A particle of mass $M$ is placed at the center of a uniform spherical shell of mass $2 M$ and radius $R$ . The gravitational potential on the surface of the shell is :

- \frac{G M}{r}

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\frac{- 3 G M}{R}

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+ \frac{G m_{1} m_{2}}{r}

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+ \frac{G m_{1} m_{2}}{R}

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Solution

The correct option is B $\frac{- 3 G M}{R}$
$\begin{matrix} p o t e n t i a l a t t h e s u r f a c e o f t h e s h e l l = p o t e n t i a l d u e t o p a r t i c l e + P o t e n t i a l d u e t o s h e l l = (\frac{- G M}{R}) + (\frac{- 2 G M}{R}) = \frac{- 3 G M}{R} \end{matrix}$
Hence, Option $B$ is correct .

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A particle of mass M is placed at the center of a uniform spherical shell of mass 2M and radius R. The gravitational potential on the surface of the shell is :

The correct option is B −3GMRpotentialatthesurfaceoftheshell=potentialduetoparticle+Potentialduetoshell=(−GMR)+(−2GMR)=−3GMRHence, Option B is correct .

A particle of mass $M$ is placed at the center of a uniform spherical shell of mass $2 M$ and radius $R$ . The gravitational potential on the surface of the shell is :