The minimum velocity (in ms^ - 1 ) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is

Circular Motion PYQ — JEE Main 2002 · Shift 0 · Q180 with Solution

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Correct answer: B

Identify the force providing centripetal force

On a flat curve, the centripetal force needed for circular motion is provided by friction.

So,

\frac{mv^2}{r} \leq f_{\max}

where

f_{\max}=\mu mg

Thus,

\frac{mv^2}{r} \leq \mu mg

Find the limiting speed

For the car to just avoid skidding,

\frac{mv^2}{r}=\mu mg

Cancelling $m$ ,

\frac{v^2}{r}=\mu g

Hence,

v=\sqrt{\mu g r}

Substitute the given values

Given:

$\mu=0.6$
$r=150\,\text{m}$
$g\approx 10\,\text{m s}^{-2}$

So,

v=\sqrt{0.6\times 10\times 150}

=\sqrt{900}

=30\,\text{m s}^{-1}

Match with the options

The required speed is

30\,\text{m s}^{-1}

So the correct option is B.

Note on wording

Physically, on a flat curve, friction sets the maximum speed to avoid skidding outward, not the minimum speed. But using the standard formula, the limiting safe speed is $30\,\text{m s}^{-1}$ .

Final Answer: $30\,\text{m s}^{-1}$ (Option B)

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