JEE Main 2025 Jan 24 Shift 2, Physics Q22: Miscellaneous Examples of SHM

For a conical pendulum, resolve the tension into vertical and horizontal components.

Vertical: $T\cos \theta =Mg$

Horizontal (provides centripetal force): $T\sin \theta =M{\omega }^{2}R$, with $R=L\sin \theta$.

Substituting $R$:

$T\sin \theta =M{\omega }^{2}L\sin \theta \Rightarrow T=M{\omega }^{2}L$

The angular frequency from $f=\frac{3}{\pi }$ rotations per second is:

$\omega =2\pi f=2\pi \cdot \frac{3}{\pi }=6\text{rad/s}$

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

Answer: 36