The rotation of the earth radius R about its axis speeds upto a value such that a man at latitude angle `60^(@)` feels weightless. The duration of the day in such case will be

Options

A. `pi sqrt((R)/(g))`
B. `(pi)/(2)sqrt((R)/(g))`
C. `(pi)/(3)sqrt((R)/(g))`
D. `pi sqrt((g)/(R))`

Explanation

To solve the problem, we need to find the duration of the day when a man at a latitude of \(60^\circ\) feels weightless due to the rotation of the Earth. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding Weightlessness Condition**:
A person feels weightless when the effective gravitational force acting on them becomes zero. The effective gravitational force (\(g'\)) at latitude \(\theta\) is given by the equation:
\[ g' = g - R \omega^2 \cos^2 \theta \]
where \(g\) is the acceleration due to gravity, \(R\) is the radius of the Earth, \(\omega\) is the angular velocity of the Earth, and \(\theta\) is the latitude.

2. **Setting Up the Equation**:
For the person to feel weightless, we set \(g' = 0\):
\[ 0 = g - R \omega^2 \cos^2(60^\circ) \]
Since \(\cos(60^\circ) = \frac{1}{2}\), we can substitute this into the equation:
\[ 0 = g - R \omega^2 \left(\frac{1}{2}\right)^2 \]
This simplifies to:
\[ 0 = g - \frac{R \omega^2}{4} \]

3. **Solving for Angular Velocity (\(\omega\))**:
Rearranging the equation gives:
\[ R \omega^2 = 4g \]
Therefore, we can express \(\omega^2\) as:
\[ \omega^2 = \frac{4g}{R} \]
Taking the square root gives us:
\[ \omega = 2\sqrt{\frac{g}{R}} \]

4. **Relating Angular Velocity to Period (T)**:
The angular velocity \(\omega\) is also related to the period \(T\) (the duration of the day) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Setting the two expressions for \(\omega\) equal gives:
\[ \frac{2\pi}{T} = 2\sqrt{\frac{g}{R}} \]

5. **Solving for the Duration of the Day (T)**:
Rearranging the equation to solve for \(T\):
\[ T = \frac{\pi}{\sqrt{\frac{g}{R}}} \]
This can be rewritten as:
\[ T = \pi \sqrt{\frac{R}{g}} \]

### Final Answer:
The duration of the day when a man at latitude \(60^\circ\) feels weightless is:
\[ T = \pi \sqrt{\frac{R}{g}} \]