Question 1 |

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 1 Explanation:

Reconnaissance\rightarrowField observations\rightarrowData analysis\rightarrowMap making

Question 2 |

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 2 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 3 |

In a survey work, three independent angle, X, Y and Z were observed with weight W_{X},W_{Y},W_{Z} respectively. The weight of the sum of angles X, Y and Z is given by:

1/(\frac{1}{W_{X}}+\frac{1}{W_{Y}}+\frac{1}{W_{Z}}) | |

(\frac{1}{W_{X}}+\frac{1}{W_{Y}}+\frac{1}{W_{Z}}) | |

W_{X}+W_{Y}+W_{Z} | |

W_{X}^{2}+W_{Y}^{2}+W_{Z}^{2} |

Question 4 |

Which of the following statements is FALSE?

Plumb line is along the direction of gravity | |

Mean Sea Level (MSL) is used as a reference surface for establishing the horizontal control | |

Mean Sea Level (MSL) is a simplification of Geoid | |

Geoid is an equi-potential surface of gravity |

Question 4 Explanation:

Mean sea level (MSL) is used as a reference surface for establishing the vertical control.

Question 5 |

The survey carried out to delineate natural features,such as hills, rivers, forests and manmade features, suchas towns, villages, buildings, roads, transmission lines
and canals is classified as

engineering survey | |

geological survey | |

land survey | |

topographic survey |

Question 5 Explanation:

Tropographic survey is done to determine the natural features of a country such as rivers, strains, lakes, wood, hills etc. and artificial features such as roads, railways, canals, towns and villages.

Question 6 |

The plan of a survey plotted to a scale of 10 m to 1 cm is reduced in such a way
that a line originally 10 cm long now measures 9 cm. The area of the reduced plan
is measured as 81 cm^{2}. The actual area (m^{2}) of the survey is

10000 | |

6561 | |

1000 | |

656 |

Question 6 Explanation:

Shrinking factor = \frac{9}{10}=0.9

Reduced plan area =(\text { Shrinkage factor })^{2}

\times Actual plan area

\Rightarrow \quad 81=(0.9)^{2} \times Actual plan area

\Rightarrow Actual plan area =100 \mathrm{cm}^{2}

\therefore \quad

Actual area of survey in \mathrm{m}^{2}

\begin{aligned} &=100 \times(10)^{2} \\ &=10000 \end{aligned}

Reduced plan area =(\text { Shrinkage factor })^{2}

\times Actual plan area

\Rightarrow \quad 81=(0.9)^{2} \times Actual plan area

\Rightarrow Actual plan area =100 \mathrm{cm}^{2}

\therefore \quad

Actual area of survey in \mathrm{m}^{2}

\begin{aligned} &=100 \times(10)^{2} \\ &=10000 \end{aligned}

Question 7 |

The type of surveying in which the curvature of the earth is taken into account
is called

Geodetic surveying | |

Plane surveying | |

Preliminary surveying | |

Topographical surveying |

Question 7 Explanation:

Geodetic surveying is that type of surveying in which the shape of the earth is taken into account.
Geodetic surveying involves spherical
trigonometry.

Question 8 |

The plan of a map was photo copied to a reduced size such that a line originally
100 mm, measures 90 mm. The original scale of the plan was 1:1000. The revised scale is

1:900 | |

1:1111 | |

1:1121 | |

1:1221 |

Question 8 Explanation:

\begin{aligned} \text { Reduction factor }&=\frac{90}{100}=0.9 \\ \text { Revised scale }&=\text { Original scale }\\ &\times \text { Reduction factor }\\ =\frac{1}{1000} \times 0.9&=\frac{1}{1000 / 0.9}=\frac{1}{1111} \end{aligned}

There are 8 questions to complete.