A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes $\left(\frac{3}{\pi}\right)$ rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is ______ ML.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Laws of Motion

A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes

(\frac{3}{π})

rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is ______ ML.

Answer: 36

The mass M performs conical-pendulum motion at the end of the string of length L. Let

θ

be the angle the string makes with the vertical, so the radius of the circular path is

R = L sin θ

.

Resolving forces on the mass: vertically,

T cos θ = M g,

and horizontally, the tension supplies the centripetal force,

T sin θ = M ω^{2} R = M ω^{2} (L sin θ) .

Dividing out

sin θ

from the second equation:

T = M ω^{2} L .

The angular velocity corresponding to

\frac{3}{π}

revolutions per second is

ω = 2 π f = 2 π \times \frac{3}{π} = 6 rad/s .

Therefore,

T = M (6)^{2} L = 36 M L .

The tension is

36 M L

, so the required integer is

36

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Integer answer