Question

A car is moving round a curve of radius 20 m without slipping on a banked road with 45°. Assuming the coefficient of friction between the
road and tyres to be 0.2. What is the
maximum speed with which the car can move?
(Take g = 10 m/s2)

(1) 10 m/s
(2) 5 m/s
(3) 15 m/s
(4) 10/3 m/s​

Answer 1

Answer:answer is in the attachment... Hope it helps!!

Actually the furmula is derived. And sry i have not writing the derivation here

Explanation:

Answer 2

Given:

• Radius of curvature = 20 m
• Angle of road = 45°
• coefficient of friction = 0.2

To find:

• Max permissible velocity of car ?

Calculation:

The general expression of maximum permissible velocity on a banked road is given as:

$\dfrac{ {v}^{2} }{rg} = \dfrac{ \mu + \tan( \theta) }{1 - \mu \tan( \theta) }$

$\implies \dfrac{ {v}^{2} }{20 \times 10} = \dfrac{ 0.2 + \tan( {45}^{ \circ} ) }{1 - (0.2) \tan( {45}^{ \circ} ) }$

$\implies \dfrac{ {v}^{2} }{200} = \dfrac{ 0.2 + \tan( {45}^{ \circ} ) }{1 - (0.2) \tan( {45}^{ \circ} ) }$

$\implies \dfrac{ {v}^{2} }{200} = \dfrac{ 0.2 + 1 }{1 - (0.2)1 }$

$\implies \dfrac{ {v}^{2} }{200} = \dfrac{ 1.2 }{0.8 }$

$\implies \dfrac{ {v}^{2} }{200} = \dfrac{ 12 }{8 }$

$\implies \dfrac{ {v}^{2} }{200} = \dfrac{ 3 }{2}$

$\implies {v}^{2} = 300$

$\implies v= 10 \sqrt{3} \: m {s}^{ - 1}$

So, max velocity allowed is 10√3 m/s (Option 4).