Giải
\(\eqalign{
& \rho = {10^3}(kg/{m^3});{\rho _{KK}} = 1,21(kg/{m^3})\cr
& {p_A} = {p_B} + {1 \over 2}{\rho _{KK}}{v^2} \cr
& {p_A} = {p_B} + \rho gh \cr
& \Rightarrow h = {{{\rho _{KK}}{v^2}} \over {2\rho g}} = {{1,{{21.15}^2}} \over {{{2.10}^3}.9,8}} \approx 0,014(m) \cr
& h = 1,4cm \cr} \) 
