The time period of a simple pendulum is T. Wh | vedclass

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(C) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$, so $T^2 = 4 \pi^2 \left(\frac{l}{g}\right) \quad ...(i)$When the

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$2 T^2=T_1^2+T_2^2$

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(C) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$, so $T^2 = 4 \pi^2 \left(\frac{l}{g}\right) \quad ...(i)$When the length is increased by $10 \ cm$, the new time period $T_1$ is given by $T_1^2 = 4 \pi^2 \left(\frac{l+10}{g}\right) \quad ...(ii)$When the length is decreased by $10 \ cm$, the new time period $T_2$ is given by $T_2^2 = 4 \pi^2 \left(\frac{l-10}{g}\right) \quad ...(iii)$Adding equations $(ii)$ and $(iii)$, we get:$T_1^2 + T_2^2 = 4 \pi^2 \left(\frac{l+10}{g}\right) + 4 \pi^2 \left(\frac{l-10}{g}\right)$$T_1^2 + T_2^2 = \frac{4 \pi^2}{g} (l + 10 + l - 10)$$T_1^2 + T_2^2 = \frac{4 \pi^2}{g} (2l)$$T_1^2 + T_2^2 = 2 \left(4 \pi^2 \frac{l}{g}\right)$Substituting equation $(i)$ into this, we get:$T_1^2 + T_2^2 = 2 T^2$

The time period of a simple pendulum is $T$. When the length is increased by $10 \ cm$, its period is $T_1$. When the length is decreased by $10 \ cm$, its period is $T_2$. Then, the relation between $T, T_1$, and $T_2$ is

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