Question

A string of length \(L\) is fixed at one end and carries a mass m at the other end. The string makes \(\frac{3}{\pi}\) revolutions per second around the vertical axis through the fixed end. The tension in the string is

• A. 8 mL
• B. 12 mL
• C. 24 mL
• D. 36 mL

Hint:

Resolve the tension: its horizontal part provides the centripetal force and R equals L sin theta.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The mass moves in a horizontal circle of radius \(R=L\sin\theta\) as a conical pendulum.

Step 2: Angular speed
\(\omega=2\pi n=2\pi\times\dfrac{3}{\pi}=6\ \text{rad/s}\).

Step 3: Horizontal component
\(T\sin\theta=m\omega^{2}R=m\omega^{2}L\sin\theta\), so \(T=m\omega^{2}L\).

Step 4: Calculation
\(T=m\times36\times L=36\,mL\).

Final Answer:
The tension in the string is 36 mL. \[ \boxed{\text{(D) }36\ \mathrm{mL}} \]