A steady flow of a liquid of density $\rho$ is shown in figure. At point $1$, the area of crosssection is $2A$ and the speed of flow of liquid is $\sqrt{2} \mathrm{m} {\mathrm{s}}^{-1}$. At point $2$, the area of crosssection is $A$. Between the points $1$ and $2$, the pressure difference is $100 \mathrm{N} {\mathrm{m}}^{-2}$ and the height difference is $10 \mathrm{cm}$. The value of $\rho$ is
(Acceleration due to gravity $=10 \mathrm{m} {\mathrm{s}}^{-2}$)