Question 1 |
If the magnetic bearing of the Sun at a place at noon is S2^{\circ}E, the magnetic declination (in degrees) at that place is
2^{\circ}E | |
2^{\circ}W | |
4^{\circ}E | |
4^{\circ}W |
Question 1 Explanation:
MB=S2^{\circ}E=180^{\circ}-2^{\circ}=178^{\circ}
TB= 180^{\circ}
Declination, \delta =TB-MB=180-178=2^{\circ} \; or \; 2^{\circ}E
TB= 180^{\circ}
Declination, \delta =TB-MB=180-178=2^{\circ} \; or \; 2^{\circ}E
Question 2 |
The bearing of a survey line is N31^{\circ}17'W. Its azimuth observed from north is
______ deg. (round off to two decimal places)
328.71 | |
458.25 | |
124.65 | |
625.25 |
Question 2 Explanation:

WCB=360-\left ( 31+\frac{17}{60} \right )^{\circ}=329.716^{\circ}
Question 3 |
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 3 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 4 |
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 4 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 5 |
Traversing is carried out for a closed traverse PQRS. The internal angles at vertices P, Q, R and S are measured as 92^{\circ} ,68^{\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 | |
196 | |
159 | |
218 |
Question 5 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}
There are 5 questions to complete.
Qno. 10 diagramn is wrong
Thanks for ur effort