The radius of a metal sphere at room temperature T is R, and the coefficient of linear expansion of the metal is α . The sphere is heated a little by a temperature ∆T so that its new temperature is T+ ∆T . The increase in the volume of the sphere is approximately

To find the increase in the volume of the metal sphere, we can use the concepts of thermal expansion.

Let the initial radius of the metal sphere at room temperature T be $R$.
The volume $V$ of a sphere of radius $R$ is given by the formula:
$V=\frac{4}{3}\pi R^3$

The coefficient of linear expansion of the metal is $\alpha$.
The coefficient of volume expansion, denoted by $\gamma$, is related to the coefficient of linear expansion $\alpha$ by the relation:
$\gamma =3\alpha$

When the sphere is heated by a small temperature change $\mathrm{\Delta T}$, the fractional change in its volume is given by:
$\frac{\mathrm{\Delta V}}{V}=\gamma \mathrm{\Delta T}$

Substituting $\gamma =3\alpha$ into the equation, we get:
$\frac{\mathrm{\Delta V}}{V}=3\alpha \mathrm{\Delta T}$

Now, we solve for the increase in volume, $\mathrm{\Delta V}$:
$\mathrm{\Delta V}=3\alpha V\mathrm{\Delta T}$

Substituting the initial volume $V=\frac{4}{3}\pi R^3$ into this expression:
$\mathrm{\Delta V}=3\alpha \left(\frac{4}{3}\pi R^3\right)\mathrm{\Delta T}$

Simplifying the expression, the factor of 3 cancels out:
$\mathrm{\Delta V}=4\pi R^3\alpha \mathrm{\Delta T}$

Thus, the increase in the volume of the sphere is approximately $4\pi R^3\alpha \mathrm{\Delta T}$.

Therefore, the correct answer is 4πR³α ∆T.