Question

A string of length 'L' fixed at one end carries a mass 'm' at the other end. The string makes \(\frac{3}{π}\) r.p.s. around the vertical axis through fixed end. The tension in the string is

• A. \(72mL\)
• B. \(36mL\)
• C. \(18mL\)
• D. \(9mL\)

Hint:

The tension provides the centripetal force, so use T = m omega squared L with omega = 2 pi f.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The mass moves in a circle of radius \(L\). Taking the string as horizontal, the tension alone gives the needed centripetal force.

Step 2: Key Formula or Approach:
\[ T = m\omega^2 L, \qquad \omega = 2\pi f \]

Step 3: Detailed Explanation:
\(f = \dfrac{3}{\pi}\) rps, so \(\omega = 2\pi\times\dfrac{3}{\pi} = 6\) rad/s.
\[ T = m(6)^2 L = 36\,mL \]

Step 4: Check the options.
72 mL would need \(\omega^2 = 72\), 18 mL needs \(\omega^2 = 18\) and 9 mL needs \(\omega^2 = 9\). Only 36 mL fits \(\omega = 6\).

Final Answer:
The tension is \(36\,mL\), option (B). \[ \boxed{36\,mL} \]