# Civil Engineering - Railways - Discussion

Discussion Forum : Railways - Section 1 (Q.No. 2)

2.

If absolute levels of rails at the consecutive axles

*A*,*B*, and*C*separated by 1.8 metres are 100.505 m, 100.530 m, and 100.525 m respectively, the unevenness of rails, isDiscussion:

31 comments Page 1 of 4.
Hardinge said:
10 months ago

@All.

U= (a+b)/2 - d.

Where a and b are the absolute levels of the 2 end axles and d is the absolute level of the middle axle.

Now the problem here is that the question here has given incorrect data. The book that this question has been copied from had actually given the middle value as 100.580 which has been misprinted as 100.530 here.

If you put the middle value as 100.580 in the above formula you will get the answer exactly.

Take a=100.505.

b=100.525.

d=100.580.

Now solve you will get 0.065 m as the answer exactly.

U= (a+b)/2 - d.

Where a and b are the absolute levels of the 2 end axles and d is the absolute level of the middle axle.

Now the problem here is that the question here has given incorrect data. The book that this question has been copied from had actually given the middle value as 100.580 which has been misprinted as 100.530 here.

If you put the middle value as 100.580 in the above formula you will get the answer exactly.

Take a=100.505.

b=100.525.

d=100.580.

Now solve you will get 0.065 m as the answer exactly.

(5)

Tariq Marwat said:
1 year ago

Please explain me the correct answer.

Naseer said:
2 years ago

Anyone, Please solve it in detail.

Aamir Ashraf said:
2 years ago

Unevenness - The longitudinal unevenness of rails is measured as a deviation in the vertical plane (depression or rise) of the middle axle of the measuring bogie with reference to the average position of the two outer axles.

The outer axles are spaced 3.6 metres apart and unevenness is recorded to a scale of 1:1 on a base of 3.6 M for left and right rail separately.

The outer axles are spaced 3.6 metres apart and unevenness is recorded to a scale of 1:1 on a base of 3.6 M for left and right rail separately.

Vivek shukla said:
3 years ago

(100.505 + 100.525 + 100.530)/3 = 100.520.

Unevenness = { (100.505 - 100.520)+ (100.530 - 100.520) + (100.525 - 100.520) } Ã· 1.8 = 0.065.

Unevenness = { (100.505 - 100.520)+ (100.530 - 100.520) + (100.525 - 100.520) } Ã· 1.8 = 0.065.

(3)

Asif wazir said:
3 years ago

Tan-1 (.025/1.8)-tan-1(.005/1.8) = .063m.

Palak kesharwani said:
3 years ago

Please give detail description.

(1)

Manju said:
4 years ago

Please anyone explain in detail.

Sattu said:
4 years ago

Take the mean of all values.

(100.505+ 100.525 + 100.530)/3 - - 100.520.

Unevenness -- 100.520/1.8.

(100.505+ 100.525 + 100.530)/3 - - 100.520.

Unevenness -- 100.520/1.8.

(1)

Amu bokato said:
5 years ago

I am not getting this. Please, anyone explain it in clearly.

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