First slide

Acceleration due to gravity and its variation

Question

If earth were to rotate on its own axis such that the weight of a person at the equator becomes half the weight at the poles, then its time period of rotation is (g=acceleration due to gravity near the poles and R is the radius of earth)

Moderate

A

$2 π \sqrt{\frac{2 R}{g}}$
B

$2 π \sqrt{\frac{3 R}{g}}$
C

$2 π \sqrt{\frac{R}{3 g}}$
D

$2 π \sqrt{\frac{4 R}{g}}$

Solution

$g_{ϕ} = g - R ω^{2} \cos^{2} ϕ \Rightarrow g_{e} = \frac{g_{P}}{2} \Rightarrow g - R ω^{2} \cos^{2} 0 = \frac{g - R ω^{2} \cos^{2} 90}{2} \Rightarrow g - R ω^{2} = \frac{g}{2} \Rightarrow g - \frac{g}{2} = R ω^{2} \Rightarrow \frac{g}{2} = R ω^{2} \Rightarrow \sqrt{\frac{g}{2 R}} = ω \Rightarrow \sqrt{\frac{g}{2 R}} = \frac{2 π}{T} \Rightarrow T = 2 π \sqrt{\frac{2 R}{g}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App