The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60° latitude becomes zero is (Radius of earth = 6400 km. At the poles g = 10 ms⁻²)

The correct option is 2.5×10⁻³ rad/sec.

Understanding the Effect of Earth's Rotation on Gravity:
The acceleration due to gravity at any latitude $\theta$ on the surface of the Earth is altered by the centrifugal force arising from the Earth's rotation. The effective acceleration due to gravity $g^{\prime }$ at latitude $\theta$ is given by the formula:

$g^{\prime }=g-{\omega }^{2}R{cos}^2\theta$

where:
• $g$ is the acceleration due to gravity at the poles (when there is no rotation effect, $g=10{\text{ms}}^{-2}$)
• $\omega$ is the angular velocity of the Earth's rotation
• $R$ is the radius of the Earth ($R=6400\text{km}=6.4\times {10}^6\text{m}$)
• $\theta$ is the latitude angle ($\theta =60°$)

Setting up the Condition for Weightlessness:
We are given that the effective acceleration due to gravity at $60°$ latitude becomes zero ($g^{\prime }=0$). Substituting this value into the equation:

$0=g-{\omega }^{2}R{cos}^2(60°)$

Rearranging the equation to solve for the angular velocity $\omega$:

${\omega }^{2}R{cos}^2(60°)=g$

${\omega }^2=\frac{g}{R{cos}^2(60°)}$

Step-by-Step Calculation:
Since $cos(60°)=\frac{1}{2}$, we have:

${cos}^2(60°)={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

Substituting this back into the expression for ${\omega }^2$:

${\omega }^2=\frac{g}{R\times \frac{1}{4}}=\frac{4g}{R}$

Taking the square root of both sides:

$\omega =2\sqrt{\frac{g}{R}}$

Now, substitute the numerical values ($g=10{\text{ms}}^{-2}$ and $R=6.4\times {10}^6\text{m}$):

$\omega =2\sqrt{\frac{10}{6.4\times {10}^6}}$

Simplify the fraction inside the square root:

$\omega =2\sqrt{\frac{1}{640000}}$

$\omega =2\times \frac{1}{800}$

$\omega =\frac{1}{400}{\text{rad s}}^{-1}$

$\omega =0.0025{\text{rad s}}^{-1}=2.5\times {10}^{-3}{\text{rad s}}^{-1}$

Thus, the required angular velocity of the Earth is $2.5\times {10}^{-3}\text{rad/sec}$.