Q101 / 150 · MHT CET 2011 · Physics
A car of mass $1500 \mathrm{~kg}$ is moving with a speed of $12.5 \mathrm{~m} / \mathrm{s}$ on a circular path of radius $20 \mathrm{~m}$ on a level road. What should be the coefficient of friction between the car and the road, so that the car does not slip?
- A$0.2$
- B$0.4$
- C$0.6$
- Correct answerD$0.8$
Solution
In this case centripetal force provides by friction
$\therefore$
$$
\frac{m v^{2}}{r}=\mu m g
$$
i. $e$,
$$
\mu=\frac{v^{2}}{r g}
$$
$$
\mu=\frac{12.5 \times 12.5}{20 \times 9.8}=0.8
$$
Hence, the correct answer is option: 'D'.