Question

An automobile negotiates a turn of radius $\frac{40}{\sqrt{3}}\text{m}$ on a road which is banked at the curve at angle ${30}^{\circ }$ to the horizontal. If coefficient of friction between road surface and automobile is $\frac{1}{\sqrt{3}}$ then maximum speed at which the car can negotiate the turn without skidding will be $(g=10{\text{m/s}}^2)$

• A. $10{\text{ms}}^{-1}$
• B. $10\sqrt{3}{\text{ms}}^{-1}$
• C. $20\sqrt{3}{\text{ms}}^{-1}$
• D. $20{\text{ms}}^{-1}$

Answer 1

The correct option is D $20{\text{ms}}^{-1}$

The maximum speed at which a vehicle can negotiate a curved road of radius $R$ and banked at an angle $\theta$ is given by,

$V_{max}=\sqrt{Rg\frac{\tan \theta +\mu }{1-\mu \tan \theta }}$

Given,

$R=\frac{40}{\sqrt{2}}\text{m}\theta ={30}^{\circ },\mu =\frac{1}{\sqrt{3}}$

$V_{max}=\sqrt{\frac{40}{\sqrt{3}}\times 10\times \frac{\tan {30}^{\circ }+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}\tan {30}^{\circ }}}$

$=\sqrt{\frac{40}{\sqrt{3}}\times 10\times \frac{\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}}}$

$=\sqrt{\frac{40}{\sqrt{3}}\times 10\times \sqrt{3}}$

$=20{\text{ms}}^{-1}$

Hence, $\left(D\right)$ is the correct answer.