Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.4 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4

1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume:

Question (a)

To find how much it can hold.

Solution:

To find how much a cylindrical tank can hold, we will find its volume.

Question (b)

Number of cement bags required to plaster it.

Solution:

To find number of cement bags required to plaster a cylindrical tank, we will find its surface area.

Question (c)

To find the number of smaller tanks that can be filled with water from it.

Solution:

To find the number of smaller tanks that can be filled with water from it, we will find volume of the tank.

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:

Radius of cylinder B is double them that of cylinder A.

∴ Volume of cylinder B should be more than that of cylinder A.

For cylinder A:

radius (r) = \(\frac{diameter}{2}\) = \(\frac{7}{2}\)

height (h) = 14 cm

Volume of cylinder A = πr²h

= \(\frac{22}{7}\) × (\(\frac{22}{7}\))^{2} × 14

= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 14

= 11 × 7 × 7

= 593 cm^{3}

For cylinder B:

radius (r) = \(\frac{diameter}{2}\) = \(\frac{14}{2}\) = 7cm and

height (h) = 7 cm

Volume of cylinder B = πr²h

= \(\frac{22}{7}\) × 7^{2} × 7

= \(\frac{22}{7}\) × 7 × 7 × 7

= 22 × 7 × 7

= 1078 cm^{3}

Total surface area:

For cylinder A:

radius (r) = \(\frac{7}{2}\) cm

height (h) = 14 cm

Total surface area of cylinder A

= 2πr (r + h)

= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\)(\(\frac{7}{2}\) +14)

= 22(\(\frac{35}{2}\))

= 11 × 35

= 385 cm^{2}

For cylinder B :

radius (r) = 7 cm, height (h) = 7 cm

Total surface area of cylinder B

= 2πr (r + h)

= 2 × \(\frac {22}{7}\) × 7(7 + 7)

= 44(14) = 616 cm^{2}

So the surface area of cylinder B is greater than that of cylinder A. Hence, the cylinder with greater volume also has greater surface area.

3. Find the height of a cuboid whose base area is 180 cm^{2} and volume is 900 cm^{3}?

Solution:

Let the height of cuboid be h cm.

Now, = Area of base × Height

∴ 900 = 180 × h

∴ h = \(\frac {900}{180}\) = 5cm

Hence, height of 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:

Volume of a cuboid = 60 × 54 × 30 cm^{3}

Volume of a cube = 6^{3} = 6 × 6 × 6 cm^{3}

∴ Number of small cubes

= \(\frac{\text { Volume of cuboid }}{\text { Volume of one cube}}\)

= \(\frac{60 \times 54 \times 30}{6 \times 6 \times 6}\)

= 10 × 9 × 5 = 450

Hence, 450 cubes can be placed in the given cuboid.

5. Find the height of the cylinder whose volume is 1.54 m^{3} and diameter of the base is 140 cm?

Solution:

For given cylinder:

Volume = 1.54 m^{3}

Radius(r) = \(\frac {diameter}{2}\) = \(\frac {140}{2}\) = 70cm = 0.7 m

Volume of cylinder = πr²h

∴ 1.54 = \(\frac {22}{7}\) × 0.7 × 0.7 × h

∴ h = \(\frac{1.54 \times 7}{22 \times 0.7 \times 0.7}\) = 1m

Hence, 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:

For given cylindrical milk tank:

Radius (r) = 1.5 m = \(\frac {15}{10}\) m

Height (h) = 7m

Volume of cylindrical milk tank

= πr²h

= \(\frac {22}{7}\) × (\(\frac {15}{10}\))^{2} × 7

= \(\frac {22}{7}\) × \(\frac {15}{10}\) × \(\frac {15}{10}\) × 7

= \(\frac{11 \times 3 \times 3}{2}=\frac{99}{2}\) = 49.5m^{3}

= 49.5 m^{3}

1 m^{3} = 1000 litres

∴ 49.5 m^{3} = 49.5 × 1000 = 49500 litres

Hence, 49,500 litres of milk can be stored in the tank.

7. If each edge of a cube is doubled:

Question (i)

How many times will its surface area increase?

Solution:

Let the edge of the original cube be x.

Then, its new edge (by doubling) = 2x

Original surface area of the cube

= 6 (side)^{2}

= 6 (x)^{2}

= 6x^{2}

New surface area of the cube

= 6 (side)^{2}

= 6 (2x)^{2}

= 6 (2x × 2x)

= 6 (4x^{2}) = 24X^{2}

= \(\frac{\text { New surface area of the cube }}{\text { Original surface area of the cube }}\) = \(\frac{24 x^{2}}{6 x^{2}}\) = 4

Hence, surface area of a cube will increase 4 times.

Question (ii)

How many times will its volume increase ?

Solution:

Original volume of the cube

= (side)^{3}

= (x)^{3}

= x^{3}

New volume of the cube = (side)^{3}

= (2x)^{3}

= (2x × 2x × 2x)

= 8x^{3}

Now,

\(\frac{\text { New volume of the cube }}{\text { Original volume of the cube }}\) = \(\frac{8 x^{3}}{x^{3}}\) = 8

Hence, volume of the cube will increase 8 times.

8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m^{3}, find the number of hours it will take to fill the reservoir.

Solution:

Volume of the cuboidal reservoir =108 m^{3} 1 m^{3} = 1000 litres

∴ 108 m^{3} = 108 × 1000 litres

= 1,08,000 litres

Water poured in a minute = 60 litres

∴ Water poured in an hour

(∵ 1 hour = 60 minutes) = 60 × 60

= 3600 litres

Time taken to fill reservoir = \(\frac{\text { volume of reservoir }}{\text { water poured in an hour }}\) = \(\frac {108000}{3600}\)

= 30 hours

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