Two spherical balls of same material and surface finish have their diameters in the ratio of 2:1. Both are heated to same temperature and allowed to cool by radiation. Rate of cooling of big ball as compared to smaller one will be in the ratio of
- A. 1:2
- B. 2:1
- C. 1:1
- D. 4:1
Answer: D. 4:1
Explanation: If 'Rate of cooling' refers to heat loss rate (Watts), $Q = \sigma A \epsilon (T^4 - T_{inf}^4) \propto A \propto D²$. Ratio $2^2:1^2 = 4:1$. If it refers to temperature drop rate $dT/dt$, it is $\propto A/V \propto 1/D$, ratio 1:2. ISRO answer implies Heat Loss Rate.