If 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:-

Solution

Physics sir · 4 steps

Inside a solid sphere the potential keeps dropping, so the centre sits at 1.5 times the surface value.

1. Surface value. On and outside its surface, a uniform sphere of mass M and radius R acts like a point mass at its centre. So on the surface V = -GM/R.
2. Inside the sphere. At a distance r from the centre, with r < R, the potential of a uniform solid sphere is given by the standard result below.

   Vᵢₙ(r) = -GM/2R³(3R² - r²)

3. Check and use. Putting r = R gives -GM/2R³(2R²) = -GM/R, which matches the surface value. Putting r = 0 gives V_(centre) = -3GM/2R.
4. So. V_(centre) = 3/2(-GM/R) = 3V/2, which is option (a).

The idea to keep. For a uniform solid sphere, V_(centre) = 3/2V_(surface). Both are negative, so the centre has the lowest potential, not zero.

Answer from the NTA final answer key. Solution by PrepWiser.