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13.Oscillations
hard
In an experiment to determine the period of a simple pendulum of length $1\, m$, it is attached to different spherical bobs of radii $r_1$ and $r_2$ . The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be $5\times10^{-4}\, s$, the difference in radii, $\left| {{r_1} - {r_2}} \right|$ is best given by .... $cm$
A
$1$
B
$0.1$
C
$0.5$
D
$0.01$
(JEE MAIN-2017)
Solution
As we know, Time-period of simple
pendulum, T $\propto \sqrt{l}$
$5 \times {10^{ – 4}} = \frac{1}{2}\frac{{{r_1} – {r_2}}}{1}$
$\because$ change in length $\Delta l=r_{1}-r_{2}$
$5 \times {10^{ – 4}} = \frac{1}{2}\frac{{{r_1} – {r_2}}}{1}$
$r_{1}-r_{2}=10 \times 10^{-4}$
$10^{-3} \mathrm{m}=10^{-1} \mathrm{cm}=0.1 \mathrm{cm}$
Standard 11
Physics
