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Thermal Properties of Matter MCQs for NEET — Physics Questions with Answers

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A black body radiates at the rate of W watts at a temperature T. If the temperature of the body is reduced to T/3, it will radiate at the rate of (in Watts)

$\frac{W}{81}$

$\frac{W}{27}$

$\frac{W}{9}$

$\frac{W}{3}$

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Explanation

(a) $P = (\frac{Q}{t}) \propto T^{4} \Rightarrow \frac{W}{P_{2}} = {(\frac{T}{T / 3})}^{4} \Rightarrow P_{2} = \frac{W}{81}$

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Star A has radius r surface temperature T while star B has radius 4r and surface temperature T/2. The ratio of the power of two stars, $P_{A}$ : $P_{B}$ is

16 : 1

1 : 16

1 : 1

1 : 4

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Explanation

$\Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{r_{2}}{r_{1}})}^{2} \times {(\frac{T_{2}}{T_{1}})}^{4} = {(\frac{4 r}{r})}^{2} \times {(\frac{T / 2}{\frac{T}{}})}^{4} = 1$

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Suppose the sun expands so that its radius becomes 100 times its present radius and its surface temperature becomes half of its present value. The total energy emitted by it then will increase by a factor of

$10^{4}$

625

256

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Explanation

(b) $\frac{Q_{2}}{Q_{1}} = {(\frac{r_{2}^{2}}{r_{1}^{2}})}^{2} \times {(\frac{T_{2}}{T_{1}})}^{4} = {(\frac{100}{1})}^{2} \times {(\frac{1}{2})}^{2} = 625$

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If the sun’s surface radiates heat at $6.3 \times 10^{7} {Wm}^{- 2}$ . Calculate the temperature of the sun assuming it to be a black body $(σ = 5.7 \times 10^{- 8} {Wm}^{- 2} K^{- 4})$

$5.8 \times 10^{3} K$

$8.5 \times 10^{3} K$

$3.5 \times 10^{8} K$

$5.3 \times 10^{8} K$

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Explanation

(a) From Stefan’s law $E = {σT}^{4}$

$T^{4} = \frac{E}{σ} = \frac{6.3 \times 10^{7}}{5.7 \times 10^{- 8}} = 1.105 \times 10^{15} = 0.1105 \times 10^{16} T = 0.58 \times 10^{4} K = 5.8 \times 10^{3} K$

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The value of Stefan’s constant is

$5.67 \times 10^{- 8} W / m^{2} - K^{4}$

$5.67 \times 10^{- 5} W / m^{2} - K^{4}$

$5.67 \times 10^{- 11} W / m^{2} - K^{4}$

None of these

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Explanation

(a) The value of stefan's constant is 5.67 $\times 10^{- 8} W / m^{2} - K^{4}$

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Rate of cooling at 600K, if surrounding temperature is 300K is R. The rate of cooling at 900K is

$\frac{16}{3} R$

$\frac{2}{3} R$

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Explanation

(a) Rate of cooling ∝ $(T^{4} - T_{0}^{4})$

$\Rightarrow \frac{R_{1}}{R_{2}} = (\frac{T_{1}^{4} - T_{0}^{4}}{T_{2}^{4} - T_{0}^{4}}) \Rightarrow \frac{R}{R_{2}} = \frac{{(600)}^{4} - {(300)}^{4}}{{(900)}^{4} - {(300)}^{4}} or R_{2} = \frac{16}{3} R$

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A black body of surface area 10 ${cm}^{2}$ is heated to 127°C and is suspended in a room at temperature 27°C. The initial rate of loss of heat from the body at the room temperature will be

2.99 W

1.89 W

1.18 W

0.99 W

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Explanation

(d) Loss of heat $∆ Q = Aεσ (T^{4} - T_{0}^{4}) t$

$\Rightarrow$ Rate of loss of heat $\frac{∆ Q}{t} = Aεσ (T^{4} - T_{0}^{4})$

$= 10 \times 10^{- 4} \times 1 \times 5.67 \times 10^{- 8} \{{(273 + 127)}^{4} - {(273 + 27)}^{4}\} = 0.99 W$

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Two identical objects A and B are at temperatures $T_{A}$ and $T_{B}$ respectively. Both objects are placed in a room with perfectly absorbing walls maintained at temperatures T( $T_{A}$ >T> $T_{B}$ )The objects A and B attain temperature T eventually which one of the following is correct statement

‘A’ only emits radiations while B only absorbs them until both attain temperature

A loses more radiations than it absorbs while B absorbs more radiations that it emits until temperature T is attained

Both A and B only absorb radiations until they attain temperature T

Both A and B only emit radiations until they attain temperature T

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Explanation

(b) According to Prevost theory every body radiate heat at all temperature (except 0 K) and also absorbs heat from surroundings.
$∵$ $T_{A}$ >T $\Rightarrow$ Object A emits radiations more than the radiations it absorbs.
and $T_{B}$ <T $\Rightarrow$ Object B absorbs more radiations than it emits.
After a certain time all bodies attains a common temperature.

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When the body has the same temperature as that of surroundings

It does not radiate heat

It radiates the same quantity of heat as it absorbs

It radiates less quantity of heat as it receives from surroundings

It radiates more quantity of heat as it receives heat from surroundings

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Explanation

(b) According to Prevost theory

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The spectral energy distribution of star is maximum at twice temperature as that of sun. The total energy radiated by star is

Twice as that of the sun

Same as that of the sun

Sixteen times as that of the sun

One sixteenth of sun

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Explanation

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Yes. You can attempt every Thermal Properties of Matter question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.