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📦 Chapter 14: Math of Space: Surface Area and Volume • Exercise Set 14.4

Verified step-by-step solutions for your class notes.

Exercise Set 14.1 7Exercise Set 14.2 4Exercise Set 14.3 4Exercise Set 14.4 4End-of-Chapter Exercises 4All Exercises 23
Q1

A ball bearing has a radius of 0.7 cm. Find its surface area.

Answer: A ball bearing is spherical. Surface Area = $4\pi r^2$.
SA = $4 \times \frac{22}{7} \times (0.7)^2 = 4 \times \frac{22}{7} \times 0.49 = 4 \times 22 \times 0.07 = 6.16$ cm².

Q2

Two solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.

Answer: Since they are made of the same metal, their weights are directly proportional to their volumes. Ratio of volumes = $5920 / 740 = 8$.
Volume ratio = $(\text{radius ratio})^3 = \left(\frac{R_{large}}{R_{small}}\right)^3 = 8 \Rightarrow \frac{R_{large}}{R_{small}} = 2$.
The radius of the smaller sphere = $5 / 2 = 2.5$ cm.
The radius of the larger sphere = $2 \times 2.5 = 5$ cm.

Q3

The diameter of the moon is approximately one fourth the diameter of the earth. Given that the moon and earth are both roughly spherical, find the ratio of their surface areas.

Answer: Let the diameter of the Earth be $D_E$. Then the diameter of the moon is $D_M = \frac{1}{4} D_E$. Their radii hold the same ratio: $r_M = \frac{1}{4} r_E$.
Ratio of Surface Areas = $\frac{4\pi r_M^2}{4\pi r_E^2} = \left(\frac{r_M}{r_E}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
The ratio is $1 : 16$.

Q5

Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions $16 \text{ cm} \times 8 \text{ cm} \times 8 \text{ cm}$. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.

Answer: Volume of the box = $16 \times 8 \times 8 = 1024$ cm³.
Volume of one sphere = $\frac{4}{3}\pi r^3 = \frac{4}{3} \times 3.14 \times (2^3) = \frac{4}{3} \times 3.14 \times 8 \approx 33.51$ cm³.
Volume of 16 spheres = $16 \times 33.51 \approx 536.16$ cm³.
Volume of preservative liquid = Volume of box - Volume of 16 spheres = $1024 - 536.16 = 487.84$ cm³.
Rounded to the nearest integer, the volume is 488 cm³.

Length = 16 cm (4 spheres × 4 cm) 8 cm
16 spheres (r = 2 cm) arranged 4 × 2 × 2 inside the 16 cm × 8 cm × 8 cm box