Gravitation

JEE Main level, single correct, medium

44 of 223

Question 44 of 223

The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point, P\mathrm{P}, located as shown, would be (nearly) :

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Answer: (C) 78GMx2\frac{7}{8} \frac{G M}{x^2}

Let mass of smaller sphere (which has to be removed) is m\mathrm{m}
Radius =R2(=\frac{\mathrm{R}}{2}( from figure ))
M43πR3=m43π(R2)3⇒m=M8\begin{aligned} & \frac{\mathrm{M}}{\frac{4}{3} \pi \mathrm{R}^3}=\frac{\mathrm{m}}{\frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^3} \\ & \Rightarrow \mathrm{m}=\frac{\mathrm{M}}{8} \end{aligned}
Mass of the left over part of the sphere
M′=M−M8=78M\mathrm{M}^{\prime}=\mathrm{M}-\frac{\mathrm{M}}{8}=\frac{7}{8} \mathrm{M}
Therefore gravitational field due to the left over part of the sphere
=GM′x2=78GMx2=\frac{\mathrm{GM}^{\prime}}{\mathrm{x}^2}=\frac{7}{8} \frac{\mathrm{GM}}{\mathrm{x}^2}

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