Gujarati
13.Oscillations
easy

The length of the second pendulum on the surface of earth is $1\, m$. The length of seconds pendulum on the surface of moon, where g is 1/6th value of $g$ on the surface of earth, is

A

$\frac{1}{6}\, m$

B

$6 \,m$

C

$\frac{1}{36}\, m$

D

$36 \,m$

Solution

(a) $T = 2\pi \sqrt {\frac{l}{g}} $

==> $\sqrt {\frac{l}{g}} $= constant

$ \Rightarrow l \propto g;$

==> $\frac{{{l_m}}}{1} = \frac{1}{6}\frac{g}{g}$

$\Rightarrow {l_m} = \frac{1}{6}\,m$

Standard 11
Physics

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