The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m with skidding is (assume g = 10 ms⁻²)

The correct option is 10 ms⁻¹.

Physical Principle:
When a car travels along a curved path on a flat road, the centripetal force required to keep the car moving in a circle is provided entirely by the static friction between the tires and the road surface. To prevent the car from skidding, this required centripetal force must not exceed the maximum possible static frictional force.

The centripetal force $F_c$ for a car of mass $m$ moving at a speed $v$ along a curve of radius $r$ is given by:

$F_c=\frac{m{v}^2}{r}$

The maximum static frictional force $f_{\mathrm{s,max}}$ is given by:

$f_{\mathrm{s,max}}=\mu N$

where $\mu$ is the coefficient of friction and $N$ is the normal force. For a flat road, the normal force is equal to the weight of the car:

$N=mg$

Therefore, the maximum force of friction is:

$f_{\mathrm{s,max}}=\mu mg$

To avoid skidding, the required centripetal force must be less than or equal to the maximum frictional force:

$\frac{m{v}^2}{r}\le \mu mg$

Dividing both sides by the mass $m$:

$\frac{v^2}{r}\le \mu g$

Solving for the maximum safe speed $v_{\max }$:

$v_{\max }=\sqrt{\mu rg}$

Given Data:
Coefficient of friction, $\mu =0.25$
Radius of the curve, $r=40$ m
Acceleration due to gravity, $g=10$ ms^-2

Step-by-Step Calculation:
Substitute the given values into the formula for maximum speed:

$v_{\max }=\sqrt{0.25\times 40\times 10}$

Simplify the terms inside the square root:

$v_{\max }=\sqrt{2.5\times 40}$

$v_{\max }=\sqrt{100}$

$v_{\max }=10$ ms^-1

Thus, the maximum speed with which the car can be driven round the curve without skidding is 10 ms^-1.