Two bodies of masses 100 kg and 10.000 kg are kept at a distance 1m apart. At which point on the line joining them a third body experience the net gravitational force of zero?
Find the distance from the first body of mass 100kg.

Given that $m_1=100\text{kg},m_2=10,000\text{kg}\text{,}\text{d =1m}$ 

The gravitational force between two bodies can be calculated as:
$F=\frac{GMm}{r^2}$
Where M and m are the masses of the objects, r is the distance between them and G is the universal gravitational constant having the value of 6.67 X 10^-11 Nm²/kg².

Let the distance of the third body from the first body be X.
The distance from the second body will be 1-X.
We have:

$F_1=F_2$

$\Rightarrow \frac{G\times 100\times m}{x^2}=\frac{G\times 1000\times m}{\left(1-x^2\right)}$

$\Rightarrow 10x^2={(1-x)}^2$

$\Rightarrow 10x=\pm (1-x)$
There are two cases:
$10x= 1-x OR 10x=-(1-x)$
Solving both, we get:
$x=\frac{1}{11} OR x=\frac{-1}{9}$
The distance cannot be negative,

$\therefore x=\frac{1}{11}\text{m}$
Therefore, option C is the correct answer