Question

A string of length 1m is fixed at one end with a bob of mass 100 g and the string makes $\frac{2}{\pi}$rev/s around a vertical axis through a fixed point. The angle of inclination of the string with vertical is-

• A. tan^–1 (5/8)
• B. tan^–1(3/5)
• C. cos^–1(8/5)
• D. cos^–1(5/8)

Answer 1

Answer: (D)

cos^–1(5/8)

Answer 2

T cos q = mg …..(i)

T sin q = mw²r …..(ii)

So tan q =$\frac{\omega^{2}r}{g}$;

̃ $\frac{\sin\theta}{\cos\theta}$ = $\frac{\omega^{2}\mathcal{l}\sin\theta}{g}$ ̃ cos q = $\frac{g}{\omega^{2}\mathcal{l}}$

̃ cos q = $\frac{10}{16 \times 1}$ = $\frac{5}{8}$