Question 31 of 223

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 $\mathrm{m}$ is placed at the centre of the arc. The gravitational force on the particle by the wire is :

A$\frac{\mathrm{GmM} \pi^2}{\mathrm{~L}^2}$B$\frac{\mathrm{GMm} \pi}{2 \mathrm{~L}^2}$C$0$D$\frac{2 \mathrm{GmM} \pi}{\mathrm{L}^2}$

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Answer: (D) $\frac{2 \mathrm{GmM} \pi}{\mathrm{L}^2}$

We have $R=\frac{L}{\pi}$
$\begin{aligned}
& \mathrm{g}_0=\frac{2 \mathrm{G} \frac{\mathrm{M}}{\mathrm{L}}}{\mathrm{R}}=\frac{2 \mathrm{GM} \pi}{\mathrm{L}^2} \\
& \therefore \mathrm{F}_{\mathrm{m}}=\mathrm{mg}_0=\frac{2 \mathrm{GM} \pi \mathrm{m}}{\mathrm{L}^2}
\end{aligned}$
Hence option (4)