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 $1\text{ cm}$.
Important Questions from Mensuration 3D (Notes)
- The height of a cylinder is 14cm and its curved surface area is 264cm². The volume of the cyclinder (in cm³) is:
($\pi=\frac{22}{7}$) - A cylindrical rod has an outer curved surface area of \(7500 \text{ cm}^2\). If the length of the rod is 92 cm, then the outer radius (in cm) of the rod, rounded off to two places of decimal, is:
\(\left(\text{Take } \pi = \frac{22}{7}\right)\) - A number of 512 identical small spheres are cast from a sphere of radius 40 cm, with the total volume of the small spheres being equal to the volume of the larger sphere. The diameter (in cm) of each of the small spheres is:
There is a wooden block in the form of a cube whose each side is 8 meters long.
The maximum possible number of cylinders with a diameter of 1 meter and a height of 4 meters were cut from this block. The cylinders are to be painted at the rate of ₹14 per square meter.
What is the total amount (in ₹) needed to paint all the cylinders if we paint the entire surface of each cylinder? (Take $\pi = \frac{22}{7}$)
- If the lateral surface area of a cylinder is $140.1 \text{ cm}^2$ and its height is $3 \text{ cm}$, then find its volume. (Use $\pi = 3.14$ and round off to two decimal places.)
