A cylindrical block of wood of base area $30\ cm^2$ , floats in a liquid of density $900\ kg/m^3$ . The block is depressed lightly and then released. The time period of the resulting oscillations of the block is equal to that of spring with block of same mass, then spring constant is equal to ........ $N/m$
Diffcult
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Lets say the displacement of cylinder from its equilibrium position is $'x'.$
$\Rightarrow \mathrm{ma}=-\rho \mathrm{g} \mathrm{Ax}$
$\Rightarrow \mathrm{a}=-\left(\frac{\rho \mathrm{g} \mathrm{A}}{\mathrm{m}}\right) \mathrm{x}$
$\Rightarrow \omega_{\text {cylinder }}=\sqrt{\frac{\rho g A}{m}} \ldots(i)$
$\Rightarrow \omega_{\text {spring -block }}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ $...(ii)$
$\Rightarrow \omega_{\text {cylinder }}=\omega_{\text {spring - block }} \ldots$ $(iii)$ $[given]$
$\Rightarrow \mathrm{K}=\rho \mathrm{g} \mathrm{A}[\text { from }(\mathrm{i}),(\mathrm{ii}),(\mathrm{iii})]$
$\mathrm{K}=27 \mathrm{N} / \mathrm{m}$
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