Question

A person stands on a spring balance at the equator. If the speed of earth's rotation increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case(approximately)?

• A. 1 hr
• B. 12 hr
• C. 2 hr
• D. 4 hr

Answer 1

The correct option is C

2 hr

Apparent weight = N = mg - m ${\omega }^2$R = mg'

Given - if $\omega$ is increased in such a way that the apparent weight is half the actual weight

$\Rightarrow mg-N=m(\omega {)}^{2}R$

N = mg - $(\omega {)}^{2}R$

$\frac{mg}{2}=mgm({\omega }^{\prime }{)}^{2}R$

M$({\omega }^{\prime }{)}^{2}R=\frac{mg}{2}$

${\omega }^{\prime }=\sqrt{\frac{g}{2R}}$

Let's assume that the length of entire day + night is T hrs. So, ${\omega }^{\prime }$ will be $\frac{2\pi }{T\times 60\times 60}$

$\Rightarrow$ $\frac{2\pi }{T\times 60\times 60}$ = $\sqrt{\frac{g}{2R}}$
$T=2hrs$