Q 11-07-058JEE MainJEE Main 2024 (4 Apr, Shift 1)Medium
A metal wire of uniform mass density having length $L$ and mass $M$ is bent to form a semicircular arc and a particle of mass $m$ is placed at the centre of the arc. The gravitational force on the particle by the wire is
Answer: (D) $\dfrac{2GMm\pi}{L^2}$
Radius of the arc: $R = \dfrac{L}{\pi}$; linear mass density $\lambda = \dfrac{M}{L}$.
Taking components along the symmetry axis, an element at angle $\theta$ contributes $\dfrac{G\lambda R\,d\theta\,m}{R^2}\sin\theta$:
$$F = \frac{G\lambda m}{R}\int_0^\pi\sin\theta\,d\theta = \frac{2G\lambda m}{R} = \frac{2GMm}{L}\cdot\frac{\pi}{L} = \frac{2GMm\pi}{L^2}$$
Solution by Sreeraj P, M.Sc Physics