Question Details

Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and length l is

Options

A

Gm/l along x-axis

B

Gm/πl along y axis

C

2πGm/l² along x axis

D

2πGm/l² along y axis

Correct Answer :

2πGm/l² along y axis

Solution :

The correct option is 2πGm/l² along y axis.

Let us derive the gravitational field at the center of the semicircle step-by-step.
Consider a thin wire of mass m and length l bent into a semicircle. Let the center of the semicircle be at the origin (0,0) of a Cartesian coordinate system, and let the semicircle lie in the upper half-plane (y > 0) symmetric about the y-axis. The ends of the wire lie on the x-axis.

The length of the semicircular wire of radius R is:
l=πR
Therefore, the radius R of the semicircle is:
R=lπ

The mass per unit length (linear mass density λ) of the wire is:
λ=ml

Let us consider a small element of the wire subtending an angle dθ at the center, located at an angle θ with respect to the positive x-axis.
The length of this small element is ds=Rdθ, and its mass dm is:
dm=λds=λRdθ

The gravitational field dE at the center due to this mass element dm is directed towards the element (along the radial direction at angle θ):
dE=GdmR2=G(λRdθ)R2=GλRdθ

Due to the symmetry of the semicircle about the y-axis, the horizontal components of the gravitational field (dEx=dEcosθ) from opposite halves of the semicircle cancel each other out. Thus, the net gravitational field along the x-axis is zero:
Ex=0

The vertical components (dEy=dEsinθ) reinforce each other and point in the positive y-direction (towards the wire since gravity is attractive).
Integrating dEy from θ=0 to θ=π:
Ey=0πdEsinθ=0πGλRsinθdθ

Ey=GλR-cosθ0π=GλR(-cosπ-(-cos0))=GλR(1+1)=2GλR

Substitute the values of λ=ml and R=lπ into the equation:
Ey=2G(ml)(lπ)=2πGml2

Therefore, the net gravitational field at the center of the semicircle is 2πGm/l² along y axis.

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