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Properties of Matter question

2010 · Q178
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Properties of Matter question

2010 · Q178

NEETPhysicsProperties of MatterMCQ+4 / −1
The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature TK is given by
  1. A
    σr2T4R2{{\sigma {r^2}{T^4}} \over {{R^2}}}R2σr2T4​
  2. B
    σr2T44πR2{{\sigma {r^2}{T^4}} \over {4\pi {R^2}}}4πR2σr2T4​
  3. C
    σr2T4R4{{\sigma {r^2}{T^4}} \over {{R^4}}}R4σr2T4​
  4. D
    4πσr2T4R2{{4\pi \sigma {r^2}{T^4}} \over {{R^2}}}R24πσr2T4​
View written solutionFree

Correct answer: A

According to the Stefan Boltzmann law, the power radiated by the star whose outer surface radiates as a black body at temperature T K is given by

P=σ4πr2T4P = \sigma 4\pi {r^2}{T^4}P=σ4πr2T4


where, r = radius of the star
σ\sigma σ = Stefan’s constant

The radiant power per unit area received at a distance R from the centre of a star is

S=P4πR2=σ4πr2T44πR2=σr2T4R2S = {P \over {4\pi {R^2}}} = {{\sigma 4\pi {r^2}{T^4}} \over {4\pi {R^2}}} = {{\sigma {r^2}{T^4}} \over {{R^2}}}S=4πR2P​=4πR2σ4πr2T4​=R2σr2T4​

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