The time period of Simple pendulum on a freely moving artificial satel

1.	The time period of Simple pendulum on a freely moving artificial satellesAFMC 2002)dyinfintyificial satellite isa) zero(b) 2 sec(c) 1 sechoomes 1/1oth its weight on the surface of earth is
Answer» Infinity!In a freely orbiting satellite which is revolving around the earth, the net value of acceleration due to gravity is zero and hence we experience zero gravity inside the spacecraft. And, the time period of a simple pendulum is is inversely proportional to the square root of acceleration due to gravity. Time period=2 ×pi× (l/g)^1/2 (where l is length of the simple pendulum and g is acceleration due to gravity) The value of g is zero inside a spacecraft, and in the equation g is in the denominator. And any number divided by zero is NOT DEFINED. As per me, either the time period is not defined or infinity. And since the time period cannot be not defined the answer is infinity which also implies that the pendulum won't oscillate.

The time period of Simple pendulum on a freely moving artificial satellesAFMC 2002)dyinfintyificial satellite isa) zero(b) 2 sec(c) 1 sechoomes 1/1oth its weight on the surface of earth is

Answer»

Infinity!In a freely orbiting satellite which is revolving around the earth, the net value of acceleration due to gravity is zero and hence we experience zero gravity inside the spacecraft.

And, the time period of a simple pendulum is is inversely proportional to the square root of acceleration due to gravity. Time period=2 ×pi× (l/g)^1/2 (where l is length of the simple pendulum and g is acceleration due to gravity)

The value of g is zero inside a spacecraft, and in the equation g is in the denominator. And any number divided by zero is NOT DEFINED.

As per me, either the time period is not defined or infinity.

And since the time period cannot be not defined the answer is infinity which also implies that the pendulum won't oscillate.

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