A body of mass m slides down an incline and reaches the bottom with a velocity v . If the same mass was in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been

Question

A body of mass m slides down an incline and reaches the bottom with a velocity v . If the same mass was in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been (A) v (B) √2v (C) (v/√2) (D) √(2/5) v

Tardigrade Admin · Accepted Answer

When body slides, v=√2 g h ....(1) When ring rolls, vâ²=√(2 g h/1 + k2 / r2)=√(2 g h/2)=(v/√2)

Q. A body of mass $m$ slides down an incline and reaches the bottom with a velocity $v$ . If the same mass was in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been

$v$

$2 v$

$\frac{v}{2}$

$\frac{2}{5} v$

When body slides, $v = 2 g h$ ....(1)
When ring rolls, $v' = \frac{2 g h}{1 + k ^{2} / r ^{2}} = \frac{2 g h}{2} = \frac{v}{2}$

Q. A body of mass m slides down an incline and reaches the bottom with a velocity v . If the same mass was in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been

v

2​v

2​v​

52​​v

When body slides, v=2gh​ ....(1) When ring rolls, v′=1+k2/r22gh​​=22gh​​=2​v​

Q. A body of mass $m$ slides down an incline and reaches the bottom with a velocity $v$ . If the same mass was in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been

$v$

$2 v$

$\frac{v}{2}$

$\frac{2}{5} v$

When body slides, $v = 2 g h$ ....(1)
When ring rolls, $v' = \frac{2 g h}{1 + k ^{2} / r ^{2}} = \frac{2 g h}{2} = \frac{v}{2}$