Question

A string of length $1m$ is fixed at one end with a bob of mass $100$ g and the string makes $\frac{2}{\pi }rev{s}^{-1}$ around a vertical axis through a fixed point. The angle of inclination of the string with vertical is

• A. ${\tan }^{-1}\left(\frac{5}{8}\right)$
• B. ${\tan }^{-1}\left(\frac{3}{5}\right)$
• C. ${\cos }^{-1}\left(\frac{8}{5}\right)$
• D. ${\cos }^{-1}\left(\frac{5}{8}\right)$

Answer 1

The correct option is D ${\cos }^{-1}\left(\frac{5}{8}\right)$

Given; $\nu =\frac{2}{\pi }rev/s$

$w=2\pi \nu =4rad/s$

Now, $mr{w}^2=Tsin\theta$ ....(1) where $r=Lsin\theta$

Also $mg=Tcos\theta$ ........(2)

Dividing the equations,

we get $cos\theta =\frac{g}{L{w}^2}$

$\cos \theta =\frac{10}{1\times 4\times 4}=\frac{5}{8}$

$\theta ={\cos }^{-1}\frac{5}{8}$