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Question

# The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of √3gl. Find the angle rotated by the string before it becomes slack.

A
cos1(13)
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B
cos1(13)
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C
cos1(23)
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D
cos1(23)
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Solution

## The correct option is B cos−1(−13)Let the string make an angle θ with the vertical finally. FBD at this position is given below. u=√3gl (given) Applying energy conservation at A and B, taking A as a reference 12mv2+mgh=12mu2 where h is the vertical height of point B. h=l+lcosθ ⇒v2=u2−2g(l+l cos θ) ⇒v2=3gl−2gl(1+cos θ) ……(1) From the FBD at point B: T+mgcosθ=mv2R Again, as tension is zero (for slack string) mv2l=mgcosθ v2=lgcosθ ……(2) From (1) and (2) equation, 3gl−2gl(1+cos θ)=glcosθ 3glcosθ=gl⇒cos θ=13 ⇒θ=cos−1(13) So, the angle rotated before the string becomes slack =180∘−cos−1(13)=cos−1(−13)

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