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Step-by-Step Solution
Step 1: Condition for Bodies to Float at the Equator
At the equator, a body floats when the gravitational force acting on it is balanced by the required centripetal force due to Earth's rotation. Mathematically, this condition is expressed as:
$mg = m \omega^{2} R$
Here,
$m$ = mass of the body,
$g$ = acceleration due to gravity,
$\omega$ = angular velocity of the Earth,
$R$ = radius of the Earth.
We can cancel $m$ from both sides and rewrite the equation as:
$g = \omega^{2} R$
Step 2: Solve for Angular Velocity $\omega$
From the condition for floating,
$ \omega = \sqrt{\frac{g}{R}}$
Step 3: Express the Duration of the Day in Terms of $\omega$
The time period $T$ (which corresponds to the length of the day when the Earth spins at this angular velocity) is given by:
$T = \frac{2 \pi}{\omega}$
Substituting $\omega = \sqrt{\frac{g}{R}}$ into the expression for $T$:
$T = 2 \pi \sqrt{\frac{R}{g}}$
Step 4: Substitute the Numerical Values
Given:
$g = 10 \, \text{m/s}^2,$
$R = 6400 \times 10^3 \, \text{m},$
$\pi = 3.14.$
Substitute these into $T = 2 \pi \sqrt{\frac{R}{g}}$:
$T = 2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}}$
$= 2 \times 3.14 \times \sqrt{640 \times 10^3}$
$= 2 \times 3.14 \times \sqrt{6.4 \times 10^5}$
$= 2 \times 3.14 \times 800$
(since $\sqrt{6.4 \times 10^5} = 800$)
$= 6.28 \times 800 = 5024\, \text{seconds}$
Step 5: Convert $T$ into Minutes
$T$ in minutes = $\frac{5024}{60} \approx 83.73 \,\text{minutes}$
Rounded off, this is approximately:
$\boxed{84 \,\text{minutes}}$
