A solid spherical body of radius R , density , and specific heat...

The rate of heat loss from the spherical body by thermal radiation is given by the Stefan-Boltzmann law:

$\dfrac{dQ}{dt} = \sigma A T^4$

where $A = 4 \pi R^2 is the surface area of the sphere.<br><br>The heat loss is also related to the rate of fall of temperature by:<br><br>$\dfrac{dQ}{dt} = -m s \dfrac{dT}{dt}$<br><br>where$m = \dfrac{4}{3} \pi R^3 \rho$is the mass of the sphere.<br><br>Equating the two expressions:<br><br>$-\left(\dfrac{4}{3} \pi R^3 \rho\right) s \dfrac{dT}{dt} = \sigma (4 \pi R^2) T^4$<br><br>$-\dfrac{dT}{dt} = \dfrac{3 \sigma}{\rho s R} T^4$<br><br>The rate of fall of temperature is proportional to$T^4$, which gives$x = 4$.<br><br>To find the time taken for the temperature to drop from$T_0 to \dfrac{T_0}{2}$, we integrate the expression:<br><br>$\int_{T_0}^{\dfrac{T_0}{2}} T^{-4} dT = -\int_{0}^{t} \dfrac{3 \sigma}{\rho s R} dt$<br><br>\left[ \dfrac{T^{-3}}{-3} \right]_{T_0}^{\dfrac{T_0}{2}} = -\dfrac{3 \sigma}{\rho s R} t$

$-\dfrac{1}{3} \left( \left(\dfrac{T_0}{2}\right)^{-3} - T_0^{-3} \right) = -\dfrac{3 \sigma}{\rho s R} t$

$\dfrac{1}{3} \left( 8 T_0^{-3} - T_0^{-3} \right) = \dfrac{3 \sigma}{\rho s R} t$

$\dfrac{7}{3} T_0^{-3} = \dfrac{3 \sigma}{\rho s R} t$

$t = \dfrac{7 \rho s R}{9 \sigma} T_0^{-3}$

From this, we can see that $t \propto \sigma^{-1} \rho^1 s^1 R^1 T_0^{-3}$.

Comparing this with $t \propto \sigma^\alpha \rho^\beta s^\gamma R^\delta T_0^{-3}$, we get:

$\alpha = -1$

$\beta = 1$

$\gamma = 1$

$\delta = 1$

Thus, the correct set of values is $x=4, \alpha=-1, \beta=1, \gamma=1, \delta=1$.

Answer: $x=4, \alpha=-1, \beta=1, \gamma=1, \delta=1$