Surveying — Part 1 MCQs with Explanations

Let's break down this surveying problem step by step! First, we need to understand what the question is asking. It's about the spacing between flight lines on a map, given a real-world area and a map scale. The area's width is 30 km, and there are 16 flight lines perpendicular to it. This means the 16 flight lines divide the 30 km width into 15 equal spaces. Therefore, the spacing between the flight lines on the *ground* is calculated as: 30 km / 15 = 2 km . Next, we need to convert this ground distance to the map distance using the scale 1:50,000. This scale means that 1 unit on the map represents 50,000 units on the ground. Convert the ground distance (2 km) to centimeters: 2 km = 2 * 1000 m = 2 * 1000 * 100 cm = 200,000 cm . Now, apply the scale to find the map distance: Map distance = Ground distance / Scale factor = 200,000 cm / 50,000 = 4 cm . Therefore, the spacing between the flight lines on the photographical map will be 4 cm. So, the correct answer is: Option D: 4 cm .

The size of a plane table is typically 600 mm × 750 mm. Plane tables are surveying instruments used for mapping and plotting, and their standard dimensions are often 600 mm × 750 mm. This size provides a suitable working area for accurate and detailed plotting of survey data. Other sizes mentioned are not standard for plane tables used in surveying.

The correct answer is Option D: 5° S latitude . Here's why: Polaris , also known as the North Star, is located very close to the North Celestial Pole . The altitude of the North Celestial Pole above the horizon is equal to the observer's latitude in the Northern Hemisphere . If you are at the Equator (0° latitude) , Polaris will be right on the horizon. As you move northward , Polaris appears higher and higher in the sky. However, Polaris can only be seen from locations in the Northern Hemisphere or very close to the equator. If you are in the Southern Hemisphere , Polaris will be below the horizon . Since 5° S latitude is in the Southern Hemisphere, Polaris will be below the horizon there. At 10° N and 50° N latitude , Polaris will be visible above the horizon. At the Equator , Polaris will be on the horizon.

The nautical mile is a unit of measurement used in navigation and is defined as one minute of latitude along any meridian. Here's the breakdown: - 1 nautical mile (NM) = 1 minute of latitude - 1 minute of latitude is equal to 1/60th of a degree of latitude - 1 degree of latitude is approximately 111.12 kilometers or 69.05 miles So, a nautical mile is approximately 1.852 kilometers or 1.15078 miles. This unit is particularly useful in navigation because it provides a consistent measure of distance that takes into account the Earth's curvature. It allows navigators to calculate distances accurately while accounting for the variations in the size of a degree of latitude at different locations on the Earth's surface. Therefore, the correct answer is Option A: 1 minute of latitude .

Correct Answer: Option C: 40° Explanation: The latitude (\(\phi\)) of a place can be determined using the formula: \[ \phi = \frac{\text{Altitude at Upper Transit} + \text{Altitude at Lower Transit}}{2} \] Given: Upper Transit Altitude = 60° 30' Lower Transit Altitude = 19° 30' \[ \phi = \frac{(60° 30') + (19° 30')}{2} \] Converting minutes: \[ \phi = \frac{(60.5 + 19.5)}{2} = \frac{80}{2} = 40° \] Thus, the latitude of the place is 40° .

Correct Answer: Option C: Equator and Zenith Explanation: Latitude (φ): Latitude measures how far north or south a location is from the Earth's equator, expressed in degrees (0° at the equator, 90° at the poles). Declination (δ): Declination is similar to latitude but for celestial objects, measuring how far a star is from the celestial equator. A star on the celestial equator has \( \delta = 0^\circ \), while a star at the celestial north pole has \( \delta = +90^\circ \). Zenith Distance (z): Zenith distance is the angle between a star and the observer's zenith (the point directly overhead). A star at the zenith has \( z = 0^\circ \), while a star on the horizon has \( z = 90^\circ \). Formula: The given formula is: \[ \phi = z - \delta \] This formula applies when the observed star is positioned between the observer's zenith and the equator . Reason: - Zenith distance is measured from the zenith downward. - Declination is measured from the celestial equator. - The subtraction only works when the star is south of the zenith, which means the star lies between the equator and the zenith. Conclusion: The correct answer is Option C: Equator and Zenith because the given formula holds only in this case.

A stadia diaphragm is a special part of a surveying instrument (like a theodolite or level). Its main purpose is to help determine distances. The diaphragm has horizontal cross wires . Typically, a stadia diaphragm contains three horizontal cross wires : one at the top, one in the middle (the central crosshair), and one at the bottom. These wires are used to read the stadia intercept on a graduated staff held at a distance, which then helps calculate the distance. Options A, B, and D are incorrect because stadia diaphragms are designed with three horizontal cross wires .

In theodolite surveying, taking both face left and face right observations is a common technique to eliminate symmetrical instrumental errors. These errors are usually equal in magnitude but opposite in sign when the instrument is reversed, so averaging both readings cancels them out. Option A: Error due to the horizontal axis not being perpendicular to the vertical axis This error is symmetrical and can be eliminated by taking both face observations. Option B: Index error due to imperfect adjustment of the vertical circle vernier This is also a symmetrical error and can be eliminated by taking face left and face right readings. Option C: Error due to non-parallelism of the axis of telescope level and the line of collimation This is an asymmetrical and inherent instrument error . It affects the sighting line itself and cannot be eliminated by reversing the face of the instrument. Option D: None of the above This is incorrect because one of the options (Option C) is indeed a valid exception. Therefore, the correct answer is: error due to non-parallelism of the axis of telescope level and line of collimation (Option C).

In surveying, the length of a parallel of latitude (λ) between two meridians is equal to the difference in longitudes multiplied by the cosine of the latitude (λ). This relationship arises from the spherical geometry of the Earth's surface. As you move away from the equator towards the poles, the parallels of latitude become shorter due to the convergence of meridians towards the poles. The cosine of the latitude accounts for this reduction in length as you move away from the equator. Therefore, the correct option is Option B: cos λ .

Understanding Reduced Bearing: The reduced bearing of a line gives its direction with respect to the North-South line. N 60° W means the line is 60° to the West of North , placing it in the North-West (NW) quadrant . Key Concept: To find the components of the line, we break it into: 1. Latitude – This is the North-South component. 2. Departure – This is the East-West component. Signs: Latitude is positive (+) if the line goes North, negative (-) if South. Departure is positive (+) if the line goes East, negative (-) if West. Formulas: Latitude = Length × cos(angle) Departure = Length × sin(angle) Given: Length = 100 m Bearing = N 60° W (angle = 60°) Step 1: Calculate Latitude Latitude = 100 × cos(60°) = 100 × 0.5 = +50 m (positive because direction is North) Step 2: Calculate Departure Departure = 100 × sin(60°) = 100 × 0.866 = 86.6 m Since the direction is West, it is negative: -86.6 m Final Answer: Latitude = +50 m Departure = -86.6 m Correct Answer: Option C: +50 m, -86.6 m

In levelling, it is essential that the staff is held vertically at the observation point. If the staff is tilted (not vertical), the horizontal distance between the instrument and the staff increases slightly. This results in the levelling instrument recording a reading higher than it would if the staff were vertical. Since the reading is higher, the calculated Reduced Level (R.L.) becomes lower than the actual or true R.L. Therefore, a non-vertical staff leads to a reduced level that is less than the true reduced level. Hence, the correct answer is Option C: less than true R.L.

The station pointer is an instrument commonly used in hydrographical surveying. It is used to plot the position of a vessel by observing angles between three known points, such as landmarks, on a chart. This method helps in determining the precise location of the survey station in relation to the known points on the shoreline or other fixed objects.

Latitude is the angular distance north or south of the equator. It is measured from the equator (0 degrees latitude) to the poles (90 degrees latitude north or south). Let's analyze each option: Option A: "Greenwich to the place" - This is incorrect because Greenwich is associated with longitude (prime meridian), not latitude. Option B: "Equator to the poles" - This is partially correct in describing latitude, but the specific measurement is from the equator to the poles, not just "to the poles." Option C: "Equator to the nearer pole" - This is correct because latitude measures the angular distance from the equator to either the North Pole (90 degrees north) or the South Pole (90 degrees south), depending on the location. Option D: "None of these" - This is incorrect as Option C accurately describes the angular distance that latitude measures. Therefore, the correct answer is Option C: Equator to the nearer pole .

In surveying, the resection by two-point problem as compared to the three-point problem gives more accurate results Resection is a method used in surveying to determine the location of an unknown point by observing known points from that location. In the two-point resection method, observations are made from two known points to determine the position of the unknown point. This method is generally more accurate because it involves observations from two distinct directions, allowing for better triangulation and reducing errors. On the other hand, the three-point resection method involves observations from three known points to determine the position of the unknown point. While it may seem more robust, it can sometimes be less accurate due to the potential for error propagation and the complexity of managing observations from three different directions. Therefore, the correct option is Option A: gives more accurate results .

Tachometric method is the correct answer to “Which of the following methods of contouring is most suitable for a hilly terrain”. This question is standard in competitive examination syllabi for Surveying.

23° 27' is the correct answer to “The angle between the plane of the equator and the plane of the ecliptic, is known as obliquity of the ecliptic and its value is”. This question is standard in competitive examination syllabi for Surveying.

0 and 100 is the correct answer to “For a tacheometer the additive and multi-plying constants are respectively”. This question is standard in competitive examination syllabi for Surveying.

f tan θ is the correct answer to “From the principal point the horizon point lies on the principal line at a distance of”. This question is standard in competitive examination syllabi for Surveying.

13° E is the correct answer to “If the magnetic bearing of the sun at a place at noon in southern hemisphere is 167°, the magnetic declination at that place is”. This question is standard in competitive examination syllabi for Surveying.

transition curve is the correct answer to “Overturning of vehicles on a curve can be avoided by using”. This question is standard in competitive examination syllabi for Surveying.

reading both verniers and taking the mean of the two is the correct answer to “The error due to eccentricity of inner and outer axes can be eliminated by”. This question is standard in competitive examination syllabi for Surveying.

all of the above is the correct answer to “Three point problem can be solved by”. This question is standard in competitive examination syllabi for Surveying.

South of observer is the correct answer to “The zenith is the point on the celestial sphere”. This question is standard in competitive examination syllabi for Surveying.

all of the above is the correct answer to “The allowable length of an offset depends upon the”. This question is standard in competitive examination syllabi for Surveying.

All the above is the correct answer to “The great circle whose plane is perpendicular to the axis of rotation of the earth, is called”. This question is standard in competitive examination syllabi for Surveying.

All the above is the correct answer to “The plane at right angle to the zenith-nadir line and passing through the centre of the earth, is called”. This question is standard in competitive examination syllabi for Surveying.

6000 m is the correct answer to “To obtain photographs of an area of 1000 m average elevation, on scale 1 : 30,000, with a camera of 30 cm focal length, the flying height is”. This question is standard in competitive examination syllabi for Surveying.

12,000 m is the correct answer to “The difference of height of two points whose parallax difference is 0.8 mm on a pair of stereo pair taken from a height ‘H’ is 100 m. If mean photo base is 95.2 mm, the flying height is”. This question is standard in competitive examination syllabi for Surveying.

+ 8".8 cos α is the correct answer to “The correction for parallax, is”. This question is standard in competitive examination syllabi for Surveying.

1° is the correct answer to “The Polaris describes a small circle round the pole whose radius is approximately”. This question is standard in competitive examination syllabi for Surveying.

radius of level tube is the correct answer to “Sensitiveness of a level tube is designated by”. This question is standard in competitive examination syllabi for Surveying.

prismatic compass is the correct answer to “Agate cap is fitted with a”. This question is standard in competitive examination syllabi for Surveying.

180° longitude is the correct answer to “International Date Line is located along”. This question is standard in competitive examination syllabi for Surveying.

outside of one handle to outside of other handle is the correct answer to “The length of a chain is measured from”. This question is standard in competitive examination syllabi for Surveying.

500 km is the correct answer to “If the equatorial distance between two meridians is 100 km, their distance at 60° latitude will be”. This question is standard in competitive examination syllabi for Surveying.

1 : 20,000 is the correct answer to “The scale of the photography taken from a height of 300 m, with a camera of focal length 15 cm, is”. This question is standard in competitive examination syllabi for Surveying.

concave is the correct answer to “In an internal focussing type of telescope, the lens provided is”. This question is standard in competitive examination syllabi for Surveying.

none of the above is the correct answer to “Which of the following is not the function of levelling head”. This question is standard in competitive examination syllabi for Surveying.

presence of magnetic substances near the instrument is the correct answer to “Local attraction in compass surveying may exist due to”. This question is standard in competitive examination syllabi for Surveying.

All the above is the correct answer to “A Nautical mile is”. This question is standard in competitive examination syllabi for Surveying.