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

T sin $\theta$ = M ${\omega }^2$ R
T sin $\theta$ = M ${\omega }^2$ L sin $\theta$
From (i) and (ii)
T = M ${\omega }^2$ L
= M 4${\pi }^2{n}^{2}L$
= M 4 ${\pi }^2\left({\frac{2}{\pi }}^2\right)L$
= 16 ML