A large cylindrical vessel contains water to a height of 10 m. If is found that the thrust acting on the curved surface is equal to that at the bottom. If atmospheric pressure can support a water column of 10 m. the radius of the vessel is

Options

A. 10 m
B. 15 m
C. 5 m
D. 25 m

Explanation

To solve the problem, we need to find the radius of a cylindrical vessel filled with water to a height of 10 m, given that the thrust acting on the curved surface is equal to that at the bottom. Let's break down the solution step by step.

### Step 1: Understand the thrust on the curved surface
The thrust (or force) acting on the curved surface of the cylinder can be calculated using the formula:
\[ F_{\text{curved}} = 2 \pi R h \left( P_0 + \frac{h}{2} \right) \]
where:
- \( R \) is the radius of the cylinder,
- \( h \) is the height of the water (10 m),
- \( P_0 \) is the atmospheric pressure.

### Step 2: Understand the thrust at the bottom
The thrust acting on the bottom of the cylinder can be calculated using:
\[ F_{\text{bottom}} = \pi R^2 (P_0 + h) \]

### Step 3: Set the thrusts equal to each other
According to the problem, the thrust on the curved surface is equal to the thrust at the bottom:
\[ 2 \pi R h \left( P_0 + \frac{h}{2} \right) = \pi R^2 (P_0 + h) \]

### Step 4: Simplify the equation
We can simplify the equation by canceling out common terms:
1. Divide both sides by \( \pi \):
\[ 2 R h \left( P_0 + \frac{h}{2} \right) = R^2 (P_0 + h) \]
2. Divide both sides by \( R \) (assuming \( R \neq 0 \)):
\[ 2 h \left( P_0 + \frac{h}{2} \right) = R (P_0 + h) \]

### Step 5: Substitute values
Given that \( h = 10 \) m and atmospheric pressure can support a water column of 10 m, we can express \( P_0 \) in terms of height:
\[ P_0 = \rho g \cdot 10 \]
where \( \rho \) is the density of water and \( g \) is the acceleration due to gravity.

### Step 6: Solve for \( R \)
Substituting \( h = 10 \) m into the equation:
\[ 2 \cdot 10 \left( P_0 + 5 \right) = R (P_0 + 10) \]
This simplifies to:
\[ 20 \left( P_0 + 5 \right) = R (P_0 + 10) \]
Now, we can solve for \( R \):
\[ R = \frac{20 \left( P_0 + 5 \right)}{P_0 + 10} \]

### Step 7: Substitute \( P_0 \)
Assuming \( P_0 \) is equal to \( \rho g \cdot 10 \):
1. Substitute \( P_0 \) into the equation.
2. After simplification, we find that \( R = 15 \) m.

### Conclusion
Thus, the radius of the vessel is:
\[ \boxed{15 \text{ m}} \]