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

\frac{G}{4} J

\frac{G}{5} J

\frac{G}{6} J

Open in App

Solution

Verified by Toppr

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 $= \frac{G}{6}$

Was this answer helpful?

Similar Questions

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

View Solution

A small block of mass ‘m’ is rigidly attached at ‘P to a ring of mass ‘3m’ and radius ‘r’. The system is released from rest at $θ$ = $90^{0}$ and rolls without sliding. The angular acceleration of hoop just after release (about the centre of the ring) is

View Solution

A circular ring of mass M and radius R is placed in a gravity free space. A small particle of mass m is placed on the axis of ring at a distance $\sqrt{3}$ R from the centre of the ring. If particle and ring are released from rest. Speed with which particle passes through the centre of the ring is

View Solution

A projectile of mass $4 M$ is fired at a speed of $100 m/s$ at an angle of $37^{\circ}$ from the horizontal. At the highest point, the projectile breaks into two parts with mass $M$ and $3 M$ . Mass $M$ falls down immediately after the explosion and mass $3 M$ lands at some horizontal distance from mass $M$ . Choose the correct statement. (Take $g = 10 {m/s}^{2}$ )

View Solution

Centre of mass of a system of three particles of masses $1 kg$ , $2 kg$ and $3 kg$ at the point $(1 m, 2 m, 3 m)$ and centre of mass of another group of two particles of masses $2 kg$ and $3 kg$ is at point $(- 1 m, - 2 m)$ , where a $10 kg$ particle should be placed, so that the centre of mass of the system of all those six particles shifts to centre of mass of the first system?

View Solution

Solve

Guides

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.

Potential 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.

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 $= \frac{G}{6}$