Q. If a cyclist moving with a speed of 4.9 $m/s$ on a level road can take a sharp circular turn of radius 4 $m,$ then coefficient of friction between the cycle tyres and road is
A) 0.41
B) 0.51
C) 0.61
D) 0.71
Explanation
(c) $\mu = \frac{v^{2}}{rg} = \frac{(4.9)^{2}}{4 \times 9.8} = 0.61$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.