Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 7 (Q.No. 12)
12.
While setting a plane table at a station it was found that the error in centering was 30 cm away from the ray of length 40 m drawn from the station. If the scale of the plan is 1 cm = 2 cm, the displacement of the end of the ray in plan from the true position will be
Discussion:
12 comments Page 2 of 2.
Shubham P said:
5 years ago
I think the answer should be 0.15 meter not 0.15 centimetre for given data.
30/2 = 15 centimetres.
If the answer is right then the scale of the plan is wrong, it should be 1cm = 2meter.
So answer will be (30/200) = 0.15 centimetres.
30/2 = 15 centimetres.
If the answer is right then the scale of the plan is wrong, it should be 1cm = 2meter.
So answer will be (30/200) = 0.15 centimetres.
(4)
Inaya said:
10 months ago
The displacement of the end of the ray in the plan from the true position will be representative fraction (RF) ×error.
D = RF×error.
Here scale.
1cm = 2m.
RF =1/200.
D = (1/200) ×30cm.
= 0. 15cm.
D = RF×error.
Here scale.
1cm = 2m.
RF =1/200.
D = (1/200) ×30cm.
= 0. 15cm.
(2)
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