Gravitation Question 112
Question: A simple pendulum has a time period $ [T_{1}] $ when on the earth’s surface and $ [T_{2}] $ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of $ [T_{2}/T_{1}] $ is[IIT-JEE 2001]
Options:
A) 1
B) $ [\sqrt{2}] $
C) 4
D) 2
Show Answer
Answer:
Correct Answer: D
Solution:
If acceleration due to gravity is g at the surface of earth then at height R it value becomes $ [g’=g{{\left( \frac{R}{R+h} \right)}^{2}}=\frac{g}{4}] $ $ [T_{1}=2\pi \sqrt{\frac{l}{g}}\text{ and },T_{2}=2\pi \sqrt{\frac{l}{g/4}}] $ $ [\frac{T_{2}}{T_{1}}=2] $
BETA
