Question

A highly conducting solid sphere of radius R, density $\rho$ and specific heat is kept in an evacuated chamber. A parallel beam of electromagnetic radiation having uniform intensity I is incident on its surface. Assume that the surface of the sphere is perfectly black . Calculate the rate of increase of temperature

• A. $\frac{3(I-2\sigma T^4)}{2R\rho s}$
• B. $\frac{3(I-4\sigma T^4)}{4R\rho s}$
• C. $\frac{4(I-4\sigma T^4)}{4R\rho s}$
• D. $\frac{3(I+4\sigma T^4)}{4R\rho s}$

Answer 1

The correct option is B $\frac{3(I-4\sigma T^4)}{4R\rho s}$

REF.Image.

Solution

we know that, $\left(\pi R^2\right)=\left(\frac{d\theta }{dt}\right)$

$d\theta =msdT$

$\Rightarrow \frac{d\theta }{dt}=ms\frac{dT}{dt}$

Here $\frac{d\theta }{dt}=\left(I\right)\left(\pi R^2\right)-\left(4\pi R^2\right)\sigma T^4$

so,$\frac{dT}{dt}=\frac{4\pi R^2[I/4-\sigma T^4]}{s\left(\frac{4\pi R^3}{3}\right)s}=\frac{3(I-4\sigma T^4)}{4R\rho s}$

option -B is correct.