When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be:

We are given that the kinetic energy of a body becomes 36 times its original value and asked to determine the percentage increase in momentum.

The relationship between kinetic energy ($$KE$$) and momentum ($$p$$) is $$KE = \frac{p^2}{2m}$$, where $$m$$ is the mass of the body, and rearranging gives $$p = \sqrt{2m \cdot KE}$$.

Letting the initial kinetic energy be $$KE$$ with corresponding momentum $$p$$ leads to $$p = \sqrt{2m \cdot KE}$$.

When the kinetic energy increases to $$KE' = 36 \, KE$$, the new momentum becomes $$p' = \sqrt{2m \cdot KE'} = \sqrt{2m \cdot 36 \, KE} = \sqrt{36} \cdot \sqrt{2m \cdot KE} = 6p$$, since $$\sqrt{36} = 6$$ and $$\sqrt{2m \cdot KE} = p$$.

The increase in momentum is then $$\Delta p = p' - p = 6p - p = 5p$$.

Therefore, the percentage increase in momentum is $$\frac{\Delta p}{p} \times 100 = \frac{5p}{p} \times 100 = 500\%$$.

The correct answer is Option (4): 500%.