JEE MAIN - Physics (2002 - No. 37)
Two spheres of the same material have radii $$1$$ $$m$$ and $$4$$ $$m$$ and temperatures $$4000$$ $$K$$ and $$2000$$ $$K$$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is
$$1:1$$
$$16:1$$
$$4:1$$
$$1:9$$
Explanation
The energy radiated per second is given by $$E = e\sigma {T^4}A$$
For same material $$e$$ is same. $$\sigma $$ is stefan's constant
$$\therefore$$ $${{{E_1}} \over {{E_2}}} = {{T_1^4{A_1}} \over {T_2^4{A_2}}} = {{T_1^44\pi r_1^2} \over {T_2^44\pi r_2^2}}$$
$$ = {{{{\left( {4000} \right)}^4} \times {1^2}} \over {{{\left( {2000} \right)}^4} \times {4^2}}} = {1 \over 1}$$
For same material $$e$$ is same. $$\sigma $$ is stefan's constant
$$\therefore$$ $${{{E_1}} \over {{E_2}}} = {{T_1^4{A_1}} \over {T_2^4{A_2}}} = {{T_1^44\pi r_1^2} \over {T_2^44\pi r_2^2}}$$
$$ = {{{{\left( {4000} \right)}^4} \times {1^2}} \over {{{\left( {2000} \right)}^4} \times {4^2}}} = {1 \over 1}$$
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