解：$(1)I=\frac {U}{R_{1}+R_{2}}=\frac {6\ \text {V}}{10 \ Ω+40 \ Ω}=0.12\ \text {A}$
$ (2)R=\frac {U}{I'}=\frac {6\ \text {V}}{0.3\ \text {A}}=20 \ Ω$
$ R_2'=R-R_1=20 \ \mathrm {Ω}-10 \ \mathrm {Ω}=10 \ \mathrm {Ω} $
$ p=\frac {F}{S}=\frac {80\ \text {N}}{0.01\ \text m^{2}}=8000\ \text {Pa}$
$ h=\frac p{ρ_{水}g}=\frac {8000\ \text {Pa}}{1.0×10^{3}\ \text {kg/m}^{3}×10\ \text {N/kg}}=0.8\ \text {m}$
$ (3)p'=ρ_{水}gh'=1.0×10^{3}\ \text {kg/m}^{3}×10\ \text {N/kg}×1\ \text {m}=1×10^{4}\ \text {Pa}$
$ F'=p'S=1×10^{4}\ \text {Pa}×0.01\ \text m^{2}=100\ \text {N}$
$ $由图知$R_2''=8 \ \mathrm {Ω}$
$ R_1'=R-R_2''=20 \ \mathrm {Ω}-8 \ \mathrm {Ω}=12 \ \mathrm {Ω} $
