A) \[\frac{H}{16}\]
B) \[\left( \frac{9}{27} \right)H\]
C) \[\left( \frac{16}{3} \right)H\]
D) \[\left( \frac{3}{16} \right)H\]
Correct Answer: D
Solution :
From Stefan?s law, the total radiant energy emitted per second per unit surface area of a black body is proportional to the fourth power of the absolute temperature of the body. That is \[E=\sigma {{T}^{4}}\] Where \[\sigma \] is Stefan?s constant. When sphere cools from 600 K to 200 K, energy radiated is \[H=\sigma \left[ {{\left( 600 \right)}^{4}}-{{\left( 200 \right)}^{4}} \right]\] Let energy radiated be \[H\] when cooled from400 K to 200 K, then \[H'=\sigma \left[ {{\left( 600 \right)}^{4}}-{{\left( 400 \right)}^{4}} \right]\] \[\therefore \] \[\frac{H}{H'}=\frac{\left[ {{\left( 600 \right)}^{4}}-{{\left( 200 \right)}^{4}} \right]}{\left[ {{\left( 600 \right)}^{4}}-{{\left( 400 \right)}^{4}} \right]}\] Using \[{{a}^{4}}-{{b}^{4}}=\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right),\] we have \[\frac{H}{H'}=\frac{\left[ {{\left( 600 \right)}^{2}}-{{\left( 200 \right)}^{2}} \right]}{\left[ {{\left( 600 \right)}^{2}}-{{\left( 400 \right)}^{2}} \right]}\times \frac{\left[ {{\left( 600 \right)}^{2}}+{{\left( 200 \right)}^{2}} \right]}{\left[ {{\left( 600 \right)}^{2}}+{{\left( 400 \right)}^{2}} \right]}\] \[\frac{H}{H'}=\frac{32}{12}\times \frac{40}{20}=\frac{16}{3}\] \[\Rightarrow \] \[H'=\frac{3}{16}H\] Note: It is important to note that the rate of radiant energy emitted does not depend upon the shape, size etc. of body. It depends only on temperature.
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