Latest Updated : November 2023

If you are in search of **NCERT solutions for Class 8 Maths Chapter** **9 Mensuration**, then you are at the right place. When it comes to **Mathematics**, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed **NCERT solutions for Class 8 Maths** of all the chapters, **exercise wise** with **updated syllabus by NCERT**. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmark **ncertforclass8.com** to boost your preparation.

## NCERT Solutions for Class 8 Maths Chapter 9 **Mensuration** Ex 9.3

**Class 8 Maths Chapter** **9 Mensuration** **Exercise 9.3**

**Exercise 9.3**

**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.**

Solution:

We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor.

When the amount of space occupied by an object such as water, milk, coffee, tea, etc., then we have to find out volume of the object.

(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 it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?**

Solution:

Yes, we can say that volume of cylinder B is greater as the radius of cylinder B is greater than that of cylinder A. Therefore greater radius will give us a greater volume.

**Volume of A-**

Diameter of cylinder A = 7 cm

Radius of cylinder A = 7/2 cm

Height of cylinder A = 14 cm

`\[Volume\ of\ cylinder\ A=\ \pi r^2h\]`

`\[=\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times14=539\ cm^3\]`

`\[Volume\ of\ cylinder\ A\ is\ 539\ cm^3\]`

**Volume of B-**

Diameter of cylinder B = 14 cm

Radius of cylinder B = 14/2 = 7 cm

Height of cylinder B = 7 cm

`\[Volume\ of\ cylinder\ B=\ \pi r^2h\]`

`\[=\frac{22}{7}\times7\times7\times7=1078\ cm^3\]`

`\[Volume\ of\ cylinder\ B\ is\ 1078\ cm^3\]`

**Surface area of cylinder A-**

`\[Surface\ area\ of\ cylinder\ A=2\pi r(r+h)\]`

`\[=2\times\frac{22}{7}\ \times\frac{7}{2}\times\left(\frac{7}{2}+14\right)=385\ cm^2\]`

**Surface area of cylinder B-**

`\[Surface\ area\ of\ cylinder\ B=2\pi r(r+h)\]`

`\[=2\times\frac{22}{7}\times7(\ 7+7)=616\ cm^2\]`

Yes, the cylinder with greater volume also has a greater surface area.

**3. Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?**

Solution:

Base area of cuboid = 180 cm2 [Given]

Volume of cuboid = 900 cm3 [Given]

We know that, Volume of cuboid = lbh

900 = 180×h

`\[h\ =\ \frac{900}{180}\ =\ 5\ cm\]`

Hence the height of the cuboid is 5 cm.

**4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?**

Solution:

Length of cuboid (l) = 60 cm [Given]

Breadth of cuboid (b) = 54 cm [Given]

Height of cuboid (h) = 30 cm [Given]

`\[We\ know\ that,\ Volume\ of\ cuboid=lbh\]`

`\[=60\times54\times30=97200\ cm^3\]`

`\[Volume\ of\ cube=(Side)^3\]`

`\[=6\times6\times6=216\ cm^3\]`

`\[Also\ ,the\ Number\ of\ small\ cubes=\frac{volume\ of\ cuboid}{volume\ of\ cube}\]`

`\[=\frac{97200}{216}=\ 450\ cubes\]`

Hence , the required number of cubes is 450.

**5. Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm ?**

Solution:

`\[Volume\ of\ cylinder=1.54\ m^3\ \ \left[Given\right]\]`

Diameter (d) of cylinder = 140 cm [Given]

Radius ( r )= d/2 = 140/2 = 70 cm [Given]

`\[Volume\ of\ cylinder=\pi r^2h\]`

`\[1.54=\frac{22}{7}\times0.7\times0.7\times h\]`

`\[h=\frac{\left(1.54\times7\right)}{(22\times0.7\times0.7)}\ =\ 1\ m\]`

Hence, the height of the cylinder is 1 m.

**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?**

Solution:

Radius of cylindrical tank (r) = 1.5 m

Height of cylindrical tank (h) = 7 m

`\[Volume\ of\ cylindrical\ \tan k\ \ \ (V)\ \ =\pi r^2h\]`

`\[=\frac{22}{7}\times1.5\times1.5\times7\ =49.5\ cm^3\]`

`\[=49.5\times1000\ liters=49500\ \ liters\ \ \ \ [∵\ 1\ m^3=1000\ litres]\]`

Hence, the required quantity of milk is 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?**

Solution:

(i) Let the edge of the cube be “ l” .

`\[Surface\ area\ of\ the\ cube\ \left(A\right)\ =6l^2\]`

`\[When\ the\ edge\ of\ the\ cube\ 'l'\ is\ doubled,\ then\]`

`\[Surface\ area\ of\ the\ cube,\ say\ A’=6(2l)^2=6\times4l^2=4\times(6l^2)\]`

`\[A’=4\ \times A\]`

`\[Hence,\ the\ surface\ area\ will\ increase\ by\ four\ times.\]`

`\[(ii)\ Volume\ of\ a\ cube\ \left(V\right)=l^3\]`

`\[When\ the\ edge\ of\ the\ cube\ 'l'\ is\ doubled,\ then\]`

`\[Volume\ of\ the\ cube\ ,\ say\ V\ ’=(2l)^3=8(l^3)\]`

`\[V\ ’=8\times V\]`

`\[Hence,\ the\ volume\ will\ increase\ 8\ times.\]`

**8. Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m3, find the number of hours it will take to fill the reservoir.**

Solution:

`\[The\ \ volume\ of\ the\ reservoir=108\ m^3\]`

`\[Rate\ of\ pouring\ water\ into\ cuboidal\ reservoir=60\ litres/\min\]`

`\[=\frac{60}{1000}m^3\ per\min\ \ \ \ \ \left[∵\ 1\ liter=\frac{1}{1000}\ m^3\right]\]`

`\[=\frac{\left(60\times60\right)}{1000}\ m^3\ per\ hour\]`

`\[Therefore,\frac{\left(60\times60\right)}{1000}\ m^3\ water\ filled\ in\ reservoir\ will\ take=1\ hour\]`

`\[Therefore\ 1\ m^3\ water\ filled\ in\ reservoir\ will\ take=\frac{1000}{60\ \times60}\ hours\]`

`\[Therefore,\ 108\ m^3\ water\ filled\ in\ reservoir\ will\ take=\frac{\left(108\times1000\right)}{(60\times60)}hours=30\ hours\]`

Hence, It will take 30 hours to fill the reservoir.

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