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Question

A bob of mass M is suspended through a massless string of length L. The horizontal velocity at position A is sufficient enough to make it reach at point B. The angle at which the speed of bob is half of that at A, satisfies


A
θ=π4
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B
π4 <θ <π2
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C
π2 <θ <3π4
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D
3π4 <θ <π
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Solution

The correct option is D 3π4 <θ <π
The bob attached to string just reaches at B that means v=0 there, hence for just completing the circular loop, velocity at bottom most point:
v=5gL ....(i)
At some position C string makes angle θ with vertical and velocity of bob is v2.


Height above bottom point A is h at position C
h=LLcosθ=L(1cosθ) ...(ii)
If an object is rising along the vertical circular path with initial velocity u, then at height h by applying mechanical energy conservation final velocity v is given by:
12mu2=12mv2+mgh
v2=u22gh ...(iii)
for position A to C
u=v, v=v2
From Eq. (ii) and (iii)
(v2)2=v22gL(1cosθ)

3v24=2gL(1cosθ)
Putting value of v from Eq. (i),
15gL4=2gL(1cosθ)
158=1cosθ
cosθ=78=0.875
It indicates that value of cosθ is ve and close to 1, so θ must be an angle of 2nd quadrant which is near to 180 or π
Hence range (3π4 <θ <π) suits most.

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