Question

A steady flow of a liquid of density \(\rho\) is shown in figure. At point \(1\), the area of cross-section is \(2A\) and the speed of flow of liquid is \(\sqrt{2}\;m s^{-1}\). At point \(2\), the area of cross-section is \(A\). Between the points \(1\) and \(2\), the pressure difference is \(100\;N m^{-2}\) and the height difference is \(10\;cm\). The value of \(\rho\) is
Take \(g=10\;m s^{-2}\).

• A. \(25\;kg\,m^{-3}\)
• B. \(30\;kg\,m^{-3}\)
• C. \(50\;kg\,m^{-3}\)
• D. \(70\;kg\,m^{-3}\)

Hint:

For fluid flow problems, first use continuity equation \(A_1v_1=A_2v_2\), then apply Bernoulli's theorem carefully with height and pressure terms.

Answer 1

The Correct Option is C

Step 1: Use equation of continuity.
For steady flow, \[ A_1v_1=A_2v_2 \] Given, \[ A_1=2A,\qquad v_1=\sqrt{2}\;m s^{-1} \] and \[ A_2=A \] So, \[ 2A\cdot \sqrt{2}=A\cdot v_2 \] \[ v_2=2\sqrt{2}\;m s^{-1} \]

Step 2: Apply Bernoulli's theorem.
Using Bernoulli's equation between points \(1\) and \(2\), \[ P_1+\frac12\rho v_1^2+\rho gh_1 = P_2+\frac12\rho v_2^2+\rho gh_2 \] Rearranging, \[ P_1-P_2 = \frac12\rho(v_2^2-v_1^2)-\rho g(h_1-h_2) \]

Step 3: Substitute the given values.
Here, \[ v_1=\sqrt{2} \] so, \[ v_1^2=2 \] Also, \[ v_2=2\sqrt{2} \] so, \[ v_2^2=8 \] Height difference is \[ h_1-h_2=10\;cm=0.1\;m \] Given pressure difference is \[ P_1-P_2=100\;N m^{-2} \] Therefore, \[ 100=\frac12\rho(8-2)-\rho(10)(0.1) \] \[ 100=3\rho-\rho \] \[ 100=2\rho \] \[ \rho=50\;kg\,m^{-3} \]

Step 4: Final conclusion.
Hence, the density of the liquid is \[ \boxed{50\;kg\,m^{-3}} \]