Area Computation And Tacheometric Surveying(Surveying ) Questions and answers for exam Preparation

Question 1. The horizontal angle subtended at the theodolite station by a subtense bar with vanes 3 m apart is 0º 10 ′40′′. Calculate the horizontal distance between the theodolite and the subtense bar?

960.00 m
966.87 m
966.78 m
906.87 m

Explanation:-

Answer: Option B. -> 966.87 m
Answer: (b).966.87 m

Question 2. The vertical angles to vanes fixed at 1 m and 3 m above the foot of the staff held vertically at a station P were – 1º 45′ and + 2º 30′, respectively. Find the horizontal distance and the reduced RL of P if the RL of the instrument axis is 110.00 m?

26.95 m, 100.177
20.95 m, 108.177
26.95 m, 108.177
26.95 m, 108.000

Explanation:-

Answer: Option C. -> 26.95 m, 108.177
Answer: (c).26.95 m, 108.177

Question 3. Following observations were taken with a tacheometer fitted with an anallactic lens having value of constant as 100.
Calculate the horizontal distance between P and Q.

149.96 m
140.26 m
141.92 m
143.56 m

Explanation:-

Answer: Option A. -> 149.96 m
Answer: (a).149.96 m

Question 4. Distance and elevation formulae for fixed hair method assuming the line of sight as horizontal and considering an external focusing type telescope is D = Ks + C. where C is _______

f/i
i/f
f + c
f – c

Explanation:-

Answer: Option C. -> f + c
Answer: (c).f + c

Question 5. The following notes refer to a traverse run by a tacheometer fitted with an anallactic lens, with constant 100 and staff held vertical. Line, Bearing, Vertical Angle, Staff Intercept -PQ, 30º 24′, + 5º 06′,1.875; QR, 300º 48′, + 3º 48′,1.445; RS, 226º 12′, − 2º 36′, 1.725 respectively. Find the length and bearing of SP.

191.930 m, 126º 47 ′47′′
190.930 m, 125º 47 ′47′′
193.930 m, 124º 47 ′47′′
192.930 m, 120º 47 ′47′′

Explanation:-

Answer: Option A. -> 191.930 m, 126º 47 ′47′′
Answer: (a).191.930 m, 126º 47 ′47′′

Question 6. Among the area calculation methods, which is more accurate?

Area by co-ordinates
Area by Simpson’s one-third rule
Area by double mean distances
Area by offsets

Explanation:-

Answer: Option B. -> Area by Simpson’s one-third rule
Answer: (b).Area by Simpson’s one-third rule

Question 7. The calculation of area by ordinate rule and Simpson’s rule will come under which category?

Area by double mean distances
Area by co-ordinates
Area by triangles
Area by offsets

Explanation:-

Answer: Option D. -> Area by offsets
Answer: (d).Area by offsets

Question 8. Find the value of number of divisions if the area is 543.89 sq. m and the summation of the co-ordinates is given as 223.98 m.

2.42 m
2.24 m
4.22 m
2.56 m

Explanation:-

Answer: Option A. -> 2.42 m
Answer: (a).2.42 m

Question 9. Which of the following indicates the formula for area by average co-ordinate method?

Δ = (L * ∑O)/(n+1)
Δ = (L * ∑O)/(n-1)
Δ = (L + ∑O)/(n+1)
Δ = (L – ∑O)/(n+1)

Explanation:-

Answer: Option A. -> Δ = (L * ∑O)/(n+1)
Answer: (a).Δ = (L * ∑O)/(n+1)

Question 10. Calculate the area by average co-ordinate rule, by using the offsets provided taken at 10m interval.
4.16, 6.34, 7.89, 6.54, 5.67, 7.76, 8.52, 5.87, 6.21

245.08m
542.08 m
524.08 m
528.04 m

Explanation:-

Answer: Option B. -> 542.08 m
Answer: (b).542.08 m