Heat Capacity of Metal Rod Heated at Constant Power

To find the heat capacity $C$ of the metal rod as a function of temperature $T$, we start from the fundamental definitions of heat capacity and power.

The rate of heat supplied to the metal rod by the current-carrying wire is given by the constant power $P$: $P = \frac{dQ}{dt}$

By definition, the heat capacity $C$ of a body relates the heat supplied $dQ$ to the resulting temperature change $dT$: $C = \frac{dQ}{dT}$

Using the chain rule, we can express the power $P$ as: $P = \frac{dQ}{dt} = \frac{dQ}{dT} \cdot \frac{dT}{dt} = C \cdot \frac{dT}{dt}$

Therefore, the heat capacity $C$ is given by: $C = \frac{P}{\frac{dT}{dt}}$

We are given the temperature variation with time as: $T(t) = T_0 \left(1 + \beta t^{1/4}\right) = T_0 + T_0 \beta t^{1/4}$

Differentiating $T(t)$ with respect to time $t$: $\frac{dT}{dt} = T_0 \beta \cdot \frac{d}{dt}\left(t^{1/4}\right) = \frac{T_0 \beta}{4} t^{-3/4} = \frac{T_0 \beta}{4 t^{3/4}}$

From the temperature equation, we can express $t^{1/4}$ in terms of $T(t)$: $T(t) - T_0 = T_0 \beta t^{1/4} \implies t^{1/4} = \frac{T(t) - T_0}{\beta T_0}$

Raising both sides to the power of $3$: $t^{3/4} = \left(\frac{T(t) - T_0}{\beta T_0}\right)^3 = \frac{(T(t) - T_0)^3}{\beta^3 T_0^3}$

Substituting $t^{3/4}$ back into the expression for $\frac{dT}{dt}$: $\frac{dT}{dt} = \frac{T_0 \beta}{4 \left( \frac{(T(t) - T_0)^3}{\beta^3 T_0^3} \right)} = \frac{\beta^4 T_0^4}{4 (T(t) - T_0)^3}$

Now, substituting $\frac{dT}{dt}$ back into the expression for $C$: $C = \frac{P}{\frac{\beta^4 T_0^4}{4 (T(t) - T_0)^3}} = \frac{4P(T(t) - T_0)^3}{\beta^4 T_0^4}$

Thus, the correct option is (A).