Principles and Classification of Surveying Questions and Answers

Answer: A

The instrument is the passometer.

A passometer is a watch-shaped instrument carried in the pocket or strapped to the leg. A pendulum inside moves with each step, and the instrument automatically registers the number of paces. Multiplying by the length of the surveyor's pace gives an approximate distance, which is useful for reconnaissance.

A pedometer works similarly but is set to the length of the pace and shows distance directly rather than the count of paces. An odometer registers the revolutions of a wheel and so measures distance travelled by a wheel. A speedometer shows speed, not paces.

Answer: A

The scale 1/1000 is 1cm=10m.

A representative fraction of 1/1000 means one unit on the map stands for 1000 of the same units on the ground. So 1 cm on paper represents 1000 cm, and

\[ 1000 \text{ cm} = 10 \text{ m} \]

giving 1 cm = 10 m.

1 cm = 100 m is the scale 1/10000. 1 cm = 100 mm is 1/10, and 1 cm = 100 cm is 1/100. Each is out by a power of ten, a common slip when converting between centimetres and metres.

Answer: A

A passometer measures the number of paces in a given distance.

The passometer counts each step automatically as the surveyor walks. Multiplying the count by the average length of the surveyor's pace gives a rough distance, which is good enough for reconnaissance and small-scale surveys.

It does not measure steps in a vertical plane, nor count obstacles. Counting the revolutions of a wheel over a distance is the job of an odometer or a perambulator (measuring wheel), not a passometer.

Answer: A

The U-fork and plumb bob are used for Centering.

In plane tabling, the U-fork (plumbing fork) has one arm placed on the drawing sheet at the plotted point and the other below the table carrying the plumb bob. Moving the table until the bob hangs over the ground station brings the plotted point exactly above it.

Levelling is done with a spirit level, orientation by back-sighting or with a trough compass, and the north direction is marked with the trough compass, not the U-fork.

Answer: C

The key gives Shrinkage factor = Calculated length / Measured length on the field.

When a map shrinks, a distance scaled off it with the original scale (the calculated length) comes out shorter than the true ground distance. The shrinkage factor is the ratio of the shrunk value to the true one, so it is less than 1: the calculated length divided by the length actually measured on the field. Dividing a calculated length by this factor recovers the true length, which is why the graphical scale, which shrinks with the paper, is printed alongside the RF.

The inverted ratio would exceed 1, and a product of the two lengths, or its reciprocal, has the dimensions of area and cannot be a dimensionless factor.

Answer: A

The most probable error is 6.745 m.

For normally distributed errors, the probable error (the value with an even chance of being exceeded) is 0.6745 times the standard deviation:

\[ \begin{aligned} E_p &= 0.6745\,\sigma \\ &= 0.6745 \times 10 = 6.745\ \text{m} \end{aligned} \]

0.6745 m is the multiplier applied to a standard deviation of 1 m instead of 10 m. 10 m is the standard deviation itself, and 20 m is twice it, roughly the 95% confidence limit, not the probable error.

Answer: A

The smallest scale is 1cm = 10 km.

A scale is 'small' when its representative fraction is small. Convert each to an RF:

• 1 cm = 10 km = 1,000,000 cm, so RF = 1/1,000,000
• 1 cm = 10 m = 1,000 cm, so RF = 1/1,000
• RF = 1/5,000
• 1 : 10,000, so RF = 1/10,000

1/1,000,000 is the smallest fraction, so 1 cm = 10 km is the smallest scale. The trap is to pick 1 : 10,000 because it has the largest number written down; the other two are larger scales still.

Answer: A

The Board's key is \(1/\sqrt{n}\), the standard result for the accidental error of the mean of n observations.

Accidental (random) errors are as likely to be positive as negative, so they partly cancel. For n independent observations of equal precision, each with standard error \(\sigma\):

\[ \begin{aligned} \sigma_{\text{sum}} &= \sigma\sqrt{n} \\ \sigma_{\text{mean}} &= \frac{\sigma\sqrt{n}}{n} = \frac{\sigma}{\sqrt{n}} \end{aligned} \]

So the accidental error in the average of the readings falls as \(1/\sqrt{n}\), which is why repeating a measurement improves it. The question does not say whether the total or the mean is meant. The error in the total of n observations grows as \(\sqrt{n}\), which is option C, and some books quote that form. The key takes the error of the result, which falls as \(1/\sqrt{n}\).

Option B (n) would suit a systematic error, which adds up in full with every reading. Option D (1/n) would make random error vanish as fast as the readings multiply, which does not happen.

Answer: D

The process is resection.

In plane tabling, resection locates on the sheet the point where the table is actually standing, by sighting known points already plotted and drawing back rays through them. The two-point and three-point problems are particular methods of resection.

Intersection does the opposite: it fixes the position of a distant, unvisited point from two known stations. Traversing runs a series of connected lines through stations. The two-point problem is one way of solving a resection, not the general name of the process.