Answer: Option D
:
D
The inner surface of B radiates at a rate of $\frac{\Delta Q}{\Delta t}=\sigma A_B{T}_A^4$,
where $\sigma \to$ stefan's constant
AB $\to$Area
TB $\to$ Temperature
Similarly, A radiates $\frac{\Delta Q_2}{\Delta t}=\sigma A_B{T}_A^4$
Now, all the radiation from inner surface of B goes inside the hollow of the shell, as B is a very poor conductor. So, this radiation falls on A, and A being a blackbody absorbs all the radiation falling on it! And it being a good conductor the heat travels to the centre almost instantly, thus making the temperature uniform throughout the ball. Similarly, all the heat radiated by A is completely absorbed by inner surface of B, and none of it is lost to the surrounding as B is a very poor conductor.
∴ Net heat lost by B = Net heat gained by A =$\sigma (A_B{T}_B^4-A_A{T}_A^4)$
= $5.67\times {10}^{-8}\times (80\times 293.1s^4-20\times 373.1s^4)\times {10}^{-4}$
= 1.15 w
Now this is equal to specific heat $\times \frac{dtemperature}{dTime}$ of both A and B
$\therefore \frac{d{T}_A}{dt}={\frac{1.15}{42}}^{\circ }c/s={0.03}^{\circ }c/s$
$\frac{d{T}_B}{dt}={\frac{1.15}{82}}^{\circ }c/s={0.01}^{\circ }c/s$