Question

The speed with the earth have to rotate on its axis so that a person on the equator would weight (3/5)th as much as present will be (Take the equilateral radius as 6400 km.)

• A. $3.28\times {10}^{-4}$ rad /sec
• B. $7.826\times {10}^{-4}$ rad /sec
• C. $3.28\times {10}^{-3}$ rad /sec
• D. $7.28\times {10}^{-3}$ rad /sec

Answer 1

The correct option is B $7.826\times {10}^{-4}$ rad /sec

Actual weight on the equator $=W=mg$

where 'm' is the mass and 'g' is the gravity$.$

According to the given condition$,$

Weight on the equator $=W^{\prime }=m{g}^{\prime }$

Weight on the equator $=W^{\prime }=3/5mg$ ........... $\left(1\right)$

We know that,$\lambda =0$ at the equator.

Now$,$

$m{g}^{\prime }=mg-mR\omega ²cos\lambda$

$3/5mg=mg-mR\omega ²cos0(\because \lambda =0)$

$3/5mg=mg-mR\omega ²$ $\left(1\right)$ $(\because cos0=1)$

$3/5mg=mg-mR\omega ²$

$mR\omega ²=mg-3/5mg$

$mR\omega ²=(1-3/5)mg$

$mR\omega ²=2/5mg$

$R\omega ²=2/5g$

$\omega ²=2g/5R$

$\omega ²=\left(2x10\right)/\left(5x6.4x10⁶\right)(\because R=RadiusofEarth=6400km=6.4x10⁶m)$

$\omega =\surd \left(6.25x10⁻⁷\right)$

$\omega =7.8x10⁻⁴rad/sec$

Hence,

option $\left(B\right)$ is correct answer.