Question

What should be the length of the day, so that the weight of a body on the equator of earth becomes zero ?

• A. $4\text{hrs}$
• B. $1.4\text{hrs}$
• C. $14\text{hrs}$
• D. $4\text{hrs}$

Answer 1

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

Weight at equator $W=M(g-R{\omega }^2)$
As $W=0$ $\Rightarrow M(g-R{\omega }^2)=0$

$\Rightarrow \omega =\sqrt{\frac{g}{R}}=\sqrt{\frac{9.8}{6400\times {10}^3}}=\frac{1}{808}\text{rad/sec}$

Now, $\frac{2\pi }{T}=\frac{1}{808}$

$T=2\pi \times 808\text{s}\approx 84.6\text{minute}$
$T=1.4\text{hr}$