Question:

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

Updated On: Jul 27, 2022
  • $\frac{Gm}{\pi{l}}$along x-axis
  • $\frac{Gm}{\pi{l}}$along y-axis
  • $\frac{2\pi Gm}{l^{2}}$along y-axis
  • $\frac{2\pi Gm}{l^{2}}$along x-axis
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The Correct Option is C

Solution and Explanation

Consider two small elements each of length $d l=r d \theta$ symmetrically. $\pi_{r}=1$ Resolve the elemental field intensities $\because\left|d E_{1}\right|=\left|d E_{2}\right|$
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Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].