Mensuration

Given radius = 5 cm and height = 10 cm.

∴ r = 5 cm, h = 10 cm.

Total surface area of a solid cylinder = 2πr(h + r) = 2π × 5 × (10 + 5) = 10π × 15 = 150π cm².

Hence, the total surface area of solid cylinder = 150π cm².

(i) Given diameter = 35 cm, height = 1.2 m = 1.2 × 100 = 120 cm.

We know,

radius = $\frac{\text{diameter}}{2} = \frac{35}{2} = 17.5$ cm.

Curved surface area = 2πrh = $2 \times \frac{22}{7} \times 120 \times 17.5 = \frac{92400}{7}$ = 13200 cm².

Hence, the outer lateral surface area of electric geyser = 13200 cm².

(ii) Volume of cylinder = πr²h.

Putting values we get,

$$\begin{aligned}\text{Volume of electric geyser } = \frac{22}{7} \times (17.5)^2 \times 120 \\[1em] = \frac{22 \times 306.25 \times 120}{7} \\[1em] = \frac{808500}{7} \\[1em] = 115500 \text{ cm}^3.\end{aligned}$$

Since 1000 cm³ = 1 litre so, 1 cm³ = $\frac{1}{1000}$ litre.

So, Volume = $115500 \times \frac{1}{1000}$ litres = 115.5 litres.

Hence, the capacity of electric geyser = 115.5 litres.

Given diameter = 7 cm, height = 12 cm.

We know,

radius = $\frac{\text{diameter}}{2} = \frac{7}{2} = 3.5$ cm.

Volume of cylinder = πr²h.

Putting values we get,

$$\begin{aligned}\text{Volume of milk in 1 glass } = \frac{22}{7} \times (3.5)^2 \times 12 \\[1em] = \frac{22 \times 12.25 \times 12}{7} \\[1em] = \frac{3234}{7} \\[1em] = 462 \text{ cm}^3.\end{aligned}$$

∴ Volume of milk in 1600 glasses = 1600 × 462 = 739200 cm³.

Since 1000 cm³ = 1 litre so, 1 cm³ = $\frac{1}{1000}$ litre.

So, Volume of milk in 1600 glasses = $739200 \times \frac{1}{1000}$ litres = 739.2 litres.

Hence, 739.2 litres of milk is required for serving 1600 students.

From figure,

The cylinder formed will have,

height = 16 cm.

and circumference of cross section = 22 cm.

We know circumference = 2πr.

$$\begin{aligned}\therefore 2πr = 22 \\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 22 \\[1em] \Rightarrow \frac{44}{7}r = 22 \\[1em] \Rightarrow r = \frac{22 \times 7}{44} \\[1em] \Rightarrow r = \frac{154}{44} \\[1em] \Rightarrow r = 3.5 \text{ cm.}\end{aligned}$$

Volume of cylinder = πr²h.

Putting values we get,

$$\begin{aligned}\text{Volume of cylinder } = \frac{22}{7} \times (3.5)^2 \times 16 \\[1em] = \frac{22 \times 12.25 \times 16}{7} \\[1em] = \frac{4312}{7} \\[1em] = 616 \text{ cm}^3.\end{aligned}$$

Hence, the volume of cylinder = 616 cm³.

A well needs to be formed which is 20 meters deep and has a diameter of 2 meters.

Height of well (h) = 20 m

Radius (r) = $\frac{\text{Diameter}}{2} = \frac{2}{2}$ = 1 m

Volume of soil to be dug out = Volume of well to be formed.

Volume of well = πr²h.

Substituting values we get,

$$\begin{aligned}\text{Volume of well }= \frac{22}{7} \times (1)^2 \times 20 \\[1em] = \frac{22}{7} \times 1 \times 20 \\[1em] = \frac{440}{7} \\[1em] = 62\frac{6}{7}.\end{aligned}$$

Hence, $62\frac{6}{7}$ m³ of soil must be dug out for the formation of well.

By formula,

Curved surface area of cylinder = 2πrh

$$\begin{aligned}\text{Curved surface area of cylinder }= 2 \times \frac{22}{7} \times 1 \times 20 \\[1em] = \frac{880}{7} m^2\end{aligned}$$

Rate of plastering = ₹ 300/m².

Cost of plastering = Curved surface area of cylinder x Rate

= $\frac{880}{7} \times 300 = \frac{264000}{7} = 37,714.28 \approx$ ₹ 37,700

Hence, the cost of plastering the inner curved surface area of the well = ₹ 37,700.

Given, diameter = 0.7 m

So radius = $\frac{\text{diameter}}{2} = \frac{0.7}{2} = 0.35 \text{ m}.$

Height = width = 1.2 m

Area covered in 1 revolution = Curved surface area of cylinder.

Curved surface area of cylinder = 2πrh

= $2 \times \frac{22}{7} \times 0.35 \times 1.2 = \frac{18.48}{7}$ = 2.64 m².

∴ Area covered in 1 revolution = Curved surface area of cylinder = 2.64 m².

Hence, the number of revolutions required to cover playground of size 120 m by 44 m

= $\frac{120 \times 44}{2.64} = \frac{5280}{2.64}$ = 2000.

Hence, the minimum number of revolutions required to cover playground of size 120 m by 44 m are 2000.

Volume of cylinder = πr²h.

Given, Volume of cylinder = 448 π cm³ and height = 7 cm.

∴ πr²h = 448 π
⇒ r²h = 448 (Dividing both sides by π)
⇒ r² × 7 = 448
⇒ r² = $\frac{448}{7}$
⇒ r² = 64
⇒ r = $\sqrt{64}$ = 8 cm.

Lateral surface area of cylinder = 2πrh = 2π × 8 × 7 = 112π cm².

Total surface area = 2πr(h + r) = 2π × 8 × (8 + 7) = 2π × 8 × 15 = 240π cm².

Hence, the lateral surface area of cylinder = 112π cm² and total surface area = 240π cm².

Given weight = 225 kg/m³

Given height = 7 m and diameter = 20 cm = $\frac{20}{100}$ = 0.2 m

Radius = $\frac{\text{Diameter}}{2} = \frac{0.2}{2}$ = 0.1 m

Volume of cylinder = πr²h.

Putting values in the formula we get,

$$\begin{aligned}\text{Volume of wooden pole } = \frac{22}{7} \times (0.1)^2 \times 7 \\[1em] = 22 \times 0.1 \times 0.1 \\[1em] = 0.22 \text{ m}^3.\end{aligned}$$

Since weight of 1 m³ of wooden pole = 225 kg.

∴ Weight of 0.22 m³ of wooden pole = 225 × 0.22 = 49.5 kg.

Hence, the weight of the wooden pole = 49.5 kg.

(i) Circumference = 2πr.

Given, circumference = 132 cm.

∴ 2πr = 132

⇒ $2 \times \frac{22}{7}r$ = 132

⇒ r = $\frac{132 \times 7}{2 \times 22} = \frac{924}{44}$ = 21 cm.

Hence, the radius of the cylinder = 21 cm.

(ii) Given height = 25 cm.

Volume of cylinder = πr²h.

Putting values we get,

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

Hence, the volume of the cylinder = 34650 cm³.

(i) Given, curved surface area of cylinder = 4400 cm².

We know that curved surface area of cylinder = 2πrh.

∴ 2πrh = 4400 .....(i)

Given, circumference of base = 110 cm.

We know that circumference = 2πr.

∴ 2πr = 110 .....(ii)

Dividing equation (i) by (ii),

$\Rightarrow \frac{2πrh}{2πr} = \frac{4400}{110}$

⇒ h = 40 cm.

Hence, the height of the cylinder = 40 cm.

(ii) We know that circumference = 2πr.

Given, circumference = 110 cm.

∴ 2πr = 110

⇒ $2 \times \frac{22}{7}r$ = 110

⇒ r = $\frac{110 \times 7}{2 \times 22} = \frac{770}{44}$ = 17.5 cm.

Volume of cylinder = πr²h.

Putting values we get,

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

Hence, the volume of the cylinder = 38500 cm³.

(i) Radius = $\frac{\text{Diameter}}{2} = \frac{20}{2}$ = 10 cm.

Curved surface area of cylinder = 2πrh.

Given, curved surface area of cylinder = 1000 cm².

∴ 2πrh = 1000

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times 10 \times h = 1000 \\[1em] \Rightarrow h = \frac{1000 \times 7}{2 \times 22 \times 10} \\[1em] \Rightarrow h = \frac{7000}{440} \\[1em] \Rightarrow h = 15.9 \text{ cm}.\end{aligned}$$

Hence, the height of the cylinder = 15.9 cm.

(ii) Volume of cylinder = πr²h.

Putting values we get,

$$\begin{aligned}\text{Volume of cylinder } = 3.14 \times (10)^2 \times 15.9 \\[1em] = 3.14 \times 100 \times 15.9 \\[1em] = 4992.6 \text{ cm}^3.\end{aligned}$$

Hence, the volume of the cylinder = 4992.6 cm³.

Height of cylindrical barrel of a pen = 7 cm.

Diameter = 5 mm = 0.5 cm. (As 10 mm = 1 cm)

Radius = $\frac{\text{Diameter}}{2} = \frac{0.5}{2} = 0.25$ cm.

Volume of cylinder = πr²h.

Putting values we get,

$$\begin{aligned}\text{Volume of barrel } = \frac{22}{7} \times (0.25)^2 \times 7 \\[1em] = 22 \times 0.0625 \\[1em] = 1.375 \text{ cm}^3.\end{aligned}$$

Ink in the bottle = One-fifth of a litre = $\frac{1}{5}$ x 1000 ml = 200 ml.

Since 1 ml = 1 cm³.

∴ 200 ml = 200 cm³.

Number of words written using 1.375 cm³ (full barrel) of ink = 310.

Number of words written using 200 cm³ (bottle) of ink = $\frac{\text{Volume of bottle}}{\text{Volume of barrel}} \times \text{Total words}$.

= $\frac{200}{1.375} \times 310$ = 145.45 × 310 = 45090.90

Rounding off to nearest 100 words = 45,100.

Hence, 45,100 words use a bottle of ink containing one-fifth litre of ink.

We know,

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

Curved surface area of cylinder = 2πrh

$$\begin{aligned}\frac{\text{Total surface area of cylinder}}{\text{Curved surface area of cylinder}} \\[1em] = \frac{2πr(h + r)}{2πrh} \\[1em] = \frac{(h + r)}{h} \\[1em] = \frac{7.5 + 3.5}{7.5} \\[1em] = \frac{11}{7.5} \\[1em] = \frac{110}{75} \\[1em] = \frac{22}{15}.\end{aligned}$$

Hence, the ratio of total surface area of cylinder to curved surface = 22 : 15.

For old cylinder,

Let Height = h and Radius = r.

So, for new cylinder,

Height = 2h and Radius = $\frac{r}{2}$.

We know that volume of cylinder = π × (radius)² × height.

∴ Volume of old cylinder = πr²h

and Volume of new cylinder = $π.\Big(\frac{r}{2}\Big)^2.2h$

Hence,

$$\begin{aligned}\Rightarrow \frac{\text{Volume of new cylinder}}{\text{Volume of old cylinder}} = \frac{π.\Big(\frac{r}{2}\Big)^2.2h}{πr^2h} \\[1em] = \frac{π.\Big(\frac{r^2}{4}\Big).2h}{πr^2h} \\[1em] = \frac{2πr^2h}{4πr^2h} \\[1em] = \frac{2}{4} \\[1em] = \frac{1}{2}.\end{aligned}$$

Hence, the ratio of the volume of the new cylinder to that of the original cylinder is 1 : 2.

Given, (h + r) = 37 cm, where h = height and r = radius of the cylinder.

Given, Total surface area = 1628 cm².

We know that Total surface area = 2πr(h + r).

∴ 2πr(h + r) = 1628

Putting values,

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r \times 37 = 1628 \\[1em] \Rightarrow r = \frac{1628 \times 7}{2 \times 22 \times 37} \\[1em] \Rightarrow r = \frac{11396}{1628} \\[1em] \Rightarrow r = 7 \text{ cm}.\end{aligned}$$

Since, h + r = 37.

⇒ h + 7 = 37
⇒ h = 37 - 7 = 30 cm.

Volume of cylinder = πr²h.

Putting values we get,

$$\begin{aligned}\text{Volume of cylinder } = \frac{22}{7}\times (7)^2 \times 30 \\[1em] = \frac{22 \times 49 \times 30}{7} \\[1em] = \frac{323400}{7} \\[1em] = 4620 \text{ cm}^3.\end{aligned}$$

Hence, the height of the cylinder = 30 cm and volume of cylinder = 4620 cm³

Given, Total surface area = 352 cm² and height = 10 cm.

We know that Total surface area = 2πr(h + r).

∴ 2πr(h + r) = 352

Putting values we get,

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r \times (10 + r) = 352 \\[1em] \Rightarrow \frac{2 \times 22 \times r \times (10 + r)}{7} = 352 \\[1em] \Rightarrow 44r(10 + r) = 352 \times 7 \\[1em] \Rightarrow 440r + 44r^2 = 2464 \\[1em]\end{aligned}$$

Dividing the complete equation by 44,

$$\begin{aligned}\Rightarrow 10r + r^2 = 56 \\[1em] \Rightarrow r^2 + 10r - 56 = 0 \\[1em] \Rightarrow r^2 + 14r - 4r - 56 = 0 \\[1em] \Rightarrow r(r + 14) - 4(r + 14) = 0 \\[1em] \Rightarrow (r - 4)(r + 14) = 0 \\[1em] \Rightarrow r - 4 = 0 \text{ or } r + 14 = 0 \\[1em] \Rightarrow r = 4 \text{ or } r = -14.\end{aligned}$$

Since radius cannot be negative hence, r ≠ -14.

∴ r = 4 cm.

Diameter = r × 2 = 4 × 2 = 8 cm.

Hence, the diameter of the base = 8 cm.

We know that, Total surface area = 2πr(h + r) and Curved surface area = 2πrh.

Given ratio between the curved surface and the total surface of a cylinder is 1 : 2.

$$\begin{aligned}\therefore \frac{2πrh}{2πr(h + r)} = \frac{1}{2} \\[1em] \Rightarrow \frac{h}{h + r} = \frac{1}{2} \\[1em] \Rightarrow 2h = h + r \\[1em] \Rightarrow 2h - h = r \\[1em] \Rightarrow h = r.\end{aligned}$$

Given, total surface area = 616 cm².

∴ 2πr(h + r) = 616

and h = r.

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

Since h = r, hence h = 7 cm.

Volume of cylinder = πr²h.

Putting values we get,

$$\begin{aligned}\text{Volume of cylinder } = \frac{22}{7}\times (7)^2 \times 7 \\[1em] = \frac{22 \times 49 \times 7}{7} \\[1em] = \frac{7546}{7} \\[1em] = 1078 \text{ cm}^3.\end{aligned}$$

Hence, the volume of cylinder = 1078 cm³.

If the amount of milk is same it means volume of milk in both jars will be equal.

Let the diameters of the jars be 3a and 4a.

Radius = $\frac{\text{Diameter}}{2}$

Hence, radius will be $\frac{3a}{2}$ and $\frac{4a}{2}$.

Let height of the two jars be h₁ and h₂.

Volume of cylinder = πr²h

Given,

Volume of 1st jar = Volume of 2nd jar.

$$\begin{aligned}\therefore π \times \Big(\frac{3a}{2}\Big)^2 \times h_1 = π \times \Big(\frac{4a}{2}\Big)^2 \times h_2 \\[1em] \Rightarrow π \times \frac{9a^2}{4} \times h_1 = π \times \frac{16a^2}{4} \times h_2\end{aligned}$$

Dividing both sides by π and multiplying by 4 we get,

$$\begin{aligned}\Rightarrow 9a^2.h_1 = 16a^2.h_2 \\[1em] \Rightarrow \frac{h_1}{h_2} = \frac{16a^2}{9a^2} \\[1em] \Rightarrow \frac{h_1}{h_2} = \frac{16}{9}\end{aligned}$$

Hence, ratio of the heights of jars = 16 : 9.

Suppose the sheet is rolled along length so the circumference of the base = 30 cm and height = 18 cm. Let radius be r₁.

or, 2πr₁ = 30

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r_1 = 30 \\[1em] \Rightarrow r_1 = \frac{30 \times 7}{2 \times 22} \\[1em] \Rightarrow r_1 = \frac{210}{44} = \frac{105}{22}.\end{aligned}$$

Suppose the sheet is rolled along breadth so the circumference of the base = 18 cm and height = 30 cm. Let radius be r₂.

or, 2πr₂ = 18

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r_2 = 18 \\[1em] \Rightarrow r_2 = \frac{18 \times 7}{2 \times 22} \\[1em] \Rightarrow r_2 = \frac{126}{44} = \frac{63}{22}.\end{aligned}$$

Volume of cylinder = πr²h

$$\begin{aligned}\therefore \frac{\text{Vol. of 1st cylinder}}{\text{Vol. of 2nd cylinder}} = \frac{π \times \Big(\frac{105}{22}\Big)^2 \times 18}{π \times \Big(\frac{63}{22}\Big)^2 \times 30} \\[1em] = \frac{105 \times 105 \times 18 \times 22^2}{63 \times 63 \times 30 \times 22^2} \\[1em] = \frac{105 \times 105 \times 18}{63 \times 63 \times 30} \\[1em] = \frac{198450}{119070} \\[1em] = \frac{50}{30} \\[1em] = \frac{5}{3}.\end{aligned}$$

Hence, the ratio of the volume of two cylinders = 5 : 3.

Internal diameter = 11.2 cm

Internal radius = $\frac{\text{Internal diameter}}{2}$

= $\frac{11.2}{2}$ = 5.6 cm.

   Thickness = External radius - Internal radius
⇒ 0.4 = External radius - Internal radius
⇒ External radius = 0.4 + Internal radius
⇒ External radius = 0.4 + 5.6 = 6.0 cm

Volume of hollow cylinder = π(R² - r²)h, where R = External radius and r = Internal radius.

Putting values we get,

$$\begin{aligned}\therefore \text{Volume of metal} = \frac{22}{7} \times (6^2 - (5.6)^2) \times 21 \\[1em] = \frac{22}{7} \times (36 - 31.36) \times 21 \\[1em] = \frac{22 \times 4.64 \times 21}{7} \\[1em] = 22 \times 4.64 \times 3 \\[1em] = 306.24 \text{ cm}^3.\end{aligned}$$

Hence, volume of metal = 306.24 cm³.

Volume of wood = π(R² - r²)h, where R = Radius of pencil and r = Radius of graphite.

Radius of pencil = $\frac{\text{Diameter of pencil}}{2}$

= $\frac{7}{2}$ = 3.5 mm

Radius of graphite = $\frac{\text{Diameter of graphite}}{2}$

= $\frac{1}{2}$ = 0.5 mm

Given height = 14 cm = 140 mm.

Putting values in formula,

$$\begin{aligned}\Rightarrow \text{ Volume of wood} = \frac{22}{7} \times ((3.5)^2 - (0.5)^2) \times 140 \\[1em] = \frac{22}{7} \times (12.25 - 0.25) \times 140 \\[1em] = \frac{22}{7} \times 12 \times 140 \\[1em] = 22 \times 12 \times 20 \\[1em] = 5280 \text{ mm}^3 \\[1em] = 5280 \times \Big(\frac{1}{10}\Big)^3 \text{ cm}^3 \\[1em] = 5.28 \text{ cm}^3.\end{aligned}$$

Volume of graphite = πr²h.

Putting values we get,

$$\begin{aligned}\text{Volume of graphite } = \frac{22}{7}\times (0.5)^2 \times 140 \\[1em] = \frac{22 \times 0.25 \times 140}{7} \\[1em] = \frac{770}{7} \\[1em] = 110 \text{ mm}^3 \\[1em] = 110 \times \Big(\frac{1}{10}\Big)^3 \\[1em] = 0.11 \text{ cm}^3\end{aligned}$$

Hence, the volume of wood = 5.28 cm³ and volume of graphite = 0.11 cm³.

Internal radius = $\frac{\text{Internal Diameter}}{2}$

= $\frac{35}{2}$ = 17.5 cm

External radius = Thickness + Internal radius = 7 + 17.5 = 24.5 cm

Height = 2 m = 2 × 100 cm = 200 cm.

Volume of hollow cylinder = π(R² - r²)h, where R = External radius and r = Internal radius.

Putting values we get,

$$\begin{aligned}\therefore \text{Volume of roller} = \frac{22}{7} \times ((24.5)^2 - (17.5)^2) \times 200 \\[1em] = \frac{22}{7} \times (600.25 - 306.25) \times 200 \\[1em] = \frac{22 \times 294 \times 200}{7} \\[1em] = 22 \times 42 \times 200 \\[1em] = 184800 \text{ cm}^3.\end{aligned}$$

Since, 1 cm³ of iron weights 8 g.

∴ 184800 cm³ weights = 184800 × 8 = 1478400 g.

Since 1 kg = 1000g or 1g = $\frac{1}{1000}kg$.

∴ 1478400 g = $1478400 \times \frac{1}{1000} = 1478.4$ kg.

Hence, the weight of the roller = 1478.4 kg.

Given, l = 10 cm and r = 7 cm.

Curved surface area of cone = πrl.

Putting values in above equation we get,

Curved surface area = $\frac{22}{7} \times 7 \times 10$ = 22 × 10 = 220 cm².

Hence, the curved surface area of right circular cone = 220 cm².

Radius of base = $\frac{\text{Diameter of base}}{2}$

= $\frac{10.5}{2}$ = 5.25 cm.

Given, l = 10 cm.

Curved surface area of cone = πrl.

Putting values in above equation we get,

Curved surface area = $\frac{22}{7} \times 5.25 \times 10$

= $\frac{1155}{7}$ = 165 cm².

Hence, the curved surface area of right circular cone = 165 cm².

(i) Given, Curved surface area of a cone = 308 cm² and slant height = 14 cm.

We know that,

Curved surface area of cone = πrl.

∴ πrl = 308

⇒ $\frac{22}{7}$ x r x 14 = 308

⇒ 22 × r × 2 = 308

⇒ 44r = 308

⇒ r = $\frac{308}{44}$ = 7 cm.

Hence, the radius of the base = 7 cm.

(ii) Total surface area of cone = πr(l + r)

Putting values in above equation we get,

Total surface area = $\frac{22}{7}$ x 7 x (14 + 7) = 22 × 1 × 21 = 462 cm².

Hence, the total surface area of right circular cone = 462 cm².

(i) Given, r = 6 cm and h = 7 cm.

Volume of cone = $\frac{1}{3}πr^2h$

Putting values in equation we get,

Volume of cone = $\frac{1}{3} \times \frac{22}{7} \times 6^2 \times 7 = \frac{22 × 36 × 7}{3 \times 7}$ = 22 × 12 = 264 cm³.

Hence, the volume of cone = 264 cm³.

(ii) Given, r = 3.5 cm and h = 12 cm.

Volume of cone = $\frac{1}{3}πr^2h$

Putting values in equation we get,

Volume of cone = $\frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 12 = \frac{22 × 12.25 × 12}{3 \times 7}$ = 22 × 1.75 × 4 = 154 cm³.

Hence, the volume of cone = 154 cm³.

(i) Given, l = 25 cm and r = 7 cm.

We know that,

    l² = r² + h²
⇒ 25² = 7² + h²
⇒ h² = 25² - 7²
⇒ h² = 625 - 49
⇒ h² = 576
⇒ h = $\sqrt{576}$ = 24 cm.

Volume of cone = $\frac{1}{3}πr^2h$

Putting values in equation we get,

Volume of cone = $\frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 24 = \frac{22 × 49 × 24}{3 \times 7}$ = 22 × 7 × 8 = 1232 cm³.

Since 1 litre = 1000 cm³ or, 1 cm³ = $\frac{1}{1000}$ litre.

∴ 1232 cm³ = $1232 \times \frac{1}{1000}$ litre = 1.232 litre.

Hence, the volume of cone = 1.232 litre.

(ii) Given, l = 13 cm and h = 12 cm.

We know that,

    l² = r² + h²
⇒ 13² = r² + 12²
⇒ r² = 13² - 12²
⇒ r² = 169 - 144
⇒ r² = 25
⇒ r = $\sqrt{25}$ = 5 cm.

Volume of cone = $\frac{1}{3}πr^2h$

Putting values in equation we get,

Volume of cone = $\frac{1}{3} \times \frac{22}{7} \times (5)^2 \times 12 = \frac{22 × 25 × 12}{3 \times 7} = \frac{6600}{21}$cm³.

Since 1 litre = 1000 cm³ or, 1 cm³ = $\frac{1}{1000}$ litre.

∴ $\frac{6600}{21}$ cm³ = $\frac{6600}{21} \times \frac{1}{1000} = \frac{66}{210} = \frac{11}{35}$ litres.

Hence, the volume of cone = $\frac{11}{35}$ litres.

Radius = $\frac{\text{Diameter}}{2} = \frac{3.5}{2} = 1.75$ m.

Height = 12 m.

Volume of cone = $\frac{1}{3}πr^2h$

Putting values in equation we get,

Volume of cone,

$= \frac{1}{3} \times \frac{22}{7} \times (1.75)^2 \times 12 = \frac{22 × 3.0625 × 12}{3 \times 7} = \frac{808.5}{21} = 38.5$ m³.

Since, 1 m³ = 1 kiloliters.

Hence, capacity of pit = 38.5 kiloliters.

We know,

Volume of cone = $\frac{1}{3}πr^2h$.

Given, Volume of cone = 48π cm³ and h = 9 cm.

$$\begin{aligned}\Rightarrow \frac{1}{3}πr^2h = 48π \\[1em] \Rightarrow r^2 = \frac{48π \times 3}{π \times h} \\[1em] \Rightarrow r^2 = \frac{48π \times 3}{π \times 9} \\[1em] \Rightarrow r^2 = 16 \\[1em] \Rightarrow r = \sqrt{16} = 4 \text{ cm}.\end{aligned}$$

Diameter = 2 × Radius = 2 × 4 = 8 cm.

Hence, the diameter of the base = 8 cm.

We know,

Volume of cone = $\frac{1}{3}πr^2h$.

Given, Volume of cone = 1570 cm³ and h = 15 cm.

$$\begin{aligned}\Rightarrow \frac{1}{3}πr^2h = 1570 \\[1em] \Rightarrow \frac{1}{3} \times 3.14 \times r^2 \times 15 = 1570 \\[1em] \Rightarrow r^2 = \frac{1570 \times 3}{15 \times 3.14} \\[1em] \Rightarrow r^2 = \frac{4710}{47.1} \\[1em] \Rightarrow r^2 = 100 \\[1em] \Rightarrow r = \sqrt{100} = 10 \text{ cm}.\end{aligned}$$

Hence, the radius of the base = 10 cm.

Cost = ₹ 210 / 100 m² = ₹ 2.1 / m².

Given, l = 25 m and diameter = 14 m.

Radius = $\frac{\text{Diameter}}{2}$ = $\frac{14}{2}$ = 7m.

Curved surface area of cone = πrl.

Putting values in equation,

Curved surface area = $\frac{22}{7} \times 7 \times 25$ = 22 × 25 = 550 m².

Cost of white washing 1 m² = ₹ 2.1

∴ Cost of white washing 550 m² = ₹ 2.1 × 550 = ₹ 1155.

Hence, the cost of white washing 550 m² = ₹ 1155.

(i) Given r = 24 m and h = 10 m.

We know that,

$l = \sqrt{r^2 + h^2}$

Putting values in the formula we get,

$$\begin{aligned}\Rightarrow l = \sqrt{(24)^2 + (10)^2} \\[1em] \Rightarrow l = \sqrt{576 + 100} \\[1em] \Rightarrow l = \sqrt{676} = 26 \text{ cm}.\end{aligned}$$

Hence, the slant height of the cone = 26 cm.

(ii) Curved surface area of cone = πrl.

Putting values in equation,

Curved surface area of tent = $\frac{22}{7} \times 24 \times 26 = \frac{13728}{7}$ m².

Cost of 1 m² of canvas = ₹ 70

∴ Cost of $\frac{13728}{7}$ m² of canvas = ₹ $\frac{13728}{7}$ × 70 = ₹ 137280.

Hence, the cost of canvas required to make tent = ₹ 137280.

Given, r = 7 cm and h = 24 cm.

We know that,

$l = \sqrt{r^2 + h^2}$

Putting values in the formula we get,

$$\begin{aligned}\Rightarrow l = \sqrt{(7)^2 + (24)^2} \\[1em] \Rightarrow l = \sqrt{49 + 576} \\[1em] \Rightarrow l = \sqrt{625} = 25 \text{ cm}.\end{aligned}$$

Curved surface area of cone = πrl.

Putting values in equation,

Curved surface area of cap = $\frac{22}{7} \times 7 \times 25$ = 22 × 25 = 550 cm².

∴ Curved surface area of 10 caps = 10 × 550 = 5500 cm².

Hence, the area of the cloth required to make 10 such caps is 5500 cm².

Let radius of cones be 3a and 4a.

Since, height of both cones is same let it be h.

We know volume of cone = $\frac{1}{3}πr^2h$.

Ratio of volume of two cones = $\frac{\text{Volume of cone 1}}{\text{Volume of cone 2}}$

$\frac{\text{Volume of cone 1}}{\text{Volume of cone 2}} = \frac{\frac{1}{3}π \times (3a)^2 \times h}{\frac{1}{3}π \times (4a)^2 \times h} = \frac{(3a)^2}{(4a)^2} = \frac{9a^2}{16a^2} = \frac{9}{16}.$

Hence, the ratio of the volume of two cones = 9 : 16.

Let height of cones be 5a and 2a and radius of cones be 2b and 5b.

We know volume of cone = $\frac{1}{3}πr^2h$.

Ratio of volume of two cones = $\frac{\text{Vol. of Cone 1}}{\text{Vol. of Cone 2}}$

$$\begin{aligned}\frac{\text{Vol. of Cone 1}}{\text{Vol. of Cone 2}} = \frac{\frac{1}{3}π \times (2b)^2 \times 5a}{\frac{1}{3}π \times (5b)^2 \times 2a} \\[1em] = \frac{4b^2 \times 5a}{25b^2 \times 2a} \\[1em] = \frac{20ab^2}{50ab^2} \\[1em] = \frac{2}{5}.\end{aligned}$$

Hence, the ratio of the volume of two cones = 2 : 5.

(i) Let the radius and height of the bigger cone be r and h respectively.

So, smaller cone's radius = $\frac{r}{2}$ and height = $\frac{h}{2}$.

We know volume of cone = $\frac{1}{3}πr^2h$.

Ratio of volume of two cones = $\frac{\text{Vol. of smaller cone}}{\text{Vol. of bigger cone}}$

$$\begin{aligned}\frac{\text{Vol. of smaller cone}}{\text{Vol. of bigger cone}} = \frac{\frac{1}{3}π \times \Big(\frac{r}{2}^2\Big) \times \frac{h}{2}}{\frac{1}{3}π \times r^2 \times h} \\[1em] = \frac{\frac{1}{3}π \times \frac{r^2}{4} \times \frac{h}{2}}{\frac{1}{3}π \times r^2 \times h} \\[1em] = \frac{\frac{1}{3}π \times r^2 \times h}{8 \times \frac{1}{3}π \times r^2 \times h} \\[1em] = \frac{1}{8}.\end{aligned}$$

Hence, the ratio of the volumes of cone = 1 : 8.

(ii) Let the slant height of bigger cone be l .

l = $\sqrt{r^2 + h^2}.$

Let slant height of smaller cone be l₁.

$$\begin{aligned}l_1 = \sqrt{\Big(\frac{r}{2}\Big)^2 + \Big(\frac{h}{2}^2\Big)} \\[1em] = \sqrt{\frac{r^2}{4} + \frac{h^2}{4}} \\[1em] = \sqrt{\frac{r^2 + h^2}{4}} \\[1em] = \frac{1}{2}\sqrt{r^2 + h^2} \\[1em] = \frac{1}{2}l.\end{aligned}$$

We know that lateral surface area of cone = π × radius × slant height.

Ratio of lateral surface area of two cones

$$\begin{aligned}= \frac{\text{Curved surface area of smaller cone}}{\text{Curved surface area of bigger cone}} \\[1em] = \frac{π \times \frac{r}{2} \times \frac{l}{2}}{π \times r \times l} \\[1em] = \frac{πrl}{4πrl} \\[1em] = \frac{1}{4}.\end{aligned}$$

Hence, the ratio of the lateral surface area of cones = 1 : 4.

Given diameter of conical tent = 20 m and l = 42 m.

Radius = $\frac{\text{Diameter}}{2} = \frac{20}{2}$ = 10 m.

We know that curved surface area of cone = π × radius × slant height.

Putting values we get,

Curved surface area of tent = $\frac{22}{7}$ x 10 x 42 = 22 × 10 × 6 = 1320 m².

10% of 1320 = $\frac{10}{100}$ x 1320 = 132 m².

Total area of canvas required for making tent = 1320 + 132 = 1452 m².

Area of rectangular cloth = l × b.

∴ l × b = 1452
⇒ l × 2 = 1452
⇒ l = $\frac{1452}{2} = 726$ m.

Since, the cost of canvas = ₹ 80/meter.

∴ The cost of 726 m of canvas = ₹ 726 × 80 = ₹ 58080.

Hence, the length of canvas required to make conical tent is 726 m and the cost of canvas = ₹ 58080.

Given, circumference of base = 44 cm.

⇒ 2πr = 44

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

Given, l = 25 cm.

We know that,

    l² = r² + h²
⇒ 25² = 7² + h²
⇒ h² = 25² - 7²
⇒ h² = 625 - 49
⇒ h² = 576
⇒ h = $\sqrt{576}$ = 24 cm.

Volume of cone = $\frac{1}{3}πr^2h$

Putting values in equation we get,

Volume of cone = $\frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 24$

= $\frac{22 × 49 × 24}{3 \times 7}$

= 22 × 7 × 8 = 1232 cm³.

Curved surface area = πrl.

Putting values in equation we get,

Curved surface area = $\frac{22}{7} \times 7 \times 25$ = 22 × 25 = 550 cm².

Given, area of base = 616 cm².

Area of base = πr².

∴ πr² = 616

$$\begin{aligned}\Rightarrow \frac{22}{7} \times r^2 = 616 \\[1em] \Rightarrow r^2 = \frac{616 \times 7}{22} \\[1em] \Rightarrow r^2 = \frac{4312}{22} \\[1em] \Rightarrow r^2 = 196 \\[1em] \Rightarrow r = \sqrt{196} = 14 \text{ cm}.\end{aligned}$$

Volume of cone = $\frac{1}{3}πr^2h$

$$\begin{aligned}\therefore \frac{1}{3}πr^2h = 9856 \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times (14)^2 \times h = 9856 \\[1em] \Rightarrow h = \frac{9856 \times 3 \times 7}{22 \times 14 \times 14} \\[1em] \Rightarrow h = \frac{206976}{4312} \\[1em] \Rightarrow h = 48 \text{ cm}.\end{aligned}$$

(i) We know that,

    l² = r² + h²
⇒ l² = 14² + 48²
⇒ l² = 196 + 2304
⇒ l² = 2500
⇒ l = $\sqrt{2500}$ = 50 cm.

Hence, the slant height of cone = 50 cm.

(ii) Total surface area of cone = πr(l + r).

Putting values in equation we get,

Total surface area of cone = $\frac{22}{7} \times 14 \times (50 + 14)$ = 22 × 2 × 64 = 2816 cm².

Hence, the total surface area of cone = 2816 cm².

Since, triangle is revolved about 8 cm side so,

h = 8 cm, r = 6 cm and l = 10 cm.

Volume of cone = $\frac{1}{3}πr^2h$

Putting values in equation we get,

Volume of cone = $\frac{1}{3} \times 3.14 \times (6)^2 \times 8$

= $\frac{904.32}{3}$

= 301.44 cm³.

Curved surface area = πrl.

Putting values in equation we get,

Curved surface area = 3.14 × 6 × 10 = 188.4 cm².

Hence, the volume of cone = 301.44 cm³ and curved surface area = 188.4 cm².

Let OAB be the given cone of height 30 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:

Then △OND ~ △OMB.

∴ $\frac{r}{R} = \frac{h}{30}$ .....(i)

According to given,

Volume of cone OCD = $\frac{1}{27}$ Volume of cone OAB

$\therefore \frac{1}{3}πr^2h = \frac{1}{27} \times \frac{1}{3}πR^2 \times 30$

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

$$\begin{aligned}\Rightarrow r^2h = \frac{30R^2}{27} \\[1em] \Rightarrow \frac{r^2}{R^2} = \frac{30}{27h} \\[1em] \Rightarrow \Big(\frac{r}{R}\Big)^2 = \frac{10}{9h}.\end{aligned}$$

Using (i)

$$\begin{aligned}\Rightarrow \Big(\frac{h}{30}\Big)^2 = \frac{10}{9h} \\[1em] \Rightarrow \frac{h^2}{900} = \frac{10}{9h} \\[1em] \Rightarrow h^3 = \frac{10 \times 900}{9} \\[1em] \Rightarrow h^3 = 10 \times 100 \\[1em] \Rightarrow h^3 = 1000 \\[1em] \Rightarrow h^3 = 10^3 \\[1em] \Rightarrow h = 10 \text{ cm}.\end{aligned}$$

The height of the cone OCD = 10 cm.

∴ The section is cut at the height of (30 - 10) cm = 20 cm.

Hence, the section cut is above 20 cm from the base.

(i) Length of the arc of the semi-circular sheet = $\frac{1}{2} \times 2πr$

= πr = $\frac{22}{7}$ x 35 = 110 cm

Let r cm be the radius of the cone, then

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

Hence, the radius of the cone is 17.5 cm.

(ii) Curved surface area of cone = area of semi-circular sheet.

$$\begin{aligned}= \frac{1}{2} \times πr^2 \\[1em] = \frac{1}{2} \times \frac{22}{7} \times (35)^2 \\[1em] = \frac{22}{14} \times 1225 \\[1em] = \frac{26950}{14} \\[1em] = 1925 \text{ cm}^2\end{aligned}$$

Hence, the curved surface area of the cone = 1925 cm².

Given, r = 14 cm.

Surface area of sphere = 4πr².

Putting values in equation we get,

Surface area of sphere =

$4 \times \frac{22}{7} \times (14)^2 = \frac{4 \times 22 \times 196}{7} = \frac{17248}{7} = 2464$ cm².

Volume of sphere = $\frac{4}{3}πr^3$

Putting values in equation we get,

Volume of sphere =

$\frac{4}{3} \times \frac{22}{7} \times (14)^3 = \frac{4 \times 22 \times 2744}{21} = \frac{241472}{21} = 11498\frac{2}{3}$ cm³.

Hence, the surface area of sphere = 2464 cm² and volume of sphere = $11498\frac{2}{3}$ cm³.

Given, diameter = 21 cm.

Radius = $\frac{\text{Diameter}}{2} = \frac{21}{2} = 10.5$ cm.

Surface area of sphere = 4πr².

Putting values in equation we get,

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

Volume of sphere = $\frac{4}{3}πr^3$

Putting values in equation we get,

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

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

Volume of sphere = $\frac{4}{3}πr^3$

Putting values in equation we get,

$$\begin{aligned}\text{Volume of sphere } = \frac{4}{3} \times \frac{22}{7} \times (4.9)^3 \\[1em] = \frac{4 \times 22 \times 117.649}{21} \\[1em] = \frac{10353.112}{21} \\[1em] = 493 \text{ cm}^3.\end{aligned}$$

Since, the density of the metal is 7.8 g per cm³.

Mass = Volume × Density.

So, the mass of the metallic sphere 493 × 7.8 = 3845441.6 g

= $\frac{3845441.6}{1000}$ = 3.845 kg.

Hence, the mass of the metallic sphere is approx 3.85 kg.

Surface area of sphere = 4πr².

Given, surface area = 154 cm².

∴ 4πr² = 154

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

Diameter = 2 × radius = 2 × 3.5 = 7 cm.

Hence, the diameter of sphere = 7 cm.

(i) Given, radius = 21 cm.

Curved surface area of hemisphere = 2πr².

Putting values in equation we get,

$$\begin{aligned}\text{Curved surface area of hemisphere } = 2 \times \frac{22}{7} \times (21)^2 \\[1em] = \frac{2 \times 22 \times 441}{7} \\[1em] = \frac{19404}{7} = 2772 \text{ cm}^2.\end{aligned}$$

Hence, the curved surface area of hemisphere = 2772 cm².

(ii) Total surface area of hemisphere = 3πr².

Putting values in equation we get,

Total surface area of hemisphere = 3 x $\frac{22}{7}$ x (21)²

$$\begin{aligned}= \frac{3 \times 22 \times 441}{7} \\[1em] = \frac{29106}{7} = 4158 \text{ cm}^2.\end{aligned}$$

Hence, the total surface area of hemisphere = 4158 cm².

Given, internal diameter = 10.5 cm.

Internal radius = $\frac{\text{Internal diameter}}{2}$

= $\frac{10.5}{2} = 5.25$ cm.

Internal curved surface area of hemispherical shell = 2πr².

Putting values in equation we get,

Internal curved surface area of hemisphere = 2 x $\frac{22}{7}$ x (5.25)²

$$\begin{aligned}= \frac{2 \times 22 \times 27.5625}{7} \\[1em] = \frac{1212.75}{7} = 173.25 \text{ cm}^2.\end{aligned}$$

The cost of tin-plating it on the inside at the rate of ₹ 16/100 cm² or ₹ 0.16/cm².

∴ Cost of tin-plating 173.25 cm² = 173.25 × 0.16 = ₹ 27.72.

Hence, the cost of tin-plating = ₹ 27.72.

Surface area of sphere = 4πr².

Given, radius in 1st case = 7 cm and in 2nd case = 14 cm.

$$\begin{aligned}\frac{\text{Surface area in 1st case}}{\text{Surface area in 2nd case}} = \frac{4 \times π \times (7)^2}{4 \times π \times (14)^2} \\[1em] = \frac{7 \times 7 }{14 \times 14} \\[1em] = \frac{49}{196} \\[1em] = \frac{1}{4}\end{aligned}$$

Hence, the ratio of the surface areas of the balloon in two cases is 1 : 4.

Let the side of the cube be a cm and let radius of sphere be r cm.

Surface area of sphere = 4πr².

Surface area of cube = 6a².

Given,
surface area of sphere = surface area of cube.

∴ 4πr² = 6a²

⇒ $\frac{r^2}{a^2} = \frac{6}{4π}$

⇒ $\frac{r}{a} = \sqrt{\frac{6}{4π}}$.

Volume of sphere = $\frac{4}{3}πr^3$.

Volume of cube = a³.

Ratio of volume of sphere to volume of cube is

$$\begin{aligned}\Rightarrow \frac{\text{Volume of sphere}}{\text{Volume of cube}} = \frac{\frac{4}{3}πr^3}{a^3} \\[1em] = \frac{4πr^3}{3a^3} \\[1em] = \frac{4π}{3} \times \frac{r^3}{a^3} \\[1em] = \frac{4π}{3} \times \Big(\frac{r}{a}\Big)^3 \\[1em] = \frac{4π}{3} \times \Big(\sqrt{\frac{6}{4π}}\Big)^3 \\[1em] = \frac{4π}{3} \times \frac{6}{4π} \times \sqrt{\frac{6}{4π}} \\[1em] = \frac{4π}{3} \times \frac{6}{4π} \times \frac{1}{2}\sqrt{\frac{6}{π}} \\[1em] = \frac{24π}{24π}\sqrt{\frac{6}{π}} \\[1em] = \sqrt{\frac{6}{π}}.\end{aligned}$$

Hence proved that the ratio is $\sqrt{6} : \sqrt{π}$.

Let the radii of two spheres be 3a and 7a.

(i) Volume of sphere = $\frac{4}{3}πr^3$.

$$\begin{aligned}\frac{\text{Vol. of Sphere 1}}{\text{Vol. of Sphere 2}} = \frac{\frac{4}{3}π(3a)^3}{\frac{4}{3}π(7a)^3} \\[1em] = \frac{\frac{4}{3}π \times 27a^3}{\frac{4}{3}π \times 343a^3} \\[1em] = \frac{27}{343}.\end{aligned}$$

Hence, the ratio of the volumes of two spheres is 27 : 343.

(ii) Surface area of sphere = 4πr².

$$\begin{aligned}\frac{\text{Surface area of Sphere 1}}{\text{Surface area of Sphere 2}} = \frac{4π(3a)^2}{4π(7a)^2} \\[1em] = \frac{4π \times 9a^2}{4π \times 49a^2} \\[1em] = \frac{9}{49}.\end{aligned}$$

Hence, the ratio of the surface areas of two spheres is 9 : 49.

Given,
ratio of the volumes of the two spheres is 125 : 64.

$$\begin{aligned}\therefore \frac{\text{Vol. of Sphere 1}}{\text{Vol. of Sphere 2}} = \frac{125}{64} \\[1em] \Rightarrow \frac{\frac{4}{3}π(r_1)^3}{\frac{4}{3}π(r_2)^3} = \frac{125}{64} \\[1em] \Rightarrow \frac{(r_1)^3}{(r_2)^3} = \frac{5^3}{4^3} \\[1em] \Rightarrow \frac{r_1}{r_2} = \frac{5}{4}.\end{aligned}$$

Surface area of sphere = 4πr².

$$\begin{aligned}\therefore \frac{\text{Surface area of Sphere 1}}{\text{Surface area of Sphere 2}} = \frac{4π(r_1)^2}{4π(r_2)^2} \\[1em] = \Big(\frac{r_1}{r_2}\Big)^2 \\[1em] = \Big(\frac{5}{4}\Big)^2 \\[1em] = \frac{25}{16}.\end{aligned}$$

Hence, the ratio of the surface areas of two spheres is 25 : 16.

We know that Surface area of sphere = 4πr².

Given,
Surface area of sphere = 154 cm².

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

Volume of sphere = $\frac{4}{3}πr^3$

Putting values in equation we get,

Volume of sphere =

$$\begin{aligned}\frac{4}{3} \times \frac{22}{7} \times (3.5)^3 \\[1em] = \frac{4 \times 22 \times 42.875}{21} \\[1em] = \frac{3773}{21} \\[1em] = \frac{539 \times 7}{3 \times 7} \\[1em] = \frac{539}{3} \\[1em] = 179\frac{2}{3} cm^3.\end{aligned}$$

Hence, the volume of sphere = $179\frac{2}{3}$ cm³.

Volume of sphere = $\frac{4}{3}πr^3$.

Given,
Volume of sphere = $179\frac{2}{3}$

$$\begin{aligned}\therefore \frac{4}{3}πr^3 = 179\frac{2}{3} \\[1em] \Rightarrow \frac{4}{3} \times \frac{22}{7} \times r^3 = \frac{539}{3} \\[1em] \Rightarrow \frac{88}{21} \times r^3 = \frac{539}{3} \\[1em] \Rightarrow r^3 = \frac{539 \times 21}{3 \times 88} \\[1em] \Rightarrow r^3 = \frac{539 \times 7}{88} \\[1em] \Rightarrow r^3 = \frac{3773}{88} \\[1em] \Rightarrow r^3 = 42.875 \\[1em] \Rightarrow r = (42.875)^{\frac{1}{3}} \\[1em] \Rightarrow r = 3.5 \text{ cm}.\end{aligned}$$

Surface area of sphere = 4πr².

Putting values in equation we get,

Surface area of sphere = $4 \times \frac{22}{7} \times (3.5)^2$

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

Hence, the radius of the sphere = 3.5 cm and surface area of sphere = 154 cm².

Volume of hemisphere = $\frac{2}{3}πr^3$.

Putting values we get,

$$\begin{aligned}\text{Volume of hemisphere} = \frac{2}{3}πr^3 \\[1em] = \frac{2}{3}π(3.5)^3 \\[1em] = \frac{2}{3} \times \frac{22}{7} \times 42.875 \\[1em] = \frac{2 \times 22 \times 42.875}{3 \times 7} \\[1em] = \frac{1886.5}{21} \\[1em] = \frac{18865}{210} \\[1em] = \frac{539}{6} \\[1em] = 89\frac{5}{6} \text{ cm}^3\end{aligned}$$

Hence, the volume of water in the hemispherical bowl = $89\frac{5}{6} \text{ cm}^3$.

Given,
surface area of the sphere = 1256 cm².

We know that, surface area of sphere = 4πr².

∴ 4πr² = 1256

$$\begin{aligned}\Rightarrow 4 \times 3.14 \times r^2 = 1256 \\[1em] \Rightarrow r^2 = \frac{1256}{3.14 \times 4} \\[1em] \Rightarrow r^2 = \frac{1256}{12.56} \\[1em] \Rightarrow r^2 = 100 \\[1em] \Rightarrow r = \sqrt{100} \\[1em] \Rightarrow r = 10 \text{ cm}.\end{aligned}$$

Total surface area of hemisphere = 3πr².

Putting values we get,

Total surface area of hemisphere = $3 \times 3.14 \times (10)^2$

$$\begin{aligned}= 3 \times 3.14 \times 100 \\[1em] = 942 \text{ cm}^2.\end{aligned}$$

Volume of hemisphere = $\frac{2}{3}πr^3$.

Volume of hemisphere = $\frac{2}{3} \times 3.14 \times 10^3$

$$\begin{aligned}= \frac{2}{3} \times 3.14 \times 1000 \\[1em] = \frac{6280}{3} \\[1em] = 2093\frac{1}{3} \text{cm}^3.\end{aligned}$$

Hence, the surface area of hemisphere = 942 cm² and volume of hemisphere = $2093\frac{1}{3}\text{ cm}^3.$

From figure,

Diameter of the biggest hole = 30 cm.

Radius (r) = $\frac{\text{Diameter}}{2} = \frac{30}{2} = 15$ cm.

Height = 70 cm.

Volume of cuboidal block = l x b x h.

Putting values we get,

Volume of cuboidal block = 70 × 30 × 30 = 63000 cm³.

Volume of cylinder = πr²h.

Putting values we get,

$$\begin{aligned}\text{Volume of cylindrical hole } = \frac{22}{7} \times 15^2 \times 70 \\[1em] = \frac{22 \times 225 \times 70}{10} \\[1em] = \frac{346500}{7} \\[1em] = 49500 \text{ cm}^3.\end{aligned}$$

Volume of wood left in the block = Volume of cuboidal block - Volume of cylindrical hole = 63000 - 49500 = 13500 cm³.

Hence, the volume of wood left in the block is 13500 cm³.

From figure,

Edge of cubical part = 28 cm.

Diameter of cylindrical part = 28 cm

radius = $\frac{\text{Diameter}}{2}$

= $\frac{\text{28}}{2}$ = 14 cm.

Height of cylinder = 28 cm.

Volume of cube = (side)³ = (28)³ = 21952 cm³.

Volume of cylinder = πr²h.

Putting values we get,

$$\begin{aligned}\text{Volume of cylinder } = \frac{22}{7} \times 14^2 \times 28 \\[1em] = \frac{22 \times 196 \times 28}{7} \\[1em] = \frac{120736}{7} \\[1em] = 17248 \text{ cm}^3.\end{aligned}$$

Total volume of trophy = Volume of cube + Volume of cylinder
= 21952 + 17248 = 39200 cm³.

Cost of 1 cm³ of glass = ₹0.75

Total cost of glass = 39200 × 0.75 = ₹29400.

Hence, the cost of making the trophy is ₹29400.

Edge of a cube = 14 cm.

Volume = (side)³ = (14)³ = 2744 cm³.

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

Diameter of the cone cut out from it = 14 cm.

Radius = $\frac{\text{Diameter}}{2}$

= $\frac{\text{14}}{2}$ = 7 cm.

Volume of cone = $\frac{1}{3}πr^2h$

Height = 14 cm.

Putting values we get,

$$\begin{aligned}\text{Volume of cone } = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 14 \\[1em] = \frac{22 \times 49 \times 14}{3 \times 7} \\[1em] = \frac{15092}{21} \\[1em] = \frac{2156}{3} \text{ cm}^3.\end{aligned}$$

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

Volume of remaining material = $2744 - \frac{2156}{3}$

$$\begin{aligned}= \frac{(3 \times 2744) - 2156}{3} \\[1em] = \frac{8232 - 2156}{3} \\[1em] = \frac{6076}{3} \\[1em] = 2025\frac{1}{3} \text{ cm}^3.\end{aligned}$$

Hence, the volume of the remaining material is $2025\frac{1}{3} \text{ cm}^3$.

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 = $\frac{\text{Diameter}}{2}$

= $\frac{10}{2}$ = 5 cm.

Height of the cone for maximum volume = 20 cm.

Volume of the cone = $\frac{1}{3}πr^2h$

Putting values we get,

Volume of cone = $\frac{1}{3} \times \frac{22}{7} \times 5^2 \times 20$

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

Volume of the remaining wood = Volume of block of wood - Volume of the cone.

Volume of remaining wood = $2000 - \frac{11000}{21}$

$$\begin{aligned}= \frac{(21 \times 2000) - 11000}{21} \\[1em] = \frac{42000 - 11000}{21} \\[1em] = \frac{31000}{21} \\[1em] = 1476\frac{4}{21} \text{ cm}^3.\end{aligned}$$

Hence, the volume of the remaining wood is $1476\frac{4}{21}$ cm³.

Volume of the box = 16 cm × 8 cm × 8 cm = 1024 cm³.

Radius of the glass sphere, r = 2 cm.

Volume of the sphere = $\frac{4}{3}πr^3$

Volume of sphere = $\frac{4}{3} \times \frac{22}{7} \times 2^3$

$$\begin{aligned}= \frac{4 \times 22 \times 8}{3 \times 7} \\[1em] = \frac{704}{21} \text{cm}^3.\end{aligned}$$

Volume of 16 spheres = 16 $\times \frac{704}{21} = \frac{11264}{21} = 536.38\text{ cm}^3$.

Volume of water filled in box = Volume of the box - Volume of 16 spheres = 1024 - 536.38 = 487.62 cm³.

Hence, the volume of the water filled in the box is approximately 487.6 cm³.

Cuboidal block of side 7 cm is surmounted by a hemisphere as shown in figure below:

Side of cuboidal block = 7 cm.

Greatest diameter of hemisphere = 7 cm.

Radius = $\frac{\text{Diameter}}{2}$

= $\frac{7}{2}$ = 3.5 cm.

Surface area of the hemisphere = $2πr^2$.

Putting values we get,

Surface area of the hemisphere = $2 \times \frac{22}{7} \times 3.5^2$

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

Surface area of the cube = 6a² = 6 x 7² = 6 × 49 = 294 cm².

Surface area of base of hemisphere = πr².

Putting values we get,

Surface area of base of hemisphere = $\frac{22}{7} \times (3.5)^2$

$$\begin{aligned}= \frac{22}{7} \times 12.25 \\[1em] = 38.5 \text{ cm}^2.\end{aligned}$$

Surface area of solid = Surface area of cube + Surface area of hemisphere - Surface area of base of hemisphere = 294 + 77 - 38.5 = 332.5 cm².

Hence, the greatest diameter that the hemisphere can have is 7 cm and surface area of the solid is 332.5 cm².

Height of the cylinder = 10 cm

Radius of the cylinder= 3.5 cm

Total surface area (T) = Curved surface area of cylinder + 2 × curved surface area of hemisphere

T = 2πrh + 2 × 2πr²
T = 2πr(h + 2r)

Putting values we get,

$$\begin{aligned}\text{T} = 2 \times \frac{22}{7} \times 3.5 \times (10 + 7) \\[1em] = \frac{154}{7} \times 17 \\[1em] = 22 \times 17 \\[1em] = 374 \text{ cm}^2.\end{aligned}$$

Hence, the total surface area of the article is 374 cm².

Given,

Diameter of solid wooden cylinder (D) = 6 cm

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

Height of solid wooden cylinder (H) = 28 cm

Diameter of cone (d) = 6 cm

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

Height of cone (h) = 10.5 cm

Volume of cylinder (V) = πr²h

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

Volume of single cone (v) = $\frac{1}{3}$ πR²H

$$\begin{aligned}v = \frac{1}{3} \times \frac{22}{7} \times 3^2 \times 10.5\\[1em] = \frac{1}{3} \times \frac{22}{7} \times 9 \times 10.5\\[1em] = 22 \times 3 \times 1.5\\[1em] = 99 \text{ cm}^2\end{aligned}$$

Volume of two conical cavities = 2 x 99 = 198 cm²

Volume of remaining solid = Volume of cylinder - Volume of 2 conical cavities = 792 - 198 = 594 cm².

Hence, volume of remaining solid = 594 cm².

Given,

Height of cone (h₁) = 3 cm.

Height of cylinder = 7 cm.

From figure,

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

∴ Volume of remaining solid = $πr^2h - \frac{1}{3}πr^2h_1 - \frac{2}{3}πr^3$

$$\begin{aligned}= πr^2\Big(h - \frac{h_1}{3} - \frac{2r}{3}\Big) \\[1em] = \frac{22}{7} \times 3 \times 3 \Big(7 - \frac{3}{3} - \frac{2 \times 3}{3}\Big) \\[1em] = \frac{198}{7} \times \Big(7 - 1 - 2\Big) \\[1em] = \frac{198}{7} \times 4 \\[1em] = \frac{792}{7} \\[1em] = 113.14 \text{ cm}^3 \approx 113 \text{ cm}^3.\end{aligned}$$

Hence, the volume of the remaining solid correct to nearest whole number is 113 cm³.

The figure of the toy in the form of a cone surmounted on a hemisphere of same radius is shown below:

Total height of the toy = 15.5 cm

Radius of the base of the conical part (r) = 3.5 cm.

Height of the cone = 15.5 - 3.5 = 12 cm.

Slant height of the cone = l.

l = $\sqrt{r^2 + h^2}$

$$\begin{aligned}l = \sqrt{3.5^2 + 12^2} \\[1em] = \sqrt{12.25 + 144} \\[1em] = \sqrt{156.25} \\[1em] = 12.5 \text{ cm}.\end{aligned}$$

Total surface area of the toy (T) = Curved surface area of cone + Curved surface area of hemisphere.

$$\begin{aligned}\therefore T = πrl + 2πr^2 \\[1em] = πr(l + 2r) \\[1em] = \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5) \\[1em] = \frac{22}{7} \times 3.5 \times 19.5 \\[1em] = 22 \times 0.5 \times 19.5 \\[1em] = 214.5 \text{ cm}^2.\end{aligned}$$

Hence, the total surface area of the toy is 214.5 cm².

The figure of the circus tent is shown below:

Diameter of cylindrical portion = 24 m

Radius of cylindrical portion (r) = $\frac{\text{Diameter}}{2}$

= $\frac{24}{2}$ = 12 m.

Height of the cylindrical part, H = 11 m.

Since vertex of cone is 16 m above the ground, height of cone, h = 16 - 11 = 5 m.

h = 5 m.

Radius of cone = 12 m.

∴ Radius of cone is also equal to r.

Slant height of the cone, l = $\sqrt{h^2 + r^2}$.

l = $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ m.

Area of canvas used to make the tent = Curved surface area of the cylindrical part + Curved surface area of the cone.

Area of the canvas used to make the tent = 2πrH + πrl = πr(2H + l).

Putting values we get,

Area of the canvas = $\frac{22}{7} \times 12 \times (2 \times 11 + 13)$

$$\begin{aligned}= \frac{22}{7} \times 12 \times 35 \\[1em] = 22 \times 12 \times 5 \\[1em] = 1320 \text{ m}^2\end{aligned}$$

Hence, the area of the canvas used to make the tent is 1320 m².

Total height of the tent = 85 m.

Height of the cylindrical part (h₁) = 50 m.

From figure,

Height of cone (h₂) = 85 - 50 = 35 m.

Diameter of the base, d = 168 m.

Radius of the base of cylindrical part, r = $\frac{d}{2} = \frac{168}{2} = 84$ m.

Slant height of the cone, l = $\sqrt{h^2 + r^2}$.

l = $\sqrt{35^2 + 84^2} = \sqrt{1225 + 7056} = \sqrt{8281} = 91$ m.

Surface area of tent (S) = Curved surface area of cylinder + Curved surface area of cone

Putting values we get,

$$\begin{aligned}S = 2πrh_1 + πrl \\[1em] = πr(2h_1 + l) \\[1em] = \frac{22}{7} \times 84 \times (2 \times 50 + 91) \\[1em] = 22 \times 12 \times 191 \\[1em] = 264 \times 191 \\[1em] = 50424 \text{ m}^2.\end{aligned}$$

Adding 20% for folds and stitches,

Area of canvas = 50424 + 20% of 50424

$$\begin{aligned}= 50424 + \frac{20}{100} \times 50424 \\[1em] = 50424 + 0.2 \times 50424 \\[1em] = 50424 + 10084.8 \\[1em] = 60508.8 \text{ m}^2 \\[1em] \approx 60509 \text{ m}^2\end{aligned}$$

Hence, the quantity of canvas required to make the tent is 60509 m².

The figure is shown below:

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

Height of the cylinder, H = 30 cm

Height of cone, h = 24 cm.

Slant height of cone, l = $\sqrt{h^2 + r^2}$.

Putting values we get,

l = $\sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$ m.

Volume of the remaining solid (V) = Volume of the cylinder - Volume of the cone.

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

Total surface area of the remaining solid (S) = Curved surface area of cylinder + Area of base of cylinder + Curved surface area of cone.

$$\begin{aligned}\therefore S = 2πrH + πr^2 + πrl \\[1em] = πr(2H + r + l) \\[1em] = \frac{22}{7} \times 7 \times (2 \times 30 + 7 + 25) \\[1em] = 22 \times (60 + 32) \\[1em] = 22 \times 92 \\[1em] = 2024 \text{ cm}^2.\end{aligned}$$

Hence, the volume of the remaining solid = 3388 cm³ and surface area = 2024 cm².

The figure of the buoy made by surmounting a right cone on a hemisphere is shown below:

Radius of base of hemisphere = Radius of cone = 3.5 m.

Volume of hemisphere (V) = $\frac{2}{3}πr^3$

Putting values,

$$\begin{aligned}V = \frac{2}{3} \times \frac{22}{7} \times \Big(3.5)^3 \\[1em] = \frac{44}{21} \times 42.875 \\[1em] = \frac{44 \times 42.875}{21} \\[1em] = \frac{1886.5}{21} \\[1em] = 89.8 \text{ m}^3\end{aligned}$$

Volume of cone = $\frac{2}{3}$ Volume of hemisphere.

∴ Volume of cone = $\frac{2}{3} \times 89.8$

$$\begin{aligned}= \frac{179.6}{3} \\[1em] = 59.87 \text{ m}^3.\end{aligned}$$

Volume of cone = $\frac{1}{3}πr^2h$.

$$\begin{aligned}\therefore \frac{1}{3}πr^2h = 59.87 \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times h = 59.87 \\[1em] \Rightarrow \frac{22}{21} \times 12.25 \times h = 59.87 \\[1em] \Rightarrow h = \frac{59.87 \times 21}{22 \times 12.25} \\[1em] \Rightarrow h = \frac{1257.27}{269.5} \\[1em] \Rightarrow h = 4.67 m.\end{aligned}$$

Slant height of cone = l = $\sqrt{h^2 + r^2}$

Putting values we get,

$$\begin{aligned}l = \sqrt{(4.67)^2 + (3.5)^2} \\[1em] l = \sqrt{21.81 + 12.25} \\[1em] l = \sqrt{34.06} \\[1em] l = 5.84 \text{ m}\end{aligned}$$

Surface area of the buoy = Curved Surface area of cone + Curved Surface area of hemisphere = πrl + 2πr².

∴ Surface area of buoy = $πr(l + 2r)$

$$\begin{aligned}= \frac{22}{7} \times 3.5 \times (5.84 + 2 \times 3.5) \\[1em] = \frac{22}{7} \times 3.5 \times (5.84 + 7) \\[1em] = \frac{22}{7} \times 3.5 \times 12.84 \\[1em] = \frac{988.68}{7} \\[1em] = 141.17 \text{ m}^2.\end{aligned}$$

Hence, the height of cone = 4.67 m and surface area of buoy is 141.17 m².

The below figure shows the building in the form of a cylinder surmounted by a hemisphere valted dome:

Let the radius of the dome be r.

∴ Internal diameter = 2r.

Given, internal diameter is equal to height.

∴ Height of building (h) = 2r.

Height of hemispherical area = r.

So, height of cylindrical area, h₁ = 2r - r = r.

Volume of building (V) = Volume of cylindrical area + Volume of hemispherical area.

$$\begin{aligned}\therefore V = πr^2h_1 + \frac{2}{3}πr^3 \\[1em] \Rightarrow V = πr^2.r + \frac{2}{3}πr^3 \\[1em] \Rightarrow V = πr^3 + \frac{2}{3}πr^3 \\[1em] \Rightarrow V = \frac{5}{3}πr^3.\end{aligned}$$

Given, V = $41\frac{19}{21} = \frac{880}{21} \text{ m}^3$

$$\begin{aligned}\therefore \frac{5}{3} \times \frac{22}{7} \times r^3 = \frac{880}{21} \\[1em] \Rightarrow r^3 = \frac{880 \times 3 \times 7}{5 \times 22 \times 21} \\[1em] \Rightarrow r^3 = \frac{18480}{2310} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = 2 \text{ m}.\end{aligned}$$

h = 2r = 2(2) = 4 m.

Hence, the height of the building is 4 m.

The below figure shows the rocket:

Given,

Height of cylindrical portion (H) = 12 cm.

Radius of cylinder and cone = $\frac{\text{Diameter}}{2} = \frac{6}{2}$ = 3 cm.

Slant height of cone (l) = 5 cm.

⇒ h² = l² - r²
⇒ h² = 5² - 3²
⇒ h² = 25 - 9
⇒ h² = 16
⇒ h = $\sqrt{16}$ = 4 cm.

Total surface area of rocket (S) = Curved surface area of cylinder + Base area of cylinder + Curved surface area of cone.

$$\begin{aligned}\therefore S = 2πrH + πr^2 + πrl \\[1em] = πr(2H + r + l) \\[1em] = 3.14 \times 3 \times (2 \times 12 + 3 + 5) \\[1em] = 9.42 \times 32 \\[1em] = 301.44 \text{ cm}^2.\end{aligned}$$

Volume of the rocket (V) = Volume of cone + Volume of cylinder.

$$\begin{aligned}\therefore V = \frac{1}{3}πr^2h + πr^2H \\[1em] = πr^2\Big(\frac{h}{3} + H) \\[1em] = 3.14 \times 3^2 \times \Big(\frac{4}{3} + 12\Big) \\[1em] = 3.14 \times 3^2 \times \Big(\frac{4 + 36}{3} \Big) \\[1em] = 3.14 \times 9 \times \frac{40}{3} \\[1em] = 376.8 \text{ cm}^3.\end{aligned}$$

Hence, the total surface area of rocket is 301.44 cm² and volume is 376.8 cm³.

Given, common radius (r) = 7 cm,

Height of cone (h) = 4 cm,

Height of cylinder (H) = 4 cm.

Volume of solid (V) = Volume of cone + Volume of cylinder + Volume of hemisphere.

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

Hence, the volume of solid = 1540 cm³.

The solid in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end is shown in the figure below:

Given,

Common Diameter = 3.5 cm,

Common Radius = $\frac{\text{Diameter}}{2} = \frac{3.5}{2}$ = 1.75 cm.

Height of cylindrical part (h₁) = 10 cm.

Height of conical part (h₂) = 6 cm.

Volume of solid (V) = Volume of cone + Volume of cylinder + Volume of hemisphere

$$\begin{aligned}V = \frac{1}{3}πr^2h_2 + πr^2h_1 + \frac{2}{3}πr^3 \\[1em] = πr^2\Big(\frac{h_2}{3} + h_1 + \frac{2r}{3}\Big) \\[1em] = 3.14 \times (1.75)^2 \times \Big(\frac{6}{3} + 10 + \frac{2 \times 1.75}{3}\Big) \\[1em] = 3.14 \times 3.0625 \times (2 + 10 + 1.167) \\[1em] = 3.14 \times 3.0625 \times (13.167) \\[1em] = 126.617 \text{ cm}^3.\end{aligned}$$

Hence, the volume of the solid is 126.62 cm³.

The toy is shown in the figure below:

Height of the cylindrical part (H) = 13 cm.

Radius = 5 cm.

Radius of cone (r) = 5 cm

Height of cone (h) = 12 cm.

Slant height of cone, l = $\sqrt{r^2 + h^2}$

Putting values we get,

$$\begin{aligned}l = \sqrt{5^2 + 12^2} \\[1em] = \sqrt{25 + 144} \\[1em] = \sqrt{169} \\[1em] = 13 \text{ cm}.\end{aligned}$$

Surface area of toy(S) = Curved surface area of cylinder + Curved surface area of hemisphere + Curved surface area of cone.

$$\begin{aligned}S = 2πrH + 2πr^2 + πrl \\[1em] = πr(2H + 2r + l) \\[1em] = \frac{22}{7} \times 5 \times (2 \times 13 + 2 \times 5 + 13) \\[1em] = \frac{110}{7} \times (26 + 10 + 13) \\[1em] = \frac{110}{7} \times 49 \\[1em] = 110 \times 7 \\[1em] = 770 \text{ cm}^2.\end{aligned}$$

Hence, the surface area of the toy is 770 cm².

(i) In the given figure,

Height of cylindrical portion (H) = 8 cm.

Radius of cylindrical portion = radius of hemispherical portion = (r) = 3 cm.

Scale = 1 : 200

∴ k = 200.

Total surface area (S) = Curved surface area of hemisphere + Curved surface area of cylinder + Area of base of cylinder

= 2πr² + 2πrH + πr²

= 3πr² + 2πrH

= πr(3r + 2H)

= 3π(3 × 3 + 2 × 8)

= 3π(9 + 16)

= 3π × 25

= 75π cm².

∴ Surface area of solid = 75π × k²

= 75π × (200)²

= 75π × 40000 cm²

= 75π × $\frac{40000}{100 \times 100}$ m²

= 300π m².

Hence, the surface area of solid = 300π m².

(ii) Volume (V) = Volume of hemisphere + Volume of cylinder

= $\frac{2}{3}πr^3 + πr^2H$.

Substituting values we get :

$$\begin{aligned}V = πr^2\Big(\frac{2}{3}r + H\Big) \\[1em] = πr^2\Big(\frac{2}{3} \times 3 + 8\Big) \\[1em] = π \times 3^2 \times (2 + 8) \\[1em] = 9π \times 10 \\[1em] = 90π \text{ cm}^3\end{aligned}$$

∴ Volume of solid = 90π × k³

= 90π × (200)³

= 90π × 8000000

= 720000000π cm³

= $\frac{720000000π}{100 \times 100 \times 100}$ m³

= 720π m³.

As, 1 m³ = 1000 litres

∴ Volume of solid = 720π × 1000 = 720000π litres.

Hence, the volume of solid = 720000π litres.

(a) 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.

(b) 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.

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.

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) = $\frac{4}{3}πr^3$

Putting values we get,

$$\begin{aligned}V = \frac{4}{3}π \times (3)^3 \\[1em] = 4π \times 3^2 \\[1em] = 36π \text{ cm}^3.\end{aligned}$$

Given, length of wire = 36 m.

So, height of cylinder = 36 m = 3600 cm.

Volume of cylinder = V = 36π cm³.

$$\begin{aligned}\therefore πr^2h = 36π \\[1em] \Rightarrow r^2 = \frac{36π}{πh} \\[1em] \Rightarrow r^2 = \frac{36}{3600} \\[1em] \Rightarrow r^2 = \frac{1}{100} \\[1em] \Rightarrow r = \sqrt{\frac{1}{100}} \\[1em] \Rightarrow r = \frac{1}{10} \text{ cm} = 1 \text{ mm}.\end{aligned}$$

Hence, the radius of the wire is 1 mm.

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

Height of cylinder (h) = 32 cm

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

Let radius of cylinder = r cm.

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

Hence, the radius of the cylinder = 3 cm.

(ii) Curved surface area of cylinder = 2πrh

Putting values we get,

Curved surface area of cylinder = 2 × 3.1 × 3 × 32 = 595.2 cm².

Hence, curved surface area of cylinder = 595.2 cm².

Radius of solid hemisphere (r) = 8 cm.

Volume of the hemisphere (V) = $\frac{2}{3}πr^3$

Radius of cone (R) = 6 cm.

Let height of cone = h cm.

Volume of cone = $\frac{1}{3}πR^2h$

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

$$\begin{aligned}\therefore \frac{1}{3}πR^2h = \frac{2}{3}πr^3 \\[1em] \Rightarrow h = \frac{2π \times 8^3 \times 3}{3 \times π \times 6^2} \\[1em] \Rightarrow h = \frac{2 \times 512}{36} \\[1em] \Rightarrow h = \frac{1024}{36} \\[1em] \Rightarrow h = \frac{256}{9} \\[1em] \Rightarrow h = 28\frac{4}{9} \text{ cm}.\end{aligned}$$

Hence, the height of the cone is $28\frac{4}{9}$ cm.

Base of water tank = 11 m × 6 m

Height of water level in rectangular water tank = 5 m

Volume of water in tank = lbh = 11 m × 6 m × 5 m = 330 m³.

Let water come upto height H in cylindrical tank.

Radius = 3.5 m

Volume of cylindrical tank = πr²H.

$$\begin{aligned}\therefore πr^2H = 330 \\[1em] \Rightarrow \frac{22}{7} \times 3.5^2 \times H = 330 \\[1em] \Rightarrow \frac{22 \times 12.25 \times H}{7} = 330 \\[1em] \Rightarrow H = \frac{330 \times 7}{22 \times 12.25} \\[1em] \Rightarrow H = \frac{2310}{269.5} \\[1em] \Rightarrow H = \frac{60}{7} = 8\frac{4}{7} \text{ m}.\end{aligned}$$

Hence, the height of water level in cylindrical tank = $8\frac{4}{7}$ m.

Diameter of the cylinder = 40 cm.

Radius (r) = $\frac{40}{2} = 20$ cm.

Height (h) = 9 cm.

∴ Volume of cylinder = πr²h = π × 20 × 20 × 9 = 3600π cm³.

Height of cone (H) = 108 cm.

Let radius of cone = R.

Volume of cone = $\frac{1}{3}πR^2H$

Given, volume of cone = volume of cylinder.

$$\begin{aligned}\therefore \frac{1}{3}πR^2H = πr^2h \\[1em] \Rightarrow R^2 = \frac{3 \times π \times 20 \times 20 \times 9}{π \times 108} \\[1em] \Rightarrow R^2 = \frac{10800}{108} \\[1em] \Rightarrow R^2 = 100 \\[1em] \Rightarrow R = \sqrt{100} \\[1em] \Rightarrow R = 10 \text{ cm}.\end{aligned}$$

Hence, the radius of the cone is 10 cm.

Given,

Radius of larger metallic sphere (R) = 6 cm

Let radius of each smaller sphere be r cm.

Given,

A solid metallic sphere of radius 6 cm is melted and recast into 64 identical solid spheres.

∴ Volume of larger metallic sphere = 64 × Volume of smaller metallic sphere

$$\begin{aligned}\Rightarrow \frac{4}{3}πR^3 = 64 \times \frac{4}{3}πr^3\\[1em] \Rightarrow R^3 = 64 \times r^3\\[1em] \Rightarrow 6^3 = 64 \times r^3\\[1em] \Rightarrow 216 = 64 \times r^3\\[1em] \Rightarrow r^3 = \frac{216}{64}\\[1em] \Rightarrow r = \sqrt[3]{\frac{216}{64}}\\[1em] \Rightarrow r = \frac{\sqrt[3]{216}}{\sqrt[3]{64}}\\[1em] \Rightarrow r = \frac{6}{4}\\[1em] \Rightarrow r = 1.5\end{aligned}$$

Hence, radius of each marble = 1.5 cm.

Diameter of hemispherical bowl = 7.2 cm

Radius of hemispherical bowl (r) = 3.6 cm

Volume of hemispherical bowl = $\frac{2}{3}πr^3$.

Radius of cone (R) = 4.8 cm.

Let height of cone = h cm.

Volume of cone = $\frac{1}{3}πR^2h$.

Volume of cone = Volume of hemispherical bowl.

$$\begin{aligned}\therefore \frac{1}{3}π \times (4.8)^2 \times h = \frac{2}{3} \times π \times (3.6)^3 \\[1em] \Rightarrow h = \frac{2π \times 46.656 \times 3}{3 \times π \times 23.04} \\[1em] \Rightarrow h = \frac{93.312}{23.04} \\[1em] \Rightarrow h = 4.05 \text{ cm}.\end{aligned}$$

Hence, the height of the cone is 4.05 cm.

Radius of the smaller sphere = r = 3 cm.

Let R be the radius of a larger new sphere.

Mass of small sphere (m₁) = 1 kg.

Mass of bigger sphere (m₂) = 7 kg.

The spheres are melted to form a new sphere.

So, the mass of the new sphere (M) = 1 + 7 = 8 kg.

Density of smaller sphere = Density of new sphere.

Let v be volume of small sphere and V be volume of new bigger sphere.

$$\begin{aligned}\frac{m_1}{v} = \frac{M}{V} \\[1em] \frac{1}{v} = \frac{8}{V} \\[1em] \frac{v}{V} = \frac{1}{8} ....(i)\end{aligned}$$

Given, radius of smaller sphere, r = 3 cm.

Volume of smaller sphere = v = $\frac{4}{3}$πr³

= $\frac{4}{3}$π(3)³

= 36π cm³.

Volume of new sphere = V = $\frac{4}{3}$πR³

Putting these values in (i),

$$\begin{aligned}\frac{36π}{\frac{4}{3}πR^3} = \frac{1}{8} \\[1em] \frac{36 \times 3}{4 \times R^3} = \frac{1}{8} \\[1em] \frac{108}{4R^3} = \frac{1}{8} \\[1em] R^3 = \frac{108 \times 8}{4} \\[1em] R^3 = 216 \\[1em] R^3 = 6^3 \\[1em] R = 6 \text{ cm}.\end{aligned}$$

Diameter = 2 × 6 = 12 cm.

Hence, the diameter of the new bigger sphere = 12 cm.

Given,

Internal radius (r) = $\frac{6}{2}$ = 3 cm.

Outer radius (R) = $\frac{10}{2}$ = 5 cm.

Given, height of the old hollow and new solid cylinder is equal let it be h.

Let the radius of new solid cylinder be r₁.

Since, old hollow cylinder is recasted into solid cylinder hence, their volume will be equal.

$\therefore π(R^2 - r^2)h = πr_1^2h$

Dividing both sides by π and h,

$$\begin{aligned}\Rightarrow R^2 - r^2 = r_1^2 \\[1em] \Rightarrow r_1^2 = 5^2 - 3^2 \\[1em] \Rightarrow r_1^2 = 25 - 9 \\[1em] \Rightarrow r_1^2 = 16 \\[1em] \Rightarrow r_1 = 4 \text{ cm}.\end{aligned}$$

Diameter = 2 × Radius = 2 × 4 = 8 cm.

Hence, the diameter of solid cylinder = 8 cm.

For sphere,

Internal radius (r) = $\frac{4}{2}$ = 2 cm,

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

For cone,

Base radius (r₁) = $\frac{8}{2}$ = 4 cm

Height = h.

Since, the hollow sphere is recasted into cone hence their volume will be equal.

$\therefore \frac{4}{3}π(R^3 - r^3) = \frac{1}{3}πr_1^2h$

Multiplying both sides by 3 and dividing by π we get,

$$\begin{aligned}\Rightarrow 4(R^3 - r^3) = r_1^2h \\[1em] \Rightarrow 4(4^3 - 2^3) = (4)^2h \\[1em] \Rightarrow 4(64 - 8) = 16h \\[1em] \Rightarrow h = \frac{4 \times 56}{16} \\[1em] \Rightarrow h = \frac{224}{16} \\[1em] \Rightarrow h = 14 \text{ cm}.\end{aligned}$$

Hence, the height of the cone = 14 cm.

Inner diameter of well = 6 m

Radius of well, r = $\frac{6}{2}$ = 3 m.

Depth (h) = 22 m.

Volume of soil dug out of well = πr²h = π × 3² × 22 = 198π m³.

Width of embankment = 5 m.

Inner radius of embankment = Inner radius of well = r = 3 m.

Outer radius of embankment (R) = inner radius + width = 3 + 5 = 8 m.

Let H be the height of soil embankment.

Volume of soil embankment (V) = π(R² - r²)H

$$\begin{aligned}V = π \times (8^2 - 3^2) \times H \\[1em] = π \times (64 - 9) \times H \\[1em] = 55πH \text{ m}^3.\end{aligned}$$

Volume of soil dug out = Volume of soil embankment.

∴ 198π = 55πH

⇒ 55H = 198 (Dividing both sides by π)

⇒ H = $\frac{198}{55}$ = 3.6 m

Hence, the height of the soil embankment is 3.6 m.

Internal diameter of cylindrical can = 21 cm.

Radius (R) = $\frac{21}{2}$ cm.

Diameter of sphere = 10.5 cm = $\frac{21}{2}$ cm.

Radius of sphere (r) = $\frac{\frac{21}{2}}{2} = \frac{21}{4}$ cm.

Let the rise in water level be h.

Rise in volume of water = Volume of sphere immersed.

$$\begin{aligned}\Rightarrow πR^2h = \frac{4}{3}πr^3 \\[1em] \Rightarrow \Big(\frac{21}{2}\Big)^2 \times h = \frac{4}{3} \times \Big(\frac{21}{4}\Big)^3 \\[1em] \Rightarrow h = \frac{4 \times 21^3 \times 2^2}{3 \times 4^3 \times 21^2} \\[1em] \Rightarrow h = \frac{4 \times 21 \times 4}{3 \times 64} \\[1em] \Rightarrow h = \frac{336}{192} \\[1em] \Rightarrow h = 1.75 \text{ cm}.\end{aligned}$$

Hence, the rise in water level is 1.75 cm.

Given, radius of glass jar = R = 8 cm

Diameter of sphere = 12 cm

Radius of sphere = r = $\frac{12}{2}$ = 6 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.

$$\begin{aligned}\Rightarrow πR^2h = \frac{4}{3}πr^3 \\[1em] \Rightarrow 8^2h = \frac{4}{3} \times 6^3 \\[1em] h = \frac{4 \times 6^3}{3 \times 8^2} \\[1em] h = \frac{4 \times 216}{3 \times 64} \\[1em] h = \frac{864}{192} \\[1em] h = 4.5 \text{ cm}.\end{aligned}$$

Hence, the height by which water level rises is 4.5 cm.

Height of cone (h) = 20 cm.

Radius of cone (r) = $\frac{16.8}{2}$ = 8.4 cm.

Volume of water in vessel = $\frac{1}{3}πr^2h$

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times (8.4)^2 \times 20 \\[1em] = \frac{22}{21} \times 70.56 \times 20 \\[1em] = \frac{31046.4}{21} \\[1em] = 1478.4 \text{ cm}^3.\end{aligned}$$

Given,

Volume of water overflown = One-third of the volume of water in the vessel = $\frac{1}{3} \times 1478.4 = 492.8$ cm³.

Volume of water overflown = Volume of two equal solid cones dropped into the vessel.

Volume of two equal solid cones dropped into the vessel = 492.8 cm³

Volume of one solid cone = $\frac{492.8}{2} = 246.4$ cm³.

Hence, the volume of each of the solid cone submerged is 246.4 cm³.

Radius of a solid cylinder (r) = 14 cm,

Height (h) = 12 cm.

Edge of cube (a) = 2 cm.

Let the no. of small cubes formed be n.

Volume of cylinder = n × Volume of each cubes.

$$\begin{aligned}\therefore πr^2h = n \times (a)^3 \\[1em] \Rightarrow \frac{22}{7} \times (14)^2 \times 12 = n \times (2)^3 \\[1em] \Rightarrow \frac{22}{7} \times 14 \times 14 \times 12 = 8n \\[1em] \Rightarrow 22 \times 2 \times 14 \times 12 = 8n \\[1em] \Rightarrow 8n = 7392 \\[1em] \Rightarrow n = \frac{7392}{8} \\[1em] \Rightarrow n = 924.\end{aligned}$$

Hence, the number of cubes formed are 924.

Shot is in the shape of sphere.

Radius of sphere (r) = $\frac{3}{2}$ = 1.5 cm.

Let the number of sphere formed = n.

Volume of cuboidal lead solid = n × Volume of each shot

$$\begin{aligned}\therefore lbh = n \times \frac{4}{3}πr^3 \\[1em] \Rightarrow 9 \times 11 \times 12 = n \times \frac{4}{3} \times \frac{22}{7} \times (1.5)^3 \\[1em] \Rightarrow 1188 = \frac{297n}{21} \\[1em] \Rightarrow n = \frac{1188 \times 21}{297} \\[1em] \Rightarrow n = \frac{24948}{297} \\[1em] \Rightarrow n = 84.\end{aligned}$$

Hence, the number of shots made from cuboidal lead of solid is 84.

Radius of a solid metallic cylinder (r) = 14 cm,

Height (h) of cylinder = 21 cm.

Radius of sphere (R) = 3.5 cm.

Let the number of spheres formed be n.

Volume of metal cylinder = n × Volume of each sphere.

$$\begin{aligned}\therefore πr^2h = n \times \frac{4}{3}πR^3 \\[1em] \Rightarrow r^2h = n \times \frac{4}{3}R^3 \text{ (Dividing both sides by π)} \\[1em] \Rightarrow (14)^2 \times 21 = n \times \frac{4}{3} \times (3.5)^3 \\[1em] \Rightarrow n = \frac{14^2 \times 3 \times 21}{4 \times (3.5)^3} \\[1em] \Rightarrow n = \frac{12348}{171.5} \\[1em] \Rightarrow n = 72.\end{aligned}$$

Hence, the number of spheres that can be made from solid cylinder are 72.

Let number of small cylinders formed be n.

Given,

Radius of cone (R) = 5 cm,

Height of cone (H) = 9 cm,

Radius of cylinder (r) = 0.5 cm,

Height of cylinder (h) = 1.5 cm

Since, cone is melted and recasted into n cylinders.

∴ Volume of cone = n × Volume of sphere

$$\begin{aligned}\Rightarrow \frac{1}{3}πR^2H = n \times πr^2h\\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 9 = n \times \frac{22}{7} \times (0.5)^2 \times 1.5\\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 25 \times 9 = n \times \frac{22}{7} \times 0.25 \times 1.5\\[1em] \Rightarrow \frac{1}{3} \times 25 \times 9 = n \times 0.25 \times 1.5\\[1em] \Rightarrow 25 \times 3 = n \times 0.25 \times 1.5\\[1em] \Rightarrow 75 = 0.375n\\[1em] \Rightarrow n = \frac{75}{0.375}\\[1em] \Rightarrow n = 200.\end{aligned}$$

Hence, number of cylinders formed = 200.

Radius of sphere (r) = 10.5 cm

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

Radius of cone (R) = 3.5 cm

Height (h) = 3 cm.

Volume of sphere = n × Volume of each cone.

$\therefore \frac{4}{3}πr^3 = n \times \frac{1}{3}πR^2h$

Dividing both sides by π and multiplying by 3,

$$\begin{aligned}\Rightarrow 4r^3 = n \times R^2h \\[1em] \Rightarrow 4 \times (10.5)^3 = n \times (3.5)^2 \times 3 \\[1em] \Rightarrow n = \frac{4 \times 1157.625}{12.25 \times 3} \\[1em] \Rightarrow n = \frac{4630.5}{36.75} \\[1em] \Rightarrow n = 126.\end{aligned}$$

Hence, the number of cones obtained from metallic sphere are 126.

Radius of each cone (r) = 2 cm

Height of cone (h) = 3 cm.

Let the number of cones required to be recast into a solid sphere of radius (R) = 6 cm be n.

n × Volume of each cone = Volume of sphere.

$\therefore n \times \frac{1}{3}πr^2h = \frac{4}{3}πR^3$

On dividing both sides by π and multiplying by 3,

$$\begin{aligned}\Rightarrow n \times r^2h = 4R^3 \\[1em] \Rightarrow n = \frac{4R^3}{r^2h} \\[1em] \Rightarrow n = \frac{4 \times (6)^3}{(2)^2 \times 3} \\[1em] \Rightarrow n = \frac{4 \times 216}{4 \times 3} \\[1em] \Rightarrow n = \frac{864}{12} \\[1em] \Rightarrow n = 72.\end{aligned}$$

Hence, the number of cones required to make a solid sphere of radius 4 cm are 72.

Radius of the top of the inverted cone (R) = 2.5 cm.

Height of cone (H) = 11 cm.

Radius of lead sphere (r) = 0.25 cm.

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

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

$$\begin{aligned}V = \frac{2}{5} \times \frac{1}{3}πR^2H \\[1em] = \frac{2}{15}π \times (2.5)^2 \times 11 \\[1em] = \frac{2π \times 6.25 \times 11}{15} \\[1em] = \frac{137.5π}{15}.\end{aligned}$$

Let the number of spheres be n.

∴ Volume of water flown out (V) = n × Volume of each lead shot.

$$\begin{aligned}\therefore \frac{137.5π}{15} = n \times \frac{4}{3}πr^3 \\[1em] \Rightarrow \frac{137.5π}{15} = n \times \frac{4}{3}π(0.25)^3 \\[1em] \Rightarrow \frac{137.5π}{15} = n \times \frac{4π \times 0.015625}{3} \\[1em] \Rightarrow n = \frac{137.5π \times 3}{15 \times 4π \times 0.015625} \\[1em] \Rightarrow n = \frac{412.5π}{0.9375π} \\[1em] \Rightarrow n = 440.\end{aligned}$$

Hence, the number of lead shots are 440.

Surface area of a metallic sphere = 616 cm².

Let the radius of this sphere be R.

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

Given, big spheres are converted into smaller spheres of diameter = 3.5 cm or radius = $\frac{3.5}{2}$.

Let the number of smaller spheres formed be n.

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

$\therefore \frac{4}{3}πR^3 = n \times \frac{4}{3}πr^3$

Dividing both sides by 4π and multiplying by 3 we get,

$$\begin{aligned}\Rightarrow R^3 = nr^3 \\[1em] \Rightarrow 7^3 = n \times \Big(\frac{3.5}{2}\Big)^3 \\[1em] \Rightarrow n = \frac{7^3 \times 2^3}{3.5^3} \\[1em] \Rightarrow n = 2^3 \times 2^3 \\[1em] \Rightarrow n = 64.\end{aligned}$$

Hence, 64 small spheres can be formed.

(i) Surface area of a metallic sphere = 1256 cm².

Let the radius of this sphere be R.

$$\begin{aligned}\therefore 4πR^2 = 1256 \\[1em] \Rightarrow R^2 = \frac{1256}{4π} \\[1em] \Rightarrow R^2 = \frac{1256}{4 \times 3.14} \\[1em] \Rightarrow R^2 = \frac{1256}{12.56} \\[1em] \Rightarrow R^2 = 100 \\[1em] \Rightarrow R = \sqrt{100} = 10 \text{ cm}.\end{aligned}$$

Hence, the radius of sphere = 10 cm.

(ii) Let the number of cones formed by recasting sphere be n.

Radius of cone (r) = 2.5 cm

Height of cone (h) = 8 cm.

Volume of sphere = n × Volume of each cone.

$\therefore \frac{4}{3}πR^3 = n \times \frac{1}{3}πr^2h$

Multiplying both sides by 3 and dividing by π.

$$\begin{aligned}\Rightarrow 4R^3 = nr^2h \\[1em] \Rightarrow 4 \times 10^3 = n \times (2.5)^2 \times 8 \\[1em] \Rightarrow n = \frac{4 \times 1000}{6.25 \times 8} \\[1em] \Rightarrow n = \frac{4000}{50} \\[1em] \Rightarrow n = 80.\end{aligned}$$

Hence, the number of cones formed by recasting sphere are 80.

(i) Radius of a cylindrical can (r) = 3.5 cm

Radius of sphere (R) = r = 3.5 cm

Height of water level in can = 7 cm.

Height of cylinder (h) = 7 cm.

Total surface area of can in contact with water (T) = Curved surface area of cylinder + base area of cylinder.

$$\begin{aligned}T = 2πrh + πr^2 \\[1em] = πr(2h + r) \\[1em] = \frac{22}{7} \times 3.5 \times (2 \times 7 + 3.5) \\[1em] = 22 \times 0.5 \times (14 + 3.5) \\[1em] = 11 \times 17.5 \\[1em] = 192.5 \text{ cm}^2.\end{aligned}$$

Hence, the surface area of can in contact with water is 192.5 cm².

(ii) Let the depth of the water before the sphere was put be d.

Volume of cylindrical can = Volume of sphere + Volume of water.

$$\begin{aligned}πr^2h = \frac{4}{3}πR^3 + πr^2d \\[1em] πr^2h = \frac{4}{3}πr^3 + πr^2d \\[1em] πr^2h = πr^2\Big(\frac{4}{3}r + d\Big) \\[1em] h = \frac{4}{3}r + d \\[1em] d = h - \frac{4}{3}r \\[1em] d = 7 - \frac{4}{3} \times 3.5 \\[1em] d = 7 - \frac{14}{3} \\[1em] d = \frac{21 - 14}{3} = \frac{7}{3} \text{ cm}.\end{aligned}$$

Hence, the depth of water before sphere was put was $\frac{7}{3}$ cm.

Given, volume of a conical tent = 462 m³ and the area of the base = 154 m²

Using formula,

area of base = πr² and volume of cone = $\frac{1}{2}$ πr²h

$$\begin{aligned}\Rightarrow \frac{1}{3}πr^2h = 462 \\[1em] \Rightarrow \frac{1}{3} \times (πr^2) \times h = 462 \\[1em] \Rightarrow \frac{1}{3} \times 154 \times h = 462 \\[1em] \Rightarrow \frac{154}{3} \times h = 462 \\[1em] \Rightarrow h = \frac{3 \times 462}{154} \\[1em] \Rightarrow h = \frac{1386}{154} \\[1em] \Rightarrow h = 9 \text{ m}.\end{aligned}$$

Hence, option 3 is the correct option.

Given, the radius of a roller = 14 cm and height = 100 cm

Curved surface area of cylinder = 2πrh

$$\begin{aligned}\text{Curved surface area of roller }= 2 \times \frac{22}{7} \times 14 \times 100 \\[1em] = 2 \times 22 \times 2 \times 100 \\[1em] = 8800 \text{ cm}^2.\end{aligned}$$

Hence, option 4 is the correct option.

Let the radius of cylinders be r₁ = 4r and r₂ = 9r.

Let heights be h₁ and h₂.

Since, cylinders have equal lateral surface area,

$\therefore 2πr_1h_1 = 2πr_2h_2$

Dividing both sides by 2π.

$$\begin{aligned}\Rightarrow r_1h_1 = r_2h_2 \\[1em] \Rightarrow 4r \times h_1 = 9r \times h_2 \\[1em] \Rightarrow \frac{h_1}{h_2} = \frac{9}{4}.\end{aligned}$$

Hence, Option 4 is the correct option.

Let radius and heights of two cylinders be r₁, h₁ and r₂, h₂.

Given, $\frac{h_1}{h_2} = \frac{5}{3} \text{ and } \frac{r_1}{r_2} = \frac{2}{3}$.

Volume of cylinder 1 = V₁ and

Volume of cylinder 2 = V₂.

$$\begin{aligned}\frac{V_1}{V_2} = \frac{πr_1^2h_1}{πr_2^2h_2} \\[1em] = \Big(\frac{r_1}{r_2}\Big)^2 \times \frac{h_1}{h_2} \\[1em] = \Big(\frac{2}{3}\Big)^2 \times \frac{5}{3} \\[1em] = \frac{4}{9} \times \frac{5}{3} \\[1em] = \frac{20}{27}.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

Radius (R) = $\frac{r}{2}$,

Slant height (L) = 2l

Total surface area of cone (S) = πR(L + R)

Putting values we get,

$$\begin{aligned}S = π \times \frac{r}{2} \times \Big(2l + \frac{r}{2}\Big) \\[1em] = \frac{πr}{2} \times 2l + \frac{πr}{2} \times \frac{r}{2} \\[1em] = πrl + \frac{πr^2}{4} \\[1em] = πr\Big(l + \frac{r}{4}\Big).\end{aligned}$$

Hence, Option 2 is the correct option.

Radius of base of cone (r) = $\frac{\text{Diameter}}{2}$

= $\frac{10}{2}$ = 5 cm.

Height (h) = 12 cm.

$$\begin{aligned}\text{l} = \sqrt{r^2 + h^2} \\[1em] = \sqrt{5^2 + 12^2} \\[1em] = \sqrt{25 + 144} \\[1em] = \sqrt{169} = 13 \text{ cm}.\end{aligned}$$

Curved surface area = πrl.

Putting values we get,

Curved surface area = π × 5 × 13 = 65π cm².

Hence, Option 2 is the correct option.

Volume of hemisphere (V) = $\frac{2}{3}πr^3$

Putting values we get,

$$\begin{aligned}V = \frac{2}{3} \times π \times (5)^3 \\[1em] = \frac{2}{3}π \times 125 \\[1em] = \frac{250π}{3} \text{ cm}^3.\end{aligned}$$

Hence, Option 1 is the correct option.

Let the diameters of two spheres be 3a and 5a.

So, their radius will be $r_1 = \frac{3a}{2} \text{ and } r_2 = \frac{5a}{2}$.

$$\begin{aligned}\text{Ratio of their surface area } = \frac{4πr_1^2}{4πr_2^2} \\[1em] = \frac{\Big(\frac{3a}{2}\Big)^2}{\Big(\frac{5a}{2}\Big)^2} \\[1em] = \frac{\Big(\frac{9a^2}{4}\Big)}{\Big(\frac{25a^2}{4}\Big)} \\[1em] = \frac{9a^2 \times 4}{25a^2 \times 4} \\[1em] = \frac{9}{25} \\[1em] = 9 : 25.\end{aligned}$$

Hence, Option 4 is the correct option.

Radius of balloon in original position (r) = 6 cm,

In pumped position radius (R) = 12 cm.

Ratio of surface areas in two situations = $\frac{4πr^2}{4πR^2}$

$$\begin{aligned}= \frac{r^2}{R^2} \\[1em] = \frac{6^2}{12^2} \\[1em] = \frac{36}{144} \\[1em] = \frac{1}{4} \\[1em] = 1 : 4.\end{aligned}$$

Hence, Option 1 is the correct option.

Two hemispheres are joined along their bases hence, the surface area of new solid will be double the surface area of hemisphere.

Curved surface area = 2πr² × 2 = 4πr².

Hence, Option 1 is the correct option.

If a solid of one shape is converted to another, then the surface area of the new solid may or may not be the same.

Hence, Option 4 is the correct option.

Edge of cube = 4.2 cm

Radius of largest cone cut out (r) = $\frac{4.2}{2}$ = 2.1 cm

Height of largest cone cut out (h) = 4.2 cm.

Volume of cone (V) = $\frac{1}{3}πr^2h$

Putting values we get,

$$\begin{aligned}V = \frac{1}{3} \times \frac{22}{7} \times (2.1)^2 \times 4.2 \\[1em] = \frac{22 \times 2.1 \times 2.1 \times 4.2}{3 \times 7} \\[1em] = \frac{407.484}{21} \\[1em] = 19.4 \text{ cm}^3\end{aligned}$$

Hence, Option 4 is the correct option.

Largest sphere that can be cut out from a cylinder of base radius 1 will have radius = 1 cm.

Volume of sphere (V) = $\frac{4}{3}πr^3 = \frac{4}{3}π(1)^3 = \frac{4}{3}π$ cm³.

Hence, Option 2 is the correct option.

Let the radius of two spheres be r₁ and r₂.

Given ratio of volumes = 64 : 27.

$$\begin{aligned}\Rightarrow \frac{\frac{4}{3}πr_1^3}{\frac{4}{3}πr_2^3} = \frac{64}{27} \\[1em] \Rightarrow \frac{r_1^3}{r_2^3} = \frac{4^3}{3^3} \\[1em] \Rightarrow \Big(\frac{r_1}{r_2}\Big)^3 = \Big(\frac{4}{3}\Big)^3 \\[1em] \Rightarrow \frac{r_1}{r_2} = \frac{4}{3}.\end{aligned}$$

Ratio of surface areas = $\frac{4πr_1^2}{4πr_2^2}$

$$\begin{aligned}= \frac{r_1^2}{r_2^2} \\[1em] = \Big(\frac{r_1}{r_2}\Big)^2 \\[1em] = \Big(\frac{4}{3}\Big)^2 \\[1em] = \frac{16}{9} = 16 : 9.\end{aligned}$$

Hence, Option 4 is the correct option.

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

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

$$\begin{aligned}\text{Ratio in their volumes } = \frac{1}{3}πr^2h : \frac{2}{3}πr^3 : πr^2h \\[1em] = \frac{1}{3} : \frac{2}{3} : 1\end{aligned}$$

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

Hence, Option 4 is the correct option.

A sphere and a cube have equal surface area.

Let a be the edge of cube and r the radius of sphere.

⇒ 4πr² = 6a²

⇒ π(2r)² = 6a²

Since, d = 2r

⇒ πd² = 6a²

$$\begin{aligned}\Rightarrow \frac{d^2}{a^2} = \frac{6}{π} \\[1em] \Rightarrow \sqrt{\frac{d^2}{a^2}} = \frac{\sqrt{6}}{\sqrt{π}} \\[1em] \Rightarrow \frac{d}{a} = \frac{\sqrt{6}}{\sqrt{π}}.\end{aligned}$$

Hence, Option 4 is the correct option.

Since, the cuboid is molded into a sphere.

Let radius of sphere be r cm.

∴ Volume of cuboid = Volume of sphere.

$$\begin{aligned}\therefore 49 \times 33 \times 24 = \frac{4}{3}πr^3 \\[1em] \Rightarrow r^3 = \frac{49 \times 33 \times 24 \times 3}{4 \times \frac{22}{7}} \\[1em] \Rightarrow r^3 = \frac{49 \times 33 \times 24 \times 3 \times 7}{4 \times 22} \\[1em] \Rightarrow r^3 = \frac{814968}{88} \\[1em] \Rightarrow r^3 = 9261 \\[1em] \Rightarrow r^3 = (21)^3 \\[1em] \Rightarrow r = 21 \text{ cm}.\end{aligned}$$

Hence, Option 1 is the correct option.

Since, cone is recasted into sphere.

∴ Volume of cone = Volume of sphere.

Given, radius of cone (R) = 6 cm, height (h) = 24 cm.

Let the radius of sphere be r.

$\therefore \frac{1}{3}πR^2h = \frac{4}{3}πr^3$

Multiplying both sides by 3 and dividing by π we get,

$$\begin{aligned}\Rightarrow R^2h = 4r^3 \\[1em] \Rightarrow 6^2 \times 24 = 4r^3 \\[1em] \Rightarrow 4r^3 = 36 \times 24 \\[1em] \Rightarrow r^3 = \frac{864}{4} \\[1em] \Rightarrow r^3 = 216 \\[1em] \Rightarrow r^3 = (6)^3 \\[1em] \Rightarrow r = 6 \text{ cm}.\end{aligned}$$

Hence, Option 2 is the correct option.

Diameter of cylinder = 15 cm

For cylinder,

Radius (r) = $\frac{15}{2} = 7.5$ cm and height (h) = 10 cm.

Let radius of sphere be R.

Volume of cylinder = Volume of sphere.

$$\begin{aligned}\Rightarrow πr^2h = \frac{4}{3}πR^3 \\[1em] \Rightarrow r^2h = \frac{4}{3}R^3 \\[1em] \Rightarrow R^3 = \frac{3 \times (7.5)^2 \times 10}{4} \\[1em] \Rightarrow R^3 = \frac{1687.5}{4} \\[1em] \Rightarrow R^3 = 421.875 \\[1em] \Rightarrow R^3 = (7.5)^3 \\[1em] \Rightarrow R = 7.5 \text{ cm}.\end{aligned}$$

Hence, Option 3 is the correct option.

Radius of sphere (R) = 10 cm.

Radius of ball (r) = 1 cm

Let number of balls formed be n.

Since a big sphere is divided into small balls.

Volume of sphere = n × Volume of each ball.

$\frac{4}{3}πR^3 = n \times \frac{4}{3}πr^3$

Multiplying both sides by $\frac{3}{4π}$.

$$\begin{aligned}\Rightarrow R^3 = nr^3 \\[1em] \Rightarrow n = \frac{R^3}{r^3} \\[1em] \Rightarrow n = \frac{10^3}{1^3} \\[1em] \Rightarrow n = 1000.\end{aligned}$$

Hence, Option 2 is the correct option.

Internal radius (r) = $\frac{4}{2}$ = 2 cm,

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

Given, spherical shell is recasted into cone.

Volume of cone = Volume of spherical shell.

Radius of cone (r₁) = $\frac{8}{2}$ = 4 cm.

Let height of cone be h.

$\frac{1}{3}πr_1^2h = \frac{4}{3}π(R^3 - r^3)$

Multiplying both sides by $\frac{3}{π}$.

$$\begin{aligned}\Rightarrow r_1^2h = 4(R^3 - r^3) \\[1em] \Rightarrow h = \frac{4(R^3 - r^3)}{r_1^2} \\[1em] = \frac{4 \times (4^3 - 2^3)}{4^2} \\[1em] = \frac{4 \times (64 - 8)}{16} \\[1em] = \frac{4 \times 56}{16} \\[1em] = \frac{224}{16} \\[1em] = 14 \text{ cm}\end{aligned}$$

Hence, Option 2 is the correct option.

Edge of cubical ice cream brick (a) = 22 cm

Volume = a³ = (22)³ = 10648 cm³.

Radius of ice cream cone (r) = 2 cm and height (h) = 7 cm.

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

= $\frac{1}{3}$ x $\frac{22}{7}$ x 2 x 2 x 7

= $\frac{22 \times 4}{3}$

= $\frac{88}{3}$ cm³.

Number of cones (n) = $\frac{\text{Vol. of ice cream brick}}{\text{Vol. of cone}}$

$$\begin{aligned}n = \frac{10648}{\frac{88}{3}} \\[1em] = \frac{10648 \times 3}{88} \\[1em] = \frac{31944}{88} \\[1em] = 363.\end{aligned}$$

Hence, Option 3 is the correct option.

Diameter of cylinder = 2 cm

Radius of cylinder (r) = $\frac{2}{2}$ = 1 cm and height (h) = 16 cm.

Given, 12 spheres are formed from this cylinder.

∴ Volume of cylinder = 12 × Volume of each sphere.

Let radius of each sphere be R.

$$\begin{aligned}\therefore πr^2h = 12 \times \frac{4}{3}πR^3 \\[1em] \Rightarrow R^3 = \frac{3π(1)^2 \times 16}{4π \times 12} \\[1em] \Rightarrow R^3 = \frac{48π}{48π} \\[1em] \Rightarrow R^3 = 1 \\[1em] \Rightarrow R = 1 \text{ cm}.\end{aligned}$$

Diameter = 2 × Radius = 2 cm.

Hence, Option 3 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.

On rotating the right angle triangle figure formed is :

Hence, Option 2 is the correct option.

When a solid sphere is cut into identical hemispheres.

Let radius of sphere be r.

Volume of a sphere = $\frac{4}{3}πr^3$

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

Volume of 2 hemisphere = $2 \times \frac{2}{3}πr^3 = \frac{4}{3}πr^3$.

The total volume of two hemispheres is equal to the volume of the original sphere.

Surface area of sphere = 4πr²

Surface area of hemisphere = 3πr²

Surface area of 2 hemisphere = 2 × 3πr² = 6πr².

The surface area of two hemispheres is not equal to the surface area of the original sphere.

Hence, Option 3 is the correct option.

Given, a sphere that can be inscribed in a hollow cube.

The maximum diameter of the sphere that can be inscribed is equal to the side of the cube (a).

The radius of the sphere is half of the diameter: r = $\frac{a}{2}$

The surface area of a sphere is given by 4πr².

So, reason(R) is false.

The surface area of a sphere = 4 x π x $\Big(\frac{a}{2}\Big)^2$

= 4 x π x $\frac{a^2}{4}$

= πa² cm².

So, assertion (A) is true.

Thus, Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

Given,

Height of cylinder,(H) = 15 cm

Diameter of cylinder, (D) = 14 cm

Radius of cylinder, r = $\frac{D}{2} = \frac{14}{2}$ = 7 cm

Height of cone,(h) = 15 cm

Radius of cone, (r) = 7 cm

The volume of the cone is given by the formula = $\frac{1}{3}$ πr²h

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 15 \\[1em] = \frac{1}{3} \times 22 \times 7 \times 15 \\[1em] = 22 \times 7 \times 3\\[1em] = 462\end{aligned}$$

So, assertion (A) is false.

The volume of a cylinder of height h and radius r is given by the formula; πr²h.

So, reason (R) is true.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given,

Height of cylinder,(H) = 15 cm

Diamter of cylinder, (D) = 14 cm

Radius of cylinder, r = $\frac{D}{2} = \frac{14}{2}$ = 7 cm

Radius of hemisphere, (r) = 7 cm

The volume of a cylinder is πr²h and the volume of a hemisphere is $\frac{2}{3}$ πr³.

So, reason (R) is true.

Volume of remaining wood = Volume of cylinder - Volume of hemisphere

$$\begin{aligned}= πR^2H - \frac{2}{3}πr^3\\[1em] = \frac{22}{7} \times 7^2 \times 15 - \frac{2}{3}\times \frac{22}{7} \times 7^3\\[1em] = 22 \times 7 \times 15 - \frac{2}{3}\times 22 \times 7^2\\[1em] = 2310 - \frac{2156}{3}\\[1em] = \frac{6930 - 2156}{3}\\[1em] = \frac{4774}{3}\\[1em] = 1591\frac{1}{3}\\[1em]\end{aligned}$$

So, assertion (A) is true.

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Since, container is open at the top hence total surface area of container (S) = curved surface area + area of base.

Radius of container = $\frac{70}{2}$ = 35 cm = 0.35 m.

$$\begin{aligned}\Rightarrow S = 2πrh + πr^2 \\[1em] = 2 \times \frac{22}{7} \times 0.35 \times 1 + \frac{22}{7} \times 0.35 \times 0.35 \\[1em] = 2 \times 22 \times 0.05 + 22 \times 0.05 \times 0.35 \\[1em] = 2.2 + 0.385 \\[1em] = 2.585 \text{ m}^2\end{aligned}$$

Cost of 1 m² sheet = ₹ 300.

Total cost = ₹ 300 × 2.585 = ₹ 775.50

Hence, the cost of tin for making the container is ₹ 775.50

Largest size of cylinder cut out of the wooden cuboid will be of diameter = 14 cm, radius = $\frac{14}{2}$ = 7 cm and height = 30 cm.

Volume of cylinder = πr²h = $\frac{22}{7} \times 7 \times 7 \times 30$ = 22 × 7 × 30 = 4620 cm³.

Volume of cuboid = lbh = 30 × 14 × 14 = 5880 cm³.

Volume of wooden wasted = Volume of cuboid - Volume of cylinder = 5880 - 4620 = 1260 cm³.

Hence, the volume of cylinder = 4620 cm³ and volume of wooden wasted = 1260 cm³.

Radius of cone = $\frac{30}{2}$ = 15 cm.

l = $\sqrt{r^2 + h^2}$

Putting values we get,

$$\begin{aligned}\Rightarrow 17 = \sqrt{15^2 + h^2} \\[1em] \Rightarrow 17^2 = 15^2 + h^2 \\[1em] \Rightarrow 289 = 225 + h^2 \\[1em] \Rightarrow h^2 = 289 - 225 \\[1em] \Rightarrow h^2 = 64 \\[1em] \Rightarrow h = \sqrt{64} = 8 \text{ cm}.\end{aligned}$$

Volume of cone (V) = $\frac{1}{3}πr^2h$

Putting values we get,

$$\begin{aligned}V = \frac{1}{3} \times 3.14 \times 15^2 \times 8 \\[1em] = \frac{3.14 \times 225 \times 8}{3} \\[1em] = 1884 \text{ cm}^3\end{aligned}$$

Total surface area of cone (S) = $πrl + πr^2 = πr(l + r)$

Putting values we get,

$$\begin{aligned}S = 3.14 \times 15 \times (17 + 15) \\[1em] = 3.14 \times 15 \times 32 \\[1em] = 1507.2 \text{ cm}^2\end{aligned}$$

Hence, the volume of cone = 1884 cm³ and surface area of cone = 1507.2 cm².

Height of cone = 8 cm.

Area of base = 156 cm².

Volume of cone (V) = $\frac{1}{3} \times$ Area of base × Height

Putting values we get,

$$\begin{aligned}V = \frac{1}{3} \times 156 \times 8 \\[1em] = \frac{1248}{3} \\[1em] = 416 \text{ cm}^3.\end{aligned}$$

Hence, the volume of cone = 416 cm³.

Circumference of the edge of bowl = 132 cm.

$$\begin{aligned}\therefore 2πr = 132 \\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 132 \\[1em] r = \frac{132 \times 7}{22 \times 2} \\[1em] r = 3 \times 7 \\[1em] r = 21 \text{ cm}.\end{aligned}$$

Volume of hemispherical bowl (V) = $\frac{2}{3}πr^3$

Putting values we get,

$$\begin{aligned}V = \frac{2}{3} \times \frac{22}{7} \times (21)^3 \\[1em] = \frac{2 \times 22 \times 21^3}{21} \\[1em] = 2 \times 22 \times 441 \\[1em] = 19404 \text{ cm}^3.\end{aligned}$$

Hence, the capacity of bowl = 19404 cm³.

Let radius of hemisphere be r cm.

Volume of hemisphere (V) = $\frac{2}{3}πr^3$

Given, V = $2425\frac{1}{2} = \frac{4851}{2}$.

$$\begin{aligned}\therefore \frac{2}{3} \times \frac{22}{7} \times r^3 = \frac{4851}{2} \\[1em] \Rightarrow r^3 = \frac{4851 \times 3 \times 7}{2 \times 22 \times 2} \\[1em] \Rightarrow r^3 = \frac{441 \times 3 \times 7}{2 \times 2 \times 2} \\[1em] \Rightarrow r^3 = \frac{9261}{8} \\[1em] \Rightarrow r^3 = \Big(\frac{21}{2}\Big)^3 \\[1em] \Rightarrow r = \frac{21}{2} \text{ cm}.\end{aligned}$$

Curved surface area = 2πr²

$$\begin{aligned}= 2 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \\[1em] = \frac{44 \times 441}{28} \\[1em] = 693 \text{ cm}^2.\end{aligned}$$

Hence, curved surface area of hemisphere = 693 cm².

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 (h₁) = 10.2 - 4.2 = 6 cm.

Volume of toy (V) = Volume of cone + Volume of hemisphere

= $\frac{1}{3}πr^2h + \frac{2}{3}πr^3$

Putting values we get,

$$\begin{aligned}V = \frac{1}{3}πr^2\Big(h + 2r\Big) \\[1em] = \frac{1}{3} \times \frac{22}{7} \times (4.2)^2 \times (6 + 2(4.2)) \\[1em] = \frac{22 \times 17.64 \times 14.4}{21} \\[1em] = \frac{5588.352}{21} \\[1em] = 266.112 \text{ cm}^3.\end{aligned}$$

Hence, the volume of the toy = 266.112 cm³.

From figure,

r = Radius of cylinder = Radius of hemisphere = $\frac{7}{2}$.

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

= 19 - $\Big(2 \times \frac{7}{2}\Big)$

= 19 - 7 = 12 cm.

Total Volume of solid (V) = 2 × Volume of hemisphere + Volume of cylinder

$$\begin{aligned}\therefore V = 2 \times \frac{2}{3}πr^3 + πr^2h \\[1em] = πr^2\Big(\frac{4r}{3} + h\Big) \\[1em] = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \Big(\frac{4}{3} \times \frac{7}{2} + 12\Big) \\[1em] = \frac{11 \times 7}{2} \times \Big(\frac{14}{3} + 12\Big) \\[1em] = \frac{77}{2} \times \Big(\frac{14}{3} + 12\Big) \\[1em] = \frac{77}{2} \times \frac{14}{3} + \frac{77}{2} \times 12 \\[1em] = \frac{539}{3} + 462 \\[1em] = \frac{539 + 1386}{3} \\[1em] = \frac{1925}{3} \\[1em] = 641\frac{2}{3} \text{ cm}^3.\end{aligned}$$

Surface area of solid (S) = 2 × Surface area of each hemisphere + Surface area of cylinder.

$$\begin{aligned}\therefore S = 2 \times 2πr^2 + 2πrh \\[1em] = πr(4r + 2h) \\[1em] = \frac{22}{7} \times \frac{7}{2} \times \Big(4 \times \frac{7}{2} + 2 \times 12\Big) \\[1em] = 11 \times (14 + 24) \\[1em] = 11 \times 38 \\[1em] = 418 \text{ cm}^2\end{aligned}$$

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

Let radius of cone (r) = 5a and height (h) = 12a.

Volume of cone (V) = $\frac{1}{3}πr^2h$

Given, V = 2512 cm³.

$$\begin{aligned}\therefore \frac{1}{3}πr^2h = 2512 \\[1em] \Rightarrow \frac{1}{3} \times 3.14 \times (5a)^2 \times 12a = 2512 \\[1em] \Rightarrow \frac{3.14 \times 25a^2 \times 12a}{3} = 2512 \\[1em] \Rightarrow \frac{942a^3}{3} = 2512 \\[1em] \Rightarrow 314a^3 = 2512 \\[1em] \Rightarrow a^3 = \frac{2512}{314} \\[1em] \Rightarrow a^3 = 8 \\[1em] \Rightarrow a^3 = (2)^3 \\[1em] \Rightarrow a = 2 \text{ cm}.\end{aligned}$$

r = 5a = 5 × 2 = 10 cm.

h = 12a = 12 × 2 = 24 cm.

$$\begin{aligned}\text{l} = \sqrt{r^2 + h^2} \\[1em] = \sqrt{10^2 + 24^2} \\[1em] = \sqrt{100 + 576} \\[1em] = \sqrt{676} \\[1em] = 26 \text{ cm}.\end{aligned}$$

Hence, the slant height of cone is 26 cm.

Let height of cone and cylinder = h.

Diameter of base of cone = 3a, (r₁) = $\frac{3a}{2}$.

Diameter of base of cylinder = 2a, (r₂) = $\frac{2a}{2}$.

Volume of cone (V₁) = $\frac{1}{3}πr_1^2h$.

Volume of cylinder (V₂) = $πr_2^2h$.

$$\begin{aligned}\therefore \frac{V_1}{V_2} = \frac{\frac{1}{3}πr_1^2h}{πr_2^2h} \\[1em] = \frac{1}{3} \times \frac{r_1^2}{r_2^2} \\[1em] = \frac{1}{3} \times \frac{\Big(\frac{3a}{2}\Big)^2}{\Big(\frac{2a}{2}\Big)^2} \\[1em] = \frac{1}{3} \times \frac{9a^2 \times 2^2}{4a^2 \times 2^2} \\[1em] = \frac{1}{3} \times \frac{9}{4} \\[1em] = \frac{3}{4} \\[1em] = 3 : 4.\end{aligned}$$

Radius of cone (r) = 9 cm.

Height of cone (h) = 10 cm.

Volume of water filled in cone = $\frac{1}{3}πr^2h$

Let h₁ be the rise in height of water in the jar.

Radius of jar (r₁) = 10.

Since, cone is submerged completely hence, volume of water level rise = volume of cone.

$$\begin{aligned}\therefore πr_1^2h_1 = \frac{1}{3}πr^2h \\[1em] \Rightarrow π(10)^2 \times h_1 = \frac{1}{3}π(9)^2 \times 10 \\[1em] \Rightarrow h_1 = \frac{\frac{1}{3}π \times 81 \times 10}{π \times (10)^2} \\[1em] \Rightarrow h_1 = \frac{\cancel{π} \times \overset{27}{\cancel{81}} \times \cancel{10}}{\cancel{3} \times \cancel{π} \times \cancel{10} \times 10} \\[1em] \Rightarrow h_1 = \frac{27}{10} \\[1em] \Rightarrow h_1 = 2.7 \text{ cm}.\end{aligned}$$

Hence, the rise in height of water in jar is 2.7 cm.

Radius of base of cone (r) = 8 cm,

Radius of cylinder (r) = 8 cm

Height of cylindrical part (h₁) = 240 cm

Height of conical part (h₂) = 36 cm.

Volume of pillar (V) = Volume of cylinder + Volume of cone.

$$\begin{aligned}V = πr^2h_1 + \frac{1}{3}πr^2h_2 \\[1em] = πr^2(h_1 + \frac{h_2}{3}) \\[1em] = \frac{22}{7} \times (8)^2 \times \Big(240 + \frac{36}{3}\Big) \\[1em] = \frac{22}{7} \times 64 \times \Big(240 + 12\Big) \\[1em] = \frac{22}{7} \times 64 \times 252 \\[1em] = 22 \times 64 \times 36 \\[1em] = 50688 \text{ cm}^3\end{aligned}$$

Weight of 1 cm³ of iron = 7.8 g

∴ Weight of 50688 cm³ of iron = 50688 × 7.8 = 395366.4 g

Converting it into Kg,

395366.4 g = $\frac{395366.4}{1000}$ Kg

= 395.3664 Kg

Hence, the weight of the pillar is 395.3664 kg.

The below figure shows the circus tent:

Diameter of cylindrical part = 126 m.

Radius of cone = Radius of cylinder = r = $\frac{126}{2}$ = 63 m.

Height of cylindrical part (h₁) = 5 m,

Total height of tent = 21 m,

Height of conical part (h) = 21 - 5 = 16 m.

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

$$\begin{aligned}= \sqrt{63^2 + 16^2} \\[1em] = \sqrt{3969 + 256} \\[1em] = \sqrt{4225} \\[1em] = 65 \text{ m}.\end{aligned}$$

Surface area of tent (S) = Surface area of cylinder + Surface area of cone.

$$\begin{aligned}S = 2πrh_1 + πrl \\[1em] = πr(2h_1 + l) \\[1em] = \frac{22}{7} \times 63 \times (2 \times 5 + 65) \\[1em] = 22 \times 9 \times (10 + 65) \\[1em] = 198 \times 75 \\[1em] = 14850 \text{ m}^2.\end{aligned}$$

Cost of one 1 m² cloth = ₹ 36.

Total cost = 14850 × 36 = ₹ 534600.

Hence, the total cost of tent is ₹ 534600.

Radius of base of cone (r₁) = 3 cm,

Height of cone (h₁) = 4 cm.

Slant height of cone (l) = $\sqrt{r_1^2 + h_1^2}$

$$\begin{aligned}\sqrt{3^2 + 4^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ cm}.\end{aligned}$$

Let height of cylinder be h₂ cm and radius be r₂ cm.

r₂ = $\frac{4}{2}$ = 2 cm.

Given, total surface area of cylinder = total surface area of cone.

⇒ 2πr(r₂ + h₂) = πr₁(l + r₁)

⇒ 2π × 2 × (2 + h₂) = π × 3 × (5 + 3)

⇒ π(8 + 4h₂) = 24π

Dividing both sides by π,

⇒ 4h₂ = 24 - 8

⇒ 4h₂ = 16

⇒ h₂ = 4 cm.

(i) Ratio between curved surface area of cone and cylinder (Ratio) = $\frac{πr_1l}{2πr_2h}$

Putting values we get,

$$\begin{aligned}\frac{\text{Curved Surface of Cone}}{\text{Curved Surface of Cylinder}} = \frac{π \times 3 \times 5}{2 \times π \times 2 \times 4} \\[1em] = \frac{15π}{16π} \\[1em] = 15 : 16.\end{aligned}$$

Hence, the ratio between curved surface area of cone and cylinder 15 : 16.

(ii) Ratio between their volumes = $\frac{\text{Vol. of Cone}}{\text{Vol. of Cylinder}}$

$$\begin{aligned}= \frac{\frac{1}{3}πr_1^2h_1}{πr_2^2h_2} \\[1em] = \frac{r_1^2h_1}{3r_2^2h_2} \\[1em] = \frac{3^2 \times 4}{3 \times 2^2 \times 4} \\[1em] = \frac{3}{2^2} \\[1em] = \frac{3}{4} \\[1em] = 3 : 4.\end{aligned}$$

Hence, the ratio between volumes of cones and cylinder is 3 : 4.

Radius of base of cone (r) = 2.1 cm and height (h) = 8.4 cm.

Let radius of sphere be R cm.

Since, cone is recasted into sphere hence, their volume will be equal.

Volume of cone = Volume of sphere.

$\therefore \frac{1}{3}πr^2h = \frac{4}{3}πR^3$

Multiplying both sides by $\frac{3}{π}$

$$\begin{aligned}\Rightarrow r^2h = 4R^3 \\[1em] \Rightarrow R^3 = \frac{r^2h}{4} \\[1em] \Rightarrow R^3 = \frac{(2.1)^2 \times 8.4}{4} \\[1em] \Rightarrow R^3 = (2.1)^2 \times (2.1) \\[1em] \Rightarrow R^3 = (2.1)^3 \\[1em] \Rightarrow R = 2.1 \text{ cm}.\end{aligned}$$

Hence, the radius of the sphere = 2.1 cm.

Given,

Radius of coin (r) = $\frac{2.5}{2}$= 1.25 cm,

Height of coin (h) = 3 mm = 0.3 cm,

Radius of cylinder (R) = 3 cm.

Height of cylinder (H) = 5 cm

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

Volume of cylinder = n × Volume of each coin.

$$\begin{aligned}\therefore πR^2H = n \times πr^2h \\[1em] n = \frac{πR^2H}{πr^2h} \\[1em] n = \frac{3^2 \times 5}{1.25^2 \times 0.3} \\[1em] n = \frac{45}{0.46875} \\[1em] n = 96.\end{aligned}$$

Hence, 96 coins are required to form a solid cylinder.

Given,

Radius of hemisphere (r) = 8 cm.

Radius of cone (R) = 6 cm.

Let height of cone be h cm.

Since, hemisphere is casted into cone,

Volume of hemisphere = Volume of cone.

$$\begin{aligned}\therefore \frac{2}{3}πr^3 = \frac{1}{3}πR^2h \\[1em] \Rightarrow 2r^3 = R^2h \\[1em] \Rightarrow h = \frac{2r^3}{R^2} \\[1em] \Rightarrow h = \frac{2 \times 8^3}{6^2} \\[1em] \Rightarrow h = \frac{1024}{36} \\[1em] \Rightarrow h = 28.44 \text{ cm}.\end{aligned}$$

Hence, the height of the cone is 28.44 cm.

Radius of hemispherical bowl (r) = 6 cm.

Radius of cylinder (R) = 4 cm.

Let h be the height of water in cylinder.

Volume of hemispherical bowl = Volume of water in cylinder.

$$\begin{aligned}\Rightarrow \frac{2}{3}πr^3 = πR^2h \\[1em] \Rightarrow \frac{2}{3}r^3 = R^2h \\[1em] \Rightarrow \frac{2}{3} \times 6^3 = 4^2 \times h \\[1em] \Rightarrow 2 \times 2 \times 36 = 16h \\[1em] \Rightarrow h = \frac{144}{16} \\[1em] \Rightarrow h = 9 \text{ cm}.\end{aligned}$$

Hence, the height of water in cylinder is 9 cm.

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

Radius of cylinder (R) = $\frac{12}{2}$ = 6 cm.

Let height of water raised be h cm.

Volume of sphere = Volume of water rise in cylinder

$$\begin{aligned}\frac{4}{3}πr^3 = πR^2h \\[1em] \frac{4}{3} \times 3^3 = 6^2 \times h \\[1em] 4 \times 3^2 = 6^2 \times h \\[1em] h = \frac{36}{36} = 1 \text{ cm}.\end{aligned}$$

Hence, the height of water raised is 1 cm.

Radius of solid sphere (r₁) = 6 cm

Volume of solid sphere (V) = $\frac{4}{3}πr_1^3$

$$\begin{aligned}= \frac{4}{3}π \times 6^3 \\[1em] = 4π \times 2 \times 6^2 \\[1em] = 288π \text{ cm}^3\end{aligned}$$

External radius of cylinder (R) = 5 cm, height (h) = 32 cm.

Let r be inner radius of cylinder.

Volume of cylinder = Volume of sphere.

$$\begin{aligned}\therefore V = π(R^2 - r^2)h \\[1em] \Rightarrow 288π = π(5^2 - r^2) \times 32 \\[1em] \Rightarrow 288 = 32(25 - r^2) \\[1em] \Rightarrow \frac{288}{32} = 25 - r^2 \\[1em] \Rightarrow 9 = 25 - r^2 \\[1em] \Rightarrow r^2 = 25 - 9 \\[1em] \Rightarrow r^2 = 16 \\[1em] \Rightarrow r = 4 \text{ cm}.\end{aligned}$$

Thickness of hollow cylinder = R - r = 5 - 4 = 1 cm.

Hence, the thickness of the cylinder = 1 cm.

When vertically placed,

Height of liquid (h) = $\frac{1 + 6}{2} = \frac{7}{2}$ cm

Diameter of base = 7 cm

Radius (r) = $\frac{7}{2}$ cm

$$\begin{aligned}\text{Area of wet surface} = πr^2 + 2πrh \\[1em] = πr(r + 2h) \\[1em] = \frac{22}{7} \times \frac{7}{2}\Big(\frac{7}{2} + 2\times \frac{7}{2}\Big) \\[1em] = 11 \times (3.5 + 7) \\[1em] = 11 \times 10.5 \\[1em] = 115.5 \text{ cm}^2.\end{aligned}$$

Hence, area of wet surface = 115.5 cm².

(a) 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.

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

= 1500 × 4 g

= 6000 g

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

Hence, mass of each box = 6 kg.