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 $$\alpha$$ is a constant, then the time taken for the sphere to cool down to temperature $$2T_0$$ will be ($$\sigma$$ is Stefan Boltzmann constant)

The sphere cools by radiation since it is in a vacuum. The net power radiated by a black body is given by the Stefan-Boltzmann law. The net power radiated per unit area is $$\sigma (T^4 - T_0^4)$$, where $$T$$ is the temperature of the sphere and $$T_0$$ is the surrounding temperature. The surface area of the sphere is $$4\pi R^2$$, so the total net power radiated is:

$$ P_{\text{net}} = 4\pi R^2 \sigma (T^4 - T_0^4) $$

This power loss equals the rate of decrease of the internal energy of the sphere. Therefore:

$$ -\frac{dU}{dt} = 4\pi R^2 \sigma (T^4 - T_0^4) $$

The internal energy $$U$$ depends on the temperature. Given the specific heat per unit mass is $$\alpha T^3$$, the heat capacity per unit mass is $$c = \alpha T^3$$. For a mass $$M$$, the change in internal energy for a temperature change $$dT$$ is:

$$ dU = M c dT = M \alpha T^3 dT $$

Thus, the rate of change of internal energy with time is:

$$ \frac{dU}{dt} = M \alpha T^3 \frac{dT}{dt} $$

Substituting into the power equation:

$$ - M \alpha T^3 \frac{dT}{dt} = 4\pi R^2 \sigma (T^4 - T_0^4) $$

Rearranging for $$dt$$:

$$ dt = -\frac{M \alpha T^3}{4\pi R^2 \sigma (T^4 - T_0^4)} dT $$

To find the time taken for the sphere to cool from $$3T_0$$ to $$2T_0$$, integrate from $$T = 3T_0$$ to $$T = 2T_0$$. Since the temperature decreases, the negative sign is handled by reversing the limits:

$$ t = \int_{2T_0}^{3T_0} \frac{M \alpha T^3}{4\pi R^2 \sigma (T^4 - T_0^4)} dT $$

Factor the denominator using $$T^4 - T_0^4 = (T^2)^2 - (T_0^2)^2 = (T^2 - T_0^2)(T^2 + T_0^2)$$. However, a substitution simplifies the integral. Let $$u = T^4 - T_0^4$$, then $$du = 4T^3 dT$$, so $$T^3 dT = \frac{du}{4}$$.

Change the limits: when $$T = 2T_0$$, $$u = (2T_0)^4 - T_0^4 = 16T_0^4 - T_0^4 = 15T_0^4$$. When $$T = 3T_0$$, $$u = (3T_0)^4 - T_0^4 = 81T_0^4 - T_0^4 = 80T_0^4$$.

Substitute:

$$ t = \int_{15T_0^4}^{80T_0^4} \frac{M \alpha}{4\pi R^2 \sigma} \cdot \frac{1}{u} \cdot \frac{du}{4} = \frac{M \alpha}{4\pi R^2 \sigma} \cdot \frac{1}{4} \int_{15T_0^4}^{80T_0^4} \frac{du}{u} $$

The integral of $$\frac{du}{u}$$ is $$\ln|u|$$, and since $$u > 0$$, it is $$\ln u$$:

$$ t = \frac{M \alpha}{16\pi R^2 \sigma} \left[ \ln u \right]_{15T_0^4}^{80T_0^4} = \frac{M \alpha}{16\pi R^2 \sigma} \left( \ln(80T_0^4) - \ln(15T_0^4) \right) $$

Simplify using logarithm properties:

$$ t = \frac{M \alpha}{16\pi R^2 \sigma} \ln \left( \frac{80T_0^4}{15T_0^4} \right) = \frac{M \alpha}{16\pi R^2 \sigma} \ln \left( \frac{80}{15} \right) $$

Reduce $$\frac{80}{15}$$ by dividing numerator and denominator by 5:

$$ \frac{80}{15} = \frac{16}{3} $$

Thus:

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

Comparing with the options, this matches Option B.

Hence, the correct answer is Option B.