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Question

A cone of height 16 cm and diameter 14 cm is mounted on a hemisphere of same diameter. What is the volume of the solid thus formed? (take π = 22/7)

This question was previously asked in
CDS II 2021 General Knowledge Previous Year Paper (14-Nov-2021)
The correct answer is

1540 cm 3

Calculating the Volume of a Composite Solid

The problem asks us to find the total volume of a solid formed by mounting a cone on top of a hemisphere. Both the cone and the hemisphere share the same diameter.

Understanding the Components and Given Information

The solid is composed of two parts:

  1. A cone
  2. A hemisphere

We are given the following dimensions:

  • Height of the cone (h) = 16 cm
  • Diameter of the cone = 14 cm
  • Diameter of the hemisphere = 14 cm
  • Value of \(\pi\) to use = 22/7

Determining the Radius

The diameter for both the cone and the hemisphere is 14 cm. The radius (r) is half of the diameter.

Radius \(r = \frac{\text{Diameter}}{2} = \frac{14 \text{ cm}}{2} = 7 \text{ cm}\)

So, the radius for both the cone and the hemisphere is 7 cm.

Calculating the Volume of the Cone

The formula for the volume of a cone is \(V_{\text{cone}} = \frac{1}{3} \pi r^2 h\).

Substitute the values: \(r = 7 \text{ cm}\), \(h = 16 \text{ cm}\), and \(\pi = \frac{22}{7}\).

\(V_{\text{cone}} = \frac{1}{3} \times \frac{22}{7} \times (7 \text{ cm})^2 \times 16 \text{ cm}\)

\(V_{\text{cone}} = \frac{1}{3} \times \frac{22}{7} \times (49 \text{ cm}^2) \times 16 \text{ cm}\)

We can cancel out one 7 from the numerator (49) with the 7 in the denominator:

\(V_{\text{cone}} = \frac{1}{3} \times 22 \times 7 \times 16 \text{ cm}^3\)

\(V_{\text{cone}} = \frac{1}{3} \times (22 \times 7) \times 16 \text{ cm}^3\)

\(V_{\text{cone}} = \frac{1}{3} \times 154 \times 16 \text{ cm}^3\)

\(V_{\text{cone}} = \frac{2464}{3} \text{ cm}^3\)

Calculating the Volume of the Hemisphere

The formula for the volume of a hemisphere is \(V_{\text{hemisphere}} = \frac{2}{3} \pi r^3\).

Substitute the values: \(r = 7 \text{ cm}\) and \(\pi = \frac{22}{7}\).

\(V_{\text{hemisphere}} = \frac{2}{3} \times \frac{22}{7} \times (7 \text{ cm})^3\)

\(V_{\text{hemisphere}} = \frac{2}{3} \times \frac{22}{7} \times (7 \times 7 \times 7) \text{ cm}^3\)

We can cancel out one 7 from the numerator with the 7 in the denominator:

\(V_{\text{hemisphere}} = \frac{2}{3} \times 22 \times (7 \times 7) \text{ cm}^3\)

\(V_{\text{hemisphere}} = \frac{2}{3} \times 22 \times 49 \text{ cm}^3\)

\(V_{\text{hemisphere}} = \frac{2}{3} \times 1078 \text{ cm}^3\)

\(V_{\text{hemisphere}} = \frac{2156}{3} \text{ cm}^3\)

Calculating the Total Volume of the Solid

The total volume of the solid is the sum of the volume of the cone and the volume of the hemisphere because the cone is mounted on the hemisphere.

\(V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}}\)

\(V_{\text{total}} = \frac{2464}{3} \text{ cm}^3 + \frac{2156}{3} \text{ cm}^3\)

Since the denominators are the same, we can add the numerators directly:

\(V_{\text{total}} = \frac{2464 + 2156}{3} \text{ cm}^3\)

\(V_{\text{total}} = \frac{4620}{3} \text{ cm}^3\)

Now, perform the division:

\(4620 \div 3\)

\(4620 / 3 = 1540\)

\(V_{\text{total}} = 1540 \text{ cm}^3\)

Conclusion

The volume of the solid formed by mounting the cone on the hemisphere is 1540 cm\(^{3}\).

Revision Table: Solid Volume Calculation

Component Shape Formula Dimensions Calculated Volume
Bottom Part Hemisphere \(\frac{2}{3} \pi r^3\) \(r=7 \text{ cm}\) \(\frac{2156}{3} \text{ cm}^3\)
Top Part Cone \(\frac{1}{3} \pi r^2 h\) \(r=7 \text{ cm}, h=16 \text{ cm}\) \(\frac{2464}{3} \text{ cm}^3\)
Composite Solid Cone on Hemisphere \(V_{\text{cone}} + V_{\text{hemisphere}}\) Combined \(\frac{2464}{3} + \frac{2156}{3} = \frac{4620}{3} = 1540 \text{ cm}^3\)

Additional Information: Volumes of Geometric Shapes

Calculating the volume of composite solids like the one in this problem involves understanding the formulas for basic geometric shapes and adding the volumes of the individual parts. Here are some related concepts:

  • Cylinder Volume: The volume of a cylinder is given by \(V = \pi r^2 h\), where \(r\) is the radius of the base and \(h\) is the height. A cone's volume is one-third of a cylinder with the same base radius and height.
  • Sphere Volume: The volume of a sphere is given by \(V = \frac{4}{3} \pi r^3\), where \(r\) is the radius. A hemisphere is half of a sphere, so its volume is half of the sphere's volume.
  • Surface Area: While this problem focuses on volume, calculating the surface area of a composite solid is often different from just adding the individual surface areas because some surfaces are hidden where the shapes join. For this solid, the total surface area would be the curved surface area of the cone plus the curved surface area of the hemisphere plus the area of the base of the hemisphere (if it's resting on a surface, but here it's joined to the cone). The base circle of the cone is hidden by the hemisphere connection.
  • Other Composite Shapes: Composite solids can be formed by combining cylinders, cubes, cuboids, pyramids, etc. The approach is similar: identify the individual shapes, calculate their volumes, and sum them up.

Understanding these basic formulas and the concept of breaking down complex shapes into simpler ones is key to solving problems involving volumes and surface areas of composite solids.

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Similar Questions

  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. The volume of a hemisphere is 155232 cm 3. What is the radius of the hemisphere?

  3. The radius and height of a right circular cone are in the ratio 3 : 7. If the volume of the cone is 528 cm 3, then what is the height of the cone? \(\left( {{\rm{Take}}\,\,{\rm{\pi }}\,{\rm{ = }}\frac{{22}}{7}} \right)\)

  4. The length, breadth and height of a cuboid are in the ratio 27 : 8 : 1. The cuboid is melted and recast into a cube. If p is the surface area of the cuboid and q is the surface area of the cube, then what is p/q equal to?

  5. A square sheet of side length 44 cm is rolled along one of its sides to form a cylinder by making opposite edges just to touch each other. What is the volume of the cylinder ? (Take π = 22/7)  

  6. A lamp shade is in the shape of a part of a cone and its top and bottom ends are circles whose circumferences are respectively 30 cm and 40 cm. The perpendicular distance between the ends is 6 cm. If the cone were to be completed, then how far would its vertex be from the top end?

  7. Three solid lead spheres of radius 6 cm, 8 cm and 10 cm are melted together and recast as a solid sphere. What is the percentage diminution of the surface area as compared to the sum of the surface areas of the three spheres ?

  8. A solid sphere of radius 3 cm is melted to form a hollow cylinder of height 4 cm and external diameter 10 cm. What is the thickness of the cylinder?

  9. A cylindrical pipe has inner diameter of 14 cm. Water flows through it at a rate of 154 litres per minute. What is the speed of water in km/hr? \(\left( {{\rm{Take}}\,\,{\rm{\pi }}\,{\rm{ = }}\frac{{22}}{7}} \right)\)

  10. What is the radius of the base of the cone ?


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. A metallic solid cuboid of dimensions 36 cm × 18 cm × 12 cm is melted and recast in the form of cubes of side 6 cm. Find the number of cubes so formed.

  3. A solid cylinder has a radius of 9 cm and a height of 25 cm. What is the ratio of its total surface area to its curved surface area?

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

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

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