Gravitation Question 330

Question: R is the radius of the earth and co is its angular velocity and $ [{g_{p}}] $ is the value of g at the poles. The effective value of g at the latitude $ [\lambda = 60{}^\circ ] $ will be equal to

Options:

A) $ [g_{p}-\frac{1}{4}R{{\omega }^{2}}] $

B) $ [g_{p}-\frac{3}{4}R{{\omega }^{2}}] $

C) $ [g_{p}-R{{\omega }^{2}}] $

D) $ [g_{p}+\frac{1}{4}R{{\omega }^{2}}] $

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Answer:

Correct Answer: A

Solution:

  • $ [g=g_{p}-R{{\omega }^{2}}{{\cos }^{2}}\lambda ] $

    $ [g=g_{p}-{{\omega }^{2}}R{{\cos }^{2}}{{60}^{o}}=g_{p}-\frac{1}{4}R{{\omega }^{2}}] $



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