Gravitation Question 253

Question: In the following four periods[AMU 2000] Time of revolution of a satellite just above the earth’s surface [(Tst)] Period of oscillation of mass inside the tunnel bored along the diameter of the earth [(Tma)] Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 N/kg [(Tsp)] Period of an infinite length simple pendulum in the earth’s real gravitational field [(Tis)]

Options:

A) [Tst>Tma]

B) [Tma>Tst]

C) [Tsp<Tis]

D) [Tst=Tma=Tsp=Tis]

Show Answer

Answer:

Correct Answer: C

Solution:

(i)[Tst=2π(R+h)3GM]

[=2πRg] [As h «R and [GM=gR2]]

(ii) [Tma=2πRg]

(iii) [Tsp=2π1g(1l+1R)=2πR2g] [As l = R] (iv) [Tis=2πRg]

[[As,,l=]]



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