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 ] $
Show Answer
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 }] $
BETA
