Q. A car turns a corner on a slippery road at a constant speed of $10\ m/s$. If the coefficient of friction is 0.5, the minimum radius of the arc in meter in which the car turns is
A) 20
B) 10
C) 5
D) 4
Explanation
(a) $v = \sqrt{\mu\ g\ r} \Rightarrow r = \frac{v^{2}}{\mu g} = \frac{100}{0.5 \times 10} = 20$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.