Q.

In the following four periods     (i) Time of revolution of a satellite just above the earth’s surface  (Tst)(ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth  (Tma)(iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 N/kg  (Tsp)(iv) Period of an infinite length simple pendulum in the earth’s real gravitational field  (Tis)

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a

Tst>Tma

b

Tma>Tst

c

Tsp

d

Tst=Tma=Tsp=Tis

answer is C.

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Detailed Solution

(i) Tst=2π(R+h)3GM=2πRg                                  [As h<
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In the following four periods     (i) Time of revolution of a satellite just above the earth’s surface  (Tst)(ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth  (Tma)(iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 N/kg  (Tsp)(iv) Period of an infinite length simple pendulum in the earth’s real gravitational field  (Tis)