Question

The reading of a pressure-meter attached to a closed pipe (with water) is \(3.5 \times 10^5 \, {Nm}^{-2}\). On opening the valve of the pipe, the reading is reduced to \(3.0 \times 10^5 \, {Nm}^{-2}\). The speed of the water flowing out the pipe is:

• A. \(10 \, {ms}^{-1}\)
• B. \(0.1 \, {ms}^{-1}\)
• C. \(1 \, {ms}^{-1}\)
• D. \(5 \, {ms}^{-1}\)

Hint:

For problems involving fluid flow and pressure changes, use Bernoulli's principle to relate the pressure and velocity at two points in the system.

Answer 1

The Correct Option is A

Using Bernoulli's principle: \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \] 
Where: 
- \(P_1 = 3.5 \times 10^5 \, {Nm}^{-2}\) (initial pressure), 
- \(P_2 = 3.0 \times 10^5 \, {Nm}^{-2}\) (final pressure), 
- \(\rho = 1000 \, {kg/m}^3\), - \(v_1 = 0 \, {m/s}\) (since the pipe is closed initially), 
- \(v_2\) is the speed of water we need to find. 
Rearranging the equation for \(v_2\): \[ v_2 = \sqrt{\frac{2(P_1 - P_2)}{\rho}} \] Substitute the known values: \[ v_2 = \sqrt{\frac{2(3.5 \times 10^5 - 3.0 \times 10^5)}{1000}} = \sqrt{\frac{10^5}{1000}} = \sqrt{100} = 10 \, {m/s} \] Thus, the speed of water flowing out of the pipe is \(10 \, {m/s}\).