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 $T^2 = 4 \pi^2 \frac{l}{g} \quad \dots (i)$When the length is i

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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 $T^2 = 4 \pi^2 \frac{l}{g} \quad \dots (i)$When the length is increased by $10 \ cm$,the new time period $T_1$ is given by $T_1^2 = 4 \pi^2 \frac{(l + 10)}{g} \quad \dots (ii)$When the length is decreased by $10 \ cm$,the new time period $T_2$ is given by $T_2^2 = 4 \pi^2 \frac{(l - 10)}{g} \quad \dots (iii)$Adding equations $(ii)$ and $(iii)$,we get$T_1^2 + T_2^2 = 4 \pi^2 \left[ \frac{l + 10}{g} + \frac{l - 10}{g} \right]$$T_1^2 + T_2^2 = 4 \pi^2 \left[ \frac{2l}{g} \right]$$T_1^2 + T_2^2 = 2 \left( 4 \pi^2 \frac{l}{g} \right)$Since $T^2 = 4 \pi^2 \frac{l}{g}$,we have$T_1^2 + T_2^2 = 2 T^2$

Column-$I$	Column-$II$
$(a)$ Land snail	$(1)$ Urea
$(b)$ Cartilaginous fishes	$(2)$ Uric acid
$(c)$ Marine fishes	$(3)$ Ammonia

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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