A wire of length l and mass m is bent in the form of a semicircle. The gravitational field intensity at the centre of semicircle is

Question

Quantrex Academy · Accepted Answer

Correct answer: C. (2 π G m)/(l^2) along y -axis

Consider two small elements each of length

d l=r d \theta

symmetrically.

\pi r=l

Resolve the elemental field intensities

\because\left|d E_1\right|=\left|d E_2\right|d E=2 d E_1 \cos \theta=\frac{2 G r d \theta}{r^2} \cos \theta

\begin{aligned} & \mathrm{G}=\text{ Gravitational constant } \\ & \lambda=\text{ Linear massdensity }=\frac{m}{l}\end{aligned}

or d E=\frac{2 \lambda G}{r} \int_0^{90^{\circ}} \cos \theta d \theta=\frac{2 \lambda G}{r}=\frac{2 \pi G m}{l^2}

A wire of length l and mass m is bent in the form of a semicircle....