Question

A bridge is in the form of a semi-circle of radius $40m$. The greatest speed with which a motor cycle can cross the bridge without leaving the ground at the highest point is $(g=10m{s}^{-2})$ (frictional force is negligibly small).

Answer 1

$h=\frac{2T\cos \theta }{rSg}$
$\left[L\right]=\frac{\left[M{T}^{-2}\right]}{\left[L\right]\left[M{L}^{-3}\right]\left[L{T}^{-2}\right]}$
$=\frac{\left[M{T}^{-2}\right]}{M{L}^{-1}{T}^2}$
$=\frac{1}{L^{-1}}=L$
Dimensionally correct $m{v}^2$ for not leave the surface.
$\frac{m{v}^2}{R}\le mg$
$\frac{v^2}{R}=g$
$V^2=Rg=40\times 10=400$
$V=\sqrt{400}=20m/s$.