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Question

# A car goes on a horizontal circular road of radius R. The speed increases at a constant rate a=dvdt. The friction coefficient between the tyre and road is μ. What is the speed at which the car will skid?

A
(μga)R2
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B
[(μ2g2a2)R2]14
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C
[(μ2g2a2)R]14
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D
(μga)R4
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Solution

## The correct option is A [(μ2g2−a2)R2]14For uniform circular motion, the resultant acceleration isaR=√a2N+a2taN=v2R,at=dvdt=aFrom FBD of car on the rough roadf=maRμN=m√(v2R)2+(dvdt)2⇒μ2N2=m2(v4R2+a2)⇒v4=μ2N2R2m2−a2R2=μ2m2g2R2m2−a2R2=μ2g2R2−a2R2v=[(μ2g2−a2)R2]14.

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