A solid spherical ball of radius $7$ cm was then immersed in the water. What would be the approximate increase in water level in the container after the ball was fully immersed?
Water Level Increase Calculation
This solution explains how to find the approximate increase in water level when a solid spherical ball is immersed into a cylindrical container partially filled with water.
Cylinder and Sphere Dimensions
- Cylindrical container radius: $r_c = 20$ cm
- Initial water height: $h_i = 25$ cm
- Solid spherical ball radius: $r_s = 7$ cm
Volume Displacement Principle
When an object is submerged in a fluid, it pushes aside (displaces) a volume of fluid equal to the object's own volume. This displaced fluid causes the fluid level to rise. To find the increase in water level, we need to equate the volume of the submerged object (the sphere) to the volume of the water that rises in the container (which takes the shape of a cylinder).
Steps to Calculate Water Level Rise
- Calculate the volume of the spherical ball. The formula for the volume of a sphere is $V_s = \frac{4}{3}\pi r_s^3$.
- Determine the volume of the displaced water. This volume is equal to the volume of the sphere since the ball is fully immersed.
- Relate displaced volume to water level rise. The displaced water occupies a cylindrical shape within the container. If the increase in water level is $\Delta h$, the volume of this rise is $V_{disp} = \pi r_c^2 \Delta h$.
- Equate volumes and solve for $\Delta h$. Since $V_{disp} = V_s$, we have $\pi r_c^2 \Delta h = \frac{4}{3}\pi r_s^3$. Solving for $\Delta h$ gives $\Delta h = \frac{4 r_s^3}{3 r_c^2}$.
Applying the Formula with Values
Now, let's substitute the given values into the formula for $\Delta h$:
- $r_s = 7$ cm
- $r_c = 20$ cm
The formula becomes:
$\Delta h = \frac{4 \times (7 \text{ cm})^3}{3 \times (20 \text{ cm})^2}$
First, calculate the cubes and squares:
- $(7 \text{ cm})^3 = 343 \text{ cm}^3$
- $(20 \text{ cm})^2 = 400 \text{ cm}^2$
Now substitute these back into the equation:
$\Delta h = \frac{4 \times 343 \text{ cm}^3}{3 \times 400 \text{ cm}^2}$
$\Delta h = \frac{1372 \text{ cm}^3}{1200 \text{ cm}^2}$
Finally, perform the division:
$\Delta h \approx 1.1433$ cm
Approximate Water Level Rise
The approximate increase in the water level in the cylindrical container after the solid spherical ball is fully immersed is approximately $1.14$ 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.)
