# Civil Engineering - UPSC Civil Service Exam Questions - Discussion

16.

The distance between two bench marks is 1000 m. If during levelling, the total error due to collimation, curvature and refraction is found to be + 0.120 m, then the magnitude of the collimation error is

 [A]. 0.00527 m [B]. 0.0527 m [C]. 0.527 m [D]. 0.673 m

Explanation:

No answer description available for this question.

 Aryan said: (Feb 11, 2014) Error due to curvature and refraction = 0.0678*(d)^2. Here, d=1 km. Error= 0.0678. Collimation error= 0.120 - 0.0678.

 Kanu said: (Mar 7, 2017) Thanks @Aryan.

 Anil said: (Mar 11, 2017) @Aryan. How can you subtract? One is in km and other is in mt.

 Taofeek said: (Apr 17, 2017) @Anil. The combined error due to curvature and refraction (ecomb) is thus given by ecomb = 0.0675 D2 m where D is the distance in km.

 Uttarayan said: (Jan 16, 2018) @Anil. The value of error due to curvature and refraction comes in m although d is taken in km.

 Mahendra Shukla said: (Dec 23, 2018) Given; Total error =collimation error + curvature error + refraction error. .120=colli.error-.07857*1+.01122*1, Colli.error = .120-.06735=.0527.

 Akash said: (Sep 26, 2019) Collimation Error = 0.12-0.0673 = 0.0527.

 Amit Singh said: (Jan 31, 2021) Curvature correction(m) = -0.06749d^2 (d is in Km), Refraction correction (m) = 0.01121d^2 (d is in Km), Combined Correction (m) = -0.06728d^2 (d is in km).