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  4. A road is 8 m wide. Its radius of curvature is 40 m. The outer edge is above the lower edge by a distance of 1.2 m. This road is most suited for a velocity of

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Q. A road is 8 m wide. Its radius of curvature is 40 m. The outer edge is above the lower edge by a distance of 1.2 m. This road is most suited for a velocity of

MGIMS WardhaMGIMS Wardha 2007

Solution:

$ \tan \theta =\frac{{{v}^{2}}}{rg} $ $ \sin \theta =\frac{{{v}^{2}}}{rg} $ $ (\sin \theta \approx \tan \theta ,for\,small\,\theta ) $ $ \therefore $ $ \frac{1.2}{8}=\frac{{{v}^{2}}}{40\times 9.8} $
$ {{v}^{2}}=1.2\times 49 $ $ v=\sqrt{1.2\times 49} $ $ =\sqrt{58.8}=7.7\,m/s $

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