A) \[4.83\times {{10}^{-3}}rad\,{{s}^{-1}}\]
B) \[5.41\times {{10}^{-3}}rad\,{{s}^{-1}}\]
C) \[7.82\times {{10}^{-4}}rad\,{{s}^{-1}}\]
D) \[8.88\times {{10}^{-14}}rad\,{{s}^{-1}}\]
Correct Answer: C
Solution :
The apparent weight of person on the equator (latitude\[\lambda =0\]) is given by \[w'=w-mR{{\omega }^{2}}\] Here,\[w'=(3/5)w=(3/5)mg\]\[[\because w=mg]\] \[\therefore \] \[(3/5)mg=mg-mR{{\omega }^{2}}\] Or \[mR{{\omega }^{2}}=mg-(3/5)mg\left( \frac{2}{5} \right)mg\] Or\[{{\omega }^{2}}=\frac{2g}{5R}\] \[\therefore \]\[\omega =\sqrt{\frac{2g}{5R}}\] Here,\[g=9.8\,m{{s}^{-2}}\]and \[R=6400\text{ }km=6400\times {{10}^{3}}m\] \[\therefore \]\[\omega =\sqrt{\left( \frac{2}{5}\times \frac{9.8}{6400\times {{10}^{3}}} \right)}rad{{s}^{-1}}\] \[=7.82\times {{10}^{-4}}rad{{s}^{-1}}\]
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