Question

If K. E. body is increased by $100\%$ . Then % change in $P$ is

• A. $50\%$
• B. $41.4\%$
• C. $10\%$
• D. $20\%$

Answer 1

Answer: (B)

$41.4\%$

We know $K.E\left(sayk\right)=\frac{p^2}{2m}$

After Increasing K.E

$2k=\frac{p_2^2}{2m}$

$\Rightarrow p_2=\sqrt{2}p$

Change in P $=\sqrt{2}p-p=0.414p$

So. Percentage change in P = $41.4\%$

Therefore, B is correct option.