Question 297 of 308

A particle of mass $m$ is suspended from a ceiling through a string of length $L$. The particle moves in a horizontal circle of radius $r$ such that $r=\dfrac{L}{\sqrt{2}}$. The speed of particle will be :

A$\sqrt{rg}$B$\sqrt{2rg}$C$\sqrt{\dfrac{rg}{2}}$D$2\sqrt{rg}$

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Answer: (A) $\sqrt{rg}$

$\sin \theta =\dfrac{1}{\sqrt{2}}$
$\theta =45^{\circ}$
$\tan \theta =\dfrac{\dfrac{m{v}^{2}}{r}}{mg}=\dfrac{{v}^{2}}{rg}=1$
${v}^{2}=rg$
$v=\sqrt{rg}$