Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 4 (Q.No. 37)
37.
The ratio of the linear displacement at the end of a line, subtended by an arc of one second to the length of the line, is
Discussion:
6 comments Page 1 of 1.
Adeel Rehman said:
5 years ago
Correct answer (A)
S=r*angle.
Or s/r = angle in radians ...eq 1
S being the linear displacement.
r being the length of the line or radius.
Here Angle = 1 sec.
So, for 360 degrees the eq 1 becomes,
s/r = 2 π in radians.
As in radians 1 sec =2pi/(360*60*60), which gives us 1sec = 1/206265 radians.
And by the relationship in eq 1 above we know that,
s/r = 1/ 206265,
Or s:r = 1: 206265 close to 206300.
S=r*angle.
Or s/r = angle in radians ...eq 1
S being the linear displacement.
r being the length of the line or radius.
Here Angle = 1 sec.
So, for 360 degrees the eq 1 becomes,
s/r = 2 π in radians.
As in radians 1 sec =2pi/(360*60*60), which gives us 1sec = 1/206265 radians.
And by the relationship in eq 1 above we know that,
s/r = 1/ 206265,
Or s:r = 1: 206265 close to 206300.
(1)
Adeel Rehman said:
5 years ago
S=r*angle.
S being the linear displacement.
R being the length of the line.
Here Angle = 1 sec.
So,l for 360 degrees s/r = 2 π in radians.
As 1 sec =2pi/(360*60*60),
So s/r = 1/ 206265,
Or s:r = 1: 206265 close to 206300.
S being the linear displacement.
R being the length of the line.
Here Angle = 1 sec.
So,l for 360 degrees s/r = 2 π in radians.
As 1 sec =2pi/(360*60*60),
So s/r = 1/ 206265,
Or s:r = 1: 206265 close to 206300.
Rahul said:
8 years ago
360 deg = 2π.
1 sec = 2π360/60/60.
1 sec = 2π360/60/60.
(2)
Shailendra said:
7 years ago
@Rahul.
It is 2π/(360*60*60).
It is 2π/(360*60*60).
Kalpesh said:
7 years ago
How? Please explain.
Shumaila said:
5 years ago
How? please explain.
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