A sphere is split in the ratio 1 : 3. The larger part is moulded into a cone having a height equal to the radius of its base, while the smaller part is moulded into a cylinder having a height equal to the radius of its base. What would be the ratio of the radius of the cone to the height of the cylinder?
This question involves understanding how the volume of a sphere is divided and then how these parts are reformed into different 3D shapes: a cone and a cylinder. The key principle here is that the volume of the material remains constant when it is moulded into a new shape. We need to use the formulas for the volume of a sphere, a cone, and a cylinder, along with the given conditions about the dimensions of the cone and cylinder.
Let's denote the original sphere's radius by $R$.
The volume of the original sphere is given by the formula:
The sphere is split into two parts in the ratio $1 : 3$. The total ratio parts are $1 + 3 = 4$.
The smaller part corresponds to the ratio 1 out of 4 total parts.
This smaller part is moulded into a cylinder.
The larger part corresponds to the ratio 3 out of 4 total parts.
This larger part is moulded into a cone.
The larger part ($V_l$) is moulded into a cone. Let the radius of the cone's base be $r_c$ and its height be $h_c$.
Given condition for the cone: height is equal to the radius of its base, so $h_c = r_c$.
The volume of the cone is given by:
Substitute $h_c = r_c$ into the volume formula:
Since the larger part of the sphere is moulded into this cone, their volumes are equal:
Now, we can solve for $r_c$ in terms of $R$:
The smaller part ($V_s$) is moulded into a cylinder. Let the radius of the cylinder's base be $r_{cy}$ and its height be $h_{cy}$.
Given condition for the cylinder: height is equal to the radius of its base, so $h_{cy} = r_{cy}$.
The volume of the cylinder is given by:
Substitute $h_{cy} = r_{cy}$ into the volume formula:
Since the smaller part of the sphere is moulded into this cylinder, their volumes are equal:
Now, we can solve for $r_{cy}$ (and thus $h_{cy}$) in terms of $R$:
Since $h_{cy} = r_{cy}$, the height of the cylinder is $h_{cy} = \frac{1}{\sqrt[3]{3}} R$.
The question asks for the ratio of the radius of the cone ($r_c$) to the height of the cylinder ($h_{cy}$).
Ratio $= \frac{r_c}{h_{cy}}$
Substitute the expressions we found for $r_c$ and $h_{cy}$ in terms of $R$:
Ratio $= \frac{\sqrt[3]{3} R}{\frac{1}{\sqrt[3]{3}} R}$
We can cancel $R$ from the numerator and denominator (assuming $R \neq 0$, which must be true for a sphere to exist).
Ratio $= \frac{\sqrt[3]{3}}{\frac{1}{\sqrt[3]{3}}} = \sqrt[3]{3} \times \sqrt[3]{3}$
Using the property of cube roots $\sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{ab}$:
Ratio $= \sqrt[3]{3 \times 3} = \sqrt[3]{9}$
So, the ratio of the radius of the cone to the height of the cylinder is $\sqrt[3]{9} : 1$.
| Shape | Original Sphere Part | Volume | Key Dimension Relation | Calculated Dimension |
|---|---|---|---|---|
| Cone | Larger (3/4 $V$) | $V_l = \pi R^3$ | $h_c = r_c$ | $r_c = \sqrt[3]{3} R$ |
| Cylinder | Smaller (1/4 $V$) | $V_s = \frac{1}{3}\pi R^3$ | $h_{cy} = r_{cy}$ | $h_{cy} = r_{cy} = \frac{1}{\sqrt[3]{3}} R$ |
Ratio of $r_c$ to $h_{cy}$:
The ratio is $\sqrt[3]{9} : 1$.
| Shape | Volume Formula | Variables |
|---|---|---|
| Sphere | $\frac{4}{3}\pi R^3$ | $R$ = radius |
| Cone | $\frac{1}{3}\pi r^2 h$ | $r$ = base radius, $h$ = height |
| Cylinder | $\pi r^2 h$ | $r$ = base radius, $h$ = height |
A fundamental concept used in this problem is the conservation of volume. When a solid object is melted, reshaped, or moulded into a new form without adding or removing material, its total volume remains the same. In this case, the volumes of the smaller and larger parts of the sphere are conserved when they are moulded into the cylinder and cone, respectively. This principle allows us to equate the volume of the sphere parts to the volumes of the resulting cylinder and cone, which is crucial for solving the problem.
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