Cylinder, Cone & Sphere (Surface Area & Volume) — Exercise 20(D)

A cone and a sphere have equal volumes. Their radii are also equal each being 10 cm; the height of the cone is :

1. 70 cm

2. 40 cm

3. 50 cm

4. 20 cm

Answer

Let radius of cone and sphere be r cm and height of cone be h cm.

Given,

A cone and a sphere have equal volumes and their radius are equal.

∴ Volume of cone = Volume of sphere

$\Rightarrow \dfrac{1}{3}πr^2h = \dfrac{4}{3}πr^3 \\[1em] \Rightarrow h = \dfrac{4 \times 3πr^3}{3πr^2} \\[1em] \Rightarrow h = 4r \\[1em] \Rightarrow h = 4 \times 10 = 40 \text{ cm}.$

Hence, Option 2 is the correct option.

A sphere and a cone have equal volumes and equal radii. The ratio between the radius and height of the cone is :

1. 3 : 4

2. 4 : 3

3. 4 : 1

4. 1 : 4

Answer

Let radius of cone and sphere be r cm and height of cone be h cm.

Given,

A cone and a sphere have equal volumes and their radius are equal.

∴ Volume of cone = Volume of sphere

$\Rightarrow \dfrac{1}{3}πr^2h = \dfrac{4}{3}πr^3 \\[1em] \Rightarrow h = \dfrac{4 \times 3πr^3}{3πr^2} \\[1em] \Rightarrow h = 4r \\[1em] \Rightarrow \dfrac{r}{h} = \dfrac{1}{4} \\[1em] \Rightarrow r : h = 1 : 4.$

Hence, Option 4 is the correct option.

The radius and the height of a cone are in the ratio 2 : 1. The ratio between the volumes of a sphere and this cone (both having equal radii) is :

1. 1 : 4

2. 4 : 1

3. 8 : 1

4. 1 : 8

Answer

Given,

The radius and the height of a cone are in the ratio 2 : 1.

Let radius of cone be 2x and height of cone be x.

Let radius f sphere be r.

Given,

Radius of cone and sphere are equal.

∴ r = 2x.

$\Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{\dfrac{4}{3}πr^3}{\dfrac{1}{3}πr^2h} \\[1em] \Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{4 \times 3 \times πr^3}{3 \times πr^2h} \\[1em] \Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{4r}{h} \\[1em] \Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{4 \times 2x}{x} \\[1em] \Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{8x}{x} \\[1em] \Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{8}{1} = 8 : 1.$

Hence, Option 3 is the correct option.

A solid metallic cylinder is melted and formed into identical solid cones each having same radius and same height as that of the cylinder. The number of cones formed is :

1. 3

2. 6

3. 9

4. 2

Answer

Given,

A solid metallic cylinder is melted and formed into identical solid cones.

Let no. of cones formed be n.

As, radius and height of cones formed is equal to the cylinder.

Let radius of cylinder and cone be r units, height of cylinder and cone be h units.

∴ Volume of cylinder = n × Volume of each cone

⇒ πr²h = n × $\dfrac{1}{3}πr^2h$

⇒ n = $\dfrac{3πr^2h}{πr^2h}$ = 3.

Hence, Option 1 is the correct option.

A solid metallic cone of radius 10 cm and height 12 cm is melted to form a wire of uniform cross-section 1 cm². The length of wire formed is :

1. 40 × π cm

2. 4 × π m

3. 400 × π m

4. 2 × π m

Answer

Given,

A solid metallic cone of radius (r) 10 cm and height (h) 12 cm is melted to form a wire of uniform cross-section 1 cm².

Let length of wire formed be l cm.

∴ Volume of cone = Volume of wire

$\Rightarrow \dfrac{1}{3}πr^2h = \text{Area of cross-section × length} \\[1em] \Rightarrow \dfrac{1}{3}π \times 10^2 \times 12 = 1 \times l \\[1em] \Rightarrow l = 4 \times π \times 10^2 \\[1em] \Rightarrow l = 400 π \text{ cm} \\[1em] \Rightarrow l = \dfrac{400π}{100} \text{ m} \\[1em] \Rightarrow l = 4 × π \text{ m}.$

Hence, Option 2 is the correct option.

A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.

Answer

Given,

Radius of the sphere (R) = 15 cm

So, the volume of sphere melted = $\dfrac{4}{3}πR^3 = \dfrac{4}{3} \times π \times 15 \times 15 \times 15$

= 4500π cm³.

Radius of each cone formed (r) = 2.5 cm

Height of each cone (h) = 8 cm

So, volume of each cone = $\dfrac{1}{3}πr^2h = \dfrac{1}{3} \times π \times 2.5 \times 2.5 \times 8$

= $\dfrac{1}{3} \times 50π = \dfrac{50}{3}π.$

Let no. of cones formed be n.

Volume of sphere = n × Volume of cone

$\Rightarrow 4500π = n \times \dfrac{50}{3}π \\[1em] \Rightarrow n = \dfrac{4500π \times 3}{50π} \\[1em] \Rightarrow n = 270.$

Hence, no. of cones formed = 270.

A hollow sphere of internal and external diameters 4 cm and 8 cm, respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.

Answer

Given,

External diameter of the hollow sphere = 8 cm

So, external radius (R) = $\dfrac{8}{2}$ = 4 cm

Internal diameter of the hollow sphere = 4 cm

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

Diameter of cone = 8 cm

Radius of cone (r₁) = 4 cm

Let height of the cone = h cm.

Since, hollow sphere is melted and recasted into cone.

∴ Volume of sphere = Volume of cone

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

Hence, the height of the cone = 14 cm.

The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid right circular cone of height 32 cm. Find the diameter of the base of the cone.

Answer

Given,

Height of the solid right circular cone (h) = 32 cm

Internal radius of metallic spherical shell (r) = 3 cm

External radius of metallic spherical shell (R) = 5 cm

Let radius of cone be r₁ cm.

As, metallic spherical shell is recasted into right circular cone.

∴ Volume of spherical shell = Volume of cone.

$\Rightarrow \dfrac{4}{3}π(R^3 - r^3) = \dfrac{1}{3}πr_1^2h \\[1em] \Rightarrow 4(R^3 - r^3) = r_1^2h \\[1em] \Rightarrow r_1^2 = \dfrac{4(R^3 - r^3)}{h} \\[1em] \Rightarrow r_1^2 = \dfrac{4 \times (5^3 - 3^3)}{32} \\[1em] \Rightarrow r_1^2 = \dfrac{4 \times (125 - 27)}{32} \\[1em] \Rightarrow r_1^2 = \dfrac{4 \times 98}{32} \\[1em] \Rightarrow r_1^2 = \dfrac{49}{4} \\[1em] \Rightarrow r_1 = \sqrt{\dfrac{49}{4}} \\[1em] \Rightarrow r_1 = \dfrac{7}{2}.$

Diameter = 2r = $2 \times \dfrac{7}{2}$ = 7 cm.

Hence, diameter = 7 cm.

Total volume of three identical cones is the same as that of a bigger cone whose height is 9 cm and diameter 40 cm. Find the radius of the base of each smaller cone, if height of each is 108 cm.

Answer

Let the radius of the smaller cone be r cm.

Given,

Height of smaller cone (h) = 108 cm

Diameter of bigger cone = 40 cm

So, radius (R) = $\dfrac{40}{2}$ = 20 cm

Height of bigger cone (H) = 9 cm.

According to question,

Volume of big cone = 3 × Volume of each smaller cone.

$\Rightarrow \dfrac{1}{3}πR^2H = 3 \times \dfrac{1}{3}πr^2h \\[1em] \Rightarrow R^2H = 3 \times r^2h \\[1em] \Rightarrow (20)^2 \times 9 = 3 \times r^2 \times 108 \\[1em] \Rightarrow r^2 = \dfrac{20^2 \times 9}{3 \times 108} \\[1em] \Rightarrow r^2 = \dfrac{3600}{324} \\[1em] \Rightarrow r^2 = \dfrac{100}{9} \\[1em] \Rightarrow r = \sqrt{\dfrac{100}{9}} \\[1em] \Rightarrow r = \dfrac{10}{3} = 3\dfrac{1}{3}\text{ cm}.$

Hence, radius of base of each cone = $3\dfrac{1}{3}$ cm.

A solid rectangular block of metal 49 cm by 44 cm by 18 cm is melted and formed into a solid sphere. Calculate the radius of the sphere.

Answer

Let radius of sphere be r cm.

Since,

A solid rectangular block of metal is melted and formed into a solid sphere.

∴ Volume of rectangular block = Volume of sphere

$\Rightarrow 49 \times 44 \times 18 = \dfrac{4}{3}πr^3 \\[1em] \Rightarrow 38808 = \dfrac{4}{3} \times \dfrac{22}{7} \times r^3 \\[1em] \Rightarrow r^3 = \dfrac{38808 \times 21}{4 \times 22} \\[1em] \Rightarrow r^3 = 9261 \\[1em] \Rightarrow r^3 = (21)^3 \\[1em] \Rightarrow r = 21 \text{ cm}.$

Hence, the radius of sphere = 21 cm.

A hemispherical bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into conical shaped small containers each of diameter 3 cm and height 4 cm. How many containers are necessary to empty the bowl?

Answer

Given,

Internal radius of hemispherical bowl (R) = 9 cm

Diameter of each conical container = 3 cm

So, radius (r) = $\dfrac{3}{2}$ = 1.5 cm and height of conical container(h) = 4 cm

Let no. of conical containers needed be n.

∴ Volume of hemispherical bowl = n × Volume of each conical shaped container

$\Rightarrow \dfrac{2}{3}πR^3 = n \times \dfrac{1}{3}πr^2h \\[1em] \Rightarrow n = \dfrac{\dfrac{2}{3}πR^3}{\dfrac{1}{3}πr^2h} \\[1em] \Rightarrow n = \dfrac{2 \times (9)^3 \times 3}{1 \times (1.5)^2 \times 4 \times 3} \\[1em] \Rightarrow n = \dfrac{2 \times 729 \times 3}{2.25 \times 4 \times 3} \\[1em] \Rightarrow n = 162.$

Hence, 162 containers are necessary to empty the bowl.

The total area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate :

(i) the radius of the solid sphere,

(ii) the number of cones recast.

[Take π = 3.14]

Answer

(i) Given,

Total area of solid metallic sphere = 1256 cm²

Let R be the radius of metallic sphere.

By formula,

Total area of solid metallic sphere = 4πR²

∴ 4πR² = 1256

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

Hence, radius of sphere = 10 cm.

(ii) Given,

Radius of cone (r) = 2.5 cm

Height of cone (h) = 8 cm.

Let no. of cones formed be n.

Since, sphere is recasted into n cones.

∴ Volume of sphere = n × Volume of cones

$\Rightarrow \dfrac{4}{3}πR^3 = n \times \dfrac{1}{3}πr^2h \\[1em] \Rightarrow 4R^3 = n \times r^2h \\[1em] \Rightarrow n = \dfrac{4R^3}{r^2h} \\[1em] \Rightarrow n = \dfrac{4 \times 10^3}{(2.5)^2 \times 8} \\[1em] \Rightarrow n = \dfrac{4 \times 1000}{6.25 \times 8} \\[1em] \Rightarrow n = \dfrac{4000}{50} \\[1em] \Rightarrow n = 80.$

Hence, no. of cones formed = 80.

The surface area of a solid metallic sphere is 2464 cm². It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate :

(i) the radius of the sphere.

(ii) the number of cones recast.

Answer

(i) Given,

Surface area of solid metallic sphere = 2464

Let radius of solid metallic sphere be R cm.

By formula,

Surface area of solid metallic sphere = 4πR²

∴ 4πR² = 2464

⇒ $4 \times \dfrac{22}{7} \times R^2 = 2464$

⇒ $R^2 = \dfrac{2464 \times 7}{22 \times 4}$

⇒ $R^2 = \dfrac{17248}{88}$

⇒ R² = 196

⇒ R = $\sqrt{196}$

⇒ R = 14 cm.

Hence, the radius of sphere = 14 cm.

(ii) Given,

Radius of cones (r) = 3.5 cm

Height of cones (h) = 7 cm.

Let no. of cones formed be n.

Since,

Sphere is melted and recasted into cones.

∴ Volume of sphere = n × Volume of cone

$\Rightarrow \dfrac{4}{3}πR^3 = \dfrac{1}{3}πr^2h \\[1em] \Rightarrow 4R^3 = n \times r^2h \\[1em] \Rightarrow n = \dfrac{4R^3}{r^2h} \\[1em] \Rightarrow n = \dfrac{4 \times (14)^3}{(3.5)^2 \times 7} \\[1em] \Rightarrow n = \dfrac{4 \times 2744}{12.25 \times 7} \\[1em] \Rightarrow n = \dfrac{10976}{85.75} \\[1em] \Rightarrow n = 128.$

Hence, no. of cones formed = 128.