pendulum, T $\propto \sqrt{l}$
$5 \times {10^{ - 4}} = \frac{1}{2}\frac{{{r_1} - {r_2}}}{1}$
$\because$ change in length $\Delta l=r_{1}-r_{2}$
$5 \times {10^{ - 4}} = \frac{1}{2}\frac{{{r_1} - {r_2}}}{1}$
$r_{1}-r_{2}=10 \times 10^{-4}$
$10^{-3} \mathrm{m}=10^{-1} \mathrm{cm}=0.1 \mathrm{cm}$
$(1)$ Amplitude $(2) $ Period $(3)$ Displacement
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