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 .......... $m/\sec$
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(d)$\frac{{{v^2}}}{{rg}} = \frac{h}{l}$
$⇒$ $v = \sqrt {\frac{{rgh}}{l}} $
$ = \sqrt {\frac{{50 \times 1.5 \times 9.8}}{{10}}} = 8.57\,m/s$
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