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Question

A right circular cone is 9.8 cm high with the radius of its base 4 cm. It is melted and recast into a right circular cone with the radius of its base 3.5 cm. Find its height (use π = \(\frac{22}{7}\))

The correct answer is

12.8 cm

Understanding Cone Volume and Recasting Problems

This problem involves melting a right circular cone and recasting it into a different right circular cone. The key principle here is that when a solid is melted and reshaped, its volume remains constant. We need to find the height of the new cone using the volume formula for a cone.

Formula for Volume of a Cone

The volume (\(V\)) of a right circular cone with radius (\(r\)) and height (\(h\)) is given by the formula:

\(V = \frac{1}{3} \pi r^2 h\)

Information Given for the First Cone

  • Height (\(h_1\)) = 9.8 cm
  • Radius (\(r_1\)) = 4 cm
  • Using \(\pi = \frac{22}{7}\)

Calculating the Volume of the First Cone

Using the volume formula:

\(V_1 = \frac{1}{3} \times \pi \times r_1^2 \times h_1\)

\(V_1 = \frac{1}{3} \times \frac{22}{7} \times (4 \, \text{cm})^2 \times 9.8 \, \text{cm}\)

\(V_1 = \frac{1}{3} \times \frac{22}{7} \times 16 \, \text{cm}^2 \times 9.8 \, \text{cm}\)

Let's simplify the calculation. \(9.8\) divided by \(7\) is \(1.4\).

\(V_1 = \frac{1}{3} \times 22 \times 16 \times 1.4 \, \text{cm}^3\)

\(V_1 = \frac{22 \times 16 \times 1.4}{3} \, \text{cm}^3\)

\(V_1 = \frac{352 \times 1.4}{3} \, \text{cm}^3\)

\(V_1 = \frac{492.8}{3} \, \text{cm}^3\)

Information Given for the Second Cone

  • Radius (\(r_2\)) = 3.5 cm
  • Height (\(h_2\)) = ? (what we need to find)
  • Using \(\pi = \frac{22}{7}\)

Setting Volumes Equal and Solving for Height

Since the first cone is melted and recast into the second cone, their volumes are equal:

\(V_1 = V_2\)

Using the volume formula for the second cone:

\(V_2 = \frac{1}{3} \pi r_2^2 h_2\)

\(V_2 = \frac{1}{3} \times \frac{22}{7} \times (3.5 \, \text{cm})^2 \times h_2\)

\(V_2 = \frac{1}{3} \times \frac{22}{7} \times (3.5 \times 3.5) \, \text{cm}^2 \times h_2\)

\(V_2 = \frac{1}{3} \times \frac{22}{7} \times 12.25 \, \text{cm}^2 \times h_2\)

Now, set \(V_1 = V_2\):

\(\frac{1}{3} \times \frac{22}{7} \times 16 \times 9.8 = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times h_2\)

We can cancel the common terms \(\frac{1}{3}\) and \(\frac{22}{7}\) from both sides:

\(16 \times 9.8 = 12.25 \times h_2\)

Calculate the left side:

\(16 \times 9.8 = 156.8\)

The equation is now:

\(156.8 = 12.25 \times h_2\)

Solve for \(h_2\):

\(h_2 = \frac{156.8}{12.25}\)

To perform the division, it might be easier to remove the decimals. Multiply both numerator and denominator by 100:

\(h_2 = \frac{156.8 \times 100}{12.25 \times 100} = \frac{15680}{1225}\)

Performing the division:

\(h_2 = 12.8\)

So, the height of the new cone is 12.8 cm.

Summary of Calculations

Item First Cone Second Cone
Height (\(h\)) 9.8 cm \(h_2\)
Radius (\(r\)) 4 cm 3.5 cm
Volume (\(V\)) \(V_1 = \frac{1}{3} \times \frac{22}{7} \times 4^2 \times 9.8\) \(V_2 = \frac{1}{3} \times \frac{22}{7} \times 3.5^2 \times h_2\)
Volume Relation \(V_1 = V_2\)
Equation \(16 \times 9.8 = 3.5^2 \times h_2\)
Result \(h_2 = 12.8\) cm

The height of the new cone is 12.8 cm.

Revision Table: Geometry Formulas

Shape Volume Formula Key Variables
Cone \(\frac{1}{3} \pi r^2 h\) \(r\) = radius, \(h\) = height
Cylinder \(\pi r^2 h\) \(r\) = radius, \(h\) = height
Sphere \(\frac{4}{3} \pi r^3\) \(r\) = radius
Cube \(s^3\) \(s\) = side length
Cuboid \(l \times w \times h\) \(l\) = length, \(w\) = width, \(h\) = height

Additional Information: Conservation of Volume

The principle of conservation of volume is fundamental in problems involving melting and recasting solids. It states that when a solid is transformed from one shape to another without any loss or addition of material, its total volume remains unchanged. This applies whether a single solid is reshaped or multiple smaller solids are combined to form a larger one, or vice versa.

  • This principle assumes the density of the material does not change during the process.
  • It is widely used in geometry and mensuration problems.
  • Knowing the volumes of different 3D shapes is essential for solving such problems.
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Important Questions from Solid Figures

  1. A cone and a hemisphere have equal bases and volumes. What is the ratio of the height of the cone to the radius of the hemisphere?

  2. If the surface area of a sphere is 64 π cm 2, then the volume of the sphere is:

  3. Find the surface area of a sphere of diameter 21 cm. (Use π = \(\frac{{22}}{7}\) )

  4. A cube is 7 cm of an edge and another cube is 14 cm on an edge. The ratios of their surface areas are

  5. Using three distinct points which of the following shapes cannot be formed?

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