Examveda

# 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

This Question Belongs to Civil Engineering >> Surveying

## Comments ( 19 )

1. Mujhe samajhane ki kripya pradhan kare

2. Mujhe samajhane ki kripya pradhan kare

3. 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

4. Harman bajwa right ans

5. D

6. How it is.?

7. 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

8. Abhijit kumar

9. I want to discuss this question

10. [30÷(16-1)]×1000×100÷50000=4

11. 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

12. 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

13. 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

14. Yes

15. 30,000/16=1875

1875/50,000 m =0.0375 Cm
~~4 cm

16. 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

17. (30*100000)/(50000*(16-1))=4cm

18. # 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

19. ??

Related Questions on Surveying