First slide

Gravitational field intensity

Question

A mass m is at a distance a from one end of a uniform rod of length l and M. The gravitational force on the mass due to the rod is

Moderate

A

$\frac{GMm}{(a + l)}$
B

$\frac{GmM}{a (l + a)}$
C

$\frac{GMm}{a}$
D

$\frac{GmM}{2 (l + a)}$

Solution

Considering mass per unit length $x = \frac{m}{l}$ mass of a small portion $d x = λ d x$ at distance x. Force due to small portion $G \frac{mλdx}{x^{2}},$
Total fore $\int_{a}^{a + l} \frac{Gmλdx}{x^{2}}$
$F = Gmλ {|\frac{- 1}{x}|}_{a}^{a + l} = Gmλ [\frac{1}{a} - \frac{1}{a + l}] = Gmλ [\frac{1}{a (a + l)}] = \frac{GMm}{a (a + l)}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App