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Asked in JEE Main 27th Aug 2nd Shift 2021 · Potential of a sphere and a shell
The field point at 25 m lies inside the shell, so the two sources must be handled differently.
The point mass at the centre contributes -(G(50))/(25)=-2G.
The shell contributes its constant interior value -(G(100))/(50)=-2G, using the shell radius, not the field-point distance.
Potentials are scalars and simply add: V=-2G-2G=-4G.
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