The acceleration due to gravity on the earth's surface at the poles is g and angular velocity of the earth about the axis passing through the pole is ω. An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is :( h

Question

The acceleration due to gravity on the earth's surface at the poles is g and angular velocity of the earth about the axis passing through the pole is ω. An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is :( h < < R , where R is the radius of the earth) (A) ( R 2 ω2/8 g ) (B) ( R 2 ω2/4 g ) (C) ( R 2 ω2/ g ) (D) ( R 2 ω2/2 g )

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f57ae3fd419432aa52e1a8ea380b88de-.png" /> ge=g-R ω2 g2=g(1-(2 h/R)) g1=g e g2=g-(2 g h/R) Now R ω2=(2 g h/R) h=(R2 ω2/2 g)

$\frac{R ^{2} ω ^{2}}{8 g}$

$\frac{R ^{2} ω ^{2}}{4 g}$

$\frac{R ^{2} ω ^{2}}{g}$

$\frac{R ^{2} ω ^{2}}{2 g}$

$g_{e} = g - R ω^{2}$
$g_{2} = g (1 - \frac{2 h}{R}) g_{1} = g e$
$g_{2} = g - \frac{2 g h}{R}$
Now $R ω^{2} = \frac{2 g h}{R}$
$h = \frac{R ^{2} ω ^{2}}{2 g}$

8gR2ω2​

4gR2ω2​

gR2ω2​

2gR2ω2​

ge​=g−Rω2 g2​=g(1−R2h​)g1​=ge g2​=g−R2gh​ Now Rω2=R2gh​ h=2gR2ω2​

$\frac{R ^{2} ω ^{2}}{8 g}$

$\frac{R ^{2} ω ^{2}}{4 g}$

$\frac{R ^{2} ω ^{2}}{g}$

$\frac{R ^{2} ω ^{2}}{2 g}$

$g_{e} = g - R ω^{2}$
$g_{2} = g (1 - \frac{2 h}{R}) g_{1} = g e$
$g_{2} = g - \frac{2 g h}{R}$
Now $R ω^{2} = \frac{2 g h}{R}$
$h = \frac{R ^{2} ω ^{2}}{2 g}$