To solve the problem of finding the relationship between the speed of water flowing in a pipe and the pressure difference, we can apply Bernoulli's principle. Bernoulli's principle states that in a streamline flow of an ideal fluid, the sum of the pressure energy, kinetic energy, and potential energy per unit volume is a constant.
The general form of Bernoulli's equation for a horizontal flow (neglecting height-related potential energy difference) is:
\(P + \frac{1}{2} \rho v^2 = \text{constant}\)
Where:
Now, when the valve is closed, the pressure is \(P_1\) and the velocity is zero because the fluid is at rest. Thus, the Bernoulli equation becomes:
\(P_1 = \text{constant}\)
When the valve is opened, the pressure drops to \(P_2\) and the water starts to flow with velocity \(v\). Applying Bernoulli's equation, we get:
\(P_2 + \frac{1}{2} \rho v^2 = \text{constant}\)
Equating the two equations, we have:
\(P_1 = P_2 + \frac{1}{2} \rho v^2\)
Simplifying for the velocity \(v\), we get:
\(\frac{1}{2} \rho v^2 = P_1 - P_2\)
\(v^2 = \frac{2 (P_1 - P_2)}{\rho}\)
\(v = \sqrt{\frac{2 (P_1 - P_2)}{\rho}}\)
This shows that the velocity \(v\) is proportional to \(\sqrt{P_1 - P_2}\). Therefore, the correct answer is \(\sqrt{P_1 - P_2}\).
$\text{The fractional compression } \left( \frac{\Delta V}{V} \right) \text{ of water at the depth of } 2.5 \, \text{km below the sea level is } \_\_\_\_\_\_\_\_\_\_ \%. \text{ Given, the Bulk modulus of water } = 2 \times 10^9 \, \text{N m}^{-2}, \text{ density of water } = 10^3 \, \text{kg m}^{-3}, \text{ acceleration due to gravity } g = 10 \, \text{m s}^{-2}.$
Given below are two statements:
Statement (I):
 
 are isomeric compounds. 
Statement (II): 
 are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below:
The effect of temperature on the spontaneity of reactions are represented as: Which of the following is correct?
