Question

A string of length $L$ is fixed at one end and carries a mass $M$ 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. ML
• B. 2 ML
• C. 4 ML
• D. 16 ML

Answer 1

The correct option is D

16 ML

$Tsin\theta =M{\omega }^{2}R$ ...........(i)

$Tsin\theta =M{\omega }^{2}Lsin\theta$ ...........(ii)

From (i) and (ii)

$T$ = $M{\omega }^{2}L$

= $M4{\pi }^2{n}^{2}L$

= $M4{\pi }^2{\left(\frac{2}{\pi }\right)}^{2}L$

= $16ML$