Two balls of same material and finish have their diameters in the ratio of 2 : 1 and both are heated to same temperature and allowed to cool by radiation. Rate of cooling by big ball as compared to smaller one will be in the ratio of
एक ही सामग्री और फिनिश की दो गेंदों का व्यास 2:1 के अनुपात में है और दोनों को एक ही तापमान पर गर्म किया जाता है और विकिरण द्वारा ठंडा होने दिया जाता है। छोटी गेंद की तुलना में बड़ी गेंद के ठंडा होने की दर किस अनुपात में होगी?
1 : 1
2 : 1
1 : 2
4 : 1
4 : 1
Rate of heat transfer 
Temperature is same for boththe sphere
Surface area of sphere, A 
Hence, the correct option is (D).
Alternate solution:
Even if it is not for radiation, let us suppose the balls are heated to a temperature "T" and from there they start cooling down, then at any moment,
Heat gained due to convection = Heat lost to surrounding
Here, (h = Coefficient of heat transfer, 
=Change in temperature, Cv= Specific heat at constant volume and m = Mass) are supposed to be constant of each balls.
Hence, the correct option is (D).
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