Question

A string of length $l$ fixed at one end carries a mass m at the other end. The string makes $\frac{2}{\pi }$ rev/s around the horizontal axis through the fixed end as shown in the figure, the tension in string is :

• A. $16\,ml$
• B. $4\,ml$
• C. $8\,ml$
• D. $2\,ml$

Answer 1

Answer: (A)

$16\,ml$

Answer 2

Horizontal component of tension balances centripetal force. The free body diagram of the given situation is shown. Taking the vertical and horizontal component of forces, we have $T \sin \theta=\frac{m v^{2}}{r} \ldots$ (i) $T \cos \theta=m g \ldots$ (ii) where linear velocity $v=r \omega$ and $\sin \theta=\frac{r}{l}$ Putting these values in (i), we get $T \times \frac{r}{l}=m \omega^{2} T$ We know $\omega=2 \pi n$, we have $\therefore T=m(2 \pi n)^{2} l$ $\Rightarrow T=m\left(2 \pi \times \frac{2}{\pi}\right)^{2} l$ $\Rightarrow T=16\, m L .$