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,
eth is the maximum allowable superelevation in (m)
Vmax 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,
eact is the actual or equilibrium superelevation
CD is Cant deficiency
Calculation:
Given,
D = 3°, CD = 76 mm = 0.076 m
eact = 10 cm = 0.1 m
As track is Broad Gauge G = 1.676 m
\({e_{th}} = {e_{act}} + CD\)
∴ eth = 0.1 + 0.076
= 0.176 m
Radius of Curvature \({\rm{R}} = \frac{{1720}}{{\rm{D}}}\)
\(∴ {\rm{\;Radius\;of\;the\;curve}},{\rm{\;R\;}} = \frac{{1720}}{3}{\rm{m}} = 573.33{\rm{\;m\;}}\)
\({e_{th}} = \frac{{G \times V_{max}^2}}{{127\;R}}\)
\(0.176 = \frac{{1.676 \times V_{max}^2}}{{127 \times 573.33}}\)
Vmax = 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.
