Answer
(a) $3.98\times 10^{-4}m^3$
$\rho_{block}=\frac{3.57}{3.98\times 10^{-4}}=8.97\times 10^3\frac{Kg}{m^3}$
(b) $820\frac{Kg}{m^3}$
Work Step by Step
(a) As $T=mg$
$\implies m=\frac{T}{g}=\frac{35.0}{9.81}=3.57Kg$
We know that
$T+\rho_w V_g-mg=0$
This can be rearranged as:
$V=\frac{mg-T}{\rho_wg}$
We plug in the known values to obtain:
$V=\frac{(3.57)(9.81)}{(1000)(9.81)-3.11}=3.98\times 10^{-4}m^3$
Now we can find the density as
$\rho_{block}=\frac{m}{V}$
We plug in the known values to obtain:
$\rho_{block}=\frac{3.57}{3.98\times 10^{-4}}=8.97\times 10^3\frac{Kg}{m^3}$
(b) We know that
$T+\rho_{0il}-mg=0$
This can be rearranged as:
$\rho_{oil}=\frac{mg-T}{Vg}$
We plug in the known values to obtain:
$\rho_{oil}=\frac{35-31.8}{(3.98\times 10^{-4})(9.81)}=820\frac{Kg}{m^3}$
