KC Sinha Mathematics Solution Class 10 Chapter 14 Surface Area and Volumes Exercise 14.2

Exercise 14.1
Exercise 14.2
Exercise 14.3
Exercise 14.4

Exercise 14.2

Question 1

A toy is in the form a cone mounted on a hemisphere of diameter 7 cm. The total height of the toy is 14.5 cm. Find the volume of the toy. $\left[\right.$ Take $\left.\pi=\frac{22}{7}\right]$

Sol :

Given: Diameter of a hemisphere = 7cm
So, Radius $=\frac{7}{2}=3.5 \mathrm{cm}$
Height of the cone, h = 143-.5 – 3.5 = 11cm

The volume of a toy = Volume of cone + Volume of a hemisphere

$=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}$

$=\frac{1}{3} \pi r^{2}(h+2 r)$

$=\frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5(11+2 \times 3.5)$

$=\frac{1}{3} \times 22 \times 0.5 \times 3.5 \times 18$

= 231cm³

Question 2

A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 2.1 cm, and the height of the cone is 4 cm. The solid is place in a cylindrical tub full of water in such a way that the whole solid is submerged in water left in the tub.
Sol :
Original volume of water in the cylindrical tub
= Volume of Cylinder
= πr²h
$=\frac{22}{7} \times 5^{2} \times 9.8$
= 22×25×1.4
= 770cm³
Given that Radius of hemisphere, R = 2.1cm
and height of cone, h = 4cm

Volume of a solid = Volume of cone + Volume of hemisphere
$=\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{h}+\frac{2}{3} \pi \mathrm{R}^{3}$
$=\frac{1}{3} \pi \mathrm{R}^{2}(\mathrm{h}+2 \mathrm{R})$
$=\frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1(4+2 \times 2.1)$
$=\frac{1}{3} \times 22 \times 0.3 \times 2.1 \times 8.2$
= 37.884cm³
∴ Volume of water displaced (removed) = 37.884cm³
Hence, the required volume of the water left in the cylindrical tub = 770 – 37.884
= 732.116 cm³

Question 3

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm, and the diameter of the capsule is 5 mm. Find its surface area.
Sol :

Given:
Diameter of cylinder = 5mm
The radius of cylinder$=\frac{\text { Diameter }}{2}=\frac{5}{2} \mathrm{mm}$
Height of cylinder = 14 – 5 = 9mm
Here, Diameter of hemisphere =. 5mm

So,

Radius of Hemisphere $=\frac{\text { Diameter }}{2}=\frac{5}{2} \mathrm{mm}$
The total area of the capsule = CSA of cylinder + CSA of 2 hemispheres
= 2πrh + 2×2πr²
$=2 \times \frac{22}{7} \times \frac{5}{2} \times 9+4 \times \frac{22}{7} \times \frac{5}{2} \times \frac{5}{2}$
$=\frac{22}{7}(45+25)$
=220 mm²

Question 4

A room in the form of a cylinder, surmounted by a hemispherical vaulted dome, contains $41 \frac{19}{21}$ m³ of air and the internal diameter of the building is equal to the height of the crown of the vault above the floor. Find the height$\left[\right.$ Take $\left.\pi=\frac{22}{7}\right]$

Sol :

Let r be the radius of hemisphere and cylinder
and height of the cylinder = h

Given: Volume of air $=\frac{880}{21} \mathrm{m}^{3}$
∴ Diameter of the building = 2r

Height of the building (H) = diameter of the building

∴ Height of the cylinder + Radius of hemispherical dome = 2r
⇒ h + r = 2r
⇒ h = 2r – r
⇒ h = r

Volume of air inside the building = Volume of hemispherical portion
+ Volume of cylindrical portion

$41 \frac{19}{21}=\frac{2}{3} \pi r^{3}+\pi r^{2} h$

$41 \frac{19}{21}=\frac{2}{3} \pi r^{3}+\pi r^{2}(r)$

$\frac{880}{21}=\frac{2}{3} \pi r^{3}+\pi r^{3}$

$\frac{880}{21}=\pi r^{3}\left(\frac{2}{3}+1\right)$

$\frac{880}{21}=\frac{22}{7} r^{3} \times \frac{5}{3}$

$\frac{880}{22 \times 5}=\mathrm{r}^{3}$
⇒ r³ = 8
⇒ r = 2
⇒ Height of building = 2r = 2×2 = 4m
Hence, the total height of the building is 4m.

Question 5

The interior of a building is in the form of a cylinder of diameter 4.3 m and height 3.8 m surmounted by a cone whose vertical angle is a right angle. Find the area of the surface and the volume of the building. [Take π=3.14]
Sol :

The diameter of a cylindrical portion BCDE of building = 4.3m

∴ The radius of a cylindrical portion$=\frac{4.3}{2}=2.15 \mathrm{m}$
Height = 3.8m

Lateral Surface Area of Cylindrical Portion BCDE = 2πrh
$=2 \times \frac{22}{7} \times 2.15 \times 3.8$
= 51.3543 m²
Let AB be the slant height of the conical portion of the building = l = AB = AC

Now, the Lateral surface of conical portion = πrl
$=\frac{22}{7} \times 2.15 \times 3.04$
= 20.5417 m²
So,
The total surface area of the building = Surface area of cylindrical portion + Surface area of the conical portion
= 51.3543 + 20.5417
= 71.8960
= 71.90 m² (approx.)

Now, In right ΔBAC,
BC² = AB² + AC²
⇒ BC² = l² + l²
⇒ (4.3)² = 2l²
⇒ 18.49 = 2l²
$\Rightarrow \mathrm{l}^{2}=\frac{18.49}{2}$
⇒ l² = 9.245
⇒ l = 3.04 m

Here, r is the radius of the cone and l is the slant height of the cone
⇒ l² = h² + r²
⇒ 9.245 = h² + (2.15)²
⇒ 9.245 = h² + 4.6225
⇒ h² = 9.245 – 4.6225
⇒ h² = 4.6225
⇒ h = 2.15 m
Now, Volume of the conical portion $=\frac{1}{3} \pi r^{2} h$
$=\frac{1}{3} \times \frac{22}{7} \times(2.15)^{2} \times 2.15$
= 10.4116 m³

Volume of cylindrical portion BCDE = πr²h
$=\frac{22}{7} \times(2.15)^{2} \times 3.8$
= 55.2059 m³
So,
The total volume of the building = Volume of the cylindrical portion
+ Volume of the conical portion
= 55.2059 + 10.4116
= 65.6175 m³
= 65.62 m³ (approx.)

Question 6

A tent of height 77 dm is in the form of a right circular cylinder of diameter 36m and height 44 dm surmounted by a right circular cone. Find the cost of the canvas at Rs 3.50 per m².

$\left[\right.$ Take $\left.\pi=\frac{22}{7}\right]$
Sol :

Given: Height of the tent = 77dm
$=\frac{77}{10}=7.7 \mathrm{m}$

The height of the conical portion = 44dm
$=\frac{44}{10}=4.4 \mathrm{m}$

∴ The height of the cylindrical portion
= Height of the tent – Height of the conical portion
= 7.7 – 4.4
= 3.3m

Given Diameter of the cylinder, d = 36m
$\therefore$ Radius $=\frac{36}{2}=18 \mathrm{m}$

CSA of cylindrical portion = 2πrh
$=2 \times \frac{22}{7} \times 18 \times 4.4$
= 497.828
= 497.83 m² (approx.)

Firstly, we have to find the slant height (l) of the conical portion
⇒ l² = h² + r²
⇒ l² = (3.3)² + (18)²
⇒ l² = 10.89 + 324
⇒ l² = 334.89
⇒ l = √334.89
⇒ l = 18.3m
∴ CSA of the conical portion = πrl
$=\frac{22}{7} \times 18 \times 18.3$
= 1035.257
= 1035.26 m² (approx.)
So,
Total Surface Area of the tent = Surface area of conical portion + surface Area of cylindrical portion
= 1035.26 + 497.83
= 1533.09 m²
∴ Canvas required to make the tent = 1533.09 m²
Cost of 1m² canvas = Rs 3.50
Cost of 1533.09m² canvas = Rs 3.50 × 1533.09
= Rs 5365.815
= Rs 5365.82

Question 7

A tent of height 3.3 m is in the form of a right circular cylinder of diameter 12m and height 2.2m, surmounted by a right circular cone of the same diameter. Find the cost of the canvas of the tent at the rate of Rs. 500 per m².

Sol :

Given: Height of the tent = 3.3 m
The height of the cylindrical portion = 2.2 m
∴ The height of the conical portion
= Height of the tent – Height of the cylindrical portion
= 3.3 – 2.2
= 1.1m
Given Diameter of the cylinder, d = 12m
$\therefore$ Radius $=\frac{12}{2}=6 \mathrm{m}$

CSA of cylindrical portion = 2πrh
$=2 \times \frac{22}{7} \times 6 \times 2.2$
= 82.971 m²

Firstly, we have to find the slant height (l) of the conical portion
⇒ l² = h² + r²
⇒ l² = (1.1)² + (6)²
⇒ l² = 1.21 + 36
⇒ l² = 37.21
⇒ l = √37.21
⇒ l = 6.1m
∴ CSA of the conical portion = πrl
$=\frac{22}{7} \times 6 \times 6.1$
= 115.029 m²
So,
Total Surface Area of the tent = Surface area of conical portion + surface Area of cylindrical portion
= 115.029 + 82.971
= 198 m²

∴ Canvas required to make the tent = 198 m²
Cost of 1m² canvas = Rs 500
Cost of 198 m² canvas = Rs 500 × 198
= Rs 99000

Question 8

A medicine capsule as shown in the given figure is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 12 mm, and the diameter of the capsule is 5 mm. Find its surface area.
Sol :

Given:
Diameter of cylinder = 5mm
Radius of cylinder $=\frac{\text { Diameter }}{2}=\frac{5}{2} \mathrm{mm}$
Height of cylinder = 12 – 5 = 7mm
Here, Diameter of hemisphere = 5mm

So,Radius of Hemisphere = $\frac{\text { Diameter }}{2}=\frac{5}{2} \mathrm{mm}$
Total area of the capsule = CSA of cylinder + CSA of 2 hemispheres
= 2πrh + 2×2πr²
$=2 \times \frac{22}{7} \times \frac{5}{2} \times 7+4 \times \frac{22}{7} \times \frac{5}{2} \times \frac{5}{2}$

$=\frac{22}{7}(35+25)$

$=\frac{22}{7} \times 60$

$=188 \frac{4}{7}$ sq. $\mathrm{mm}$

Question 9

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm, and the total height of the vessel is 13cm. Find the inner surface area of the vessel.

Sol :

Given: Diameter of the hemisphere = 14cm
∴ Radius of the hemisphere$=\frac{14}{2}=7 \mathrm{cm}$
Curved Surface Area of Hemisphere = 2πr²
$=2 \times \frac{22}{7} \times 7 \times 7$
= 308cm²
Now,
Radius of cylinder = radius of hemisphere = 7cm
Height of cylinder = Total height – Radius of hemisphere
= 13 – 7
= 6cm
So, CSA of cylinder = 2πrh
$=2 \times \frac{22}{7} \times 7 \times 6$
= 264cm²
The inner Surface area of the vessel
= CSA of hemisphere + CSA of cylinder
= 308 + 264
= 572cm²
Hence, Inner surface area of vessel is 572cm²

Question 10

From a circular cylinder of base diameter 10cm and height 12cm, a conical cavity with the same base and height is carved out. Find the volume of the remaining solid.
Sol :
Given that Diameter of circular cylinder = 10cm
∴ Radius of the cylinder$=\frac{10}{2}=5 \mathrm{cm}$
and Height of the cylinder = 12cm

So,
Volume of the cylinder = πr²h
$=\frac{22}{7} \times 5 \times 5 \times 12$
= 942.857 cm³

Volume of the cone $=\frac{1}{3} \pi r^{2} h$
$=\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12$
= 314.285 cm³

Remaining Volume = Volume of the cylinder – Volume of cone
= 942.857 – 314.285
= 628.572 cm³

Question 11

An ice-cream cone consists of a right circular cone of height 14cm and diameter of the circular top is 5 cm. It has hemisphere on the top with the same diameter as of circular top. Find the volume of ice-cream in the cone.
Sol :
Given: Height of the cone = 14cm
and diameter of the circular top = 5cm
$\therefore$ Radius $=\frac{5}{2} \mathrm{cm}$

Now,
Volume of the cone $=\frac{1}{3} \pi r^{2} h$
$=\frac{1}{3} \times \frac{22}{7} \times \frac{5}{2} \times \frac{5}{2} \times 14$
= 91.6666cm³

Now,
Volume of the hemisphere $=\frac{2}{3} \pi r^{3}$
$=\frac{2}{3} \times \frac{22}{7} \times\left(\frac{5}{2}\right)^{3}$
= 32.738 cm³

So,
The volume of ice cream in the cone
= Volume of cone + Volume of a hemisphere
= 91.6666 + 32.738
= 124.404cm³

Question 12

A student was asked to make a model in his workshop, which shaped like a cylinder with two cones attached at its two ends, using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 10 cm. If each cone has a height of 2cm, find the volume of air contained in the model. (Consider the outer and inner dimensions of the model to be nearly the same).
Sol :

Given that diameter of cylinder = 3cm
So, Radius $=\frac{3}{2}=1.5 \mathrm{cm}$
Given: Height of cone = 2cm
and
Height of the cylinder + height of cone + height of cone = 10cm
Height of the cylinder + 2 + 2 = 10
Height of the cylinder = 6cm
So,
Volume of the cylinder = πr²h
$=\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 6$
= 42.428 cm³
= 42.43 cm³ (approx.)

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

$=\frac{1}{3} \times \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 2$

= 4.71 cm³
Volume of 1^st cone = Volume of 2^nd cone = 4.71cm³
So, Total volume of the model
= Volume of cylinder + volume of 1^st cone + volume of 2^nd cone
= 42.43 + 4.71 + 4.71
= 51.85cm³

Question 13

A decorative block as shown in the given figure is made of a cube and a hemisphere. The edge of the cube is 10cm, and the radius of the hemisphere attached on the top is 3.5. Find the cost of painting the block at the rate of 50 paise per sq.cm.

Sol :

Given that side of a cube = 10cm
Total Surface Area of cube = 6(side)²
= 6 × 10 × 10
= 600cm²

Now, Radius of hemisphere = 3.5cm
Curved Surface Area of hemisphere = 2πr²

$=2 \times \frac{22}{7} \times 3.5 \times 3.5$
= 77cm²

Base area of hemisphere = Area of circle = πr²
$=\frac{22}{7} \times 3.5 \times 3.5$
= 38.5cm²
Therefore,
Total Surface Area of block = Total Surface area of a cube + CSA of the hemisphere - Base area of a hemisphere
= 600 + 77 – 38.5
= 638.5cm²

Cost of painting the block of 1cm² = Rs 0.50
Cost of painting the block of 638.5cm² = Rs 0.50 × 638.5
= Rs 319.25

Question 14

A godown building is in the form as shown in the adjoining figure. The vertical cross-section parallel to the width side of the building is a rectangle of size 7 m x 3 m mounted by a semicircle of radius 3.5 m. The inner measurement of the cuboidal portion are 10 m x 7 m x 3 m. Find the

(i) the volume of the godown, and
(ii) the total internal surface area excluding the floor.

Sol :
(i) Volume of godown = Volume of cuboid +$\frac{1}{2}$volume of cylinder
Volume of cuboid = l × b × h = 10 × 7 × 3 = 210 m³
Volume of cylinder = πr²h $=\frac{22}{7} \times(3.5)^{2} \times 10$

= 385 m³
∴ Volume of godown

$=210+\frac{1}{2} \times 384.65$
= 210 + 192.5
= 402.5 cm³

(ii) Internal Surface Area of a godown excluding the floor
$=\mathrm{CSA}$ of cuboid $+\frac{1}{2}$the total surface area of the cylinder
$=2 \mathrm{h}(1+\mathrm{b})+\frac{1}{2} \times 2 \pi \mathrm{r}(\mathrm{r}+\mathrm{h})$
$=2 \times 3(10+7)+\frac{22}{7} \times 3.5(3.5+10)$
= 6 × 17 + 22 × 0.5 × 13.5
= 102 + 148.5
= 250.5 cm²
Hence, internal surface area of godown excluding floor is 250.5cm²

Question 15

A solid iron pole is having a cylindrical portion 110 cm high and of base diameter, 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that mass of 1 cm³ of iron is 8 g.
Sol :

Given: Diameter of cone = 12cm
So, Radius of cone = 6cm
and Height of cone = 9cm

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

$=\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 9$

= 339.428 cm³

Radius of the cylinder, R = Radius of cone = 6cm
Height of the cylinder, H = 110cm

Volume of cylinder = πR²H
$=\frac{22}{7} \times 6 \times 6 \times 110$
= 12445.714cm³
Hence, the volume of the pole

= Volume of conical part + Volume of the cylindrical part
= 339.43 + 12445.714
= 12785.142 cm³

Required mass of pole = 8 × 12785.142
= 102281.13 gm
= 102.281kg

Question 16

A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in the given figure. The height of the entire rocket is 26 cm, while the height of the conical part is 6cm. The base of the conical portion has a diameter of 5cm, while the base diameter of the cylindrical portion is 3cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. [Take π=3.14]

Sol :

The area to be painted orange
= CSA of cone + Base area of Cone – Base area of the cylinder

Curved Surface Area of Cone
Given that diameter of conical portion = 5cm

The radius of conical portion

$=\frac{5}{2}=2.5 \mathrm{cm}$
Height of the conical part = h = 6cm
We need to find the ‘l’ first
We know that
l² = h² + r²
⇒ l² = 6² + (2.5)²
⇒ l² = 36 + (6.25)
⇒ l² = 42.25
⇒ l = √42.25
⇒ l = 6.5cm

So, CSA of conical portion = πrl
=3.14×2.5×6.5
= 51.025 cm²

Base area of the cone = πr²
=3.14×2.5×2.5
=19.625 cm²

Diameter of the cylinder = 3cm
So, Radius of the cylinder = 1.5cm
Base area of the cylinder = π(r’)²
=3.14×1.5×1.5
=7.065 cm²

So, the area to be painted orange = 51.025 + 19.625 – 7.065
= 63.585 cm²

Now, the area to be painted yellow
= CSA of the cylinder + Area of one bottom base of the cylinder
= 2πr’h’ + π(r’)²
= 2 × 3.14 × 1.5 × 20 + 7.065
= 188.4 + 7.065
= 195.465 cm²

Question 17

The inner diameter of glass is 7cm, and it has a raise portion in the bottom in the shape of a hemisphere as shown in the figure. If the height of the glass is 16 cm, find the apparent capacity and the actual capacity of the glass. $\left[\right.$ Take $\left.\pi=\frac{22}{7}\right]$

Sol :
Given the inner diameter of the glass = 7cm
So, the radius of the glass = $r=\frac{7}{2}=3.5 \mathrm{cm}$
Height of the glass = 16cm

The volume of the cylindrical glass = πr²h
$=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 16$
= 616cm³

Now, radius of the hemisphere = Radius of the cylinder =r = 3.5cm

Volume of the hemisphere$=\frac{2}{3} \pi r^{3}$
$=\frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5$
= 89.83 cm³

Now,
Apparent capacity of the glass = Volume of cylinder = 616cm³

The actual capacity of the glass
= Total volume of cylinder – Volume of hemisphere
= 616 – 89.83
= 526.17cm³

Hence,
Apparent Capacity of the glass = 616cm³
and actual capacity of the glass = 526.17cm³

S.no	Chapters	Links
1	Real numbers	Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4
2	Polynomials	Exercise 2.1 Exercise 2.2 Exercise 2.3
3	Pairs of Linear Equations in Two Variables	Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5
4	Trigonometric Ratios and Identities	Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4
5	Triangles	Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5
6	Statistics	Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
7	Quadratic Equations	Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5
8	Arithmetic Progressions (AP)	Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4
9	Some Applications of Trigonometry: Height and Distances	Exercise 9.1
10	Coordinates Geometry	Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4
11	Circles	Exercise 11.1 Exercise 11.2
12	Constructions	Exercise 12.1
13	Area related to Circles	Exercise 13.1
14	Surface Area and Volumes	Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4
15	Probability	Exercise 15.1