In a B.G. railway track, the specified ruling gradent is 1 in 250. The horizontal curve of 3° on a gradient of 1 in 250 will have the permissible gradient of
A. 1 in 257
B. 1 in 357
C. 1 in 457
D. 1 in 512
Answer: Option B
Solution (By Examveda Team)
When a train moves on a curve, additional resistance acts on it because the wheels experience extra friction and lateral force while turning.If the track is also on a gradient, the engine must overcome both:
1. Gradient resistance
2. Curve resistance
To reduce the extra resistance caused by the curve, railway engineers slightly flatten the gradient on curves. This process is called grade compensation.
The purpose of grade compensation is to ensure that the total resistance on a curved gradient does not exceed the resistance on a straight ruling gradient.
Given:
Ruling Gradient = 1 in 250
Degree of Curve = 3°
First, convert the ruling gradient into percentage form:
Gradient % = (1 / 250) × 100
= 0.4%
For a B.G. (Broad Gauge) railway track, the standard grade compensation is:
Grade Compensation = 0.04% per degree of curve
Therefore:
Grade Compensation = 0.04 × 3
= 0.12%
Now calculate the compensated or permissible gradient:
Permissible Gradient = Ruling Gradient − Grade Compensation
= 0.4% − 0.12%
= 0.28%
Now convert 0.28% into the form 1 in N:
N = 100 / 0.28
N ≈ 357
Thus, the permissible gradient becomes:
1 in 357
Therefore, the correct answer is Option B: 1 in 357.

Please slove the question
Ruling Gradient = 1/250 = 0.4℅
Grade compensation for BG = 0.04% per curve =0.04*3 = 0.12%
Permissible gradient = 0.4 - 0.12 = 0.28% = 1/357
For BG grade compensation =min of (70/r or .04 per) therefore GC=.12 ,per gradient =.4-.12 which is Equal to 1 in 357 ,also 1in 250 = .4.
explain how ?