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Physics MCQs for NEET — Practice Questions with Answers

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

$\frac{3}{2} λ_{m}$

$\frac{2}{3} λ_{m}$

$\frac{4}{9} λ_{m}$

$\frac{9}{4} λ_{m}$

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Explanation

(b) $λ_{m_{2}} = \frac{T_{1}}{T_{2}} \times λ_{m_{1}} = \frac{2000}{3000} \times λ_{m_{1}} = \frac{2}{3} λ_{m_{1}} = \frac{2}{3} λ_{m}$

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A black body at a temperature of 1640 K has the wavelength corresponding to maximum emission equal to 1.75 $μm$ . Assuming the moon to be a perfectly black body, the temperature of the moon, if the wavelength corresponding to maximum emission is 14.35 $μ$ m is

100 K

150 K

200 K

250 K

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Explanation

(c) $\frac{T_{2}}{T_{1}} = \frac{λ_{m_{1}}}{λ_{m_{2}}} = \frac{1.75}{14.35} \Rightarrow T_{2} = \frac{1.75}{14.35} \times 1640 = 200 K$

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A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} K$ . The wavelength in nanometers at which its radiation becomes maximum is -
(b = 0.0029 mK)

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Explanation

(b) According to Wein’s displacement law

$λ_{m} T = b or λ_{m} = \frac{b}{T} = \frac{0.0029}{5 \times 10^{4}} = 58 \times 10^{- 9} m = 58 nm$

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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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