**Concept**

For the Indian railway's track

Theoretical Cant is given by:

\({e_{th}} = \frac{{G \times V_{max}^2}}{{127\;R}}\)

Actual cant or equilibrium cant provided is given by

\({e_{act}} = \frac{{G \times V_{avg}^2}}{{127 \times R\;}}\)

Where,

eth is the maximum allowable super-elevation in (m)

Vmax is the maximum speed on the curve in kmph,

V_{avg} is the average speed on the curve

G is the gauge length of the track in m and

R is the radius of the curve in m where

D is the curvature of the curve in degree.

,\({\rm{R}} = \frac{{1720}}{{\rm{D}}}\),

\({e_{th}} = {e_{act}} + CD\)

Where,

eact is the actual or equilibrium super-elevation

CD is Cant deficiency

__Calculation:__

Given,

Degree of curve = 2°

Type of Gauge is Broad Gauge (BG) G = 1750 mm

V_{max} = 100 kmph, V_{avg} = 80 kmph

Radius of curve \(R = \frac{{1720}}{D}\) = 1720/2 = 875 m

Taking V_{max} theoretical cant is

\({e_{th}} = \frac{{G \times V_{max}^2}}{{127\;R}}\)

\({e_{th}} = \frac{{1.750 \times {{100}^2}}}{{127 \times 875\;}}\) = 157.480 mm

Taking Veq equilibrium cant is

\({e_{act}} = \frac{{G \times V_{avg}^2}}{{127 \times R\;}}\)

\({e_{eq}} = \frac{{1.750 \times {{80}^2}}}{{127 \times 875\;}}\) = 100.787 mm

∴ Cant Deficiency = 157.480 - 100.787 = 56.693 mm

** ∴ Cant Deficiency is 57 mm** as in Question it is asked to roundoff to integer.