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…

Case-I

In equilibrium

$${F}_{\mathrm{up}}=\mathrm{mg}$$Fup​=mg

$${V}_{i}{\rho }_{L}g=mg$$Vi​ρL​g=mg

$$Ah{\rho }_{L}g=AL{\rho }_{s}g\ldots \ldots \ldots (1)$$AhρL​g=ALρs​g………(1)

Case-II

Let block is pressed by x more in liquid

$${F}_{net}={F}_{up}^{1}-mg$$Fnet​=Fup1​−mg

$$={V}_{i}^{1}{\rho }_{L}g-\mathrm{mg}$$=Vi1​ρL​g−mg

$${F}_{\mathrm{net}}=\mathrm{A }(h+x){\rho }_{L}g-\mathrm{AL}{\rho }_{s}g$$Fnet​=A(h+x)ρL​g−ALρs​g

${F}_{\mathrm{net}}=\mathrm{Ah}{\rho }_{L}g+\mathrm{Ax}{\rho }_{L}g-\mathrm{AL}{\rho }_{s}g$Fnet​=AhρL​g+AxρL​g−ALρs​g. . . . . . . . . . (2)

from equation (1) & (2)

${F}_{\mathrm{net}}=\mathrm{Ax}{\rho }_{L}g$Fnet​=AxρL​grestoring force in upward direction

$$\alpha =\frac{{F}_{net}}{m}=(\frac{A{\rho }_{L}g}{A{\rho }_{s}L})x$$α=mFnet​​=(Aρs​LAρL​g​)x

$$\alpha ={\omega }^{2}\mathrm{x for SHM}$$α=ω2xforSHM

and ${\omega }^{2}=\frac{g}{l}$ω2=lg​

so $\frac{g}{l}=\frac{{\rho }_{L}g}{{\rho }_{s}L}$lg​=ρs​LρL​g​

so $l=\frac{{\rho }_{s}L}{{\rho }_{L}}=\frac{650\times 54}{900}=39\mathrm{cm}$l=ρL​ρs​L​=900650×54​=39cm