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Question: A circular racetrack of radius 300 m is banked at an angle of \(15 ^ { \circ }\) . The coefficient ...

A circular racetrack of radius 300 m is banked at an angle of 15∘15 ^ { \circ } . The coefficient of friction between the wheels of a race car and the road is 0.2. The optimum speed of the race car to avoid wear and tear on its tyres is (Take tan⁡15∘=0.27, g=10 m s−2\tan 15 ^ { \circ } = 0.27 , \mathrm {~g} = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 } )

A

103 m s−110 \sqrt { 3 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

B

910 m s−19 \sqrt { 10 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

C

10 m s−1\sqrt { 10 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

D

210 m s−12 \sqrt { 10 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

Answer

910 m s−19 \sqrt { 10 } \mathrm {~m} \mathrm {~s} ^ { - 1 }

Explanation

Solution

Here R=300 m,θ=15∘,g=10 ms=2,μ=0.2R = 300 \mathrm {~m} , \theta = 15 ^ { \circ } , g = 10 \mathrm {~ms} ^ { = 2 } , \mu = 0.2

The optimum speed of the car to avoid wear and tear is given by

v=Rgtan⁡θ=300×10×tan⁡15∘v = \sqrt { R g \tan \theta } = \sqrt { 300 \times 10 \times \tan 15 ^ { \circ } } =810=910 ms−1= \sqrt { 810 } = 9 \sqrt { 10 } \mathrm {~ms} ^ { - 1 }