Calculating Water Level Rise in a Flask

This problem involves finding the increase in water level within a rectangular flask after adding spherical marbles. We need to calculate the volume of the marbles and determine how this volume affects the water height based on the flask's base dimensions.

1. Calculate the Volume of a Single Marble

• The formula for the volume of a sphere is $V_{\text{sphere}} = \frac{4}{3}\pi r^3$.
• Given the radius $r = 1\text{ cm}$.
• Volume of one marble: $V_{\text{marble}} = \frac{4}{3}\pi (1\text{ cm})^3 = \frac{4}{3}\pi \text{ cm}^3$.

2. Calculate the Total Volume of Marbles

• There are $21$ marbles.
• Total volume $V_{\text{total marbles}} = 21 \times V_{\text{marble}} = 21 \times \frac{4}{3}\pi \text{ cm}^3$.
• $V_{\text{total marbles}} = 7 \times 4\pi \text{ cm}^3 = 28\pi \text{ cm}^3$.
• This is the volume of water that will be displaced.

3. Calculate the Base Area of the Flask

• The flask has a rectangular base with length $L = 11\text{ cm}$ and width $W = 8\text{ cm}$.
• Base Area $A_{\text{base}} = L \times W = 11\text{ cm} \times 8\text{ cm} = 88\text{ cm}^2$.

4. Calculate the Rise in Water Level

• The displaced water volume ($V_{\text{total marbles}}$) causes the water level to rise by $\Delta h$.
• The volume added is equal to the base area multiplied by the height change: $V_{\text{displaced}} = A_{\text{base}} \times \Delta h$.
• So, $28\pi \text{ cm}^3 = 88\text{ cm}^2 \times \Delta h$.
• Solving for $\Delta h$: $\Delta h = \frac{28\pi}{88} \text{ cm}$.
• Simplifying the fraction: $\Delta h = \frac{7\pi}{22} \text{ cm}$.
• Using the approximation $\pi \approx \frac{22}{7}$: $\Delta h = \frac{7 \times (\frac{22}{7})}{22} \text{ cm} = \frac{22}{22} \text{ cm} = 1\text{ cm}$.

5. Final Check

• The initial water level is $5\text{ cm}$.
• The rise in water level is $1\text{ cm}$.
• The new water level is $5\text{ cm} + 1\text{ cm} = 6\text{ cm}$.
• This is less than the flask height ($20\text{ cm}$), so the water does not overflow.

The rise in water level is $1\text{ cm}$.