The kinetic energy of a particle depends on the square of speed of the particle. If error in measurement of speed is $30\%$, the error in the measurement of kinetic energy will be $n\%$. The value of $n\%$ is ________

Hint : 

Error in kinetic energy is $\frac{\Delta K}{K}$

$\text{Step1:Relation between kinetic energy and speed}$

Kinetic Energy of particle is $\frac{1}{2}m{v}^2$

$\Rightarrow$$KE\propto v^2$

$\Rightarrow$$\frac{{\mathrm{K}\mathrm{E}}_2}{{\mathrm{K}\mathrm{E}}_1}=\left(\frac{{\mathrm{v}}_1^2}{{\mathrm{v}}_2^2}\right)$

$\text{Step2:Relation between initial and final kinetic energies}$

Given that the error in measure of speed is  $30\%$

$\therefore \frac{{\mathrm{K}\mathrm{E}}_2}{{\mathrm{K}\mathrm{E}}_1}={\left(\frac{100+30}{100}\right)}^2$

$\Rightarrow \frac{{\mathrm{K}\mathrm{E}}_2}{{\mathrm{K}\mathrm{E}}_1}=1.69$

$\Rightarrow {\mathrm{K}\mathrm{E}}_2=1.69{\mathrm{K}\mathrm{E}}_1$

$\text{step3:Percentage change in kinetic energy}$

Percentage change in kinetic energy is 

$=\frac{{\mathrm{K}\mathrm{E}}_2-{\mathrm{K}\mathrm{E}}_1}{{\mathrm{K}\mathrm{E}}_1}\times 100$

$=\frac{1.69{\mathrm{K}\mathrm{E}}_1-{\mathrm{K}\mathrm{E}}_1}{{\mathrm{K}\mathrm{E}}_1}$

$=69\%$

Therefore the value of $n\%$ is $69\%$.