1. Tardigrade
  2. Question
  3. Physics
  4. A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is √ x ( u 2- gL ). The value of x is

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Q. A stone tide to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is $\sqrt{ x \left( u ^{2}- gL \right)}$. The value of $x$ is

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Solution:

$v=\sqrt{u^{2}-2 g L}$
$\Delta v=\sqrt{u^{2}+v^{2}}$
$\Delta v=\sqrt{u^{2}+v^{2}-2 g L}$
$\Delta v=\sqrt{2 u^{2}-2 g L}$
$\Delta v=\sqrt{2\left(u^{2}-g L\right)} x=2$