Ch 12: SA & Volumes (Old Syllabus)
Covers the "Conversion of Solids" (Melt & Recast) and "Frustum of a Cone" from Old Exercises 13.3 & 13.4.
Exercise 13.3 (Conversion of Solids)
💡 Trick: When a solid is melted and recast into another, Volume of Initial Solid = Volume of Final Solid.
Q1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Volume of Sphere = Volume of Cylinder
$\frac{4}{3}\pi R^3 = \pi r^2 h$
$\frac{4}{3} \times (4.2)^3 = (6)^2 \times h$
$\frac{4}{3} \times 4.2 \times 4.2 \times 4.2 = 36 \times h$
$4 \times 1.4 \times 4.2 \times 4.2 = 36 \times h$
$h = \frac{98.784}{36} =$ 2.744 cm.
Q3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Volume of earth dug out = Volume of cylindrical well
$V = \pi r^2 h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 = 22 \times 3.5 \times 10 = \mathbf{770 \text{ m}^3}$.
This earth is spread to form a cuboidal platform.
Volume of platform = Length $\times$ Breadth $\times$ Height
$770 = 22 \times 14 \times h$
$h = \frac{770}{22 \times 14} = \frac{35}{14} = \frac{5}{2} =$ 2.5 m.
Q4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Radius of well ($r$) = $3/2 = 1.5 \text{ m}$. Depth ($h$) = $14 \text{ m}$.
Volume of earth dug = $\pi r^2 h = \pi \times (1.5)^2 \times 14 = \mathbf{31.5\pi \text{ m}^3}$.
The embankment is a hollow cylinder (ring) around the well.
Inner radius ($r$) = $1.5 \text{ m}$.
Outer radius ($R$) = $1.5 + 4 (\text{width}) = \mathbf{5.5 \text{ m}}$.
Volume of embankment = Area of ring $\times$ height ($H$)
$V = \pi (R^2 - r^2) \times H$
$31.5\pi = \pi ((5.5)^2 - (1.5)^2) \times H$
$31.5 = (30.25 - 2.25) \times H$
$31.5 = 28 \times H \implies H = \frac{31.5}{28} =$ 1.125 m.
Exercise 13.4 (Frustum of a Cone)
💡 Formulas: Vol = $\frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)$ | CSA = $\pi l (r_1 + r_2)$
Q1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
Height ($h$) = $14 \text{ cm}$, $r_1 = 4/2 = 2 \text{ cm}$, $r_2 = 2/2 = 1 \text{ cm}$.
Capacity (Volume) = $\frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)$
= $\frac{1}{3} \times \frac{22}{7} \times 14 \times (2^2 + 1^2 + 2\times1)$
= $\frac{1}{3} \times 44 \times (4 + 1 + 2)$
= $\frac{1}{3} \times 44 \times 7 = \frac{308}{3} =$ 102.67 cm³.
Q3. A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
Solution:
Open side radius ($r_1$) = $10 \text{ cm}$. Upper closed side radius ($r_2$) = $4 \text{ cm}$. Slant height ($l$) = $15 \text{ cm}$.
Area of material = CSA of frustum + Area of upper closed circular base
= $\pi l (r_1 + r_2) + \pi r_2^2$
= $\frac{22}{7} \times 15 \times (10 + 4) + \frac{22}{7} \times (4)^2$
= $\frac{22}{7} \times 15 \times 14 + \frac{22}{7} \times 16$
= $22 \times 30 + \frac{352}{7} = 660 + 50.28 =$ 710.28 cm².