If a man at the equator would weigh (35)thof his weight, the angular speed of the earth is
A
√25gR
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B
√gR
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C
√Rg
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D
√25Rg
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Solution
The correct option is A√25gR Weight of the body at equator =35 of initial weight ∴g′=35g(becausemassremainsconstant) g′=g−ω2Rcos2λ⇒35g=g−ω2Rcos2(0∘) ⇒ω2=2g5R⇒ω=√2g5R