Q2GATE 2024MCQ

For a reconnaissance survey, it is necessary to obtain vertical aerial photographs of a terrain at an average scale of 1:13000 using a camera. If the permissible flying height is assumed as 3000m above a datum and the average terrain elevation is 1050 m above the datum, the required focal length (in $\mathrm{mm}$ ) of the camera is

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📖 Explanation

$\text { Scale }=\frac{1}{13000}$ Flying height $(\mathrm{H})=3000 \mathrm{~m}$ Elevation $(h)=1050 m$

\begin{aligned} S & =\frac{f}{H-h} \\ \frac{1}{13000} & =\frac{f}{3000-1050} \\ \therefore \quad f & =0.15 m=150 \mathrm{~mm} \end{aligned}

Q3GATE 2023NAT

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).

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📖 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, $S=1: 20000$

Q4GATE 2022MCQ

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

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📖 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}

Q5GATE 2019MCQ

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?

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📖 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}

Q6GATE 2018MCQ

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

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📖 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}

Q7GATE 2018NAT

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 ______

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📖 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}

Q8GATE 2017NAT

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_____

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📖 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}

Q9GATE 2017MCQ

The number of spectral bands in the Enhanced Thematic Mapper sensor on the remote sensing satellite Landsat-7 is

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Q10GATE 2016MCQ

Optimal flight planning for a photogrammetric survey should be carried out considering

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Q11GATE 2016MCQ

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

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Q12GATE 2016MCQ

The minimum number of satellites needed for a GPS to determine its position precisely is

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Q13GATE 2016NAT

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 _____.

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📖 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}

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