# Statement I : A cyclist is moving on an unbanked road with a speed

Question:

Statement I : A cyclist is moving on an unbanked road with a speed of $7 \mathrm{kmh}^{-1}$ and takes a sharp circular turn along a path of radius of $2 \mathrm{~m}$ without reducing the speed. The static friction coefficient is $0.2$. The

cyclist will not slip and pass the curve $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$

Statement II : If the road is banked at an angle of $45^{\circ}$, cyclist can cross the curve of $2 \mathrm{~m}$ radius with the speed of $18.5 \mathrm{kmh}^{-1}$ without slipping.

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

1. (1) Statement I is incorrect and statement II is correct

2. (2) Statement I is correct and statement II is incorrect

3. (3) Both statement I and statement II are false

4. (4) Both statement I and statement II are true

Correct Option: , 4

Solution:

(4)

Statement I :

$\mathrm{v}_{\max }=\sqrt{\mu \mathrm{Rg}}=\sqrt{(0.2) \times 2 \times 9.8}$

$\mathrm{v}_{\max }=1.97 \mathrm{~m} / \mathrm{s}$

$7 \mathrm{~km} / \mathrm{h}=1.944 \mathrm{~m} / \mathrm{s}$

Speed is lower than $\mathrm{v}_{\max }$, hence it can take safe turn.

Statement II

$\mathrm{v}_{\max }=\sqrt{\mathrm{Rg}\left[\frac{\tan \theta+\mu}{1-\mu \tan \theta}\right]}$

$=\sqrt{2 \times 9.8\left[\frac{1+0.2}{1-0.2}\right]}=5.42 \mathrm{~m} / \mathrm{s}$

$18.5 \mathrm{~km} / \mathrm{h}=5.14 \mathrm{~m} / \mathrm{s}$

Speed is lower than $\mathbf{v}_{\max }$, hence it can take safe turn.