Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 1 (Q.No. 38)
38.
If the radius of the earth is 600 km the height of a mountain above sea level at the top of which a beat seconds pendulum at sea level, looses 27 seconds a day, is
Discussion:
3 comments Page 1 of 1.
Mahadev said:
8 years ago
Applied correction in question take the radius of earth=6400km.
dn/n=(-H/R).
dn=No. of seconds the pendulum will lose in one day=(-27)
n=no. of seconds in one day or 24 hours=24*60*60=86400
H=Height of a mountain in km=?
R=radius of the earth=6400km.
Using above formula and details you will get the correct answer which is 2000 metre(2km).
dn/n=(-H/R).
dn=No. of seconds the pendulum will lose in one day=(-27)
n=no. of seconds in one day or 24 hours=24*60*60=86400
H=Height of a mountain in km=?
R=radius of the earth=6400km.
Using above formula and details you will get the correct answer which is 2000 metre(2km).
Subha Annadurai said:
8 years ago
Explain please.
Rajesh kumar arya said:
9 years ago
g' = g[1+h/Re]^ - 2.
And g' = [2π/T]^2*L.
T(ideal) - loss time per second = T(actual).
T' = 2 - 27/3600.
=1.9925.
g' = [2*3. 14/1. 9925]^2*.0. 994 =9. 874.
H = [{9. 81/9. 874)^1/2 - 1]*600000 => 2000.
And g' = [2π/T]^2*L.
T(ideal) - loss time per second = T(actual).
T' = 2 - 27/3600.
=1.9925.
g' = [2*3. 14/1. 9925]^2*.0. 994 =9. 874.
H = [{9. 81/9. 874)^1/2 - 1]*600000 => 2000.
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