A cylindrical vessel of radius r is filled with a homogeneous liquid upto a height h. It is found that the force exerted by the liquid column on the bottom of the cylinder is equal to the force exerted by the liquid on the sides of the cylinder. What is the relation between r and h?

Options

A. `r=h/2`
B. r=h
C. r=2h
D. `r=3/2h`

Explanation

To solve the problem, we need to find the relationship between the radius \( r \) of a cylindrical vessel and the height \( h \) of the liquid it contains, given that the force exerted by the liquid column on the bottom of the cylinder is equal to the force exerted by the liquid on the sides of the cylinder.

### Step-by-Step Solution:

**Step 1: Calculate the force exerted by the liquid column on the bottom of the cylinder.**

The force exerted by the liquid column on the bottom can be calculated using the formula:
\[ F_{\text{bottom}} = \text{Pressure} \times \text{Area} \]
The pressure at the bottom due to the liquid column is given by:
\[ P = \rho g h \]
where:
- \( \rho \) = density of the liquid,
- \( g \) = acceleration due to gravity,
- \( h \) = height of the liquid column.

The area \( A \) of the bottom of the cylinder (which is circular) is:
\[ A = \pi r^2 \]
Thus, the force on the bottom becomes:
\[ F_{\text{bottom}} = \rho g h \cdot \pi r^2 \]

**Step 2: Calculate the force exerted by the liquid on the sides of the cylinder.**

The force exerted by the liquid on the sides of the cylinder can be calculated by considering the pressure at a height \( y \) from the bottom, which is:
\[ P(y) = \rho g y \]
The differential force \( dF \) acting on a small strip of the side of the cylinder of height \( dy \) and circumference \( 2\pi r \) is:
\[ dF = P(y) \cdot \text{Area of the strip} = \rho g y \cdot (2 \pi r \cdot dy) \]
To find the total force on the side, we integrate this from \( y = 0 \) to \( y = h \):
\[ F_{\text{sides}} = \int_0^h \rho g y (2 \pi r) \, dy \]
Calculating the integral:
\[ F_{\text{sides}} = 2 \pi r \rho g \int_0^h y \, dy = 2 \pi r \rho g \left[ \frac{y^2}{2} \right]_0^h = 2 \pi r \rho g \cdot \frac{h^2}{2} = \pi r \rho g h^2 \]

**Step 3: Set the forces equal to each other.**

According to the problem, these two forces are equal:
\[ F_{\text{bottom}} = F_{\text{sides}} \]
Substituting the expressions we derived:
\[ \rho g h \cdot \pi r^2 = \pi r \rho g h^2 \]

**Step 4: Simplify the equation.**

We can cancel out \( \pi \), \( \rho \), and \( g \) from both sides (assuming they are non-zero):
\[ h r^2 = r h^2 \]

**Step 5: Rearranging the equation.**

Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ h r = h^2 \]
Now, dividing both sides by \( h \) (assuming \( h \neq 0 \)):
\[ r = h \]

### Final Relation:
Thus, the relationship between the radius \( r \) and the height \( h \) is:
\[ r = h \]