NCERT Solutions Class 8 Maths Chapter 9 Mensuration Exercise 9.3

NCERT Solutions Class 8 Maths Chapter 9 , Mensuration, helps students understand how to calculate surface areas and volumes of different 3D shapes. In NCERT Solutions Class 8 Maths Chapter 9 Exercise 9.3, the focus is on finding the volume of cubes and cuboids. This is useful in real life when we need to know how much space an object can hold — like a box, tank, or room. 

These solutions provided an easy-to-understand explanation of the method and followed the format of the NCERT textbook. Students who practice using these solutions will be able to perform better on tests and develop their calculation skills.

1.0Download NCERT Solutions Class 8 Maths Chapter 9 Mensuration Exercise 9.3: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 9 Mensuration Exercise 9.3 PDF with clear, step-by-step answers to help you understand volume concepts easily and score well.

2.0Key Concepts in Exercise 9.3 of Class 8 Maths Chapter 9

Exercise 9.3 focuses on finding the volume and capacity of solid shapes like cube, cuboid, and cylinder. This exercise helps students understand how much space a 3D object occupies and introduces the concept of measuring the quantity a container can hold.

Key Concepts Covered in Exercise 9.3:

• Volume of a Cube
  A cube has all sides equal. The formula to calculate volume is:
  $Volume=a^3$
• where aa is the length of a side.
• Volume of a Cuboid
  A cuboid has different lengths, breadths, and heights. The volume is calculated using:
  Volume=l×b×h
  where ll, bb, and hh are length, breadth, and height.
• Volume of a Cylinder
  A cylinder's volume is calculated using the formula:
  $Volume=\pi r^{2}h$
  where rr is the radius and hh is the height.
• Capacity and Volume Relation
  Capacity refers to how much liquid a container can hold. 1,000 cm³ = 1 litre. Students learn to convert between volume (cm³ or m³) and capacity (litres).
• Units and Conversions
  Understanding and using correct units like cubic centimetres (cm³), cubic metres (m³), and litres is emphasized.

3.0NCERT Class 8 Maths Chapter 9: Other Exercises

4.0NCERT Class 8 Maths Chapter 9 Exercise 9.3: Detailed Solutions

1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume. (a) To find how much it can hold. (b) Number of cement bags required to plaster it. (c) To find the number of smaller tanks that can be filled with water from it. Sol. (a) Volume (b) Surface area (c) Volume
2. Diameter of cylinder $A$ is 7 cm , and the height is 14 cm . Diameter of cylinder $B$ is 14 cm and height is 7 cm . Without doing any calculations can you suggest whose volume is greater? Verify by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
   
   Sol. The heights and diameters of these cylinders A and B are interchanged. We know that, Volume of cylinder $=\pi r^{2}h$ Radius of cylinder $\mathrm{A}=\frac{7}{2}\mathrm{cm}$ Radius of cylinder $\mathrm{B}\left(\frac{14}{2}\right)\mathrm{cm}=7\mathrm{cm}$ As the radius of cylinder B is greater, therefore, the volume of cylinder $B$ will be greater. Let us verify it by calculating the volume of both the cylinders. Volume of cylinder $\mathrm{A}=\pi {\mathrm{r}}^2\mathrm{h}$ $=\left(\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 14\right){\mathrm{cm}}^3=539{\mathrm{cm}}^3$ Volume of cylinder $B=\pi r^{2}h$ $=\left(\frac{22}{7}\times 7\times 7\times 7\right){\mathrm{cm}}^3=1078{\mathrm{cm}}^3$ Volume of cylinder B is greater. $=$ Surface area of cylinder $\mathrm{A}=2\pi \mathrm{r}(\mathrm{r}+\mathrm{h})$ $=\left[2\times \frac{22}{7}\times \frac{7}{2}\left(\frac{7}{2}+14\right)\right]{\mathrm{cm}}^2$ $=\left[22\times \left(\frac{7+28}{2}\right)\right]{\mathrm{cm}}^2=\left(22\times \frac{35}{2}\right){\mathrm{cm}}^2$ $=385{\mathrm{cm}}^2$ Surface area of cylinder B $=2\pi r(r+h)$ $=\left[2\times \frac{22}{7}\times 7\times (7+7)\right]{\mathrm{cm}}^2$ $=(44\times 14){\mathrm{cm}}^2$ $=616{\mathrm{cm}}^2$ Thus, the surface area of cylinder B is also greater than the surface area of cylinder A.
3. Find the height of a cuboid whose base area is $180{\mathrm{cm}}^2$ and volume is $900{\mathrm{cm}}^3$ ? Sol. Volume of the cuboid $=900{\mathrm{cm}}^3$ $\Rightarrow$ (Area of the base) $\times$ Height $=900{\mathrm{cm}}^3$ $\Rightarrow 180\times$ Height $=900$ $\Rightarrow$ Height $=\frac{900}{180}=5$ Hence, the height of the cuboid is 5 cm .
4. A cuboid is of dimensions $60\mathrm{cm}\times 54\mathrm{cm}\times$ 30 cm . How many small cubes with side 6 cm can be placed in the given cuboid? Sol. Volume of cuboid $=(60\times 54\times 30){\mathrm{cm}}^3$ $=97200{\mathrm{cm}}^3$ Volume of cube $=(6\times 6\times 6){\mathrm{cm}}^3$ $=216{\mathrm{cm}}^3$ No. of cubes $=\frac{\text{ Volume of cuboid }}{\text{ Volume of cube }}$ $=\frac{97200}{216}=450$
5. Find the height of the cylinder whose volume is $1.54{\mathrm{m}}^3$ and diameter of the base is 140 cm ? Sol. Let $h$ be the height of cylinder whose radius, $\mathrm{r}=\frac{140}{2}\mathrm{cm}=70\mathrm{cm}=\frac{70}{100}\mathrm{m}=0.7\mathrm{m}$ and volume $=1.54{\mathrm{m}}^3$. $\because$ volume $=\pi r^{2}h$ $\Rightarrow \pi r^2\mathrm{h}=1.54$ $\Rightarrow \frac{22}{7}\times 0.7\times 0.7\times \mathrm{h}=1.54$ $\Rightarrow \mathrm{h}=\frac{1.54\times 7}{22\times 0.7\times 0.7}=1\mathrm{m}$ Hence, the height of cylinder is 1 metre.
6. A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m . Find the quantity of milk in litres that can be stored in the tank.
   
   Sol. Quantity of milk that can be stored in the tank = Volume of the tank $=\pi r^{2}h$, where $r=1.5$ and $h=7\mathrm{m}$ $=\left(\frac{22}{7}\times 1.5\times 1.5\times 7\right){\mathrm{m}}^3=49.5{\mathrm{m}}^3$ $=(49.5\times 1000)$ litres $\left[\because 1{\mathrm{m}}^3=1000\mathrm{ltr}.\right]$ $=49500$ litres.
7. If each edge of a cube is doubled, (i) How many times will its surface area increase? (ii) How many times will its volume increase? Sol. Let x units be the edge of the cube. Then, its surface area $=6x^2$ and its volume $=x^3$. When its edge is doubled, (i) Its surface area $=6(2\mathrm{x}{)}^2$ $=6\times 4x^2=24x^2$ $\Rightarrow$ The surface area of the new cube will be 4 times that of the original cube. (ii) Its volume $=(2\mathrm{x}{)}^3=8{\mathrm{x}}^3$ $\Rightarrow$ The volume of the new cube will be 8 times that of the original cube.
8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is $108{\mathrm{m}}^3$, find the number of hours it will take to fill the reservoir.
   
   Sol. Volume of the reservoir $=108{\mathrm{m}}^3$ $=108\times 1000$ litres $=108000$ litres Since water is pouring into reservoir at the rate of 60 litres per minute. $\therefore$ Time taken to fill the reservoir $=\frac{108000}{60\times 60}$ hours $=30$ hours

5.0Key Features and Benefits of Class 8 Maths Chapter 9 Exercise 9.3

• Useful in Daily Life: These concepts are useful for determining how much space is available in boxes, tanks, rooms, etc. 
• Volume of Cubes and Cuboids: Exercise 9.3 teaches students how to calculate the space inside 3D shapes like cubes and cuboids using basic volume formulas. 
• Clear and Easy Steps: The questions are simple to follow, with direct formulas and examples that help students solve problems step by step.
• Develops Spatial Understanding: Students who practice this exercise better understand space and measurements and are able to visualize solid shapes. 
• NCERT-Based Practice: The exercise strictly follows the NCERT textbook, making it ideal for exams, homework, and concept revision.