KCET 2017 Physics Solved Paper

Time period of oscillations of a simple pendulum is given as $T=2\sqrt{\frac{l}{g}}$ where l is length of pendulum. For pendulum A: Given, completes 10 oscillations is 20 s. Therefore, time period of single oscillation $=\frac{20}{10}=2$ $\Rightarrow T_A=2$ $\Rightarrow 2\pi\sqrt{\frac{I_A}{g}}=2$ → (1) For pendulum B: Given, completes 8 oscillations in 10 s. Therefore, time period of single oscillation $=\frac{10}{8}=\frac{5}{4}$ $\Rightarrow T_{B}=\frac{5}{4}$ $\Rightarrow 2\pi\sqrt{\frac{l_{B}}{g}}=\frac{5}{4}$ →(2) Divide Eq. (1) by Eq. (2), we get $\frac{2\pi\sqrt{\frac{l_A}{g}}}{2\pi\sqrt{\frac{l_B}{g}}}=\frac{\frac{2}{\frac{1}{5}}}{4}$ $\Rightarrow \sqrt{\frac{l_A}{l_B}}=\frac{2\times4}{5}$ $\Rightarrow \sqrt{\frac{l_A}{l_B}}=\frac{8}{5}$ $\Rightarrow \frac{l_A}{l_B}=\left(\frac{8}{5}\right)^{2}=\frac{64}{25}$ Therefore, ratio of lengths of pendulums A and B is $\frac{64}{25}$