NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 14 Math of Space Surface Area and Volume Exercise 14.4

Exercise Set 14.4

Question 1. A ball bearing has a radius of 0.7 cm. Find its surface area.
Answer:

Surface area = 4πr² = 4 × \( \frac{22}{7} \) × 0.7 × 0.7 = 4 × 22 × 0.1 × 0.7 = 6.16 cm²
The surface area of the ball bearing is 6.16 cm².

Question 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.
Answer:

The spheres are made of the same metal, so their weights are in the same ratio as their volumes.
Radius of the smaller sphere r = 5 ÷ 2 = 2.5 cm; let the radius of the larger one be R.
\( \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = \frac{5920}{740} \;\Rightarrow\; \frac{R^3}{r^3} = 8 \)
\( \frac{R}{r} = \sqrt[3]{8} = 2 \), so R = 2 × 2.5 = 5 cm.
The radius of the larger sphere is 5 cm.

Teacher's Note:
1. Same metal means weight is proportional to volume, so the weights can be used in place of volumes.
2. A weight ratio of 8 gives a radius ratio of only ∛8 = 2, not 8.

Question 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.
Answer:

Let the earth's radius be R. The moon's diameter is a quarter of the earth's, so the moon's radius is \( \frac{R}{4} \).
\( \frac{\text{surface area of moon}}{\text{surface area of earth}} = \frac{4\pi \left(\frac{R}{4}\right)^2}{4\pi R^2} = \frac{R^2/16}{R^2} = \frac{1}{16} \)
Surface area of the moon : surface area of the earth = 1 : 16.

Question 4. Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
Answer:

(i) CSA = 2πr² = 2 × \( \frac{22}{7} \) × 21 × 21 = 2 × 22 × 3 × 21 = 2772 cm²
(ii) TSA = curved part + flat circle = 2πr² + πr² = 3πr² = 3 × \( \frac{22}{7} \) × 21 × 21 = 3 × 22 × 3 × 21 = 4158 cm²

Question 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.
Answer:

Each sphere has diameter 4 cm, so exactly \( \frac{16}{4} \) × \( \frac{8}{4} \) × \( \frac{8}{4} \) = 4 × 2 × 2 = 16 spheres fit in the box.
Volume of the box = 16 × 8 × 8 = 1024 cm³
Volume of 16 spheres = 16 × \( \frac{4}{3} \) π × 2³ = \( \frac{512}{3} \) π ≈ \( \frac{512}{3} \) × 3.1416 ≈ 536.17 cm³
Volume of liquid = volume of box − volume of spheres ≈ 1024 − 536.17 ≈ 487.83 cm³
The volume of the liquid is about 488 cm³ (the same rounded answer comes with π ≈ 3.14 or π = \( \frac{22}{7} \)).

Teacher's Note:
1. The liquid fills whatever the spheres leave empty, so subtract the spheres' volume from the box's volume.
2. Round only at the last step to avoid rounding errors.

Question 6. The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.
Answer:
Let the original radius be r, so the original volume is V = \( \frac{4}{3} \) πr³.
New radius = r + 10% of r = 1.1r.
New volume = \( \frac{4}{3} \) π(1.1r)³ = 1.331 × \( \frac{4}{3} \) πr³ = 1.331V
Percentage increase = \( \frac{1.331V - V}{V} \) × 100% = 0.331 × 100% = 33.1%
Hence the volume increases by about 33.1%. Because volume depends on the cube of the radius, a 10% larger radius gives far more than 10% extra volume.

Question 7. The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.
Answer:
Let the original radius be r; the new radius is \( \left(1 + \frac{x}{100}\right)r \).
The new volume is 100% + 72.8% = 172.8% of the old one, i.e. 1.728 times it:
\( \frac{4}{3}\pi \left(1 + \frac{x}{100}\right)^3 r^3 = 1.728 \times \frac{4}{3}\pi r^3 \)
\( \left(1 + \frac{x}{100}\right)^3 = 1.728 \;\Rightarrow\; 1 + \frac{x}{100} = \sqrt[3]{1.728} = 1.2 \) (since 1.2³ = 1.728)
\( \frac{x}{100} \) = 0.2, so x = 20
x = 20; the radius was increased by 20%.

Question 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.

Answer:
Let the radius of the base of the dome be r.
2πr = 35.2, i.e. 2 × \( \frac{22}{7} \) × r = 35.2, so r = \( \frac{35.2 \times 7}{44} \) = 5.6 m.
Only the curved surface of the dome is painted:
CSA = 2πr² = 2 × \( \frac{22}{7} \) × 5.6 × 5.6 = 2 × 22 × 0.8 × 5.6 = 197.12 m²
Change to cm²: 1 m² = 100 × 100 = 10000 cm², so 197.12 m² = 19,71,200 cm².
Number of 100 cm² units = 1971200 ÷ 100 = 19712
Cost = 19712 × ₹ 10 = ₹ 1,97,120
The cost of painting the dome is ₹ 1,97,120.

Teacher's Note:
1. The rate is per 100 cm², so the area in m² must first be converted to cm².
2. Only the curved surface is painted; the flat base is the floor of the dome.