CBSE · Class 10 · Mathematics · Surface Areas and VolumesA solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. (Take $\pi = 3.14$). Also, find the volume of the circumscribing cylinder. Find the difference between the volumes of the cylinder and the toy.

Step-by-Step Solution:

1. Given Data:

• Diameter of the base ($d$) = 4 cm
• Radius of the base ($r$) = $\frac{4}{2} = 2$ cm
• Height of the cone ($h$) = 2 cm
• Radius of the hemisphere = Radius of the cone ($r$) = 2 cm

2. Formulae Used:

• Volume of cone = $\frac{1}{3} \pi r^2 h$
• Volume of hemisphere = $\frac{2}{3} \pi r^3$
• Volume of toy = Volume of cone + Volume of hemisphere

3. Calculation of Toy's Volume:

• Volume of cone = $\frac{1}{3} \times 3.14 \times (2)^2 \times 2$ $= \frac{1}{3} \times 3.14 \times 4 \times 2$ $= \frac{25.12}{3} = 8.373$ $cm^3$

• Volume of hemisphere = $\frac{2}{3} \times 3.14 \times (2)^3$ $= \frac{2}{3} \times 3.14 \times 8$ $= \frac{50.24}{3} = 16.747$ $cm^3$

• Volume of toy = $8.373 + 16.747 = 25.12$ $cm^3$

4. Calculation of Circumscribing Cylinder's Volume:

• Height of cylinder ($H$) = Height of cone + Radius of hemisphere = $2 + 2 = 4$ cm
• Radius of cylinder ($r$) = 2 cm
• Volume of cylinder = $\pi r^2 H = 3.14 \times (2)^2 \times 4$ $= 3.14 \times 4 \times 4 = 50.24$ $cm^3$

5. Difference in Volumes:

• Difference = Volume of cylinder - Volume of toy $= 50.24 - 25.12 = 25.12$ $cm^3$

Answer:\nThe volume of the toy is $25.12$ $cm^3$ and the difference between the volumes of the cylinder and the toy is also $25.12$ $cm^3$.