# 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
0.02 cm
0.15 cm
02 cm
0.1 cm
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

Shubham P said:   4 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.
(3)

Karthi said:   5 years ago
@Hemanth.

1cm = 2cm is given Not 2m.

Karthi said:   5 years ago
60/4000 = 0.015.
Not .15.
(1)

Hemanth said:   6 years ago
It is reduction scale,
The scale should be 1cm = 2 m,
Then (1/200)*30 =0.15 cm.
(Cross multiplication)
(1)

Civillian said:   6 years ago
Here, 1.5/100=0.015.

Salamuddin said:   7 years ago
1cm=2cm given.

So 30*2= 60cm,
60/4000=0.15.
(1)

G1 rathod said:   7 years ago
(1)

Arun karthik said:   7 years ago
1cm=2cm given.

So 30*2= 60cm,
60/40=1.5,
1.5/100=0.15.
(1)

Swathi Jagatheesan said:   7 years ago
Is there any formula for this? Please explain.

Gagan said:   8 years ago
30 * 2 = 60/4000 = 0.15.