JEE MAIN - Physics (2023 - 11th April Evening Shift - No. 17)
If $$\mathrm{V}$$ is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:-
$$\frac{3 \mathrm{~V}}{2}$$
$$\frac{\mathrm{V}}{2}$$
$$\frac{4}{3} \mathrm{~V}$$
$$\mathrm{V}$$
Explanation
The gravitational potential (V) due to a sphere of uniform density at a distance r from its center is given by:
$$ V(r) = \frac{GM}{2R^3} \left(3R^2 - r^2\right) $$
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):
$$ V(0) = \frac{GM}{2R^3} \left(3R^2 - 0^2\right) = \frac{3GM}{2R} $$
The gravitational potential at the center of the sphere is $$\frac{3}{2}$$ times the potential at the surface of the sphere. Therefore:
$$ V(0) = \frac{3V}{2} $$
$$ V(r) = \frac{GM}{2R^3} \left(3R^2 - r^2\right) $$
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):
$$ V(0) = \frac{GM}{2R^3} \left(3R^2 - 0^2\right) = \frac{3GM}{2R} $$
The gravitational potential at the center of the sphere is $$\frac{3}{2}$$ times the potential at the surface of the sphere. Therefore:
$$ V(0) = \frac{3V}{2} $$
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