Heat Transfer and Radiation PYQs: MHT CET, JEE Main, NEET

Q1MHT CET 2026 · 13 APR · SHIFT 2Easy

Heat Conduction · Thermal conductivity & rate of heat flow

The coefficient of thermal conductivity of a metal rod depends on its

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Answer: (c)

In $\frac{Q}{t} = \frac{KA\Delta T}{L}$ , $K$ is a property of the material only.
Area, mass and length change the heat flow, not $K$ .
So the answer is (c) material of the rod.

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Q2MHT CET 2026 · 13 MAY · SHIFT 1Easy

Heat Conduction · Composite rods, slabs & junction temperature

Two rods A and B are of equal lengths. Their ends are kept between the same temperatures and their area cross-sections are $A_1$ and $A_2$ , thermal conductivities are $K_1$ and $K_2$ . The rate of heat transmission in the two rods will be equal if

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Answer: (b)

$H = \frac{KA\Delta T}{L}$ with the same $L$ and $\Delta T$ for both rods.
$H_1 = H_2 \Rightarrow K_1A_1 = K_2A_2$
So the answer is (b).

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Q3MHT CET 2026 · 18 APR · SHIFT 1Easy

Heat Conduction · Composite rods, slabs & junction temperature

The ratio of thermal conductivity of two rods of different materials is $4 : 3$ . The two rods have same area of cross-section and same thermal resistance. The lengths of rods are in the ratio

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Answer: (b)

Thermal resistance $R = \frac{L}{KA}$ ; same $R$ and $A$ gives $L \propto K$ .
$\frac{L_1}{L_2} = \frac{K_1}{K_2} = \frac{4}{3}$
So the answer is (b) $4:3$ .

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Q4MHT CET 2026 · 15 APR · SHIFT 2Easy

Heat Conduction · Composite rods, slabs & junction temperature

The ratio of the thermal conductivity of two rods of different materials is 6:5. The two rods of same area of cross section and same thermal resistance will have the length in the ratio

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Answer: (b)

Thermal resistance $R = \frac{L}{KA}$ ; same $R$ and $A$ gives $L \propto K$ .
$\frac{L_1}{L_2} = \frac{K_1}{K_2} = \frac{6}{5}$
So the answer is (b) $6:5$ .

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Q5MHT CET 2026 · 16 APR · SHIFT 1Medium

Heat Conduction · Thermal conductivity & rate of heat flow

Rate of flow of heat through a cylindrical rod is ' $H_1$ '. The temperature at the ends of the rod are $T_1$ and $T_2$ . If all the dimensions of the rod become double and the temperature difference remains the same and if the rate of flow of heat becomes ' $H_2$ ' then $H_2 =$

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Answer: (c)

$H = \frac{KA\Delta T}{L}$ , with $A = \pi r^2$ .
Doubling all dimensions: $A \to 4A$ , $L \to 2L$ .
$H_2 = \frac{4}{2}H_1 = 2H_1$ , so the answer is (c).

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Q6MHT CET 2026 · 12 MAY · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

Three rods made of the same material and the same cross-sectional areas have been joined as shown in the figure. Each rod is of the same length. The temperature of the junction is

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Answer: (a)

All rods are identical, so each has the same resistance. Heat in from the two hot rods equals heat out through the cold rod.
$2(90-\theta) = \theta - 0$
$3\theta = 180 \Rightarrow \theta = 60^\circ\text{C}$ , so the answer is (a).

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Q7MHT CET 2026 · 20 APR · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

Two rods of same length, radius and material transfer a given amount of heat in 't' second when they are joined as shown in fig (1). But when they are joined as shown in fig (2) then they will transfer same heat in same condition in time

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Answer: (c)

Let each rod have resistance $R$ . Fig (1) is parallel: $R_1 = \frac{R}{2}$ . Fig (2) is series: $R_2 = 2R$ .
For the same heat and $\Delta T$ , time $t \propto$ thermal resistance.
$t_2 = \frac{2R}{R/2}\,t = 4t$ , so the answer is (c).

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Q8MHT CET 2026 · 17 APR · SHIFT 2Medium

Heat Conduction · Composite rods, slabs & junction temperature

In a composite slab there are two materials having coefficients of thermal conductivity K and 2K, thickness x and 4x repectively. The temperature of the two outer surfaces of a composite slab are $T_2$ and $T_1\ (T_2 > T_1).\ T_2$ is on side K and $T_1$ is on side 2K. The rate of heat transfer through the slab in a steady state is $\left[\frac{A(T_2-T_1)K}{x}\right] \cdot f$ , where f is equal to

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Answer: (d)

Series resistances: $R = \frac{x}{KA} + \frac{4x}{2KA} = \frac{3x}{KA}$
$H = \frac{T_2-T_1}{R} = \frac{A(T_2-T_1)K}{x}\cdot\frac{1}{3}$
So $f = \frac{1}{3}$ , answer (d).

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Q9MHT CET 2026 · 17 APR · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

Consider two rods 1 and 2 of same length. They have different specific heats $(C_1, C_2)$ , thermal conductivities $(K_1, K_2)$ and area of cross-section $(A_1, A_2)$ respectively. Both the rods have temperatures $(T_1, T_2)$ at their ends. If their rate of loss of heat due to conduction is equal, then

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Answer: (b)

Conduction rate $H = \frac{KA\Delta T}{L}$ does not involve specific heat.
Same $L$ and $\Delta T$ : $H_1 = H_2 \Rightarrow A_1K_1 = A_2K_2$
So the answer is (b).

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Q10JEE Main 2026 · 22 JAN · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

Rods $x$ and $y$ of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $A$ and $F$ are maintained at $100^\circ\text{C}$ and $40^\circ\text{C}$ respectively. Given the thermal conductivity of $\text{rod } x$ is three times of that of $\text{rod } y$ , the temperature at junction points $B$ and $E$ are (close to):

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Answer: (c)

$K_x = 3K_y$ , so take resistance of an $x$ rod $= r$ and a $y$ rod $= 3r$ .
B to E: path BCE $= 6r$ in parallel with path BDE $= 2r$ , giving $\frac{6 \times 2}{8}r = 1.5r$ .
Total $= r + 1.5r + 3r = 5.5r$ ; current $H = \frac{60}{5.5r} \approx \frac{10.9}{r}$
$T_B = 100 - 10.9 \approx 89^\circ\text{C}$ , $T_E = 40 + 3(10.9) \approx 73^\circ\text{C}$
So the answer is (c).

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Q11MHT CET 2026 · 11 APR · SHIFT 1Difficult

Heat Conduction · Composite rods, slabs & junction temperature

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the side have thermal conductivity $2K$ while that in the middle has thermal conductivity $K$ . The left end of the combination is maintained at temperature $3T$ and the right end at $T$ . The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_1$ and that at the right junctions is $T_2$ . The ratio $T_1/T_2$ is

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Answer: (c)

Resistance of a $2K$ rod $= R$ , of the $K$ rod $= 2R$ ; total $= 4R$ for a drop of $3T - T = 2T$ .
Drop per $R$ $= \dfrac{2T}{4} = \dfrac{T}{2}$
$T_1 = 3T - \dfrac{T}{2} = \dfrac{5T}{2}$ , $T_2 = T + \dfrac{T}{2} = \dfrac{3T}{2}$
So the answer is (c) $\dfrac{T_1}{T_2} = \dfrac{5}{3}$ .

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Q12MHT CET 2025 · 22 APR · SHIFT 2Easy

Heat Conduction · Thermal conductivity & rate of heat flow

The two ends of a rod of length ' $x$ ' and uniform cross-sectional area ' A ' are kept at temperatures ' $T_1$ ' and ' $T_2$ ' respectively ( $T_1 > T_2$ ). If the rate of heat transfer is ' $Q/t$ ', through the rod in steady state, then the coefficient of thermal conductivity ' K ' is

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Answer: (b)

Fourier's law: $\frac{Q}{t} = \frac{KA(T_1-T_2)}{x}$
Rearranging: $K = \frac{xQ}{tA(T_1-T_2)}$
So the answer is (b).

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Q13JEE Main 2025 · 29 JAN · SHIFT 2Easy

Heat Conduction · Thermal conductivity & rate of heat flow

The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have

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Answer: (a)

A thermoelectric material must keep a temperature difference across it, so it needs low thermal conductivity.
It must carry the generated current easily, so it needs high electrical conductivity.
So the answer is (a).

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Q14MHT CET 2025 · 19 APR · SHIFT 2Medium

Heat Conduction · Composite rods, slabs & junction temperature

A composite slab consists of two materials having coefficients of thermal conductivity K and $2K$ , thickness $x$ and $4x$ respectively. The temperatures of two outer surfaces of a composite slab are $T_2$ and $T_1$ respectively $(T_2 > T_1)$ . The rate of heat transfer through the slab in a steady state is $\left[\frac{A(T_2-T_1)K}{x}\right] f$ , where $f$ is equal to

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Answer: (d)

Series resistances: $R = \frac{x}{KA} + \frac{4x}{2KA} = \frac{3x}{KA}$
$H = \frac{T_2-T_1}{R} = \frac{A(T_2-T_1)K}{x}\cdot\frac{1}{3}$
So $f = \frac{1}{3}$ , answer (d).

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Q15NEET 2025Medium

Heat Conduction · Composite rods, slabs & junction temperature

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity $2K$ while that in the middle has thermal conductivity $K$ . The left end of the combination is maintained at temperature $3T$ and the right end at $T$ . The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_1$ and that at the right junction is $T_2$ . The ratio $T_1/T_2$ is

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Answer: (a)

$R = \frac{L}{KA}$ : the $2K$ rods have resistance $r$ , the middle $K$ rod has $2r$ ; total $4r$ .
Total drop $3T - T = 2T$ , so the drop per $r$ is $\frac{T}{2}$ .
$T_1 = 3T - \frac{T}{2} = \frac{5T}{2}$ , $T_2 = \frac{5T}{2} - T = \frac{3T}{2}$
$\frac{T_1}{T_2} = \frac{5}{3}$ , so the answer is (a).

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Q16JEE Main 2025 · 7 APR · SHIFT 2Medium

Heat Conduction · Composite rods, slabs & junction temperature

Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratios of lengths, radii and thermal conductivities of these rods are: $\frac{L_A}{L_B} = \frac{1}{2},\ \frac{r_A}{r_B} = 2$ and $\frac{K_A}{K_B} = \frac{1}{2}$ . The free ends of rods A and B are maintained at 400 K, 200 K, respectively. The temperature of rods interface is ______________ $K$ , when equilibrium is established.

Numerical answer type. Work it out, then open the answer.

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Answer: 360

Thermal resistance $R = \dfrac{L}{K\pi r^2}$ , so $\dfrac{R_A}{R_B} = \dfrac{L_A}{L_B}\cdot\dfrac{K_B}{K_A}\cdot\dfrac{r_B^2}{r_A^2} = \dfrac{1}{2}\cdot 2\cdot\dfrac{1}{4} = \dfrac{1}{4}$
In series the temperature drop divides in ratio of resistances: drop across A $= 200\times\dfrac{1}{5} = 40$ K
Interface temperature $= 400 - 40 = 360$ K

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Q17JEE Main 2025 · 22 JAN · SHIFT 1Difficult

Heat Conduction · Thermal conductivity & rate of heat flow

Three conductors of same length having thermal conductivity $k_1, k_2$ and $k_3$ are connected as shown in figure.

Area of cross sections of $1$ st and $2$ nd conductor are same and for $3$ rd conductor it is double of the $1$ st conductor. The temperatures are given in the figure. In steady state condition, the value of $\theta$ is _________ $^\circ\text{C}$ . (Given: $k_1 = 60\text{ J s}^{-1}\text{ m}^{-1}\text{ K}^{-1},\ k_2 = 120\text{ J s}^{-1}\text{ m}^{-1}\text{ K}^{-1},\ k_3 = 135\text{ J s}^{-1}\text{ m}^{-1}\text{ K}^{-1}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 40

Rods 1 and 2 are in parallel between $100^\circ$ C and $\theta$ ; rod 3 (area $2A$ ) carries the same total heat from $\theta$ to $0^\circ$ C.
$\dfrac{(k_1+k_2)A(100-\theta)}{l} = \dfrac{k_3(2A)\theta}{l}$
$180(100-\theta) = 270\theta \Rightarrow 450\theta = 18000$
$\theta = 40^\circ$ C

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Q18NEET 2024Easy

Heat Conduction · Thermal conductivity & rate of heat flow

Given below are two statements: One is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$ .

Assertion A: Houses made of concrete roofs overlaid with foam keep the room hotter during summer.

Reason R: The layer of foam insulation prohibits heat transfer, as it contains air pockets. In the light of the above statements, choose the correct answer from the options given below.

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Answer: (b)

Foam with trapped air is a poor conductor, so it blocks heat flow: R is true.
Blocking heat entering from the hot roof keeps the room cooler in summer, not hotter: A is false.
So the answer is (b) A is false but R is true.

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Q19MHT CET 2024 · 15 MAY · SHIFT 2Medium

Heat Conduction · Thermal conductivity & rate of heat flow

A cylindrical rod is having temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $Q\text{ J/s}$ . All the linear dimensions of the rod are doubled by keeping the temperature constant. The new rate of flow of heat is

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Answer: (b)

$Q = \frac{KA\Delta\theta}{L}$ , with $A = \pi r^2$ .
Doubling all linear dimensions: $A \to 4A$ , $L \to 2L$ .
New rate $= \frac{4}{2}Q = 2Q$ , so the answer is (b).

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Q20MHT CET 2024 · 2 MAY · SHIFT 2Medium

Heat Conduction · Thermal conductivity & rate of heat flow

Rate of flow of heat through a cylindrical rod is ' $H_1$ '. The temperature at the ends of the rod are ' $T_1$ ' and ' $T_2$ '. If all the dimensions of the rod become double and the temperature difference remains the same, the rate of flow of heat becomes ' $H_2$ '. Then

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Answer: (b)

$H = \frac{KA\Delta T}{L}$ , with $A = \pi r^2$ .
Doubling all dimensions: $A \to 4A$ , $L \to 2L$ .
$H_2 = 2H_1$ , so the answer is (b).

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Q21MHT CET 2024 · 4 MAY · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

Two rods of same length \& material transfer a given amount of heat in 12 s when they are joined end to end. But when they are joined length wise parallel to each other they will transfer same amount of heat in same condition in time

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Answer: (b)

Each rod has resistance $R$ . End to end (series): $2R$ . Side by side (parallel): $\frac{R}{2}$ .
Time $\propto$ resistance, so the time drops by a factor of 4.
$t = \frac{12}{4} = 3\ \text{s}$ , so the answer is (b).

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Q22MHT CET 2023 · 13 MAY · SHIFT 2Easy

Heat Conduction · Thermal conductivity & rate of heat flow

Which one of the following is based on convection?

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Answer: (b)

A room heater warms air near it; the hot air rises and cool air sinks, setting up convection currents.
Utensil and iron rod heat by conduction; Sun to Earth is radiation.
So the answer is (b) heating a room by heater.

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Q23MHT CET 2023 · 13 MAY · SHIFT 1Easy

Heat Conduction · Thermal conductivity & rate of heat flow

The temperature gradient in a rod of length $75\text{ cm}$ is $40^\circ\text{C/m}$ . If the temperature of cooler end of the rod is $10^\circ\text{C}$ , then the temperature of hotter end is

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Answer: (b)

Temperature difference $= \text{gradient} \times \text{length} = 40 \times 0.75 = 30^\circ\text{C}$
Hot end $= 10 + 30 = 40^\circ\text{C}$
So the answer is (b) $40^\circ\text{C}$ .

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Q24MHT CET 2023 · 11 MAY · SHIFT 1Easy

Heat Conduction · Thermal conductivity & rate of heat flow

The rate of flow of heat through a metal rod with temperature difference $40^\circ\text{C}$ is $1600\text{ cal/s}$ . The thermal resistance of metal rod in $^\circ\text{C s/cal}$ is

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Answer: (a)

Thermal resistance $R = \frac{\Delta T}{H}$
$R = \frac{40}{1600} = 0.025\ ^\circ\text{C s/cal}$
So the answer is (a) 0.025.

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Q25MHT CET 2023 · 12 MAY · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

A composite slab consists of two materials having coefficient of thermal conductivity $K$ and $2K$ , thickness $x$ and $4x$ respectively. The temperature of the two outer surfaces of a composite slab are $T_2$ and $T_1\ (T_2 > T_1)$ . The rate of heat transfer through the slab in a steady state is $\left[\frac{A(T_2-T_1)K}{x}\right] \cdot f$ where ' $f$ ' is equal to

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Answer: (d)

Series resistances: $R = \frac{x}{KA} + \frac{4x}{2KA} = \frac{3x}{KA}$
$H = \frac{T_2-T_1}{R} = \frac{A(T_2-T_1)K}{x}\cdot\frac{1}{3}$
So $f = \frac{1}{3}$ , answer (d).

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Q26JEE Main 2023 · 13 APR · SHIFT 2Medium

Heat Conduction · Composite rods, slabs & junction temperature

Two plates $\text{A}$ and $\text{B}$ have thermal conductivities $84\text{ W m}^{-1}\text{ K}^{-1}$ and $126\text{ W m}^{-1}\text{ K}^{-1}$ respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of $\text{A}$ and $\text{B}$ are kept at $100^\circ\text{C}$ and $0^\circ\text{C}$ respectively, then the temperature of the surface of contact in steady state is _____________ $^\circ\text{C}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 40

Same area and thickness: $R \propto \dfrac{1}{K}$ , so $\dfrac{R_A}{R_B} = \dfrac{126}{84} = \dfrac{3}{2}$
Drop across A $= 100\times\dfrac{3}{5} = 60^\circ$ C
Contact temperature $= 100 - 60 = 40^\circ$ C

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Q27JEE Main 2022 · 24 JUN · SHIFT 1Medium

Heat Conduction · Thermal conductivity & rate of heat flow

Two metallic blocks M1 and M2 of same area of cross-section are connected to each other (as shown in figure). If the thermal conductivity of M2 is K then the thermal conductivity of M1 will be: [Assume steady state heat conduction]

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Answer: (b)

Same heat current through both blocks: $\frac{K_1A(100-80)}{16} = \frac{KA(80-0)}{8}$
$\frac{20K_1}{16} = 10K$
$K_1 = 8K$ , so the answer is (b).

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Q28NEET 2022 · PHASE 2Medium

Heat Conduction · Composite rods, slabs & junction temperature

Two rods one made of copper and other made of steel of same length and same cross sectional area are joined together. The thermal conductivity of copper and steel are 385 J s ${}^{-1}$ K ${}^{-1}$ m ${}^{-1}$ and 50 J s ${}^{-1}$ K ${}^{-1}$ m ${}^{-1}$ respectively. The free ends of copper and steel are held at 100 $^\circ$ C and 0 $^\circ$ C respectively. The temperature at the junction is, nearly:

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Answer: (a)

Same heat current in both rods (same $L$ , $A$ ): $385(100-\theta) = 50(\theta - 0)$
$435\theta = 38500$
$\theta \approx 88.5^\circ\text{C}$ , so the answer is (a).

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Q29JEE Main 2022 · 27 JUL · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

If $K_1$ and $K_2$ are the thermal conductivities, $L_1$ and $L_2$ are the lengths and $A_1$ and $A_2$ are the cross sectional areas of steel and copper rods respectively such that $\frac{K_2}{K_1} = 9,\ \frac{A_1}{A_2} = 2,\ \frac{L_1}{L_2} = 2$ . Then, for the arrangement as shown in the figure, the value of temperature $T$ of the steel - copper junction in the steady state will be:

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Answer: (c)

Steel: $R_s = \frac{L_1}{K_1A_1}$ . Copper: $R_c = \frac{L_2}{K_2A_2} = \frac{L_1/2}{9K_1 \cdot A_1/2} = \frac{R_s}{9}$
Same heat current: $\frac{450 - T}{R_s} = \frac{T - 0}{R_s/9}$
$450 - T = 9T \Rightarrow T = 45^\circ\text{C}$ , so the answer is (c).

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Q30JEE Main 2022 · 29 JUN · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 120 cm2 for each plate. The equivalent thermal conductivity of the compound plate is $\left(1 + \frac{5}{\alpha}\right)$ K, then the value of $\alpha$ will be ______________.

Numerical answer type. Work it out, then open the answer.

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Answer: 21

Plates in series: $\dfrac{L_1+L_2}{K_{eq}} = \dfrac{L_1}{K_1} + \dfrac{L_2}{K_2}$
$\dfrac{6.5}{K_{eq}} = \dfrac{4}{K} + \dfrac{2.5}{2K} = \dfrac{5.25}{K} \Rightarrow K_{eq} = \dfrac{6.5}{5.25}K = \dfrac{26}{21}K$
$\dfrac{26}{21} = 1 + \dfrac{5}{21}$
So $\alpha = 21$

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Q31JEE Main 2022 · 26 JUL · SHIFT 2Difficult

Heat Conduction · Thermal conductivity & rate of heat flow

An ice cube of dimensions $60\text{ cm} \times 50\text{ cm} \times 20\text{ cm}$ is placed in an insulation box of wall thickness $1\text{ cm}$ . The box keeping the ice cube at $0^\circ\text{C}$ of temperature is brought to a room of temperature $40^\circ\text{C}$ . The rate of melting of ice is approximately:

(Latent heat of fusion of ice is $3.4 \times 10^{5}\text{ J kg}^{-1}$ and thermal conducting of insulation wall is $0.05\text{ W m}^{-1}\,{}^\circ\text{C}^{-1}$ )

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Answer: (b)

Wall area $= 2(0.6\times0.5 + 0.5\times0.2 + 0.6\times0.2) = 1.04\ \text{m}^2$
$H = \dfrac{KA\Delta T}{d} = \dfrac{0.05\times1.04\times40}{0.01} = 208$ W
Rate of melting $= \dfrac{H}{L} = \dfrac{208}{3.4\times10^{5}} \approx 6.1\times10^{-4}$ kg/s
So the answer is (b) $61\times10^{-5}$ kg s $^{-1}$ .

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Q32MHT CET 2021 · 24 SEP · SHIFT 1Easy

Heat Conduction · Thermal conductivity & rate of heat flow

The temperature difference between two sides of an iron plate, $1.8\text{ cm}$ thick is $9^\circ\text{C}$ . Heat is transmitted through the plate $10\text{ kcal/s m}^2$ at steady state. The thermal conductivity of iron is

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Answer: (a)

Heat flux $\frac{H}{A} = \frac{K\Delta T}{d} \Rightarrow K = \frac{(H/A)\,d}{\Delta T}$
$K = \frac{10 \times 0.018}{9} = 0.02\ \text{kcal m}^{-1}\text{s}^{-1}\,{}^\circ\text{C}^{-1}$
So the answer is (a).

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Q33MHT CET 2021 · 22 SEP · SHIFT 2Easy

Heat Conduction · Thermal conductivity & rate of heat flow

The rate of flow of heat through a copper rod with temperature difference $28^\circ\text{C}$ is $1400\text{ cal s}^{-1}$ . The thermal resistance of copper rod will be

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Answer: (b)

Thermal resistance $R = \frac{\Delta T}{H}$
$R = \frac{28}{1400} = 0.02\ ^\circ\text{C s/cal}$
So the answer is (b).

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Q34MHT CET 2021 · 23 SEP · SHIFT 2Medium

Heat Conduction · Thermal conductivity & rate of heat flow

A cylindrical rod has temperatures ' $T_1$ ' and ' $T_2$ ' at its ends. The rate of flow of heat is ' $Q_1$ ' cal $\text{s}^{-1}$ . If length and radius of the rod are doubled keeping temperature constant, then the rate of flow of heat ' $Q_2$ ' will be

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Answer: (d)

$Q = \frac{K\pi r^2\Delta T}{L} \Rightarrow Q \propto \frac{r^2}{L}$
$\frac{Q_2}{Q_1} = \frac{(2)^2}{2} = 2$
So the answer is (d) $Q_2 = 2Q_1$ .

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Q35MHT CET 2021 · 22 SEP · SHIFT 1Medium

Heat Conduction · Thermal conductivity & rate of heat flow

A cylindrical rod is having temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is ' $Q$ ' $\text{J s}^{-1}$ . All the linear dimensions of the rod are doubled by keeping the temperatures constant. What is the new rate of flow of heat?

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Answer: (c)

$Q = \frac{K\pi r^2\Delta\theta}{L} \Rightarrow Q \propto \frac{r^2}{L}$
Doubling $r$ and $L$ : factor $\frac{4}{2} = 2$
New rate $= 2Q$ , so the answer is (c).

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Q36MHT CET 2021 · 21 SEP · SHIFT 1Medium

Heat Conduction · Thermal conductivity & rate of heat flow

The temperature difference bewtween two sides of metal plate, $3\text{ cm}$ thick is $15^\circ\text{C}$ . Heat is transmitted through plate at the rate of $900\text{ kcal}$ per minute per $\text{m}^2$ at steady state. The thermal conductivity of metal is

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Answer: (c)

Heat flux $= \frac{900}{60} = 15\ \text{kcal s}^{-1}\text{m}^{-2}$
$K = \frac{(H/A)\,d}{\Delta T} = \frac{15 \times 0.03}{15}$
$K = 3 \times 10^{-2}\ \text{kcal m}^{-1}\text{s}^{-1}\,{}^\circ\text{C}^{-1}$ , so the answer is (c).

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Q37MHT CET 2021 · 21 SEP · SHIFT 2Medium

Heat Conduction · Composite rods, slabs & junction temperature

Two rods of same length and material are joined end to end. They transfer heat in 8 second. When they are joined in parallel they transfer same amount of heat in same conditions in time

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Answer: (b)

Each rod has resistance $R$ . Series: $2R$ ; parallel: $\frac{R}{2}$ .
Time $\propto$ resistance, so the time drops by a factor of 4.
$t = \frac{8}{4} = 2\ \text{s}$ , so the answer is (b).

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Q38JEE Main 2021 · 17 MAR · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

Two identical metal wires of thermal conductivities K1 and K2 respectively are connected in series. The effective thermal conductivity of the combination is:

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Answer: (a)

Identical wires: same $L$ and $A$ . In series, resistances add: $\frac{2L}{K_{eq}A} = \frac{L}{K_1A} + \frac{L}{K_2A}$
$\frac{2}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2}$
$K_{eq} = \frac{2K_1K_2}{K_1+K_2}$ , so the answer is (a).

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Q39JEE Main 2021 · 26 FEB · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

The temperature $\theta$ at the junction of two insulating sheets, having thermal resistances R1 and R2 as well as top and bottom temperatures $\theta$ 1 and $\theta$ 2 (as shown in figure) is given by:

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Answer: (a)

Same heat current through both sheets: $\frac{\theta_2 - \theta}{R_2} = \frac{\theta - \theta_1}{R_1}$
$\theta_2R_1 - \theta R_1 = \theta R_2 - \theta_1R_2$
$\theta = \frac{\theta_1R_2 + \theta_2R_1}{R_1+R_2}$ , so the answer is (a).

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Q40JEE Main 2021 · 27 AUG · SHIFT 1Medium

Heat Conduction · Composite rods, slabs & junction temperature

A rod CD of thermal resistance 10.0 KW ${}^{-1}$ is joined at the middle of an identical rod AB as shown in figure. The end A, B and D are maintained at 200 $^\circ$ C, 100 $^\circ$ C and 125 $^\circ$ C respectively. The heat current in CD is P watt. The value of P is ..................

Numerical answer type. Work it out, then open the answer.

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Answer: 2

C is the middle of AB, so $R_{AC} = R_{CB} = 5$ K/W and $R_{CD} = 10$ K/W.
Heat balance at C: $\dfrac{200-T}{5} + \dfrac{100-T}{5} + \dfrac{125-T}{10} = 0$
$400 - 2T + 200 - 2T + 125 - T = 0 \Rightarrow T = 145^\circ$ C
$P = \dfrac{145-125}{10} = 2$ W

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Q41MHT CET 2021 · 20 SEP · SHIFT 2Difficult

Heat Conduction · Thermal conductivity & rate of heat flow

A conducting rod of length $1\text{ m}$ has area of cross-section $10^{-3}\text{ m}^2$ . One end is immersed in boiling water $(100^\circ\text{C})$ and the other end in Ice $(0^\circ\text{C})$ . If coefficient of thermal conductivity of $\text{rod}$ is $96\text{ cal/s m}^\circ\text{C}$ and latent heat for ice is $8 \times 10^{4}\text{ cal/kg}$ then the amount of ice which will melt in one minute is

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Answer: (b)

Rate of heat flow $H = \dfrac{KA\Delta T}{L} = \dfrac{96\times10^{-3}\times100}{1} = 9.6$ cal/s
Heat in 60 s: $Q = 9.6\times60 = 576$ cal
$m = \dfrac{Q}{L_f} = \dfrac{576}{8\times10^{4}} = 7.2\times10^{-3}$ kg
So the answer is (b) $7.2\times10^{-3}$ kg.

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Q42JEE Main 2021 · 31 AUG · SHIFT 2Difficult

Heat Conduction · Composite rods, slabs & junction temperature

Two thin metallic spherical shells of radii r1 and r2 (r1 < r2) 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 :-

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Answer: (a)

Thin spherical shell: $H = K(4\pi r^2)\dfrac{d\theta}{dr}$ , constant through all shells.
Integrate from $r_1$ to $r_2$ : $H\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right) = 4\pi K(\theta_2-\theta_1)$
$H = \dfrac{4\pi K r_1 r_2(\theta_2-\theta_1)}{r_2-r_1}$
So the answer is (a).

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Q43JEE Main 2020 · 6 SEP · SHIFT 2Medium

Heat Conduction · Composite rods, slabs & junction temperature

Three rods of identical cross-section and lengths are made of three different materials of thermal conductivity K1, K2 and K3, respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at 100oC and the other at 0oC (see figure). If the joints of the rod are at 70oC and 20oC in steady state and there is no loss of energy from the surface of the rod, the correct relationship between K1, K2 and K3 is:

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Answer: (a)

Same $L$ , $A$ and heat current: $K_1(100-70) = K_2(70-20) = K_3(20-0)$
$30K_1 = 50K_2 = 20K_3$
$\frac{K_1}{K_3} = \frac{20}{30} = \frac{2}{3}$ , so the answer is (a) $K_1:K_3 = 2:3$ .

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Q44JEE Main 2019 · 10 JAN · SHIFT 1Easy

Heat Conduction · Composite rods, slabs & junction temperature

A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 WK–1m–1, the energy flux through it in the steady state is -

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Answer: (d)

Temperatures are $10^3\ \text{K}$ and $10^2\ \text{K}$ , so $\Delta T = 900\ \text{K}$ .
Energy flux $\frac{H}{A} = \frac{K\Delta T}{d} = \frac{0.1 \times 900}{1}$
$= 90\ \text{W m}^{-2}$ , so the answer is (d).

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Q45JEE Main 2019 · 9 APR · SHIFT 2Medium

Heat Conduction · Composite rods, slabs & junction temperature

Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness 'd' and '3d', respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ' $\theta$ 2' and ' $\theta$ 1' respectively, ( $\theta$ 2 > $\theta$ 1). The temperature at the interface is :-

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Answer: (a)

$R = \frac{L}{KA}$ : $R_1 = \frac{d}{3KA}$ , $R_2 = \frac{3d}{KA}$ , ratio $1:9$ .
The drop across the $3K$ layer is $\frac{1}{10}(\theta_2 - \theta_1)$ .
$\theta = \theta_2 - \frac{\theta_2-\theta_1}{10} = \frac{\theta_1}{10} + \frac{9\theta_2}{10}$
So the answer is (a).

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Q46NEET 2019Difficult

Heat Conduction · Thermal conductivity & rate of heat flow

A deep rectangular pond of surface area $A$ , containing water (density = $\rho$ ), specific heat capacity = $s$ ), is located in a region where the outside air temperature is at a steady value of $-26^\circ\text{C}$ . The thickness of the frozen ice layer in this pond, at a certain instant is $x$ .

Taking the thermal conductivity of ice as $K$ , and its specific latent heat of fusion as $L$ , the rate of increase of the thickness of ice layer, at this instant, would be given by

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Answer: (d)

Water below the ice is at $0^\circ$ C, top of ice at $-26^\circ$ C; heat conducted through the ice layer: $\dfrac{KA(26)}{x}$
This heat comes from freezing a layer $dx$ : $\rho A\,dx\,L = \dfrac{26KA}{x}dt$
$\dfrac{dx}{dt} = \dfrac{26K}{\rho x L}$
So the answer is (d).

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Q47JEE Main 2019 · 9 JAN · SHIFT 1Difficult

Heat Conduction · Thermal conductivity & rate of heat flow

Temperature difference of 120oC is maintained between ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length $\frac{3L}{2},$ is connected across AB (see figure). In steady state, temperature difference between P and Q will be close to:

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Answer: (a)

Same material and area, so resistance $\propto$ length. AP $= QB = L/2$ ; between P and Q: straight part $L$ in parallel with bent rod $3L/2$ .
$R_{PQ} = \dfrac{L\cdot\frac{3L}{2}}{L+\frac{3L}{2}} = \dfrac{3L}{5}$ (in units of $\frac{1}{KA}$ )
Total $= \dfrac{L}{2} + \dfrac{3L}{5} + \dfrac{L}{2} = \dfrac{8L}{5}$
$\Delta T_{PQ} = 120\times\dfrac{3/5}{8/5} = 45^\circ$ C
So the answer is (a) $45^\circ$ C.

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Q48JEE Main 2019 · 12 JAN · SHIFT 1Difficult

Heat Conduction · Composite rods, slabs & junction temperature

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 K1 and the of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is:

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Answer: (b)

Heat flows along the length: the two parts are in parallel, $K_{eq}A_{total} = K_1A_1 + K_2A_2$
$A_1 = \pi R^2$ , $A_2 = \pi(4R^2 - R^2) = 3\pi R^2$ , $A_{total} = 4\pi R^2$
$K_{eq}(4\pi R^2) = K_1\pi R^2 + 3K_2\pi R^2$
So the answer is (b) $\dfrac{K_1+3K_2}{4}$ .

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Q49NEET 2017Medium

Heat Conduction · Composite rods, slabs & junction temperature

Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are K1 and K2. The thermal conductivity of the composite will be

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Answer: (d)

The rods are side by side between $T_1$ and $T_2$ , so they are in parallel; each has area $A$ , total $2A$ .
$\frac{K(2A)\Delta T}{d} = \frac{K_1A\Delta T}{d} + \frac{K_2A\Delta T}{d}$
$K = \frac{K_1+K_2}{2}$ , so the answer is (d).

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Q50JEE Main 2014Difficult

Heat Conduction · Composite rods, slabs & junction temperature

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 $100^\circ\text{C}$ where as ends of brass and steel are kept at $0^\circ\text{C}$ . Lengths of the copper, brass and steel rods are $46,\ 13$ and $12\text{ cms}$ respectively. The rods are thermally insulated from surroundings excepts 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:

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Answer: (c)

Let junction temperature be $T$ . Heat in through copper = heat out through brass + steel (same area).
$\dfrac{0.92(100-T)}{46} = \dfrac{0.26T}{13} + \dfrac{0.12T}{12} \Rightarrow 0.02(100-T) = 0.03T$
$T = 40^\circ$ C
$H_{Cu} = \dfrac{0.92\times4\times60}{46} = 4.8$ cal/s
So the answer is (c) 4.8 cal/s.

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Q51JEE Main 2009 · AIEEEEasy

Heat Conduction · Thermal conductivity & rate of heat flow

A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $\theta$ along the length $x$ of the bar from its hot end is best described by which of the following figures?

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Answer: (a)

In steady state with no side loss, $H = -KA\frac{d\theta}{dx}$ is the same at every section.
So $\frac{d\theta}{dx}$ is constant: $\theta$ falls linearly with $x$ .
So the answer is (a), a straight line sloping down.

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Q52JEE Main 2007 · AIEEEMedium

Heat Conduction · Composite rods, slabs & junction temperature

One end of a thermally insulated rod is kept at a temperature $T_1$ and the other at $T_2$ . The rod is composed of two sections of length $L_1$ and $L_2$ and thermal conductivities $K_1$ and $K_2$ respectively. The temperature at the interface of the two section is

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Answer: (d)

Same heat current: $\frac{K_1A(T_1 - T)}{L_1} = \frac{K_2A(T - T_2)}{L_2}$
$K_1L_2(T_1 - T) = K_2L_1(T - T_2)$
$T = \frac{K_1L_2T_1 + K_2L_1T_2}{K_1L_2 + K_2L_1}$ , so the answer is (d).

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Q53JEE Main 2005 · AIEEEMedium

Heat Conduction · Composite rods, slabs & junction temperature

The figure shows a system of two concentric spheres of radii $r_1$ and $r_2$ are kept at temperatures $T_1$ and $T_2$ , respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to

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Answer: (d)

A thin shell of radius $r$ and thickness $dr$ : $H = -K(4\pi r^2)\dfrac{dT}{dr}$ , same $H$ through every shell.
$H\int_{r_1}^{r_2}\dfrac{dr}{r^2} = 4\pi K(T_1-T_2) \Rightarrow H\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right) = 4\pi K(T_1-T_2)$
$H = \dfrac{4\pi K r_1 r_2 (T_1-T_2)}{r_2-r_1}$
So the answer is (d) $H \propto \dfrac{r_1 r_2}{r_2-r_1}$ .

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Q54JEE Main 2004 · AIEEEMedium

Heat Conduction · Composite rods, slabs & junction temperature

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and $2K$ and thickness $x$ and $4x,$ respectively, are $T_2$ and $T_1\ (T_2 > T_1).$ The rate of heat transfer through the slab, in a steady state is $\left(\frac{A(T_2-T_1)K}{x}\right) f,$ with $f$ equal to

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Answer: (d)

Slabs in series: thermal resistances add, $R = \dfrac{x}{KA} + \dfrac{4x}{2KA} = \dfrac{3x}{KA}$
$H = \dfrac{T_2-T_1}{R} = \dfrac{A(T_2-T_1)K}{x}\cdot\dfrac{1}{3}$
So the answer is (d) $f = \dfrac{1}{3}$ .

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Q55NEET 1995 · AIPMTMedium

Heat Conduction · Thermal conductivity & rate of heat flow

Heat is flowing through two cylindrical rods of the same material. The diameters of the rods are in the ratio $1 : 2$ and the lengths in the ratio $2 : 1$ . If the temperature difference between the ends is same, then ratio of the rate of flow of heat through them will be

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Answer: (d)

$H \propto \frac{A}{L} \propto \frac{d^2}{L}$ (same material and $\Delta T$ ).
$\frac{H_1}{H_2} = \left(\frac{1}{2}\right)^2 \times \frac{1}{2} = \frac{1}{8}$
So the answer is (d) $1:8$ .

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Q56MHT CET 2026 · 13 MAY · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

If $150$ J of energy is incident on area $2\text{ m}^2$ and amount of heat reflected is $15$ J, coefficient of absorption is $0.6$ the amount of energy transmitted is

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Answer: (b)

$a + r + t = 1$ ; absorbed $= 0.6\times150 = 90$ J
Transmitted $= 150 - 90 - 15 = 45$ J
So the answer is (b) 45 J.

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Q57MHT CET 2026 · 13 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

Heat energy is incident on the surface at the rate of $1000$ J/min. If coefficient of absorption is $0.8$ and coefficient of reflection is $0.1$ then heat energy transmitted by the surface in 5 minute is

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Answer: (b)

$t = 1 - a - r = 1 - 0.8 - 0.1 = 0.1$
Energy incident in 5 min $= 5000$ J
Transmitted $= 0.1\times5000 = 500$ J
So the answer is (b) 500 J.

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Q58MHT CET 2026 · 19 APR · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

When $170$ J of energy is incident on a surface of a body, $17$ J of energy is reflected by it. If the coefficient of absorption is $0.7$ , then the amount of energy transmitted will be

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Answer: (c)

$r = \dfrac{17}{170} = 0.1$ , $a = 0.7$ , so $t = 1 - 0.7 - 0.1 = 0.2$
Transmitted $= 0.2\times170 = 34$ J
So the answer is (c) 34.0 J.

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Q59MHT CET 2026 · 17 APR · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

' $x$ ' joule of heat is incident on a body. Out of it, the total heat reflected and transmitted by it is ' $y$ ' joule. The absorption coefficient of the body is

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Answer: (b)

Heat absorbed $= x - y$ (rest is reflected or transmitted).
$a = \dfrac{\text{absorbed}}{\text{incident}} = \dfrac{x-y}{x}$
So the answer is (b).

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Q60MHT CET 2026 · 19 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A perfectly black body radiates power at temperature $T_1$ K. If it is to radiate 16 times this power, its temperature should be increased by

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Answer: (d)

Stefan's law: $P \propto T^4$
$\dfrac{T_2}{T_1} = 16^{1/4} = 2 \Rightarrow T_2 = 2T_1$
Increase $= 2T_1 - T_1 = T_1$
So the answer is (d) $(T_1)$ K.

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Q61MHT CET 2026 · 15 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

The temperature of a black body is increased by $50\%$ then the percentage increase of radiation is nearly

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Answer: (c)

$E \propto T^4$ ; $T' = 1.5T$
$\dfrac{E'}{E} = (1.5)^4 = 5.06$
Increase $= (5.06 - 1)\times100 \approx 406\% \approx 400\%$
So the answer is (c) 400%.

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Q62MHT CET 2026 · 19 APR · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A black body is at temperature $827^\circ$ C. The rate at which it emits energy is proportional to

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Answer: (c)

Stefan's law uses absolute temperature: $T = 827 + 273 = 1100$ K
$E \propto T^4 = (1100)^4$
So the answer is (c).

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Q63MHT CET 2026 · 20 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Temperature of two stars are in the ratio $3 : 2$ . If wavelength of maximum intensity of first star is $4200$ Å, the corresponding wavelength for second star is

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Answer: (c)

Wien's law: $\lambda_m T =$ constant, so $\lambda_m \propto \dfrac{1}{T}$
$\lambda_2 = \lambda_1\dfrac{T_1}{T_2} = 4200\times\dfrac{3}{2} = 6300$ Å
So the answer is (c) 6300 Å.

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Q64MHT CET 2026 · 20 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Identify the INCORRECT statement about the black body radiation

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Answer: (a)

A black body emits all wavelengths, and its spectrum has a peak at $\lambda_m$ (Wien's law), falling off at longer wavelengths.
So intensity is not the same for all wavelengths.
So the incorrect statement is (a).

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Q65MHT CET 2026 · 20 APR · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Two stars A and B radiate maximum energy at wavelength $3.9 \times 10^{-7}$ m and $5.2 \times 10^{-7}$ m respectively. The ratio of the temperature of star A to that of star B will be

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Answer: (d)

Wien's law: $\lambda_m T = b$ , so $T \propto \dfrac{1}{\lambda_m}$ .
$\dfrac{T_A}{T_B} = \dfrac{\lambda_B}{\lambda_A} = \dfrac{5.2 \times 10^{-7}}{3.9 \times 10^{-7}} = \dfrac{4}{3}$
So the answer is (d) $4 : 3$ .

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Q66MHT CET 2026 · 19 APR · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Out of the following, which statement is NOT true about black body radiation?

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Answer: (b)

A black body emits all wavelengths, but the intensity varies with $\lambda$ (the Planck curve has a peak at $\lambda_m$ ).
Intensity is therefore not the same for all wavelengths.
So the statement that is NOT true is (b).

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Q67MHT CET 2026 · 17 APR · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A black body emits radiation of maximum intensity of wavelength ' $\lambda$ ' at temperature 'T' K. Its corresponding wavelength at temperature $(2.5T)$ K will be

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Answer: (b)

Wien's law: $\lambda_m \propto \dfrac{1}{T}$ .
$\lambda' = \lambda \dfrac{T}{2.5T} = \dfrac{\lambda}{2.5} = \dfrac{2\lambda}{5}$
So the answer is (b) $\dfrac{2\lambda}{5}$ .

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Q68MHT CET 2026 · 15 APR · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

About black body radiation, which one of the following is a WRONG statement?

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Answer: (a)

A black body emits all wavelengths, but the power radiated is different at different wavelengths, with a peak at $\lambda_m$ .
So 'power radiated is same for all wavelengths' is false.
So the wrong statement is (a).

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Q69MHT CET 2026 · 20 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

Two bodies 'X' and 'Y' at temperatures $T_1$ K and $T_2$ K respectively have the same dimensions. Their emissivities $e_1$ and $e_2$ are in the ratio of $1 : 6$ . If their emissive powers are same, the relation between their temperatures is

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Answer: (c)

Emissive power $E = e\sigma T^4$ ; equal $E$ gives $e_1 T_1^4 = e_2 T_2^4$ .
$\left(\dfrac{T_1}{T_2}\right)^4 = \dfrac{e_2}{e_1} = 6$
$\dfrac{T_1}{T_2} = 6^{1/4}$
So the answer is (c) $6^{1/4} : 1$ .

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Q70MHT CET 2026 · 13 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

Rate of radiation by a black body is 'R' at temperature 'T'. Another body has same area but emissivity is 0.2 and temperature 3T. Its rate of radiation is

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Answer: (c)

Rate of radiation $P = e\sigma A T^4$ ; for the black body $R = \sigma A T^4$ .
$P' = 0.2\,\sigma A (3T)^4 = 0.2 \times 81\,\sigma A T^4$
$P' = 16.2R$
So the answer is (c) $16.2R$ .

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Q71MHT CET 2026 · 11 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

A wire of length $L$ and diameter $d$ is used in a bulb. The temperature of wire is $T$ and power radiated by the wire is $P$ . Its emissivity is ( $\sigma$ = Stefan's constant) (Assume that emissivity of wire material is same at all wavelength)

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Answer: (a)

Radiating area of the wire (curved surface) $A = \pi d L$ .
$P = e\sigma A T^4 = e\sigma \pi d L T^4$
$e = \dfrac{P}{\sigma T^4 \pi d L}$
So the answer is (a).

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Q72MHT CET 2026 · 14 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

At $273^\circ$ C, the emissive power of a perfectly black body is $R$ . The value at $0^\circ$ C is

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Answer: (b)

Stefan's law: $E \propto T^4$ (T in kelvin).
$273^\circ\text{C} = 546$ K, $0^\circ\text{C} = 273$ K
$\dfrac{E'}{R} = \left(\dfrac{273}{546}\right)^4 = \dfrac{1}{16}$
So the answer is (b) $\dfrac{R}{16}$ .

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Q73MHT CET 2026 · 13 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

The radiation emitted by a star A is $10000$ times that of the sun. If surface temperature of sun and star A are $6000$ K and $2000$ K respectively, the ratio of radii of star A and sun is

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Answer: (b)

Power of a star $P = 4\pi r^2 \sigma T^4$ , so $P \propto r^2 T^4$ .
$\dfrac{P_A}{P_S} = \left(\dfrac{r_A}{r_S}\right)^2 \left(\dfrac{2000}{6000}\right)^4 = 10000$
$\left(\dfrac{r_A}{r_S}\right)^2 = 10000 \times 81 \Rightarrow \dfrac{r_A}{r_S} = 900$
So the answer is (b) $900 : 1$ .

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Q74MHT CET 2026 · 13 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

The total radiant energy per unit area normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature 'T' K is ( $\sigma$ = Stefan's constant)

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Answer: (a)

Total power of the star $P = 4\pi r^2 \sigma T^4$ .
At distance $R$ this spreads over a sphere of area $4\pi R^2$ .
Energy per unit area per unit time $= \dfrac{4\pi r^2 \sigma T^4}{4\pi R^2} = \dfrac{\sigma r^2 T^4}{R^2}$
So the answer is (a).

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Q75MHT CET 2026 · 20 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A sphere having temperature 600 K is losing heat due to radiation. At this temperature its rate of cooling is $R$ . The rate of cooling of this sphere at 400 K is $\frac{x}{243}$ . The value of $x$ is (Temperature of surrounding is 300 K)

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Answer: (d)

Net rate of cooling $\propto T^4 - T_0^4$ with $T_0 = 300$ K.
At 600 K: $600^4 - 300^4 = 15 \times 300^4 = 1215 \times 100^4$
At 400 K: $400^4 - 300^4 = (256 - 81) \times 100^4 = 175 \times 100^4$
$R' = R \times \dfrac{175}{1215} = \dfrac{35R}{243}$
So $x = 35R$ , answer (d).

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Q76MHT CET 2026 · 20 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A sphere is at temperature $600$ K. In an external environment of $200$ K, its cooling rate is R. When the temperature of the sphere falls to $400$ K then cooling rate R' will become

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Answer: (d)

Net rate of cooling $\propto T^4 - T_0^4$ with $T_0 = 200$ K.
At 600 K: $600^4 - 200^4 = (81 - 1) \times 200^4 = 80 \times 200^4$
At 400 K: $400^4 - 200^4 = (16 - 1) \times 200^4 = 15 \times 200^4$
$R' = \dfrac{15}{80}R = \dfrac{3}{16}R$
So the answer is (d).

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Q77MHT CET 2026 · 18 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Two spherical black bodies of radii $R_1$ and $R_2$ having surface temperature $T_1$ and $T_2$ respectively, radiate the same power. The ratio of $R_1$ to $R_2$ will be

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Answer: (c)

Power $P = 4\pi R^2 \sigma T^4$ ; equal power gives $R_1^2 T_1^4 = R_2^2 T_2^4$ .
$\dfrac{R_1}{R_2} = \left(\dfrac{T_2}{T_1}\right)^2$
So the answer is (c).

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Q78MHT CET 2026 · 17 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

In an external environment of temperature ( $T$ ) kelvin, a sphere at temperature ( $3T$ ) kelvin has cooling rate $R_1$ . When the temperature of that sphere falls to ( $2T$ ) kelvin, the cooling rate $R_2$ of the sphere will become

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Answer: (d)

Net rate of cooling $\propto T_s^4 - T^4$ .
At $3T$ : $(81 - 1)T^4 = 80T^4$ ; at $2T$ : $(16 - 1)T^4 = 15T^4$
$R_2 = \dfrac{15}{80}R_1 = \dfrac{3}{16}R_1$
So the answer is (d).

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Q79MHT CET 2026 · 13 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

What will be the ratio of rate of radiation of metal sphere at two different temperatures $T_1 = 527^\circ$ C and $T_2 = 127^\circ$ C

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Answer: (b)

Stefan's law: rate of radiation $\propto T^4$ (T in kelvin).
$T_1 = 800$ K, $T_2 = 400$ K
$\dfrac{P_1}{P_2} = \left(\dfrac{800}{400}\right)^4 = 16$
So the answer is (b) $16 : 1$ .

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Q80MHT CET 2026 · 14 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

The wavelength of maximum intensity of radiation emitted by a star is $289.8$ nm. The radiation intensity of the star is (Stefan's constant = $5.67 \times 10^{-8}$ SI unit, Wein's constant $2898\,\mu$ mk)

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Answer: (a)

Wien's law: $T = \frac{b}{\lambda_m} = \frac{2898 \times 10^{-6}}{289.8 \times 10^{-9}} = 10^4\text{ K}$
Intensity $= \sigma T^4 = 5.67 \times 10^{-8} \times 10^{16}$
$= 5.67 \times 10^{8}\text{ W/m}^2$ , so the answer is (a).

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Q81MHT CET 2026 · 12 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A rectangular black body of temperature $27^\circ\text{C}$ has surface area $6$ cm $\times 3$ cm and rate of radiation $E$ . If its temperature is increased by $300^\circ\text{C}$ and its linear dimensions are reduced to $\left(\frac{1}{3}\right)^{rd}$ of their initial value then the rate of radiation is

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Answer: (b)

$P = \sigma A T^4$ , so $P \propto A T^4$ .
$T_1 = 300\text{ K}$ , $T_2 = 600\text{ K}$ : factor $2^4 = 16$ .
Linear dimensions become $\frac{1}{3}$ , so area becomes $\frac{1}{9}$ .
$P' = \frac{16}{9}E$ , so the answer is (b).

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Q82MHT CET 2026 · 20 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A black body is at temperature of 5780 K. The energy of radiation emitted by the body at wavelength 300 nm is $U_1$ , at wavelength 500 nm is $U_2$ and that at 900 nm is $U_3$ respectively. Wien's constant $b = 2.89 \times 10^{6}$ nmK. This shows that

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Answer: (c)

Wien's law: $\lambda_m = \frac{b}{T} = \frac{2.89 \times 10^6}{5780} = 500\text{ nm}$ .
Energy is maximum at 500 nm and falls on either side.
So $U_2$ is the largest: $U_1 < U_2 > U_3$ , answer (c).

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Q83MHT CET 2026 · 16 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A black rectangular surface of area 'A' emits energy 'E' per second at $27^\circ$ C. If length and breadth is reduced to half of its initial value and temperature is raised to $327^\circ$ C then energy emitted per second becomes

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Answer: (b)

$E \propto A T^4$ .
Length and breadth halved: $A \to \frac{A}{4}$ .
$T: 300\text{ K} \to 600\text{ K}$ , factor $16$ .
$E' = \frac{16}{4}E = 4E$ , so the answer is (b).

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Q84MHT CET 2026 · 15 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A black rectangular surface of area A emits energy E per second at $127^\circ$ C. If length and breadth is reduced to half of initial value and temperature is raised to $527^\circ$ C then energy emitted becomes

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Answer: (c)

$E \propto A T^4$ .
Length and breadth halved: $A \to \frac{A}{4}$ .
$T: 400\text{ K} \to 800\text{ K}$ , factor $16$ .
$E' = \frac{16}{4}E = 4E$ , so the answer is (c).

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Q85MHT CET 2026 · 13 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A black rectangular surface of area A emits energy E per unit time at $27^\circ$ C. If length and breadth is reduced to $\left(\frac{1}{4}\right)^{th}$ of initial value and temperature is raised to $327^\circ$ C, then energy emitted per unit time becomes

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Answer: (a)

$E \propto A T^4$ .
Length and breadth become $\frac{1}{4}$ : $A \to \frac{A}{16}$ .
$T: 300\text{ K} \to 600\text{ K}$ , factor $16$ .
$E' = \frac{16}{16}E = E$ , so the answer is (a).

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Q86MHT CET 2026 · 19 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

The power radiated by a black body is $P$ and it radiates maximum energy at wavelength $\lambda_0$ . If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $0.75\lambda_0$ , the power radiated by it becomes $nP$ . The value $n$ is

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Answer: (b)

Wien: $T \propto \frac{1}{\lambda_m}$ , so $\frac{T'}{T} = \frac{\lambda_0}{0.75\lambda_0} = \frac{4}{3}$ .
Stefan: $P \propto T^4 \Rightarrow n = \left(\frac{4}{3}\right)^4 = \frac{256}{81}$
So the answer is (b) $\frac{256}{81}$ .

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Q87MHT CET 2026 · 14 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

The power radiated by a black body is P and it radiates maximum energy at wavelength $\lambda_0$ . If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $\frac{3\lambda_0}{4}$ , the power radiated by it becomes nP. The value of n is

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Answer: (a)

Wien: $T \propto \frac{1}{\lambda_m}$ , so $\frac{T'}{T} = \frac{4}{3}$ .
Stefan: $P \propto T^4 \Rightarrow n = \left(\frac{4}{3}\right)^4 = \frac{256}{81}$
So the answer is (a) $\frac{256}{81}$ .

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Q88MHT CET 2026 · 18 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

A black body radiates maximum energy at wavelength ' $\lambda$ ' at temperature $T_1$ and its emissive power is E. When the temperature of the body is changed to $T_2$ , it radiates maximum energy at wavelength $\frac{2\lambda}{3}$ , then the emissive power will become

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Answer: (b)

Wien: $T \propto \frac{1}{\lambda_m}$ , so $\frac{T_2}{T_1} = \frac{\lambda}{2\lambda/3} = \frac{3}{2}$ .
Stefan: $E \propto T^4 \Rightarrow E' = \left(\frac{3}{2}\right)^4 E = \frac{81}{16}E$
So the answer is (b).

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Q89MHT CET 2026 · 16 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

The energy spectrum of a black body exhibits a maximum around a wavelength ' $\lambda$ '. The temperature of the black body is now changed such that the energy is maximum around a wavelength $2\lambda/3$ . The power radiated by the black body will now increase by a factor

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Answer: (b)

Wien: $T \propto \frac{1}{\lambda_m}$ , so $\frac{T'}{T} = \frac{3}{2}$ .
Stefan: $P \propto T^4 \Rightarrow$ factor $= \left(\frac{3}{2}\right)^4 = \frac{81}{16}$
So the answer is (b) $\frac{81}{16}$ .

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Q90MHT CET 2026 · 15 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

The energy spectrum of a black body exhibits a maximum around a wavelength $\lambda$ . The temperature of a black body is now changed such that the energy is maximum at wavelength $2\lambda/3$ . The power radiated by the black body will now increase by a factor of

Show answer and solution

Answer: (b)

Wien: $T \propto \frac{1}{\lambda_m}$ , so $\frac{T'}{T} = \frac{3}{2}$ .
Stefan: $P \propto T^4 \Rightarrow$ factor $= \left(\frac{3}{2}\right)^4 = \frac{81}{16}$
So the answer is (b) $\frac{81}{16}$ .

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Q91MHT CET 2026 · 15 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

A black disc has a radius 'R' and wavelength corresponding to the maximum intensity is ' $\lambda$ '. The emissive power (E) for the different radii and maximum wavelengths is directly proportional to

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Answer: (b)

Power from the disc: $E = \sigma A T^4$ with $A \propto R^2$ .
Wien: $T = \frac{b}{\lambda} \Rightarrow T^4 \propto \lambda^{-4}$ .
$E \propto R^2 \lambda^{-4}$ , so the answer is (b).

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Q92MHT CET 2026 · 11 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

The energy spectrum of a black body exhibits a maximum of wavelength $\lambda_0$ . The temperature of the black body is now changed such that the energy is maximum at wavelength $\frac{3\lambda_0}{2}$ . The ratio of power radiated by the black body will be

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Answer: (b)

Wien: $T \propto \frac{1}{\lambda_m}$ , so $\frac{T'}{T} = \frac{\lambda_0}{3\lambda_0/2} = \frac{2}{3}$ .
Stefan: $\frac{P'}{P} = \left(\frac{2}{3}\right)^4 = \frac{16}{81}$
So the answer is (b) $\frac{16}{81}$ .

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Q93MHT CET 2026 · 12 MAY · SHIFT 1Difficult

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

Three spheres, P, Q and R, having radii $4$ m, $5$ m and $6$ m respectively, are coated with lamp black on their outer surfaces. The wavelengths corresponding to maximum intensity (radiated) are $400$ nm, $500$ nm and $600$ nm respectively. If the powers radiated by them are $P_P, P_Q$ and $P_R$ respectively, then $P_P : P_Q : P_R$ is

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Answer: (b)

$P = \sigma (4\pi r^2) T^4$ and $T \propto \frac{1}{\lambda_m}$ , so $P \propto \frac{r^2}{\lambda_m^4}$ .
$P_P : P_Q : P_R = \frac{4^2}{4^4} : \frac{5^2}{5^4} : \frac{6^2}{6^4}$
$= \frac{1}{16} : \frac{1}{25} : \frac{1}{36}$
So the answer is (b).

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Q94MHT CET 2025 · 22 APR · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

If 120 J of thermal energy is incident on area $3\text{ m}^{2}$ , the amount of heat transmitted is 12 J, coefficient of absorption is 0.6, then the amount of heat reflected is

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Answer: (c)

Absorbed $= 0.6\times120 = 72$ J
Reflected $= 120 - 72 - 12 = 36$ J
So the answer is (c) 36 J.

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Q95MHT CET 2025 · 20 APR · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

For a perfectly black body, coefficient of emission is

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Answer: (b)

A perfectly black body absorbs all radiation ( $a = 1$ ), and by Kirchhoff's law $e = a$ .
So the answer is (b) unity.

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Q96MHT CET 2025 · 19 APR · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

The co-efficient of absorption and the coefficient of reflection of a thin uniform plate are 0.77 and 0.17 respectively. If 250 kcal of heat is incident on the surface of the plate, the quantity of heat transmitted is

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Answer: (c)

$t = 1 - a - r = 1 - 0.77 - 0.17 = 0.06$
Transmitted $= 0.06\times250 = 15$ kcal
So the answer is (c) 15 kcal.

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Q97MHT CET 2025 · 19 APR · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Black sphere of radius R radiates power P at certain temperature $T$ . If the temperature is doubled, the radius gets doubled. Now the power radiated would be

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Answer: (d)

$P = \sigma(4\pi R^2)T^4 \Rightarrow P \propto R^2T^4$
$\dfrac{P'}{P} = 2^2\times2^4 = 64$
So the answer is (d) 64 P.

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Q98MHT CET 2025 · 5 MAY · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A black body has maximum wavelength ' $\lambda_{m}$ ' at temperature 2000 K. Its maximum wavelength at 3000 K will be

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Answer: (d)

Wien's law: $\lambda_m T =$ constant.
$\lambda' = \lambda_m \times \dfrac{2000}{3000} = \dfrac{2}{3}\lambda_m$
So the answer is (d) $\dfrac{2}{3}\lambda_m$ .

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Q99MHT CET 2025 · 25 APR · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A black body emits radiation of maximum intensity at wavelength ' $\lambda$ ' at temperature T K. Its corresponding wavelength at temperature 1.5 T K will be

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Answer: (a)

Wien's law: $\lambda_m \propto \dfrac{1}{T}$ .
$\lambda' = \lambda \dfrac{T}{1.5T} = \dfrac{2\lambda}{3}$
So the answer is (a) $\dfrac{2\lambda}{3}$ .

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Q100MHT CET 2025 · 20 APR · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

The following graph represents the radiant power versus wavelength of the black body. The area under the curve represents

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Answer: (d)

The y-axis is energy emitted per unit time per unit area per unit wavelength, $E_\lambda$ .
Area under the curve $= \int E_\lambda\, d\lambda$ adds this over all wavelengths.
This is the total energy emitted per unit time per unit area, $E = \sigma T^4$ .
So the answer is (d).

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Q101MHT CET 2025 · 22 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

Two bodies A and B at temperatures ' $T_1$ ' K and ' $T_2$ ' K respectively have the same dimensions. Their emissivities are in the ratio $16 : 1$ . At $T_1 = xT_2$ , they radiate the same amount of heat per unit area per unit time. The value of $x$ is

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Answer: (d)

Same heat per unit area per unit time: $e_1 T_1^4 = e_2 T_2^4$ .
$\left(\dfrac{T_1}{T_2}\right)^4 = \dfrac{e_2}{e_1} = \dfrac{1}{16}$
$\dfrac{T_1}{T_2} = \dfrac{1}{2} \Rightarrow x = 0.5$
So the answer is (d) 0.5.

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Q102JEE Main 2025 · 7 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is $1727^\circ\text{C}$ and power radiated by the wire is 94.2 W. Its emissivity is $\frac{x}{8}$ where $x =$ _________. (Given $\sigma = 6.0 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}, \pi = 3.14$ and assume that the emissivity of wire material is same at all wavelength.)

Numerical answer type. Work it out, then open the answer.

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Answer: 5

$P = e\sigma A T^4$ with $A = \pi d L$ and $T = 1727 + 273 = 2000$ K.
$A = 3.14 \times 0.5 \times 10^{-3} \times 0.1 = 1.57 \times 10^{-4}\ \text{m}^2$
$\sigma A T^4 = 6 \times 10^{-8} \times 1.57 \times 10^{-4} \times 16 \times 10^{12} = 150.72$ W
$e = \dfrac{94.2}{150.72} = 0.625 = \dfrac{5}{8}$
So $x = 5$ .

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Q103MHT CET 2025 · 5 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

The outer surface of star in the form of a sphere radiates heat as a black body at temperature ' T '. The total radiant energy per unit area, normal to the direction of incidence, received at a distance ' $R$ ' from the centre of a star of radius ' $r$ ' is $(R > r)(\sigma =$ Stefan's constant $)$

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Answer: (a)

Total power of the star $P = 4\pi r^2 \sigma T^4$ .
At distance $R$ it spreads over area $4\pi R^2$ .
Energy per unit area per unit time $= \dfrac{4\pi r^2 \sigma T^4}{4\pi R^2} = \dfrac{\sigma r^2 T^4}{R^2}$
So the answer is (a).

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Q104MHT CET 2025 · 26 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Two spherical black bodies have radii ' $R_1$ ' and ' $R_2$ '. Their surface temperatures are ' $T_1$ ' and ' $T_2$ '. If they radiate same power, then $\frac{R_2}{R_1}$ is

Show answer and solution

Answer: (d)

Power $P = 4\pi R^2 \sigma T^4$ ; equal power gives $R_1^2 T_1^4 = R_2^2 T_2^4$ .
$\dfrac{R_2}{R_1} = \left(\dfrac{T_1}{T_2}\right)^2$
So the answer is (d).

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Q105MHT CET 2025 · 26 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A black sphere has radius $R$ whose rate of radiation is E at temperature T. If radius is made R / 2 and temperature 3T, the rate of radiation will be

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Answer: (c)

Rate of radiation $E = 4\pi R^2 \sigma T^4 \propto R^2 T^4$ .
$\dfrac{E'}{E} = \left(\dfrac{1}{2}\right)^2 (3)^4 = \dfrac{81}{4}$
So the answer is (c) $\dfrac{81E}{4}$ .

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Q106MHT CET 2025 · 23 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Two spherical black bodies have radii ' $R_1$ ' and ' $R_2$ '. Their surface temperatures are $T_1 K$ and $T_2 K$ respectively. If they radiate the same power, the ratio $\frac{R_1}{R_2}$ is

Show answer and solution

Answer: (d)

Power $P = 4\pi R^2 \sigma T^4$ ; equal power gives $R_1^2 T_1^4 = R_2^2 T_2^4$ .
$\dfrac{R_1}{R_2} = \left(\dfrac{T_2}{T_1}\right)^2$
So the answer is (d).

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Q107MHT CET 2025 · 21 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

If a black body at 400 K surrounded by atmosphere at 300 K has rate of cooling ' $R_0$ ', the same body at 900 K, surrounded by same atmosphere, will have rate of cooling nearly

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Answer: (c)

Net rate of cooling $\propto T^4 - T_0^4$ with $T_0 = 300$ K.
At 400 K: $(256 - 81) \times 100^4 = 175 \times 100^4$
At 900 K: $(6561 - 81) \times 100^4 = 6480 \times 100^4$
$\dfrac{R}{R_0} = \dfrac{6480}{175} \approx 37$ , nearest option $36R_0$
So the answer is (c) $36R_0$ .

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Q108JEE Main 2025 · 22 JAN · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K. If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding temperature to be negligible),

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Answer: (c)

$P = \sigma e (4\pi r^2) T^4 \Rightarrow P \propto r^2 T^4$ (same material, same $e$ ).
$\frac{P_{big}}{P_{small}} = \left(\frac{0.8}{0.2}\right)^2 \left(\frac{400}{800}\right)^4 = 16 \times \frac{1}{16} = 1$
So the answer is (c) $E$ .

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Q109MHT CET 2025 · 25 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A rectangular black body of temperature $127^\circ\text{C}$ has surface area $4\text{ cm} \times 2\text{ cm}$ and rate of radiation is E. If its temperature is increased by $400^\circ\text{C}$ and surface area is reduced to half of the initial value then the rate of radiation is

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Answer: (d)

$E \propto A T^4$ .
$T: 400\text{ K} \to 800\text{ K}$ , factor $2^4 = 16$ .
Area halved: factor $\frac{1}{2}$ .
$E' = 16 \times \frac{1}{2}E = 8E$ , so the answer is (d).

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Q110MHT CET 2025 · 22 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Black bodies A and B radiate maximum energy with wavelength difference $4\,\mu\text{m}$ . The absolute temperature of body A is 3 times that of B. The wavelength at which body $B$ radiates maximum energy is

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Answer: (b)

Wien's law: $\lambda_m T = b$ , so $\lambda_A = \frac{\lambda_B}{3}$ since $T_A = 3T_B$ .
$\lambda_B - \frac{\lambda_B}{3} = 4\,\mu\text{m} \Rightarrow \frac{2}{3}\lambda_B = 4\,\mu\text{m}$
$\lambda_B = 6\,\mu\text{m}$ , so the answer is (b).

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Q111MHT CET 2025 · 21 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A rectangular block of surface area A emits energy E per second at $27^\circ\text{C}$ . If length and breadth is reduced to half of initial value and temperature is raised to $327^\circ\text{C}$ then energy emitted per second becomes

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Answer: (b)

$E \propto A T^4$ .
Length and breadth halved: $A \to \frac{A}{4}$ .
$T: 300\text{ K} \to 600\text{ K}$ , factor $16$ .
$E' = \frac{16}{4}E = 4E$ , so the answer is (b).

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Q112MHT CET 2025 · 26 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

Two black spheres $P \,\&\, Q$ have radii in the ratio $4 : 3$ . The wavelength of maximum intensity of radiation are in the ratio $4 : 5$ respectively. The ratio of radiated power by P to Q is

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Answer: (a)

$P \propto r^2 T^4$ and Wien gives $T \propto \frac{1}{\lambda_m}$ , so $P \propto \frac{r^2}{\lambda_m^4}$ .
$\frac{P_P}{P_Q} = \left(\frac{4}{3}\right)^2 \left(\frac{5}{4}\right)^4 = \frac{16}{9} \times \frac{625}{256}$
$= \frac{625}{144}$ , so the answer is (a).

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Q113MHT CET 2024 · 9 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

Three identical metal spheres (of same surface area) have red, black and white colors and they are heated up to same temperature. They are allowed to cool. Arrange them from maximum rate of cooling to minimum rate of cooling

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Answer: (a)

Good absorbers are good emitters: emissivity order black > red > white.
Same area and temperature, so rate of cooling $\propto e$ .
So the answer is (a) black, red, white.

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Q114MHT CET 2024 · 3 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

Given that ' $x$ ' joule of heat is incident on a body. Out of that, total heat reflected and transmitted is ' $y$ ' joule. The absorption coefficient of body is

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Answer: (c)

Absorbed heat $= x - y$
$a = \dfrac{x-y}{x}$
So the answer is (c).

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Q115MHT CET 2024 · 11 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

Two spheres $S_1$ and $S_2$ have same radii but temperatures $T_1$ and $T_2$ respectively. Their emissive power is same and emissivity in the ratio 1:4. Then the ratio $T_1 : T_2$ is

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Answer: (b)

Same emissive power: $e_1 T_1^4 = e_2 T_2^4$ .
$\left(\dfrac{T_1}{T_2}\right)^4 = \dfrac{e_2}{e_1} = 4$
$\dfrac{T_1}{T_2} = 4^{1/4} = \sqrt{2}$
So the answer is (b) $\sqrt{2} : 1$ .

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Q116MHT CET 2024 · 10 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

Rate of radiation by a black body is ' R ' at temperature 'T'. Another body has same area but emissivity is 0.2 and temperature 3T. Its rate of radiation is

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Answer: (c)

Rate of radiation $P = e\sigma A T^4$ ; for the black body $R = \sigma A T^4$ .
$P' = 0.2\,\sigma A (3T)^4 = 0.2 \times 81 R$
$P' = 16.2R$
So the answer is (c) $(16.2)R$ .

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Q117MHT CET 2024 · 10 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A black sphere has radius $R$ whose rate of radiation is E at temperature T. If radius is made half and temperature 4 T, the rate of radiation will be

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Answer: (a)

Rate of radiation $E \propto R^2 T^4$ .
$\dfrac{E'}{E} = \left(\dfrac{1}{2}\right)^2 (4)^4 = \dfrac{256}{4} = 64$
So the answer is (a) $64E$ .

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Q118MHT CET 2024 · 4 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Two bodies ' X ' and ' Y ' at temperatures ' $T_1$ ' K and ' $T_2$ ' K respectively have the same dimensions. If their emissive powers are same, the relation between their temperatures is

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Answer: (a)

Emissive power $E = e\sigma T^4$ ; equal $E$ gives $e_1 T_1^4 = e_2 T_2^4$ .
The emissivity ratio is missing in the question; in the original version $e_1 : e_2 = 1 : 3$ .
$\left(\dfrac{T_1}{T_2}\right)^4 = \dfrac{e_2}{e_1} = 3 \Rightarrow \dfrac{T_1}{T_2} = 3^{1/4}$
So the answer works out to (c) $3^{1/4} : 1$ .

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Q119MHT CET 2024 · 4 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A sphere is at temperature 600 K. In an external environment of 200 K, its cooling rate is ' $R$ ' When the temperature of the sphere falls to 400 K, then cooling rate ' $R'$ ' will become

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Answer: (a)

Net rate of cooling $\propto T^4 - T_0^4$ with $T_0 = 200$ K.
At 600 K: $(81 - 1) \times 200^4 = 80 \times 200^4$
At 400 K: $(16 - 1) \times 200^4 = 15 \times 200^4$
$R' = \dfrac{15}{80}R = \dfrac{3}{16}R$
So the answer is (a).

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Q120MHT CET 2024 · 3 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Two spherical black bodies of radii ' $R_1$ ' and ' $R_2$ ' and with surface temperature ' $T_1$ ' and ' $T_2$ ' respectively radiate the same power. The ratio of ' $R_1$ ' to ' $R_2$ ' will be

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Answer: (b)

Power $P = 4\pi R^2 \sigma T^4$ ; equal power gives $R_1^2 T_1^4 = R_2^2 T_2^4$ .
$\dfrac{R_1}{R_2} = \left(\dfrac{T_2}{T_1}\right)^2$
So the answer is (b).

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Q121MHT CET 2024 · 10 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Ordinary bodies P and Q radiate maximum energy with wavelength difference $3\,\mu\text{m}$ . The absolute temperature of body P is four times that of Q. The wavelength at which body Q radiates maximum energy is

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Answer: (b)

Wien's law: $\lambda_m \propto \frac{1}{T}$ , so $\lambda_P = \frac{\lambda_Q}{4}$ .
$\lambda_Q - \frac{\lambda_Q}{4} = 3\,\mu\text{m} \Rightarrow \frac{3}{4}\lambda_Q = 3\,\mu\text{m}$
$\lambda_Q = 4\,\mu\text{m}$ , so the answer is (b).

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Q122MHT CET 2024 · 2 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

The frequency ' $\nu_{\text{m}}$ ' corresponding to which the energy emitted by a black body is maximum may vary with the temperature ' $T$ ' of the body as shown by the curves ' A ', ' B ', ' C ' and ' D ' in the figure. Which one of these represents the correct variation?

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Answer: (c)

Wien's law: $\lambda_m T = b$ and $\nu_m = \frac{c}{\lambda_m}$ .
$\nu_m = \frac{c}{b}T \Rightarrow \nu_m \propto T$
The graph of $\nu_m$ against $T$ is a straight line through the origin.
So the answer is (c) straight line B.

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Q123MHT CET 2024 · 16 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

The power radiated by a black body is P and it radiates maximum energy around the wavelength $\lambda_0$ . Now the temperature of the black body is changed so that it radiates maximum energy around wavelength $\left(\frac{\lambda_0}{2}\right)$ . The power radiated by it will now increase by a factor of

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Answer: (c)

Wien: $\lambda_m \to \frac{\lambda_0}{2}$ means $T \to 2T$ .
Stefan: $P \propto T^4 \Rightarrow$ factor $2^4 = 16$ .
So the answer is (c) 16.

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Q124MHT CET 2024 · 10 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

The black discs $x, y$ and z have radii $1\text{ m}, 2\text{ m}$ and 3 m respectively. The wavelengths corresponding to maximum intensity are $200\text{ nm}, 300\text{ nm}$ and 400 nm respectively. The relation between emissive power $E_x, E_y$ and $E_z$ is

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Answer: (a)

Power $E \propto R^2 T^4 \propto \frac{R^2}{\lambda_m^4}$ .
Taking $\lambda$ in units of 100 nm: $E_x \propto \frac{1}{2^4} = 0.0625$ , $E_y \propto \frac{4}{3^4} = 0.049$ , $E_z \propto \frac{9}{4^4} = 0.035$ .
So $E_x > E_y > E_z$ , answer (a).

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Q125MHT CET 2024 · 9 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

A black body radiates power ' P ' and maximum energy is radiated by it at a wavelength $\lambda_0$ . The temperature of the black body is now so changed that it radiates maximum energy at the wavelength $\frac{\lambda_0}{4}$ . The power radiated by it at new temperature is

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Answer: (b)

Wien: $\lambda_m \to \frac{\lambda_0}{4}$ means $T \to 4T$ .
Stefan: $P \propto T^4 \Rightarrow P' = 4^4 P = 256P$
So the answer is (b) $256P$ .

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Q126MHT CET 2023 · 14 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

A body is said to be opaque to the radiation if (a, r and t are coefficient of absorption, reflection and transmission respectively)

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Answer: (a)

Opaque body transmits nothing: $t = 0$ .
From $a + r + t = 1$ : $a + r = 1$
So the answer is (a).

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Q127MHT CET 2023 · 10 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

Heat energy is incident on the surface at the rate of X J/min. If ' $a$ ' and ' $r$ ' represent coefficient of absorption and reflection respectively then the heat energy transmitted by the surface in ' $t$ ' minutes is

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Answer: (a)

$a + r + t = 1 \Rightarrow t = 1 - (a+r)$
Energy incident in $t$ minutes $= Xt$
Transmitted $= [1-(a+r)]Xt$
None of the options matches exactly; (a) $(a+r)Xt$ is the energy absorbed plus reflected, not transmitted.

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Q128MHT CET 2023 · 11 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

If the temperature of a hot body is increased by $50\%$ , then the increase in the quantity of emitted heat radiation will be approximately

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Answer: (d)

$E \propto T^4$ ; $\dfrac{E'}{E} = (1.5)^4 \approx 5.06$
Increase $\approx 406\% \approx 400\%$
So the answer is (d) 400%.

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Q129MHT CET 2023 · 12 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

About black body radiation, which of the following is the wrong statement?

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Answer: (a)

A black body emits all wavelengths, but with different intensities; the curve rises to a peak at $\lambda_m$ and falls on either side.
So intensity is not the same for all wavelengths.
So the wrong statement is (a).

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Q130MHT CET 2023 · 11 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

Two bodies $A$ and $B$ at temperatures ' $T_1$ ' $\text{K}$ and ' $T_2$ ' $\text{K}$ respectively have the same dimensions. Their emissivities are in the ratio $1 : 3$ . If they radiate the same amount of heat per unit area per unit time, then the ratio of their temperatures $(T_1 : T_2)$ is

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Answer: (b)

Same heat per unit area per unit time: $e_1 T_1^4 = e_2 T_2^4$ .
$\left(\dfrac{T_1}{T_2}\right)^4 = \dfrac{e_2}{e_1} = 3$
$\dfrac{T_1}{T_2} = 3^{1/4}$
So the answer is (b) $3^{1/4} : 1$ .

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Q131MHT CET 2023 · 14 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A sphere and a cube, both of copper have equal volumes and are black. They are allowed to cool at same temperature and in same atmosphere. The ratio of their rate of loss of heat will be

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Answer: (c)

Same material, same temperature and surroundings, so rate of heat loss $\propto$ surface area $A$ .
Equal volumes: $\frac{4}{3}\pi r^3 = a^3$ , so $a = \left(\frac{4\pi}{3}\right)^{1/3} r$ .
$\frac{A_s}{A_c} = \frac{4\pi r^2}{6a^2} = \frac{2\pi/3}{(4\pi/3)^{2/3}}$
Cubing: $\left(\frac{A_s}{A_c}\right)^3 = \frac{8\pi^3/27}{16\pi^2/9} = \frac{\pi}{6}$
So the answer is (c) $\left(\frac{\pi}{6}\right)^{1/3}$ .

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Q132MHT CET 2023 · 14 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A sphere, a cube and a thin circular plate all made of same material and having the same mass are heated to same temperature of $200^\circ\text{C}$ . When these are left in a room.

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Answer: (c)

Same mass and material, so same heat capacity $mc$ ; rate of cooling $\frac{dT}{dt} \propto \frac{A}{mc}$ .
For a given volume, the sphere has the least surface area and a thin plate has the largest.
Largest area means fastest heat loss, so the plate cools fastest.
So the answer is (c) the circular plate reaches room temperature fast.

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Q133MHT CET 2023 · 13 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A black body at temperature $127^\circ\text{C}$ radiates heat at the rate of $5\text{ cal/cm}^{2}\text{ s}$ . At a temperature $927^\circ\text{C}$ , its rate of emission in units of $\text{cal/cm}^{2}\text{ s}$ will be

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Answer: (a)

Stefan's law: $E \propto T^4$ (T in kelvin).
$T_1 = 400\text{ K}$ , $T_2 = 1200\text{ K}$ , so $\frac{E_2}{E_1} = 3^4 = 81$ .
$E_2 = 81 \times 5 = 405\text{ cal/cm}^2\text{ s}$
So the answer is (a) 405.

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Q134MHT CET 2023 · 12 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Compare the rate of loss of heat from a metal sphere at $627^\circ\text{C}$ with the rate of loss of heat from the same sphere at $327^\circ\text{C}$ , if the temperature of the surrounding is $27^\circ\text{C}$ . (nearly)

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Answer: (b)

Net rate of heat loss $\propto T^4 - T_0^4$ , with $T_0 = 300\text{ K}$ .
$T_1 = 900\text{ K} = 3T_0$ , $T_2 = 600\text{ K} = 2T_0$ .
$\frac{Q_1}{Q_2} = \frac{3^4 - 1}{2^4 - 1} = \frac{80}{15} \approx 5.3$
So the answer is (b) 5.3.

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Q135MHT CET 2023 · 12 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A black sphere has radius ' $R$ ' whose rate of radiation is ' $E$ ' at temperature ' $T$ '. If radius is made $R/3$ and temperature ' $3T$ ', the rate of radiation will be

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Answer: (d)

Radiated power $P = \sigma (4\pi R^2) T^4$ , so $P \propto R^2 T^4$ .
$\frac{P'}{E} = \left(\frac{1}{3}\right)^2 (3)^4 = \frac{81}{9} = 9$
So the answer is (d) $9E$ .

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Q136MHT CET 2023 · 11 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Two spherical black bodies of radii ' $r_1$ ' and ' $r_2$ ' at temperature ' $T_1$ ' and ' $T_2$ ' respectively radiate power in the ratio $1 : 2$ Then $r_1 : r_2$ is

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Answer: (b)

$P \propto r^2 T^4$ , so $\frac{P_1}{P_2} = \frac{r_1^2 T_1^4}{r_2^2 T_2^4} = \frac{1}{2}$ .
$\left(\frac{r_1}{r_2}\right)^2 = \frac{1}{2}\left(\frac{T_2}{T_1}\right)^4$
$\frac{r_1}{r_2} = \frac{1}{\sqrt{2}}\left(\frac{T_2}{T_1}\right)^2$
So the answer is (b).

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Q137MHT CET 2023 · 12 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

A black body radiates maximum energy at wavelength ' $\lambda$ ' and its emissive power is ' $E$ '. Now due to a change in temperature of that body, it radiates maximum energy at wavelength $\frac{\lambda}{3}$ . At that temperature emissive power is

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Answer: (c)

Wien: $\lambda_m \to \frac{\lambda}{3}$ means $T \to 3T$ .
Stefan: $E \propto T^4 \Rightarrow E' = 3^4 E = 81E$
Ratio new : old $= 81 : 1$ , so the answer is (c).

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Q138MHT CET 2023 · 10 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

A black body radiates maximum energy at wavelength ' $\lambda$ ' and its emissive power is 'E' Now due to change in temperature of that body, it radiates maximum energy at wavelength $\frac{2\lambda}{3}$ . At that temperature emissive power is

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Answer: (a)

Wien: $\lambda_m \to \frac{2\lambda}{3}$ means $T \to \frac{3}{2}T$ .
Stefan: $E \propto T^4 \Rightarrow E' = \left(\frac{3}{2}\right)^4 E = \frac{81}{16}E$
So the answer is (a) $\frac{81}{16}$ (times $E$ ).

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Q139MHT CET 2023 · 10 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

A black body radiates maximum energy at wavelength ' $\lambda$ ' and its emissive power is $E$ . Now due to change in temperature of that body, it radiates maximum energy at wavelength $\frac{2\lambda}{3}$ . At that temperature emissive power is

Show answer and solution

Answer: (b)

Wien's law: $\lambda_m T = b$ , so $T \propto \frac{1}{\lambda_m}$ .
$\lambda_m' = \frac{2\lambda}{3} \Rightarrow T' = \frac{3}{2}T$
Stefan's law: $E \propto T^4 \Rightarrow E' = \left(\frac{3}{2}\right)^4 E = \frac{81}{16}E$
So the answer is (b) $\frac{81E}{16}$ .

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Q140MHT CET 2022 · 11 AUG · SHIFT 2Difficult

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

Three discs $x, y$ and $z$ having radii $2\text{ m}, 3\text{ m}$ and $6\text{ m}$ respectively are coated on outer surfaces. The wavelength corresponding to maximum intensity are $300\text{ nm}, 400\text{ nm}$ and $500\text{ nm}$ respectively. If $P_x, P_y$ and $P_z$ are power radiated by them respectively then

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Answer: (b)

$P = \sigma A T^4$ with $A \propto r^2$ and $T \propto \frac{1}{\lambda_m}$ , so $P \propto \frac{r^2}{\lambda_m^4}$ .
$P_x \propto \frac{2^2}{3^4} = \frac{4}{81} \approx 0.049$
$P_y \propto \frac{3^2}{4^4} = \frac{9}{256} \approx 0.035$ , $P_z \propto \frac{6^2}{5^4} = \frac{36}{625} \approx 0.058$
So the answer is (b) $P_z$ is maximum.

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Q141MHT CET 2021 · 23 SEP · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Absorptivity, reflectivity & emissivity

For a perfectly black body, coefficient of emission is

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Answer: (c)

A perfectly black body has $a = 1$ ; by Kirchhoff's law its emissivity $e = a = 1$ .
So the answer is (c) unity.

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Q142MHT CET 2021 · 23 SEP · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

The temperature of a black body is increased by $50\%$ , then the percentage increase in the rate of radiation by the body is approximated

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Answer: (c)

$E \propto T^4$ ; $\dfrac{E'}{E} = (1.5)^4 \approx 5.06$
Increase $\approx 406\% \approx 400\%$
So the answer is (c) 400%.

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Q143MHT CET 2021 · 22 SEP · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

The emissive power of sphere of area $0.04\text{ m}^{2}$ is $0.7\text{ k cal s}^{-1}\text{ m}^{-2}$ . The amount of heat radiated in 20 second is

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Answer: (d)

Heat radiated $Q = E\times A\times t$
$Q = 0.7\times0.04\times20 = 0.56$ kcal
So the answer is (d) 0.56 (the unit should be kcal; the unit printed in the options is wrong).

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Q144MHT CET 2021 · 20 SEP · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A perfectly black body emits a radiation at temperature 'T $_{1}$ ' K. If it is to radiate at 16 times this power, its temperature 'T $_{2}$ ' K should be

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Answer: (c)

$P \propto T^4$ ; $\dfrac{P_2}{P_1} = 16$
$\dfrac{T_2}{T_1} = 16^{1/4} = 2$
So the answer is (c) $2T_1$ .

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Q145MHT CET 2021 · 24 SEP · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Assume that for solar radiation, surface temperature of the sun is $6000\text{ K}$ . If Wien's constant 'b' is $2.897 \times 10^{-3}\text{ mK}$ , the value of maximum wavelength will be

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Answer: (a)

Wien's law: $\lambda_m = \dfrac{b}{T}$ .
$\lambda_m = \dfrac{2.897 \times 10^{-3}}{6000} = 4.828 \times 10^{-7}\ \text{m}$
$= 4828\ \mathring{A}$
So the answer is (a) $4828\ \mathring{A}$ .

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Q146MHT CET 2021 · 21 SEP · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A black body has maximum wavelength ' $\lambda_{m}$ ' at temperature $2000\text{ K}$ . Its corresponding wavelength at temperature $3000\text{ K}$ will be

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Answer: (b)

Wien's law: $\lambda_m T =$ constant.
$\lambda' = \lambda_m \times \dfrac{2000}{3000} = \dfrac{2\lambda_m}{3}$
So the answer is (b) $\dfrac{2\lambda_m}{3}$ .

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Q147MHT CET 2021 · 20 SEP · SHIFT 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Two stars 'P' and 'Q' emit yellow and blue light respectively. The relation between their temperatures $(T_{P}$ and $T_{Q})$ is

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Answer: (d)

Wien's law: $T \propto \dfrac{1}{\lambda_m}$ .
Yellow light has a longer wavelength than blue, so $\lambda_P > \lambda_Q$ .
Hence $T_P < T_Q$ , so the answer is (d).

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Q148MHT CET 2021 · 22 SEP · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

If the temperature of the sun is doubled, the rate of energy received by the earth will be increased by a factor

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Answer: (d)

Power from the sun $P \propto T^4$ ; energy received by the earth is proportional to $P$ .
$T \to 2T \Rightarrow P \to 2^4 P = 16P$
So the answer is (d) 16.

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Q149MHT CET 2021 · 21 SEP · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A black reactangular surface of area ' $a$ ' emits energy ' $E$ ' per second at $27^\circ\text{C}$ . If length and breadth is reduced to $\left(\frac{1}{3}\right)$ rd of initial value and temperature is raised to $327^\circ\text{C}$ then energy emitted per second becomes

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Answer: (a)

$E \propto A T^4$ .
Length and breadth become $\frac{1}{3}$ : $A \to \frac{A}{9}$ .
$T: 300\text{ K} \to 600\text{ K}$ , factor $16$ .
$E' = \frac{16E}{9}$ , so the answer is (a).

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Q150MHT CET 2021 · 20 SEP · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

A black rectangular surface of area ' $A$ ' emits energy ' $E$ ' per second at $27^\circ\text{C}$ . If length and breadth is reduced to $(1/3)$ rd of its initial value and temperature is raised to $327^\circ\text{C}$ then energy emitted per second becomes

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Answer: (c)

$E \propto A T^4$ .
Length and breadth become $\frac{1}{3}$ : $A \to \frac{A}{9}$ .
$T: 300\text{ K} \to 600\text{ K}$ , factor $16$ .
$E' = \frac{16E}{9}$ , so the answer is (c).

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Q151NEET 2020 · PHASE 2Easy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Three stars $A, B, C$ have surface temperatures $T_{A}, T_{B}, T_{C}$ respectively. Star $A$ appears bluish, star $B$ appears reddish and star $C$ yellowish. Hence,

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Answer: (a)

Wien's law: $T \propto \dfrac{1}{\lambda_m}$ .
Wavelengths: blue < yellow < red, so $\lambda_A < \lambda_C < \lambda_B$ .
Hence $T_A > T_C > T_B$ , so the answer is (a).

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Q152MHT CET 2020 · 16 OCT · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Two spherical black bodies of radius $r_1$ and $r_2$ with surface temperature $T_1$ and $T_2$ respectively, radiate same power, then $r_1 : r_2$ is

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Answer: (b)

$P \propto r^2 T^4$ ; equal power gives $r_1^2 T_1^4 = r_2^2 T_2^4$ .
$\frac{r_1}{r_2} = \left(\frac{T_2}{T_1}\right)^2$
So the answer is (b).

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Q153MHT CET 2019 · 3 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

The SI unit and dimensions of Stefan's constant $\sigma$ in case of Stefan's law of radiation is

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Answer: (d)

$E = \sigma T^4 \Rightarrow \sigma = \dfrac{E}{T^4}$ , with $E$ in J m $^{-2}$ s $^{-1}$
Unit: $\dfrac{\text{J}}{\text{m}^2\,\text{s}\,\text{K}^4}$ (W m $^{-2}$ K $^{-4}$ )
Dimensions: $\dfrac{[ML^2T^{-2}]}{[L^2][T][K^4]} = [M^1L^0T^{-3}K^{-4}]$
So the answer is (d).

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Q154MHT CET 2019 · 2 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

The maximum wavelength of radiation emitted by a star is 289.8 nm. Then intensity of radiation for the star is (Given: Stefan's constant $= 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}$ , Wien's constant, $b = 2898\,\mu\text{m K}$ )

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Answer: (c)

Wien's law: $T = \frac{b}{\lambda_m} = \frac{2898 \times 10^{-6}}{289.8 \times 10^{-9}} = 10^4\text{ K}$
Intensity $= \sigma T^4 = 5.67 \times 10^{-8} \times 10^{16}$
$= 5.67 \times 10^{8}\text{ W m}^{-2}$ , so the answer is (c).

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Q155NEET 2018Medium

Thermal Radiation: Black Body, Stefan & Wien · Stefan + Wien combined

The power radiated by a black body is P and it radiates maximum energy at wavelength, $\lambda$ 0. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $\frac{3}{4}\lambda_0$ , the power radiated by it becomes nP. The value of n is

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Answer: (c)

Wien's law: $T \propto \frac{1}{\lambda_m}$ , so $\lambda_m' = \frac{3}{4}\lambda_0 \Rightarrow T' = \frac{4}{3}T$ .
Power $P \propto T^4 \Rightarrow nP = \left(\frac{4}{3}\right)^4 P$
$n = \frac{256}{81}$
So the answer is (c) $\frac{256}{81}$ .

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Q156JEE Main 2014 · 19 APR · SHIFT 1Difficult

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A black coloured solid sphere of radius R and mass M is inside a cavity with vacuum inside. The walls of the cavity are maintained at a temperature T0. The initial temperature of the sphere is 3T0. If the specific heat of the material of the sphere varies as $\alpha$ T3 per unit mass with the temperature T of the sphere, where $\alpha$ is a constant, then the time taken for the sphere to cool down to temperature 2T0 will be ( $\sigma$ is Stefan Boltzmann constant)

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Answer: (c)

Net loss by radiation: $-Mc\frac{dT}{dt} = \sigma (4\pi R^2)(T^4 - T_0^4)$ with $c = \alpha T^3$ .
$\frac{M\alpha}{4\pi R^2 \sigma} \int_{3T_0}^{2T_0} \frac{-T^3\,dT}{T^4 - T_0^4} = \int_0^t dt$
$\int \frac{T^3 dT}{T^4 - T_0^4} = \frac{1}{4}\ln(T^4 - T_0^4)$ , so $t = \frac{M\alpha}{16\pi R^2\sigma}\ln\frac{81T_0^4 - T_0^4}{16T_0^4 - T_0^4}$
$\frac{80}{15} = \frac{16}{3} \Rightarrow t = \frac{M\alpha}{16\pi R^2\sigma}\ln\left(\frac{16}{3}\right)$
So the answer is (c).

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Q157JEE Main 2006 · AIEEEMedium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Assuming the Sun to be a spherical body of radius $R$ at a temperature of $TK$ , evaluate the total radiant power incident on Earth at a distance $r$ from the Sun

Where r0 is the radius of the Earth and $\sigma$ is Stefan's constant.

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Answer: (b)

Power of the sun: $P = 4\pi R^2 \sigma T^4$ .
Intensity at distance $r$ : $I = \frac{P}{4\pi r^2} = \frac{R^2 \sigma T^4}{r^2}$ .
Earth intercepts this over its cross-section $\pi r_0^2$ .
$P_E = \pi r_0^2 R^2 \sigma \frac{T^4}{r^2}$ , so the answer is (b).

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Q158JEE Main 2004 · AIEEEMedium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

If the temperature of the sun were to increase from $T$ to $2T$ and its radius from $R$ to $2R$ , then the ratio of the radiant energy received on earth to what it was previously will be

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Answer: (d)

Energy received on earth $\propto$ power of the sun $\propto R^2 T^4$ .
Ratio $= 2^2 \times 2^4 = 4 \times 16 = 64$
So the answer is (d) 64.

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Q159JEE Main 2003 · AIEEEEasy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

The earth radiates in the infra-red region of the spectrum. The spectrum is correctly given by

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Answer: (d)

Earth's temperature is about 300 K, so by Wien's law $\lambda_m = \dfrac{b}{T} \approx \dfrac{2.9 \times 10^{-3}}{300} \approx 10\ \mu\text{m}$ , which is in the infra-red.
The key uses Wien's law to explain why the earth's radiation lies in the infra-red.
So the answer (as per key) is (d) Wien's law.

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Q160JEE Main 2002 · AIEEEEasy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Infrared radiation is detected by

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Answer: (b)

A pyrometer measures temperature from the thermal (infra-red) radiation a body emits.
Spectrometer analyses spectra, photometer measures visible light; nanometer is a unit, not an instrument.
So the answer is (b) pyrometer.

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Q161JEE Main 2002 · AIEEEEasy

Thermal Radiation: Black Body, Stefan & Wien · Wien's displacement law

Which of the following is more close to a black body?

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Answer: (a)

A black body absorbs nearly all radiation falling on it.
Lamp-black / black board paint absorbs about 96-99% of incident radiation; leaves and roses reflect a lot of light.
So the answer is (a) black board paint.

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Q162JEE Main 2002 · AIEEEMedium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

Two spheres of the same material have radii $1\text{ m}$ and $4\text{ m}$ and temperatures $4000\text{ K}$ and $2000\text{ K}$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

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Answer: (a)

$P \propto r^2 T^4$ for the same material.
$\frac{P_1}{P_2} = \left(\frac{1}{4}\right)^2 \left(\frac{4000}{2000}\right)^4 = \frac{1}{16} \times 16 = 1$
So the answer is (a) $1 : 1$ .

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Q163NEET 1998 · AIPMTMedium

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

The radiant energy from the sun, incident normally at the surface of earth is $20\text{ kcal/m}^{2}\text{ min}$ . What would have been the radiant energy, incident normally on the earth, if the sun had a temperature, twice of the present one?

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Answer: (c)

Energy received on earth $\propto$ power of the sun $\propto T^4$ .
$T \to 2T$ gives a factor $2^4 = 16$ .
$16 \times 20 = 320\text{ kcal/m}^2\text{ min}$
So the answer is (c) $320\text{ kcal/m}^2\text{ min}$ .

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Q164NEET 1997 · AIPMTEasy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

A black body is at a temperature of 500 K. It emits energy at a rate which is proportional to

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Answer: (b)

Stefan's law: $E \propto T^4$ with $T$ in kelvin.
$E \propto (500)^4$
So the answer is (b).

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Q165NEET 1993 · AIPMTEasy

Thermal Radiation: Black Body, Stefan & Wien · Stefan-Boltzmann law & radiated power

If the temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of

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Answer: (d)

Power emitted by the sun $\propto T^4$ ; energy received on earth is a fixed fraction of it.
$\dfrac{E'}{E} = 2^4 = 16$
So the answer is (d) 16.

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Q166MHT CET 2026 · 13 MAY · SHIFT 2Medium

Newton's Law of Cooling

Two hot bodies $B_1$ and $B_2$ have temperature $100^\circ$ C and $80^\circ$ C respectively at $t = 0$ . The temperature of the surroundings is $40^\circ$ C. The ratio of the respective rates of cooling $R_1$ and $R_2$ of these two bodies at $t = 0$ will be

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Answer: (a)

Newton's law of cooling: rate $\propto (T - T_0)$ .
$\frac{R_1}{R_2} = \frac{100 - 40}{80 - 40} = \frac{60}{40}$
So the answer is (a) $R_1 : R_2 = 3 : 2$ .

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Q167MHT CET 2026 · 18 APR · SHIFT 2Medium

Newton's Law of Cooling

A body cools from a temperature $3\theta$ to $2\theta$ in 10 minute. The room temperature is $\theta$ . The temperature of the body at the end of the next 10 minute is ' $x$ '. Assuming that Newton's law of cooling is applicable, the value of ' $x$ ' will be

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Answer: (c)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
First 10 min: $\frac{3\theta - 2\theta}{10} = k(2.5\theta - \theta) \Rightarrow k = \frac{1}{15}$ per min
Next 10 min: $\frac{2\theta - x}{10} = \frac{1}{15}\left(\frac{2\theta + x}{2} - \theta\right) = \frac{x}{30}$
$2\theta - x = \frac{x}{3} \Rightarrow x = \frac{3}{2}\theta$
So the answer is (c) $\frac{3}{2}\theta$ .

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Q168MHT CET 2026 · 16 APR · SHIFT 2Medium

Newton's Law of Cooling

A bowl filled with very hot water cools from $98^\circ$ C to $86^\circ$ C in 2 minutes when the room temperature is $22^\circ$ C. How long it will take to cool from $75^\circ$ C to $69^\circ$ C?

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Answer: (b)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
Case 1: $\frac{98 - 86}{2} = k(92 - 22) \Rightarrow 6 = 70k$
Case 2: $\frac{75 - 69}{t} = k(72 - 22) = 50k$
Dividing: $\frac{6/t}{6} = \frac{50}{70} \Rightarrow t = 1.4$ min
So the answer is (b) $1.4$ minute.

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Q169MHT CET 2026 · 15 APR · SHIFT 1Medium

Newton's Law of Cooling

An ordinary body cools from $4\theta$ to $3\theta$ in 't' minutes. The temperature of the body after next 't' minutes is (Assume Newton's law of cooling and room temperature as $\theta$ )

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Answer: (b)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - \theta\right)$ .
First $t$ : $\frac{\theta}{t} = k(3.5\theta - \theta) = 2.5k\theta$
Next $t$ : $\frac{3\theta - T}{t} = k\left(\frac{3\theta + T}{2} - \theta\right) = k\frac{\theta + T}{2}$
Dividing: $\frac{3\theta - T}{\theta} = \frac{\theta + T}{5\theta} \Rightarrow 15\theta - 5T = \theta + T \Rightarrow T = \frac{7\theta}{3}$
So the answer is (b) $\frac{7\theta}{3}$ .

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Q170MHT CET 2026 · 11 APR · SHIFT 2Medium

Newton's Law of Cooling

A body cools in $7$ minutes from $60^\circ$ C to $40^\circ$ C. What time (in minutes) does it take to cool from $40^\circ$ C to $28^\circ$ C, if its surrounding temperature is $10^\circ$ C? (Newton's law of cooling holds good)

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Answer: (d)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
Case 1: $\frac{20}{7} = k(50 - 10) = 40k \Rightarrow k = \frac{1}{14}$
Case 2: $\frac{12}{t} = \frac{1}{14}(34 - 10) = \frac{24}{14}$
$t = 7$ min
So the answer is (d) $7$ .

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Q171MHT CET 2025 · 25 APR · SHIFT 1Medium

Newton's Law of Cooling

A body cools from $80^\circ\text{C}$ to $50^\circ\text{C}$ in 5 min. In the next time of ' t ' in, the body continues to cool from $50^\circ\text{C}$ to $30^\circ\text{C}$ . The total time taken by the body to cool from $80^\circ\text{C}$ to $30^\circ\text{C}$ is [The temperature of the surroundings is $20^\circ\text{C}$ .]

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Answer: (d)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
$80 \to 50$ : $\frac{30}{5} = k(65 - 20) \Rightarrow k = \frac{2}{15}$
$50 \to 30$ : $\frac{20}{t} = \frac{2}{15}(40 - 20) \Rightarrow t = 7.5$ min
Total time $= 5 + 7.5 = 12.5$ min
So the answer is (d) 12.5 min.

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Q172MHT CET 2025 · 21 APR · SHIFT 2Medium

Newton's Law of Cooling

Hot water cools from $80^\circ\text{C}$ to $60^\circ\text{C}$ in 1 minutes. In cooling from $60^\circ\text{C}$ to $50^\circ\text{C}$ it will take (room temperature $= 30^\circ\text{C}$ )

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Answer: (a)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
Case 1: $\frac{20}{1} = k(70 - 30) \Rightarrow k = \frac{1}{2}$ per min
Case 2: $\frac{10}{t} = \frac{1}{2}(55 - 30) = 12.5 \Rightarrow t = 0.8$ min
$0.8$ min $= 48$ s
So the answer is (a) 48 s.

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Q173MHT CET 2025 · 20 APR · SHIFT 2Medium

Newton's Law of Cooling

A metal sphere cools at a rate of $1.5^\circ\text{C/min}$ when its temperature is $80^\circ\text{C}$ . When the temperature of the sphere is $40^\circ\text{C}$ , its rate of cooling is $0.3^\circ\text{C/min}$ . The temperature of the surrounding $(\theta_0)$ is

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Answer: (a)

Newton's law: rate $\propto (T - \theta_0)$ .
$\frac{1.5}{0.3} = \frac{80 - \theta_0}{40 - \theta_0} = 5$
$80 - \theta_0 = 200 - 5\theta_0 \Rightarrow \theta_0 = 30^\circ$ C
So the answer is (a) $30^\circ$ C.

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Q174MHT CET 2025 · 20 APR · SHIFT 1Medium

Newton's Law of Cooling

A body cools from $60^\circ\text{C}$ to $40^\circ\text{C}$ in 6 minutes. After next 6 minutes its temperature will be (Temperature of the surroundings is $10^\circ\text{C}$ )

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Answer: (b)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
First 6 min: $\frac{20}{6} = k(50 - 10) \Rightarrow k = \frac{1}{12}$
Next 6 min: $\frac{40 - T}{6} = \frac{1}{12}\left(\frac{40 + T}{2} - 10\right) \Rightarrow 40 - T = \frac{T + 20}{4}$
$160 - 4T = T + 20 \Rightarrow T = 28^\circ$ C
So the answer is (b) $28^\circ$ C.

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Q175JEE Main 2025 · 29 JAN · SHIFT 2Medium

Newton's Law of Cooling

A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

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Answer: (a)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
$90 \to 80$ : $\frac{10}{t} = k(85 - 20) = 65k$
$80 \to 60$ : $\frac{20}{t'} = k(70 - 20) = 50k$
Dividing: $\frac{t'}{t} = \frac{20}{10}\times\frac{65}{50} = \frac{13}{5}$
So the answer is (a) $\frac{13}{5}t$ .

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Q176JEE Main 2025 · 24 JAN · SHIFT 2Medium

Newton's Law of Cooling

The temperature of a body in air falls from $40^\circ\text{C}$ to $24^\circ\text{C}$ in 4 minutes. The temperature of the air is $16^\circ\text{C}$ . The temperature of the body in the next 4 minutes will be:

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Answer: (b)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
First 4 min: $\frac{16}{4} = k(32 - 16) \Rightarrow k = \frac{1}{4}$
Next 4 min: $\frac{24 - T}{4} = \frac{1}{4}\left(\frac{24 + T}{2} - 16\right) \Rightarrow 24 - T = \frac{T - 8}{2}$
$48 - 2T = T - 8 \Rightarrow T = \frac{56}{3}^\circ$ C
So the answer is (b) $\frac{56}{3}{}^\circ$ C.

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Q177MHT CET 2025 · 26 APR · SHIFT 2Difficult

Newton's Law of Cooling

A calorimeter contains 10 g of water at $20^\circ\text{C}$ . The temperature fall to $15^\circ\text{C}$ in 10 min. When calorimeter contains 20 g of water at $20^\circ\text{C}$ , it takes 15 min. for the temperature to become $15^\circ\text{C}$ . The water equivalent at the calorimeter is

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Answer: (c)

Same calorimeter and same temperature range, so the rate of heat loss is the same in both cases.
$\frac{(m_1 + W)\Delta T}{t_1} = \frac{(m_2 + W)\Delta T}{t_2}$
$\frac{10 + W}{10} = \frac{20 + W}{15} \Rightarrow 150 + 15W = 200 + 10W$
$W = 10$ g
So the answer is (c) 10 g.

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Q178MHT CET 2024 · 16 MAY · SHIFT 1Medium

Newton's Law of Cooling

A bucket full of hot water is kept in a room. If it cools from $75^\circ\text{C}$ to $70^\circ\text{C}$ in $t_1$ minutes, from $70^\circ\text{C}$ to $65^\circ\text{C}$ in $t_2$ minutes and $65^\circ\text{C}$ to $60^\circ\text{C}$ in $t_3$ minutes, then

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Answer: (a)

Newton's law: rate of cooling $\propto$ (body temperature $-$ room temperature).
Each step is the same $5^\circ$ C drop, but the average excess temperature gets smaller each time.
Smaller rate means more time for the same drop.
So the answer is (a) $t_1 < t_2 < t_3$ .

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Q179MHT CET 2024 · 10 MAY · SHIFT 2Medium

Newton's Law of Cooling

An ordinary body cools from ' $4\theta'$ ' to ' $3\theta'$ ' in ' t ' minutes. The temperature of that body after next 't' minutes is (Assume Newton's law of cooling and room temperature is $\theta$ )

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Answer: (d)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - \theta\right)$ .
First $t$ : $\frac{\theta}{t} = k(3.5\theta - \theta) = 2.5k\theta$
Next $t$ : $\frac{3\theta - T}{t} = k\frac{\theta + T}{2}$
Dividing: $15\theta - 5T = \theta + T \Rightarrow T = \frac{7\theta}{3}$
So the answer is (d) $\frac{7\theta}{3}$ .

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Q180MHT CET 2024 · 9 MAY · SHIFT 2Medium

Newton's Law of Cooling

The temperature of a liquid falls from 365 K to 359 K in 3 minutes. The time during which temperature of this liquid falls from 342 K to 338 K is [Let the room temperature be 296 K ]

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Answer: (c)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
Case 1: $\frac{6}{3} = k(362 - 296) = 66k \Rightarrow k = \frac{1}{33}$
Case 2: $\frac{4}{t} = \frac{1}{33}(340 - 296) = \frac{44}{33}$
$t = 3$ min
So the answer is (c) 3 min.

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Q181MHT CET 2023 · 14 MAY · SHIFT 2Medium

Newton's Law of Cooling

A metal rod cools at the rate of $4^\circ\text{C/min}$ when its temperature is $90^\circ\text{C}$ and the rate of $1^\circ\text{C/m}$ in when its temperature is $30^\circ\text{C}$ . The temperature of the surrounding is

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Answer: (c)

Newton's law: rate $\propto (T - \theta_0)$ .
$\frac{4}{1} = \frac{90 - \theta_0}{30 - \theta_0}$
$120 - 4\theta_0 = 90 - \theta_0 \Rightarrow \theta_0 = 10^\circ$ C
So the answer is (c) $10^\circ$ C.

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Q182JEE Main 2022 · 27 JUL · SHIFT 1Medium

Newton's Law of Cooling

Read the following statements:

A. When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice.

B. Two bodies $P$ and $Q$ having equal surface areas are maintained at temperature $10^\circ\text{C}$ and $20^\circ\text{C}$ . The thermal radiation emitted in a given time by $P$ and $Q$ are in the ratio $1 : 1.15$ . C. A Carnot Engine working between $100\text{ K}$ and $400\text{ K}$ has an efficiency of $75\%$ .

D. When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice.

Choose the correct answer from the options given below:

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Answer: (a)

A: Newton's law, rate $\propto \Delta T$ , so doubling $\Delta T$ doubles the rate. True.
B: $\frac{E_P}{E_Q} = \left(\frac{283}{293}\right)^4 \approx \frac{1}{1.15}$ . True.
C: $\eta = 1 - \frac{100}{400} = 0.75 = 75\%$ . True.
D: quadrupling $\Delta T$ makes the rate four times, not twice. False.
So the answer is (a) A, B, C only.

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Q183MHT CET 2021 · 24 SEP · SHIFT 2Medium

Newton's Law of Cooling

A metal sphere cools at the rate of $1.5^\circ\text{C/min}$ when its temperature is $80^\circ\text{C}$ . At what rate will it cool when its temperature falls to $50^\circ\text{C}$ . [Temperature of surrounding is $30^\circ\text{C}$ ]

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Answer: (b)

Newton's law: rate $\propto (T - \theta_0)$ .
$\frac{R}{1.5} = \frac{50 - 30}{80 - 30} = \frac{2}{5}$
$R = 0.6^\circ$ C/min
So the answer is (b) $0.6^\circ$ C/min.

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Q184NEET 2021Medium

Newton's Law of Cooling

A cup of coffee cools from 90 $^\circ$ C to 80 $^\circ$ C in t minutes, when the room temperature is 20 $^\circ$ C. The time taken by a similar cup of coffee to cool from 80 $^\circ$ C to 60 $^\circ$ C at a room temperature same at 20 $^\circ$ C is:

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Answer: (c)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
$90 \to 80$ : $\frac{10}{t} = k(85 - 20) = 65k$
$80 \to 60$ : $\frac{20}{t'} = k(70 - 20) = 50k$
Dividing: $\frac{t'}{t} = \frac{20}{10}\times\frac{65}{50} = \frac{13}{5}$
So the answer is (c) $\frac{13}{5}t$ .

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Q185JEE Main 2019 · 8 APR · SHIFT 1Medium

Newton's Law of Cooling

Two identical beakers A and B contain equal volumes of two different liquids at 60°C each and left to cool down. Liquid in A has density of 8 × 102 kg/m3 and specific heat of 2000 J kg–1 K–1 while liquid in B has density of 103 kg m–3 and specific heat of 4000 J kg–1 K–1. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

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Answer: (b)

Same volume $V$ , same area and emissivity, so the heat loss rate is the same for both at the same temperature.
Rate of fall $\frac{dT}{dt} = \frac{\text{heat loss rate}}{\rho V c} \propto \frac{1}{\rho c}$
A: $\rho c = 800 \times 2000 = 1.6\times10^6$ ; B: $\rho c = 1000 \times 4000 = 4\times10^6$
A has smaller $\rho c$ , so A cools faster and its curve lies below B's at all times.
So the answer is (b).

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