Question 174 of 183

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

A$2$B$3$C$4$D$1$

Show answer and solution▾

Answer: (A) $2$

Applying conservation of mechanical energy at the point $A$ and $B$, we get$\dfrac{1}{2}m{u}^{2}+0=\dfrac{1}{2}m{v}^{2}+mgL$
$\Rightarrow v=\sqrt{{u}^{2}-2gL}$
Now in vector form. ${\vec{v}}_{i}=u\hat{i}$ and ${\vec{v}}_{f}=v\hat{j}$
Therefore, change in velocity will be,$∆\vec{V}=v\hat{j}-u\hat{i}$ and $\left|\Delta \vec{V}\right|=\sqrt{{u}^{2}+{v}^{2}}$
$\Rightarrow \Delta V=\sqrt{{u}^{2}+\left({u}^{2}-2gL\right)}$
$=\sqrt{2\left({u}^{2}-gL\right)}$
Hence, $x=2$