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} ,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 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}

Question 6 |

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 6 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 7 |

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 7 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 8 |

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 8 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}

Question 9 |

An open traverse PQRST is surveyed using theodolite and the consecutive coordinates
obtained are given in the table

If the independent coordinates (Northing, Easting) of station P are (400 m, 200 m) the independent coordinates (in m) of station T, are

If the independent coordinates (Northing, Easting) of station P are (400 m, 200 m) the independent coordinates (in m) of station T, are

194.7, 370.1 | |

205.3, 429.9 | |

394.7, 170.1 | |

405.3, 229.9 |

Question 9 Explanation:

\begin{aligned} \Delta L &=-5.3 \\ \Delta D&=-29.9 \\ T, \; \text{Northing}&\{400+(-5.3)\}=394.7 \\ T, \; \text{Easting}&\{200+(-29.9)\}=170.1\\ &T[394.7m, 170.1m] \end{aligned}

Question 10 |

The data from a closed traverse survey PQRS (run in the clockwise direction) are given in the table

The closing error for the traverse PQRS (in degrees) is ___

The closing error for the traverse PQRS (in degrees) is ___

2 | |

3 | |

4 | |

5 |

Question 10 Explanation:

Assuming it as anticlockwise traverse.

Mathematically sum of interior angle for a closed traverse =(2n-4)\times 90 = (2 \times 4 -4) \times 90 = 360^{\circ}

Given sum of interior angles, = 88 + 92 + 94 + 89 = 363^{\circ}

Then error in interior angle = 363 - 360 = 3^{\circ}

Note: In this question as per clockwise traverse included angle should be taken as exterior angle. But if we take exterior angle then we get all interior angles more than 180^{\circ}.

Mathematically sum of interior angle for a closed traverse =(2n-4)\times 90 = (2 \times 4 -4) \times 90 = 360^{\circ}

Given sum of interior angles, = 88 + 92 + 94 + 89 = 363^{\circ}

Then error in interior angle = 363 - 360 = 3^{\circ}

Note: In this question as per clockwise traverse included angle should be taken as exterior angle. But if we take exterior angle then we get all interior angles more than 180^{\circ}.

There are 10 questions to complete.

Qno. 10 diagramn is wrong

Thanks for ur effort