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 :

Gravitation PYQ — JEE Main 2024 · 4 Apr · Shift 1 · Q63 with Solution

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Correct answer: D

Given data

Length of wire $= L$
Mass of wire $= M$
It is bent into a semicircular arc.
A particle of mass $m$ is placed at the centre of the arc.

We need the net gravitational force on the particle due to the semicircular wire.

Find the radius of the semicircle

The length of a semicircular arc is

L = \pi R

So,

R = \frac{L}{\pi}

Linear mass density of the wire

Since the wire is uniform,

\lambda = \frac{M}{L}

Take a small element of the arc

Consider a small element of length $dl$ at angle $\theta$ . Its mass is

dm = \lambda \, dl

For a circular arc,

dl = R\, d\theta

Hence,

dm = \lambda R \, d\theta

The distance of this element from the particle at the centre is $R$ . So the gravitational force due to this element is

dF = \frac{Gm\,dm}{R^2}

Substitute $dm$ :

dF = \frac{Gm}{R^2}(\lambda R\, d\theta) = \frac{Gm\lambda}{R} d\theta

Resolve components

By symmetry, horizontal components cancel out. Only the components along the symmetry axis of the semicircle add up.

For the element at angle $\theta$ ,

dF_{\text{net axis}} = dF\sin\theta

Taking $\theta$ from $0$ to $\pi$ ,

F = \int_0^\pi dF\sin\theta = \frac{Gm\lambda}{R} \int_0^\pi \sin\theta \, d\theta

Now,

\int_0^\pi \sin\theta \, d\theta = 2

Therefore,

F = \frac{Gm\lambda}{R} \cdot 2 = \frac{2Gm\lambda}{R}

Substitute $\lambda = \dfrac{M}{L}$ and $R = \dfrac{L}{\pi}$ :

F = \frac{2Gm}{R}\cdot \frac{M}{L} = 2Gm \cdot \frac{M}{L} \cdot \frac{\pi}{L}

So,

F = \frac{2\pi GmM}{L^2}

Match with options

F = \frac{2\pi GmM}{L^2}

This matches Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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View written solutionFree

Correct answer: D

Given data

Length of wire $= L$
Mass of wire $= M$
It is bent into a semicircular arc.
A particle of mass $m$ is placed at the centre of the arc.

We need the net gravitational force on the particle due to the semicircular wire.

Find the radius of the semicircle

The length of a semicircular arc is

L = \pi R

So,

R = \frac{L}{\pi}

Linear mass density of the wire

Since the wire is uniform,

\lambda = \frac{M}{L}

Take a small element of the arc

Consider a small element of length $dl$ at angle $\theta$ . Its mass is

dm = \lambda \, dl

For a circular arc,

dl = R\, d\theta

Hence,

dm = \lambda R \, d\theta

The distance of this element from the particle at the centre is $R$ . So the gravitational force due to this element is

dF = \frac{Gm\,dm}{R^2}

Substitute $dm$ :

dF = \frac{Gm}{R^2}(\lambda R\, d\theta) = \frac{Gm\lambda}{R} d\theta

Resolve components

By symmetry, horizontal components cancel out. Only the components along the symmetry axis of the semicircle add up.

For the element at angle $\theta$ ,

dF_{\text{net axis}} = dF\sin\theta

Taking $\theta$ from $0$ to $\pi$ ,

F = \int_0^\pi dF\sin\theta = \frac{Gm\lambda}{R} \int_0^\pi \sin\theta \, d\theta

Now,

\int_0^\pi \sin\theta \, d\theta = 2

Therefore,

F = \frac{Gm\lambda}{R} \cdot 2 = \frac{2Gm\lambda}{R}

Substitute $\lambda = \dfrac{M}{L}$ and $R = \dfrac{L}{\pi}$ :

F = \frac{2Gm}{R}\cdot \frac{M}{L} = 2Gm \cdot \frac{M}{L} \cdot \frac{\pi}{L}

So,

F = \frac{2\pi GmM}{L^2}

Match with options

F = \frac{2\pi GmM}{L^2}

This matches Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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