In case of rod:-

1st method

Velocity at the top of circle(U)=0

length covered when it reaches at bottommost POINT=2L

acceleration=g

final velocity=v(say)

From Newton's third equation,we can write

v^2=u^2+2as

v^2=0+2×g×2l

v=√4gl

2nd method

At the top point, Velocity of block will be zero,and potential energy will be at it's maximum

using work energy theorem,we can write

mv^2/2 =mg×2l

v=√4gl

In case of string:-

Required velocity at the top(to complete the circle)=√gl(minimum)

Velocity at the bottom=v(say)

acceleration=g

distance to travel=2l

using third equation of MOTION we get

v^2=(√gl)^2+2×g×2l

v=√5gl

Doubts:-

1) Why velocity at top(in case of string) is not taken as zero?

Ans:-String is weak(not compressible),and if at the top, velocity of block will become zero then the block will fall directly at the bottom without completing the circle.

2)Why √gl is PARTICULARLY taken?

Ans:-It is mathematically proved that,√5gl is the minimum velocity required to give to the block attached by a string to complete the circle .So at the top velocity will be √gl