Chapter 10 of 12

Updated 6 August 2026

NCERT Solutions

Surface Areas and Volumes — NCERT Solutions

Madhya Pradesh Board · Class 9 · Mathematics

NCERT Solutions for Surface Areas and Volumes, Madhya Pradesh Board Class 9 Mathematics: 36 textbook questions solved step by step.

30 questions20 flashcards4 formulas & key relations5 concepts

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Illustrates how a right-angled triangle, when rotated about one of its perpendicular sides, forms a right circular cone. Shows the progression from a flat triangle to a 3D cone. — Learn better with visuals Super Tutor pairs illustrations like this with notes and quizzes for Surface Areas and Volumes.

36 Questions Solved · 4 Sections

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Exercise 11.1

1Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.Show solution

Given: Diameter = 10.5 cm, so radius $r = \dfrac{10.5}{2} = 5.25$ cm; slant height $l = 10$ cm.

Formula: Curved Surface Area of cone $= \pi r l$

Calculation:
$\text{CSA} = \frac{22}{7} \times 5.25 \times 10$
$= \frac{22}{7} \times 52.5$
$= 22 \times 7.5$
$= 165 \text{ cm}^2$

Answer: The curved surface area of the cone is $\mathbf{165 \text{ cm}^2}$ .

2Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.Show solution

Given: Slant height $l = 21$ m; diameter = 24 m, so radius $r = 12$ m.

Formula: Total Surface Area $= \pi r l + \pi r^2 = \pi r(l + r)$

Calculation:
$\text{TSA} = \frac{22}{7} \times 12 \times (21 + 12)$
$= \frac{22}{7} \times 12 \times 33$
$= \frac{22 \times 396}{7}$
$= \frac{8712}{7}$
$= 1244.57 \text{ m}^2 \text{ (approx.)}$

Answer: The total surface area of the cone is approximately $\mathbf{1244.57 \text{ m}^2}$ .

3Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.Show solution

Given: Curved Surface Area $= 308$ cm²; slant height $l = 14$ cm.

(i) Finding the radius:

$\pi r l = 308$
$\frac{22}{7} \times r \times 14 = 308$
$44r = 308$
$r = \frac{308}{44} = 7 \text{ cm}$

Radius of the base = 7 cm.

(ii) Total Surface Area:

$\text{TSA} = \pi r l + \pi r^2 = \pi r(l + r)$
$= \frac{22}{7} \times 7 \times (14 + 7)$
$= 22 \times 21$
$= 462 \text{ cm}^2$

Answer: (i) Radius $= 7$ cm; (ii) Total surface area $= \mathbf{462 \text{ cm}^2}$ .

4A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent. (ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹ 70.Show solution

Given: Height $h = 10$ m; radius $r = 24$ m; cost per m² $= ₹70$ .

(i) Slant height:
$l = \sqrt{r^2 + h^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \text{ m}$

Slant height $= 26$ m.

(ii) Cost of canvas:

Canvas required = Curved Surface Area of the cone
$\text{CSA} = \pi r l = \frac{22}{7} \times 24 \times 26 = \frac{22 \times 624}{7} = \frac{13728}{7} = 1961.14 \text{ m}^2 \text{ (approx.)}$

$\text{Cost} = 1961.14 \times 70 = ₹\,137279.80 \approx ₹\,137280$

Answer: (i) Slant height $= 26$ m; (ii) Cost of canvas $\approx ₹\,1,37,280$ .

5What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).Show solution

Given: Height $h = 8$ m; radius $r = 6$ m; width of tarpaulin $= 3$ m; extra length $= 20$ cm $= 0.20$ m; $\pi = 3.14$ .

Step 1: Find slant height.
$l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m}$

Step 2: Find curved surface area of the cone.
$\text{CSA} = \pi r l = 3.14 \times 6 \times 10 = 188.4 \text{ m}^2$

Step 3: Find length of tarpaulin.
$\text{Area of tarpaulin} = \text{length} \times \text{width}$
$\text{Length} = \frac{\text{CSA}}{\text{width}} = \frac{188.4}{3} = 62.8 \text{ m}$

Step 4: Add extra length for wastage.
$\text{Total length} = 62.8 + 0.20 = 63 \text{ m}$

Answer: The required length of tarpaulin is $\mathbf{63 \text{ m}}$ .

6The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m².Show solution

Given: Slant height $l = 25$ m; diameter $= 14$ m, so radius $r = 7$ m; rate $= ₹210$ per $100$ m².

Step 1: Curved Surface Area.
$\text{CSA} = \pi r l = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \text{ m}^2$

Step 2: Cost of white-washing.
$\text{Cost} = \frac{550}{100} \times 210 = 5.5 \times 210 = ₹\,1155$

Answer: The cost of white-washing the curved surface of the conical tomb is $\mathbf{₹\,1155}$ .

7A joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.Show solution

Given: Radius $r = 7$ cm; height $h = 24$ cm; number of caps $= 10$ .

Step 1: Find slant height.
$l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm}$

Step 2: Curved surface area of one cap.
$\text{CSA} = \pi r l = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \text{ cm}^2$

Step 3: Area for 10 caps.
$\text{Total area} = 10 \times 550 = 5500 \text{ cm}^2$

Answer: The area of the sheet required to make 10 caps is $\mathbf{5500 \text{ cm}^2}$ .

8A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)Show solution

Given: Base diameter $= 40$ cm $= 0.40$ m, so radius $r = 0.20$ m; height $h = 1$ m; number of cones $= 50$ ; cost $= ₹12$ per m²; $\pi = 3.14$ ; $\sqrt{1.04} = 1.02$ .

Step 1: Find slant height.
$l = \sqrt{r^2 + h^2} = \sqrt{(0.20)^2 + 1^2} = \sqrt{0.04 + 1} = \sqrt{1.04} = 1.02 \text{ m}$

Step 2: Curved surface area of one cone.
$\text{CSA} = \pi r l = 3.14 \times 0.20 \times 1.02 = 3.14 \times 0.204 = 0.64056 \text{ m}^2$

Step 3: Total curved surface area of 50 cones.
$\text{Total CSA} = 50 \times 0.64056 = 32.028 \text{ m}^2$

Step 4: Cost of painting.
$\text{Cost} = 32.028 \times 12 = ₹\,384.336 \approx ₹\,384.34$

Answer: The cost of painting all 50 cones is approximately $\mathbf{₹\,384.34}$ .

Exercise 11.2

1Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cmShow solution

Formula: Surface area of a sphere $= 4\pi r^2$

(i) $r = 10.5$ cm:
$= 4 \times \frac{22}{7} \times 10.5 \times 10.5 = 4 \times \frac{22}{7} \times 110.25 = \frac{4 \times 22 \times 110.25}{7} = \frac{9702}{7} = 1386 \text{ cm}^2$

(ii) $r = 5.6$ cm:
$= 4 \times \frac{22}{7} \times 5.6 \times 5.6 = 4 \times \frac{22}{7} \times 31.36 = \frac{4 \times 22 \times 31.36}{7} = \frac{2759.68}{7} = 394.24 \text{ cm}^2$

(iii) $r = 14$ cm:
$= 4 \times \frac{22}{7} \times 14 \times 14 = 4 \times 22 \times 2 \times 14 = 4 \times 22 \times 28 = 2464 \text{ cm}^2$

Answers: (i) $1386 \text{ cm}^2$ , (ii) $394.24 \text{ cm}^2$ , (iii) $2464 \text{ cm}^2$ .

2Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 mShow solution

Formula: Surface area $= 4\pi r^2$ , where $r = \dfrac{d}{2}$ .

(i) Diameter $= 14$ cm, $r = 7$ cm:
$= 4 \times \frac{22}{7} \times 7 \times 7 = 4 \times 22 \times 7 = 616 \text{ cm}^2$

(ii) Diameter $= 21$ cm, $r = 10.5$ cm:
$= 4 \times \frac{22}{7} \times 10.5 \times 10.5 = \frac{4 \times 22 \times 110.25}{7} = \frac{9702}{7} = 1386 \text{ cm}^2$

(iii) Diameter $= 3.5$ m, $r = 1.75$ m:
$= 4 \times \frac{22}{7} \times 1.75 \times 1.75 = \frac{4 \times 22 \times 3.0625}{7} = \frac{269.5}{7} = 38.5 \text{ m}^2$

Answers: (i) $616 \text{ cm}^2$ , (ii) $1386 \text{ cm}^2$ , (iii) $38.5 \text{ m}^2$ .

3Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)Show solution

Given: Radius $r = 10$ cm; $\pi = 3.14$ .

Formula: Total surface area of hemisphere $= 3\pi r^2$

$= 3 \times 3.14 \times 10 \times 10$
$= 3 \times 314$
$= 942 \text{ cm}^2$

Answer: Total surface area of the hemisphere $= \mathbf{942 \text{ cm}^2}$ .

4The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.Show solution

Given: Initial radius $r_1 = 7$ cm; final radius $r_2 = 14$ cm.

Formula: Surface area of sphere $= 4\pi r^2$

$\frac{\text{Surface area}_1}{\text{Surface area}_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{r_1^2}{r_2^2} = \frac{7^2}{14^2} = \frac{49}{196} = \frac{1}{4}$

Answer: The ratio of surface areas of the balloon in the two cases is $\mathbf{1 : 4}$ .

5A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 cm².Show solution

Given: Inner diameter $= 10.5$ cm, so inner radius $r = 5.25$ cm; rate $= ₹16$ per $100$ cm².

Step 1: Curved surface area of hemisphere (inner side).
$\text{CSA} = 2\pi r^2 = 2 \times \frac{22}{7} \times 5.25 \times 5.25$
$= 2 \times \frac{22}{7} \times 27.5625$
$= \frac{2 \times 22 \times 27.5625}{7}$
$= \frac{1212.75}{7} = 173.25 \text{ cm}^2$

Step 2: Cost of tin-plating.
$\text{Cost} = \frac{173.25}{100} \times 16 = 1.7325 \times 16 = ₹\,27.72$

Answer: The cost of tin-plating the bowl on the inside is $\mathbf{₹\,27.72}$ .

6Find the radius of a sphere whose surface area is 154 cm².Show solution

Given: Surface area $= 154$ cm².

Formula: $4\pi r^2 = 154$

$r^2 = \frac{154}{4\pi} = \frac{154}{4 \times \frac{22}{7}} = \frac{154 \times 7}{4 \times 22} = \frac{1078}{88} = \frac{49}{4}$

$r = \frac{7}{2} = 3.5 \text{ cm}$

Answer: The radius of the sphere is $\mathbf{3.5 \text{ cm}}$ .

7The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.Show solution

Given: Let diameter of earth $= d$ , then diameter of moon $= \dfrac{d}{4}$ .

So radius of earth $= \dfrac{d}{2}$ and radius of moon $= \dfrac{d}{8}$ .

Formula: Surface area $= 4\pi r^2$

$\frac{\text{Surface area of moon}}{\text{Surface area of earth}} = \frac{4\pi \left(\frac{d}{8}\right)^2}{4\pi \left(\frac{d}{2}\right)^2} = \frac{\frac{d^2}{64}}{\frac{d^2}{4}} = \frac{d^2}{64} \times \frac{4}{d^2} = \frac{4}{64} = \frac{1}{16}$

Answer: The ratio of the surface area of the moon to that of the earth is $\mathbf{1 : 16}$ .

8A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.Show solution

Given: Inner radius $= 5$ cm; thickness $= 0.25$ cm.

Outer radius $r = 5 + 0.25 = 5.25$ cm.

Outer curved surface area:
$= 2\pi r^2 = 2 \times \frac{22}{7} \times 5.25 \times 5.25$
$= 2 \times \frac{22}{7} \times 27.5625$
$= \frac{1212.75}{7}$
$= 173.25 \text{ cm}^2$

Answer: The outer curved surface area of the bowl is $\mathbf{173.25 \text{ cm}^2}$ .

9A right circular cylinder just encloses a sphere of radius r (see Fig. 11.10). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).Show solution

Given: A right circular cylinder just encloses a sphere of radius $r$ .

Since the cylinder just encloses the sphere:

Radius of cylinder $= r$
Height of cylinder $= 2r$ (diameter of sphere)

(i) Surface area of the sphere:
$= 4\pi r^2$

(ii) Curved surface area of the cylinder:
$= 2\pi r h = 2\pi r \times 2r = 4\pi r^2$

(iii) Ratio of surface area of sphere to curved surface area of cylinder:
$= \frac{4\pi r^2}{4\pi r^2} = \frac{1}{1}$

Answer: (i) $4\pi r^2$ ; (ii) $4\pi r^2$ ; (iii) The ratio is $\mathbf{1 : 1}$ .

Exercise 11.3

1Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cmShow solution

Formula: Volume of cone $= \dfrac{1}{3}\pi r^2 h$

(i) $r = 6$ cm, $h = 7$ cm:
$V = \frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 7$
$= \frac{1}{3} \times 22 \times 36$
$= \frac{792}{3} = 264 \text{ cm}^3$

(ii) $r = 3.5$ cm, $h = 12$ cm:
$V = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 12$
$= \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 12$
$= \frac{1}{3} \times \frac{22 \times 147}{7}$
$= \frac{1}{3} \times 22 \times 21$
$= \frac{462}{3} = 154 \text{ cm}^3$

Answers: (i) $264 \text{ cm}^3$ , (ii) $154 \text{ cm}^3$ .

2Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm

Formula: Volume of cone $= \dfrac{1}{3}\pi r^2 h$ ; $1000 \text{ cm}^3 = 1$ litre.

(i) $r = 7$ cm, $l = 25$ cm:

First find height: $h = \sqrt{l^2 - r^2} = \sqrt{625 - 49} = \sqrt{576} = 24$ cm.

$V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 = \frac{1}{3} \times 22 \times 7 \times 24 = \frac{3696}{3} = 1232 \text{ cm}^3$

$= \frac{1232}{1000} = 1.232 \text{ litres}$

(ii) $h = 12$ cm, $l = 13$ cm:

First find radius: $r = \sqrt{l^2 - h^2} = \sqrt{169 - 144} = \sqrt{25} = 5$ cm.

$V = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 = \frac{1}{3} \times \frac{22 \times 300}{7} = \frac{6600}{21} = \frac{2200}{7} \approx 314.28 \text{ cm}^3$

$= \frac{2200}{7000} = \frac{11}{35} \approx 0.314 \text{ litres}$

Answers: (i) $1.232$ litres, (ii) $\dfrac{11}{35}$ litres $\approx 0.314$ litres.

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3The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Given: Height $h = 15$ cm; Volume $= 1570$ cm³; $\pi = 3.14$ .

Formula: $V = \dfrac{1}{3}\pi r^2 h$

$1570 = \frac{1}{3} \times 3.14 \times r^2 \times 15$
$1570 = 3.14 \times 5 \times r^2$
$1570 = 15.7 \times r^2$
$r^2 = \frac{1570}{15.7} = 100$
$r = 10 \text{ cm}$

Answer: The radius of the base of the cone is $\mathbf{10 \text{ cm}}$ .

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4If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

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5A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Given: Top diameter $= 3.5$ m, so radius $r = 1.75$ m; depth (height) $h = 12$ m.

Formula: Volume of cone $= \dfrac{1}{3}\pi r^2 h$

$V = \frac{1}{3} \times \frac{22}{7} \times 1.75 \times 1.75 \times 12$
$= \frac{1}{3} \times \frac{22}{7} \times 3.0625 \times 12$
$= \frac{1}{3} \times \frac{22 \times 36.75}{7}$
$= \frac{1}{3} \times \frac{808.5}{7}$
$= \frac{808.5}{21} = 38.5 \text{ m}^3$

Since $1 \text{ m}^3 = 1$ kilolitre:
$\text{Capacity} = 38.5 \text{ kilolitres}$

Answer: The capacity of the conical pit is $\mathbf{38.5 \text{ kilolitres}}$ .

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6The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone

Given: Volume $= 9856$ cm³; diameter $= 28$ cm, so radius $r = 14$ cm.

(i) Height of the cone:
$V = \frac{1}{3}\pi r^2 h$
$9856 = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times h$
$9856 = \frac{1}{3} \times \frac{22 \times 196}{7} \times h$
$9856 = \frac{1}{3} \times 616 \times h$
$9856 = \frac{616h}{3}$
$h = \frac{9856 \times 3}{616} = \frac{29568}{616} = 48 \text{ cm}$

(ii) Slant height:
$l = \sqrt{r^2 + h^2} = \sqrt{14^2 + 48^2} = \sqrt{196 + 2304} = \sqrt{2500} = 50 \text{ cm}$

(iii) Curved surface area:
$\text{CSA} = \pi r l = \frac{22}{7} \times 14 \times 50 = 22 \times 2 \times 50 = 2200 \text{ cm}^2$

Answers: (i) Height $= 48$ cm; (ii) Slant height $= 50$ cm; (iii) CSA $= 2200 \text{ cm}^2$ .

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7A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Given: Right triangle with sides 5 cm, 12 cm, 13 cm (hypotenuse). Revolved about the side of 12 cm.

When revolved about the side 12 cm, the triangle generates a cone where:

Height $h = 12$ cm (axis of revolution)
Radius $r = 5$ cm (the other leg)
Slant height $l = 13$ cm (hypotenuse)

Volume of cone:
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12$
$= \frac{1}{3} \times \frac{22}{7} \times 300$
$= \frac{6600}{21} = \frac{2200}{7} \approx 314.28 \text{ cm}^3$

Answer: The volume of the solid obtained is $\dfrac{2200}{7} \approx \mathbf{314.28 \text{ cm}^3}$ .

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8If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Given: Same right triangle revolved about the side 5 cm.

When revolved about the side 5 cm:

Height $h = 5$ cm
Radius $r = 12$ cm

Volume of cone:
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 12 \times 12 \times 5$
$= \frac{1}{3} \times \frac{22}{7} \times 720$
$= \frac{15840}{21} = \frac{5280}{7} \approx 754.28 \text{ cm}^3$

Ratio of volumes (Q7 : Q8):
$= \frac{\frac{2200}{7}}{\frac{5280}{7}} = \frac{2200}{5280} = \frac{5}{12}$

Answer: Volume $= \dfrac{5280}{7} \approx 754.28 \text{ cm}^3$ ; Ratio of volumes (Q7 to Q8) $= \mathbf{5 : 12}$ .

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9A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Given: Diameter $= 10.5$ m, so radius $r = 5.25$ m; height $h = 3$ m.

Step 1: Volume of the heap.
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 3$
$= \frac{22}{7} \times 5.25 \times 5.25$
$= \frac{22}{7} \times 27.5625$
$= \frac{606.375}{7} = 86.625 \text{ m}^3$

Step 2: Slant height for canvas area.
$l = \sqrt{r^2 + h^2} = \sqrt{(5.25)^2 + 3^2} = \sqrt{27.5625 + 9} = \sqrt{36.5625} = 6.046 \text{ m (approx.)}$

Step 3: Curved surface area (canvas required).
$\text{CSA} = \pi r l = \frac{22}{7} \times 5.25 \times 6.046$
$= \frac{22 \times 5.25 \times 6.046}{7}$
$= \frac{22 \times 31.7415}{7}$
$= \frac{698.313}{7} \approx 99.76 \text{ m}^2$

Answer: Volume of the heap $\approx 86.625 \text{ m}^3$ ; Area of canvas required $\approx \mathbf{99.76 \text{ m}^2}$ .

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Exercise 11.4

1Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m

Formula: Volume of sphere $= \dfrac{4}{3}\pi r^3$

(i) $r = 7$ cm:
$V = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7$
$= \frac{4}{3} \times 22 \times 49$
$= \frac{4312}{3} = 1437\frac{1}{3} \text{ cm}^3$

(ii) $r = 0.63$ m:
$V = \frac{4}{3} \times \frac{22}{7} \times 0.63 \times 0.63 \times 0.63$
$= \frac{4}{3} \times \frac{22}{7} \times 0.250047$
$= \frac{4 \times 22 \times 0.250047}{21}$
$= \frac{22.00414}{21} \approx 1.0478 \text{ m}^3$

More precisely: $\dfrac{4}{3} \times \dfrac{22}{7} \times (0.63)^3 = \dfrac{4 \times 22 \times 0.250047}{21} \approx 0.9477 \text{ m}^3$

Let us recalculate: $(0.63)^3 = 0.250047$
$V = \frac{4}{3} \times \frac{22}{7} \times 0.250047 = \frac{88}{21} \times 0.250047 = 4.19 \times 0.250047 \approx 1.0478$

Using exact fractions: $r = \dfrac{63}{100}$ m
$V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{63}{100}\right)^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{250047}{1000000}$
$= \frac{88}{21} \times \frac{250047}{1000000} = \frac{22004136}{21000000} \approx 1.0478 \text{ m}^3$

Note: Standard answer is $\approx 1.05 \text{ m}^3$ .

Answers: (i) $\dfrac{4312}{3} \approx 1437.33 \text{ cm}^3$ ; (ii) $\approx 1.05 \text{ m}^3$ .

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2Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m

Formula: Volume of sphere $= \dfrac{4}{3}\pi r^3$ (water displaced = volume of sphere)

(i) Diameter $= 28$ cm, $r = 14$ cm:
$V = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14$
$= \frac{4}{3} \times \frac{22}{7} \times 2744$
$= \frac{4 \times 22 \times 2744}{21}$
$= \frac{241472}{21} = \frac{34496}{3} \approx 11498.67 \text{ cm}^3$

(ii) Diameter $= 0.21$ m, $r = 0.105$ m:
$V = \frac{4}{3} \times \frac{22}{7} \times 0.105 \times 0.105 \times 0.105$
$= \frac{4}{3} \times \frac{22}{7} \times 0.001157625$
$= \frac{88}{21} \times 0.001157625$
$\approx 0.004851 \text{ m}^3$

Answers: (i) $\approx 11498.67 \text{ cm}^3$ ; (ii) $\approx 0.004851 \text{ m}^3$ .

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3The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

Given: Diameter $= 4.2$ cm, so radius $r = 2.1$ cm; density $= 8.9$ g/cm³.

Step 1: Volume of the ball.
$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1$
$= \frac{4}{3} \times \frac{22}{7} \times 9.261$
$= \frac{4 \times 22 \times 9.261}{21}$
$= \frac{814.968}{21} = 38.808 \text{ cm}^3$

Step 2: Mass of the ball.
$\text{Mass} = \text{Volume} \times \text{Density} = 38.808 \times 8.9 = 345.39 \text{ g (approx.)}$

Answer: The mass of the metallic ball is approximately $\mathbf{345.39 \text{ g}}$ .

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4The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Given: Diameter of moon $= \dfrac{1}{4} \times$ diameter of earth.

Let radius of earth $= R$ , then radius of moon $= \dfrac{R}{4}$ .

Formula: Volume of sphere $= \dfrac{4}{3}\pi r^3$

$\frac{\text{Volume of moon}}{\text{Volume of earth}} = \frac{\frac{4}{3}\pi \left(\frac{R}{4}\right)^3}{\frac{4}{3}\pi R^3} = \frac{\frac{R^3}{64}}{R^3} = \frac{1}{64}$

Answer: The volume of the moon is $\dfrac{1}{64}$ of the volume of the earth.

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5How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Given: Diameter $= 10.5$ cm, so radius $r = 5.25$ cm.

Formula: Volume of hemisphere $= \dfrac{2}{3}\pi r^3$

$V = \frac{2}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 5.25$
$= \frac{2}{3} \times \frac{22}{7} \times 144.703125$
$= \frac{2 \times 22 \times 144.703125}{21}$
$= \frac{6366.9375}{21} = 303.1875 \text{ cm}^3$

Converting to litres ( $1000 \text{ cm}^3 = 1$ litre):
$= \frac{303.1875}{1000} \approx 0.303 \text{ litres}$

Answer: The hemispherical bowl can hold approximately $\mathbf{0.303}$ litres of milk.

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6A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Given: Inner radius $r = 1$ m; thickness $= 1$ cm $= 0.01$ m.

Outer radius $R = 1 + 0.01 = 1.01$ m.

Volume of iron used = Volume of outer hemisphere $-$ Volume of inner hemisphere:
$V = \frac{2}{3}\pi R^3 - \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(R^3 - r^3)$
$= \frac{2}{3} \times \frac{22}{7} \times \left[(1.01)^3 - (1)^3\right]$
$= \frac{2}{3} \times \frac{22}{7} \times [1.030301 - 1]$
$= \frac{2}{3} \times \frac{22}{7} \times 0.030301$
$= \frac{44}{21} \times 0.030301$
$= \frac{1.333244}{21} \approx 0.06348 \text{ m}^3$

Answer: The volume of iron used to make the tank is approximately $\mathbf{0.06348 \text{ m}^3}$ .

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7Find the volume of a sphere whose surface area is 154 cm².

Given: Surface area $= 154$ cm².

Step 1: Find radius.
$4\pi r^2 = 154$
$r^2 = \frac{154}{4 \times \frac{22}{7}} = \frac{154 \times 7}{88} = \frac{1078}{88} = \frac{49}{4}$
$r = \frac{7}{2} = 3.5 \text{ cm}$

Step 2: Find volume.
$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5$
$= \frac{4}{3} \times \frac{22}{7} \times 42.875$
$= \frac{4 \times 22 \times 42.875}{21}$
$= \frac{3773}{21} = \frac{539}{3} = 179\frac{2}{3} \text{ cm}^3$

Answer: The volume of the sphere is $\mathbf{179\dfrac{2}{3} \text{ cm}^3}$ .

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8A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 4989.60. If the cost of white-washing is ₹ 20 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome.

Given: Total cost $= ₹4989.60$ ; rate $= ₹20$ per m².

(i) Inside surface area:
$\text{Surface area} = \frac{\text{Total cost}}{\text{Rate}} = \frac{4989.60}{20} = 249.48 \text{ m}^2$

(ii) Volume of air inside the dome:

Curved surface area of hemisphere $= 2\pi r^2 = 249.48$
$r^2 = \frac{249.48}{2 \times \frac{22}{7}} = \frac{249.48 \times 7}{44} = \frac{1746.36}{44} = 39.69$
$r = \sqrt{39.69} = 6.3 \text{ m}$

$V = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 6.3 \times 6.3 \times 6.3$
$= \frac{2}{3} \times \frac{22}{7} \times 250.047$
$= \frac{2 \times 22 \times 250.047}{21}$
$= \frac{11002.068}{21} = 523.908 \text{ m}^3 \approx 523.9 \text{ m}^3$

Answers: (i) Inside surface area $= 249.48 \text{ m}^2$ ; (ii) Volume of air $\approx \mathbf{523.9 \text{ m}^3}$ .

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9Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the (i) radius r' of the new sphere, (ii) ratio of S and S'.

Given: 27 spheres each of radius $r$ and surface area $S$ are melted to form a new sphere of radius $r'$ and surface area $S'$ .

(i) Radius $r'$ of the new sphere:

Volume of 27 small spheres = Volume of new sphere
$27 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi r'^3$
$27r^3 = r'^3$
$r' = \sqrt[3]{27r^3} = 3r$

Radius of new sphere $r' = 3r$ .

(ii) Ratio of $S$ and $S'$ :
$S = 4\pi r^2$
$S' = 4\pi r'^2 = 4\pi (3r)^2 = 4\pi \times 9r^2 = 36\pi r^2$

$\frac{S}{S'} = \frac{4\pi r^2}{36\pi r^2} = \frac{1}{9}$

Answer: (i) $r' = 3r$ ; (ii) $S : S' = \mathbf{1 : 9}$ .

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10A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Given: Diameter $= 3.5$ mm, so radius $r = 1.75$ mm.

Formula: Volume of sphere $= \dfrac{4}{3}\pi r^3$

$V = \frac{4}{3} \times \frac{22}{7} \times 1.75 \times 1.75 \times 1.75$
$= \frac{4}{3} \times \frac{22}{7} \times 5.359375$
$= \frac{4 \times 22 \times 5.359375}{21}$
$= \frac{471.625}{21}$
$= 22.458 \text{ mm}^3 \approx 22.46 \text{ mm}^3$

Answer: The amount of medicine needed to fill the capsule is approximately $\mathbf{22.46 \text{ mm}^3}$ .

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