NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.4#
Sphere: $\text{SA} = 4\pi r^2$, $V = \dfrac{4}{3}\pi r^3$
Hemisphere: $\text{CSA} = 2\pi r^2$, $\text{TSA} = 3\pi r^2$, $V = \dfrac{2}{3}\pi r^3$
1. A ball bearing has a radius of 0.7 cm. Find its surface area. #
Solution
$$\text{SA} = 4 \times \dfrac{22}{7} \times 0.7^2$$
$$\text{SA} = 4 \times \dfrac{22}{7} \times 0.49$$
$$\text{SA} = 6.16 \text{ cm}^2$$
*2. 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. #
Solution
Same metal, so weight is proportional to volume, which is proportional to $r^3$:
$$\dfrac{R^3}{r^3} = \dfrac{5920}{740} = 8$$$$\Rightarrow \dfrac{R}{r} = 2$$
The smaller radius is $\dfrac{5}{2} = 2.5$ cm, so:
$$R = 2 \times 2.5 = 5 \text{ cm}$$3. 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. #
Solution
$$\dfrac{\text{SA}_{\text{moon}}}{\text{SA}_{\text{earth}}} = \dfrac{4\pi r^2}{4\pi (4r)^2} = \dfrac{1}{16}$$
∴ The ratio is 1 : 16.
4. Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm. #
Solution
$$\text{CSA} = 2 \times \dfrac{22}{7} \times 21^2 = 2772 \text{ cm}^2$$
$$\text{TSA} = 3 \times \dfrac{22}{7} \times 21^2 = 4158 \text{ cm}^2$$
5. Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 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. #
Solution
$$V_{\text{box}} = 16 \times 8 \times 8 = 1024 \text{ cm}^3$$$$V_{\text{spheres}} = 16 \times \dfrac{4}{3}\pi \times 2^3$$
$$V_{\text{spheres}} = \dfrac{512\pi}{3} \approx 536.17 \text{ cm}^3$$$$V_{\text{liquid}} = 1024 - 536.17 \approx 488 \text{ cm}^3$$
∴ About 488 cm³ of liquid. (Using $\pi = \dfrac{22}{7}$ gives about 488 cm³ as well.)
6. The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%. #
Solution
The new radius is $1.1r$:
$$V_{\text{new}} = \dfrac{4}{3}\pi (1.1r)^3$$$$V_{\text{new}} = 1.331 \times \dfrac{4}{3}\pi r^3$$
The volume becomes 1.331 times the original, an increase of $0.331$, that is, 33.1%.
*7. The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x. #
Solution
$$\left(1 + \dfrac{x}{100}\right)^3 = 1.728$$
$$\Rightarrow 1 + \dfrac{x}{100} = 1.2$$
$$\Rightarrow x = 20$$
(since $1.2^3 = 1.728$). ∴ The radius increased by 20%.
8. The hemispherical dome of a building needs to be painted (see Fig. 14.16). If the circumference of the base of the dome is 35.2 m, find the cost of painting it, given the cost of painting is ₹10 per 100 cm². #
Solution
$$2\pi r = 35.2$$
$$\Rightarrow r = \dfrac{35.2 \times 7}{44}$$
$$\Rightarrow r = 5.6 \text{ m}$$
The curved surface is painted:
$$\text{CSA} = 2 \times \dfrac{22}{7} \times 5.6^2$$$$\text{CSA} = 197.12 \text{ m}^2$$
$$\text{CSA} = 1971200 \text{ cm}^2$$
The cost is ₹10 per 100 cm²:
$$\text{Cost} = \dfrac{1971200}{100} \times 10$$$$\text{Cost} = ₹197120$$
∴ Painting the dome costs ₹1,97,120.