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 meter in which the car turns is

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Solution

The correct option is A
20

$v = \sqrt{μ g r} \Rightarrow r = \frac{v^{2}}{μ g} = \frac{100}{0.5 \times 10} = 20$

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

The correct option is A 20 v=√μgr⇒r=v2μg=1000.5×10=20

The correct option is A
20

$v = \sqrt{μ g r} \Rightarrow r = \frac{v^{2}}{μ g} = \frac{100}{0.5 \times 10} = 20$