Question

Two spheres of the same material have radii \(1\,m\) and \(2\,m\), and temperatures \(2000\,K\) and \(1000\,K\) respectively. What is the ratio of energy radiated per second by the first sphere to that by the second?

• A. \(1:2\)
• B. \(2:1\)
• C. \(1:4\)
• D. \(4:1\)

Hint:

Thermal radiation power is \[ P\propto r^2T^4. \] A small increase in temperature greatly increases radiation because of the fourth-power dependence on \(T\).

Answer 1

The Correct Option is D

Concept: According to Stefan–Boltzmann law, \[ P=e\sigma AT^4 \] where \[ A=4\pi r^2. \] Since both spheres are made of the same material, \[ e=\text{constant}. \] Thus, \[ P\propto r^2T^4. \]

Step 1: Write the ratio of powers. \[ \frac{P_1}{P_2} = \frac{r_1^2T_1^4}{r_2^2T_2^4}. \] Substituting values, \[ \frac{P_1}{P_2} = \frac{(1)^2(2000)^4}{(2)^2(1000)^4}. \]

Step 2: Simplify. \[ = \frac{1}{4} \left(\frac{2000}{1000}\right)^4 \] \[ = \frac{1}{4}(2)^4 \] \[ = \frac{16}{4} \] \[ =4. \] Therefore, \[ P_1:P_2=4:1. \] \[\begin{aligned} \boxed{4:1} \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.