Category: Surveying Objective Questions

Surveying Multiple Choice Questions & Answers on “Interpolation of Contours”.

1. What is the process of spacing the contours proportional between the plotted ground points established by indirect methods?
a) Interpolation
b) Tacheometric method
c) Cross section method
d) By squares method
Answer: a
Clarification: Interpolation of the contours is the process of spacing the contours proportionately between the plotted ground points established by indirect methods.

2. The methods of interpolation are based on the assumption that the slope of ground between the two points is uniform.
a) True
b) False
Answer: a
Clarification: Interpolation of the contours is the process of spacing the contours proportionately between the plotted ground points established by indirect methods. The methods of interpolation are based on the assumption that the slope of ground between the two points is uniform.

3. Which of the following is not a chief method of Interpolation?
a) By estimation
b) By arithmetic calculation
c) By graphical method
d) By cross sections
Answer: d
Clarification: The methods of interpolation are based on the assumption that the slope of ground between the two points is uniform. The chief methods of Interpolation are by estimation, by arithmetic calculations, by graphical method.

4. Which of the following methods is extremely rough and is used for small scale work only?
a) By estimation
b) By arithmetic calculation
c) By graphical method
d) By cross sections
Answer: a
Clarification: Estimation method is extremely rough and is used for small scale work only. The positions of contour points between the guide points are located by estimation.

5. The positions of contour points between the guide points are located by _______
a) Estimation
b) Arithmetic calculations
c) Graphical method
d) Cross sections
Answer: a
Clarification: Estimation method is extremely rough and is used for small scale work only. The positions of contour points between the guide points are located by estimation.

6. Which of the following Interpolation method is accurate and time consuming?
a) by estimation
b) by Arithmetic method
c) by graphical method
d) by squares
Answer: b
Clarification: Arithmetic method is accurate but time consuming. The positions of contour between the guide points are located by arithmetic calculation.

7. In which of the following methods Interpolation is done with the help of a tracing paper?
a) by estimation
b) by Arithmetic method
c) by graphical method
d) by squares
Answer: c
Clarification: In the graphical method, the Interpolation is done with the help of a tracing paper. Sometimes tracing cloth is also used.

8. In cross sections method, cross sections are run traverse to the centre line of a road, railway or canal etc.
a) True
b) False
Answer: a
Clarification: In this method, cross sections are run transverse to the centre line of a road, railway or canal etc. The method is most suitable for railway route surveys.

9. In the case of hilly terrain, the tacheometric method may be used with advantages.
a) True
b) False
Answer: a
Clarification: In the case of hilly terrain, the tacheometric method may be used with advantages. A tacheometer is a theodolite fitted with stadia diaphragm so that staff readings against all the three hairs may be taken.

10. In cross sections method, If there are irregularities in the surface between two cross lines, additional guide points may be located on intermediate cross lines.
a) True
b) False
Answer: a
Clarification: If there are irregularities in the surface between two cross lines, additional guide points may be located on intermediate cross lines. If required, some of the cross lines may also be chosen at any inclination other 90° to the main line.

Surveying Multiple Choice Questions on “Plane Table Surveying – Graphic Triangulation”.

1. Which among the following is not a method of plane table surveying?
a) Radiation
b) Trisection
c) Intersection
d) Resection
Answer: b
Clarification: The methods which are adopted in case of plane table surveying involve radiation, Intersection, resection and traversing, which are used based on the type of output required.

2. The method of radiation is used when distances are small.
a) True
b) False
Answer: a
Clarification: In the method of radiation, a ray is drawn from the instrument station towards the point, the distance is measured between the instrument station and that point by which, the point is located. This method is more suitable when the distances are small.

3. Method of Intersection is used when _____________
a) Distance between point and instrument is very small
b) Distance between point and instrument is infinity
c) Distance between point and instrument is small
d) Distance between point and instrument is large
Answer: d
Clarification: Method of intersection is used when the distance between the point and the instrument station is either too large or cannot be measured accurately due to some conditions. The location of an object is determined by sighting the object from two plane table stations.

4. Which of the following methods required two instrument stations?
a) Radiation
b) Intersection
c) Resection
d) Traversing
Answer: b
Clarification: The location of an object is determined by sighting the object from two plane table stations and drawing the rays. The intersection of these rays will give a position of the object, because of this it is essential to have at least two instrument stations for locating any point.

5. Which among the following methods of plane table is most commonly used?
a) Intersection
b) Resection
c) Traversing
d) Radiation
Answer: a
Clarification: The method of intersection usually involves locating a station point and marking the remaining points, followed by changing the instrument to another point and computing the remaining points from which the required distance between the points can be found out with the minimal effort.

6. Which of the following set indicates the location of details of the survey?
a) Resection, intersection
b) Radiation, resection
c) Radiation, intersection
d) Traversing, resection
Answer: c
Clarification: The methods radiation and intersection usually involves collecting or locating the required details which accessed for computing the survey. Traversing and resection involve the location of plane table stations.

7. Method of intersection is also known as _____________
a) Resection
b) Graphical triangulation
c) Radiation
d) Traversing
Answer: b
Clarification: Method of intersection is also known as graphical triangulation because it involves the formation of triangles in computed traverse and moreover this process is a graphical method.

8. The process of determining the plotted position of the station occupied by the plane table is known as _________
a) Trisection
b) Radiation
c) Intersection
d) Resection
Answer: d
Clarification: By the resection process, the plane table station points are located in which they are taken as reference for locating the remaining points of the traverse.

9. Orientation might cause a huge problem in method of resection.
a) True
b) False
Answer: a
Clarification: If the table is not correctly oriented at the station in the method of resection, the intersection of the two resectors will not give the correct location of the station. The problem, therefore, lies in the orienting table at the stations.

10. The observations made in method of traversing can be used as ___________
a) Bisecting point
b) Traverse point
c) Instrument stations
d) Survey point
Answer: c
Clarification: The main difference between method of traversing and method of radiation is that in the case of radiation the observations are taken to those points which are to be detailed while in the case of traversing the observations are made to those points which will be subsequently used as instrument stations.

Surveying Multiple Choice Questions on “Tacheometric Surveying – Stadia Method Principle”.

1. Horizontal distances are measured by direct methods, i.e. laying of chains or tapes on the ground.
a) True
b) False
Answer: a
Clarification: Generally, horizontal distances are measured by direct methods, i.e. laying of chains or tapes on the ground. These methods are not always convenient if the ground is undulating, rough, difficult and inaccessible.

2. If the ground is undulating, rough, difficult and inaccessible. Under these circumstances _________ methods are used to obtain distances.
a) Direct methods
b) Indirect methods
c) Chain surveying
d) Tacheometry
Answer: b
Clarification: Generally, horizontal distances are measured by direct methods, i.e. laying of chains or tapes on the ground. These methods are not always convenient if the ground is undulating, rough, difficult and inaccessible. Under these circumstances, indirect methods are used to obtain distances.

3. Which of the following is an indirect method of surveying?
a) Chain surveying
b) Tacheometry
c) Countouring
d) All of the mentioned
Answer: b
Clarification: Generally, horizontal distances are measured by direct methods, i.e. laying of chains or tapes on the ground. These methods are not always convenient if the ground is undulating, rough, difficult and inaccessible. Under these circumstances, indirect methods are used to obtain distances. One such method is “Tacheometry”.

4. Using tacheometric methods, elevations can also be determined.
a) True
b) False
Answer: a
Clarification: Using tacheometric methods, elevations can also be determined. It is in fact a branch of angular surveying in which both the horizontal and vertical positions of points are determined from the instrumental observations, the chain surveys being entirely eliminated.

5. Which of the following is the branch of angular surveying in which both the horizontal and vertical positions of points are determined from the instrumental observations, the chain surveys being entirely eliminated?
a) Tacheometry
b) Contouring
c) Ranging
d) Random line method
Answer: a
Clarification: Using tacheometric methods, elevations can also be determined. It is in fact a branch of angular surveying in which both the horizontal and vertical positions of points are determined from the instrumental observations, the chain surveys being entirely eliminated.

6. Tacheometer has ______ number of horizontal hairs.
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: A tacheometer is similar to an ordinary transit theodolite, generally a vernier theodolite itself, fitted with two stadia wires in addition to the central cross-hair. The stadia diaphragm has three horizontal hairs viz., a central horizontal hair and upper and lower stadia hairs.

7. Which of horizontal hairs are equivalent in stadia diaphragm of tacheometer?
a) Upper and central
b) Central and lower
c) Upper and lower
d) Lower, central and upper
Answer: c
Clarification: The stadia diaphragm has three horizontal hairs viz., a central horizontal hair and upper and lower stadia hairs. The upper and lower stadia hairs are equidistant from the central horizontal hair. Stadia hairs are sometimes called stadia lines.

8. The magnification of the telescope in tacheometer should be at least _______ to _______ diameters.
a) 10 to 20
b) 10 to 30
c) 20 to 30
d) 20 to 40
Answer: c
Clarification: For the purpose of tacheometry, even though an ordinary transit can be employed, accuracy and speed are increased if the instrument is specially designed for the work. The magnification of the telescope in tacheometer should be at least 20 to 30 diameters.

9. What should be the aperture required for tacheometer?
a) 30mm
b) 40mm
c) 25mm
d) 15mm
Answer: b
Clarification: The magnification of the telescope in tacheometer should be at least 20 to 30 diameters, with an aperture of at least 40 mm for a sufficiently bright image.

10. The magnifying power of the eyepiece is also smaller than for an ordinary transit to produce a clearer image of a staff held far away.
a) True
b) False
Answer: a
Clarification: The magnifying power of the eyepiece is also greater than for an ordinary transit to produce a clearer image of staff held far away.

11. For short sights of about _____ m or less, an ordinary levelling staff may be used. For long sights, special staff called stadia rod is generally used.
a) 50
b) 100
c) 150
d) 200
Answer: b
Clarification: For short sights of about 100 m or less, an ordinary levelling staff may be used. For long sights, special staff called stadia rod is generally used. The graduations are in bold type (face about 50 mm to 150 mm wide and 15 mm to 60 mm thick) and the stadia rod is 3 m to 5 m long. To keep the staff or stadia rod vertical, a small circular spirit level is fitted on its backside. It is hinged to fold up.

12. In fixed hair method, the distance between _______ hair and__________ hair are fixed.
a) Upper and central
b) Central and lower
c) Upper and lower
d) Lower, central and upper
Answer: c
Clarification: In this method, the distance between the upper hair and lower hair, i.e. stadia interval i, on the diaphragm of the lens system is fixed. The staff intercepts, therefore, changes according to the distance D and vertical angle θ.

13. Distance and elevation formulae for fixed hair method assuming line of sight as horizontal and considering an external focusing type telescope is D = Ks + C. where K is _______
a) f/i
b) i/f
c) f + c
d) f – c
Answer: a
Clarification: The constant K is equal to (f /i). It is called multiplying constant of the tacheometer and is generally kept as 100. The constant C is equal to (f + c). It is called additive constant whose value ranges from 30 cm to 50 cm for external focusing telescopes and 10 cm to 20 cm for internal focusing telescopes.

14. For anallactic lens in D = Ks + C, which of the following is zero?
a) D
b) K
c) C
d) S
Answer: c
Clarification: The constant C is equal to (f + c). It is called additive constant whose value ranges from 30 cm to 50 cm for external focusing telescopes and 10 cm to 20 cm for internal focusing telescopes. For telescopes fitted with anallactic lens, C equals zero.

15. 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 _______
a) f/i
b) i/f
c) f + c
d) f – c
Answer: c
Clarification: The constant K is equal to (f /i). It is called multiplying constant of the tacheometer and is generally kept as 100. The constant C is equal to (f + c). It is called additive constant whose value ranges from 30 cm to 50 cm for external focusing telescopes and 10 cm to 20 cm for internal focusing telescopes.

Surveying Multiple Choice Questions on “Ideal Transition Curve”.

1. A curve which is having a varying radius is called _____________
a) Simple curve
b) Compound curve
c) Transition curve
d) Reverse curve
Answer: c
Clarification: A transition curve is having a varying radius which is introduced in between the branches of different curves. It is used in areas where a straight line and a curve have to be connected.

2. In order to prevent the case of overturning, which of the following is provided?
a) Super elevation
b) Reverse curve
c) Simple curve
d) Compound curve
Answer: a
Clarification: Super elevation is provided in such a way that there is a rise at a side of the curve which helps in avoiding overturning, which is given as B/gR which helps in determining the amount of elevation needed to be provided.

3. The provision of equilibrium cant can be seen in case of__________
a) Curves
b) Mountains
c) Highways
d) Railways
Answer: d
Clarification: The provision of equilibrium cant can be seen in case of railways, which is capable of providing the same amount of load on either side of the track, which makes the passenger not to lean in any direction.

4. Side friction factor is used in which of the following cases?
a) Reverse curve
b) Transition curve
c) Simple curve
d) Compound curve
Answer: b
Clarification: Side friction factor is the force transferred by friction to the pavement, which can be used in case of highways while providing super elevation to avoid overturning of vehicles.

5. With increase in super elevation there must be subsequent increase in centrifugal force for balancing it.
a) True
b) False
Answer: a
Clarification: While considering super elevation, the weight of the vehicle and centrifugal force must be taken into consideration. So if there is an increase in super elevation, centrifugal force must be increased so that it balances the motion with less frictional force.

6. Length of transition curve can be given as __________
a) L = r tan(δ/2)
b) L = ne
c) L = n – e
d) L = n + e
Answer: b
Clarification: Length of transition curve is given as L = ne in which, e describes about the amount of super elevation provided which should be at a required rate and n varies in between 300 to 1200.

7. Which of the following methods doesn’t describe the method of finding length of transition curve?
a) Arbitrary gradient
b) Time rate
c) Bisection of arcs
d) Rate of Change of radial acceleration
Answer: c
Clarification: The length of transition curve depends upon amount of super elevation provided which must be at a required rate. It can be determined by an arbitrary gradient, time rate and change of radial acceleration.

8. Among the methods available for determining the length of the curve, which is commonly used?
a) Bisection of arcs
b) Time rate
c) Arbitrary method
d) Rate of change of radial acceleration
Answer: d
Clarification: In the rate of change of radial acceleration, time rate methods length is directly proportional to which makes both of them useful. But, in accuracy perspective, rate of change of radial acceleration method is used most commonly.

9. Clothoid is also known as___________
a) Glover’s spiral
b) Froude’s equation
c) Cartesian curve
d) Cubic spiral
Answer: a
Clarification: A clothoid is an ideal transition curve in which, length is inversely proportional to the radius of the curve. It is also known as Glover’s spiral.

10. Froude’s transition is also known as___________
a) Clothoid
b) Cubic parabola
c) Cubic spiral
d) Glover’s spiral
Answer: b
Clarification: Cubic parabola is also known as Froude’s transition curve. It uses Cartesian co-ordinates which are required for setting the curve. It is the most commonly used transition curve, which can use rectangular co-ordinates too.

11. Determine the super elevation, if the width of the road can be given as 2.96m, radius of curve as 62.96m, vehicle speed = 56m/s.
a) 51.303m
b) 51.03m
c) 15.03m
d) 15.3m
Answer: c
Clarification: Super elevation or cant can be derived by,
e = Bv²/g*R. on substitution, we get
e = 2.96*56² / 9.81*62.96
e = 15.03m.

12. If the super elevation for a road is given as 24m, find the length of transition curve.
a) 15.6 km
b) 15.6 m
c) 15.6 cm
d) 15.6 mm
Answer: a
Clarification: The length of transition curve can be determined by,
L = n*e, the value of n varies in between 300-1200. Assume the value of n = 650. On substitution, we get
L = 650*24
L = 15600 m. = 15.6 km.

13. Using intrinsic equation, find the value of the length of curve between two points of a 45m transition curve having radius 24.76m with an inclination of 8˚43ꞌ.
a) 319.37m
b) 913.37m
c) 139.73m
d) 139.37m
Answer: d
Clarification: From the intrinsic equation,
( l = sqrt{2RLθ}.) On substitution, we get
(l = sqrt{2*24.76*45*8˚43ꞌ})
l = 139.37m.

14. If the radius of curvature of a curve being 45.42m with an inclination of 7˚52ꞌ. The value of s and L corresponds to 2 and 56m respectively, find the total tangent length of a transition curve.
a) 03.02m
b) 30.02m
c) 2.3m
d) 3.2m
Answer: b
Clarification: The total length of a transition curve can be determined by,
(R+s)*tan (θ/2) + L*(1 – (frac{s}{R})) / 2. On substitution, we get
= (45.42+2)*tan (7˚52ꞌ/2) + 56*(1 – (frac{2}{45.42})) / 2
= 30.02m.

15. In a cubic parabola, if the value of x co-ordinate is 7, radius of the curve is given as 42.69m and the length of the curve as 24m. Find the y co-ordinate.
a) 2.76
b) 1.05
c) 0.05
d) 5
Answer: c
Clarification: The value of y co-ordinate can be determined by using the formula,
y = x³ / 6*R*L. on substitution, we get
y = 7³ / 6*42.69*24
y = 0.05.

Surveying Multiple Choice Questions on “Field Astronomy – Astronomical Corrections”.

1. Which of the following doesn’t belong to the set of corrections applied in astronomical corrections?
a) Correction of parallax
b) Correction of sag
c) Correction of refraction
d) Correction of semi-diameter
Answer: b
Clarification: Astronomical corrections are applied in the case of celestial bodies. They generally include corrections for parallax, for refraction, for semi-diameter, for the dip of horizon. All these are not applied at once but are applied based on the necessity.

2. Magnitude of refraction depends upon which of the following factors?
a) Density
b) Surface tension
c) Reflection
d) Polarisation
Answer: a
Clarification: Due to the curvature of the earth surface, the layers of the atmosphere can be thinner as its distance from surface increases. It may cause the deviation in the angle of ray which is equal to refraction angle. The magnitude of refraction depends on density of air, temperature, pressure of barometer, altitude.

3. Correction to the dip is always_____________
a) Zero
b) Multiplicative
c) Subtractive
d) Additive
Answer: c
Clarification: Angle of dip can be assumed as the angle between true and visible horizon. Due to the curvature of the earth, the dip angle must be subtracted from the observed angle, which makes the correction as subtractive.

4. Determine the index error for face right, if the face left and face right readings were given as 18˚24ꞌ52ꞌꞌ and 18˚23ꞌ24ꞌꞌ.
a) +24ꞌꞌ
b) +49ꞌꞌ
c) +4ꞌꞌ
d) +44ꞌꞌ
Answer: d
Clarification: For determining the correction for index error, calculate the mean reading = (18˚24ꞌ52ꞌꞌ + 18˚23ꞌ24ꞌꞌ) / 2 = 18˚24ꞌ8ꞌꞌ
Index error for face right can be given as 18˚24ꞌ8ꞌꞌ – 18˚23ꞌ24ꞌꞌ = +44ꞌꞌ.

5. Find the altitude correction for semi-diameter which is having index error 25˚46ꞌ21ꞌꞌ and semi-diameter 0˚26ꞌ21ꞌꞌ.
a) 26˚12ꞌ42ꞌꞌ
b) 62˚12ꞌ42ꞌꞌ
c) 26˚21ꞌ42ꞌꞌ
d) 26˚12ꞌ24ꞌꞌ
Answer: a
Clarification: The altitude correction for the semi-diameter can be given as the summation of index error with the semi-diameter mentioned i.e.,
25˚46ꞌ21ꞌꞌ + 0˚26ꞌ21ꞌꞌ = 26˚12ꞌ42ꞌꞌ.

6. Determine the correction for refraction if the angle of azimuth is given as 62˚21ꞌ24ꞌꞌ.
a) 10˚11ꞌ50.74ꞌꞌ
b) 0˚1ꞌ50.74ꞌꞌ
c) 0˚1ꞌ0.74ꞌꞌ
d) 50˚1ꞌ50.74ꞌꞌ
Answer: b
Clarification: The correction for refraction can be given as 58ꞌꞌ tan z. Where, z is the Azimuthal angle. On substitution, we get
58ꞌꞌ*tan (62˚21ꞌ24ꞌꞌ) = 0˚1ꞌ50.74ꞌꞌ.

7. Determine the correction for parallax which has to be applied to 29˚42ꞌ31ꞌꞌ for obtaining altitude of the sun, which is given as 32˚41ꞌ15ꞌꞌ.
a) 53˚51ꞌ
b) 53˚15ꞌ
c) 35˚51ꞌ
d) 5˚51ꞌ
Answer: a
Clarification: Correction for parallax can be given as
Correction for parallax = horizontal parallax * cos apparent latitude
Correction for parallax = 8*8cos α. On substitution, we get
Correction for parallax = 53˚51ꞌ, this has to applied for 29˚42ꞌ31ꞌꞌ in order to have the correct altitude of the sun.

8. Determine the corrected azimuth value if the azimuth is given as 54˚32ꞌ15ꞌꞌhaving b = +23ꞌꞌ with vertical angle 29˚42ꞌ31ꞌꞌ.
a) 67˚39ꞌ40ꞌꞌ
b) 67˚30ꞌ39.78ꞌꞌ
c) 67˚39ꞌ39.78ꞌꞌ
d) 76˚39ꞌ39.78ꞌꞌ
Answer: c
Clarification: The corrected azimuth can be determined by the summation of azimuth with correction applied.
Correction = b*tan α = 23*tan 29˚42ꞌ31ꞌꞌ = 13˚7ꞌ24.8ꞌꞌ. On summation,
Corrected azimuth = 54˚32ꞌ15ꞌꞌ + 13˚7ꞌ24.8ꞌꞌ
Corrected azimuth = 67˚39ꞌ39.78ꞌꞌ.

9. Which of the following is always subtractive?
a) Correction for reflection
b) Correction for dip
c) Correction for parallax
d) Correction for polarization
Answer: b
Clarification: The correction applied for the dip is always subtractive. Dip is the angle between true and the visible horizon, the observations are taken from the sextant, the altitude of the star can be measured by this. Correction applied for the parallax is always additive.

10. Apply correction for dip angle if the height of the observer at the sea level is given as 54m.
a) 23˚13ꞌ45.05ꞌꞌ
b) 0˚3ꞌ45.05ꞌꞌ
c) 0˚31ꞌ45.05ꞌꞌ
d) 0˚13ꞌ45.05ꞌꞌ
Answer: d
Clarification: The correction for dip is found out by using the formula,
Tan β = (sqrt{2h/R}). Where, R is the radius of earth.
Tan β = (sqrt{2*54/(6370*1000)})
β = 0˚13ꞌ45.05ꞌꞌ.

Surveying Multiple Choice Questions on “Remote Sensing – Electromagnetic Spectrum”.

1. Energy of the discrete particles can be given by_______
a) Photons
b) Protoplasm
c) Electrons
d) Neutrons
Answer: a
Clarification: Photons act as a constituent particle in case of electromagnetic wave. These possess certain properties like energy and momentum which is able to deliver it to the wave and helps in certain classifications.

2. Which among the following is having more wavelengths?
a) X-rays
b) Cosmic waves
c) Radio waves
d) Gamma rays
Answer: c
Clarification: Radio waves possess the longest wavelength in the electromagnetic spectrum, which makes it suitable in the usage of classified radars for operating in a particular region.

3. Which among the following wave is not employed in case of remote sensing?
a) X-ray
b) Visible ray
c) Thermal IR
d) Radio waves
Answer: a
Clarification: Gamma rays, X-rays and UV rays will be absorbed by the atmosphere so that the sensor which is mounted on the satellite can’t use the facilities which can be provided by these rays and also these possess very less wavelength.

4. Optical mechanical scanner is used in which type of electromagnetic waves?
a) X-rays
b) Cosmic waves
c) Radio waves
d) Thermal IR
Answer: d
Clarification: Infrared region has been classified into thermal IR and reflected IR. Thermal IR is equipped with the optical mechanical scanner and a special system which is free from a film produces images by using wavelengths of range 3-5 micrometer.

5. Radio waves are having the longest wavelength among all the electromagnetic waves.
a) False
b) True
Answer: b
Clarification: In an electromagnetic spectrum, certain classifications were made to determine the wave properties. Among them, radio and television waves stand at last having a long wavelength parameter. This makes it suitable for usage of radars.

6. Gamma rays are having a wavelength of _________
a) Zero
b) Greater than 0.03nm
c) Less than 0.03nm
d) Equal to 0.03nm
Answer: c
Clarification: In an electromagnetic spectrum Gamma rays are having second least wavelength. It can be clearly observed that these rays are absorbed by the atmosphere which makes it possess least wavelength, which is less than 0.03nm.

7. Which of the following waves can be used in case of remote sensing?
a) UV rays
b) X-rays
c) Gamma rays
d) Visible rays
Answer: d
Clarification: Electromagnetic spectrum consists of a wide range of classifications among those some of them are absorbed by the atmosphere and most of them are used in remote sensing those include visible rays, IR rays, Radar waves, Radio waves etc.

8. Which of the following indicates the correct set of combination in radio waves?
a) Shorter wavelength – high frequency
b) Longer wavelength – less frequency
c) Shorter wavelength – less frequency
d) Longer wavelength – high frequency
Answer: a
Clarification: From the relation it can be clearly seen that wavelength and frequency are inversely proportional to each other which makes it form a relation that is shorter wavelength must possess high frequency which makes it suitable in case of cosmic rays and radio waves.

9. How much wave length is reflected back by the earth surface from the absorbed sun radiation?
a) 0.5meter
b) 0.5 micrometer
c) 0.5 centimeter
d) 0.5 decimeter
Answer: b
Clarification: In general, the radiation received from sun is distributed to all over the world in a manner that the entire radiation is useful. But some of this is reflected back, which is 0.5 micrometer, a minute one.

10. EM waves varies from ______ to ________
a) Meters to nano-meters
b) Meters to micro-meters
c) Nano to micro-meters
d) Centimeters to nano-meters
Answer: a
Clarification: The EM wave is continuum of energy, which under goes certain propagations. During this course, it may undergo some undulations which result in decrease of the wave length capacity. Generally, these vary from meters to nano-meters.

11. The formula of energy produced from the body can be given as _________
a) Q = h- c / λ
b) Q = h*c * λ
c) Q = h+ c / λ
d) Q = h*c / λ
Answer: d
Clarification: The energy propagated from the body can be determined by
Q = h*c / λ. From this, the wave length can be determined by having an energy value. This help in the determination of the wavelength by using plank’s constant.