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Laws of Motion question

2016 · Q143
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Laws of Motion question

2016 · Q143

NEETPhysicsLaws of MotionMCQ+4 / −1
A car is negotiating a curved road of radius R. The road is banked at an angle θ\thetaθ. The coefficient of friction between the tyres of the car and the road is μ\muμs. The maximum safe velocity on this road is
  1. A
    gRμs+tan⁡θ1−μstan⁡θ\sqrt {{g \over R}{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}}Rg​1−μs​tanθμs​+tanθ​​
  2. B
    gR2μs+tan⁡θ1−μstan⁡θ\sqrt {{g \over {{R^2}}}{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}}R2g​1−μs​tanθμs​+tanθ​​
  3. C
    gR2μs+tan⁡θ1−μstan⁡θ\sqrt {g{R^2}{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}}gR21−μs​tanθμs​+tanθ​​
  4. D
    gRμs+tan⁡θ1−μstan⁡θ\sqrt {gR{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}}gR1−μs​tanθμs​+tanθ​​
View written solutionFree

Correct answer: D

On a banked road,
Vmax⁡2Rg=(μs+tan⁡θ1−μstan⁡θ){{V_{\max }^2} \over {Rg}} = \left( {{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}} \right)RgVmax2​​=(1−μs​tanθμs​+tanθ​)

Maximum safe velocity of a car on the banked road
Vmax⁡=Rg[μs+tan⁡θ1−μstan⁡θ]{V_{\max }} = \sqrt {Rg\left[ {{{{\mu _s} + \tan \theta } \over {1 - {\mu _s}\tan \theta }}} \right]} Vmax​=Rg[1−μs​tanθμs​+tanθ​]​

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