Question

If the speed of the earth’s rotation is increased such that the apparent weight at the equator is one third the initial value, then what will be the length of the day in this case? (Take radius of Earth =$6.4\times {\text{1}0}^6\text{m and}g=10{\text{m/s}}^2$)

• A. $\text{3.28 hrs}$
• B. $\text{87.06 hrs}$
• C. $\text{10.54 hrs}$
• D. $\text{1.7 hrs}$

Answer 1

The correct option is D $\text{1.7 hrs}$

Initial apparent weight when earth rotates at angular velocity ${\omega }_o$,
$\begin{array}{}{W}_{app}=m(g-{\omega }_o^{2}R)& \text{(1)}\end{array}$
Final apparent weight when speed of earth increases from ${\omega }_o$ to $\omega$,
$W_{app}^{\prime }=m(g-{\omega }^{2}R)$
Since $W_{app}^{\prime }=\frac{W_{app}}{3}$
$\begin{array}{}\frac{W_{app}}{3}=m(g-{\omega }^{2}R)& \text{(2)}\end{array}$
Dividing equation $\left(1\right)$ and $\left(2\right)$,
$3=\frac{g-{\omega }_o^{2}R}{g-{\omega }^{2}R}$
$\Rightarrow \omega =\sqrt{\frac{2g+{\omega }_o^{2}R}{3R}}$
Here ${\omega }_o=\frac{2\pi }{24\times 60\times 60}$
Putting these values,
$\omega =1.02\times {10}^{-3}\text{rad/s}$
Hence new time period, $T=\frac{2\pi }{1.02\times {10}^{-3}}=6151.036\text{s}=1.7\text{hrs}$