Question 1: Palette 1/81

DPT MCQ

Concept: Thermal Stress and Tension in rods. When a rod is fixed between two points and its temperature changes, thermal expansion or contraction is prevented, leading to the development of thermal stress. The formula for the force (tension) generated is $F = Y A \alpha \Delta T$, where $Y$ is Young's modulus, $A$ is the cross-sectional area, $\alpha$ is the coefficient of linear expansion, and $\Delta T$ is the change in temperature.

A wire of cross-sectional area $3 \text{ mm}^2$ is first stretched between two fixed points at a temperature of $20^\circ\text{C}$. Determine the tension when the temperature falls to $10^\circ\text{C}$. Coefficient of linear expansion $\alpha = 10^{-5} \text{ }^\circ\text{C}^{-1}$ and $Y = 2 \times 10^{11} \text{N/m}^2$.

A

$30\text{ N}$

B

$60\text{ N}$

C

$90\text{ N}$

D

$120\text{ N}$

Question 2: Palette 2/81

DPT MCQ

Concept: Thermal Expansion of a Pendulum. When temperature increases, the length of a pendulum increases, which increases its time period ($T = 2\pi\sqrt{l/g}$). This causes the clock to run slow and lose time. The time lost or gained is given by $\Delta t = \frac{1}{2} \alpha \Delta \theta t$, where $\alpha$ is the coefficient of linear expansion, $\Delta \theta$ is the change in temperature, and $t$ is the total time elapsed.

Estimate the time lost or gained by a pendulum clock at the end of a week when the atmospheric temperature rises to 40°C. The clock is known to give correct time at 15°C and the pendulum is of steel. (Coefficient of linear expansion of steel is $12 \times 10^{-6} / ^\circ\text{C}$).

A

90.72 s lost

B

90.72 s gained

C

45.36 s lost

D

45.36 s gained

Question 3: Palette 3/81

DPT MCQ

Concept: The moment of inertia $I$ for a rigid body is related to its dimensions (e.g., length $l$). For a rod of length $l$ and mass $m$, $I = \frac{m}{3}l^2$. When temperature increases, the length changes according to the linear expansion formula: $l = l_0(1 + \alpha\Delta\theta)$. Substituting this into the moment of inertia formula and using the binomial approximation for small changes, the fractional change in moment of inertia is related to the fractional change in length. Since $\frac{\Delta l}{l} = \alpha\Delta\theta$, the relation becomes $\frac{\Delta I}{I} = 2(\alpha\Delta\theta)$.

If temperature is increase by 60°C find % increase in moment of inertia. ($\alpha = 10^{-5} /^\circ\text{C}$)

A

0.06%

B

0.12%

C

0.18%

D

0.24%

Question 4: Palette 4/81

DPT MCQ

Concept: The amount of heat $Q$ required to change the temperature of a substance is given by the formula $Q = mS\Delta T$, where $m$ is the mass of the body, $S$ is the specific heat capacity, and $\Delta T$ is the change in temperature. For water, the specific heat capacity $S = 1 \text{ cal/gm}^\circ\text{C}$.

Find the amount of heat required to raise the temperature of 50 gm water at 20°C to 60°C.

A

1000 cal

B

2000 cal

C

3000 cal

D

4000 cal

Question 5: Palette 5/81

DPT MCQ

Concept: The amount of heat $Q$ released or absorbed when the temperature of a substance changes is given by the formula $Q = mS\Delta T$, where $m$ is the mass, $S$ is the specific heat capacity, and $\Delta T$ is the change in temperature. For water, the specific heat capacity $S = 1 \text{ cal/gm}^\circ\text{C}$.

Find the amount of heat released from 50 gm water at 50°C to 50 gm water 20°C.

A

1200 cal

B

1500 cal

C

1800 cal

D

2000 cal

Question 6: Palette 6/81

DPT MCQ

Concept: The amount of heat $Q$ required to change the temperature of a substance is given by the formula $Q = mS\Delta T$, where $m$ is the mass, $S$ is the specific heat capacity, and $\Delta T$ is the change in temperature. For ice, the specific heat capacity $S_{ice} = 0.5 \text{ cal/gm}^\circ\text{C}$ (or $\frac{1}{2} \text{ cal/gm}^\circ\text{C}$).

Find the amount of heat required to raise the temperature of 50 gm of ice from -20°C to -5°C.

A

275 cal

B

375 cal

C

425 cal

D

500 cal

Question 7: Palette 7/81

DPT MCQ

Concept: When specific heat capacity $C$ depends on temperature $T$, the total heat $Q$ required to change the temperature of a mass $m$ from $T_1$ to $T_2$ is found by integrating the heat capacity over the temperature range: $Q = \int_{T_1}^{T_2} m C \, dT$.

The specific heat of a substance is given by $C = a + bT$, where $a = 1.12 \text{ kJ kg}^{-1}\text{K}^{-1}$ and $b = 0.016 \text{ kJ kg}^{-1}\text{K}^{-2}$. The amount of heat required to raise the temperature of 1.2 kg of the material from 280 K to 312 K is:

A

215 kJ

B

225 kJ

C

235 kJ

D

245 kJ

Question 8: Palette 8/81

DPT MCQ

Concept: The amount of heat $Q$ required to change the state of a substance from solid to liquid at its melting point without changing its temperature is given by the formula $Q = mL$, where $m$ is the mass of the substance and $L$ is the latent heat of fusion. For water/ice, the latent heat of fusion is $L = 80 \text{ cal/g}$.

Find the amount of heat required to melt 20 g of ice at 0°C.

A

800 cal

B

1200 cal

C

1600 cal

D

2000 cal

Question 9: Palette 9/81

DPT MCQ

Concept: The heat $\Delta Q$ required to change the state of a substance from liquid to gas at its boiling point without a change in temperature is given by the formula $\Delta Q = m L_v$, where $m$ is the mass of the substance and $L_v$ is the latent heat of vaporization. For water, the latent heat of vaporization is typically taken as 540 cal/gm.

Find the amount of heat required to change 10 gm of water at 100°C to 10 gm of vapor at 100°C.

A

540 cal

B

2700 cal

C

5400 cal

D

10800 cal

Question 10: Palette 10/81

DPT MCQ

Concept: The amount of heat $\Delta Q$ required to change the state of a liquid to a gas at its boiling point without a change in temperature is given by the formula $\Delta Q = m L_v$, where $m$ is the mass of the liquid that converts into gas and $L_v$ is the latent heat of vaporization. For water, the latent heat of vaporization is $L_v = 540 \text{ cal/gm}$.

Find the amount of heat required to change 10 gm of water at $100^\circ\text{C}$ to 8 gm of water and 2 gm of vapour at $100^\circ\text{C}$.

A

540 cal

B

1080 cal

C

5400 cal

D

10800 cal

Question 11: Palette 11/81

DPT MCQ

Concept: The total heat required involves two steps: first, raising the temperature of the entire mass of water from its initial temperature to the boiling point using $\Delta Q_1 = mS\Delta T$; second, changing the phase of a specific mass of water to vapour using $\Delta Q_2 = m'L_v$. The total heat is the sum of these processes: $\Delta Q = \Delta Q_1 + \Delta Q_2$.

Find the amount of heat required to change 10 gm water at $20^\circ\text{C}$ to 6 gm water and 4 gm vapour at $100^\circ\text{C}$.

A

2160 cal

B

2960 cal

C

3760 cal

D

6200 cal

Question 12: Palette 12/81

DPT MCQ

Concept: The total heat required involves two steps: first, raising the temperature of the entire mass of ice from its initial temperature to its melting point ($0^\circ\text{C}$) using $\Delta Q_1 = m S_{ice} \Delta T$; second, changing the phase of a specific mass of the ice to water at $0^\circ\text{C}$ using $\Delta Q_2 = m' L_f$. The total heat is the sum of these processes: $\Delta Q = \Delta Q_1 + \Delta Q_2$.

Find the amount of heat required to change 10 gm ice at $-20^\circ\text{C}$ to 6 gm water + 4 gm ice at $0^\circ\text{C}$.

A

380 cal

B

480 cal

C

580 cal

D

680 cal

Question 13: Palette 13/81

DPT MCQ

Concept: The total heat required involves three stages: first, raising the temperature of the ice to its melting point (0°C) using $\Delta Q_1 = m S_{ice} \Delta T_1$; second, melting the ice into water at 0°C using $\Delta Q_2 = m L_f$; third, raising the temperature of the resulting water to the final temperature of 50°C using $\Delta Q_3 = m S_w \Delta T_2$. The net heat required is the sum of these parts: $\Delta Q_{net} = \Delta Q_1 + \Delta Q_2 + \Delta Q_3$.

Find the amount of heat required to change 10 gm of ice at -20°C to 10 gm of water at 50°C.

A

1200 cal

B

1300 cal

C

1400 cal

D

1500 cal

Question 14: Palette 14/81

DPT MCQ

Concept: The total heat required involves four stages: first, raising the temperature of the entire mass of ice to its melting point (0°C) using $Q_1 = m S_{ice} \Delta T_1$; second, melting all the ice into water at 0°C using $Q_2 = m L_f$; third, raising the temperature of the resulting water to the boiling point (100°C) using $Q_3 = m S_w \Delta T_2$; and fourth, changing the phase of a specific mass of the water to vapour at 100°C using $Q_4 = m' L_v$. The net heat required is the sum of these parts: $Q_{net} = Q_1 + Q_2 + Q_3 + Q_4$.

Find the amount of heat required to change 10 gm ice at -20°C to 4 gm vapour + 6 gm water at 100°C.

A

3060 cal

B

4060 cal

C

5060 cal

D

6060 cal

Question 15: Palette 15/81

DPT MCQ

Concept: The total heat required involves five stages: first, heating the ice to its melting point using $Q_1 = m s_{ice} \Delta T_1$; second, melting the ice into water at 0°C using $Q_2 = m L_f$; third, heating the water to its boiling point using $Q_3 = m s_w \Delta T_2$; fourth, boiling the water into steam at 100°C using $Q_4 = m L_v$; and fifth, heating the steam to the final temperature using $Q_5 = m s_{steam} \Delta T_3$. The net heat required is the sum of these parts: $Q_{net} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5$.

Find the amount of heat required if 100 g ice at -10°C is converted into 100 g steam at 120°C.

A

71.52 kcal

B

72.42 kcal

C

73.42 kcal

D

74.42 kcal

Question 16: Palette 16/81

DPT MCQ

Liquid A with mass $m_1$, specific heat $s_1$, and temperature $T_1$ is mixed with liquid B with mass $m_2$, specific heat $s_2$, and temperature $T_2$. Find the final temperature of the mixture.

A

$\frac{m_1 s_1 T_1 + m_2 s_2 T_2}{m_1 s_1 + m_2 s_2}$

B

$\frac{m_1 s_1 T_1 - m_2 s_2 T_2}{m_1 s_1 + m_2 s_2}$

C

$\frac{m_1 s_1 T_2 + m_2 s_2 T_1}{m_1 s_1 + m_2 s_2}$

D

$\frac{m_1 T_1 + m_2 T_2}{m_1 + m_2}$

Question 17: Palette 17/81

DPT MCQ

A calorimeter of heat capacity $100\text{ J/K}$ is at room temperature of $30^{\circ}\text{C}$. $100\text{ g}$ of water at $40^{\circ}\text{C}$ of specific heat $4200\text{ J/kg}\cdot\text{K}$ is poured into the calorimeter. What is the temperature of water in calorimeter?

A

$35.42^{\circ}\text{C}$

B

$38.07^{\circ}\text{C}$

C

$36.50^{\circ}\text{C}$

D

$39.12^{\circ}\text{C}$

Question 18: Palette 18/81

DPT MCQ

10 g Ice at $0^{\circ}\text{C}$ is mixed with 10 g water at $60^{\circ}\text{C}$. Find the final temperature of the mixture.

A

$5^{\circ}\text{C}$

B

$10^{\circ}\text{C}$

C

$0^{\circ}\text{C}$

D

$15^{\circ}\text{C}$

Question 19: Palette 19/81

DPT MCQ

Steam at $100^{\circ}\text{C}$ is passed into $1.1\text{ kg}$ of water contained in a calorimeter of water equivalent $0.02\text{ kg}$ at $15^{\circ}\text{C}$ till the temperature of the calorimeter and its contents rises to $80^{\circ}\text{C}$. What is the mass of steam condensed? (Latent heat of steam = $536\text{ cal/g}$)

A

$0.130\text{ kg}$

B

$0.065\text{ kg}$

C

$0.260\text{ kg}$

D

$0.115\text{ kg}$

Question 20: Palette 20/81

DPT MCQ

An iron block of mass 2 kg falls from a height 10 m. After colliding with the ground, it loses 25% energy to surroundings. Then find the temperature rise of the block. (Take specific heat of iron $470\text{ J/kg}\cdot^{\circ}\text{C}$)

A

$0.159^{\circ}\text{C}$

B

$0.213^{\circ}\text{C}$

C

$0.425^{\circ}\text{C}$

D

$0.053^{\circ}\text{C}$

Question 21: Palette 21/81

DPT MCQ

540 g of ice at $0^{\circ}\text{C}$ is mixed with 540 g of water at $80^{\circ}\text{C}$. The final temperature of the mixture is (Given latent heat of fusion of ice = $80\text{ cal/g}$ and specific heat capacity of water = $1\text{ cal/g}\cdot^{\circ}\text{C}$)

A

$0^{\circ}\text{C}$

B

$40^{\circ}\text{C}$

C

$80^{\circ}\text{C}$

D

less than $0^{\circ}\text{C}$

Question 22: Palette 22/81

DPT MCQ

Concept: According to the principle of calorimetry, when two or more bodies at different temperatures are mixed, heat lost by the hotter bodies equals the heat gained by the colder bodies, assuming no heat exchange with the surroundings. The heat exchanged is given by $Q = mc\Delta T$, where $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature.

Illustration 31. Three liquids P, Q and R are given. It is observed that 4 kg of P at 60 °C and 1 kg of R at 50 °C, when mixed produce a resultant temperature of 55 °C. A mixture of 1 kg of P at 60 °C and 1 kg of Q at 50 °C shows a temperature of 55 °C. Find the resulting temperature when 1 kg of Q at 60 °C is mixed with 1 kg of R at 50 °C.

A

50 °C

B

52 °C

C

55 °C

D

58 °C

Question 23: Palette 23/81

DPT MCQ

Concept: Principle of Calorimetry (Heat Lost = Heat Gained)

10 g of water at $70\text{ }^\circ\text{C}$ is mixed with 5 g of water at $30\text{ }^\circ\text{C}$. Find the temperature of the mixture in equilibrium.

A

$45\text{ }^\circ\text{C}$

B

$50\text{ }^\circ\text{C}$

C

$56.7\text{ }^\circ\text{C}$

D

$60\text{ }^\circ\text{C}$

Question 24: Palette 24/81

DPT MCQ

Concept: Principle of Calorimetry (Heat Lost = Heat Gained)

The temperatures of equal masses of three different liquids A, B and C are $12\text{ }^\circ\text{C}$, $19\text{ }^\circ\text{C}$ and $28\text{ }^\circ\text{C}$ respectively. The temperature when liquids A and B are mixed is $16\text{ }^\circ\text{C}$ and when liquids B and C are mixed is $23\text{ }^\circ\text{C}$. What should be the temperature when liquids A and C are mixed?

A

$18.2\text{ }^\circ\text{C}$

B

$20.3\text{ }^\circ\text{C}$

C

$22.0\text{ }^\circ\text{C}$

D

$24.5\text{ }^\circ\text{C}$

Question 25: Palette 25/81

DPT MCQ

Concept: Conservation of Energy and Calorimetry

A copper cube of mass 200 g slides down a rough inclined plane of inclination $37^\circ$ at a constant speed. Assuming that the loss in mechanical energy goes into the copper block as thermal energy, find the increase in temperature of the block as it slides down through 60 cm. Specific heat capacity of copper is equal to $420\text{ J kg}^{-1}\text{ K}^{-1}$. Take $g = 10\text{ m s}^{-2}$.

A

$4.3 \times 10^{-3}\text{ }^\circ\text{C}$

B

$8.6 \times 10^{-3}\text{ }^\circ\text{C}$

C

$1.2 \times 10^{-2}\text{ }^\circ\text{C}$

D

$5.7 \times 10^{-3}\text{ }^\circ\text{C}$

Question 26: Palette 26/81

DPT MCQ

Concept: Principle of Calorimetry and Phase Change

1 kg of ice at $0\text{ }^\circ\text{C}$ is mixed with 1 kg of steam at $100\text{ }^\circ\text{C}$. Find the equilibrium temperature and the final composition of the mixture. Given that latent heat of fusion of ice is $3.36 \times 10^5\text{ J kg}^{-1}$ and latent heat of vaporization of water is $2.27 \times 10^6\text{ J kg}^{-1}$ and specific heat of water is $4200\text{ J kg}^{-1}\text{ }^\circ\text{C}^{-1}$.

A

$100\text{ }^\circ\text{C}$, $0.66\text{ kg}$ steam and $1.34\text{ kg}$ water

B

$100\text{ }^\circ\text{C}$, $0.34\text{ kg}$ steam and $1.66\text{ kg}$ water

C

$80\text{ }^\circ\text{C}$, $2.0\text{ kg}$ water

D

$100\text{ }^\circ\text{C}$, $0.42\text{ kg}$ steam and $1.58\text{ kg}$ water

Question 27: Palette 27/81

DPT MCQ

Concept: Conversion of Mechanical Energy into Heat and Calorimetry

A lead bullet just melts when stopped by an obstacle. Assuming 25% heat to be absorbed by the obstacle, find the velocity of the bullet if its initial temperature is $27\text{ }^\circ\text{C}$. Given melting point of lead is $327\text{ }^\circ\text{C}$, $c_{\text{lead}}$ is $0.03\text{ cal g}^{-1}\text{ }^\circ\text{C}^{-1}$, $L_{\text{bullet}} = 6\text{ cal g}^{-1}$ and $J = 4.2\text{ joule cal}^{-1}$.

A

$410\text{ m s}^{-1}$

B

$320\text{ m s}^{-1}$

C

$280\text{ m s}^{-1}$

D

$500\text{ m s}^{-1}$

Question 28: Palette 28/81

DPT MCQ

Concept: Principle of Calorimetry and Phase Change

5 g ice at $0\text{ }^\circ\text{C}$ is mixed with 5 g of steam at $100\text{ }^\circ\text{C}$. What is the final temperature?

A

$50\text{ }^\circ\text{C}$

B

$80\text{ }^\circ\text{C}$

C

$100\text{ }^\circ\text{C}$

D

$40\text{ }^\circ\text{C}$

Question 29: Palette 29/81

DPT MCQ

Concept: The rate of heat flow ($dQ/dt$) through a conductor is given by the formula:
$\frac{dQ}{dt} = \frac{KA \Delta T}{l}$
where $K$ is the coefficient of thermal conductivity, $A$ is the cross-sectional area, $\Delta T$ is the temperature difference, and $l$ is the length of the rod.
The rate of heat required for melting ice is:
$\frac{dQ}{dt} = L \frac{dm}{dt}$
where $L$ is the latent heat of fusion of ice and $dm/dt$ is the rate of melting of ice.

Find the rate of flow of heat and the rate of melting of ice for the given rod with $T_1 = 100^\circ C$ and $T_2 = 0^\circ C$ (melting ice), length $l = 10 m$, cross-sectional area $A = 2 m^2$, and thermal conductivity $K = 100$ in SI units.

A

Rate of heat flow = $2000 J/sec$, Rate of melting of ice = $2000/L kg/sec$

B

Rate of heat flow = $1000 J/sec$, Rate of melting of ice = $1000/L kg/sec$

C

Rate of heat flow = $200 J/sec$, Rate of melting of ice = $200/L kg/sec$

D

Rate of heat flow = $4000 J/sec$, Rate of melting of ice = $4000/L kg/sec$

Question 30: Palette 30/81

DPT MCQ

Concept: In a steady state, the rate of heat flow ($\frac{dQ}{dt}$) is constant across any cross-section of the rod. The rate of heat flow is given by:
$\frac{dQ}{dt} = \frac{KA \Delta T}{l}$
where $K$ is thermal conductivity, $A$ is the cross-sectional area, $\Delta T$ is the temperature difference, and $l$ is the length. By equating the heat flow through the first segment of length $4 m$ and the total length of $10 m$, we can solve for the unknown temperature $T$.

Find the temperature at a distance $4 m$ from the hot end ($100^\circ C$) for a rod of total length $10 m$ with the other end at $0^\circ C$.

A

$40^\circ C$

B

$50^\circ C$

C

$60^\circ C$

D

$70^\circ C$

Question 31: Palette 31/81

DPT MCQ

Concept: For a uniform rod in a steady state with no heat accumulation, the temperature decreases uniformly with length. The temperature gradient ($\frac{\Delta T}{\Delta x}$) remains constant throughout the rod. Therefore, the temperature drop is directly proportional to the length of the section.

Find temperature at C for a uniform rod of length 10 m with its ends held at $80^\circ\text{C}$ and $20^\circ\text{C}$, at a distance of 6 m from the hotter end.

A

$36^\circ\text{C}$

B

$44^\circ\text{C}$

C

$50^\circ\text{C}$

D

$60^\circ\text{C}$

Question 32: Palette 32/81

DPT MCQ

Concept: In a steady state, the rate of heat flow $\frac{dQ}{dt}$ is the same through both rods connected in series because there is no accumulation of heat at the junction. The rate of heat flow is given by:
$\frac{dQ}{dt} = \frac{KA \Delta T}{l}$
where $K$ is the thermal conductivity, $A$ is the cross-sectional area, $\Delta T$ is the temperature difference across the rod, and $l$ is the length of the rod.

Two different rods are connected in series as shown in the figure, with the first rod having properties ($L, K, A$) and the second rod having properties ($2L, 3K, A$). If the temperatures at the free ends are $100^\circ C$ and $0^\circ C$ respectively, find the temperature at the junction $B$.

A

$30^\circ C$

B

$40^\circ C$

C

$50^\circ C$

D

$60^\circ C$

Question 33: Palette 33/81

DPT MCQ

Concept: In a steady state, the rate of heat flow $\frac{dQ}{dt}$ is consistent across all cross-sections of the composite rod because there is no accumulation of heat anywhere. The heat flow rate through each rod is given by:
$\frac{dQ}{dt} = \frac{KA \Delta T}{l}$
where $K$ is the thermal conductivity, $A$ is the cross-sectional area, $\Delta T$ is the temperature difference, and $l$ is the length of the rod.

Find the temperature $T$ at the junction of two rods connected in series as shown in the figure, where the first rod has properties ($L, K, A$) and the second rod has properties ($L, 2K, 4A$). The temperatures at the free ends are $80^\circ C$ and $20^\circ C$ respectively.

A

$200/9^\circ C$

B

$80/3^\circ C$

C

$100/3^\circ C$

D

$40^\circ C$

Question 34: Palette 34/81

DPT MCQ

Concept: In a steady state, the rate of heat flow (heat current, i) through series-connected rods is constant. The thermal resistance of a rod of length l, thermal conductivity K, and cross-sectional area A is given by the formula:
$R_{th} = \frac{l}{KA}$
For rods in series, the total thermal resistance is the sum of the individual thermal resistances:
$R_{total} = R_1 + R_2$
The heat current is driven by the temperature difference across the ends:
$i = \frac{\Delta T}{R_{total}}$

Two cylindrical rods of the same cross-sectional area A are joined end-to-end. The first rod has length 2L, thermal conductivity K, and its free end is maintained at 100 degrees Celsius. The second rod has length 6L, thermal conductivity 2K, and its free end is maintained at 0 degrees Celsius. Find the heat current and the temperature T' at the junction.

A

$i = \frac{10KA}{L}$, $T' = 50^\circ\text{C}$

B

$i = \frac{20KA}{L}$, $T' = 60^\circ\text{C}$

C

$i = \frac{15KA}{L}$, $T' = 45^\circ\text{C}$

D

$i = \frac{30KA}{L}$, $T' = 80^\circ\text{C}$

Question 35: Palette 35/81

DPT MCQ

Concept: In steady state, the rate of heat flow (heat current, i) through series-connected rods is constant. The thermal resistance of a rod is given by:
$R_{th} = \frac{length}{K \times A}$
For multiple rods in series, the equivalent thermal resistance is the sum of individual resistances:
$R_{eq} = R_1 + R_2 + R_3$
The heat current is calculated using:
$i = \frac{\Delta T}{R_{eq}}$

Three cylindrical rods are joined end-to-end in series. The first rod has length l, thermal conductivity K, and cross-sectional area A. The second rod has length 2l, thermal conductivity 2K, and cross-sectional area A. The third rod has length 4l, thermal conductivity 2K, and cross-sectional area A. If the free ends are maintained at $100^\circ\text{C}$ and $20^\circ\text{C}$ respectively, find the steady-state heat current flowing through the system.

A

$i = \frac{10KA}{l}$

B

$i = \frac{15KA}{l}$

C

$i = \frac{20KA}{l}$

D

$i = \frac{25KA}{l}$

Question 36: Palette 36/81

DPT MCQ

Concept: In steady state, the rate of heat flow (heat current, i) remains constant through series-connected rods. The thermal resistance of a rod is given by:
$R_{th} = \frac{length}{K \times A}$
For multiple rods in series, the total equivalent thermal resistance is the sum of the individual resistances:
$R_{eq} = R_1 + R_2 + R_3 + R_4$
The heat current is calculated using:
$i = \frac{\Delta T}{R_{eq}}$

Four cylindrical rods are joined end-to-end in series. The first rod has length l, thermal conductivity K, and cross-sectional area A. The second rod has length 4l, thermal conductivity 2K, and cross-sectional area A. The third rod has length 6l, thermal conductivity 3K, and cross-sectional area A. The fourth rod has length 16l, thermal conductivity 4K, and cross-sectional area A. If the outer ends are maintained at temperatures $T_1$ and $T_2$ respectively, find the steady-state heat current flowing through the system.

A

$i = \frac{KA(T_1 - T_2)}{5l}$

B

$i = \frac{KA(T_1 - T_2)}{7l}$

C

$i = \frac{KA(T_1 - T_2)}{9l}$

D

$i = \frac{KA(T_1 - T_2)}{11l}$

Question 37: Palette 37/81

DPT MCQ

Concept: When rods are connected in parallel, the temperature difference ($\Delta T$) across each rod is the same. The total heat current ($i_{net}$) is the sum of the individual heat currents passing through each rod:
$i_{net} = i_1 + i_2 + i_3$
The thermal resistance of a rod is given by:
$R_{th} = \frac{length}{K \times Area}$
For parallel combinations, the equivalent thermal resistance ($R_{eq}$) satisfies:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
The total rate of heat flow is:
$i_{net} = \frac{\Delta T}{R_{eq}}$

Find the net rate of heat flow through a combination of three cylindrical rods connected in parallel between two reservoirs at temperatures $100^\circ\text{C}$ and $0^\circ\text{C}$. The dimensions and thermal conductivities of the rods are:
Rod 1: length l, thermal conductivity K, cross-sectional area A
Rod 2: length l, thermal conductivity K/4, cross-sectional area 2A
Rod 3: length l, thermal conductivity K/6, cross-sectional area 3A

A

$i_{net} = \frac{KA \times 100}{2l}$

B

$i_{net} = \frac{2KA \times 100}{l}$

C

$i_{net} = \frac{3KA \times 100}{2l}$

D

$i_{net} = \frac{5KA \times 100}{2l}$

Question 38: Palette 38/81

DPT MCQ

Concept: In steady state, the rate of heat flow (heat current, $i = \frac{dQ}{dt}$) through series-connected rods remains constant. The thermal resistance of a rod is given by:
$R_{th} = \frac{length}{K \times A}$
For multiple rods in series, the total equivalent thermal resistance is the sum of the individual resistances:
$R_{eq} = R_1 + R_2 + R_3$
The heat current is calculated using:
$i = \frac{\Delta T}{R_{eq}}$

Three cylindrical rods are joined end-to-end in series. The first rod has length l, thermal conductivity K, and cross-sectional area A. The second rod has length 2l, thermal conductivity K/2, and cross-sectional area A. The third rod has length 3l, thermal conductivity 3K, and cross-sectional area A. If the free ends are maintained at $100^\circ\text{C}$ and $20^\circ\text{C}$ respectively, find the steady-state temperatures $T_1$, $T_2$ at the junctions and the rate of heat flow $\frac{dQ}{dt}$.

A

$T_1 = \frac{260}{3}^\circ\text{C}$, $T_2 = \frac{100}{3}^\circ\text{C}$, $\frac{dQ}{dt} = \frac{40KA}{3l}$

B

$T_1 = \frac{260}{3}^\circ\text{C}$, $T_2 = \frac{100}{3}^\circ\text{C}$, $\frac{dQ}{dt} = \frac{80KA}{6l}$

C

$T_1 = \frac{520}{6}^\circ\text{C}$, $T_2 = \frac{400}{6}^\circ\text{C}$, $\frac{dQ}{dt} = \frac{80KA}{6l}$

D

$T_1 = \frac{520}{6}^\circ\text{C}$, $T_2 = \frac{400}{6}^\circ\text{C}$, $\frac{dQ}{dt} = \frac{40KA}{3l}$

Question 39: Palette 39/81

DPT MCQ

Concept: According to Kirchhoff's current law applied to thermal systems, the net heat current entering or leaving a junction in a steady state is zero:
$\sum i = 0$
The thermal resistance of a rod is given by:
$R_{th} = \frac{\text{length}}{K \times \text{area}}$
The rate of melting of ice ($\frac{dm}{dt}$) is related to the heat current reaching the ice ($i_{ice}$) by:
$i_{ice} = L_f \left(\frac{dm}{dt}\right)$

Find the temperature at the junction $T'$ and the rate of melting of ice for three cylindrical rods joined at a common junction as shown in the figure. The first rod has length L, thermal conductivity K, cross-sectional area A, and its free end is at $40^\circ\text{C}$. The second rod has length L, thermal conductivity K/4, cross-sectional area A, and its free end is at $100^\circ\text{C}$. The third rod has length 4L, thermal conductivity 2K, cross-sectional area A, and its free end is connected to ice at $0^\circ\text{C}$. (Take the latent heat of fusion of ice as $L_f$).

A

$T' = \frac{260}{7}^\circ\text{C}$, Rate of melting = $\frac{2KA}{4LL_f}\left(\frac{260}{7}\right)$

B

$T' = \frac{100}{7}^\circ\text{C}$, Rate of melting = $\frac{KA}{2LL_f}\left(\frac{100}{7}\right)$

C

$T' = \frac{180}{7}^\circ\text{C}$, Rate of melting = $\frac{3KA}{4LL_f}\left(\frac{180}{7}\right)$

D

$T' = \frac{200}{7}^\circ\text{C}$, Rate of melting = $\frac{KA}{4LL_f}\left(\frac{200}{7}\right)$

Question 40: Palette 40/81

DPT MCQ

Concept: The heat current through any rod is given by:
$i = \frac{\Delta T}{R_{th}}$
where $R_{th} = \frac{\text{length}}{K \times \text{area}}$.
If the heat current in rod BC is zero, it implies there is no temperature difference across rod BC. Therefore, the temperature at junction B ($T_1$) must be equal to the temperature at end C ($T_0$). Consequently, all the heat current coming from rod AB must flow entirely into rod BD.

If heat current in rod BC is zero, find $T_C$ (or $T_0$) given the system of rods connected at junction B as shown in the figure. Rod AB has length 2l, thermal conductivity K, cross-sectional area A, and its end is at $100^\circ\text{C}$. Rod BD has length l, thermal conductivity K/3, cross-sectional area A, and its end is at $40^\circ\text{C}$. Rod BC has length l, thermal conductivity 5K, cross-sectional area A.

A

$76^\circ\text{C}$

B

$60^\circ\text{C}$

C

$80^\circ\text{C}$

D

$100^\circ\text{C}$

Question 41: Palette 41/81

DPT MCQ

Concept: The thermal resistance of a rod is given by:
$R_{th} = \frac{\text{length}}{K \times \text{area}}$
For a series-parallel combination, first find the equivalent resistance ($R'$) of the parallel section:
$\frac{1}{R'} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$
Then, find the total equivalent resistance ($R_{eq}$) of the system by adding the series resistance:
$R_{eq} = R_1 + R'$
The total heat current ($\frac{dQ}{dt}$) flowing through the first rod is:
$i = \frac{\Delta T}{R_{eq}}$
The temperature at the junction $T'$ can be found using the potential drop across the first rod:
$i = \frac{100 - T'}{R_1}$

Find $T'$ and net rate of flow of heat ($\frac{dQ}{dt}$) through rod '1' for the combination of cylindrical rods shown in the figure. The first rod has length L, thermal conductivity K, cross-sectional area A, and its left end is maintained at $100^\circ\text{C}$. It is connected to three parallel rods, each of length L and cross-sectional area A, having thermal conductivities K/4, K/8 (with cross-sectional area 2A), and K/2 respectively, whose right ends are maintained at $0^\circ\text{C}$.

A

$T' = 40^\circ\text{C}$, $\frac{dQ}{dt} = \frac{60KA}{l}$

B

$T' = 50^\circ\text{C}$, $\frac{dQ}{dt} = \frac{50KA}{l}$

C

$T' = 60^\circ\text{C}$, $\frac{dQ}{dt} = \frac{40KA}{l}$

D

$T' = 70^\circ\text{C}$, $\frac{dQ}{dt} = \frac{30KA}{l}$

Question 42: Palette 42/81

DPT MCQ

Concept: Thermal resistance with variable thermal conductivity.

Find the thermal resistance $R_{AB}$ for a rod of cross-sectional area $A$ and length $l$ where the thermal conductivity varies with length $x$ as $K = 2x + 3$ from $x = 0$ to $x = l$.

A

$\frac{1}{2A} \ln\left(\frac{2l + 3}{3}\right)$

B

$\frac{1}{A} \ln\left(\frac{2l + 3}{3}\right)$

C

$\frac{1}{2A} \ln(2l + 3)$

D

$\frac{2}{A} \ln\left(\frac{2l + 3}{3}\right)$

Question 43: Palette 43/81

DPT MCQ

Concept: Thermal resistance and rate of heat flow for variable thermal conductivity.

Find the equivalent thermal resistance $R_{eq}$ and the rate of heat flow through a solid cylinder where the thermal conductivity varies as $K = 2x + 3$, the length of the rod is $10\text{ m}$, the radius is $1\text{ m}$, and the end temperatures are maintained at $100^\circ\text{C}$ and $20^\circ\text{C}$.

A

$R_{eq} = \frac{1}{2\pi} \ln\left(\frac{23}{3}\right)\text{ K/W}$, $\text{Rate of heat flow} = \frac{160\pi}{\ln(23/3)}\text{ W}$

B

$R_{eq} = \frac{1}{\pi} \ln\left(\frac{23}{3}\right)\text{ K/W}$, $\text{Rate of heat flow} = \frac{80\pi}{\ln(23/3)}\text{ W}$

C

$R_{eq} = \frac{1}{2\pi} \ln(23)\text{ K/W}$, $\text{Rate of heat flow} = \frac{160\pi}{\ln(23)}\text{ W}$

D

$R_{eq} = \frac{1}{2\pi} \ln\left(\frac{20}{3}\right)\text{ K/W}$, $\text{Rate of heat flow} = \frac{160\pi}{\ln(20/3)}\text{ W}$

Question 44: Palette 44/81

DPT MCQ

Concept: Effective thermal conductivity in a series combination.

Two identical metal rods of thermal conductivities $K_1$ and $K_2$ respectively are connected in series. The effective thermal conductivity of the combination is:

A

$\frac{K_1 + K_2}{2}$

B

$\frac{2K_1 K_2}{K_1 + K_2}$

C

$\frac{K_1 K_2}{K_1 + K_2}$

D

$\frac{K_1 + K_2}{K_1 K_2}$

Question 45: Palette 45/81

DPT MCQ

Concept: Analysis of heat flow through a combination of rods using the electrical analogy of thermal resistance.

Three rods of material x and three of material y are connected as shown in the figure. All the rods are identical in length and cross-sectional area. If the end A is maintained at $60^\circ\text{C}$ and the junction E at $10^\circ\text{C}$, calculate the temperature of the junction B. The thermal conductivity of x is $800\text{ W m}^{-1}\text{ }^\circ\text{C}^{-1}$ and that of y is $400\text{ W m}^{-1}\text{ }^\circ\text{C}^{-1}$.

A

$30^\circ\text{C}$

B

$40^\circ\text{C}$

C

$45^\circ\text{C}$

D

$50^\circ\text{C}$

Question 46: Palette 46/81

DPT MCQ

Concept: Thermal resistance and heat flow rate through a tapered rod (frustum of a cone).

Find the rate of heat flow through a cross section of the rod shown in figure ($\theta_2 > \theta_1$). Thermal conductivity of the material of the rod is $K$.

A

$\frac{K \pi r_1 r_2 (\theta_2 - \theta_1)}{L}$

B

$\frac{K \pi (r_1 + r_2)^2 (\theta_2 - \theta_1)}{4L}$

C

$\frac{K \pi (r_2^2 - r_1^2) (\theta_2 - \theta_1)}{L}$

D

$\frac{K \pi r_1 r_2 (\theta_2 + \theta_1)}{L}$

Question 47: Palette 47/81

DPT MCQ

Concept: Thermal resistance and heat flow rate in parallel and series combinations of conductors.

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 $100^\circ\text{C}$ whereas ends of brass and steel are kept at $0^\circ\text{C}$. Lengths of the copper, brass and steel rods are $46\text{ cm}$, $13\text{ cm}$ and $12\text{ cm}$ respectively. The rods are thermally insulated from surrounding except at ends. Thermal conductivities of copper, brass and steel are $0.92$, $0.26$ and $0.12\text{ CGS units}$ respectively. Rate of heat flow through copper rod is:

A

$1.2\text{ cal/s}$

B

$2.4\text{ cal/s}$

C

$4.8\text{ cal/s}$

D

$6.0\text{ cal/s}$

Question 48: Palette 48/81

DPT MCQ

Concept: Radial heat flow through a spherical shell and its thermal resistance.

Two thin metallic spherical shells of radii $r_1$ and $r_2$ ($r_1 < r_2$) are placed with their centres coinciding. A material of thermal conductivity $K$ is filled in the space between the shells. The inner shell is maintained at temperature $\theta_1$ and the outer shell at temperature $\theta_2$ ($\theta_1 > \theta_2$). The rate at which heat flows radially through the material is:

A

$\frac{4\pi K r_1 r_2 (\theta_1 - \theta_2)}{r_2 - r_1}$

B

$\frac{4\pi K (\theta_1 - \theta_2)}{r_2 - r_1}$

C

$\frac{4\pi K r_1 r_2 (\theta_1 - \theta_2)}{r_1 + r_2}$

D

$\frac{K r_1 r_2 (\theta_1 - \theta_2)}{4\pi (r_2 - r_1)}$

Question 49: Palette 49/81

DPT MCQ

Concept: Junction temperature in steady-state heat conduction through layers in series.

The temperature $\theta$ at the junction of two insulating sheets, having thermal resistances $R_1$ and $R_2$ as well as top and bottom temperatures $\theta_2$ and $\theta_1$ respectively, is given by:

A

$\frac{\theta_1 R_1 + \theta_2 R_2}{R_1 + R_2}$

B

$\frac{\theta_1 R_2 + \theta_2 R_1}{R_1 + R_2}$

C

$\frac{\theta_1 R_1 - \theta_2 R_2}{R_1 + R_2}$

D

$\frac{\theta_1 R_2 - \theta_2 R_1}{R_1 + R_2}$

Question 50: Palette 50/81

DPT MCQ

Concept: Steady-state heat conduction through multiple layers.

Consider a lake of depth $L$. The temperature of the base is constant at $4^\circ\text{C}$ and the atmosphere temperature is $-10^\circ\text{C}$. Thermal conductivity of water is $K_1$ and for ice is $K_2$. Find the expression for the maximum thickness of ice that can be formed theoretically.

A

$\frac{5K_2 L}{2K_1 + 5K_2}$

B

$\frac{2K_1 L}{2K_1 + 5K_2}$

C

$\frac{5K_1 L}{5K_1 + 2K_2}$

D

$\frac{2K_2 L}{5K_1 + 2K_2}$

Question 51: Palette 51/81

DPT MCQ

Concept: Newton's Law of Cooling

A beaker of boiled water cools from $80^\circ\text{C}$ to $40^\circ\text{C}$ in $6\text{ minutes}$. What is the time taken to cool from $40^\circ\text{C}$ to $30^\circ\text{C}$ if the temperature of the surroundings is $20^\circ\text{C}$?

A

$3\text{ minutes}$

B

$4\text{ minutes}$

C

$5\text{ minutes}$

D

$6\text{ minutes}$

Question 52: Palette 52/81

DPT MCQ

Concept: Newton's Law of Cooling

A body takes 4 minutes to cool from $100^\circ\text{C}$ to $70^\circ\text{C}$. If the temperature of the surroundings is $20^\circ\text{C}$, the time taken by it to cool from $70^\circ\text{C}$ to $50^\circ\text{C}$ is:

A

$\frac{13}{3}\text{ minutes}$

B

$5\text{ minutes}$

C

$\frac{10}{3}\text{ minutes}$

D

$4\text{ minutes}$

Question 53: Palette 53/81

DPT MCQ

Concept: Thermal conduction through series and parallel combinations of conductors, and the relation of thermal resistance to heat transfer rate:
$Q = \frac{\Delta T}{R_{eq}} \cdot t$

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 $k$ and the other $2k$. The temperature difference between the ends along the $x$-axis is the same in both the configurations. It takes 9s 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 :-

A

2.0 s

B

3.0 s

C

4.5 s

D

6.0 s

Question 54: Palette 54/81

DPT MCQ

Concept: Thermal resistance and rate of heat transfer. The rate of heat flow through a conductor is given by:
$H = \frac{dQ}{dt} = \frac{\Delta T}{R}$
where the thermal resistance of a rod of length $L$, cross-sectional area $A$, and thermal conductivity $K$ is:
$R = \frac{L}{KA}$

A rod of length $l$ and cross-section $A$ is used to melt a piece of ice as shown. Now if the rod is broken into two equal parts and is arranged as shown, the time taken to melt ice in the second case becomes:

A

Half

B

One-fourth

C

Twice

D

Four times

Question 55: Palette 55/81

DPT MCQ

Concept: When rods of the same length $d$ and cross-sectional area $A$ are connected in series, the equivalent thermal resistance is the sum of individual thermal resistances:
$R_{eq} = R_1 + R_2 + R_3$
The thermal resistance $R$ of a rod is given by:
$R = \frac{d}{KA}$
The equivalent thermal conductivity $K_{eq}$ for the composite rod of total length $3d$ is:
$R_{eq} = \frac{3d}{K_{eq}A}$

A composite rod made of three rods of equal length and cross-section as shown in the fig. The thermal conductivities of the materials of the rods are $K/2$, $5K$ and $K$ respectively. The end $A$ and end $B$ are at constant temperatures. All heat entering the end $A$ goes out of the end $B$, there being no loss of heat from the sides of the bar. The effective thermal conductivity of the bar is

A

$15K/16$

B

$6K/13$

C

$5K/16$

D

$2K/13$

Question 56: Palette 56/81

DPT MCQ

Concept: Junction temperature in thermal conduction. According to the principle of conservation of energy, under steady-state conditions, the net rate of heat entering a junction must equal the net rate of heat leaving the junction. The rate of heat flow is given by:
$H = \frac{dQ}{dt} = \frac{\Delta T}{R}$
Since the three rods are made of the same material, have the same cross-section, and the same length, their thermal resistances $R$ are identical:
$R = \frac{L}{KA}$

Three rods made of same material and having the same cross section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at $0^\circ\text{C}$ and $90^\circ\text{C}$ respectively. The temperature of the junction of the three rods will be :

A

$45^\circ\text{C}$

B

$60^\circ\text{C}$

C

$30^\circ\text{C}$

D

$20^\circ\text{C}$

Question 57: Palette 57/81

DPT MCQ

Concept: According to Newton's Law of Cooling, the rate of cooling of a body is directly proportional to the difference in temperature between the body and its surroundings. For a small temperature interval, the average rate of cooling is given by:
$\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$
where $T_1$ is the initial temperature, $T_2$ is the final temperature, $t$ is the time interval, $T_s$ is the surrounding temperature, and $K$ is a positive constant.

A body cools in 7 minutes from $60^\circ\text{C}$ to $40^\circ\text{C}$. The temperature of the surrounding is $10^\circ\text{C}$. The temperature of the body after the next 7 minutes will be:

A

$32^\circ\text{C}$

B

$30^\circ\text{C}$

C

$28^\circ\text{C}$

D

$34^\circ\text{C}$

Question 58: Palette 58/81

DPT MCQ

Concept: According to Newton's Law of Cooling, the rate of cooling of a body is directly proportional to the difference in temperature between the body and its surroundings. For a small temperature interval, the average rate of cooling is given by:
$\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$
where $T_1$ is the initial temperature, $T_2$ is the final temperature, $t$ is the time interval, $T_s$ is the surrounding temperature, and $K$ is a positive constant.

A body cools from $80^\circ\text{C}$ to $60^\circ\text{C}$ in 5 minutes. The temperature of the surrounding is $20^\circ\text{C}$. The time it takes to cool from $60^\circ\text{C}$ to $40^\circ\text{C}$ is:

A

500 s

B

25/3 s

C

450 s

D

420 s

Question 59: Palette 59/81

DPT MCQ

Concept: In a series combination of slabs, the steady-state heat current $H$ flowing through each slab is the same. The heat current is given by:
$H = \frac{\Delta T_i}{R_i}$
where $\Delta T_i$ is the temperature difference across the $i$-th slab, and $R_i$ is its thermal resistance.
For slabs of the same thickness $d$ and cross-sectional area $A$, the thermal resistance is:
$R = \frac{d}{kA}$
Thus, $R \propto \frac{1}{k}$.
The total temperature difference $\Delta T_{total}$ across the entire wall is the sum of the temperature differences across individual slabs:
$\Delta T_{total} = \Delta T_1 + \Delta T_2 + \Delta T_3$
Since $H$ is constant through all slabs in series:
$\Delta T_i = H \cdot R_i$
This implies:
$\Delta T_1 : \Delta T_2 : \Delta T_3 = R_1 : R_2 : R_3 = \frac{1}{k_1} : \frac{1}{k_2} : \frac{1}{k_3}$

Figure shows three different arrangements of materials 1, 2 and 3 to form a wall. Thermal conductivities are $k_1 > k_2 > k_3$. The left side of the wall is $20^\circ\text{C}$ higher than the right side. Temperature difference $\Delta T$ across the material 1 has following relation in three cases :

A

$\Delta T_a > \Delta T_b > \Delta T_c$

B

$\Delta T_a = \Delta T_b = \Delta T_c$

C

$\Delta T_a = \Delta T_b > \Delta T_c$

D

$\Delta T_a = \Delta T_b$ $\Delta T_c$

Question 60: Palette 60/81

DPT MCQ

Concept: The total heat required $Q$ is the sum of three heat exchange processes:
1. Heating ice from $-20^\circ\text{C}$ to $0^\circ\text{C}$ using $Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T_1$.
2. Phase conversion of ice into water at $0^\circ\text{C}$ using $Q_2 = m \cdot L_{\text{ice}}$.
3. Heating water from $0^\circ\text{C}$ to $20^\circ\text{C}$ using $Q_3 = m \cdot c_{\text{water}} \cdot \Delta T_2$.

Find the quantity of heat required to convert $40\text{ g}$ of ice at $-20^\circ\text{C}$ into water at $20^\circ\text{C}$. Given $L_{\text{ice}} = 0.336 \times 10^6\text{ J/kg}$, specific heat capacity of ice $c_{\text{ice}} = 2100\text{ J/kg}\cdot\text{K}$, and specific heat capacity of water $c_{\text{water}} = 4200\text{ J/kg}\cdot\text{K}$.

A

$15120\text{ J}$

B

$18480\text{ J}$

C

$16800\text{ J}$

D

$20160\text{ J}$

Question 61: Palette 61/81

DPT MCQ

Concept: The total heat supplied to the system over time raises the temperature of the ice and container to $0^\circ\text{C}$, melts the ice into water, and then raises the temperature of the resulting water and container to a final temperature $T$.
Total heat supplied:
$\Delta Q = \text{Rate} \cdot t$
Heat absorbed during heating phase:
$Q = m \cdot c \cdot \Delta T$
Heat absorbed during phase change:
$Q = m \cdot L$

An aluminium container of mass $100\text{ g}$ contains $200\text{ g}$ of ice at $-20^\circ\text{C}$. Heat is added to the system at the rate of $100\text{ cal/s}$. Find the temperature of the system after $4\text{ minutes}$. (Given specific heat capacity of ice $c_{\text{ice}} = 0.5\text{ cal/g}\cdot^\circ\text{C}$, latent heat of fusion $L = 80\text{ cal/g}$, specific heat capacity of aluminium $c_{\text{Al}} = 0.2\text{ cal/g}\cdot^\circ\text{C}$, and specific heat capacity of water $c_{\text{water}} = 1.0\text{ cal/g}\cdot^\circ\text{C}$)

A

$20^\circ\text{C}$

B

$25.45^\circ\text{C}$

C

$28.5^\circ\text{C}$

D

$32^\circ\text{C}$

Question 62: Palette 62/81

DPT MCQ

Concept: According to the principle of calorimetry, the heat lost by the condensing steam and cooling water must equal the heat gained by the cold water to reach the final equilibrium temperature.
Heat gained by water:
$Q_{\text{gained}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_f - T_i)$
Heat released by steam:
$Q_{\text{lost}} = m_{\text{steam}} \cdot L_v + m_{\text{steam}} \cdot c_{\text{water}} \cdot (100 - T_f)$

The temperature of $100\text{ g}$ of water is to be raised from $24^\circ\text{C}$ to $90^\circ\text{C}$ by adding steam at $100^\circ\text{C}$ to it. Calculate the mass of the steam required for this purpose. (Given latent heat of vaporization of steam $L_v = 540\text{ cal/g}$ and specific heat capacity of water $c_{\text{water}} = 1\text{ cal/g}\cdot^\circ\text{C}$)

A

$10\text{ g}$

B

$15\text{ g}$

C

$12\text{ g}$

D

$14\text{ g}$

Question 63: Palette 63/81

DPT MCQ

Concept: The heat lost by the block as it cools down to $0^\circ\text{C}$ is absorbed by the ice at $0^\circ\text{C}$ to melt into water.
Heat lost by the block:
$Q = m_{\text{block}} \cdot c \cdot \Delta T$
Heat gained by the ice to melt:
$Q = m_{\text{ice}} \cdot L$
By the principle of calorimetry:
$m_{\text{block}} \cdot c \cdot \Delta T = m_{\text{ice}} \cdot L$

A block of mass $2.5\text{ kg}$ is heated to a temperature of $500^\circ\text{C}$ and placed on a large ice block. What is the maximum amount of ice that can melt (approx.)? (Given specific heat capacity of the body $c = 0.1\text{ cal/g}\cdot^\circ\text{C}$ and latent heat of fusion of ice $L = 80\text{ cal/g}$)

A

$1.25\text{ kg}$

B

$2.00\text{ kg}$

C

$1.56\text{ kg}$

D

$0.85\text{ kg}$

Question 64: Palette 64/81

DPT MCQ

Concept: For a substance whose specific heat capacity $S(T)$ varies with temperature, the heat energy $dQ$ required to change the temperature of mass $m$ by an infinitesimal amount $dT$ is given by $dQ = m \cdot S(T) \cdot dT$. The total heat energy $\Delta Q$ is calculated by integrating this expression between the temperature limits $T_1$ and $T_2$:
$\Delta Q = \int_{T_1}^{T_2} m \cdot S(T) \cdot dT$

The specific heat of a metal at low temperatures varies according to $S = a T^3$, where $a$ is a constant and $T$ is the absolute temperature. The heat energy needed to raise unit mass of the metal from $T = 1\text{ K}$ to $T = 2\text{ K}$ is:

A

$\frac{3a}{4}$

B

$\frac{9a}{4}$

C

$\frac{15a}{4}$

D

$\frac{7a}{2}$

Question 65: Palette 65/81

DPT MCQ

Concept: Heat capacity $C$ of a body is given by $C = m \cdot c = \rho \cdot V \cdot c$, where $m$ is mass, $\rho$ is density, $V$ is volume, and $c$ is specific heat capacity. When heat capacities of two bodies are equal, their products $\rho \cdot V \cdot c$ are equal.

The density of a material A is $1500\text{ kg/m}^3$ and that of another material B is $2000\text{ kg/m}^3$. It is found that the heat capacity of $8$ volumes of A is equal to the heat capacity of $12$ volumes of B. The ratio of specific heats of A and B will be:

A

$1 : 2$

B

$3 : 4$

C

$2 : 1$

D

$4 : 3$

Question 66: Palette 66/81

DPT MCQ

Concept: In an insulated container, the principle of conservation of energy states that total heat gained by cold substances equals total heat lost by hot substances:
$Q_{\text{gained}} = Q_{\text{lost}}$
Heat gained by ice to melt and then heat up to final temperature $T_f$:
$Q_{\text{gained}} = m_{\text{ice}} \cdot L_f + m_{\text{ice}} \cdot c_{\text{water}} \cdot (T_f - 0)$
Heat lost by steam to condense and then cool down to final temperature $T_f$:
$Q_{\text{lost}} = M \cdot L_v + M \cdot c_{\text{water}} \cdot (100 - T_f)$

$M\text{ grams}$ of steam at $100^\circ\text{C}$ is mixed with $200\text{ g}$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^\circ\text{C}$ [heat of vaporization of water is $540\text{ cal/g}$ and heat of fusion of ice is $80\text{ cal/g}$], the value of $M$ is:

A

$30\text{ g}$

B

$50\text{ g}$

C

$40\text{ g}$

D

$60\text{ g}$

Question 67: Palette 67/81

DPT MCQ

Concept: When a bullet hits a target, its kinetic energy $K = \frac{1}{2} m v^2$ is converted into heat energy. Since half of this energy is absorbed by the bullet, the heat absorbed by the bullet is given by:
$Q = \frac{1}{2} K = \frac{1}{2} \left( \frac{1}{2} m v^2 \right)$
This heat energy causes a rise in the bullet's temperature, described by:
$Q = m \cdot s \cdot \Delta T$
Equating the heat absorbed to the thermal energy formula:
$\frac{1}{4} m v^2 = m \cdot s \cdot \Delta T \implies \Delta T = \frac{v^2}{4 s}$

A bullet of mass $5\text{ g}$, traveling with a speed of $210\text{ m/s}$, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is $0.030\text{ cal/g}\cdot^\circ\text{C}$ ($1\text{ cal} = 4.2 \times 10^7\text{ ergs} = 4.2\text{ J}$) is close to:

A

$84.5^\circ\text{C}$

B

$42.2^\circ\text{C}$

C

$169.0^\circ\text{C}$

D

$116.2^\circ\text{C}$

Question 68: Palette 68/81

DPT MCQ

Concept: When the temperature increases, the total free thermal expansion of the two rods is constrained by the three springs connected in series.
1. Total free expansion of the rods:
$\Delta L_{\text{total}} = \alpha \cdot L \cdot \Delta T + \alpha \cdot \frac{L}{2} \cdot \Delta T = \frac{3}{2} \alpha L \Delta T$
2. The tension $T$ in each spring connected in series must be equal:
$T = K x_1 = 2K x_2 = 3K x_3$
3. The sum of the compression of all three springs equals the total thermal expansion of the rods:
$x_1 + x_2 + x_3 = \Delta L_{\text{total}}$

The system shown consists of $3$ springs and two rods. If the temperature of the rods is increased by $\Delta T$, calculate the force exerted by springs on wall. Neglect friction and thermal stress and take the coefficient of linear expansion of the material of rods equal to $\alpha$. (Rod lengths are $L$ and $L/2$, spring constants are $K$, $2K$, and $3K$).

A

$\frac{9}{11} K \alpha L \Delta T$

B

$\frac{6}{11} K \alpha L \Delta T$

C

$\frac{3}{11} K \alpha L \Delta T$

D

$\frac{18}{11} K \alpha L \Delta T$

Question 69: Palette 69/81

DPT MCQ

Concept: 1. Center of Mass Position: Since there are no external horizontal forces acting on the system of two rods, the center of mass of the combined system does not move. Therefore:
$M \cdot S_1 + 2M \cdot S_2 = 0 \implies S_1 + 2 S_2 = 0$
where $S_1$ and $S_2$ are the displacements of the centers of mass of rod 1 and rod 2, respectively.
2. Relative Displacement of Point of Contact: The displacement of the point of contact $P$ relative to the center of mass of each rod can be determined by the thermal expansion of the half-length of that rod:
$S_P - S_1 = \frac{L}{2} \cdot (2\alpha) \cdot \Delta T = L \alpha \Delta T$
$S_P - S_2 = -\frac{2L}{2} \cdot \alpha \cdot \Delta T = -L \alpha \Delta T$

Two separate thin rods of mass $M$ and $2M$ are kept on a smooth horizontal surface such that they are just touching each other. If the lengths of the rods are $L$ and $2L$, and their coefficients of linear expansion are $2\alpha$ and $\alpha$ respectively, find the displacement of their point of contact when both of them are heated such that their temperature rises by $\Delta T$.

A

$\frac{1}{3} L \alpha \Delta T$

B

$\frac{4}{3} L \alpha \Delta T$

C

$\frac{2}{3} L \alpha \Delta T$

D

$\frac{1}{2} L \alpha \Delta T$

Question 70: Palette 70/81

DPT MCQ

Concept: Thermal expansion of a rod is given by $\Delta L = L \cdot \alpha \cdot \Delta T$.
Thermal strain when expansion is restricted is $\text{strain} = \frac{\text{restricted expansion}}{L}$.
Thermal stress is calculated using Young's modulus: $\text{stress} = Y \cdot \text{strain}$.

A rod of length $2\text{ m}$ is at a temperature of $20^\circ\text{C}$. Find the free expansion of the rod if the temperature is increased to $50^\circ\text{C}$, and find the thermal stresses produced when the rod is (i) fully prevented from expanding and (ii) permitted to expand by $0.4\text{ mm}$. (Given $Y = 2 \times 10^{11}\text{ N/m}^2$ and $\alpha = 15 \times 10^{-6}\text{ /}^\circ\text{C}$)

A

Free expansion $= 0.9\text{ mm}$, (i) Stress $= 9 \times 10^7\text{ N/m}^2$, (ii) Stress $= 5 \times 10^7\text{ N/m}^2$

B

Free expansion $= 0.6\text{ mm}$, (i) Stress $= 6 \times 10^7\text{ N/m}^2$, (ii) Stress $= 3 \times 10^7\text{ N/m}^2$

C

Free expansion $= 0.9\text{ mm}$, (i) Stress $= 4.5 \times 10^7\text{ N/m}^2$, (ii) Stress $= 2.5 \times 10^7\text{ N/m}^2$

D

Free expansion $= 1.2\text{ mm}$, (i) Stress $= 9 \times 10^7\text{ N/m}^2$, (ii) Stress $= 4 \times 10^7\text{ N/m}^2$

Question 71: Palette 71/81

DPT MCQ

Concept: Thermal expansion increases the total length of the steel rail from $L$ to $L + \Delta L$, where $\Delta L = L \cdot \alpha \cdot \Delta T$.
When clamped at both ends, the deformed rail buckles, forming a triangle with two equal hypotenuses of length $\frac{L + \Delta L}{2}$ and base $\frac{L}{2}$.
Using Pythagoras' theorem, the central vertical displacement $x$ is:
$x = \sqrt{\left(\frac{L + \Delta L}{2}\right)^2 - \left(\frac{L}{2}\right)^2} \approx \sqrt{\frac{L \cdot \Delta L}{2}}$

A rail track made of steel having length $10\text{ m}$ is clamped on a railway line at its two ends. On a summer day due to a rise in temperature by $20^\circ\text{C}$, it is deformed as shown in the figure. Find $x$ (displacement of the centre in $\text{cm}$) if $\alpha_{\text{steel}} = 1.2 \times 10^{-5}\text{ /}^\circ\text{C}$. Round off to the nearest integer.

A

$11\text{ cm}$

B

$15\text{ cm}$

C

$8\text{ cm}$

D

$22\text{ cm}$

Question 72: Palette 72/81

DPT MCQ

Concept: According to the principle of calorimetry, in an isolated system, the heat lost by the hotter body equals the heat gained by the colder body. The heat transferred is given by $Q = m c \Delta T$, where $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the temperature change.

Three liquids $A$, $B$, and $C$ of equal mass are at temperatures $12\text{ °C}$, $19\text{ °C}$, and $28\text{ °C}$ respectively. When liquids $A$ and $B$ are mixed, the resultant temperature is $16\text{ °C}$. When liquids $B$ and $C$ are mixed, the resultant temperature is $23\text{ °C}$. Find the resulting temperature when liquids $A$ and $C$ are mixed.

A

$18.2\text{ °C}$

B

$20.3\text{ °C}$

C

$22.0\text{ °C}$

D

$24.5\text{ °C}$

Question 73: Palette 73/81

DPT MCQ

Concept: According to Stefan-Boltzmann Law, the net rate of heat loss or net power emitted by a body at temperature $T$ surrounded by an environment at temperature $T_0$ is given by:
$P = e \sigma A (T^4 - T_0^4)$
where:
$e$ is the emissivity of the body
$\sigma$ is Stefan's constant ($\approx 5.67 \times 10^{-8}\text{ W/m}^2\text{K}^4$)
$A$ is the surface area of the body
$T$ is the absolute temperature of the body in Kelvin
$T_0$ is the absolute temperature of the surroundings in Kelvin

A body of emissivity ($e = 0.75$), surface area of $300\text{ cm}^2$ and temperature $227^\circ\text{C}$ is kept in a room at temperature $27^\circ\text{C}$. Calculate the initial value of net power emitted by the body.

A

$34.7\text{ Watt}$

B

$52.0\text{ Watt}$

C

$69.4\text{ Watt}$

D

$138.8\text{ Watt}$

Question 74: Palette 74/81

DPT MCQ

Concept: 1. Rate of heat loss by radiation is given by $\frac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$, which is directly proportional to the surface area $A$.
2. Rate of cooling is given by $\frac{dT}{dt} = \frac{e \sigma A (T^4 - T_0^4)}{m s} = \frac{e \sigma A (T^4 - T_0^4)}{\rho V s}$, which is directly proportional to the ratio of surface area to volume $\left(\frac{A}{V}\right)$.

A sphere of radius $R$ and a cube of side length $R$ are made of the same material, kept at the same temperature, and placed in identical surroundings with the same surface finish. What are the ratios of their rate of heat loss and rate of cooling (Sphere : Cube), respectively?

A

$\frac{2\pi}{3}\text{ and }\frac{1}{2}$

B

$\frac{4\pi}{3}\text{ and }2$

C

$\frac{\pi}{3}\text{ and }\frac{1}{4}$

D

$\frac{2\pi}{3}\text{ and }2$

Question 75: Palette 75/81

DPT MCQ

Concept: According to Stefan-Boltzmann Law, the rate of heat radiation emitted per unit area by a body is directly proportional to the fourth power of its absolute temperature:
$\frac{dQ}{dt} = e A \sigma T^4$
Therefore, for a given body, the rate of heat radiation $E$ is proportional to $T^4$:
$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$
where $T$ represents the absolute temperature in Kelvin ($K = {^\circ\text{C}} + 273$).

If the rate of heat radiation from a body at temperature $273^\circ\text{C}$ is $E$, find the rate of heat radiation when the body is at $546^\circ\text{C}$.

A

$\frac{16}{81} E$

B

$\frac{81}{16} E$

C

$2 E$

D

$16 E$

Question 76: Palette 76/81

DPT MCQ

Concept: According to Wien's Displacement Law, the wavelength $\lambda_m$ corresponding to maximum energy emission by a black body is inversely proportional to its absolute temperature $T$:
$\lambda_m T = b$
where:
$\lambda_m$ is the peak wavelength
$T$ is the absolute temperature in Kelvin
$b$ is Wien's displacement constant ($2.9 \times 10^{-3}\text{ m K}$)

Light from the sun is found to have maximum intensity near a wavelength of $470\text{ nm}$. Assuming the surface of the sun acts as a black body, find the temperature of the sun. (Take Wien's constant $b = 2.9 \times 10^{-3}\text{ m K}$)

A

$5250\text{ K}$

B

$6170\text{ K}$

C

$7400\text{ K}$

D

$4800\text{ K}$

Question 77: Palette 77/81

DPT MCQ

Concept: According to Wien's Displacement Law, the wavelength $\lambda_m$ corresponding to maximum energy emission is inversely proportional to the absolute temperature $T$:
$\lambda_m \propto \frac{1}{T} \implies \lambda_m T = \text{Constant}$
Therefore, for two radiating bodies:
$\frac{(\lambda_m)_1}{(\lambda_m)_2} = \frac{T_2}{T_1}$

The maximum in the energy distribution spectrum of the sun is at $4753Å$ and its temperature is $6050\text{ K}$. What will be the temperature of a star whose energy distribution shows a maximum at $9506Å$?

A

$12100\text{ K}$

B

$3025\text{ K}$

C

$4753\text{ K}$

D

$6050\text{ K}$

Question 78: Palette 78/81

DPT MCQ

Concept: 1. Stefan-Boltzmann Law states that the total power emitted by a spherical blackbody star is given by:
$P = \sigma A T^4 = \sigma (4\pi R^2) T^4$
where $R$ is the radius and $T$ is the absolute temperature.
2. Wien's Displacement Law states that the peak wavelength $\lambda$ is inversely proportional to the absolute temperature $T$:
$\lambda T = \text{constant} \implies \frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A}$

Two spherical stars A and B emit blackbody radiation. The radius of A is $400$ times that of B and A emits $10^4$ times the power emitted from B. The ratio of their wavelengths $\frac{\lambda_A}{\lambda_B}$ at which the peaks occur in their respective radiation curves is:

A

$\frac{1}{2}$

B

$1$

C

$2$

D

$4$

Question 79: Palette 79/81

DPT MCQ

Concept: 1. Wien's Displacement Law:
The wavelength $\lambda_m$ corresponding to maximum intensity is inversely proportional to absolute temperature $T$:
$\lambda_m \propto \frac{1}{T} \implies \frac{T_2}{T_1} = \frac{\lambda_1}{\lambda_2}$
2. Stefan-Boltzmann Law:
The total energy radiated per unit area per second $E$ is directly proportional to the fourth power of absolute temperature $T$:
$E \propto T^4 \implies \frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$

A hot black body emits energy at the rate of $16\text{ J m}^{-2}\text{ s}^{-1}$ and its most intense radiation corresponds to $20,000Å$. When the temperature of this body is further increased such that its most intense radiation corresponds to $10,000Å$, find the value of energy radiated in $\text{J m}^{-2}\text{ s}^{-1}$.

A

$64\text{ J m}^{-2}\text{ s}^{-1}$

B

$128\text{ J m}^{-2}\text{ s}^{-1}$

C

$256\text{ J m}^{-2}\text{ s}^{-1}$

D

$512\text{ J m}^{-2}\text{ s}^{-1}$

Question 80: Palette 80/81

DPT MCQ

Concept: The area under the blackbody radiation spectrum curve ($E_\lambda$ vs $\lambda$) represents the total emissive power $E$ radiated by the black body.
According to Stefan-Boltzmann Law, total emissive power is directly proportional to the fourth power of absolute temperature:
$E \propto T^4 \implies A \propto T^4$
Therefore, the ratio of the areas under the curves at two temperatures $T_1$ and $T_2$ is:
$\frac{A_2}{A_1} = \frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4$

The spectra of a black body at temperatures $273^\circ\text{C}$ and $546^\circ\text{C}$ are shown in the figure. If $A_1$ and $A_2$ are the areas under the two curves corresponding to $273^\circ\text{C}$ and $546^\circ\text{C}$ respectively, the value of $\frac{A_2}{A_1}$ is:

A

$\frac{16}{81}$

B

$\frac{81}{16}$

C

$\frac{1}{16}$

D

$\frac{81}{8}$

Question 81: Palette 81/81

DPT MCQ

Concept: 1. Stefan-Boltzmann Law:
The rate of radiated energy is given by $\frac{dQ}{dt} = e \sigma A T^4$.
2. Wien's Displacement Law:
The peak wavelength is inversely proportional to temperature: $\lambda T = b = \text{constant}$.

Two bodies A and B have thermal emissivities of $0.01$ and $0.81$ respectively. The outer surface areas of the two bodies are the same. The two bodies radiate energy at the same rate. The wavelength $\lambda_B$, corresponding to the maximum spectral radiancy in the radiation from B, is shifted from the wavelength corresponding to the maximum spectral radiancy in the radiation from A by $1.00\text{ }\mu\text{m}$. If the temperature of A is $5802\text{ K}$, calculate the wavelength $\lambda_B$.

A

$1.50\text{ }\mu\text{m}$

B

$0.50\text{ }\mu\text{m}$

C

$2.00\text{ }\mu\text{m}$

D

$3.00\text{ }\mu\text{m}$