Question

Two points A and B are taken on opposite banks of a river in reciprocal levelling. When the level was set up near A, the staff readings on A and B were 1.225 and 2.375. When the level was set up near B, the respective readings were 0.945 and 2.115. The true differences of level between A and B is

• A. 1.150 m fall from A to B
• B. 1.170 m rise from A to B
• C. 1.160 m fall from A to B
• D. 1.160 m rise from A to B

Hint:

Reciprocal levelling is designed to cancel out errors.
The true difference in level is simply the average of the two apparent differences calculated from each setup.
$\Delta H = \frac{(\text{Reading}_B - \text{Reading}_A)_{\text{Setup 1}} + (\text{Reading}_B - \text{Reading}_A)_{\text{Setup 2}}}{2}$

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the true difference in level between two points, A and B, using the data from a reciprocal levelling operation, which is used to cancel out errors due to instrument collimation and earth's curvature.

Step 2: Key Formula or Approach:
Let $h_A$ and $h_B$ be the staff readings on A and B when the instrument is near A.
Let $h'_A$ and $h'_B$ be the staff readings on A and B when the instrument is near B.
The true difference in level, $\Delta H$, between A and B is the average of the two apparent differences in level.
\[ \Delta H = \frac{(h_B - h_A) + (h'_B - h'_A)}{2} \] A positive result for $\Delta H$ means B is lower than A (a fall). A negative result means B is higher than A (a rise).

Step 3: Detailed Explanation:

Case 1: Instrument near A
- Staff reading on A ($h_A$) = 1.225 m
- Staff reading on B ($h_B$) = 2.375 m
- Apparent difference = $h_B - h_A = 2.375 - 1.225 = 1.150$ m

Case 2: Instrument near B
- Staff reading on A ($h'_A$) = 2.115 m
- Staff reading on B ($h'_B$) = 0.945 m
- Apparent difference = $h'_B - h'_A = 0.945 - 2.115 = -1.170$ m
Wait, the readings are given as "respective readings", so when the instrument is near B, the reading on A is 0.945 and on B is 2.115. Let's re-read carefully: "When the level was set up near B, the respective readings were 0.945 and 2.115". This is ambiguous. Let's assume the first reading is always A and the second is B. - $h'_A = 0.945$ m - $h'_B = 2.115$ m - Apparent difference = $h'_B - h'_A = 2.115 - 0.945 = 1.170$ m
Now, calculate the true difference in level:
\[ \Delta H = \frac{(1.150) + (1.170)}{2} \] \[ \Delta H = \frac{2.320}{2} = 1.160 \text{ m} \] Since the result is positive, it means that point B is lower than point A. This is a

fall from A to B.

Step 4: Final Answer:
The true difference of level is a 1.160 m fall from A to B.