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Question

# A body crosses the topmost point of a vertical circle at critical speed. What will be its centripetal acceleration (in m/s2) when the string becomes horizontal?

A
g
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B
2g
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C
3g
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D
5g
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Solution

## The correct option is C 3g At highest point, if body is having critical speed v then tension T=0 i.e condition for string becoming just slacked. From newton's 2nd law towards centre ′o′: mg+T=mv2R.....(i) Putting T=0 in Eq.(i) ⇒mv2R=mg ∴v=√Rg Applying law of conservation of energy between point B & C, PEB+KEB=PEC+KEC ⇒mg(2R)+12mv2=mg(R)+12mv2c......(ii) Substituting v=√gR ∴vc=√3gR Now centripetal acceleration at horizontal position C: ac=v2cR=(√3gR)2R=3g ∴ac=3g m/s2 Hence option C is the correct answer

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