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 the string is

• A. 16 ml
• B. 4 ml
• C. 8 ml
• D. 2 ml

Answer 1

Answer: (A)

16 ml

Answer 2

By the diagram T sin $\theta = \frac{ mv^2 }{ r }$ \hspace20mm ..(i) T cos $\theta = mg$ \hspace20mm ..(ii) where linear velocity v = r $\omega$ and sin $\theta = \frac{ r}{ l }$ Putting these values in E (i), we get T $\times \frac{ r }{ l } = m \omega^2 r$ T = $m \omega^2 \, l$ We know $\omega$ = 2 $\pi$n, we have $\therefore T = m ( 2 \pi n )^2 \, l$ $\Rightarrow T = m \bigg( 2 \pi \times \frac{ 2 }{ \pi } \bigg)^2 \, l$ $\Rightarrow T = 16 \, ml$