Question

What should be the angular velocity of earth due to rotation about its own axis so that the weight at equator becomes \((\frac{3}{5})^{th}\) of initial value? (\(g =\) acceleration due to gravity, R = radius of earth)

• A. \((\frac{R}{3g})^{\frac{1}{2}}\)
• B. \((\frac{2g}{5R})^{\frac{1}{2}}\)
• C. \((\frac{5g}{9R})^{\frac{1}{2}}\)
• D. \((\frac{3g}{2R})^{\frac{1}{2}}\)

Hint:

Apparent g at the equator is g - omega^2 R.

Answer 1

The Correct Option is B

Step 1: Forces at the equator:
Net inward force $=mg'=mg-m\omega^2R$ for the apparent weight $mg'$.

Step 2: Fraction:
$mg'=\frac35mg$ means the centrifugal term is $\frac25mg$, so $m\omega^2R=\frac25mg$.

Step 3: Solve:
$\omega^2=\dfrac{2g}{5R}$. Option (B).

Final Answer:
The centrifugal term must be 2/5 of mg. \[ \boxed{B} \]