Answer
0.74 m
Work Step by Step
We can write, Pressure at the bottom of the container P,
$P=P_{atm}+h_{w}\rho_{w}g+h_{m}\rho_{m}g-(1)$
Given that,
$P=2P_{atm}$ &
$h_{w}+h_{m} = 1\space m=>h_{w}=1-h_{m}$
(1)=>
$2P_{atm}=P_{atm}+(1-h_{m})\rho_{w}g+h_{m}\rho _{m}g$
$P_{atm}-\rho_{w}g=h_{m}g(\rho_{m}-\rho_{w})$
$h_{m}=\frac{P_{atm}-\rho_{w}g}{g(\rho_{m}-\rho_{w})}$ ; Let's plug known values into this equation.
$h_{m}=\frac{1.01\times10^{5}\space Pa-1000\space kg/m^{3}\times9.8\space m/s^{2}}{9.8\space m/s^{2}(13600\space kg/m^{3}-1000\space kg/m^{3})}=0.74\space m$
