Category: Surveying Objective Questions

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.

Surveying Multiple Choice Questions on “GPS Surveying Techniques”.

1. Which of the following can be affected by atmospheric path disturbances?
a) Modern GPS surveying
b) Conventional GPS
c) Absolute positioning
d) Resection method
Answer: a
Clarification: Modern GPS surveying can be affected by atmospheric conditions. It can produce more accuracy in its output when compared to pseudo ranging method. This method costs more and is effective for engineering applications.

2. Which among the following can be described as an application of pseudo ranging?
a) Computation of distance between satellite and user
b) Computation of distance between GPS antenna and satellite
c) Computation of distance between GPS antenna and user
d) Computation of distance between satellite and object
Answer: b
Clarification: Application of pseudo ranging involves in computing distance between GPS antenna and satellite by correlation between transmitted code and reference code. It needs synchronization between transmitter and receiver clock signal.

3. By using pseudo ranging method, two dimensional and three dimensional GPS positions can be located.
a) True
b) False
Answer: a
Clarification: In the process of pseudo ranging, at least four observations are taken. By using these, the solution of 2D and 3D GPS positions can be determined. But in general, only three are required for 2D GPS positions.

4. Which of the following error occurs due to atmospheric conditions?
a) Natural error
b) User error
c) Propagation error
d) Signal multipath error
Answer: d
Clarification: Signal multipath is an error in GPS tracking, which is generated by atmospheric interference errors. It can be expressed in terms of the reflected signal from geographic based buildings, high rocks.

5. Which of the following is not used in the tracking system?
a) Multiple frequency
b) Dual frequency
c) Single frequency
d) Military navigation
Answer: a
Clarification: The GPS surveying includes tracking system, off which the civilian navigation, military navigation, single frequency and the dual frequency are used. These possess carrier waves which can be used for further processing of the function.

6. Which of the following doesn’t belong to the relative positioning techniques?
a) Real-time kinematic technique
b) Viscous GPS technique
c) Kinematic GPS surveying technique
d) Differential GPS technique
Answer: b
Clarification: The relative positioning technique involves static and kinematic GPS surveying technique, differential GPS technique, and real-time kinematic technique. These are to be identified before conducting the GPS surveying method.

7. Which of the following classes of positioning technique possess high precision?
a) GPS
b) Viscous technique
c) Real time technique
d) Kinematic technique
Answer: d
Clarification: The real time kinematic technique is having a high precision, which use carrier phase measurements in the instantaneous positioning mode. It is considered as the most powerful GPS positioning technology.

8. Which among the following indicates the correct set of static GPS surveying technology classification?
a) Long and normal base lines
b) Medium and short baselines
c) Long and short baselines
d) Normal and short base lines
Answer: c
Clarification: The classification of the static GPS surveying involves long baseline GPS technique and Short baseline GPS technique. Both are determined based on the length of the base line taken into consideration.

9. What will be the length of the base line in case of short baseline method of GPS surveying?
a) Less than 50km
b) Greater than 50km
c) Less than 2km
d) Greater than 100km
Answer: a
Clarification: In the case of short baseline of GPS surveying, baselines would be typically less than 50km. they can support the control network applications with data processing packages.

10. Which of the following is considered as modern GPS technology?
a) GIS
b) GPS mode
c) Instantaneous mode
d) Kinematic positioning technique
Answer: d
Clarification: Among the different GPS surveying techniques available, the kinematic positioning technique is considered as the modern GPS surveying technology. It requires the usage of all the specialized hardware and software as well as new filed procedures.

Surveying Multiple Choice Questions on “Introduction – Errors and Mistakes in Chaining”.

1. The length of a line measured with a 20 m chain was found to be 250 m. Calculate the true length of the line if the chain was 10 cm too long.
a) 252.25 m
b) 251.25 m
c) 225.25 m
d) 221.25 m
Answer: b
Clarification: Incorrect length of the chain is 20 + 10/100, ie 20.1 m. Measured length is 250, hence true length of the line is 250 × (20.1/20)=251.25 m.

2. The length of a survey line was measured with a 20 m chain and was found to be equal to 1200 m. If the length again measured with 25 m chain it is 1212 m. On comparing the 20 m chain with the test gauge, it was found to be 1 decimeter too long. Find the actual length of 25 m chain used.
a) 22.25 m
b) 21.64 m
c) 24.25 m
d) 24. 88 m
Answer: d
Clarification: Incorrect length of 20 m line is 20+0.10 = 20.10 m. True length of line = 1200×(20.10/20) = 1206 m. Actual or True length of 25 m chain = (1206×25)/1212 = 24.88 m.

3. A surveyor measured the distance between two points on the plan drawn to a scale of 1 cm is equal 40 m and the result was 468 m. But, actual scale is 1 cm = 20 m. Find the true distance between the two points.
a) 992 m
b) 936 m
c) 987 m
d) 967 m
Answer: b
Clarification: Distance between two points measured with a scale of 1 cm to 20 m is 468/20 = 23.4 cm. Actual scale of a plan is 1 cm = 40 m. True distance between the points is 23.4 × 40 = 936 m.

4. If L is true length of chain and L’ is incorrect length of the chain the correction to area A is _________
(Where ∆L/L = e, e is small and A’ is measured area)
a) 1+2e A’
b) (1+2e)/A’
c) (1+2e) x A’
d) (1+ e)xA’
Answer: c
Clarification: By using A=A'(L’/L)² and L’/L=(L+∆L)/L=1+e where e = ∆L/L.

5. If L is true length of chain and L’ is incorrect length of the chain the correction to Volume V is _______
(Where ∆L/L = e, e is small and V’ is measured area)
a) 1+3e)+ V’
b) (1+3e)/V’
c) (1+3e)xV’
d) (1+ e) ×V’
Answer: c
Clarification: By using V = V'(L’/L)³, e = ∆L/L and L’/L = (L+∆L)/L = 1+e. Then V = V’ (1+e)³ here e is small so V = (1+3e)xV’.

6. The difference between a measurement and the true value of the quantity measured is _____
a) True error
b) Discrepancy
c) Limit of error
d) Accuracy
Answer: a
Clarification: The difference between a measurement and the true value of the quantity measured is the true error of the measurement. The important function of a surveyor is to secure measurements that are correct within a certain limit of error prescribed by the nature and purpose of a particular survey.

7. The difference between the two measured values of the same quantity is ______
a) Precision
b) Accuracy
c) Discrepancy
d) Error
Answer: c
Clarification: A discrepancy is a difference between two measured values of the same quantity. A discrepancy may be small, yet the error may be great if each of the two measurements contains an error that may be large.

8. Which of the following are not sources of errors?
a) Instrumental
b) Personal
c) Natural
d) Artificial
Answer: d
Clarification: Error may arise from three sources namely instrumental, personal and natural.

9. A tape may be too long or an angle measuring instrument may be out of adjustment. Then such type of error comes under which source of error?
a) Instrumental
b) Personal
c) Natural
d) Artificial
Answer: a
Clarification: Error may arise due to imperfection or faulty adjustment of the instrument with which measurement is being taken comes under an instrumental source of error.

10. Investigation of observations of various types shows that accidental errors follow a definite law. This law is called ______
a) Law of probability
b) Law of recurrence
c) Law of precise
d) Law of accuracy
Answer: a
Clarification: This law defines the occurrence of errors and can be expressed in the form of the equation which is used to compute the probable value or the probable precision of a quantity. This is also termed as a theory of probability.

Surveying Questions and Answers for Experienced people on “Traversing by Direct Observation of Angles”.

1. In which of the following transverse method angles are measured by theodolite?
a) By fast needle
b) By direct observation of angles
c) By locating details with transit and tape
d) By free needle
Answer: b
Clarification: In transversing by direct observation of angles, angles between the lines are directly measured by a theodolite. The method is therefore accurate in comparison to the previous three methods.

2. In transversing by direct observation of angles, magnetic bearing of any one line can also be measured if required.
a) True
b) False
Answer: a
Clarification: The magnetic bearings of any one line can be measured and magnetic bearing of other lines can be calculated. The angles measured at different stations may be either included angle and deflection angle.

3. Transversing by included angles comes under which of the following?
a) Transversing by fast needle
b) Transversing by free needle
c) Transversing by direct observation of angles
d) Transversing by chain and compass
Answer: c
Clarification: Transversing by included angles and transversing by deflection angles comes under transversing by direct observation of angles.

4. __________ at a station is either of the two angles by the two survey lines meeting there.
a) Included angle
b) Deflection angle
c) Transverse angle
d) Deviated angle
Answer: a
Clarification: An included angle at a station is either of the two angles formed by the two survey lines meeting there. The method consists simply in measuring each angle directly from a backsight on the preceding station.

5. Included angles can be measured _________
a) Clockwise
b) Counter clockwise
c) Clockwise and counterclockwise
d) Clockwise or counterclockwise
Answer: d
Clarification: Included angles can be measured either clockwise or counterclockwise. But it is better to measure all angles clockwise.

6. All angles are preferred to measure clockwise because of the graduations of theodolite circle increase in this direction.
a) True
b) False
Answer: a
Clarification: It is better to measure included angles clockwise. It is because of the graduations of theodolite circle increase in this direction.

7. A deflection angle is an angle in which a survey line makes with prolongation of back sight.
a) True
b) False
Answer: b
Clarification: A deflection angle is an angle in which a survey line makes with the prolongation of the preceding line.

8. Transversing by deflection angles is more suitable for surveys of roads railways, pipe lines etc.
a) True
b) False
Answer: a
Clarification: Transversing by deflection angles is more suitable for surveys of roads railways, pipe lines etc, where the survey lines make small deflection angles.

9. Deflection angle may vary from __________ to __________
a) 0° to 90°
b) 90° to 180°
c) 0° to 180°
d) 0° to 270°
Answer: a
Clarification: A deflection angle is an angle in which a survey line makes with the prolongation of the preceding line. It may vary from 0° to 180°.

10. In following figure deflection angle at Q is teta L.
a) True
b) False
Answer: b
Clarification: The deflection angle at Q is alpha R and that at R is teta L.

To practice all areas of Surveying for Experienced people,