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A car of mass 800 kg takes a turn of radius 40 m at 20 m/s on a level road. The minimum coefficient of static friction needed is
A0.82
B1.02
C0.51
D0.25
Answer & Solution
Correct answer: B. 1.02
1. For circular motion on a level road friction provides the centripetal force.
2. μ_s ≥ v²/(rg) = (20)²/(40×9.8) = 400/392.
3. μ_s ≈ 1.02 (greater than 1, so this turn is not safe without banking).
_Source: Maharashtra Balbharati Std XII Physics, Ch 1 "Rotational Dynamics" §1.3_
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