The gravitational force of attraction between two objects decreases by 19 % when the distance between them increased by 4m. Find the original distance between them.

Let the original force between the two objects be:
$F_1=F$
Force after 19% decrease will be:
$F_2=\frac{81}{100}F$
Let the original distance be r.
$r_1=r$
The distance after 4m increase will be:
$r_2=r+4$ 
We know that when the masses of both bodies are constant,
$F\propto \frac{1}{r^2}$

$\Rightarrow \frac{F_1}{F_2}=\frac{r_2^2}{r_1^2}$
Substituting the values,

$\Rightarrow \frac{100F}{81F}={\left(\frac{r+4}{r}\right)}^2$

$\Rightarrow \frac{10}{9}=\frac{r+4}{r}$

$10r=9r+36$

$r=36m$
Therefore, option C is the correct answer.