Question

A heavy particle hanging from a string of length/ is projected horizontally with speed $\sqrt{gl}$. The speed of particle at the point where the tension in the string equals weight of the particle is

• A. $\sqrt{2gl}$
• B. $\sqrt{3gl}$
• C. $\sqrt{gll2}$
• D. $\sqrt{gl/3}$

Answer 1

The correct option is D $\sqrt{gl/3}$

Let T=mg

$h=l(1-cos\theta )$

using energy conservation

$\frac{1}{2}m(U^2-V^2)=mgh$

U=gl, V= speed of particle at B

$V^2=V^2-2gh$

$T-mgcos\theta =\frac{m{V}^2}{l}$

$mg-mgcos\theta =\frac{m{V}^2}{l}$

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

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

$cos\theta =\frac{2}{3}$

$\therefore V=\sqrt{\frac{gl}{3}}$