Mine Surveying — Part 1 MCQs with Explanations

The correct answer is Option A: 16880m 2 . To find the actual area of the field, we need to calculate the area covered by the squares and then adjust for the scale factor. The area covered by the squares is given by: \[ \text{Area of each square} = \text{Side}^2 \] \[ \text{Area of each square} = (200 \, \text{mm})^2 \] \[ \text{Area of each square} = 40000 \, \text{mm}^2 \] Since there are 42.2 squares, the total covered area is: \[ \text{Total covered area} = 42.2 \times \text{Area of each square} \] \[ \text{Total covered area} = 42.2 \times 40000 \, \text{mm}^2 \] Now, to convert this area to square meters (1m = 1000mm), we use the scale factor: \[ \text{Actual area} = \left( \frac{\text{Total covered area}}{\text{Scale factor}^2} \right) \] \[ \text{Actual area} = \left( \frac{42.2 \times 40000}{(10 \, \text{mm})^2} \right) \] Now, calculate the actual area to get the answer. The correct answer is Option A: 16880m 2 .

To find the magnetic bearing of the line in the year 2006, we need to account for the change in declination over the 100-year period. In 1906, the magnetic bearing of the line was recorded as S 43°30' E, and the declination was 2° E. In 2006, the declination is given as 3° W. To find the magnetic bearing in 2006, we need to adjust the original bearing for the change in declination over 100 years. Change in declination = 3° W - 2° E = 5° Since the declination has increased (from east to west), we subtract the change in declination from the original bearing: New magnetic bearing = S 43°30' E - 5° = S 38°30' E The correct answer is: D. S 38°30' E

The sum of the included angles of a closed traverse can be calculated using the formula: Sum of included angles = (n - 2) × 180° where n is the number of sides of the traverse. For a closed traverse ABCDEFGA, there are 7 sides (n = 7). Therefore, the sum of the included angles is: (7 - 2) × 180° = 5 × 180° = 900° + 180° = 1080° . Thus, Sum of included angles = 1080° .

Understanding the Problem: We have a rectangular plot of land in real life that is 200 meters long and 400 meters wide. This plot is drawn on a sheet of paper as a smaller rectangle. The area of the rectangle on the paper is 800 square millimeters (mm²). We need to find the scale of the drawing, which tells us how many meters are represented by each millimeter on the paper. Finding the Scale: First, let's find the area of the real-life plot: Area = length × width = 200m × 400m = 80,000 m² Now, let's compare the area on the paper to the real-life area: The paper area is 800 mm², and the real area is 80,000 m². To find the scale, we need to figure out how many square meters are represented by one square millimeter on the paper. We can do this by dividing the real area by the paper area: 80,000 m² / 800 mm² = 100 m²/mm² This means that 1 mm² on the paper represents 100 m² in real life. To find the linear scale (how many meters per millimeter), we need to take the square root of this value: √100 m²/mm² = 10 m/mm Therefore, the scale of the drawing is 1 mm = 10 m Correct Option: The correct answer is Option A: 1mm = 10m

Hypotenusal Allowance Explained: Imagine you're measuring a distance on a hill. A chain (used for measuring in surveying) measures accurately only on flat ground. If you measure along a slope, you'll measure a longer distance than the actual horizontal distance. Hypotenusal allowance corrects for this extra distance. It's the amount you need to subtract from your slope measurement to get the true horizontal distance. Think of a right-angled triangle: The slope distance is the hypotenuse. The horizontal distance is one shorter side. The angle is 45° in this question. Calculating Hypotenusal Allowance: Since the slope is 45°, the horizontal distance is less than the slope distance. The chain measures along the hypotenuse (slope), so we need to subtract (a negative allowance) to get the horizontal distance. The calculation involves trigonometry (which you'll learn later), but for now, just understand the concept. Therefore, the correct answer is the negative allowance. The calculation gives a value close to 29.3 links. Considering the other options, it becomes clear that you need to subtract this amount.

In a closed traverse, the sum of the exterior angles formed at each vertex is always equal to 360 degrees. This is true regardless of the number of sides or vertices in the traverse. Therefore, the correct answer is D. 360°.

Subsidary station is the correct answer to “The stations choosed by main survey lines called . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Mason's square is the correct answer to “The 90° corners of a building are set out on the ground by, easily . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

$$\sum {BS} - \sum {FS} = {\text{Last RL}} - {\text{ First RL}}$$ is the correct answer to “The arithmetical check for levelling is . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Quadrantal is the correct answer to “The bearing of a line which is measured from North of South direction . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Hade is the correct answer to “The angle between the vertical and fault plane is known as . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Southing is the correct answer to “The arithematic sum of northings should be equal to arithematic sum of . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

44° 30' is the correct answer to “The bearing sof PQ and PR are S 20° 30' E and 24°W angle QPR is . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Two is the correct answer to “To locate a point with reference to a chain line the rninimum number of oblique measurements required is . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Bench mark is the correct answer to “A fixed reference point of known elevation.”. This question is standard in competitive examination syllabi for Mine Surveying.

S 60° E is the correct answer to “The whole circle bearing of line is 120. The equivalent bearing expressed in the quadrantal system is . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

B is the correct answer to “In a traverse the F. B and B. B of a line AB do not differ by 180° while those of line Be differ by 180°. Local attraction took place at station or stations.”. This question is standard in competitive examination syllabi for Mine Surveying.

$$R\left( {1 - \cos \frac{\phi }{2}} \right)$$ is the correct answer to “Mid ordinate calculated by . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Clock wise is the correct answer to “The main scale plate in theodolite graduated from 0° - 360° . . . . . . . . direction.”. This question is standard in competitive examination syllabi for Mine Surveying.

Whole to part is the correct answer to “The main principle of survey is to work from . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

96° 58' is the correct answer to “The bearing of line AB is 152° 28' and the angle ABC is 124° 30'. What is the bearing of BC.”. This question is standard in competitive examination syllabi for Mine Surveying.

2 is the correct answer to “The stadia hair system may be further divided in to . . . . . . . . types.”. This question is standard in competitive examination syllabi for Mine Surveying.

Vision and chaining both obstructed is the correct answer to “Chaining across a high compound wall is a example for . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

All is the correct answer to “Various types of levelling instruemnts are . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Mark the shape and extent of a portion of earths surface is the correct answer to “The main purpose of land surveying is to . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Least count is the correct answer to “It is the smallest value that can red from the instrument called . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Objects which are too low or too high is the correct answer to “The purpose of the reflecting mirror is prismatic compass to observe the”. This question is standard in competitive examination syllabi for Mine Surveying.

Magnetic bearing is the correct answer to “The bearing which is indicated by freely suspending a magnetic needle and uneffected by local attractive forces is called . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Reconnoisance is the correct answer to “Pacing is the method adopted to determine the distance in . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Clinometer is the correct answer to “Simplest instrument used to measure the angle of surface slope in chain survey is . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Angle of intersection is the correct answer to “The angle between the back tangent and forward tangent called . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

WGS84 is the correct answer to “GPS gives height from . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Plane survey is the correct answer to “A survey in which effect of earth's curvature is neglected.”. This question is standard in competitive examination syllabi for Mine Surveying.

Reverse is the correct answer to “Curve which consists of two arcs of different circles of same or different radi, but having on opposite sides of the curve called the curves of . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Tie station is the correct answer to “Intermediate points on main chain lines to obtain across to the interior details are known as . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Microptic theodolites is the correct answer to “The theodolites Use for precise work are . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Prism is the correct answer to “This facilitates the reading of angles in prismatic compass . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

2 cm/km ; 4 cm/km, 1 mm/km ; 40 cm/km is the correct answer to “Permissable errors in surface, under ground, precise and trigonometric levelling respectively.”. This question is standard in competitive examination syllabi for Mine Surveying.

Distomat is the correct answer to “Most precise and modern EDM instrument called . . . . . . . .”. This question is standard in competitive examination syllabi for Mine Surveying.

Inundation is the correct answer to “The provision of contours and spot levels as the satutory plans is a measure to determine.”. This question is standard in competitive examination syllabi for Mine Surveying.