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 $\left(\frac{2}{\pi }\right)\text{rev/s}$ around a vertical axis through a fixed point. The tension in the string is

• A. $\frac{5}{8}\text{N}$
• B. $\frac{8}{5}\text{N}$
• C. $\frac{50}{8}\text{N}$
• D. $\frac{80}{5}\text{N}$

Answer 1

The correct option is B $\frac{8}{5}\text{N}$

From FBD of the bob:


$T\cos \theta =mg$ ... (1)
$T\sin \theta =mr{\omega }^2$ ... (2)
Since, $r=l\sin \theta$ and using this in (2),
$T=ml{\omega }^2$ ... (3)
Here, $\omega =\frac{2}{\pi }\times 2\pi =4\text{rad/s}$
From (2) and (3) we get
$\cos \theta =\frac{g}{l{\omega }^2}=\frac{5}{8}$

From equation (1),
$T=\frac{mg}{\cos \theta }=\frac{0.1\times 10}{5/8}=\frac{8}{5}\text{N}$