Volume & Surface Area of Solids — Exercise 21(A)

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.