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(D) The length of the semicircular arc is $L = \pi R$,so the radius is $R = \frac{L}{\pi}$.Consider an infinitesimal element of the wire of length $dl

Vedclass · Accepted Answer

$\frac{2 GmM \pi}{L^2}$

Vedclass · Answer

(D) The length of the semicircular arc is $L = \pi R$,so the radius is $R = \frac{L}{\pi}$.Consider an infinitesimal element of the wire of length $dl = R d	heta$ at an angle $	heta$. The mass of this element is $dm = \lambda dl = \frac{M}{L} R d	heta$.The gravitational force exerted by this element on the particle of mass $m$ at the center is $dF = \frac{G m dm}{R^2} = \frac{G m (M/L) R d	heta}{R^2} = \frac{GMm}{LR} d	heta$.Due to symmetry,the horizontal components of the force cancel out,and the vertical components add up.The net force is $F = \int_{-\pi/2}^{\pi/2} dF \sin	heta = \frac{GMm}{LR} \int_{-\pi/2}^{\pi/2} \sin	heta d	heta$. Wait,the standard result for the field at the center of a semicircular arc is $E = \frac{2GM}{LR}$.Substituting $R = L/\pi$,we get $F = mE = m \left( \frac{2GM}{L(L/\pi)} ight) = \frac{2GMm\pi}{L^2}$.Thus,the correct option is $D$.

$A$ 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:

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