Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $$u = \frac{U}{V} \propto T^4$$ and pressure $$p = \frac{1}{3}\left(\frac{U}{V}\right)$$. If the shell now undergoes an adiabatic expansion the relation between T and R is:

We are told that the photon gas inside the spherical cavity behaves like an ideal gas of photons. For such radiation we have two experimentally established relations:

$$u=\frac{U}{V}\propto T^{4} \qquad\text{and}\qquad p=\frac13\left(\frac{U}{V}\right).$$

Let us put a constant of proportionality so that the equations can be handled algebraically. We write

$$\frac{U}{V}=aT^{4}\quad\Longrightarrow\quad U=aT^{4}V,$$

where $$a$$ is the radiation constant.

The process described is adiabatic, so by the First Law of Thermodynamics the infinitesimal form is

$$dU+pdV=0.$$

First we find $$dU$$. Using $$U=aT^{4}V$$ we differentiate:

$$dU=d(aT^{4}V)=a\left(4T^{3}dT\right)V+aT^{4}dV.$$

Now we substitute $$dU$$ and $$p=\dfrac13 aT^{4}$$ into the adiabatic condition:

$$a\left(4T^{3}VdT+T^{4}dV\right)+\frac13 aT^{4}dV=0.$$

Every term contains the constant $$a$$, so we divide through by $$a$$ and simplify:

$$4T^{3}VdT+T^{4}dV+\frac13T^{4}dV=0.$$

The two $$dV$$ terms combine:

$$4T^{3}VdT+\left(1+\frac13\right)T^{4}dV=0 \;\Longrightarrow\; 4T^{3}VdT+\frac43T^{4}dV=0.$$

Dividing every term by $$T^{3}$$ gives

$$4V\,dT+\frac43T\,dV=0.$$

Now divide by $$4$$ to make the coefficients simpler:

$$V\,dT+\frac13T\,dV=0.$$

We separate the variables by bringing all $$T$$-terms to one side and all $$V$$-terms to the other side:

$$\frac{dT}{T}=-\frac13\frac{dV}{V}.$$

Integrating both sides, we have

$$\int\frac{dT}{T}=-\frac13\int\frac{dV}{V}$$

which yields

$$\ln T=-\frac13\ln V+\text{constant}.$$

Exponentiating converts the logarithms back to algebraic form:

$$T\,V^{\;1/3}=\text{constant}.$$

So we can write the adiabatic relation for black-body radiation as

$$T\propto V^{-1/3}.$$

The cavity is a sphere of radius $$R$$, so its volume is $$V=\dfrac43\pi R^{3}$$. Because the factor $$\dfrac43\pi$$ is a constant, we use the proportionality $$V\propto R^{3}$$. Substituting this into the temperature-volume relation gives

$$T\propto\left(R^{3}\right)^{-1/3}=R^{-1}.$$

Thus the temperature varies inversely with the radius:

$$T\propto\frac1R.$$

Among the given options this corresponds to Option D.

Hence, the correct answer is Option D.