An object with uniform density rho is attache | vedclass

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(B) For the equilibrium of the block hung from the spring,the sum of the spring force and the buoyant force must equal the weight of the block.Let $V$

Vedclass · Accepted Answer

$\rho=\frac{\rho_1 x_2-\rho_2 x_1}{x_2-x_1}$

Vedclass · Answer

(B) For the equilibrium of the block hung from the spring,the sum of the spring force and the buoyant force must equal the weight of the block.Let $V$ be the volume of the object and $k$ be the spring constant.The weight of the object is $W = \rho V g$.The buoyant force in a liquid of density $\rho_f$ is $F_B = \rho_f V g$.The spring force is $F_s = kx$.For equilibrium: $kx + \rho_f V g = \rho V g$.In the first liquid (density $\rho_1$): $k x_1 + \rho_1 V g = \rho V g \quad \dots(i)$In the second liquid (density $\rho_2$): $k x_2 + \rho_2 V g = \rho V g \quad \dots(ii)$From $(i)$,$k = \frac{(\rho - \rho_1) V g}{x_1}$.From $(ii)$,$k = \frac{(\rho - \rho_2) V g}{x_2}$.Equating the two expressions for $k$:$\frac{(\rho - \rho_1) V g}{x_1} = \frac{(\rho - \rho_2) V g}{x_2}$$(\rho - \rho_1) x_2 = (\rho - \rho_2) x_1$$\rho x_2 - \rho_1 x_2 = \rho x_1 - \rho_2 x_1$$\rho (x_2 - x_1) = \rho_1 x_2 - \rho_2 x_1$$\rho = \frac{\rho_1 x_2 - \rho_2 x_1}{x_2 - x_1}$

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