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 |
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 2 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 3 |
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 3 Explanation:
Reconnaissance\rightarrowField observations\rightarrowData analysis\rightarrowMap making
Question 4 |
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 4 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 5 |
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 6 |
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 6 Explanation:
Mean sea level (MSL) is used as a reference surface for establishing the vertical control.
Question 7 |
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 7 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 8 |
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 8 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 9 |
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 9 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 10 |
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 10 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 10 questions to complete.