Mensuration — Exercise 17.1

Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of π.

Answer

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.

An electric geyser is cylindrical in shape, having a diameter of 35 cm and height 1.2 m. Neglecting the thickness of its walls, calculate

(i) its outer lateral surface area

(ii) its capacity in litres.

Answer

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

We know,

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

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

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

(ii) Volume of cylinder = πr²h.

Putting values we get,

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

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

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

Hence, the capacity of electric geyser = 115.5 litres.

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk upto a height of 12 cm, find how many litres of milk is needed to serve 1600 students.

Answer

Given diameter = 7 cm, height = 12 cm.

We know,

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

Volume of cylinder = πr²h.

Putting values we get,

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

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

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

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

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

In the given figure, a rectangular tin foil of size 22 cm by 16 cm is wrapped around to form a cylinder of height 16 cm. Find the volume of the cylinder.

Answer

From figure,

The cylinder formed will have,

height = 16 cm.

and circumference of cross section = 22 cm.

We know circumference = 2πr.

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

Volume of cylinder = πr²h.

Putting values we get,

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

Hence, the volume of cylinder = 616 cm³.

How many cubic metres of soil must be take out to make a well 20 metres deep and 2 metres in diameter?

Answer

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) = $\dfrac{\text{Diameter}}{2} = \dfrac{2}{2}$ = 1 m

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

Volume of well = πr²h.

Substituting values we get,

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

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

If the inner curved surface of the well in part (i) above is to be plastered at the rate of ₹ 300 per m², find the cost of plastering (rounded to the nearest hundred rupees).

Answer

By formula,

Curved surface area of cylinder = 2πrh

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

Rate of plastering = ₹ 300/m².

Cost of plastering = Curved surface area of cylinder x Rate

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

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

A roadroller (in the shape of the cylinder) has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must make in order to level a playground of size 120 m by 44 m.

Answer

Given, diameter = 0.7 m

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

Height = width = 1.2 m

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

Curved surface area of cylinder = 2πrh

= $2 \times \dfrac{22}{7} \times 0.35 \times 1.2 = \dfrac{18.48}{7}$ = 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

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

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

If the volume of a cylinder of height 7 cm is 448 π cm³, find its lateral surface area and total surface area.

Answer

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 = $\dfrac{448}{7}$
⇒ 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.

A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m³.

Answer

Given weight = 225 kg/m³

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

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

Volume of cylinder = πr²h.

Putting values in the formula we get,

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

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.

The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the

(i) radius of the cylinder.

(ii) volume of the cylinder.

Answer

(i) Circumference = 2πr.

Given, circumference = 132 cm.

∴ 2πr = 132

⇒ $2 \times \dfrac{22}{7}r$ = 132

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

Hence, the radius of the cylinder = 21 cm.

(ii) Given height = 25 cm.

Volume of cylinder = πr²h.

Putting values we get,

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

Hence, the volume of the cylinder = 34650 cm³.

The area of the curved surface of a cylinder is 4400 cm², and the circumference of its base is 110 cm. Find :

(i) the height of the cylinder.

(ii) the volume of the cylinder.

Answer

(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),

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

⇒ h = 40 cm.

Hence, the height of the cylinder = 40 cm.

(ii) We know that circumference = 2πr.

Given, circumference = 110 cm.

∴ 2πr = 110

⇒ $2 \times \dfrac{22}{7}r$ = 110

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

Volume of cylinder = πr²h.

Putting values we get,

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

Hence, the volume of the cylinder = 38500 cm³.

A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm². Find

(i) the height of the cylinder correct to one decimal place.

(ii) the volume of the cylinder correct to one decimal place. (Take π = 3.14)

Answer

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

Curved surface area of cylinder = 2πrh.

Given, curved surface area of cylinder = 1000 cm².

∴ 2πrh = 1000

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

Hence, the height of the cylinder = 15.9 cm.

(ii) Volume of cylinder = πr²h.

Putting values we get,

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

Hence, the volume of the cylinder = 4992.6 cm³.

The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre ?

Answer correct to the nearest 100 words.

Answer

Height of cylindrical barrel of a pen = 7 cm.

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

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

Volume of cylinder = πr²h.

Putting values we get,

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

Ink in the bottle = One-fifth of a litre = $\dfrac{1}{5}$ 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 = $\dfrac{\text{Volume of bottle}}{\text{Volume of barrel}} \times \text{Total words}$.

= $\dfrac{200}{1.375} \times 310$ = 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.

Find the ratio between the total surface area of a cylinder to its curved surface area given that its height and radius are 7.5 cm and 3.5 cm.

Answer

We know,

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

Curved surface area of cylinder = 2πrh

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

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

The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder ?

Answer

For old cylinder,

Let Height = h and Radius = r.

So, for new cylinder,

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

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

∴ Volume of old cylinder = πr²h

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

Hence,

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

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

The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm². Find the height and the volume of the cylinder.

Answer

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,

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

Since, h + r = 37.

⇒ h + 7 = 37
⇒ h = 37 - 7 = 30 cm.

Volume of cylinder = πr²h.

Putting values we get,

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

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

The total surface area of cylinder is 352 cm². If its height is 10 cm, then find the diameter of the base.

Answer

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,

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

Dividing the complete equation by 44,

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

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.

The ratio between the curved surface and the total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm².

Answer

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.

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

Given, total surface area = 616 cm².

∴ 2πr(h + r) = 616

and h = r.

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

Since h = r, hence h = 7 cm.

Volume of cylinder = πr²h.

Putting values we get,

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

Hence, the volume of cylinder = 1078 cm³.

Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.

Answer

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

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

Let height of the two jars be h₁ and h₂.

Volume of cylinder = πr²h

Given,

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

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

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

$\Rightarrow 9a^2.h_1 = 16a^2.h_2 \\[1em] \Rightarrow \dfrac{h_1}{h_2} = \dfrac{16a^2}{9a^2} \\[1em] \Rightarrow \dfrac{h_1}{h_2} = \dfrac{16}{9}$

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

A rectangular sheet of tin foil of size 30 cm × 18 cm can be rolled to form a cylinder in two ways along length and along breadth. Find the ratio of volumes of the two cylinder thus formed.

Answer

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

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

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

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

Volume of cylinder = πr²h

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

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

A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the volume of the metal.

Answer

Internal diameter = 11.2 cm

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

= $\dfrac{11.2}{2}$ = 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,

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

Hence, volume of metal = 306.24 cm³.

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Answer

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

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

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

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

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

Given height = 14 cm = 140 mm.

Putting values in formula,

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

Volume of graphite = πr²h.

Putting values we get,

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

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

A cylindrical roller made of iron is 2 m long. Its inner diameter is 35 cm and the thickness is 7 cm all round. Find the weight of the roller in kg, if 1 cm³ of iron weights 8 g.

Answer

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

= $\dfrac{35}{2}$ = 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,

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

Since, 1 cm³ of iron weights 8 g.

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

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

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

Hence, the weight of the roller = 1478.4 kg.