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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the pendulum and $g$ is the acceleration

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$64/25$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.For pendulum $A$: It completes $10$ oscillations in $20 \ s$.Therefore,the time period $T_A = \frac{20}{10} = 2 \ s$.$2\pi \sqrt{\frac{l_A}{g}} = 2 \quad (1)$For pendulum $B$: It completes $8$ oscillations in $10 \ s$.Therefore,the time period $T_B = \frac{10}{8} = \frac{5}{4} \ s$.$2\pi \sqrt{\frac{l_B}{g}} = \frac{5}{4} \quad (2)$Dividing equation $(1)$ by equation $(2)$:$\frac{2\pi \sqrt{l_A/g}}{2\pi \sqrt{l_B/g}} = \frac{2}{5/4}$$\sqrt{\frac{l_A}{l_B}} = \frac{2 	imes 4}{5} = \frac{8}{5}$Squaring both sides:$\frac{l_A}{l_B} = \left(\frac{8}{5}ight)^2 = \frac{64}{25}$

Two simple pendulums $A$ and $B$ are made to oscillate simultaneously. It is found that $A$ completes $10$ oscillations in $20 \ s$ and $B$ completes $8$ oscillations in $10 \ s$. The ratio of the lengths of $A$ and $B$ is

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