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Physics · Heat Transfer

JEE Main 2025 — 22 January, Morning Shift — Question 49

Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of bigger body is 400 K . If the energy radiate from the smaller body is E , the energy radiated from the bigger body is (assume, effect of the surrounding to be negligible)

  1. Option A:

    256 E

  2. Option B:

    E

    Correct
  3. Option C:

    64 E

  4. Option D:

    16 E

Answer: B

Step-by-step solution

dθdt=σAT4⇒P∝AT4\frac{\mathrm{d} \theta}{\mathrm{dt}}=\sigma \mathrm{AT}^{4} \Rightarrow \mathrm{P} \propto \mathrm{AT}^{4}

Psmaller Plarger =(0.2)2×8004(0.8)2×4004\frac{\mathrm{P}_{\text {smaller }}}{\mathrm{P}_{\text {larger }}}=\frac{(0.2)^{2} \times 800^{4}}{(0.8)^{2} \times 400^{4}}

116×16=1\frac{1}{16} \times 16=1

∴Plarger =Psmaller \therefore \mathrm{P}_{\text {larger }}=\mathrm{P}_{\text {smaller }}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
22 January, Morning Shift
Subject
Physics
Chapter
Heat Transfer
Topic
Convection and Radiation

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