A simple pendulum of length l_1 has time peri | vedclass

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(C) For a simple pendulum,the time period is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.Thus,$T_1 = 2 \pi \sqrt{\frac{l_1}{g}}$ and $T_2 = 2 \pi \sqrt{\f

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$\sqrt{T_1^2 - T_2^2}$

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(C) For a simple pendulum,the time period is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.Thus,$T_1 = 2 \pi \sqrt{\frac{l_1}{g}}$ and $T_2 = 2 \pi \sqrt{\frac{l_2}{g}}$.Squaring both sides,we get $T_1^2 = 4 \pi^2 \frac{l_1}{g}$ and $T_2^2 = 4 \pi^2 \frac{l_2}{g}$.From these,we can express the lengths as $l_1 = \frac{T_1^2 g}{4 \pi^2}$ and $l_2 = \frac{T_2^2 g}{4 \pi^2}$.Now,for a pendulum of length $(l_1 - l_2)$,the time period $T$ is given by $T = 2 \pi \sqrt{\frac{l_1 - l_2}{g}}$.Substituting the expressions for $l_1$ and $l_2$:$T = 2 \pi \sqrt{\frac{1}{g} \left( \frac{T_1^2 g}{4 \pi^2} - \frac{T_2^2 g}{4 \pi^2} \right)}$.$T = 2 \pi \sqrt{\frac{g}{g \cdot 4 \pi^2} (T_1^2 - T_2^2)}$.$T = 2 \pi \sqrt{\frac{1}{4 \pi^2} (T_1^2 - T_2^2)}$.$T = 2 \pi \cdot \frac{1}{2 \pi} \sqrt{T_1^2 - T_2^2}$.$T = \sqrt{T_1^2 - T_2^2}$.

$A$ simple pendulum of length $l_1$ has time period $T_1$. Another simple pendulum of length $l_2$ $(l_1 > l_2)$ has time period $T_2$. Then the time period of the pendulum of length $(l_1 - l_2)$ will be

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