Answer
(a) The density of the liquid is $1.21\times 10^3~kg/m^3$
(b) $\rho = \frac{(M-M_a)~\rho_o}{M}$
Work Step by Step
(a) Let $M$ be the mass of the aluminum. Let $\rho_a$ be the density of aluminum. We can find the volume $V$ of the aluminum as:
$V = \frac{M}{\rho_a}$
$V = \frac{3.80~kg}{2.70\times 10^3~kg/m^3}$
$V = 1.41\times 10^{-3}~m^3$
Let $M_a$ be the apparent mass of the aluminum. Let $\rho$ be the density of the liquid. Note that the apparent weight is equal to the true weight of the aluminum minus the buoyant force provided by the liquid. Therefore;
$M_a~g = M~g-F_B$
$F_B = M~g-M_a~g$
$\rho~V~g = M~g-M_a~g$
$\rho = \frac{M-M_a}{V}$
$\rho = \frac{3.80~kg-2.10~kg}{1.41\times 10^{-3}~m^3}$
$\rho = 1.21\times 10^3~kg/m^3$
The density of the liquid is $1.21\times 10^3~kg/m^3$.
(b) Let $M$ be the mass of the object and let $M_a$ be the apparent mass. Let $\rho_o$ be the density of the object. Let $\rho$ be the density of the liquid.
$M_a~g = M~g-F_B$
$F_B = M~g-M_a~g$
$\rho~V~g = M~g-M_a~g$
$\rho = \frac{M-M_a}{V}$
$\rho = \frac{M-M_a}{\frac{M}{\rho_o}}$
$\rho = \frac{(M-M_a)~\rho_o}{M}$
