The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30 ° banked road at maximum possible speed without skidding is ____________ × 103 kg m/s2. [Given cos30 ° = 0.87, μ s = 0.2] (a) 12.4 (b) 7.2 (c) 6.96 (d) 10.2

Answer: (d). θNμNmgcentre of turn ← At maximum speed friction μs N acts down the slope. Vertical balance: Ncos θ - μs Nsin θ = mg. N = (mg)/(cos θ - μssin θ) = (800 × 9.8)/(0.87 - 0.2 × 0.5) = (7840)/(0.77) N ≈ 10.2 × 10³ kg m/s ² So the answer is (d) 10.2.

The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a… | JEE Main 2021 (20 Jul, Shift 1)

Q13JEE Main 2021 · 20 JUL · SHIFT 1Difficult

Banking of Roads & Bridges

The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30 $^\circ$ banked road at maximum possible speed without skidding is ____________ $\times$ 103 kg m/s2. [Given cos30 $^\circ$ = 0.87, $\mu$ s = 0.2]

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Answer: (d)

At maximum speed friction $\mu_s N$ acts down the slope.
Vertical balance: $N\cos\theta - \mu_s N\sin\theta = mg$ .
$N = \frac{mg}{\cos\theta - \mu_s\sin\theta} = \frac{800 \times 9.8}{0.87 - 0.2 \times 0.5} = \frac{7840}{0.77}$
$N \approx 10.2 \times 10^3$ kg m/s $^2$
So the answer is (d) 10.2.

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