Two solid spheres are heated to the same temperature and allowed to cool under identical conditions. Assume that all the surfaces have the same emissivity and ratios of their radii, specific heats and densities are respectively $1 : α, 1 : β, 1 : γ$ . Then,

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The ratio of initial rate of fall of temperature is $α β γ : 1$

The ratio of initial rate of fall of temperature is $β : α γ$

The ratio of initial rate of loss of heat is $1 : α^{2}$

The ratio of initial rate of loss of heat is $1 : α^{3}$

answer is A, C.

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Detailed Solution

Given $R_{1} : R_{2} = 1 : α, S_{1} : S_{2} = 1 : β, ρ_{1} : ρ_{2} = 1 : γ$
$e_{1} = e_{2}, \frac{d θ}{d t} = \frac{- σ A e}{m s} (T^{4} - T_{0}^{4})$
$\frac{{(\frac{d θ}{d t})}_{1}}{{(\frac{d θ}{d t})}_{2}} = \frac{(\frac{A_{1}}{m_{1} s_{1}})}{(\frac{A_{2}}{m_{2} s_{2}})} = \frac{A_{1}}{A_{2}} \times \frac{m_{2}}{m_{1}} \times \frac{S_{2}}{S_{1}}$
$= \frac{4 π R_{1}^{2}}{4 π R_{2}^{2}} \times \frac{ρ_{2}}{ρ_{1}} \times \frac{\frac{4}{3} π R_{2}^{3}}{\frac{4}{3} π R_{1}^{3}} \times \frac{S_{2}}{S_{1}}$
$= {(\frac{R_{1}}{R_{2}})}^{2} \times {(\frac{R_{2}}{R_{1}})}^{3} \times \frac{ρ_{2}}{ρ_{1}} \times \frac{s_{2}}{s_{1}} = (\frac{R_{2}}{R_{1}}) (\frac{ρ_{2}}{ρ_{1}}) (\frac{s_{2}}{s_{1}}) = α β γ : 1$
$\frac{d Q}{d t} = σ A e (T^{4} - T_{0}^{4}) \Rightarrow \frac{{(\frac{d Q}{d t})}_{1}}{{(\frac{d Q}{d t})}_{2}} = \frac{A_{1}}{A_{2}} = {(\frac{R_{1}}{R_{2}})}^{2} = \frac{1}{α^{2}} = 1 : α^{2}$