On a planet a freely falling body takes 2 ,sec when it is dropped from a height of 8 ,m , time perio

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13.Oscillations

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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 ..... $\sec$

A

$3.14 $

B

$16.28$

C

$1.57$

D

None of these

Solution

(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.

Standard 11

Physics

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