A particle of mass

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 :

 

 

  1. $\sqrt{1 g}$

  2. $\sqrt{2 \mathrm{rg}}$

  3. $2 \sqrt{\mathrm{rg}}$

  4. $\sqrt{\frac{\mathrm{rg}}{2}}$


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

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