A car turns a corner on a slippery road at a constant speed of $10$ m $/$ s. If the coefficient of friction is $0.5$ , the minimum radius of the arc in metre in which the car turns is?

20

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10

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5

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4

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Solution

The correct option is B $20$
Given, $v = 10 m / s μ = 0.5$
We know that $v_{m a x} = \sqrt{μ g R}$
On squaring both sides, we get
$R = \frac{v^{2}}{μ g}$
$= \frac{100}{0.5 \times 10} = 20$ .

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A car turns a corner on a slippery road at a constant speed of 10m/s. If the coefficient of friction is 0.5, the minimum radius of the arc in metre in which the car turns is?

The correct option is B 20Given, v=10m/sμ=0.5We know that vmax=√μgROn squaring both sides, we getR=v2μg=1000.5×10=20.

A car turns a corner on a slippery road at a constant speed of $10$ m $/$ s. If the coefficient of friction is $0.5$ , the minimum radius of the arc in metre in which the car turns is?

The correct option is B $20$
Given, $v = 10 m / s μ = 0.5$
We know that $v_{m a x} = \sqrt{μ g R}$
On squaring both sides, we get
$R = \frac{v^{2}}{μ g}$
$= \frac{100}{0.5 \times 10} = 20$ .