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Question

A level, when set up $20$ m from peg A and $70$ m from peg B, reads $0.750$ m on a staff held on A and $2.065$ m on a staff held on B, keeping the bubble at its centre while reading. If the reduced levels of A and B are $100.500$ m and $101.800$ m respectively, what is the collimation error per $100.0$ m?

The correct answer is

$0.030$ m

Calculating Collimation Error

This solution explains how to determine the collimation error per 100.0 m in a leveling survey based on the provided readings and reduced levels.

Identify Given Data

  • Reduced Level of Peg A ($RL_A$): $100.500$ m
  • Reduced Level of Peg B ($RL_B$): $101.800$ m
  • Distance from level to Peg A ($d_A$): $20$ m
  • Distance from level to Peg B ($d_B$): $70$ m
  • Staff reading on Peg A (Backsight, $BS_{obs}$): $0.750$ m
  • Staff reading on Peg B (Foresight, $FS_{obs}$): $2.065$ m

Determine Collimation Error Rate

The collimation error ($e$) per meter is calculated using the differences in reduced levels, staff readings, and distances. A common formula derived from relating the true elevation change to the observed readings is:

$ e = \frac{(RL_B - RL_A) - (FS_{obs} - BS_{obs})}{d_A - d_B} $

First, calculate the components:

  • Difference in Reduced Levels: $RL_B - RL_A = 101.800 \text{ m} - 100.500 \text{ m} = 1.300 \text{ m}$
  • Difference in Staff Readings (FS - BS): $FS_{obs} - BS_{obs} = 2.065 \text{ m} - 0.750 \text{ m} = 1.315 \text{ m}$
  • Difference in Distances: $d_A - d_B = 20 \text{ m} - 70 \text{ m} = -50 \text{ m}$

Now, substitute these values into the formula:

$ e = \frac{1.300 \text{ m} - 1.315 \text{ m}}{-50 \text{ m}} = \frac{-0.015 \text{ m}}{-50 \text{ m}} = 0.0003 \text{ m/m} $

This value, $e = 0.0003$ m/m, represents the collimation error for each meter the instrument is set up from the staff.

Calculate Error per 100.0 m

The question asks for the collimation error per $100.0$ m. To find this, multiply the error rate per meter by $100.0$ m:

Error per $100.0$ m $= e \times 100.0 \text{ m}$

Error per $100.0$ m $= 0.0003 \text{ m/m} \times 100.0 \text{ m} = 0.030 \text{ m}$

Conclusion

The collimation error is $0.030$ m per $100.0$ m.

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Important Questions from Levelling

  1. The expression for sensitivity of the bubble tube (α) can be taken as, ______

    Where n = No. of divisions, s = Net staff reading, D = Distance, R = Radius of curvature, l = Length of one division

  2. A vertical line which is perpendicular to the level line is called:

  3. In levelling between two points A and B on the opposite sides of a river, the level was first set up near A and the staff readings on A and B were 2.645 m and 2.30 m respectively. The level was then moved near B and set up; the respective staff readings then were 1.085 m and 1.665 m on A and B respectively. What is the true difference of level between A and B?

  4. In levelling back, sight is also called as ______.

  5. A 'level line' is a-

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