Question

A car is negotiating a curve of radius R = 20 m on a banked road with banking angle $\theta$ and coefficient of friction $\mu$. Take speed of car as v.

$\begin{array}{cccccc}& \text{Column 1}& & \text{Column 2}& & \text{Column 3}\\ \left(I\right)& v=10\sqrt{2}m/s& \left(i\right)& \mu =0.2& \left(p\right)& \theta ={30}^{\circ }\\ \left(II\right)& v=20m/s& \left(ii\right)& \mu =0.4& \left(Q\right)& \theta ={45}^{\circ }\\ \left(III\right)& v=50m/s& \left(iii\right)& \mu =0.6& \left(R\right)& \theta ={37}^{\circ }\\ \left(IV\right)& v=5m/s& \left(iv\right)& \mu =0.8& \left(S\right)& \theta ={60}^{\circ }\end{array}$

In which case will the friction be zero ?

• A. $\left(I\right)\left(i\right)\left(Q\right)$
• B. $\left(II\right)\left(ii\right)\left(R\right)$
• C. $\left(II\right)\left(iii\right)\left(R\right)$
• D. $\left(I\right)\left(iii\right)\left(R\right)$

Answer 1

The correct option is A

$\left(I\right)\left(i\right)\left(Q\right)$

For zero friction, the centripetal acceleration is provided by the horizontal component of normal reaction by the road.