Question

A particle suspended by a thread of length ${}^{\prime }{l}^{\prime }$ is projected horizontally with a velocity $\sqrt{3gl}$ at the lowest point. The height from the bottom at which the tension in the string becomes zero is :

• A. $\frac{4l}{3}$
• B. $\frac{2l}{3}$
• C. $\frac{5l}{3}$
• D. $\frac{l}{3}$

Answer 1

The correct option is B $\frac{2l}{3}$

When no external force is acting on the particle then,

$\Delta K.E.+\Delta P.E.=0$

Let at height$h$ the tension in thread becomes zero and the radial force acting at this height is given as and the height at which particle makes angle$\theta$is given as

$h=l+l\cos \theta$

$F_{net}=T+mg\cos \theta$

Let$v_B$be speed of the particle when$T=0$,

$F_r=\frac{m{v}_{{}_B}^2}{l}$

Now on equating above two equations we get

$v_{{}_B}^2=gl\cos \theta$

By using law of conservation we get

$\frac{1}{2}mgl\cos \theta +\frac{3}{2}mgl-mgl\left(1+\cos \theta \right)=0$

$\cos \theta =-\frac{1}{3}$

By substituting$\cos \theta =1$in$h=l+l\cos \theta$we get

$h=\frac{2l}{3}$