Question 176 of 194

A cylindrical block of wood (density = $650kg{m}^{-3}$), of base area $30{cm}^{2}$ and height $54cm$, floats in a liquid of density $900kg{m}^{-3}$. 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 (nearly) :

A$52cm$B$26cm$C$39cm$D$65cm$

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Answer: (C) $39cm$

Case-I

In equilibrium

${F}_{up}=mg$

${V}_{i}{\rho }_{L}g=mg$

$Ah{\rho }_{L}g=AL{\rho }_{s}g\ldots \ldots \ldots \left(1\right)$



Case-II

Let block is pressed by x more in liquid

${F}_{net}={F}_{up}^{1}-mg$

$={V}_{i}^{1}{\rho }_{L}g-mg$

${F}_{net}=A\left(h+x\right){\rho }_{L}g-AL{\rho }_{s}g$

${F}_{net}=Ah{\rho }_{L}g+Ax{\rho }_{L}g-AL{\rho }_{s}g$. . . . . . . . . . (2)

from equation (1) & (2)

${F}_{net}=Ax{\rho }_{L}g$restoring force in upward direction

$\alpha =\dfrac{{F}_{net}}{m}=\left(\dfrac{A{\rho }_{L}g}{A{\rho }_{s}L}\right)x$

$\alpha ={\omega }^{2}x for SHM$

and ${\omega }^{2}=\dfrac{g}{l}$

so $\dfrac{g}{l}=\dfrac{{\rho }_{L}g}{{\rho }_{s}L}$

so $l=\dfrac{{\rho }_{s}L}{{\rho }_{L}}=\dfrac{650\times 54}{900}=39cm$