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Question

A ball of mass 2 kg is on a frictionless horizontal surface. It is attached to one end of a string and rotates about a fixed other end at an angular velocity 2 rev/sec. If the length of the string is doubled and angular velocity is halved, the tension in the string is equal to (Let initial tension be T0 )

A
T0
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B
T02
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C
4T0
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D
8T0
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Solution

The correct option is B T02
Given, Initial tension =T0

Since tension provides centripetal force T=mlω2.

The ratio of the tension before and after will be,

T1T2=l1ω21l2ω22,
where T1 is the initial tension=T0.

Length of the string is doubled l2=2l1
Angular velocity is halved ω2=ω12
T1T2=l12l1×(2ω2ω2)2
T1T2=12×4=2
T2=T12=T02

Hence, the new tension will be half of the initial tension.

Option B is the correct answer

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