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

Question

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

Tardigrade Admin · Accepted Answer

m g'=m g-m R ω2 cos 2 φ Now, (3/5) m g=m g-m R ω2 or m R ω2=m g-(3/5) m g=(2/5) m g or ω=√(2 g/5 R)

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

$\frac{g}{3 R}$

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

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

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

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

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

3Rg​​

3R2g​​

5R2g​​

7R2g​​