Question:
Locations \(P\) and \(Q\) are separated by a wide valley. The difference in levels of locations \(P\) and \(Q\) measured by a levelling instrument stationed near \(P\) is 3.0 m. The same instrument stationed near \(Q\) measures the difference in levels of locations \(P\) and \(Q\) as \(-1.0\) m. Assume that the atmospheric refraction is the same during the measurements. The true difference in levels (in m) of locations \(P\) and \(Q\) is
Locations \(P\) and \(Q\) are separated by a wide valley. The difference in levels of locations \(P\) and \(Q\) measured by a levelling instrument stationed near \(P\) is 3.0 m. The same instrument stationed near \(Q\) measures the difference in levels of locations \(P\) and \(Q\) as \(-1.0\) m. Assume that the atmospheric refraction is the same during the measurements. The true difference in levels (in m) of locations \(P\) and \(Q\) is
Show Hint
Use the reciprocal levelling rule: true difference = mean of the two apparent (reciprocal) observations.
Updated On: Jul 17, 2026
- 1.0
- 1.5
- 2.0
- 4.0
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Understand why reciprocal levelling is used.
A wide valley means the instrument cannot be set up midway between \(P\) and \(Q\), so the two sight lengths (to \(P\) and to \(Q\)) end up very different. A long sight picks up more error from curvature of the earth and atmospheric refraction than a short sight does. This error shows up in the observed difference of level and needs to be removed to get the true value.
Step 2: Set up the two observed (apparent) differences.
When the instrument stands near \(P\), the sight to \(P\) is short (almost no error) and the sight to \(Q\) is long (carries an error, call it \(e\)). The apparent difference in level recorded this way is \(h_1 = 3.0\) m.
When the instrument stands near \(Q\), the sight to \(Q\) is now short (almost no error) and the sight to \(P\) is long, so the same error \(e\) enters the reading but with the opposite effect on the computed difference. The apparent difference recorded this way is \(h_2 = -1.0\) m.
Since both readings are taken and reported in the same sense, we can write
\[ h_1 = h_{true} + e \]
\[ h_2 = h_{true} - e \]
Step 3: Add the two equations to cancel the error.
\[ h_1 + h_2 = 2h_{true} \]
\[ h_{true} = \frac{h_1 + h_2}{2} \]
This is the standard reciprocal levelling result: the true difference in level is the mean of the two apparent (reciprocal) observations, because the curvature and refraction error is the same size \(e\) each time and cancels out on averaging.
Step 4: Substitute the given values.
\[ h_{true} = \frac{3.0 + (-1.0)}{2} = \frac{2.0}{2} = 1.0 \text{ m} \]
Step 5: Check the other options.
Option (C) 2.0 m comes from wrongly dropping the negative sign and computing \(\frac{3.0+1.0}{2}\).
Option (D) 4.0 m comes from taking the difference of the two readings, \(3.0-(-1.0)=4.0\), instead of the mean.
Option (B) 1.5 m does not follow from either reading using the correct averaging rule, so it is incorrect.
Final Answer:
The true difference in levels of \(P\) and \(Q\) is \(1.0\) m, option (A).
\[ \boxed{h_{true} = 1.0 \text{ m}} \]
A wide valley means the instrument cannot be set up midway between \(P\) and \(Q\), so the two sight lengths (to \(P\) and to \(Q\)) end up very different. A long sight picks up more error from curvature of the earth and atmospheric refraction than a short sight does. This error shows up in the observed difference of level and needs to be removed to get the true value.
Step 2: Set up the two observed (apparent) differences.
When the instrument stands near \(P\), the sight to \(P\) is short (almost no error) and the sight to \(Q\) is long (carries an error, call it \(e\)). The apparent difference in level recorded this way is \(h_1 = 3.0\) m.
When the instrument stands near \(Q\), the sight to \(Q\) is now short (almost no error) and the sight to \(P\) is long, so the same error \(e\) enters the reading but with the opposite effect on the computed difference. The apparent difference recorded this way is \(h_2 = -1.0\) m.
Since both readings are taken and reported in the same sense, we can write
\[ h_1 = h_{true} + e \]
\[ h_2 = h_{true} - e \]
Step 3: Add the two equations to cancel the error.
\[ h_1 + h_2 = 2h_{true} \]
\[ h_{true} = \frac{h_1 + h_2}{2} \]
This is the standard reciprocal levelling result: the true difference in level is the mean of the two apparent (reciprocal) observations, because the curvature and refraction error is the same size \(e\) each time and cancels out on averaging.
Step 4: Substitute the given values.
\[ h_{true} = \frac{3.0 + (-1.0)}{2} = \frac{2.0}{2} = 1.0 \text{ m} \]
Step 5: Check the other options.
Option (C) 2.0 m comes from wrongly dropping the negative sign and computing \(\frac{3.0+1.0}{2}\).
Option (D) 4.0 m comes from taking the difference of the two readings, \(3.0-(-1.0)=4.0\), instead of the mean.
Option (B) 1.5 m does not follow from either reading using the correct averaging rule, so it is incorrect.
Final Answer:
The true difference in levels of \(P\) and \(Q\) is \(1.0\) m, option (A).
\[ \boxed{h_{true} = 1.0 \text{ m}} \]
Was this answer helpful?
0
0
Top GATE CE Surveying Questions
If the size of the ground area is $6 \, \text{km} \times 3 \, \text{km}$ and the corresponding photo size in the aerial photograph is $30 \, \text{cm} \times 15 \, \text{cm}$, then the scale of the photograph is $1$
- A 30 m long tape is standardized at \(25^{\circ}\text{C}\). It was used to measure the length of a line which came out to be 200 m. The temperature during the measurement was \(35^{\circ}\text{C}\). The coefficient of expansion of the tape was \(11 \times 10^{-6}\) per \(^{\circ}\text{C}\). The correction in the measured length (in mm) due to change in temperature is (in integer).
- The height of the plane of collimation of a levelling instrument is 100.000 m from a datum. The levelling instrument measures a Back Sight of 1.500 m on a vertically held staff at a ground point \(P\). The reduced level (in m) of the ground point \(P\) with respect to the datum is (rounded off to three decimal places).
- The following consecutive staff readings (in m) were taken with a dumpy level and a levelling staff at a common interval of 20 m: 0.385; 1.030; 1.925; 2.825; 3.730; 4.850; 1.045; 2.005; 3.330; 4.580. The dumpy level was shifted after taking the sixth reading. The gradient (in %) of the ground between the first and the last location of the readings is ______ (in integer).
- Which one of the following statements is TRUE with respect to levelling survey?
View More Questions
Top GATE CE Levels and Levelling Questions
- The height of the plane of collimation of a levelling instrument is 100.000 m from a datum. The levelling instrument measures a Back Sight of 1.500 m on a vertically held staff at a ground point \(P\). The reduced level (in m) of the ground point \(P\) with respect to the datum is (rounded off to three decimal places).
- The following consecutive staff readings (in m) were taken with a dumpy level and a levelling staff at a common interval of 20 m: 0.385; 1.030; 1.925; 2.825; 3.730; 4.850; 1.045; 2.005; 3.330; 4.580. The dumpy level was shifted after taking the sixth reading. The gradient (in %) of the ground between the first and the last location of the readings is ______ (in integer).
- Which one of the following statements is TRUE with respect to levelling survey?
- For the traverse given below, the benchmark is at ground point L. The Reduced Level (RL) of L is 150.000 m. The direction of the traverse is from L to N. The readings on levelling staff using the same levelling instrument kept at Stations S1 and S2 during the traversing are shown in the figure.
Column 1 Column 2 (P) 3.300 m (I) Sum of Back Sights (Q) 2.500 m (II) Sum of Fore Sights (R) 150.300 m (III) RL of Point N (S) 3.000 m (T) 150.500 m
Matching the information given under Column 1 and Column 2, which one of the following options is CORRECT?
Top GATE CE Questions
- If \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}....\frac{1}{29} =\frac{N}{29!}\) Find the remainder when N is divided by 19.
- What will be the relative density?
M = 0.5 percentage of air voids
γmin = 1.3 gm/cc
γmax = 1.8 gm/cc - Find RL of R?
RL of P [BM = 200.000Point Staff Readings
Back sideStaff Readings
Fore sideRemarks P -2.050 - 200.000 Q 1.050 0.95 Change Point R - -1.655 - - What will the angle of the failure plane from horizontal for sand be?
σ3=300
σ1=300+100=400 - Which of the following having negative skamptam Pore water coefficient?
View More Questions