JEE Main 2014 — Physics Thermodynamics

2014mcqhard

A black coloured solid sphere of radius $R$ and mass $M$ is inside a cavity with a vacuum inside. The walls of the cavity are maintained at temperature ${T}_{0}$ . The initial temperature of the sphere is $3{T}_{0}$ . If the specific heat of the material of the sphere varies as $\alpha {T}^{3}$ per unit mass with the temperature $T$ of the sphere, where $\alpha$ is a constant, then the time taken for the sphere to cool down to temperature $2{T}_{0}$ will be

( $\sigma$ is Stefan Boltzmann constant)

Official previous-year question

Held on 19 Apr 2014 · Verified 6 Jul 2026.

Options

A
$\frac{M\alpha }{16\pi {R}^{2}\sigma }\mathrm{ln}(\frac{3}{2})$
B
$\frac{M\alpha }{16\pi {R}^{2}\sigma }\mathrm{ln}(\frac{16}{3})$
C
$\frac{M\alpha }{4\pi {R}^{2}\sigma }\mathrm{ln}(\frac{3}{2})$
D
$\frac{M\alpha }{4\pi {R}^{2}\sigma }\mathrm{ln}(\frac{16}{3})$

Did you get this right?

Solution

$\frac{d\theta }{dt}=-\frac{\sigma Ae}{Ms}({\theta }^{4}-{\theta }_{0}^{4})$

JEE Main 2014 Physics, Thermal Properties of Matter — solution figure

$\frac{d\theta }{dt}=-\frac{\sigma Ae}{M\alpha {\theta }^{3}}({\theta }^{4}-{\theta }_{0}^{4})$

$\int \frac{{\theta }^{3}}{({\theta }^{4}-{\theta }_{0}^{4})}d\theta =-\frac{\sigma Ae}{M\alpha }\int dt$

Let $({\theta }^{4}-{\theta }_{0}^{4})=x$

$4{\theta }^{3}d\theta =dx$

$\int \frac{dx}{4x}=-\frac{\sigma Ae}{M\alpha }{[t]}_{0}^{t}$

$\frac{1}{4}[\mathrm{ln}x]=\frac{M\sigma Ae}{M\alpha }{[t]}_{0}^{t}$

$\frac{1}{4}[\mathrm{ln}({\theta }^{4}-{\theta }_{0}^{4})]\begin{matrix}2{T}_{0} \\ 3{T}_{0}\end{matrix}=-\frac{\sigma Ae}{M\alpha }t$

$\frac{1}{4}\mathrm{ln}(\frac{{(2{T}_{0})}^{4}-{T}_{0}^{4}}{{(3{T}_{0})}^{4}-{T}_{0}^{4}})=-\frac{\sigma Ae}{M\alpha }t$

$t=-\frac{\alpha M}{4\sigma Ae}\mathrm{ln}(\frac{15{T}_{0}^{4}}{80{T}_{0}^{4}})$

$=-\frac{\alpha M}{4\sigma Ae}\mathrm{ln}(\frac{3}{16})$

$=-\frac{\alpha M}{4\sigma (4\pi {R}^{2})e}\mathrm{ln}(\frac{3}{16})$ where $e=1$

$t=\frac{\alpha M}{16\pi \sigma {R}^{2}}\mathrm{ln}(\frac{16}{3})$

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Options

$\frac{M\alpha }{16\pi {R}^{2}\sigma }\mathrm{ln}(\frac{3}{2})$

$\frac{M\alpha }{16\pi {R}^{2}\sigma }\mathrm{ln}(\frac{16}{3})$

$\frac{M\alpha }{4\pi {R}^{2}\sigma }\mathrm{ln}(\frac{3}{2})$

$\frac{M\alpha }{4\pi {R}^{2}\sigma }\mathrm{ln}(\frac{16}{3})$

Solution

$\frac{d\theta }{dt}=-\frac{\sigma Ae}{Ms}({\theta }^{4}-{\theta }_{0}^{4})$

JEE Main 2014 Physics, Thermal Properties of Matter — solution figure

$\frac{d\theta }{dt}=-\frac{\sigma Ae}{M\alpha {\theta }^{3}}({\theta }^{4}-{\theta }_{0}^{4})$

$\int \frac{{\theta }^{3}}{({\theta }^{4}-{\theta }_{0}^{4})}d\theta =-\frac{\sigma Ae}{M\alpha }\int dt$

Let $({\theta }^{4}-{\theta }_{0}^{4})=x$

$4{\theta }^{3}d\theta =dx$

$\int \frac{dx}{4x}=-\frac{\sigma Ae}{M\alpha }{[t]}_{0}^{t}$

$\frac{1}{4}[\mathrm{ln}x]=\frac{M\sigma Ae}{M\alpha }{[t]}_{0}^{t}$

$\frac{1}{4}[\mathrm{ln}({\theta }^{4}-{\theta }_{0}^{4})]\begin{matrix}2{T}_{0} \\ 3{T}_{0}\end{matrix}=-\frac{\sigma Ae}{M\alpha }t$

$\frac{1}{4}\mathrm{ln}(\frac{{(2{T}_{0})}^{4}-{T}_{0}^{4}}{{(3{T}_{0})}^{4}-{T}_{0}^{4}})=-\frac{\sigma Ae}{M\alpha }t$

$t=-\frac{\alpha M}{4\sigma Ae}\mathrm{ln}(\frac{15{T}_{0}^{4}}{80{T}_{0}^{4}})$

$=-\frac{\alpha M}{4\sigma Ae}\mathrm{ln}(\frac{3}{16})$

$=-\frac{\alpha M}{4\sigma (4\pi {R}^{2})e}\mathrm{ln}(\frac{3}{16})$ where $e=1$

$t=\frac{\alpha M}{16\pi \sigma {R}^{2}}\mathrm{ln}(\frac{16}{3})$

JEE Main 2014 — Physics Thermodynamics

Options

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Solution

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JEE Main 2014 — Physics Thermodynamics

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics