Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 5 (Q.No. 49)
49.
A sewer is laid from a manhole A to a manhole B, 250 m away along a gradient of 1 in 125. If the reduced level of the invert at A is 205.75 m and the height of the boning rod is 3 m, the reduced level of the sight rail at B, is
Discussion:
16 comments Page 1 of 2.
Praveen said:
8 years ago
Can you explain it?
Vinay said:
8 years ago
Answer C. 206.75.
205.75 - 1*250/125 = 206.75 m.
205.75 - 1*250/125 = 206.75 m.
Atul Patel said:
8 years ago
205.75 - 1 * 205/125 = 203.75,
203.75 + 3 = 206.75.
203.75 + 3 = 206.75.
Arun said:
8 years ago
Why added 3 to 203.75?
Karthik said:
8 years ago
@Arun
3 is the height of boring rod.
3 is the height of boring rod.
Piku said:
8 years ago
205.75-(1 * 250/150)+3 = 206.75.
Gaurav said:
8 years ago
Gradient is downward.
Chadha said:
8 years ago
205.75 - (250*1/125-3)= 206.75.
Rajib kumar said:
7 years ago
Please explain the solution.
Mahesh said:
6 years ago
RL of invert at B.
= 205.75 - (250 ÷ 125) = 203.75.
RL of sight at B = 203.75+3 = 206.75.
= 205.75 - (250 ÷ 125) = 203.75.
RL of sight at B = 203.75+3 = 206.75.
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