Volume & Surface Area of Solids — Analytical & Application Based Questions

A famous sweet shop “Madanlal Sweets” sells tinned rasgullas. The tin container is cylindrical in shape with diameter 14 cm, height 16 cm, and it can hold 20 spherical rasgullas of diameter 6 cm and sweetened liquid such that the can is filled and then sealed. Find out how much sweetened liquid the can contains. Take π = 3.14.

Answer

Given,

Diameter of container = 14 cm

Radius (R) = $\dfrac{\text{Diameter}}{2} = \dfrac{14}{2}$ = 7 cm

Height of container (H) = 16 cm

Diameter of rasgulla = 6 cm

Radius of rasgulla (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{6}{2}$ = 3 cm

Volume of container = Volume of rasgullas + Volume of liquid

⇒ πR²H = $20 \times \dfrac{4}{3}$ πr³ + Volume of liquid

⇒ Volume of liquid = πR²H - $\dfrac{80}{3}$ πr³

⇒ Volume of liquid = π(R²H - $\dfrac{80}{3}r^3$)

⇒ Volume of liquid = 3.14 × (7² × 16 - $\dfrac{80}{3} \times 3^3$)

⇒ Volume of liquid = 3.14 × (784 - 720)

⇒ Volume of liquid = 3.14 × 64

⇒ Volume of liquid = 200.96 cm³.

Hence, volume of sweetened liquid in the container = 200.96 cm³.

The ratio of the radius and the height of a solid metallic right circular cylinder is 7 : 27. This is melted and made into a cone of diameter 14 cm and slant height 25 cm. Find the height of the :

(a) cone

(b) cylinder

Answer

Given,

Diameter of cone = 14 cm

Radius of cone (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{14}{2}$ = 7 cm

Slant height (l) = 25 cm

Ratio of the radius and the height of a solid metallic right circular cylinder is 7 : 27.

Radius of cylinder (R) = 7x

Height of cylinder (H) = 27x

(a) Let height of cone be h cm.

By formula,

⇒ l² = r² + h²

⇒ 25² = 7² + h²

⇒ 625 = 49 + h²

⇒ h² = 625 - 49

⇒ h² = 576

⇒ h = $\sqrt{576}$ = 24 cm.

Hence, height of cone = 24 cm.

(b) Given,

A solid metallic right circular cylinder is melted and made into a cone.

∴ Volume of cylinder = Volume of cone

⇒ πR²H = $\dfrac{1}{3}πr^2h$

⇒ R²H = $\dfrac{1}{3}r^2h$

⇒ (7x)² × 27x = $\dfrac{1}{3} \times 7^2 \times 24$

⇒ 1323x³ = $\dfrac{1176}{3}$

⇒ 1323x³ = 392

⇒ x³ = $\dfrac{392}{1323}$

⇒ x³ = $\dfrac{8}{27}$

⇒ x³ = $\Big(\dfrac{2}{3}\Big)^3$

⇒ x = $\dfrac{2}{3}$

Height of cylinder (H) = 27x = $27 \times \dfrac{2}{3}$ = 18 cm.

Hence, height of cylinder = 18 cm.

The curved surface area of a right circular cone is half of another right circular cone. If the ratio of their slant heights is 2 : 1 and that of their volumes is 3 : 1, find ratio of their:

(a) radii

(b) heights

Answer

(a) Let radius of smaller and larger cone be r and R, respectively.

Let height of smaller and larger cone be h and H, respectively.

Let slant height of smaller and larger cone be l and L respectively.

Given,

Ratio of their slant heights is 2 : 1.

∴ l : L = 2 : 1.

Given,

Curved surface area of a right circular cone is half of the other right circular cone.

∴ πrl = $\dfrac{1}{2}πRL$

⇒ rl = $\dfrac{1}{2}RL$

⇒ $\dfrac{r}{R} = \dfrac{L}{2l} = \dfrac{1}{2} \times \dfrac{1}{2}$

⇒ $\dfrac{r}{R} = \dfrac{1}{4}$

⇒ r : R = 1 : 4

Hence, ratio of the radii = 1 : 4.

(b) Let volume of cone with smaller curved surface area be v and that with larger curved surface area be V.

Given,

Volumes are in the ratio 3 : 1.

∴ v : V = 3 : 1

$\Rightarrow \dfrac{\text{Volume of larger CSA cone}}{\text{Volume of smaller CSA cone}} = \dfrac{V}{v} \\[1em] \Rightarrow \dfrac{\dfrac{1}{3}πR^2H}{\dfrac{1}{3}πr^2h} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{4^2}{1^2} \times \dfrac{H}{h} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{16}{1} \times \dfrac{H}{h} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{H}{h} = \dfrac{1}{3 \times 16} \\[1em] \Rightarrow \dfrac{H}{h} = \dfrac{1}{48} \\[1em] \Rightarrow \dfrac{h}{H} = \dfrac{48}{1}.$

Hence, ratio of heights = 48 : 1.

A mathematics teacher uses certain amount of terracotta clay to form different shaped solids. First, she turned it into a sphere of radius 7 cm and then she made a right circular cone with base radius 14 cm. Find the height of the cone so formed. If the same clay is turned to make a right circular cylinder of height $\dfrac{7}{3}$ cm, then find the radius of the cylinder so formed. Also, compare the total surface areas of sphere and cylinder so formed.

Answer

First, a sphere of radius (r) 7 cm is formed.

Volume of sphere = $\dfrac{4}{3}πr^3$

$= \dfrac{4}{3} \times \dfrac{22}{7} \times 7^3 \\[1em] = \dfrac{4}{3} \times 22 \times 7^2 \\[1em] = \dfrac{4312}{3} \text{ cm}^3.$

Next, a right circular cone with radius (r₁) 14 cm is formed. Let height of cone be h cm.

Since same amount of clay is used to make cone and sphere.

∴ Volume of cone = Volume of sphere

$\Rightarrow \dfrac{1}{3}πr_1^2h = \dfrac{4312}{3} \\[1em] \Rightarrow πr_1^2h = 4312 \\[1em] \Rightarrow \dfrac{22}{7} \times 14^2 \times h = 4312 \\[1em] \Rightarrow 22 \times 2 \times 14 \times h = 4312 \\[1em] \Rightarrow h = \dfrac{4312}{22 \times 2 \times 14} \\[1em] \Rightarrow h = \dfrac{4312}{616} = 7 \text{ cm}.$

Given,

The same clay is used to make a right circular cylinder of height (h₁) $\dfrac{7}{3}$ cm. Let its radius be r₂.

Since same amount of clay is used to make cylinder and sphere.

∴ Volume of cylinder = Volume of sphere

$\Rightarrow πr_2^2h_1 = \dfrac{4312}{3} \\[1em] \Rightarrow \dfrac{22}{7} \times r_2^2 \times \dfrac{7}{3} = \dfrac{4312}{3} \\[1em] \Rightarrow 22 \times r_2^2 = 4312 \\[1em] \Rightarrow r_2^2 = \dfrac{4312}{22} \\[1em] \Rightarrow r_2^2 = 196 \\[1em] \Rightarrow r_2 = \sqrt{196} = 14 \text{ cm}.$

Total surface area of sphere = 4πr²

Total surface area of cylinder = 2πr₂(r₂ + h₁)

$\Rightarrow \dfrac{\text{TSA sphere}}{\text{TSA cylinder}} = \dfrac{4πr^2}{2πr_2(r_2 + h_1)} \\[1em] = \dfrac{2r^2}{r_2(r_2 + h_1)} \\[1em] = \dfrac{2 \times 7^2}{14(14 + \dfrac{7}{3})} \\[1em] = \dfrac{98}{14 \times \dfrac{49}{3}} \\[1em] = \dfrac{98 \times 3}{14 \times 49} \\[1em] = \dfrac{3}{7}.$

Hence, height of cone = 7 cm, radius of cylinder = 14 cm and Total surface area of sphere : Total surface area of cylinder = 3 : 7.