MCQ
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 ..... $\sec$
- ✓$3.14 $
- B$16.28$
- C$1.57$
- DNone of these
$ \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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