Question

Question Paper
PHYSICS
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A small ring of mass 'm' is attached at one end of a light string of length 1 = 0.6m. The other end of the string is tied to a small
block B of mass '2m'. The ring is free to move on a fixed smooth horizontal rod. The block B is released from rest from the position
shown. The velocity (in ms-1) of the ring when the string becomes vertical is (g = 10 ms2)
cm
M
{= 0.6 m
(A)
4
(B) 3
(C) 3.5
(D) 2.5
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Answer 1

Answer: The correct answer is option (A) 4  $\frac{m}{s}$ .

Explanation:

The ring can move freely on the string and when the string becomes vertical then let us take that the velocity of the ring is ' v ' .

∴ Deriving expression for velocity,

Now from the conservation of energy we can write that ,

$\frac{1}{2}$ $mv^2$ = mgl + $\frac{1}{3}$ mgl

$\frac{1}{2}$ $mv^2$ = $\frac{4}{3}$ mgl

$\frac{1}{2}$$v^2$ = $\frac{4}{3}$ g.l

$v^2$ = $\frac{8}{3}$.g.l

v = $\sqrt{\frac{8gl}{3 }$

Substituting values given in the question,

v = $\sqrt{\frac{8 X 10 X 0.6}{3} }$

v = $\sqrt{16}$

v = 4  $\frac{m}{s}$ .

The velocity (in ms-1) of the ring when the string becomes vertical is 4 $\frac{m}{s}$ .

Answer 2

Given:

Mass of ring = m

Length of string = 0.6m

Mass of Block B = 2m

To find:

Velocity of the ring

Solution:

Let the velocity of the ring is "v"

According to question, the ring can move freely on the string and the becomes vertical.

So, velocity can be calculated as follows.

law of conservation of mass can be written as-

$\frac{1}{2}mv^{2} = mgI + \frac{1}{3}mgI$

$\frac{1}{2}mv^{2} = \frac{4}{3}mgI$

$\frac{1}{2}v^{2} = \frac{4}{3}gI$

$v^{2} = \frac{8}{3}gI$

$v = \sqrt{\frac{8 \cdot g \cdot I}{3} }$

by putting all the values in the above equation,

$v = \sqrt{\frac{8 \times 10 \times 0.6}{3} }$

v = √16

v = 4 m/s

Hence, when string becomes vertical the velocity of the ring is 4m/s.