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 $=10 \mathrm{m} {\mathrm{s}}^{-2}$)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.