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

$160 kcal / m^{2} \min$

$40 kcal / m^{2} \min$

$320 kcal / m^{2} \min$

$80 kcal / m^{2} \min$

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Explanation

(c) $\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow \frac{E_{2}}{20} = {(\frac{2 T}{T})}^{4} = 16 \Rightarrow E_{2} = 320 kcal / m^{2} \min$

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

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Explanation

(d) Amount of energy radiated ∝ $(Temperature)^{4}$

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The ratio of energy of emitted radiation of a black body at 27 $° C$ and 927 $° C$ is

1 : 4

1 : 16

1 : 64

1 : 256

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Explanation

(d) $\frac{Q_{1}}{Q_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {(\frac{273 + 27}{273 + 927})}^{4} = {(\frac{1}{4})}^{4} = \frac{1}{256}$

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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. Then the ratio of $r_{1}$ and $r_{2}$ will be

${(\frac{T_{2}}{T_{1}})}^{2}$

${(\frac{T_{2}}{T_{1}})}^{4}$

${(\frac{T_{1}}{T_{2}})}^{2}$

${(\frac{T_{1}}{T_{2}})}^{4}$

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Explanation

(a) For black body, $P = {AεσT}^{4}$ . For same power $A \propto \frac{1}{T^{4}}$

$\Rightarrow {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow \frac{r_{1}}{r_{2}} = {(\frac{T_{2}}{T_{1}})}^{2}$

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A black body is at a temperature 300 K. It emits energy at a rate, which is proportional to

300

${(300)}^{2}$

${(300)}^{3}$

${(300)}^{4}$

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Explanation

(d) $E \propto T^{4}$

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Two identical metal balls at temperature 200 $° C$ and 400 $° C$ kept in air at 27 $° C$ . The ratio of net heat loss by these bodies is

1/4

1/2

1/16

$\frac{473^{4} - 300^{4}}{673^{4} - 300^{4}}$

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Explanation

(d) If temperature of surrounding is considered then
net loss of energy of a body by radiation

$Q = Aεσ (T^{4} - T_{0}^{4}) t \Rightarrow Q \propto (T^{4} - T_{0}^{4}) \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{T_{1}^{4} - T_{0}^{4}}{T_{2}^{4} - T_{0}^{4}} = \frac{{(273 + 200)}^{4} - {(273 + 27)}^{4}}{{(273 + 400)}^{4} - {(273 + 27)}^{4}} = \frac{{(473)}^{2} - {(300)}^{4}}{{(673)}^{4} - {(300)}^{4}}$

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A black body radiates 20 W at temperature 227 $° C$ . If temperature of the black body is changed to 727 $° C$ then its radiating power will be

120W

240 W

320 W

360 W

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Explanation

$\Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} \Rightarrow \frac{P_{2}}{20} = {(\frac{273 + 727}{273 + 227})}^{4} \Rightarrow \frac{P_{2}}{20} = 2^{4} \Rightarrow P_{2} = 320 W$

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The radiation emitted by a star A is 10,000 times that of the sun. If the surface temperatures of the sun and the star A are 6000 K and 2000 K respectively, the ratio of the radii of the star A and the sun is

300 : 1

600 : 1

900 : 1

1200 : 1

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Explanation

(c) $Q \propto {AT}^{4} \propto r^{2} T^{4} \frac{Q_{star}}{Q_{sun}} = \frac{r_{star}^{2} . T_{star}^{4}}{r_{sun}^{2} . T_{sun}^{4}} \Rightarrow \frac{10000}{1} = \frac{r_{star}^{2}}{r_{sun}^{2}} \times {(\frac{6000}{2000})}^{4} \Rightarrow \frac{r_{star}}{r_{sun}} = \frac{100 \times 9}{1} = \frac{900}{1}$

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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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