#
Multiple Choice Questions
One and More than One Correct
1. Assume that a drop of liquid evaporates by decrease in its surface energy. So, that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is $T$, density of liquid is $\rho$ and $L$ is its latent heat of vaporisation. [JEE Main 2013]
a. $\frac{T}{\rho L}$
b. $\frac{L}{T \rho}$
c. $\frac{T}{L \rho}$
d. $\frac{2T}{\rho L}$
Answer: (d)
Solution:
Surface energy of the drop $= 4\pi R^2 T$
Let $\delta m$ be the mass evaporated.
$$\delta m = \delta V \cdot \rho = 4\pi R^2 \delta R \cdot \rho$$Heat required for evaporation $= \delta m \cdot L = 4\pi R^2 \delta R \cdot \rho L$
Equating decrease in surface energy to heat required for evaporation:
$$8\pi R \delta R \cdot T = 4\pi R^2 \delta R \cdot \rho L \implies R = \frac{2T}{\rho L}$$2. The figure shows the variation of specific heat capacity $c$ of a solid as a function of temperature ($T$). The temperature is increased continuously from $0$ to $500\text{ K}$ at a constant rate. Ignoring any volume change, the following statement (s) is (are) correct to a reasonable approximation. [JEE Adv. 2013 P-II]
a. The rate at which heat is absorbed in the range 0-100 K varies linearly with temperature T
b. Heat absorbed in increasing the temperature from 0–100 K is less than the heat required for increasing the temperature from 400–500 K
c. There is no change in the rate of heat absorption in range 400 – 500 K
d. The rate of heat absorption increases in the range 200–300 K
Answer: (a, b, c, d)
Solution:
The graph between $0-100\text{ K}$ appears to be a straight line up to a reasonable approximation. Area under the curve in the temperature range $0-100\text{ K}$ is less than in range $400-500\text{ K}$. The graph of $C$ versus $T$ is constant in the temperature range $400-500\text{ K}$. In the temperature range $200-300\text{ K}$, the specific heat capacity increases with temperature.
3. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the coefficient of linear expansion $\alpha$ of the metal of the pendulum shaft are respectively: [JEE Main 2016]
a. $25^\circ\text{C}, 1.85 \times 10^{-5}\text{ /}^\circ\text{C}$
b. $60^\circ\text{C}, 1.85 \times 10^{-4}\text{ /}^\circ\text{C}$
c. $30^\circ\text{C}, 1.85 \times 10^{-3}\text{ /}^\circ\text{C}$
d. $55^\circ\text{C}, 1.85 \times 10^{-2}\text{ /}^\circ\text{C}$
Answer: (a)
Solution:
$$\Delta t_1 = \frac{1}{2} \alpha (40 - T_0) \times 86400 = -12\text{ s} \quad \text{--(i)}$$ $$\Delta t_2 = \frac{1}{2} \alpha (20 - T_0) \times 86400 = +4\text{ s} \quad \text{--(ii)}$$Dividing (i) by (ii):
$$\frac{40 - T_0}{20 - T_0} = \frac{-12}{4} = -3 \implies 40 - T_0 = -60 + 3T_0 \implies 4T_0 = 100 \implies T_0 = 25^\circ\text{C}$$Substituting $T_0$ in equation (ii):
$$\frac{1}{2} \alpha (20 - 25) \times 86400 = 4 \implies \alpha = 1.85 \times 10^{-5}\text{ /}^\circ\text{C}$$4. Consider an expanding sphere of instantaneous radius $R$ whose total mass remains constant. The expansion is such that the instantaneous density $\rho$ remains uniform throughout the volume. The rate of fractional change in density $\frac{1}{\rho}\frac{d\rho}{dt}$ is constant. The velocity $v$ of any point on the surface of the expanding sphere is proportional to [JEE Adv. 2017 P-II]
a. $R$
b. $\frac{1}{R}$
c. $R^3$
d. $R^{2/3}$
Answer: (a)
Solution:
$$M = \frac{4}{3}\pi R^3 \rho = \text{constant}$$Differentiating with respect to time $t$:
$$\frac{1}{\rho}\frac{d\rho}{dt} + \frac{3}{R}\frac{dR}{dt} = 0 \implies \frac{dR}{dt} = -\frac{R}{3} \left(\frac{1}{\rho}\frac{d\rho}{dt}\right)$$Since $\frac{1}{\rho}\frac{d\rho}{dt}$ is constant, we have:
$$v = \frac{dR}{dt} \propto R$$5. An unknown metal of mass 192 g heated to a temperature of 100ºC was immersed into a brass calorimeter of mass 128 g containing 240 g of water a temperature of 8.4ºC. Calculate the specific heat of the unknown metal if water temperature stabilises at 21.5ºC (Specific heat of brass is $394\text{ J kg}^{-1}\text{ K}^{-1}$) [JEE Main 2019]
a. $1232\text{ J kg}^{-1}\text{ K}^{-1}$
b. $458\text{ J kg}^{-1}\text{ K}^{-1}$
c. $654\text{ J kg}^{-1}\text{ K}^{-1}$
d. $916\text{ J kg}^{-1}\text{ K}^{-1}$
Answer: (d)
Solution:
Heat lost by metal = Heat gained by calorimeter + Heat gained by water
$$0.192 \times S \times (100 - 21.5) = [0.128 \times 394 \times (21.5 - 8.4)] + [0.240 \times 4200 \times (21.5 - 8.4)]$$ $$S = 916\text{ J kg}^{-1}\text{ K}^{-1}$$6. A heat source at $T = 10^3\text{ K}$ is connected to another heat reservoir at $T = 10^2\text{ K}$ by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is $0.1\text{ W K}^{-1}\text{ m}^{-1}$, the energy flux through it in the steady state is: [JEE Main 2019]
a. $90\text{ W m}^{-2}$
b. $200\text{ W m}^{-2}$
c. $65\text{ W m}^{-2}$
d. $120\text{ W m}^{-2}$
Answer: (a)
Solution:
$$\text{Energy flux} = \frac{1}{A}\frac{dQ}{dt} = \frac{k \Delta T}{\ell} = \frac{0.1 \times (10^3 - 10^2)}{1} = \frac{0.1 \times 900}{1} = 90\text{ W m}^{-2}$$7. Ice at –20° C is added to 50 g of water at 40°C. When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2 J/g/°C, Specific heat of Ice = 2.1 J/g/°C, Heat of fusion of water at 0°C = 334 J/g) [JEE Main 2019]
a. 50 g
b. 40 g
c. 60 g
d. 100 g
Answer: (b)
Solution:
Let the amount of ice added be $m\text{ g}$. Since $20\text{ g}$ of ice is unmelted at $0^\circ\text{C}$, the mass of ice that melted is $(m - 20)\text{ g}$.
Heat taken by ice = Heat given by water
$$m \times 2.1 \times [0 - (-20)] + (m - 20) \times 334 = 50 \times 4.2 \times 40$$ $$42\text{m} + 334\text{m} - 6680 = 8400$$ $$376\text{m} = 15080 \implies m \approx 40.1\text{ g} \approx 40\text{ g}$$8. Two rods A and B of identical dimensions are at temperature 30°C. If A is heated up to 180°C and B up to T°C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is: [JEE Main 2019]
a. 270ºC
b. 230ºC
c. 250ºC
d. 200ºC
Answer: (b)
Solution:
$$\ell_A = \ell_B \implies \ell_0 [1 + \alpha_A \Delta T_A] = \ell_0 [1 + \alpha_B \Delta T_B]$$ $$\alpha_A (180 - 30) = \alpha_B (T - 30) \implies \frac{\alpha_A}{\alpha_B} \times 150 = T - 30$$ $$\frac{4}{3} \times 150 = T - 30 \implies 200 = T - 30 \implies T = 230^\circ\text{C}$$9. When 100 g of a liquid A at 100°C is added to 50 g of a liquid B at temperature 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100 g of liquid A at 100°C is added to 50 g of liquid B at 50°C, will be: [JEE Main 2019]
a. 80°C
b. 60°C
c. 70°C
d. 85°C
Answer: (a)
Solution:
Using principle of calorimetry for the first case:
$$100 \times S_A \times (100 - 90) = 50 \times S_B \times (90 - 75)$$ $$1000 S_A = 750 S_B \implies \frac{S_A}{S_B} = \frac{3}{4}$$For the second case, let the final temperature be $T$:
$$100 \times S_A \times (100 - T) = 50 \times S_B \times (T - 50)$$ $$2 \times \frac{3}{4} \times (100 - T) = T - 50 \implies 300 - 3T = 2T - 100 \implies 5T = 400 \implies T = 80^\circ\text{C}$$10. A thermometer graduated according to a linear scale reads a value $X_0$ when in contact with boiling water, and $X_0/3$ when in contact with ice. What is the temperature of an object in 0 ºC, if this thermometer in the contact with the object reads $X_0/2$? [JEE Main 2019]
a. 35
b. 25
c. 60
d. 40
Answer: (b)
Solution:
$$\frac{X - X_{\text{ice}}}{X_{\text{steam}} - X_{\text{ice}}} = \frac{T - 0}{100 - 0}$$ $$\frac{X_0/2 - X_0/3}{X_0 - X_0/3} = \frac{T}{100} \implies \frac{X_0/6}{2X_0/3} = \frac{T}{100} \implies \frac{1}{4} = \frac{T}{100} \implies T = 25^\circ\text{C}$$11. A metal ball of mass 0.1 kg is heated up to 500°C and dropped into a vessel of heat capacity $800\text{ J K}^{-1}$ and containing 0.5 kg water. The initial temperature of water and vessel is 30°C. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, $4200\text{ J kg}^{-1}\text{ K}^{-1}$ and $400\text{ J kg}^{-1}\text{ K}^{-1}$] [JEE Main 2019]
a. 30%
b. 20%
c. 25%
d. 15%
Answer: (b)
Solution:
$$0.1 \times 400 \times (500 - T) = [0.5 \times 4200 + 800] \times (T - 30)$$ $$40(500 - T) = 2900(T - 30)$$ $$20000 - 40T = 2900T - 87000 \implies 2940T = 107000 \implies T \approx 36.39^\circ\text{C}$$ $$\text{Percentage increment} = \frac{36.39 - 30}{30} \times 100 \approx 21.33\% \approx 20\%$$12. A cylinder of radius $R$ is surrounded by a cylindrical shell of inner radius $R$ and outer radius $2R$. The thermal conductivity of the material of the inner cylinder is $K_1$ and that of the outer cylinder is $K_2$. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is: [JEE Main 2019]
a. $K_1 + K_2$
b. $\frac{K_1 + K_2}{2}$
c. $\frac{2K_1 + 3K_2}{5}$
d. $\frac{K_1 + 3K_2}{4}$
Answer: (d)
Solution:
For heat flowing parallel to the length, the cylinders act in parallel:
$$K_{\text{eq}} A_{\text{total}} = K_1 A_1 + K_2 A_2$$ $$A_1 = \pi R^2, \quad A_2 = \pi (2R)^2 - \pi R^2 = 3\pi R^2, \quad A_{\text{total}} = 4\pi R^2$$ $$K_{\text{eq}} (4\pi R^2) = K_1 (\pi R^2) + K_2 (3\pi R^2) \implies K_{\text{eq}} = \frac{K_1 + 3K_2}{4}$$Integer and Subjective
13. In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Then, find the final temperature of the mixture given...
Solution:
Heat lost by steam to convert into water and cool down equals heat required by ice to warm up, melt, and warm up to the final temperature. The final equilibrium temperature is found to be $0^\circ\text{C}$ with partial melting/freezing.
14. A metal rod AB of length $10x$ has its one end A in ice at 0°C and the other end B in water at 100°C. If a point P on the rod is maintained at 400°C, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is $540\text{ cal/g}$ and latent heat of melting of ice is $80\text{ cal/g}$. If the point P is at a distance of $\lambda x$ from the ice end A, find the value of $\lambda$. [JEE 2009 P-II]
Solution:
$$\frac{dm_{\text{vapour}}}{dt} = \frac{dm_{\text{ice}}}{dt}$$ $$\frac{k S (400 - 0)}{L_v (\lambda x)} = \frac{k S (400 - 100)}{L_f (10x - \lambda x)}$$ $$\frac{400}{540 \lambda} = \frac{300}{80 (10 - \lambda)} \implies \lambda = 9$$15. A piece of ice of mass $m$ grams is at $-5^\circ\text{C}$ at atmospheric pressure. It is given $420\text{ J}$ of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that $1\text{ gm}$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is: [JEE 2010 P-I]
Solution:
$$Q = m \cdot S_{\text{ice}} \cdot \Delta T + m_{\text{melted}} \cdot L_f$$ $$420 = m \times 2100 \times [0 - (-5)] \times 10^{-3} + 1 \times 3.36 \times 10^2$$ $$420 = 10.5m + 336 \implies 10.5m = 84 \implies m = 8\text{ g}$$16. A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_i$ (in Kelvin) and the final temperature is $a T_i$, the value of $a$ is: [JEE 2010 P-II]
Solution:
For adiabatic process, $TV^{\gamma - 1} = \text{constant}$. For a diatomic gas, $\gamma = \frac{7}{5}$.
$$T_i V_i^{\frac{7}{5} - 1} = T_f V_f^{\frac{7}{5} - 1} \implies T_i V_i^{2/5} = (a T_i) \left(\frac{V_i}{32}\right)^{2/5}$$ $$1 = a \left(\frac{1}{32}\right)^{2/5} = a \left(\left(\frac{1}{2}\right)^5\right)^{2/5} = a \left(\frac{1}{2}\right)^2 = \frac{a}{4} \implies a = 4$$
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.