A cylindrical block of wood (density = 650, kg, m^{-3} ), of base area 30,cm^2 and height 54, cm , f

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13.Oscillations

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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)

A

$52$

B

$65$

C

$39$

D

$26$

(JEE MAIN-2015)

Solution

Required equivalent length

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

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

$\Rightarrow l=0.39 \mathrm{m}=39 \mathrm{cm}$

Standard 11

Physics

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