JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Laws of Motion

A car of 800 kg is taking turn on a banked road of radius 300 m and angle of banking

3 0^{\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}, 3 = 1.73)

(A) $264 m / s$
(B) $51.4 m / s$ ✓ Correct
(C) $70.4 m / s$
(D) $102.8 m / s$

Answer: (B)

Solution

For a banked road with friction, the maximum safe speed is

v_{ma x} = R g (\frac{tan θ + μ _{s}}{1 - μ _{s} tan θ})

With

R = 300 m

g = 10 m/s^{2}

θ = 3 0^{\circ}

(

tan 3 0^{\circ} \approx 0.577

) and

μ_{s} = 0.2

v_{ma x} = 3000 \times \frac{0.577 + 0.2}{1 - 0.2 ( 0.577 )} = 3000 \times \frac{0.777}{0.885} \approx 51.4 m/s

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option

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