5) A uniform solid sphere of radius R has a mass M.A spherical position of the sphere centred at distance R/2 from the centre C of the sphere and of radius R/2 is cut away and removed. The force of gravity due to the remaining mass of the sphere on unit mass at a point P at a distance 2R from C along CO will be (in magnitude)

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$\frac{7 GM}{8 R^{2}}$

$\frac{7 GM}{32 R^{2}}$

$\frac{7 GM}{36 R^{2}}$

$\frac{7 GM}{16 R^{2}}$

answer is B.

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Detailed Solution

Mass of cut-out area : M/8
F₁ = force due to complete disc $= \frac{GM}{(2 R)^{2}} = \frac{GM}{4 R^{2}}$
F₂ = Force due to cavity $= \frac{G (M / 8)}{{(2 R - R / 2)}^{2}} = \frac{GM}{18 R^{2}}$
F₂ = Required force $= F_{1} - F_{2} = \frac{7 GM}{36 R^{2}}$