Question

How many hours would make a day if the earth were rotating at such a high speed that the weight of a body on the equator were zero (Take radius of Earth, $R=6400\text{km}$ and Acceleration due to gravity, $g=10{\text{ms}}^{-2}$).

• A. $6.2\text{hours}$
• B. $1.4\text{hours}$
• C. $12\text{hours}$
• D. $24\text{hours}$

Answer 1

The correct option is B $1.4\text{hours}$

The apparent weight of a body of mass $m$ at the Equator is given by $$m{g}^{\prime }=m(g-{\omega }^{2}R)$$Where $\omega$ is the angular velocity of earth and $R$ is its radius.

Since, the weight of the body becomes zero.i.e., $m{g}^{\prime }=0$

$\Rightarrow mg(1-{\omega }^{2}R)=0$

$\Rightarrow mg=m{\omega }^{2}R$

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

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

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

Given,
$R=6400\text{km}=64\times {10}^{5}m.$
$g=10{\text{ms}}^{-2}$

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

$\Rightarrow T=2\pi \times 8\times {10}^2$

$\Rightarrow T=1600\pi \text{sec}.$

$\Rightarrow T=\frac{1600\pi }{60\times 60}\text{hr}$

$\therefore T=1.4\text{hr}$

Hence, option (b) is the correct answer.

Why this question? To familiarize students with the change in acceleration due to gravity due to the rotation of the earth.