Question

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 the 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 \, \text{m} $
• B. $ 0.030 \, \text{m} $
• C. $ 0.045 \, \text{m} $
• D. $ 0.060 \, \text{m} $

Hint:

To find collimation error: - Find difference in observed and actual elevation differences. - Divide by distance to get error per meter. - Multiply by 100 for error per 100 m.

Answer 1

The Correct Option is D

Step 1: Understand the concept of collimation error.
Collimation error occurs when the line of sight of a leveling instrument is not perfectly horizontal. This causes incorrect readings depending on the distance from the instrument. We use the formula: $$ e = \frac{(R_A - R_B) - (RL_A - RL_B)}{D_B - D_A} $$ Where:
$ R_A = 2.847 \, \text{m} $, $ R_B = 3.462 \, \text{m} $
$ RL_A = 283.665 \, \text{m} $, $ RL_B = 284.295 \, \text{m} $
$ D_A = 25 \, \text{m} $, $ D_B = 50 \, \text{m} $
Step 2: Compute difference in observed and actual levels.
Observed difference: $$ R_A - R_B = 2.847 - 3.462 = -0.615 \, \text{m} $$ Actual difference: $$ RL_A - RL_B = 283.665 - 284.295 = -0.630 \, \text{m} $$ Error: $$ \text{Error} = (-0.615) - (-0.630) = 0.015 \, \text{m} $$
Step 3: Calculate collimation error per 100 m.
Total distance between A and B = $ 50 - 25 = 25 \, \text{m} $ So, error over 25 m is $ 0.015 \, \text{m} $ Then, error per 100 m: $$ \frac{0.015}{25} \times 100 = 0.060 \, \text{m} $$ $$ \boxed{0.060 \, \text{m}} $$