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Physics Question on Thermodynamics

Two metallic spheres S1S_1, and S2_2 are made of the same material and have got identical surface finish. The mass of S1_1 is thrice that of S2_2. Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of S1S_1 to that of S2_2 is

A

13\frac{1}{3}

B

13\frac{1}{\sqrt 3}

C

31\frac{\sqrt 3}{1}

D

(13)1/3\bigg(\frac{1}{3}\bigg)^{1/3}

Answer

(13)1/3\bigg(\frac{1}{3}\bigg)^{1/3}

Explanation

Solution

The rate at which energy radiates from the object is
              ΔQΔt=eσAT4\, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{\Delta Q}{\Delta t}=e \sigma AT^4

Since,       ΔQ=mcΔT, \, \, \, \, \, \, \Delta Q=mc \Delta T,we get
                ΔTΔt=eσAT4mc\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{\Delta T}{\Delta t}=\frac{e \sigma AT^4}{mc}
Also, since m=43πr3ρ\frac{4}{3}\pi r^3 \rho for a sphere, we get
            A=4πr2=4π(3m4πρ)2/3\, \, \, \, \, \, \, \, \, \, \, \, A=4\pi r^2=4\pi \bigg(\frac{3m}{4\pi \rho}\bigg)^{2/3}
Hence, ΔTΔt=eσT4mc[4π(3m4πρ)2/3]\frac{\Delta T}{\Delta t} =\frac{e \sigma T^4}{mc}\bigg[4\pi \bigg(\frac{3m}{4\pi \rho}\bigg)^{2/3}\bigg]
              =K(1m)1/3\, \, \, \, \, \, \, \, \, \, \, \, \, \, =K\bigg(\frac{1}{m}\bigg)^{1/3}
For the given two bodies

(ΔT/Δt)1(ΔT/Δt)2=(m2m1)1/3=(13)1/3\frac{(\Delta T / \Delta t)_1}{(\Delta T / \Delta t )_2}=\bigg(\frac{m_2}{m_1}\bigg)^{1/3}=\bigg( \frac{1}{3}\bigg)^{1/3}