Question 1 |
If the size of the ground area is 6 \mathrm{~km} \times 3 \mathrm{~km} and the corresponding photo size in the aerial photograph is 30 \mathrm{~cm} \times 15 \mathrm{~cm}, then the scale of the photograph is 1 : ____ (in integer).
15000 | |
20000 | |
25000 | |
30000 |
Question 1 Explanation:
Scale of Photograph (\mathrm{S})=\frac{\text { Photo distance }}{\text { Ground distance }}
For Length scale :
\begin{aligned} S & =\frac{30 \mathrm{~cm}}{6 \mathrm{~cm}} \\ & =\frac{30 \mathrm{~cm}}{60 \times 10^{4} \mathrm{~cm}} \end{aligned}
\Rightarrow \quad \mathrm{S}=\frac{1}{20000}
For Breadth Scale :
\begin{aligned} S & =\frac{15 \mathrm{~cm}}{3 \mathrm{Km}} \\ & =\frac{15 \mathrm{~cm}}{30 \times 10^{4} \mathrm{~cm}} \end{aligned}
S=\frac{1}{20000}
\therefore Scale of photograph, S=1: 20000
For Length scale :
\begin{aligned} S & =\frac{30 \mathrm{~cm}}{6 \mathrm{~cm}} \\ & =\frac{30 \mathrm{~cm}}{60 \times 10^{4} \mathrm{~cm}} \end{aligned}
\Rightarrow \quad \mathrm{S}=\frac{1}{20000}
For Breadth Scale :
\begin{aligned} S & =\frac{15 \mathrm{~cm}}{3 \mathrm{Km}} \\ & =\frac{15 \mathrm{~cm}}{30 \times 10^{4} \mathrm{~cm}} \end{aligned}
S=\frac{1}{20000}
\therefore Scale of photograph, S=1: 20000
Question 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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}
There are 5 questions to complete.