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:

A wire of length L and mass M is bent into a semicircular arc. The radius of the semicircle is determined by $$\pi R = L \Rightarrow R = \frac{L}{\pi}$$.

The linear mass density is $$\lambda = \frac{M}{L}$$. A particle of mass m is placed at the centre, and by symmetry the net force is directed toward the midpoint of the arc along the axis of symmetry.

Consider an element $$d\theta$$ at angle $$\theta$$ from the axis of symmetry. Its mass is $$dm = \lambda R\,d\theta = \frac{M}{L} \cdot \frac{L}{\pi}\,d\theta = \frac{M}{\pi}\,d\theta$$. The gravitational force from this element is $$dF = \frac{Gm\,dm}{R^2}$$, and the component along the axis of symmetry is $$dF\cos\theta$$.

Integrating from $$-\pi/2$$ to $$\pi/2$$ gives the total force:

$$ F = \int_{-\pi/2}^{\pi/2} \frac{Gm}{R^2} \cdot \frac{M}{\pi}\cos\theta\,d\theta = \frac{GmM}{\pi R^2}[\sin\theta]_{-\pi/2}^{\pi/2} = \frac{GmM}{\pi R^2}(2) = \frac{2GmM}{\pi R^2} $$

Substituting $$R = L/\pi$$:

$$ F = \frac{2GmM}{\pi(L/\pi)^2} = \frac{2GmM}{\pi \cdot L^2/\pi^2} = \frac{2GmM\pi}{L^2} $$

The correct answer is Option 4: $$\frac{2GmM\pi}{L^2}$$.