The over-bridge of a river is in the form of a circular arc of radius of curvature 10 m. If g = 10 ms^-2. The highest speed (in m/s) at which a motor cyclist can cross the bridge without leaving the ground is

The motor cyclist can leave the ground only at the highest point on the bridge. At this point, the centripetal force is mv /R. He will not leave the ground if the centripetal force equals the weight mg.
Thus $\frac{{\mathrm{mv}}^2}{\mathrm{R}}=\mathrm{mg}\text{ or }\mathrm{v}=\sqrt{\mathrm{gR}}=\sqrt{10\times 10}=10 {\mathrm{ms}}^{-1}.$