There are two bodies of masses 100 kg and 10000 kg separated by a distance 1m. At what distance from the smaller body, the intensity of gravitational field will be zero

The correct answer is Option 3: 1/11 m.

Step-by-Step Explanation:
Let us analyze the given information:
- Mass of the smaller body, $m_1=100\text{kg}$
- Mass of the larger body, $m_2=10000\text{kg}$
- Distance between the two bodies, $d=1\text{m}$

Let $P$ be the point between the two bodies where the net gravitational field intensity is zero. Let this point be at a distance $x$ from the smaller body of mass $m_1$.
Consequently, the distance of the point $P$ from the larger body of mass $m_2$ will be $(d-x)=(1-x)\text{m}$.

The gravitational field intensity $E$ due to a body of mass $M$ at a distance $r$ is given by the formula:
$E=\frac{GM}{r^2}$
where $G$ is the universal gravitational constant.

For the net gravitational field intensity at point $P$ to be zero, the magnitude of the gravitational field intensity due to $m_1$ must be equal to the magnitude of the gravitational field intensity due to $m_2$:
$E_1=E_2$

Substituting the expressions, we get:
$\frac{G{m}_1}{x^2}=\frac{G{m}_2}{{(1-x)}^2}$

We can cancel the gravitational constant $G$ from both sides:
$\frac{m_1}{x^2}=\frac{m_2}{{(1-x)}^2}$

Now, substitute the values of $m_1$ and $m_2$:
$\frac{100}{x^2}=\frac{10000}{{(1-x)}^2}$

Simplify the equation by dividing both sides by 100:
$\frac{1}{x^2}=\frac{100}{{(1-x)}^2}$

Taking the square root on both sides:
$\frac{1}{x}=\frac{10}{1-x}$

Cross-multiply to solve for $x$:
$1-x=10x$
$1=10x+x$
$1=11x$
$x=\frac{1}{11}\text{m}$

Therefore, the intensity of the gravitational field is zero at a distance of 1/11 m from the smaller body.