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.
120o
135o
150o
30o
A
30o
B
120o
C
150o
D
135o
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Solution
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As shown in figure, h=L+sinϕ
By energy conservation,
12mv2=mgh+12mv′2
12×(√7gl2)2=g(L+Lsinϕ)+12v′2
v′2=−2gL−2gLsinϕ+72gL
=32gL−2gLsinϕ
If string slack then,
mgsinϕ=mv′2L
gsinϕ=1L(v′)2=1L(32gL−2gLsinϕ)
gsinϕ=32g−2gsinϕ
3gsinϕ=32g
sinϕ=12
ϕ=π6
θ=π2+π6=4π6=120∘
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