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(D) The time period $T$ of $SHM$ for a floating body undergoing small vertical oscillations is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the

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$2\pi \sqrt{\frac{da}{g}}$

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(D) The time period $T$ of $SHM$ for a floating body undergoing small vertical oscillations is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the depth of the object submerged in the liquid at equilibrium.According to the law of floatation,the weight of the object equals the weight of the displaced liquid:$V_{total} \cdot \rho_{object} \cdot g = V_{submerged} \cdot \rho_{water} \cdot g$$(abc) \cdot d \cdot g = (b \cdot c \cdot l) \cdot 1 \cdot g$Simplifying the equation:$abc \cdot d = bcl$$l = da$Substituting $l$ into the time period formula:$T = 2\pi \sqrt{\frac{da}{g}}$

$A$ cuboidal piece of wood has dimensions $a, b$ and $c$. Its relative density is $d$. It is floating in a large body of water such that side $a$ is vertical. It is pushed down a bit and released. The time period of $SHM$ executed by it is:

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