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Multiple Choice Questions

One and More than One Correct

1. In a dark room with ambient temperature a black body is kept at a temperature T. Keeping the temperature of the black body constant (at T), sunrays are allowed to fall on the black body through a hole in the roof of the dark room. Assuming that there is no change in the ambient temperature of the room, which of the following statement(s) is/are correct? [JEE 2006]

a. The quantity of radiation absorbed by the black body in unit time will increase

b. Since emissivity = absorptivity, hence the quantity of radiation emitted by black body in unit time will increase.

c. Black body radiates more energy in unit time in the visible spectrum

d. The reflected energy in unit time by the black body remains same

Answer: (a, b, c, d)

Solution:

a. When sunrays fall on the black body, additional radiant energy is incident upon it per unit time, leading to an increase in the total quantity of radiation absorbed by it per unit time.

b. To keep its temperature constant at $T$ despite absorbing additional solar energy, the black body must emit energy at an increased rate. By Kirchhoff's law, emissivity equals absorptivity ($\varepsilon = a$).

c. Sunrays contain high-frequency visible spectrum radiation. Upon absorption and emission, the total power emitted increases, including an increase in energy radiated in the visible spectrum.

d. An ideal black body absorbs all incident radiation and reflects zero energy. Thus, the reflected energy per unit time remains zero (same).

2. A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then in steady state [JEE 2011 P-I]

a. heat flow through A and E slabs are same

b. heat flow through slab E is maximum

c. temperature difference across slab E is smallest

d. heat flow through C = heat flow through B + heat flow through D

Answer: (a, c, d)

Solution:

Slabs A and E are in series with the entire combination, so the total heat current passing through slab A, through the parallel block combination (B, C, D), and through slab E must be equal ($H_A = H_E = H_{\text{total}}$). Thus, heat flow through A and E is the same and maximum.

Treating slabs B, C, and D as a joint parallel block, the equivalent thermal resistance of slab E is smallest because of its large conductivity ($6K$) and short length ($L$), resulting in the smallest temperature difference across slab E ($\Delta T = H \cdot R$).

For parallel slabs B, C, and D, the sum of individual heat currents equals the total current entering them:

$$H_C = H_B + H_D$$

3. Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e., second) plate under steady state condition is: [JEE 2012 P-I]

a. $\left(\frac{65}{2}\right)^{1/4} T$

b. $\left(\frac{97}{4}\right)^{1/4} T$

c. $\left(\frac{97}{2}\right)^{1/4} T$

d. $\left(\frac{97}{4}\right) T$

Answer: (c)

Solution:

Consider unit area of each plate. Let the steady-state temperature of plate II be $T'$.

Plate II emits radiation from both of its surfaces at a total rate of:

$$E_{\text{emitted}} = 2 \sigma (T')^4$$

Amount of radiation incident on and absorbed by plate II from plates I (at $2T$) and III (at $3T$):

$$E_{\text{absorbed}} = \sigma (2T)^4 + \sigma (3T)^4$$

By steady-state condition ($E_{\text{emitted}} = E_{\text{absorbed}}$):

$$2 \sigma (T')^4 = \sigma (2T)^4 + \sigma (3T)^4$$ $$2 (T')^4 = 16 T^4 + 81 T^4 = 97 T^4$$ $$(T')^4 = \frac{97}{2} T^4$$ $$\Rightarrow T' = \left(\frac{97}{2}\right)^{1/4} T$$

4. Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity $\kappa$ and the other $2\kappa$. The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuration II is: [JEE Adv. 2013 P-I]

a. 2.0 s

b. 3.0 s

c. 4.5 s

d. 6.0 s

Answer: (a)

Solution:

For Configuration I (Series connection):

$$R_I = R_1 + R_2 = \frac{L}{\kappa A} + \frac{L}{2\kappa A} = \frac{3L}{2\kappa A}$$

For Configuration II (Parallel connection):

$$\frac{1}{R_{II}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{\kappa A}{L} + \frac{2\kappa A}{L} = \frac{3\kappa A}{L}$$ $$\Rightarrow R_{II} = \frac{L}{3\kappa A}$$

Comparing resistances:

$$\frac{R_{II}}{R_I} = \frac{L / (3\kappa A)}{3L / (2\kappa A)} = \frac{2}{9}$$

The rate of heat transfer is $H = \frac{Q}{t} = \frac{\Delta T}{R} \implies t \propto R$ for the same amount of heat $Q$ and temperature difference $\Delta T$:

$$\frac{t_{II}}{t_I} = \frac{R_{II}}{R_I} \implies t_{II} = \left(\frac{2}{9}\right) \times 9\text{ s} = 2.0\text{ s}$$

5. The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is: [JEE Main 2013]

a. $p_0 v_0$

b. $\left(\frac{13}{2}\right) p_0 v_0$

c. $\left(\frac{11}{2}\right) p_0 v_0$

d. $4 p_0 v_0$

Answer: (b)

Solution:

Heat is extracted from the source during the expansion paths where temperature increases ($\Delta Q > 0$). For an ideal monatomic gas, $C_V = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$.

Summing heat supplied along the constant volume heating step and constant pressure expansion step:

$$Q_{\text{in}} = n C_V \Delta T_1 + n C_p \Delta T_2 = \frac{3}{2} V \Delta P + \frac{5}{2} P \Delta V$$ $$Q_{\text{in}} = \frac{3}{2} (p_0 v_0) + \frac{5}{2} (2 p_0 v_0) = \frac{13}{2} p_0 v_0$$

6. If a piece of metal is heated to temperature $\theta$ and then allowed to cool in a room which is at temperature $\theta_0$, the graph between the temperature T of the metal and time t will be closed to [JEE Main 2013]

Answer: (c)

Solution:

By Newton’s law of cooling, the rate of cooling is proportional to the temperature difference between the body and its surroundings:

$$\frac{dT}{dt} = -k(T - \theta_0)$$

Integrating gives an exponential decay curve of temperature $T$ with time $t$ asymptoting towards $\theta_0$ as $t \to \infty$. Thus, the slope decreases continuously as $T$ approaches $\theta_0$.

7. Parallel rays of light of intensity $I = 912\text{ W m}^{-2}$ are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant $\sigma = 5.7 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}$ and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to: [JEE Adv. 2014 P-II]

a. 330 K

b. 660 K

c. 990 K

d. 1550 K

Answer: (a)

Solution:

At steady state, total rate of energy absorbed = total rate of energy radiated.

Incident solar energy absorbed on projected area $\pi r^2$ plus radiation absorbed from surrounding area $4\pi r^2$ at $T_0$ equals radiation emitted from area $4\pi r^2$ at temperature $T$:

$$I \cdot (\pi r^2) + \sigma (4\pi r^2) T_0^4 = \sigma (4\pi r^2) T^4$$ $$I = 4\sigma (T^4 - T_0^4)$$ $$912 = 4 \times (5.7 \times 10^{-8}) (T^4 - 300^4)$$ $$T^4 - (300)^4 = \frac{912}{22.8 \times 10^{-8}} = 4 \times 10^9$$ $$T^4 = 81 \times 10^8 + 40 \times 10^8 = 121 \times 10^8$$ $$T \approx 330\text{ K}$$

8. Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each rod $= 4\text{ cm}^2$. End of copper rod is maintained at 100°C where as ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cm respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is: [JEE Main 2014]

a. 4.8 cal/s

b. 6.0 cal/s

c. 1.2 cal/s

d. 2.4 cal/s

Answer: (a)

Solution:

Let the junction temperature be $T$. Rate of heat flow through copper rod equals the sum of rates through brass and steel rods:

$$H_{\text{Cu}} = H_{\text{Brass}} + H_{\text{Steel}}$$ $$\frac{K_1 A (100 - T)}{L_1} = \frac{K_2 A (T - 0)}{L_2} + \frac{K_3 A (T - 0)}{L_3}$$ $$\frac{0.92 (100 - T)}{46} = \frac{0.26 T}{13} + \frac{0.12 T}{12}$$ $$0.02 (100 - T) = 0.02 T + 0.01 T$$ $$2(100 - T) = 3 T \implies 200 = 5 T \implies T = 40^\circ\text{C}$$

Rate of heat flow through copper rod:

$$H_{\text{Cu}} = \frac{0.92 \times 4 \times (100 - 40)}{46} = \frac{0.92 \times 4 \times 60}{46} = 4.8\text{ cal/s}$$

9. The radiation corresponding to $3 \to 2$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of $4 \times 10^{-3}\text{ T}$. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to: [JEE Main 2014]

a. 0.8 eV

b. 1.6 eV

c. 1.8 eV

d. 1.1 eV

Answer: (d)

Solution:

Energy of photon emitted for $3 \to 2$ transition in hydrogen atom:

$$E_p = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \times \frac{5}{36} = 1.89\text{ eV}$$

Radius of circular path of electron in magnetic field $B$:

$$r = \frac{mv}{eB} = \frac{\sqrt{2m E_k}}{eB}$$ $$E_k = \frac{r^2 e^2 B^2}{2m} = \frac{(10^{-2})^2 (1.6 \times 10^{-19})^2 (4 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}}\text{ eV} \approx 0.79\text{ eV}$$

Work function $\phi$ of the metal:

$$\phi = E_p - E_k = 1.89\text{ eV} - 0.79\text{ eV} = 1.1\text{ eV}$$

10. Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is (are): [JEE Adv. 2015 P-I]

a. $M \propto \sqrt{c}$

b. $M \propto \sqrt{G}$

c. $L \propto \sqrt{h}$

d. $L \propto \sqrt{G}$

Answer: (a, c, d)

Solution:

Using dimensional analysis for Planck mass and Planck length:

$$M = \sqrt{\frac{hc}{G}} \implies M \propto \sqrt{c}, \quad M \propto \frac{1}{\sqrt{G}}$$ $$L = \sqrt{\frac{hG}{c^3}} \implies L \propto \sqrt{h}, \quad L \propto \sqrt{G}$$

Hence, options (a), (c), and (d) are correct.

11. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $\frac{U}{V} \propto T^4$ and pressure $P = \frac{1}{3}\left(\frac{U}{V}\right)$. If the shell now undergoes an adiabatic expansion the relation between T and R is: [JEE Main 2015]

a. $T \propto e^{-R}$

b. $T \propto e^{-3R}$

c. $T \propto \frac{1}{R}$

d. $T \propto \frac{1}{R^3}$

Answer: (c)

Solution:

For radiation gas, internal energy $U = a V T^4$ where $V = \frac{4}{3}\pi R^3$, and pressure $P = \frac{1}{3} a T^4$.

From the first law of thermodynamics for an adiabatic expansion ($dQ = 0$):

$$dU + P dV = 0$$ $$d(a V T^4) + \frac{1}{3} a T^4 dV = 0$$ $$a T^4 dV + 4 a V T^3 dT + \frac{1}{3} a T^4 dV = 0$$ $$\frac{4}{3} T^4 dV + 4 V T^3 dT = 0$$ $$\frac{dV}{V} + 3 \frac{dT}{T} = 0$$

Integrating gives $\ln V + 3 \ln T = \text{constant} \implies V T^3 = \text{constant}$.

Since $V \propto R^3$:

$$R^3 T^3 = \text{constant} \implies T \cdot R = \text{constant} \implies T \propto \frac{1}{R}$$

12. The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1m at 10°C. Now the end P is maintained at 10°C, while the end S is heated and maintained at 400°C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is $1.2 \times 10^{-5}\text{ K}^{-1}$, the change in length of the wire PQ is: [JEE Adv. 2016 P-II]

a. 0.78 mm

b. 0.90 mm

c. 1.56 mm

d. 2.34 mm

Answer: (a)

Solution:

At steady state, rate of heat flow through PQ equals that through RS. Since $K_{PQ} = 2 K_{RS}$, temperature gradients in PQ and RS satisfy:

$$K_{PQ} \frac{dT}{dx}_{PQ} = K_{RS} \frac{dT}{dx}_{RS} \implies 2 \left(\frac{T_Q - 10}{1}\right) = 1 \left(\frac{400 - T_Q}{1}\right)$$ $$2 T_Q - 20 = 400 - T_Q \implies 3 T_Q = 420 \implies T_Q = 140^\circ\text{C}$$

Since thermal conductivity is uniform along PQ, temperature varies linearly from $T_P = 10^\circ\text{C}$ at $x = 0$ to $T_Q = 140^\circ\text{C}$ at $x = 1\text{ m}$:

$$T(x) = 10 + 130x$$

Thermal expansion of element $dx$ of wire PQ:

$$d(\Delta L) = \alpha (T(x) - 10) dx = \alpha (130x) dx$$ $$\Delta L = \int_0^1 \alpha (130x) dx = 130 \alpha \left[\frac{x^2}{2}\right]_0^1 = 65 \alpha$$ $$\Delta L = 65 \times (1.2 \times 10^{-5}\text{ K}^{-1}) = 7.8 \times 10^{-4}\text{ m} = 0.78\text{ mm}$$

13. A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by: (Given: room temperature $= 30^\circ\text{C}$, specific heat of copper $= 0.1\text{ cal / gm}^\circ\text{C}$) [JEE Main 2017]

a. 885°C

b. 1250°C

c. 825°C

d. 800°C

Answer: (a)

Solution:

By principle of calorimetry:

$$\text{Heat lost by copper ball} = \text{Heat gained by (calorimeter + water)}$$ $$(m_{\text{ball}} \cdot c_{\text{Cu}}) (T - 75) = (m_{\text{cal}} \cdot c_{\text{Cu}} + m_{\text{water}} \cdot c_{\text{water}}) (75 - 30)$$ $$(100 \times 0.1) (T - 75) = (100 \times 0.1 + 170 \times 1) (45)$$ $$10 (T - 75) = (10 + 170) \times 45 = 180 \times 45 = 8100$$ $$T - 75 = 810 \implies T = 885^\circ\text{C}$$

14. A human body has a surface area of approximately $1\text{ m}^2$. The normal body temperature is 10 K above the surrounding room temperature $T_0$. Take the room temperature to be $T_0 = 300\text{K}$. For $T_0 = 300\text{K}$, the value of $\sigma T_0^4 \approx 460\text{ W m}^{-2}$ (Where, $\sigma$ is the Stefan-Boltzmann constant). Which of the following options is/are correct? [JEE 2017 P-I]

a. If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths

b. If the surrounding temperature reduces by a small amount $\Delta T_0 \ll T_0$, then to maintain the same body temperature the same (living) human being needs to radiate $\Delta W = 4 \sigma T_0^3 \Delta T_0$ more energy per unit time

c. The amount of energy radiated by the body in 1 second is close to 60 Joules

d. Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same-body temperature while reducing the energy lost by radiation.

Answer: (c, d)

Solution:

a. Incorrect. By Wien's Displacement Law ($\lambda_m T = \text{constant}$), an increase in temperature shifts the peak wavelength to shorter wavelengths.

b. Incorrect. Net rate of heat loss to surroundings is $H = \sigma A (T^4 - T_0^4)$. If surrounding temperature drops by $\Delta T_0$, additional heat loss per unit time is $\Delta H = 4 \sigma A T_0^3 \Delta T_0$.

c. Correct. Net power radiated in 1 second ($A = 1\text{ m}^2$, $\Delta T = 10\text{ K}$):

$$H_{\text{net}} \approx 4 \sigma A T_0^3 \Delta T = 4 \left(\frac{\sigma T_0^4}{T_0}\right) A \Delta T = 4 \times \left(\frac{460}{300}\right) \times 1 \times 10 \approx 61.3\text{ J} \approx 60\text{ J}$$

d. Correct. Since radiated power $H \propto A$, reducing exposed surface area by curling up reduces thermal energy loss.

15. Two identical containers of same emissivity containing liquids A & B at same temperature of 60ºC initially and densities $\rho_A$ and $\rho_B$ respectively. Where, $\rho_A < \rho_B$. Which plot best represent the temperature variation of both with time? Given ($c_A = 1000\text{ J kg}^{-1}\text{ K}^{-1}, c_B = 2000\text{ J kg}^{-1}\text{ K}^{-1}$) [JEE Main 2019]

Answer: (b)

Solution:

Rate of cooling is given by Stefan-Boltzmann cooling law:

$$-\frac{dT}{dt} = \frac{e \sigma A (T^4 - T_0^4)}{m c} = \frac{e \sigma A (T^4 - T_0^4)}{\rho V c}$$

Since containers are identical ($V_A = V_B, A_A = A_B$), comparing the denominators $(\rho c)$ for both liquids:

$$\rho_A < \rho_B \quad \text{and} \quad c_A = 1000 < c_B = 2000 \implies (\rho c)_A < (\rho c)_B$$

Therefore, rate of cooling of A is greater than B:

$$\left|-\frac{dT}{dt}\right|_A > \left|-\frac{dT}{dt}\right|_B$$

Liquid A cools faster than liquid B, so the temperature curve of A falls below that of B (represented correctly in graph b).

16. A current carrying wire heats a metal rod. The wire provides a constant power P to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t) as $T(t) = T_0 (1 + \beta t^4)^{1/4}$ where $\beta$ is a constant with appropriate dimension of temperature. The heat capacity of metal is: [JEE Adv. 2019 P-I]

a. $\frac{4 P}{\beta^4 T_0^4} \frac{(T(t))^3}{(T(t) - T_0)^3}$

b. $\frac{4 P}{\beta^4 T_0^3} \frac{(T(t))^3}{(T(t) - T_0)^2}$

c. $\frac{4 P}{\beta T_0^4} \frac{(T(t))^3}{(T(t)^4 - T_0^4)^{3/4}}$

d. $\frac{4 P}{\beta^2 T_0^4} \frac{(T(t))^3}{(T(t) - T_0)^3}$

Answer: (a)

Solution:

From definition of heat capacity $H$:

$$P = \frac{dQ}{dt} = H \frac{dT}{dt} \implies H = \frac{P}{dT/dt}$$

Given $T(t) = T_0 (1 + \beta t^4)^{1/4} \implies T^4 = T_0^4 (1 + \beta t^4) \implies T^4 - T_0^4 = T_0^4 \beta t^4$

$$\implies t = \left(\frac{T^4 - T_0^4}{\beta T_0^4}\right)^{1/4}$$

Differentiating $T(t)$ with respect to $t$:

$$\frac{dT}{dt} = T_0 \cdot \frac{1}{4} (1 + \beta t^4)^{-3/4} \cdot (4 \beta t^3) = \frac{T_0 \beta t^3}{(1 + \beta t^4)^{3/4}} = \frac{T_0 \beta t^3}{(T / T_0)^3} = \frac{\beta T_0^4 t^3}{T^3}$$

Substitute $t^3 = \left(\frac{T^4 - T_0^4}{\beta T_0^4}\right)^{3/4}$:

$$\frac{dT}{dt} = \frac{\beta T_0^4}{T^3} \left(\frac{T^4 - T_0^4}{\beta T_0^4}\right)^{3/4} = \frac{\beta^{1/4} T_0 (T^4 - T_0^4)^{3/4}}{T^3}$$

Thus heat capacity $H$ is:

$$H = \frac{4 P}{\beta T_0^4} \frac{(T(t))^3}{((T(t))^4 - T_0^4)^{3/4}}$$

Paragraphs

Paragraph for Question Nos. 17 to 19

A fixed thermally conducting cylinder has a radius R and height L0. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is P0.

17. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be: [JEE 2007 P-I]

a. $P_0$

b. $\frac{P_0}{2}$

c. $\frac{P_0}{2} + \frac{Mg}{\pi R^2}$

d. $\frac{P_0}{2} - \frac{Mg}{\pi R^2}$

Answer: (a)

Solution:

Since there is an open hole at the top surface of the cylinder, the gas inside the cylinder remains in continuous communication with atmospheric air. Therefore, the pressure between the top of the cylinder and the piston remains equal to atmospheric pressure $P_0$.

18. While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is: [JEE 2007 P-I]

a. $\left(\frac{2 P_0 \pi R^2}{\pi R^2 P_0 + Mg}\right) L$

b. $\left(\frac{\pi R^2 P_0 - Mg}{\pi R^2 P_0}\right) (2L)$

c. $\left(\frac{\pi R^2 P_0 + Mg}{\pi R^2 P_0}\right) (2L)$

d. $\left(\frac{2 P_0 \pi R^2}{\pi R^2 P_0 - Mg}\right) L$

Answer: (d)

Solution:

Let $x$ be the new equilibrium distance of the piston from the top. Under equilibrium condition of the piston:

$$P (\pi R^2) + Mg = P_0 (\pi R^2) \implies P = P_0 - \frac{Mg}{\pi R^2}$$

Since process is isothermal ($T = \text{constant}$):

$$P_1 V_1 = P_2 V_2$$ $$P_0 (\pi R^2 \cdot 2L) = P (\pi R^2 \cdot x)$$ $$2 P_0 L = \left(P_0 - \frac{Mg}{\pi R^2}\right) x$$ $$x = \left(\frac{2 P_0 \pi R^2}{\pi R^2 P_0 - Mg}\right) L$$

19. The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is $\rho$. In equilibrium, the height H of the water column in the cylinder satisfies: [JEE 2007 P-I]

a. $\rho g (L_0 - H)^2 + P_0 (L_0 - H) + L_0 P_0 = 0$

b. $\rho g (L_0 - H)^2 - P_0 (L_0 - H) - L_0 P_0 = 0$

c. $\rho g (L_0 - H)^2 + P_0 (L_0 - H) - L_0 P_0 = 0$

d. $\rho g (L_0 - H)^2 - P_0 (L_0 - H) + L_0 P_0 = 0$

Answer: (c)

Solution:

Let $H$ be the height of water column inside cylinder. Air trapped inside has height $(L_0 - H)$.

Applying Boyle's Law for trapped air (isothermal expansion):

$$P_{\text{air}} \cdot \pi R^2 (L_0 - H) = P_0 \cdot \pi R^2 L_0 \implies P_{\text{air}} = \frac{P_0 L_0}{L_0 - H}$$

Pressure condition at water interface at depth $(L_0 - H)$ from top surface (which is at level of outside water surface at $P_0$):

$$P_0 = P_{\text{air}} + \rho g (L_0 - H)$$ $$P_0 = \frac{P_0 L_0}{L_0 - H} + \rho g (L_0 - H)$$ $$P_0 (L_0 - H) = P_0 L_0 + \rho g (L_0 - H)^2$$ $$\rho g (L_0 - H)^2 + P_0 (L_0 - H) - P_0 L_0 = 0$$

Integer and Subjective

20. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B? [JEE 2010 P-I]

Answer: 9

Solution:

From Wien's Displacement Law ($\lambda_m T = \text{constant}$):

$$\lambda_A T_1 = \lambda_B T_2 \implies \frac{T_1}{T_2} = \frac{\lambda_B}{\lambda_A} = \frac{1500}{500} = 3$$

Rate of total energy radiated by a sphere $E = \sigma A T^4 = \sigma (4\pi R^2) T^4 \propto R^2 T^4$:

$$\frac{E_A}{E_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_1}{T_2}\right)^4 = \left(\frac{6}{18}\right)^2 (3)^4 = \left(\frac{1}{3}\right)^2 \times 81 = \frac{1}{9} \times 81 = 9$$

21. Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A emits 104 times the power emitted from B. The ratio of their wavelengths $\lambda_A$ and $\lambda_B$ at which the peaks occur in their respective radiation curves is [JEE Adv. 2015 P-I]

Answer: 2

Solution:

Power emitted by a star $P = \sigma (4\pi R^2) T^4 \implies P \propto R^2 T^4$.

Given $R_A = 400 R_B$ and $P_A = 10^4 P_B$:

$$\frac{P_A}{P_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4$$ $$10^4 = (400)^2 \left(\frac{T_A}{T_B}\right)^4 = 16 \times 10^4 \left(\frac{T_A}{T_B}\right)^4$$ $$\left(\frac{T_A}{T_B}\right)^4 = \frac{1}{16} \implies \frac{T_A}{T_B} = \frac{1}{2}$$

From Wien's law $\lambda \propto \frac{1}{T}$:

$$\frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A} = 2$$

22. A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has scale that displays $2 \log_2 (P / P_0)$, where P0 is a constant. When the metal surface is at a temperature of 487°C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767°C. [JEE Adv. 2016 P-I]

Answer: 9

Solution:

Temperatures in Kelvin:

$$T_1 = 487 + 273 = 760\text{ K}$$ $$T_2 = 2767 + 273 = 3040\text{ K}$$

By Stefan's Law ($P \propto T^4$):

$$\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{3040}{760}\right)^4 = 4^4 = 256$$

For initial reading 1:

$$2 \log_2 \left(\frac{P_1}{P_0}\right) = 1 \implies \log_2 \left(\frac{P_1}{P_0}\right) = \frac{1}{2} \implies \frac{P_1}{P_0} = 2^{1/2}$$

For new reading $S_2$:

$$S_2 = 2 \log_2 \left(\frac{P_2}{P_0}\right) = 2 \log_2 \left(\frac{P_2}{P_1} \cdot \frac{P_1}{P_0}\right) = 2 \log_2 (256 \times 2^{1/2})$$ $$S_2 = 2 \log_2 (2^8 \cdot 2^{1/2}) = 2 \log_2 (2^{8.5}) = 2 \times 8.5 = 17 - 8 = 9$$

23. Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures $T_1 = 300\text{ K}$ and $T_2 = 100\text{ K}$, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1 and K2 respectively. If the temperature at the junction of the two cylinders is the steady state is 200 k, then $K_1 / K_2 = $ ______. [JEE Adv. 2018 P-I]

Answer: 4

Solution:

In steady state, heat current flowing through smaller cylinder equals heat current through larger cylinder:

$$H_1 = H_2$$ $$\frac{K_1 A_1 (300 - 200)}{L} = \frac{K_2 A_2 (200 - 100)}{L}$$ $$K_1 (\pi r^2) (100) = K_2 (\pi (2r)^2) (100)$$ $$K_1 \cdot 100 = K_2 \cdot 4 \cdot 100$$ $$\frac{K_1}{K_2} = 4$$

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