JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 5)
A stone tide to a spring of length L is whirled in a vertical circle with the other end of the spring 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 $$\sqrt {x({u^2} - gL)} $$. The value of x is -
3
2
1
5
Explanation
$$\overrightarrow v = \sqrt {{u^2} - 2gL} \widehat j$$
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$$\overrightarrow u = u\widehat i$$
$$\therefore$$ $$\left| {\overrightarrow v - \overrightarrow u } \right| = \sqrt {({u^2} - 2gL) + {u^2}} $$
$$ = \sqrt {2u - 2gL} $$
$$\therefore$$ $$x = 2$$
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