The speed with which the earth would have to rotate about its axis so that a person on the equator would…

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 ( $\mathrm{g}=$ gravitational acceleration, $R=$ equatorial radius of the earth)
  1. $\sqrt{\frac{2 \mathrm{~g}}{5 \mathrm{R}}}$
  2. $\sqrt{\frac{3 \mathrm{~g}}{5 \mathrm{R}}}$
  3. $\sqrt{\frac{5 \mathrm{R}}{3 \mathrm{~g}}}$
  4. $\sqrt{\frac{3}{5}} \mathrm{gR}$

Solution

$\begin{array}{ll} & g^{\prime}=g-R \omega^2 \cos ^2 \theta \\ \therefore \quad & g^{\prime}=g-R \omega^2 \\ & \text { Given } g^{\prime}=\frac{3 g}{5} \\ \therefore \quad & \omega^2=\frac{2 g}{5 R} \text { or } \omega=\sqrt{\frac{2 g}{5 R}}\ldots(\because \theta=0)\end{array}$

Asked in: MHT CET 2024 (16 May Shift 1)

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