Question

A string of length 'L' fixed at one end carries a mass 'm' at the other end. The string makes \(\frac{3}{2\pi }\) revolutions per second around the vertical axis through fixed end. The tension in the string is

• A. \(36mL\)
• B. \(18mL\)
• C. \(9mL\)
• D. \(4.5mL\)

Hint:

The horizontal component of tension gives the centripetal force. Use the angular speed from the frequency.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The mass moves in a horizontal circle, so the string forms a conical pendulum.
Let the string make angle \(\theta\) with the vertical axis; the circle has radius \(L\sin\theta\).

Step 2: Angular speed
\(\omega=2\pi f=2\pi\times\dfrac{3}{2\pi}=3\) rad/s

Step 3: Force along the radius
\(T\sin\theta=m\omega^{2}L\sin\theta\)
The factor \(\sin\theta\) cancels, so \(T=m\omega^{2}L\).

Step 4: Result
\(T=m\times 9\times L=9mL\), option (C).

Final Answer:
The tension in the string is \(9mL\). \[ \boxed{\text{(C) }9mL} \]