Physics · Gravitation

JEE Main 2024 — 4 April, Shift 1 — Question 37

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:

Option A:
$\frac{\mathrm{GMm} \pi}{2 \mathrm{~L}^{2}}$
Option B:
zero
Option C:
$\frac{G m M \pi^{2}}{L^{2}}$
Option D:
$\frac{2 \mathrm{GmM} \pi}{\mathrm{L}^{2}}$
Correct

Answer: D

Step-by-step solution

We have

$\mathrm{R}=\frac{\mathrm{L}}{\pi}$ $\mathrm{g}_{0}=\frac{2 \mathrm{G} \frac{\mathrm{M}}{\mathrm{L}}}{\mathrm{R}}=\frac{2 \mathrm{GM} \pi}{\mathrm{L}^{2}}$

$\therefore \mathrm{F}_{\mathrm{m}}=\mathrm{mg}_{0}=\frac{2 \mathrm{GM} \pi \mathrm{m}}{\mathrm{L}^{2}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Physics
Chapter: Gravitation
Topic: Gravitational Field and Gravity