Electrostatics

JEE Main level, single correct, medium

177 of 311

Question 177 of 311

If VV is the gravitational potential due to sphere of uniform density on its surface, then its value at the centre of sphere will be:

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Answer: (D) 3V2\dfrac{3V}{2}

The magnitude of gravitational potential of a solid sphere at a distance rr from the centre is
V=−GM2R3(3R2−r2)V=\dfrac{-GM}{2{R}^{3}}(3{R}^{2}-{r}^{2})
At r=Rr=R, V=−GMRV=\dfrac{-GM}{R}.
At r=0r=0, V′=−GM2R3(3R2)=−3GM2RV'=\dfrac{-GM}{2{R}^{3}}(3{R}^{2})=\dfrac{-3GM}{2R}
Clearly, V′=3V2V'=\dfrac{3V}{2}.

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