A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes (\frac3\pi) rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is \underline\hspace2cm ML.

Circular Motion PYQ — JEE Main 2025 · 24 Jan · Shift 2 · Q73 with Solution

View written solutionFree

Correct answer: 36

The mass executes a conical pendulum motion.

\omega = 2\pi f = 2\pi\left(\frac{3}{\pi}\right)=6\ \text{rad/s}

If the string makes angle $\theta$ with the vertical, then radius of circular motion is

r=L\sin\theta

The horizontal component of tension provides centripetal force:

T\sin\theta = M\omega^2 r

Substitute $r=L\sin\theta$ :

T\sin\theta = M\omega^2 (L\sin\theta)

Since $\sin\theta \neq 0$ , cancel it:

T = M\omega^2 L

T = M(6)^2L = 36ML

The question asks: tension is $\underline{\hspace{2cm}}\, ML$

So the required integer is

\boxed{36}

Stored correct answer = $36$

This matches the derived answer.