A road is 10 m wide. Its radius of curvature is 50 m. The outer edge is above the lower edge by a distance of 1.5 m. This road is most suited for the velocity

2.5 m/sec

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4.5 m/sec

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6.5 m/sec

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8.5 m/sec

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Solution

The correct option is D
8.5 m/sec

$\frac{v^{2}}{r g} = \frac{h}{l} \Rightarrow v = \sqrt{\frac{r g h}{l}} = \sqrt{\frac{50 \times 1.5 \times 9.8}{10}} = 8.57 m / s$

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A road is 10 m wide. Its radius of curvature is 50 m. The outer edge is above the lower edge by a distance of 1.5 m. This road is most suited for the velocity

The correct option is D 8.5 m/sec v2rg=hl⇒v=√rghl=√50×1.5×9.810=8.57 m/s

The correct option is D
8.5 m/sec

$\frac{v^{2}}{r g} = \frac{h}{l} \Rightarrow v = \sqrt{\frac{r g h}{l}} = \sqrt{\frac{50 \times 1.5 \times 9.8}{10}} = 8.57 m / s$