The acceleration due to gravity on the earth's surface at the poles is g an and angular velocity of the earth about the axis passing through the pole is $\omega$. An object is weight 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 : ( $\mathrm{h} \ll \mathrm{R}$, where R is the radius of the earth)

(A) $\frac{R^{2} \omega^{2}}{2 g}$

(B) $\frac{R^{2} \omega^{2}}{g}$

(C) $\frac{R^{2} \omega^{2}}{8 g}$

(D) $\frac{R^{2} \omega^{2}}{4 g}$

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Gravitation

The acceleration due to gravity on the earth's surface at the poles is g an and angular velocity of the earth about the axis passing through the pole is

ω

. An object is weight 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) $\frac{R ^{2} ω ^{2}}{2 g}$ ✓ Correct
(B) $\frac{R ^{2} ω ^{2}}{g}$
(C) $\frac{R ^{2} ω ^{2}}{8 g}$
(D) $\frac{R ^{2} ω ^{2}}{4 g}$

Answer: (A)

Solution

At the equator the spin lowers the measured gravity to

g_{e} = g - R ω^{2}

. At a height

h

above a pole the value falls with altitude as

g_{h} = g (1 - \frac{2 h}{R})

for

h ≪ R

. Equal spring-balance weights mean equal effective gravities, so

g - R ω^{2} = g - \frac{2 g h}{R}

. This gives

R ω^{2} = \frac{2 g h}{R}

, hence

h = \frac{R ^{2} ω ^{2}}{2 g}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Gravitation

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