Question:
Statement I : A cyclist is moving on an unbanked road with a speed of 7 km h-1 and takes a sharp circular turn along a path of radius of 2 m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (\(g = 9.8\ \mathrm{m/s^2}\))
Statement II : If the road is banked at an angle of \(45^\circ\), cyclist can cross the curve of 2 m radius with the speed of 18.5 km h-1 without slipping.
In the light of the above statements, choose the correct answer from the options given below.
Statement I : A cyclist is moving on an unbanked road with a speed of 7 km h-1 and takes a sharp circular turn along a path of radius of 2 m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (\(g = 9.8\ \mathrm{m/s^2}\))
Statement II : If the road is banked at an angle of \(45^\circ\), cyclist can cross the curve of 2 m radius with the speed of 18.5 km h-1 without slipping.
In the light of the above statements, choose the correct answer from the options given below.
Statement II : If the road is banked at an angle of \(45^\circ\), cyclist can cross the curve of 2 m radius with the speed of 18.5 km h-1 without slipping.
In the light of the above statements, choose the correct answer from the options given below.
Show Hint
Check each statement with the limiting-friction condition: on a flat road \(v^2/(rg)\le\mu\); on a banked road the safe speed range is between \(v_{min}\) and \(v_{max}\) given by \(v^2=rg\,\frac{\tan\theta\pm\mu}{1\mp\mu\tan\theta}\).
Updated On: Nov 4, 2026
- Statement I is incorrect and statement II is correct
- Both statement I and statement II are true
- Statement I is correct and statement II is incorrect
- Both statement I and statement II are false
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Statement I: \(v = 7\ \mathrm{km/h} = 7\times\frac{5}{18} = 1.94\ \mathrm{m/s}\), so \(v^2 = 3.78\ \mathrm{m^2/s^2}\).
Friction must supply the centripetal force: \(\mu_{req} = \frac{v^2}{rg} = \frac{3.78}{2\times 9.8} = 0.193\).
Since \(0.193 \lt 0.2\), the cyclist does not slip. Statement I is true.
Statement II: banked road, \(\theta = 45^\circ\), \(\tan\theta = 1\), \(r = 2\ \mathrm{m}\), \(\mu = 0.2\).
\(v_{max}^2 = rg\,\frac{\tan\theta+\mu}{1-\mu\tan\theta} = 19.6\times\frac{1.2}{0.8} = 29.4\), so \(v_{max} = 5.42\ \mathrm{m/s} = 19.5\ \mathrm{km/h}\).
\(v_{min}^2 = rg\,\frac{\tan\theta-\mu}{1+\mu\tan\theta} = 19.6\times\frac{0.8}{1.2} = 13.07\), so \(v_{min} = 3.61\ \mathrm{m/s} = 13.0\ \mathrm{km/h}\).
The given speed 18.5 km/h lies between 13.0 km/h and 19.5 km/h, so the cyclist does not slip. Statement II is true.
Answer: (2) Both statement I and statement II are true
Friction must supply the centripetal force: \(\mu_{req} = \frac{v^2}{rg} = \frac{3.78}{2\times 9.8} = 0.193\).
Since \(0.193 \lt 0.2\), the cyclist does not slip. Statement I is true.
Statement II: banked road, \(\theta = 45^\circ\), \(\tan\theta = 1\), \(r = 2\ \mathrm{m}\), \(\mu = 0.2\).
\(v_{max}^2 = rg\,\frac{\tan\theta+\mu}{1-\mu\tan\theta} = 19.6\times\frac{1.2}{0.8} = 29.4\), so \(v_{max} = 5.42\ \mathrm{m/s} = 19.5\ \mathrm{km/h}\).
\(v_{min}^2 = rg\,\frac{\tan\theta-\mu}{1+\mu\tan\theta} = 19.6\times\frac{0.8}{1.2} = 13.07\), so \(v_{min} = 3.61\ \mathrm{m/s} = 13.0\ \mathrm{km/h}\).
The given speed 18.5 km/h lies between 13.0 km/h and 19.5 km/h, so the cyclist does not slip. Statement II is true.
Answer: (2) Both statement I and statement II are true
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