Surface Areas and Volumes Test

Result

Surface Areas and Volumes Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Use the following information to answer the next question.

The dimensions of a box are 32 cm × 28 cm × 24 cm. It is to be painted from outside.

What is the area to be painted?

A
2336 cm²

B
4672 cm²

C
6742 cm²

D
7634 cm²

Solution

Since the box is to be painted from outside, the area to be painted is the total surface area of the box.

The length, breadth, and height of the box are given as 32 cm, 28 cm, and 24 cm respectively.

It is known that, total surface area of a cuboid = 2(lb + bh + hl) where l, b, and h are length, breadth, and height of the cuboid respectively.

Therefore, total surface area of the box = 2(32 × 28 + 28 × 24 + 24 × 32) cm²

= 2(896 + 672 + 768) cm²

= 2 × 2336 cm²

= 4672 cm²

Hence, the area to be painted is 4672 cm².

The correct answer is B.
Question 2
1 / -0

Use the following information to answer the next question.

A cylindrical container of radius 21 cm contains some water. Three iron balls of diameter 14 cm are immersed in the cylindrical container.

What is the approximate rise in the level of water in the container?

A
2.21 cm

B
3.11 cm

C
4.06 cm

D
4.27 cm

Solution

Let h be the rise in the water level.

Diameter of each iron ball = 14 cm

⇒ Radius of each iron ball

The volume of water which rises in the container is equal to the volume of the three iron balls.

∴ Volume of each iron ball

Radius of the container = 21 cm

∴ Volume of water which rises in the container = π × (21 cm)² × h

Thus, the water rises by 3.11 cm due to the immersion of iron balls in the container.

The correct answer is B.