If a cyclist moving with a speed of 4.9 m /s on a level road can take a sharp circular turn of radius 4m, then coefficient of friction between the cycle tyres and road is

To find the minimum coefficient of friction between the bicycle tyres and the road, we analyze the forces acting on the cyclist during the circular turn on a level road.

When a cyclist takes a turn of radius $r$ with a velocity $v$, the necessary centripetal force is provided by the frictional force between the tyres and the road.

The frictional force $f$ must be less than or equal to the maximum static friction, which is given by:
$f\le \mu N$
where:
$\mu$ is the coefficient of static friction,
$N$ is the normal reaction force. Since the road is level, the normal force balances the weight of the cyclist and the cycle:
$N=mg$

The centripetal force required to keep the cyclist in a circular path of radius $r$ is:
$F_c=\frac{m{v}^2}{r}$

Equating the centripetal force to the force of static friction for the limiting case (maximum safe speed or minimum required coefficient of friction):
$\frac{m{v}^2}{r}=\mu mg$

By simplifying the equation, we can cancel the mass ($m$) from both sides:
$\mu =\frac{v^2}{rg}$

Now, we substitute the given values into the formula:
Speed, $v=4.9\text{m/s}$
Radius, $r=4\text{m}$
Acceleration due to gravity, $g=9.8{\text{m/s}}^2$

Calculation:
$\mu =\frac{{\left(4.9\right)}^2}{4\times 9.8}$

Evaluating the terms:
$\mu =\frac{24.01}{39.2}$

Dividing the values:
$\mu \approx 0.6125$

Therefore, the coefficient of friction between the cycle tyres and the road is approximately 0.61.