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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 =6.4×106 m and g=10 m/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 D 1.7 hrs
Initial apparent weight when earth rotates at angular velocity ωo,
Wapp=m(gω2oR)(1)
Final apparent weight when speed of earth increases from ωo to ω,
Wapp=m(gω2R)
Since Wapp=Wapp3
Wapp3=m(gω2R)(2)
Dividing equation (1) and (2),
3=gω2oRgω2R
ω=2g+ω2oR3R
Here ωo=2π24×60×60
Putting these values,
ω=1.02×103 rad/s
Hence new time period, T=2π1.02×103=6151.036 s=1.7 hrs

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