Q135 / 150 · MHT CET 2021 (22 Sep Shift 1) · Physics
A car of mass ' $m$ ' moving with velocity ' $u$ ' on a straight road in a straight line, doubles its velocity in time $t$. the power delivered by the engine of a car for doubling the velocity is
- Correct answerA$\frac{3 \mathrm{mu}^2}{2 \mathrm{t}}$
- B$\frac{\mathrm{mu}^2}{2 \mathrm{t}}$
- C$\frac{2 \mathrm{mu}^2}{\mathrm{t}}$
- D$\frac{3 \mathrm{mu}^2}{\mathrm{t}}$
Solution
Initial kinetic energy $\mathrm{k}_1=\frac{1}{2} \mathrm{mu}^2$
Final kinetic energy $\mathrm{k}_2=\frac{1}{2} \mathrm{mu}_2^2=\frac{1}{2} \mathrm{~m}(2 \mathrm{u})^2=\frac{1}{2}\left(4 \mathrm{mu}^2\right)$
$$
\therefore \mathrm{k}_2-\mathrm{k}_1=\frac{3}{2} \mathrm{mu}^2
$$
Change in K.E. is work done
Power, $\mathrm{P}=\frac{\text { workdone }}{\mathrm{t}}=\frac{3 \mathrm{mu}^2}{2 \mathrm{t}}$
Hence, the correct answer is option: 'A'.