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NEET PHYSICSThermal Properties of MatterMedium

Question

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}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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