Question

A long container has a square base of side $L$. The height to which water should be filled so that the force on one of its side surface is the same as that at its bottom is (ignore atmospheric pressure)

• A. $L/4$
• B. $L$
• C. $L/2$
• D. $2L$

Answer 1

Answer: (D)

2L

Answer 2

Let $H$ be the height to which water is filled in the container. Let $\rho$ be the density of water and $g$ be the acceleration due to gravity. The side of the square base is $L$.

1. Force on the bottom surface ($F_B$):

The pressure at the bottom of the container is uniform and given by $P_B = \rho g H$. The area of the square base is $A_B = L \times L = L^2$. The force on the bottom surface is: $F_B = P_B \times A_B = (\rho g H) \times L^2 = \rho g H L^2$

2. Force on one of the side surfaces ($F_S$):

A side surface is a rectangle with width $L$ and height $H$. The pressure on the side surface varies linearly with depth, from $0$ at the water surface to $\rho g H$ at the bottom. To find the total force, we can use the average pressure, as the pressure varies linearly. The average pressure on the side surface is: $P_{avg} = \frac{\text{Pressure at top} + \text{Pressure at bottom}}{2} = \frac{0 + \rho g H}{2} = \frac{1}{2} \rho g H$ The area of one side surface is $A_S = L \times H$. The force on one side surface is: $F_S = P_{avg} \times A_S = \left(\frac{1}{2} \rho g H\right) \times (L H) = \frac{1}{2} \rho g H^2 L$

3. Equating the forces:

According to the problem statement, the force on one of its side surfaces is the same as that at its bottom: $F_B = F_S$ $\rho g H L^2 = \frac{1}{2} \rho g H^2 L$

To solve for $H$, we can cancel common terms from both sides. Assuming $H \neq 0$, $L \neq 0$, $\rho \neq 0$, and $g \neq 0$: Divide both sides by $\rho g H L$: $L = \frac{1}{2} H$ Multiply both sides by 2: $H = 2L$

Thus, the height to which water should be filled is $2L$.