The length of the seconds pendulum is decreas | vedclass

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(B) For a seconds pendulum,the time period $T = 2 \ s$.The formula for the time period is $T = 2 \pi \sqrt{\frac{l}{g}}$.Substituting $T = 2$,we get $

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$978$

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(B) For a seconds pendulum,the time period $T = 2 \ s$.The formula for the time period is $T = 2 \pi \sqrt{\frac{l}{g}}$.Substituting $T = 2$,we get $2 = 2 \pi \sqrt{\frac{l}{g}}$,which simplifies to $1 = \pi \sqrt{\frac{l}{g}}$.Squaring both sides,we get $1 = \pi^2 \frac{l}{g}$,so $l = \frac{g}{\pi^2}$.At place $A$,$g_A = 981 \ cm/s^2$ and $\pi^2 = 10$,so $l_A = \frac{981}{10} = 98.1 \ cm$.At place $B$,the length is decreased by $0.3 \ cm$,so $l_B = 98.1 - 0.3 = 97.8 \ cm$.Since it is still a seconds pendulum,$T = 2 \ s$ at place $B$ as well.Using $T = 2 \pi \sqrt{\frac{l_B}{g_B}}$,we have $1 = \pi^2 \frac{l_B}{g_B}$.Therefore,$g_B = \pi^2 	imes l_B = 10 	imes 97.8 = 978 \ cm/s^2$.

The length of the seconds pendulum is decreased by $0.3 \ cm$ when it is shifted from place $A$ to place $B$. If the acceleration due to gravity at place $A$ is $981 \ cm/s^2$,the acceleration due to gravity at place $B$ is $($ Take $\pi^2 = 10$ $)$ (in $cm/s^2$)

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