Question

A particle is attached to massless string whose on end is attached to a fixed point. The particle is moved to a position such that the string is horizontal and straight and then released from this position. The angle made by the string with vertical when the tension in string is equal to the weight of the particle is :

• A. ${\cos }^{-1}\frac{1}{2}$
• B. ${\cos }^{-1}\frac{1}{3}$
• C. ${45}^o$
• D. ${\cos }^{-1}\frac{1}{\sqrt{3}}$

Answer 1

The correct option is D ${\cos }^{-1}\frac{1}{\sqrt{3}}$

Given,

Tension in the string is equal to the weight of the body, $T=mg$

Kinetic energy = potential energy

$\frac{1}{2}m{v}^2=mg\left(rcos\theta \right)$ …… (1)

Tension is the string, $T=\frac{m{v}^2}{r}+mg\cos \theta$

$mg=\frac{2}{r}\left(\frac{m{v}^2}{2}\right)+mg\cos \theta$

from equation (1)

$\Rightarrow mg=\frac{2}{r}\left(mgr\cos \theta \right)+mg\cos \theta$

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

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{3}\right)$