If gravitational field due to uniform thin hemispherical shell at point $P$ is $I$ , then the magnitude of gravitational field at $Q$ is (Mass of hemisphere is $M$ , radius $R$ ).

\frac{G M}{2 R^{2}} - I

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\frac{G M}{2 R^{2}} + I

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\frac{G M}{4 R^{2}} - I

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2 I - \frac{G M}{2 R^{2}}

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Solution

The correct option is A $\frac{G M}{2 R^{2}} - I$
Assume another thin hemispherical shell of same mass and radius as shown in the figure
Due to symmetry

|Gravitational field due to $2^{n d}$ at P| = |Gravitational field due to $1^{s t}$ at Q|

At P gravitational field due to $1^{s t}$ + gravitational field due to $2^{n d} = \frac{G (2 M)}{4 R^{2}} = \frac{G M}{2 R^{2}}$
$I + I_{2 n d} = \frac{G M}{2 R^{2}}$
Gravitational field due to $2^{n d}$ at $P = \frac{G M}{2 R^{2}} - I$

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