Gravitation Question 43
Question: The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60o latitude becomes zero is (Radius of earth = 6400 km. At the poles $ [g=10m{{s}^{-2}})] $ [EAMCET 2000]
Options:
A) $ [2.5\times {{10}^{-3}}rad/s] $
B) $ [5.0\times {{10}^{-1}}rad/s] $
C) $ [10\times 10^{1}rad/s] $
D) $ [7.8\times {{10}^{-2}}rad/s] $
Show Answer
Answer:
Correct Answer: A
Solution:
$ [{g}’=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda ] $
Þ $ [0=g-{{\omega }^{2}}R{{\cos }^{2}}60^{o}] $
$ [0=g-\frac{{{\omega }^{2}}R}{4}\Rightarrow \omega =2\sqrt{\frac{g}{R}}=\frac{1}{400}\frac{rad}{\sec }] $
$ [=2.5\times {{10}^{-3}}\frac{rad}{sec}] $
