Mensuration — Multiple Choice Questions

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

1. 15 m

2. 12 m

3. 9 m

4. 24 m

Answer

Given, volume of a conical tent = 462 m³ and the area of the base = 154 m²

Using formula,

area of base = πr² and volume of cone = $\dfrac{1}{2}$ πr²h

$\Rightarrow \dfrac{1}{3}πr^2h = 462 \\[1em] \Rightarrow \dfrac{1}{3} \times (πr^2) \times h = 462 \\[1em] \Rightarrow \dfrac{1}{3} \times 154 \times h = 462 \\[1em] \Rightarrow \dfrac{154}{3} \times h = 462 \\[1em] \Rightarrow h = \dfrac{3 \times 462}{154} \\[1em] \Rightarrow h = \dfrac{1386}{154} \\[1em] \Rightarrow h = 9 \text{ m}.$

Hence, option 3 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

Given, the radius of a roller = 14 cm and height = 100 cm

Curved surface area of cylinder = 2πrh

$\text{Curved surface area of roller }= 2 \times \dfrac{22}{7} \times 14 \times 100 \\[1em] = 2 \times 22 \times 2 \times 100 \\[1em] = 8800 \text{ cm}^2.$

Hence, option 4 is the correct option.

If two cylinders of the same lateral surface have their radii in the ratio 4 : 9, then the ratio of their heights is

1. 2 : 3

2. 3 : 2

3. 4 : 9

4. 9 : 4

Answer

Let the radius of cylinders be r₁ = 4r and r₂ = 9r.

Let heights be h₁ and h₂.

Since, cylinders have equal lateral surface area,

$\therefore 2πr_1h_1 = 2πr_2h_2$

Dividing both sides by 2π.

$\Rightarrow r_1h_1 = r_2h_2 \\[1em] \Rightarrow 4r \times h_1 = 9r \times h_2 \\[1em] \Rightarrow \dfrac{h_1}{h_2} = \dfrac{9}{4}.$

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 volumes is

1. 10 : 17

2. 20 : 27

3. 17 : 27

4. 20 : 37

Answer

Let radius and heights of two cylinders be r₁, h₁ and r₂, h₂.

Given, $\dfrac{h_1}{h_2} = \dfrac{5}{3} \text{ and } \dfrac{r_1}{r_2} = \dfrac{2}{3}$.

Volume of cylinder 1 = V₁ and

Volume of cylinder 2 = V₂.

$\dfrac{V_1}{V_2} = \dfrac{πr_1^2h_1}{πr_2^2h_2} \\[1em] = \Big(\dfrac{r_1}{r_2}\Big)^2 \times \dfrac{h_1}{h_2} \\[1em] = \Big(\dfrac{2}{3}\Big)^2 \times \dfrac{5}{3} \\[1em] = \dfrac{4}{9} \times \dfrac{5}{3} \\[1em] = \dfrac{20}{27}.$

Hence, Option 2 is the correct option.

The total surface area of a cone whose radius is $\dfrac{r}{2}$ and slant height 2l is

1. 2πr(l + r)

2. $πr \Big(l + \dfrac{r}{4}\Big)$

3. πr(l + r)

4. 2πrl

Answer

Given,

Radius (R) = $\dfrac{r}{2}$,

Slant height (L) = 2l

Total surface area of cone (S) = πR(L + R)

Putting values we get,

$S = π \times \dfrac{r}{2} \times \Big(2l + \dfrac{r}{2}\Big) \\[1em] = \dfrac{πr}{2} \times 2l + \dfrac{πr}{2} \times \dfrac{r}{2} \\[1em] = πrl + \dfrac{πr^2}{4} \\[1em] = πr\Big(l + \dfrac{r}{4}\Big).$

Hence, Option 2 is the correct option.

If the diameter of the base of cone is 10 cm and its height is 12 cm, then its curved surface area is

1. 60π cm²

2. 65π cm²

3. 90π cm²

4. 120π cm²

Answer

Radius of base of cone (r) = $\dfrac{\text{Diameter}}{2}$

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

Height (h) = 12 cm.

$\text{l} = \sqrt{r^2 + h^2} \\[1em] = \sqrt{5^2 + 12^2} \\[1em] = \sqrt{25 + 144} \\[1em] = \sqrt{169} = 13 \text{ cm}.$

Curved surface area = πrl.

Putting values we get,

Curved surface area = π × 5 × 13 = 65π cm².

Hence, Option 2 is the correct option.

If the radius of a hemisphere is 5 cm, then its volume is

1. $\dfrac{250}{3}π$ cm³

2. $\dfrac{500}{3}π$ cm³

3. 75π cm³

4. $\dfrac{125}{3}π$ cm³

Answer

Volume of hemisphere (V) = $\dfrac{2}{3}πr^3$

Putting values we get,

$V = \dfrac{2}{3} \times π \times (5)^3 \\[1em] = \dfrac{2}{3}π \times 125 \\[1em] = \dfrac{250π}{3} \text{ cm}^3.$

Hence, Option 1 is the correct option.

If the ratio of the diameters of the two spheres is 3 : 5, then the ratio of their surface areas is

1. 3 : 5

2. 5 : 3

3. 27 : 125

4. 9 : 25

Answer

Let the diameters of two spheres be 3a and 5a.

So, their radius will be $r_1 = \dfrac{3a}{2} \text{ and } r_2 = \dfrac{5a}{2}$.

$\text{Ratio of their surface area } = \dfrac{4πr_1^2}{4πr_2^2} \\[1em] = \dfrac{\Big(\dfrac{3a}{2}\Big)^2}{\Big(\dfrac{5a}{2}\Big)^2} \\[1em] = \dfrac{\Big(\dfrac{9a^2}{4}\Big)}{\Big(\dfrac{25a^2}{4}\Big)} \\[1em] = \dfrac{9a^2 \times 4}{25a^2 \times 4} \\[1em] = \dfrac{9}{25} \\[1em] = 9 : 25.$

Hence, Option 4 is the correct option.

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is

1. 1 : 4

2. 1 : 3

3. 2 : 3

4. 2 : 1

Answer

Radius of balloon in original position (r) = 6 cm,

In pumped position radius (R) = 12 cm.

Ratio of surface areas in two situations = $\dfrac{4πr^2}{4πR^2}$

$= \dfrac{r^2}{R^2} \\[1em] = \dfrac{6^2}{12^2} \\[1em] = \dfrac{36}{144} \\[1em] = \dfrac{1}{4} \\[1em] = 1 : 4.$

Hence, Option 1 is the correct option.

If two solid hemispheres of same base radius r are joined together along with their bases, then the curved surface of this new solid is

1. 4πr²

2. 6πr²

3. 3πr²

4. 8πr²

Answer

Two hemispheres are joined along their bases hence, the surface area of new solid will be double the surface area of hemisphere.

Curved surface area = 2πr² × 2 = 4πr².

Hence, Option 1 is the correct option.

If a solid of one shape is converted to another, then the surface area of the new solid

1. remains same

2. increases

3. decreases

4. can't say

Answer

If a solid of one shape is converted to another, then the surface area of the new solid may or may not be the same.

Hence, Option 4 is the correct option.

The volume of the largest right circular cone that can be carved out from a cube of edge 4.2 cm is

1. 9.7 cm³

2. 77.6 cm³

3. 58.2 cm³

4. 19.4 cm³

Answer

Edge of cube = 4.2 cm

Radius of largest cone cut out (r) = $\dfrac{4.2}{2}$ = 2.1 cm

Height of largest cone cut out (h) = 4.2 cm.

Volume of cone (V) = $\dfrac{1}{3}πr^2h$

Putting values we get,

$V = \dfrac{1}{3} \times \dfrac{22}{7} \times (2.1)^2 \times 4.2 \\[1em] = \dfrac{22 \times 2.1 \times 2.1 \times 4.2}{3 \times 7} \\[1em] = \dfrac{407.484}{21} \\[1em] = 19.4 \text{ cm}^3$

Hence, Option 4 is the correct option.

The volume of the greatest sphere cut off from a circular cylindrical wood of base radius 1 cm and height 6 cm is

1. 288π cm³

2. $\dfrac{4}{3}π$ cm³

3. 6π cm³

4. 4π cm³

Answer

Largest sphere that can be cut out from a cylinder of base radius 1 will have radius = 1 cm.

Volume of sphere (V) = $\dfrac{4}{3}πr^3 = \dfrac{4}{3}π(1)^3 = \dfrac{4}{3}π$ cm³.

Hence, Option 2 is the correct option.

The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is

1. 3 : 4

2. 4 : 3

3. 9 : 16

4. 16 : 9

Answer

Let the radius of two spheres be r₁ and r₂.

Given ratio of volumes = 64 : 27.

$\Rightarrow \dfrac{\dfrac{4}{3}πr_1^3}{\dfrac{4}{3}πr_2^3} = \dfrac{64}{27} \\[1em] \Rightarrow \dfrac{r_1^3}{r_2^3} = \dfrac{4^3}{3^3} \\[1em] \Rightarrow \Big(\dfrac{r_1}{r_2}\Big)^3 = \Big(\dfrac{4}{3}\Big)^3 \\[1em] \Rightarrow \dfrac{r_1}{r_2} = \dfrac{4}{3}.$

Ratio of surface areas = $\dfrac{4πr_1^2}{4πr_2^2}$

$= \dfrac{r_1^2}{r_2^2} \\[1em] = \Big(\dfrac{r_1}{r_2}\Big)^2 \\[1em] = \Big(\dfrac{4}{3}\Big)^2 \\[1em] = \dfrac{16}{9} = 16 : 9.$

Hence, Option 4 is the correct option.

If a cone, a hemisphere and a cylinder have equal bases and have same height, then the ratio of their volumes is

1. 1 : 3 : 2

2. 2 : 3 : 1

3. 2 : 1 : 3

4. 1 : 2 : 3

Answer

Let the common radius of shapes be r and height be h.

Ratio in their volumes = Volume of cone : Volume of hemisphere : Volume of cylinder.

$\text{Ratio in their volumes } = \dfrac{1}{3}πr^2h : \dfrac{2}{3}πr^3 : πr^2h \\[1em] = \dfrac{1}{3} : \dfrac{2}{3} : 1$

On multiplying by 3, ratio = 1 : 2 : 3.

Hence, Option 4 is the correct option.

If a sphere and a cube have equal surface areas, then the ratio of the diameter of the sphere to the edge of the cube is

1. 1 : 2

2. 2 : 1

3. $\sqrt{π} : \sqrt{6}$

4. $\sqrt{6} : \sqrt{π}$

Answer

A sphere and a cube have equal surface area.

Let a be the edge of cube and r the radius of sphere.

⇒ 4πr² = 6a²

⇒ π(2r)² = 6a²

Since, d = 2r

⇒ πd² = 6a²

$\Rightarrow \dfrac{d^2}{a^2} = \dfrac{6}{π} \\[1em] \Rightarrow \sqrt{\dfrac{d^2}{a^2}} = \dfrac{\sqrt{6}}{\sqrt{π}} \\[1em] \Rightarrow \dfrac{d}{a} = \dfrac{\sqrt{6}}{\sqrt{π}}.$

Hence, Option 4 is the correct option.

A solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm is molded to form a sphere. The radius of the sphere is

1. 21 cm

2. 23 cm

3. 25 cm

4. 19 cm

Answer

Since, the cuboid is molded into a sphere.

Let radius of sphere be r cm.

∴ Volume of cuboid = Volume of sphere.

$\therefore 49 \times 33 \times 24 = \dfrac{4}{3}πr^3 \\[1em] \Rightarrow r^3 = \dfrac{49 \times 33 \times 24 \times 3}{4 \times \dfrac{22}{7}} \\[1em] \Rightarrow r^3 = \dfrac{49 \times 33 \times 24 \times 3 \times 7}{4 \times 22} \\[1em] \Rightarrow r^3 = \dfrac{814968}{88} \\[1em] \Rightarrow r^3 = 9261 \\[1em] \Rightarrow r^3 = (21)^3 \\[1em] \Rightarrow r = 21 \text{ cm}.$

Hence, Option 1 is the correct option.

If a solid right circular cone of height 24 cm and base radius 6 cm is melted and recast in the shape of sphere, then the radius of the sphere is

1. 4 cm

2. 6 cm

3. 8 cm

4. 12 cm

Answer

Since, cone is recasted into sphere.

∴ Volume of cone = Volume of sphere.

Given, radius of cone (R) = 6 cm, height (h) = 24 cm.

Let the radius of sphere be r.

$\therefore \dfrac{1}{3}πR^2h = \dfrac{4}{3}πr^3$

Multiplying both sides by 3 and dividing by π we get,

$\Rightarrow R^2h = 4r^3 \\[1em] \Rightarrow 6^2 \times 24 = 4r^3 \\[1em] \Rightarrow 4r^3 = 36 \times 24 \\[1em] \Rightarrow r^3 = \dfrac{864}{4} \\[1em] \Rightarrow r^3 = 216 \\[1em] \Rightarrow r^3 = (6)^3 \\[1em] \Rightarrow r = 6 \text{ cm}.$

Hence, Option 2 is the correct option.

If a solid circular cylinder of iron whose diameter is 15 cm and height 10 cm is melted and recasted into a sphere, then the radius of the sphere is

1. 15 cm

2. 10 cm

3. 7.5 cm

4. 5 cm

Answer

Diameter of cylinder = 15 cm

For cylinder,

Radius (r) = $\dfrac{15}{2} = 7.5$ cm and height (h) = 10 cm.

Let radius of sphere be R.

Volume of cylinder = Volume of sphere.

$\Rightarrow πr^2h = \dfrac{4}{3}πR^3 \\[1em] \Rightarrow r^2h = \dfrac{4}{3}R^3 \\[1em] \Rightarrow R^3 = \dfrac{3 \times (7.5)^2 \times 10}{4} \\[1em] \Rightarrow R^3 = \dfrac{1687.5}{4} \\[1em] \Rightarrow R^3 = 421.875 \\[1em] \Rightarrow R^3 = (7.5)^3 \\[1em] \Rightarrow R = 7.5 \text{ cm}.$

Hence, Option 3 is the correct option.

The number of balls of radius 1 cm that can be made from a sphere of radius 10 cm is

1. 100

2. 1000

3. 10000

4. 100000

Answer

Radius of sphere (R) = 10 cm.

Radius of ball (r) = 1 cm

Let number of balls formed be n.

Since a big sphere is divided into small balls.

Volume of sphere = n × Volume of each ball.

$\dfrac{4}{3}πR^3 = n \times \dfrac{4}{3}πr^3$

Multiplying both sides by $\dfrac{3}{4π}$.

$\Rightarrow R^3 = nr^3 \\[1em] \Rightarrow n = \dfrac{R^3}{r^3} \\[1em] \Rightarrow n = \dfrac{10^3}{1^3} \\[1em] \Rightarrow n = 1000.$

Hence, Option 2 is the correct option.

A metallic spherical shell of internal and external diameters 4 cm and 8 cm respectively is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is

1. 12 cm

2. 14 cm

3. 15 cm

4. 18 cm

Answer

Internal radius (r) = $\dfrac{4}{2}$ = 2 cm,

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

Given, spherical shell is recasted into cone.

Volume of cone = Volume of spherical shell.

Radius of cone (r₁) = $\dfrac{8}{2}$ = 4 cm.

Let height of cone be h.

$\dfrac{1}{3}πr_1^2h = \dfrac{4}{3}π(R^3 - r^3)$

Multiplying both sides by $\dfrac{3}{π}$.

$\Rightarrow r_1^2h = 4(R^3 - r^3) \\[1em] \Rightarrow h = \dfrac{4(R^3 - r^3)}{r_1^2} \\[1em] = \dfrac{4 \times (4^3 - 2^3)}{4^2} \\[1em] = \dfrac{4 \times (64 - 8)}{16} \\[1em] = \dfrac{4 \times 56}{16} \\[1em] = \dfrac{224}{16} \\[1em] = 14 \text{ cm}$

Hence, Option 2 is the correct option.

A cubical ice cream brick of edge 22 cm is to be distributed among some children by filling ice cream cones of radius 2 cm and height 7 cm up to its brim. The number of children who will get the ice cream cones is

1. 163

2. 263

3. 363

4. 463

Answer

Edge of cubical ice cream brick (a) = 22 cm

Volume = a³ = (22)³ = 10648 cm³.

Radius of ice cream cone (r) = 2 cm and height (h) = 7 cm.

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

= $\dfrac{1}{3}$ x $\dfrac{22}{7}$ x 2 x 2 x 7

= $\dfrac{22 \times 4}{3}$

= $\dfrac{88}{3}$ cm³.

Number of cones (n) = $\dfrac{\text{Vol. of ice cream brick}}{\text{Vol. of cone}}$

$n = \dfrac{10648}{\dfrac{88}{3}} \\[1em] = \dfrac{10648 \times 3}{88} \\[1em] = \dfrac{31944}{88} \\[1em] = 363.$

Hence, Option 3 is the correct option.

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

1. 4 cm

2. 3 cm

3. 2 cm

4. 6 cm

Answer

Diameter of cylinder = 2 cm

Radius of cylinder (r) = $\dfrac{2}{2}$ = 1 cm and height (h) = 16 cm.

Given, 12 spheres are formed from this cylinder.

∴ Volume of cylinder = 12 × Volume of each sphere.

Let radius of each sphere be R.

$\therefore πr^2h = 12 \times \dfrac{4}{3}πR^3 \\[1em] \Rightarrow R^3 = \dfrac{3π(1)^2 \times 16}{4π \times 12} \\[1em] \Rightarrow R^3 = \dfrac{48π}{48π} \\[1em] \Rightarrow R^3 = 1 \\[1em] \Rightarrow R = 1 \text{ cm}.$

Diameter = 2 × Radius = 2 cm.

Hence, Option 3 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.

A right angle triangle shaped piece of hard board is rotated completely about its hypotenuse, as shown in the diagram. The solid so formed is always :

(i) a single cone

(ii) a double cone

Which of the statement is valid ?

1. only (i)

2. only (ii)

3. both (i) and (ii)

4. neither (i) nor (ii)

Answer

On rotating the right angle triangle figure formed is :

Hence, Option 2 is the correct option.

A solid sphere is cut into identical hemispheres.

Statement 1 : The total volume of two hemispheres is equal to the volume of the original sphere.

Statement 2 : The total surface area of two hemispheres together is equal to the surface area of the original sphere.

Which of the following is valid ?

1. Both the statements are true

2. Both the statements are false

3. Statement 1 is true, and statement 2 is false

4. Statement 1 is false, and statement 2 is true

Answer

When a solid sphere is cut into identical hemispheres.

Let radius of sphere be r.

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

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

Volume of 2 hemisphere = $2 \times \dfrac{2}{3}πr^3 = \dfrac{4}{3}πr^3$.

The total volume of two hemispheres is equal to the volume of the original sphere.

Surface area of sphere = 4πr²

Surface area of hemisphere = 3πr²

Surface area of 2 hemisphere = 2 × 3πr² = 6πr².

The surface area of two hemispheres is not equal to the surface area of the original sphere.

Hence, Option 3 is the correct option.