Question 1 |

The error in measuring the radius of a 5 cm circular rod was 0.2%. If the
cross-sectional area of the rod was calculated using this measurement, then the
resulting absolute percentage error in the computed area is______.
(round off to two decimal places)

0.25 | |

0.40 | |

0.67 | |

0.83 |

Question 1 Explanation:

\begin{aligned}
r&=5 \\
e_r&=\frac{0.2}{100} \times 5=0.01cm \\
A&=\pi r^2 \\
e_A&=2 \pi r.e_r
\end{aligned}

Absolute perecentage error in computed area

\begin{aligned} &=\frac{e_A}{A} \times 100 \\ &=\frac{2 \pi r.e_r}{\pi r^2} \times 100\\ &=2 \times \left ( \frac{e_r}{r} \times 100 \right )\\ &=2 \times 0.2=0.4 \end{aligned}

Absolute perecentage error in computed area

\begin{aligned} &=\frac{e_A}{A} \times 100 \\ &=\frac{2 \pi r.e_r}{\pi r^2} \times 100\\ &=2 \times \left ( \frac{e_r}{r} \times 100 \right )\\ &=2 \times 0.2=0.4 \end{aligned}

Question 2 |

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 2 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 3 |

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 3 Explanation:

WCB=360-\left ( 31+\frac{17}{60} \right )^{\circ}=329.716^{\circ}

Question 4 |

A line between stations P and Q laid on a slope of 1 in 5 was measured as 350 m using a 50 m tape. The tape is known to be short by 0.1 m.

The corrected horizontal length (in m) of the line PQ will be

The corrected horizontal length (in m) of the line PQ will be

342.52 | |

349.3 | |

356.2 | |

350.7 |

Question 4 Explanation:

Horizontal distance of line

PQ=350 \cos \theta =\frac{350 \times 5}{\sqrt{26}}=343.20m

Tape is 0.1 m short.

Nominal length of tape, l = 50 m

Actual length of tape, l' = 50 - 0.1 = 49.9 m

Corrected horizontal length of line PQ

\begin{aligned} &=\left ( \frac{l'}{l} \right ) \times 343.2\\ &=\left ( \frac{49.9}{50} \right )\times 343.2\\ &=342.517\simeq 342.52m \end{aligned}

Question 5 |

An aerial photograph is taken from a flight at a height of 3.5 km above mean
sea level, using a camera of focal length 152 mm. If the average ground
elevation is 460 m above mean sea level, then the scale of the photograph is

1 : 20000 | |

01:20 | |

1 : 100000 | |

1.986111111 |

Question 5 Explanation:

\begin{aligned}
H &=3.5Km=3500m\\
f&=152mm \\
h_{avg}&=460m \\
Scale&=\frac{f}{H-h_{avg}} \\
&= \frac{152 \times 10^{-3}}{3500-460}\\
&=\frac{1}{20000}
\end{aligned}

Question 6 |

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 6 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 7 |

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 7 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 8 |

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 8 Explanation:

Reconnaissance\rightarrowField observations\rightarrowData analysis\rightarrowMap making

Question 9 |

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 9 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 10 |

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 10 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.

There are 10 questions to complete.

Thankyou very much for providing Pyq s with neat explanation…