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 :

Correct answer: D. 2 GmM π L^2 We have R=(L)/(π) aligned & g_0= 2 G M L R = 2 GM π L^2 & F_ m =mg_0= 2 GM π m L^2 aligned Hence option (4)

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Physics
Chapter: Gravitation

JEE Main 2024 · 4 April · Morning Shift · Physics · Gravitation

A metal wire of uniform mass density having length L and mass M is...

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 metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc and a particle of

A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc and a particle of

Options

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

Correct answer

D. $\frac{2 \mathrm{GmM} \pi}{\mathrm{L}^2}$

Step-by-step solution

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)

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