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NEET PHYSICSThermal Properties of MatterMedium

Question

The total radiant energy per unit area per unit time, normal to the direction of incidence, received at a distance RR from the centre of a star of radius rr, whose outer surface radiates as a black body at a temperature T KT\text{ K} is given by (where σ\sigma is Stefan's constant):

A

σr2T4R2\frac{\sigma r^2 T^4}{R^2}

B

σr2T44πr2\frac{\sigma r^2 T^4}{4\pi r^2}

C

σr4T4r4\frac{\sigma r^4 T^4}{r^4}

D

4πσr2T4R2\frac{4\pi \sigma r^2 T^4}{R^2}

Step-by-Step Solution

According to the Stefan-Boltzmann law, the total radiant power PP emitted by a star of radius rr and temperature TT is P=σAT4=σ(4πr2)T4P = \sigma A T^4 = \sigma (4\pi r^2) T^4. This energy spreads uniformly over a spherical region as it travels outwards. At a distance RR from the center of the star, the energy crosses a spherical surface of area 4πR24\pi R^2. Therefore, the radiant energy received per unit area per unit time (intensity II) at distance RR is given by: I=P4πR2=σ(4πr2)T44πR2=σr2T4R2I = \frac{P}{4\pi R^2} = \frac{\sigma (4\pi r^2) T^4}{4\pi R^2} = \frac{\sigma r^2 T^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.

PHYSICSThermal Properties of Matterradiantenergynormaldirectionincidence

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