Gravitation Question 38

Question: What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km [AMU 1999]

Options:

A) $ [7.4\times {{10}^{-4}},rad/\sec ] $

B) $ [6.7\times {{10}^{-4}},rad/\sec ] $

C) $ [7.8\times {{10}^{-4}},rad/\sec ] $

D) $ [8.7\times {{10}^{-4}},rad/\sec ] $

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Answer:

Correct Answer: C

Solution:

Weight of the body at equator = $ [\frac{3}{5}] $

of initial weight $ [g’=\frac{3}{5}g] $

(because mass remains constant) $ [g’=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda ] $

Þ $ [\frac{3}{5}g=g-{{\omega }^{2}}R{{\cos }^{2}}(0{}^\circ )] $

Þ $ [{{\omega }^{2}}=\frac{2g}{5R}] $

Þ $ [\omega =\sqrt{\frac{2g}{5R}}=\sqrt{\frac{2\times 10}{5\times 6400\times 10^{3}}}] $ = $ [7.8\times {{10}^{-4}}\frac{rad}{\sec }] $



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