Gravitation Question 253
Question: In the following four periods[AMU 2000] Time of revolution of a satellite just above the earth’s surface $ [(T_{st})] $ Period of oscillation of mass inside the tunnel bored along the diameter of the earth $ [(T_{ma})] $ Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 N/kg $ [(T_{sp})] $ Period of an infinite length simple pendulum in the earth’s real gravitational field $ [(T_{is})] $
Options:
A) $ [T_{st}>T_{ma}] $
B) $ [T_{ma}>T_{st}] $
C) $ [T_{sp}<T_{is}] $
D) $ [T_{st}=T_{ma}=T_{sp}=T_{is}] $
Show Answer
Answer:
Correct Answer: C
Solution:
(i)$ [T_{st}=2\pi \sqrt{\frac{{{(R+h)}^{3}}}{GM}}] $
$ [=2\pi \sqrt{\frac{R}{g}}] $ [As h «R and $ [GM=gR^{2}]] $
(ii) $ [T_{ma}=2\pi \sqrt{\frac{R}{g}}] $
(iii) $ [T_{sp}=2\pi \sqrt{\frac{1}{g\left( \frac{1}{l}+\frac{1}{R} \right)}}=2\pi \sqrt{\frac{R}{2g}}] $ [As l = R] (iv) $ [T_{is}=2\pi \sqrt{\frac{R}{g}}] $
$ [[As,,l=\infty ]] $
BETA
