Mensuration

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 (around 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.

Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.

Answer

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

Curved surface area of cone = πrl.

Putting values in above equation we get,

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

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

Diameter of the base of a cone is 10.5 cm and slant height is 10 cm. Find its curved surface area.

Answer

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

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

Given, l = 10 cm.

Curved surface area of cone = πrl.

Putting values in above equation we get,

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

= $\dfrac{1155}{7}$ = 165 cm².

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

Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find :

(i) radius of the base

(ii) total surface area of the cone.

Answer

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

We know that,

Curved surface area of cone = πrl.

∴ πrl = 308

⇒ $\dfrac{22}{7}$ x r x 14 = 308

⇒ 22 × r × 2 = 308

⇒ 44r = 308

⇒ r = $\dfrac{308}{44}$ = 7 cm.

Hence, the radius of the base = 7 cm.

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

Putting values in above equation we get,

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

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

Find the volume of the right circular cone with

(i) radius 6 cm and height 7 cm

(ii) radius 3.5 cm and height 12 cm.

Answer

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

Volume of cone = $\dfrac{1}{3}πr^2h$

Putting values in equation we get,

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

Hence, the volume of cone = 264 cm³.

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

Volume of cone = $\dfrac{1}{3}πr^2h$

Putting values in equation we get,

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

Hence, the volume of cone = 154 cm³.

Find the capacity in litres of a conical vessel with

(i) radius 7 cm, slant height 25 cm

(ii) height 12 cm, slant height 13 cm.

Answer

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

We know that,

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

Volume of cone = $\dfrac{1}{3}πr^2h$

Putting values in equation we get,

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

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

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

Hence, the volume of cone = 1.232 litre.

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

We know that,

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

Volume of cone = $\dfrac{1}{3}πr^2h$

Putting values in equation we get,

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

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

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

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

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters ?

Answer

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

Height = 12 m.

Volume of cone = $\dfrac{1}{3}πr^2h$

Putting values in equation we get,

Volume of cone,

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

Since, 1 m³ = 1 kiloliters.

Hence, capacity of pit = 38.5 kiloliters.

If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

Answer

We know,

Volume of cone = $\dfrac{1}{3}πr^2h$.

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

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

Diameter = 2 × Radius = 2 × 4 = 8 cm.

Hence, the diameter of the base = 8 cm.

The height of the cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Answer

We know,

Volume of cone = $\dfrac{1}{3}πr^2h$.

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

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

Hence, the radius of the base = 10 cm.

The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface area at the rate of ₹210 per 100 m².

Answer

Cost = ₹ 210 / 100 m² = ₹ 2.1 / m².

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

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

Curved surface area of cone = πrl.

Putting values in equation,

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

Cost of white washing 1 m² = ₹ 2.1

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

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

A conical tent is 10 m high and the radius of its base is 24 m. Find :

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹ 70.

Answer

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

We know that,

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

Putting values in the formula we get,

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

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

(ii) Curved surface area of cone = πrl.

Putting values in equation,

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

Cost of 1 m² of canvas = ₹ 70

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

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

A Joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the cloth required to make 10 such caps.

Answer

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

We know that,

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

Putting values in the formula we get,

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

Curved surface area of cone = πrl.

Putting values in equation,

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

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

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

The ratio of the base radii of two right circular cones of the same height is 3 : 4. Find the ratio of their volumes.

Answer

Let radius of cones be 3a and 4a.

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

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

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

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

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

The ratio of the heights of two right circular cones is 5 : 2 and that of their base radii is 2 : 5. Find the ratio of their volumes.

Answer

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

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

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

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

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

The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone. Find :

(i) the ratio of their volumes.

(ii) the ratio of their lateral surface areas.

Answer

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

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

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

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

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

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

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

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

Let slant height of smaller cone be l₁.

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

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

Ratio of lateral surface area of two cones

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

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

Find what length of canvas 2 m in width is required to make a conical tent 20 m in diameter and 42 m in slant height allowing 10% for folds and the stitching. Also find the cost of the canvas at the rate of ₹80 per meter.

Answer

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

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

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

Putting values we get,

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

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

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

Area of rectangular cloth = l × b.

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

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

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

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

The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.

Answer

Given, circumference of base = 44 cm.

⇒ 2πr = 44

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

Given, l = 25 cm.

We know that,

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

Volume of cone = $\dfrac{1}{3}πr^2h$

Putting values in equation we get,

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

= $\dfrac{22 × 49 × 24}{3 \times 7}$

= 22 × 7 × 8 = 1232 cm³.

Curved surface area = πrl.

Putting values in equation we get,

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

The volume of a right circular cone is 9856 cm³ and the area of its base is 616 cm². Find

(i) the slant height of the cone.

(ii) total surface area of the cone.

Answer

Given, area of base = 616 cm².

Area of base = πr².

∴ πr² = 616

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

Volume of cone = $\dfrac{1}{3}πr^2h$

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

(i) We know that,

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

Hence, the slant height of cone = 50 cm.

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

Putting values in equation we get,

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

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

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed. (Take π = 3.14)

Answer

Since, triangle is revolved about 8 cm side so,

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

Volume of cone = $\dfrac{1}{3}πr^2h$

Putting values in equation we get,

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

= $\dfrac{904.32}{3}$

= 301.44 cm³.

Curved surface area = πrl.

Putting values in equation we get,

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

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

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be $\dfrac{1}{27}$ 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 30 cm and base radius R cm. Let this cone be cut by the plane CND (parallel to the base plane AMB) to obtain cone OCD with height h cm and base radius r cm as shown in the figure below:

Then △OND ~ △OMB.

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

According to given,

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

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

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

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

Using (i)

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

The height of the cone OCD = 10 cm.

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

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

A semi-circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find :

(i) the radius of the cone.

(ii) the (lateral) surface area of the cone.

Answer

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

= πr = $\dfrac{22}{7}$ x 35 = 110 cm

Let r cm be the radius of the cone, then

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

Hence, the radius of the cone is 17.5 cm.

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

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

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

Find the surface area and volume of a sphere of radius 14 cm.

Answer

Given, r = 14 cm.

Surface area of sphere = 4πr².

Putting values in equation we get,

Surface area of sphere =

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

Volume of sphere = $\dfrac{4}{3}πr^3$

Putting values in equation we get,

Volume of sphere =

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

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

Find the surface area and volume of a sphere of diameter 21 cm.

Answer

Given, diameter = 21 cm.

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

Surface area of sphere = 4πr².

Putting values in equation we get,

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

Volume of sphere = $\dfrac{4}{3}πr^3$

Putting values in equation we get,

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

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

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³, find the mass of the shot-put.

Answer

Volume of sphere = $\dfrac{4}{3}πr^3$

Putting values in equation we get,

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

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

Mass = Volume × Density.

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

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

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

Find the diameter of a sphere whose surface area is 154 cm².

Answer

Surface area of sphere = 4πr².

Given, surface area = 154 cm².

∴ 4πr² = 154

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

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

Hence, the diameter of sphere = 7 cm.

Find

(i) the curved surface area.

(ii) the total surface area of a hemisphere of radius 21 cm.

Answer

(i) Given, radius = 21 cm.

Curved surface area of hemisphere = 2πr².

Putting values in equation we get,

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

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

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

Putting values in equation we get,

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

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

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

A hemispherical brass bowl has inner-diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 cm².

Answer

Given, internal diameter = 10.5 cm.

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

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

Internal curved surface area of hemispherical shell = 2πr².

Putting values in equation we get,

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

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

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

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

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

The radius of a spherical balloon increases from 7 cm to 14 cm as air is pumped into it. Find the ratio of the surface areas of the balloon in two cases.

Answer

Surface area of sphere = 4πr².

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

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

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

A sphere and a cube have the same surface. Show that the ratio of the volume of the sphere to that of the cube is $\sqrt{6} : \sqrt{π}$.

Answer

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

Surface area of sphere = 4πr².

Surface area of cube = 6a².

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

∴ 4πr² = 6a²

⇒ $\dfrac{r^2}{a^2} = \dfrac{6}{4π}$

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

Volume of sphere = $\dfrac{4}{3}πr^3$.

Volume of cube = a³.

Ratio of volume of sphere to volume of cube is

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

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

If the ratio of the radii of two spheres is 3 : 7, find :

(i) the ratio of their volumes.

(ii) the ratio of their surface areas.

Answer

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

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

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

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

(ii) Surface area of sphere = 4πr².

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

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

If the ratio of the volumes of the two spheres is 125 : 64, find the ratio of their surface areas.

Answer

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

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

Surface area of sphere = 4πr².

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

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

Find the volume of a sphere whose surface area is 154 cm².

Answer

We know that Surface area of sphere = 4πr².

Given,
Surface area of sphere = 154 cm².

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

Volume of sphere = $\dfrac{4}{3}πr^3$

Putting values in equation we get,

Volume of sphere =

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

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

If the volume of a sphere is $179\dfrac{2}{3}$ cm³, find its radius and the surface area.

Answer

Volume of sphere = $\dfrac{4}{3}πr^3$.

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

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

Surface area of sphere = 4πr².

Putting values in equation we get,

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

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

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

A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain ?

Answer

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

Putting values we get,

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

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

The surface area of a solid sphere is 1256 cm². It is cut into two hemispheres. Find the total surface area and the volume of a hemisphere. Take π = 3.14

Answer

Given,
surface area of the sphere = 1256 cm².

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

∴ 4πr² = 1256

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

Total surface area of hemisphere = 3πr².

Putting values we get,

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

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

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

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

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

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

The adjoining figure shows a cuboidal block of wood through which a circular cylindrical hole of the biggest size is drilled. Find the volume of the wood left in the block.

Answer

From figure,

Diameter of the biggest hole = 30 cm.

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

Height = 70 cm.

Volume of cuboidal block = l x b x h.

Putting values we get,

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

Volume of cylinder = πr²h.

Putting values we get,

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

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

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

The adjoining figure shows a solid trophy made of shining glass. If one cubic centimeter of glass costs ₹ 0.75, find the cost of the glass for making the trophy.

Answer

From figure,

Edge of cubical part = 28 cm.

Diameter of cylindrical part = 28 cm

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

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

Height of cylinder = 28 cm.

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

Volume of cylinder = πr²h.

Putting values we get,

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

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

Cost of 1 cm³ of glass = ₹0.75

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

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

From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.

Answer

Edge of a cube = 14 cm.

Volume = (side)³ = (14)³ = 2744 cm³.

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

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

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

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

Volume of cone = $\dfrac{1}{3}πr^2h$

Height = 14 cm.

Putting values we get,

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

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

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

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

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

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 remaining wood.

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

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

Height of the cone for maximum volume = 20 cm.

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

Putting values we get,

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