QJEEM23Pheb2e: JEE Main 2023 Physics - Class 11 JEEM23Pheb2e Solution: If V is the gravitational potential due to sphere of uniform…

The gravitational potential (V) due to a sphere of uniform density at a distance r from its center is given by:

$$V(r)=\frac{GM}{2R^3}(3R^2-r^2)$$

At the surface of the sphere (r = R), the gravitational potential is:

$$V=\frac{GM}{R}$$

Now, let's find the gravitational potential at the center of the sphere (r = 0):

$$V(0)=\frac{GM}{2R^3}(3R^2-0^2)=\frac{3GM}{2R}$$

The gravitational potential at the center of the sphere is $$\frac{3}{2}$$ times the potential at the surface of the sphere. Therefore:

$$V(0)=\frac{3V}{2}$$