Question

A spherical black body of radius \[r\] radiated power\[P\], and its rate of cooling is\[R\]
A. \[P \propto r\]
B. \[P \propto {r^2}\]
C. \[R \propto {r^2}\]
D.\[R \propto \left( {\dfrac{1}{r}} \right)\]

Answer 1

Hint: First of all, we will apply the concept in which the power is directly proportional to area of the surface area of the sphere and the square of radius of the sphere. After that we will find out the expression of power which relates mass, rate of change of temperature. We will put the values of mass in the equation and rearrange it further. We will carry out further manipulation and obtain the result.

Complete step by step answer:
In the given question, we are given a spherical black of radius \[r\] and which radiated power of magnitude \[P\] .
We are asked to find the rate of cooling of the black body.
To begin with, we will first find the expression of power which varies according to the area of the sphere and the radius of the square.
The power at which the body radiates is directly proportional to area:
\[
  P \propto A \\
\Rightarrow P \propto {r^2} \\
 \]
We know that power is related to mass, rate of change of temperature, etc. by the following expression, as given below:
\[P = mC\dfrac{{dT}}{{dt}}\]
Put the value of \[m\] and \[C\] in the above equation:
\[P = \dfrac{4}{3}\pi {r^3}DS\dfrac{{dT}}{{dt}}\]
i.e.
\[
\Rightarrow \dfrac{4}{3}\pi {r^3}DS\dfrac{{dT}}{{dt}} \propto {r^{\text{2}}} \\
\Rightarrow \dfrac{{dT}}{{dt}} \propto \dfrac{1}{{RDS}} \\
 \]
i.e. Rate of fall of temperature is \[R \propto \dfrac{1}{r}\]

Hence the correct option is (D) \[R \propto \dfrac{1}{r}\]

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