Question 1 |
For a given traverse, latitudes and departures are calculated and it is found that sum of latitudes is equal to +2.1 m and the sum of departures is equal to -2.8 m. The length and bearing of the closing error, respectively, are
3.50 \mathrm{~m} \text { and } 53^{\circ} 7^{\prime} 48^{\prime \prime} \mathrm{NW} | |
2.45 \mathrm{~m} \text { and } 53^{\circ} 7^{\prime} 48^{\prime \prime} \text { NW } | |
0.35 \mathrm{~m} \text { and } 53.13^{\circ} \mathrm{SE} | |
3.50 \mathrm{~m} \text { and } 53.13^{\circ} \mathrm{SE} |
Question 1 Explanation:
\begin{aligned} e_{L} &=+2.1 \mathrm{~m} \\ e_{D} &=-2.8 \mathrm{~m} \\ e &=\sqrt{e_{L}^{2}+e_{D}^{2}} \\ &=\sqrt{(2.1)^{2}+(2.8)^{2}}=3.5 \mathrm{~m} \\ \text { Bearing of closing error } &=\tan ^{-1}\left(\frac{e_{D}}{e_{L}}\right) \\ &=\tan ^{-1}\left(\frac{-2.8}{2.1}\right)=-53.13^{\circ} \\ &=53^{\circ} 7^{\prime} 48^{\prime \prime} \mathrm{NW} \end{aligned}
Question 2 |
A horizontal angle \theta is measured by four different surveyors multiple times and the values reported are given below.
\begin{array}{|c|c|c|} \hline \text { Surveyor } & \text { Angle } \theta & \text { Number of observations } \\ \hline 1 & 36^{\circ} 30^{\prime} & 4 \\ \hline 2 & 36^{\circ} 00^{\prime} & 3 \\ \hline 3 & 35^{\circ} 30^{\prime} & 8 \\ \hline 4 & 36^{\circ} 30^{\prime} & 4 \\ \hline \end{array}
he most probable value of the angle \theta ( in degree, round off to two decimal placesis ________
\begin{array}{|c|c|c|} \hline \text { Surveyor } & \text { Angle } \theta & \text { Number of observations } \\ \hline 1 & 36^{\circ} 30^{\prime} & 4 \\ \hline 2 & 36^{\circ} 00^{\prime} & 3 \\ \hline 3 & 35^{\circ} 30^{\prime} & 8 \\ \hline 4 & 36^{\circ} 30^{\prime} & 4 \\ \hline \end{array}
he most probable value of the angle \theta ( in degree, round off to two decimal placesis ________
12 | |
28 | |
36 | |
44 |
Question 2 Explanation:
\begin{aligned} \mathrm{MPV} &=\frac{\left(36^{\circ} 30^{\prime} \times 4\right)+\left(36^{\circ} \times 3\right)+\left(35^{\circ} 30^{\prime} \times 8\right)+\left(36^{\circ} 30^{\prime} \times 4\right)}{4+3+8+4} \\ &=36^{\circ} \end{aligned}
Question 3 |
In general, the CORRECT sequence of surveying operations is
Field observations\rightarrow
Reconnaissance\rightarrow
Data analysis\rightarrow
Map making | |
Data analysis\rightarrow
Reconnaissance\rightarrow
Field observations \rightarrow
Map making | |
Reconnaissance\rightarrow
Field observations \rightarrow
Data analysis \rightarrow
Map making | |
Reconnaissance\rightarrow
Data analysis \rightarrow
Field observations \rightarrow
Map making |
Question 3 Explanation:
Reconnaissance\rightarrowField observations\rightarrowData analysis\rightarrowMap making
Question 4 |
Traversing is carried out for a closed traverse PQRS. The internal angles at vertices P, Q, R and S are measured as 92^{\circ} ,62^{\circ} ,123^{\circ} and 77^{\circ} , respectively. If fore bearing of line PQ is 27^{\circ} , fore bearing of line RS (in degrees, in integer) is _________
258 | |
753 | |
159 | |
218 |
Question 4 Explanation:
\begin{array}{l} Q=\left[\begin{array}{l} B B \text { of } P Q=27^{\circ}+180^{\circ}=207^{\circ} \\ F B \text { of } Q R=207^{\circ}-68^{\circ}=139^{\circ} \end{array}\right. \\ R=\left[\begin{array}{l} B B \text { of } Q R=139^{\circ}+180^{\circ}=319^{\circ} \\ F B \text { of } R S=319^{\circ}-123^{\circ}=196^{\circ} \end{array}\right. \end{array}
\begin{array}{l} Q=\left[\begin{array}{l} B B \text { of } P Q=27^{\circ}+180^{\circ}=207^{\circ} \\ F B \text { of } Q R=207^{\circ}+68^{\circ}=275^{\circ} \end{array}\right. \\ R=\left[\begin{array}{l} B B \text { of } Q R=275^{\circ}-180^{\circ}=95^{\circ} \\ F B \text { of } R S=95^{\circ}+123^{\circ}=218^{\circ} \end{array}\right. \end{array}
Question 5 |
Which of the following is/are correct statement(s)?
Back Bearing of a line is equal to Fore Bearing \pm 180^{\circ} | |
If the whole circle bearing of a line is 270^{\circ}, its reduced bearing is 90^{\circ} \mathrm{NW} | |
The boundary of water of a calm water pond will represent contour line | |
In the case of fixed hair stadia tachometry, the staff intercept will be larger, when the staff is held nearer to the observation point |
Question 5 Explanation:
The principal of fixed hair tacheometry is that distances are proportional to staff intercept.
As distance increase, staff intercept also increases.
As distance increase, staff intercept also increases.
Question 6 |
Which of the following is NOT a correct statement?
The first reading from a level station is a 'Fore Sight' | |
Basic principle of surveying is to work from whole to parts | |
Contours of different elevations may intersect each other in case of an overhanging cliff | |
Planimeter is used for measuring 'area' |
Question 6 Explanation:
First reading from level station is called BS.
Question 7 |
A theodolite is set up at station A. The RL of instrument axis is 212.250 m. The angle
of elevation to the top of a 4 m long staff, held vertical at station B, is 7^{\circ}. The horizontal
distance between station A and B is 400 m. Neglecting the errors due to curvature of
earth and refraction, the RL (in m, round off to three decimal places) of station B is
__________
257.363 | |
145.126 | |
472.156 | |
324.422 |
Question 7 Explanation:
\begin{aligned} V&=400 \tan 7^{\circ}\\ &=49.113\\ x&=(49.113-4)=45.113\\ RL_B&=212.25+45.113\\ &=257.363m \end{aligned}
Question 8 |
The diameter and height of a right circular cylinder are 3 cm and 4 cm, respectively.
The absolute error in each of these two measurements is 0.2 cm. The absolute error in
the computed volume (in cm^3, round off to three decimal places), is _______.
1.65 | |
7.52 | |
3.25 | |
5.18 |
Question 8 Explanation:
Let diameter, x = 3 and height = y = 4 and error= \pm 0.2
\begin{aligned} V&= \pi\left ( \frac{x}{2} \right )^2y=\frac{\pi x^2 y}{4}\\ V&=f(x,y) \\ dV&= \left ( \frac{\partial V}{\partial x} \right )dx +\left ( \frac{\partial V}{\partial y} \right )dy\\ dV&= \left ( \frac{1}{2}\pi xy \right )dx +\left ( \frac{\pi x^2}{4} \right )dy\\ &= \frac{1}{2} \pi \times 3 \times 4 \times (0.2)+\frac{\pi}{4} \times (3)^2 \times (0.2)\\ &= 1.65 \pi\\ &=1.65 \times 3.14=5.18 \end{aligned}
i.e., absolute error = |5.18| = 5.18
\begin{aligned} V&= \pi\left ( \frac{x}{2} \right )^2y=\frac{\pi x^2 y}{4}\\ V&=f(x,y) \\ dV&= \left ( \frac{\partial V}{\partial x} \right )dx +\left ( \frac{\partial V}{\partial y} \right )dy\\ dV&= \left ( \frac{1}{2}\pi xy \right )dx +\left ( \frac{\pi x^2}{4} \right )dy\\ &= \frac{1}{2} \pi \times 3 \times 4 \times (0.2)+\frac{\pi}{4} \times (3)^2 \times (0.2)\\ &= 1.65 \pi\\ &=1.65 \times 3.14=5.18 \end{aligned}
i.e., absolute error = |5.18| = 5.18
Question 9 |
A theodolite was set up at a station P. The angle of depression to a vane 2 m above
the foot of a staff held at another station Q was 45^{\circ}. The horizontal distance between
stations P and Q is 20 m. The staff reading at a benchmark S of RL 433.050 m
is 2.905 m. Neglecting the errors due to curvature and refraction, the RL of the station Q (in m), is
413.05 | |
413.955 | |
431.05 | |
435.955 |
Question 9 Explanation:
\begin{aligned} \frac{x}{20}&=\tan 45^{\circ}\\ x&=20m\\ RL \; of \; Q&=433.05+2.905-x-2\\ &=433.05+2.905-20-2\\ &=413.955m \end{aligned}
Question 10 |
The length and bearings of a traverse PQRS are:
The length of line segment SP (in m, round off to two decimal places), is ________.
The length of line segment SP (in m, round off to two decimal places), is ________.
33.07 | |
25.36 | |
47.78 | |
44.79 |
Question 10 Explanation:
\begin{aligned} \Delta L&= 40 \cos 80^{\circ}+50 \cos 10^{\circ}+30 \cos 210^{\circ}\\ &=30.20 \\ \Delta D&=40 \sin 80^{\circ}+50 \sin 10^{\circ}+30 \sin 210^{\circ} \\ &=33.07 \\ \text{Length, \; SP}&=\sqrt{\Delta L^2+\Delta D^2}\\ &=44.79m \end{aligned}
There are 10 questions to complete.
