A spherical ball of lead, 3 cm in diameter, is melted and recast into three spherical balls. The diameters of two of these balls are \(\frac{3}{2}\) cm and 2 cm, respectively. Find the diameter of the third ball.
2.5 cm
Understanding the Problem: Melting and Recasting Spheres
The question describes a scenario where a single large spherical ball made of lead is melted down and then remolded into three smaller spherical balls. A key principle in such problems is the conservation of volume. When a material is melted and recast, its total volume remains unchanged, assuming no material is lost or added.
We are given the diameter of the original large ball and the diameters of two of the three smaller balls. We need to find the diameter of the third smaller ball.
Key Concept: Volume of a Sphere
The volume of a sphere is calculated using the formula:
\[ V = \frac{4}{3}\pi r^3 \]where \(V\) is the volume and \(r\) is the radius of the sphere.
Since the diameter \(D\) is twice the radius \(r\), i.e., \(D = 2r\), the radius can be expressed as \(r = \frac{D}{2}\). Substituting this into the volume formula, we get:
\[ V = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 = \frac{4}{3}\pi \frac{D^3}{8} = \frac{\pi D^3}{6} \]However, it's often easier to work with radii directly.
Step-by-Step Calculation
Let's denote the diameter of the original large ball as \(D\), and its radius as \(R\). The diameters of the three smaller balls are \(D_1\), \(D_2\), and \(D_3\), with corresponding radii \(r_1\), \(r_2\), and \(r_3\).
According to the problem:
- Diameter of the original ball, \(D = 3\) cm.
- Diameter of the first small ball, \(D_1 = \frac{3}{2}\) cm.
- Diameter of the second small ball, \(D_2 = 2\) cm.
- Let the diameter of the third small ball be \(D_3\).
First, calculate the radius of each ball from its diameter:
- Original ball radius, \(R = \frac{D}{2} = \frac{3}{2}\) cm.
- First small ball radius, \(r_1 = \frac{D_1}{2} = \frac{3/2}{2} = \frac{3}{4}\) cm.
- Second small ball radius, \(r_2 = \frac{D_2}{2} = \frac{2}{2} = 1\) cm.
- Third small ball radius, \(r_3 = \frac{D_3}{2}\) cm.
Next, calculate the volume of the original large ball:
\[ V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \left(\frac{3}{2}\right)^3 = \frac{4}{3}\pi \left(\frac{27}{8}\right) = \frac{9\pi}{2} \text{ cm}^3 \]Then, calculate the volumes of the two known smaller balls:
\[ V_1 = \frac{4}{3}\pi r_1^3 = \frac{4}{3}\pi \left(\frac{3}{4}\right)^3 = \frac{4}{3}\pi \left(\frac{27}{64}\right) = \frac{9\pi}{16} \text{ cm}^3 \] \[ V_2 = \frac{4}{3}\pi r_2^3 = \frac{4}{3}\pi (1)^3 = \frac{4\pi}{3} \text{ cm}^3 \]Let the volume of the third small ball be \(V_3\). According to the principle of conservation of volume:
\[ V = V_1 + V_2 + V_3 \]Substitute the calculated volumes:
\[ \frac{9\pi}{2} = \frac{9\pi}{16} + \frac{4\pi}{3} + V_3 \]We can divide the entire equation by \(\pi\) to simplify:
\[ \frac{9}{2} = \frac{9}{16} + \frac{4}{3} + \frac{V_3}{\pi} \]Now, isolate the term with \(V_3\):
\[ \frac{V_3}{\pi} = \frac{9}{2} - \frac{9}{16} - \frac{4}{3} \]Find a common denominator for the fractions (the least common multiple of 2, 16, and 3 is 48):
\[ \frac{V_3}{\pi} = \frac{9 \times 24}{2 \times 24} - \frac{9 \times 3}{16 \times 3} - \frac{4 \times 16}{3 \times 16} \] \[ \frac{V_3}{\pi} = \frac{216}{48} - \frac{27}{48} - \frac{64}{48} \] \[ \frac{V_3}{\pi} = \frac{216 - 27 - 64}{48} = \frac{216 - 91}{48} = \frac{125}{48} \]So, \(V_3 = \frac{125\pi}{48}\) cm\(^3\).
Now we use the volume formula for the third ball, \(V_3 = \frac{4}{3}\pi r_3^3\), to find \(r_3\):
\[ \frac{125\pi}{48} = \frac{4}{3}\pi r_3^3 \]Divide both sides by \(\frac{4}{3}\pi\):
\[ \frac{125\pi}{48} \times \frac{3}{4\pi} = r_3^3 \] \[ \frac{125 \times 3}{48 \times 4} = r_3^3 \] \[ \frac{125 \times 1}{16 \times 4} = r_3^3 \] \[ \frac{125}{64} = r_3^3 \]To find \(r_3\), take the cube root of both sides:
\[ r_3 = \sqrt[3]{\frac{125}{64}} = \frac{\sqrt[3]{125}}{\sqrt[3]{64}} = \frac{5}{4} \text{ cm} \]Finally, find the diameter of the third ball, \(D_3 = 2r_3\):
\[ D_3 = 2 \times \frac{5}{4} = \frac{10}{4} = \frac{5}{2} = 2.5 \text{ cm} \]Thus, the diameter of the third spherical ball is 2.5 cm.
| Ball | Diameter (cm) | Radius (cm) | Volume (cm\(^3\)) |
|---|---|---|---|
| Original | 3 | \(\frac{3}{2}\) | \(\frac{4}{3}\pi \left(\frac{3}{2}\right)^3 = \frac{9\pi}{2}\) |
| Small 1 | \(\frac{3}{2}\) | \(\frac{3}{4}\) | \(\frac{4}{3}\pi \left(\frac{3}{4}\right)^3 = \frac{9\pi}{16}\) |
| Small 2 | 2 | 1 | \(\frac{4}{3}\pi (1)^3 = \frac{4\pi}{3}\) |
| Small 3 | \(D_3\) | \(\frac{D_3}{2}\) | \(\frac{4}{3}\pi \left(\frac{D_3}{2}\right)^3 = \frac{\pi D_3^3}{6}\) |
Volume conservation: \(\frac{9\pi}{2} = \frac{9\pi}{16} + \frac{4\pi}{3} + \frac{\pi D_3^3}{6}\)
Divide by \(\pi\): \(\frac{9}{2} = \frac{9}{16} + \frac{4}{3} + \frac{D_3^3}{6}\)
Multiply by 48 (LCM of 2, 16, 3, 6):
\(48 \times \frac{9}{2} = 48 \times \frac{9}{16} + 48 \times \frac{4}{3} + 48 \times \frac{D_3^3}{6}\)
\(216 = 27 + 64 + 8D_3^3\)
\(216 = 91 + 8D_3^3\)
\(216 - 91 = 8D_3^3\)
\(125 = 8D_3^3\)
\(D_3^3 = \frac{125}{8}\)
\(D_3 = \sqrt[3]{\frac{125}{8}} = \frac{\sqrt[3]{125}}{\sqrt[3]{8}} = \frac{5}{2} = 2.5\)
So, the diameter of the third ball is 2.5 cm.
Comparing with Options
The calculated diameter of the third ball is 2.5 cm, which matches one of the given options.
Revision Table: Sphere Volume and Recasting
| Concept | Description | Formula |
|---|---|---|
| Volume of Sphere | Amount of space a sphere occupies, depends on its radius or diameter. | \(V = \frac{4}{3}\pi r^3\) or \(V = \frac{\pi D^3}{6}\) |
| Melting and Recasting | Process where a solid is melted into liquid and reshaped into a new solid. | Total Volume (before) = Total Volume (after) |
| Diameter and Radius | Diameter is the distance across the center of a circle/sphere; radius is half the diameter. | \(D = 2r\) or \(r = \frac{D}{2}\) |
Additional Information: Applications of Volume Conservation
The principle of conservation of volume is fundamental in various problems involving melting, casting, or reshaping materials. It applies whenever a material changes form but its total quantity remains constant. Here are a few examples:
- Melting a cube into a sphere: The volume of the original cube equals the volume of the new sphere.
- Drawing a wire from a cylinder: The volume of the original cylinder equals the volume of the resulting cylindrical wire.
- Converting coins into a cuboid: The sum of the volumes of all coins equals the volume of the resulting cuboid.
- Filling a container: If liquid from one container is poured into another, the volume of the liquid remains constant, changing only its shape.
In this problem, we dealt with spheres, using their volume formula and the conservation principle to find an unknown dimension.
Similar Questions
If the total surface area of a cube is 24 sq.units, then what is the volume of the cube?
The volume of a cone is 73920 cm3. If the height of the cone is 160 cm, then find the diameter of its base.
Ranu carries water to school in a cylindrical flask with diameter 12 cm and height 21 cm. Determine the amount of water that she can carry in the flask. (Use π = \(\frac{22}{7}\))
The volume of a cone with height equal to radius, and slant height 5 cm is :
What is the whole surface area of a cone of base radius 6 cm and height 8 cm?
What is the volume of a cube if the perimeter of one face of the cube is 40 cm?
A conical tent of height 10 m and base diameter 48 m was erected by a company in a park. Find the curved surface area of the tent (In m2).
If the surface area of a cube is 5046 cm2, then the volume of the cube is:
The volume of a cuboid is twice that of a cube. If the dimensions of the cuboid are (8 m × 8 m ×16 m), the total surface area of the cube is:
If the slant height of a cone is 60 cm and the radius of its base is 21 cm, then find its curved surface area. (use π = \({22 \ {} \over 7}\))
Important Questions from Solid Figures
The area of the floor of a cubical room is 192 m 2. The length of the longest rod that can be kept in that room is :
A solid metallic rectangular block of dimensions 112 cm × 44 cm × 25 cm is melted and recast into a cylinder of radius 35 cm. The curved surface area (in cm 2) of the cylinder is: (Take π = 22/7)
If the volume of a cube is 175616 cm 3, what is its side?
The volume of a right circular cone is 1232 cm 3. If the height of the cone is 24 cm, then what will be the radius of its base?
A right triangle contains the right angle between the sides 5 cm and 7 cm. A cone is generated by revolving about the side 5 cm. The volume of this cone is:
