RBSE Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 11 Mensuration Ex 11.4

Question 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.

Answer:
(a) Volume
(b) Surface Area
(c) Volume.

Question 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?

Cylinder B has greater surface area.
Surface area of Cylinder A
= (2πrh + 2πr²) sq. unit
(Where r = \(\frac{7}{2}\) cm and h = 14 cm)
= (2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × 14 + 2 × \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)) cm²
= (308 + 77) cm² = 385 cm²
Surface area of cylinder B
= (2πrh + 2πr) sq. units
(Where r = 7 cm, h = 7 cm)
= \(\left(2 \times \frac{22}{7} \times 7 \times 7+2 \times \frac{22}{7} \times 7 \times 7\right)\) cm²
= (308 + 308) cm²
= 616 cm²
Cylinder B has greater surface area.
Volume of cylinder A = πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times\) 14 cm³
= 539 cm³
Volume of cylinder B = \(\left(\frac{22}{7} \times 7 \times 7 \times 7\right)\) cm³
= 1078 cm³
Cylinder B has greater volume.
Hence cylinder haying greater volume also has greater surface area.

Question 3.
Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³?
Answer:
Volume of the cuboid = 900 cm³
or (Area of the base) × Height = 900 cm³
or 180 × Height = 900
or Height = \(\frac{900}{180}\) = 5
Hence, the height of the cuboid is 5 cm.

Question 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 Answer:
Volume of cuboid = 60 × 54 × 30 = 97,200 cm²
∴ Volume of cube = (6 cm)³ = 216 cm³
Number of small cubes
= \(\frac{\text { Volume of cuboid }}{\text { Volume of each cube }}\)
= \(\frac{97200}{216}\) = 450 cubes.

Question 5.
Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm ?
Answer:
Let the h be the height of cylinder 140
whose radius, r = \(\frac{140}{2}\)cm = 70 cm = 0.7 m and volume = 154 m³
Now Volume = 1.54 m³
or πr²h = 1.54
or \(\frac{22}{7}\) × 0.7 × 0.7 × h = 1.54
or h = \(\frac{1.54 \times 7}{22 \times 0.7 \times 0.7}\) = 1
Hence, the height of cylinder is 1 metre.

Question 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?

Answer:
Quantity of milk that can be stored in the tank = Volume of the tank
= πr²h, where r = 1.5 m and h = 7 m
= (\(\frac{22}{7}\) × 1.5 × 1.5 × 7)m³ = 49.5 m³
= (49.5 × 1000) litre
[∵ 1 m³ = 1000 liter]
= 49,500 litre

Question 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?
Answer:
Let x units be the edge of the cube. Then, its surface area = 6x² and its volume = x³
When its edge is doubled,
(i) Its surface area = 6(2x)² = 6 × 4x² = 24x²
⇒ The surface area of the new cube will be 4 times that of the original cube.
(ii) Its volume = (2x)³ = 8x³
⇒ The volume of the new cube will be 8 times that of the original cube.

Question 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.

Answer:
Volume of reservoir = 108 m³
= 108 × 1000 litres
∵ 1 m³ = 1000 litres
= 1,08,000 litres
Since water is pouring into reservoir @ 60 litres per minute
∴ Time taken to fill the reservoir
= \(\frac{1,08,000}{60 \times 60}\) hours
= 30 hours