Answer
1.46 m
Work Step by Step
The pressure at the bottom of the right container $P_{R},$
$P_{R}=P_{atm}+h_{1}\rho_{m}g-(1)$
The pressure at the bottom of the left container $P_{L},$
$P_{L}=P_{atm}+h_{w}\rho_{w}g+h_{2}\rho_{m}g-(2)$
(2)=(1)=>
$P_{atm}+h_{w}\rho_{w}g+h_{2}\rho_{m}g=P_{atm}+h_{1}\rho_{m}g$
$h_{w}\rho_{w}g+(h_{2}-h_{1})\rho_{m}g=0-(3)$
Given that,
$h_{w}=1\space m$ & $h_{1}+h_{2}=1\space m$
(3)=>
$h_{w}\rho_{w}g+(2h_{2}-1)\rho_{m}g=0$
$h_{2}=\frac{1}{2}(1-\frac{\rho_{w}}{\rho_{m}})$ ; Let's plug known values into this equation.
$h_{2}=\frac{1}{2}(1-\frac{1000\space kg/m^{3}}{13600\space kg/m^{3}})=0.46\space m$
The fluid level in the = $h_{w}+h_{2}=1\space m+0.46\space m=1.46\space m$
left container
