This set of Surveying Multiple Choice Questions & Answers (MCQs) focuses on “Area by Planimeter”.

1. Planimeter is an instrument which is used for__________

a) Locating co-ordinates

b) Transferring point from paper to ground

c) Measuring area of plan

d) Sighting parallel and perpendicular points to station

View Answer

Explanation: By the use of planimeter, the area of the land can be measured which is used for processing. Plumb bob is used for transferring point from paper to ground and alidade for sighting parallel and perpendicular points to station.

2. The formula for finding area by the use of planimeter is______________

a) Δ = M(F-I±10N+C)

b) Δ = M(F+I±10N+C)

c) Δ = M(F-I±10N-C)

d) Δ = M(F-I±10N±C)

View Answer

Explanation: The area of the obtained figure can be calculated by, Δ = M(F-I±10N+C)

Where, F = Final reading, I = Initial reading, N = number of times the zero mark of the dial passes the fixed index mark, M = A multiplying constant, C= Constant of the instrument which when multiplied by M.

3. Which of the following methods will give the best output for area?

a) Area by double mean distances

b) Area by triangles

c) Area by co-ordinates

d) Area by planimeter

View Answer

Explanation: The area calculated by forming triangles will be able to give the best output because it involves formation of frame work.

4. Multiplier constant(M) is also known as___________

a) Planimeter constant

b) Tacheometric constant

c) Meridian constant

d) Simpson’s constant

View Answer

Explanation: From the formula of area by planimeter, Δ = M (F-I±10N+C) the variable M represents multiplier constant which is required for further calculations.

5. Multiplier constant is equal to__________

a) A+nꞌ

b) A*nꞌ

c) A/nꞌ

d) nꞌ/A

View Answer

Explanation: The value of multiplier constant is given as M= A/nꞌ in which,

A = known area, nꞌ = change in wheel readings.

6. Calculate the area if I = 8.257, M = 150 sq.cm, F = 4.143, C = 31.155.

a) 6255.15

b) 2565.15

c) 2655.15

d) 2556.15

View Answer

Explanation: The area can given by, A=M (F- I ± 10N + C)

On substituting, I = 8.257, M = 150 sq.cm, F = 4.143, C = 31.155 and N = -1

A = 150 (4.143- 8.257- 10 + 31.155) = 2556.15sq.cm.

7. Which of the following mathematical operations can be used for area computation?

a) Euler’s equation

b) Simpson’s one-third rule

c) Quadratic equation

d) Simultaneous differential equation

View Answer

Explanation: Simpson’s one-third rule assumes that the short length of the boundary between the ordinates is parabolic arc and this method is more useful when the boundary line departs considerably from the straight line.

8. Simpson’s rule is capable of producing more accurate results.

a) True

b) False

View Answer

Explanation: The results obtained by the use of Simpson’s rule are more accurate in all cases. The result obtained is smaller than those obtained by using the trapezoidal rule.

9. Find the value of multiplier constant if the length of the arm is given as 45.78 m and diameter as 2.54 m.

a) 643.86 sq. m

b) 436.86 sq. m

c) 346.68 sq. m

d) 364.86 sq. m

View Answer

Explanation: The value of multiplier constant can be given as M = L * circumference

Circumference = π * D = π *2.54 = 7.97 m. On substitution, we get

M = 45.78 * 7.97 = 364.86 sq. m.

10. If length of the arm = 23.31 m, distance between wheel and pivot = 2 m, D = 3m. Find the value of constant C.

a) 4.63 sq. m

b) 6.64 sq. m

c) 6.46 sq. m

d) 4.95 sq. m

View Answer

Explanation: The value of the constant C can be given as, \(C = \frac{π (L^2-2aL+R^2)}{M}\). The value of M can be given as M = L* π*D = 219.691 sq. m. On substitution, we get

C = π (23.31

^{2}-2*2*23.31+1.5

^{2}) / 219.691 = 6.46 sq. m.

11. If the area of the traverse is 645.32 sq. m and the change in the wheel readings can be given as 10, find the value of multiplier constant.

a) 64.532 sq. m

b) 6453.2 sq. m

c) 6.4532 sq. m

d) 0.65432 sq. m

View Answer

Explanation: If the values of area and the change in wheel readings are given then the multiplier constant can be given as M = A/nꞌ, where A is the area and nꞌ is the change in wheel readings. On substitution, we get

M = 645.32 / 10 = 64.532 sq. m.

12. If the area of traverse is drawn to a scale 1ꞌꞌ = 23 ft, find the change in area if the original area is 497.76 sq. in.

a) 6.04 sq. m

b) 6.04 m

c) 6.04 sq. in

d) 6.04 acres

View Answer

Explanation: Here, the scale is given as 1ꞌꞌ = 23 ft. So, 1 sq. in = 23*23 sq. ft

And the area of the field can be given as (23*23*497.76) / 43560 = 6.044 acres.

13. Find the value of I if the area of the field is given as 234.315 sq. m, M = 22.15 sq. m, F = 3.256, N = 1, C = 26.43.

a) 21.907 sq. m

b) 29.107 sq. m

c) 29.701 sq. m

d) 23.071 sq. m

View Answer

Explanation: The area of the field can be given as A=M (F- I ± 10N + C). On substitution of the given values we get,

234.315 = 22.15 (3.256-I+10+26.43)

I = 29.107 sq. m.

**Sanfoundry Global Education & Learning Series – Surveying.**

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