Question

A spherical black body of radius r radiates power P, and its rate of cooling is R.
A. $P \propto r$
B. $P \propto {r^2}$
C. $R \propto r$
D. $R \propto \dfrac{1}{r}$

Answer 1

Hint: We will apply the concept in which the power is directly proportional to area of the surface area of the sphere area of the sphere and the square of radius of the sphere. We will put values of mass in the equation and rearrange it further. We will carry out further manipulation and obtain the results.

Complete step by step answer:
From the above question, we are given that the radius of the spherical block is r and it is radiating the power is P.
We are asked to find the rate of cooling of the black body.
We will find the expression of power which varies according to the area of the sphere and the radius of the square.
The power of the body which radiates is directly proportional to area
$P \propto A$
$P \propto {r^2}$
We know that the relationship between power and rate of change of temperature etc, by the following expression, as given by
$P = mC\dfrac{{dT}}{{dt}}$
Substitute the value of m and C in the above equation
$P = \dfrac{4}{3}\pi {r^3}DS\dfrac{{dT}}{{dt}}$
Substitute the values of m and C in the above equation
$P = \dfrac{4}{3}\pi {r^3}\dfrac{{dT}}{{dt}} \propto {r^2}$
$\dfrac{{dT}}{{dt}} \propto \dfrac{1}{{RDS}}$
Therefore rate of fall in temperature is given by
$R \propto \dfrac{1}{r}$

So, the correct answer is “Option D”.

Note:
An object that absorbs all the radiation falling on it is called a black body at all wavelengths. When a black body material maintains at a constant temperature, its emission has a characteristic frequency distribution and it depends on temperature-dependent. Its emission is called radiation of the black-body.