If the momentum of a body is increased by 100 %, then the percentage increase in the kinetic energy is

The correct option is 300 %.

Let us understand the step-by-step relationship between the momentum and the kinetic energy of a body.

The kinetic energy ($K$) of a body of mass $m$ and velocity $v$ is given by:
$K=\frac{1}{2}m{v}^2$

The linear momentum ($p$) of the body is defined as:
$p=mv$

We can express the kinetic energy in terms of momentum by multiplying and dividing the kinetic energy equation by mass ($m$):
$K=\frac{m^2{v}^2}{2m}=\frac{p^2}{2m}$
This shows that kinetic energy is directly proportional to the square of momentum ($K\propto p^2$) for a constant mass.

Let the initial momentum of the body be $p$ and the initial kinetic energy be $K$.

According to the problem, the momentum of the body is increased by 100 %. Therefore, the new momentum ($p^{\prime }$) is:
$p^{\prime }=p+100\%\text{of}p=p+p=2p$

Now, let us find the new kinetic energy ($K^{\prime }$) using the new momentum:
$K^{\prime }=\frac{{\left(2p\right)}^2}{2m}=4·\frac{p^2}{2m}=4K$

The percentage increase in the kinetic energy is calculated as:
$\text{Percentage Increase}=\frac{K^{\prime }-K}{K}$

$\text{Percentage Increase}=\frac{4K-K}{K}$

Thus, the kinetic energy of the body increases by 300 %.