JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Gravitation

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:

Answer: (D)

Solution

The wire (length

L

, mass

M

) is bent into a semicircle, so its radius satisfies

π R = L \Rightarrow R = \frac{L}{π}

, and the linear mass density is

λ = \frac{M}{L}

.

The gravitational field at the centre of a semicircular arc (only the component along the symmetry axis survives) is

g = \int \frac{G d m}{R ^{2}} sin θ = \frac{G λ}{R} \int_{0}^{π} sin θ d θ = \frac{2 G λ}{R} .

Substituting

λ = \frac{M}{L}

and

R = \frac{L}{π}

g = \frac{2 G ( M / L )}{L / π} = \frac{2 π GM}{L ^{2}} .

Force on the particle:

F = m g = \frac{2 π GM m}{L ^{2}} .

Option D.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option