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MHT CET 2026 (20 Apr, Shift 2) · Thermal Radiation: Black Body, Stefan & Wien

Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2026 (20 Apr, Shift 2)

Q20MHT CET 2026 · 20 APR · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien

A sphere having temperature 600 K is losing heat due to radiation. At this temperature its rate of cooling is RR. The rate of cooling of this sphere at 400 K is x243\frac{x}{243}. The value of xx is (Temperature of surrounding is 300 K)

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Answer: (d)

  1. Net rate of cooling ∝T4−T04\propto T^4 - T_0^4 with T0=300T_0 = 300 K.
  2. At 600 K: 6004−3004=15×3004=1215×1004600^4 - 300^4 = 15 \times 300^4 = 1215 \times 100^4
  3. At 400 K: 4004−3004=(256−81)×1004=175×1004400^4 - 300^4 = (256 - 81) \times 100^4 = 175 \times 100^4
  4. R′=R×1751215=35R243R' = R \times \dfrac{175}{1215} = \dfrac{35R}{243}
  5. So x=35Rx = 35R, answer (d).
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