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

• A

  $\frac{G m}{\pi l}$ along $x$-axis
• B

  $\frac{G m}{\pi l}$ along $y$-axis
• Correct answer
  C

  $\frac{2 \pi G m}{l^2}$ along $y$-axis
• D

  $\frac{2 \pi G m}{l^2}$ along $x$-axis

Solution

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

Hence, the correct answer is option: 'C'.