Neglecting side friction, super-elevation for V=60 km/h and radius R=200 m is

Why is this the answer?

IRC superelevation formula (neglecting lateral friction): e = V^2 / (225*R), where V = design speed (km/h) and R = radius of horizontal curve (m). Derived from IRC equilibrium equation: e + f = V^2/(127R); setting f = 0 gives this formula. Max superelevation: 0.07 (plain/rolling terrain), 0.10 (hill roads) per IRC 38:2019. e=V²/(225R)=60²/(225×200)=0.080.

Related Transportation Engineering MCQs

More solved Transportation Engineering questions for AE/JE Civil Engineering exams.

• 'Road safety audits' (RSA) in India are conducted as per:
• The 'gauge' of a railway track in India (broad gauge) is:
• The 'saturation flow rate' at a signalised intersection is the flow that would cross...
• The 'PMGSY' (Pradhan Mantri Gram Sadak Yojana) standard road uses which pavement type?
• 'Pavement Management System' (PMS) in highways involves:
• 'Channelisation' at a traffic intersection improves safety by: