MHT CET · Physics · Gravitation
The radius of earth is \(6400 \mathrm{~km}\) and acceleration due to gravity \(\mathrm{g}=10 \mathrm{~ms}^{-2}\). For the weight of body of mass \(5 \mathrm{~kg}\) to be zero on equator, rotational velocity of the earth must be (in rad/s)
- A \(\frac{1}{80}\)
- B \(\frac{1}{400}\)
- C \(\frac{1}{800}\)
- D \(\frac{1}{1600}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{800}\)
Step-by-step Solution
Detailed explanation
At equator, for the weight to be zero, the gravitational force must be equal to centrifugal force.
\( \begin{aligned} & \mathrm{mR} \omega^2=\mathrm{mg} \\ & \omega^2=\frac{\mathrm{g}}{\mathrm{R}} \\ & \omega=\sqrt{\frac{\mathrm{g}}{\mathrm{R}}} \\ & \omega=\sqrt{\frac{10}{6.4 \times 10^6}} \\ & \omega=\frac{1}{800} \frac{\mathrm{rad}}{\mathrm{s}} \end{aligned} \)
\( \begin{aligned} & \mathrm{mR} \omega^2=\mathrm{mg} \\ & \omega^2=\frac{\mathrm{g}}{\mathrm{R}} \\ & \omega=\sqrt{\frac{\mathrm{g}}{\mathrm{R}}} \\ & \omega=\sqrt{\frac{10}{6.4 \times 10^6}} \\ & \omega=\frac{1}{800} \frac{\mathrm{rad}}{\mathrm{s}} \end{aligned} \)
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