lass 10 Maths Chapter 12 Surface Areas and Volumes – Exercise 12.3 NCERT Solutions

Class 10 Maths Chapter 12 Surface Areas and Volumes – Exercise 12.3 NCERT Solutions

Introduction

Exercise 12.3 focuses on conversion of solids from one shape to another. These problems involve calculating the volume of a solid and then using it to determine dimensions or number of new solids formed by melting or reshaping. This exercise strengthens your ability to apply volume formulas in practical situations.

Formula Used

• Cube:

$$V=a^3$$

• Cuboid:

$$V=l\cdot b\cdot h$$

• Cylinder:

$$V=\pi r^{2}h$$

• Sphere:

$$V=\frac{4}{3}\pi r^3$$

• Hemisphere:

$$V=\frac{2}{3}\pi r^3$$

• Cone:

$$V=\frac{1}{3}\pi r^{2}h$$

NCERT Questions with Solutions (10)

Q1. A metallic sphere of radius 7 cm is melted and recast into smaller spheres of radius 1 cm. Find number of spheres.

$$V_{\text{big}}=\frac{4}{3}\pi (7^3)=1436{\text{cm}}^3$$

$$V_{\text{small}}=\frac{4}{3}\pi (1^3)=4.19{\text{cm}}^3$$

$$\text{Number}=\frac{1436}{4.19}=343$$

Q2. A solid sphere of radius 6 cm is melted and recast into cones of radius 3 cm and height 4 cm. Find number of cones.

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

$$V_{\text{cone}}=\frac{1}{3}\pi (3^2)(4)=37.68{\text{cm}}^3$$

$$\text{Number}=\frac{904.32}{37.68}=24$$

Q3. A solid sphere of radius 3 cm is melted and recast into smaller spheres of radius 1 cm. Find number of spheres.

$$V_{\text{sphere}}=\frac{4}{3}\pi (3^3)=113.1{\text{cm}}^3$$

$$V_{\text{small}}=\frac{4}{3}\pi (1^3)=4.19{\text{cm}}^3$$

$$\text{Number}=\frac{113.1}{4.19}=27$$

Q4. A solid sphere of radius 12 cm is melted and recast into cubes of side 3 cm. Find number of cubes.

$$V_{\text{sphere}}=\frac{4}{3}\pi ({12}^3)=7238{\text{cm}}^3$$

$$V_{\text{cube}}=3^3=27{\text{cm}}^3$$

$$\text{Number}=\frac{7238}{27}=268$$

Q5. A solid sphere of radius 7 cm is melted and recast into cones of radius 7 cm and height 9 cm. Find number of cones.

$$V_{\text{sphere}}=\frac{4}{3}\pi (7^3)=1436{\text{cm}}^3$$

$$V_{\text{cone}}=\frac{1}{3}\pi (7^2)(9)=462{\text{cm}}^3$$

$$\text{Number}=\frac{1436}{462}\approx 3$$

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

$$V_{\text{big}}=\frac{4}{3}\pi ({14}^3)=11494{\text{cm}}^3$$

$$V_{\text{small}}=\frac{4}{3}\pi (7^3)=1436{\text{cm}}^3$$

$$\text{Number}=\frac{11494}{1436}=8$$

Q7. A solid sphere of radius 5 cm is melted and recast into cones of radius 5 cm and height 10 cm. Find number of cones.

$$V_{\text{sphere}}=\frac{4}{3}\pi (5^3)=523.6{\text{cm}}^3$$

$$V_{\text{cone}}=\frac{1}{3}\pi (5^2)(10)=261.8{\text{cm}}^3$$

$$\text{Number}=\frac{523.6}{261.8}=2$$

Q8. A solid sphere of radius 9 cm is melted and recast into cubes of side 3 cm. Find number of cubes.

$$V_{\text{sphere}}=\frac{4}{3}\pi (9^3)=3053{\text{cm}}^3$$

$$V_{\text{cube}}=3^3=27{\text{cm}}^3$$

$$\text{Number}=\frac{3053}{27}=113$$

Q9. A solid sphere of radius 10 cm is melted and recast into cones of radius 5 cm and height 12 cm. Find number of cones.

$$V_{\text{sphere}}=\frac{4}{3}\pi ({10}^3)=4188.8{\text{cm}}^3$$

$$V_{\text{cone}}=\frac{1}{3}\pi (5^2)(12)=314.2{\text{cm}}^3$$

$$\text{Number}=\frac{4188.8}{314.2}\approx 13$$

Q10. A solid sphere of radius 8 cm is melted and recast into smaller spheres of radius 2 cm. Find number of spheres.

$$V_{\text{big}}=\frac{4}{3}\pi (8^3)=2144.7{\text{cm}}^3$$

$$V_{\text{small}}=\frac{4}{3}\pi (2^3)=33.5{\text{cm}}^3$$

$$\text{Number}=\frac{2144.7}{33.5}=64$$

FAQs (10)

1. What is conversion of solids? Changing one solid shape into another by melting or reshaping.

2. What remains constant in conversion? Volume remains constant.

3. Can surface area change in conversion? Yes, surface area changes.

4. Formula for sphere volume?

$$V=\frac{4}{3}\pi r^3$$

5. Formula for cone volume?

$$V=\frac{1}{3}\pi r^{2}h$$

6. Formula for cube volume?

$$V=a^3$$

7. Formula for cuboid volume?

$$V=l\cdot b\cdot h$$

8. Formula for hemisphere volume?

$$V=\frac{2}{3}\pi r^3$$

9. Why is conversion important? Used in manufacturing and design.

10. Why is this exercise important? It builds skills in applying volume formulas to practical

Conclusion

Exercise 12.3 has 10 solved questions and 10 FAQs that strengthen your understanding of volumes of solid figures. This builds the foundation for advanced problems in Class 10 Maths.

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New Syllabus-Class 10 Maths Chapter 12 Surface Areas and Volumes – Exercise 12.3 NCERT Solutions