Volume & Surface Area of Solids

Find the curved surface area and the total surface area of the cylinder for which:

(i) h = 16 cm, r = 10.5 cm

(ii) h = 5 cm, r = 21 cm

(iii) h = 20 cm, r = 14 cm

(iv) h = 1 m, r = 1.4 cm

Answer

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

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

Curved surface area of cylinder = 2πrh

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

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

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

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

Curved surface area of cylinder = 2πrh

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

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

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

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

Curved surface area of cylinder = 2πrh

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

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

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

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

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

Curved surface area of cylinder = 2πrh

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

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

Find the volume of the cylinder in which:

(i) Height = 21 cm and Base Radius = 5 cm

(ii) Diameter = 28 cm and Height = 40 cm

Answer

By formula, Volume of cylinder = πr²h

(i) Given,

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

Volume of cylinder = πr²h

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

Hence, volume of a cylinder = 1650 cm³.

(ii) Given,

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

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

Volume of cylinder = πr²h

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

Hence, volume of a cylinder = 24640 cm³.

Find the weight of a solid cylinder of radius of radius 10.5 cm and height 60 cm, if the material of the cylinder weighs 5 grams per cu. cm

Answer

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

By formula,

Volume of cylinder = πr²h

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

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

⇒ 1 gram = $\dfrac{1}{1000}$ kg

⇒ 5 gram = $\dfrac{5}{1000}$ kg

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

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

A cylindrical tank has a capacity of 6160 m³. Find its depth, if its radius is 14 m. Also, find the cost of painting its curved surface at ₹ 30 per m².

Answer

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

By formula,

Volume of cylinder = πr²h

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

Curved surface area of cylinder = 2πrh

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

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

⇒ 880 × 30 = ₹ 26,400.

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

(i) The curved surface area of a cylinder is 4400 cm² and the circumference of its base is 110 cm. Find the height and the volume of the cylinder.

(ii) The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the radius of the cylinder and also its volume.

Answer

Given, curved surface area of a cylinder = 4400 cm²

(i) By formula,

Curved surface area of cylinder = 2πrh

∴ 2πrh = 4400 ....(1)

Given, circumference of base = 110 cm

We know that circumference = 2πr

∴ 2πr = 110 ....(2)

Dividing eq. (1) by (2),

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

From eq.(1), we have,

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

Volume of cylinder = πr²h

Putting values we get,

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

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

(ii) Given, circumference of base = 132 cm

We know that circumference = 2πr

∴ 2πr = 132

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

Given, height (h) = 25 cm.

Volume of cylinder = πr²h

Putting values we get,

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

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

The total surface area of a solid cylinder is 462 cm² and its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

Answer

Given,

Total surface area = 462 cm²

⇒ 2πr(h + r) = 462

⇒ πr(h + r) = $\dfrac{462}{2}$

⇒ πr(h + r) = 231 ....(1)

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

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

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

⇒ 3h = h + r

⇒ 3h - h = r

⇒ 2h = r

⇒ h = $\dfrac{\text{r}}{2}$

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

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

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

Volume of cylinder = πr²h

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

Hence, volume of cylinder is 539 cm³.

The sum of the radius of the base and the height of a solid cylinder is 37 m. If the total surface area of the cylinder be 1628 m², find its volume.

Answer

Given, r + h = 37 m

Total surface area = 1628 m²

⇒ 2πr(h + r) = 1628

⇒ πr(h + r) = $\dfrac{1628}{2}$

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

⇒ r + h = 37

⇒ 7 + h = 37

⇒ h = 37 - 7

⇒ h = 30 m.

Volume of cylinder = πr²h

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

Hence, volume of solid cylinder is 4620 m³.

Find the height of a solid circular cylinder having total surface area of 660 cm² and radius 5 cm.

Answer

Total surface area = 660 cm²

⇒ 2πr(h + r) = 660

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

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

Find the total surface area of a hollow cylinder open at both ends, if its length is 12 cm, external diameter is 8 cm and the thickness is 2 cm.

Answer

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

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

Let internal radius be r cm.

Thickness = (R - r)

⇒ 2 = 4 - r

⇒ r = 4 - 2

⇒ r = 2 cm.

By formula,

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

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

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

A solid metallic cylinder is cut into two identical halves along its height. The diameter of the cylinder is 7 cm and the height is 10 cm. Find :

(i) The total surface area (both the halves).

(ii) The total cost of painting the two halves at the rate of ₹ 30 per cm².

$\Big(\text{use } \pi = \dfrac{22}{7}\Big)$

Answer

(i) Given,

Diameter of cylinder (d) = 7 cm

Radius of cylinder (r) = $\dfrac{d}{2} = \dfrac{7}{2}$ = 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)]

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

= (2 × 22 × 0.5 × 13.5) + 140

= 297 + 140

= 437 cm².

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

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

= 437 × 30

= ₹ 13,110.

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

Water is flowing at the rate of 3 km/hr through a circular pipe of 20 cm internal diameter into a circular cistern of diameter 10 m and depth 2 m. In how much time will the cistern be filled?

Answer

Let the cistern be filled in x hours.

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

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

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

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

= πr² × rate of flow of water

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

Given, diameter of cistern = 10 m

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

Depth of cistern = 2 m

Volume of cistern = πr²h

$= \dfrac{22}{7} \times 5^2 \times 2 \\[1em] = \dfrac{22}{7} \times 25 \times 2 \\[1em] = \dfrac{1100}{7}$

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

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

= 1 hour 40 min.

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

A swimming pool 70 m long, 44 m wide and 3 m deep, is filled by water issuing from a pipe of diameter 35 cm, at 6 m per second. How many hours does it take to fill the pool?

Answer

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

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

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

= πr² × 6 m/s

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

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

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

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

Water is flowing at the rate of 8 m per second through a circular pipe whose internal diameter is 2 cm, into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase in the water level in 30 minutes.

Answer

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

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

= πr² × 8 m/s

= πr² × 8 × 100 cm/s

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

Volume of cylindrical tank = πR²h

$= \dfrac{22}{7} \times 40^2 \times \text{h} \\[1em] = \dfrac{22}{7} \times 1600 \times \text{h} \\[1em] = \dfrac{35200}{7} \times \text{h}$

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

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

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

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

A 20 m deep well with diameter 7 m is dug up and the earth from digging is spread evenly to form a platform 22 m × 14 m. Determine the height of the platform.

Answer

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

By formula,

Volume of a cylinder = πr²h

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

Let height of platform be H.

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

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

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

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

Hence, the height of the platform is 2.5 m.

Find the mass of a metallic hollow cylindrical pipe 24 cm long with internal diameter 10 cm and made of 5 mm thick metal, if 1 cm³ of the metal weighs 7.5 grams.

Answer

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

Given,

Thickness = 5 mm = 0.5 cm

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

⇒ R = r + thickness

⇒ R = 5.5 cm

Length of the pipe (h) = 24 cm

External volume = πR²h

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

Internal volume = πr²h

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

Volume of metal = External volume - Internal volume

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

Given, 1 cm³ of the metal weighs 7.5 grams.

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

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

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

A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.

Answer

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

Depth of the well (h) = 8.4 m

Volume of the earth dug out from the well = πr²h

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

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

The embankment forms a hollow cylinder around the well.

Let height of the embankment be H.

∴ Volume of embankment = πR²H - πr²H

= πH(R² - r²)

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

Volume of earth dug out = Volume of the embankment

$\Rightarrow 660 = 412.5 \text{H} \\[1em] \Rightarrow \text{H} = \dfrac{660}{412.5} \\[1em] \Rightarrow \text{H} = 1.6 \text{ m.}$

Hence, the height of the embankment is 1.6 m.

The height of a right circular cone is 24 cm and the radius of its base is 7 cm. Calculate :

(i) the slant height of the cone

(ii) the lateral surface area of the cone

(iii) the total surface area of the cone

(iv) the volume of the cone

Answer

Given, h = 24 cm and r = 7 cm

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

Hence, slant height of the cone is 25 cm.

(ii) Lateral surface area = πrl

$= \dfrac{22}{7} \times 7 \times 25 \\[1em] = \dfrac{3850}{7} \\[1em] = 550 \text{ cm}^2$

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

(iii) Total surface area = πr(l + r)

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

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

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

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

Hence, volume of the cone is 1232 cm³.

The height of a right circular cone is 8 cm and the diameter of its base is 12 cm. Calculate :

(i) the slant height of the cone

(ii) the total surface area of the cone

(iii) the volume of the cone

Answer

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

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

Hence, slant height of the cone is 10 cm.

(ii) Total surface area = πr(l + r)

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

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

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

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

Hence, volume of the cone is 301.7 cm³.

The slant height of a cone is 17 cm and the radius of its base is 15 cm. Find:

(i) the height of the cone

(ii) the volume of the cone

(iii) the total surface area of the cone

Answer

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

(i) l² = r² + h²

⇒ h² = l² - r²

⇒ h² = 17² - 15²

⇒ h² = 289 - 225

⇒ h² = 64

⇒ h = $\sqrt{64} = 8 \text{ cm.}$

Hence, height of the cone is 8 cm.

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

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

Hence, volume of the cone is 1885.7 cm³.

(iii) Total surface area = πr(l + r)

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

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

The volume of a right circular cone is 660 cm³ and diameter of its base is 12 cm. Calculate:

(i) the height of the cone

(ii) the slant height of the cone

(ii) the total surface area of the cone

Answer

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

Volume of cone = 660 cm³

(i) By formula,

Volume of cone = $\dfrac{1}{3}$ πr²h

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

Hence, the height of the cone is 17.5 cm.

(ii) By formula,

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

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

Hence, slant height of the cone is 18.5 cm.

(iii) Total surface area = πr(l + r)

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

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

The total surface of a right circular cone of slant height 20 cm is 384π cm². Calculate:

(i) its radius in cm

(ii) its volume in cm³, in terms of π

Answer

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

(i) By formula,

Total surface area = πr(l + r)

⇒ 384π = πr(20 + r)

⇒ 384 = 20r + r²

⇒ r² + 20r - 384 = 0

⇒ r² + 32r - 12r - 384 = 0

⇒ r(r + 32) - 12(r + 32) = 0

⇒ (r + 32) = 0 or (r - 12) = 0

⇒ r = - 32 or r = 12

Since, radius cannot be negative.

∴ r = 12 cm.

Hence, radius of the cone is 12 cm.

(ii) l² = r² + h²

⇒ h² = l² - r²

⇒ h² = 20² - 12²

⇒ h² = 400 - 144

⇒ h² = 256

⇒ h = $\sqrt{256} = 16 \text{ cm.}$

Volume of cone = $\dfrac{1}{3}$ πr²h

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

Hence, volume of the cone is 768 π cm³.

The radius and the height of a right circular cone are in the ratio of 5 : 12 and its volume is 2512 cm³. Find:

(i) the radius and height of the cone

(ii) the curved surface area of the cone

(iii) the total surface area of the cone

(Take π = 3.14)

Answer

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

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

Volume of cone = $\dfrac{1}{3}$ πr²h

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

⇒ r = 5x = 5 × 2 = 10 cm

⇒ h = 12x = 12 × 2 = 24 cm

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

(ii) Curved surface area = πrl

l² = r² + h²

⇒ l² = 10² + 24²

⇒ l² = 100 + 576

⇒ l² = 676

⇒ l = $\sqrt{676}$ = 26 cm

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

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

(iii) Total surface area = πr(l + r)

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

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

A right circular cone of radius 20 cm has its volume 8800 cm³. Find its:

(i) height

(ii) curved surface area.

Give your answer to the nearest whole number.

Answer

Let height of cone be h cm.

Given,

Radius of cone (r) = 20 cm

(i) Given,

Volume = 8800 cm³

By formula,

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

∴ $\dfrac{1}{3}\pi r^2 h = 8800$

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

Hence, height of cone = 21 cm.

(ii) By formula,

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

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

By formula,

Curved Surface Area of cone = πrl

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

⇒ $\dfrac{12760}{7}$

⇒ 1822.857 ≈ 1823 cm².

Hence, curved surface area of cone = 1823 cm².

How many metres of canvas 1.25 m will be needed to make a conical tent whose base radius is 17.5 m and height 6 m?

Answer

Given, r = 17.5 m and h = 6 m

l² = r² + h²

⇒ l² = 17.5² + 6²

⇒ l² = 306.25 + 36

⇒ l² = 342.25

⇒ l = $\sqrt{342.25}$ = 18.5 m

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

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

Width of the canvas used = 1.25 m

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

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

A circus tent is cylindrical to a height of 3 meters and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 2.5 m wide to make the required tent.

Answer

From figure,

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

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

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

Slant height (l) = 53 m

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

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

Width of the canvas used = 2.5 m

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

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

An iron pillar consists of a cylindrical portion, 2.8 m high and 20 cm in diameter and a cone 42 cm high is surrounding it. Find the weight of the pillar, given that 1 cm³ of iron weighs 7.5 g.

Answer

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

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

Height of the conical portion, H = 42 cm

From figure,

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

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

= πr²h + $\dfrac{1}{3}$ πr²H

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

Given,

Weight of 1 cm³ of iron = 7.5 gm.

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

Hence, the weight of the pillar is 693 kg.

Water flows at the rate of 10 m per minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm?

Answer

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

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

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

Height of the conical portion, H = 24 cm

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

Volume of water that flows in 1 min = πr²h

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

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

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

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

$= \dfrac{\dfrac{70400}{7}}{\dfrac{1375}{7}} \\[1em] = \dfrac{70400}{7} \times \dfrac{7}{1375} \\[1em] = \dfrac{70400}{1375} \\[1em] = 51.2 \text{ min.}$

= 51 min 12 sec.

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

A conical tent is to accommodate 11 persons. Each person must have 4 m² of the space on the ground and 20m³ of air to breathe. Find the height of the cone.

Answer

Given,

Each person must have 20 m³ of air to breathe.

∴ 11 persons need 11 × 20 m³ = 220 m³

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

∴ 11 persons need 11 × 4 m² = 44 m²

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

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

Let height of the conical tent be h meters.

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

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

Hence, height of the tent is 15 m..

A conical tent is to accommodate 77 persons. Each person must have 16 m³ of air to breathe. Given the radius of the tent as 7 m, find the height of the tent and also its curved surface area.

Answer

Given,

Each person must have 16 m³ of air to breathe.

∴ 77 persons need 77 × 16 m³ = 1232 m³

Radius of the tent (r) = 7 m

Let height of the conical tent be h meters.

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

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

By formula,

l² = r² + h²

⇒ l² = 7² + 24²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$ = 25 m

Curved surface area of the tent = πrl

$= \dfrac{22}{7} \times 7 \times 25 \\[1em] = 22 \times 25 \\[1em] = 550 \text{ m}^2$

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

A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm. It is melted and recast into a right circular cone with radius of its base as 1.2 cm. Find its height.

Answer

Radius of cone, r = 1.6 cm

Height of the cone, h = 3.6 cm

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

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

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

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

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

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

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

Hence, height of the cone is 6.4 cm.

A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to form cones, each of height 1 cm and base radius 1 mm. Find the number of cones.

Answer

For larger cylinder,

Height (H) = 5 cm

Radius (R) = 3 cm

For smaller cones,

Height (h) = 1 cm

Radius (r) = 1 mm = 0.1 cm

Let no. of smaller cones formed be n.

Volume of larger cylinder = n × Volume of smaller cones

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

Hence, the number of cones formed = 13500.

A conical vessel, whose internal radius is 12 cm and height 50 cm, is full of liquid. The contents are emptied into a cylindrical vessel with internal radius 10 cm. Find the height to which the liquid rises in the cylindrical vessel.

Answer

For cylindrical vessel,

Let height be H cm

Radius (R) = 10 cm

For conical vessel,

Height (h) = 50 cm

Radius (r) = 12 cm

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

∴ Volume of cylinder = Volume of conical vessel

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

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

The height of a cone is 40 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be $\dfrac{1}{64}$ of the volume of the given cone, at what height above the base is the section cut?

Answer

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

From figure,

∠NOD = ∠MOB (Common angle)

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

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

∴ △OND ~ △OMB (By AA similarity)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

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

According to given,

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

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

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

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

Using eq.(1),

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

The height of the cone OCD = 10 cm

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

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

A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal is 0.5 cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone, correct to one decimal place.

Answer

Internal radius (r) = 3 cm

Height (h) = 21 cm

Thickness = External radius (R) - Internal radius

⇒ 0.5 = External radius - 3 cm

⇒ External radius = 0.5 + 3 = 3.5 cm

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

Putting values we get,

∴ Volume of metal = π(3.5² - 3²) × 21

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

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

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

∴ 214.5 = $\dfrac{1}{3}$ πr²H

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

Hence, radius of the cone is 5.4 cm.

From a circular cylinder of diameter 10 cm and height 12 cm, a conical cavity of the same base radius and of the same height is hollowed out. Find the volume and the whole surface of the remaining solid. Leave the answer in π

Answer

Given,

Height of the cylinder (H) = 12 cm

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

Height of the cone (h) = 12 cm

Radius of the cone (r) = 5 cm

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

= πR²H - $\dfrac{1}{3}$ πr²h

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

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

By formula,

l² = r² + h²

⇒ l² = 5² + 12²

⇒ l² = 25 + 144

⇒ l² = 169

⇒ l = $\sqrt{169}$ = 13 cm

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

= 2πRH + πrl + πR²

= π(2RH + rl + R²)

= π(2 × 5 × 12 + 5 × 13 + 5²)

= π(120 + 65 + 25)

= 210 π cm².

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

From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the cone and of the remaining material, each correct to one place of decimal.

Answer

Edge of a cube = 14 cm

Volume = side³ = 14³ = 2744 cm³.

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

Diameter of the cone cut out from it = 14 cm

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

Height, h = 14 cm

Volume of cone = $\dfrac{1}{3}$ πr²h

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

Rounding off to one decimal place = 718.8 cm³

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

= 2744 - 718.67

= 2025.33 cm³

Rounding off to one decimal place = 2025.3 cm³

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

A cone of maximum volume is carved out of a block of wood of size 20 cm × 10 cm × 10 cm. Find the volume of the cone carved out, correct to one decimal place.

Answer

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

Diameter of the cone for maximum volume = 10 cm

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

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

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

Volume of cone = $\dfrac{1}{3}$ πr²h

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

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

From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cones are also of 6 cm and height 10.5 cm. Find the volume of the remaining solid.

Answer

Given,

Diameter of solid wooden cylinder (D) = 6 cm

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

Height of solid wooden cylinder (H) = 28 cm

Diameter of cone (d) = 6 cm

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

Height of the cone (h) = 10.5 cm

Volume of cylinder = πR²H

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

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

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

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

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

= 792 - 198

= 594 cm³.

Hence, volume of the remaining solid = 594 cm³.

Find the volume and surface area of a sphere of radius :

(i) 10.5 cm

(ii) 4.2 cm

Answer

(i) Given, r = 10.5 cm

Surface area of sphere = 4πr²

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

Volume of sphere = $\dfrac{4}{3}$ πr³

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

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

(ii) Given, r = 4.2 cm

Surface area of sphere = 4πr²

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

Volume of sphere = $\dfrac{4}{3}$ πr³

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

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

The volume of a sphere is 288 π cm³. Calculate its radius and hence its surface area to nearest cm².

Answer

Given, volume of a sphere is 288 π cm³.

Let radius be r cm.

Volume of sphere = $\dfrac{4}{3}$ πr³

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

Surface area of sphere = 4πr²

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

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

The curved surface area of a sphere is 2826 cm². Calculate its radius and hence its volume .

(Take π = 3.14)

Answer

Given, curved surface area of a sphere is 2826 cm²

Curved surface area of sphere = 4πr²

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

Volume of sphere = $\dfrac{4}{3}$ πr³

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

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

Find the curved surface area and the total surface area of a hemisphere of diameter 10 cm.

Answer

Given, diameter = 10 cm

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

Curved surface area of hemisphere = 2πr²

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

Total surface area of hemisphere = 3πr²

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

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

(i) How many bullets can be made out of a cube of lead whose edge measures 22 cm, each bullet being 2 cm in diameter?

(ii) How many lead shots each 3 mm in diameter can be made from a cuboid of dimensions 9 cm × 11 cm × 12 cm?

Answer

(i)

Edge of a cube = 22 cm

Volume of cube = side³ = 22³ = 10648 cm³.

Bullets are spherical in shape.

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

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

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

Let the number of bullets formed be n.

∴ Volume of cube = n × Volume of each bullet

$\Rightarrow 10648 = \text{n} \times \dfrac{88}{21} \\[1em] \Rightarrow \text{n} = \dfrac{10648 \times 21}{88} \\[1em] \Rightarrow \text{n} = \dfrac{223608}{88} \\[1em] \Rightarrow \text{n} = 2541$

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

(ii)

Shots is in the shape of sphere.

(3 mm = 0.3 cm)

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

Let the number of sphere formed be n.

Volume of cuboid = n × Volume of each shot

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

Hence, there are 84000 lead shots.

How many bullets, each of diameter 1.5 cm, can be made by melting a cylinder of lead having radius of the base 5 cm and height 18 cm ?

Answer

Bullet is in shape of sphere.

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

Radius of cylinder, R = 5 cm

Height of cylinder, h = 18 cm

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

∴ Volume of cylinder = n × Volume of each bullet

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

Hence, 800 bullets are made.

How many lead balls, each of radius 1.5 cm can be made by melting a bigger ball of radius 9 cm?

Answer

Given,

Radius of bigger ball, R = 9 cm

Radius of smaller ball, r = 1.5 cm

Let the number of smaller lead balls formed be n.

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

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

Hence, 216 lead balls can be made.

A manufacturing company prepares spherical ball bearings, each of radius 7 mm and mass 4 gm. These ball bearings are packed into boxes. Each box can have maximum of 2156 cm³ of ball bearings. Find the :

(i) maximum number of ball bearings that each box can have.

(ii) mass of each box of ball bearings in kg.

$\Big(\text{use } \pi = \dfrac{22}{7}\Big)$

Answer

(i) Given,

Radius of ball bearings = 7 mm

Volume of box = 2156 cm³ = 2156 × 10³ mm³

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

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

Substituting values we get :

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

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

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

= 1500 × 4 gm

= 6000 gm

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

Hence, mass of each box = 6 kg.

A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained.

Answer

Radius of sphere, r = 10.5 cm

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

Radius of cone, R = 3.5 cm

Height, h = 3 cm

Volume of sphere = n × Volume of each cone

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

Hence, the number of cones obtained is 126.

The surface area of a solid metallic sphere is 616 cm² . It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained?

Answer

Surface area of a metallic sphere = 616 cm²

Let the radius of this sphere be R cm.

∴ 4πR² = 616

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

Given,

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

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

Let the number of smaller spheres formed be n.

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

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

Hence, 64 small spheres can be formed.

A copper sphere having a radius of 6 cm is melted and then drawn into a cylindrical wire of radius 2 mm. Calculate the length of the wire.

Answer

Let the length of the wire be h cm

Radius of the sphere, R = 6 cm

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

Given, copper sphere is melted into wire.

∴ Volume of the wire = Volume of the sphere

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

Hence, the length of the wire is 72 m.

A hemisphere of lead of radius 12 cm is melted and cast into a right circular cone of height 54 cm. Find the radius of the base of the cone.

Answer

Radius of hemisphere, r = 12 cm

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

Radius of the cone = R cm

Height of the cone, h = 54 cm

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

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

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

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

A spherical metallic ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 2.5 cm 2 cm respectively. Find the radius of the third ball.

Answer

Radius of larger spherical metallic ball, R = 3 cm

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

Given,

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

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

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