Volume & Surface Area of Solids — Multiple Choice Questions

The weight of a metallic right circular cylinder with base radius 10.5 cm and height 60 cm, it being given that 1 cm³ of metal weighs 5 gm, is :

1. 48.75 kg

2. 97.65 kg

3. 102.45 kg

4. 103.95 kg

Answer

Radius of cylinder, r = 10.5 cm

Height of cylinder, h = 60 cm

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, 1 cm³ of metal weighs 5 gm.

Toatl weight = 5 × 20790 = 103950 g = 103950 × $\dfrac{1}{1000}$ = 103.95 kg.

Hence, option 4 is the correct option.

If the diameter of the base of a closed right circular cylinder be equal to its height h, then its whole surface area is :

1. πh²

2. $\dfrac{3}{2}$ πh²

3. $\dfrac{4}{3}$ πh²

4. 2 πh²

Answer

Let radius of cylinder be r cm.

Given, diameter = height = h cm

Radius = $\dfrac{\text{diameter}}{2} = \dfrac{\text{h}}{2}$

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

$= 2 π \dfrac{\text{h}}{2} \Big(\dfrac{\text{h}}{2} + \text{h}\Big) \\[1em] = π \text{h} \times \Big(\dfrac{\text{h} + 2 \text{h}}{2}\Big) \\[1em] = π \text{h} \times \dfrac{3\text{h}}{2} \\[1em] = \dfrac{3}{2} π \text{h}^2$

Hence, option 2 is the correct option.

The radius of a roller 100 cm long is 14 cm. The curved surface area of the roller is :

1. 13200 cm²

2. 15400 cm²

3. 4400 cm²

4. 8800 cm²

Answer

Length of the roller, h = 100 cm

Radius of the roller, r = 14 cm

Roller is in the shape of a cylinder.

∴ Curved surface area of roller = 2πrh

= $2 \times \dfrac{22}{7} \times 14 \times 100$

= 2 × 22 × 2 × 100

= 8800 cm².

Hence, option 4 is the correct option.

The ratio between the radius of the base and the height of a cylinder is 2 : 3. If its volume is 1617 cm³ the total surface area of the cylinder is :

1. 308 cm²

2. 462 cm²

3. 540 cm²

4. 770 cm²

Answer

Let radius of cylinder, r = 2a and height, h = 3a.

Given,

Volume of cylinder = 1617 cm³

By formula,

Volume of cylinder = πr²h

$\Rightarrow 1617 = \dfrac{22}{7} \times (\text{2a})^2 \times \text{3a} \\[1em] \Rightarrow 1617 \times \dfrac{7}{22} = 4(\text{a})^2 \times \text{3a} \\[1em] \Rightarrow \dfrac{11319}{22} = 12\text{a}^3 \\[1em] \Rightarrow \dfrac{514.5}{12} = \text{a}^3 \\[1em] \Rightarrow \text{a}^3 = 42.875 \\[1em] \Rightarrow \text{a} = \sqrt[3]{42.875} \\[1em] \Rightarrow \text{a} = 3.5$

∴ Radius = 2a = 2 × 3.5 = 7 cm

Height = 3a = 3 × 3.5 = 10.5 cm

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

$= 2 \times \dfrac{22}{7} \times 7 (7 + 10.5) \\[1em] = 2 \times 22 \times 17.5 \\[1em] = 770 \text{cm}^2.$

Hence, option 4 is the correct option.

A solid cylinder has a total surface area of 231 cm². If its curved surface area is two-thirds of the total surface area, the volume of the cylinder is :

1. 269.5 cm³

2. 308 cm³

3. 363.4 cm³

4. 385 cm³

Answer

Given,

Total surface area of cylinder = 231 cm²

⇒ 2πr² + 2πrh = 231 ...(1)

Curved surface area = $\dfrac{2}{3}$ (Total surface area)

= $\dfrac{2}{3} \times 231 = 2 \times 77$ = 154 cm²

By formula,

Curved surface area of cylinder = 2πrh

⇒ 2πrh = 154 ...(2)

Substituting eq.(2) in eq.(1), we have :

⇒ 2πr² + 154 = 231

⇒ 2πr² = 231 - 154

⇒ 2πr² = 77

$\Rightarrow \text{r}^2 = \dfrac{77}{2π} \\[1em] \Rightarrow \text{r}^2 = \dfrac{77}{2 \times \dfrac{22}{7}} \\[1em] \Rightarrow \text{r}^2 = \dfrac{77 \times 7}{2 \times 22} \\[1em] \Rightarrow \text{r}^2 = \dfrac{539}{44} \\[1em] \Rightarrow \text{r}^2 = 12.25 \\[1em] \Rightarrow \text{r} = \sqrt{12.25} \\[1em] \Rightarrow \text{r} = 3.5 \text{ cm.}$

Substituting value of r in eq.(2), we get:

$\Rightarrow 2 \times \dfrac{22}{7} \times 3.5 \times \text{h} = 154 \\[1em] \Rightarrow 2 \times 22 \times 0.5 \times \text{h} = 154 \\[1em] \Rightarrow 22\text{h} = 154 \\[1em] \Rightarrow \text{h} = \dfrac{154}{22} \\[1em] \Rightarrow \text{h} = 7 \text{ cm.}$

Volume of cylinder = πr²h

$= \dfrac{22}{7} \times 3.5^2 \times 7 \\[1em] = 22 \times 12.25 \\[1em] = 269.5 \text{ cm}^3.$

Hence, option 1 is the correct option.

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, its volume is :

1. 3180 m³

2. 4620 m³

3. 5240 m³

4. None of these

Answer

Let radius be r m and height be h m.

Given,

r + h = 37 m ...(1)

Totals surface area of cylinder = 1628 m²

⇒ 2πr(r + h) = 1628

⇒ 2πr × 37 = 1628

$\Rightarrow 74 \times \dfrac{22}{7} \times \text{r} = 1628 \\[1em] \Rightarrow \dfrac{1628}{7} \times \text{r} = 1628 \\[1em] \Rightarrow \text{r} = \dfrac{1628 \times 7}{1628} \\[1em] \Rightarrow \text{r} = 7 \text{ m.}$

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

⇒ r + h = 37

⇒ 7 + h = 37

⇒ h = 37 - 7

⇒ h = 30 m.

Volume of cylinder = πr²h

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

Hence, option 2 is the correct option.

The curved surface area of a cylinder is 4400 cm² and the circumference of its base is 110 cm. The volume of the cylinder (in cm³) is :

1. 36000

2. 38500

3. 40150

4. 42250

Answer

Given, curved surface area of cylinder = 4400 cm²

We know that curved surface area of cylinder = 2πrh

∴ 2πrh = 4400 .....(1)

Given, circumference of base = 110 cm

We know that circumference = 2πr

∴ 2πr = 110 .....(2)

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

Dividing eq.(1) by (2), we get:

$\dfrac{2π\text{rh}}{2π\text{r}} = \dfrac{4400}{110}$

⇒ h = 40 cm

Volume of cylinder = πr²h

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

Hence, option 2 is the correct option.

A rectangular sheet of paper of size 11 cm x 7 cm is first rotated about the side 11 cm and then about the side 7 cm to form a cylinder, as shown in the diagram. The ratio of their curved surface areas is:

1. 1 : 1

2. 7 : 11

3. 11 : 7

4. $\dfrac{11π}{7} : \dfrac{7π}{11}$

Answer

In first case :

Height of cylinder (h) = 7 cm

Let radius be r cm

⇒ 2πr = 11

⇒ r = $\dfrac{11}{2π}$

In second case :

Height of cylinder (H) = 11 cm

Let radius be R cm

⇒ 2πR = 7

⇒ R = $\dfrac{7}{2π}$

$\therefore \dfrac{\text{CSA of 1st cylinder}}{\text{CSA of 2nd cylinder}} = \dfrac{2πrh}{2πRH} \\[1em] = \dfrac{rh}{RH} \\[1em] = \dfrac{\dfrac{11}{2π} \times 7}{\dfrac{7}{2π} \times 11} \\[1em] = \dfrac{\dfrac{77}{2π}}{\dfrac{77}{2π}} \\[1em] = \dfrac{1}{1} \\[1em] = 1 : 1.$

Hence, Option 1 is the correct option.

Two steel sheets each of length a₁ and breadth a₂ are used to prepare the surface of two right circular cylinders - one having volume V₁ and height a₂ and the other having volume V₂ and height a₁. Then :

1. V₁ = V₂

2. a₁ V₁ = a₂ V₂

3. a₂ V₁ = a₁ V₂

4. $\dfrac{\text{V}_1}{\text{a}_2} = \dfrac{\text{V}_2}{\text{a}_1}$

Answer

For cylinder 1,

Height of cylinder, h = a₂

Radius of cylinder be r cm

Circumference of base = 2πr = a₁

$\Rightarrow 2 \times π \times \text{r} = \text{a}_1 \\[1em] \Rightarrow \text{r} = \dfrac{\text{a}_1}{2π}$

Volume of cylinder 1, V₁ = πr²h

$= π \times \Big(\dfrac{\text{a}_1}{2π}\Big)^2 \times \text{a}_2 \\[1em] = π \times \dfrac{\text{a}_1^2}{4π^2} \times \text{a}_2 \\[1em] = \dfrac{\text{a}_1^2 \text{a}_2 }{4π}$

For cylinder 2,

Height of cylinder, H = a₁

Radius of cylinder be R

Circumference of base = 2πR = a₂

$\Rightarrow 2 \times π \times \text{R} = \text{a}_2 \\[1em] \Rightarrow \text{R} = \dfrac{\text{a}_2}{2π} \\[1em] \Rightarrow \text{R} = \dfrac{\text{a}_2}{2π}$

Volume of cylinder 2, V₂ = πR²H

$= π \times \Big(\dfrac{\text{a}_2}{2π}\Big)^2 \times \text{a}_1 \\[1em] = π \times \dfrac{\text{a}_2^2}{4π^2} \times \text{a}_1 \\[1em] = \dfrac{\text{a}_2^2 \text{a}_1}{4π}$

Ratio of the volumes of the two cylinders:

$\Rightarrow \dfrac{\text{Volume of cylinder 1}}{\text{Volume of cylinder 2}} = \dfrac{\dfrac{\text{a}_1^2 \text{a}_2 }{4π}}{\dfrac{\text{a}_2^2 \text{a}_1}{4π}} \\[1em] \Rightarrow \dfrac{\text{V}_1}{\text{V}_2} = \dfrac{\text{a}_1^2 \text{a}_2 }{4π} \times \dfrac{4π}{\text{a}_2^2 \text{a}_1} \\[1em] \Rightarrow \dfrac{\text{V}_1}{\text{V}_2} = \dfrac{\text{a}_1}{\text{a}_2} \\[1em] \Rightarrow \text{V}_1 \text{a}_2 = \text{V}_2 \text{a}_1$

Hence, option 3 is the correct option.

Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. The ratio of their radii is :

1. 1 : $\sqrt{2}$

2. $\sqrt{2}$ : 1

3. 1 : 2

4. 1 : 4

Answer

Let radius and heights of two cylinders be r, h and R, H.

Given,

$\dfrac{\text{h}}{\text{H}} = \dfrac{1}{2}$

Volume of cylinder 1 = v

Volume of cylinder 2 = V

⇒ v = V

$\Rightarrow π\text{r}^2\text{h} = π\text{R}^2\text{H} \\[1em] \text{Divide by π on both sides, we get:} \\[1em] \Rightarrow \text{r}^2 = \text{R}^2 \times \dfrac{\text{H}}{\text{h}} \\[1em] \Rightarrow \dfrac{\text{r}^2}{\text{R}^2} = \dfrac{2}{1} \\[1em] \Rightarrow \Big(\dfrac{\text{r}}{\text{R}}\Big)^2 = \dfrac{2}{1} \\[1em] \Rightarrow \dfrac{\text{r}}{\text{R}} = \sqrt{\dfrac{2}{1}} \\[1em] \Rightarrow \dfrac{\text{r}}{\text{R}} = \dfrac{\sqrt{2}}{1} \\[1em]$

∴ r : R = $\sqrt{2}$ : 1

Hence, option 2 is the correct option.

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. If the total surface area is 616 cm², the volume of the cylinder is :

1. 1232 cm³

2. 1078 cm³

3. 1848 cm³

4. 1548 cm³

Answer

Total surface area = 616 cm²

⇒ 2πr(h + r) = 616

⇒ πr(h + r) = $\dfrac{616}{2}$

⇒ πr(h + r) = 308 ....(1)

Ratio between its curved surface area and total surface area = 1 : 2

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

⇒ 2h = h + r

⇒ 2h - h = r

⇒ h = r

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

⇒ πr(r + r) = 308

⇒ πr × 2r = 308

⇒ 2πr² = 308

$\Rightarrow 2 \times \dfrac{22}{7} \times \text{r}^2 = 308 \\[1em] \Rightarrow \dfrac{44}{7} \times \text{r}^2 = 308 \\[1em] \Rightarrow \text{r}^2 = \dfrac{308 \times 7}{44} \\[1em] \Rightarrow \text{r}^2 = \dfrac{2156}{44} \\[1em] \Rightarrow \text{r}^2 = 49 \\[1em] \Rightarrow \text{r} = \sqrt{49} \\[1em] \Rightarrow \text{r} = 7 \text{ cm.}$

⇒ h = 7 cm

Volume of cylinder = πr²h

= $\dfrac{22}{7}$ × 7² × 7

= 22 × 49

= 1078 cm³.

Hence, option 2 is the correct option.

The number of coins, 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm, is :

1. 380

2. 450

3. 472

4. 540

Answer

Given,

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

Height of coin, h = 0.2 cm

Radius of cylinder, R = $\dfrac{\text{diameter}}{2} = \dfrac{4.5}{2}$ = 2.25 cm

Height of cylinder, H = 10 cm

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

Volume of cylinder = n × Volume of each coin

∴ πR²H = n × πr²h

$\Rightarrow \text{n} = \dfrac{π\text{R}^2\text{H}}{π\text{r}^2\text{h}} \\[1em] \Rightarrow \text{n} = \dfrac{2.25^2 \times 10}{0.75^2 \times 0.2} \\[1em] \Rightarrow \text{n} = \dfrac{5.0625 \times 10}{0.5625 \times 0.2} \\[1em] \Rightarrow \text{n} = \dfrac{50.625}{0.1125} \\[1em] \Rightarrow \text{n} = 450$

Hence, option 2 is the correct option.

A rectangular tin sheet is 12 cm long and 5 cm broad. It is rolled along its length to form a cylinder by making the opposite edges just touch each other. The volume of the cylinder (in cm³) is :

1. $\dfrac{60}{π}$

2. $\dfrac{100}{π}$

3. $\dfrac{120}{π}$

4. $\dfrac{180}{π}$

Answer

For cylinder, rolled along its length:

Height of cylinder, h = 5 cm

Radius of cylinder be r cm

Circumference of base = 2πr = 12

$\Rightarrow 2 \times π \times \text{r} = 12 \\[1em] \Rightarrow \text{r} = \dfrac{12}{2π} \\[1em] \Rightarrow \text{r} = \dfrac{6}{π} \text{ cm.}$

Volume of cylinder = πr²h

$= π \times \Big(\dfrac{6}{π}\Big)^2 \times 5 \\[1em] = π \times \Big(\dfrac{36}{π^2}\Big) \times 5 \\[1em] = \dfrac{180}{π} \text{ cm}^3$

Hence, option 4 is the correct option.

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their curved surface areas is :

1. 2 : 5

2. 8 : 7

3. 10 : 9

4. 16 : 9

Answer

Given,

r : R = 2 : 3

Let r = 2x and R = 3x

h : H = 5 : 3

Let h = 5y and H = 3y

$\dfrac{\text{CSA of 1st cylinder}}{\text{CSA of 2nd cylinder}} = \dfrac{2π\text{rh}}{2π\text{RH}} \\[1em] = \dfrac{\text{rh}}{\text{RH}} \\[1em] = \dfrac{\text{2x} \times \text{5y}}{\text{3x} \times {\text{3y}}} \\[1em] = \dfrac{10\text{xy}}{9\text{xy}} \\[1em] = \dfrac{10}{9} \\[1em] = 10 : 9$

Hence, option 3 is the correct option.

If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the new cylinder to that of the original one?

1. 1 : 2

2. 1 : 4

3. 1 : 8

4. 4 : 1

Answer

For old cylinder,

Let height = h and Radius = r

So, for new cylinder,

Height = h and radius = $\dfrac{\text{r}}{2}$

We know that volume of cylinder = π × radius² × height

∴ Volume of old cylinder = πr²h

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

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

Hence, option 2 is the correct option.

A cylindrical metallic wire is stretched to double its length. Which of the following will NOT change for the wire after stretching?

1. Its curved surface area

2. Its total surface area

3. Its volume

4. Its radius

Answer

On changing the shape of a container, its volume remains same.

Hence, option 3 is the correct option.

Two cylindrical vessels with radii 15 cm and 10 cm and heights 35 cm and 15 cm respectively are filled with water. If this water when poured into a cylindrical vessel, 15 cm in height, fills it completely then the radius of the vessel is :

1. 17.5 cm

2. 18 cm

3. 20 cm

4. 25 cm

Answer

Given,

For cylinder 1,

Radius, r = 15 cm

Height, h = 35 cm

For cylinder 2,

Radius, R = 10 cm

Height, H = 15 cm

By formula, Volume of cylinder = πr²h

Volume of cylinder 1 = v

$= \dfrac{22}{7} \times 15^2 \times 35 \\[1em] = 22 \times 225 \times 5 \\[1em] = 24750 \text{ cm}^3.$

Volume of cylinder 2 = V

$= \dfrac{22}{7} \times 10^2 \times 15 \\[1em] = \dfrac{22}{7} \times 100 \times 15 \\[1em] = \dfrac{33000}{7} \text{ cm}^3.$

Given, water from cylinder 1 and 2 is poured into cylinder 3.

Volume of cylinder 3 = Volume of cylinder 1 + Volume of cylinder 2

= 24750 + $\dfrac{33000}{7}$

= $\dfrac{173250 + 33000}{7} = \dfrac{206250}{7} \text{ cm}^3$

For cylinder 3,

Radius be a cm

Height = 15 cm

Volume of cylinder 3 = πa² × 15

$\Rightarrow \dfrac{206250}{7} = \dfrac{22}{7} \times \text{a}^2 \times 15 \\[1em] \Rightarrow \dfrac{206250}{7} \times \dfrac{7}{22 \times 15} = \text{a}^2 \\[1em] \Rightarrow \dfrac{206250}{330} = \text{a}^2 \\[1em] \Rightarrow \text{a}^2 = 625 \\[1em] \Rightarrow \text{a} = \sqrt{625} \\[1em] \Rightarrow \text{a} = 25 \text{ cm.}$

Hence, option 4 is the correct option.

A hollow garden roller 63 cm wide with a girth of 440 cm is made of iron 4 cm thick. The volume of the iron used is :

1. 154982 cm³

2. 106372 cm³

3. 107812 cm³

4. 107712 cm³

Answer

Length of the roller (h) = 63 cm

Let external radius be R cm and internal radius be r cm.

Girth of the roller = Circumference of roller = 440 cm

⇒ 2πR = 440

$\Rightarrow 2 \times \dfrac{22}{7} \text{R} = 440 \\[1em] \Rightarrow \dfrac{44}{7} \text{R} = 440 \\[1em] \Rightarrow \text{R} = 440 \times \dfrac{7}{44} \\[1em] \Rightarrow \text{R} = \dfrac{3080}{44} \\[1em] \Rightarrow \text{R} = 70 \text{ cm.}$

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

⇒ 4 = 70 - r

⇒ r = 70 - 4 = 66 cm

External volume = πR²h

$= \dfrac{22}{7} \times (70)^2 \times 63 \\[1em] = 22 \times 4900 \times 9 \\[1em] = 970200$

Internal volume = πr²h

$= \dfrac{22}{7} \times (66)^2 \times 63 \\[1em] = 22 \times 4356 \times 9 \\[1em] = 862488$

Volume of iron = External volume - Internal volume

= 970200 - 862488

= 107712 cm³

Hence, option 4 is the correct option.

If the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base, then its volume is :

1. $\dfrac{1}{3}$ πr³

2. $\dfrac{2}{3}$ πr³

3. 3πr³

4. 9πr³

Answer

Given, radius = 3r and height(h) = 3r

Volume of cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times π \times (3\text{r})^2 \times 3\text{r} \\[1em] = π \times 9\text{r}^2 \times \text{r} \\[1em] = 9 π\text{r}^3$

Hence, option 4 is the correct option.

A right circular cone has the radius of the base equal to the height of the cone. If the volume of the cone is 9702 cu. cm, then the diameter of the base of the cone is :

1. 21 cm

2. $2\sqrt{7}$ cm

3. 42 cm

4. $21\sqrt{7}$ cm

$\Big[\text{Use } \pi = \dfrac{22}{7}\Big]$

Answer

Given,

Height of cone (h) = Radius of cone (r) = a cm (let)

Given,

Volume = 9702 cm³

$\therefore \dfrac{1}{3}πr^2h = 9702 \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times a^2 \times a = 9702 \\[1em] \Rightarrow a^3 = \dfrac{9702 \times 7 \times 3}{22} \\[1em] \Rightarrow a^3 = 441 \times 21 \\[1em] \Rightarrow a^3 = 9261 \\[1em] \Rightarrow a = \sqrt[3]{9261} = 21 \text{ cm}.$

Diameter = 2 × radius = 2 × 21 = 42 cm.

Hence, option 3 is the correct option.

The ratio of diameters of two right circular cones is 3 : 7 and that of their heights is 14 : 9, then their volumes are in ratio:

1. 3 : 7

2. 2 : 7

3. 3 : 2

4. 9 : 49

Answer

Let the diameters of the two right circular cones be d₁ and d₂, their heights be h₁ and h₂ and their radius be r₁ and r₂.

Given,

⇒ d₁ : d₂ = 3 : 7

Let d₁ and d₂ be 3x and 7x respectively.

⇒ r₁ = $\dfrac{3x}{2}$

⇒ r₂ = $\dfrac{7x}{2}$

Given,

⇒ h₁ : h₂ = 14 : 9

Let h₁ and h₂ be 14a and 9a respectively.

⇒ h₁ : h₂ = 14a : 9a

By formula,

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

$\Rightarrow V_1 : V_2 = \dfrac{1}{3}\pi r_1^2h_1:\dfrac{1}{3}\pi r_2^2h_2 \\[1em] = \dfrac{\dfrac{1}{3}\pi r_1^2h_1}{\dfrac{1}{3} \pi r_2^2h_2} \\[1em] = \dfrac{r_1^2h_1}{r_2^2h_2} \\[1em] = \dfrac{\Big(\dfrac{3x}{2}\Big)^2 \times 14a}{\Big(\dfrac{7x}{2}\Big)^2 \times 9a} \\[1em] = \dfrac{\Big(\dfrac{9x^2}{4}\Big) \times 14}{\Big(\dfrac{49x^2}{4}\Big) \times 9} \\[1em] = \dfrac{9x^2 \times 14 \times 4}{49x^2 \times 9 \times 4} \\[1em] = \dfrac{9 \times 14}{49 \times 9} \\[1em] = \dfrac{14}{49} \\[1em] = \dfrac{2}{7} = 2 : 7.$

Hence, option 2 is the correct option.

If the volumes of two cones are in the ratio of 1 : 4 and their diameters are in the ratio 4 : 5, then the ratio of their heights is :

1. 1 : 5

2. 5 : 4

3. 5 : 16

4. 25 : 64

Answer

Let Volume of cones be V and v respectively.

⇒ V : v = 1 : 4

Diameters of cones be D and d respectively.

⇒ D = 4b and d = 5b

Radius of 1st cone = $\dfrac{\text{diameter}}{2} = \dfrac{4\text{b}}{2}$ = 2b

Radius of 2nd cone = $\dfrac{\text{diameter}}{2} = \dfrac{5\text{b}}{2}$ = 2.5b

Let the height of cones be h and H respectively.

By formula,

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

$\therefore \dfrac{\text{V}}{\text{v}} = \dfrac{\dfrac{1}{3}π (\text{2b})^2 \text{h}}{\dfrac{1}{3}π (\text{2.5b})^2 \text{H}} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{\dfrac{1}{3}π \times \text{4b}^2 \text{h}}{\dfrac{1}{3}π \times 6.25\text{b}^2 \text{H}} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{4 \text{h}}{6.25\text{H}} \\[1em] \Rightarrow \dfrac{\text{H}}{\text{h}} = \dfrac{4 \times 4}{6.25} \\[1em] \Rightarrow \dfrac{\text{H}}{\text{h}} = \dfrac{16}{6.25} \\[1em] \Rightarrow \dfrac{\text{H}}{\text{h}} = \dfrac{16}{6.25} \\[1em]$

Multipy and divide by 100 on R.H.S

$\Rightarrow \dfrac{\text{H}}{\text{h}} = \dfrac{16 \times 100}{6.25 \times 100} \\[1em] \Rightarrow \dfrac{\text{H}}{\text{h}} = \dfrac{1600}{625} \\[1em] \Rightarrow \dfrac{\text{H}}{\text{h}} = \dfrac{64}{25}$

∴ h : H = 25 : 64

Hence, option 4 is the correct option.

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

(Take π = 3.14)

1. 14 cm

2. 16 cm

3. 24 cm

4. 26 cm

Answer

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

Let r = 5x and h = 12x

Volume of cone = $\dfrac{1}{3}$ πr²h

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

⇒ r = 5x = 5 × 2 = 10 cm

⇒ h = 12x = 12 × 2 = 24 cm

Curved surface area = πrl

l² = r² + h²

⇒ l² = 10² + 24²

⇒ l² = 100 + 576

⇒ l² = 676

⇒ l = $\sqrt{676}$ = 26 cm

Hence, option 4 is the correct option.

How many metres of cloth 2.5 m wide will be required to make a conical tent whose base radius is 7 m and height is 24 m?

1. 120 m

2. 180 m

3. 220 m

4. 550 m

Answer

Given, r = 7 m and h = 24 m

l² = r² + h²

⇒ l² = 7² + 24²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$ = 25 m

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

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

Width of the cloth used = 2.5 m

Length of canvas = $\dfrac{\text{area of canvas}}{\text{width of canvas}} = \dfrac{550}{2.5}$ = 220 m.

Hence, option 3 is the correct option.

The length of the canvas, 1.1 m wide required to build a conical tent of height 14 m and floor area 346.5 m², is :

1. 490 m

2. 525 m

3. 665 m

4. 860 m

Answer

Given,

Height of cone, h = 14 m

Area of base floor = 346.5 m²

$\Rightarrow π\text{r}^2 = 346.5 \\[1em] \Rightarrow \dfrac{22}{7} \times \text{r}^2 = 346.5 \\[1em] \Rightarrow \text{r}^2 = \dfrac{346.5 \times 7}{22} \\[1em] \Rightarrow \text{r}^2 = \dfrac{2425.5}{22} \\[1em] \Rightarrow \text{r}^2 = 110.25 \\[1em] \Rightarrow \text{r} = \sqrt{110.25} \\[1em] \Rightarrow \text{r} = 10.5 \text{ m.}$

By formula,

l² = r² + h²

⇒ l² = 10.5² + 14²

⇒ l² = 110.25 + 196

⇒ l² = 306.25

⇒ l = $\sqrt{306.25}$ = 17.5 m

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

$= \dfrac{22}{7} \times 10.5 \times 17.5 \\[1em] = \dfrac{4042.5}{7} \\[1em] = 577.5 \text{ m}^2$

Let length of canvas be a m.

Area of canvas = Curved surface area of cone

⇒ a × b = 577.5

⇒ a × 1.1 = 577.5

⇒ a = $\dfrac{577.5}{1.1}$

⇒ a = 525 m

Hence, option 2 is the correct option.

The volume of conical tent is 462 m³ and the area of the base is 154 m². The height of the cone is :

1. 15 m

2. 12 m

3. 9 m

4. 24 m

Answer

Given, Area of base = 154 m²

⇒ πr² = 154

$\Rightarrow \dfrac{22}{7} \text{r}^2 = 154 \\[1em] \Rightarrow \text{r}^2 = 154 \times \dfrac{7}{22} \\[1em] \Rightarrow \text{r}^2 = \dfrac{1078}{22} \\[1em] \Rightarrow \text{r}^2 = 49 \\[1em] \Rightarrow \text{r} = \sqrt{49} \\[1em] \Rightarrow \text{r} = 7 \text{ m.}$

Volume of cone = 462 m³

By formula,

Volume of cone = $\dfrac{1}{3}$ πr²h

$\Rightarrow 462 = \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times \text{h} \\[1em] \Rightarrow 462 = \dfrac{22}{21} \times 49 \times \text{h} \\[1em] \Rightarrow \text{h} = \dfrac{462 \times 21}{22 \times 49} \\[1em] \Rightarrow \text{h} = \dfrac{9702}{1078} \\[1em] \Rightarrow \text{h} = 9 \text{ m}.$

Hence, option 3 is the correct option.

A conical tent is to accomodate 11 persons such that each person occupies 4 m² space on the ground and has 20 m³ of air to breathe. The height of the cone is :

1. 14 m

2. 15 m

3. 16 m

4. 20 m

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, option 2 is the correct option.

The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4 the ratio of their curved surface areas is:

1. 4 : 5

2. 5 : 4

3. 16 : 25

4. 25 : 16

Answer

Given, ratio of slant height = 5 : 4

Let slant height of 1st cone be 5a and 2nd cone be 4a cm.

For 1st cone,

⇒ Diameter = d

⇒ Radius = r

⇒ Slant height, l = 5a

For 2nd cone,

⇒ Diameter = D

⇒ Radius = R

⇒ Slant height, L = 4a

Given,

⇒ d = D

∴ r = R

$\Rightarrow \dfrac{\text{CSA of 1st cone}}{\text{CSA of 2nd cone}} = \dfrac{π\text{rl}}{π\text{RL}} \\[1em] = \dfrac{\text{l}}{\text{L}} \\[1em] = \dfrac{5\text{a}}{4\text{a}} \\[1em] = \dfrac{5}{4} \\[1em] = 5 : 4$

Hence, option 2 is the correct option.

If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the new cone to that of the original cone?

1. 1 : 2

2. 1 : 3

3. 1 : 4

4. 2 : 3

Answer

For older cone,

⇒ Radius = r

⇒ Height = h

⇒ Volume = v

For new cone,

⇒ Radius = R = $\dfrac{\text{r}}{2}$

⇒ Height = h

⇒ Volume = V

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

$\Rightarrow \dfrac{\text{V}}{\text{v}} = \dfrac{\dfrac{1}{3}π \text{R}^2 \text{h}}{\dfrac{1}{3}π \text{r}^2 \text{h}} \\[1em] = \dfrac{\Big(\dfrac{\text{r}}{2}\Big)^2}{\text{r}^2} \\[1em] = \dfrac{\dfrac{\text{r}^2}{4}}{\text{r}^2} \\[1em] = \dfrac{\text{r}^2}{4\text{r}^2} \\[1em] = \dfrac{1}{4}.$

Hence, option 3 is the correct option.

A cone of height 7 cm and base radius 3 cm is carved from a rectangular block of wood 10 cm × 5 cm × 2 cm. The percentage of wood wasted is:

1. 34%

2. 46%

3. 54%

4. 66%

Answer

Volume of rectangular block = 10 cm × 5 cm × 2 cm = 100 cm³

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

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 3^2 \times 7 \\[1em] = \dfrac{1}{3} \times 22 \times 9 \\[1em] = 22 \times 3 \\[1em] = 66 \text{ cm}^3.$

Volume of wood wasted = Volume of rectangular block - Volume of cone

= 100 - 66

= 34 cm³

Percentage of wood wasted

= $\dfrac{\text{Volume of wood wasted}}{\text{Total volume of rectangular block}} \times 100 = \dfrac{34}{100} \times 100$ = 34%

Hence, option 1 is the correct option.

An edge of a cube measures 10 cm. If the largest possible right circular cone is cut out of this cube, then the volume of the cone is:

1. 260 cm³

2. 260.9 cm³

3. 261.9 cm³

4. 262.7 cm³

Answer

Edge of a cube = 10 cm

Volume = side³ = 10³ = 1000 cm³.

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

Diameter of the cone cut out from it = 10 cm

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

Height, h = 10 cm

Volume of cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 5^2 \times 10 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 25 \times 10 \\[1em] = \dfrac{5500}{21} \\[1em] = 261.9 \text{ cm}^3.$

Hence, option 3 is the correct option.

If the height of a cone is doubled, then its volume is increased by :

1. 100%

2. 200%

3. 300%

4. 400%

Answer

For older cone,

⇒ Radius = r

⇒ Height = h

⇒ Volume = v

For new cone,

⇒ Radius = r

⇒ Height = 2h

⇒ Volume = V

Volume of cone = $\dfrac{1}{3}π \times \text{(radius)}^2 \times \text{height}$

Increase in the volume of cone = $\dfrac{\text{Volume of new cone - Volume of old cone}}{\text{Volume of old cone}} \times 100$

$= \dfrac{\text{V - v}}{\text{v}} \times 100 \\[1em] = \dfrac{(\dfrac{1}{3}π \text{r}^2 \times \text{2h}) - (\dfrac{1}{3}π \text{r}^2 \text{h})}{\dfrac{1}{3}π \text{r}^2 \text{h}} \times 100 \\[1em] = \dfrac{\dfrac{1}{3}π \text{r}^2 \text{h}(2 - 1)}{\dfrac{1}{3}π \text{r}^2 \text{h}} \times 100 \\[1em] = 100$

Hence, option 1 is the correct option.

If the height of two cones are in the ratio of 1 : 4 and the radii of their bases are in the ratio 4 : 1, then the ratio of their volumes is:

1. 1 : 2

2. 2 : 3

3. 3 : 4

4. 4 : 1

Answer

Let height of cones be 1a and 4a and radius of the cones be 4b and 1b.

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

Volume of 1st cone, V = $\dfrac{1}{3}π (\text{4b})^2 \text{1a} = \dfrac{1}{3}π \times 16\text{b}^2 \times \text{a}$

Volume of 2nd cone, v = $\dfrac{1}{3}π (\text{1b})^2 \text{4a} = \dfrac{1}{3}π \times \text{b}^2 \times 4\text{a}$

$\Rightarrow \dfrac{\text{V}}{\text{v}} = \dfrac{\dfrac{1}{3}π \times 16\text{b}^2 \times \text{a}}{\dfrac{1}{3}π \times \text{b}^2 \times 4\text{a}} \\[1em] = \dfrac{16}{4} \\[1em] = \dfrac{4}{1}.$

= 4 : 1

Hence, option 4 is the correct option.

The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. Then their volumes are in the ratio:

1. 3 : 4

2. 4 : 3

3. 8 : 9

4. 9 : 8

Answer

For cylinder,

Radius = 3a

Height = 2b

Volume of cylinder, v = πr²h = π × (3a)² × 2b = π × 9a² × 2b = 18πa²b

For cone,

Radius = 4a

Height = 3b

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

$= \dfrac{1}{3}π (\text{4a})^2 \text{3b} \\[1em] = π \times 16\text{a}^2 \text{b} \\[1em]$

$\therefore \dfrac{\text{v}}{\text{V}} = \dfrac{π \times 18\text{a}^2 \text{b}}{π \times 16\text{a}^2 \text{b}} \\[1em] = \dfrac{18}{16} \\[1em] = \dfrac{9}{8}$

= 9 : 8

Hence, option 4 is the correct option.

A right cylindrical vessel is full with water. How many cones having the same diameter and height as those of the right cylinder will be needed to store that water?

1. 2

2. 3

3. 4

4. 5

Answer

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

Volume of cylinder = πr²h

Let the number of cones required be n.

∴ Volume of cylinder = n × Volume of cone

⇒ πr²h = n × $\dfrac{1}{3}π \text{r}^2 \text{h}$

⇒ 3 × πr²h = n × πr²h

∴ n = 3

Hence, option 2 is the correct option.

A conical vessel whose internal radius is 10 cm and height 48 cm is full of water. If this water is poured into a cylindrical vessel with internal radius 20 cm, the height to which water rises in it is:

(Take π = 3.14)

1. 3 cm

2. 4 cm

3. 5 cm

4. 6 cm

Answer

Given, radius of cone, R = 10 cm

Height of cone, H = 48 cm

Height of water in cylinder be h cm

Radius of cylinder, r = 20 cm

Since, water from conical vessel is poured into cylindrical vessel.

∴ Volume of cone = Volume of water in cylinder

$\Rightarrow \dfrac{1}{3}π \text{R}^2 \text{H} = π \text{r}^2 \text{h} \\[1em] \Rightarrow \dfrac{1}{3} \times 10^2 \times 48 = 20^2 \times \text{h} \\[1em] \Rightarrow 100 \times 16 = 400 \times \text{h} \\[1em] \Rightarrow \text{h} = \dfrac{100 \times 16}{400} \\[1em] \Rightarrow \text{h} = \dfrac{1600}{400} \\[1em] \Rightarrow \text{h} = 4 \text{ cm.}$

Hence, option 2 is the correct option.

A cylindrical vessel 32 cm high and 18 cm as the radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, the radius of its base is :

1. 12 cm

2. 24 cm

3. 36 cm

4. 48 cm

Answer

Given, radius of conical heap be R cm

Height of cone, H = 24 cm

Height of cylinder, h = 32 cm

Radius of cylinder, r = 18 cm

Since, sand from cylindrical vessel is poured to form conical heap.

∴ Volume of sand in cone = Volume of cylinder

$\Rightarrow \dfrac{1}{3}π \text{R}^2 \text{H} = π \text{r}^2 \text{h} \\[1em] \Rightarrow \dfrac{1}{3} \text{R}^2 \text{H} = \text{r}^2 \text{h} \\[1em] \Rightarrow \dfrac{1}{3} \times \text{R}^2 \times 24 = 18^2 \times 32 \\[1em] \Rightarrow \text{R}^2 \times 8 = 324 \times 32 \\[1em] \Rightarrow \text{R}^2 \times 8 = 10368 \\[1em] \Rightarrow \text{R}^2 = \dfrac{10368}{8} \\[1em] \Rightarrow \text{R}^2 = 1296 \\[1em] \Rightarrow \text{R} = \sqrt{1296} \\[1em] \Rightarrow \text{R} = 36 \text{ cm.}$

Hence, option 3 is the correct option.

The volume of a sphere is 38808 cu. cm. The curved surface area of the sphere (in cm²) is:

1. 1386

2. 4158

3. 5544

4. 8316

Answer

Given,

Volume of sphere = 38808 cm³

Let the radius of the sphere be r cm.

By formula,

Volume of sphere = $\dfrac{4}{3}$ πr³

$\Rightarrow \dfrac{4}{3} \times \dfrac{22}{7} \times \text{r}^3 = 38808 \\[1em] \Rightarrow \text{r}^3 = \dfrac{38808 \times 7 \times 3}{22 \times 4} \\[1em] \Rightarrow \text{r}^3 = \dfrac{814968}{88} \\[1em] \Rightarrow \text{r}^3 = 9261 \\[1em] \Rightarrow \text{r} = \sqrt[3]{9261} \\[1em] \Rightarrow \text{r} = 21 \text{ cm}.$

Curved surface area of sphere = 4πr²

= 4 × $\dfrac{22}{7}$ × 21 × 21

= 4 × 22 × 3 × 21

= 5544 cm².

Hence, option 3 is the correct option.

If the ratio of volumes of two spheres is 1 : 8, then the ratio of their surface areas is :

1. 1 : 2

2. 1 : 4

3. 1 : 8

4. 1 : 16

Answer

Let radius of two spheres be r and R.

Given,

Ratio of volume of the two spheres is 1 : 8.

Volume of sphere = $\dfrac{4}{3}$ π.(radius)³

$\therefore \dfrac{\text{Volume of Sphere 1}}{\text{Volume of Sphere 2}} = \dfrac{1}{8} \\[1em] \Rightarrow \dfrac{\dfrac{4}{3} \times π \times \text{r}^3}{\dfrac{4}{3} \times π \times \text{R}^3} = \dfrac{1}{8} \\[1em] \Rightarrow \dfrac{\text{r}^3}{\text{R}^3} = \dfrac{1}{8} \\[1em] \Rightarrow \Big(\dfrac{\text{r}}{\text{R}}\Big)^3 = \dfrac{1}{8} \\[1em] \Rightarrow \dfrac{\text{r}}{\text{R}} = \sqrt[3]{\Big(\dfrac{1}{8}\Big)} \\[1em] \Rightarrow \dfrac{\text{r}}{\text{R}} = \dfrac{1}{2}.$

Surface area of sphere = 4π.(radius)²

$\therefore \dfrac{\text{Surface area of Sphere 1}}{\text{Surface area of Sphere 2}} = \dfrac{1}{8} \\[1em] = \dfrac{4 \times π \times \text{r}^2}{4 \times π \times \text{R}^2} \\[1em] = \Big(\dfrac{\text{r}}{\text{R}}\Big)^2 \\[1em] = \Big(\dfrac{1}{2}\Big)^2 \\[1em] = \dfrac{1}{4} \\[1em] = 1 : 4.$

Hence, option 2 is the correct option.

The surface area of a sphere is 154 cm². The volume of the sphere is :

1. 359 $\dfrac{1}{3}$ cm³

2. 179 $\dfrac{2}{3}$ cm³

3. 736 $\dfrac{1}{3}$ cm³

4. 1437 $\dfrac{1}{3}$ cm³

Answer

Let radius of sphere be r cm.

Given, surface area of a sphere = 154 cm²

⇒ 4πr² = 154

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

Volume of sphere = $\dfrac{4}{3}$ πr³

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

Hence, option 2 is the correct option.

Volume of a cylinder of height 3 cm is 48π cm³. Radius of the cylinder is :

1. 48 cm

2. 16 cm

3. 4 cm

4. 24 cm

Answer

Given, height of cylinder, h = 3 cm

Radius of cylinder be r cm

Volume of cylinder = 48π cm³

By formula,

Volume of cylinder = πr²h

⇒ πr²h = 48π

⇒ r² × 3 = 48

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

Hence, option 3 is the correct option.

Three solid spherical beads of radii 3 cm, 4 cm and 5 cm are melted and recast into a single spherical bead. Its radius is :

1. 6 cm

2. 7 cm

3. 8 cm

4. 9 cm

Answer

Given, three spherical beads of radii 3 cm, 4 cm and 5 cm.

Volume of sphere = $\dfrac{4}{3}$ πr³

Volume of single spherical bead = Volume of bead of radius 3 cm + Volume of bead of radius 4 cm + Volume of bead of radius 5 cm

$\Rightarrow \dfrac{4}{3}π \times \text{r}^3 = \dfrac{4}{3}π \times 3^3 + \dfrac{4}{3}π \times 4^3 + \dfrac{4}{3}π \times 5^3 \\[1em] \Rightarrow \text{r}^3 = (3^3 + 4^3 + 5^3) \\[1em] \Rightarrow \text{r}^3 = 27 + 64 + 125 \\[1em] \Rightarrow \text{r}^3 = 216 \\[1em] \Rightarrow \text{r} = \sqrt[3]{216} \\[1em] \Rightarrow \text{r} = 6 \text{ cm.}$

Hence, option 1 is the correct option.

If a solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then the surface area of each ball (in cm²) is :

1. 50 π

2. 60 π

3. 75 π

4. 100 π

Answer

Let radius of small spherical balls be r cm.

Radius of sphere, R = 10 cm

Volume of solid sphere = 8 × Volume of small spherical balls

$\Rightarrow \dfrac{4}{3}π \times \text{R}^3 = 8 \times \dfrac{4}{3}π \times \text{r}^3 \\[1em] \Rightarrow 10^3 = 8 \times \text{r}^3 \\[1em] \Rightarrow 1000 = 8 \times \text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \dfrac{1000}{8} \\[1em] \Rightarrow \text{r} = \sqrt[3]{\dfrac{1000}{8}} \\[1em] \Rightarrow \text{r} = \dfrac{10}{2} \\[1em] \Rightarrow \text{r} = 5 \text{ cm.}$

Surface area of small ball = 4 πr²

= 4 × π × 5 × 5

= 100 π cm².

Hence, option 4 is the correct option.

The number of solid spheres, each of diameter 6 cm that can be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm, is :

1. 3

2. 4

3. 5

4. 6

Answer

Given,

Diameter of sphere = 6 cm

Radius of sphere, r = $\dfrac{\text{diameter}}{2} = \dfrac{6}{2}$ = 3 cm

Height of cylinder, h = 45 cm

Diameter of cylinder = 4 cm

Radius of cylinder, R = $\dfrac{\text{diameter}}{2} = \dfrac{4}{2}$ = 2 cm

Number of solid spheres required be n.

Since, the number of solid spheres, are moulded to form a solid metal cylinder.

∴ Volume of cylinder = n × Volume of solid sphere

$\Rightarrow π\text{R}^2\text{h} = \text{n} \times \dfrac{4}{3}π \times \text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \text{n} \times \dfrac{4}{3} \times \text{r}^3 \\[1em] \Rightarrow 2^2 \times 45 = \text{n} \times \dfrac{4}{3} \times 3^3 \\[1em] \Rightarrow 4 \times 45 = \text{n} \times \dfrac{4}{3} \times 27 \\[1em] \Rightarrow 180 = \text{n} \times 4 \times 9 \\[1em] \Rightarrow 180 = \text{n} \times 36 \\[1em] \Rightarrow \text{n} = \dfrac{180}{36} \\[1em] \Rightarrow \text{n} = 5.$

Hence, option 3 is the correct option.

If the height and diameter of a right circular cylinder are 32 cm and 6 cm respectively, then the radius of the sphere whose volume is equal to the volume of the cylinder is :

1. 3 cm

2. 4 cm

3. 4.5 cm

4. 6 cm

Answer

Given,

Height of cylinder, h = 32 cm

Diameter of cylinder = 6 cm

Radius of cylinder, R = $\dfrac{\text{diameter}}{2} = \dfrac{6}{2}$ = 3 cm

Let radius of sphere be r cm.

Since, volume of sphere is equal to the volume of the cylinder.

∴ Volume of cylinder = Volume of solid sphere

$\Rightarrow π\text{R}^2\text{h} = \dfrac{4}{3}π \times \text{r}^3 \\[1em] \Rightarrow \text{R}^2\text{h} = \dfrac{4}{3} \times \text{r}^3 \\[1em] \Rightarrow 3^2 \times 32 = \dfrac{4}{3} \times \text{r}^3 \\[1em] \Rightarrow 9 \times 32 \times 3 = 4 \times \text{r}^3 \\[1em] \Rightarrow 864 = 4 \times \text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \dfrac{864}{4} \\[1em] \Rightarrow \text{r}^3 = 216 \\[1em] \Rightarrow \text{r} = \sqrt[3]{216} \\[1em] \Rightarrow \text{r} = 6 \text{ cm.}$

Hence, option 4 is the correct option.

If the volume of a sphere is twice that of the other, then the ratio of their radii is :

1. 2 : 1

2. 4 : 1

3. $\sqrt{2}$ : 1

4. $\sqrt[3]{2}$ : 1

Answer

Let the radius of sphere 1 be r cm and radius of sphere 2 be R cm.

Volume of sphere = $\dfrac{4}{3}π \times \text{r}^3$

Given,

Volume of sphere 1 = 2 × Volume of sphere 2

$\Rightarrow \dfrac{4}{3}π \times \text{r}^3 = 2 \times \dfrac{4}{3}π \times \text{R}^3 \\[1em] \Rightarrow \text{r}^3 = 2 \times \text{R}^3 \\[1em] \Rightarrow \dfrac{\text{r}^3}{\text{R}^3} = \dfrac{2}{1} \\[1em] \Rightarrow \Big(\dfrac{\text{r}}{\text{R}}\Big)^3 = \dfrac{2}{1} \\[1em] \Rightarrow \dfrac{\text{r}}{\text{R}} = \sqrt[3]{\dfrac{2}{1}} \\[1em] \Rightarrow \dfrac{\text{r}}{\text{R}} = \dfrac{\sqrt[3]{2}}{1}$

Hence, option 4 is the correct option.

If the volume of two spheres is in the ratio 27: 64, then the ratio of their radii is:

1. 3 : 4

2. 4 : 3

3. 9 : 16

4. 16 : 9

Answer

Given, the volume of two spheres = 27: 64

By formula, volume of a sphere = $\dfrac{4πr^3}{3}$, where r is the radius of the sphere.

Let r and R be the radii of the two spheres.

$\Rightarrow \dfrac{\dfrac{4πr^3}{3}}{\dfrac{4πR^3}{3}} = \dfrac{27}{64} \\[1em] \Rightarrow \dfrac{r^3}{R^3} = \dfrac{27}{64} \\[1em] \Rightarrow \dfrac{r}{R} = \dfrac{\sqrt[3]{27}}{\sqrt[3]{64}} \\[1em] \Rightarrow \dfrac{r}{R} = \dfrac{3}{4} \\[1em] \Rightarrow r : R = 3 : 4.$

Hence, option 1 is the correct option.

If a sphere just fits in a right circular cylinder, then the ratio of the volume of sphere to the volume of the cylinder is :

1. 1 : 3

2. 1 : 2

3. 2 : 3

4. 1 : 4

Answer

Let r be the radius of the sphere

The radius of cylinder is also r cm and height of the cylinder is 2r cm

By formula,

Volume of cylinder = πr²h

= πr²(2r)

= 2πr³

By formula,

Volume of sphere = $\dfrac{4}{3}π \times \text{r}^3$

Ratio of the volume of sphere to the volume of the cylinder:

$= \dfrac{\dfrac{4}{3}π \times \text{r}^3}{2π \times \text{r}^3} \\[1em] = \dfrac{4}{3} \times \dfrac{1}{2} \\[1em] = \dfrac{4}{6} \\[1em] = \dfrac{2}{3}$

Hence, option 3 is the correct option.

A sphere of radius 6 cm is dropped into a cylindrical vessel, partly filled with water. The radius of the vessel is 8 cm. If the sphere is submerged completely, then the surface of the water rises by :

1. 2 cm

2. 3 cm

3. 4 cm

4. 4.5 cm

Answer

Radius of sphere, r = 6 cm

Radius of cylinder, R = 8 cm

Let the rise in water level be x cm.

∴ Volume of water that rises by x cm in the cylindrical vessel = Volume of sphere submerged

$\Rightarrow π\text{R}^2\text{x} = \dfrac{4}{3} π\text{r}^3 \\[1em] \Rightarrow 8^2\text{x} = \dfrac{4}{3} \times 6^3 \\[1em] \Rightarrow 64\text{x} = \dfrac{4}{3} \times 216 \\[1em] \Rightarrow 64\text{x} = 4 \times 72 \\[1em] \Rightarrow 64\text{x} = 288 \\[1em] \Rightarrow \text{x} = \dfrac{288}{64} \\[1em] \Rightarrow \text{x} = 4.5 \text{ cm.}$

Hence, option 4 is the correct option.

If a cylindrical rod of iron whose length is 12 times its radius is melted and cast into spherical balls of the same radius, then the number of balls will be :

1. 3

2. 6

3. 9

4. 27

Answer

Let radius of the cylinder be r cm.

Length of cylindrical rod, h = 12r

Radius of spherical balls be r cm

Number of spherical balls required be n.

Since, a cylindrical rod of iron is melted and cast into spherical balls of the same radius.

∴ Volume of cylindrical rod = n × Volume of spherical ball

$\Rightarrow π\text{r}^2\text{h} = \text{n} \times \dfrac{4}{3} π\text{r}^3 \\[1em] \Rightarrow \text{r}^2 \times \text{12r} = \text{n} \times \dfrac{4}{3} \text{r}^3 \\[1em] \Rightarrow 12\text{r}^3 = \text{n} \times \dfrac{4}{3} \text{r}^3 \\[1em] \Rightarrow 12 = \text{n} \times \dfrac{4}{3} \\[1em] \Rightarrow \text{n} = \dfrac{12 \times 3}{4} \\[1em] \Rightarrow \text{n} = \dfrac{36}{4} \\[1em] \Rightarrow \text{n} = 9.$

Hence, option 3 is the correct option.

If a solid sphere of radius r is melted and recast into the shape of a solid cone of height r then the radius of the base of the cone is :

1. r

2. 2r

3. 3r

4. 4r

Answer

Given,

Radius of sphere = r cm

Height of cone, h = r cm

Let radius of cone be R cm

Since, sphere is melted and recasted into cone.

∴ Volume of sphere = Volume of cone

$\Rightarrow \dfrac{4}{3} π\text{r}^3 = \dfrac{1}{3} π\text{R}^2\text{h} \\[1em] \Rightarrow \dfrac{4}{3} \text{r}^3 = \dfrac{1}{3} \text{R}^2 \times \text{r} \\[1em] \Rightarrow 3 \times \dfrac{4}{3} \times \text{r}^3 = \text{R}^2 \times \text{r} \\[1em] \Rightarrow 4 \times \dfrac{\text{r}^3}{\text{r}} = \text{R}^2 \\[1em] \Rightarrow 4 \times \text{r}^2 = \text{R}^2 \\[1em] \Rightarrow \text{R} = \sqrt{4 \times \text{r}^2} \\[1em] \Rightarrow \text{R} = 2\text{r}$

Hence, option 2 is the correct option.

If the volume and the surface area of a sphere are numerically the same, then its radius is:

1. 1 unit

2. 2 units

3. 3 units

4. 4 units

Answer

Let radius of sphere be r cm.

Given,

Volume of sphere = Surface area of sphere

$\Rightarrow \dfrac{4}{3} π\text{r}^3 = 4π\text{r}^2 \\[1em] \Rightarrow 4π\text{r}^3 = 3 \times 4π\text{r}^2 \\[1em] \Rightarrow 4\text{r}^3 = 12 \times \text{r}^2 \\[1em] \Rightarrow \text{r}^3 = \dfrac{12}{4} \times \text{r}^2 \\[1em] \Rightarrow \dfrac{\text{r}^3}{\text{r}^2} = 3 \\[1em] \Rightarrow \text{r} = 3 \text{ units.}$

Hence, option 3 is the correct option.

The diameter of a copper sphere is 6 cm. The sphere is melted and drawn into a long wire of uniform circular cross section. If the length of the wire is 36 cm, then its radius is :

1. 0.5 cm

2. 1 cm

3. 1.2 cm

4. 1.5 cm

Answer

Given,

Let the wire's radius be a.

Given, sphere is melted into the wire.

The wire formed is a cylinder, hence the volume of wire will be equal to the volume of sphere.

Radius of sphere, r = $\dfrac{\text{diameter}}{2} = \dfrac{6}{2}$ = 3 cm

Volume of sphere, V = $\dfrac{4}{3} π\text{r}^3$

$= \dfrac{4}{3}π \times 3^3 \\[1em] = \dfrac{4}{3}π \times 27 \\[1em] = 4 \times 9π \\[1em] = 36 π \text{ cm}^3$

Given, length of wire = 36 cm

So, height of cylinder = 36 cm

Volume of cylinder, V = 36 π cm³

∴ πr²h = 36 π

$\Rightarrow \text{r}^2 \times 36 = 36 \\[1em] \Rightarrow \text{r}^2 = \dfrac{36}{36} \\[1em] \Rightarrow \text{r}^2 = 1 \\[1em] \Rightarrow \text{r} = \sqrt{1} \\[1em] \Rightarrow \text{r} = 1 \text{ cm.}$

Hence, option 2 is the correct option.

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. The radius of the third ball is :

1. 0.5 cm

2. 1 cm

3. 1.5 cm

4. 2.5 cm

Answer

Radius of larger spherical metallic ball, R = 3 cm

Radius of smaller spherical balls are 1.5 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 1.5 cm + Volume of ball of radius 2 cm + Volume of ball of radius r cm

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

Hence, option 4 is the correct option.

A solid sphere with a radius of 4 cm is cut into 4 identical pieces by two mutually perpendicular planes passing through its center. Find the total surface area of one-quarter piece.

1. 24π

2. 32π

3. 48π

4. 64π

Answer

Total surface area of semi-hemisphere = 2πr²

= 2π × 4²

= 2π × 16

= 32π.

Hence, option 2 is the correct option.

Two identical solid hemispheres are kept in contact to form a sphere. The ratio of the total surface areas of two hemispheres to the surface area of the sphere formed is :

1. 1 : 1

2. 3 : 2

3. 2 : 3

4. 2 : 1

Answer

Let radius of hemisphere be r.

Total surface area of hemisphere = 3πr²

Total surface area of two hemisphere = 2 × 3πr² = 6πr².

Total surface area of sphere = 4πr²

Total surface area of two hemisphere : Total surface area of sphere = 6πr² : 4πr²

= 6 : 4

= 3 : 2.

Hence, option 2 is the correct option.

A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. The radius of the base of the cone is :

1. 2 cm

2. 2.1 cm

3. 3 cm

4. 6.3 cm

Answer

Radius of sphere, r = $\dfrac{\text{diameter}}{2} = \dfrac{12.6}{2} = 6.3 \text{ cm.}$

Volume of sphere = $\dfrac{4}{3}π\text{r}^3$

Radius of the cone = R cm

Height of the cone, h = 25.2 cm

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

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

$\therefore \dfrac{1}{3}π\text{R}^2 \text{h} = \dfrac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \dfrac{1}{3}\text{R}^2 \text{h} = \dfrac{4}{3}\text{r}^3 \\[1em] \Rightarrow \text{R}^2 = \dfrac{4 \times 3}{3 \times \text{h}} \times \text{r}^3 \\[1em] \Rightarrow \text{R}^2 = \dfrac{12}{3 \times 25.2} \times 6.3^3 \\[1em] \Rightarrow \text{R}^2 = \dfrac{12}{75.6} \times 250.047 \\[1em] \Rightarrow \text{R}^2 = \dfrac{3000.564}{75.6} \\[1em] \Rightarrow \text{R}^2 = 39.69 \\[1em] \Rightarrow \text{R} = \sqrt{39.69} \\[1em] \Rightarrow \text{R} = 6.3 \text{ cm.}$

Hence, option 4 is the correct option.

How many lead shots each 0.3 cm in diameter can be made from a cuboid of dimensions 9 cm × 11 cm × 12 cm?

1. 7200

2. 8400

3. 72000

4. 84000

Answer

Shots is in the shape of sphere.

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

Let the number of spheres formed be n.

Volume of cuboid = n × Volume of each lead 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, option 4 is the correct option.

A metallic sphere of radius 10.5 cm in melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. The number of cones formed is :

1. 21

2. 63

3. 126

4. 130

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

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