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MHT CET 2026 (20 Apr, Shift 1) · Thermal Radiation: Black Body, Stefan & Wien

Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2026 (20 Apr, Shift 1)

Q21MHT CET 2026 · 20 APR · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien

A sphere is at temperature 600600 K. In an external environment of 200200 K, its cooling rate is R. When the temperature of the sphere falls to 400400 K then cooling rate R' will become

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Answer: (d)

  1. Net rate of cooling ∝T4−T04\propto T^4 - T_0^4 with T0=200T_0 = 200 K.
  2. At 600 K: 6004−2004=(81−1)×2004=80×2004600^4 - 200^4 = (81 - 1) \times 200^4 = 80 \times 200^4
  3. At 400 K: 4004−2004=(16−1)×2004=15×2004400^4 - 200^4 = (16 - 1) \times 200^4 = 15 \times 200^4
  4. R′=1580R=316RR' = \dfrac{15}{80}R = \dfrac{3}{16}R
  5. So the answer is (d).
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