JEE Advanced
Physics
Thermal Properties of Matter
2021
JEE Advanced 2021 (Paper 1)
JEE Advanced Physics Question (2021) — Solution
Question
A small object is placed at the centre of a large evacuated hollow spherical container. Assume the container is maintained at 0   K . At time t = 0 , the temperature of the object is 200   K . The temperature of the object becomes 100   K at t = t 1 and 50   K at t = t 2 . Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature. The ratio t 2 t 1 _________ .
Step-by-step solution
Using Stefan-Boltzmann law, P = d Q d t = σ e A T B 4 − T S 4 ,             ∵ T S = 0   K So, d Q d t = e σ A T 4 ,         consider ,   T = T B also, d Q d t = - m s d T d t = e σ A T 4 - ∫ 200 T d T T 4 = ∫ 0 t k d t ,   where   k = σ A m s ⇒ 1 T 3 − 1 200 3 = 3 k t Now, For case 1 1 100 3 − 1 200 3 = 3 k t 1               . . . 1 For case 2 1 50 3 − 1 200 3 = 3 k t 2               . . . 2 Divide equation 2 by 1 , 1 50 3 − 1 200 3 1 100 3 − 1 200 3 = t 2 t 1 So, t 2 t 1 = 9
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