Question

At two points on a horizontal tube of varying cross-section the radii are \(1 \text{ cm}\) and \(0.4 \text{ cm}\), velocities of fluid are \(V_1, V_2\) and pressure difference \((P_1 - P_2)\) between these points is \(4.9 \text{ cm of water}\). The value of \( \sqrt{V_2^2 - V_1^2} \) is given, where \( g = 980 \text{ cm/s}^2 \).

• A. $9.8 \text{ cm/s}$
• B. $98 \text{ cm/s}$
• C. $980 \text{ cm/s}$
• D. $0.98 \text{ cm/s}$

Hint:

- Use $P = \rho g h$ for fluid column - Combine with Bernoulli for velocity relation

Answer 1

The Correct Option is B

Concept: For horizontal flow (Bernoulli’s equation): \[ P_1 - P_2 = \frac{1}{2}\rho (V_2^2 - V_1^2) \] Also, pressure difference in terms of height: \[ P_1 - P_2 = \rho g h \]

Step 1: Equate both expressions.
\[ \rho g h = \frac{1}{2}\rho (V_2^2 - V_1^2) \]

Step 2: Cancel $\rho$.
\[ g h = \frac{1}{2}(V_2^2 - V_1^2) \]

Step 3: Substitute values.
\[ h = 4.9\ \text{cm}, \quad g = 980\ \text{cm/s}^2 \] \[ 980 \times 4.9 = \frac{1}{2}(V_2^2 - V_1^2) \]

Step 4: Solve.
\[ V_2^2 - V_1^2 = 2 \times 980 \times 4.9 = 9604 \] \[ \sqrt{V_2^2 - V_1^2} = \sqrt{9604} = 98\ \text{cm/s} \]