Civil Engineering - Surveying - Discussion

The answer should be A. 1/2m.

With ref of Punmia : the second term in slope correction shall be neglected for slopes flatter than about 1 in 25 (i.e slope should be lesser than 0.04).

The slope is Vertical distance/Horizontal distance.
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Here Vertical distance is h and the horizondal distance is √((L)^2 - (h)^2).

L = 20 given in the problem. We need to find, for which value of h the slope will be less than 0.04.

Sub h = 0.5, we get slope = 0.025;
Sub h = 1, we get slope = 0.050;
Sub h = 2, we get slope = 0.100;
Sub h = 3, we get slope = 0.150;

Therefore Option A (h=0.5m) is the correct answer.

If the value of h is maximum then h^4 value is greater than the other value. So how is possible to this? Value goes nearest to zero right?

ASSUMING L=100m then ,Cs=(400/(2*100))+(8000/(8*100^3)= 2.001m.
Which less than 3m.

How will get 3.3?

The question is wrong it should be h2/2l -h4/8l3 instead of h4/8l3.

How will get 3.3?

The formula (h^2/2L)+(h^4/8L^3) is for slope that ranges from 20%-30%.

If (h^4/8L^3) is to be neglected, h^2/2L is left and if we are going to equate it with 20% the answer would be 2.8~3 and if h^2/2L is to be equated to 30% the answer would be 3.4~3 therefore the answer is 3.

Correct @Imran.

As we decrease the value of h less than 3, the second term impact becomes negligence so 3 is the answer.

H should be less than 3.

Even if it is substituted in the formula, the second term will get negligible (very small number-even neglecting the term will not affect the answer).

H^4/8L^3 = 0 or near to 0 which is neglected to measure, when I put h<3 (1/2,1,2), results are below 1mm, which is hard to measurable. So, 3 is correct.