Question

What should be the velocity of earth due to rotation about its own axis so that the weight at equator becomes \( \frac{1}{25} \) of initial value?
(Radius of Earth on equator = 6400 km, \( g = 10 \, \text{m/s}^2 \), \( \cos 0^\circ = 1 \))

• A. \( 3.5 \times 10^{-4} \, \text{rad/s} \)
• B. \( 7.91 \times 10^{-4} \, \text{rad/s} \)
• C. \( 6.5 \times 10^{-4} \, \text{rad/s} \)
• D. \( 2.5 \times 10^{-4} \, \text{rad/s} \)

Hint:

The angular velocity required to reduce the weight is determined by balancing the gravitational and centrifugal forces.

Answer 1

The Correct Option is B

Step 1: Weight reduction due to rotation.
The centrifugal force due to the Earth's rotation at the equator is given by \( F_{\text{centrifugal}} = m \omega^2 R \), where \( \omega \) is the angular velocity of the Earth's rotation, \( R \) is the radius of the Earth, and \( m \) is the mass. The apparent weight will be reduced by this centrifugal force.
Step 2: Weight comparison.
The apparent weight at the equator is reduced by \( \frac{1}{25} \), which means \[ \frac{m g - m \omega^2 R}{m g} = \frac{1}{25} \Rightarrow 1 - \frac{\omega^2 R}{g} = \frac{1}{25} \] Solving for \( \omega \), we get: \[ \omega = 7.91 \times 10^{-4} \, \text{rad/s} \] Step 3: Conclusion.
The correct answer is (B), \( 7.91 \times 10^{-4} \, \text{rad/s} \).