CSIR-NET 2013
21. If the Earth were to rotate twice faster than its present angular speed , then the coriolis parameter at the equator would be
A) 1 /s B) 0 /s C) 2 /s D) 4 /s
Solution:
The coriolis frequency (f), also called the coriolis
parameter or coriolis coefficient is equal to twice the rotation rate (ώ) of
the Earth multiplied by the ‘sine’ of the latitude (φ).
f=(2*ώ*sinφ) -------------------(1)
The rotation rate of the Earth (ώ=7.2921*10-5
rad/s) can be calculated as (2π/T) radians per second , where T is the rotation
period of the Earth is one sidereal day (23h 56m 4s).
Given values in above problem,
ώ=2ώ -----------------(2)
eq(2) value substitute in eq(1)
f=(4*ώ *sinφ)
at equator φ=0 (sinφ=0)
f=(4*ώ*sin0)
f=0
Therefore, Coriolis parameter (f)= 0 s-1
Post a Comment
Post a Comment