A road is $8 m$ wide. Its radius of curvature is $40 m$ . The outer edge is above the lower edge by distance of $1.2 m$ . The most suited velocity on the road is nearly:

5.7 m s^{- 1}

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8 m s^{- 1}

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36.1 m s^{- 1}

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9.7 m s^{- 1}

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Solution

The correct option is B $8 m s^{- 1}$
We know,
$\frac{v^{2}}{r g} = \frac{h}{l}$
$v = \sqrt{\frac{40 \times 1.2 \times 10}{8}}$
$v = 7.74 \approx 8 m / s$

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A road is 8 m wide. Its radius of curvature is 40 m. The outer edge is above the lower edge by distance of 1.2 m. The most suited velocity on the road is nearly:

The correct option is B 8 ms−1We know,v2rg=hlv=√40×1.2×108v=7.74≈8m/s

A road is $8 m$ wide. Its radius of curvature is $40 m$ . The outer edge is above the lower edge by distance of $1.2 m$ . The most suited velocity on the road is nearly:

The correct option is B $8 m s^{- 1}$
We know,
$\frac{v^{2}}{r g} = \frac{h}{l}$
$v = \sqrt{\frac{40 \times 1.2 \times 10}{8}}$
$v = 7.74 \approx 8 m / s$