Questions and answers for exam Preparation

MCQs

Total Questions : 45 | Page 1 of 5 pages

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

Discuss Question

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

Discuss Question

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

Discuss Question

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

Discuss Question

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′′

Discuss Question

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

Discuss Question

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

Discuss Question

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

Discuss Question

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)

Discuss Question

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