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Question

A bob attached to a light string of length 10 m is given an initial velocity of 20 m/s at the bottom-most point to perform vertical circular motion. What will be the velocity of bob just before slacking of string?
(Take g=10 m/s2)

A
0 m/s
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B
1023 m/s
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C
203 m/s
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D
103 m/s
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Solution

The correct option is B 1023 m/s
The FBD of the bob at the position where the slackness happen is shown in the figure

The slackness of the string happens when the tension in the string becomes zero.
Hence the component of mg along the string (mgcosθ) will balance the centrifugal force which is mv2R in the radially outward direction.

mgcosθ=mv2R ...(i),
Height of the bob, H=R+Rcosθ

Now by the conservation of mechanical energy we get,
12mu2+0=12mv2+mg(R+Rcosθ) ...(ii), where u is the initial velocity given to the bob at the bottom most point and v is the velocity at the point where the slackness happen.

Putting the value of , cosθ=v2Rg from equation (i) in the equation (ii) we get

v2=u22gR3 , where u=20 m/s, R=10 m

v=2003=1023 m/s

Hence option B is the correct answer

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