Answer
(a) $2.34\times 10^{-4}m^3$
(b) $8.70\times 10^3\frac{Kg}{m^3}$
Work Step by Step
(a) We know that
$\Sigma F=T_w+T_B-W=0$
$\implies F_B=\rho_w Vg=W-T_w$
This can be rearranged as:
$V=\frac{W-T}{\rho_wg}$
We plug in the known values to obtain:
$V=\frac{20.0-17.7}{1000(9.81)}=2.34\times 10^{-4}m^3$
(b) As $\rho=\frac{m}{V}$
$\rho=\frac{W}{gV}$
$\implies \rho=\frac{20.0}{(9.81)(2.345\times 10^{-4})}=8.70\times 10^3\frac{Kg}{m^3}$
