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: Understanding the Question:
This is a case of a conical pendulum. The tension $T$ in the string provides the horizontal centripetal force and balances the weight vertically.
Step 2: Key Formula or Approach:
Let $T$ be tension, $\theta$ be the angle with vertical.
1. $T \cos \theta = mg$
2. $T \sin \theta = m \omega^2 r$, where $r = l \sin \theta$ is the radius of the circular path.
Step 3: Detailed Explanation:
From the second equation:
\[ T \sin \theta = m \omega^2 (l \sin \theta) \]
Dividing both sides by $\sin \theta$ (assuming $\theta \neq 0$):
\[ T = m \omega^2 l \]
Given:
Frequency $f = \frac{1}{\pi}$ rev/s
Angular velocity $\omega = 2\pi f = 2\pi \left(\frac{1}{\pi}\right) = 2$ rad/s
Length $= l$
Mass $= m$
Substitute $\omega = 2$ into the tension formula:
\[ T = m \cdot (2)^2 \cdot l = 4ml \]
Step 4: Final Answer:
The tension in the string is 4 ml.