If 16 flight lines are run perpendicular to an area 30 km wide, their spacings on a photographical map on scale 1 : 50,000 , will be
A. 1 cm
B. 2 cm
C. 3 cm
D. 4 cm
Answer: Option D
Solution (By Examveda Team)
Let's break down this surveying problem step by step!First, we need to understand what the question is asking.
It's about the spacing between flight lines on a map, given a real-world area and a map scale.
The area's width is 30 km, and there are 16 flight lines perpendicular to it. This means the 16 flight lines divide the 30 km width into 15 equal spaces.
Therefore, the spacing between the flight lines on the *ground* is calculated as: 30 km / 15 = 2 km.
Next, we need to convert this ground distance to the map distance using the scale 1:50,000.
This scale means that 1 unit on the map represents 50,000 units on the ground.
Convert the ground distance (2 km) to centimeters: 2 km = 2 * 1000 m = 2 * 1000 * 100 cm = 200,000 cm.
Now, apply the scale to find the map distance: Map distance = Ground distance / Scale factor = 200,000 cm / 50,000 = 4 cm.
Therefore, the spacing between the flight lines on the photographical map will be 4 cm.
So, the correct answer is: Option D: 4 cm.
This Question Belongs to Civil Engineering >> Surveying
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Solution to the problem:
1. Calculate the spacing of flight lines on the ground:
The total width of the area is 30 km, and there are 16 flight lines. The spacing between flight lines on the ground can be calculated as:
Spacing (ground) = Total width / Number of flight lines
Spacing (ground) = 30km / 16 = 1.875
2. Convert the ground spacing to centimeters: Since the map scale is given in centimeters, convert the ground spacing from kilometers to centimeters: 1km = 1,00,000 cm
Spacing (ground) = 1.875 km x 10,00,000 cm/km = 187,500 cm
3. Calculate the spacing on the map using the given scale:
The map scale is 1:50,000, meaning 1 unit on the map represents 50,000 units on the ground. Spacing (map) = Spacing (ground) / Scale denominator Spacing (map) = 187,500 cm / 50,000 = 3.75 cm
4. Compare with the given options: The calculated spacing on the map is 3.75 cm. Looking at the options, 4 cm is the closest and likely intended answer if rounding is applied or if the number of segments is considered (15 segments for 16 lines). However, strictly calculating, 3.75 cm is the result. Given the options, and typical multiple-choice rounding, 4 cm is the most plausible choice. Therefore, the spacing on the photographical map will be approximately 4 cm.
Mujhe samajhane ki kripya pradhan kare
Mujhe samajhane ki kripya pradhan kare
Hi guys I am prathamesh ahire
And I am trying to solve this question
Width- 30 km
Flight lines - 16
Therefore total flights - 15
Then width of 1 (one) flight - 30/15 = 2 km
As given scale is 1:50000 ( means 1cm = 50000 cm )
We know that , 1 km = 100000 cm
Therefore , for 2km scale will be 4 cm
Harman bajwa right ans
D
How it is.?
16 lines divide the area to 15 spacings, so
Distance / spacing = 30/15.
Equate it with (map/ground)
=>1/50,000= x/ (3/15)
X= 4cm
Abhijit kumar
I want to discuss this question
[30÷(16-1)]×1000×100÷50000=4
30 km = 30 x 1000 x 100 cm
Scale on Map = 1: 50,000
Area on Map = (30 x 1000 x 100)/50000 = 60
IF 16 lines divide the area into 15 parts, then 60/15 = 4
i.e. Distance between two flights is 4 cm
Flight lines = (Width(coverage)/spacing)
16 = (30,000/x)
x = 30,000 / 16
x = 1875
scale = map/ ground
1/50,000 = x / 1875
x = 0.0375 m = 3.75 cm ~~ 4cm
Flight lines = (Width(coverage)/spacing)
16 = (30,000/x)
x = 30,000 / 16
x = 1875
scale = map/ ground
1/50,000 = x / 1875
x = 0.0375 m = 3.75 cm ~~ 4cm
Yes
30,000/16=1875
1875/50,000 m =0.0375 Cm
~~4 cm
reason??# of Flight lines = (Width(coverage)/spacing)
16 = (30,000/x)
x = 30,000 / 16
x = 1875
scale = map/ ground
1/50,000 = x / 1875
x = 0.0375 m = 3.75 cm ~~ 4cm
(30*100000)/(50000*(16-1))=4cm
# of Flight lines = (Width(coverage)/spacing)
16 = (30,000/x)
x = 30,000 / 16
x = 1875
scale = map/ ground
1/50,000 = x / 1875
x = 0.0375 m = 3.75 cm ~~ 4cm
??