Question

A string of length $1m$ is fixed at one end and a mass of $100gm$ is attached at the other end. The string makes $2/\pi$ rev/sec around a vertical axis through the fixed point. The angle of inclination of the string with the vertical is

($g = 10\mspace{6mu} m/\sec^{2}{}$)

• A. $\tan^{- 1}\frac{5}{8}$
• B. $\tan^{- 1}\frac{8}{5}$
• C. $\cos^{- 1}\frac{8}{5}$
• D. $\cos^{- 1}\frac{5}{8}$

Answer 1

Answer: (D)

$\cos^{- 1}\frac{5}{8}$

Answer 2

For the critical condition, in equilibrium

$T\sin\theta = m\omega^{2}r$ and $T\cos\theta = mg\therefore\tan\theta = \frac{\omega^{2}r}{g}$

⇒ $\frac{4\pi^{2}n^{2}r}{g} = \frac{4\pi^{2}(2/\pi)^{2}.1}{10} = \frac{8}{5}$