Question

A string of length $l=1\text{m}$ fixed at one end carries a mass $m=1\text{kg}$ at the other end. The string is rotating at a rate of $\frac{2}{\pi }\text{rev/s}$ about the axis through the fixed end as shown in the figure. Then the tension in the string is

• A. $11\text{N}$
• B. $12\text{N}$
• C. $13\text{N}$
• D. $16\text{N}$

Answer 1

The correct option is D $16\text{N}$


From the FBD of mass attached to string,
$T\sin \theta =m{\omega }^{2}r....\left(i\right)$
Here,
$r=l\sin \theta ,\omega =2\pi f$
$f=\text{frequency of revolution}=\frac{2}{\pi }\text{rev/s}$
From Eq. $\left(i\right)$,
$\Rightarrow T\sin \theta =m\left(l\sin \theta \right)(2\pi f{)}^2$
$T\sin \theta =m\times l\sin \theta \times {(2\pi \times \frac{2}{\pi })}^2$
$\therefore T=16ml=(16\times 1\times 1)=16\text{N}$