A stone is tied to a string of length l and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at the lowest position nad has a speed u. The magnitude of change in velocity as it reaches the position where the string is horizontal is ? (g being the accelration due to gravity)

Let the direction of revolution be counter-clockwise

Initial velocity = u i^ (at bottom)

i^ ( i-cap suggests that the direction of initial velocity is towards positive x-axis)

--

When the string is horizontal, the body has covered a quarter of revolution, its velocity is v j^

j^ (j-cap suggests that the direction of its velocity is towards positive y-axis when the string is horizontal)

-

Change in velocity = vj^ - ui^

magnitude of change in velocity = √ v2 + u2

-

Using conservation of energy

Kinetic Energy of body when it is at bottom = 1/2 mu2

When it goes to the quarter position, it does some work against gravity. So a part of its kinetic energy is spent in doing work equal to mgh and remaining kinetic energy is 1/2 mv2

1/2 mu2 = 1/2 mv2 + mgh

here h = l

1/2 mu2 = 1/2 mv2 + mgl

v2 - u2 = -2gl

v2 - u2 + 2u2 = 2u2 - 2gl

v2 + u2 = 2 (u2 - gl)

Therefore, change in magnitude of velocity = √ v2 + u2 = √2 (u2 - gl)