Question 1 |

The magnetic bearing of the sun for a location at noon is 183^{\circ} 30^{\prime}. If the sun is exactly on the geographic meridian at noon, the magnetic declination of the location is ___

3^{\circ} 30^{\prime} \mathrm{W} | |

3^{\circ} 30^{\prime} \mathrm{E} | |

93^{\circ} 30^{\prime} \mathrm{W} | |

93^{\circ} 30^{\prime} \mathrm{E} |

Question 1 Explanation:

At any place 12 O'clock sun will be exactly over the true meridian of that place.

Declination =183^{\circ} 30^{\prime}-180^{\circ}=3^{\circ} 30^{\prime} \mathrm{W}

Magnetic declination will be 3^{\circ} 30 \mathrm{~W}.

Declination =183^{\circ} 30^{\prime}-180^{\circ}=3^{\circ} 30^{\prime} \mathrm{W}

Magnetic declination will be 3^{\circ} 30 \mathrm{~W}.

Question 2 |

A student is scanning his 10 inch \times 10 inch certificate at 600 dots per inch (dpi) to convert it to raster. What is the percentage reduction in number of pixels if the same certificate is scanned at 300 dpi?

62 | |

88 | |

75 | |

50 |

Question 2 Explanation:

DPI = Dots Per Inch, it should be noted that DPI is not dots per square inch.

\% reduction in number of pixels

=\frac{10 \times 600 \times 10 \times 600-100 \times 300 \times 10 \times 300}{10 \times 600 \times 10 \times 600} \times 100

=\frac{600^{2}-300^{2}}{600^{2}} \times 100

=75 \%

\% reduction in number of pixels

=\frac{10 \times 600 \times 10 \times 600-100 \times 300 \times 10 \times 300}{10 \times 600 \times 10 \times 600} \times 100

=\frac{600^{2}-300^{2}}{600^{2}} \times 100

=75 \%

Question 3 |

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 3 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 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 |

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 5 Explanation:

Reconnaissance\rightarrowField observations\rightarrowData analysis\rightarrowMap making

There are 5 questions to complete.