QJEEAdv08Ph95de: JEE Advanced 2008 Physics - Class 11 JEEAdv08Ph95de Solution: A bob of mass M is suspended by a massless string of length L.…

The bob of mass m requires velocity to reach at a point with vertical circular motion.

$V = \sqrt {5gl}$

Then, $h = L - L\cos \theta = L(1 - \cos \theta )$

Applying law of conservation of energy at point A & C

Kinetic energy (K.E)_A + Potential energy (U_A) = Kinetic energy (K.E_C) + Potential energy (U_C)

$\Rightarrow {1 \over 2}m{V^2} + 0 = {1 \over 2}m{\left( {{{\sqrt {5gl} } \over 2}} \right)^2} + mgL(1 - \theta )$

$\Rightarrow {{5L} \over 2} = {{5L} \over 8} + L(1 - \cos \theta )$ ( $\because$ $V = \sqrt {5gl}$ )

$\Rightarrow {5 \over 2} - {5 \over 8} = 1 - \cos \theta$

$\Rightarrow {{20 - 5} \over 8} = 1 - \cos \theta$

$\Rightarrow \cos \theta = 1 - {{15} \over 8}$

$\Rightarrow \cos \theta = {{ - 7} \over 8}$

$\Rightarrow \theta = 152^\circ$

$\therefore$ {{3\pi } \over 4} < \theta < \pi