On a planet a freely falling body takes $2 \,sec$ when it is dropped from a height of $8 \,m$, the time period of simple pendulum of length $1\, m$ on that planet is ..... $\sec$
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(a) On a planet, if a body dropped initial velocity $(u = 0)$ from a height h and takes time t to reach the ground then $h = \frac{1}{2}{g_P}{t^2}$
$ \Rightarrow {g_P} = \frac{{2\,h}}{{{t^2}}} = \frac{{2 \times 8}}{4} = 4\,m/{s^2}$
Using $T = 2\pi \sqrt {\frac{l}{g}} $ ==> $T = 2\,\pi \sqrt {\frac{1}{4}} = \pi = 3.14\,$sec.
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