What should be the velocity of earth due to rotation about its own axis so that the weight at equator…

What should be the velocity of earth due to rotation about its own axis so that the weight at equator becomes $\left(\frac{3}{5}\right)^{\text {th }}$ of initial value? (Radius of earth on equator $=6400 \mathrm{~km}, \mathrm{~g}=10 \frac{\mathrm{m}}{\mathrm{s}^{2}}, \cos 0^{\circ}=1$ )
  1. $3 \cdot 5 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}$
  2. $7 \cdot 91 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}$
  3. $6 \cdot 5 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}$
  4. $2 \cdot 5 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}$

Solution

True weight at equator, $\mathrm{W}=\mathrm{mg}$ Observed weight at equator, $\mathrm{W}^{\prime}=\mathrm{mg}^{\prime}=\frac{3}{5} \mathrm{mg}$ At equator, latitude $\lambda=0$ Using the formula, $\begin{aligned} & \mathrm{mg}^{\prime}=\mathrm{mg}-\mathrm{mR} \omega^2 \cos ^2 \lambda \\ & =\frac{3}{5} \mathrm{mg}=\mathrm{mg}-\mathrm{mR}^2 \omega^2 \cos ^2 \mathrm{o}=\mathrm{mg}-\mathrm{mR} \omega^2 \\ & \Rightarrow \mathrm{mR} \omega^2=\mathrm{mg}-\frac{3}{5} \mathrm{mg}=\frac{2}{5} \mathrm{mg} \\ & \therefore \omega=\left(\frac{2 \mathrm{~g}}{5 \mathrm{R}}\right)^{1 / 2} \\ & =\left(\frac{2 \times 9.8}{5 \times 6.4 \times 10^6}\right)^{1 / 2}=7.8 \times 10^{-4} \mathrm{rad} / \mathrm{s} . \end{aligned}$ ~

Asked in: MHT CET 2020 (15 Oct Shift 2)

Practice more Gravitation questions on Aicharya