Question:
A particle of mass $\mathrm{m}$ is suspended from a ceiling through a string of length $L$. The particle moves in
a horizontal circle of radius $r$ such that $r=\frac{L}{\sqrt{2}}$. The
speed of particle will be :
Correct Option: 1
Solution:
$\mathrm{r}=\frac{\ell}{\sqrt{2}}$
$\sin \theta=\frac{\mathrm{r}}{\ell}=\frac{1}{\sqrt{2}}$
$\theta=45^{\circ}$
$T \sin \theta=\frac{m v^{2}}{r}$
$\mathrm{T} \cos \theta=\mathrm{mg}$
$\tan \theta=\frac{v^{2}}{r g} \Rightarrow v=\sqrt{r g}$
Ans. 1