Question

Statement I : A cyclist is moving on an unbanked road with a speed of $7{\text{kmh}}^{-1}$ and takes a sharp circular turn along a path of radius of $2\text{m}$ without reducing the speed. The static friction coefficient is $\mathrm{0.2.}$ The cyclist will not slip and pass the curve.
$(g=10{\text{m/s}}^2)$

Statement II : If the road is banked at an angle of ${45}^{\circ }$, cyclist can cross the curve of $2\text{m}$ radius with the speed of $15{\text{kmh}}^{-1}$ 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

Statement 1 :
On a horizontal ground,
$V_{max}=\sqrt{\mu Rg}=\sqrt{0.2\times 2\times 10}=2\text{m/s}=2\times \frac{18}{5}$
$=7.2\text{km/h}$ > given velocity.
Hence, it can turn safely.

Statement - 2 :
Given, $\theta ={45}^{\circ }\Rightarrow \tan \theta =\tan {45}^{\circ }=1$
Also, $V_{max}=\sqrt{gr\left(\frac{\tan \theta +\mu }{1-\mu \tan \theta }\right)}=\left(\sqrt{2\times 10\times \frac{1.2}{0.8}}\right)\times \frac{18}{5}=19.72\text{km/h}$

And, $V_{min}=\sqrt{rg\left(\frac{\tan \theta -\mu }{1+\mu \tan \theta }\right)}=\left(\sqrt{2\times 10\times \frac{0.8}{1.2}}\right)\times \frac{18}{5}=13.14\text{km/h}$

Since, given velocity is between $V_{max}$ and $V_{min}$, cyclist can safely turn.