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The gravitational field, due to the leftover part of a uniform sphere from which a part as shown has been removed, at a very far off point P located as shown in the figure, would be nearly

Asked in JEE Main Online 2013 · Field inside a cavity

Figure: Field inside a cavity
Answer: (3) 7/8(GM)/(x²)

Step-by-step solution

P is very far off, so where the hole sits stops mattering and only the mass that is left counts.

The removed piece is a sphere of half the radius, so it carries (1/2)³=1/8 of the mass.

Leftover mass is M-M/8=(7M)/8.

From far away it acts as a point mass: E=(G((7M)/8))/(x²)=7/8(GM)/(x²).

Why the other options are wrong

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