Question

An inextensible string of length 'l' fixed at one end, carries a mass 'm' at the other end. If the string makes $\frac{1}{\pi}$ revolutions per second around the vertical axis through the fixed end, the tension in the string is [The string makes an angle $\theta$ with the vertical]

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

Hint:

In a conical pendulum, the tension $T$ equals $ml\omega^2$ because $T \sin\theta$ provides the centripetal force.

Answer 1

The Correct Option is C

Step 1: Conical Pendulum Analysis
Horizontal component of Tension $T \sin \theta = m r \omega^2$
Since $r = l \sin \theta$, then $T \sin \theta = m (l \sin \theta) \omega^2$
Step 2: Solving for Tension
$T = m l \omega^2$
Step 3: Frequency to Angular Velocity
$\omega = 2 \pi f = 2 \pi (\frac{1}{\pi}) = 2 \text{ rad/s}$
Step 4: Calculation
$T = m \cdot l \cdot (2)^2 = 4 m l$
Final Answer:(C)