A rod of length `l` whose lower end is fixed along the horizontal plane starts from the vertical position. The velocity of the upper end of the rod when it hits the ground is

Explanation

When upper end of the rod hits the gound, the distance through which centre of gravity of the rod falls `= L//2`.
`:.` Loss of potantial energy `= Mg (L//2)`, where `M` is mass of the rod.
Gain in rotational `K.E. = (1)/(2)I omega^(2)`
Where `I = ML^(2)//3 and omega = v//L`
`:.` Gain in rotational `K.E.`
`= (1)/(2) ((ML^(2))/(3)) ((v)/(L))^(2) = (M v^(2))/(6)`
According to the law of conservation of energy
`(M v^(2))/(6) = Mg (L)/(2)`
`v = sqrt(3gL)`