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

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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

$v^2=u^2-2gh$

$T-mg \cos \theta=\large\frac{mv^2}{l}$

$mg-mg \cos \theta=\large\frac{mv^2}{l}$

$v^2=gl(1-\cos \theta)$-----(1)

$gl(1-\cos \theta)=gl-2gl(1-\cos \theta)$

$\cos \theta= \large\frac{2}{3}$ , Substituting in (1)

Therefore $v=\sqrt {\large\frac{gl}{3}}$

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