Category: Surveying Objective Questions

Surveying Multiple Choice Questions on “Plane Table Surveying Accessories”.

1. Which of the following does not belong to the instruments of plane table?
a) Spirit level
b) Plumb bob
c) Compass
d) Theodolite
Answer: d
Clarification: Spirit level is used for levelling the table at the station point. Plumb bob is used to transfer the point in sheet to the ground. Compass is used to indicate the magnetic north. Theodolite is not used in this case as the entire process is done manually without any instrumental usage.

2. In plane table surveying, plotting and recording of values are done simultaneously.
a) True
b) False
Answer: a
Clarification: Plane table surveying is a graphical method, in which the field observations and plotting process is done simultaneously. It involves making a map in the field while the ground can be seen by the topographer and without intermediate steps of recording in field notes.

3. What are the dimensions of Johnson table?
a) 60*45 cm
b) 45*60 cm
c) 40*60 cm
d) 45*65 cm
Answer: b
Clarification: The Johnson board usually consists of 45*60 cm dimension, which is used while doing important field works which needs precise values.

4. For obtaining high precision values, which among the following is used?
a) Wooden table
b) Johnson table
c) Coast survey table
d) Traverse table
Answer: c
Clarification: Coast survey table can be able to obtain more precise values than the remaining plane tables. Due to more accuracy in levelling, the ability of producing precise values increases.

5. Alidade is used for___________
a) Sighting
b) Levelling
c) Transferring point to ground
d) Drawing lines
Answer: a
Clarification: The presence of eye vane of object vane allows the alidade to bisect the required points which are near by the station.

6. Orientation of table involves which among the following?
a) Traversing
b) Fore sighting
c) Back sighting
d) Measuring bearings
Answer: c
Clarification: Orientation by back sighting is a process that involves setting up of the plane table when there is no possibility over the station mark.

7. Orientation by trough compass is done for obtaining precise values.
a) True
b) False
Answer: b
Clarification: Orientation by trough compass is low accurate method which is done when speed of the work is more important than obtaining accurate values. Orientation by means of back sighting involves in obtaining more precise values.

8. Which of the following indicates the correct set of alidade?
a) Plain, reverse
b) Complex, telescopic
c) Plain, theodolite
d) Plain, telescopic
Answer: d
Clarification: In general, two types of alidades are used. They include plain, telescopic alidades. Now-a-days telescopic alidade is fitted on the plane table which is making the work from complex to simple.

9. By plotting and recording values simultaneously, there is a possibility to occur more errors.
a) True
b) False
Answer: a
Clarification: It might be possible to note the wrong value while processing simultaneously, but it is one of the most accurate methods which gives the best output.

10. In which of the following cases, orientation by back sighting is used?
a) When approximate levelling is required
b) When second point is available for orientation
c) When it is not possible to set instrument over station mark
d) When speed is important than accuracy
Answer: c
Clarification: Orientation by back sighting is used when it is not possible to set instruments over station mark and also for obtaining more precise and accurate values.

Surveying Multiple Choice Questions on “Area Computation – Ordinate Rule”.

1. Which of the following represents the correct set of ordinate rules used?
a) Average ordinate rule, Trapezoidal rule
b) Mid-ordinate rule, Mean ordinate rule
c) Mid-ordinate rule, Average ordinate rule
d) Trapezoidal rule, Mean ordinate rule
Answer: c
Clarification: The area from offsets can be calculated by using certain rules which include Mid-ordinate rule, Average ordinate rule, Trapezoidal rule, Simpson’s one-third rule, of which the mid-ordinate and average ordinate rules come under ordinate rules set.

2. Find the length of the base line if the number of divisions are 4 and d = 1.5m.
a) 2 m
b) 6 m
c) 2.5 m
d) 8 m
Answer: b
Clarification: The value of length of the base can be found out by using the formula,
L = n*d and on substitution, we get
L = 4*1.5 = 6 m.

3. Ordinate rule is based on which of the following assumptions?
a) Boundaries of the offsets are straight lines
b) Boundaries of the offsets are perpendicular
c) Boundaries of the offsets are curves
d) Boundaries of the offsets are parabolic
Answer: a
Clarification: The ordinate rule is used with the assumption that the boundaries between the extremities of the ordinates are straight lines. The base line is divided into number of divisions and the ordinates are measured at the midpoints of each division.

4. The area of the figure from ordinate rule can be determined as__________
a) A = average ordinate * perimeter
b) A = average ordinate * breadth
c) A = average ordinate * area
d) A = average ordinate * length of base
Answer: d
Clarification: The formula for area of figure from ordinate rule can be given as,
A = average ordinate * length of base
Where, L can be determined by no. of divisions*distance of each division.

5. Calculate the area by mid-ordinate rule if the value of d = 2m and the ordinates are given as 24.69m, 42.96 m, 26.74m.
a) 188.87 sq. m
b) 881.78 sq. m
c) 188.78 sq. m
d) 198.78 sq. m
Answer: c
Clarification: The formula for the area of the mid-ordinate can be given as
A = d*∑O. On substitution, we get
A = 2 * (24.69+42.96+26.74)
A = 188.78 sq. m.

6. Among the area calculation methods, which is more accurate?
a) Area by co-ordinates
b) Area by Simpson’s one-third rule
c) Area by double mean distances
d) Area by offsets
Answer: b
Clarification: Though the area calculated by dividing into triangles helps in determining the area of the figure, the area calculated by using Simpson’s rule helps in providing accurate results than the previously mentioned process.

7. 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
a) 245.08m
b) 542.08 m
c) 524.08 m
d) 528.04 m
Answer: b
Clarification: We have, Δ = (L * ∑O) / (n+1)
Here n = number of divisions = 8; n + 1 = number of ordinates = 8 + 1 = 9
L = Length of base = 10 x 8 = 80 m
∑O = 4.16+6.34+7.89+6.54+5.67+7.76+8.52+5.87+6.21 = 58.96m
Δ = (80*58.96)/9 = 524.089m.

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.
a) 2.42 m
b) 2.24 m
c) 4.22 m
d) 2.56 m
Answer: a
Clarification: We know that, the area by mid-ordinate can be given as, A = d*∑O. from this, the value of d can be calculated as,
d = A/∑O
d = 543.89 / 223.98
d = 2.42m.

9. The calculation of area by ordinate rule and Simpson’s rule will come under which category?
a) Area by double mean distances
b) Area by co-ordinates
c) Area by triangles
d) Area by offsets
Answer: d
Clarification: The area by offset method is suitable for long narrow strips of land. The offsets are measured from the boundary of the base line or a survey line at regular intervals. This method can also be applied to a plotted plan from which the offsets to a line can be scaled off.

10. Which of the following indicates the formula for area by average co-ordinate method?
a) Δ = (L * ∑O)/(n+1)
b) Δ = (L * ∑O)/(n-1)
c) Δ = (L + ∑O)/(n+1)
d) Δ = (L – ∑O)/(n+1)
Answer: a
Clarification: The area by average co-ordinate method is given as,
Δ = Average ordinate * length of base
Δ = (L * ∑O) / (n+1) where, ∑O = O1+O2+………+O_n

Surveying Multiple Choice Questions on “Setting out Compound Curve”.

1. The observations made for setting a compound curve must be equal to_________
a) 180-(Δ/2)
b) 180-(Δ1/2)
c) 180-(Δ2/2)
d) 180+(Δ/2)
Answer: a
Clarification: The points which are plotted for setting a compound curve involve a check at the last for determining the errors. The check can be given as 180-(Δ/2). Here, Δ is the deflection angle at P.I.

2. A compound curve can be set by which of the following methods?
a) Two-theodolite method
b) Deflection angles
c) Bisection of arcs
d) Tacheometric method
Answer: b
Clarification: Since a compound curve involves a combination of both long and short curves, it is best to adopt the deflection angles method so that work can be completed with accurate values.

3. In a compound curve, the point at which both the long curve and short curve will meet is called_________
a) Point of radius
b) Point of curvature curve
c) Point of compound curve
d) Point of deflection curve
Answer: c
Clarification: The point at which both the long curve and short curve will meet is determined as point of compound curve or P.C.C. It is the point where point of intersection lies perpendicular to P.C.C.

4. For setting a compound curve, the theodolite is first placed at______________
a) P.I
b) P.C.C
c) P.T
d) P.C
Answer: d
Clarification: Generally, a compound curve can be set by the method of deflection angles. In this method, the theodolite is placed at point of curve (P.C) and later at P.C.C for obtaining long curve.

5. Find the value of radius of curvature, if the degree of the curve is given as 7˚.
a) 136.71 m
b) 163.17 m
c) 163.71 m
d) 613.71 m
Answer: c
Clarification: The radius of curvature can be determined by
R = 1146 / D. On substitution, we get
R = 1146 / 7˚
R = 163.71 m.

6. Determine the value of long curve length of a compound curve, if the radius of curvature is given as 56.87m and the deflection angle is given as 65˚43ꞌ.
a) 56.22m
b) 65.22m
c) 65.44m
d) 69.22m
Answer: b
Clarification: The long curve length can be determined by using the formula,
t = R*θ*π/180. On substitution, we get
t = 56.87*65˚43ꞌ*π/180
t = 65.22m.

7. Find the value of the chainage of T1 by using the chainage of P.I and the tangent distance is given as 1024.31m and 707.57m.
a) 616.74m
b) 313.74m
c) 613.74m
d) 316.74m
Answer: d
Clarification: The chainage of T1 can be calculated as,
Chainage of T1 = chainage of P.I – T_s. On substitution, we get
Chainage of T1 = 1024.31-707.57
Chainage of T1 = 316.74m.

8. Determine the short curve tangent distance by using the short curve and long curve tangent lengths given as 54.98m and 89.32m. The deflection angles at P.I and at long curve are given as 86˚45ꞌ and 43˚31ꞌ.
a) 154.5 m
b) 145.5 m
c) 514.5 m
d) 451.5 m
Answer: a
Clarification: The short curve tangent distance can be determined by using the formula,
(T = t_s + (t_s + t_l)* frac{sin θ_2}{sin θ}.) On substitution, we get
(T = 54.98 + (54.98 + 89.32)* frac{sin 43˚31ꞌ}{sin 86˚45ꞌ})
T = 154.5 m.

9. Find the chainage of second tangent point, if the chainage at the point of compound curve is given as 2345.87m and the curve length can be given as 568.54m.
a) 2941.41 m
b) 9214.41 m
c) 2914.14 m
d) 2914.41 m
Answer: d
Clarification: The chainage at T₂ can be given as,
Chainage at T₂ = chainage at P.C.C + curve length. On substitution, we get
Chainage at T₂ = 2345.87 + 568.54
Chainage at T₂ = 2914.41 m.

10. All the necessary calculations will be done before setting the curve.
a) True
b) False
Answer: b
Clarification: The determination of tangent length, curve length, tangent distance will be done after setting of the curve, but not before. The curve setting process actually involves in the determination of these. By using the required data further process can be done.

Surveying Multiple Choice Questions on “Field Astronomy – Relation Between the Co-ordinates”.

1. The relation between altitude and latitude of the observer can be given as__________
a) Equal
b) Minimum
c) Maximum
d) Not equal
Answer: a
Clarification: Relation between the co-ordinates must be established in such a way that they might form a relation. In case of altitude and latitude, they must be equal so that they can satisfy the required relation in case of solving problem.

2. The sign for the deflection angle depends upon _______________
a) Altitude
b) Zenith
c) Celestial body
d) Horizon
Answer: c
Clarification: The placement of star defines the sign of the deflection angle. If the star is below the equator, then negative sign is allocated and if it is above then positive sign is allocated. By the allocation of sign to the deflection angle it might help in the determination of relation between latitude and declination.

3. Determine the latitude of the observer if the altitude of the pole is given as 23˚41ꞌ.
a) 32˚14ꞌ
b) 32˚41ꞌ
c) 23˚14ꞌ
d) 23˚41ꞌ
Answer: d
Clarification: The relation between latitude and the altitude can be given as
θ = α. So, the value of latitude of the observer can be given as
θ = 23˚41ꞌ.

4. Find the latitude of the observer if the declination is about 54˚32ꞌ having a meridian zenith of about 10m.
a) 64˚2ꞌ
b) 64˚32ꞌ
c) 46˚32ꞌ
d) 64˚23ꞌ
Answer: b
Clarification: The latitude in case of declination and zenith can be given as
θ = δ + z. On substitution, we get
θ = 54˚32ꞌ +10
θ = 64˚32ꞌ.

5. Find the difference of longitude between A = 30˚E and B = 160˚E.
a) 130˚
b) 310˚
c) 13˚
d) 30˚
Answer: a
Clarification: The difference can be calculated by
B – A = 160˚- 30˚ = 130˚. In case of change in direction, the angle must be subtracted or multiplied with 180. This will help in change in its direction.

6. If longitudes of A and B are given as 32˚12ꞌ W, 44˚2ꞌ W having latitude 29˚24ꞌ. Find the distance in km between the points A and B.
a) 19.54km
b) 91.1km
c) 11.9km
d) 19.1km
Answer: d
Clarification: The distance can be calculated by
Distance = difference of longitude * cos latitude. On substitution, we get
Distance = (44˚2ꞌ W – 32˚12ꞌ W)*cos29˚24ꞌ * 1.852
Distance = 19.1km.

7. Which of the following indicates the formula for hour angle of equinox?
a) Hour angle of star- R.A of the star
b) Hour angle of star+ R.A of the star
c) Hour angle of star / R.A of the star
d) Hour angle of star* R.A of the star
Answer: b
Clarification: The relation between right ascension and hour angle can be determined by using the hour angle of equinox, which can be given as hour angle of equinox = hour angle of star+ R.A of the star.

8. The distance between the points in a celestial body can be determined by using __________
a) Napier’s rule
b) Celestial rule
c) Zenithal rule
d) Obligate rule
Answer: a
Clarification: Napier’s rule is used in the determination of the distance between two points. It also involves in the measurement of altitude and hour angle if the declination and latitude are known.

9. Determine the zenith distance if the declination of star is given as 74˚32ꞌ and the latitude of the observer as 54˚21ꞌ.
a) 0˚12ꞌ
b) 2˚11ꞌ
c) 20˚11ꞌ
d) 20˚15ꞌ
Answer: c
Clarification: Here, declination of a star is greater than the latitude the,
Zenith distance = ZP – AP
Zenith distance = (90-θ)- (90-δ)
Zenith distance = 90-54˚21ꞌ-90+74˚32ꞌ
Zenith distance = 20˚11ꞌ.

10. If the zenith distance is given as 26˚57ꞌ, find the altitude of the star at upper culmination.
a) 63˚30ꞌ
b) 36˚3ꞌ
c) 3˚36ꞌ
d) 63˚3ꞌ
Answer: d
Clarification: Altitude of the star at upper culmination can be given as
= 90 – zenith distance
= 90 – 26˚57ꞌ
= 63˚3ꞌ.

Surveying Multiple Choice Questions on “Remote Sensing – Basic Principles”.

1. The relation between velocity, wavelength and frequency can be given as _________
a) λ = c / r
b) λ = c / f
c) λ = c / h
d) λ = h*c / f
Answer: b
Clarification: Photons are the main constituent particles in the electromagnetic energy. The relation between velocity, wavelength and frequency is determined as λ = c / f, where λ represents wavelength, f is the frequency of the wave and c represents the velocity of the wave, which is equal to speed of light.

2. Remote sensing uses which of the following waves in its procedure?
a) Electric field
b) Sonar waves
c) Gamma- rays
d) Electro-magnetic waves
Answer: d
Clarification: Electro-magnetic waves are used in case of remote sensing. The different waves present in this spectrum enables us to use a variety of waves based on the condition present and can be able have a better output.

3. Which of the following is not a principle of remote sensing?
a) Interaction of energy with satellite
b) Electromagnetic energy
c) Electro-magnetic spectrum
d) Interaction of energy with atmosphere
Answer: a
Clarification: Remote sensing involves certain principles which are applied for having a good result of the desired output. The principles are electromagnetic energy, electro-magnetic spectrum, interaction of energy with atmosphere etc.

4. Which among the following waves is having less wavelength range?
a) 0.03mm
b) 0.03nm
c) 0.03m
d) 0.03km
Answer: b
Clarification: A wide range of waves are present in case of electromagnetic spectrum, off which the gamma-rays are having a nano level wave length capacity i.e., less than 0.03nm.

5. In visible region, the blue light is having a wave length range of __________
a) 0.42-0.52 micrometer
b) 0.24-0.52 micrometer
c) 0.42-0.92 micrometer
d) 0.22-0.32 micrometer
Answer: a
Clarification: Visible region consist of three color waves red, blue and green remaining are the combination of those. The blue light is having a wavelength range of 0.42-0.52 micrometer.

6. Which of the following field is used by the EM waves?
a) Solar field
b) Polarized field
c) Electric field
d) Micro field
Answer: c
Clarification: EM waves used two major sources of fields i.e., electric and magnetic fields. Both are placed orthogonal to each other in a wave pattern. The electric components are placed in vertical manner and magnetic components in horizontal manner.

7. Among the following, which describes Stefan- Boltzmann formula?
a) M = σ/T⁴
b) M = σ-T⁴
c) M = σ+T⁴
d) M = σ*T⁴
Answer: d
Clarification: Stefan- Boltzmann law is based on the radiation produced and emitted by the body. This can be mathematically represented by M = σ*T⁴. Here, α is the Stefan- Boltzmann constant, T is the absolute temperature, M is the spectral existence of the body.

8. Which of the following is not a classification of scattering principle?
a) Faraday scattering
b) Rayleigh scattering
c) Mie scattering
d) Non-selective scattering
Answer: a
Clarification: Scattering involves in distribution of the light ray in more than two directions. It can be further classified as Rayleigh scattering, Mie scattering, non-selective scattering.

9. Which of the following can act as an example for air-borne platform?
a) LISS-III
b) Dakota
c) MOS
d) LISS-II
Answer: b
Clarification: At present, the air-borne platforms in use are Dakota, AVRO and beach-craft. A sensor is mounted on them and is placed at an altitude which can be able to access the specified object.

10. Polar orbiting satellites are generally placed at an altitude range of __________
a) 7-15km
b) 7000-15000km
c) 700-1500km
d) 70-150km
Answer: c
Clarification: Polar orbiting satellites are also known as sun-synchronous satellites, which are generally placed at an altitude range of 700-1500km from the ground level. These are able to deliver accurate information about the object which we need access to.