Question

A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2/π 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

Answer: (D)

16 ML

Answer 2

$T\sin\theta = M\omega^{2}R$ ..... (i)

$T\sin\theta = M\omega^{2}L\sin\theta$..... (ii)

From (i) and (ii)

$T = M\omega^{2}L = M4\pi^{2}n^{2}L$ $= M4\pi^{2}\left( \frac{2}{\pi} \right)^{2}L = 16ML$