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  • A stone tide to a string of length is whirled in a vertical circle with the other end of the string a

A stone tide to a string of length \mathrm{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 \mathrm{u}. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is \mathrm{\sqrt{x\left(u^{2}-g L\right)}}. The value of \mathrm{x} is-

Option: 1

3


Option: 2

2


Option: 3

1


Option: 4

5


Answers (1)

best_answer



\mathrm{\overrightarrow{\Delta v}= v\hat{j}-u\hat{i}}
\mathrm{\left | \overrightarrow{\Delta v} \right |= \sqrt{v^{2}+u^{2}}}

Applying mechanical energy conservation. at 1 & 2.
\mathrm{\frac{1}{2}mu^{2}+0= \frac{1}{2}mv^{2}+mgl}
              \mathrm{v^{2}= u^{2}-2gl}

\mathrm{\therefore \Delta v= \sqrt{u^{2}-2gl+u^{2}}= \sqrt{2\left ( u^{2}-2gl \right )}}

The correct answer is (2)
 

Posted by

Ritika Harsh

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