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

<p>The relationship between kinetic energy (K.E) and momentum (p) of a body can be expressed through their respective definitions. Kinetic energy is given by $K.E = \frac{1}{2} mv^2$ where $m$ is the mass of the body and $v$ is its velocity. The momentum (p) of a body is given by $p = mv$. To express kinetic energy in terms of momentum, we can manipulate the expression for momentum as follows:</p> <p>$p = mv \implies v = \frac{p}{m}$</p> <p>Substituting $v$ in the kinetic energy formula, we get</p> <p>$K.E = \frac{1}{2} m\left(\frac{p}{m}\right)^2 = \frac{1}{2} \frac{p^2}{m}$</p> <p>Therefore, we see that kinetic energy is directly proportional to the square of the momentum $(K.E \propto p^2)$.</p> <p>Now, given that the kinetic energy of a body becomes 36 times its original value, we can set up the proportionality as</p> <p>$\frac{K.E_{\text{final}}}{K.E_{\text{original}}} = 36$</p> <p>Since $K.E_{\text{final}} = 36 \times K.E_{\text{original}}$ and knowing $K.E \propto p^2$, we can express this relationship through the squares of the initial and final momentum:</p> <p>$\frac{p_{\text{final}}^2}{p_{\text{original}}^2} = 36$</p> <p>Taking the square root of both sides to find the ratio of final to initial momentum, we have</p> <p>$\frac{p_{\text{final}}}{p_{\text{original}}} = \sqrt{36} = 6$</p> <p>This indicates that the final momentum is 6 times the original momentum. To find the percentage increase in the momentum, we calculate the increase from the original to the final, subtracting the original momentum (which is considered 1 times itself):</p> <p>$$\text{Percentage increase} = \left(\frac{p_{\text{final}} - p_{\text{original}}}{p_{\text{original}}}\right) \times 100\% = \left(\frac{6p - p}{p}\right) \times 100\% = \left(6 - 1\right) \times 100\% = 5 \times 100\% = 500\%$$</p> <p>Therefore, the correct answer is Option B: 500%.</p>