Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A car of mass $m$ is moving on a level circular track of radius $R$ . If $\mu_s$ represents the static friction between the road and tyres of the car, then the maximum speed of the car in circular motion is given by:

$\sqrt{\mu_s mRg}$

$\sqrt{\frac{Rg}{\mu_s}}$

$\sqrt{\frac{mRg}{\mu_s}}$

$\sqrt{\mu_s Rg}$

Step-by-Step Solution

Forces on a Level Road: On a level circular road, the vertical forces (weight $mg$ and normal reaction $N$ ) balance each other, so $N = mg$ . The necessary centripetal force required for the car to move in a circle of radius $R$ is provided solely by the static friction ( $f_s$ ) between the tyres and the road.
Centripetal Force Condition: The centripetal force is given by $F_c = \frac{mv^2}{R}$ . This force must be less than or equal to the maximum static friction ( $f_{s,max}$ ) to prevent skidding. $\frac{mv^2}{R} \le f_{s,max}$ $f_{s,max} = \mu_s N = \mu_s mg$
Derivation: $\frac{mv^2}{R} \le \mu_s mg$ $v^2 \le \mu_s Rg$ $v \le \sqrt{\mu_s Rg}$ Therefore, the maximum speed is $v_{max} = \sqrt{\mu_s Rg}$ . (Reference: NCERT Class 11, Physics Part I, Chapter 5, Section 5.10, Eq. 4.18).

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started