Question 1 |
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 1 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 2 |
A camera with a focal length of 20 cm fitted in an aircraft is used for taking vertical aerial photographs of a terrain. The average elevation of the terrain is 1200 m above mean sea level (MSL). What is the height above MSL at which an aircraft must fly in order to get the aerial photographs at a scale of 1:8000?
2600 m | |
2800 m | |
3000 m | |
3200 m |
Question 2 Explanation:
Given focal length = 20 cm
as we know scale of vertical photograph =\frac{f}{H-h_{avg}}
its given as 1 : 8000
Hence,
\begin{aligned} \frac{f}{H-h_{avg}} &=\frac{1}{8000} \\ \frac{20 cm}{(H-1200) \times 100cm}&=\frac{1}{8000} \\ \Rightarrow \;\;H &=2800m \end{aligned}
Question 3 |
An aerial photograph of a terrain having an average elevation of 1400 m is taken at a scale of 1:7500. The focal length of the camera is 15 cm. The altitude of the flight above mean sea level (in m, up to one decimal place) is ______
2680 | |
2460 | |
2525 | |
3620 |
Question 3 Explanation:
\begin{aligned} h &=1400 \mathrm{m} \\ \text { Scale } &=1: 7500 \\ f &=15 \mathrm{cm} \\ \text { Scale } &=\frac{f}{H-h} \\ \Rightarrow\quad \frac{1}{7500} &=\frac{15 \times 10^{-2}}{H-1400} \\ \Rightarrow\quad H &=2525 \mathrm{m} \end{aligned}
Question 4 |
A square area (on the surface of the earth) with side 100 m and uniform height, appears as 1 cm^{2} on a vertical aerial photograph. The topographic map shows that a contour of 650 m passes through the area. If focal length of the camera lens is 150 mm, the height from which the aerial photograph was taken, is
800 m | |
1500 m | |
2150 m | |
3150 m |
Question 4 Explanation:
\begin{aligned} A&=100 \times 100 \mathrm{m}^{2}\\ \text { Area on photo, }\\ a &=1 \mathrm{cm}^{2} \\ \text { Scale } \quad 1 \mathrm{cm} &=100 \mathrm{m} \\ f &=150 \mathrm{mm} \\ h &=650 \mathrm{m} \\ \text { Scale } \quad &=\frac{1}{100}=\frac{1}{100 \times 10^{2}}\\ &=\frac{1}{10000}=\frac{f}{H-h}\\ \Rightarrow \quad \frac{1}{10000}&=\frac{150 \times 10^{-3}}{H-650} \\ \Rightarrow \quad H&=2150 \mathrm{m} \end{aligned}
Question 5 |
Two towers A and B, standing vertically on a horizontal ground, appear in a vertical aerial
photograph as shown in the figure.
The length of the image of the tower A on the photograph is1.5 cm and of the tower B is 2.0 cm. The distance of the top the tower A (as shown by the arrowhead) is 4.0 cm and the distance of the top of the tower B is 6.0 cm, as measured form the principal point p of the photograph. If the height of the tower B is 80 m, the height (in meters) of the tower A is_____
The length of the image of the tower A on the photograph is1.5 cm and of the tower B is 2.0 cm. The distance of the top the tower A (as shown by the arrowhead) is 4.0 cm and the distance of the top of the tower B is 6.0 cm, as measured form the principal point p of the photograph. If the height of the tower B is 80 m, the height (in meters) of the tower A is_____
90 | |
240 | |
30 | |
4 |
Question 5 Explanation:
For Tower A:
Reliet displacement,
\begin{array}{l} d_{A}=1.5 \mathrm{cm} \\ r_{A}=4 \mathrm{cm} \\ h_{A}=? \end{array}
For Tower B:
Relier displacement.
\begin{array}{l} d_{B}=2.0 \mathrm{cm} \\ r_{B}=6 \mathrm{cm} \\ h_{B}=80\mathrm{m} \end{array}
Reliet displacement.
\begin{aligned} D&=\frac{h.r}{\left(H-h_{\text{avg}} \right )}\\ \text{For A:}d_{A}&=\frac{h_{A}.r_{A}}{\left(H-h_{\text{avg}} \right )}\\ 1.5&=\frac{h_{A}\times 4}{240}\\ h_{A}&=90m\\ \text{For B: }d_{B}&=\frac{h_{B}.r_{B}}{\left(H-h_{\text{avg}} \right )}\\ \Rightarrow \quad 2&=\frac{80\times 6}{\left(H-h_{\text{avg}} \right )}\\ \Rightarrow \left(H-h_{\text{avg}} \right )&=240m \end{aligned}
Reliet displacement,
\begin{array}{l} d_{A}=1.5 \mathrm{cm} \\ r_{A}=4 \mathrm{cm} \\ h_{A}=? \end{array}
For Tower B:
Relier displacement.
\begin{array}{l} d_{B}=2.0 \mathrm{cm} \\ r_{B}=6 \mathrm{cm} \\ h_{B}=80\mathrm{m} \end{array}
Reliet displacement.
\begin{aligned} D&=\frac{h.r}{\left(H-h_{\text{avg}} \right )}\\ \text{For A:}d_{A}&=\frac{h_{A}.r_{A}}{\left(H-h_{\text{avg}} \right )}\\ 1.5&=\frac{h_{A}\times 4}{240}\\ h_{A}&=90m\\ \text{For B: }d_{B}&=\frac{h_{B}.r_{B}}{\left(H-h_{\text{avg}} \right )}\\ \Rightarrow \quad 2&=\frac{80\times 6}{\left(H-h_{\text{avg}} \right )}\\ \Rightarrow \left(H-h_{\text{avg}} \right )&=240m \end{aligned}
Question 6 |
The number of spectral bands in the Enhanced Thematic Mapper sensor on the remote sensing satellite Landsat-7 is
64 | |
10 | |
8 | |
15 |
Question 6 Explanation:
Total number of spectrum band in Enhanced
Thematic Mapper sender =8
Question 7 |
A tall tower was photographed from an elevation of 700 m above the datum. The radial distances of the top and bottom of the tower from the principal points are 112.50 mm and 82.40 mm, respectively. If the bottom of the tower is at an elevation 250 m above the datum, then the height (expressed in m) of the tower is _____.
120.4 | |
100.8 | |
50.4 | |
60.2 |
Question 7 Explanation:
Given
H=700 \mathrm{m}_{1} h_{\text {avg }}=250
Relief distance,
\begin{aligned} d&=f_{1}=1 \\ &=112.5-82.40=30.1 \mathrm{mm}\\ \therefore\quad d&=\frac{h_{1}}{H=h_{a v}}\\ &[\text{where his height of tower}]\\ \Rightarrow \quad 30.1&=\frac{h \times 112.5}{100=250} \\ \therefore \quad h&=120.4 \mathrm{mm} \end{aligned}
Question 8 |
Optimal flight planning for a photogrammetric survey should be carried out considering
only side-lap | |
only end-lap | |
either side-lap or end-lap | |
both side-lap as well as end-lap |
Question 9 |
The system that uses the Sun as a source of electromagnetic energy and records the naturally radiated and reflected energy from the object is called
Geographical Information System | |
Global Positioning System | |
Passive Remote Sensing | |
Active Remote Sensing |
Question 9 Explanation:
The sun provide a very convenient source of energy for remote sensing. The Sun e energy is either refleoted, as it is tor visible wavelengths, of absorbed and then re-emitted, as it is for themal intrared wavelengths, Remote sensing system which measure energy le. naturally available are called as passive remote sensing.
Question 10 |
The minimum number of satellites needed for a GPS to determine its position precisely is
2 | |
3 | |
4 | |
24 |
Question 10 Explanation:
At a minimum, four satellite must be in view of the receiver tor it to compute four unknown quantity (three position eoordinate and one for clock deviation trom satellite time).
There are 10 questions to complete.