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A heavy particle hanging from fixed point by light inextensible string of length l is projected horizontally with speed $\sqrt {gl}$. Find speed of particle at the instant of the motion when tension in the string is equal to weight of particle

\[(a)\;v=\sqrt {\frac{g}{3l}} \quad (b)\;v=\sqrt {\frac{gl}{3}} \quad (c)\;v=\frac{gl}{3}\quad (d)\;v= \frac{g}{3l}\]

1 Answer

Let $T=mg$ when angle is $\theta$
$h=l(1-\cos \theta)$
Using concervation of energy at A and B
$\large\frac{1}{2} $$m(u^2-v^2)=mgh$
$u=gl,v=$speed of particle at B
$T-mg \cos \theta=\large\frac{mv^2}{l}$
$mg-mg \cos \theta=\large\frac{mv^2}{l}$
$v^2=gl(1-\cos \theta)$-----(1)
$gl(1-\cos \theta)=gl-2gl(1-\cos \theta)$
$\cos \theta= \large\frac{2}{3}$ , Substituting in (1)
Therefore $v=\sqrt {\large\frac{gl}{3}}$
answered Jul 18, 2013 by meena.p

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