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Standard XII

Physics

Finding Average Velocity

A particle of...

Question

A particle of mass 1kg is placed at a distance of 4m from the centre and on the axis of a uniform ring of mass 5kg and radius 3m. The work done to increase the distance of the particle from 4m to $\sqrt{3} m$ is.

\frac{G}{3} J

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\frac{G}{4} J

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\frac{G}{5} J

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\frac{G}{6} J

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Solution

The correct option is D $\frac{G}{6} J$
Potential of a ring, $V = \frac{- G M}{y}$
where, $y = \sqrt{x^{2} + r^{2}}$
So initial, energy is $\frac{- 5 G \times 1}{5}$
Final energy is $\frac{- 5 G \times 1}{6}$
So the difference $=$ Final $-$ Initial $= \frac{G}{6}$

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A particle of mass 1kg is placed at a distance of 4m from the centre and on the axis of a uniform ring of mass 5kg and radius 3m. The work done to increase the distance of the particle from 4m to √3m is.

The correct option is D G6JPotential of a ring, V=−GMywhere, y=√x2+r2So initial, energy is −5G×15Final energy is −5G×16So the difference = Final − Initial =G6

A particle of mass 1kg is placed at a distance of 4m from the centre and on the axis of a uniform ring of mass 5kg and radius 3m. The work done to increase the distance of the particle from 4m to $\sqrt{3} m$ is.

The correct option is D $\frac{G}{6} J$
Potential of a ring, $V = \frac{- G M}{y}$
where, $y = \sqrt{x^{2} + r^{2}}$
So initial, energy is $\frac{- 5 G \times 1}{5}$
Final energy is $\frac{- 5 G \times 1}{6}$
So the difference $=$ Final $-$ Initial $= \frac{G}{6}$