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

The Correct Option is A

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 $