Question

Statement I: A cyclist is moving on an unbanked road with a speed of $7km{h}^{-1}$and takes a sharp circular turn along a path of radius of$2m$ without reducing the speed. The static friction coefficient is $0.2$. The cyclist will not slip and pass the curve.

($g=9.8m/s^2$)

Statement-II: If the road is banked at an angle of $45°$, cyclists can cross the curve of $2m$ radius with the speed of $18.5km{h}^{-1}$ without slipping.

In the light of the above statements, choose the correct answer from the options given below:

• A. Both statement I and statement II are false
• B. Both statement I and statement II are true
• C. Statement I is correct and statement II is incorrect
• D. Statement I is incorrect and statement II is correct

Answer 1

The correct option is B

Both statement I and statement II are true

Step 1. Given data

For unbanked road

speed of cyclist, $v_1=7km/hr$

radius of path taken $r=2m$

coefficient of friction $\mu =0.2$

For banked road

angle on bank, $\theta =45°$

speed of cyclist, $v_1=18.5km/hr$

radius of path taken $r=2m$

coefficient of friction $\mu =0.2$

Step 2. Solving for safe speed

For Unbanked road that is horizontal path, the maximum safe velocity is

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

where $\mu$ is coefficient of friction $g$ is acceleration due to gravity and $r$ is radius of path. we have

$$\begin{gathered} v_{max}=\sqrt{0.2\times 9.8\times 2}=1.97m/s \\ =7.2km/hr \end{gathered}$$

Since the speed of cyclist is less than maximum safe velocity, he can turn safely

For banked road, the maximum safe velocity is

$v_{max}=\sqrt{gr\frac{\tan \theta +\mu }{1-\mu \tan \theta }}=\sqrt{9.8\times 2\times \frac{1+0.2}{1-0.2\times 1}}=19.5km/hr$

the minimum safe velocity is

$v_{min}=\sqrt{gr\frac{\tan \theta -\mu }{1+\mu \tan \theta }}=\sqrt{9.8\times 2\times \frac{1-0.2}{1+0.2\times 1}}=12.0km/hr$

since the speed of cyclist is $18.5km{h}^{-1}$, which is in range $12.0to19.5km/hr$ of safe velocity, therefore he can turn safely.

Hence, option B is correct