The temperature of a metallic sphere of radius R is increased by a...

The initial volume of the metallic sphere is $V = \dfrac{4}{3}\pi R^3$.

The coefficient of volume expansion $\gamma$ is related to the linear coefficient of thermal expansion $\alpha$ as $\gamma = 3\alpha$.

The increase in volume $\Delta V$ due to a temperature increase $\Delta T$ is given by $\Delta V = V \gamma \Delta T$.

Substituting the values of $V$ and $\gamma$, we get:

$\Delta V = \left(\dfrac{4}{3}\pi R^3\right)(3\alpha)\Delta T$

$\Delta V = 4\pi R^3\alpha\Delta T$

Answer: $4\pi R^3\alpha\Delta T$