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

Practice free Thermal Properties of Matter (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the north star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperature of the sun and north star is

1.46

0.69

1.21

0.83

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Explanation

(b) $\frac{T_{S}}{T_{N}} = \frac{{(λ_{N})}_{\max}}{{(λ_{S})}_{\max}} = \frac{350}{510} = 0.69$

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The amount of radiation emitted by a perfectly black body is proportional to

Temperature on ideal gas scale

Fourth root of temperature on ideal gas scale

Fourth power of temperature on ideal gas scale

Source of temperature on ideal gas scale

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Explanation

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A black body radiates energy at the rate of E $W / m^{2}$ at a high temperature TK. When the temperature is reduced to $\frac{T}{2} K$ , the radiant energy will be

$\frac{E}{16}$

$\frac{E}{4}$

$4 E$

$16 E$

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Explanation

(a) $E \propto T^{4} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{T^{4}}{T^{4}} \times 2^{4} \Rightarrow E_{2} = \frac{E}{16}$

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An object is at a temperature of $400 ° C$ . At what temperature would it radiate energy twice as fast? The temperature of the surroundings may be assumed to be negligible .

$200 ° C$

200 K

$800 ° C$

800 K

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Explanation

(d) $\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow \frac{2}{1} = {(\frac{T}{400 + 273})}^{4} = {(\frac{T}{673})}^{4} \Rightarrow T = 2^{1 / 4} \times 673 = 800 K$

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A black body at a temperature of 227 $° C$ radiates heat energy at the rate of 5 $cal / {cm}^{2} - \sec$ . At a temperature of $727 ° C$ , the rate of heat radiated per unit area in $cal / {cm}^{2}$ will be

160

250

500

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Explanation

(a) $\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} = (\frac{273 + 727}{237 + 227}) = \frac{{(1000)}^{4}}{{(500)}^{4}} = 16 \Rightarrow E_{2} = 80$

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Energy is being emitted from the surface of a black body at 127 $° C$ temperature at the rate of $1.0 \times 10^{6} J / \sec - m^{2}$ . Temperature of the black body at which the rate of energy emission is $16.0 \times 10^{6} J / \sec - m^{2}$ will be -

254 $° C$

508 $° C$

527 $° C$

727 $° C$

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Explanation

(c) $\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow T_{2} = {(\frac{E_{2}}{E_{1}})}^{1 / 4} \times T_{1} = {(16)}^{1 / 4} \times (273 + 127) \Rightarrow T_{2} = 800 k = 527 ° C$

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If temperature of a black body increases from $7 ° C$ to $287 ° C$ , then the rate of energy radiation increases by

${(\frac{287}{7})}^{4}$

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Explanation

(b) For a block body rate of energy $\frac{Q}{t} = P = {AσT}^{4}$

$P \propto T^{4} \Rightarrow \frac{P_{1}}{P_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {\{\frac{(273 + 7)}{(273 + 287)}\}}^{4} = \frac{1}{16}$

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The area of a hole of heat furnace is $10^{- 4} m^{2}$ . It radiates $1.58 \times 10^{5}$ calories of heat per hour. If the emissivity of the furnace is 0.80, then its temperature is

1500 K

2000 K

2500 K

3000 K

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Explanation

$\Rightarrow \frac{1.58 \times 10^{5} \times \times 4.2}{60 \times 60} = 5.6 \times 10^{- 8} \times 10^{- 4} \times 0.8 \times T^{4} T \approx 2500 K$

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Two spheres P and Q, of same colour having radii 8 cm and 2 cm are maintained at temperatures 127 $° C$ and 527 $° C$ respectively. The ratio of energy radiated by P and Q is

0.054

0.0034

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Explanation

$\Rightarrow Q \propto {AT}^{4} \propto r^{2} T^{4} (∵ A = 4 {πr}^{2}) \Rightarrow \frac{Q_{P}}{Q_{Q}} = {(\frac{r_{P}}{r_{Q}})}^{2} {(\frac{T_{P}}{T_{Q}})}^{4} = {(\frac{8}{2})}^{2} {\{\frac{(273 + 127)}{(273 + 527)}\}}^{4} = 1$

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A body radiates energy 5W at a temperature of 127 $° C$ . If the temperature is increased to 927 $° C$ , then it radiates energy at the rate of

410 W

81 W

405 W

200 W

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Explanation

$\Rightarrow \frac{P_{1}}{P_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {(\frac{927 + 273}{127 + 273})}^{4} \Rightarrow P_{1} = 405 W$

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