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:

Concept:

Rate of cooling is given by the expression ϵσA i.e. Q = ϵσ(πr2)

where ϵ is the emissivity, σ is the Stefan-Boltzmann constant = 5.67 × 10-8 W/m2K4, A = area, T = absolute temperature

The ratio of the diameters of two spherical balls is \(\frac{d_1}{d_2}=\frac{2}{1}\)

Let us assume the diameter of 1st ball is d1 = 2d and diameter of 2nd ball is d2 = d

So, the ratio of the rate of cooling of the big ball as compared to the smaller ball =

\(\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{A_1}}}{{{A_2}}} = \frac{{π d_1^2}}{{π d_2^2}} = {\left( {\frac{{2d}}{d}} \right)^2} = \frac{4}{1}\)

so, the ratio is 4 : 1