MHT CET · Physics · Gravitation
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)
- A \(\sqrt{\frac{2 \mathrm{~g}}{5 \mathrm{R}}}\)
- B \(\sqrt{\frac{3 \mathrm{~g}}{5 \mathrm{R}}}\)
- C \(\sqrt{\frac{5 \mathrm{R}}{3 \mathrm{~g}}}\)
- D \(\sqrt{\frac{3}{5}} \mathrm{gR}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\frac{2 \mathrm{~g}}{5 \mathrm{R}}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & g^{\prime}=g-R \omega^2 \cos ^2 \theta \\ \therefore \quad & g^{\prime}=g-R \omega^2 \\ & \text { Given } g^{\prime}=\frac{3 g}{5} \\ \therefore \quad & \omega^2=\frac{2 g}{5 R} \text { or } \omega=\sqrt{\frac{2 g}{5 R}}\ldots(\because \theta=0)\end{array}\)
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