Bob `B` of the pendulum `AB` is given an initial velocity `sqrt(3Lg)` in horizontal direction. Final the maximum height of the bob from the starting point,

(a) if `AB` is a massless rod,
(b) if `AB` is a massless staring.

Answer

A, B, C, D

Explanation

Applying conservation of energy
`mgh=(1)/(2)m(sqrt(3Lg))^(2)`
o`h=(3L)/(2)`
(b) Since, `sqrt(3Lg)` lies between `sqrt(2Lg)` and `sqrt(5Lg)` , the stringt will slack in upper half of the circle. Assuming that string slack when it makes an angle `theta` with horizontal. We have
`mgsintheta=(mv^(2))/(L)` ...(i)
`v^(2)=(sqrt(3gL)^(2))-2gL(1+sintheta)` ...(ii)
Solving Eq. `(i)` and Eq. `(ii)` , we get
`sintheta=(1)/(3)`
and `v^(2)=(gL)/(3)`
Maximum height of the bob from starting point,

`H=L(1+sintheta)+(v^(2)sin^(2)(90^(@)-theta))/(2g)`
`=(4L)/(3)+((gL)/(6g))cos^(2)theta(4L)/(3)+(4L)/(27)`
`=(40)/(27)L`