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Standard XII

Physics

Gravitational Field Due to a Ring

A uniform rin...

Question

A uniform ring and uniform sphere are placed as shown in the figure. The gravitational attraction between the sphere and ring ( $d = \sqrt{3} R$ ) is.

\frac{8 G M^{2}}{R^{2}}

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

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\frac{3 G M^{2}}{2 R^{2}}

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\frac{\sqrt{3} G M^{2}}{R^{2}}

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Solution

The correct option is D $\frac{\sqrt{3} G M^{2}}{R^{2}}$
Gravitational field due to ring at a distance
$d = \sqrt{3}$ R on its axis is $E = \frac{G M d}{(R^{2} + d^{2})^{3 / 2}} = \frac{\sqrt{3} G M}{8 R^{2}}$
Force on the sphere = mass of sphere × E

$= 8 M E = \frac{\sqrt{3} G M^{2}}{R^{2}}$

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A uniform ring and uniform sphere are placed as shown in the figure. The gravitational attraction between the sphere and ring (d=√3R) is.

The correct option is D √3GM2R2 Gravitational field due to ring at a distance d=√3 R on its axis is E=GMd(R2+d2)3/2 =√3GM8R2 Force on the sphere = mass of sphere × E =8ME=√3GM2R2

A uniform ring and uniform sphere are placed as shown in the figure. The gravitational attraction between the sphere and ring ( $d = \sqrt{3} R$ ) is.

The correct option is D $\frac{\sqrt{3} G M^{2}}{R^{2}}$
Gravitational field due to ring at a distance
$d = \sqrt{3}$ R on its axis is $E = \frac{G M d}{(R^{2} + d^{2})^{3 / 2}} = \frac{\sqrt{3} G M}{8 R^{2}}$
Force on the sphere = mass of sphere × E

$= 8 M E = \frac{\sqrt{3} G M^{2}}{R^{2}}$