the gas to the reservoir at temperature T₂ equals the work done on the goes.
i.e., W_{3 → 4};= μRT₂ log_e \(\frac{V_4}{V_3}\) = -μRT₁ log_e \(\frac{V_3}{V_4}\)

Step 4 → 1: Adiabatic compression of the gas from (P₄, V₄, T₂) to (P₁, V₁, T₁). It is shown in curve (d) Work done on the gas is given by.
W_{4 → 1} = μR \(\frac{\left(T_2-T_1\right)}{(r-1)}=-\mu R \frac{\left(T_1-T_2\right)}{(r-1)}\)
Total work done by the gas in one complete cycle is
W = W_{1 → 2} + W_{2 → 3} + W_{3 → 4} + W_{4 → 1}
= \(\mu R T_1 \log _e \frac{V_2}{V_1}+\frac{\mu R\left(T_1-T_2\right)}{(r-1)}-\mu R T_2 \log _e \frac{V_3}{V_4}-\frac{\mu R\left(T_1-T_2\right)}{(r-1)}\)
∴ W = μRT₁ log_e \(\frac{V_2}{V_1}\) – μRT₂ log_e \(\frac{3}{V_4}\) = Q₁ – Q₂ …………… (2)
The efficiency of the cannot engine is
η = \(\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}\) ………… (3)
Now since step 2 → 3 is an adiabatic process,
T₁V₂^{r – 1} ≈ T₂V₃^{r – 1}
Similarly, since step 4 → 1 is an adiabatic process,
T₂V₂^{r – 1} = T₁V₁^{r – 1} ⇒ T₁V₁^{r – 1} = T₂V₂^{r – 1}

Question 2.
State second law of thermodynamics. How is heat engine different form a refrigerator. [TS – Mar. ’18, ’16; AP – Mar. ’15, ’13]
Answer:
Second law of thermodynamics gives the direction of flow of heat. Second law consists two statements.

1. Kelvin – Planck statement: “No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work”.
   (or)
   “It is impossible to derive a continuous supply of engines in cooling a body below the coldest of its surroundings1′.
2. Clausius statement : “No process is possible whose sole result is the transfer of heat from a colder object to a hotter object”.
   (or)
   “Heat can not itself flow from cold body to hot body”.

Heat engine : A device which converts heat energy into work is called heat engine.
A heat engine consists of the essential parts.

1. Source or hot reservoir: It is maintained at a very high temperature T₁. Heat is extracted from this body.
2. Working substance : In a steam engine working substance is steam. In a diesel engine working substance is mixture of fuel vapour and air.
3. Sink or cold reservoir: It is maintained at a very low temperature T₂. It absorbs heat energy, rejected by working substance.

Work done: The difference of heat absorbed from source and heat rejected to sink is equal to work done by the engine, i.e., W = Q₁ – Q₂.
Efficiency : The ratio of workdone (W) by the engine to the amount of heat absorbed (Q₁) by the engine is called efficiency.
i.e., η = \(\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}=1-\frac{Q_2}{Q_1}\)
Refrigerator : Refrigerator is a heat pump which is the reverse of a heat engine.

In refrigerator the working substance extracts (Q₂) an amount of heat from sink of lower temperature and a external work W is done on the working substance and finally amount of heat (Q₁) is transfered to source.
The coefficient of performance (α) of a refrigerator is given by
α = \(\frac{\mathrm{Q}_2}{\mathrm{~W}}=\frac{\mathrm{Q}_2}{\mathrm{Q}_1-\mathrm{Q}_2}\) (∴ heat released Q₁ = Q₂ + W)
For heat engine η can never exceed 1. For refregirator a can be greater than 1.
Therefore working of refrigerator is opposite to that of heat engine.

Problems

Question 1.
If a monoatomic ideal gas of volume 1 litre at N.T.P. is compressed (i) adia- batically to half of its volume, find the work done on the gas. Also find (ii) the work done if the cdmpression is iso- thermal, (γ = 5/3)
Solution:
i) During an adiabatic process
T₁V₁^{r – 1} = T₂V₂^{r – 1}
Here T₁ = 273 K
V₂ = \(\frac{V_1}{2}\)
r = \(\frac{5}{3}\)
T₂ = \(\left[\frac{V_1}{V_2}\right]^{\frac{5}{3}-1}=273\left[\frac{V_1}{V_1}\right]^{\frac{2}{3}}\)
T₂ = (2)^{\(\frac{2}{3}\)} = 273 = 431.6 K
Number of moles = \(\frac{1 \text { litre }}{22.4 \text { litre }}=\frac{1}{22.4}\)
Work done = \(\frac{\mu R}{(r-1)}\) [T₁ – T₂]
= \(\frac{8.314}{22.4\left[\frac{5}{3}-1\right]}\) × (273 – 431.6)
= \(\frac{8.314 \times 3}{22.4 \times 2}\) (-158.6)
= -89 J

ii) Work done during isothermal compression is
w = 2.3026 μRT log₁₀\(\frac{V_2}{V_1}\)
μ = Number of moles = \(\frac{1}{22.4}\)
T = 273K
R = 8.314 Jmol^-1K^-2
\(\frac{V_2}{V_1}=\frac{1}{2}\) = 0.5
∴ W = \(\frac{2.3026 \times 8.314 \times 273 \log _{10}(0.5)}{22.4}\)
Or w = -70J

Question 2.
Five moles of hydrogren when heated through 20 K expand by an amount of 8.3 × 10^-3m³ under a constant pressure of 10⁵ N/m². If C_v = 20.1/mole K, find C_p.
Solution:
Mayer’s relation C_p – C_v = R
Multiplying throughout by μ∆T
μC_p∆T – μC_v∆T = μ R∆T
μ ∆T(C_p – C_v) = P∆T [∴ μ R∆T = P∆V]
5 × 20 (C_p – 20) = 105 × 8.3 × (10 – 3)
[∴ μ = 5, ∆T = 20 K, P = 1 × 10⁵ N/m² C_v = 20 J/mole K and ∆V = 8.3 × 10³ M³]
C_p – 20 = 8.3
∴ C_p = 28.3 J/mole-K

Additional Problems

Question 1.
A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operate on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 10⁴ J/g ?
Solution:
Here volume of water heated = 3.0 lit/min.
Mass of water heated m = 3000 g/min
Rise in temperature ∆T = 77 – 27 = 50°C
Sp. heat of water C = 4.2 Jg^-1C^-1
Amount of heat used
∆Q = mc∆T = 3000 × 4.2 × 50
= 63 × 10⁴ J/min
Heat of combustion = 4 × 10⁴ J/g
Rate of combustion of fuel = \(\frac{63 \times 10^4}{4 \times 10^4}\)
= 15.75 g/min

Question 2.
What amount of heat must be supplied t 2.0 × 10^-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N₂ = 28; R = 8.3 J mol^-1 K^-1.)
Solution:
Here mass of gas m = 2 × 10^-2kg = 20g
Rise in temperature ∆T = 45°C
Heat required ∆Q = ?
Molecular mass M = 28
Number of moles n = \(\frac{m}{M}=\frac{20}{28}\) = 0.714
As nitrogen is a diatomic gas, molecular specific heat of constant pressure is
C_p = \(\frac{7}{2}\) R = \(\frac{7}{2}\) × 8.3 J mole^-1 K^-1
As ∆Q = nC_p∆T
∆Q = 0.714 × \(\frac{7}{2}\) × 8.3 × 45 J = 933.4 J

Question 3.
Explain why
a) Two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂)/2.
b) The coolant in a chemical or nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
c) Air pressure in a car tyre increases during driving.
d) The climate of a harbour town is more temperate then that of a town in a desert at the same latitude.
Solution:
a) In thermal contact heat flows from the body at higher temp to the body at lower temperature till temperatures becomes equal. The final temperature can be the mean temp (T₁ + T₂)/2 only when thermal capacities of the two bodies are equal.

b) This is because heat absorbed by a substance is directly proportional to the specific heat of the substance.

c) During, driving the temperature of air inside the tyre increases due to motion. According to Charles law, P α T. Therefore, air pressure inside the tyre increases.

d) This is because in a harbour town, the relative humidity is more than in a desert town. Hence the climate of a harbour town is without extremes of hot and cold.

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ?
Solution:
As no heat is allowed to be exchanged, the process is adiabatic.
∴ P₂V₂^r = P₁V₁^r (or) \(\frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^r\)
As V₂ = \(\frac{V_1}{2}\)
∴ P₂/P₁ = \(\left(\frac{V_1}{\frac{1}{2} V_1}\right)^{1 .4}\)
= 2^1.4 = 2.64

Question 5.
In changing the state of a gas adiabatically form an equilibrium state A to another equilibrium state B, and amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal. how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J)
Solution:
Here, when the change is adiabatic, ∆Q = 0, ∆w = -22.3 J
If ∆u is change in internal energy of the system, then as
∆Q = ∆u + ∆w
O = ∆u – 22.3 (or) ∆u = 22.3 J
In the second case ∆Q = 9.35 cal
= 9.35 × 4.2 J = 39.3 J
∆w = ?
As ∆u + ∆w = ∆Q
∆w = ∆Q – ∆u
= 39.3 – 22.3 = 17.0 J

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :
a) What is the final pressure of the gas in A and B ?
b) What is the change in internal energy of the gas ?
c) What is the change in the temperature of the gas ?
d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ?
Solution:
a) When the stopcock is suddenly opened, the volume available to the gas at 1 atm pressure will become two times. Therefore pressure will decrease to one-half i.e., 0.5 atm.

b) There will be no change in the internal energy of the gas as no work is done on/ by the gas.

c) Also there will be no change in temp of the gas as gas does no work in expansion.

d) No, because the process called free expansion is rapid and cannot be controlled. The intermediate states are non equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.

Question 7.
A steam engine delivers 5.4 × 10⁸J . of work per minute and services 3.6 × 10⁹J of heat per minute form its boiler, what is the efficiency of the engine ? How much heat is wasted per minute ? .
Solution:
Output i.e., useful workdone per min
= 5.4 × 10⁸ J
Input i.e., heat absorbed per min
= 3.6 × 10⁹ J
Efficiency = \(\frac{\text { Output }}{\text { Input }}=\frac{5.4 \times 10^8}{3.6 \times 10^9}\) = 0.15
= 0.15 × 100% = 15%
heat energy wasted / minute = Heat absorbed / minute – useful work done / minute
= 3.6 × 10⁹ – 5.4 × 10⁸ = 109 (3.6 – 0.54)
= 3.06 × 10⁹

Question 8.
An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing ?
Solution:
Heat supplied, ∆Q = 100 w = 100 J/s
Useful workdone ∆w = 75 J/s
Increase in internal energy/ sec, ∆u = ?
As ∆Q = ∆u + ∆w
∴ ∆u = ∆Q – ∆w
= 100 – 75 = 25 J/S

Question 9.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig.

Its volume is then reduced to the original value form E to F by an isobaric process. Calculate the total work done by the gas from D to E to F
Solution:
A is clear from fig.
Change in pressure dp = EF = 5.0 – 2.0
= 3.0 atm = 3.0 × 10¹⁵ Nm^-2
Change in volume dv = DF = 600 – 300
= 300 cc = 300 × 10^-6 m³
Workdone by the gas from D to E to F = area of ∆DEF
w = \(\frac{1}{2}\) × DF × EF
= \(\frac{1}{2}\) × (300 × 10^-6 ) × (3.0 × 10⁵ ) = 45 J

Question 10.
A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C. calculate the coefficient of performance.
Solution:
Here T₁ = 36°C = 36 + 273 = 309 K
T₂ = 10°C = 10 + 273 = 283 K
COP = \(\frac{T_2}{T_1-T_2}=\frac{283}{309-283}\)
= \(\frac{283}{26}\) = 10.9

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter

Very Short Answer Questions

Question 1.
Distinguish between heat and temperature. [T.S. Mar. 15]
Answer:
Heat

1. It is a form of energy.
2. It is the cause on the body.
3. It is determined with calorimeter.
4. It’s unit is Joule and Calories.

Temperature

1. It is a degree of hotness or coldness of a body.
2. It is an effect on a body.
3. It is measured with thermometer.
4. It’s unit is degree’Celsius or Kelvin or degree Fahrenheit.

Question 2.
What are the lower and upper fixing points in Celsius and Fahrenheit scales ? [T.S. Mar. 16]
Answer:
In Celsius scale, lower fixed point is ice point or 0°C and upper fixed point is steam point or 100°C. In Fahrenheit scale, lower fixed point is 32°F and upper fixed point is 212°F.

Question 3.
Do the values of coefficients of expansion differ, when the temperatures are measured on Centigrade scale or on Fahrenheit scale ?
Answer:
Yes. The coefficients of expansion depends on scale of temperature because
\(\frac{\alpha}{{ }^{\circ} \mathrm{C}}=\frac{9}{5} \alpha /{ }^{\circ} \mathrm{F}\)

Question 4.
Can a substance contract on heating ? Give an example.
Answer:
Yes. Rubber, type metal, cast iron contract on heating.

Question 5.
Why gaps are left between rails on a railway track ? [A.P. Mar. 17, 16]
Answer:
The length of the rails increases in summer due to high temperature. Therefore a gap is left to allow this expansion.

Question 6.
Why do liquids have no linear and areal expansions ?
Answer:
Liquids occupy the same shape of vessel. They do not have individual length and area. Hence, liquids have no linear and areal expansions.

Question 7.
What is latent heat of fusion ?
Answer:
The amount of heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure is called the latent heat of fusion (L_f).

Question 8.
What is latent heat of vapourisation ? [Mar. 13]
Answer:
The amount of heat per unit mass required to change a substance from the liquid to the vapour state at the same temperature and pressure is called the latent heat of vaporisation (L_v).

Question 9.
What is specific gas constant ? Is it same for all gases ?
Answer:
Universal gas constant per molecular mass is called specific gas constant.
i.e., r = \(\frac{\mathrm{R}}{\mathrm{m}}\). No, it is different for different gases.

Question 10.
What are the units and dimensions of specific gas constant ?
Answer:
The units of specific gas constant is J/Kg/K. Dimensional formula is (m⁰.L²t^-2K¹).

Question 11.
Why utensils are coated black ? Why the bottom of the utensils are made of copper ?
Answer:

1. Utensils are coated black, because it is a good absorber of heat.
2. Copper is a good conductor of heat. So, copper is used at the bottom of cooking utensils.

Question 12.
State Weins displacement law. [A.P. Mar. 17]
Answer:
The wavelength (λ_m) corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature
i.e., λ_m ∝ \(\frac{1}{T}\)

Question 13.
Ventilators are provided in rooms just below the roof. Why ? [Mar. 14]
Answer:
Ventilators are provided in rooms just below the roof, because the hot air escapes out and fresh air enters into the rooms due to convection.

Question 14.
Does a body radiate heat at 0 K ? Does it radiate heat at 0°C ?
Answer:

1. No. A body does not radiate heat at 0k.
2. Yes. A body radiate heat at 0°C.

Question 15.
State the different modes of transmission of heat. Which of these modes require medium ?
Answer:
The different modes of transmission are : 1) conduction 2) convection and 3) radiation.
Among three modes conduction and convection require medium.

Question 16.
Define coefficient of thermal conductivity and temperature gradient.
Answer:
Coefficient of thermal conductivity : It is defined as the quantity of heat energy flowing per second between the opposite faces of cube of unit side, which are maintained at unit temperature difference.
Temperature gradient: The change of temperature per unit length of conductor is called temperature gradient.

Question 17.
What is thermal resistance of a conductor ? On what factors does it depend ?
Answer:
The resistance offered by the conductor for the flow of heat is called thermal resistance (R) of the conductor. R = \(\frac{\mathrm{d}}{\mathrm{kA}}\). It depends on

1. The nature of the material, i.e., thermal conductivity (K).
2. d; length of conductor along which heat flows and
3. A; area of cross section of the conductor.

Question 18.
State the units and dimensions of coefficient of convection.
Answer:
The unit of coefficient of convection is, wm^-2k^-1.
Dimensional formula (m.L⁰T³θ^-1).

Question 19.
Define emissive power and emissivity ?
Answer:
Emissive power : The energy flux emitted by unit surface area of a radiating body is known as emissive power (e_λ).

Missivity (e): It is defined as the ratio of the emissive power of the body to that of black body at the same temperature.

Question 20.
What is greenhouse effect ? Explain global warming. [A.P. Mar. 15, 13]
Answer:
Greenhouse effect: When the earth receives sun light, it gets heated up and emits infrared radiation into air. CO₂, CH₄, N₂O, O₃, Chlorofluoro carbon (green house gases) present in the air absorbs the heat content of infrared radiation and keeps the earth warm. This is called green house effect.

Global warming : As CO₂ content increases, more heat is retained in the atmosphere and the temperatures all over the world increases. This is called global warming.

Effects : a) Polar ice caps melt and fed more water in rivers and seas results in severe floods, b) In some areas, water resources dry up leading to drought conditions.

Question 21.
Define absorptive power of a body. What is the absorptive power of a perfect black body ?
Answer:
Absorptive power : At a given temperature and wavelength, the ratio of the amount of radiant energy absorbed to the amount of radiant energy incident in a wavelength range is called the absorptive power at that temperature and wavelength. Absorptive power of a perfect black body is 1.

Question 22.
State Newton’s law of cooling. [A.P. Mar. 18, 16; T.S. Mar. 18]
Answer:
The rate of loss of heat is directly proportional to the difference in temperature between the body and its surroundings provided the temperature difference is small. i.e., – \(\frac{\mathrm{dQ}}{\mathrm{dt}} \alpha\left(\mathrm{T}_B-\mathrm{T}_{\mathrm{S}}\right)\)

Question 23.
State the conditions under which Newton’s law of cooling is applicable.
Answer:

1. Loss of heat is negligible by conduction and only when it is due to convection.
2. Loss of heat occurs in a streamlined flow of air i.e., forced convection.
3. Temperature of the body is uniformly distributed over it.

Question 24.
The roof of buildings are often painted white during summer. Why ? [T.S. Mar. 17, 15]
Answer:
The roof of buildings are painted white during summer, because it reflects more heat radiations and absorbs less heat radiations. Hence we feel comfort inside the house.

Short Answer Questions

Question 1.
Explain Celsius and Fahrenheit scales of temperature. Obtain the relation between Celsius and Fahrenheit scales of temperature.
Answer:
Centigrade (Celsius) scale of temperature: In the Celsius scale of temperature, the lower fixed point is called the ice point and is assigned the value 0°C. The upper fixed point is called the steam point and is assigned the value 100°C. The interval between these two points (i.e., 100°C – 0 = 100°C) is subdivided into 100 equal parts each one corresponding to 1°C.

Fahrenheit scale of temperature : In the Fahrenheit scale of temperature, the lower fixed point is the ice point and is assigned a value 32°F and the upper fixed point is the staempoint and is assigned a value 212°F. The interval between these two points (i.e., 212°F – 32°F = 180°F) is subdivided into 180 equal parts, each one corresponding to 1°F.

Fahrenheit scale of temperature : In the Fahrenheit scale of temperature, the lower fixed point is the ice point and is assigned a value 32°F and the upper fixed point is the staempoint and is assigned a value 212°F. The interval between these two points (i.e., 212°F – 32°F = 180°F) is subdivided into 180 equal parts, each one corresponding to 1°F.

Relation between Celsius and Fahreinheit scales of temperature:
Difference of 100 Celsius degrees = Difference of 180 Fahrenheit degress
When the temperature of a body is measured on both the Celsius and Fahrenheit scales, let the readings be t_C and t_F respectively. Then
\(\frac{t_c-0}{t^{100}}=\frac{t_F-32}{180} \Rightarrow \frac{t_C}{5}=\frac{t_F-32}{9}\)
C = \(\frac{5}{9}\) (F – 32)

Question 2.
Two identical rectangular strips, one of copper and the other of steel, are riveted together to form a compound bar. What will happen on heating ?
Answer:
Bimetallic strip:

Two similar strips of copper and steel are revitted side by side as a compound strip called bimetallic strip. It is straight at ordinary temperatures. When the bimetallic strip is heated, copper expands more than steel. So, it bends with copper on convex side. When the bimetallic strip is cooled below room temperature, copper contracts more than steel and it bends with copper on concave side. Bimetallic strip is used in refrigerators, automatic iron, fire alarm etc.

Question 3.
Pendulum clocks generally go fast in winter and slow in summer. Why ?
Answer:
The time period of a pendulum at t₁°C is T₁ = 2π \(\sqrt{\frac{L_1}{g}}\) ………….. (1)
Where L₁ is length of pendulum at t₁°C.
If L₂ is length of pendulum at t₂°C,
then T₂ = 2π \(\sqrt{\frac{L_2}{g}}\) ………………. (2)
\(\frac{(2)}{(1)}\) gives \(\frac{T_2}{T_1}=\sqrt{\frac{L_2}{L_1}}=\sqrt{\frac{L_1(1+\alpha t)}{L_1}}\)
Where a is the linear coefficient of expansion of pendulum clock. Where t = t₂ – t₁
\(\frac{T_2}{T_1}=\sqrt{(1+\alpha t)}=(1+\alpha t)^{\frac{1}{2}}\)
\(\frac{T_2}{T_1}=1+\frac{\alpha t}{2}\)
\(\frac{T_2}{T_1}-1=\frac{\alpha t}{2}\)
\(\frac{T_2-T_1}{T_1}=\frac{\alpha t}{2}\)
\(\frac{T_2-T_1}{T_1}\) = time lost by pendulum per second.
Due to expansion in length, during summer, time period increases or the clock looses time in summer. In winter due to fall in temperature, the length contracts, time period decreases, hence clock gains time.

Question 4.
In what way is the anomalous behaviour of water advantageous to acquatic animals ? [A.P. Mar. 18, Mar. 14]
Answer:
In cold countries, as atmospheric temperature decreases, the upper layers of the lakes, rivers etc., cool, contract and sink to the bottom (fig). This goes on until the whole of the water reaches the temperature of 4°C. When the top layers cool further temperature falls below 4°C, it expands and becomes lighter. It does not sink downwards and remains at the top. With further cooling the top layer gradually form ice at the top (fig). Ice and water are bad conductors of heat. So the lower layers are protected against freezing by the layers of ice and cold water at 1°C, 2°C and 3°C. This results in water remaining at the bottom at 4°C. So that aquatic animals survive in those layers of water.

Question 5.
Explain conduction, convection and radiation with examples. [TS – Mar. ’18, ’16, ’15; AP – Mar. ’15]
Answer:
The heat is transmitted in three types. They are 1) Conduction 2) Convection 3) Radiation.

1. Conduction : The process of transmission of heat from one place to other without actual movement of the particles of the medium is called conduction.
   Ex : When long iron rod is heated at one end, heat transmits to the other end.
2. Convection : The process of transmission of heat from one place to another by the actual
   movement of the particles is called convection.
   Ex. : If water in a beaker is heated, the particles of water at the bottom receive the heat first. These particles expand, become lighter and rise up. At the same time colder and denser particles reach the bottom. They get in their turn heated and move up. This process is known as convection.
3. Radiation : The process of transmission of heat from one place to another without any intervening medium is called radiation.
   Ex. : Earth receives heat radiations from the sun.

Long Answer Questions

Question 1.
State Boyle’s law and charle’s law. Hence, derive ideal gas equation which of two laws is better for the purpose of Thermometry and why ?
Answer:
P ∝ \(\frac{1}{\mathrm{V}}\)
⇒ PV = constants ⇒ P₁V₁ – P₂V₂
Charles law at constant volume : At constant volume the pressure of given mass of a gas is directly proportional to absolute temperature of gas.
P ∝ T
⇒ \(\frac{P}{T}\) = constant ⇒ \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)
Charles law at constant pressure : At constant pressure the volume of given mass of gas is directly proportional to absolute temperature.
V ∝ T
⇒ \(\frac{V}{T}\) = constant ⇒ \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)
Ideal gas equation : A gas which obeys all the gas laws at all temperature and all pressures is called an ideal gas.

Consider are mole of a gas initially at a pressure P₁ volume V₁ and temperature T₁ and Let P₂, V₂ and T₂ be the final pressure, volume and absolute temperature T₂.
From Boyle’s law, P₁V₁ = P₂V₂
⇒ V = \(\frac{P_1 V_1}{P_2}\) ………….. (1)
Now the gas is heated at constant pressure P₂, then its volume changes from V to V₂ and temperature changes from T₁ and T₂.
From charles law, \(\frac{V}{T_1}=\frac{V_2}{T_2} \Rightarrow V=\frac{V_2 T_1}{T_2}\) ………….. (2)
From (1) & (2) \(\frac{P_1 V_1}{P_2}=\frac{V_2 T_1}{T_2} \Rightarrow \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)
⇒ \(\frac{P V}{T}\) = constant ⇒ \(\frac{P V}{T}\) = R
Where ‘R’ is coniversal gas constant.
⇒ PV = RT
From ‘n’ gram moles PV = nRT. This law is called ideal gas equation.
Out of Boyle’s law, charles law is better for the purpose of thermometry because with increasing temperature, pressure and volume of gas also increase.

Question 2.
Explain thermal conductivity and coefficient of thermal conductivity. A copper bar of thermal conductivity 401 W/(mK) has one end at 104°C and the other end at 24°C. The length of the bar is 0.10 m and the cross – sectional area is 1.0 × 10^-6 m^-2. What is the rate of heat conduction, along the bar ?
Answer:
Thermal conductivity : Thermal conductivity of a solid body is ability to conduct heat in the body. Coefficient of thermal conductivity: The coefficient of thermal conductivity of a material may be defined as the quantity of heat energy flowing per second between the opposite faces of the cube of unit side, which are maintained at unit temperature difference.

Explanation of conduction : In steady state, conduction between the opposite faces which are at temperatures θ₁°C and θ₂°C separated by distance d is

1. Directly proportional to area of cross section of the rod.
   Directly proportional to temperature difference (θ₂ – θ₁) between the opposite faces.
2. Time of flow of heat, t
3. Inversely proportional to the separation of faces ‘d’.
   
   ∴ Q ∝ \(\frac{\mathrm{A}\left(\theta_2-\theta_1\right) t}{\mathrm{~d}}\)
   or Q = \(\frac{K A\left(\theta_2-\theta_1\right) t}{d}\)
   K is called coefficient of thermal conductivity of the material.
   When A = 1; (θ₂ – θ₁) = 1; t = 1; d = 1
   ∴ K = Q

Problem:
K_c = 401 W/mK, θ₂ = 104°C, θ₂ = 24°C, d = 0.10 m, A = 1.6 × 10^-6 m^-2
Rate of heat conduction = \(\frac{Q}{t}=K_c A \frac{\left(\theta_2-\theta_1\right)}{d}\)
= 401 × 1 × 10^-6 × (104-24) = 0.32 W

Question 3.
State and explain Newton’s law of cooling. State the conditions under which Newton’s law of cooling is applicable. A body cools down from 60°C to 50°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surroundings.
Answer:
Expression for Newton’s law of cooling : Consider a hot body at temperature T. Let T₀ be the temperature of its surroundings. According to Newton’s law of cooling,
Rate of loss of heat ∝ Temperature difference between the body and its surroundings.
\(\frac{-\mathrm{dQ}}{\mathrm{dt}}\) ∝ (T – T₀)
\(\frac{-\mathrm{dQ}}{\mathrm{dt}}\) = k (T – T₀) ⇒ (1) where k is proportionality constant. Let m be the mass and C the specific heat of the body at temperature T. If the temperature of the body falls by small amount dT in time dt, then the amount of heat lost is
dQ = mc dT
∴ Rate of loss of heat is given by
\(\frac{\mathrm{dQ}}{\mathrm{dt}}\) = mc \(\frac{\mathrm{dT}}{\mathrm{dt}}\) ………….. (2)
putting (2) in (1) ⇒
-mc \(\frac{\mathrm{dT}}{\mathrm{dt}}\) = k(T – T₀)
\(\frac{\mathrm{dT}}{\mathrm{dt}}=\frac{\mathrm{k}}{\mathrm{mc}}\) (T – T₀) = -K(T – T₀)
where K = \(\frac{\mathrm{k}}{\mathrm{mc}}\) is another constant.
⇒ \(\frac{d T}{\left(T-T_0\right)}\) = – K dt
on integrating both sides, we get
\(\int \frac{d T}{\left(T-T_0\right)}=-K \int d t\)
log_e (T – T₀) = – Kt + c …………… (3)
T – T₀ = e^{-Kt + c}
T = T₀ + e^c e^-Kt
T = T₀ + Ce^-Kt ……………… (4)
Where c is integration constant and C = e^c
equations (1), (2), (3) and (4) are the different expressions for Newton’s law of cooling.

Explanation of Graphs:
1) If we plot a graph by taking equation (4) different values of temperature difference ∆T = T – T₀ along y – axis and the corresponding values of t along X – axis, we get a curve of the form shown in figure. It clearly shows that the rate of cooling is higher initially and then decreases as the temperature of the body falls.
Curve showing cooling of hot water with time.

2) More over, the equation (3) is of the form y = mx + c. So if we plot a graph, by taking log_e(T – T₀) along Y – axis and time t along x – axis, we must get a straight line, as shown in figure. It has a negative slope equal to – K and intercept on Y – axis equal to C.
In both of the above situations, Newton’s law of cooling stands verified.

Newton’s law of cooling is applicable :

1. Loss of heat is negligible by conduction and only when it is due to convection.
2. Loss of heat occurs in a stream lined flow of air i.e., forced convection.
3. Temperature of the body is uniformly distributed over it.
4. Temperature differences are moderate i.e., upto 30 K, however if heat body is due to forced convection the law is valid for large differences of temperature also.
   
   ∴ Surrounding temperature, θ₀ = \(\frac{85^{\circ} \mathrm{C}}{3}\) = 28.33°C

Problems

Question 1.
What is the temperature for which the readings on Kelvin and Fahrenheit scales are same?
Answer:
The relation between Kelvin and Fahrenheit scales is given by
\(\frac{K-273.15}{100}=\frac{F-32}{180}\)
but K = F
\(\frac{F-273.15}{100}=\frac{F-32}{180}\)
F – 273.15 = \(\frac{5}{9}\) F – \(\frac{160}{9}\)
F – \(\frac{5}{9}\) F = 273.15 – \(\frac{160}{9}\)
= 273.15 – 17.77
\(\frac{4F}{9}\) = 255.38
∴ F = \(\frac{9}{4}\) (255.38)
= 574.6°F

Question 2.
Find the increase in temperature of aluminium rod if its length ¡s to be increased by 1%. (a for aluminium = 25 × 10^-6/0 C). [A.P (Mar. ‘15)]
Answer:
Percentage increase in length = \(\frac{l_2-l_1}{l_1}\) × 100
= α(t₂ – t₁) × 100 [∵ \(\frac{l_2-l_1}{l_1}\) = α(t₂ – t₁)]
Here, percentage increase in length = 1,
α = 25 × 10^-6/ °C
1 = 25 × 10^-6(t₂ – t₁) × 100
t₂ – t₁ = \(\frac{1}{25 \times 10^{-4}}=\frac{10^4}{25}\) = 400°C

Question 3.
How much steam at 100°C is to be passed into water of mass 100 g at 20°C to raise its temperature by 5°C ? (Latent heat of steam is 540 cal/g and specific heat of water is 1 cal/g°C).
Answer:
In the method of mixtures,
Heat lost by steam = Heat gained by water
m_SL_S + m_SS(100 – 1) = m_WS (t – 20)
Where m_S is the mass of steam, L_S is the latent heat of steam, S is the specific heat of steam and m_W is the mass of water.
Here, L_S = 540 cal/g
S = 1 cal/g°C;
m_W = 100 g
t = 20 + 5 = 25°C
m_S × 540 + m_S × 1 × (100 – 25) = 100 × 1 × (25 – 20)
615 m_S = 500
m_S = \(\frac{500}{615}\)
= 0.813 g.

Question 4.
2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313 K. If the specific heat of air at constant volume is 0.718 kJ/kgK, find the amount of heat absorbed in kJ and kcal.
(J = 4.2 kJ/kcal).
Answer:
Here, M = 2 kg
dT = 313- 293 = 20 K
C_V = 0.718 × 10³ J/Kg – K
J = 4.2 KJ/Kcal
C_V = \(\frac{1}{\mathrm{~m}} \frac{\mathrm{d} \theta}{\mathrm{dt}}\)
dθ = mC_VdT
= 0.718 × 10³ × 2 × 20
= 28720 J
dθ = 28.72 KJ
dθ = \(\frac{28720}{4.2 \times 10^3}\)
= 6.838 Kcal.

Question 5.
A dock, with a brass pendulum, keeps correct time at 20°C, but loses 8.212 s per day, when the temperature rises to 30°C. Calculate the coefficient of linear expansion of brass.
Answer:
Here t₁ = 20°C, t₂ = 30°C
Loss of time per day = 8.212 sec
Loss of time per day = \(\frac{1}{2}\) α (t₂ – t₁) × 86,400
8.212 = \(\frac{1}{2}\) α (30 – 20) × 86,400
∴ α = \(\frac{16.424}{864 \times 10^3}\)
= 19 × 10^-6/C°

Question 6.
A body cools from 60°C to 40°C in 7 minutes. What will be its temperature after next 7 minutes if the temperature of its surroundings is 10°C ?
Answer:
Rate of cooling \(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = K(θ_av – θ₀)
Case (i) :
Given, .
θ₁ = 60°C, θ₂ = 40°C, θ₀ = 10°C, t = 7 min
⇒ \(\frac{60^{\circ} \mathrm{C}-40^{\circ} \mathrm{C}}{7}=K\left[\frac{60^{\circ} \mathrm{C}+40^{\circ} \mathrm{C}}{2}-10^{\circ} \mathrm{C}\right]\)
\(\frac{20}{7}\) = K[50 – 10]
⇒ \(\frac{20}{7}\) = K × 40
∴ K = \(\frac{1}{14}\) ……………. (1)

Case (ii) :
Given,
θ₁ = 40°C, θ₀ = 10°C, t = 7 min, θ₂ = ?
⇒ \(\frac{40-\theta_2}{7}=K\left[\frac{40+\theta_2}{2}-10\right]\)
\(\frac{40-\theta_2}{7}=\frac{1}{14}\left[\frac{40+\theta_2}{2}-10\right]\)
80 – 2θ₂ = \(\frac{40+\theta_2-20}{2}\)
160 – 4θ₂ = 20 + θ₂
5θ₂ = 140
∴ θ₂ = \(\frac{140}{5}\) = 28°C

Question 7.
If the maximum intensity of radiation for a black body is found at 2.65 pm what is the temperature of the radiating body ? (Wien’s constant = 2.9 × 10^-3 mK)
Answer:
λ_max = 2.65 pm = 2.65 × 10^-6m
Wien’s constant (b) = 2.90 × 10^-3 mk
λ_max T = b(constant)
T = \(\frac{b}{\lambda_{\max }}=\frac{2.9 \times 10^{-3}}{2.65 \times 10^{-6}}\)
= 1094 K.

Additional Problems

Question 1.
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Answer:
Relation between kelvin scale and Celsius scale is T_C = T_K – 273.15
Where T_C.T_K = temperature on Celsius and kelvin scales respectively
For Neon T_C = 24.57 – 273.15 = – 248.58°C
For CO₂ T_C = 216.55 – 273.15 = -56.60°C
Relation between Kelvin and Fahrenheit scales is
\(\frac{T_F-32}{180}=\frac{T_K-273.15}{100}\)
T_F = \(\frac{180}{100}\) (T_K – 273.15) + 32
FQ or Neon T_F
= \(\frac{180}{100}\)(24.57 – 273.15) + 32 = – 415.44° F
FQ or CO₂ T_F = \(\frac{180}{100}\) (216.55 – 273.15) + 32
= – 69.88° F

Question 2.
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B ?
Answer:
Given triple point of water on scale A = 200
A triple point of water on scale B = 350 B
As per question 200 A = 350, B = 273.16 K
(Or)
IA = \(\frac{273.16}{200}\) K and IB = \(\frac{273.16}{350}\) K
If T_A and T_B represent the triple point of water on two scales A and B then
\(\frac{273.16}{200}\) T_A = \(\frac{273.16}{350}\)T_B (Or) \(\frac{T_A}{T_B}=\frac{200}{350}=\frac{4}{7}\)
(Or) T_A = \(\frac{4}{7}\) T_B

Question 3.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = R₀[1 + α(T – T₀)]
The resistance is 101.6 Ω at the triple-point of water 273.16 K and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?
Answer:
Here, R₀ = 101. 6 Ω, T₀ = 273.16 K
Case (i) R₁ = 165.5 Ω, T₁ = 600.5 K
Case (ii) R₂ = 123.4 Ω; T₂ = ?
Using the relation R = R₀(1 + α(T – T₀))
Case (i):
165.5 = 101.6(1 + α(600.5-273.16))
α = \(\frac{165.5-101.6}{101.6 \times(600.5-273.16)}\)
= \(\frac{63.9}{101.6 \times 327.34}\)

Case (ii): 123.4 = 101.6(1 + α(T₂ – 273.16))
(Or)
123.4 = 101.6
(1 + \(\frac{63.9}{101.6 \times 327.34}\) (T₂ – 273.16))
= 101.6 + \(\frac{63.9}{327.34}\) (T₂ – 273.16)
T₂ = \(\frac{(123.4-101.6) \times 327.34}{63.9}\) + 273.16
= 111.67 + 273.16 = 384.83 K

Question 4.
Answer the following :
a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ?
b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ?
c) The absolute temperature (Kelvin scale) T is related to the temperature t_C on the Celsius scale by
t_c = T – 273.15
Why do we have 273.15 in this relation and not 273.16 ?
Answer:
a) This is on account of the fact that the triple point of water has a unique value i.e., 273.16 K at a unique point, where exists unique values of pressure and volume. On the other hand, the melting point of ice and boiling point of water do not have unique set of value as they change in pressure and volume.

b) The other fixed point on the absolute scale is the absolute zero itself.

c) On Celsius 0°C corresponds to melting points of ice of normal pressure. The corresponding value of absolute temperature is 273.15 K. The temperature 273.16 K corresponds to the triple point of water.
From the given relation the corresponding value of triple point of water on Celsius scale
= 273.16 – 273. 15 = 0.01°C.

Question 5.
Two. ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
b) What do you think is the reason behind the slight difference in answers of thermoineters A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ?
Answer:
a). Let T be the melting point of sulphur, for water T_tr = 273.16 K
For thermometer A, T = P/P_tr × 273.16
= \(\frac{1.797 \times 10^5}{1.250 \times 10^5}\) × 273.16 = 392.69 K
For thermometer B, T = P/P_tr × 273.16
= \(\frac{0.287 \times 10^5}{0.200 \times 10^5}\) × 273.16 = 391.98 K

b) The cause of slight different answers is that the oxygen and hydrogen gases are not perfectly ideal. To reduce this discrepancy, the readings should be at lower and lower pressure as in that case, the gases approach to the ideal gas behaviour.

Question 6.
A steel tape 1 m long is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0°C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0°C ? Coefficient of linear expansion of steel = 1.20 × 10^-5 K^-1.
Answer:
Length of steell tape at 27°C is 100 cm i.e.
L = 100 cm and T = 27°C
The length of steel tape at 45°C is L¹ = L + ∆L
= L + αL∆T
= 100 + (1.20 × 10^-5) × 100 × (45° – 27)
= 100.0216 cm
Length of 1 cm mark at 27°C on this scale, at 45°C
= 100.0216/100 cm
Length of 63 cm measured’ by this tape at 45°C will be
= \(\frac{100.0216}{100}\) × 63 = 63.0136 cm
Length of the same steel rod on a day when the temperature is 27°C = 63 × 1 = 63 cm.

Question 7.
A large steel wheel is to be fitted on to a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft ? Assume coefficient of linear expansion of the steel to be constant over the required temperature range :
α_steel = 1.20 × 10^-5K^-1.
Answer:
Here T₁ = 27°C = 27 + 273 = 300 K
Length at temperature, T₁K = L_T1 = 8.70 cm
Length at temperature, T₂K = L_T2 – 8.69 cm
Change in length = L_T2 – L_T1 = L_T2 α(T₂ – T₂)
Or 8.69 – 8.70 = 8.70 × (1.20 × 10^-5) × (T₂ – 300)
Or T₂ – 300 = \(\frac{0.01}{8.70 \times 1.2 \times 10^{-5}}\) = -95.8
Or T₂ = 300 – 95.8 = 204.2 K = -68.8°C

Question 8.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C ? Coefficient of linear expansion of copper = 1.70 × 10^-5K^-1.
Answer:
In this problem superfical expansion of copper sheet will be involved on heating. Here, area of hole at 227°C, then area of the hole at 227°C,
S₂ = \(\frac{\pi \mathrm{D}_2^2}{4}\)cm².
Coefficient of superficial expansion of copper is β = 2α = 2 × 1.70 × 10^-5 = 3.4 × 10^-5 c¹
Increase in area = S₂ – S₁ = βS₁ ∆T (or)
S₂ = S₁ + βS₁ ∆T
= S₁(1 + β∆T)
(Or)
\(\frac{\pi \mathrm{D}_2^2}{4}=\frac{\pi}{4}\) (4.24)² [1 + 3.4 × 10^-5 (228 – 27)]
(Or)
D₂² = (4.24)² × 1.0068
D₂ = 4.2544 cm
Change in diameter = D₂ – D₁
= 4.2544 – 4.24
= 0.0144 cm.

Question 9.
A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 × 10^-5K^-1; Young’s modulus of brass = 0.91 × 10¹¹ Pa.
Answer:
Here L = 1.8m, T₁ = 27°C, T₂ = -39°C, r = 1
mm = 10^-3m, F = ?
α = 2 × 10^-5C^-1, y = 0.91 × 10¹¹ N/m²
From y = \(\frac{\mathrm{FL}}{a \Delta L}\), ⇒ ∆L = \(\frac{\mathrm{FL}}{\mathrm{ay}}\)
Also ∆L = αL∆T = ∴ \(\frac{\mathrm{FL}}{\mathrm{ay}}\) = αL∆T
(Or) F = α∆Tay = α(T₂ – T₁) πr²y
= 2 × 10^-5 × (-39 – 27) × (10^-3)² × 0.91 × 10¹¹
= -3.77 × 10^-2 N
Negative sign indicates that the force is inward due to contraction of the wire.

Question 10.
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C ? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2,0 × 10^-5K^-1, steel = 1.2 × 10^-5K^-1.
Answer:
∆L₁ = L₁α₁∆T = 50 × (2.10 × 10^-5) (250 – 40) = 0.2205 cm
∆L₂ = L₂α₂∆T
= 50 × (1.2 × 10^-5) (250 – 40) = 0.216 cm
∴ Change in length of combined rod
= ∆L₁ + ∆L₂
= 0.220 + 0.126 = 0.346 cm

Question 11.
The coefficient of volume expansion of glycerin is 49 × 10^-5K^-1. What is the fractional change in its density for a 30°C rise in temperature ?
Answer:
Here r = 49 × 10^-5C^-1, ∆T = 30°C .
As V¹ = V + ∆V = V(1 + r∆T)
∴ V¹ = V(1 + 49 + 10^-5 × 30) = 1.0147 V
As P = \(\frac{m}{V}\), p¹ = \(\frac{m}{V^1}\) = \(\frac{\mathrm{m}}{1.0147 \mathrm{~V}}\)= 09855 P
Fractional change in density = \(\frac{\rho-\rho^{\prime}}{\rho}\)
= \(\frac{\rho-0.9855 \rho}{\rho}\)
= 0.0145

Question 12.
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 Jg^-1K^-1.
Answer:
Here p = 10 kw = 10⁵w,
Mass m = 8.0 kg = 8 × 10³ g
Rise in temp; ∆T = ?
time t = 2.5 min = 2.5 × 60 sec
Sp. heat C = 0.91 Jg^-1°C^-1
Total energy = p × t = 10⁴ × 150
= 15 × 10⁵ J
As 50% of energy is lost
∴ Energy available ∆Q = \(\frac{1}{2}\) × 15 × 10⁵
= 7.5 × 10⁵ J
As ∆Q = mc ∆T
∴ ∆T = \(\frac{\Delta Q}{\mathrm{mc}}=\frac{7.5 \times 10^5}{8 \times 10^3 \times 0.91}\) = 103°C

Question 13.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt ? (Specific heat of copper = 0.39 Jg^-1K^-1 heat of fusion of water = 335 Jg^-1.
Answer:
Here, mass of copper block m = 2.5 kg = 2500 kg
Fall in temperature ∆T = 500 – 0 = 500°C
Specific heat of copper C = 0.39 Jg^-1°C^-1
Latent heat of fusion L = 335 Jg^-1
Let the mass of ice melted by m¹
As heat gained by ice = Heat lost by copper
m^-1L = Mc∆T
m^-1 = \(\frac{\mathrm{mC} \Delta \mathrm{T}}{\mathrm{L}}\)
m¹ = \(\frac{2500 \times 0.39 \times 500}{335}\)
= 1500 g = 1.5 kg

Question 14.
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?
Answer:
Here, mass of metal, m = 0.20 kg = 200 g
Fall in temperature of metal ∆T = 150 – 40
= 110°C
If L is specific heat of metal, then heat lost by the metal
∆Q = mc∆T = 200 × L × 110
Volume of water = 150 C.C
Mass of water m¹ = 150 g
Water equivalent of calorimeter, w = 0.025 kg = 25 kg
Rise in temp of water and calorimeter
∆T¹ = 40 – 27 = 13°C
Heat gained by water and calorimeter,
∆Q¹ = (m¹ + w)T¹
= (150 + 25) × 13 = 175 × 13
As ∆Q = ∆Q¹
∴ From (i) and (ii)
200 × C × 110 = 175 × 13
C = \(\frac{175 \times 13}{200 \times 110}\) ≈ 0.1
(Or)
If some heat is lost to the sorroundings, value of L is so obtained will be less than the actual value of L.

Question 15.
Given below are observations on molar specific heats at room temperature of some common gases.

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?
Answer:
The gases which are listed in the above table are diatomic gases and not mono atomic gases. For diatomic gases, molar specific heat
= \(\frac{5}{2}\) R = \(\frac{5}{2}\) × 1.98 = 4.95, which agrees fairly well with all observations listed in the . table except for chlorine. A monoatomic gas molecules has only the transnational motion. A diatomic gas molecules, a part from translation motion. The vibrational as well as rotational motion is also possible. Therefore to raise the temperature of 1 mole of diatomic gas through 1°C heat is also to be supplied increase not only translation energy but also rotational and vibrational energies. Hence molar specific heat of diatomic gas is greater than that for monoatomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. Shows that for chlorine molecule, at room temperature vibrational motion. Also occurs along with translational and rotational motions. Where as other diatomic molecules at room temperature usually have rotational motion apart from their translational motion. This the reason that chlorine has some what larger value of molar specific heat.

Question 16.
Answer the following questions based on the P-T phase diagram of carbon dioxide :
a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂ co-exist in equilibrium ?
b) What is the effect of decrease of pressure on the fusion and boiling point of CO₂ ?
c) What are the critical temperature and pressure for CO₂ ? What is their significance ?
d) Is CO₂ solid, liquid or gas at (a) – 70°C under 1 atm, (b) -60°C under 10 atm, (c) 15°C under 56 atm ?
Answer:
a) The solid, liquid and vapour phase of carbon dioxide exist in equilibrium at the point i.e., temperature = -56.6°C and pressure = 5.11 atm.

b) With the decrease in pressure, both the fusion and boiling point of carbondioxide will decrease.

c) For carbondioxide, the critical tempera-ture is 31,1°C and critical pressure is 73.0 atm. If the temp of CO₂ is more than 31.1°C. It cannot be liquified, however large pressure we may apply.

d) Carbondioxide will be (a) a vapour at – 70°C under 1 atm (b) a solid at 6°C under 10 atm (c) a liquid at 15°C under 56 atm.

Question 17.
Answer the following questions based on the P – T phase diagram of CO₂ :
a) CO₂ at 1 atm pressure and temperature – 60°C is compressed isothermally. Does it go through a liquid phase ?
b) What happens when CO₂ at 4 atm pressure is cooled from room tem-perature at constant pressure ?
c) Describe qualitatively the changes in a given mass of solid CO₂ at 10 atm pressure and temperature – 65°C as it is heated up to room temperature at constant pressure.
d) CO₂ is heated to a temperature 70°C and compressed isothermally. What changes in its properties, do you expect to observe ?
Answer:
a) Since the temp – 60°C lies to the left of 56.6°C on the curve i.e. lies in the region of vapour and solid phase. So CO₂ will condense directly into solid with out becoming liquid.

b) Since the pressure 4 atm is less than 5.11 atm. The carbondioxde will condenses directly into solid without becoming liquid.

c) When a solid CO₂ at 10 atm. Pressure and -65°C temp is heated, It is first converted into a liquid. A further increase in temperature brings it to the vapour phase. At P = 10 atm. If a horizontal line is drawn parallel to the T-axis. Then the points of intersection of this line with the fusion and vapourisation curve will give the fu-sion and boiling point of CO₂ at 10 atm.

d) Since 70°C is higher than the critical temperature of CO₂. So the CO₂ gas can not be converted into liquid state on being compressed isothermally at 70°C. It will remain in the vapour state. Nowever the gas will depart more and more now its perfect gas behavious with the increase in pressure.

Question 18.
A child running a temperature of 101 °F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580 cal g^-1.
Answer:
Here fall in temp = ∆T = 101 – 98 = 3°F
= 3 × \(\frac{5}{3}\)°C = \(\frac{5}{3}\)°C
Mass of child M = 30 kg
Sp. heat of human body = Sp heat of water
C = 1000 cal.kg^-1C^-1
∴ Heat last by the body of child, ∆Q = mC∆T
= 30 × 1000 × \(\frac{5}{3}\) = 5000 calories
If m be the mass of water evapourated in 20 min then m¹L = ∆Q
m¹ = \(\frac{\Delta Q}{L}=\frac{5000}{580}\) = 86.2 g
∴ Average rate of extra evapouration = \(\frac{86.2}{20}\)
= 4.31 gmin^-1

Question 19.
A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C and co-efficient of thermal conductivity of thermacole is 0.01 Js^-1m^-1 K^-1. [Heat of fusion of wafer = 335 × 10³ J kg^-1]
Answer:
Here length of each side l = 30 cm = 0.3 m
Thickness of each side ∆x = 5 cm = 0.05 m
Total surface area through which heat enters into the box
A = 6 l² = 6 × 0.3 × 0.3 = 0 Jum²
Temp.diff ∆T = 45 – 0 = 45°C,
K = 0.01 JS^-1m^-1°C^-1
Time ∆T = 6 hrs = 6 × 60 × 60 S
Latent heat of fusion L = 335 × 10³ J/kg
Let m be the mass of ice melted in this time
∆Q = mL = KA \(\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)\) ∆t
m = kA \(\left(\frac{\Delta T}{\Delta \mathrm{T}}\right) \frac{\Delta \mathrm{t}}{\mathrm{L}}\) = 0.01 × 0.54 × \(\frac{45}{0.05}\) × \(\frac{6 \times 60 \times 60}{335 \times 10^3}\) = 0.313 kg
Mass of ice left = 4 – 0.313 = 3.687 kg

Question 20.
A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 Js^-1m^-1 K^-1: Heat of vaporisation of water = 2256 × 10³ Jkg^-1.
Answer:
Were A = 0.15 m² ∆x = 1.0 m = 10^-2 m
\(\frac{\Delta Q}{\Delta t}=\frac{6 \times 10^3 \times 2256}{60}\) JS^-1 = 2256 × 10² JS^-1
K = 609 JS^-1m^-1°C^-1, ∆T = (t – 100)
From \(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\mathrm{KA}\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)\)
2256 × 10² = 609 × 0.15 \(\frac{(t-100)}{10^{-2}}\)
t – 100 = \(\frac{2256}{609 \times 0.15}\) = 2470 t
t = 24.70

Question 21.
Explain why :
a) a body with large reflectivity is a poor emitter.
b) a brass tumbler feels much colder than a wooden tray on a chilly day.
c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.
d) the earth without its atmosphere would be inhospitably cold.
e) healting systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
Answer:
a) This is because a body with large refelectivity is a poor absorber of heat and poor absorbers of heat are poor imilters.

b) When we touch a brass tumbler on a chill day; heat flows from our body to the tumbler quickly cas thermal conductivity or brass is very high and as a result, it appears colder. On the other hand as the wood is bad conductor, heat does not flow to the wooden tray from our body, on touching it.

c) When the red hot iron pieces is in the oven, its temperature TK is the given by the relation.
E = αT⁴ But if the red hot iron piece is in the open having the surrounding temperature T₀, its energy is radiated according to relation E¹ = α(T⁴ – T⁴₀). As the working principle of optical pysometer is based on the fact that the brightness of – a glowing surface of a body depends , upon its temperature. Therefore, pysometes gives too low a value for the temperature of red iron in the open.

d) The lower layer’s of Earth’s atmosphere reflect infrared radiations from earth back to the surface of the earth. Thus the heat radiation recieved by the earth from the sun during the day are kept trapped by the atm. If atmosphere of earth were not there, its surface would become too cold to live.

e) Steam at 100°C possess more heat than the same mass of water at 100°C possess 540 calories of heat more than possessed by 1 gm of ware at 100°C. That is why heating systems based on circulation of stream are more efficient than those based on circulation of hot water.

Question 22.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.
Answer:
\(\frac{\mathrm{d} \theta}{\mathrm{dt}}\) = K [θ_av – θ₀]
\(\frac{80-50}{5}\) = K(65 – 20)
⇒ \(\frac{30}{5}\) = K × 45 = K = \(\frac{2}{15}\)
\(\frac{60-30}{t}=\frac{2}{15}[45-20]\)
\(\frac{30}{t}=\frac{2}{15} \times 15\)
t = 5 min
Time of cooling is given by t = \(\frac{2.3026}{k}\) log₁₀
\(\frac{T_1-T_0}{T_2-T_0}\)
As per question condition (i) T₁ = 80°C, T₂ = 50°C, T₀ = 20°C, t = 5 min
5 × 60 = \(\frac{2.3026}{K} \log _{10} \frac{80-20}{50-20}\)
= \(\frac{2.3026}{k} \log _{10^2}\) ……………. (2)
As per question condition (i) T₁ = 60°C, T₂ = 30°C, T₀ = 20°C, t = ?
t = \(\frac{2.3026}{K} \log _{10} \frac{60-20}{30-20}\)
= \(\frac{2.3026}{k} \log _{10^4}\) ……………. (3)
Div (3) by (2) we get
\(\frac{t}{5 \times 60}=\frac{\log _{10^4}}{\log _{10^2}}=\frac{0.6021}{0.3010}\) = 2
(Or) t = 5 × 60 × 2 = 10 × 60 s
= 10 min

Textual Examples

Question 1.
Show that the coefficient of areal expansions. (∆A/A)/∆T, of rectangular sheet of the solid is twice its linear expansivity, α₁.
Answer:
∆A₃ = (∆a) (∆b)

When the temperature increases by ∆T, a increases by ∆a = α₁a ∆T and b increases by ∆b = α₁b ∆T. From fig. the increase in area
∆A = ∆A₁ + ∆A₂ + ∆A₃
∆A = a ∆b + b ∆a + (∆a) (∆b)
= a α₁b ∆T + b α₁a ∆T + (α₁)² ab(∆T)²
= α₁ab ∆T(2 + α₁∆T)
= α₁A ∆T(2 + α₁∆T)
Since α₁ = 10^-5 K^-1, from Table 12.1, the product α₁∆T for fractional temperature is small in comparision with 2 and may be neglected.
Hence,
\(\left(\frac{\Delta \mathrm{A}}{\mathrm{A}}\right) \frac{1}{\Delta \mathrm{T}}\) ≈ 2α₁

Question 2.
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diametbr of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27°C. To what temperature should the ring be heated so as to fit the rim of the wheel ?
Answer:
Given,
T₁ = 27°C
L_T1 = 5.231 m
L_T2 = 5.243 m
So,
L_T2 = L_T1 [1 + α₁(T₂ – T₁)]
5.243 m = 5.231 m[1 + 1.20 × 10^-5 K^-1 (T₂ – 27°C)]
or T₂ = 218°C

Question 3.
A sphere of aluminium of 0.047 kg is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100°C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg of water at 20°C. The temperature of water rises and attains a steady state at 23°C. Calculate the specific heat capacity of aluminium.
Answer:
Mass of aluminium sphere (m.,) = 0.047 kg
Initial temp, of aluminium sphere = 100°C
Final temp. = 23°C
Change in temp. (∆T) = (100°C – 23°C)
= 77°C
Let specific heat capacity of aluminium be s_Al. The amount of heat lost by the aluminium sphere = m₁s_Al ∆T = 0.047 kg × s_Al × 77°C s ……………. (i)
Mass of water (m₂) = 0.25 kg
Mass of calorimeter (m₃) = 0.14 kg
Initial temp, of water and calorimeter = 20°C
Final temp, of the mixture = 23°C
Change in temp. (∆T₂) = 23°C – 20°C = 3°C
Specific heat capacity of water (s_w) from table
12.3 = 4.18 × 10³ J kg^-1 K^-1
Specific heat capacity of copper calorimeter = 0.386 × 10³ J kg^-1 K^-1
The amount of heat gained by water and calorimeter = m₂s_w ∆T₂ + m₃s_cu ∆T₂ = (m₂s_w + m₃s_cu) (∆T₂)
= (0.25 kg × 4.18 × 10³ J kg^-1 K^-1 + 0.14 kg × 0.386 × 10³ J kg^-1 K^-1) (23°C – 20°C) ………………… (ii)
In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter.
So, from (i) and (ii)
0.047 kg × s_Al × 77°C = (0.25 kg × 4.18 × 10³ J kg^-1 K^-1 + 0.14 kg × 0.386 × 10³ J kg^-1 K^-1) (3°C)
s_Al = 0.911 kJ kg^-1 K^-1

Question 4.
When 0.15 kg of ice of 0°C mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Calculate the heat of fusion of ice.
(s_water = 4186 J kg^-1 K^-1)
Answer:
Heat lost by water = ms_w (θ_f – θ_i)_w
= (0.30 kg) (4186 kg^-1 K^-1) (50.0°C – 6.7°C)
= 54376.14 J
Heat required to melt ice = m₂L_f = (0.15 kg) L_f
Heat required to raise temperature of ice water to final temperature
= m_Is_w (θ_f – θ_i)_I
= (0.15 kg) (4186 J kg^-1 K^-1) (6.7°C – 0°C)
= 4206.93 J
Heat lost = heat gained
54376.14 J = (0.15 Kg)L_f + 4206.93 J
L_f = 3.34 × 10⁵ J kg^-1

Question 5.
Calculate the heat required to convert 3 kg of ice at -12°C kept in a calorimeter-to steam at 100°C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg^-1 K^-1, specific heat capacity of water = 4186 J kg^-1 K^-1, latent heat of fusion of ice = 3.35 × 10⁵ J kg^-1 and latent heat of steam = 2.256 × 10⁶ J kg^-1.
Answer:
We have
Mass of the ice, m = 3 kg
Specific heat capacity of ice, s_ice
= 2100 J kg^-1 K^-1
Specific heat capacity of water, s_water
= 4186 J kg^-1 K^-1
Latent heat of fusion ice, L_{f ice}
= 3.35 × 10⁵ J kg^-1
Latent heat of steam, L_steam
= 2.256 × 10⁶ J kg^-1
Now,
Q = Heat required to convert 3 kg of ice at -12°C to steam at 100°C.
Q₁ = Heat required to convert ice at – 12°C to ice at 0°C. .
ms_ice ∆T₁ = (3 kg) (2100 J kg^-1 K^-1) [0 – (-12)]°C = 75600 J
Q₂ = Heat required to melt ice at – 0°C to water at 0°C.
mL_{f ice} = (3 kg) (3.35 × 10⁵ J kg^-1) = 1005000 J
Q₃ = Heat required to convert water at 0°C to water at 100°C.
ms_W ∆T₂ = (3 kg) (4186 J kg^-1 K^-1) (100°C)
= 1255800 J
Q₄ = Heat required to convert water at 100°C to steam at 100°C.
mL_steam = (3 kg) (2-256 × 10⁶ J kg^-1)
= 6768000 J
So,
Q = Q₁ + Q₂ + Q₃ + Q₄
= 75600 J + 1005000 J + 1255800 J + 6768000 J
= 9.1 × 10⁶ J

Question 6.
What is the temperature of the steel- copper junction in the steady state of the system shown in fig. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace = 300°C, temperature of the other end = 0°C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = 50.2 J s^-1m^-1K^-1 and of copper = 385 J s^-1m^-1K^-1].

Answer:
Let T be the temperature of the steel-copper junction in the steady state. Then,
\(\frac{K_1 A_1(300-T)}{L_2}=\frac{K_2 A_2(T-0)}{L_2}\)
Where 1 and 2 refer to the steel and copper rod respectively. For A₁ = 2, A₂, L₁ = 15.0 cm, L₂ = 10.0 cm, K₁ = 50.2 J s^-1m^-1K^-1,
K₂ = 385 J s^-1m^-1K^-1, we have
\(\frac{50.2 \times 2(300-\mathrm{T})}{15}=\frac{385 \mathrm{~T}}{10}\)
which gives T = 44.4°C

Question 7.
An iron bar (L₁ = 0.1 m, A₁ = 0.02 m², K₁ = 79 W m^-1 K^-1) and a brass bar (L₂ = 0.1 m, A₂ = 0.02 m², K₂ = 109 W m^-1K^-1) are soldered end to end as shown in fig. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar and (iii) the heat current through the compound bar.

Answer:
Given,
L₁ = L₂ = L = 0.1 m, A₁ = A₂ = A = 0.02 m², K₁ = 79 W m^-1 K^-1, K₂ = 109 W m^-1K^-1, T₁ = 373 K and T₂ = 273 K.
Under steady state condition, the heat current (H₁) through iron bar is equal to the heat current (H₂) through brass bar.
So, H = H₁ = H₂
= \(\frac{K_1 A_1\left(T_1-T_0\right)}{L_1}=\frac{K_2 A_2\left(T_0-T_2\right)}{L_2}\)
For A₁ = A₂ = A and L₁ = L₂ = L this equation leads to
K₁(T₁ – T₀) = K₂(T₀ – T₂)
Thus the junction temperature T₀ of the two bars is T₀ = \(\frac{\left(K_1 T_1+K_2 T_2\right)}{\left(K_1+K_2\right)}\) ………….. (a)
Using this equation, the heat current H through either bar is

Using these equations, the heat current H’ through the compound bar of length L₁ + L₂ = 2L and the equivalent thermal conductivity K, of the compound bar are given by

Question 8.
A pan filled with hot food cools from 94°C to 86°C in 2 minutes when the room temperature is at 20°C. How long will it take to cool from 71°C to 69°C ?
Answer:
The average temperature of 94°C and 86°C is 90°C, which is 70°C above the room temperature. Under these conditions the pan cools 8°C in 2 minutes.
Using \(\frac{\mathrm{dT}_2}{\mathrm{~T}_2-\mathrm{T}_1}=-\frac{\mathrm{k}}{\mathrm{ms}}\) dt = – K dt we have
The average of 69°C and 71°C is 70°C, which is 50°C above room temperature. K is the same for this situation as for the original.

The average of 69°C and 71°C is 70°C, which is 50°C above room temperature. K is the same for this situation as for the original.
\(\frac{2^{\circ} \mathrm{C}}{\text { Time }}\) = K(50°C)
When we divide above two equations, we have
\(\frac{8^{\circ} \mathrm{C} / 2 \min }{2^{\circ} \mathrm{C} / \text { time }}=\frac{K\left(70^{\circ} \mathrm{C}\right)}{\mathrm{K}\left(50^{\circ} \mathrm{C}\right)}\)
Time = 0.7 min = 42 s

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids

Very Short Answer Questions

Question 1.
Define average pressure. Mention it’s unit and dimensional formula. Is it a scalar or a vector ? [A.P. Mar. 17]
Answer:
Average pressure (P_av) : Average power is defined as the normal force acting per unit area.
P_av = \(\frac{F}{A}\)
units → N/m² (or) pascal
Dimensional formula → [ML^-1 T^-2]
Pressure is a scalar quantity.

Question 2.
Define Viscosity. What are it’s units and dimensions ?
Answer:
Viscosity : The property of a liquid which opposes the relative motion between its layers is called viscosity. .
G.G.S unit poise
S.I unit → Nm^-2s
Dimensional formula = [M¹L^-1T^-1]

Question 3.
What is the principle behind the carburetor of an automobile ? [AP – Mar. 15; TS – Mar. ’18, ’17]
Answer:
The carburetor of automobile has a venturi channel (nozzle) through which air flows with a large speed. The pressure is then lowered at the narrow neck and the petrol is sucked up in the chamber to provide the correct mixture of air to fuel necessary for combustion. ”

Question 4.
What is magnus effect ? [T.S. Mar. 16; A.P. Mar. 15]
Answer:
The difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called magnus effect.

Question 5.
Why are drops and bubbles spherical ? [A.P. Mar. 18, 17, 16, 14]
Answer:
The surface tension of a liquid tends to have minimum surface area. For a given volume, the surface area is minimum for a sphere. Hence rain drops are spherical shape.

Question 6.
Give the expression for the excess pressure in a liquid drop. [T.S. Mar. 17]
Answer:
Excess pressure in a liquid drop, p_i – p₀ = \(\frac{2 s}{r}\)
where s = Surface tension; r = Radius of the liquid drop.

Question 7.
Give the expression for the excess pressure in an air bubble inside the liquid.
Answer:
Excess pressure in an air bubble inside the liquid, p_i – p₀ = \(\frac{2 s}{r}\)
where s = Surface tension
r = Radius of the air bubble.
Air bubble forms inside the liquid, hence it has one liquid surface.

Question 8.
Give the expression for the excess pressure soap bubble in air. [T.S. Mar. 16]
Answer:
Soap bubble have two interfaces, hence excess pressure inside a soap bubble is p_i – p₀ = \(\frac{4 s}{r}\)
where s = Surface tension
r = Radius of the soap bubble.

Question 9.
What are water proofing agents and water wetting agents ? What do they do ?
Answer:
Water proofing agents are added, to create a large angle of contact between water and fibres.
Soaps, detergents and dying substances are wetting agents. When they are added, the angle of contact becomes small. So that they may penetrate well and become effective.

Question 10.
What is angle of contact ? [A.P. Mar. 16]
Answer:
The angle between tangent to the liquid surface at the point of contact and solid surface inside the liquid is termed as angle of contact (0).

Question 11.
Mention any two examples that obey Bernoullis theorem and justify them. [A.P. Mar. 18]
Answer:

1. In heavy winds house roof’s are blown off. When the velocity of the wind is greater on the roof top than inside the house, then the pressure on the roof top becomes less than that inside the house. This pressure difference causes the dynamic lift.
2. When a fan is rotating, papers are blown off from .the table top. The velocity of wind on the paper increases due to fan and hence pressure decreases. Due to this pressure difference papers are blown off.

Question 12.
When water flows through a pipe, which of the layers moves fastest and slowest ?
Answer:
Water flows through a pipe, the layers near the axis of the tube are fastest and at the walls of the tube are slowest

Question 13.
“Terminal velocity is more if surface area of the body is more”. Give reasons in support of your answer.
Answer:
Surface area (A) = 4πr² and terminal velocity (υ_t) αr²
As surface area increases, r2 is also increases. Then terminal velocity is also increases.
∴ Terminal velocity is more, if surface area of the body is more.

Short Answer Questions

Question 1.
What is atmospheric pressure and how is it determined using Barometer ?
Answer:
Atmospheric pressure : Atmospheric pressure at any point is equal to the weight of a column of air of unit cross sectional area extending from that point to the top of the earth’s atmosphere.
1 atm = 1.013 × 10⁵ pa
Determination of atmospheric pressure using barometer: A long glass tube closed at one end and filled with mercury is inverted into a trough of mercury. This device is known as mercury barometer.

The space above the mercury column in the tube contains only mercury vapour whose pressure Pis so small, that it may be neglected. The pressure inside the column at point A must equal the pressure at B.
∴ Pressure at B = Atmospheric pressure = P_a
P_a = ρgh = Pressure at A …………………… (1)
Where ρ is density of mercury and h is the height of the mercury column in the tube. In the experiment it is found that the mercury column in the barometer has a height of about 76cm at the sea level equivalent to one atmosphere.

Question 2.
What is guage pressure and how is a manometer used for measuring pressure differences ?
Answer:
Guage pressure : The guage pressure is the difference of the actual pressure and the atmospheric pressure. P_g = P – P_a
measurement of pressure differences :

1. The manometer consists of a U-shaped tube, which is filled with a low density liquid (oil) for measuring small pressure difference and high density liquid
   (mercury) for measuring large pressure difference.
2. One end of the tube is connected to the vessel D whose pressure of air measure and the other end of the tube is open.
3. If pressure of air in vessel D is more than the earth’s atmosphere, the level of liquid in arm I will go down up to point A and level of liquid in arm II rises up to C.
4. Then the pressure of air in vessel is equal to pressure at point A.
   
5. Note the difference of liquid levels in the two arms of U-tube.(Say h). p be density of the liquid. Pa is the atmospheric pressure.
6. Pressure at point A (P_A) = Pressure at point B = Pressure at point C + Pressure due to column of liquid.
   P_A = P_c + hρg (or) P_A – P_c = hρg
   Here P_c = P_a, P_A = P, ∴ P – P_a = hρg
   P – P_a = P_g = guage pressure = hρg

Question 3.
State Pascal’s law and verify it with the help of an experiment.
Answer:
Pascal’s law : It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest is same.
Proof :

1. Imagine a circular cylinder of uniform cross-sectional area A, such that points C and D lie on flat faces of the cylinder.
2. The liquid inside the cylinder is in equilibrium under the action of forces exerted by the liquid outside the cylinder.
3. These forces are acting every where perpendicular to the surface of the cylinder.
4. Thus the forces on the flat faces of the cylinder at C and D will be perpendicular to the forces on the curved surface of the cylinder.
5. Since liquid is in equilibrium, the sum of the forces acting on the curved surface of the cylinder must be zero.
   
6. If P₁ and P₂ are the pressures at points C and D respectively. F₁ and F₂ are the forces acting on the flat surfaces of the cylinder due to liquid, then
   F₁ = P₁ A and F₂ = P₂A Since liquid is in equilibrium, therefore
   F₁ = F₂
   P₁ A = P₂A
   (or) P₁ = P₂
   It means the pressures at C and D are the same. This proves the pascal’s law.

Question 4.
Explain hydraulic lift and hydraulic brakes.
Answer:
Hydraulic lift and hydraulic brakes are based on the Pascal’s law.

Hydraulic lift : Here C and D are two cylinders of different areas of cross section. They are connected to each other with a pipe E. Each cylinder is provided with airtight frictionless piston. Let a, A be the area’s of cross-sections of the piston at C and D (a < < A)

The cylinders are filled with an incompressible liquid.
Let f be the applied force at C. Pressure exerted on the liquid
P = \(\frac{f}{a}\) ……………. (1)
According to pascal’s law, this pressure is transmitted to piston of cylinder D. Upward force at D will be
F = PA = \(\frac{f}{a}\) A = f \(\frac{A}{a}\) …………… (2)
As A > > a
∴ F > > f .
∴ Heavy load placed on the larger piston is easily lifted.

Hydraulic Brakes :
When the brake pedal is pressed, the piston (P) of the master cylinder is pushed inwards. There will be increased pressure on liquid at P, which is transmitted equally to P₁ and P₂ of wheel cylinder in accordance with pascal’s law. Due to which P₁ and P₂ move outwards. Breakshoes to move away from each other which in turn press against the inner rim of the wheel. The brake becomes operative.

Question 5.
What is hydrostatic paradox ?
Answer:
Consider three vessels A, B and C of different shapes. They are connected at the bottom by a horizontal pipe. On filling with the level in the three vessels is the same, though they hold different amounts of water. This is so because water at the bottom has the same pressure below each section of the vessel. It means the liquid pressure at a point is independent of the quantity of liquid but depends upon the depth of point below the liquid surface. This is known as hydrostatic paradox.

Question 6.
Explain how pressure varies with depth.
Answer:
Consider a fluid at rest in a container. In figure point 1 is at a height h above a point 2. The pressure at points 1 and 2 are P₁ and P₂. As the fluid is at rest, the horizontal forces should be zero. The resultant vertical forces should balance the weight of the element. Pressure at top (P₁A) acting downward, pressure at bottom (P₂A) acting upward.

(P₂ – P₁) A = mg ………………… (1)
Mass of the fluid (m) = ρv = ρhA
(P₂ – P₁) = ρgh ………………. (2)
Pressure difference depends on the vertical distance h.
If the point 1 under discussion is shifted to the top of the fluid (water), which is open to the atmosphere, P₁ may be replaced by atmospheric pressure (P_a) and P₂ by P. Then eq (2) gives
P – P_a = ρgh
P = P_a + ρgh ……………… (3)
Thus the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh.

Question 7.
What is Torricelli’s law ? Explain how the speed of efflux is determined with an experiment.
Answer:
Torricelli’s theorem : The velocity of efflux i.e., the velocity with which the liquid flows out of an orifice (i.e., a narrow hole) is equal to that which is freely falling body would acquire in falling through a vertical distance equal to the depth of orifice below the free surface of liquid.

Consider ideal liquid of density p contained in a tank provided with a narrow hole.
Let h = Height of free surface of liquid above O.
P = Atmospheric pressure
v = Velocity of efflux
Applying Bernoulli’s theorem at A and O
(P + ρgh + O)_atA = [P + 0 + \(\frac{1}{2}\)ρv²]_atO
P + ρgh = P + \(\frac{1}{2}\) ρv² = ρgh = \(\frac{1}{2}\)ρv²
v = \(\sqrt{2 g h}\)

Question 8.
What is Venturi-meter? Explain how it is used.
Answer:
Venturi meter: The Venturi-meter is a device to measure the flow speed of incompressible fluid.

1. It consists of a tube with a broad diameter and a small constriction at the middle.
2. A manometer in the form of a U-tube is also attached to it, with one of arm at the broad neck point of the tube and the other at constriction as shown in figure.
   
3. The manometer contains a liquid of density ρ_m.
4. The pressure difference causes the fluid in the U-tube connected at the narrow neck to rise in comparison to the other arm.
5. Filter pumps, sprayers used for perfumes, carburetor of automobile has used on this principle.

Question 9.
What is Reynolds number ? What is it’s significance ?
Answer:
Reynolds number : “Reynold number is a pure number which determines the nature of flow of liquid through a pipe”.
Reynold number (R_e) = \(\frac{\rho \mathrm{vd}}{\eta}\)
Where ρ is density of the fluid
v is speed of the fluid
d is diameter of the pipe

1. If the flow is stream line (or) laminor R_e < 1000
2. If the flow is turbulent, R_e > 2000
3. If the flow becomes unsteady, 1000 < R_e < 2000

Physical significance of Reynolds number : Reynold’s number describes the ratio of the inertial force per unit area to the viscous force per unit area for a flowing fluid.

Question 10.
Explain dynamic lift with examples.
Answer:
Dynamic lift: Dynamic lift is the force that act on a body, by virtue of its motion through a fluid.

e.g.1 : Fig (a) shows, ball moving without spin. Stream lines are equally distributed above and below the ball. The velocity above and below the ball is same resulting zero pressure difference. There no upward (or) downward force on the ball.

Fig (b) shows, ball moving with spin stream lines are more crowded above the ball than below. The velocity of air above the ball is large (v + v_r) and below it is smaller (v – v_r). As a result, there is a pressure difference between lower and upper faces. Pressure is less at top of the ball and pressure is morebelow the ball. There is a net upward force on the ball.
e.g. 2 : Dynamic lift also acts on the an aeroplane wing.

Question 11.
Explain Surface Tension and Surface energy. [Mar. 13]
Answer:
Surface tension (S): The force acting per unit length of an imaginary line drawn on the surface of a liquid, normal to it and parallel to the surface is called surface tension.
T = \(\frac{\mathrm{F}}{l}\)
S.l unit → N/m
D.F → [MT^-2]
Surface energy (E): The additional potential energy due to molecular forces per unit surface area is called surface tension.
Surface energy = \(\)
S.l Unit → J/m²
D.F → (MT^-2 )

Consider a horizontal liquid film ending in bar free to slide over parallel guides. We move the bar by a small distance d. The area of the surface increases, the system now has more energy, this means – that some work has to be done against an internal force F.
Work done (W) = F.d
If the surface energy of the film is S per unit area, the extra area is 2dl. A film has two sides and the liquid in between them.
So there are two surfaces and the extra energy is
S (2dl) = Fd
S = \(\frac{\mathrm{F}}{2 l}\)
Surface tension is equal to the surface energy and is also equal to the force per unit length exerted by the fluid on the movable bar.

Question 12.
Explain how surface tension can be measured experimentally.
Answer:
A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of the balance. The plate is balanced by weights on the otherside, with its horizontal edge just over water. The vessel is raised slightly till the liquid just touches the glass plate and pulls it down a little because of surface tension. Weights are added till the plate just clears water. Suppose the additional weight requires is W.

Surface tension of liquid air interface is
S_la = \(\frac{\mathrm{W}}{2 l}=\frac{\mathrm{mg}}{2 l}\)
Where l is length of the plate edge, m is extra mass.

Long Answer Questions

Question 1.
State Bernoulli’s principle. From conservation of energy in a fluid flow through a tube, arrive at Bernoulli’s equation. Give an application of Bernoulli’s theorem.
Answer:
Bernoulli’s principle : Bernoulli principle states that in a stream line flow, the sum of the pressure, the K.E per unit volume and the P.E per unit volume remains a constant.
P + \(\frac{1}{2}\) ρv² + ρgh = constant

1. Consider a non-viscous, incompressible fluid is flowing through the pipe in a steady flow.
2. Consider the flow at two regions BC and DE. Initially the fluid lying between B and D.
   
3. During short time interval At, this fluid would have moved. Suppose V₁ is the speed at B and V₂ is the speed at D.
4. In time ∆t, distance moved from B to C is V₁ ∆t in the same interval (∆t) distance moved from D to E is V₂∆t.
5. Let P₁ and P₂ be pressure act at area’s of cross¬sections A₁ and A₂ of the two regions.
6. The work done on the fluid at left end (BC)
   = Force × displacement
   = Pressure × Area × displacement
   = P₁A₁ × V₁∆t (∆V = A₁V₁∆t) = P₁∆V ………………. (1)
7. Similarly work done by the fluid at right end (DE)
   = P₂A₂ × V₂∆t = P₂∆V …………………. (2)
8. Work done on the fluid is taken as positive and workdone by the fluid is taken as negative.
   ∴ Total work done (W) = (P₁ – P₁) ∆V ………………. (3)
   Part of this work goes into changing the K.E of the fluid and part goes into changing gravitational P.E.
9. Mass of the fluid(∆m) passing through the pipe in time (∆t) is ∆m = ρA₁V₁∆t
   where ρ is the density of the fluid.
   ∆m = ρ∆V ………………….. (4)
10. Gravitational P.E = ρg∆V (h₂ – h₁) ………………… (5)
    Change in K.E (∆K) = \(\frac{1}{2}\) ρ∆V (V₂² – V₁²) …………….. (6)
11. According to law of conservation of energy
    (P₁ – P₂) ∆V = \(\frac{1}{2}\) ρ ∆V (V₂² – V₁²) + ρg∆V (h₂ – h₁)
    P₁ – P₂ = \(\frac{1}{2}\) ρ (V₂² – V₁²) + ρg (h₂ – h₁)
    P₁ + \(\frac{1}{2}\) ρ V₁² + ρgh₁ = P₁ + \(\frac{1}{2}\) ρ V₂² + ρgh₂
    P + \(\frac{1}{2}\) ρ v² + ρgh = constant ………………. (7)

∴ Sum of the pressure, K.E per unit volume and P.E per unit volume remains constant. Application of Bernoulli’s theorem :

1. It explains the dynamic lift on the wings of aeroplanes.
2. It explains the dynamic lift experienced by a spinning cricket ball.

Question 2.
Define coefficient of viscosity. Explain Stoke’s law and explain the conditions under which a rain drop attains terminal velocity, υ_t. Give the expression for υ_t.
Answer:
Coefficient of viscosity (η): The coefficient of viscosity is defined as the tangential force per unit area of the layer, required to maintain unit velocity gradient.
η = \(\frac{F}{A\left(\frac{\Delta V}{\Delta x}\right)}\)
S.l. unit → Nm^-2 s (or) P_aS
C.G.S unit → Poise
Dimensional formula = [ML^-1T^-1]
Stoke’s law : According to this law the viscous force acting on a moving body which is spherical in shape is directly proportional to

1. Coefficient of viscosity of fluid (η)
2. Radius of the spherical body (r)
3. Velocity of the body (v)

∴ F ∝ ηrv
F = Kηrv
Where K is a constant of proportionality. Experimentally it was found to be 6π.
∴ F = 6πηrv
When rain drops falling through air from the clouds reach the surface with almost constant speed through they are moving under gravitational force. This velocity is called terminal velocity. After attaining the terminal velocity, net force acting on the rain drop is zero.
According to stokes law, F ∝ ηrv
F = 6πηrv (∵ 6π = K = Proportionality constant)
Let ρ, r be the density and radius of the sphere.
The fluid density be σ.
The forces acting on the sphere are

1. Weight of the sphere W = mg
   W = Vρg = \(\frac{4}{3}\)πr³ρg …………….. (1)
2. The force of buoyancy (B) = V σ g = \(\frac{4}{3}\)πr³σg ……………. (2)
3. Viscous force (f) = 6πηv ………………… (3)
   When the sphere attains terminal velocity (v_t), the net force on the body becomes zero.
   
   ∴ At terminal velocity
   Net downward force = Net upward force
   W = B + f, W – B = f
   \(\frac{4}{3}\)πr³ρg – \(\frac{4}{3}\)πr³σg = 6πηrv_t;
   ∴ \(\frac{4}{3}\)πr³g (ρ – σ) = 6πηrv_t
   V_t = \(\frac{2}{9} \frac{r^2 g(\rho-\sigma)}{\eta}\) ………………. (4)

Problems

Question 1.
Calculate the work done in blowing a soap bubble of diameter 0.6 cm against the surface tension force. (Surface tension of soap solution = 2.5 × 10² Nm^-1).
Solution:
D = 0.6 cm = 0.6 × 10² m
r = \(\frac{D}{2}=\frac{0.6 \times 10^{-2}}{2}\) = 0.3 × 10² m
S = 2.5 × 10² N/m
W = 8πr²s
= 8 × 3.14 × (0.3 × 10^-2)₂ × 2.5 × 10^-2
W = 5.652 × 10^-6 J

Question 2.
How high does methyl alcohol rise in a glass tube of diameter 0.06 cm ? (Surface tension of methyl alcohol = 0.023 Nm^-1 and density = 0.8 gmcm^-3. Assume that the angle of contact is zero)
Solution:
D = 0.06cm, θ = 0°
r = \(\frac{D}{2}=\frac{0.06}{2}\) = 0.03 cm = 0.03 × 10^-2 m
= 3 × 10^-4 m
S = 0.023 N/m, Density = 0.8 gm/c.c.
= 0.8 × \(\frac{10^{-3}}{10^{-6}}\)
ρ = 0.8 × 10³ kg/m³
S = \(\frac{\mathrm{hr \rho g}}{2 \cos \theta}\)
h = \(\frac{2 \cos \theta}{r \rho g}\)
= \(\frac{2 \times 0.023}{3 \times 10^{-4} \times 0.8 \times 10^3 \times 9.8}\) (∵ cos 0° = 1)
= 0.0019 × 10^-1 m
≈ 0.002 m
h = 2 cm .

Question 3.
What should be the radius of a capillay tube if water has to rise to a height pf 6 cm in it ? (Surface tension of water = 7.2 × 10^-2 Nm^-1).
Solution:
h = 6 × 10^-2 m
S = 7.2 × 10^-2 N/m
Density of water (ρ) = 10³ kg/m³
S = \(\frac{\mathrm{hr \rho g}}{2}\)
r = \(\frac{2 \mathrm{~S}}{\mathrm{h \rho g}}=\frac{2 \times 7.2 \times 10^{-3}}{6 \times 10^{-2} \times 10^3 \times 9.8}\)
r = \(\frac{14.4}{58.8} \) × 10^-3
r = 0.24 × 10^-3 m
r = 0.24 mm

Question 4.
Find the depression of the meniscus in the capillary tube of diameter 0.4 mm dipped in a beaker containing , mercury. (Density of mercury = 13.6 × 10³ Kg m^-3 and surface tension of mercury = 0.49 Nm^-1 and angle of contact = 135°).
Solution:
D = 4 m.m
r = \(\frac{D}{2}=\frac{4}{2}\) = 2m.m = 2 × 10^-3 m
ρ = 13.6 × 10³ kg/m³
θ = 135°, S = 0.49 Nm
cosθ = cos 135°
= – sin 45° = – \(\frac{1}{\sqrt{2}}\)
S = \(\frac{\mathrm{hr \rho g}}{2 \cos \theta}\)
h = \(\frac{2 \mathrm{~s} \cos \theta}{\mathrm{r \rho g}}\)
= 2 × 0.49 \(\left(\frac{-1}{\sqrt{2}}\right) \frac{1}{2 \times 10^{-3} \times 13.6 \times 10^3 \times 9.8}\)
h = \(\frac{-0.49}{13.6 \times 9.8 \times \sqrt{2}}\)
h = -0.024m.

Question 5.
If the diameter of a soap bubble is 10 mm and its surface tension is 0.04 Nm^-1, find the excess pressure inside the bubble. [Mar. 14]
Solution:
D = 10 m.m
r = \(\frac{D}{2}=\frac{10}{2}\) = 5 m.m
= 5 × 10^-3
S = 0.04 N/m
P_i – P₀ = \(\frac{4 S}{r}=\frac{4 \times 0.04}{5 \times 10^{-3}}\)
= 0.032 × 10³
P_i – P₀ = 32 N/m² (or) Pascal.

Question 6.
If work done by an agent to form a bubble of radius R is W, then how much energy is required to increase its radius to 2R ?
Solution:
R₁ = R, R₂ = 2R
Initial work done (W) = 8πR²S
Final work done (W) = 8π[R₂² – R₁²]S
= 8π [4R² – R²]S
= 3 × 8π R²S
W = 3W.

Question 7.
If two soap bubbles of radii R₁ and R₂ (in vacuum) coalasce under isothermal conditions, what is the radius of the new bubble. Take T as the surface tension of soap solution.
Solution:
R₁, R₂ and R be the radii of first, second and resultant bubble. The soap bubbles coalesce in vacuum, so surface energy do not change
E = E₁ + E₂
8π R²T = 8π R₁²T + 8π R₂² T
R² = R₁² + R₂²
R = \(\sqrt{R_1^2+R_2^2}\)

Additional Problems

Question 1.
Explain why
a) The blood pressure in humans is greater at the feet than at the brain.
b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km.
c) Hydrostatic pressure is a scalar quantities even though pressure is force divided by area.
Answer:
a) The height of the blood column in the human body is more at feet than at the brain. That is why, the blood exerts more pressure at the feet than at the brain (∴ pressure = hρg)

b) We know that the density of air is maximum near the surface of earth and decreases rapidly with height and at a height of about 6 km, it decreases to near by half its value at the sea level. Beyond 6 km height the density of air decreases very slowly with height. Due to this reason, the atmospheric pressure at a height of about 6 km decreases to nearby half of its value at sea level.

c) Since due to applied force on liquid, the pressure is transmitted equally in all directions inside the liquid. That is why there is no fixed direction for the pressure due to liquid. Hence hydrostatic pressure is a scalar quantity.

Question 2.
Explain why
a) The angle of contact of mercurywith glass is obtuse, while that of water with glass is acute.
b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.) [T.S. Mar. 15]
c) Surface tension of a liquid is independent of the area of the surface.
d) Water with detergent disolved in it should have small angles of contact.
e) A drop of liquid under no external forces is always spherical in shape.
Answer:
a) When a small quantity of liquid is poured on solid, three interfaces, namely liquid- air, solid-air and solid-liquid are formed. The surface tensions corresponding to these three interfaces are S_LA, S_SA and S_SL respectively. Let 0 be the angle of contact between the liquid and solid. The molecules in the region, where the three interfaces meet are in equilibrium. It means net force acting on them is zero. For the molecule at 0 to be in equilibrium, we have.

In case of mercury glass, S_SA < S_SL, therefore cosθ is negative or θ > 90° i.e. obtuse. In case of water-glass, S_SA > S_SL, therefore cosθ is positive or θ < 90° i,e. acute.

b) For mercury glass, angle of contact is obtuse. In order to achieve this obtuse value of angle of contact, the mercury tends to form a drop. In case of water glass, the angle of contact is acute. To achieve this acute value of angle of contact, the water tends to spread.

c) Surface tension of liquid is the force acting per unit length on a line drawn tangentially to the liquid surface at rest. Since this force is independent of the area of liquid surface, therefore surface tension is also independent of the area of the liquid surface.

d) We know that the cloth has narrow spaces in the form of capillaries. The rise of liquid in a capillary tude is directly proportional to cos0. if 0 is small cos0 will be large. Due to which capillary rise will be more and so the detergent will penetrate more in cloth.

e) In the absence of external forces, the surface of the liquid drop tends to acquire the minimum surface area due to surface tension. Since for a given volume, the surface area of sphere is least, hence the liquid drop takes the spherical shape.

Question 3.
Fill in the blanks using the word (S) from the list appended with each statement ;
a) Surface tension of liquids generally … with temperatures (increases / decreases)
b) Viscosity of gases… with temperature, whereas viscosity of liquids … with temperature (increases / decreases)
c) For solids with elastic modulus of rigidity, the shearing force is proportional to…. while for fluids it is proportional to … (shear strain / rate of shear strain)
d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle)
e) For the model of a plane in a wind tunnel, turbulence occurs at a… speed for turbulence for an actual plane (greater / smaller)
Answer:
a) Decreases
b) increases; decreases
c) Shear strain; rate of shear strain.
d) Conservation of mass; Bernoullis principle.
e) Greater.

Question 4.
Explain why
a) To keep a piece of paper horizontal, you should blow over, not under, it.
b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
e) A spinning cricket ball in air does not follow a parabolic trajectory.
Answer:
a) When we blow over the paper, the velocity of air blow increases and hence pressure of air on it decreases (according to Bernoullis theorem), whereas pressure of air below the paper is atmosphere. Hence the paper stays horizontal.

b) By doing so the area of the outlet of water jet is reduced, so velocity of water increases according to equation of continuity av = a constant.

c) When a fluid is flowing out of a small hole in a vessel it acquires a large velocity and hence possesses large momentum. Since no external force is acting on the system, a backward velocity must be attained by the vessel (according to law of conservation of momentum). As a result of it, impulse (backward thrust) is experienced by the vessel.

d) There, size of the needle controls velocity of flow and thumb pressure controls pressure. According to Bernoulli’s theorem, P + ρgh + \(\frac{1}{2}\) ρV² = a constant shows that P occurs with power one and V occurs with power two, hence the velocity has more influence. That is why the needle has a better control over flow.

e) If the ball is spinning well as moving linearly, the streamlines at the top of ball due to two types of motion are opposed to each other and those below are in the same direction. As a result of it, the velocity of air flow is greater below than above the ball. Now, according to Bernoullis principle, the pressure on the upper side of the ball becomes more than the pressure on the lower side of ball. Due to it, a resultant force F acts upon the ball at right angle to linear motion in the downward, direction, resulting the ball to move along a curved path.

Question 5.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor ?
Answer:
Here, m = 50kg; r =D/2 = 1/2 cm = \(\frac{1}{200}\) m
Pressure = \(\frac{\text { Force }}{\text { area }}=\frac{m g}{\pi r^2}=\frac{50 \times 9.8}{(22 / 7)(1 / 200)^2}\)
= 6.24 × 10⁶ Nm^-2.

Question 6.
Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmosphere pressure.
Answer:
P = 0.76 × (13.6 × 10³) × 9.8
= h × 984 × 9.8 or
h = \(\frac{0.76 \times 13.6 \times 10^3 \times 9.8}{984 \times 9.8}\)
= 10.5 m.

Question 7.
A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean ? Take the depth of the ocean to be roughly 3 km and ignore ocean currents.
Answer:
Here, maximum stress = 10⁹Pa,
h = 3km = 3 × 10³m;
ρ(water) = 10³kg/m³ and g = 9.8 m/s²
The structure will be suitable for putting upon top of an oil well provided the pressure exerted by sea water is less than the maximum stress it can bear.
Pressure due to sea water, P = hρg
= 3 × 10³ × 10³ × 9.8
= 2.94 × 10⁷ Pa
Since the pressure of sea water is less than the maximum stress of 10⁹ Pa, the structure will be suitable for putting upon top of the oil well.

Question 8.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear ?
Answer:
The maximum force, which the bigger piston can bear,
F = 3000 kgf = 3000 × 9.8 N
∴ Area of piston, A = 425 cm²
= 425 × 10^-4 m²
∴ maximum pressure on the bigger piston.
P = \(\frac{F}{A}=\frac{3000 \times 9.8}{425 \times 10^{-4}}\) = 6.92 × 10⁵ Pa
Since the liquid transmits pressure equally, therefore the maximum pressure the smaller piston can bear is 6.92 × 10⁵ Pa.

Question 9.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit ?
Answer:
For water column in one arm of U-tube,
h₁ = 10.0 cm; ρ₁ (density) = 1g cm^-3
For spirit column in other arm of U-tube, h₂ = 12.5 cm; ρ₁ = ?
As the mercury column in the two arms of U-tube are in level, therefore pressure exerted by each is equal. Hence h₁ρ₁g = h₂ρ₂g or
P₂ = \(\frac{h_1 \rho_1}{h_2}=\frac{10 \times 1}{12.5}\) = 0.8 g cm^-3
Therefore, relative density of spirit = ρ₂/ρ₁
= \(\frac{0.8}{1}\) = 0.8

Question 10.
In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific, gravity of mercury = 13.6)
Answer:
On pouring 15.0 cm of water and spirit each into the respective arms of U-tube, the mercury level will rise in the arm containing spirit. Let h be the difference in the levels of mercury in two arms of U-tube and p be the density of mercury.
∴ The pressure exerted by h cm of mercury column = difference in pressure exerted by water and spirit.
∴ h₁ρ₁g = h₂ρ₂g ……………. (1)
Here h = ? ρ =13.6 g cm^-3
ρ₁ =1 cm^-3
ρ₂ = 0.8 g cm^-3
h₁ = 15 + 10 = 25 cm
h₂ = 15 + 12.5 = 27.5 cm
Putting values in (i) we get h × 13.6 × g
= 25 × 1 × g – 27.5 × 0.8 × g = 3g
or h = 3/13.6 = 0.22cm

Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain.
Answer:
No, Bernoulli’s theorem is used only for stream – line flow.

Question 12.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain.
Answer:
No, it does not matter if one uses gauge instead of absolute pressures in applying Bernoulli’s’ equation, provided the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.

Question 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10^-3 kg s^-1, what is the pressure difference between the two ends of the tube ? (Density of glycerine = 1.3 × 10³ kg m^-3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].
Answer:
Here, l = 1.5 m, r = 1.0 cm = 10^-2 m, ρ = 1; 1.3 × 10^-3 kg/m3; η = 0.83 Nsm^-2.
Mass of glycerine flouring per sec, M = 4 × 10^-3 kg/s
Volume of glycerine flouring per sec, V = \(\frac{M}{\rho}\)
V = \(\frac{4 \times 10^{-3}}{1.3 \times 10^3} \mathrm{~m}^3 \mathrm{~s}^{-1}\) m³s^-1
If ρ is the difference of pressure between two ends of the tube,then using poisevilles formula we have
V = \(\frac{\pi \rho r^4}{8 \eta l}\) or P = \(\frac{V \times 8 \eta l}{\pi r^4}\)
P = \(\left(\frac{4 \times 10^{-3}}{1.3 \times 10^3}\right) \times \frac{8 \times 0.83 \times 1.5}{3.142 \times\left(10^{-2}\right)^4}\)
= 975.37 Pa

Question 14.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^-1 and 63 m s^-1 respectively. What is the lift on the wing if its area is 2.5 m² ? Take the density of air to be 1.3 kg m^-3.
Answer:
Let V₁, V₂ be the speeds on the upper and lower surfaces of the wing of aeroplane and P₁ and P₂ be the pressures on upper and lower surfaces of the wing respectively.
Then V₁ = 70ms^-1; V₂ = 63ms^-1,
P = 1.3kg m^-3.
This difference of pressure provides the lift to the aeroplane.
\(\frac{P_1}{\rho}\) + gh + \(\frac{1}{2} V_1^2\) = \(\frac{P_2}{\rho}\) + gh + \(\frac{1}{2} V_2^2\)
∴ \(\frac{P_1}{\rho}-\frac{P_2}{\rho}=\frac{1}{2}\left(V_2^2-V_1^2\right)\)
or P₁ – P₂ = \(\frac{1}{2} \rho\left(V_2^2-V_1^2\right)\)
= \(\frac{1}{2}\) × 1.3 [(70)² – (63)²]
= 605.15 Pa
This difference of pressure provides the lift to the aeroplane.
So, lift on the aeroplane = Pressure difference × area of wings
= 605.15 × 2.5
= 1512.875 N
= 1.51 × 10³N.

Question 15.
Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?

Answer:
Figure a is incorrect. According to equation of continuity i.e. av = a constant, where area of cross – section of tube is less, the velocity of liquid flow is more than the other portion of tube. According to Bernoulli’s theorem,
P + \(\frac{1}{2}\) ρv² = a constant i.e. where V is more P is less and avice versa.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm² one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min^-1, what is the speed of ejection of the liquid through the holes ?
Answer:
Area of cross – section of tube, a₁ = 8.0 cm² = 8 × 10^-4 m²
No. of holes = 40, Diameter of each hole, D = 1 mm = 10^-3 m
∴ Radius of hole, r = \(\frac{D}{2}=\frac{1}{2}\) × 10^-3 m
= 5 × 10^-4 m
Area of cross – section of each hole = πr²
= π(5 × 10^-4)²m²
Total area of cross – section of 40 holes,
a₂ = 40 × π (5 × 10^-4)²m²
Speed of liquid inside the tube,
V₁ = 1.5m/min
= \(\frac{1.5}{60}\) ms^-1
If V₂ is the velocity of ejection of the liquid through the holes, then
a₁V₁ = a₂V₂ or V₂ = \(\frac{a_1 V_1}{a_2}\)
V₂ = \(\frac{\left(8 \times 10^{-4}\right) \times 1.5}{60 \times 40 \times \pi \times\left(5 \times 10^4\right)^2}\)
= 0.637 ms^-1

Question 17.
A U-shaped wire is dipped in a soap solution and removed. The thin shaped film formed between the surfaces and the light slider supports a weight of 1.5 × 10^-2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film ?
Answer:
We know that soap film has two free surfaces, so total length of the film to be supported,
= 2l = 2 × 30 = 60 cm
= 0.6 m
Total force on the slider due to surface tension will be,
F = S × 2l
= S × 0.6 N
In equilibrium position, the force F on slider due to surface tension must balance the weight mg
(1.5 × 10^-2)N i.e. F = 1.5 × 10^-2
= S × 0.6 Or
S = \(\frac{1.5 \times 10^{-2}}{0.6}\)
2.5 × 10^-2Nm^-1

Question 18.
Figure (a) shows a thin liquid film supporting a small weight = 4.5 × 10^-2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c) ? Explain your answer physically.

Answer:
a) Here, length of film supporting the weight
= 40cm = 0.4 m
Total liquid film supported (or force)
= 4.5 × 10^-2 N
film has two free surfaces, ∴ surface tension,
S = \(\frac{4.5 \times 10^{-2}}{2 \times 0.4}\)
S = 5.625 × 10^-2 Nm^-1
Since the liquid is same for all the cases (a), (b), (c) and temperature is also same, therefore surface tension for cases (b) and (c) will also be the same = 5.625 × 10^-2. In figure (b) and (c) the length of the film supporting the weight is also the same as that of(a), hence the total weight supported in each case is 4.5 × 10^-2 N.

Question 19.
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20°C) is 4.65 × 10^-1 Nm^-1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.
Answer:
Here, r = 3.0 mm = 3 × 10^-3 m;
S = 4.65 × 10^-1 Nm^-1;
P = 1.01 × 10⁵ Pa
Excess of pressure inside the drop of mercury is given by
P = \(\frac{2 S}{r}=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}\)
= 310 Pa
Total pressure inside the drop = P + ρ
= 1.01 × 10⁵ + 310
= 1.01 31 × 10⁵ Pa

Question 20.
What is the excess pressure inside a bubble of soap solution of radius 5.0 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 × 10^-2 Nm^-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 10⁵ Pa).
Answer:
Here, S = 2.5 × 10^-2 Nm^-1, r = 5.00 mm = 5 × 10^-3 m.
Density of soap solution, ρ = 1.2 × 10³ kg m^-3
Excess pressure inside the soap bubble,
P = \(\frac{4 s}{r}=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\) = 20 Pa
Excess pressure inside the air bubble, P’ = \(\frac{2S}{r}\)
= \(\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\) = 10 Pa
Total pressure inside the air bubble at depth h in soap solution – ρ’ + atmospheric pressure + hρg
= 10 + 1.01 × 10⁵ + 0.4 × 1.2 × 10³ × 9.8
= 1.06 × 10³ Pa

Question 21.
A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm². The tank is filled with water in one compartment and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.
Answer:
For compartment containing water,
h₁ = 4m, ρ₁ = 10³ kg m^-3
The pressure exerted by water at the door provided at bottom

P₁ = h₁ρ₁g
= 4 × 10³ × 9.8
= 3.92 × 10⁴ Pa
For compartment containing acid,
h₂ = 4m,
ρ₂ = 1.7 × 1.0³ kg/m³
The pressure exerted by acid at the door provided at bottom.
P₂ = h₂ρ₂g
= 4 × 1.7 × 10³ × 9.8
= 6.664 × 10⁴ Pa
∴ Difference of pressure = P₂P₁
= 6.664 × 10⁴ – 3.92 × 10⁴
= 2.774 × 10⁴Pa
Given, area of door, A = 20cm²
= 20 × 10^-4m²
Force on the door = difference in pressure × area
= (P₂ – P₁) × A
= (2.774 × 10⁴) × (20 × 10⁴)
= 54.88N
≈ 55N
To keep the door closed, the force equal to 55 N should be applied horizontally on the door from compartment containing water to that containing acid.

Question 22.
A manometer reads the pressure of a gas in an enclosure as shown in Fig. (a) when a pump removes some of the gas, the manometer reads as in Fig. (b) The liquid used in the manometers is mercury and the atmo-spheric pressure is 76 cm of mercury.
a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and b) in units of cm of mercury.
b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury, are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

Answer:
a) Here, atomospheric presure, p = 76 cm of mercury
In Fig (a) Pressure head
h = + 20 cm
∴ Absolute pressure = p + h
= 76 + 20
= 96 cm of mercury
Gauge pressure = h = 20 cm of mercury.
In fig (b) pressure head, h = -18 cm
Absolute pressure = p + h
= 76+ (-18)
= 58 cm of mercury
Gauge pressure = h = -18cm of mercury.

b) Here 13.6 cm of water added in right limb is equiralent to
\(\frac{13.6}{13.6}\) = 1 cm of mercury column.
i.e., h¹ = 1 cm of mercury column.
Now pressure at A, PA = P + h¹ = 76 + 1
= 77 cm
Let the difference in mercury levels in the two lumbs be h₁, then pressure at B,
P_B = 58 + h₁ or
As P_A = P_B = 77 = 58 + h₁
h₁ = 77 – 58 = 19 cm of mercury column.

Question 23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular comon height. Is the force exerted by the water on the base of the vessel the same in the two cases ? If so, why do the vessels filled with water to that same height give different readings on a weighing scale ?
Answer:

Since the pressure depends upon the height of water column and the height of the water column in the two vessels of different shapes is the same.hence there will be same pressure due to water on the base of each vessel. As the base area of each vessel is same, hence there will be equal force acting on the two base areas due to water pressure. The water exerts force on the walls of the vessel also. In case, the walls of the vessel are not perpendicular to base, the force exerted by water on the walls has a net non-zero vertical component which is more in first vessel than that of second vessel. That is why, the two vessel:; filled with water to same vertical height show different readings on a weighing machine.

Question 24.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein ? Use the density of whole blood from Table 1.
Answer:
h = \(\frac{p}{p g}=\frac{200}{1.06 \times 10^3}\) × 9.8 = 0.1925 m
The blood may just enter the vein if the height at which the blood container be kept must be slightly greater than 0.1925 m i.e. 0.2 m.

Question 25.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.
a) What is the largest average velocity of blood flow in an artery of diameter 2 × 10^-3 m if the flow must remain laminar ?
b) Do the dissipative forces become more important as the fluid velocity increases ? Discuss qualitatively.
Answer:
a) If dissipative forces are present, then some forces in liquid flow due to pressure different is spent against dissipative forces. Due to which the pressure drop becomes large.

b) The dissipative forces become more important with increasing flow velocity, because of tubulence.

Question 26.
a) What is the largest average velocity of blood flow in an artery of radius 2 × 10^-3m if the flow must remain lanimar ?
Answer:
Here, r = 2 × 10^-3m ;
D = 2r = 2 × 2 × 10^-3 = 4 × 10^-3m
η = 2.084 × 10^-3 Pa s;
p = 1.06 × 10³ kgm³
For flow tobe laminar, N_r = 2000
a) Now, V_c = \(\frac{N_r \eta}{\rho D}\)
= \(\frac{2000 \times\left(2.084 \times 10^{-3}\right)}{\left(1.06 \times 10^3\right) \times\left(4 \times 10^{-3}\right)}\) = 0.98m/s.

b) What is the corresponding flow rate ? (Take viscosity of blood to be 2.084 × 10^-3 Pa s).
Answer:
Volume flowing per second = πr²Vc
= \(\frac{22}{7}\) × (2 × 10^-3)² × 0.98
= 1.23 × 10^-5 m³s^-1

Question 27.
A plane is in level flight at constant speed and each of its two wings has an area of 25m². If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m^-3)
Answer:
Here, V₁ = 180 km/h = 50m/s, V₂ = 234 km/ h = 65 m/s;
A = 2 × 25 = 50m²; P = 1kg/m³
P₁ – P₂ = \(\frac{1}{2}\) p (V₂² – V₁²)
= \(\frac{1}{2}\) × 1 × (65² – 50²)
Upward force = (P₁ – P₂) A = \(\frac{1}{2}\) × (65² – 50²) × 50N
As the plane is in level flight, so
mg = (P₁ – P₂)A
or m = \(\frac{\left(P_1-P_2\right) A}{g}=\frac{1 \times\left(65^2-50^2\right) \times 50}{2 \times 9.8}\)
= 4.4 × 10³N

Question 28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10^-5 m and density 1.2 × 10³ kg m^-3. Take the viscosity of air at the temperature of the experiment to be 1.8 × 10⁵ Pa s. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air.
Answer:
Here, r = 2.0 × 10^-5m; ρ = 1.2 × 10³ kg m^-3;
η = 1.8 × 10⁵ Ns m^-2
P₀ = 0, V = ?, F = ?
Terminal velocity V = \(\frac{2 r^2\left(\rho-\rho_0\right) g}{9 \eta}\)
= \(\frac{2 \times\left(2.0 \times 10^{-5}\right)^2 \times\left(1.2 \times 10^3-0\right) \times 9.8}{9 \times 1.8 \times 10^{-5}}\)
= 5.8 × 10^-2ms^-1 = 5.8 cms^-1
Viscous force on the drop, F = 6πηrv
= 6 × \(\frac{22}{7}\) × (1.8 × 10^-5) × (2.0 × 10^-5) × (5.8 × 10^-2)
= 3.93 × 10^-10N.

Question 29.
Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.456 N m^-1. Density of mercury = 1.36 × 10³ kg m^-3.
Answer:
Here, θ = 140°, r = 1 × 10^-3 m;
S = 0.465 Nm^-1, ρ = 13.6 × 10³ kg, h = ?]
Cos = 140° = – cos40° = -0.7660
Now h = \(\frac{2 S \cos \theta}{r \rho g}\)
= \(\frac{2 \times 0.465 \times \cos 140^{\circ}}{10^{-3} \times 13.6 \times 10^3 \times 9.8}\)
= \(\frac{2 \times 0.465 \times(-0.7660)}{10^{-3} \times 13.6 \times 10^3 \times 9.8}\)
= -5.34 × 10^-3m
= -5.34mm
Negative value of h shows that the mercury levels is depressed in the tube.

Question 30.
Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to forma a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube ? Surface tension of water at the temperature of the experiment is 7.3 × 10^-2 N m^-1. Take the angle of contact to be zero and density of water to be 1.0 × 10³ kg m^-3 (g = 9.8 m^-2).
Answer:
Here, S = 7.3 ^-2 Nm^-1, ρ = 1.0 × 10³ kg m^-3; θ = 0°
For narrow tube, 2r₁ = 3.00 mm = 3 × 10^-3 m or r₁ =1.5 × 10^-3 m .
For wider tube, 2r₂ = 6.00 mm = 6 × 10^-3 m or r₂ = 3 × 10^-3 m
let h₁, h₂ be the heights to which water rises in narrow tube and wider tube respectively.
Then, h₁ = \(\frac{2 s \cos \theta}{r_1 \rho g}\) and h₂ = \(\frac{2 s \cos \theta}{r_2 \rho g}\)
∴ Difference in levels of water in two limbs of U-tube is

Question 31.
a) It is known that density of air decreases with height y as
ρ = ρ₀e^-y/y₀
Where p₀ = 1.25 kg m^-3 is the density at sea level, and y₀ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.
Answer:
We know that the rate of decrease of density p of air is directly proportional to height y i.e.
\(\frac{-d \rho}{d y}\) ∝ p or \(\frac{d \rho}{d y}\) = – Kρ
Where K is a constant of proportionality. Here – ve sign shows that ρ decreases as y increases.
\(\frac{d y}{\rho}\) = – Kρ
Integrating it with in the conditions, as y changes fromotoy density changes from ρ₀ to ρ, we have
\(\left.\int_{P_0}^p \frac{d P}{P}=-\int_0^y k d y=\left[\log e^p\right)\right]_{\rho_0}^p=k y\)
\(\frac{\rho}{\rho_0}=e^{-k y}=\rho=\rho_0 e^{-\mathrm{y} / y_0}\)

b) A large He balloon of volume 1425 m³ is used to lift a payload of 400 kg. Assum that the balloon maintains constant radius as it rises. How high does it rise ?
(Take y₀ = 8000 m and P_He = 0.18 kgm^-3)
Answer:
The balloon will rise to aheight, where its density becomes equal to the air at that height.

y = 8000 × 1 = 8000 m
= 8 km.

Textual Examples

Question 1.
The two thigh bones (femurs), each of cross-sectional area 10 cm² mass 40 kg. Estimate the average pressure sustained by the femurs.
Answer:
Total cross-sectional area of the femurs is A = 2 × 10 cm² 20 × 10^-4 m². The force acting on them is F = 40 kg wt = 400 N (taking g 10 ms^-2). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
P_av = \(\frac{F}{A}\) = 2 × 10⁵ N m^-2

Question 2.
What is the pressure on a swimmer 10m below the surface of a lake ?
Answer:
Here
h = 10 m and p = 1000 kg m^-3.
Take g 10 m S^-2
From Equation = P = P_a + ρgh
P = P_a + ρgh
= 1.01 × 10⁵ Pa + 1000 kgm^-3 × 10m S^-2 × 10 m
= 2.10 × 10⁵ Pa
≈ 2 atm
This is a 100% increase in pressure from surface level. At a depth of 1 km the increase in pressure is 100 atm! Submarines are designed to withstand such enormous pressures.

Question 3.
The density of the atmosphere at sea level is 1.29 kg/m³. Assume that it does not change with altitude. Then how high would the atmosphee extend ?
Answer:
We use P – P_a = ρgh
ρgh = 1.29 kg m^-3 × 9.8 ms² × hm
= 1.01 × 10⁵ pa
∴ h = 7989 m ≈ 8 km
In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing press-ure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm.

Question 4.
At a depth of 1000 m in an ocean
(a) What is the absolute pressure ?
b) What is the guage Pressure ?
c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sealevel atmospheric pressure. (The density of sea water is 1.03 × 10³ kg m^-3, g = 10ms^-2.
Answer:
Here h = 1000 m and ρ = 1.03 × 10³ kg m^-3
a) From P₂ – P₁ = ρgh absolute pressure
P = P_a + ρgh
= 1.01 × 10⁵ Pa + 1.03 × 10³ kg m^-3 × 10 m s^-2 × 1000 m
= 104.01 × 10⁵ Pa .
= 104 atm

b) Gauge pressure is P – P_a = ρgh = P_g
P_g = 1.03 × 10³ kg m^-3 × 10 m s² × 1000 m
= 103 × 10⁵ Pa
≈ 103 atm

c) The pressure outside the submarine is P = P_a + ρgh and the pressure inside it is P_a. Hence, the net pressure acting on the window is gauge pressure, P_g = ρgh. Since the area of the window is A = 0.04 m², the force acting on it is
F = P_gA = 103 × 10⁵ Pa × 0.04m²
= 4.12 × 10⁵ N
≈ 103 atm

Question 5.
Two syringes of different cross sections A₁, A₂ and lengths L₁, L₂ (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively.
a) Find the force exerted on the larger piston when a force of ION is applied to the smaller piston, b) if the smaller piston is pushed in through 6.0 cm,’how much does the larger piston move out ?
Answer:
a) Since pressure is transmitted undiminished throughout the fluid.
F₂ = \(\frac{A_2}{A_1} F_1=\frac{\pi\left(3 / 2 \times 10^{-2} m\right)^2}{\pi\left(1 / 2 \times 10^{-2} \mathrm{~m}\right)^2} \times 10 \mathrm{~N}\)
= 90 N

b) Water is considered to be perfectly incompressible. Volume covered by the move-ment of smaller piston inwards is equal to volume moved outwards due to the larger piston.
L₁A₁ = L₂A₂
L₂ = \(\frac{A_1}{A_2} L_1=\frac{\pi\left(1 / 2 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(3 / 2 \times 10^{-2} \mathrm{~m}\right)^2}\)
= × 6 × 10^-2m
≃ 0.67 × 10^-2m = 0.67 cm.
Note, atmospheric pressure is common to both pistons and has been ignored.

Question 6.
In a car lift compressed air exerts a force F₁ on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston or radius 15 cm (Fig). If the mass of the car to be lifted is 1350 kg, calculate F₁. What is the pressure necessary to accomplish this task ? (g = 9.8 ms^-2).

Answer:
Since pressure is transmitted undiminished throughout the fluid.
F₁ = \(\frac{A_1}{A_2} F_2=\frac{\pi\left(5 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(15 \times 10^{-2} \mathrm{~m}\right)^2}\)
= (1350 N × 9.8m s^-2) = 1470 N
= 1.5 × 10³N
The air pressure that will produce this force is
P = \(\frac{F_1}{A_1}=\frac{1.5 \times 10^{-3} \mathrm{~N}}{p\left(5 \times 10^{-2}\right)^2 \mathrm{~m}}\) = 1.9 × 10⁵ Pa
This is almost double the atmospheric pressure.

Question 7.
Blooc velocity : The flow of blood in a large artery of an anesthetised dog is diverted through a Venturi meter. The wider part of the meter has a cross-sectional area equal to that of the artery. A = 8 mm². The harrower part has an area = 4mn². The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery ?
Answer:
We take the density of blood from whole blood to be 1.06 × 10³ kg m³. The ratio of the areas is \(\left(\frac{\mathrm{A}}{\mathrm{a}}\right)\) = 2
Using v₁ = \(\left(\sqrt{\frac{2 \rho_m g h}{\rho}}\right)\left(\left(\frac{A}{2}\right)^2-1\right)^{-1 / 2}\)
we obtain
v₁ = \(\sqrt{\frac{2 \times 24 \mathrm{pA}}{1060 \mathrm{~kg} \mathrm{~m}^{-3} \times\left(2^2-1\right)}=0.125 \mathrm{~m} \mathrm{~s}^{-1}}\)

Question 8.
A fully loaded Boeing aircraft has a mass of 3.3 × 10⁵ kg. Its total wing area is 500 m2. It is in level flight with a speed of 960 km/h. a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ = 1.2 kgm^-3].
Answer:
a) The weight of’the Boeing aircraft is balanced by the upward force due to the pressure difference
∆P × A – 3.3 × 10⁵ kg × 9.8 = mg.
∆P = (3.3 × 10⁵ kg × 9.8 m s^-2) / 500 m²
= 6.6 × 10³ N m^-2

b) We ignore the small height difference between the top and bottom sides in
P₁ + (\(\frac{1}{2}\)) ρv₁² + ρgh₁ = P₂ + (\(\frac{1}{2}\))ρv₂² + ρgh₂
The pressure difference between them is then
∆P = \(\frac{\rho}{2}\left(v_2^2-v_1^2\right)\)
Where v₂ is the speed of air over the upper surface and v₁ is the speed under the bottom surface.
(v₂ – v₁) = \(\frac{2 \Delta P}{\rho\left(v_2+v_1\right)}\)
Taking the average speed v_av (v₂ + v₁)/2 – 960 km/h = 267 m s^-1, we have
(v₂ – v₁)/v_av = \(\frac{\Delta \mathrm{P}}{\rho v_{\mathrm{av}}^2}\) ≈ 0.08
The speed of air above the wing needs to be only 8% higher than that below.

Question 9.
A metal block of area 0.10 m² is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released, the block moves to the right with a constant speed of 0.085 ms^-1. Find the co-efficient of viscosity of the liquid.

Answer:
The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus the shear force F is
F = T = mg = 0.010 kg × 9.8 ms^-2
= 9.8 × 10^-2 N
Shear stress on the fluid = F/A = \(\frac{9.8 \times 10^{-2}}{0.10}\)
Strain rate = \(\frac{v}{l}=\frac{0.085 \mathrm{~ms}^{-1}}{0.3 \times 10^{-3} \mathrm{~m}}\)
η = \(\frac{\text { stress }}{\text { strain rate }}\)
= \(\frac{\left(9.8 \times 10^{-2} \mathrm{~N}\right)\left(0.30 \times 10^{-3} \mathrm{~m}\right)}{\left(0.085 \mathrm{~ms}^{-1}\right)\left(0.10 \mathrm{~m}^2\right)}\)
= 3.45 × 10^-3 Pa s

Question 10.
The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20°C is 6.5 cm s^-1. Compute the viscosity of the oil at 20°C. Density of oil is 1.5 × 10³ kg m^-3, density of copper is 8.9 × 10³ kg m^-3.
Answer:
We have v_t = 6.5 × 10^-2 ms^-1,
a = 2 × 10^-3m,
g = 9.8 ms^-2, ρ = 8.9 × 10³ kg m^-3,
σ = 1.5 × 10³ kg m^-3. From
v_t = \(\frac{2 a^2(\rho-\sigma) g}{(9 \eta)}\)
η = \(\frac{2}{9} \times \frac{\left(2 \times 10^{-3}\right) \mathrm{m}^2 \times 9.8 \mathrm{~ms}^{-2}}{6.5 \times 10^{-2} \mathrm{~ms}^{-1}}\) × 7.4 × 10³ kg m^-3
= 9.9 × 10^-1 kg m^-1s^-1

Question 11.
a) The flow rate of water from a tap of diameter 1.25 cm is L7min. The co-efficient of viscosity of water is 10^-3 Pa s.
b) After sometime the flow rate is increased to 3L / min. Characterise the flow for both the flow rates.
Answer:
a) Let the speed of the flow be v and the diameter of the tap be d = 1.25 cm. The volume of the water flowing out per second is
Q = v × π d²/4
v = 4Q / d² π
We then estimate the Reynolds number to be
R_e = 4vQ / πdη
= 4 × 10³ kg m^-3 × Q / (3.14 × 1.25 × 10^-2 m × 10^-3 Pa S)
= 1.019 × 10⁸ m^-3 SQ
Since initially (a)
Q = 0.48 L/min = 8cm³/s
= 8 × 10^-6 m³ s^-1, we obtain,
R_e = 815
Since this is below 1000, the flow is steady.
After some time

b) When Q = 3L/ min = 50 cm³, s = 5 × 10^-5 m³ s^-1 we obtain,
R_e = 5095
The flow will be turbulent. You may carry out an experiment in your washbasin to determine the ransition from laminar to turbulent flow.

Question 12.
The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water ? The surface tension of water at temperature of the experiments is 7.30 × 10^-2 Nm¹. 1 atmospheric pressure = 1.01 × 10⁵ Pa, density of water = 1000 kg/m³, g = 9.8 × ms^-2. Also calculate the excess pressure.
Answer:
The excess pressure in a bubble of gas in a liquid is given by 2S / r, where S is the surface tension of the liquid gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is 4S / r.) The radius of the bubble is r. Now the pressure outside the bubble, within water, P₀ equals atmospheric pressure plus the pressure due to 8.00 cm of water column. That is
P₀ = (1.01 × 10⁵ Pa + 0.08 m × 1000 kg m^-3 × 9.80 m s^-2)
= 1.01784 × 10⁵ Pa. .
Therefore, the pressure inside the bubble is
P₁ = P₀ + 2S / r (as r = 1 mm)
= 1.01784 × 10⁵ Pa + (2 × 7.3 × 10^-2 Pa m/10^-3 m)
= (1.01784 + 0.00146) × 10⁵ Pa
= 1.02 × 10⁵ Pa
where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical ! (The answer has been rounded off to three significant figures). The excess pressure in the bubble is 146 Pa from (1.0178 + 0.00146) × 10⁵ Pa.

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids

Very Short Answer Questions

Question 1.
State Hooke’s law of elasticity.
Answer:
“With in the elastic limit stress directly proportional to the strain”.
Stress ∝ strain
Stress = k × strain
k = \(\frac{\text { Stress }}{\text { Strain }}\)
Where k is modulus of elasticity.

Question 2.
State the units and dimensions of stress.
Answer:

1. Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{A}\)
   S.l units → N/m² (or) Pascal
2. Dimensional formula
   Stress = \(\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}\) = [ML^-1T^-2].

Question 3.
State the units and dimensions of modulus of elasticity.
Answer:
Modulus of elasticity (k) = \(\frac{\text { Stress }}{\text { Strain }}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2]

Question 4.
State the units and dimensions of Young’s modulus.
Answer:
Young’s modules (y) = \(\frac{\text { LongitudinalStress }}{\text { Longitudinal Strain }}=\frac{\frac{F}{A}}{\frac{e}{L}}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2]

Question 5.
State the units and dimensions of modulus of rigidity.
Answer:
Modulus of rigidity (G) = \(\frac{F}{A \theta}=\frac{\text { Shearing Stress }}{\text { Shearing Strain }}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2].

Question 6.
State the units and dimensions of Bulk modulus.
Answer:
Bulk modulus (B) = \(\frac{\text { Bulk Stress }}{\text { Bulk Strain }}=\frac{-P V}{\Delta V}\)
Units → N/m² (or) Pascal
Dimensional formula → [ML^-1T^-2].

Question 7.
State the examples of nearly perfect elastic and plastic bodies.
Answer:

• Nearly perfect elastic bodies are quartz fibre.
• Nearly perfect plastic bodies are dough and day.

Short Answer Questions

Question 1.
Define Hooke’s Law of elasticity, proportionality, permanent set and breaking stress.
Answer:
Hooke’s law : “With in the elastic limit stress is directly proportional to the strain”.
Stress ∝ strain
Stress = k × strain
Where k is modulus of elasticity.
Proportionality limit: The maximum stress developed in a body till it obeys Hookes law is called proportionality limit.
Permanent Set : Permanent deformation produced when a body is stretched beyond its elastic limit.
Breaking stress : The maximum stress a body can bear before it breaks.

Question 2.
Define modulus of elasticity, stress, strain and Poisson’s ratio.
Answer:
Modulus of elasticity : It is the ratio stress applied on a body to the strain produced in the body.
k = \(\frac{\text { Stress }}{\text { Strain }}\)
S.I unit → N/m² (or) Pascal
Stress : When a body is subjected to an external force, the force per unit area is called stress.
Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{A}\)
S.I unit → N/m² (or) Pascal
Strain : When deforming forces act on a body, the fractional deformation produced in the body. It has no units

Poisson’s ratio (σ) : The ratio between lateral strain to longitudinal strain of a body is called poisson’s ratio.
σ = \(\frac{\text { Lateral Strain }}{\text { Longitudinal Strain }}=\frac{\frac{-\Delta \mathrm{r}}{\mathrm{r}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}\)

Question 3.
Define Young’s modulus. Bulk modulus and Shear modulus.
Answer:
Young’s modulus (y) : With in the elastic limit, the ratio of longitudinal stress to longitudinal strain is called young’s modulus.
y = \(\frac{\text { Longitudinal Stress }}{\text { Longitudinal Strain }}=\frac{\frac{F}{A}}{\frac{e}{L}}\)
y = \(\frac{\mathrm{FL}}{\mathrm{A} \cdot \mathrm{e}}\)
S.I unit → N/m² (or) Pascal

Bulk modulus (B) : With in the elastic limit, it is defined as the ratio of Bulk stress to Bulk strain
B = \(\frac{\text { Bulk Stress }}{\text { Bulk Strain }}\)
B = \(\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{-\Delta \mathrm{V}}{\mathrm{V}}}=\frac{-\mathrm{PV}}{\Delta \mathrm{V}}\) (∵ -ve sign indicates volume decreases)
S.I unit → N/m² (or) Pascal

Rigidity modulus (G) : With in the elastic limit, it is defined as the ratio of shearing stress to shearing strain.
G = \(\frac{\text { Shearing Stress }}{\text { Shearing Strain }}\)
G = \(\frac{\frac{F}{A}}{\theta}=\frac{F}{A \theta}\)
S.I unit → N/m² (or) Pascal

Question 4.
Define stress and explain the types of stress. [T.S. Mar. 16]
Answer:
Stress : The restoring force per unit area is called stress
∴ Stress = \(\frac{\text { Restoring Force }}{\text { Area }}=\frac{F}{A}\)
Stress is classified into three types.

1. Longitudinal stress
2. Volume (or) Bulk stress
3. Tangential (or) shearing stress

1. Longitudinal stress (or) Linear stress : When a normal stress changes the length of a body, then it is called longitudinal stress.
Longitudinal stress = \(\frac{F}{A}\)

2. Volume (or) Bulk stress : When a normal stresschanges the volume of a body, then it is called volume stress.
Volume stress = \(\frac{\text { Force }}{\text { Area }}\) = pressure.

3. Tangential (or) shearing stress : When the stress is tangential to the surface due to the application of forces parallel to the surface, then the stress is called tangential stress.
Tangential stress = \(\frac{F}{A}\).

Question 5.
Define strain and explain the types of strain.
Answer:
Strain : It is the ratio of change in dimension to its original dimension.
Strain = \(\frac{\text { Changes in dimension }}{\text { Original dimension }}\)
Strain is of three types.
1. Longitudinal strain : It is the ratio of change in length to its original length.
Longitudinal strain = \(\frac{\text { Changes in length }}{\text { Original length }}=\frac{e}{L}\)

2. Shearing strain (or) Tangential strain : When simultaneous compression and extension in mutually perpendicular direction takes place in a body, the change of shape it under goes, is called shearing strain.

Shearing strain (θ) = \(\frac{1}{L}\)

3. Bulk (or) volume strain : It is the ratio of change in volume to its original volume is called bulk strain. It is called Bulk (or) volume strain.
Bulk strain = \(\frac{\text { Change in Volume }}{\text { Original Volume }}=\frac{\Delta V}{V}\)

Question 6.
Define strain energy and derive the equation for the same. [Mar. 14]
Answer:
The potential energy stored in a body when stretched is called strain energy.
Let us consider a wire of length L and cross – sectional area A. Let x be the change in length of the wire by the application of stretching force F.

Strain energy per unit volume = \(\frac{1}{2} \times \frac{F}{A} \cdot \frac{x}{L}\)
= \(\frac{1}{2}\) × stress × strain.

Question 7.
Explain why steel is preferred to copper, brass, aluminium in heavy-duty machines and in structural designs.
Answer:
The elastic behavior of materials plays an important role in everyday life. Designing of buildings, the structural design of the columns, beams and supports require knowledge of strength of material used. The elasticity of the material is due to stress developed with in the body, when extenal force acts on it. A material is of more elastic nature if it develops more stress (or) restoring force. Steel develops more stress than copper, brass, aluminium for same strain. So steel is more elastic.
y = \(\frac{\text { Stress }}{\text { Strain }}\)

Question 8.
Describe the behaviour of a wire under gradually increasing load. [A.P. – Mar. ’18, ’16, ’15; TS – Mar. ’18, ’15, ’13]
Answer:
When the load is increased in steps, a graph is drawn between stress on y-axis and corresponding strain on x-axis.

1. Proportionality limit : In the linear position OA, stress is proportional to strain, i.e. Hookes law is obeyed by the wire upto point A. The graph is a straight lint. When ever the stretching force at A is removed, the wire regains its original length.
A is called proportionality limit.

2. Elastic limit : In the graph B is the elastic limit.
Through the wire doesnot obey Hooke’s law at B. The wire regains its original length after removing the stretching force at B. upto point B the wire is under elastic behaviour.

3. Permanent set (or) yield point: In the graph c is the yield point. If the stretching force at c is removed, the wire doesnot regain its original length and the length of the wire changes permanently. In this position the wire flows like a viscous liquid. After the point c, the wire is under plastic behavior, c is called permanent set (or) yield point.

4. Breaking point: When the stress increased, the wire becomes thinner and thinner. When the stress increases to a certain limit the wire breaks. The stress at which the wire breaks is called breaking stress and the point D is called breaking point.

5. Elastic fatigue : The state of temperary loss of elastic nature of a body due to continuous strain is called elastic fatigue. When a body is subjected to continuous strain with in the elastic limit, it appears to have lost Hastic property temporarily to some extent and becomes weak.

Question 9.
Two identical solid balls, one of ivory and the other of wet-day are dropped from the same height onto the floor. Which one will rise to greater height after striking the floor and why ?
Answer:
Ivory ball rise to greater height after striking the floor. The ivory ball regain its original shape after striking the floor. The elastic property of ivory ball is more. Where as wet-day ball does not regain its original shape after striking the floor.

So wet-day ball acts like plastic body.

Question 10.
While constructing buildings and bridges a pillar with distributed ends is preferred to a pillar with rounded ends. Why ?
Answer:
Use of pillars (or) columns is also very common in buildings and bridges. A pillar with rounded ends supports less load than that with a distributed shape at the ends. The precise design of a bridge (or) a building has to take into account the conditions under which it will function, the cost and long period, reliability of usable materials.

Question 11.
Explain why the maximum height of a mountain on earth is approximately 10 km ?
Answer:
The maximum height of a mountain on earth is 10 km, can also be provided by considering the elastic properties of rocks. A mountain base is not under uniform compression and this provides some shearing stress to rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow.

At the bottom of a mountain of height h, the force per unit area due to the weight of the mountain is hρg. Where ρ is the density of the mountain. The material at the bottom experiences this force in the vertical direction and the sides of the mountain are free.
There is a shear component, approximately hρg itself.
Elastic limit for 3 typical rock is 30 × 10⁷ N/m²
hρg = 30 × 10⁷ (ρ = 3 × 10³ kg/m³)
h = \(\frac{30 \times 10^7}{3 \times 10^3 \times 10}\)
h = 10 km.

Question 12.
Explain the concept of Elastic Potential Energy in a stretched wire and hence obtain the expression for it.
Answer:
“When a wire is put under a tensile stress, work is done against the inter-atomic forces. The work is stored in the wire in the form of elastic potential energy”.

Expression for elastic potential energy : Consider a wire of length L and area of cross section A is subjected to a deforming force F along the length of the wire. Let the length of the wire is elongated by l.
Young’s modulus (y) = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
F = \(\frac{\mathrm{yAl}}{\mathrm{L}}\) ……………. (1)
Work done due to further elongation of small length dl
Work done (dw) = F × dl = (\(\frac{\mathrm{yAl}}{\mathrm{L}}\))dl ……………… (2)
Total work done in increasing the length of the wire from L to (L + l)
w = \(\int_0^1 \frac{\mathrm{yAl} }{\mathrm{L}} \mathrm{dl}=\frac{\mathrm{yA}}{2} \times \frac{l^2}{\mathrm{~L}}\)
w = \(\frac{1}{2} \times \mathrm{y} \times\left(\frac{l}{\mathrm{~L}}\right)^2 \times \mathrm{Al}\)
= \(\frac{1}{2}\) y × stress² × volume of the wire
w = \(\frac{1}{2}\) × stress × strain × volume of the wire.
This work is stored in the wire in elastic potential energy (u).

Long Answer Question

Question 1.
Define Hooke’s law of elasticity and describe an experiment to determine the Young’s modulus of the material of a wire.
Answer:
Hooke’s law : With in the elastic limit, stress is directly proportional to the strain.
Stress ∝ strain
Stress = k × strain
Where k is modulus of elasticity.
Determination of young’s modulus of the material of a wire:

Young’s Modulus of the Material of a wire

1. It consists of two long straight wires of same length and same area of cross-section suspended side by side from a rigid .support.
2. The wire A (reference wire) carries a metre scale M and a pan to place a weight.
3. The wire B (experimental wire) carries a pan in which known weights can be placed.
4. A vernier scale v is attached to a pointer at the bottom of the experimental wire B and the main scale M is fixed to the wire A.
5. The weights placed in the pan, the elongation of the wire is measured by the vernier arrangement.
6. The reference wire is used to compensate for any change in length that may occur due to change in room temperature.
7. Both the reference and experimental wires are given an initial small load to keep the wires straight and the vernier reading is noted.
8. Now the experimental wire is gradually loaded with more weights, the vernier reading is noted again.
9. The difference between two vernier readings gives the elongation produced in the wire.
10. Let r and L be the radius and initial length of the experimental wire. Let M be the mass that produced an elongation ∆L in the wire.
    Young’s modulus of the material of the experimental wire is given by
    y = \(\frac{\text { Longitudinal Stress }}{\text { Longitudinal Strain }}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\frac{\Delta \mathrm{L}}{\mathrm{L}}}\)
    y = \(\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}\)
    y = \(\frac{\mathrm{MgL}}{\pi r^2 \times \Delta \mathrm{L}}\)
    From above equation young’s modulus of the material of the wire is determined.

Problems

Question 1.
A copper wire of 1mm diameter is stretched by applying a force of 10 N. Find the stress in the wire.
Solution:
D = 1 m.m = 10^-3m, r = \(\frac{D}{2}\) = 0.5 × 10^-3 m.
F = 10 N
Stress = \(\frac{F}{A}=\frac{F}{\pi r^2}\)
= \(\frac{10}{3.14 \times\left(0.5 \times 10^{-3}\right)^2}\)
= 1.273 × 10⁷ N/m².

Question 2.
A tungsten wire of length 20 cm is stretched by 0.1 cm. Find the strain on the wire.
Solution:
L = 20 × 10^-2 m, ∆L = 0.1 × 10^-2 m
Strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{0.1 \times 10^{-2}}{20 \times 10^{-2}}\) = 0.005.

Question 3.
If an iron wire is stretched by 1 %, what is the strain on the Wire ?
Solution:
Strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}\) = 1 %
Strain = \(\frac{1}{100}\) = 0.01

Question 4.
A brass wire of diameter 1mm and length 2 m is streched by applying a force of 20N. If the increase in length is 0.51 mm. find
(i) the stress,
(ii) the strain and
(iii) the Young’s modulus of the wire.
Solution:
D = 1 m.m, r = \(\frac{D}{2}\) = 0.5 × 10^-3 m
L = 2 m, F = 20 N, ∆L = 0.51 m.m = 0.51 × 10^-3 m

= 9.984 × 10¹⁰ N/m²

Question 5.
A copper wire and an aluminium wire have lengths in the ratio 3 : 2, diameters in the ratio 2 : 3 and forces applied in the ratio 4: 5. Find the ratio of increase in length of the two wires. (Y_cu = 1.1 × 10¹¹ Nm^-2, Y_Al = 0.7 × 10¹¹ Nm^-2).
Solution:

= \(\frac{4}{5} \times \frac{3}{2} \times\left(\frac{0.7 \times 10^{11}}{1.1 \times 10^{11}}\right) \times\left(\frac{3}{2}\right)^2\)
\(\frac{\Delta \mathrm{L}_1}{\Delta \mathrm{L}_2}=\frac{189}{110}\)

Question 6.
A brass wire of cross-sectional area 2 mm2 is suspended from a rigid support and a body of volume 100 cm3 is attached to its other end. If the decrease in the length of the wire is 0.11 mm, when the body is completely immersed in water, find the natural length of the wire.
(Y_brass = 0.91 × 10¹¹ Nm^-2, ρ_water = 10³ kg m^-3).
Solution:
A = πr² = 2 × 10^-6 m², V = 100 × 10^-6 = 10^-4 m³
∆L = 0.11 × 10^-3 m, y_Brass = 0.91 × 10¹¹ N/m², ρ = 10³ kg/m³
y = \(\frac{M g L}{A \times \Delta L}=\frac{v \rho g L}{A \times \Delta L}\)
L = \(\frac{\mathrm{yA} \Delta \mathrm{L}}{\mathrm{v \rho g}}=\frac{0.91 \times 10^{11} \times 2 \times 10^{-6} \times 0.11 \times 10^{-3}}{10^{-4} \times 10^3 \times 9.8}\)
L = 2.04 m.

Question 7.
There are two wires of same material. Their radii and lengths are both in the ratio 1:2. If the extensions produced are equal, what is the ratio of loads ?
Solution:

Question 8.
Two wires of different material have same lengths and areas of cross¬section. What is the ratio of their increase in length when forces applied are the same ?
(Y₁ = 0.9 × 10¹¹ Nm^-2, Y₂ = 3.6 × 10¹¹ Nm^-2)
Solution:
y₁ = 0.9 × 10¹¹ Nm^-2
y₂ = 3.6 × 10¹¹ Nm^-2
y = \(\frac{F L}{A \times \Delta L}\)
∆L ∝ \(\frac{1}{y}\)
\(\frac{(\Delta L)_1}{(\Delta L)_2}=\frac{y_2}{y_1}=\frac{3.6 \times 10^{11}}{0.9 \times 10^{11}}=\frac{4}{1}\)

Question 9.
A metal wire of length 2.5 m and area of cross-section 1.5 × 10^-6 m² is stretched through 2 mm. if its Young’s modulus is 1.25 × 10¹¹ N.m², find the tension in the wire.
Solution:
L = 2.5 m, A = 1.5 × 10^-6 m²
∆L = 2 × 10^-9 m
y = 1.25 × 10¹¹ N.m²
y = \(\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}\)
F = \(\frac{\mathrm{yA} \Delta \mathrm{L}}{\mathrm{L}}\)
= \(\frac{1.25 \times 10^{11} \times 1.5 \times 10^{-6} \times 2 \times 10^{-3}}{2.5}\)
F = 150 N

Question 10.
An aluminium wire and a steel wire of the same length and cross-section are joined end-to-end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 1.35 mm, find the ratio of the
(i) stress in the two wires and
(ii) strain in the two wires.
(Y_Al = 0.7 × 10¹¹ N.m², Y_Steel = 2 × 10¹¹ Nm²).
Solution:
Y_Al = 0.7 × 10¹¹ N.m², Y_Steel = 2 × 10¹¹ Nm²
∆L₁ + ∆L₂ = 1.35 mm ……………… (1)

Question 11.
A 2 cm cube of some substance has its upper face displaced by 0.15 cm due to a tangential force of 0.3 N while keeping the lower face fixed, Calculate the rigidity modulus of the substance.
Solution:
L = 2 × 10^-2 m, A = L² = 4 × 10^-4 m²
∆x = 0.15 × 10^-2 m
F = 0.3 N
G = \(\frac{\frac{F}{A}}{\frac{\Delta x}{L}}=\frac{F L}{A \Delta x}\) (∵ θ = \(\frac{\Delta x}{L}\))
G = \(\frac{0.3 \times 2 \times 10^{-2}}{4 \times 10^{-4} \times 0.15^6 \times 10^{-2}}\)
G = 10⁴ N/m²

Question 12.
A spherical ball of volume 1000 cm³ is subjected to a pressure of 10 Atmosphere. The change in volume is 10^-2 cm³. If the ball is made of iron, find its bulk modulus.
(1 atmosphere = 1 × 10⁵ Nm^-2).
Solution:
v = 1000 cm³ = 1000 × 10^-6 = 10^-3 m³
p = 1 atm = 1 × 10⁵ = 10⁵ N/m²
-∆v = 10^-2 cm³ = 10^-2 × 10^-6 = 10^-8 m³
Bulk modulus (B) = \(\frac{-p v}{\Delta v}\)
= \(\frac{10^5 \times 10^{-3}}{10^{-8}}\)
B = 10¹⁰ N/m².

Question 13.
A copper cube of side of length 1 cm is subjected to a pressure of 100 atmosphere. Find the change in its volume if the bulk modulus of copper is 1.4 × 10¹¹ Nm^-2. (1 atm = 1 × 10⁵ Nm^-2).
Solution:
l = 1 cm = 10^-2 m
V = Volume of the cube = l³ = 1cm³
= 10^-6 m³
P = 100 atm = 100 × 10⁵ = 10⁷ N/m²
B = 1.4 × 10¹¹ N/m²
B = \(\frac{-P V}{\Delta V}\)
-∆V = \(\frac{P V}{B}=\frac{10^7 \times 10^{-6}}{1.4 \times 10^{11}}\)
-∆V = 0.7143 × 10^-10 m³.

Question 14.
Determine the pressure required to reduce the given volume of water by 2%. Bulk modulus of water is 2.2 × 10⁹ Nm^-2.
Solution:
\(\frac{-\Delta V}{V}\) = 2 % = \(\frac{2}{100}\)
B = 2.2 × 10⁹ Nm²
B = \(\frac{-P V}{\Delta V}\)
P = -B × \(\frac{\Delta V}{V}\)
= 2.2 × 10⁹ × \(\frac{2}{100}\)
P = 4.4 × 10⁷ N/m².

Question 15.
A steel wire of length 20 cm is stretched to increase its length by 0.2 cm. Find the lateral strain in the wire if the Poisson’s ratio for steel is 0.19.
Solution:
L = 20 cm = 20 × 10^-2 m
∆L = 0.2 × 10^-2 m
σ = 0.19
σ = \(\frac{\text { Lateral strain }}{\text { Longitudinal strain }\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)}\)
Lateral strain = σ × \(\frac{\Delta L}{L}\)
= \(\frac{0.19 \times 0.2 \times 10^{-2}}{20 \times 10^{-2}}\)
= 0.0019

Additional Problems

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10^-5 m²,
Solution:
Given, for steel wire, a₁ = 3.0 × 10^-5 m², l₁ = 4.7 m, ∆l₁ = ∆l, F₁ = F
For copper wire, a₂ = 4.0 × 10^-5 m², l₂ = 3.5 m, ∆l₂ = ∆l, F₂ = F .
Let y₁, y₂ be the young modulus of steel wire and copper wire respectively.
∴ y₁ = \(\frac{F_1}{a_1} \times \frac{l_1}{\Delta l_2}=\frac{F}{3.0 \times 10^{-5}} \times \frac{4.7}{\Delta l}\) ………….. (i)
and y₂ = \(\frac{F_2 \times l_2}{a_2 \times \Delta l_2}=\frac{F \times 3.2}{4 \times 10^{-5} \times \Delta l}\)
\(\frac{\mathrm{y}_1}{\mathrm{y}_2}=\frac{4.7 \times 4 \times 10^{-5}}{3.5 \times 3.0 \times 10^{-5}}\) = 1.8
Here y₁ : y₁ = 1.8 : 1.

Question 2.
Figure shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material ?

Solution:
a) From graph, for stress = 150 × 10⁶ Nm^-2 the corresponding strain = 0.002
young’s modulus y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{150 \times 10^6}{0.002}\)
= 7.5 × 10¹⁰ Nm^-2

b) Approximate yeild strength will be equal to the maximum stress it can substain with out crossing the elastic limit. Therefore, the approximate yeild strength
= 300 × 10⁶ Nm^-2
= 3 × 10⁸ Nm^-2

Question 3.
The stress-strain graphs for materials A and B are shown in Fig.

The graphs are drawn to the same scale.
a) Which of the materials has the greater Young’s modulus ?
b) Which of the two is the stronger material ?
Solution:
a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence young’s modulus (= stress/ strain) is greater for A than that of B.

b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.

Question 4.
Read the following two statements below carefully and state, with reasons, if it is true or false.
a) The Young’s modulus of rubber is greater than that of steel;
b) The stretching of a coil is determined by its shear modulus.
Solution:
a) False, because for a given stress there is more strain in rubber than steel and modulus of elasticity is inversly proportional to strain.

b) True because the strecting of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elasticity is involved.

Question 5.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of steel wire is 1.5 m and that of brass Wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Solution:
For steel wire, total force on steel wire,
F₁ = 4 + 6 = 10 kg, f = 10 × 9.8 N
l₁ = 1.5 m, ∆l₁ = ?, 2r₁ = 0.25cm or r₁ =(0.25/2)cm = 0.125 × 10^-2 m
y₁ = 2.0 × 10¹¹ pa
For brass wire, F₂ = 6.0 kg, f = 6 × 9.8 N
2r₂ = 0.25 cm or r₂ = (0.25/2) cm = 0.125 × 10^-2 m,
y₂ = 0.91 × 10¹¹ pa, l₂ = 1.0 m, ∆l₂ = ?

Question 6.
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a verticle wall. A mass of 100 kdis then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Solution:
A = 0.10 × 0.10 = 10^-2 m², F = mg = 100 × 10 N
Shearing strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{\left(\frac{\mathrm{F}}{\mathrm{A}}\right)}{\eta}\)
or ∆L = \(\frac{F L}{A \eta}\)
= \(\frac{(100 \times 10) \times(0.10)}{10^{-2} \times\left(25 \times 10^9\right)}\) = 4 × 10^-7 m.

Question 7.
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the corn pressional strain of each column.
Solution:
Load on each column, F = \(\frac{50,000}{4}\) kgwt
= \(\frac{50,000 \times 9.8}{4}\) N
A = π(r₂² – r₁²) = \(\frac{22}{7}\)(0.60)² – (0.30)²]
= \(\frac{22}{7}\) 0.27 m²
Compression strain = \(\frac{\frac{F}{A}}{y}=\frac{F}{A y}\)
= \(\frac{50,000 \times 9.8}{4 \times\left(\frac{22}{7} \times 0.27\right) \times 2.0 \times 10^{11}}\)
= 7.21 × 10^-7.

Question 8.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Solution:
Here, A = 15.2 × 19.2 × 10^-6 m², F = 44,
500 N, η = 42 × 10⁹ Nm^-2
Strain = \(\frac{\text { Stress }}{\text { Modulus of elasticity }}\)
= \(\frac{\frac{F}{A}}{\eta}=\frac{F}{A \eta}=\frac{44500}{\left(15.2 \times 19.2 \times 10^{-6}\right) \times 42 \times 10^9}\)
= 3.65 × 10^-3.

Question 9.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress Is not to exceed 10⁸ N m^-2, what is the maximum load the cable can support ?
Solution:
Maximum load macimum stress × area of cross-section
= 10⁸πr²
= 10⁸ × \(\frac{22}{7}\) × (1.5 × 10^-2)²
= 7.07 × 104 N.

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
Solution:
As each wire has same tension F, so each wire has same extansion due to mass of rigid bar. As each wire is of same length, hence each wire has same strain, if D is the diameter of wire, then
y = \(\frac{4 \mathrm{~F} / \pi \mathrm{D}^2}{\text { Strain }}\) or D² ∝ 1/y
\(\frac{D_{\mathrm{cu}}}{\mathrm{D}_{\mathrm{iron}}}=\sqrt{\frac{\mathrm{y}_{\mathrm{iron}}}{\mathrm{y}_{\mathrm{cu}}}}\)
= \(\sqrt{\frac{190 \times 10^9}{110 \times 10^9}}=\sqrt{\frac{19}{11}}\) = 1.31.

Question 11.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
Solution:
Here, m = 14.5 kg, l = r = 1m, v = 2 rps, A = 0.065 × 10^-4 m²
Total pulling force on mass, when it is at the lowest position of the vertical circle is
F = mg + mrω²
= mg + mr4πv²
= 14.5 × 9.8 + 14.5 × 1 × 4 × (\(\frac{22}{7}\))² × 2²
= 142.1 +2291.6
= 2433.7 N
y = \(\frac{F}{A} \times \frac{l}{\Delta l}\) or ∆l = \(\frac{F l}{A y}\)
= \(\frac{2433.7 \times 1}{\left(0.065 \times 10^{-4}\right) \times\left(2 \times 10^{11}\right)}\)
= 1.87 × 10^-3 m
= 1.87 mm.

Question 12.
Compute the bulk modulus of water from the following data : Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Solution:
Here, V = 100 lit = 100 × 10^-3 m³, P = 100 atm = 100 × 1.013 × 10⁵ Pa
V + ∆V = 100.5 litre or ∆V= (V + ∆V) – V
= 100.5 – 100
= 0.5 litre = 0.5 × 10^-3 m³
We known that bulk modulus, B = \(\frac{\mathrm{PV}}{\Delta \mathrm{V}}\)
= \(\frac{100 \times 1.013 \times 10^5 \times 100 \times 10^{-3}}{0.5 \times 10^{-3}}\)
= 2.026 × 10⁹ Pa
Bulk modulus of air = 1.0 × 10⁵ Pa
\(\frac{\text { Bulk modulus of water }}{\text { Bulk modulus of air }}=\frac{2.026 \times 10^9}{1.0 \times 10^5}\)
= 2.026 × 10¹⁴.
It is so because gases are much more compressible than those of liquids. The molecules in gases are very poorly coupled to their neighbours as compared to those of gases.

Question 13.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^-3 ?
Solution:
Here, P = 80.0 atm = 80.0 × 1.013 × 10⁵ pa,
compressibility = \(\left(\frac{1}{B}\right)\) = 45.8 × 10^-11 pa^-1
Density of water at surface,
ρ = 1.03 × 10³ kg m^-3
Let p be the density of water at the given depth, if v and v’ are volumes of certain mass M of ocean water at surface and at a given

Putting this value in (i) we get
1 – \(\frac{1.03 \times 10^3}{\rho^{\prime}}\) = 3.712 × 10^-3 or
ρ’ = \(\frac{1.03 \times 10^3}{1-3.712 \times 10^{-3}}\) = 1.034 × 10³ kg m^-3.

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Solution:
Here, P = 10 atm = 10 × 1.013 × 10⁵ pa,
B = 37 × 10⁹ Nm^-2
Volumetric strain = \(\frac{\Delta V}{V}=\frac{P}{B}\)
= \(\frac{10 \times 1.013 \times 10^5}{37 \times 10^9}\) = 2.74 × 10^-5
∴ Fractional change in volume = \(\frac{\Delta V}{V}\)
= 2.74 × 10^-5

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa.
Solution:
Here, L = 10 cm = 0.10m; P = 7 × 10⁶ pa B = 140 Gpa = 140 × 10⁹ pa
As B = \(\frac{\mathrm{PV}}{\Delta \mathrm{V}}=\frac{\mathrm{Pl}^3}{\Delta \mathrm{V}}\) or ∆V = \(\frac{\mathrm{Pl}^3}{\mathrm{~B}}\)
= \(\frac{\left(7 \times 10^6\right) \times(0.10)^3}{140 \times 10^9}\) = 5 × 10^-8 m³
= 5 × 10^-2 mm³

Question 16.
How much should be pressure on a litre of water be changed to compress it by 0.10% ?
Solution:
Here, V = 1 litre = 10^-3m³;
∆V/V = 0,10/100 = 10^-3
B = \(\frac{P V}{\Delta V}\) or P = B \(\frac{\Delta V}{V}\)
= (2.2 × 10⁹) × 10^-3 = 2.2 × 10⁶pa

Question 17.
Anvils made of single crystals of diamond, with the shape as shown in Fig. are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil ?

Solution:
Here, D = 0.5 mm = 0.5 × 10^-3m = 5 × 10^-4m
F = 50,000 N = 5 × 10⁴N
Pressure at the tip of anvil.
P = \(\frac{F}{\pi D^2 / 4}=\frac{4 F}{\pi D^2}\)
P = \(\frac{4 \times\left(5 \times 10^4\right)}{(22 / 7) \times\left(5 \times 10^{-4}\right)^2}\) = 2.5 × 10¹¹pa.

Question 18.
A rod of length 1.05 m having negligible mass is supported at its end by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm², respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Solution:
For Steel wire A, l₁ = l, A₁ = 1 mm²
Y₁ = 2 × 10¹¹Nm^-2
For aluminium wire B, l₂ = l;
A₂ = 2mm²; y₂ = 7 × 10¹⁰ Nm^-2
a) Let mass m be suspended from the rod at the distance × from the end where wires A is connected. Let F₁ and F₂ be the tension in two wires and there is equal stress in two wires, then
\(\frac{F_1}{A_1}=\frac{F_2}{A_2} \text { or } \frac{F_1}{F_2}=\frac{A_1}{A_2}=\frac{1}{2}\) …………………. (i)
Taking moment of forces about the point of suspension of mass from the rod, we have
F₁x = F₂ (1.05 – x) or \(\frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{1}{2}\)
or 2.10 – 2x = x or x = 0.70m = 70 cm

b) Let mass m be supended from the rod at distance × from the end where wire A is connected. Let F₁ and F₂ be the tension in the wires and there is equal strain in the two wires i.e.
\(\frac{F_1}{A_1 Y_1}=\frac{F_2}{A_2 Y_2}\) or \(\frac{F_1}{F_2}=\frac{A_1}{A_2} \frac{Y_1}{Y_2}\)
= \(\frac{1}{2} \times \frac{2 \times 10^{11}}{7 \times 10^{10}}=\frac{10}{7}\)
As the rod is stationary, so F₁x = F₂(1.05 – x)
or \(\frac{1.05-x}{x}=\frac{F_1}{F_2}=\frac{10}{7}\)
or 10 x = 7.35 – 7x
or x = 0.4324 m
x = 43.2cm.

Question 19.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10^-2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the- mid-point.
Solution:
Refer the figure, let x be the depression at the mid point i.e CD = x
In fig. AC = CB = Z = 0.5m
m = 100g = 0.100 kg
AD = BD = (l² + x²)^1/2
Increase in length, ∆l = AD + DB – AB
= 2 AD – AB
= 2 (l² + x²)^1/2 – 2l
= 2l(1 + \(\frac{x^2}{l^2}\))^1/2 – 2l
= 2l (1 + \(\frac{x^2}{2 l^2}\)) – 2l = \(\frac{x^2}{l}\)
Strain = \(\frac{\Delta l}{2 l}=\frac{x^2}{2 l^2}\)
If T is the tension in the wire, then 2T cos θ
= mg or T = \(\frac{\mathrm{mg}}{2 \cos \theta}\)

Question 20.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 10⁷ Pa ? Assume that each rivet is to carry one quarter of the load.
Solution:
Here, r = 6/2 = 3mm = 3 × 10^-3 m, max.
stress = 6.9 × 10⁷ Pa
Max . load on a rivet = Max stress × area of cross section
= 6.9 × 10⁷ × (22/7) × (3 × 10^-3)²
∴ Maximum tension
= 4 (6.9 × 10⁷ × \(\frac{22}{7}\) × 9 × 10^-6)
= 7.8 × 10³N.

Question 21.
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom ?
Solution:
Here, P = 1.1 × 10⁸ Pa, V = 0.32 m³,
B = 16 × 10¹¹Pa
∆V = \(\frac{\mathrm{PV}}{\mathrm{B}}\)
= \(\frac{\left(1.1 \times 10^8\right) \times 0.32}{1.6 \times 10^{11}}\)
= 2.2 × 10^-4m³

Textual Examples

Question 1.
A structural steel rod has a radius of 10 mm and a length of 10 m. A 100 KN force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel ¡s 2.0 × 10¹¹ Nm².
Answer:
a) Given Stress = \(\frac{F}{A}=\frac{F}{\pi r^2}\)
= \(\frac{100 \times 10^3 \mathrm{~N}}{3.14 \times\left(10^{-2} \mathrm{~m}\right)^2}\) = 3.18 × 10⁸ Nm^-2

b) The elongation
∆L = \(\frac{(\mathrm{F} / \mathrm{A}) \mathrm{L}}{\mathrm{Y}}\)
= \(\frac{\left(3.18 \times 10^8 \mathrm{Nm}^2\right)(1 \mathrm{~m})}{2 \times 10^{11} \mathrm{Nm}^{-2}}\)
= 1.59 × 10^-3 m
= 1.59 mm

c) The strain is given by
Strain = ∆L/L = (1.59 × 10^-3) km
= 1.59 × 10^-3 = 0.16%

Question 2.
A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.
Answer:
From y = \(\frac{\sigma}{\varepsilon}\)
we have stress = strain × Young’s modulus.
W/A = Y_c × (∆L_c/L_c) = Y_s × (∆L_s/L_s)
where the subscripts c and s refer to copper and stainless steel respectively,
∆L_c/∆L_s = (Y_s/Y_s) () (L_c/L_s)
Given L_c = 2.2 m, L_s = 1.6 m,
Y_c = 1.1 × 10¹¹ N.m^-2 and Y_s = 2.0 × 10¹¹ N.m^-2.
∆L_c/∆L_s = \(\frac{2.0 \times 10^{11}}{1.1 \times 10^{11}}=\frac{2.2}{1.6}\) = 2.5
∆L_c + ∆L_s = 7.0 × 10^-4 m
Solving the above equations,
∆L_c = 5.0 × 10^-4 m and ∆L_s = 2.0 × 10^-4 m.
∴ W = (A × Y_c × ∆L_c)L_c
= π(1.5 × 10^-3)² × \(\left[\frac{\left(5.0 \times 10^{-4} \times 1.1 \times 10^{11}\right)}{2.2}\right]\) = 1.8 × 10² N.

Question 3.
In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in Fig.) The combined mass of all the persons performing the act and the tables, planks etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load.

Answer:
Total mass of all the performers, tables, plaques = 280 kg
Mass of the performer = 60 kg
Mass supported by the legs of the performer at the bottom of the pyramid = 280 – 60
= 220 kg
Weight of this supported mass = 220 kg wt.
= 220 × 9.8 N = 2156 N
Weight supported by each thighbone of the performer = \(\frac{1}{2}\) (2156) N = 1078 N.
The Young’s modulus for bone is Y = 9.4 × 10⁹ Nm^-2 (compressive)
Length of each thighbone L = 0.5 m the radius of thigbone = 2.0 cm
Thus the cross-sectional area of the thighbone
A = π × (2 × 10^-2)² m²
= 1.26 × 10^-3 m²
Using \(\frac{(F \times L)}{(A \times \Delta L)}\) the compression in each thigbone (∆L) can be computed as
∆L = \(\frac{F \times L}{Y \times A}\)
= [(1078 × 0.5)/(9.4 9 × 10⁹ × 1.26 × 10^-3)]
= 4.55 × 10^-5 m or 4.55 × 10^-3 cm.
This is a very small change ! The fractional decrease in the thighbone is ∆L/L = 0.000091 or 0.0091%.

Question 4.
A square slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × 10⁴ N. The lower edge is riveted to the floor. How much will the upper edge be displaced ?
Answer:
The area (A) = 50 cm × 10 cm
= 0.5 m × 0.1 m
= 0.05 m²
Therefore, the stress appIid is
= (9.4 × 10⁴N/0.05 m²)
= 1.80 × 10⁶ N.m²
We know that shearing strain = (∆x/L)
= Stress/G.
Therefore the displacement
∆x = (Stress × L)/G = \(\frac{\left(1.8 \times 10^6 \mathrm{Nm}^{-2} \times 0.5 \mathrm{~m}\right)}{5.6 \times 10^9 \mathrm{Nm}^{-2}}\)
= 1.6 × 10^-4 m = 0.16 mm.

Question 5.
The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression. ∆V/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 Nm^-2. (Take g = 10 ms^-2)
Answer:
The pressure exerted by a 3000 m column of water on the bottom layer
ρ = hρg
= 3000 m × 1000 kg m^-3× 10 ms^-2
= 3 × 10⁷ Nm^-2
Fractional compression ∆V/V, is
∆V/V = stress \(\frac{\left(3 \times 10^7 \mathrm{Nm}^{-2}\right)}{2.2 \times 109 \mathrm{Nm}^{-2}}\)
= 1.36 × 10^-2 or 1.36%

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 9th Lesson Gravitation Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 9th Lesson Gravitation

Very Short Answer Questions

Question 1.
State the unit and dimension of the universal gravitational constant (G).
Answer:
F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}^2}\)
units of G = Nm² Kg^-2
dimensional formula of G = \(\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{[\mathrm{M}][\mathrm{M}]}\) = [M^-1 L³ T^-2]

Question 2.
State the vector form of Newtons’s law of gravitation.
Answer:
Vector form of Newton’s law of gravitation is
F = \(\frac{-G m_1 m_2}{r^3} \hat{r}\) where \(\hat{r}\) is unit vector.

Question 3.
If the gravitational force of Earth on the Moon is F, what is the gravitational force of moon on earth ? Do these forces to attraction-reaction pair ?
Answer:
F. Yes, they form action and reaction pair.

Question 4.
What would be the change in acceleration due to gravity (g) at the surface, if the radius of Earth decreases by 2% keeping the mass of Earth constant ?
Answer:
g₁r₁², l = g₂r₂², r₂ = \(\frac{98}{100}\) r₁
\(\frac{g_2}{g_1}=\frac{r_1^2}{r_2^2}=\frac{r_1^2}{\left(\frac{98}{100}\right) r_1^2}=\frac{100 \times 100}{98 \times 98}\)
\(\frac{g_2}{g_1}\) = 1.04
\(\frac{g_2}{g_1}\) – 1 = 1.04 – 1
\(\frac{g_2-g_1}{g_1}\) = 0.04

Question 5.
As we go from one planet to another, how will
a) the mass and
b) the weight of a body change ?
Answer:
a) The mass remains constant.
b) The weight (w = mg), changes from one planet to another planet.

Question 6.
Keeping the length of a simple pendulum constant, will the time period be the same on all planets ? Support your answer with reason.
Answer:
No, Time period depends on acceleration due to gravity (g).
T = 2π \(\sqrt{\frac{l}{g}}\)
g value varies from planet to planet. So time period changes.

Question 7.
Give the equation for the value of g at a depth ‘d’ from the surface of Earth. What is the value of ‘g’ at the centre of Earth ?
Answer:

1. g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\)) where d = Depth
   R = Radius of the Earth
2. At the centre of the Earth g = 0.

Question 8.
What are the factors that make ‘g’ the least at the equator and maximum at the poles ? > .
Answer:

1. g value is maximum at poles due to
   a) Rotation of the Earth
   b) Earth is flattened at the poles
   c) The equatorial radius is less at the poles.
2. g value minimum at equator due to
   a) Rotation of the earth
   b) Bulging near the equator.

Question 9.
“Hydrogen is in abundance around the sun but not around Earth”. Explain.
Answer:
The escape velocity on the sun is 620 km/s and escape velocity on the Earth is 11.2 km/s. The r.m.s velocities of hydrogen (2 km/s) is less than escape velocity on the Sun. So hydrogen is more abundant around the Sun and less around the Earth.

Question 10.
What is the time period of revolution of a geostationary satellite ? Does it rorate from West to East or from East to West ?
Answer:
Time period of geo-stationary satellite is 24 hours. It can rotate from west to east.

Question 11.
What are polar satellites ?
Answer:
Polar satellites are low altitude satellites (500 to 800 km), but they go around the poles of the earth in a north-south direction. Its time period is around 100 minutes.

Short Answer Questions

Question 1.
State Kepler’s laws of planetary motion.
Answer:
The three laws of Kepler can be stated as follows.

1. Law of orbits : All planets move in elliptical orbits with the sun situated at one of the foci.
   
2. Law of areas : The line that joins any planet to the sun sweeps equal areas in equal intervals of time.
3. Law of periods : The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.
   T² ∝ R³

Question 2.
Derive the relation between acceleration due to gravity (g) at the surface of a planet and Gravitational constant (G).
Answer:
Consider a body of mass m on the surface of the planet. Let R be the radius of the Earth and M be the mass of the Earth.
Force acting on the body due to gravitational pull of the planet is
F = m g → (1)
According to Newton’s gravitational law, Force on the body is F = \(\frac{\mathrm{GMm}}{\mathrm{R}^2}\) → (2)

From eq’s (1) and (2), we have
m g = \(\frac{\mathrm{GMm}}{\mathrm{R}^2}\)
g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
This is the relation between g and G
Mass of the earth (M) = Volume × density of the earth
M = \(\frac{4}{3}\) π R² × ρ
g = \(\frac{4}{3}\) π G R ρ

Question 3.
How does the acceleration due to gravity (g) change for the same values of height(h) and depth (d).
Answer:
a) g_h = g(1 – \(\frac{2 \mathrm{~h}}{\mathrm{R}}\)), g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\))
Same values of height and depth, h = d
g_h = g (1 – \(\frac{2 \mathrm{~d}}{\mathrm{R}}\)) and g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\))
∴ g_d > g_h

b) For large height and large depth
g_h = \(\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}\) and g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\))
If h = d = R
g_h = \(\frac{\mathrm{g}}{\left(1+\frac{\mathrm{R}}{\mathrm{R}}\right)^2}\) and g_d = g(1 – \(\frac{\mathrm{R}}{\mathrm{R}}\)) = 0
∴ g_h > g_d

Question 4.
What is orbital velocity ? Obtain an expression for it. [Mar. 14]
Answer:
Orbital velocity (V₀) : The horizontal velocity required for an object to revolve around a planet in a circular orbit is called orbital velocity.

Expression for orbital velocity :
Consider a body (satellite) of mass m, revolves round the earth in a circular orbit. Let h be the height of the satellite from the surface of the earth. Then (R + h) is the radius of the orbit.
The Gravitational force of attraction of the earth on the body is given by F = \(\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^2}\) ………….. (1)
Where M = Mass of the earth, R = Radius of the earth, G = universal gravitational constant. If V₀ is the orbital velocity of the body.
The centripetal force on the body is given by F = \(\frac{\mathrm{mv}_{\mathrm{o}}^2}{(\mathrm{R}+\mathrm{h})}\) …………… (2)
In order to make the body revolve in the same orbit, its centripetal force must be equal to the gravitational force

Question 5.
What is escape velocity ? Obtain an expression for it.
Answer:
Escape velocity : It is the minimum velocity with which a body should be projected, so that it moves into the space by overcoming the earth’s gravitational field.

Expression for escape velocity :
Consider a body of mass m thrown with a velocity v₂
Then K.E = \(\frac{1}{2}\) m v_e² …………. (1)
The gravitational force of attraction of the earth of mass M and Radius R on a body of mass m at its surface is F = \(\frac{\mathrm{GMm}}{\mathrm{R}^2}\) ……………… (2)
Gravitational P. E. = work done on the body
∴ P. E. = F × R = \(\frac{\mathrm{GMm}}{\mathrm{R}^2}\) × R
P.E. = \(\frac{\mathrm{GMm}}{\mathrm{R}}\) …………….. (3)
A body just escapes when its K. E. = P. E
\(\frac{1}{2}\) m v_e² = \(\frac{\mathrm{GMm}}{\mathrm{R}^2}\)
v_e² = \(\frac{2 \mathrm{GM}}{\mathrm{R}}\) (∵ g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\))
v_e = \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\)
v_e = \(\sqrt{2 g R}\) (gR = \(\frac{\mathrm{GM}}{\mathrm{R}}\))
v_e = \(\sqrt{2} \times \sqrt{g R}\) (∵ v₀ = \(\sqrt{g R}\))
v_e = \(\sqrt{2}\) × v₀
∴ Escape velocity is \(\sqrt{2}\) times the orbital velocity.

Question 6.
What is a geostationary satellite ? State its uses. [T.S. Mar. 18, 15; A.P. Mar. 16]
Answer:
Geo-stationary satellite : If the period of revolution of an artificial satellite is equal to the period of rotation of earth, then such a satellite is called geo-stationary satellite.
Time period of geo-stationary satellite is 24 hours.
Uses :

1. Study the upper layers of atmosphere
2. Forecast the changes in atmosphere
3. Know the shape and size of the earth.
4. Identify the minerals and natural resources present inside and on the surface of the earth.
5. Transmit the T. V. programmes to distant objects
6. Under take space research i.e. to know about the planets, satellites, comets etc.

Question 7.
If two places are at the same height from the mean sea level; One is a mountain and other is in air at which place will ‘g’ be greater ? State the reason for your answer.
Answer:
The acceleration due to gravity on mountain is greater than that of air.
g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) ………….. (1)
Mass (M) = volume × density (ρ)
M = \(\frac{4}{3}\)π R³ × ρ
g = \(\frac{\mathrm{G}}{\mathrm{R}^2}\) × \(\frac{4}{3}\) π R³ ρ
g = – \(\frac{4}{3}\) π R G ρ …………….. (2)
g ∝ ρ
So density is more at mountains. So g is more on mountain.

Question 8.
The weight of an object is more at the poles than at the equator. At which of these can we get more sugar for the same weight ? State the reason for your answer.
Answer:
Weight of the object at poles = m_p g_p (∵ w = mg)
Weight of the object at equator = m_e g_e
Given weight of the object at poles > weight of the object at equator
m_p g_p > m_eg_e
We know that g_p > g_e
Then m_p < m_e
Hence we can get more sugar at equator.

Question 9.
If a nut becomes loose and gets detached form a satellite revolving around the earth, will it fall down to earth or will it revolve around earth ? Give reasons for your answer.
Answer:
When a nut is detached from a satellite revolving around the earth, the nut is also moving with the speed of the satellite as the orbit of a satellite does not depend upon its mass. Hence nut is moving in the same orbit under centripetal force.

Question 10.
An object projected with a velocity greater than or equal to 11.2 kms it will not return to earth. Explain the reason.
Answer:
The escape velocity on the surface of the earth (v_e) = 11.2 km/s. Any object projected with the velocity greater then (or) equal to 11.2 km/s it will not come back. Because it has overcome the earth’s gravitational pull.
So an object never come back to earth.

Long Answer Questions

Question 1.
Define gravitational potential energy and derive an expression for it associated with two particles of masses m₁ and m₂.
Answer:
Gravitational potential energy : Gravitational potential energy of a body at a point in a gravitational field of another body is defined as the amount of work done in brining the given body from infinity to that point without acceleration.

Expression for gravitational potential energy : Consider a gravitational field due to earth of mass M, radius R. The mass of the earth can be supposed to be concentrated at its centre 0. Let us calculate the gravitational the potential energy of the body of mass m placed at point p in the gravitational field, where OP = r and r > R. Let OA = x and AB = dx.
The gravitational force on the body at A will be
F = \(\frac{\mathrm{GMm}}{\mathrm{X}^2}\) ……………… (1)
Small amount of work done in bringing the body without acceleration through a small distance dx is given by
dw = Force × displacement
dw = F × dx
dw = \(\frac{\mathrm{GMm}}{\mathrm{X}^2}\) × dx ……………… (2)
Total work done in bringing the body from infinity to point P is given by

This work done is stored in the body as its gravitational potential energy (u)
∴ Gravitational potential energy (u) = \(\frac{\mathrm{GMm}}{\mathrm{r}}\) ……………….. (4)
Gravitational potential energy associated with two particles of masses m, and m2 separated by a distance r is given by
u = –\(\frac{G m_1 m_2}{r}\) ……………….. (5) (if we choose u = 0 as r → ∞).

Question 2.
Derive an expression for the variation of acceleration due to gravity (a) above and (b) below the surface of the Earth.
Answer:
i) Variation of g with height:
When an object is on the surface of the earth, it will be at a distance r = R radius of the earth, then we have g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
Where G = universal gravitational constant, M = Mass of the earth
When the object is at a height h above the surface of the earth, Then r = R + h

g value decreases with altitude.

ii) Variation of g with depth :
Let us assume that the earth to be a homogeneous uniform sphere of radius R, mass M and of uniform density ρ.
We know that g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) = \(\frac{4}{3}\) π ρ G R ………………… (1)
Consider a body of mass m be placed at a depth d.

The value of g decreases with depth.

Question 3.
State Newton’s Universal Law of gravitation. Explain how the value of the Gravitational constant (G) can be determined by Cavendish method.
Answer:
Newton’s law of gravitation :
“Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversly proportional to the square of the distance between them”
Determination of G value by cavendish method :

1. In 1798 Henry Cavendish determined the value of G experimentally.
2. The bar AB has two small lead spheres attached at its ends.
3. The bar is suspended from a rigid support by a fine wire.
4. Two large lead spheres are brought close to the small ones but on opposite sides as shown in figure.
5. The big spheres attract the nearby small ones by equal and opposite force as shown in figurer.
6. There is no net force on the bar but only a torque which is clearly equal to F times the length of the bar. When F is the force of attraction between a big sphere and its neighbouring small sphere.
7. Due to this torque, the suspended wire gets twisted till such time as the restoring torque of the wire equals the gravitational torque.
   Restoring torque = τ θ ………………… (1)
   Where τ is restoring couple per unit twist 0 is the angle
8. If d is the seperation between big and small balls having masses M and m.
   Gravitational force (F) = \(\frac{\mathrm{GMm}}{\mathrm{d}^2}\) ……………… (2)
   ix) If L is the length of the bar A B, then the torque arising out of F is F multiplied by L. At equilibrium, this is equal to the restoring torque.
   \(\frac{\mathrm{GMm}}{\mathrm{d}^2}\) = τ θ
   observations of θ thus enables one to calculate G.
   The measurement of G = 6.67 × 10^-11 Nm²/ Kg²

Problems

(Gravitational Constant ‘G’ = 6.67 × 10^-11 Nm²/ Kg^-2; Radius of earth ‘R’ = 6400 km; Mass of earth ‘ME’ = 6 × 10²⁴ kg)

Question 1.
Two spherical balls each of mass 1 kg are placed 1 cm apart. Find the gravitational force of attraction between them.
Solution:
m₁ = m₂ = 1 kg, d = 1 cm = 1 × 10^-2 m
F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}^2}\)
F = \(\frac{6.67 \times 10^{-11} \times 1 \times 1}{\left(1 \times 10^{-2}\right)^2}\) = 6.67 × 10^-7N

Question 2.
The mass of a ball is four times the mass of another ball. When these balls are separated by a distance of 10 cm, the force of gravitation between them is 6.67 × 10^-7 N. Find the masses of the two balls.
Solution:
m₁ = m, m₂ = 4m, d = 10 cm = 10 × 10^-2 m,
F = 6.67 × 10^-7 N
G = 6.67 × 10^-11 Nm²/kg ²
F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}^2}\)
6.67 × 10^-7 = \(\frac{6.67 \times 10^{-11} \times \mathrm{m} \times 4 \mathrm{~m}}{\left(10 \times 10^{-2}\right)^2}\)
4 m² = 10²
m² = \(\frac{100}{4}\) = 25
m = 5 kg
∴ m₁ = m = 5 kg
m₂ = 4m = 4 × 5 = 20 kg

Question 3.
Three spherical balls of masses 1 kg, 2kg and 3 kg are placed at the corners of an equilateral triangle of side 1 m. Find the magnitude of gravitational force exerted by the 2 kg and 3kg masses on the 1 kg mass.
Solution:
The force of attraction at 2 kg on the 1 kg particle
F₂ = \(\frac{\mathrm{Gmn}{\mathrm{~d}^2}\) = \(\frac{\mathrm{G} \times 1 \times 2}{1^2}\)
F₂ = 2 G

Question 4.
At a certain height above the earth’s surface, the acceleration due to gravity is 4% of its value at the surface of earth. Determine the height.
Solution:
g_h = 4% of g = \(\frac{4}{100}\)g, R = 6400 km
g_h = \(\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}\)
\(\frac{4 \mathrm{~g}}{100}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}\)
\(\left(1+\frac{h}{R}\right)^2=\frac{100}{4}\) = 25
1 + \(\frac{h}{R}\) = 5
\(\frac{h}{R}\) = 4
h = 4 × R = 4 × 6400 = 25,600 km.

Question 5.
A satellite is orbiting the earth at a height of 1000km. Find its orbital speed.
Solution:
h = 1000 km
Oribital velocity (v₀) = \(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}\)
G = 6.67 × 10^-11 Nm²/kg ², M = 6 × 10²⁴ kg
R + h = 6400 + 1000 = 7400 km
= 7400 × 10³m
v₀ = \(\sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{7400 \times 10^3}}\)
v₀ = \(\sqrt{0.5408 \times 10^{10}}\) = 73.54 × 10³ m/s
v₀ = 7.354km/s

Question 6.
A satellite orbits the earth at a height equal to the radius of earth. Find it’s
(i) orbital speed and
(ii) Period of revolution
Solution:
Height h = R

Question 7.
The gravitational force of attraction between two objects decreases by 36% when the distance between them is increased by 4 m. Find the original distance between them.
Solution:
F₁ = F, F₂ = \(\frac{64}{100}\) F
d₁ = d, d₂ = (d + 4) m

5d = 4d + 16
d = 16 m.

Question 8.
Four identical masses of m are kept at the corners of a square of side a. Find the gravitational force exerted on one of the masses by the other masses.
Solution:

Question 9.
Two spherical balls of 1 kg and 4kg are separated by a distance of 12 cm. Find the distance of a point from the 1 kg mass at which the gravitational force on any mass becomes zero.
Solution:
m₁ = 1 kg, m₂ = 4kg, r = 12 cm
∴ x = \(\frac{r}{\sqrt{\frac{m_2}{m_1}}+1}\) from m₁
= \(\frac{12}{\sqrt{\frac{4}{1}}+1}=\frac{12}{2+1}=\frac{12}{3}\) = 4 cm
At x = 4 cm the gravitational force is zero.

Question 10.
Three uniform spheres each of mass m and radius R are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any one of the spheres due to the other two.
Solution:

Question 11.
Two satellites are revolving round the earth at different heights. The ratio of their orbital speeds ¡s 2 : 1. If one of them is at a height of 100 km, what is the height of the other satellite ?
Solution:

4R + 400 = R + h₂
h₂ = 3R + 400 = 3 × 6400 + 400
= 19200 + 400
h₂ = 19,600 km.

Question 12.
A satellite is revolving round in a circular orbit with a speed of 8 km s-1 at a height where the value of acceleration due to gravity is 8 m s-2. How high is the satellite from the Earth’s surface ? (Radius of planet = 6000 km)
Solution:
v₀ = 8 km/s = 8000 m/s
g_h = 8 m/s², R = 6000 km = 6000 × 10³ m
∴ v₀ = \(\sqrt{\frac{G M}{R+h}}=\sqrt{g(R+h)}\)
v₀² = g(R + h)
(8000)² = 8(6000 × 10³ + h)
6000 × 10³ + h = 8 × 10⁶
h = (8 – 6) 10⁶
h = 2 × 10⁶m
h = 2000 × 10³ = 2000 km.

Question 13.
(a) Calculate the escape velocity of a body from the Earth’s surface, (b) If. the Earth were made of wood, its mass would be 10% of its current mass. What would be the escape velocity, if the Earth were made of wood ?
Solution:
R = 6400 × 10³m,

Additional Problems

Question 1.
Answer the following :
a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?
b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ?
c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull, (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ?
Solution:
a) We cannot shield a body from the gravitational influence of nearby matter because the gravitational force on the body due to near by matter is independent of the presence of other matter, whereas it is not so in the case of electrical forces it means the gravitational screens are not possible.

b) Yes, if the size of the spaceship orbiting around the earth is large enough, an astronaut inside the spaceship can detect the variation in g.

c) Tidal effect depends inversly on the cube of the distance, unlike force which depends inversly on the square of the distance. Since the distance of moon from the ocean water is very small as compared to the distance of sun from the ocean water on earth. Therefore, the tidal effect of moon’s pull is greater than the tidal effect of the sun.

Question 2.
Choose the correct alternative :
a) Acceleration due to gravity increase^ decreases with increasing altitude.
b) Acceleration due to gravity increases/decreases with increas¬ing depth (assume the earth to be a sphere of uniform density).
c) Acceleration due to gravity is independent of mass of the earth/ mass of the body.
d) The formula – G Mm (1/r₂ – 1/r₁) is more/less accurate than the formula mg (r₂ – r₁) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth.
Solution:
a) decreases
b) decreases
c) mass of the body
d) more

Question 3.
Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ?
Solution:
Here, T_e = 1 year; T_p = \(\frac{T_c}{2}=\frac{1}{2}\) year; r_e = 1
A.U.; r_p = ?
Using Kepler’s third law, we have
r_p = r_e\(\left(\frac{T_p}{T_e}\right)^{2 / 3}\) = \(1\left(\frac{1 / 2}{1}\right)^{2 / 3}\)
= 0.63 AU

Question 4.
Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10⁸m. Show that the mass of Jupiter is about-one-thousandth that of the sun.
Solution:
For a satellite of Jupiter, orbital period,
T₁ = 1.769 days = 1.769 × 24 × 60 × 60 s
Radius of the orbit of satellite,
r₁ = 4.22 × 10⁸ m
mass of Jupiter, M₁ is given by M₁
= \(\frac{4 \pi^2 \times\left(4.22 \times 10^8\right)^3}{G \times(1.769 \times 24 \times 60 \times 60)^2}\)
= \(\frac{4 \pi^2 r_1^3}{\mathrm{GT}_1^2}\) ……………. (1)
We know that the orbital period of earth around the sun,
T = 1 year = 365.25 × 24 × 60 × 60 s
Oribital radius, r = 1 A.U = 1.496 × 10¹¹ m

Question 5.
Let us assume that our galaxy consists of 2.5 × 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be 105 ly.
Solution:
Here, r = 50,000 ly =50,000 × 9.46 × 10¹⁵m
= 4.73 × 10²⁰m.
M = 2.5 × 10¹¹ solar, mass = 2.5 × 10¹¹ × (2 × 10³⁰) kg
= 5.0 × 10⁴¹ kg
We know that, M = \(\frac{4 \pi^2 r^3}{\mathrm{GT}^2}\)
or T = \(\left(\frac{4 \pi^2 r^3}{G M}\right)^{1 / 2}\)
= \(\left[\frac{4 \times(22 / 7)^2 \times\left(4.73 \times 10^{20}\right)^3}{\left(6.67 \times 10^{11}\right) \times\left(5.0 \times 10^{41}\right)}\right]^{1 / 2}\)
= 1.12 × 10¹⁶S.

Question 6.
Choose the correct alternative :
a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potentia! energy.
b) The energy required to launch an orbiting .satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.
Solution:
a) Kinetic energy
b) Less.

Question 7.
Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the Ideation from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched ?
Solution:
The escape velocity is independent of mass of the body and the direction of projection it depends upon the gravitational potential at the point from where the body is launched. Since this potential depends slightly on the latitude and height of the point, therefore, the escape velocity depends slightly on these factors.

Question 8.
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit ? Neglect any mass loss of the comet when it comes very close to the Sun.
Solution:
A comet while going on elliptical orbit around the sun has constant angular momentum and total energy at all locations but other quantities vary with locations.

Question 9.
Which of the following symptoms is likely to afflict an astronaut in space
(a) swollen feet,
(b) swollen face,
(c) headache,
(d) orientational problem.
Solution:
a) We know that the legs carry the weight of the body in the normal position due to gravity pull. The astronaut in space is in weightless state. Hence, swollen feet may not affect his working.

b) In the conditions of weightless, the face of the astronaut is expected to get more supply. Due to it, the astronaut may develop swollen face. As eyes, ears, nose, mouth etc. are all embedded in the face, hence, swollen face may affect to great extent the seeing / hearing / eating / smelling capabilities of the astronaut in space.

c) Headache is due to metal strain it will persist whether a person is an astronaut in space or he is on earth it means headache will have the same effect on the astronaut in space as on a person on earth.

d) Space also has orientation. We also have the frames of reference in space. Hence, orientational problem will affect the astronaut in space.

Question 10.
In the following two exercises, choose the correct answer from among the given ones : The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig) (i) a, (ii) b, (iii) c, (iv) 0.

Solution:
We know that the gravitational potential is constant at all points upside a spherical shell. Therefore, the gravitational potential gradient at all points inside the spherical shell is zero [i.e as v is constant, \(\frac{\mathrm{dv}}{\mathrm{dr}}\) = 0].

Since gravitational intensity is equal to negative of the gravitational potential gradient, hence the gravitational intensity is zero at all points inside a hollow spherical shell. This indicates that the gravitational forces acting on a particle at any point inside a spherical shell, will be symmetrically placed. Therefore if we remove the upper hemispherical shell, the net gravitational forces acting on the particle at the centre Q or at some other point P will be acting downwards which will also be the direction of gravitational intensity it is so because, the gravitational intensity at a point is the gravitational force per unit mass at that point. Hence the gravitational intensity at the centre Q will be along c, i.e., option (iii) is correct.

Question 11.
For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g. .
Solution:
As per explanation given in the answer of Q. 10, the direction of gravitational intensity at P will be along e i.e., option (ii) is correct.

Question 12.
A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun = 2 × 10³⁰ kg, mass of the earth 6 × 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius 1.5 × 10¹¹ m).
Solution:
Here M_s = 2 × 10³⁰ kg ; M_c = 6 × 10²⁴ kg ; r = 1.5 × 10¹¹ m .
Let x be the distance of a point from the earth where gravitational forces on the rocket due to sun and earth become equal and opposite. Then distance of rocket from the sun
= (r – x). If m is the mass of rocket then
\(\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{m}}{(\mathrm{r}-\mathrm{x})^2}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{x}^2} \text { or } \frac{(\mathrm{r}-\mathrm{x})^2}{\mathrm{x}^2}=\frac{\mathrm{M}_{\mathrm{s}}}{\mathrm{M}_{\mathrm{e}}}\)

Question 13.
How will you ‘weigh the sun1, that is estimate its mass ? The mean orbital radius of the earth around the sun is 1.5 × 10⁸ km.
Solution:
To estimate the mass of the sun, we require, the time period of revolution T of one of its planets (say the earth). Let M_s, M_e be the masses of sun and earth respectively and r be the mean orbital radius of the earth around the sun. The gravitational force acting on earth due to sum is
F = \(\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{M}_{\mathrm{e}}}{\mathrm{r}^2}\)
Let, the earth be moving in circular orbit around the sun, with a uniform angular velocity ω, the centripetal force acting on earth is.
F¹ = M_erω² = M_er \(\frac{4 \pi^2}{T^2}\)
As this centripetal force is provided by the gravitational pull of sun on earth, So
\(\frac{\mathrm{GM}_{\mathrm{s}} \mathrm{M}_{\mathrm{e}}}{\mathrm{r}^2}=\mathrm{M}_{\mathrm{e}} \mathrm{r} \frac{4 \pi^2}{\mathrm{~T}^2} \text { or } \mathrm{M}_{\mathrm{s}}=\frac{4 \pi^2 \mathrm{r}^3}{G \mathrm{~T}^2}\)
Knowing r and T, mass M_s of the sun can be estimated.
In this Question, we are given, r = 1.5 × 10⁸ km
= 1.5 × 10¹¹ m
T = 365 days = 365 × 24 × 60 × 60 s
∴ M_s = \(\frac{4 \times(22 / 7)^2 \times\left(1.5 \times 10^{11}\right)^3}{\left(6.67 \times 10^{-11}\right) \times(365 \times 24 \times 60 \times 60)^2}\)
= 2 × 10³⁰ kg.

Question 14.
A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 10⁸ km away from the sun ?
Solution:
Here, T_s = 29.5 T_e; R_e = 1.5 × 10⁸ km; R_s =?
Using the relation, \(\frac{\mathrm{T}_{\mathrm{s}}^2}{\mathrm{R}_{\mathrm{s}}^3}=\frac{\mathrm{T}_{\mathrm{e}}^2}{\mathrm{R}_{\mathrm{e}}^3}\)
or R = Re \(\left(\frac{\mathrm{T}_{\mathrm{s}}}{\mathrm{T}_{\mathrm{e}}}\right)^{2 / 3}\)
= 1.5 × 10⁸ \(\left(\frac{29.5 \mathrm{~T}_{\mathrm{e}}}{\mathrm{T}_{\mathrm{e}}}\right)^{2 / 3}\)
= 1.43 × 10⁹ km.

Question 15.
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Solution:
Weight of the body = mg = 63N
At height h, the value of g is given by
g’ = \(\frac{g R^2}{(R+h)^2}=\frac{g R^2}{(R+R / 2)^2}\) = 4/9 g
Gravitational force on body at height h is
F = mg’ = m × \(\frac{4}{9}\) g = \(\frac{4}{9}\) mg
= \(\frac{4}{9}\) × 63 = 28N

Question 16.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ?
Solution:
wt. of body at a depth d = mg¹
= m × g \(\left(1-\frac{d}{R}\right)\)
= 250 \(\left(1-\frac{R / 2}{R}\right)\)
= 125 N

Question 17.
A rocket is fired vertically with a speed of 5 km s^-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^-11 N m² kg^-2.
Solution:
Let the rocket be fired with velocity v from the surface of earth and it reaches a height h from the surface of earth where its velocity becomes zero.
Total energy of rocket at the surface of energy

or h = \(\frac{\mathrm{Rv}^2}{2 \mathrm{gR}-\mathrm{v}^2}\)
= \(\frac{\left(6.4 \times 10^6\right) \times\left(5 \times 10^3\right)^2}{2 \times 9.8 \times\left(6.4 \times 10^6\right)-\left(5 \times 10^3\right)^2}\)
= 1.6 × 10⁶m

Question 18.
The escape speed of a projectile on the earth’s surface is 11.2 km s-1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth ? Ignore the presence of the sun and other planets.
Solution:
Here, v_e = 11.2 kms^-1, velocity of projection of the body v = 3ve. Let m be the mass of the projectile and v₀ be the velocity of the projectile when far away from the earth (i.e) out of gravitational field of earth) then from the law of conservation of energy
\(\frac{1}{2}\) mv₀² = \(\frac{1}{2}\) mv² – \(\frac{1}{2}\) mv_e²
or v₀ = \(\sqrt{v^2-v_e^2}\)
= \(\sqrt{(3 v e)^2-v_e^2}\)
= \(\sqrt{8} v_e=\sqrt{8}\) × 11.2 = 31.68 kms^-1

Question 19.
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence ? Mass of the satellite = 200 kg; mass of the earth = 6.0 × 1024 kg; radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^-11 N m² kg^-2.
Solution:
Total energy of orbiting satellite at a hight h.
= – \(\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}+\frac{1}{2} \mathrm{mv}^2\)
= – \(\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}+\frac{1}{2} m \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}\)
= \(\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}\)
energy expended to rocket the satellite out of the earth’s gravitational field.
= – (total energy of orbiting satellite)
= \(\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}\)
= \(\frac{\left(6.67 \times 10^{-11}\right) \times\left(6 \times 10^{24}\right) \times 200}{2\left(6.4 \times 10^6+4 \times 10^5\right)}\)
= 5.9 × 10⁹J

Question 20.
Two stars each of one solar mass (= 2 × 10^30< kg) are approaching each other for a head on collision. When they are a distance i09 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10^4< km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Solution:
Here, mass of each star, M = 2 × 10^30< kg
initial distance between two stars, r = 10^9<
km = 10^12< m.
initial potential energy of the system = – \(\frac{\text { GMM }}{r}\)
Total K.E. of the stars = \(\frac{1}{2}\) mv^2< + \(\frac{1}{2}\) mv^2<
= Mv^2<
Where v is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres, r^1< = 2R.
∴ Final potential energy of two starts = \(\frac{-\mathrm{GMM}}{2 \mathrm{R}}\)
since gain in K.E. is at the cost of loss in P.E

Question 21.
Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium ? If so, is the equilibrium stable or unstable ?
Solution:
Gravitational field at the mid – point of the line joining the centres of the two spheres.
= \(\frac{\mathrm{GM}}{(r / 2)^2}(-\hat{r})+\frac{\mathrm{GM}}{(r / 2)^2} \hat{r}=0\)
Gravitational potential at the mid point of the list joining the centres of the two spheres is
v = \(\frac{-\mathrm{GM}}{r / 2}+\left(\frac{-\mathrm{GM}}{r / 2}\right)=\frac{-4 \mathrm{GM}}{r}\)
\(\frac{-4 \times 6.67 \times 10^{-11} \times 100}{1.0}\) = -2.7 × 10^-8< J/kg
As the effective force on the body placed at mid-point is zero, so the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its initial position of equilibrium. Hence, the body is in unstable equilibrium.

Question 22.
As you have learnt in the text, a geo-stationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 × 10²⁴ kg, radius = 6400 km.
Solution:
Gravitational potential at height h from the surface of earth is
v = \(\frac{-\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}\)
= \(\frac{-6.67 \times 10^{-11} \times\left(6 \times 10^{24}\right)}{\left(6.4 \times 10^6+36 \times 10^6\right)}\)
= -9.4 × 10⁶ J/kg.

Question 23.
A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity ? (mass of the sun = 2 × 10^30< kg).
Solution:
The object will remain struck to the surface of star due to gravity, if the accerlation due to gravity is more than the centrifugal accerlation due to its rotation.
Accerlation due to gravity, g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
= \(\frac{6.67 \times 10^{-11} \times 2.5 \times 2 \times 10^{30}}{(12000)^2}\)
= 2.3 × 10¹² m/s²
centrifugal accerlation = rw²
= r(2πv)²
= 12000 (2π × 1.5)²
= 1.1 × 10⁶ ms^-2
since g > rω² , therefore the body will remain struck with the surface of star.

Question 24.
A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun = 2 × 10³⁰ kg; mass of mars = 6.4 × 10^23< kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 10^8< km; G = 6.67 × 10^-11< N m² kg ² .
Solution:
Let R, be the radius of the orbit of mars and R be the radius of the mars. M be the mass of the sun and M’ be the mass of mars. If m is the mass of the space ship, then potential energy of space-ship due to gravitational attraction of the sun = \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
potential energy of space – ship due to gravitational attraction of mars = – \(\frac{\mathrm{GM}^1 \mathrm{~m}}{\mathrm{R}^1}\)
since K.E of space ship is zero, therefore total energy of spaceship
= \(\frac{-\mathrm{GMm}}{\mathrm{R}}\) – \(\frac{\mathrm{GM}^1 \mathrm{~m}}{\mathrm{R}^1}\)
= – Gm \(\left(\frac{M}{R}+\frac{M^1}{R^1}\right)\)
∴ energy required to rocket out the spaceship from the solar system = – (total energy of space ship)

Question 25.
A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s^-1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it ? Mass of mars = 6.4 × 10^23< kg; radius of mars = 3395 km; G = 6.67 × 10^-11< N m² kg^-2.
Solution:
Let m = mass of the rocket, M = mass of the mars and
R = radius of mars. Let v be the initial velocity of rocket.
Initial K.E = \(\frac{1}{2}\) mv²; Initial P.E = – \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
Total initial energy = \(\frac{1}{2}\) mv² – \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
since 20% of K.E is lost, only 80% is left behind to reach the height. Therefore
Total energy available = \(\frac{80}{100} \times \frac{1}{2}\) mv²
– \(\frac{-\mathrm{GMm}}{\mathrm{R}}\) = 0.4 mv² – \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
If the rocket reaches the higher point which is at a height h from the surface of Mars, its
K.E. is zero and P.E. = \(\frac{-\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}\)
using principle of conservation of energy, we have
0.4 mv² – \(\frac{\mathrm{GMm}}{\mathrm{R}}=-\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}\)
or \(\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}=\frac{\mathrm{GM}}{\mathrm{R}}\) – 0.4 v²

Textual Examples

Question 1.
Let the speed of the planet at the perihelion P in Fig. be υ_p and the Sun- planet distance SP be r_p. Relate {r_p, υ_p} to the corresponding quantities at the aphelion {r_A, υ_A}. Will the planet take equal times to traverse BAC and CPB ?

(a) An ellipse traced out by a planet around the sun. The colsest point is P and the farthest point is A. P is called the perihelion and A the aphelion. The semimajor axis (a) is half the distance AP
Answer:
The magnitude of the angular momentum at P is L_p = m_p r_p υ_p. Similarly, L_A = m_p r_A υ_A. From angular momentum conservation
m_p r_p υ_p = m_p r_A υ_A
or \(\frac{v_p}{v_A}=\frac{r_A}{r_p}\)

Question 2.
Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. (a) What is the force acting on a mass,2m placed at the centroid O of the triangle ? (b) What is the force if the mass at the vertex A is doubled ?
Take AO = BO = CO = 1 m (see Fig)

Three equal masses are placed at the three vertices of the ∆ABC. A mass 2m is placed at the centroid O.
Answer:
(a) The angle between OC and the positive x- axix is 30° and so is the angle between OB and the negative x-axis. The individual forces a vector notation are

From the principle of superposition and the law of vector addition, the resultant gravitational force F_R on (2m) at O is
F_R = F_OA + F_OB + F_OC
F_R = 2Gm² \(\hat{\mathrm{j}}\) + 2Gm² \(-\hat{\mathrm{i}}\) cos 30° – \(\hat{\mathrm{j}}\) sin 30°) + 2Gm² (\(\hat{\mathrm{i}}\) cos 30° – \(\hat{\mathrm{j}}\) sin 30°) = 0
Alternatively, one expects on the basis of symmetry that the resultant force ought to be zero.

(b) By symmetry the x-component of the force cancels out. The y-component survives.
F_R = 4Gm² \(\hat{\mathrm{j}}\) – 2Gm² \(\hat{\mathrm{j}}\) = 2Gm² \(\hat{\mathrm{j}}\)

Question 3.
Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square.
Answer:
We have four mass pairs at distance l and two diagonal pairs at distance \(\sqrt{2}\)1 Hence,

= \(\frac{2 \mathrm{Gm}}{1}\left(2+\frac{1}{\sqrt{2}}\right)\) = -5.41 \(\frac{\mathrm{Gm}^2}{l}\)
The gravitational potential U(r) at the centre of the square (r = \(\sqrt{2}\) l / 2) is
U(r) = \(-4 \sqrt{2} \frac{\mathrm{Gm}}{\mathrm{l}}\)

Question 4.
Two uniform solid spheres of equal radii R, but mass M and 4 M have a centre to centre separation 6 R, as shown in Fig. The two spheres are held fixed A projeetile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.

Answer:
If ON = r, we have
\(\frac{\mathrm{GMm}}{\mathrm{r}^2}=\frac{4 \mathrm{GMm}}{\left(6 \mathrm{R}-\mathrm{r}^2\right)}\)
(6R – r)² = 4r²
6R – r = ±2r
r = 2R or – 6R.
The neutral point r = -6R does not concern us in this example. Thus ON = r = 2R.
Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface of M is
E_i = \(\frac{1}{2} \mathrm{~m} v^2-\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{4 \mathrm{GMm}}{5 \mathrm{R}}\)
The mechanical energy at N is purely potential.
E_N = \(-\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{4 \mathrm{GMm}}{4 \mathrm{R}}\)
From the principle of conservation of mechanical energy
\(\frac{1}{2} v^2-\frac{G M}{R}-\frac{4 G M}{5 R}=-\frac{G M}{2 R}-\frac{G M}{R}\)
υ² = \(\frac{2 G M}{R}\left(\frac{4}{5}-\frac{1}{2}\right)\)
υ² = \(\left(\frac{3 \mathrm{GM}}{5 R}\right)^{1 / 2}\)

Question 5.
The planet Mars has two moons, phobos and delmos. (i) phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 × 10³ km. Calculate the mass of Mars, (ii) Assume that Earth and Mars move in circular orbits around the sun, with the Martian orbits being 1.52 times the orbital radius of the earth. What is the length of the Martain year in days ?
Answer:
(i) We employ T² = K (R_E + h)³ (where K = 4π² / GM_E) with the Earth’s mass replaced by the Martian mass M_m

(ii) Once again Kepler’s third law comes to our aid,
\(\frac{T_M^2}{T_E^2}=\frac{R_{M S}^3}{R_{E S}^3}\)
Where R_MS is the Mars-Sun distance and R_ES is the Earth-Sun distance.
∴ T_M = (1.52)^3/2 × 365
= 684 days
For example. the ratio of the semi-minor to semi-major axis for our Earth is, b/a = 0.99986.

Question 6.
Weighing the Earth : You are given the following data g = 9.81 ms² R_E = 6.37 x106 m the distance to the moon R = 3.4 × 10⁸ m and the time period of the moons revolution is 27.3 days. Obtain the mass of the Earth M_E in two different ways.
Answer:
(1) From g = \(\frac{F}{m}=\frac{G M_E}{R_E^2}\)
M_E = \(\frac{g R_E^2}{G}\)
= \(\frac{9.81 \times\left(6.37 \times 10^6\right)^2}{6.67 \times 10^{-11}}\)
= 5.97 × 10²⁴kg. (by Method – 1)

(2) The moon is a satellite of the Earth. From the derivation of Kepler’s third law

= 6.02 × 10²⁴kg (by Method – 2)
Both methods yield almost the same answer the difference between them being less than 1%.

Question 7.
Express the constant k T² = K (R_E + h)² where K = 4π²/GM_E of in days and kilometres. Given k = 10^-13 s² m^-3. The moon is at a distance of 3.84 × 10⁵ km from the earth. Obtain its time period of revolution in days.
Answer:
Given
k = 10^-13 s² m^-3 (d = day)
= 10^-13 \(\left[\frac{1}{(24 \times 60 \times 60)^2} d^2\right]\)
\(\left[\frac{1}{(1 / 1000)^3 \mathrm{~km}^3}\right]\) = 1.33 × 10^-14 d² km^-3
Using T² = K (R_E + h)³ (where k = 4π²/ GM_E) and the given value of k the time period of the moon is
T² = (1.33 × 10^-14) (3.84 × 10⁵)³
T = 27.3 d