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 =6.4×106m and g=10m/s2)
A
3.28 hrs
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B
87.06 hrs
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C
10.54 hrs
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D
1.7 hrs
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Solution
The correct option is D1.7 hrs Initial apparent weight when earth rotates at angular velocity ωo, Wapp=m(g−ω2oR)(1) Final apparent weight when speed of earth increases from ωo to ω, W′app=m(g−ω2R) Since W′app=Wapp3 Wapp3=m(g−ω2R)(2) Dividing equation (1) and (2), 3=g−ω2oRg−ω2R ⇒ω=√2g+ω2oR3R Here ωo=2π24×60×60 Putting these values, ω=1.02×10−3rad/s Hence new time period, T=2π1.02×10−3=6151.036s=1.7hrs