The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as at present weight is ($g=$ gravitational acceleration,$R=$ equatorial radius of the earth).
- A$\sqrt{\frac{2g}{5R}}$
- B$\sqrt{\frac{3g}{5R}}$
- C$\sqrt{\frac{5R}{3g}}$
- D$\sqrt{\frac{3}{5}gR}$
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