R is the radius of the Earth and ω is its angular velocity and gp is the value of g at poles. The effective value of ' g ' at the latitude λ=60° will be equal to

Question

R is the radius of the Earth and ω is its angular velocity and gp is the value of g at poles. The effective value of ' g ' at the latitude λ=60° will be equal to (A) gp-(1/4) R ω2 (B) gp-(3/4) R ω2 (C) gp-R ω2 (D) gp+(1/4) R ω2

Tardigrade Admin · Accepted Answer

g=gp-ω2 R cos 2 λ Here λ=60° ⇒ g=gp-ω2 R cos 2 60°=gp-(ω2 R/4)

Q. $R$ is the radius of the Earth and $ω$ is its angular velocity and $g_{p}$ is the value of $g$ at poles. The effective value of ' $g$ ' at the latitude $λ = 6 0^{\circ}$ will be equal to

$g_{p} - \frac{1}{4} R ω^{2}$

$g_{p} - \frac{3}{4} R ω^{2}$

$g_{p} - R ω^{2}$

$g_{p} + \frac{1}{4} R ω^{2}$

$g = g_{p} - ω^{2} R cos^{2} λ$
Here $λ = 6 0^{\circ}$
$\Rightarrow g = g_{p} - ω^{2} R cos^{2} 6 0^{\circ} = g_{p} - \frac{ω ^{2} R}{4}$

Q. R is the radius of the Earth and ω is its angular velocity and gp​ is the value of g at poles. The effective value of ' g ' at the latitude λ=60∘ will be equal to

gp​−41​Rω2

gp​−43​Rω2

gp​−Rω2

gp​+41​Rω2