A) \[\frac{{{\omega }^{2}}}{R}\]
B) \[R{{\omega }^{2}}\]
C) \[{{R}^{2}}{{\omega }^{2}}\]
D) \[\frac{{{\omega }^{2}}}{{{R}^{2}}}\]
Correct Answer: B
Solution :
Effective value of acceleration due to gravity \[({{g}_{\phi }})\] is given by \[{{g}_{\phi }}=g\left( 1-\frac{{{\omega }^{2}}R}{g}{{\cos }^{2}}\phi \right)\]where\[\omega \]is angular velocity. At poles: \[\phi =90{}^\circ \] \[\therefore \] \[{{g}_{\phi }}={{g}_{p}}=g\] ?. (i) At equator \[\phi =0\] \[{{g}_{\phi }}={{g}_{e}}=g\left( 1-\frac{{{\omega }^{2}}R}{g} \right)=g-{{\omega }^{2}}R\] ?.. (ii) From Eqs. (i) and (ii), we get \[\therefore \] \[{{g}_{p}}-{{g}_{e}}=g-(g-{{\omega }^{2}}R)={{\omega }^{2}}R\]
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