Question

A heavy particle hangs from a point O, by a string of length a. It is projected horizontally with a velocity $v=\sqrt{(2+\sqrt{3})ag}$.
The angle with the upward vertical, string makes where string becomes slack is

• A. $\theta =si{n}^{-1}\left(\frac{-1}{\sqrt{3}}\right)$
• B. $\theta =co{s}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• C. $\theta =co{s}^{-1}\left(\frac{1}{\sqrt{2}}\right)$
• D. $\theta =si{n}^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Answer 1

The correct option is B $\theta =co{s}^{-1}\left(\frac{1}{\sqrt{3}}\right)$



$mgcos\theta =\frac{m{v}^2}{a}\Rightarrow v^2=agcos\theta$
$\frac{1}{2}m(2+\sqrt{3})ag=mga(1+cos\theta )$
$+\frac{1}{2}mgacos\theta$
$\Rightarrow 1+\frac{\sqrt{3}}{2}=1+\frac{3cos\theta }{2}\Rightarrow cos\theta =\frac{1}{\sqrt{3}}$
$\therefore \theta =co{s}^{-1}\left(\frac{1}{\sqrt{3}}\right)$ with upward vertical.