Question

If all objects on the equator of earth feel weightless, then the duration of the day will nearly become
(Take $g=10{\text{m/s}}^2$)

• A. $6.2\text{hr}$
• B. $4.4\text{hr}$
• C. $2.2\text{hr}$
• D. $1.4\text{hr}$

Answer 1

The correct option is D $1.4\text{hr}$

The weight at the equator is given by

$m{g}^{\prime }=m(g-{\omega }^{2}R)$

Since, the object feel weightless, that means weight of body becomes zero.

$\therefore m{g}^{\prime }=0$

$\Rightarrow m(g-{\omega }^{2}R)=0$

$\Rightarrow g={\omega }^{2}R$

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

Now time period, $T=\frac{2\pi }{\omega }$

$\Rightarrow T=2\pi \sqrt{\frac{R}{g}}$

Here, $R=6400\text{km}=64\times {10}^5\text{m}$
$g=10{\text{m/s}}^2$

$\Rightarrow T=2\pi \sqrt{\frac{64\times {10}^5}{10}}=2\pi \times 8\times {10}^2$

$\Rightarrow T=1600\pi \text{s}=1.4\text{hr}$

Hence, option $\left(d\right)$ is correct.

Why this question: To familiarize students with the change in acceleration due to gravity due to the rotation of the earth. Key concepts: The acceleration due to gravity at the equator is given by $g^{\prime }=(g-{\omega }^{2}R)$, where $\omega$ is the angular velocity of the earth and $R$ the radius of the earth.