Back to Directory
NEET PHYSICSMedium

Assuming the sun to have a spherical outer surface of radius rr, radiating like a black body at temperature tCt^\circ \text{C}, the power received by a unit surface of the earth (normal to the incident rays) at a distance RR from the centre of the sun will be: (where σ\sigma is Stefan's constant)

A

4πr2σt4R2\frac{4\pi r^2\sigma t^4}{R^2}

B

r2σ(t+273)44πR2\frac{r^2\sigma(t+273)^4}{4\pi R^2}

C

16π2r2σt4R2\frac{16\pi^2 r^2\sigma t^4}{R^2}

D

r2σ(t+273)4R2\frac{r^2\sigma(t+273)^4}{R^2}

Step-by-Step Solution

According to the Stefan-Boltzmann law, the total power radiated by the sun acting as a black body is P=σAT4P = \sigma A T^4. The absolute temperature is T=(t+273) KT = (t + 273) \text{ K} and the surface area of the sun is A=4πr2A = 4\pi r^2. Thus, P=σ(4πr2)(t+273)4P = \sigma (4\pi r^2)(t+273)^4. This radiated power spreads uniformly over all directions. At a distance RR from the centre of the sun, this power is distributed over a spherical surface area of 4πR24\pi R^2. The power received per unit area (intensity) at the earth is given by: I=P4πR2=σ(4πr2)(t+273)44πR2=r2σ(t+273)4R2I = \frac{P}{4\pi R^2} = \frac{\sigma (4\pi r^2)(t+273)^4}{4\pi R^2} = \frac{r^2\sigma(t+273)^4}{R^2}.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started