A) moving up with uniform velocity
B) moving up with uniform acceleration
C) moving down with uniform acceleration
D) moving around the earth in geostationary orbit
Correct Answer: A
Solution :
Time period (T) of a seconds pendulum of length I, acceleration due to gravity g, is given \[T=2\pi \sqrt{\frac{l}{g}}\] \[\Rightarrow \] \[T\propto \frac{1}{\sqrt{g}}\] Given that time period of a seconds pendulum decreases, since \[T\propto \frac{1}{\sqrt{g}}\] it implies that effective value of g is increasing. Hence, it means that rocket is accelerating upwards, with uniform velocity.
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