JEE Main 2014 · 19 Apr · Shift 1 · Physics Q11

← JEE Main 2014 · 19 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 11 of 90

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 temperature $T_0$ . The initial temperature of the sphere is $3T_0$ . 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 α is a constant, then the time taken for the sphere to cool down to temperature $2T_0$ will be (σ is Stefan Boltzmann constant)

\dfrac{M\alpha}{4\pi R^2\sigma}\ln\left(\dfrac32\right)

\dfrac{M\alpha}{4\pi R^2\sigma}\ln\left(\dfrac{16}3\right)

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

\dfrac{M\alpha}{16\pi R^2\sigma}\ln\left(\dfrac32\right)

Answer and worked solution

Official answer: C

$M\alpha T^3\dfrac{dT}{dt}=-4\pi R^2\sigma\left(T^4-T_0^4\right)$ . Integrating from $3T_0$ to $2T_0$ :
$t=\frac{M\alpha}{4\pi R^2\sigma}\cdot\frac14\ln\frac{81T_0^4-T_0^4}{16T_0^4-T_0^4}=\frac{M\alpha}{16\pi R^2\sigma}\ln\frac{16}3.$

Topics:Radiation and Stefan's law Calorimetry and specific heat

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JEE Main 2014 · 19 Apr · Shift 1, Physics question 11