Question

On a dry road, the maximum speed of a vehicle along a circular path is $V$. When the road becomes wet, the maximum speed becomes $\frac{V}{2}$. If the coefficient of friction of the dry road is $\mu$, then that of the wet road is

• A. $\frac{2\mu}{3}$
• B. $\frac{\mu}{4}$
• C. $\frac{\mu}{3}$
• D. $\frac{3\mu}{4}$

Hint:

Physics Tip: {square root} of the friction coefficient, reducing the speed by half requires the friction to decrease to one-fourth of its original value.

Answer 1

The Correct Option is B

Concept: Physics (Circular Motion) - Centripetal Force and Friction.

Step 1: State the formula for maximum velocity on a level curve. The maximum speed $V$ at which a vehicle can round a curve of radius $r$ without skidding is governed by the coefficient of friction $\mu$: $$V = \sqrt{\mu rg} \text{ --- (i) }$$

Step 2: Set up the equation for the wet road. Let $\mu'$ be the coefficient of friction for the wet road where the maximum speed is $V/2$: $$\frac{V}{2} = \sqrt{\mu' rg} \text{ --- (ii)}$$

Step 3: Compare the two equations. Dividing equation (i) by equation (ii): $$\frac{V}{V/2} = \frac{\sqrt{\mu rg}}{\sqrt{\mu' rg}} \text{ }$$ $$2 = \sqrt{\frac{\mu}{\mu'}} \text{ }$$

Step 4: Solve for $\mu'$. Squaring both sides: $$4 = \frac{\mu}{\mu'} \implies \mu' = \frac{\mu}{} \text{ }$$ $$ \therefore \text{The coefficient of friction of the wet road is } \frac{\mu}{4}. \text{ } $$