$A$ mass falls from a height $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of the Earth,it is found that $t = 2T$. The entire setup is taken to the surface of another planet whose mass is half that of the Earth and radius is the same. The same experiment is repeated and corresponding times are noted as $t'$ and $T'$.
- A$t' = \sqrt{2} T'$
- B$t' > 2 T'$
- C$t' < 2 T'$
- D$t' = 2 T'$
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