Gravitation Question 23
Question: R is the radius of the earth and $ [\omega ] $ 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[MP PMT 1999]
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}}] $
Show Answer
Answer:
Correct Answer: A
Solution:
$ [g=g_{p}-R{{\omega }^{2}}{{\cos }^{2}}\lambda ] $ =$ [g_{p}-{{\omega }^{2}}R{{\cos }^{2}}60{}^\circ ] $ =$ [g_{p}-\frac{1}{4}R{{\omega }^{2}}] $
