Correct Answer - Option 1 : 87.6 km/h

__Concept: __

For the Indian railways track

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

Where,

e_{th} is the maximum allowable superelevation in (m)

V_{max} is the maximum speed on the curve in kmph,

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,

e_{act }is the actual or equilibrium superelevation

CD is Cant deficiency

__Calculation: __

Given,

D = 3°, CD = 76 mm = 0.076 m

e_{act }= 10 cm = 0.1 m

As track is Broad Gauge G = 1.676 m

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

∴ e_{th} = 0.1 + 0.076

= 0.176 m

Radius of Curvature \({\rm{R}} = \frac{{1720}}{{\rm{D}}}\)

\(∴ {\rm{\;Radius\;of\;the\;curve}},{\rm{\;R\;}} = \frac{{1720}}{3}{\rm{m}} = 573.33{\rm{\;m\;}}\)

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

\(0.176 = \frac{{1.676 \times V_{max}^2}}{{127 \times 573.33}}\)

V_{max} = 87.6 kmph

∴ The maximum speed of the train is approximately **87.6 kmph.**

__Important Point: __

In the formula of maximum allowable super elevation \({e_{th}} = \frac{{G \times V_{max}^2}}{{127\;R}}\) all the variables should be replaced in the particular unit mentioned above, otherwise the constant term ‘127’ is not valid.