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Question

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?

The correct answer is \(\sqrt[3]{9}{\rm{}}:1\)

Understanding the Sphere Splitting and Moulding Problem

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.

Step-by-Step Solution: Calculating Volumes and Radii

Let's denote the original sphere's radius by $R$.

The volume of the original sphere is given by the formula:

  • Volume of Sphere ($V$) $= \frac{4}{3}\pi R^3$

The sphere is split into two parts in the ratio $1 : 3$. The total ratio parts are $1 + 3 = 4$.

Volume of the Smaller Part

The smaller part corresponds to the ratio 1 out of 4 total parts.

  • Volume of Smaller Part ($V_s$) $= \frac{1}{4} \times V = \frac{1}{4} \times \frac{4}{3}\pi R^3 = \frac{1}{3}\pi R^3$

This smaller part is moulded into a cylinder.

Volume of the Larger Part

The larger part corresponds to the ratio 3 out of 4 total parts.

  • Volume of Larger Part ($V_l$) $= \frac{3}{4} \times V = \frac{3}{4} \times \frac{4}{3}\pi R^3 = \pi R^3$

This larger part is moulded into a cone.

Analysing the Cone and Cylinder Dimensions

The 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:

  • Volume of Cone ($V_{cone}$) $= \frac{1}{3}\pi r_c^2 h_c$

Substitute $h_c = r_c$ into the volume formula:

  • $V_{cone} = \frac{1}{3}\pi r_c^2 (r_c) = \frac{1}{3}\pi r_c^3$

Since the larger part of the sphere is moulded into this cone, their volumes are equal:

  • $V_{cone} = V_l$
  • $\frac{1}{3}\pi r_c^3 = \pi R^3$

Now, we can solve for $r_c$ in terms of $R$:

  • Divide both sides by $\pi$: $\frac{1}{3} r_c^3 = R^3$
  • Multiply both sides by 3: $r_c^3 = 3R^3$
  • Take the cube root of both sides: $r_c = \sqrt[3]{3R^3} = \sqrt[3]{3} \times \sqrt[3]{R^3} = \sqrt[3]{3} R$

The Cylinder

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:

  • Volume of Cylinder ($V_{cyl}$) $= \pi r_{cy}^2 h_{cy}$

Substitute $h_{cy} = r_{cy}$ into the volume formula:

  • $V_{cyl} = \pi r_{cy}^2 (r_{cy}) = \pi r_{cy}^3$

Since the smaller part of the sphere is moulded into this cylinder, their volumes are equal:

  • $V_{cyl} = V_s$
  • $\pi r_{cy}^3 = \frac{1}{3}\pi R^3$

Now, we can solve for $r_{cy}$ (and thus $h_{cy}$) in terms of $R$:

  • Divide both sides by $\pi$: $r_{cy}^3 = \frac{1}{3}R^3$
  • Take the cube root of both sides: $r_{cy} = \sqrt[3]{\frac{1}{3}R^3} = \sqrt[3]{\frac{1}{3}} \times \sqrt[3]{R^3} = \frac{1}{\sqrt[3]{3}} R$

Since $h_{cy} = r_{cy}$, the height of the cylinder is $h_{cy} = \frac{1}{\sqrt[3]{3}} R$.

Finding the Required Ratio

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 $\sqrt[3]{9} : 1$.

Summary of Volumes and Dimensions

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 Calculation Steps

Ratio of $r_c$ to $h_{cy}$:

  • $\frac{r_c}{h_{cy}} = \frac{\sqrt[3]{3} R}{\frac{1}{\sqrt[3]{3}} R}$
  • $= \frac{\sqrt[3]{3}}{\frac{1}{\sqrt[3]{3}}}$
  • $= \sqrt[3]{3} \times \sqrt[3]{3}$
  • $= \sqrt[3]{3 \times 3}$
  • $= \sqrt[3]{9}$

The ratio is $\sqrt[3]{9} : 1$.

Revision Table: Geometric Volumes and Formulas

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

Additional Information: Conservation of Volume

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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Important Questions from Solid Figures

  1. A cylindrical tube, open at both ends, is made of a metal sheet which is 0.5 cm thick. Its outer radius is 4 cm and length is 2 m. How much metal (in cm 3) has been used in making the tube?

  2. The volume of a right circular cone is 308 cm 3 and the radius of its base is 7 cm. What is the curved surface area (in cm 2) of the cone? (Take π =  \(\frac{22}{7} \) )

  3. The slant height and radius of a right circular cone are in the ratio 29 ∶ 20. If its volume is 4838.4 π cm 3, then its radius is: 

  4. Six cubes, each of edge 2 cm, are joined end to end. What is the total surface area of the resulting cuboid in cm 2?

  5. A solid cube of side 8 cm is dropped into a rectangular container of length 16 cm, breadth 8 cm and height 15 cm which is partly filled with water. If the cube is completely submerged, then the rise of water level (in cm) is:

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