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?
2 hr
Apparent weight = N = mg - m ω2R = mg'
Given - if ω is increased in such a way that the apparent weight is half the actual weight
⇒mg−N=m(ω)2R
N = mg - (ω)2R
mg2=mgm(ω′)2R
M(ω′)2R=mg2
ω′=√g2R
Let's assume that the length of entire day + night is T hrs. So, ω′ will be 2πT×60×60
⇒ 2πT×60×60 = √g2R
T=2hrs