Two identical thin rings each of radius R are coaxially placed at a distance R. If mass of rings are m1, m2 respectively, then the work done in moving a mass m from centre of one ring to that of the other is

Question

Two identical thin rings each of radius R are coaxially placed at a distance R. If mass of rings are m1, m2 respectively, then the work done in moving a mass m from centre of one ring to that of the other is (A) (Gm(m1+m2)/R)(√2+1) (B) (Gm(m1-m2)/√2R)(√2-1) (C) (Gm√2/R)(m1+m2) (D) zero

Tardigrade Admin · Accepted Answer

The situation is as shown in the figure.

Gravitational potential at the centre of the first ring (i.e., at

O_{1}

) is

V_{1} = \frac{G m _{1}}{R} - \frac{G M _{2}}{R ^{2} + R ^{2}}

= - \frac{G m _{1}}{R} - \frac{G m _{2}}{2 R}

Gravitational potential at the centre of the second ring

(i . e . a t O_{2})

is

V_{2} = - \frac{G m _{2}}{R} - \frac{G m _{1}}{R ^{2} + R ^{2}}

= - \frac{G m _{2}}{R} - \frac{G m _{1}}{2 R}

Work done in moving a mass

m

from

O_{1}

to

O_{2}

is

W = m (V_{2} - V_{1})

= m [- \frac{G m _{2}}{R} - \frac{G m _{1}}{2 R} - (- \frac{G m _{1}}{R} - \frac{G m _{2}}{2 R})]

= m [- \frac{G m _{2}}{R} - \frac{G m _{1}}{2 R} + \frac{G m _{1}}{R} + \frac{G m _{2}}{2 R}]

= \frac{G m}{R} [\frac{- 2 m _{2} - m _{1} + 2 m _{1} + m _{2}}{2}]

= \frac{G m}{2 R} [2 (m_{1} - m_{2}) - (m_{1} - m_{2})

= \frac{G m ( m _{1} - m _{2} )}{2 R} (2 - 1)

Q. Two identical thin rings each of radius $R$ are coaxially placed at a distance $R$ . If mass of rings are $m_{1}$ , $m_{2}$ respectively, then the work done in moving a mass $m$ from centre of one ring to that of the other is

$\frac{G m ( m _{1} + m _{2} )}{R} (2 + 1)$

$\frac{G m ( m _{1} - m _{2} )}{2 R} (2 - 1)$

$\frac{G m 2}{R} (m_{1} + m_{2})$

zero

Q. Two identical thin rings each of radius R are coaxially placed at a distance R. If mass of rings are m1​, m2​ respectively, then the work done in moving a mass m from centre of one ring to that of the other is

RGm(m1​+m2​)​(2​+1)

2​RGm(m1​−m2​)​(2​−1)

RGm2​​(m1​+m2​)

zero

Q. Two identical thin rings each of radius $R$ are coaxially placed at a distance $R$ . If mass of rings are $m_{1}$ , $m_{2}$ respectively, then the work done in moving a mass $m$ from centre of one ring to that of the other is

$\frac{G m ( m _{1} + m _{2} )}{R} (2 + 1)$

$\frac{G m ( m _{1} - m _{2} )}{2 R} (2 - 1)$

$\frac{G m 2}{R} (m_{1} + m_{2})$