JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Laws of Motion

The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a

3 0^{\circ}

banked road at maximum possible speed without skidding is

____

\times 1 0^{3} kg m / s^{2}

.
[Given

cos 3 0^{\circ} = 0.87, μ_{5} = 0.2

]

(A)6.96
(B)10.2✓ Correct
(C)12.4
(D)7.2

Answer: (B)

Solution

At the greatest speed the vehicle is on the verge of sliding up the bank, so friction acts down the slope with magnitude

μ N

. Resolving along the vertical, where there is no acceleration,

N cos θ = m g + μ N sin θ ⟹ N = \frac{m g}{cos θ - μ sin θ}

With

cos 3 0^{\circ} = 0.87

sin 3 0^{\circ} = 0.5

and

μ = 0.2

the denominator is

0.87 - 0.1 = 0.77

, so

N = \frac{800 \times 9.8}{0.77} = 1.02 \times 1 0^{4} kg m/ s^{2}

The normal reaction is

10.2 \times 1 0^{3}

kg m/s

^{2}

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option

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