Question

Consider a particle of mass m on the axis of a ring of mass M & radius r, at a distance r from the centre of ring, this particle moving under the gravitational attraction of the ring, reaches the centre of the ring. The velocity of the particle at the centre of the ring will be –

• A.
• B.
• C. $\sqrt { \frac { 2 \mathrm { GM } } { \mathrm { r } ( \sqrt { 2 } + 1 ) } }$
• D. $\sqrt { \frac { 2 \mathrm { GM } } { \mathrm { r } } \left( 1 - \frac { 1 } { \sqrt { 2 } } \right) }$

Answer 1

Answer: (D)

$\sqrt { \frac { 2 \mathrm { GM } } { \mathrm { r } } \left( 1 - \frac { 1 } { \sqrt { 2 } } \right) }$

Answer 2

Gravitational P.E. = $- \frac { \mathrm { GMm } } { \sqrt { \mathrm { R } ^ { 2 } + \mathrm { x } ^ { 2 } } }$

(T.E.)_A = (T.E.)_B

Let velocity at B is V so

K.E. + P.E. = constant

0 + = $\frac { 1 } { 2 }$ mv² +

v = $\sqrt { \frac { 2 \mathrm { GM } } { \mathrm { r } } \left( 1 - \frac { 1 } { \sqrt { 2 } } \right) }$