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Rotational Motion question

2003 · Q160
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Rotational Motion question

2003 · Q160

NEETPhysicsRotational MotionMCQ+4 / −1
A stone is tied to a string of length lll 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 its lowest position and has a speed u. The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is
  1. A
    2(μ2−gl)\sqrt {2\left( {{\mu ^2} - gl} \right)}2(μ2−gl)​
  2. B
    u2−gl\sqrt {{u^2} - gl}u2−gl​
  3. C
    u−u2−2glu - \sqrt {{u^2} - 2gl}u−u2−2gl​
  4. D
    2gl\sqrt {2gl}2gl​
View written solutionFree

Correct answer: A

The total energy at A = the total energy at B

⇒12mu2=12mv2+mgl \Rightarrow {1 \over 2}m{u^2} = {1 \over 2}m{v^2} + mgl⇒21​mu2=21​mv2+mgl

⇒v=u2−2gl\Rightarrow v = \sqrt {{u^2} - 2gl}⇒v=u2−2gl​

The change in magnitude of velocity = u2+v2\sqrt {{u^2} + {v^2}} u2+v2​

=2(u2−gl)= \sqrt {2\left( {{u^2} - gl} \right)}=2(u2−gl)​

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