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A bob of mas M is suspended by a massles...

A bob of mas `M` is suspended by a massless string of length `L`. The horizontal velocity `v` at position `A` is just sufficient to make it reach the point `B`. The angle `theta` at which the speed of the bob is half of that at `A`, satisfies.
.

A

`theta = (pi)/(4)`

B

`(pi)/(4) lt theta lt (pi)/(2)`

C

`(pi)/(2) lt theta lt (3pi)/(4)`

D

`(3pi)/(4) lt theta lt pi`

Text Solution

Verified by Experts

The correct Answer is:
D
(d) This is the case of vertical motion when the body just
completes the circle. Here
`v = sqrt(5gL) …..(i)`
Applying energy conservation,
Total energy at A = Total energy at P
`(1)/(2)mv^2 = )1)/(2)m ((v)/(2))^2 + mgh.
`rArr h = (3v^2)/(8g)`
`= (3)/(8g)xx5gL = (15L)/(8)....(ii)`

In `Delta OPM, cos theta = (L -h)/(L) = ((L - (15L)/(8)/(L) = (-7)/(8)`
Therefore, the value of `theta` lies in the range `(3pi)/(4) ltthetaltpi.`
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