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. $2ML$
• C. $4ML$
• D. $16ML$

Answer 1

The correct option is D $16ML$

From the figure, we find that
$T\sin \theta =M{\omega }^{2}R$
$=M{\omega }^{2}L\sin \theta$
$\therefore T=M{\omega }^{2}L=M(2\pi n{)}^{2}L$
$=M4{\pi }^2{\left(\frac{2}{\pi }\right)}^{2}L=16ML$
Hence, the correct answer is option (d).