A curved road of diameter $1.8 k m$ is banked so that no friction is required at a speed of $30 m / s$ . What is the banking angle?

6^{\circ}

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16^{\circ}

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26^{\circ}

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{0.6}^{\circ}

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Solution

The correct option is A $6^{\circ}$
Radius of curved road $r = \frac{D}{2} = \frac{1.8 k m}{2} = 900 m$
Speed $v = 30 m / s$
So, banking angle $tan θ = \frac{v^{2}}{r g}$
$∴$ $tan θ = \frac{(30)^{2}}{900 \times 9.8} = 0.1$
$⟹ θ = t a n^{- 1} (0.1) = 6^{o}$

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A curved road of diameter 1.8 km is banked so that no friction is required at a speed of 30 m/s. What is the banking angle?

The correct option is A 6∘Radius of curved road r=D2=1.8 km2=900 mSpeed v=30 m/sSo, banking angle tanθ=v2rg∴ tanθ=(30)2900×9.8=0.1⟹ θ=tan−1(0.1)=6o

A curved road of diameter $1.8 k m$ is banked so that no friction is required at a speed of $30 m / s$ . What is the banking angle?

The correct option is A $6^{\circ}$
Radius of curved road $r = \frac{D}{2} = \frac{1.8 k m}{2} = 900 m$
Speed $v = 30 m / s$
So, banking angle $tan θ = \frac{v^{2}}{r g}$
$∴$ $tan θ = \frac{(30)^{2}}{900 \times 9.8} = 0.1$
$⟹ θ = t a n^{- 1} (0.1) = 6^{o}$