Q76 / 90 · JEE Main 2024 (06 Apr Shift 2) · Physics
When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be:
- A$6 \%$
- B$600 \%$
- C$60 \%$
- Correct answerD$500 \%$
Solution
Kinetic energy $(K)=\frac{P^2}{2 m}$
$\Rightarrow \mathrm{P}=\sqrt{2 \mathrm{mK}}$
If $K_f=36 K_i$ So, $\mathrm{P}_{\mathrm{f}}=6 \mathrm{P}_{\mathrm{i}}$ $\begin{aligned} \% \text { increase in momentum } & =\frac{\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\ & =\frac{6 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\ & =500 \% \end{aligned}$
If $K_f=36 K_i$ So, $\mathrm{P}_{\mathrm{f}}=6 \mathrm{P}_{\mathrm{i}}$ $\begin{aligned} \% \text { increase in momentum } & =\frac{\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\ & =\frac{6 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \% \\ & =500 \% \end{aligned}$
Hence, the correct answer is option: 'D'.