A cylindrical block of wood (density \(= 650\, kg\, m^{-3}\)), of base area \(30\,cm^2\) and height \(54\, cm\), floats in a liquid of density \(900\, kg\, m^{-3}\) . The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length ..... \(cm\) (nearly)

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Correct Answer

(C) \(39\)

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Detailed explanation

Required equivalent length

\(=\frac{\rho_{\text {wood }}}{\rho_{\text {liquid }}} \times\) height of block

\(=\frac{650}{900} \times 54 \times 10^{-2}\)

\(\Rightarrow l=0.39 \mathrm{m}=39 \mathrm{cm}\)

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