Question 120 of 126

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)

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

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Answer: (B) $\dfrac{M\alpha }{16\pi {R}^{2}\sigma }\ln \left(\dfrac{16}{3}\right)$

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



$\dfrac{d\theta }{dt}=-\dfrac{\sigma Ae}{M\alpha {\theta }^{3}}\left({\theta }^{4}-{\theta }_{0}^{4}\right)$
$\int \dfrac{{\theta }^{3}}{\left({\theta }^{4}-{\theta }_{0}^{4}\right)}d\theta =-\dfrac{\sigma Ae}{M\alpha }\int dt$
Let $\left({\theta }^{4}-{\theta }_{0}^{4}\right)=x$
$4{\theta }^{3}d\theta =dx$
$\int \dfrac{dx}{4x}=-\dfrac{\sigma Ae}{M\alpha }{\left[t\right]}_{0}^{t}$
$\dfrac{1}{4}\left[\ln x\right]=\dfrac{M\sigma Ae}{M\alpha }{\left[t\right]}_{0}^{t}$
$\dfrac{1}{4}\left[\ln \left({\theta }^{4}-{\theta }_{0}^{4}\right)\right]\begin{matrix}2{T}_{0} \\ 3{T}_{0}\end{matrix}=-\dfrac{\sigma Ae}{M\alpha }t$
$\dfrac{1}{4}\ln \left(\dfrac{{\left(2{T}_{0}\right)}^{4}-{T}_{0}^{4}}{{\left(3{T}_{0}\right)}^{4}-{T}_{0}^{4}}\right)=-\dfrac{\sigma Ae}{M\alpha }t$
$t=-\dfrac{\alpha M}{4\sigma Ae}\ln \left(\dfrac{15{T}_{0}^{4}}{80{T}_{0}^{4}}\right)$
$=-\dfrac{\alpha M}{4\sigma Ae}\ln \left(\dfrac{3}{16}\right)$
$=-\dfrac{\alpha M}{4\sigma \left(4\pi {R}^{2}\right)e}\ln \left(\dfrac{3}{16}\right)$ where $e=1$
$t=\dfrac{\alpha M}{16\pi \sigma {R}^{2}}\ln \left(\dfrac{16}{3}\right)$