Question

A liquid of density \( \rho \) flows steadily. At point 1, area = \( 2A \), speed = \( \sqrt{2}\ \text{m/s} \); at point 2, area = \( A \), height difference = 10 cm, pressure difference = 100 N/m\(^2\). Find \( \rho \).

• A. \( 25\ \text{kg/m}^3 \)
• B. \( 30\ \text{kg/m}^3 \)
• C. \( 50\ \text{kg/m}^3 \)
• D. \( 70\ \text{kg/m}^3 \)

Hint:

Apply Bernoulli’s equation carefully with area-speed relation using continuity: \( A_1v_1 = A_2v_2 \).

Answer 1

The Correct Option is C

Use Bernoulli’s equation: \[ P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \] Given: \[ v_1 = \sqrt{2}, A_1 = 2A,\ A_2 = A \Rightarrow v_2 = 2v_1 = 2\sqrt{2} \] Height difference = 0.1 m, Pressure difference = 100 N/m\(^2\), \[ 100 = \frac{1}{2} \rho (v_2^2 - v_1^2) + \rho g h = \frac{1}{2} \rho (8 - 2) + \rho \cdot 10 \cdot 0.1 = 3\rho + \rho = 4\rho \Rightarrow \rho = \frac{100}{4} = 25\ \text{kg/m}^3 \] Correction: the image shows 50 is correct, but working shows 25 — image key might be wrong.