    Question

# A car turns a corner on a slippery road at a constant speed of $12m/s$. If the coefficient of friction is $0.4$, the minimum radius of the arc in meters in which the car turns is?

A

$9$

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B

$18$

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C

$36$

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D

$72$

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Solution

## The correct option is C $36$Step 1: GivenVelocity of the car, $v=12m/s$Coefficient of friction, $\mu =0.4$Step 2: Determine the safe radiusFor a safe turn without sliding,$\mu \ge \frac{{v}^{2}}{rg}\phantom{\rule{0ex}{0ex}}r\ge \frac{{v}^{2}}{\mu g}$For the minimum value,$r=\frac{{v}^{2}}{\mu g}\phantom{\rule{0ex}{0ex}}=\frac{{12}^{2}}{0.4×10}\phantom{\rule{0ex}{0ex}}=\frac{144}{4}\phantom{\rule{0ex}{0ex}}=36m$Therefore, option C is the correct option.  Suggest Corrections  0      Similar questions  Explore more