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There is a small hole at the bottom of t...

There is a small hole at the bottom of tank filled with water. If total pressure at the bottom is `3 atm(1 atm=10^(5)Nm^(-2))`, then find the velocity of water flowing from hole.

A

`sqrt(400)ms^(-1)`

B

`sqrt(600)ms^(-1)`

C

`sqrt(60)ms^(-1)`

D

none of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the velocity of water flowing from a hole at the bottom of a tank filled with water, we can use Bernoulli's equation and the concept of pressure difference. Here’s a step-by-step solution: ### Step 1: Understand the given data - The total pressure at the bottom of the tank is given as \(3 \, \text{atm}\). - The atmospheric pressure is \(1 \, \text{atm}\). - We need to convert these pressures into SI units (Pascals). ### Step 2: Convert pressures to SI units 1 atm = \(10^5 \, \text{N/m}^2\) - Pressure at the bottom of the tank: \[ P_{\text{bottom}} = 3 \, \text{atm} = 3 \times 10^5 \, \text{N/m}^2 \] - Pressure at the uppermost position (atmospheric pressure): \[ P_{\text{top}} = 1 \, \text{atm} = 1 \times 10^5 \, \text{N/m}^2 \] ### Step 3: Calculate the pressure difference The pressure difference (\(\Delta P\)) between the bottom and the top of the tank is: \[ \Delta P = P_{\text{bottom}} - P_{\text{top}} = (3 \times 10^5) - (1 \times 10^5) = 2 \times 10^5 \, \text{N/m}^2 \] ### Step 4: Use Bernoulli's equation According to Bernoulli's principle, the velocity of fluid flowing out of a hole can be derived from the pressure difference: \[ \Delta P = \frac{1}{2} \rho V^2 \] Where: - \(\Delta P\) is the pressure difference, - \(\rho\) is the density of water (approximately \(1000 \, \text{kg/m}^3\)), - \(V\) is the velocity of water flowing out. ### Step 5: Rearranging the equation to solve for \(V\) Rearranging the equation gives: \[ V^2 = \frac{2 \Delta P}{\rho} \] ### Step 6: Substitute the values Substituting the known values into the equation: \[ V^2 = \frac{2 \times (2 \times 10^5)}{1000} \] \[ V^2 = \frac{4 \times 10^5}{1000} = 400 \] ### Step 7: Calculate \(V\) Taking the square root of both sides: \[ V = \sqrt{400} = 20 \, \text{m/s} \] ### Final Answer The velocity of water flowing from the hole is: \[ \boxed{20 \, \text{m/s}} \]
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