Volume & Surface Area of Solids

(i) Total surface area of cylinder = 2πr(h + r)

$$\begin{aligned}= 2 \times \frac{22}{7} \times 10.5(10.5 + 16) \\[1em] = 2 \times \frac{22}{7} \times 10.5(26.5) \\[1em] = 2 \times \frac{22}{7} \times 278.25 \\[1em] = 1749 \text{ cm}^2\end{aligned}$$

Curved surface area of cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 10.5 \times 16 \\[1em] = \frac{7392}{7} \\[1em] = 1056 \text{ cm}^2\end{aligned}$$

Hence, the curved surface area = 1056 cm² and the total surface area of the cylinder = 1749 cm².

(ii) Total surface area of cylinder = 2πr(h + r)

$$\begin{aligned}= 2 \times \frac{22}{7} \times 21(21 + 5) \\[1em] = 2 \times \frac{22}{7} \times 21(26) \\[1em] = \frac{24024}{7} \\[1em] = 3432 \text{ cm}^2\end{aligned}$$

Curved surface area of cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 21 \times 5 \\[1em] = \frac{4620}{7} \\[1em] = 660 \text{ cm}^2\end{aligned}$$

Hence, the curved surface area = 660 cm² and the total surface area of the cylinder = 3432 cm².

(iii) Total surface area of cylinder = 2πr(h + r)

$$\begin{aligned}= 2 \times \frac{22}{7} \times 14(14 + 20) \\[1em] = 2 \times \frac{22}{7} \times 14(34) \\[1em] = \frac{20944}{7} \\[1em] = 2992 \text{ cm}^2\end{aligned}$$

Curved surface area of cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 14 \times 20 \\[1em] = \frac{12320}{7} \\[1em] = 1760 \text{ cm}^2\end{aligned}$$

Hence, the curved surface area = 1760 cm² and the total surface area of the cylinder = 2992 cm².

(iv) Given, h = 1 m = 100 cm and r = 1.4 cm

Total surface area of cylinder = 2πr(h + r)

$$\begin{aligned}= 2 \times \frac{22}{7} \times 1.4 \times (1.4 + 100) \\[1em] = 2 \times \frac{22}{7} \times 1.4 \times (101.4) \\[1em] = \frac{6246.24}{7} \\[1em] = 892.32 \text{ cm}^2\end{aligned}$$

Curved surface area of cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 1.4 \times 100 \\[1em] = \frac{6160}{7} \\[1em] = 880 \text{ cm}^2\end{aligned}$$

Hence, the curved surface area = 880 cm² and the total surface area of the cylinder = 892.32 cm².

By formula, Volume of cylinder = πr²h

(i) Given,

Height (h) = 21 cm and base radius (r) = 5 cm

Volume of cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 5^2 \times 21 \\[1em] = \frac{22}{7} \times 25 \times 21 \\[1em] = \frac{11550}{7} \\[1em] = 1650 \text{ cm}^3.\end{aligned}$$

Hence, volume of a cylinder = 1650 cm³.

(ii) Given,

Height (h) = 40 cm and Diameter (d) = 28 cm

Radius (r) = $\frac{\text{Diameter}}{2} = \frac{28}{2} = 14 \text{ cm}$

Volume of cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 14^2 \times 40 \\[1em] = \frac{22}{7} \times 196 \times 40 \\[1em] = \frac{172480}{7} \\[1em] = 24640 \text{ cm}^3.\end{aligned}$$

Hence, volume of a cylinder = 24640 cm³.

Given, radius (r) = 10.5 cm and height (h) = 60 cm

By formula,

Volume of cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times (10.5)^2 \times 60 \\[1em] = \frac{22}{7} \times 110.25 \times 60 \\[1em] = \frac{145530}{7} \\[1em] = 20790 \text{ cm}^3.\end{aligned}$$

Given, the material of the cylinder weighs 5 grams per cu.cm

⇒ 1 gram = $\frac{1}{1000}$ kg

⇒ 5 gram = $\frac{5}{1000}$ kg

∴ Total weight of the cylinder = 20790 × $\frac{5}{1000} = \frac{103950}{1000} = 103.95$ kg.

Hence, the weight of a solid cylinder is 103.95 kg.

Let radius (r) = 14 m and depth or height = h meters

By formula,

Volume of cylinder = πr²h

$$\begin{aligned}\Rightarrow 6160 = \frac{22}{7} \times 14^2 \times \text{h} \\[1em] \Rightarrow 6160 = \frac{22}{7} \times 196 \times \text{h} \\[1em] \Rightarrow \text{h} = \frac{6160 \times 7}{22 \times 196} \\[1em] \Rightarrow \text{h} = \frac{43120}{4312} \\[1em] \Rightarrow \text{h} = 10 \text{ m}.\end{aligned}$$

Curved surface area of cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 14 \times 10 \\[1em] = \frac{6160}{7} \\[1em] = 880 \text{ m}^2\end{aligned}$$

Given, the cost of painting its curved surface is ₹ 30 per m².

⇒ 880 × 30 = ₹ 26,400.

Hence, depth of the cylindrical tank is 10 m and the cost of painting its curved surface is ₹ 26,400.

Given, curved surface area of a cylinder = 4400 cm²

(i) By formula,

Curved surface area of cylinder = 2πrh

∴ 2πrh = 4400 ....(1)

Given, circumference of base = 110 cm

We know that circumference = 2πr

∴ 2πr = 110 ....(2)

Dividing eq. (1) by (2),

$$\begin{aligned}\Rightarrow \frac{2π\text{rh}}{2π\text{r}} = \frac{4400}{110} \\[1em] \Rightarrow \text{h} = 40 \text{ cm.}\end{aligned}$$

From eq.(1), we have,

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times \text{r} = 110 \\[1em] \Rightarrow \text{r} = \frac{110 \times 7}{22 \times 2} \\[1em] \Rightarrow \text{r} = \frac{770}{44} \\[1em] \Rightarrow \text{r} = 17.5 \text{ cm.}\end{aligned}$$

Volume of cylinder = πr²h

Putting values we get,

$$\begin{aligned}= \frac{22}{7} \times (17.5)^2 \times 40 \\[1em] = \frac{22}{7} \times 306.25 \times 40 \\[1em] = \frac{269500}{7} \\[1em] = 38500 \text{ cm}^3.\end{aligned}$$

Hence, the height of the cylinder is 40 cm and volume of the cylinder = 38500 cm³.

(ii) Given, circumference of base = 132 cm

We know that circumference = 2πr

∴ 2πr = 132

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times \text{r} = 132 \\[1em] \Rightarrow \text{r} = \frac{132 \times 7}{22 \times 2} \\[1em] \Rightarrow \text{r} = \frac{924}{44} \\[1em] \Rightarrow \text{r} = 21 \text{ cm.}\end{aligned}$$

Given, height (h) = 25 cm.

Volume of cylinder = πr²h

Putting values we get,

$$\begin{aligned}= \frac{22}{7} \times (21)^2 \times 25 \\[1em] = \frac{22}{7} \times 441 \times 25 \\[1em] = \frac{242550}{7} \\[1em] = 34650 \text{ cm}^3.\end{aligned}$$

Hence, radius of the cylinder is 21 cm and volume of the cylinder is 34650 cm³.

Given,

Total surface area = 462 cm²

⇒ 2πr(h + r) = 462

⇒ πr(h + r) = $\frac{462}{2}$

⇒ πr(h + r) = 231 ....(1)

Given, curved surface area is one-third of its total surface area.

⇒ Curved surface area = $\frac{1}{3}$ × total surface area

$$\begin{aligned}\Rightarrow \frac{\text{curved surface area}}{\text{total surface area}} = \frac{1}{3} \\[1em] \Rightarrow \frac{2π\text{rh}}{2π\text{r(h + r)}} = \frac{1}{3} \\[1em] \Rightarrow \frac{\text{h}}{\text{(h + r)}} = \frac{1}{3} \\[1em]\end{aligned}$$

⇒ 3h = h + r

⇒ 3h - h = r

⇒ 2h = r

⇒ h = $\frac{\text{r}}{2}$

Substituting value of h in eq.(1), we have:

$$\begin{aligned}\Rightarrow π\text{r(h + r)} = 231 \\[1em] \Rightarrow \frac{22}{7} \times \text{r}\Big(\frac{\text{r}}{2} + \text{r}\Big) = 231 \\[1em] \Rightarrow \frac{22}{7} \times \text{r}\Big(\frac{\text{r} + 2\text{r}}{2}\Big) = 231 \\[1em] \Rightarrow \frac{22}{7} \times \text{r} \times \frac{3\text{r}}{2} = 231 \\[1em] \Rightarrow \frac{22}{7} \times \frac{3\text{r}^2}{2} = 231 \\[1em] \Rightarrow \text{r}^2 = \frac{7 \times 2 \times 231}{22 \times 3} \\[1em] \Rightarrow \text{r}^2 = \frac{3234}{66} \\[1em] \Rightarrow \text{r}^2 = 49 \\[1em] \Rightarrow \text{r} = \sqrt{49} \\[1em] \Rightarrow \text{r} = 7 \text{ cm}.\end{aligned}$$

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

Volume of cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 7^2 \times 3.5 \\[1em] = \frac{22}{7} \times 49 \times 3.5 \\[1em] = \frac{3773}{7} \\[1em] = 539 \text{ cm}^3.\end{aligned}$$

Hence, volume of cylinder is 539 cm³.

Given, r + h = 37 m

Total surface area = 1628 m²

⇒ 2πr(h + r) = 1628

⇒ πr(h + r) = $\frac{1628}{2}$

$$\begin{aligned}\Rightarrow \frac{22}{7} \times \text{r} \times 37 = 814 \\[1em] \Rightarrow \text{r} = \frac{814 \times 7}{22 \times 37} \\[1em] \Rightarrow \text{r} = \frac{5698}{814} \\[1em] \Rightarrow \text{r} = 7 \text{ m.}\end{aligned}$$

⇒ r + h = 37

⇒ 7 + h = 37

⇒ h = 37 - 7

⇒ h = 30 m.

Volume of cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 7^2 \times 30 \\[1em] = \frac{22}{7} \times 49 \times 30 \\[1em] = \frac{32340}{7} \\[1em] = 4620 \text{ m}^3.\end{aligned}$$

Hence, volume of solid cylinder is 4620 m³.

Total surface area = 660 cm²

⇒ 2πr(h + r) = 660

$$\begin{aligned}\Rightarrow πr(h + r) = \frac{660}{2} \\[1em] \Rightarrow \frac{22}{7} \times 5(\text{h} + 5) = \frac{660}{2} \\[1em] \Rightarrow \text{h} + 5 = 330 \times \frac{7}{22 \times 5} \\[1em] \Rightarrow \text{h} + 5 = \frac{2310}{110} \\[1em] \Rightarrow \text{h} + 5 = 21 \\[1em] \Rightarrow \text{h} = 21 - 5 \\[1em] \Rightarrow \text{h} = 16 \text{ cm.}\end{aligned}$$

Hence, the height of a solid circular cylinder is 16 cm.

Given, external diameter = 8 cm and h = 12 cm

External radius (R) = $\frac{\text{diameter}}{2} = \frac{8}{2} = 4$ cm.

Let internal radius be r cm.

Thickness = (R - r)

⇒ 2 = 4 - r

⇒ r = 4 - 2

⇒ r = 2 cm.

By formula,

Total surface area of hollow cylinder = 2π(Rh + rh + R² - r²)

$$\begin{aligned}= 2 \times \frac{22}{7} (4 \times 12 + 2 \times 12 + 4^2 - 2^2) \\[1em] = \frac{44}{7} (48 + 24 + 16 - 4) \\[1em] = \frac{44}{7} \times 84 \\[1em] = \frac{3696}{7} \\[1em] = 528 \text{ cm}^2.\end{aligned}$$

Hence, the total surface area of the given hollow cylinder open at both ends is 528 cm².

(i) Given,

Diameter of cylinder (d) = 7 cm

Radius of cylinder (r) = $\frac{d}{2} = \frac{7}{2}$ = 3.5 cm

Height of cylinder (h) = 10 cm

Total surface area (both the halves) = Total surface area of cylinder + Area of two rectangles

= [2πr(h + r)] + [2 × (l × b)]

= [2πr(h + r)] + [2 × (h × d)]

= $\Big[2 \times \frac{22}{7} \times 3.5 \times (3.5 + 10) \Big] + [2 \times 10 \times 7]$

= (2 × 22 × 0.5 × 13.5) + 140

= 297 + 140

= 437 cm².

Hence, total surface area of both the halves = 437 cm².

(ii) Total cost of painting the two halves = Total surface area × Rate

= 437 × 30

= ₹ 13,110.

Hence, total cost of painting the two halves = ₹ 13,110.

Let the cistern be filled in x hours.

Water column forms a cylinder of radius (r) = $\frac{\text{diameter}}{2} = \frac{20}{2} \times \frac{1}{100} = \frac{1}{10} \text{ m}.$

Given, water is flowing at the rate of 3 km/hr through a circular pipe.

= $\frac{3 \times 1000}{60 \times 60} = \frac{5}{6}$ m/s

Volume of water that flows in 1 second = Area of cross section of pipe × rate of flow of water

= πr² × rate of flow of water

$$\begin{aligned}= \frac{22}{7} \times \Big(\frac{1}{10}\Big)^2 \times \frac{5}{6} \\[1em] = \frac{22}{7} \times \frac{1}{100} \times \frac{5}{6} \\[1em] = \frac{110}{4200} \\[1em] = \frac{11}{420}\end{aligned}$$

Given, diameter of cistern = 10 m

Radius (r) = $\frac{\text{diameter}}{2} = \frac{10}{2} = 5$ m

Depth of cistern = 2 m

Volume of cistern = πr²h

$$\begin{aligned}= \frac{22}{7} \times 5^2 \times 2 \\[1em] = \frac{22}{7} \times 25 \times 2 \\[1em] = \frac{1100}{7}\end{aligned}$$

Required time = $\frac{\text{Volume of cistern}}{\text{volume of water flows in 1 second}}$

$$\begin{aligned}= \frac{\frac{1100}{7}}{\frac{11}{420}} \\[1em] = \frac{1100}{7} \times \frac{420}{11} \\[1em] = \frac{462000}{77} \text{ seconds}\\[1em] = \frac{462000}{77} \times \frac{1}{60 \times 60}\text{ hours} \\[1em] = \frac{5}{3} \\[1em] = 1\frac{2}{3} \text{ hours.}\end{aligned}$$

= 1 hour 40 min.

Hence, cistern will be filled in 1 hour 40 min.

Volume of the swimming pool = 70 × 44 × 3 = 9240 m³

Radius of the pipe = $\frac{\text{Diameter}}{2} = \frac{35}{2} \times \frac{1}{100} = \frac{35}{200} = 0.175$ m

Volume of water flowing out of the pipe per second = Area of cross section of the pipe × rate of flow of water

= πr² × 6 m/s

$$\begin{aligned}= \frac{22}{7} \times (0.175)^2 \times 6 \\[1em] = \frac{22}{7} \times 0.030625 \times 6 \\[1em] = \frac{4.0425}{7} \\[1em] = 0.5775 \text{ m}^3\end{aligned}$$

Required time = $\frac{\text{Volume of swimming pool}}{\text{volume of water flow per second}}$

$$\begin{aligned}= \frac{9240}{0.5775} \\[1em] = 16000 \text{ seconds} \\[1em] = \frac{16000}{60 \times 60} \\[1em] = \frac{40}{9} \\[1em] = 4\frac{4}{9} \text{ hours.}\end{aligned}$$

Hence, time required to fill the pool is $4\frac{4}{9}$ hours.

Internal radius of circular pipe = $\frac{\text{Diameter}}{2} = \frac{2}{2} = 1 \text{ cm.}$

Volume of water flowing through pipe per second = Area of cross section of pipe × rate of flow of water

= πr² × 8 m/s

= πr² × 8 × 100 cm/s

$$\begin{aligned}= \frac{22}{7} \times (1)^2 \times 800 \\[1em] = \frac{22}{7} \times 800 \\[1em] = \frac{17600}{7} \text{ cm}^3/s\end{aligned}$$

Volume of cylindrical tank = πR²h

$$\begin{aligned}= \frac{22}{7} \times 40^2 \times \text{h} \\[1em] = \frac{22}{7} \times 1600 \times \text{h} \\[1em] = \frac{35200}{7} \times \text{h}\end{aligned}$$

Volume of water flowing through pipe in 30 minutes = Volume of cylindrical tank

(1 minute = 60 second so, 30 minutes = 30 × 60 = 1800 s)

$$\begin{aligned}\Rightarrow 1800 \times \frac{17600}{7} = \frac{35200}{7} \times \text{h} \\[1em] \Rightarrow \frac{31680000}{7} = \frac{35200}{7} \times \text{h} \\[1em] \Rightarrow \text{h} = \frac{31680000}{7} \times \frac{7}{35200} \\[1em] \Rightarrow \text{h} = 900 \text{ cm.} \\[1em] \Rightarrow \text{h} = 9 \text{ m.}\end{aligned}$$

Hence, the increase in the water level in 30 minutes is 9 m.

The shape of the deep well will be cylindrical with radius (r) and height (h) = 20 m.

By formula,

Volume of a cylinder = πr²h

r = $\frac{\text{diameter}}{2} = \frac{7}{2} = 3.5$ m.

Let height of platform be H.

Volume of earth is spread evenly to form a rectangular platform (cuboid).

Volume of platform = 22 m × 14 m × height(H)

Volume of earth dug from well = Volume of earth spread evenly to form a platform

$$\begin{aligned}\Rightarrow π\text{r}^2 \text{ h} = 22 \times 14 \times \text{H} \\[1em] \Rightarrow \frac{22}{7} \times (3.5)^2 \times 20 = 308 \times \text{H} \\[1em] \Rightarrow \frac{22}{7} \times 12.25 \times 20 = 308 \times \text{H} \\[1em] \Rightarrow \frac{5390}{7 \times 308} = \text{H} \\[1em] \Rightarrow \text{H} = \frac{5390}{2156} \\[1em] \Rightarrow \text{H} = 2.5 \text{ m.}\end{aligned}$$

Hence, the height of the platform is 2.5 m.

Internal radius of the pipe, r = $\frac{\text{diameter}}{2} = \frac{10}{2} = 5$ cm

Given,

Thickness = 5 mm = 0.5 cm

Thickness = External radius (R) - Internal radius (r)

⇒ R = r + thickness

⇒ R = 5.5 cm

Length of the pipe (h) = 24 cm

External volume = πR²h

$$\begin{aligned}= \frac{22}{7} \times (5.5)^2 \times 24 \\[1em] = \frac{22}{7} \times 30.25 \times 24 \\[1em] = \frac{15972}{7}\end{aligned}$$

Internal volume = πr²h

$$\begin{aligned}= \frac{22}{7} \times (5)^2 \times 24 \\[1em] = \frac{22}{7} \times 25 \times 24 \\[1em] = \frac{13200}{7}\end{aligned}$$

Volume of metal = External volume - Internal volume

$$\begin{aligned}= \frac{15972}{7} - \frac{13200}{7} \\[1em] = \frac{15972 - 13200}{7} \\[1em] = \frac{2772}{7} \\[1em] = 396 \text{ cm}^3.\end{aligned}$$

Given, 1 cm³ of the metal weighs 7.5 grams.

(1g = $\frac{1}{1000} \text{kg}$)

Total weight of the pipe = $396 \times 7.5 \times \frac{1}{1000} = \frac{2970}{1000} = 2.97$ kg

Hence, the mass of a metallic hollow cylindrical pipe is 2.97 kg.

Radius of the well, r = $\frac{\text{diameter}}{2} = \frac{10}{2} = 5$ m

Depth of the well (h) = 8.4 m

Volume of the earth dug out from the well = πr²h

$$\begin{aligned}= \frac{22}{7} \times 5^2 \times 8.4 \\[1em] = \frac{22}{7} \times 25 \times 8.4 \\[1em] = \frac{4620}{7} \\[1em] = 660 \text{ m}^3\end{aligned}$$

External radius, R = radius of the well + 7.5 = 5 + 7.5 = 12.5 m

The embankment forms a hollow cylinder around the well.

Let height of the embankment be H.

∴ Volume of embankment = πR²H - πr²H

= πH(R² - r²)

$$\begin{aligned}= \frac{22}{7} \times \text{H} \times [(12.5)^2 - 5^2] \\[1em] = \frac{22}{7} \times \text{H} \times (156.25 - 25) \\[1em] = \frac{22}{7} \times \text{H} \times 131.25 \\[1em] = \frac{2887.5}{7} \times \text{H} \\[1em] = 412.5 \text{ H} \text{ m}^3\end{aligned}$$

Volume of earth dug out = Volume of the embankment

$$\begin{aligned}\Rightarrow 660 = 412.5 \text{H} \\[1em] \Rightarrow \text{H} = \frac{660}{412.5} \\[1em] \Rightarrow \text{H} = 1.6 \text{ m.}\end{aligned}$$

Hence, the height of the embankment is 1.6 m.

Given, h = 24 cm and r = 7 cm

(i) Slant height, l = $\sqrt{\text{h}^2 + \text{r}^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm.}$

Hence, slant height of the cone is 25 cm.

(ii) Lateral surface area = πrl

$$\begin{aligned}= \frac{22}{7} \times 7 \times 25 \\[1em] = \frac{3850}{7} \\[1em] = 550 \text{ cm}^2\end{aligned}$$

Hence, lateral surface area of the cone is 550 cm².

(iii) Total surface area = πr(l + r)

$$\begin{aligned}= \frac{22}{7} \times 7(25 + 7) \\[1em] = \frac{154}{7} \times (32) \\[1em] = \frac{4928}{7} \\[1em] = 704 \text{ cm}^2\end{aligned}$$

Hence, total surface area of the cone is 704 cm².

(iv) Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 \\[1em] = \frac{22}{21} \times 49 \times 24 \\[1em] = \frac{25872}{21} \\[1em] = 1232 \text{ cm}^3\end{aligned}$$

Hence, volume of the cone is 1232 cm³.

Given, h = 8 cm and r = $\frac{\text{diameter}}{2} = \frac{12}{2} = 6$ cm

(i) Slant height, l = $\sqrt{\text{h}^2 + \text{r}^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm.}$

Hence, slant height of the cone is 10 cm.

(ii) Total surface area = πr(l + r)

$$\begin{aligned}= \frac{22}{7} \times 6(10 + 6) \\[1em] = \frac{132}{7} \times (16) \\[1em] = \frac{2112}{7} \\[1em] = 301.7 \text{ cm}^2\end{aligned}$$

Hence, total surface area of the cone is 301.7 cm².

(iii) Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 6^2 \times 8 \\[1em] = \frac{22}{21} \times 36 \times 8 \\[1em] = \frac{6336}{21} \\[1em] = 301.7 \text{ cm}^3\end{aligned}$$

Hence, volume of the cone is 301.7 cm³.

Given, slant height, l = 17 cm and radius, r = 15 cm

(i) l² = r² + h²

⇒ h² = l² - r²

⇒ h² = 17² - 15²

⇒ h² = 289 - 225

⇒ h² = 64

⇒ h = $\sqrt{64} = 8 \text{ cm.}$

Hence, height of the cone is 8 cm.

(ii) Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 15^2 \times 8 \\[1em] = \frac{22}{21} \times 225 \times 8 \\[1em] = \frac{39600}{21} \\[1em] = 1885.7 \text{ cm}^3\end{aligned}$$

Hence, volume of the cone is 1885.7 cm³.

(iii) Total surface area = πr(l + r)

$$\begin{aligned}= \frac{22}{7} \times 15(17 + 15) \\[1em] = \frac{330}{7} \times (32) \\[1em] = \frac{10560}{7} \\[1em] = 1508.6 \text{ cm}^2\end{aligned}$$

Hence, total surface area of the cone is 1508.6 cm².

Given, radius, r = $\frac{\text{diameter}}{2} = \frac{12}{2} = 6 \text{ cm.}$

Volume of cone = 660 cm³

(i) By formula,

Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}\Rightarrow 660 = \frac{1}{3} \times \frac{22}{7} \times 6^2 \times \text{h} \\[1em] \Rightarrow 660 = \frac{22}{21} \times 36 \times \text{h} \\[1em] \Rightarrow \text{h} = \frac{660 \times 21}{22 \times 36} \\[1em] \Rightarrow \text{h} = \frac{13860}{792} \\[1em] \Rightarrow \text{h} = 17.5 \text{ cm}.\end{aligned}$$

Hence, the height of the cone is 17.5 cm.

(ii) By formula,

Slant height (l) = $\sqrt{\text{h}^2 + \text{r}^2}$

$= \sqrt{(17.5)^2 + 6^2} = \sqrt{306.25 + 36} = \sqrt{342.25} = 18.5 \text{ cm.}$

Hence, slant height of the cone is 18.5 cm.

(iii) Total surface area = πr(l + r)

$$\begin{aligned}= \frac{22}{7} \times 6 \times (18.5 + 6) \\[1em] = \frac{132}{7} \times (24.5) \\[1em] = \frac{3234}{7} \\[1em] = 462 \text{ cm}^2\end{aligned}$$

Hence, total surface area of the cone is 462 cm².

Given, slant height, l = 20 cm and total surface area of cone = 384π cm²

(i) By formula,

Total surface area = πr(l + r)

⇒ 384π = πr(20 + r)

⇒ 384 = 20r + r²

⇒ r² + 20r - 384 = 0

⇒ r² + 32r - 12r - 384 = 0

⇒ r(r + 32) - 12(r + 32) = 0

⇒ (r + 32) = 0 or (r - 12) = 0

⇒ r = - 32 or r = 12

Since, radius cannot be negative.

∴ r = 12 cm.

Hence, radius of the cone is 12 cm.

(ii) l² = r² + h²

⇒ h² = l² - r²

⇒ h² = 20² - 12²

⇒ h² = 400 - 144

⇒ h² = 256

⇒ h = $\sqrt{256} = 16 \text{ cm.}$

Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times π \times 12^2 \times 16 \\[1em] = \frac{1}{3} \times π \times 144 \times 16 \\[1em] = \frac{2304}{3} π \\[1em] = 768 π \text{ cm}^3\end{aligned}$$

Hence, volume of the cone is 768 π cm³.

Given, radius(r) : height(h) = 5 : 12

(i) Let r = 5x and h = 12x

Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}\Rightarrow 2512 = \frac{1}{3} \times 3.14 \times (5\text{x})^2 \times 12\text{x} \\[1em] \Rightarrow 2512 = 3.14 \times 25\text{x}^2 \times 4\text{x} \\[1em] \Rightarrow \text{x}^3 = \frac{2512}{3.14 \times 25 \times 4} \\[1em] \Rightarrow \text{x}^3 = \frac{2512}{314} \\[1em] \Rightarrow \text{x}^3 = 8 \\[1em] \Rightarrow \text{x} = \sqrt[3]{8} \\[1em] \Rightarrow \text{x} = 2\end{aligned}$$

⇒ r = 5x = 5 × 2 = 10 cm

⇒ h = 12x = 12 × 2 = 24 cm

Hence, radius of the cone is 10 cm and height of the cone is 24 cm.

(ii) Curved surface area = πrl

l² = r² + h²

⇒ l² = 10² + 24²

⇒ l² = 100 + 576

⇒ l² = 676

⇒ l = $\sqrt{676}$ = 26 cm

$$\begin{aligned}= 3.14 \times 10 \times 26 \\[1em] = 816.4 \text{ cm}^2\end{aligned}$$

Hence, curved surface area of the cone is 816.4 cm².

(iii) Total surface area = πr(l + r)

$$\begin{aligned}= 3.14 \times 10(26 + 10) \\[1em] = 3.14 \times 10 \times 36 \\[1em] = 1130.4 \text{ cm}^2\end{aligned}$$

Hence, total surface area of the cone is 1130.4 cm².

Let height of cone be h cm.

Given,

Radius of cone (r) = 20 cm

(i) Given,

Volume = 8800 cm³

By formula,

Volume of cone = $\frac{1}{3}\pi r^2 h$

∴ $\frac{1}{3}\pi r^2 h = 8800$

$$\begin{aligned}\Rightarrow \frac{1}{3} \times \frac{22}{7} \times 20^2 \times h = 8800 \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 400 \times h = 8800 \\[1em] \Rightarrow h = \frac{8800 \times 7 \times 3}{400 \times 22} \\[1em] \Rightarrow h = 7 \times 3 \\[1em] \Rightarrow h = 21 \text{ cm}.\end{aligned}$$

Hence, height of cone = 21 cm.

(ii) By formula,

Slant Height of cone (l) = $\sqrt{(r^2 + h^2)}$

$$\begin{aligned}l = \sqrt{(20)^2 + (21)^2} \\[1em] = \sqrt{400 + 441} \\[1em] = \sqrt{841} \\[1em] = 29 \text{ cm}\end{aligned}$$

By formula,

Curved Surface Area of cone = πrl

Curved Surface Area of cone = $\frac{22}{7} \times 20 \times 29$

⇒ $\frac{12760}{7}$

⇒ 1822.857 ≈ 1823 cm².

Hence, curved surface area of cone = 1823 cm².

Given, r = 17.5 m and h = 6 m

l² = r² + h²

⇒ l² = 17.5² + 6²

⇒ l² = 306.25 + 36

⇒ l² = 342.25

⇒ l = $\sqrt{342.25}$ = 18.5 m

So, the total curved surface area of the tent = πrl

$$\begin{aligned}= \frac{22}{7} \times 17.5 \times 18.5 \\[1em] = \frac{7122.5}{7} \\[1em] = 1017.5 \text{ m}^2\end{aligned}$$

Width of the canvas used = 1.25 m

Length of canvas = $\frac{\text{area of canvas}}{\text{width of canvas}} = \frac{1017.5}{1.25}$ = 814 m.

Hence, 814 metres of canvas will be needed to make a conical tent.

From figure,

Radius of conical part = radius of cylindrical part = r.

Radius of the cylindrical part of the tent (r) = $\frac{\text{diameter}}{2} = \frac{105}{2} \text{ m}$

Radius of the conical part (r) = $\frac{105}{2} \text{ m}$

Slant height (l) = 53 m

So, the total curved surface area of the tent = 2πrh + πrl

$$\begin{aligned}= 2 \times \frac{22}{7} \times \frac{105}{2} \times 3 + \frac{22}{7} \times \frac{105}{2} \times 53 \\[1em] = \frac{13860}{14} + \frac{122430}{14} \\[1em] = 990 + 8745 \\[1em] = 9735 \text{ m}^2.\end{aligned}$$

Width of the canvas used = 2.5 m

Length of canvas = $\frac{\text{area of canvas}}{\text{width of canvas}} = \frac{9735}{2.5}$ = 3894 m.

Hence, length of the canvas required to make tent is 3894 m.

Radius of cylindrical portion, r = $\frac{\text{diameter}}{2} = \frac{20}{2}$ = 10 cm

Height of the cylindrical portion, h = 2.8 m = 2.8 × 100 = 280 cm

Height of the conical portion, H = 42 cm

From figure,

Radius of the conical part = radius of the cylindrical portion = r = 10 cm

Volume of iron pole = Volume of cylindrical portion + Volume of conical portion

= πr²h + $\frac{1}{3}$ πr²H

$$\begin{aligned}= \frac{22}{7} \times 10^2 \times 280 + \frac{1}{3} \times \frac{22}{7} \times 10^2 \times 42 \\[1em] = \frac{22}{7} \times 100 \times 280 + \frac{1}{3} \times \frac{22}{7} \times 100 \times 42 \\[1em] = \frac{22}{7} \times 100 \times 280 + \frac{22}{7} \times 100 \times 14 \\[1em] = \frac{616000}{7} + \frac{30800}{7} \\[1em] = \frac{616000 + 30800}{7} \\[1em] = \frac{646800}{7} \\[1em] = 92400 \text{ cm}^3.\end{aligned}$$

Given,

Weight of 1 cm³ of iron = 7.5 gm.

Total weight = 92400 × 7.5 = 693000 gm = $\frac{693000}{1000}$ kg = 693 kg.

Hence, the weight of the pillar is 693 kg.

Radius of cylindrical pipe, r = $\frac{\text{diameter}}{2} = \frac{0.5}{2}$ = 0.25 cm

Given, water flows at the rate of 10 m per minute.

Length of the cylindrical portion, h = 10 m = 10 × 100 = 1000 cm

Height of the conical portion, H = 24 cm

Radius of conical pipe, R = $\frac{\text{diameter}}{2} = \frac{40}{2}$ = 20 cm

Volume of water that flows in 1 min = πr²h

$$\begin{aligned}= \frac{22}{7} \times (0.25)^2 \times 1000 \\[1em] = \frac{22}{7} \times 0.0625 \times 1000 \\[1em] = \frac{1375}{7}\end{aligned}$$

Volume of the conical vessel = $\frac{1}{3}$ πR²H

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 20^2 \times 24 \\[1em] = \frac{22}{7} \times 400 \times 8 \\[1em] = \frac{70400}{7}\end{aligned}$$

Required time = $\frac{\text{Volume of conical vessel}}{\text{Volume of water that flows in 1 min}}$

$$\begin{aligned}= \frac{\frac{70400}{7}}{\frac{1375}{7}} \\[1em] = \frac{70400}{7} \times \frac{7}{1375} \\[1em] = \frac{70400}{1375} \\[1em] = 51.2 \text{ min.}\end{aligned}$$

= 51 min 12 sec.

Hence, time required to fill a conical vessel is 51 min 12 sec.

Given,

Each person must have 20 m³ of air to breathe.

∴ 11 persons need 11 × 20 m³ = 220 m³

Each person must have 4 m² of the space on the ground.

∴ 11 persons need 11 × 4 m² = 44 m²

Base of the conical tent = area of the circle = πr²

$$\begin{aligned}\Rightarrow 44 = \frac{22}{7} \times \text{r}^2 \\[1em] \Rightarrow \text{r}^2 = \frac{7 \times 44}{22} \\[1em] \Rightarrow \text{r}^2 = \frac{308}{22} \\[1em] \Rightarrow \text{r}^2 = 14 \text{ m.}\end{aligned}$$

Let height of the conical tent be h meters.

Since, conical tent needs to accomodate 11 persons, so its volume will be equal to volume of air required for 11 persons.

$$\begin{aligned}\Rightarrow \frac{1}{3}π \text{r}^2 \text{h} = 220 \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 14 \times \text{h} = 220 \\[1em] \Rightarrow \text{h} = \frac{220 \times 7 \times 3}{22 \times 14} \\[1em] \Rightarrow \text{h} = \frac{4620}{308} \\[1em] \Rightarrow \text{h} = 15 \text{ m.}\end{aligned}$$

Hence, height of the tent is 15 m..

Given,

Each person must have 16 m³ of air to breathe.

∴ 77 persons need 77 × 16 m³ = 1232 m³

Radius of the tent (r) = 7 m

Let height of the conical tent be h meters.

Since, conical tent needs to accomodate 77 persons, so its volume will be equal to volume of air required for 77 persons.

$$\begin{aligned}\Rightarrow \frac{1}{3}π \text{r}^2 \text{h} = 1232 \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 7^2 \times \text{h} = 1232 \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 49 \times \text{h} = 1232 \\[1em] \Rightarrow \text{h} = \frac{1232 \times 7 \times 3}{22 \times 49} \\[1em] \Rightarrow \text{h} = \frac{25872}{1078} \\[1em] \Rightarrow \text{h} = 24 \text{ m.}\end{aligned}$$

By formula,

l² = r² + h²

⇒ l² = 7² + 24²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$ = 25 m

Curved surface area of the tent = πrl

$$\begin{aligned}= \frac{22}{7} \times 7 \times 25 \\[1em] = 22 \times 25 \\[1em] = 550 \text{ m}^2\end{aligned}$$

Hence, height of the tent is 24 m and curved surface area of the tent is 550 m².

Radius of cone, r = 1.6 cm

Height of the cone, h = 3.6 cm

Volume of circular cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times (1.6)^2 \times 3.6 \\[1em] = \frac{1}{3} \times \frac{22}{7} \times 2.56 \times 3.6 \\[1em] = \frac{202.752}{21}\end{aligned}$$

Volume of cone of radius (R) = 1.2 cm and height (H)

Volume of circular cone = $\frac{1}{3}$ πR²H

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times (1.2)^2 \times \text{H} \\[1em] = \frac{1}{3} \times \frac{22}{7} \times 1.44 \times \text{H} \\[1em] = \frac{31.68}{21} \text{H}\end{aligned}$$

Since, cone is melted and recasted into right circular cone of radius 1.2 cm, the volume remains the same.

$$\begin{aligned}\therefore \frac{202.752}{21} = \frac{31.68}{21} \text{H} \\[1em] \Rightarrow \text{H} = \frac{202.752 \times 21}{31.68 \times 21} \\[1em] \Rightarrow \text{H} = 6.4 \text{ cm.}\end{aligned}$$

Hence, height of the cone is 6.4 cm.

For larger cylinder,

Height (H) = 5 cm

Radius (R) = 3 cm

For smaller cones,

Height (h) = 1 cm

Radius (r) = 1 mm = 0.1 cm

Let no. of smaller cones formed be n.

Volume of larger cylinder = n × Volume of smaller cones

$$\begin{aligned}\Rightarrow π\text{R}^2\text{H} = \text{n} \times \frac{1}{3}π\text{r}^2\text{h} \\[1em] \Rightarrow \frac{22}{7} \times (3)^2 \times 5 = \text{n} \times \frac{1}{3} \times \frac{22}{7} \times (0.1)^2 \times 1 \\[1em] \Rightarrow \frac{22}{7} \times 9 \times 5 = \text{n} \times \frac{1}{3} \times \frac{22}{7} \times 0.01 \times 1 \\[1em] \Rightarrow \frac{990}{7} = \text{n} \times \frac{0.22}{21} \\[1em] \Rightarrow \text{n} = \frac{990 \times 21}{0.22 \times 7} \\[1em] \Rightarrow \text{n} = \frac{20790}{1.54} \\[1em] \Rightarrow \text{n} = 13500\end{aligned}$$

Hence, the number of cones formed = 13500.

For cylindrical vessel,

Let height be H cm

Radius (R) = 10 cm

For conical vessel,

Height (h) = 50 cm

Radius (r) = 12 cm

Since, contents in conical vessel are emptied into a cylindrical vessel, hence there volume will be same.

∴ Volume of cylinder = Volume of conical vessel

$$\begin{aligned}\Rightarrow π\text{R}^2\text{H} = \frac{1}{3}π\text{r}^2\text{h} \\[1em] \Rightarrow \text{R}^2\text{H} = \frac{1}{3}\text{r}^2\text{h} \\[1em] \Rightarrow 10^2 \times \text{H} = \frac{1}{3} \times 12^2 \times 50 \\[1em] \Rightarrow 100 \times \text{H} = \frac{1}{3} \times 144 \times 50 \\[1em] \Rightarrow \text{H} = \frac{144 \times 50}{3 \times 100} \\[1em] \Rightarrow \text{H} = \frac{7200}{300} \\[1em] \Rightarrow \text{H} = 24 \text{ cm.}\end{aligned}$$

Hence, height to which the liquid rises in the cylindrical vessel is 24 cm.

Let OAB be the given cone of height 40 cm and base radius R cm. Let this cone be cut by the plane CND (parallel to the base plane AMB) to obtain cone OCD with height h cm and base radius r cm as shown in the figure below :

From figure,

∠NOD = ∠MOB (Common angle)

∠OND = ∠OMB = 90° (Heights are perpendicular to radii)

∠ODN = ∠OBM (Corresponding angles since ND || MB)

∴ △OND ~ △OMB (By AA similarity)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

∴ $\frac{\text{r}}{\text{R}} = \frac{\text{h}}{40}$ ...(1)

According to given,

Volume of cone OCD = $\frac{1}{64}$ Volume of cone OAB

∴ $\frac{1}{3}$ πr²h = $\frac{1}{64} \times \frac{1}{3}$ πR² × 40

Dividing both sides by π and multiplying by 3 we get,

$$\begin{aligned}\Rightarrow \text{r}^2 \text{h} = \frac{40}{64} \times \text{R}^2 \\[1em] \Rightarrow \frac{\text{r}^2}{\text{R}^2} = \frac{5}{8 \text{h}} \\[1em] \Rightarrow \Big(\frac{\text{r}}{\text{R}}\Big)^2 = \frac{5}{8 \text{h}} \\[1em]\end{aligned}$$

Using eq.(1),

$$\begin{aligned}\Rightarrow \Big(\frac{\text{h}}{40}\Big)^2 = \frac{5}{8 \text{h}} \\[1em] \Rightarrow \frac{\text{h}^2}{1600} = \frac{5}{8 \text{h}} \\[1em] \Rightarrow \text{h}^3 = \frac{5 \times 1600}{8} \\[1em] \Rightarrow \text{h}^3 = 5 \times 200 \\[1em] \Rightarrow \text{h}^3 = 1000 \\[1em] \Rightarrow \text{h} = \sqrt[3]{1000} \\[1em] \Rightarrow \text{h} = 10 \text{ cm.}\end{aligned}$$

The height of the cone OCD = 10 cm

∴ The section is cut at the height of 40 - 10 = 30 cm.

Hence, the section is cut above 30 cm from the base.

Internal radius (r) = 3 cm

Height (h) = 21 cm

Thickness = External radius (R) - Internal radius

⇒ 0.5 = External radius - 3 cm

⇒ External radius = 0.5 + 3 = 3.5 cm

Volume of hollow cylinder = π(R² - r²)h,

Putting values we get,

∴ Volume of metal = π(3.5² - 3²) × 21

$$\begin{aligned}= \frac{22}{7} \times (12.25 - 9) \times 21 \\[1em] = 22 \times 3.25 \times 3 \\[1em] = 214.5 \text{ cm}^3\end{aligned}$$

Given the tube is melted and cast into a right circular cone of height (H) 7 cm.

So, the volume of metal and volume of cone will be same.

∴ 214.5 = $\frac{1}{3}$ πr²H

$$\begin{aligned}\Rightarrow 214.5 = \frac{1}{3} \times \frac{22}{7} \times \text{r}^2 \times 7 \\[1em] \Rightarrow 214.5 = \frac{1}{3} \times 22 \times \text{r}^2 \\[1em] \Rightarrow \text{r}^2 = \frac{214.5 \times 3}{22} \\[1em] \Rightarrow \text{r}^2 = 29.25 \\[1em] \Rightarrow \text{r} = \sqrt{29.25} \\[1em] \Rightarrow \text{r} = 5.4 \text{ cm.}\end{aligned}$$

Hence, radius of the cone is 5.4 cm.

Given,

Height of the cylinder (H) = 12 cm

Radius of the base of the cylinder (R) = $\frac{\text{diameter}}{2} = \frac{10}{2}$ = 5 cm

Height of the cone (h) = 12 cm

Radius of the cone (r) = 5 cm

Volume of the remaining part = Volume of cylinder - Volume of cone

= πR²H - $\frac{1}{3}$ πr²h

$$\begin{aligned}= π \times 5^2 \times 12 - π \times \frac{1}{3} \times 5^2 \times 12 \\[1em] = π(25 \times 12 - 25 \times 4) \\[1em] = π(300 - 100) \\[1em] = 200 π \text{ cm}^3\end{aligned}$$

∴ The volume of the remaining solid is 200 π cm³.

By formula,

l² = r² + h²

⇒ l² = 5² + 12²

⇒ l² = 25 + 144

⇒ l² = 169

⇒ l = $\sqrt{169}$ = 13 cm

Total surface area of remaining solid = Curved surface area of cylinder + curved surface area of cone + base area of cylinder

= 2πRH + πrl + πR²

= π(2RH + rl + R²)

= π(2 × 5 × 12 + 5 × 13 + 5²)

= π(120 + 65 + 25)

= 210 π cm².

Hence, the volume of the remaining solid is 200 π cm³ and total surface area of remaining solid is 210 π cm².

Edge of a cube = 14 cm

Volume = side³ = 14³ = 2744 cm³.

Cone of maximum size is carved out as shown in figure,

Diameter of the cone cut out from it = 14 cm

Radius, r = $\frac{\text{diameter}}{2} = \frac{14}{2}$ = 7 cm

Height, h = 14 cm

Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 14 \\[1em] = \frac{1}{3} \times 22 \times 49 \times 2 \\[1em] = \frac{2156}{3} \\[1em] = 718.67 \text{ cm}^3.\end{aligned}$$

Rounding off to one decimal place = 718.8 cm³

Volume of the remaining material = Volume of the cube - Volume of the cone

= 2744 - 718.67

= 2025.33 cm³

Rounding off to one decimal place = 2025.3 cm³

Hence, the volume of the cone is 718.8 cm³ and of the remaining material is 2025.3 cm³.

Volume of block of wood = 20 cm × 10 cm × 10 cm = 2000 cm³

Diameter of the cone for maximum volume = 10 cm

Cone of maximum volume is carved out as shown in figure,

Radius, r = $\frac{\text{diameter}}{2} = \frac{10}{2}$ = 5 cm.

Height of the cone for maximum volume, h = 20 cm

Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 20 \\[1em] = \frac{1}{3} \times \frac{22}{7} \times 25 \times 20 \\[1em] = \frac{11000}{21} \\[1em] = 523.8 \text{ cm}^3.\end{aligned}$$

Hence, the volume of the cone carved out is 523.8 cm³.

Given,

Diameter of solid wooden cylinder (D) = 6 cm

Radius of solid wooden cylinder (R) = $\frac{6}{2}$ = 3 cm

Height of solid wooden cylinder (H) = 28 cm

Diameter of cone (d) = 6 cm

Radius of cone (r) = $\frac{6}{2}$ = 3 cm

Height of the cone (h) = 10.5 cm

Volume of cylinder = πR²H

$$\begin{aligned}= \frac{22}{7} \times 3^2 \times 28 \\[1em] = 22 \times 9 \times 4 \\[1em] = 792 \text{ cm}^3.\end{aligned}$$

Volume of single cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 3^2 \times 10.5 \\[1em] = \frac{22}{7} \times 9 \times 3.5 \\[1em] = \frac{693}{7} \\[1em] = 99 \text{ cm}^3.\end{aligned}$$

Volume of two conical cavities = 2 × 99 = 198 cm³

Volume of remaining solid = Volume of cylinder - Volume of 2 conical cavities

= 792 - 198

= 594 cm³.

Hence, volume of the remaining solid = 594 cm³.

(i) Given, r = 10.5 cm

Surface area of sphere = 4πr²

$$\begin{aligned}= 4 \times \frac{22}{7} \times 10.5^2 \\[1em] = 4 \times \frac{22}{7} \times 110.25 \\[1em] = \frac{9702}{7} \\[1em] = 1386 \text{ cm}^2.\end{aligned}$$

Volume of sphere = $\frac{4}{3}$ πr³

$$\begin{aligned}= \frac{4}{3} \times \frac{22}{7} \times 10.5^3 \\[1em] = \frac{4}{3} \times \frac{22}{7} \times 1157.625 \\[1em] = \frac{101871}{21} \\[1em] = 4851 \text{ cm}^3.\end{aligned}$$

Hence, volume of the sphere is 4851 cm³ and surface area of a sphere is 1386 cm².

(ii) Given, r = 4.2 cm

Surface area of sphere = 4πr²

$$\begin{aligned}= 4 \times \frac{22}{7} \times 4.2^2 \\[1em] = 4 \times \frac{22}{7} \times 17.64 \\[1em] = \frac{1552.32}{7} \\[1em] = 221.76 \text{ cm}^2.\end{aligned}$$

Volume of sphere = $\frac{4}{3}$ πr³

$$\begin{aligned}= \frac{4}{3} \times \frac{22}{7} \times 4.2^3 \\[1em] = \frac{4}{3} \times \frac{22}{7} \times 74.088 \\[1em] = \frac{6519.744}{21} \\[1em] = 310.464 \text{ cm}^3.\end{aligned}$$

Hence, volume of the sphere is 310.464 cm³ and surface area of a sphere is 221.76 cm².

Given, volume of a sphere is 288 π cm³.

Let radius be r cm.

Volume of sphere = $\frac{4}{3}$ πr³

$$\begin{aligned}\Rightarrow 288 π = \frac{4}{3} \times π \times \text{r}^3 \\[1em] \text{Dividing by π on both sides, we get:} \\[1em] \Rightarrow 288 = \frac{4}{3} \times \text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \frac{288 \times 3}{4} \\[1em] \Rightarrow \text{r}^3 = \frac{864}{4} \\[1em] \Rightarrow \text{r}^3 = 216 \\[1em] \Rightarrow \text{r} = 6 \text{ cm.}\end{aligned}$$

Surface area of sphere = 4πr²

$$\begin{aligned}= 4 \times \frac{22}{7} \times 6^2 \\[1em] = 4 \times \frac{22}{7} \times 36 \\[1em] = \frac{3168}{7} \\[1em] = 452.57 \approx 453 \text{ cm}^2.\end{aligned}$$

Hence, radius is 6 cm and its surface area is 453 cm².

Given, curved surface area of a sphere is 2826 cm²

Curved surface area of sphere = 4πr²

$$\begin{aligned}\Rightarrow 2826 = 4 \times 3.14 \times \text{r}^2 \\[1em] \Rightarrow 2826 = 12.56 \times \text{r}^2 \\[1em] \Rightarrow \text{r}^2 = \frac{2826}{12.56} \\[1em] \Rightarrow \text{r}^2 = 225 \\[1em] \Rightarrow \text{r} = \sqrt{225} \\[1em] \Rightarrow \text{r} = 15 \text{ cm.}\end{aligned}$$

Volume of sphere = $\frac{4}{3}$ πr³

$$\begin{aligned}= \frac{4}{3} \times 3.14 \times 15^3 \\[1em] = \frac{4}{3} \times 3.14 \times 3375 \\[1em] = \frac{42390}{3} \\[1em] = 14130 \text{ cm}^3.\end{aligned}$$

Hence, radius is 15 cm and volume of sphere is 14130 cm³.

Given, diameter = 10 cm

Radius, r = $\frac{\text{diameter}}{2} = \frac{10}{2}$ = 5 cm.

Curved surface area of hemisphere = 2πr²

$$\begin{aligned}= 2 \times \frac{22}{7} \times 5^2 \\[1em] = 2 \times \frac{22}{7} \times 25 \\[1em] = \frac{1100}{7} \\[1em] = 157.1 \text{ cm}^2.\end{aligned}$$

Total surface area of hemisphere = 3πr²

$$\begin{aligned}= 3 \times \frac{22}{7} \times 5^2 \\[1em] = 3 \times \frac{22}{7} \times 25 \\[1em] = \frac{1650}{7} \\[1em] = 235.71 \text{ cm}^2.\end{aligned}$$

Hence, the curved surface area is 157.1 cm² and the total surface area is 235.71 cm².

(i)

Edge of a cube = 22 cm

Volume of cube = side³ = 22³ = 10648 cm³.

Bullets are spherical in shape.

Radius of bullet, r = $\frac{\text{diameter}}{2} = \frac{2}{2} = 1 \text{cm.}$

Volume of each bullet = $\frac{4}{3}$ πr³

$$\begin{aligned}= \frac{4}{3} \times \frac{22}{7} \times 1^3 \\[1em] = \frac{88}{21} \text{ cm}^3.\end{aligned}$$

Let the number of bullets formed be n.

∴ Volume of cube = n × Volume of each bullet

$$\begin{aligned}\Rightarrow 10648 = \text{n} \times \frac{88}{21} \\[1em] \Rightarrow \text{n} = \frac{10648 \times 21}{88} \\[1em] \Rightarrow \text{n} = \frac{223608}{88} \\[1em] \Rightarrow \text{n} = 2541\end{aligned}$$

Hence, 2541 bullets can be made out of a cube of lead.

(ii)

Shots is in the shape of sphere.

(3 mm = 0.3 cm)

Radius of sphere, r = $\frac{\text{diameter}}{2} = \frac{0.3}{2} = 0.15 \text{cm.}$

Let the number of sphere formed be n.

Volume of cuboid = n × Volume of each shot

$$\begin{aligned}\therefore \text{lbh} = \text{n} \times \frac{4}{3} π \text{r}^3 \\[1em] \Rightarrow 9 \times 11 \times 12 = \text{n} \times \frac{4}{3} \times \frac{22}{7} \times 0.15^3 \\[1em] \Rightarrow 1188 = \text{n} \times \frac{4}{3} \times \frac{22}{7} \times 0.003375 \\[1em] \Rightarrow 1188 = \text{n} \times \frac{0.297}{21} \\[1em] \Rightarrow \text{n} = \frac{21 \times 1188}{0.297} \\[1em] \Rightarrow \text{n} = \frac{24948}{0.297} \\[1em] \Rightarrow \text{n} = 84000.\end{aligned}$$

Hence, there are 84000 lead shots.

Bullet is in shape of sphere.

Radius of bullet, r = $\frac{\text{diameter}}{2} = \frac{1.5}{2}$ = 0.75 cm

Radius of cylinder, R = 5 cm

Height of cylinder, h = 18 cm

Given, bullets are made by melting a cylinder of lead.

∴ Volume of cylinder = n × Volume of each bullet

$$\begin{aligned}\Rightarrow π \text{R}^2 \text{h} = \text{n} \times \frac{4}{3} π \text{r}^3 \\[1em] \text{Divide by π on both sides, we get:} \\[1em] \Rightarrow \text{R}^2 \text{h} = \text{n} \times \frac{4}{3} \text{r}^3 \\[1em] \Rightarrow 5^2 \times 18 = \text{n} \times \frac{4}{3} \times 0.75^3 \\[1em] \Rightarrow 25 \times 18 = \text{n} \times \frac{4}{3} \times 0.421875 \\[1em] \Rightarrow 450 = \text{n} \times 0.5625 \\[1em] \Rightarrow \text{n} = \frac{450}{0.5625} \\[1em] \Rightarrow \text{n} = 800.\end{aligned}$$

Hence, 800 bullets are made.

Given,

Radius of bigger ball, R = 9 cm

Radius of smaller ball, r = 1.5 cm

Let the number of smaller lead balls formed be n.

∴ Volume of big ball = n × Volume of each small ball

$$\begin{aligned}\Rightarrow \frac{4}{3}π\text{R}^3 = \text{n} \times \frac{4}{3}π\text{r}^3 \\[1em] \text{Dividing both sides by 4π and multiplying by 3, we get :} \\[1em] \Rightarrow \text{R}^3 = \text{n} \times \text{r}^3 \\[1em] \Rightarrow 9^3 = \text{n} \times 1.5^3 \\[1em] \Rightarrow 729 = \text{n} \times 3.375 \\[1em] \Rightarrow \text{n} = \frac{729}{3.375} \\[1em] \Rightarrow \text{n} = 216.\end{aligned}$$

Hence, 216 lead balls can be made.

(i) Given,

Radius of ball bearings = 7 mm

Volume of box = 2156 cm³ = 2156 × 10³ mm³

Number of ball bearings that each box can have (N)

= $\frac{\text{Volume of box}}{\text{Volume of each ball}}$

Substituting values we get :

$$\begin{aligned}N = \frac{2156 \times 10^3}{\frac{4}{3} \times \frac{22}{7} \times 7^3} \\[1em] = \frac{2156 \times 10^3}{\frac{88}{3} \times 7^2} \\[1em] = \frac{2156 \times 3 \times 1000}{88 \times 49} \\[1em] = \frac{6468000}{4312} \\[1em] = 1500.\end{aligned}$$

Hence, maximum no. of ball bearings in a box = 1500.

(ii) Mass of each box = No. of balls × Mass of each ball

= 1500 × 4 gm

= 6000 gm

= $\frac{6000}{1000}$ = 6 kg.

Hence, mass of each box = 6 kg.

Radius of sphere, r = 10.5 cm

Let the number of cones formed by recasting metallic sphere be n.

Radius of cone, R = 3.5 cm

Height, h = 3 cm

Volume of sphere = n × Volume of each cone

$$\begin{aligned}\Rightarrow \frac{4}{3}π\text{r}^3 = \text{n} \times \frac{1}{3}π\text{R}^2 \text{h} \\[1em] \text{Dividing both sides by π and multiplying by 3, we get :} \\[1em] \Rightarrow 4\text{r}^3 = \text{n} \times \text{R}^2 \text{h} \\[1em] \Rightarrow 4 \times 10.5^3 = \text{n} \times 3.5^2 \times 3 \\[1em] \Rightarrow 4 \times 1157.625 = \text{n} \times 12.25 \times 3 \\[1em] \Rightarrow 4630.5 = \text{n} \times 36.75 \\[1em] \Rightarrow \text{n} = \frac{4630.5}{36.75} \\[1em] \Rightarrow \text{n} = 126.\end{aligned}$$

Hence, the number of cones obtained is 126.

Surface area of a metallic sphere = 616 cm²

Let the radius of this sphere be R cm.

∴ 4πR² = 616

$$\begin{aligned}\Rightarrow 4 \times \frac{22}{7} \times \text{R}^2 = 616 \\[1em] \Rightarrow \frac{88}{7} \times \text{R}^2 = 616 \\[1em] \Rightarrow \text{R}^2 = \frac{616 \times 7}{88} \\[1em] \Rightarrow \text{R}^2 = \frac{4312}{88} \\[1em] \Rightarrow \text{R}^2 = 49 \\[1em] \Rightarrow \text{R} = \sqrt{49} \\[1em] \Rightarrow \text{R} = 7 \text{ cm.}\end{aligned}$$

Given,

Big sphere is melted and recast into smaller spheres of diameter 3.5 cm.

Radius, r = $\frac{\text{diameter}}{2} = \frac{3.5}{2}$ = 1.75 cm

Let the number of smaller spheres formed be n.

Volume of big sphere = n × Volume of each small sphere

$$\begin{aligned}\Rightarrow \frac{4}{3}π\text{R}^3 = \text{n} \times \frac{4}{3}π\text{r}^3 \\[1em] \text{Dividing both sides by 4π and multiplying by 3, we get :} \\[1em] \Rightarrow \text{R}^3 = \text{n} \times \text{r}^3 \\[1em] \Rightarrow 7^3 = \text{n} \times 1.75^3 \\[1em] \Rightarrow 343 = \text{n} \times 5.359375 \\[1em] \Rightarrow \text{n} = \frac{343}{5.359375} \\[1em] \Rightarrow \text{n} = 64.\end{aligned}$$

Hence, 64 small spheres can be formed.

Let the length of the wire be h cm

Radius of the sphere, R = 6 cm

Radius of the wire, r = 2 mm = 0.2 cm

Given, copper sphere is melted into wire.

∴ Volume of the wire = Volume of the sphere

$$\begin{aligned}\Rightarrow π\text{r}^2 \text{h} = \frac{4}{3}π\text{R}^3 \\[1em] \Rightarrow \text{r}^2 \text{h} = \frac{4}{3}\text{R}^3 \\[1em] \Rightarrow 0.2^2 \times \text{h} = \frac{4}{3} \times 6^3 \\[1em] \Rightarrow 0.04 \times \text{h} = \frac{4}{3} \times 216 \\[1em] \Rightarrow \text{h} = \frac{4 \times 216}{3 \times 0.04} \\[1em] \Rightarrow \text{h} = \frac{864}{0.12} \\[1em] \Rightarrow \text{h} = 7200 \text{ cm.} \\[1em] \Rightarrow \text{h} = 72 \text{ m}.\end{aligned}$$

Hence, the length of the wire is 72 m.

Radius of hemisphere, r = 12 cm

Volume of hemisphere = $\frac{2}{3}π\text{r}^3$

Radius of the cone = R cm

Height of the cone, h = 54 cm

Volume of cone = $\frac{1}{3}π\text{R}^2 \text{h}$

Since, hemisphere is melted and recasted into a cone, the volume remains the same.

$$\begin{aligned}\therefore \frac{1}{3}π\text{R}^2 \text{h} = \frac{2}{3}π\text{r}^3 \\[1em] \Rightarrow \frac{1}{3}\text{R}^2 \text{h} = \frac{2}{3}\text{r}^3 \\[1em] \Rightarrow \text{R}^2 = \frac{2 \times 3}{3 \times \text{h}} \times \text{r}^3 \\[1em] \Rightarrow \text{R}^2 = \frac{2 \times 3}{3 \times 54} \times 12^3 \\[1em] \Rightarrow \text{R}^2 = \frac{6}{162} \times 1728 \\[1em] \Rightarrow \text{R}^2 = \frac{10368}{162} \\[1em] \Rightarrow \text{R}^2 = 64 \\[1em] \Rightarrow \text{R} = \sqrt{64} \\[1em] \Rightarrow \text{R} = 8 \text{ cm.}\end{aligned}$$

Hence, the radius of the base of the cone is 8 cm.

Radius of larger spherical metallic ball, R = 3 cm

Radius of smaller spherical balls are 2.5 cm, 2 cm and r cm

Given,

A spherical metallic ball of radius 3 cm is melted and recast into three spherical balls.

∴ Volume of larger spherical ball = Volume of ball of radius 2.5 cm + Volume of ball of radius 2 cm + Volume of ball of radius r cm

$$\begin{aligned}\Rightarrow \frac{4}{3}π\text{R}^3 = \frac{4}{3}π \times 2.5^3 + \frac{4}{3}π \times 2^3 + \frac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \frac{4}{3}π\text{R}^3 = \frac{4}{3}π(2.5^3 + 2^3 + \text{r}^3) \\[1em] \Rightarrow \text{R}^3 = (2.5^3 + 2^3 + \text{r}^3) \\[1em] \Rightarrow 3^3 = 15.625 + 8 + \text{r}^3 \\[1em] \Rightarrow \text{r}^3 = 27 - 15.625 - 8 \\[1em] \Rightarrow \text{r}^3 = 3.375 \\[1em] \Rightarrow \text{r} = \sqrt[3]{3.375} \\[1em] \Rightarrow \text{r} = 1.5 \text{ cm.}\end{aligned}$$

Hence, the radius of the third ball is 1.5 cm.

(i) Radius of the metallic sphere, R = 6 cm

Height of the cylinder, h = 32 cm

Volume of cylinder = Volume of metallic sphere (As sphere is melted and formed into a cylinder)

$$\begin{aligned}\therefore π\text{r}^2\text{h} = \frac{4}{3}π\text{R}^3 \\[1em] \Rightarrow \text{r}^2 = \frac{4}{3} \times \frac{\text{R}^3}{\text{h}} \\[1em] \Rightarrow \text{r}^2 = \frac{4}{3} \times \frac{6^3}{32} \\[1em] \Rightarrow \text{r}^2 = \frac{4}{3} \times \frac{216}{32} \\[1em] \Rightarrow \text{r}^2 = \frac{864}{96} \\[1em] \Rightarrow \text{r}^2 = 9 \\[1em] \Rightarrow \text{r} = \sqrt{9} \\[1em] \Rightarrow \text{r} = 3 \text{ cm.}\end{aligned}$$

Hence, radius of the cylinder is 3 cm.

(ii) Curved surface area of cylinder = 2πrh

= 2 × 3.1 × 3 × 32

= 595.2 cm²

Hence, curved surface area of the cylinder is 595.2 cm².

Given,

Radius of spherical vessel (r) = 28 cm

Volume of spherical vessel (v) = $\frac{4}{3}πr^3$

Volume of oil in vessel = $\frac{3}{4}v$

Substituting values we get :

$$\begin{aligned}\Rightarrow v = \frac{4}{3} \times \frac{22}{7} \times 28^3 \\[1em] \Rightarrow \frac{3}{4}v = \frac{3}{4} \times \frac{4}{3} \times \frac{22}{7} \times 28^3 \\[1em] \Rightarrow \frac{3}{4}v = \frac{22}{7} \times 28^3.\end{aligned}$$

Radius of cylindrical vessel (R) = 21 cm

Let height of oil in cylindrical vessel be h cm.

Volume of oil = Volume of cylinder upto which oil is filled (πR²h)

$$\begin{aligned}\Rightarrow \frac{22}{7} \times 28^3 = \frac{22}{7} \times 21^2 \times h \\[1em] \Rightarrow 28^3 = 21^2 \times h \\[1em] \Rightarrow h = \frac{28^3}{21^2}\\[1em] \Rightarrow h = \frac{21952}{441} \\[1em] \Rightarrow h = 49.77 ≈ 50 \text{ cm}.\end{aligned}$$

Hence, height of the oil in the cylindrical vessel = 50 cm.

Given,

Internal radius of hemispherical bowl, R = $\frac{\text{diameter}}{2} = \frac{36}{2}$ = 18 cm

Radius of cylindrical bottles, r = 3 cm

Height of the cylindrical bottles, h = 6 cm

Let number of cylindrical bottles needed be n.

∴ Volume of hemispherical bowl = n × Volume of each cylindrical bottle

$$\begin{aligned}\Rightarrow \frac{2}{3}π\text{R}^3 = \text{n} \times π\text{r}^2\text{h} \\[1em] \Rightarrow \frac{2}{3}\text{R}^3 = \text{n} \times \text{r}^2\text{h} \\[1em] \Rightarrow \frac{2}{3} \times 18^3 = \text{n} \times 3^2 \times 6 \\[1em] \Rightarrow \frac{2}{3} \times 5832 = \text{n} \times 9 \times 6 \\[1em] \Rightarrow \frac{11664}{3} = \text{n} \times 54 \\[1em] \Rightarrow \text{n} = \frac{11664}{3 \times 54} \\[1em] \Rightarrow \text{n} = \frac{11664}{162} \\[1em] \Rightarrow \text{n} = 72\end{aligned}$$

Hence, 72 bottles are required to empty the bowl.

Radius of the sphere, r = $\frac{\text{diameter}}{2} = \frac{36}{2}$ = 18 cm

Radius of cylinder, R = $\frac{\text{diameter}}{2} = \frac{60}{2}$ = 30 cm

Let height of water raised be h cm.

Volume of water rise in cylinder = Volume of sphere

$$\begin{aligned}\Rightarrow π\text{R}^2\text{h} = \frac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \frac{4}{3}\text{r}^3 \\[1em] \Rightarrow \text{h} = \frac{4}{3} \times \frac{\text{r}^3}{\text{R}^2} \\[1em] \Rightarrow \text{h} = \frac{4}{3} \times \frac{18^3}{30^2} \\[1em] \Rightarrow \text{h} = \frac{4}{3} \times \frac{5832}{900} \\[1em] \Rightarrow \text{h} = \frac{23328}{2700} \\[1em] \Rightarrow \text{h} = 8.64 \text{ cm.}\end{aligned}$$

Hence, the height by which water level raised is 8.64 cm.

Given, radius of glass jar, R = 12.5 cm

Diameter of sphere = 15 cm

Radius of sphere, r = $\frac{\text{diameter}}{2} = \frac{15}{2}$ = 7.5 cm

When the sphere is removed from the jar, volume of water decreases.

Let h be the height by which water level decrease.

Volume of water decreased = Volume of sphere

$$\begin{aligned}\Rightarrow π\text{R}^2\text{h} = \frac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \frac{4}{3}\text{r}^3 \\[1em] \Rightarrow \text{h} = \frac{4}{3} \times \frac{\text{r}^3}{\text{R}^2} \\[1em] \Rightarrow \text{h} = \frac{4}{3} \times \frac{7.5^3}{12.5^2} \\[1em] \Rightarrow \text{h} = \frac{4}{3} \times \frac{421.875}{156.25} \\[1em] \Rightarrow \text{h} = \frac{1687.5}{468.75} \\[1em] \Rightarrow \text{h} = 3.6 \text{ cm.}\end{aligned}$$

Hence, the height by which water level decrease is 3.6 cm.

Let the radius of the sphere be r cm.

Radius of cylinder, R = 12 cm

SInce, a spherical iron ball is dropped into the tub and is fully immersed in it.

Height of water raised by 6.75 cm.

∴ h = 6.75 cm.

Volume of water rise in cylinder = Volume of sphere

$$\begin{aligned}\Rightarrow π\text{R}^2\text{h} = \frac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \frac{4}{3}\text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \frac{3}{4} \times \text{R}^2 \times \text{h} \\[1em] \Rightarrow \text{r}^3 = \frac{3}{4} \times 12^2 \times 6.75 \\[1em] \Rightarrow \text{r}^3 = \frac{3}{4} \times 144 \times 6.75 \\[1em] \Rightarrow \text{r}^3 = \frac{2916}{4} \\[1em] \Rightarrow \text{r}^3 = 729 \\[1em] \Rightarrow \text{r} = \sqrt[3]{729} \\[1em] \Rightarrow \text{r} = 9 \text{ cm.}\end{aligned}$$

Hence, the radius of the ball is 9 cm.

Given,

Diameter of lead spheres = 6 cm

Radius, r = $\frac{\text{diameter}}{2} = \frac{6}{2}$ = 3 cm

Diameter of beaker = 18 cm

Radius of beaker, R = $\frac{\text{diameter}}{2} = \frac{18}{2}$ = 9 cm

Increase in water level, h = 40 cm

Let n spheres are dropped.

∴ Volume of water increased in beaker = n × Volume of one sphere

$$\begin{aligned}\Rightarrow π\text{R}^2 \text{h} = \text{n} \times \frac{4}{3} π\text{r}^3 \\[1em] \text{Divide by π on both sides, we get:} \Rightarrow \text{R}^2 \text{h} = \text{n} \times \frac{4}{3} \text{r}^3 \\[1em] \Rightarrow 9^2 \times 40 = \text{n} \times \frac{4}{3} \times 3^3 \\[1em] \Rightarrow 81 \times 40 = \text{n} \times \frac{4}{3} \times 27 \\[1em] \Rightarrow 3240 = \text{n} \times 36 \\[1em] \Rightarrow \text{n} = \frac{3240}{36} \\[1em] \Rightarrow \text{n} = 90.\end{aligned}$$

Hence, the number of lead spheres dropped into the beaker are 90.

Radius of the top of the inverted cone, R = 2.5 cm

Height of the cone, H = 11 cm

Radius of lead shot, r = 0.25 cm

When lead shots are dropped into vessel, $\frac{2}{5}$ of water flows out.

∴ Volume of water flown out = $\frac{2}{5}$ Volume of cone.

$$\begin{aligned}= \frac{2}{5} \times \frac{1}{3}π\text{R}^2 \text{H} \\[1em] = \frac{2}{15} \times π\times 2.5^2 \times 11 \\[1em] = \frac{2}{15} \times π\times 6.25 \times 11 \\[1em] = \frac{137.5}{15} π\end{aligned}$$

Let the number of spheres be n.

∴ Volume of water flown out = n × Volume of each lead shot

$$\begin{aligned}\Rightarrow \frac{137.5}{15} π = \text{n} \times \frac{4}{3} \times π\text{r}^3 \\[1em] \text{Divide by π on both sides, we get:} \Rightarrow \frac{137.5}{15} = \text{n} \times \frac{4}{3} \times 0.25^3 \\[1em] \Rightarrow \frac{137.5}{15} = \text{n} \times \frac{4}{3} \times 0.015625 \\[1em] \Rightarrow \frac{137.5}{15} = \text{n} \times \frac{0.0625}{3} \\[1em] \Rightarrow \text{n} = \frac{137.5 \times 3}{0.0625 \times 15} \\[1em] \Rightarrow \text{n} = \frac{412.5}{0.9375} \\[1em] \Rightarrow \text{n} = 440.\end{aligned}$$

Hence, the number of lead shots are 440.

Given,

Height of the solid right circular cylinder, h = 37 cm

Internal radius of metallic spherical shell, r = $\frac{\text{diameter}}{2} = \frac{18}{2}$ = 9 cm

External radius of metallic spherical shell, R = $\frac{\text{diameter}}{2} = \frac{24}{2}$ = 12 cm

Let the radius of cylinder be a cm.

As, metallic spherical shell is recasted into right circular cylinder.

∴ Volume of spherical shell = Volume of cylinder

$$\begin{aligned}\Rightarrow \frac{4}{3} π(\text{R}^3 - \text{r}^3) = π\text{a}^2\text{h} \\[1em] \text{Divide by π on both sides, we get:} \Rightarrow \frac{4}{3} (\text{R}^3 - \text{r}^3) = \text{a}^2\text{h} \\[1em] \Rightarrow \frac{4}{3} \times (12^3 - 9^3) = \text{a}^2 \times 37 \\[1em] \Rightarrow \frac{4}{3} \times (1728 - 729) = \text{a}^2 \times 37 \\[1em] \Rightarrow \frac{4}{3} \times 999 = \text{a}^2 \times 37 \\[1em] \Rightarrow \text{a}^2 = \frac{4 \times 999}{3 \times 37} \\[1em] \Rightarrow \text{a}^2 = \frac{3996}{111} \\[1em] \Rightarrow \text{a}^2 = 36 \\[1em] \Rightarrow \text{a} = \sqrt{36} \\[1em] \Rightarrow \text{a} = 6 \text{ cm.}\end{aligned}$$

Diameter = 2a = 2 × 6 = 12 cm.

Hence, diameter of the base of the cylinder is 12 cm.

Given,

The solid wooden toy is in the shape of a right circular cone mounted on a hemisphere.

Radius of hemisphere, r = 4.2 cm

Total height, h = 10.2 cm

Height of conical part, H = 10.2 - 4.2 = 6 cm

(i) Volume of wood used in toy = Volume of cone + Volume of hemisphere

$$\begin{aligned}= \frac{1}{3} π\text{r}^2\text{H} + \frac{2}{3} π\text{r}^3 \\[1em] = \frac{1}{3} \times \frac{22}{7} \times 4.2^2 \times 6 + \frac{2}{3} \times \frac{22}{7} \times 4.2^3 \\[1em] = \frac{1}{3} \times \frac{22}{7} \times 17.64 \times 6 + \frac{2}{3} \times \frac{22}{7} \times 74.088 \\[1em] = \frac{2328.48}{21} + \frac{3259.872}{21} \\[1em] = \frac{5588.352}{21} \\[1em] = 266.11 \text{cm}^3\end{aligned}$$

Hence, the volume of wood used in the toy is 266.11 cm³.

(ii) By formula,

l² = r² + h²

⇒ l² = 4.2² + 6²

⇒ l² = 17.64 + 36

⇒ l² = 53.64

⇒ l = $\sqrt{53.64}$ = 7.32 cm

Total surface area of toy = Curved surface area of cone + curved surface area of hemisphere

$$\begin{aligned}= π\text{rl} + 2π\text{r}^2 \\[1em] = \frac{22}{7} \times 4.2 \times 7.32 + 2 \times \frac{22}{7} \times 4.2^2 \\[1em] = \frac{676.368}{7} + 2 \times \frac{22}{7} \times 17.64 \\[1em] = \frac{676.368}{7} + \frac{776.16}{7} \\[1em] = \frac{676.368 + 776.16}{7} \\[1em] = \frac{1452.528}{7} \\[1em] = 207.56 \text{cm}^2\end{aligned}$$

Hence, the total surface area of the toy is 207.56 cm².

Given,

Radius of cylindrical portion = Radius of hemispherical portion = r = 3.5 m

Height of cylinder, h = 7 m

(i) Area of internal surface = Surface area of cylinder + Surface area of hemisphere

= 2πrh + 2πr²

= 2πr(h + r)

= 2 × $\frac{22}{7}$ × 3.5(7 + 3.5)

= 2 × $\frac{22}{7}$ × 36.75

= 2 × 22 × 5.25

= 231 m²

Hence, the total area of the internal surface, excluding the base is 231 m².

(ii) Internal volume of container = Volume of hemisphere + Volume of cylinder

$$\begin{aligned}= \frac{2}{3} π\text{r}^3 + π\text{r}^2\text{h} \\[1em] = \frac{2}{3} \times \frac{22}{7} \times 3.5^3 + \frac{22}{7} \times 3.5^2 \times 7 \\[1em] = \frac{2}{3} \times \frac{22}{7} \times 42.875 + \frac{22}{7} \times 12.25 \times 7 \\[1em] = \frac{2}{3} \times 22 \times 6.125 + 269.5 \\[1em] = 89.83 + 269.5 \\[1em] = 359.33 \\[1em] = 359 \frac{1}{3} \text{m}^3\end{aligned}$$

Hence, the volume of container is 359.33 m³.

Given, common radius, r = 3.5 cm

Height of cylinder, H = 6.5 cm

Height of hemisphere = radius of hemisphere = 3.5 cm

Height of cone, h = Total height of the solid - height of cylinder - height of hemisphere = 12.8 - 6.5 - 3.5 = 2.8 cm

Volume of solid = Volume of cone + Volume of cylinder + Volume of hemisphere

$$\begin{aligned}= \frac{1}{3} π\text{r}^2\text{h} + π\text{r}^2\text{H} + \frac{2}{3} π\text{r}^3 \\[1em] = π\text{r}^2 \Big(\frac{1}{3} \text{h} + \text{H} + \frac{2}{3} \text{r}\Big) \\[1em] = \frac{22}{7} \times (3.5)^2 \times \Big(\frac{1}{3} \times 2.8 + 6.5 + \frac{2}{3} \times 3.5 \Big) \\[1em] = \frac{22}{7} \times 12.25 \times \Big(\frac{2.8}{3} + 6.5 + \frac{7}{3} \Big) \\[1em] = 22 \times 1.75 \times \Big(\frac{2.8 + 19.5 + 7}{3} \Big) \\[1em] = 38.5 \times \frac{29.3}{3} \\[1em] = \frac{1128.05}{3} \\[1em] = 376.02 \approx 376 \text{ cm}^3.\end{aligned}$$

Hence, volume of solid is 376 cm³.

Given, common radius, r = 7 cm

Height of cone, h = 4 cm

Height of cylinder, H = 4 cm

Volume of solid = Volume of cone + Volume of cylinder + Volume of hemisphere

$$\begin{aligned}= \frac{1}{3} π\text{r}^2\text{h} + π\text{r}^2\text{H} + \frac{2}{3} π\text{r}^3 \\[1em] = π\text{r}^2 \Big(\frac{1}{3} \text{h} + \text{H} + \frac{2}{3} \text{r}\Big) \\[1em] = \frac{22}{7} \times 7^2 \Big(\frac{1}{3} \times 4 + 4 + \frac{2}{3} \times 7 \Big) \\[1em] = \frac{22}{7} \times 49 \Big(\frac{4}{3} + 4 + \frac{14}{3} \Big) \\[1em] = 22 \times 7 \Big(\frac{4 + 12 + 14}{3} \Big) \\[1em] = 154 \times \frac{30}{3} \\[1em] = 154 \times 10 \\[1em] = 1540 \text{ cm}^3.\end{aligned}$$

Hence, volume of solid is 1540 cm³.

Radius of hemisphere, r = 7 cm

Height of cone, h = 4 cm

Radius of cylinder, R = 14 cm

Let the rise in water level be x cm.

∴ Volume of water that rises by x cm in the cylindrical container = Volume of hemisphere submerged + Volume of cone submerged

$$\begin{aligned}\Rightarrow π\text{R}^2\text{x} = \frac{2}{3} π\text{r}^3 + \frac{1}{3} π\text{r}^2\text{h} \\[1em] \Rightarrow \text{R}^2\text{x} = \frac{1}{3} (2\text{r}^3 + \text{r}^2\text{h}) \\[1em] \Rightarrow 3 \times 14^2\text{x} = 2 \times 7^3 + 7^2 \times 4 \\[1em] \Rightarrow 3 \times 196 \text{x} = 2 \times 343 + 49 \times 4 \\[1em] \Rightarrow 588 \text{x} = 686 + 196 \\[1em] \Rightarrow 588 \text{x} = 882 \\[1em] \Rightarrow \text{x} = \frac{882}{588} \\[1em] \Rightarrow \text{x} = 1.5 \text{ cm.}\end{aligned}$$

Hence, rise in water level is 1.5 cm.

From figure,

Radius of cylinder = Radius of hemisphere = r = $\frac{\text{diameter}}{2} = \frac{7}{2}$ = 3.5 cm

Height of cylinder, h = Total height - (2 × Radius of hemisphere)

= 19 - 2 × 3.5

= 19 - 7

= 12 cm.

Total volume of solid = 2 × Volume of hemisphere + Volume of cylinder

$$\begin{aligned}= 2 \times \frac{2}{3} π\text{r}^3 + π\text{r}^2\text{h} \\[1em] = π\text{r}^2 (\frac{4}{3} \text{r} + \text{h}) \\[1em] = \frac{22}{7} \times 3.5^2 (\frac{4}{3} \times 3.5 + 12) \\[1em] = \frac{22}{7} \times 12.25 (\frac{14}{3} + 12) \\[1em] = \frac{269.5}{7} \times \frac{14 + 36}{3} \\[1em] = \frac{269.5}{7} \times \frac{50}{3} \\[1em] = \frac{13475}{21} \\[1em] = \frac{1925}{3} \\[1em] = 641.67 \text{ cm}^3.\end{aligned}$$

Surface area of solid = 2 × 2πr² + 2πrh

= πr(4r + 2h)

$$\begin{aligned}= \frac{22}{7} \times 3.5 (4 \times 3.5 + 2 \times 12) \\[1em] = 22 \times 0.5 (14 + 24) \\[1em] = 11 \times 38 \\[1em] = 418 \text{ cm}^2.\end{aligned}$$

Hence, the volume of solid is $641\frac{2}{3}$ cm³ and surface area of solid is 418 cm².

Given,

Radius, r = 3 cm

Height of cone, h = 3 cm

Height of cylinder, H = 7 cm

From figure,

Volume of remaining solid = Volume of cylinder - Volume of cone - Volume of hemisphere

∴ Volume of remaining solid = πr²H - $\frac{1}{3}$ πr²h - $\frac{2}{3}$ πr³

$$\begin{aligned}= π\text{r}^2(\text{H} - \frac{1}{3}\text{h} - \frac{2}{3} \text{r}) \\[1em] = \frac{22}{7} \times 3^2(7 - \frac{1}{3} \times 3 - \frac{2}{3} \times 3) \\[1em] = \frac{22}{7} \times 9(7 - 1 - 2) \\[1em] = \frac{22}{7} \times 9 \times 4 \\[1em] = \frac{792}{7} \\[1em] = 113.14 \approx 113 \text{ cm}^3.\end{aligned}$$

Hence, the volume of the remaining solid is 113 cm³.

Let the common radius of shapes be r and height be h.

Ratio of their volumes = Volume of cylinder : Volume of hemisphere : Volume of cone

Since, a cylinder, a hemisphere and a cone have equal base diameters and have the same height.

⇒ They share a same radius r.

For hemisphere, the height is the distance from the centre of its base to its heighest point, which is equal to the radius.

⇒ h = r

$$\begin{aligned}= π\text{r}^2\text{h} : \frac{2}{3} π\text{r}^3 : \frac{1}{3} π\text{r}^2\text{h} \\[1em] = π\text{r}^2\text{h} : \frac{2}{3} π\text{r}^2 \text{r} : \frac{1}{3} π\text{r}^2\text{h} \\[1em] = π\text{r}^2\text{h} : \frac{2}{3} π\text{r}^2 \text{h} : \frac{1}{3} π\text{r}^2\text{h} \\[1em] = 1 : \frac{2}{3} : \frac{1}{3}\end{aligned}$$

On multiplying by 3, ratio = 3 : 2 : 1

Hence, proved that their volumes are in the ratio 3 : 2 : 1.

Let original radius be r units and new radius be R units.

Given, radius of a sphere is doubled.

∴ R = 2 × r = 2r

Let the original surface area be s and new surface area be S.

By formula,

Percentage increase in surface area = $\frac{\text{S - s}}{\text{s}} \times 100$

$$\begin{aligned}= \frac{4 π\text{R}^2 - 4 π\text{r}^2}{4π\text{r}^2} \times 100 \\[1em] = \frac{4 π(\text{R}^2 - \text{r}^2)}{4π\text{r}^2} \times 100 \\[1em] = \frac{(\text{R}^2 - \text{r}^2)}{\text{r}^2} \times 100 \\[1em] = \frac{(\text{(2r)}^2 - \text{r}^2)}{\text{r}^2} \times 100 \\[1em] = \frac{(4\text{r}^2 - \text{r}^2)}{\text{r}^2} \times 100 \\[1em] = \frac{3\text{r}^2}{\text{r}^2} \times 100 \\[1em] = 300 %.\end{aligned}$$

Hence, percentage increase in surface area is 300%.

Let original radius be r units and new radius be R units.

Given, radius of a sphere is increased by 50%.

∴ R = r + $\frac{50}{100}$ × r = r + $\frac{1}{2}$ r = r + 0.5 r = 1.5 r

Let the original volume be v and new volume be V.

By formula,

Percentage increase in volume = $\frac{\text{V - v}}{\text{v}} \times 100$

$$\begin{aligned}= \frac{\frac{4}{3} π\text{R}^3 - \frac{4}{3} π\text{r}^3}{\frac{4}{3} π\text{r}^3} \times 100 \\[1em] = \frac{\frac{4}{3} π(\text{R}^3 - \text{r}^3)}{\frac{4}{3} π\text{r}^3} \times 100 \\[1em] = \frac{(\text{R}^3 - \text{r}^3)}{\text{r}^3} \times 100 \\[1em] = \frac{(\text{(1.5r)}^3 - \text{r}^3)}{\text{r}^3} \times 100 \\[1em] = \frac{(3.375\text{r}^3 - \text{r}^3)}{\text{r}^3} \times 100 \\[1em] = \frac{2.375\text{r}^3}{\text{r}^3} \times 100 \\[1em] = 237.5 %\end{aligned}$$

Hence, percentage increase in volume is 237.5%.

Given,

Ratio of the volumes of the two spheres is 1 : 8

$$\begin{aligned}\therefore \frac{\text{Volume of sphere 1}}{\text{Volume of sphere 2}} = \frac{1}{8} \\[1em] \Rightarrow \frac{\frac{4}{3} π\text{R}^3}{\frac{4}{3} π\text{r}^3} = \frac{1}{8} \\[1em] \Rightarrow \frac{\text{R}^3}{\text{r}^3} = \frac{1^3}{2^3} \\[1em] \Rightarrow \frac{\text{R}}{\text{r}} = \frac{1}{2}\end{aligned}$$

Surface area of sphere = 4πr²

$$\begin{aligned}\therefore \frac{\text{Surface area of sphere 1}}{\text{Surface area of sphere 2}} = \frac{4π\text{R}^2}{4π\text{r}^2} \\[1em] = \Big(\frac{\text{R}}{\text{r}}\Big)^2 \\[1em] = \Big(\frac{1}{2}\Big)^2 \\[1em] = \frac{1}{4}.\end{aligned}$$

Hence, the ratio of the surface areas of two spheres is 1 : 4.

From figure,

Radius of cylindrical block = radius of hemispheres = r = 3.5 cm

Height of cylindrical block = h = 14 cm.

By formula,

Total surface area of cylinder = 2πr(h + r)

Total surface area of a hemisphere = 3πr²

Total surface area = Total surface area of cylinder + Total surface area of 2 hemispheres

= 2πr(h + r) + 2 × 3πr²

= 2πrh + 2πr² + 6πr²

= 2πrh + 8πr²

= 2πr(h + 4r).

Substituting values, we get :

$$\begin{aligned}\text{Total surface area} = 2 \times \frac{22}{7} \times 3.5 \times (14 + 4 \times 3.5) \\[1em] = 2 \times 22 \times 0.5 \times (14 + 14) \\[1em] = 22 \times 28 \\[1em] = 616 \text{ cm}^2.\end{aligned}$$

Hence, the total surface area = 616 cm².

Given,

External radius of hollow sphere (R) = 5 cm, internal radius of hollow sphere (r) = 3 cm, height of cone (h) = 8 cm.

By formula,

Volume of metal in the hollow sphere V = $\frac{4}{3} \pi (R^3 − r^3)$

Substituting values we get :

$$\begin{aligned}\Rightarrow V_{\text{sphere}}=\frac{4}{3} \times \pi \times \Big(5^{3}-3^{3}\Big) \\[1em] \Rightarrow V_{\text{sphere}} =\frac{4}{3} \times \pi \times (125-27) \\[1em] \Rightarrow V_{\text{sphere}} =\frac{4}{3} \times \pi \times 98 \\[1em] \Rightarrow V_{\text{sphere}} =\frac{392}{3} \pi\end{aligned}$$

By formula,

Volume of cone = $\frac{1}{3} \pi \times \text{radius}^2h$

Let radius of cone formed be m cm.

$V_{\text{cone}}=\frac{1}{3}\pi m^{2}\times 8=\frac{8}{3}\pi m^{2}$

Since, the sphere is melted to form, the cone the volume of both objects will be equal.

$$\begin{aligned}\Rightarrow \frac{392}{3}\pi=\frac{8}{3}\pi m^{2} \\[1em] \Rightarrow 392 = 8m^{2} \\[1em] \Rightarrow m^2 = \frac{392}{8} \\[1em] \Rightarrow m^{2} = 49 \\[1em] \Rightarrow m = 7 \text{ cm}.\end{aligned}$$

Hence, the radius of the cone formed = 7 cm.

Radius of cylinder, r = 10.5 cm

Height of cylinder, h = 60 cm

Volume of cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 10.5^2 \times 60 \\[1em] = \frac{22}{7} \times 110.25 \times 60 \\[1em] = \frac{145530}{7} \\[1em] = 20790 \text{ cm}^3\end{aligned}$$

Given, 1 cm³ of metal weighs 5 gm.

Toatl weight = 5 × 20790 = 103950 g = 103950 × $\frac{1}{1000}$ = 103.95 kg.

Hence, option 4 is the correct option.

Let radius of cylinder be r cm.

Given, diameter = height = h cm

Radius = $\frac{\text{diameter}}{2} = \frac{\text{h}}{2}$

Total surface area of cylinder = 2πr(r + h)

$$\begin{aligned}= 2 π \frac{\text{h}}{2} \Big(\frac{\text{h}}{2} + \text{h}\Big) \\[1em] = π \text{h} \times \Big(\frac{\text{h} + 2 \text{h}}{2}\Big) \\[1em] = π \text{h} \times \frac{3\text{h}}{2} \\[1em] = \frac{3}{2} π \text{h}^2\end{aligned}$$

Hence, option 2 is the correct option.

Length of the roller, h = 100 cm

Radius of the roller, r = 14 cm

Roller is in the shape of a cylinder.

∴ Curved surface area of roller = 2πrh

= $2 \times \frac{22}{7} \times 14 \times 100$

= 2 × 22 × 2 × 100

= 8800 cm².

Hence, option 4 is the correct option.

Let radius of cylinder, r = 2a and height, h = 3a.

Given,

Volume of cylinder = 1617 cm³

By formula,

Volume of cylinder = πr²h

$$\begin{aligned}\Rightarrow 1617 = \frac{22}{7} \times (\text{2a})^2 \times \text{3a} \\[1em] \Rightarrow 1617 \times \frac{7}{22} = 4(\text{a})^2 \times \text{3a} \\[1em] \Rightarrow \frac{11319}{22} = 12\text{a}^3 \\[1em] \Rightarrow \frac{514.5}{12} = \text{a}^3 \\[1em] \Rightarrow \text{a}^3 = 42.875 \\[1em] \Rightarrow \text{a} = \sqrt[3]{42.875} \\[1em] \Rightarrow \text{a} = 3.5\end{aligned}$$

∴ Radius = 2a = 2 × 3.5 = 7 cm

Height = 3a = 3 × 3.5 = 10.5 cm

Total surface area of cylinder = 2πr(r + h)

$$\begin{aligned}= 2 \times \frac{22}{7} \times 7 (7 + 10.5) \\[1em] = 2 \times 22 \times 17.5 \\[1em] = 770 \text{cm}^2.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

Total surface area of cylinder = 231 cm²

⇒ 2πr² + 2πrh = 231 ...(1)

Curved surface area = $\frac{2}{3}$ (Total surface area)

= $\frac{2}{3} \times 231 = 2 \times 77$ = 154 cm²

By formula,

Curved surface area of cylinder = 2πrh

⇒ 2πrh = 154 ...(2)

Substituting eq.(2) in eq.(1), we have :

⇒ 2πr² + 154 = 231

⇒ 2πr² = 231 - 154

⇒ 2πr² = 77

$$\begin{aligned}\Rightarrow \text{r}^2 = \frac{77}{2π} \\[1em] \Rightarrow \text{r}^2 = \frac{77}{2 \times \frac{22}{7}} \\[1em] \Rightarrow \text{r}^2 = \frac{77 \times 7}{2 \times 22} \\[1em] \Rightarrow \text{r}^2 = \frac{539}{44} \\[1em] \Rightarrow \text{r}^2 = 12.25 \\[1em] \Rightarrow \text{r} = \sqrt{12.25} \\[1em] \Rightarrow \text{r} = 3.5 \text{ cm.}\end{aligned}$$

Substituting value of r in eq.(2), we get:

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times 3.5 \times \text{h} = 154 \\[1em] \Rightarrow 2 \times 22 \times 0.5 \times \text{h} = 154 \\[1em] \Rightarrow 22\text{h} = 154 \\[1em] \Rightarrow \text{h} = \frac{154}{22} \\[1em] \Rightarrow \text{h} = 7 \text{ cm.}\end{aligned}$$

Volume of cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 3.5^2 \times 7 \\[1em] = 22 \times 12.25 \\[1em] = 269.5 \text{ cm}^3.\end{aligned}$$

Hence, option 1 is the correct option.

Let radius be r m and height be h m.

Given,

r + h = 37 m ...(1)

Totals surface area of cylinder = 1628 m²

⇒ 2πr(r + h) = 1628

⇒ 2πr × 37 = 1628

$$\begin{aligned}\Rightarrow 74 \times \frac{22}{7} \times \text{r} = 1628 \\[1em] \Rightarrow \frac{1628}{7} \times \text{r} = 1628 \\[1em] \Rightarrow \text{r} = \frac{1628 \times 7}{1628} \\[1em] \Rightarrow \text{r} = 7 \text{ m.}\end{aligned}$$

Substituting value of r in eq.(1), we have:

⇒ r + h = 37

⇒ 7 + h = 37

⇒ h = 37 - 7

⇒ h = 30 m.

Volume of cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 7^2 \times 30 \\[1em] = \frac{22}{7} \times 49 \times 30 \\[1em] = 22 \times 7 \times 30 \\[1em] = 4620 \text{ m}^3.\end{aligned}$$

Hence, option 2 is the correct option.

Given, curved surface area of cylinder = 4400 cm²

We know that curved surface area of cylinder = 2πrh

∴ 2πrh = 4400 .....(1)

Given, circumference of base = 110 cm

We know that circumference = 2πr

∴ 2πr = 110 .....(2)

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times \text{r} = 110 \\[1em] \Rightarrow \text{r} = \frac{110 \times 7}{22 \times 2} \\[1em] \Rightarrow \text{r} = \frac{770}{44} \\[1em] \Rightarrow \text{r} = 17.5 \text{ cm.}\end{aligned}$$

Dividing eq.(1) by (2), we get:

$\frac{2π\text{rh}}{2π\text{r}} = \frac{4400}{110}$

⇒ h = 40 cm

Volume of cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 17.5^2 \times 40 \\[1em] = \frac{22}{7} \times 306.25 \times 40 \\[1em] = \frac{269500}{7} \\[1em] = 38500 \text{ cm}^3.\end{aligned}$$

Hence, option 2 is the correct option.

In first case :

Height of cylinder (h) = 7 cm

Let radius be r cm

⇒ 2πr = 11

⇒ r = $\frac{11}{2π}$

In second case :

Height of cylinder (H) = 11 cm

Let radius be R cm

⇒ 2πR = 7

⇒ R = $\frac{7}{2π}$

$$\begin{aligned}\therefore \frac{\text{CSA of 1st cylinder}}{\text{CSA of 2nd cylinder}} = \frac{2πrh}{2πRH} \\[1em] = \frac{rh}{RH} \\[1em] = \frac{\frac{11}{2π} \times 7}{\frac{7}{2π} \times 11} \\[1em] = \frac{\frac{77}{2π}}{\frac{77}{2π}} \\[1em] = \frac{1}{1} \\[1em] = 1 : 1.\end{aligned}$$

Hence, Option 1 is the correct option.

For cylinder 1,

Height of cylinder, h = a₂

Radius of cylinder be r cm

Circumference of base = 2πr = a₁

$$\begin{aligned}\Rightarrow 2 \times π \times \text{r} = \text{a}_1 \\[1em] \Rightarrow \text{r} = \frac{\text{a}_1}{2π}\end{aligned}$$

Volume of cylinder 1, V₁ = πr²h

$$\begin{aligned}= π \times \Big(\frac{\text{a}_1}{2π}\Big)^2 \times \text{a}_2 \\[1em] = π \times \frac{\text{a}_1^2}{4π^2} \times \text{a}_2 \\[1em] = \frac{\text{a}_1^2 \text{a}_2 }{4π}\end{aligned}$$

For cylinder 2,

Height of cylinder, H = a₁

Radius of cylinder be R

Circumference of base = 2πR = a₂

$$\begin{aligned}\Rightarrow 2 \times π \times \text{R} = \text{a}_2 \\[1em] \Rightarrow \text{R} = \frac{\text{a}_2}{2π} \\[1em] \Rightarrow \text{R} = \frac{\text{a}_2}{2π}\end{aligned}$$

Volume of cylinder 2, V₂ = πR²H

$$\begin{aligned}= π \times \Big(\frac{\text{a}_2}{2π}\Big)^2 \times \text{a}_1 \\[1em] = π \times \frac{\text{a}_2^2}{4π^2} \times \text{a}_1 \\[1em] = \frac{\text{a}_2^2 \text{a}_1}{4π}\end{aligned}$$

Ratio of the volumes of the two cylinders:

$$\begin{aligned}\Rightarrow \frac{\text{Volume of cylinder 1}}{\text{Volume of cylinder 2}} = \frac{\frac{\text{a}_1^2 \text{a}_2 }{4π}}{\frac{\text{a}_2^2 \text{a}_1}{4π}} \\[1em] \Rightarrow \frac{\text{V}_1}{\text{V}_2} = \frac{\text{a}_1^2 \text{a}_2 }{4π} \times \frac{4π}{\text{a}_2^2 \text{a}_1} \\[1em] \Rightarrow \frac{\text{V}_1}{\text{V}_2} = \frac{\text{a}_1}{\text{a}_2} \\[1em] \Rightarrow \text{V}_1 \text{a}_2 = \text{V}_2 \text{a}_1\end{aligned}$$

Hence, option 3 is the correct option.

Let radius and heights of two cylinders be r, h and R, H.

Given,

$\frac{\text{h}}{\text{H}} = \frac{1}{2}$

Volume of cylinder 1 = v

Volume of cylinder 2 = V

⇒ v = V

$$\begin{aligned}\Rightarrow π\text{r}^2\text{h} = π\text{R}^2\text{H} \\[1em] \text{Divide by π on both sides, we get:} \\[1em] \Rightarrow \text{r}^2 = \text{R}^2 \times \frac{\text{H}}{\text{h}} \\[1em] \Rightarrow \frac{\text{r}^2}{\text{R}^2} = \frac{2}{1} \\[1em] \Rightarrow \Big(\frac{\text{r}}{\text{R}}\Big)^2 = \frac{2}{1} \\[1em] \Rightarrow \frac{\text{r}}{\text{R}} = \sqrt{\frac{2}{1}} \\[1em] \Rightarrow \frac{\text{r}}{\text{R}} = \frac{\sqrt{2}}{1} \\[1em]\end{aligned}$$

∴ r : R = $\sqrt{2}$ : 1

Hence, option 2 is the correct option.

Total surface area = 616 cm²

⇒ 2πr(h + r) = 616

⇒ πr(h + r) = $\frac{616}{2}$

⇒ πr(h + r) = 308 ....(1)

Ratio between its curved surface area and total surface area = 1 : 2

$$\begin{aligned}\Rightarrow \frac{\text{Curved surface area}}{\text{Total surface area}} = \frac{1}{2} \\[1em] \Rightarrow \frac{2π\text{rh}}{2π\text{r(h + r)}} = \frac{1}{2} \\[1em] \Rightarrow \frac{\text{h}}{\text{(h + r)}} = \frac{1}{2} \\[1em]\end{aligned}$$

⇒ 2h = h + r

⇒ 2h - h = r

⇒ h = r

Substituting value of h in eq.(1), we get:

⇒ πr(r + r) = 308

⇒ πr × 2r = 308

⇒ 2πr² = 308

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times \text{r}^2 = 308 \\[1em] \Rightarrow \frac{44}{7} \times \text{r}^2 = 308 \\[1em] \Rightarrow \text{r}^2 = \frac{308 \times 7}{44} \\[1em] \Rightarrow \text{r}^2 = \frac{2156}{44} \\[1em] \Rightarrow \text{r}^2 = 49 \\[1em] \Rightarrow \text{r} = \sqrt{49} \\[1em] \Rightarrow \text{r} = 7 \text{ cm.}\end{aligned}$$

⇒ h = 7 cm

Volume of cylinder = πr²h

= $\frac{22}{7}$ × 7² × 7

= 22 × 49

= 1078 cm³.

Hence, option 2 is the correct option.

Given,

Radius of coin, r = $\frac{\text{diameter}}{2} = \frac{1.5}{2}$ = 0.75 cm

Height of coin, h = 0.2 cm

Radius of cylinder, R = $\frac{\text{diameter}}{2} = \frac{4.5}{2}$ = 2.25 cm

Height of cylinder, H = 10 cm

Let no. of coins required to be melted to form cylinder be n.

Volume of cylinder = n × Volume of each coin

∴ πR²H = n × πr²h

$$\begin{aligned}\Rightarrow \text{n} = \frac{π\text{R}^2\text{H}}{π\text{r}^2\text{h}} \\[1em] \Rightarrow \text{n} = \frac{2.25^2 \times 10}{0.75^2 \times 0.2} \\[1em] \Rightarrow \text{n} = \frac{5.0625 \times 10}{0.5625 \times 0.2} \\[1em] \Rightarrow \text{n} = \frac{50.625}{0.1125} \\[1em] \Rightarrow \text{n} = 450\end{aligned}$$

Hence, option 2 is the correct option.

For cylinder, rolled along its length:

Height of cylinder, h = 5 cm

Radius of cylinder be r cm

Circumference of base = 2πr = 12

$$\begin{aligned}\Rightarrow 2 \times π \times \text{r} = 12 \\[1em] \Rightarrow \text{r} = \frac{12}{2π} \\[1em] \Rightarrow \text{r} = \frac{6}{π} \text{ cm.}\end{aligned}$$

Volume of cylinder = πr²h

$$\begin{aligned}= π \times \Big(\frac{6}{π}\Big)^2 \times 5 \\[1em] = π \times \Big(\frac{36}{π^2}\Big) \times 5 \\[1em] = \frac{180}{π} \text{ cm}^3\end{aligned}$$

Hence, option 4 is the correct option.

Given,

r : R = 2 : 3

Let r = 2x and R = 3x

h : H = 5 : 3

Let h = 5y and H = 3y

$$\begin{aligned}\frac{\text{CSA of 1st cylinder}}{\text{CSA of 2nd cylinder}} = \frac{2π\text{rh}}{2π\text{RH}} \\[1em] = \frac{\text{rh}}{\text{RH}} \\[1em] = \frac{\text{2x} \times \text{5y}}{\text{3x} \times {\text{3y}}} \\[1em] = \frac{10\text{xy}}{9\text{xy}} \\[1em] = \frac{10}{9} \\[1em] = 10 : 9\end{aligned}$$

Hence, option 3 is the correct option.

For old cylinder,

Let height = h and Radius = r

So, for new cylinder,

Height = h and radius = $\frac{\text{r}}{2}$

We know that volume of cylinder = π × radius² × height

∴ Volume of old cylinder = πr²h

and Volume of new cylinder = π $(\frac{\text{r}}{2})^2$ h

$$\begin{aligned}\therefore \frac{\text{Volume of new cylinder}}{\text{Volume of old cylinder}} = \frac{π(\frac{\text{r}}{2})^2\text{h}}{π\text{r}^2\text{h}} \\[1em] = \frac{\frac{\text{r}^2}{4}}{\text{r}^2} \\[1em] = \frac{\text{r}^2}{4\text{r}^2} \\[1em] = \frac{1}{4}\end{aligned}$$

Hence, option 2 is the correct option.

On changing the shape of a container, its volume remains same.

Hence, option 3 is the correct option.

Given,

For cylinder 1,

Radius, r = 15 cm

Height, h = 35 cm

For cylinder 2,

Radius, R = 10 cm

Height, H = 15 cm

By formula, Volume of cylinder = πr²h

Volume of cylinder 1 = v

$$\begin{aligned}= \frac{22}{7} \times 15^2 \times 35 \\[1em] = 22 \times 225 \times 5 \\[1em] = 24750 \text{ cm}^3.\end{aligned}$$

Volume of cylinder 2 = V

$$\begin{aligned}= \frac{22}{7} \times 10^2 \times 15 \\[1em] = \frac{22}{7} \times 100 \times 15 \\[1em] = \frac{33000}{7} \text{ cm}^3.\end{aligned}$$

Given, water from cylinder 1 and 2 is poured into cylinder 3.

Volume of cylinder 3 = Volume of cylinder 1 + Volume of cylinder 2

= 24750 + $\frac{33000}{7}$

= $\frac{173250 + 33000}{7} = \frac{206250}{7} \text{ cm}^3$

For cylinder 3,

Radius be a cm

Height = 15 cm

Volume of cylinder 3 = πa² × 15

$$\begin{aligned}\Rightarrow \frac{206250}{7} = \frac{22}{7} \times \text{a}^2 \times 15 \\[1em] \Rightarrow \frac{206250}{7} \times \frac{7}{22 \times 15} = \text{a}^2 \\[1em] \Rightarrow \frac{206250}{330} = \text{a}^2 \\[1em] \Rightarrow \text{a}^2 = 625 \\[1em] \Rightarrow \text{a} = \sqrt{625} \\[1em] \Rightarrow \text{a} = 25 \text{ cm.}\end{aligned}$$

Hence, option 4 is the correct option.

Length of the roller (h) = 63 cm

Let external radius be R cm and internal radius be r cm.

Girth of the roller = Circumference of roller = 440 cm

⇒ 2πR = 440

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \text{R} = 440 \\[1em] \Rightarrow \frac{44}{7} \text{R} = 440 \\[1em] \Rightarrow \text{R} = 440 \times \frac{7}{44} \\[1em] \Rightarrow \text{R} = \frac{3080}{44} \\[1em] \Rightarrow \text{R} = 70 \text{ cm.}\end{aligned}$$

Thickness = External radius (R) - internal radius (r)

⇒ 4 = 70 - r

⇒ r = 70 - 4 = 66 cm

External volume = πR²h

$$\begin{aligned}= \frac{22}{7} \times (70)^2 \times 63 \\[1em] = 22 \times 4900 \times 9 \\[1em] = 970200\end{aligned}$$

Internal volume = πr²h

$$\begin{aligned}= \frac{22}{7} \times (66)^2 \times 63 \\[1em] = 22 \times 4356 \times 9 \\[1em] = 862488\end{aligned}$$

Volume of iron = External volume - Internal volume

= 970200 - 862488

= 107712 cm³

Hence, option 4 is the correct option.

Given, radius = 3r and height(h) = 3r

Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times π \times (3\text{r})^2 \times 3\text{r} \\[1em] = π \times 9\text{r}^2 \times \text{r} \\[1em] = 9 π\text{r}^3\end{aligned}$$

Hence, option 4 is the correct option.

Given,

Height of cone (h) = Radius of cone (r) = a cm (let)

Given,

Volume = 9702 cm³

$$\begin{aligned}\therefore \frac{1}{3}πr^2h = 9702 \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times a^2 \times a = 9702 \\[1em] \Rightarrow a^3 = \frac{9702 \times 7 \times 3}{22} \\[1em] \Rightarrow a^3 = 441 \times 21 \\[1em] \Rightarrow a^3 = 9261 \\[1em] \Rightarrow a = \sqrt[3]{9261} = 21 \text{ cm}.\end{aligned}$$

Diameter = 2 × radius = 2 × 21 = 42 cm.

Hence, option 3 is the correct option.

Let the diameters of the two right circular cones be d₁ and d₂, their heights be h₁ and h₂ and their radius be r₁ and r₂.

Given,

⇒ d₁ : d₂ = 3 : 7

Let d₁ and d₂ be 3x and 7x respectively.

⇒ r₁ = $\frac{3x}{2}$

⇒ r₂ = $\frac{7x}{2}$

Given,

⇒ h₁ : h₂ = 14 : 9

Let h₁ and h₂ be 14a and 9a respectively.

⇒ h₁ : h₂ = 14a : 9a

By formula,

Volume of cone = $\frac{1}{3}\pi r^2h$

$$\begin{aligned}\Rightarrow V_1 : V_2 = \frac{1}{3}\pi r_1^2h_1:\frac{1}{3}\pi r_2^2h_2 \\[1em] = \frac{\frac{1}{3}\pi r_1^2h_1}{\frac{1}{3} \pi r_2^2h_2} \\[1em] = \frac{r_1^2h_1}{r_2^2h_2} \\[1em] = \frac{\Big(\frac{3x}{2}\Big)^2 \times 14a}{\Big(\frac{7x}{2}\Big)^2 \times 9a} \\[1em] = \frac{\Big(\frac{9x^2}{4}\Big) \times 14}{\Big(\frac{49x^2}{4}\Big) \times 9} \\[1em] = \frac{9x^2 \times 14 \times 4}{49x^2 \times 9 \times 4} \\[1em] = \frac{9 \times 14}{49 \times 9} \\[1em] = \frac{14}{49} \\[1em] = \frac{2}{7} = 2 : 7.\end{aligned}$$

Hence, option 2 is the correct option.

Let Volume of cones be V and v respectively.

⇒ V : v = 1 : 4

Diameters of cones be D and d respectively.

⇒ D = 4b and d = 5b

Radius of 1st cone = $\frac{\text{diameter}}{2} = \frac{4\text{b}}{2}$ = 2b

Radius of 2nd cone = $\frac{\text{diameter}}{2} = \frac{5\text{b}}{2}$ = 2.5b

Let the height of cones be h and H respectively.

By formula,

Volume of cone = $\frac{1}{3}π \text{r}^2 \text{h}$

$$\begin{aligned}\therefore \frac{\text{V}}{\text{v}} = \frac{\frac{1}{3}π (\text{2b})^2 \text{h}}{\frac{1}{3}π (\text{2.5b})^2 \text{H}} \\[1em] \Rightarrow \frac{1}{4} = \frac{\frac{1}{3}π \times \text{4b}^2 \text{h}}{\frac{1}{3}π \times 6.25\text{b}^2 \text{H}} \\[1em] \Rightarrow \frac{1}{4} = \frac{4 \text{h}}{6.25\text{H}} \\[1em] \Rightarrow \frac{\text{H}}{\text{h}} = \frac{4 \times 4}{6.25} \\[1em] \Rightarrow \frac{\text{H}}{\text{h}} = \frac{16}{6.25} \\[1em] \Rightarrow \frac{\text{H}}{\text{h}} = \frac{16}{6.25} \\[1em]\end{aligned}$$

Multipy and divide by 100 on R.H.S

$$\begin{aligned}\Rightarrow \frac{\text{H}}{\text{h}} = \frac{16 \times 100}{6.25 \times 100} \\[1em] \Rightarrow \frac{\text{H}}{\text{h}} = \frac{1600}{625} \\[1em] \Rightarrow \frac{\text{H}}{\text{h}} = \frac{64}{25}\end{aligned}$$

∴ h : H = 25 : 64

Hence, option 4 is the correct option.

Given, radius(r) : height(h) = 5 : 12

Let r = 5x and h = 12x

Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}\Rightarrow 2512 = \frac{1}{3} \times 3.14 \times (5\text{x})^2 \times 12\text{x} \\[1em] \Rightarrow 2512 = 3.14 \times 25\text{x}^2 \times 4\text{x} \\[1em] \Rightarrow \text{x}^3 = \frac{2512}{3.14 \times 25 \times 4} \\[1em] \Rightarrow \text{x}^3 = \frac{2512}{314} \\[1em] \Rightarrow \text{x}^3 = 8 \\[1em] \Rightarrow \text{x} = \sqrt[3]{8} \\[1em] \Rightarrow \text{x} = 2 \text{ cm.}\end{aligned}$$

⇒ r = 5x = 5 × 2 = 10 cm

⇒ h = 12x = 12 × 2 = 24 cm

Curved surface area = πrl

l² = r² + h²

⇒ l² = 10² + 24²

⇒ l² = 100 + 576

⇒ l² = 676

⇒ l = $\sqrt{676}$ = 26 cm

Hence, option 4 is the correct option.

Given, r = 7 m and h = 24 m

l² = r² + h²

⇒ l² = 7² + 24²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$ = 25 m

So, the total curved surface area of the tent = πrl

$$\begin{aligned}= \frac{22}{7} \times 7 \times 25 \\[1em] = 22 \times 25 \\[1em] = 550 \text{ m}^2\end{aligned}$$

Width of the cloth used = 2.5 m

Length of canvas = $\frac{\text{area of canvas}}{\text{width of canvas}} = \frac{550}{2.5}$ = 220 m.

Hence, option 3 is the correct option.

Given,

Height of cone, h = 14 m

Area of base floor = 346.5 m²

$$\begin{aligned}\Rightarrow π\text{r}^2 = 346.5 \\[1em] \Rightarrow \frac{22}{7} \times \text{r}^2 = 346.5 \\[1em] \Rightarrow \text{r}^2 = \frac{346.5 \times 7}{22} \\[1em] \Rightarrow \text{r}^2 = \frac{2425.5}{22} \\[1em] \Rightarrow \text{r}^2 = 110.25 \\[1em] \Rightarrow \text{r} = \sqrt{110.25} \\[1em] \Rightarrow \text{r} = 10.5 \text{ m.}\end{aligned}$$

By formula,

l² = r² + h²

⇒ l² = 10.5² + 14²

⇒ l² = 110.25 + 196

⇒ l² = 306.25

⇒ l = $\sqrt{306.25}$ = 17.5 m

So, the total curved surface area of the tent = πrl

$$\begin{aligned}= \frac{22}{7} \times 10.5 \times 17.5 \\[1em] = \frac{4042.5}{7} \\[1em] = 577.5 \text{ m}^2\end{aligned}$$

Let length of canvas be a m.

Area of canvas = Curved surface area of cone

⇒ a × b = 577.5

⇒ a × 1.1 = 577.5

⇒ a = $\frac{577.5}{1.1}$

⇒ a = 525 m

Hence, option 2 is the correct option.

Given, Area of base = 154 m²

⇒ πr² = 154

$$\begin{aligned}\Rightarrow \frac{22}{7} \text{r}^2 = 154 \\[1em] \Rightarrow \text{r}^2 = 154 \times \frac{7}{22} \\[1em] \Rightarrow \text{r}^2 = \frac{1078}{22} \\[1em] \Rightarrow \text{r}^2 = 49 \\[1em] \Rightarrow \text{r} = \sqrt{49} \\[1em] \Rightarrow \text{r} = 7 \text{ m.}\end{aligned}$$

Volume of cone = 462 m³

By formula,

Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}\Rightarrow 462 = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times \text{h} \\[1em] \Rightarrow 462 = \frac{22}{21} \times 49 \times \text{h} \\[1em] \Rightarrow \text{h} = \frac{462 \times 21}{22 \times 49} \\[1em] \Rightarrow \text{h} = \frac{9702}{1078} \\[1em] \Rightarrow \text{h} = 9 \text{ m}.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Each person must have 20 m³ of air to breathe.

∴ 11 persons need 11 × 20 m³ = 220 m³

Each person must have 4 m² of the space on the ground.

∴ 11 persons need 11 × 4 m² = 44 m²

Base of the conical tent = area of the circle = πr²

$$\begin{aligned}\Rightarrow 44 = \frac{22}{7} \times \text{r}^2 \\[1em] \Rightarrow \text{r}^2 = \frac{7 \times 44}{22} \\[1em] \Rightarrow \text{r}^2 = \frac{308}{22} \\[1em] \Rightarrow \text{r}^2 = 14 \text{ m.}\end{aligned}$$

Let height of the conical tent be h meters.

Since, conical tent needs to accomodate 11 persons, so its volume will be equal to volume of air required for 11 persons.

$$\begin{aligned}\Rightarrow \frac{1}{3}π \text{r}^2 \text{h} = 220 \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 14 \times \text{h} = 220 \\[1em] \Rightarrow \text{h} = \frac{220 \times 7 \times 3}{22 \times 14} \\[1em] \Rightarrow \text{h} = \frac{4620}{308} \\[1em] \Rightarrow \text{h} = 15 \text{ m.}\end{aligned}$$

Hence, option 2 is the correct option.

Given, ratio of slant height = 5 : 4

Let slant height of 1st cone be 5a and 2nd cone be 4a cm.

For 1st cone,

⇒ Diameter = d

⇒ Radius = r

⇒ Slant height, l = 5a

For 2nd cone,

⇒ Diameter = D

⇒ Radius = R

⇒ Slant height, L = 4a

Given,

⇒ d = D

∴ r = R

$$\begin{aligned}\Rightarrow \frac{\text{CSA of 1st cone}}{\text{CSA of 2nd cone}} = \frac{π\text{rl}}{π\text{RL}} \\[1em] = \frac{\text{l}}{\text{L}} \\[1em] = \frac{5\text{a}}{4\text{a}} \\[1em] = \frac{5}{4} \\[1em] = 5 : 4\end{aligned}$$

Hence, option 2 is the correct option.

For older cone,

⇒ Radius = r

⇒ Height = h

⇒ Volume = v

For new cone,

⇒ Radius = R = $\frac{\text{r}}{2}$

⇒ Height = h

⇒ Volume = V

Volume of cone = $\frac{1}{3}π \text{r}^2 \text{h}$

$$\begin{aligned}\Rightarrow \frac{\text{V}}{\text{v}} = \frac{\frac{1}{3}π \text{R}^2 \text{h}}{\frac{1}{3}π \text{r}^2 \text{h}} \\[1em] = \frac{\Big(\frac{\text{r}}{2}\Big)^2}{\text{r}^2} \\[1em] = \frac{\frac{\text{r}^2}{4}}{\text{r}^2} \\[1em] = \frac{\text{r}^2}{4\text{r}^2} \\[1em] = \frac{1}{4}.\end{aligned}$$

Hence, option 3 is the correct option.

Volume of rectangular block = 10 cm × 5 cm × 2 cm = 100 cm³

Volume of cone = $\frac{1}{3}π \text{r}^2 \text{h}$

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 3^2 \times 7 \\[1em] = \frac{1}{3} \times 22 \times 9 \\[1em] = 22 \times 3 \\[1em] = 66 \text{ cm}^3.\end{aligned}$$

Volume of wood wasted = Volume of rectangular block - Volume of cone

= 100 - 66

= 34 cm³

Percentage of wood wasted

= $\frac{\text{Volume of wood wasted}}{\text{Total volume of rectangular block}} \times 100 = \frac{34}{100} \times 100$ = 34%

Hence, option 1 is the correct option.

Edge of a cube = 10 cm

Volume = side³ = 10³ = 1000 cm³.

Cone of maximum size is carved out as shown in figure,

Diameter of the cone cut out from it = 10 cm

Radius, r = $\frac{\text{diameter}}{2} = \frac{10}{2}$ = 5 cm

Height, h = 10 cm

Volume of cone = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 10 \\[1em] = \frac{1}{3} \times \frac{22}{7} \times 25 \times 10 \\[1em] = \frac{5500}{21} \\[1em] = 261.9 \text{ cm}^3.\end{aligned}$$

Hence, option 3 is the correct option.

For older cone,

⇒ Radius = r

⇒ Height = h

⇒ Volume = v

For new cone,

⇒ Radius = r

⇒ Height = 2h

⇒ Volume = V

Volume of cone = $\frac{1}{3}π \times \text{(radius)}^2 \times \text{height}$

Increase in the volume of cone = $\frac{\text{Volume of new cone - Volume of old cone}}{\text{Volume of old cone}} \times 100$

$$\begin{aligned}= \frac{\text{V - v}}{\text{v}} \times 100 \\[1em] = \frac{(\frac{1}{3}π \text{r}^2 \times \text{2h}) - (\frac{1}{3}π \text{r}^2 \text{h})}{\frac{1}{3}π \text{r}^2 \text{h}} \times 100 \\[1em] = \frac{\frac{1}{3}π \text{r}^2 \text{h}(2 - 1)}{\frac{1}{3}π \text{r}^2 \text{h}} \times 100 \\[1em] = 100%.\end{aligned}$$

Hence, option 1 is the correct option.

Let height of cones be 1a and 4a and radius of the cones be 4b and 1b.

Volume of cone = $\frac{1}{3}π \text{r}^2 \text{h}$

Volume of 1st cone, V = $\frac{1}{3}π (\text{4b})^2 \text{1a} = \frac{1}{3}π \times 16\text{b}^2 \times \text{a}$

Volume of 2nd cone, v = $\frac{1}{3}π (\text{1b})^2 \text{4a} = \frac{1}{3}π \times \text{b}^2 \times 4\text{a}$

$$\begin{aligned}\Rightarrow \frac{\text{V}}{\text{v}} = \frac{\frac{1}{3}π \times 16\text{b}^2 \times \text{a}}{\frac{1}{3}π \times \text{b}^2 \times 4\text{a}} \\[1em] = \frac{16}{4} \\[1em] = \frac{4}{1}.\end{aligned}$$

= 4 : 1

Hence, option 4 is the correct option.

For cylinder,

Radius = 3a

Height = 2b

Volume of cylinder, v = πr²h = π × (3a)² × 2b = π × 9a² × 2b = 18πa²b

For cone,

Radius = 4a

Height = 3b

Volume of cone, V = $\frac{1}{3}π \text{r}^2 \text{h}$

$$\begin{aligned}= \frac{1}{3}π (\text{4a})^2 \text{3b} \\[1em] = π \times 16\text{a}^2 \text{b} \\[1em]\end{aligned}$$

$$\begin{aligned}\therefore \frac{\text{v}}{\text{V}} = \frac{π \times 18\text{a}^2 \text{b}}{π \times 16\text{a}^2 \text{b}} \\[1em] = \frac{18}{16} \\[1em] = \frac{9}{8}\end{aligned}$$

= 9 : 8

Hence, option 4 is the correct option.

Volume of cone = $\frac{1}{3}π \text{r}^2 \text{h}$

Volume of cylinder = πr²h

Let the number of cones required be n.

∴ Volume of cylinder = n × Volume of cone

⇒ πr²h = n × $\frac{1}{3}π \text{r}^2 \text{h}$

⇒ 3 × πr²h = n × πr²h

∴ n = 3

Hence, option 2 is the correct option.

Given, radius of cone, R = 10 cm

Height of cone, H = 48 cm

Height of water in cylinder be h cm

Radius of cylinder, r = 20 cm

Since, water from conical vessel is poured into cylindrical vessel.

∴ Volume of cone = Volume of water in cylinder

$$\begin{aligned}\Rightarrow \frac{1}{3}π \text{R}^2 \text{H} = π \text{r}^2 \text{h} \\[1em] \Rightarrow \frac{1}{3} \times 10^2 \times 48 = 20^2 \times \text{h} \\[1em] \Rightarrow 100 \times 16 = 400 \times \text{h} \\[1em] \Rightarrow \text{h} = \frac{100 \times 16}{400} \\[1em] \Rightarrow \text{h} = \frac{1600}{400} \\[1em] \Rightarrow \text{h} = 4 \text{ cm.}\end{aligned}$$

Hence, option 2 is the correct option.

Given, radius of conical heap be R cm

Height of cone, H = 24 cm

Height of cylinder, h = 32 cm

Radius of cylinder, r = 18 cm

Since, sand from cylindrical vessel is poured to form conical heap.

∴ Volume of sand in cone = Volume of cylinder

$$\begin{aligned}\Rightarrow \frac{1}{3}π \text{R}^2 \text{H} = π \text{r}^2 \text{h} \\[1em] \Rightarrow \frac{1}{3} \text{R}^2 \text{H} = \text{r}^2 \text{h} \\[1em] \Rightarrow \frac{1}{3} \times \text{R}^2 \times 24 = 18^2 \times 32 \\[1em] \Rightarrow \text{R}^2 \times 8 = 324 \times 32 \\[1em] \Rightarrow \text{R}^2 \times 8 = 10368 \\[1em] \Rightarrow \text{R}^2 = \frac{10368}{8} \\[1em] \Rightarrow \text{R}^2 = 1296 \\[1em] \Rightarrow \text{R} = \sqrt{1296} \\[1em] \Rightarrow \text{R} = 36 \text{ cm.}\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Volume of sphere = 38808 cm³

Let the radius of the sphere be r cm.

By formula,

Volume of sphere = $\frac{4}{3}$ πr³

$$\begin{aligned}\Rightarrow \frac{4}{3} \times \frac{22}{7} \times \text{r}^3 = 38808 \\[1em] \Rightarrow \text{r}^3 = \frac{38808 \times 7 \times 3}{22 \times 4} \\[1em] \Rightarrow \text{r}^3 = \frac{814968}{88} \\[1em] \Rightarrow \text{r}^3 = 9261 \\[1em] \Rightarrow \text{r} = \sqrt[3]{9261} \\[1em] \Rightarrow \text{r} = 21 \text{ cm}.\end{aligned}$$

Curved surface area of sphere = 4πr²

= 4 × $\frac{22}{7}$ × 21 × 21

= 4 × 22 × 3 × 21

= 5544 cm².

Hence, option 3 is the correct option.

Let radius of two spheres be r and R.

Given,

Ratio of volume of the two spheres is 1 : 8.

Volume of sphere = $\frac{4}{3}$ π.(radius)³

$$\begin{aligned}\therefore \frac{\text{Volume of Sphere 1}}{\text{Volume of Sphere 2}} = \frac{1}{8} \\[1em] \Rightarrow \frac{\frac{4}{3} \times π \times \text{r}^3}{\frac{4}{3} \times π \times \text{R}^3} = \frac{1}{8} \\[1em] \Rightarrow \frac{\text{r}^3}{\text{R}^3} = \frac{1}{8} \\[1em] \Rightarrow \Big(\frac{\text{r}}{\text{R}}\Big)^3 = \frac{1}{8} \\[1em] \Rightarrow \frac{\text{r}}{\text{R}} = \sqrt[3]{\Big(\frac{1}{8}\Big)} \\[1em] \Rightarrow \frac{\text{r}}{\text{R}} = \frac{1}{2}.\end{aligned}$$

Surface area of sphere = 4π.(radius)²

$$\begin{aligned}\therefore \frac{\text{Surface area of Sphere 1}}{\text{Surface area of Sphere 2}} = \frac{1}{8} \\[1em] = \frac{4 \times π \times \text{r}^2}{4 \times π \times \text{R}^2} \\[1em] = \Big(\frac{\text{r}}{\text{R}}\Big)^2 \\[1em] = \Big(\frac{1}{2}\Big)^2 \\[1em] = \frac{1}{4} \\[1em] = 1 : 4.\end{aligned}$$

Hence, option 2 is the correct option.

Let radius of sphere be r cm.

Given, surface area of a sphere = 154 cm²

⇒ 4πr² = 154

$$\begin{aligned}\Rightarrow 4 \times \frac{22}{7} \times \text{r}^2 = 154 \\[1em] \Rightarrow \text{r}^2 = \frac{154 \times 7}{22 \times 4} \\[1em] \Rightarrow \text{r}^2 = \frac{1078}{88} \\[1em] \Rightarrow \text{r}^2 = 12.25 \\[1em] \Rightarrow \text{r} = \sqrt{12.25} \\[1em] \Rightarrow \text{r} = 3.5 \text{ cm.}\end{aligned}$$

Volume of sphere = $\frac{4}{3}$ πr³

$$\begin{aligned}= \frac{4}{3} \times \frac{22}{7} \times 3.5^3 \\[1em] = \frac{4}{3} \times \frac{22}{7} \times 42.875 \\[1em] = \frac{3773}{21} \\[1em] = 179\frac{2}{3} \text{ cm}^3.\end{aligned}$$

Hence, option 2 is the correct option.

Given, height of cylinder, h = 3 cm

Radius of cylinder be r cm

Volume of cylinder = 48π cm³

By formula,

Volume of cylinder = πr²h

⇒ πr²h = 48π

⇒ r² × 3 = 48

$$\begin{aligned}\Rightarrow \text{r}^2 = \frac{48}{3} \\[1em] \Rightarrow \text{r}^2 = 16 \\[1em] \Rightarrow \text{r} = \sqrt{16} \\[1em] \Rightarrow \text{r} = 4 \text{ cm.}\end{aligned}$$

Hence, option 3 is the correct option.

Given, three spherical beads of radii 3 cm, 4 cm and 5 cm.

Volume of sphere = $\frac{4}{3}$ πr³

Volume of single spherical bead = Volume of bead of radius 3 cm + Volume of bead of radius 4 cm + Volume of bead of radius 5 cm

$$\begin{aligned}\Rightarrow \frac{4}{3}π \times \text{r}^3 = \frac{4}{3}π \times 3^3 + \frac{4}{3}π \times 4^3 + \frac{4}{3}π \times 5^3 \\[1em] \Rightarrow \text{r}^3 = (3^3 + 4^3 + 5^3) \\[1em] \Rightarrow \text{r}^3 = 27 + 64 + 125 \\[1em] \Rightarrow \text{r}^3 = 216 \\[1em] \Rightarrow \text{r} = \sqrt[3]{216} \\[1em] \Rightarrow \text{r} = 6 \text{ cm.}\end{aligned}$$

Hence, option 1 is the correct option.

Let radius of small spherical balls be r cm.

Radius of sphere, R = 10 cm

Volume of solid sphere = 8 × Volume of small spherical balls

$$\begin{aligned}\Rightarrow \frac{4}{3}π \times \text{R}^3 = 8 \times \frac{4}{3}π \times \text{r}^3 \\[1em] \Rightarrow 10^3 = 8 \times \text{r}^3 \\[1em] \Rightarrow 1000 = 8 \times \text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \frac{1000}{8} \\[1em] \Rightarrow \text{r} = \sqrt[3]{\frac{1000}{8}} \\[1em] \Rightarrow \text{r} = \frac{10}{2} \\[1em] \Rightarrow \text{r} = 5 \text{ cm.}\end{aligned}$$

Surface area of small ball = 4 πr²

= 4 × π × 5 × 5

= 100 π cm².

Hence, option 4 is the correct option.

Given,

Diameter of sphere = 6 cm

Radius of sphere, r = $\frac{\text{diameter}}{2} = \frac{6}{2}$ = 3 cm

Height of cylinder, h = 45 cm

Diameter of cylinder = 4 cm

Radius of cylinder, R = $\frac{\text{diameter}}{2} = \frac{4}{2}$ = 2 cm

Number of solid spheres required be n.

Since, the number of solid spheres, are moulded to form a solid metal cylinder.

∴ Volume of cylinder = n × Volume of solid sphere

$$\begin{aligned}\Rightarrow π\text{R}^2\text{h} = \text{n} \times \frac{4}{3}π \times \text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \text{n} \times \frac{4}{3} \times \text{r}^3 \\[1em] \Rightarrow 2^2 \times 45 = \text{n} \times \frac{4}{3} \times 3^3 \\[1em] \Rightarrow 4 \times 45 = \text{n} \times \frac{4}{3} \times 27 \\[1em] \Rightarrow 180 = \text{n} \times 4 \times 9 \\[1em] \Rightarrow 180 = \text{n} \times 36 \\[1em] \Rightarrow \text{n} = \frac{180}{36} \\[1em] \Rightarrow \text{n} = 5.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Height of cylinder, h = 32 cm

Diameter of cylinder = 6 cm

Radius of cylinder, R = $\frac{\text{diameter}}{2} = \frac{6}{2}$ = 3 cm

Let radius of sphere be r cm.

Since, volume of sphere is equal to the volume of the cylinder.

∴ Volume of cylinder = Volume of solid sphere

$$\begin{aligned}\Rightarrow π\text{R}^2\text{h} = \frac{4}{3}π \times \text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \frac{4}{3} \times \text{r}^3 \\[1em] \Rightarrow 3^2 \times 32 = \frac{4}{3} \times \text{r}^3 \\[1em] \Rightarrow 9 \times 32 \times 3 = 4 \times \text{r}^3 \\[1em] \Rightarrow 864 = 4 \times \text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \frac{864}{4} \\[1em] \Rightarrow \text{r}^3 = 216 \\[1em] \Rightarrow \text{r} = \sqrt[3]{216} \\[1em] \Rightarrow \text{r} = 6 \text{ cm.}\end{aligned}$$

Hence, option 4 is the correct option.

Let the radius of sphere 1 be r cm and radius of sphere 2 be R cm.

Volume of sphere = $\frac{4}{3}π \times \text{r}^3$

Given,

Volume of sphere 1 = 2 × Volume of sphere 2

$$\begin{aligned}\Rightarrow \frac{4}{3}π \times \text{r}^3 = 2 \times \frac{4}{3}π \times \text{R}^3 \\[1em] \Rightarrow \text{r}^3 = 2 \times \text{R}^3 \\[1em] \Rightarrow \frac{\text{r}^3}{\text{R}^3} = \frac{2}{1} \\[1em] \Rightarrow \Big(\frac{\text{r}}{\text{R}}\Big)^3 = \frac{2}{1} \\[1em] \Rightarrow \frac{\text{r}}{\text{R}} = \sqrt[3]{\frac{2}{1}} \\[1em] \Rightarrow \frac{\text{r}}{\text{R}} = \frac{\sqrt[3]{2}}{1}\end{aligned}$$

Hence, option 4 is the correct option.

Given, the volume of two spheres = 27: 64

By formula, volume of a sphere = $\frac{4πr^3}{3}$, where r is the radius of the sphere.

Let r and R be the radii of the two spheres.

$$\begin{aligned}\Rightarrow \frac{\frac{4πr^3}{3}}{\frac{4πR^3}{3}} = \frac{27}{64} \\[1em] \Rightarrow \frac{r^3}{R^3} = \frac{27}{64} \\[1em] \Rightarrow \frac{r}{R} = \frac{\sqrt[3]{27}}{\sqrt[3]{64}} \\[1em] \Rightarrow \frac{r}{R} = \frac{3}{4} \\[1em] \Rightarrow r : R = 3 : 4.\end{aligned}$$

Hence, option 1 is the correct option.

Let r be the radius of the sphere

The radius of cylinder is also r cm and height of the cylinder is 2r cm

By formula,

Volume of cylinder = πr²h

= πr²(2r)

= 2πr³

By formula,

Volume of sphere = $\frac{4}{3}π \times \text{r}^3$

Ratio of the volume of sphere to the volume of the cylinder:

$$\begin{aligned}= \frac{\frac{4}{3}π \times \text{r}^3}{2π \times \text{r}^3} \\[1em] = \frac{4}{3} \times \frac{1}{2} \\[1em] = \frac{4}{6} \\[1em] = \frac{2}{3}\end{aligned}$$

Hence, option 3 is the correct option.

Radius of sphere, r = 6 cm

Radius of cylinder, R = 8 cm

Let the rise in water level be x cm.

∴ Volume of water that rises by x cm in the cylindrical vessel = Volume of sphere submerged

$$\begin{aligned}\Rightarrow π\text{R}^2\text{x} = \frac{4}{3} π\text{r}^3 \\[1em] \Rightarrow 8^2\text{x} = \frac{4}{3} \times 6^3 \\[1em] \Rightarrow 64\text{x} = \frac{4}{3} \times 216 \\[1em] \Rightarrow 64\text{x} = 4 \times 72 \\[1em] \Rightarrow 64\text{x} = 288 \\[1em] \Rightarrow \text{x} = \frac{288}{64} \\[1em] \Rightarrow \text{x} = 4.5 \text{ cm.}\end{aligned}$$

Hence, option 4 is the correct option.

Let radius of the cylinder be r cm.

Length of cylindrical rod, h = 12r

Radius of spherical balls be r cm

Number of spherical balls required be n.

Since, a cylindrical rod of iron is melted and cast into spherical balls of the same radius.

∴ Volume of cylindrical rod = n × Volume of spherical ball

$$\begin{aligned}\Rightarrow π\text{r}^2\text{h} = \text{n} \times \frac{4}{3} π\text{r}^3 \\[1em] \Rightarrow \text{r}^2 \times \text{12r} = \text{n} \times \frac{4}{3} \text{r}^3 \\[1em] \Rightarrow 12\text{r}^3 = \text{n} \times \frac{4}{3} \text{r}^3 \\[1em] \Rightarrow 12 = \text{n} \times \frac{4}{3} \\[1em] \Rightarrow \text{n} = \frac{12 \times 3}{4} \\[1em] \Rightarrow \text{n} = \frac{36}{4} \\[1em] \Rightarrow \text{n} = 9.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Radius of sphere = r cm

Height of cone, h = r cm

Let radius of cone be R cm

Since, sphere is melted and recasted into cone.

∴ Volume of sphere = Volume of cone

$$\begin{aligned}\Rightarrow \frac{4}{3} π\text{r}^3 = \frac{1}{3} π\text{R}^2\text{h} \\[1em] \Rightarrow \frac{4}{3} \text{r}^3 = \frac{1}{3} \text{R}^2 \times \text{r} \\[1em] \Rightarrow 3 \times \frac{4}{3} \times \text{r}^3 = \text{R}^2 \times \text{r} \\[1em] \Rightarrow 4 \times \frac{\text{r}^3}{\text{r}} = \text{R}^2 \\[1em] \Rightarrow 4 \times \text{r}^2 = \text{R}^2 \\[1em] \Rightarrow \text{R} = \sqrt{4 \times \text{r}^2} \\[1em] \Rightarrow \text{R} = 2\text{r}\end{aligned}$$

Hence, option 2 is the correct option.

Let radius of sphere be r cm.

Given,

Volume of sphere = Surface area of sphere

$$\begin{aligned}\Rightarrow \frac{4}{3} π\text{r}^3 = 4π\text{r}^2 \\[1em] \Rightarrow 4π\text{r}^3 = 3 \times 4π\text{r}^2 \\[1em] \Rightarrow 4\text{r}^3 = 12 \times \text{r}^2 \\[1em] \Rightarrow \text{r}^3 = \frac{12}{4} \times \text{r}^2 \\[1em] \Rightarrow \frac{\text{r}^3}{\text{r}^2} = 3 \\[1em] \Rightarrow \text{r} = 3 \text{ units.}\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Let the wire's radius be a.

Given, sphere is melted into the wire.

The wire formed is a cylinder, hence the volume of wire will be equal to the volume of sphere.

Radius of sphere, r = $\frac{\text{diameter}}{2} = \frac{6}{2}$ = 3 cm

Volume of sphere, V = $\frac{4}{3} π\text{r}^3$

$$\begin{aligned}= \frac{4}{3}π \times 3^3 \\[1em] = \frac{4}{3}π \times 27 \\[1em] = 4 \times 9π \\[1em] = 36 π \text{ cm}^3\end{aligned}$$

Given, length of wire = 36 cm

So, height of cylinder = 36 cm

Volume of cylinder, V = 36 π cm³

∴ πr²h = 36 π

$$\begin{aligned}\Rightarrow \text{r}^2 \times 36 = 36 \\[1em] \Rightarrow \text{r}^2 = \frac{36}{36} \\[1em] \Rightarrow \text{r}^2 = 1 \\[1em] \Rightarrow \text{r} = \sqrt{1} \\[1em] \Rightarrow \text{r} = 1 \text{ cm.}\end{aligned}$$

Hence, option 2 is the correct option.

Radius of larger spherical metallic ball, R = 3 cm

Radius of smaller spherical balls are 1.5 cm, 2 cm and r cm

Given,

A spherical metallic ball of radius 3 cm is melted and recast into three spherical balls.

∴ Volume of larger spherical ball = Volume of ball of radius 1.5 cm + Volume of ball of radius 2 cm + Volume of ball of radius r cm

$$\begin{aligned}\Rightarrow \frac{4}{3}π\text{R}^3 = \frac{4}{3}π \times 1.5^3 + \frac{4}{3}π \times 2^3 + \frac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \frac{4}{3}π\text{R}^3 = \frac{4}{3}π(1.5^3 + 2^3 + \text{r}^3) \\[1em] \Rightarrow \text{R}^3 = (1.5^3 + 2^3 + \text{r}^3) \\[1em] \Rightarrow 3^3 = 3.375 + 8 + \text{r}^3 \\[1em] \Rightarrow \text{r}^3 = 27 - 3.375 - 8 \\[1em] \Rightarrow \text{r}^3 = 15.625 \\[1em] \Rightarrow \text{r} = \sqrt[3]{15.625} \\[1em] \Rightarrow \text{r} = 2.5 \text{ cm.}\end{aligned}$$

Hence, option 4 is the correct option.

Total surface area of semi-hemisphere = 2πr²

= 2π × 4²

= 2π × 16

= 32π.

Hence, option 2 is the correct option.

Let radius of hemisphere be r.

Total surface area of hemisphere = 3πr²

Total surface area of two hemisphere = 2 × 3πr² = 6πr².

Total surface area of sphere = 4πr²

Total surface area of two hemisphere : Total surface area of sphere = 6πr² : 4πr²

= 6 : 4

= 3 : 2.

Hence, option 2 is the correct option.

Radius of sphere, r = $\frac{\text{diameter}}{2} = \frac{12.6}{2} = 6.3 \text{ cm.}$

Volume of sphere = $\frac{4}{3}π\text{r}^3$

Radius of the cone = R cm

Height of the cone, h = 25.2 cm

Volume of cone = $\frac{1}{3}π\text{R}^2 \text{h}$

Since, sphere is melted and recasted into a cone, the volume remains the same.

$$\begin{aligned}\therefore \frac{1}{3}π\text{R}^2 \text{h} = \frac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \frac{1}{3}\text{R}^2 \text{h} = \frac{4}{3}\text{r}^3 \\[1em] \Rightarrow \text{R}^2 = \frac{4 \times 3}{3 \times \text{h}} \times \text{r}^3 \\[1em] \Rightarrow \text{R}^2 = \frac{12}{3 \times 25.2} \times 6.3^3 \\[1em] \Rightarrow \text{R}^2 = \frac{12}{75.6} \times 250.047 \\[1em] \Rightarrow \text{R}^2 = \frac{3000.564}{75.6} \\[1em] \Rightarrow \text{R}^2 = 39.69 \\[1em] \Rightarrow \text{R} = \sqrt{39.69} \\[1em] \Rightarrow \text{R} = 6.3 \text{ cm.}\end{aligned}$$

Hence, option 4 is the correct option.

Shots is in the shape of sphere.

Radius of sphere, r = $\frac{\text{diameter}}{2} = \frac{0.3}{2} = 0.15 \text{cm.}$

Let the number of spheres formed be n.

Volume of cuboid = n × Volume of each lead shot

$$\begin{aligned}\therefore \text{lbh} = \text{n} \times \frac{4}{3} π \text{r}^3 \\[1em] \Rightarrow 9 \times 11 \times 12 = \text{n} \times \frac{4}{3} \times \frac{22}{7} \times 0.15^3 \\[1em] \Rightarrow 1188 = \text{n} \times \frac{4}{3} \times \frac{22}{7} \times 0.003375 \\[1em] \Rightarrow 1188 = \text{n} \times \frac{0.297}{21} \\[1em] \Rightarrow \text{n} = \frac{21 \times 1188}{0.297} \\[1em] \Rightarrow \text{n} = \frac{24948}{0.297} \\[1em] \Rightarrow \text{n} = 84000.\end{aligned}$$

Hence, option 4 is the correct option.

Radius of sphere, r = 10.5 cm

Let the number of cones formed by recasting metallic sphere be n.

Radius of cone, R = 3.5 cm

Height, h = 3 cm

Volume of sphere = n × Volume of each cone

$$\begin{aligned}\Rightarrow \frac{4}{3}π\text{r}^3 = \text{n} \times \frac{1}{3}π\text{R}^2 \text{h} \\[1em] \text{Dividing both sides by π and multiplying by 3, we get :} \\[1em] \Rightarrow 4\text{r}^3 = \text{n} \times \text{R}^2 \text{h} \\[1em] \Rightarrow 4 \times 10.5^3 = \text{n} \times 3.5^2 \times 3 \\[1em] \Rightarrow 4 \times 1157.625 = \text{n} \times 12.25 \times 3 \\[1em] \Rightarrow 4630.5 = \text{n} \times 36.75 \\[1em] \Rightarrow \text{n} = \frac{4630.5}{36.75} \\[1em] \Rightarrow \text{n} = 126.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Internal radius of hemispherical bowl, R = 9 cm

Radius of cylindrical bottles, r = $\frac{\text{diameter}}{2} = \frac{3}{2}$ = 1.5 cm

Height of the cylindrical bottles, h = 4 cm

Let number of cylindrical bottles needed be n.

∴ Volume of hemispherical bowl = n × Volume of each cylindrical bottle

$$\begin{aligned}\Rightarrow \frac{2}{3}π\text{R}^3 = \text{n} \times π\text{r}^2\text{h} \\[1em] \Rightarrow \frac{2}{3}\text{R}^3 = \text{n} \times \text{r}^2\text{h} \\[1em] \Rightarrow \frac{2}{3} \times 9^3 = \text{n} \times 1.5^2 \times 4 \\[1em] \Rightarrow \frac{2}{3} \times 729 = \text{n} \times 2.25 \times 4 \\[1em] \Rightarrow 2 \times 243 = \text{n} \times 9 \\[1em] \Rightarrow \text{n} = \frac{486}{9} \\[1em] \Rightarrow \text{n} = 54.\end{aligned}$$

Hence, option 3 is the correct option.

Let the common radius of shapes be r and height be h.

Ratio of their volumes = Volume of cone : Volume of hemisphere : Volume of cylinder

$$\begin{aligned}= \frac{1}{3} π\text{r}^2\text{h} : \frac{2}{3} π\text{r}^3 : π\text{r}^2\text{h} \\[1em] = \frac{1}{3} : \frac{2}{3} : 1\end{aligned}$$

On multiplying by 3, ratio = 1 : 2 : 3

Hence, option 1 is the correct option.

In hollow cylindrical drum,

Height, H = 1 m = 100 cm

Radius, R = $\frac{\text{Diameter}}{2} = \frac{30}{2}$ = 15 cm

For each cylindrical boxes,

Height, h = 10 cm

Radius, r = $\frac{\text{Diameter}}{2} = \frac{10}{2}$ = 5 cm

Let the maximum number of cylindrical boxes that can be packed be n.

Volume of hollow cylinder = n × Volume of cylindrical box

$$\begin{aligned}\therefore π\text{R}^2\text{H} = \text{n} \times π\text{r}^2\text{h} \\[1em] \Rightarrow \text{R}^2\text{H} = \text{n} \times \text{r}^2\text{h} \\[1em] \Rightarrow 15^2 \times 100 = \text{n} \times 5^2 \times 10 \\[1em] \Rightarrow 225 \times 100 = \text{n} \times 25 \times 10 \\[1em] \Rightarrow 22500 = \text{n} \times 250 \\[1em] \Rightarrow \text{n} = \frac{22500}{250} \\[1em] \Rightarrow \text{n} = 90.\end{aligned}$$

Hence, option 4 is the correct option.

In ice-cream cylinder,

Radius of cylinder, R = $\frac{\text{diameter}}{2} = \frac{35}{2}$ = 17.5 cm

Height of cylinder, H = 32 cm

In each ice-cream cone,

Radius of cone part, r = $\frac{\text{diameter}}{2} = \frac{4}{2}$ = 2 cm

Height of cone, h = 7 cm

Let the number of children who get ice-cream cone be n.

Volume of cylinder = n × Volume of ice-cream cone

$$\begin{aligned}\therefore π\text{R}^2\text{H} = \text{n} \times \frac{1}{3}π\text{r}^2\text{h} \\[1em] \Rightarrow 3 \times \text{R}^2\text{H} = \text{n} \times \text{r}^2\text{h} \\[1em] \Rightarrow 3 \times 17.5^2 \times 32 = \text{n} \times 2^2 \times 7 \\[1em] \Rightarrow 3 \times 306.25 \times 32 = \text{n} \times 4 \times 7 \\[1em] \Rightarrow 29400 = \text{n} \times 28 \\[1em] \Rightarrow \text{n} = \frac{29400}{28} \\[1em] \Rightarrow \text{n} = 1050.\end{aligned}$$

Hence, option 3 is the correct option.

Let the radius of the sphere be r cm.

Radius of cylinder, R = $\frac{\text{diameter}}{2} = \frac{14}{2}$ = 7 cm

Since, a spherical iron ball is dropped into the vessel.

Height of water raised by 9 $\frac{1}{3} \text{cm} = \frac{28}{3}$ cm

Volume of water rise in cylinder = Volume of sphere

$$\begin{aligned}\Rightarrow π\text{R}^2\text{h} = \frac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \frac{4}{3}\text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \frac{3}{4} \times \text{R}^2 \times \text{h} \\[1em] \Rightarrow \text{r}^3 = \frac{3}{4} \times 7^2 \times \frac{28}{3} \\[1em] \Rightarrow \text{r}^3 = \frac{3}{4} \times 49 \times \frac{28}{3} \\[1em] \Rightarrow \text{r}^3 = \frac{4116}{12} \\[1em] \Rightarrow \text{r}^3 = 343 \\[1em] \Rightarrow \text{r} = \sqrt[3]{343} \\[1em] \Rightarrow \text{r} = 7 \text{ cm.}\end{aligned}$$

Hence, option 2 is the correct option.

From figure,

Height of cylinder be h cm

Given, Diameter of cylinder = $\frac{1}{4}$ h

Radius of cylinder = Radius of hemisphere = r = $\frac{\text{diameter}}{2} = \frac{\frac{1}{4} \times \text{h}}{2} = \frac{\text{h}}{8}$

Height of cylinder, h = Total height - (2 × Radius of hemisphere)

$$\begin{aligned}\Rightarrow \text{h} = 35 - 2 \times \frac{\text{h}}{8} \\[1em] \Rightarrow \text{h} = 35 - \frac{\text{h}}{4} \\[1em] \Rightarrow \text{h} + \frac{\text{h}}{4} = 35 \\[1em] \Rightarrow \frac{4 \text{h} + \text{h}}{4} = 35 \\[1em] \Rightarrow \frac{5 \text{h}}{4} = 35 \\[1em] \Rightarrow 5\text{h} = 35 \times 4 \\[1em] \Rightarrow 5\text{h} = 140 \\[1em] \Rightarrow \text{h} = \frac{140}{5} \\[1em] \Rightarrow \text{h} = 28 \text{ cm.}\end{aligned}$$

∴ r = $\frac{\text{h}}{8} = \frac{28}{8} = 3.5 \text{ cm}$

Surface area of solid = 2 × 2πr² + 2πrh

= πr(4r + 2h)

$$\begin{aligned}= \frac{22}{7} \times 3.5 (4 \times 3.5 + 2 \times 28) \\[1em] = 22 \times 0.5 (14 + 56) \\[1em] = 11 \times 70 \\[1em] = 770 \text{ cm}^2.\end{aligned}$$

Hence, option 4 is the correct option.

By formula,

Volume of sphere = $\frac{4}{3} \pi r^3$

Given,

A solid sphere is cut into two identical hemispheres.

Volume of hemisphere = $\frac{2}{3} \pi r^3$

Volume of two identical hemispheres = $2 \times \frac{2}{3} \pi r^3$

= $\frac{4}{3} \pi r^3$

Thus, volume of a sphere = volume of two identical hemispheres.

∴ Statement 1 is true.

We know that,

Surface area of sphere = 4πr²

When a sphere is cut into two hemispheres, two new flat circular surfaces are created,

Surface area of a single hemisphere = Curved surface area + Area of its flat circular face

= 2πr² + πr²

= 3πr²

Total surface area of two hemispheres = 2 × 3πr² = 6πr².

Thus, the surface area of the original sphere ≠ the total surface area of the two hemispheres.

∴ Statement 2 is false.

Hence, option 3 is the correct option.

Directions:

At an NCC camp, several tents were installed. Each tent is cylindrical to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m.

Based on this information, answer the following questions:

67. The slant height of the conical portion of the tent is :

(a) 16.5 m
(b) 17.5 m
(c) 18.5 m
(d) 19.5 m

68. The cost of cloth required to make each tent at the rate of ₹ 80 per square meter is :

(a) ₹ 76560
(b) ₹ 80140
(c) ₹ 82720
(d) ₹ 85960

69. If each cadet requires 8 m² of floor space and there are 15 tents in all how many cadets can be accommodated in the camp?

(a) 960
(b) 1155
(c) 1320
(d) 1440

70. If a tent has maximum number of cadets that it can accommodate as calculated in the above questions, what is the volume of air available to each cadet to breathe?

(a) 48 m³
(b) 52 m³
(c) 55 m³
(d) 77 m³

67. Given,

Height of cylinder, h = 3 m

Total height of tent, T = 13.5 m

Height of cone, H = T - h = 13.5 - 3 = 10.5 m

Radius of base of cylinder = Radius of cone = r = 14 m

Slant height of cone be l m.

l² = r² + H²

⇒ l² = 14² + 10.5²

⇒ l² = 196 + 110.25

⇒ l² = 306.25

⇒ l = $\sqrt{306.25}$ = 17.5 m

Hence, Option (b) is the correct option.

68. Curved surface area of tent = Curved surface area of cone + Curved surface area of cylinder

= 2πrh + πrl

= πr(2h + l)

$$\begin{aligned}= \frac{22}{7} \times 14 (2 \times 3 + 17.5) \\[1em] = 22 \times 2(6 + 17.5) \\[1em] = 44 \times 23.5 \\[1em] = 1034 \text{ m}^2.\end{aligned}$$

Given, cost of cloth required to make each tent is ₹ 80 per square meter.

⇒ Total cost = 80 × 1034 = ₹ 82720

Hence, Option (c) is the correct option.

69. The floor space of a tent is base area of cylinder.

∴ Area of base = πr²

= $\frac{22}{7}$ × 14 × 14

= 22 × 2 × 14

= 616 m²

Given each cadet requires 8 m² of floor space.

The number of cadets per tent = $\frac{\text{Area of base}}{\text{space required per cadet}} = \frac{616}{8}$ = 77 cadets.

Given, there are 15 tents.

∴ Total number of cadets = 77 × 15 = 1155 cadets.

Hence, Option (b) is the correct option.

70. Volume of air in each tent = Volume of air in cylinder + Volume of air in cone

$$\begin{aligned}= π\text{r}^2\text{h} + \frac{1}{3}π\text{r}^2\text{H} \\[1em] = π\text{r}^2 (\text{h} + \frac{1}{3}\text{H}) \\[1em] = \frac{22}{7} \times 14^2 (3 + \frac{1}{3} \times 10.5) \\[1em] = \frac{22}{7} \times 196 (3 + 3.5) \\[1em] = 22 \times 28 \times 6.5 \\[1em] = 4004 \text{ m}^3.\end{aligned}$$

Volume of air available to each cadet to breathe = $\frac{\text{Volume of air per tent}}{\text{Cadets per tent}} = \frac{4004}{77}$ = 52 m³.

Hence, Option (b) is the correct option.

Directions:

From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out.

Based on this information, answer the following questions:

71. The volume of the remaining solid is :

(a) 2856 cm³
(b) 3388 cm³
(c) 3672 cm³
(d) 4620 cm³

72. The total surface area of the remaining solid is :

(a) 1870 cm²
(b) 2024 cm²
(c) 2178 cm²
(d) 2332 cm²

73. The slant height of the cut out cone is :

(a) 18 cm
(b) 25 cm
(c) 26 cm
(d) 32 cm

74. The total surface area of the cut out cone is :

(a) 550 cm²
(b) 704 cm²
(c) 858 cm²
(d) 616 cm²

71. Radius of solid cylinder = Radius of cone = r = 7 cm

Height of the cylinder, H = 30 cm

Height of cone, h = 24 cm

Volume of remaining solid = Volume of cylinder - Volume of cone

$$\begin{aligned}= π\text{r}^2\text{H} - \frac{1}{3}π\text{r}^2\text{h} \\[1em] = π\text{r}^2(\text{H} - \frac{\text{h}}{3}) \\[1em] = \frac{22}{7} \times 7^2 \times (30 - \frac{24}{3}) \\[1em] = \frac{22}{7} \times 49 \times (30 - 8) \\[1em] = 22 \times 7 \times 22 \\[1em] = 3388 \text{ cm}^3.\end{aligned}$$

Hence, Option (b) is the correct option.

72. Slant height of cone, l = $\sqrt{\text{h}^2 + \text{r}^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625}$ = 25 cm

Total surface area of reamining solid = Curved surface area of cylinder + Area of base of cylinder + Curved surface area of cone

= 2πrH + πr² + πrl

= πr(2H + r + l)

= $\frac{22}{7}$ × 7 (2 × 30 + 7 + 25)

= 22 × (60 + 32)

= 22 × 92

= 2024 cm².

Hence, Option (b) is the correct option.

73. Slant height of cone, l = $\sqrt{\text{h}^2 + \text{r}^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625}$ = 25 cm

Hence, Option (b) is the correct option.

74. Total surface area of cut out cone = πr² + πrl

= πr(r + l)

= $\frac{22}{7}$ × 7 × (7 + 25)

= 22 × 32

= 704 cm²

Hence, Option (b) is the correct option.

Directions:

The surface area of a solid metallic sphere is 900 π cm².

Based on this information, answer the following questions:

75. If the given sphere is melted and recast into 3 smaller spheres of equal volumes, then the radius of each smaller sphere is :

(a) 5 cm
(b) 5 $\sqrt{3}$ cm
(c) 5 $\sqrt[3]{3}$ cm
(d) 5 $\sqrt[3]{9}$ cm

76. If the given sphere is cut into two hemispheres, then how much does the total surface area get increased? (Take π = 3.14) :

(a) no change
(b) 706.5 cm²
(c) 1015 cm²
(d) 1413 cm²

77. If the given sphere is melted and recast into solid right cones, each of radius 2.5 cm and height 8 cm, how many cones are formed?

(a) 135
(b) 270
(c) 405
(d) 540

78. If the given sphere is melted and recast into small spheres each of radius 0.5 cm, then the number of spheres formed is :

(a) 1350
(b) 2700
(c) 13500
(d) 27000

75. Given,

Let radius of sphere be R cm.

Surface area of a solid metallic sphere = 900 π cm²

∴ 4πR² = 900 π

⇒ 4R² = 900

⇒ R² = $\frac{900}{4}$

⇒ R² = 225

⇒ R = $\sqrt{225}$

⇒ R = 15 cm

Let radius of small spheres be r cm.

Since, the given sphere is melted and recast into 3 smaller spheres of equal volumes.

Volume of sphere = 3 × Volume of small spheres

$$\begin{aligned}\therefore \frac{4}{3}π\text{R}^3 = 3 \times \frac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \text{R}^3 = 3 \times \text{r}^3 \\[1em] \Rightarrow 15^3 = 3 \times \text{r}^3 \\[1em] \Rightarrow 3375 = 3 \times \text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \frac{3375}{3} \\[1em] \Rightarrow \text{r}^3 = 1125 \\[1em] \Rightarrow \text{r} = \sqrt[3]{1125} \\[1em] \Rightarrow \text{r} = \sqrt[3]{125 \times 9} \\[1em] \Rightarrow \text{r} = 5\sqrt[3]{9} \text{ cm.}\end{aligned}$$

Hence, Option (d) is the correct option.

76. When a sphere is cut into two hemispheres, two new circular faces are exposed.

∴ Radius of hemisphere = R = 15 cm.

The increase in total surface area = Area of two circular faces

= 2 × πR²

= 2 × 3.14 × 15²

= 2 × 3.14 × 225

= 1413 cm²

Hence, Option (d) is the correct option.

77. Let, radius of cone be a = 2.5 cm

Height of cone, h = 8 cm

Let solid right cones formed be n.

Since, the given sphere is melted and recast into solid right cones.

∴ Volume of sphere = n × Volume of cone

$$\begin{aligned}\therefore \frac{4}{3}π \text{R}^3 = \text{n} \times \frac{1}{3}π\text{a}^2 \text{h} \\[1em] \Rightarrow 4 \text{R}^3 = \text{n} \times \text{a}^2 \text{h} \\[1em] \Rightarrow \text{n} = \frac{4 \text{R}^3}{\text{a}^2 \text{h}} \\[1em] \Rightarrow \text{n} = \frac{4 \times 15^3}{2.5^2 \times 8} \\[1em] \Rightarrow \text{n} = \frac{4 \times 3375}{6.25 \times 8} \\[1em] \Rightarrow \text{n} = \frac{13500}{50} \\[1em] \Rightarrow \text{n} = 270.\end{aligned}$$

Hence, Option (b) is the correct option.

78. Let radius of small spheres be b = 0.5 cm.

Let number of small spheres formed be n.

Since, the given sphere is melted and recast into small spheres.

∴ Volume of sphere = n × Volume of small spheres

$$\begin{aligned}\therefore \frac{4}{3}π \text{R}^3 = \text{n} \times \frac{4}{3}π \text{b}^3 \\[1em] \Rightarrow \text{R}^3 = \text{n} \times \text{b}^3 \\[1em] \Rightarrow \text{n} = \frac{\text{R}^3}{\text{b}^3} \\[1em] \Rightarrow \text{n} = \frac{15^3}{0.5^3} \\[1em] \Rightarrow \text{n} = \frac{3375}{0.125} \\[1em] \Rightarrow \text{n} = 27000.\end{aligned}$$

Hence, Option (d) is the correct option.

Given, radius r = 3 cm and height h = 4 cm.

By formula, l² = h² + r² = 4² + 3² = 16 + 9 = 25

⇒ l = $\sqrt{25}$ = 5 cm.

So, the slant height is 5 cm, not (4 + 3) = 7 cm.

∴ Assertion (A) is false.

The curved surface area of a cone of radius r and slant height l is πrl.

∴ Reason (R) is true.

Hence, option 4 is the correct option.

The cone of maximum volume that can be carved out of a solid hemisphere of radius r has the same base radius r and the same height r.

Volume of cone = $\frac{1}{3}$ πr²h = $\frac{1}{3}$ πr² × r = $\frac{1}{3}$ πr³.

∴ Assertion (A) is true.

For a cone of radius r and height h, slant height l = $\sqrt{h^2 + r^2}$.

∴ Reason (R) is true.

Reason (R) is a correct statement, but it states the slant-height formula, which is not used to establish the maximum volume in the Assertion. So, R is not the correct explanation of A.

Hence, option 2 is the correct option.

Given, radius r = 5 cm and slant height l = 13 cm.

Total surface area of cone = πr(l + r) = π × 5 × (13 + 5) = π × 5 × 18 = 90π cm².

∴ Assertion (A) is true.

The expression πr(l + r) is the formula for the total surface area of a cone, whereas the curved surface area of a cone is πrl. So, Reason (R) wrongly calls πr(l + r) the curved surface area.

∴ Reason (R) is false.

Hence, option 3 is the correct option.

Total volume of the two spheres = $\frac{4}{3}$ π(2)³ + $\frac{4}{3}$ π(4)³ = $\frac{4}{3}$ π(8 + 64) = $\frac{4}{3}$ π × 72 = 96π cm³.

Let the height of the cone be h cm. Radius of cone = 6 cm.

Volume of cone = $\frac{1}{3}$ π(6)²h = 12πh.

On recasting, the volume remains the same.

⇒ 12πh = 96π ⇒ h = 8 cm.

∴ Assertion (A) is true.

When one solid is converted (melted and recast) into another, the volume remains unchanged, and this is exactly the principle used to find the height.

∴ Reason (R) is true and is the correct explanation of A.

Hence, option 1 is the correct option.

By formula,

Volume of sphere = $\frac{4}{3} \pi r^3$

Given,

A solid sphere is cut into two identical hemispheres.

Volume of hemisphere = $\frac{2}{3} \pi r^3$

Volume of two identical hemispheres = $2 \times \frac{2}{3} \pi r^3$

= $\frac{4}{3} \pi r^3$

Thus, volume of a sphere = volume of two identical hemispheres.

So assertion (A) is true.

We know that,

Surface area of sphere = 4πr²

When a sphere is cut into two hemispheres, two new flat circular surfaces are created,

Surface area of a single hemisphere = Curved surface area + Area of its flat circular face

= 2πr² + πr²

= 3πr²

Total surface area of two hemispheres = 2 × 3πr² = 6πr².

Thus, the surface area of the original sphere ≠ the total surface area of the two hemispheres.

So reason (R) is false.

(A) is true, (R) is false.

Hence, option 3 is the correct option.

Given,

Diameter of container = 14 cm

Radius (R) = $\frac{\text{Diameter}}{2} = \frac{14}{2}$ = 7 cm

Height of container (H) = 16 cm

Diameter of rasgulla = 6 cm

Radius of rasgulla (r) = $\frac{\text{Diameter}}{2} = \frac{6}{2}$ = 3 cm

Volume of container = Volume of rasgullas + Volume of liquid

⇒ πR²H = $20 \times \frac{4}{3}$ πr³ + Volume of liquid

⇒ Volume of liquid = πR²H - $\frac{80}{3}$ πr³

⇒ Volume of liquid = π(R²H - $\frac{80}{3}r^3$)

⇒ Volume of liquid = 3.14 × (7² × 16 - $\frac{80}{3} \times 3^3$)

⇒ Volume of liquid = 3.14 × (784 - 720)

⇒ Volume of liquid = 3.14 × 64

⇒ Volume of liquid = 200.96 cm³.

Hence, volume of sweetened liquid in the container = 200.96 cm³.

Given,

Diameter of cone = 14 cm

Radius of cone (r) = $\frac{\text{Diameter}}{2} = \frac{14}{2}$ = 7 cm

Slant height (l) = 25 cm

Ratio of the radius and the height of a solid metallic right circular cylinder is 7 : 27.

Radius of cylinder (R) = 7x

Height of cylinder (H) = 27x

(a) Let height of cone be h cm.

By formula,

⇒ l² = r² + h²

⇒ 25² = 7² + h²

⇒ 625 = 49 + h²

⇒ h² = 625 - 49

⇒ h² = 576

⇒ h = $\sqrt{576}$ = 24 cm.

Hence, height of cone = 24 cm.

(b) Given,

A solid metallic right circular cylinder is melted and made into a cone.

∴ Volume of cylinder = Volume of cone

⇒ πR²H = $\frac{1}{3}πr^2h$

⇒ R²H = $\frac{1}{3}r^2h$

⇒ (7x)² × 27x = $\frac{1}{3} \times 7^2 \times 24$

⇒ 1323x³ = $\frac{1176}{3}$

⇒ 1323x³ = 392

⇒ x³ = $\frac{392}{1323}$

⇒ x³ = $\frac{8}{27}$

⇒ x³ = $\Big(\frac{2}{3}\Big)^3$

⇒ x = $\frac{2}{3}$

Height of cylinder (H) = 27x = $27 \times \frac{2}{3}$ = 18 cm.

Hence, height of cylinder = 18 cm.

(a) Let radius of smaller and larger cone be r and R, respectively.

Let height of smaller and larger cone be h and H, respectively.

Let slant height of smaller and larger cone be l and L respectively.

Given,

Ratio of their slant heights is 2 : 1.

∴ l : L = 2 : 1.

Given,

Curved surface area of a right circular cone is half of the other right circular cone.

∴ πrl = $\frac{1}{2}πRL$

⇒ rl = $\frac{1}{2}RL$

⇒ $\frac{r}{R} = \frac{L}{2l} = \frac{1}{2} \times \frac{1}{2}$

⇒ $\frac{r}{R} = \frac{1}{4}$

⇒ r : R = 1 : 4

Hence, ratio of the radii = 1 : 4.

(b) Let volume of cone with smaller curved surface area be v and that with larger curved surface area be V.

Given,

Volumes are in the ratio 3 : 1.

∴ v : V = 3 : 1

$$\begin{aligned}\Rightarrow \frac{\text{Volume of larger CSA cone}}{\text{Volume of smaller CSA cone}} = \frac{V}{v} \\[1em] \Rightarrow \frac{\frac{1}{3}πR^2H}{\frac{1}{3}πr^2h} = \frac{1}{3} \\[1em] \Rightarrow \frac{4^2}{1^2} \times \frac{H}{h} = \frac{1}{3} \\[1em] \Rightarrow \frac{16}{1} \times \frac{H}{h} = \frac{1}{3} \\[1em] \Rightarrow \frac{H}{h} = \frac{1}{3 \times 16} \\[1em] \Rightarrow \frac{H}{h} = \frac{1}{48} \\[1em] \Rightarrow \frac{h}{H} = \frac{48}{1}.\end{aligned}$$

Hence, ratio of heights = 48 : 1.

First, a sphere of radius (r) 7 cm is formed.

Volume of sphere = $\frac{4}{3}πr^3$

$$\begin{aligned}= \frac{4}{3} \times \frac{22}{7} \times 7^3 \\[1em] = \frac{4}{3} \times 22 \times 7^2 \\[1em] = \frac{4312}{3} \text{ cm}^3.\end{aligned}$$

Next, a right circular cone with radius (r₁) 14 cm is formed. Let height of cone be h cm.

Since same amount of clay is used to make cone and sphere.

∴ Volume of cone = Volume of sphere

$$\begin{aligned}\Rightarrow \frac{1}{3}πr_1^2h = \frac{4312}{3} \\[1em] \Rightarrow πr_1^2h = 4312 \\[1em] \Rightarrow \frac{22}{7} \times 14^2 \times h = 4312 \\[1em] \Rightarrow 22 \times 2 \times 14 \times h = 4312 \\[1em] \Rightarrow h = \frac{4312}{22 \times 2 \times 14} \\[1em] \Rightarrow h = \frac{4312}{616} = 7 \text{ cm}.\end{aligned}$$

Given,

The same clay is used to make a right circular cylinder of height (h₁) $\frac{7}{3}$ cm. Let its radius be r₂.

Since same amount of clay is used to make cylinder and sphere.

∴ Volume of cylinder = Volume of sphere

$$\begin{aligned}\Rightarrow πr_2^2h_1 = \frac{4312}{3} \\[1em] \Rightarrow \frac{22}{7} \times r_2^2 \times \frac{7}{3} = \frac{4312}{3} \\[1em] \Rightarrow 22 \times r_2^2 = 4312 \\[1em] \Rightarrow r_2^2 = \frac{4312}{22} \\[1em] \Rightarrow r_2^2 = 196 \\[1em] \Rightarrow r_2 = \sqrt{196} = 14 \text{ cm}.\end{aligned}$$

Total surface area of sphere = 4πr²

Total surface area of cylinder = 2πr₂(r₂ + h₁)

$$\begin{aligned}\Rightarrow \frac{\text{TSA sphere}}{\text{TSA cylinder}} = \frac{4πr^2}{2πr_2(r_2 + h_1)} \\[1em] = \frac{2r^2}{r_2(r_2 + h_1)} \\[1em] = \frac{2 \times 7^2}{14(14 + \frac{7}{3})} \\[1em] = \frac{98}{14 \times \frac{49}{3}} \\[1em] = \frac{98 \times 3}{14 \times 49} \\[1em] = \frac{3}{7}.\end{aligned}$$

Hence, height of cone = 7 cm, radius of cylinder = 14 cm and Total surface area of sphere : Total surface area of cylinder = 3 : 7.