Class 9 Maths Chapter 11 Surface Areas and Volumes – Exercise 11.8 NCERT Solutions

Introduction

Exercise 11.8 focuses on problems involving conversion of one solid shape into another. Students learn how to apply the principle of conservation of volume when solids are melted and recast into different shapes such as spheres, cones, or cylinders. This exercise is highly practical and connects geometry with real‑life applications.

Key Concept

• Volume Conservation Principle: When a solid is melted and recast, its volume remains unchanged.

$$V_{\text{original}}=V_{\text{new}}$$

Common Mistakes

• Forgetting to cube the radius in volume formulas.

• Using diameter instead of radius.

• Confusing surface area with volume.

• Not converting units consistently (cm${}^3$, m${}^3$, litres).

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. A metallic sphere of radius 4.2 cm is melted and recast into smaller spheres of radius 2.1 cm. Find number of spheres formed.

$$\text{Number}=\frac{(4.2{)}^3}{(2.1{)}^3}=\frac{74.088}{9.261}=8$$

Q2. A solid sphere of radius 6 cm is melted and recast into cones of radius 6 cm and height 12 cm. Find number of cones formed.

$$V_{sphere}=\frac{4}{3}\pi (6{)}^3=904.32$$

$$V_{cone}=\frac{1}{3}\pi (6{)}^2\times 12=452.16$$

$$\text{Number}=\frac{904.32}{452.16}=2$$

Q3. A solid sphere of radius 7 cm is melted and recast into cones of radius 7 cm and height 24 cm. Find number of cones formed.

$$V_{sphere}=\frac{4}{3}\pi (7{)}^3=1437.3$$

$$V_{cone}=\frac{1}{3}\pi (7{)}^2\times 24=1232$$

$$\text{Number}\approx 1$$

Q4. A solid sphere of radius 3.5 cm is melted and recast into cones of radius 3.5 cm and height 7 cm. Find number of cones formed.

$$V_{sphere}=\frac{4}{3}\pi (3.5{)}^3=179.6$$

$$V_{cone}=\frac{1}{3}\pi (3.5{)}^2\times 7=89.6$$

$$\text{Number}=2$$

Q5. A solid sphere of radius 21 cm is melted and recast into smaller spheres of radius 7 cm. Find number of spheres formed.

$$\text{Number}=\frac{(21{)}^3}{(7{)}^3}=\frac{9261}{343}=27$$

Q6. A solid sphere of radius 14 cm is melted and recast into smaller spheres of radius 7 cm. Find number of spheres formed.

$$\text{Number}=\frac{(14{)}^3}{(7{)}^3}=\frac{2744}{343}=8$$

Q7. A solid sphere of radius 10.5 cm is melted and recast into cones of radius 3.5 cm and height 7 cm. Find number of cones formed.

$$V_{sphere}=\frac{4}{3}\pi (10.5{)}^3=4851$$

$$V_{cone}=\frac{1}{3}\pi (3.5{)}^2\times 7=89.6$$

$$\text{Number}=\frac{4851}{89.6}\approx 54$$

Q8. A solid sphere of radius 7 cm is melted and recast into smaller spheres of radius 1 cm. Find number of spheres formed.

$$\text{Number}=\frac{(7{)}^3}{(1{)}^3}=343$$

Q9. A solid sphere of radius 3.5 cm is melted and recast into smaller spheres of radius 0.7 cm. Find number of spheres formed.

$$\text{Number}=\frac{(3.5{)}^3}{(0.7{)}^3}=\frac{42.875}{0.343}=125$$

Q10. A solid sphere of radius 4.2 cm is melted and recast into cones of radius 2.1 cm and height 4.2 cm. Find number of cones formed.

$$V_{sphere}=\frac{4}{3}\pi (4.2{)}^3=310.3$$

$$V_{cone}=\frac{1}{3}\pi (2.1{)}^2\times 4.2=19.4$$

$$\text{Number}=\frac{310.3}{19.4}\approx 16$$

FAQs (10)

FAQ1. What is principle of conservation of volume? Volume remains constant when solid is melted and recast.

FAQ2. What is formula for volume of sphere? $\frac{4}{3}\pi r^3$.

FAQ3. What is formula for volume of cone? $\frac{1}{3}\pi r^{2}h$.

FAQ4. Why use radius instead of diameter? Formulas require radius.

FAQ5. What is unit of volume? Cubic units (cm${}^3$, m${}^3$).

FAQ6. What is real‑life example of recasting solids? Melting metal to form new shapes.

FAQ7. Why cube radius in volume formula? Volume depends on three dimensions.

FAQ8. What is difference between TSA and volume? TSA measures surface covering, volume measures capacity.

FAQ9. Why is Exercise 11.8 important? It applies mensuration to practical recasting problems.

FAQ10. What is practical use of these problems? Used in manufacturing, packaging, and design.

Conclusion

Exercise 11.8 covers conversion of solids from one shape to another with solved examples and FAQs. Mastering these problems helps students apply conservation of volume in practical contexts.

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