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MHT CET 2024 (4 May, Shift 1) · Thermal Radiation: Black Body, Stefan & Wien

Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2024 (4 May, Shift 1)

Q64MHT CET 2024 · 4 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien

A sphere is at temperature 600 K. In an external environment of 200 K, its cooling rate is ' $R$ ' When the temperature of the sphere falls to 400 K, then cooling rate ' $R'$ ' will become

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Answer: (a)

Net rate of cooling $\propto T^4 - T_0^4$ with $T_0 = 200$ K.
At 600 K: $(81 - 1) \times 200^4 = 80 \times 200^4$
At 400 K: $(16 - 1) \times 200^4 = 15 \times 200^4$
$R' = \dfrac{15}{80}R = \dfrac{3}{16}R$
So the answer is (a).

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