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 ]] $



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