Class 9 Maths Chapter 11 Surface Areas and Volumes – Exercise 11.5 NCERT Solutions

Class 9 Maths Chapter 11 Surface Areas and Volumes – Exercise 11.5 NCERT Solutions

Introduction

Exercise 11.5 focuses on surface areas and volumes of spheres and hemispheres in practical problems. Students apply formulas to real‑life contexts such as balls, globes, and containers. This exercise strengthens problem‑solving skills and connects geometry with everyday applications.

Key Formulas

1. Surface Area of Sphere:

$$\text{Surface Area}=4\pi r^2$$

2. Volume of Sphere:

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

3. Surface Area of Hemisphere:

   • Curved surface area:

$$2\pi r^2$$

• Total surface area:

$$3\pi r^2$$

4. Volume of Hemisphere:

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

Common Mistakes

• Confusing curved surface area with total surface area.

• Forgetting to cube radius in volume formula.

• Using diameter directly instead of halving to get radius.

• Not converting units consistently.

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

Q1. A spherical ball of radius 7 cm is melted to form small balls of radius 1.4 cm. Find number of balls formed.

$$\text{Volume of big ball}=\frac{4}{3}\pi (7{)}^3=\frac{4}{3}\pi \cdot 343$$

$$\text{Volume of small ball}=\frac{4}{3}\pi (1.4{)}^3$$

$$\text{Number of balls}=\frac{343}{(1.4{)}^3}=343\mathrm{/}2.744\approx 125$$

Q2. A spherical ball of diameter 12 cm is melted to form small balls of diameter 3 cm. Find number of balls formed. Radius of big ball = 6 cm, small ball = 1.5 cm.

$$\text{Number of balls}=\frac{6^3}{{1.5}^3}=\frac{216}{3.375}=64$$

Q3. A hemispherical bowl of radius 7 cm is filled with soup. How much soup can it hold?

$$V=\frac{2}{3}\pi r^3=\frac{2}{3}\times \frac{22}{7}\times 343=718.67{\text{cm}}^3$$

Q4. Find surface area of hemisphere of radius 10 cm.

$$\text{TSA}=3\pi r^2=3\times \frac{22}{7}\times 100=942.86{\text{cm}}^2$$

Q5. Find volume of hemisphere of radius 10 cm.

$$V=\frac{2}{3}\pi r^3=\frac{2}{3}\times \frac{22}{7}\times 1000=2094.3{\text{cm}}^3$$

Q6. A spherical ball of radius 3.5 cm is melted to form cones of radius 3.5 cm and height 7 cm. Find number of cones formed.

$$\text{Volume of sphere}=\frac{4}{3}\pi (3.5{)}^3$$

$$\text{Volume of cone}=\frac{1}{3}\pi (3.5{)}^2\times 7$$

$$\text{Number of cones}=\frac{\frac{4}{3}\pi (42.875)}{\frac{1}{3}\pi (85.75)}=\frac{171.5}{85.75}=2$$

Q7. A spherical ball of radius 7 cm is melted to form cones of radius 7 cm and height 24 cm. Find number of cones formed.

$$\text{Volume of sphere}=\frac{4}{3}\pi (343)=1437.3{\text{cm}}^3$$

$$\text{Volume of cone}=\frac{1}{3}\pi (49)\times 24=1232{\text{cm}}^3$$

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

Q8. A spherical ball of radius 21 cm is melted to form small spheres of radius 7 cm. Find number of spheres formed.

$$\text{Number of spheres}=\frac{{21}^3}{7^3}=\frac{9261}{343}=27$$

Q9. A spherical ball of radius 14 cm is melted to form small spheres of radius 7 cm. Find number of spheres formed.

$$\text{Number of spheres}=\frac{{14}^3}{7^3}=\frac{2744}{343}=8$$

Q10. A hemispherical tank of radius 3.5 m is full of water. Find capacity in litres.

$$V=\frac{2}{3}\pi r^3=\frac{2}{3}\times \frac{22}{7}\times (3.5{)}^3=89.6{\text{m}}^3$$

$$1{\text{m}}^3=1000\text{litres}\Rightarrow 89.6\times 1000=89600\text{litres}$$

FAQs (10)

FAQ1. What is formula for volume of sphere? $\frac{4}{3}\pi r^3$.

FAQ2. What is formula for volume of hemisphere? $\frac{2}{3}\pi r^3$.

FAQ3. What is TSA of hemisphere? $3\pi r^2$.

FAQ4. What is CSA of hemisphere? $2\pi r^2$.

FAQ5. Why cube radius in volume formula? Volume depends on three dimensions.

FAQ6. What is unit of volume? Cubic units (cm${}^3$, m${}^3$).

FAQ7. What is unit of surface area? Square units (cm${}^2$, m${}^2$).

FAQ8. What is real‑life example of hemisphere? Bowls, tanks, domes.

FAQ9. Why is Exercise 11.5 important? It connects geometry with practical problems.

FAQ10. What is practical use of sphere/hemisphere formulas? Used in packaging, storage, and design.

Conclusion

Exercise 11.5 covers surface areas and volumes of spheres and hemispheres in practical contexts with solved examples and FAQs. Mastering these problems helps students apply mensuration to real‑life situations.

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