A bob of mass M is suspended by a massless string of length L. The horizonta 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

Options

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`

Answer

D

Explanation

As the body just reaches the topmost point `B`,
therefore, `upsilon_(A) = sqrt(5gL)` and `upsilon_(B) = sqrt(gL)`
Let at point `P`having angular displacement `theta` the speed becomes half of the initial value
`upsilon_(A)`. Using the law of conservation of energy
Energy at A = Energy at `B`
`(1)/(2) m upsilon^(2) A=(1)/(2) m upsilon^(2) P +mgL(1-cos theta)`
`(1)/(2)m(upsilon_(A)^(2)-upsilon_(P)^(2)) = mgL (1 - cos theta)`
`(1)/(2)m(5gL -(5gL)/(4)) =mgL (1 - cos theta)`
`(15)/(8) =(1-cos theta) , cos theta = -(7)/(8)`
Which means `(3 pi)/(4) lt theta lt pi`.