What should be the angular velocity of rotation of Earth about its own axis so that the weight of a body at the equator reduces to (3/5) or r its present value? (Take R as the radius of the Earth)

Question

What should be the angular velocity of rotation of Earth about its own axis so that the weight of a body at the equator reduces to (3/5) or r its present value? (Take R as the radius of the Earth) (A) √(g/3R) (B) √(2g/3R) (C) √(2g/5R) (D) √(2g/7R)

Tardigrade Admin · Accepted Answer

mg' = mg - mRω2 cos2 φ Now, (3/5) mg = mg - mRω2 or mRω2 = mg - (3/5) mg or mRω2 = (3/5) mg or ω = √(2g/5R)

Q. What should be the angular velocity of rotation of Earth about its own axis so that the weight of a body at the equator reduces to $\frac{3}{5}$ or r its present value? (Take $R$ as the radius of the Earth)

$\frac{g}{3 R}$

$\frac{2 g}{3 R}$

$\frac{2 g}{5 R}$

$\frac{2 g}{7 R}$

$m g^{'} = m g - m R ω^{2} co s^{2} ϕ$

Now, $\frac{3}{5} m g = m g - m R ω^{2}$

or $m R ω^{2} = m g - \frac{3}{5} m g$

or $m R ω^{2} = \frac{3}{5} m g$ or $ω = \frac{2 g}{5 R}$

Q. What should be the angular velocity of rotation of Earth about its own axis so that the weight of a body at the equator reduces to 53​ or r its present value? (Take R as the radius of the Earth)

3Rg​​

3R2g​​

5R2g​​

7R2g​​