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# A particle suspended from a fixed point, by a light inextensible thread of length L is projected horizontally from its lowest position with velocity √7gL2. The thread will slack after swinging through an angle θ, such that θ equal.120o135o150o30o

A
120o
B
150o
C
30o
D
135o
Solution
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#### As shown in figure, h=L+sinϕBy energy conservation,12mv2=mgh+12mv′212×(√7gl2)2=g(L+Lsinϕ)+12v′2v′2=−2gL−2gLsinϕ+72gL=32gL−2gLsinϕIf string slack then,mgsinϕ=mv′2Lgsinϕ=1L(v′)2=1L(32gL−2gLsinϕ)gsinϕ=32g−2gsinϕ3gsinϕ=32gsinϕ=12ϕ=π6θ=π2+π6=4π6=120∘

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