Answer
$(a)\space 5.97\space cm $
$(b)\space 7.12\space cm$
Work Step by Step
When the system is in equilibrium, the buoyant force is equal to the weight of the tube.
$F_{B}=W_{T}-(1)$
According to the principle of Archimedes', the magnitude of the buoyant force is equal to the weight of the displaced fluid. Therefore
$F_{B}=V\rho g=(hA)\rho g-(2)$
(1),(2)=>
$(hA)\rho g=W_{T}=> h=\frac{W_{T}}{A\rho g}$
(a) $ h_{acid}=\frac{W_{T}}{A\rho_{acid} g}$ ; Let's plug known values into this equation.
$h_{acid}=\frac{5.88\times10^{-2}N}{(1280\space kg/m^{3})(7.85\times10^{-5}m^{2})(9.8\space m/s^{2})}=5.97\space cm$
The height that the acid rises = 5.97 cm
(b) $ h_{af}=\frac{W_{T}}{A\rho_{af} g}$ ; Let's plug known values into this equation.
$h_{af}=\frac{5.88\times10^{-2}N}{(1073\space kg/m^{3})(7.85\times10^{-5}m^{2})(9.8\space m/s^{2})}=7.12\space cm$
The height of antifreeze rises = 7.12 cm
