Kindly elaborate. Ans-1
Two metallic spheres A₁ and A₂ are made of the same material and have identical surface finish. The mass of A₁ is eight times the mass of A₂. Both the spheres are heated to same temperature and placed in same room having lower temperature but thermally insulated from each other. The ratio of initial rate of cooling of A₁ to that of A₂ is
(1) 1/2 (2) 1/4
(3) 4/1 (4) 1/8

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Solution

$D e a r s t u d e n t, L e t r_{1} a n d r_{2} b e t h e r a d i i o f S_{1} a n d S_{2} a n d ρ b e d e n s i t y o f t h e m a t e r i a l . T h e m a s s e s o f S_{1} a n d S_{2} a r e, m_{1} = \frac{4}{3} {πr}^{3}_{1} ρ m_{2} = \frac{4}{3} {πr}^{3}_{2} ρ m_{1} = 8 m_{2} so we get, r_{1} = 2 r_{2} Let T be temperature of two spheres and T_{o} be room temperature . The rate of radiation heat loss by S_{1} and S_{2} are, \frac{{dQ}_{1}}{dt} = 4 {πσer}_{1}^{2} (T^{4} - {T_{o}}^{4}) \frac{{dQ}_{2}}{dt} = 4 {πσer}_{2}^{2} (T^{4} - {T_{o}}^{4}) T h e t e m p e r a t u r e o f t h e s p h e r e r e d u c e s d u e t o r a d i a t i o n h e a t l o s s . T h e r a t e o f t e m p e r a t u r e c h a n g e f o r S_{1} a n d S_{2} a r e g i v e n b y, \frac{{dQ}_{1}}{dt} = - m_{1} S (\frac{d T_{1}}{d t}) \frac{{dQ}_{2}}{dt} = - m_{2} S (\frac{d T_{2}}{d t}) s o, d i v i n i d i n g t h e s e t w o e q u a t i o n a n d p u t t i n g t h e v a l u e s d e r i v e d a b o v e w e g e t \frac{{r^{2}}_{1}}{{r^{2}}_{2}} = \frac{m_{1}}{m_{2}} \times \frac{(\frac{d T_{1}}{d t})}{(\frac{d T_{2}}{d t})} \frac{4 {r^{2}}_{2}}{{r^{2}}_{2}} = \frac{8 m_{2}}{m_{2}} \times \frac{(\frac{d T_{1}}{d t})}{(\frac{d T_{2}}{d t})} \Rightarrow \frac{(\frac{d T_{1}}{d t})}{(\frac{d T_{2}}{d t})} = \frac{1}{2} R e g a r d s$

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