Civil Engineering - Surveying - Discussion

8. 

A level when set up 25 m from peg A and 50 m from peg B reads 2.847 on a staff held on A and 3.462 on a staff held on B, keeping bubble at its centre while reading. If the reduced levels of A and B are 283.665 m and 284.295 m respectively, the collimation error per 100 m is

[A]. 0.015 m
[B]. 0.030 m
[C]. 0.045 m
[D]. 0.060 m

Answer: Option D

Explanation:

No answer description available for this question.

Idowu S A said: (Oct 6, 2015)  
Actual height difference between A and B = 284.295 - 283.665 = 0.63.

Height difference between A and B based on staff readings = 3.462 - 2.847 = 0.615.

Distance between A and B = 50-25 = 25 m.

Therefore collimation error per 100 m.

= (0.63 - 0.615)*(100/25).

= 0.060 m.

Deependra Jaiswal said: (Mar 4, 2018)  
Thank you @Idowu.

Naveen Nayak said: (Mar 28, 2018)  
The Least count of staff 5mm.

Civillian said: (Apr 9, 2018)  
The distance between A and B is 75 possible then why taking only 25 by subtracting? Please explain in detail.

Ghalib said: (May 1, 2018)  
I can prove that .060 isn't it's right answer?

Try to use it as a check and see after applying this correction you wouldn't get (.63) as δh rather use 4.980 as error per 100 meters you will get the correct answer.

Awinash said: (Jun 11, 2018)  
No, The Right answer is 5. Agree @Naveen.

Amal Singha said: (Nov 12, 2019)  
hi = 283.665+2.847 = 286.512m.
Rl of point B = 286.512-3.462 = 283.048m.
Collimaton error = 284.295-283.048 = 1.247m in 25m.
Collimation error for 100m = 1.247*4 = 4.988m.

Iceryi said: (Jun 13, 2020)  
Yes, the distance between a and b is 75.

Dhirendra said: (Feb 24, 2021)  
Yes, 0.6 is correct. I agree with the given option.

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