Volumes of Combinations of Solids for CBSE Class 10 Mathematics

While calculating the surface area of combined solids requires careful tracking of which boundaries are exposed and which are glued together, calculating volume is refreshingly straightforward. Volume represents the three-dimensional space occupied by matter—a scalar quantity that obeys the law of physical superposition.

In CBSE Class 10 Mathematics, Chapter 12 (Surface Areas and Volumes), computing the volumes of combinations of solids involves straightforward addition or subtraction: when solids are joined together, their volumes add; when cavities or depressions are carved out, their volumes subtract.

What You Will Learn

• The universal additive rule for composite volumes: $\text{Total Volume} = V_1 + V_2 + \dots$
• Volume formula reference table (cubes, cylinders, cones, spheres, hemispheres)
• Problem Type 1: Solid cone standing on a hemisphere
• Problem Type 2: The 45 Gulab Jamun Sugar Syrup Problem (NCERT Classic)
• Problem Type 3: Cuboidal wooden pen stand with conical depressions
• Step-by-step solved CBSE board examination problems and arithmetic shortcuts

1. The Fundamental Volume Rule

Unlike surface areas (where internal contact faces vanish), volume depends solely on the amount of material present:

    Case 1: Joining Solids Together
    [ Total Volume ] = [ Volume of Solid 1 ] + [ Volume of Solid 2 ]

    Case 2: Scooping / Carving Cavities Out
    [ Remaining Volume ] = [ Volume of Original Solid ] - [ Volume of Carved Cavities ]

The Core Rule: <u>Volume is ALWAYS additive when solids are merged, and subtractive when cavities are carved out! Contact surfaces between joined components have zero volume and do not affect the total space occupied.</u>

2. Volume Quick-Reference Table

Notice the elegant relationship: The volume of a cone is exactly one-third the volume of a cylinder having the same base radius and height!

3. High-Yield Solved Board Examination Problems

Solved Example 1: Cone on a Hemisphere in Terms of $\pi$ (NCERT Classic)

Problem: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\text{ cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.

Solution:

1. Analyze Dimensions:
   • Radius of hemisphere: $r = 1\text{ cm}$.
   • Radius of cone: $r = 1\text{ cm}$.
   • Height of cone: $h = r = 1\text{ cm}$.
2. Formulate Total Volume: $\mathbf{\text{Total Volume} = \text{Volume of Cone} + \text{Volume of Hemisphere}}$ $\text{Total Volume} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$
3. Factor Out Common Terms: $\text{Total Volume} = \frac{1}{3}\pi r^2 (h + 2r)$
4. Substitute $r = 1$ and $h = 1$: $\text{Total Volume} = \frac{1}{3}\pi (1)^2 [1 + 2(1)] = \frac{1}{3}\pi (1) [3] = \mathbf{\pi\text{ cm}^3}$
5. Therefore, <u>the volume of the solid is $\pi\text{ cm}^3$</u>.

Solved Example 2: The 45 Gulab Jamun Sugar Syrup Problem (CBSE 5-Mark Classic)

Problem: A gulab jamun contains sugar syrup up to about $30\%$ of its volume. Find approximately how much syrup would be found in $45$ gulab jamuns, each shaped like a cylinder with two hemispherical ends with length $5\text{ cm}$ and diameter $2.8\text{ cm}$. (Use $\pi = 22/7$).

                      Hemisphere         Cylinder         Hemisphere
                        (-----[=============================]-----)
                        < 1.4 > <---------- 2.2 cm --------> < 1.4 >
                        <----------------- 5.0 cm ------------------>

Solution:

1. Analyze Dimensions of ONE Gulab Jamun:
   • Diameter $d = 2.8\text{ cm} \implies$ Radius $r = 1.4\text{ cm}$.
   • The two hemispherical ends each occupy a length equal to radius: $1.4\text{ cm}$.
   • Length (height) of the cylindrical part: $h = 5.0 - (1.4 + 1.4) = 5.0 - 2.8 = \mathbf{2.2\text{ cm}}$
2. Calculate Volume of ONE Gulab Jamun: $\text{Volume of 1 Gulab Jamun} = \text{Volume of Cylinder} + 2 \times (\text{Volume of Hemisphere})$ Notice that two hemispheres equal one complete sphere: $\text{Volume} = \pi r^2 h + \frac{4}{3}\pi r^3 = \mathbf{\pi r^2 \left(h + \frac{4}{3}r\right)}$
3. Substitute Values for One Piece: $\text{Volume} = \frac{22}{7} \times (1.4)^2 \times \left[2.2 + \frac{4}{3}(1.4)\right]$ $\text{Volume} = \frac{22}{7} \times 1.4 \times 1.4 \times \left[2.2 + \frac{5.6}{3}\right] = 22 \times 0.2 \times 1.4 \times \left[\frac{6.6 + 5.6}{3}\right]$ $\text{Volume} = 6.16 \times \frac{12.2}{3} = \mathbf{\frac{75.152}{3}\text{ cm}^3}$
4. Calculate Volume of All 45 Gulab Jamuns: $\text{Total Volume of 45 pieces} = 45 \times \left(\frac{75.152}{3}\right) = 15 \times 75.152 = \mathbf{1127.28\text{ cm}^3}$
5. Calculate Quantity of Sugar Syrup ($30\%$ of Total Volume): $\text{Syrup Volume} = 30\% \times 1127.28 = \frac{30}{100} \times 1127.28 = 0.3 \times 1127.28 = 338.184\text{ cm}^3 \approx \mathbf{338\text{ cm}^3}$
6. Therefore, <u>approximately $338\text{ cm}^3$ of sugar syrup is found in the $45$ gulab jamuns</u>.

Solved Example 3: Wooden Pen Stand with Conical Depressions

Problem: A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15\text{ cm}$ by $10\text{ cm}$ by $3.5\text{ cm}$. The radius of each of the depressions is $0.5\text{ cm}$ and the depth is $1.4\text{ cm}$. Find the volume of wood in the entire stand.

Solution:

1. Calculate Volume of the Wooden Cuboid: $V_{\text{cuboid}} = l \times b \times h = 15 \times 10 \times 3.5 = 150 \times 3.5 = \mathbf{525\text{ cm}^3}$
2. Calculate Volume of the 4 Conical Depressions:
   • Radius of depression $r = 0.5\text{ cm}$, depth (height) $h = 1.4\text{ cm}$. $V_{\text{one cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4 = \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4$ $V_{\text{one cone}} = \frac{1}{3} \times 22 \times 0.25 \times 0.2 = \frac{1.1}{3}\text{ cm}^3$
   • Volume of 4 depressions: $V_{\text{4 cones}} = 4 \times \frac{1.1}{3} = \frac{4.4}{3} = \mathbf{1.47\text{ cm}^3}$
3. Calculate Remaining Volume of Wood: $V_{\text{wood}} = V_{\text{cuboid}} - V_{\text{4 cones}} = 525 - 1.47 = \mathbf{523.53\text{ cm}^3}$
4. Therefore, <u>the volume of wood in the entire stand is $523.53\text{ cm}^3$</u>.

4. Summary and Examination Tips

Exam Tip: In the Gulab Jamun problem, DO NOT divide $\frac{75.152}{3}$ into recurring decimals early on! Keep the denominator $3$ in place until you multiply by $45$; the $45$ and $3$ divide cleanly ($45/3 = 15$), eliminating all messy decimal divisions!

Common Mistake: Confusing cone height with slant height. In volume calculations, you must ALWAYS use the vertical height $h$, NOT the slant height $l$! Slant height is strictly for curved surface area ($\pi r l$).