A car of 800 ~kg is taking turn on a banked road of radius 300 ~m and angle of banking 30^\circ . If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: (g=10 ~m / s^2, \sqrt3=1.73)

Circular Motion PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q60 with Solution

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Correct answer: A

For maximum speed on a banked road

At the highest safe speed, the car tends to slip up the bank, so static friction acts down the slope.

Let:

mass $m = 800\,\text{kg}$
radius $r = 300\,\text{m}$
banking angle $\theta = 30^\circ$
coefficient of static friction $\mu = 0.2$
$g = 10\,\text{m/s}^2$

Forces acting on the car

वजन: $mg$ vertically downward
Normal reaction: $N$ perpendicular to the road
Friction: $f = \mu N$ down the slope

Resolve forces into:

horizontal direction toward the center
vertical direction

Equations of motion

For maximum speed:

Horizontal centripetal force:

N\sin\theta + f\cos\theta = \frac{mv^2}{r}

Since $f=\mu N$ ,

N\sin\theta + \mu N\cos\theta = \frac{mv^2}{r}

N(\sin\theta + \mu\cos\theta)=\frac{mv^2}{r}

Vertical equilibrium:

N\cos\theta - f\sin\theta = mg

N\cos\theta - \mu N\sin\theta = mg

N(\cos\theta - \mu\sin\theta)=mg

Eliminate $N$

Divide the first equation by the second:

\frac{\dfrac{mv^2}{r}}{mg} = \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}

\frac{v^2}{rg} = \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}

So,

v^2 = rg\cdot \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}

Substitute values

Using

\sin 30^\circ = 0.5, \qquad \cos 30^\circ = \frac{\sqrt 3}{2} = \frac{1.73}{2}=0.865

Then

\sin\theta + \mu\cos\theta = 0.5 + 0.2(0.865)=0.5+0.173=0.673

\cos\theta - \mu\sin\theta = 0.865 - 0.2(0.5)=0.865-0.1=0.765

Thus,

v^2 = 300\times 10 \times \frac{0.673}{0.765}

v^2 = 3000 \times 0.8797 \approx 2639.2

v \approx \sqrt{2639.2} \approx 51.4\,\text{m/s}

Match with options

\boxed{v_{\max} = 51.4\,\text{m/s}}

So the correct option is A.

Comparison with stored answer

Stored correct answer: A

My derived answer also gives A, so they agree.

More from Circular Motion

View written solutionFree

Correct answer: A

For maximum speed on a banked road

At the highest safe speed, the car tends to slip up the bank, so static friction acts down the slope.

Let:

mass $m = 800\,\text{kg}$
radius $r = 300\,\text{m}$
banking angle $\theta = 30^\circ$
coefficient of static friction $\mu = 0.2$
$g = 10\,\text{m/s}^2$

Forces acting on the car

वजन: $mg$ vertically downward
Normal reaction: $N$ perpendicular to the road
Friction: $f = \mu N$ down the slope

Resolve forces into:

horizontal direction toward the center
vertical direction

Equations of motion

For maximum speed:

Horizontal centripetal force:

N\sin\theta + f\cos\theta = \frac{mv^2}{r}

Since $f=\mu N$ ,

N\sin\theta + \mu N\cos\theta = \frac{mv^2}{r}

N(\sin\theta + \mu\cos\theta)=\frac{mv^2}{r}

Vertical equilibrium:

N\cos\theta - f\sin\theta = mg

N\cos\theta - \mu N\sin\theta = mg

N(\cos\theta - \mu\sin\theta)=mg

Eliminate $N$

Divide the first equation by the second:

\frac{\dfrac{mv^2}{r}}{mg} = \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}

\frac{v^2}{rg} = \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}

So,

v^2 = rg\cdot \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}

Substitute values

Using

\sin 30^\circ = 0.5, \qquad \cos 30^\circ = \frac{\sqrt 3}{2} = \frac{1.73}{2}=0.865

Then

\sin\theta + \mu\cos\theta = 0.5 + 0.2(0.865)=0.5+0.173=0.673

\cos\theta - \mu\sin\theta = 0.865 - 0.2(0.5)=0.865-0.1=0.765

Thus,

v^2 = 300\times 10 \times \frac{0.673}{0.765}

v^2 = 3000 \times 0.8797 \approx 2639.2

v \approx \sqrt{2639.2} \approx 51.4\,\text{m/s}

Match with options

\boxed{v_{\max} = 51.4\,\text{m/s}}

So the correct option is A.

Comparison with stored answer

Stored correct answer: A

My derived answer also gives A, so they agree.

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