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Mathematics · Surface Areas and Volumes · NCERT Examples

Example 7

A complete, independently verified solution for NCERT Example 7.

Mathematics · Chapter 12 · NCERT Examples

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NCERT Example

Example 7

toy volume · circumscribing cylinder

A toy is a hemisphere surmounted by a right circular cone. The cone is 2 cm high and the common base diameter is 4 cm. Find the toy's volume. A right circular cylinder circumscribes the toy; find the difference between the cylinder's volume and the toy's volume. Take π = 3.14.

Cone over a hemisphere circumscribed by a right circular cylinder, with points A to H, O and P labelled.
Figure 12-14 from the supplied NCERT chapter.

Solution

Board-exam working:

r=42=2 cm,hcone=2 cmr=\frac42=2\text{ cm},\qquad h_{cone}=2\text{ cm}

Volume of the toy:

Vt=23πr3+13πr2hV_t=\frac23\pi r^3+\frac13\pi r^2h
=23(3.14)(2)3+13(3.14)(2)2(2)=\frac23(3.14)(2)^3+\frac13(3.14)(2)^2(2)
=25.12 cm3=25.12\text{ cm}^3

The circumscribing cylinder has radius 2 cm and height 2 + 2 = 4 cm.

Vc=πr2H=3.14×22×4=50.24 cm3V_c=\pi r^2H=3.14\times2^2\times4=50.24\text{ cm}^3

Required difference:

Vc−Vt=50.24−25.12=25.12 cm3V_c-V_t=50.24-25.12=25.12\text{ cm}^3

Final answer

Vt=25.12 cm3,Vc−Vt=25.12 cm3V_t=25.12\text{ cm}^3,\qquad V_c-V_t=25.12\text{ cm}^3