SATHEE: Laws of Motion - Result Question 55

Laws of Motion - Result Question 55

55. A car is negotiating a curved road of radius $R$ . The road is banked at an angle $θ$ . the coefficient of friction between the tyres of the car and the road is $μ_{s}$ . The maximum safe velocity on this road is :

[2016]

(a) $\sqrt{g R^{2} (\frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ})}$

(b) $\sqrt{g R (\frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ})}$

(d) $\sqrt{\frac{g}{R^{2}} (\frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ})}$

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Answer:

Correct Answer: 55. (b)

Solution:

(b) On a banked road,

$\frac{V_{max}^{2}}{R g} = (\frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ})$

Maximum safe velocity of a car on the banked road

$V_{max} = \sqrt{R g [\frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ}]}$

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Laws of Motion - Result Question 55

55. A car is negotiating a curved road of radius R. The road is banked at an angle θ. the coefficient of friction between the tyres of the car and the road is μs. The maximum safe velocity on this road is :

Answer:

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55. A car is negotiating a curved road of radius $R$ . The road is banked at an angle $θ$ . the coefficient of friction between the tyres of the car and the road is $μ_{s}$ . The maximum safe velocity on this road is :