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The angular speed of earth is `"rad s"^(-1)`, so that the object on equator may appear weightless, is (radius of earth = 6400 km)A. `1.23 xx 10^(-3)`B. `6.20 xx 10^(-3)`C. `1.56`D. `1.23 xx 10^(-5)` |
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Answer» Correct Answer - A At equator, `g=g-R omega^(2)rArr 0=g-Romega^(2)` `:. omega=sqrt((g)/(R))=sqrt((9.8)/(6400xx10^(3)))=1.23xx10^(-3) "rad s"^(-1)`. |
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