Oil is filled in a cylindrical container up to height 4m. A small hole of area 'p' is punched in the wall of the container at a height 1.52m from the bottom. The cross sectional area of the container is Q. If`p/q = 0.1` then v is (where vis the velocity of oil coming out of the hole)

Options

A. `5 sqrt2`
B. `6sqrt3`
C. `8sqrt2`
D. `7sqrt5`

Answer

To solve the problem, we will use Bernoulli's equation and the principle of conservation of mass. Here’s a step-by-step solution:

### Step 1: Identify the heights involved
- The height of the oil column is \( h_1 = 4 \, \text{m} \).
- The height of the hole from the bottom is \( h_2 = 1.52 \, \text{m} \).
- The difference in height \( h = h_1 - h_2 = 4 - 1.52 = 2.48 \, \text{m} \).

### Step 2: Apply Bernoulli's equation
According to Bernoulli's theorem, for two points in a fluid flow, we have:

\[ P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2 \]

Since the hole is open to the atmosphere, the pressure at both points can be considered equal (let's denote it as \( P \)). Therefore, we can simplify the equation:

\[ \rho g h_1 + \frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2} \rho v_2^2 \]

### Step 3: Assume initial conditions
Assuming the velocity of the oil in the container at the surface is negligible (i.e., \( v_1 \approx 0 \)), we can simplify further:

\[ \rho g h_1 = \rho g h_2 + \frac{1}{2} \rho v_2^2 \]

### Step 4: Cancel out the density
Since the density \( \rho \) is common on both sides, we can cancel it out:

\[ g h_1 = g h_2 + \frac{1}{2} v_2^2 \]

### Step 5: Rearranging the equation
Rearranging gives us:

\[ g (h_1 - h_2) = \frac{1}{2} v_2^2 \]

### Step 6: Substitute the values
Substituting the values we have:

\[ g = 9.8 \, \text{m/s}^2 \]
\[ h_1 - h_2 = 2.48 \, \text{m} \]

Thus,

\[ 9.8 \times 2.48 = \frac{1}{2} v_2^2 \]

### Step 7: Solve for \( v_2 \)
Calculating the left side:

\[ 24.304 = \frac{1}{2} v_2^2 \]

Multiplying both sides by 2:

\[ 48.608 = v_2^2 \]

Taking the square root:

\[ v_2 = \sqrt{48.608} \approx 6.96 \, \text{m/s} \]

### Step 8: Relate \( v \) to \( V \)
From the problem, we know that:

\[ \frac{p}{Q} = 0.1 \implies v = 10V \]

Thus, the velocity \( v \) of the oil coming out of the hole is:

\[ v \approx 6.96 \, \text{m/s} \]

### Final Answer:
The velocity of oil coming out of the hole is approximately \( 6.96 \, \text{m/s} \).

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