A cylindrical block of wood (density = 650 k | vedclass

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(C) The time period of oscillation for a floating body is given by $T = 2\pi \sqrt{\frac{h'}{g}}$,where $h'$ is the depth of the submerged part of the

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$39$

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(C) The time period of oscillation for a floating body is given by $T = 2\pi \sqrt{\frac{h'}{g}}$,where $h'$ is the depth of the submerged part of the block in equilibrium.For a floating body,the weight of the body equals the weight of the displaced liquid: $A \cdot h \cdot ho_{	ext{wood}} \cdot g = A \cdot h' \cdot ho_{	ext{liquid}} \cdot g$.Thus,$h' = h \cdot \frac{ho_{	ext{wood}}}{ho_{	ext{liquid}}}$.Substituting the given values: $h' = 54 \ cm 	imes \frac{650}{900}$.$h' = 54 	imes \frac{13}{18} = 3 	imes 13 = 39 \ cm$.The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.Comparing the two expressions,the equivalent length $l$ of the simple pendulum is equal to the submerged depth $h'$.Therefore,$l = 39 \ cm$.

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