MCQs on Land Surveying

Correct Answer: B. 150 cm

Solution:

According to the Indian Standard IS: 1562-1962, which specifies the details for drafting scales, the 'Diagonal Scale – A' is designed with a total graduated length of 150 cm.

Correct Answer: C. 2

Solution:

The fundamental principle of surveying is to work from the whole to the part. To locate a new point, you need at least two known reference points (control points) to establish its position accurately through methods like triangulation or intersection.

Correct Answer: D. Engineer's scale and representative fraction

Solution:

Both an Engineer's scale (e.g., 1 cm = 10 m) and a Representative Fraction (e.g., 1/1000) are numerical scales. If the paper shrinks, these scales will no longer be accurate. A graphical scale, which is a drawn line on the map, shrinks along with the paper, thus maintaining the correct proportions for measurement.

Correct Answer: D. straight; parallel

Solution:

In plane surveying, the curvature of the Earth is ignored. A level line (a line of constant elevation) is therefore considered a straight line, and plumb lines (lines indicating the direction of gravity) at different points are considered to be parallel to each other.

Correct Answer: C. Diagonal scale

Solution:

A diagonal scale is used to measure three dimensions: a unit, its tenth, and its hundredth. For example, meters, decimeters, and centimeters.

Correct Answer: D. 400 grads

Solution:

The centesimal system divides a full circle (circumference) into 400 grads. Each grad is further divided into 100 centigrads, and each centigrad into 100 milligrads.

Correct Answer: B. 1 / 50000

Solution:

To find the Representative Fraction (RF), both distances must be in the same unit.
1 dm = 0.1 m
5 km = 5000 m
RF = (Distance on map) / (Distance on ground) = 0.1 / 5000 = 1 / 50000.

Correct Answer: C. ii and iii

Solution:

Engineer's scale and Representative Fraction are numerical scales that become inaccurate if the paper shrinks. Plane and Diagonal scales are graphical scales drawn on the map, so they shrink along with the paper, maintaining correct proportions.

Correct Answer: D. i and iii

Solution:

Working from whole to part involves establishing a network of high-precision control points first, and then filling in the details with less precise measurements. This method localizes and prevents the accumulation of errors.

Correct Answer: D. at least two, already fixed points of reference

Solution:

To accurately determine the location of a new point, measurements must be taken from at least two existing, known (fixed) points. This provides the necessary geometric constraints.

Correct Answer: B. whole to part.

Solution:

This is a fundamental principle of all surveying. It ensures that errors are contained within smaller sections and do not accumulate across the entire survey area.

Correct Answer: D. 0.4

Solution:

Ground distance = Map distance × Scale factor = 10 cm × 4000 = 40000 cm.
To convert cm to km, divide by 100,000 (since 1 km = 100,000 cm).
40000 / 100000 = 0.4 km.

Correct Answer: D. 1: 250000

Solution:

Convert both sides to the same unit (cm).
1 km = 100,000 cm, so 2.5 km = 250,000 cm.
The scale is 1 cm = 250,000 cm, so the RF is 1:250000.

Correct Answer: C. 1: 400000

Solution:

Area on map = 1.5 cm × 0.5 cm = 0.75 cm².
Area on ground = 12 km².
1 km = 100,000 cm, so 1 km² = (100,000)² cm² = 10¹⁰ cm².
Area on ground = 12 × 10¹⁰ cm².
Area Scale = Area on map / Area on ground = 0.75 / (12 × 10¹⁰).
Linear Scale = √(Area Scale) = √(0.75 / (12 × 10¹⁰)) = √(1 / (16 × 10¹⁰)) = 1 / (4 × 10⁵) = 1 / 400000.

Correct Answer: D. 0.97

Solution:

The shrinkage factor is the ratio of the shrunk length to the original length.

Shrinkage Factor = Shrunk Length / Original Length

Shrinkage Factor = 9.7 cm / 10 cm = 0.97