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: Find the speed.
Angular speed $\omega = 2\pi f = 6$ rad/s, so the linear speed is $v = \omega L = 6L$.

Step 2: Use centripetal force.
$T = \dfrac{mv^2}{L} = \dfrac{m(6L)^2}{L}$.

Step 3: Simplify.
\[ T = 36\,mL \]

Final Answer:
Option (B). \[ \boxed{36\,mL} \]