Banking Of Roads

On a frictionless banked road the horizontal component of the normal force alone supplies the centripetal force needed for circular motion, while its vertical component balances gravity. Dividing these two equilibrium relations gives the design condition $\tan \theta =\frac{v^2}{rg}$. Substituting the given data, $\tan \theta =\frac{(20{)}^2}{80\times 10}=\frac{400}{800}=0.50$. The value 0.25 would correspond to half the required centripetal demand, perhaps from using $v$ instead of $v^2$ incorrectly. The value 0.75 overstates the angle and does not satisfy the force balance for the given speed and radius. The value 1.00 would demand a 45-degree bank, which corresponds to a much higher design speed than 20 m/s on this radius. An important conceptual feature is that the design relation contains no mass term, so the same banking angle works for a light motorcycle and a heavy lorry alike, since both the centripetal requirement and the supporting normal force scale with mass. This is the NCERT frictionless banking relation central to circular dynamics, and including friction would merely widen the band of safe speeds about this ideal value. As a plausibility check, a moderate tangent of 0.5 corresponds to roughly a 27-degree bank, a realistic slope for a highway curve at this speed.