1. Condition for bodies to start floating at the equator

At the equator, if bodies just start floating, the normal reaction becomes zero. So the required centripetal acceleration is provided entirely by gravity:

$\omega^2 R = g$

Thus,

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

2. Substitute the given values

Given:

$g = 10\ \text{m s}^{-2}, \qquad R = 6400 \times 10^3\ \text{m} = 6.4 \times 10^6\ \text{m}$

So,

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

$\omega = \sqrt{1.5625 \times 10^{-6}}$

$\omega = 1.25 \times 10^{-3}\ \text{rad s}^{-1}$

3. Find the time period of rotation

The duration of the day is the time period:

$T = \frac{2\pi}{\omega}$

Using $\pi = 3.14$,

$T = \frac{2 \times 3.14}{1.25 \times 10^{-3}}$

$T = \frac{6.28}{1.25 \times 10^{-3}}$

$T = 5024\ \text{s}$

4. Convert into minutes

$T = \frac{5024}{60} \approx 83.7\ \text{minutes}$

So the duration of the day is approximately

$\boxed{84\ \text{minutes}}$

5. Check options

• A: 84 minutes ✅
• B: 1200 minutes ❌
• C: 60 minutes ❌
• D: does not change ❌

Hence, the correct option is A.