Surface Areas and Volumes

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 $\mathbf{165 \text{ cm}^2}$.

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 $\mathbf{1244.57 \text{ m}^2}$.

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 $= \mathbf{462 \text{ cm}^2}$.

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 $\approx ₹\,1,37,280$.

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 $\mathbf{63 \text{ m}}$.

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 $\mathbf{₹\,1155}$.

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 $\mathbf{5500 \text{ cm}^2}$.

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 $\mathbf{₹\,384.34}$.

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) $2464 \text{ cm}^2$.

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) $38.5 \text{ m}^2$.

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 $= \mathbf{942 \text{ cm}^2}$.

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 $\mathbf{1 : 4}$.

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 $\mathbf{₹\,27.72}$.

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 $\mathbf{3.5 \text{ cm}}$.

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 $\mathbf{1 : 16}$.

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 $\mathbf{173.25 \text{ cm}^2}$.

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 $\mathbf{1 : 1}$.

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) $154 \text{ cm}^3$.

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.

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 $\mathbf{10 \text{ cm}}$.

Given: Height $h = 9$ cm; Volume $= 48\pi$ cm³.

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

$$48\pi = \frac{1}{3} \times \pi \times r^2 \times 9$$
$$48\pi = 3\pi r^2$$
$$r^2 = \frac{48}{3} = 16$$
$$r = 4 \text{ cm}$$

$$\text{Diameter} = 2r = 8 \text{ cm}$$

Answer: The diameter of the base is $\mathbf{8 \text{ cm}}$.

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 $\mathbf{38.5 \text{ kilolitres}}$.

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 $= 2200 \text{ cm}^2$.

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 $\dfrac{2200}{7} \approx \mathbf{314.28 \text{ cm}^3}$.

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) $= \mathbf{5 : 12}$.

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 $\approx \mathbf{99.76 \text{ m}^2}$.

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) $\approx 1.05 \text{ m}^3$.

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) $\approx 0.004851 \text{ m}^3$.

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 $\mathbf{345.39 \text{ g}}$.

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.

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.

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 $\mathbf{0.06348 \text{ m}^3}$.

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 $\mathbf{179\dfrac{2}{3} \text{ cm}^3}$.

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 $\approx \mathbf{523.9 \text{ m}^3}$.

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) $S : S' = \mathbf{1 : 9}$.

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 $\mathbf{22.46 \text{ mm}^3}$.