Question

Mass $M$ is distributed uniformly along a line of length $2L$. A particle of mass $m$ is at a point ($\text{P}$) at distance $a$ above the centre of the line on its perpendicular bisector as shown in figure. The gravitational force that the mass distribution along the line exerts on the particle is

• A. $\frac{2GMm}{a\sqrt{L^2+a^2}}$
• B. $\frac{GMm}{a\sqrt{L^2+a^2}}$
• C. $\frac{GMm}{\sqrt{L^2+a^2}}$
• D. $\frac{3GMm}{a\sqrt{L^2+a^2}}$

Answer 1

The correct option is B $\frac{GMm}{a\sqrt{L^2+a^2}}$

Consider an element of length $dx$ at a distance $x$ from the center of the rod.


The mass of the element, $dm=\frac{M}{2L}dx$

Gravitational field at point $\text{P}$ due the element,

$dE=\frac{Gdm}{(\sqrt{a^2+x^2}{)}^2}$

$\Rightarrow dE=\frac{M}{2L}\frac{Gdx}{(a^2+x^2)}$

Now consider a symmetrical element on the other side at a distance $x$ from the centre.

From the figure we see that the component $dE\sin \theta$ will get cancelled.

So, the net field will be along the perpendicular of the rod.

$E_{net}={\int }_0^{L}2dE\cos \theta$

Substituting the value of $dE$ and $\cos \theta$,

$\Rightarrow E_{net}={\int }_0^L\frac{2MG}{2L}\frac{dx}{a^2+x^2}\left(\frac{a}{\left(\sqrt{a^2+x^2}\right)}\right)$

$\Rightarrow E_{net}=\frac{MGa}{L}{\int }_0^L\frac{dx}{(a^2+x^2{)}^{3/2}}$

From the figure,
$[x=a\tan \theta \therefore dx=a{\sec }^2\theta d\theta ]$

$\Rightarrow E_{net}=\frac{MGa}{L}{\int }_0^L\frac{a{\sec }^2\theta d\theta }{(a^2+a^2{\tan }^2\theta {)}^{3/2}}$

$\Rightarrow E_{net}=\frac{MG}{La}{\int }_0^L\frac{{\sec }^2\theta d\theta }{(1+{\tan }^2\theta {)}^{3/2}}$

$\Rightarrow E_{net}=\frac{MG}{La}{\int }_0^L\frac{{\sec }^2\theta d\theta }{({\sec }^2\theta {)}^{3/2}}[\because (1+{\tan }^2\theta )={\sec }^2\theta ]$

$\Rightarrow E_{net}=\frac{MG}{La}{\int }_0^L\cos \theta d\theta$

$\Rightarrow E_{net}=\frac{MG}{La}[\sin \theta {]}_0^L$

From the figure, value of $\sin \theta =\frac{x}{\sqrt{x^2+a^2}}$

$\Rightarrow E_{net}=\frac{MG}{aL}{\left[\frac{x}{\sqrt{x^2+a^2}}\right]}_0^L$

$\Rightarrow E_{net}=\frac{MG}{aL}\frac{L}{\sqrt{L^2+a^2}}$

$\therefore E_{net}=\frac{MG}{a\sqrt{a^2+L^2}}$

So, the force acting on the particle of mass $m$ placed at point $\text{P}$,

$F=m{E}_{net}=\frac{GMm}{a\sqrt{a^2+L^2}}$

Hence, option (b) is correct answer.

Why this question: To make students practice field calculation due to continuous mass distribution. Let's try: Notice the similarity between the given question with value of force due to uniform finite line charge at a point along its perpendicular.