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JEE Main 2014 (19 Apr, Shift 1) · Thermal Radiation: Black Body, Stefan & Wien

Thermal Radiation: Black Body, Stefan & Wien question from JEE Main 2014 (19 Apr, Shift 1)

Q101JEE Main 2014 · 19 APR · SHIFT 1Difficult

Thermal Radiation: Black Body, Stefan & Wien

A black coloured solid sphere of radius R and mass M is inside a cavity with vacuum inside. The walls of the cavity are maintained at a temperature T0. The initial temperature of the sphere is 3T0. If the specific heat of the material of the sphere varies as α\alphaT3 per unit mass with the temperature T of the sphere, where α\alpha is a constant, then the time taken for the sphere to cool down to temperature 2T0 will be (σ\sigma is Stefan Boltzmann constant)

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Answer: (c)

  1. Net loss by radiation: −McdTdt=σ(4πR2)(T4−T04)-Mc\frac{dT}{dt} = \sigma (4\pi R^2)(T^4 - T_0^4) with c=αT3c = \alpha T^3.
  2. Mα4πR2σ∫3T02T0−T3 dTT4−T04=∫0tdt\frac{M\alpha}{4\pi R^2 \sigma} \int_{3T_0}^{2T_0} \frac{-T^3\,dT}{T^4 - T_0^4} = \int_0^t dt
  3. ∫T3dTT4−T04=14ln⁡(T4−T04)\int \frac{T^3 dT}{T^4 - T_0^4} = \frac{1}{4}\ln(T^4 - T_0^4), so t=Mα16πR2σln⁡81T04−T0416T04−T04t = \frac{M\alpha}{16\pi R^2\sigma}\ln\frac{81T_0^4 - T_0^4}{16T_0^4 - T_0^4}
  4. 8015=163⇒t=Mα16πR2σln⁡(163)\frac{80}{15} = \frac{16}{3} \Rightarrow t = \frac{M\alpha}{16\pi R^2\sigma}\ln\left(\frac{16}{3}\right)
  5. So the answer is (c).
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