Question

The velocity of a body increases by $10\%$. The percentage increase in its kinetic energy is

• A. $10\%$
• B. $21\%$
• C. $40\%$
• D. $38\%$

Answer 1

Answer: (B)

Here, $v'=v+\frac{10}{100}v=\frac{11}{10}v$
$\therefore \frac{k'}{k}=\frac{\frac{1}{2}mv'^{2}}{\frac{1}{2}mv^{2}}$ or $\frac{k'}{k}=\left(\frac{v'}{v}\right)^{2}=\left(\frac{11}{10}\right)^{2}$
Now, $\%$ increase in kinetic energy
$=\left(\frac{k'-k}{k}\right) \times100$
$=\left(\frac{k'}{k}-1\right) \times100$
$=\left(\frac{121}{100}-1\right) \times 100$
$=21\%$