Radius of curved road is ' R ', width of road is ' b '. The outer edge of road is raised by ' h ' with respect to inner edge so that a car with velocity ' V ' can pass safe over it, then value of ' h ' is ( g= acceleration due to gravity)

Step-by-step solution

For a car moving with velocity V on a banked road of radius R , the horizontal component of the normal force provides the centripripetal force, while the vertical component balances gravity. For a banking angle : N = mV^2 R N = mg Dividing these expressions yields = V^2 Rg . From the road geometry with width b and outer edge raised by h , = h b . Equating gives: h b = V^2 Rg Solving for the height difference: h = V^2 b Rg , which corresponds to option A .