Question

A string of length $L$ is fixed at one end and carries a mass $\text{M}$ at the other end. The string makes $\frac{2}{\pi }$ revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is

• A. $\text{ML}$
• B. $2\text{ML}$
• C. $4\text{ML}$
• D. $16\text{ML}$

Answer 1

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

$T\sin \theta =M{\omega }^{2}R...\left(i\right)$
$T\sin \theta =M{\omega }^{2}L\sin \theta ...\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$
$T=M{\omega }^{2}L$
$=M4{\pi }^2{n}^{2}L$
$=M4{\pi }^2{\left(\frac{2}{\pi }\right)}^{2}L$
$=16ML$

Why this question? Note:For the conical pendulum, the horizontal component of tension provides the centripetal force while the vertical component balances the weight.