Question

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

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

Answer 1

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

FBD of the bob:


$T\cos \theta =mg$ ... (1)
$T\sin \theta =mr{\omega }^2$ ... (2)
From the diagram, radius of circular path $r=l\sin \theta$
$\therefore T\sin \theta =m\left(l\sin \theta \right){\omega }^2$
$\Rightarrow T=ml{\omega }^2$ ... (3)
$\therefore \cos \theta =\frac{mg}{ml{\omega }^2}$ (from (1) and (3))
$\omega =2\pi f=2\pi \times \frac{2}{\pi }\text{rad/s}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{10}{1\times {(2\pi \times \frac{2}{\pi })}^2}\right)$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{5}{8}\right)$