A uniform ring of mass m and radius r is placed directly above a uniform sphere of mass M and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance $r \sqrt{3}$ as shown in the figure. The gravitational force exerted by the sphere on the ring will be.

\frac{G M m}{8 r^{2}}

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\frac{G M m}{4 r^{2}}

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\sqrt{3} \frac{G M m}{8 r^{2}}

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\frac{G M m}{8 r^{3} \sqrt{3}}

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Solution

The correct option is C $\sqrt{3} \frac{G M m}{8 r^{2}}$
R.E.F image
On a differential part
of ring , the gravitational
force (dF) by sphere $= \frac{G M d m (2) \to r}{(2 R^{2}) (2 R)}$
$\Rightarrow d \to F = \frac{4 M d m \to r}{4 R^{2} (R)} = \frac{G M d m \to r}{4 R^{3}}$
Total gravitational $(F) = \int_{0}^{m} \frac{G M d m \to r}{4 R^{3}}$
$\to r$ is resolved into vectors are
with magnitude $(\frac{\sqrt{3} \to r}{2})$ directed perpendicular
to the plane of ring and
other radially with magnitude $(\frac{| \to r |}{2})$
Let them be $r_{⊥}$ and $_{c}$ receptively.
Then,
$\to r =_{⊥} + \to r c$
So, $d F = \frac{G M d m \to r}{4 R^{3}} = \frac{G M d m}{4 R^{3}} (_{⊥} +_{c})$
$\to F = \int_{0}^{m} \to d f = \int_{0}^{m} \frac{G M d m (_{⊥})}{4 R^{3}} + \int_{0}^{m} \frac{G M d m_{c}}{4 R^{3}}$
$\to F = \int_{0}^{m} \frac{G M d m}{4 R^{3}} (\frac{\sqrt{3}}{2}) R_{⊥} + \int_{0}^{m} \frac{G M d m}{4 R^{3}} (\frac{R}{2})_{c}$
$\to F = \frac{\sqrt{3} G M}{8 R^{2}} \int_{0}^{m} d m_{⊥} + \frac{G M}{8 R^{2}} \int_{0}^{m} d m_{c}$
At every point $_{⊥}$ is same but
$_{c}$ changes accordingly such that
$\int_{0}^{m} d m_{⊥} = M_{⊥}$ and $\int_{0}^{m}_{c} = 0$
So, $\to F = \frac{\sqrt{3} G M}{8 R^{2}} m_{⊥}$
option C is correct.

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