Volume & Surface Area of Solids

Given:

Length of the cuboid (l) = 24 cm

Breadth of the cuboid (b) = 16 cm

Height of the cuboid (h) = 7.5 cm

Volume of a cuboid = l × b × h

= 24 × 16 × 7.5 cm³

= 2880 cm³

As we know, the total surface area of the cuboid = 2(l × b + b × h + h × l)

= 2(24 × 16 + 16 × 7.5 + 7.5 × 24) cm²

= 2(384 + 120 + 180) cm²

= 2 × 684 cm²

= 1368 cm²

As we know, the lateral surface area of the cuboid = 2(l + b) × h

= 2(24 + 16) × 7.5 cm²

= 2 × 40 × 7.5 cm²

= 600 cm²

Hence, the volume is 2880 cm³, the total surface area is 1368 cm² and the lateral surface area is 600 cm².

Given:

Length of the cuboid (l) = 10 m

Breadth of the cuboid (b) = 35 cm = 0.35 m

Height of the cuboid (h) = 1.2 m

Volume of a cuboid = l × b × h

= 10 × 0.35 × 1.2 m³

= 4.2 m³

As we know, the total surface area of the cuboid = 2(l × b + b × h + h × l)

= 2(10 × 0.35 + 0.35 × 1.2 + 1.2 × 10) m²

= 2(3.5 + 0.42 + 12) m²

= 2 × 15.92 m²

= 31.84 m²

As we know, the lateral surface area of the cuboid = 2(l + b) × h

= 2(10 + 0.35) × 1.2 m²

= 2 × 10.35 × 1.2 m²

= 24.84 m²

Hence, the volume is 4.2 m³, the total surface area is 31.84 m² and the lateral surface area is 24.84 m².

Given:

Length of the cistern (l) = 6 m

Breadth of the cistern (b) = 2.5 m

Depth of the cistern (h) = 1.4 m

Capacity of the cistern = Volume of the cistern = l × b × h

= 6 × 2.5 × 1.4 m³

= 21 m³

Area of the iron sheet required = Total surface area of the cistern

As we know, the total surface area of the cuboid = 2(l × b + b × h + h × l)

= 2(6 × 2.5 + 2.5 × 1.4 + 1.4 × 6) m²

= 2(15 + 3.5 + 8.4) m²

= 2 × 26.9 m²

= 53.8 m²

Hence, the capacity of the cistern is 21 m³ and the area of the iron sheet required is 53.8 m².

Given:

Length of the wall = 13.5 m = 1350 cm

Width of the wall = 60 cm

Height of the wall = 1.6 m = 160 cm

Volume of the wall = length × width × height

= 1350 × 60 × 160 cm³

= 12960000 cm³

Length of the brick = 22.5 cm

Width of the brick = 12 cm

Height of the brick = 8 cm

Volume of one brick = length × width × height

= 22.5 × 12 × 8 cm³

= 2160 cm³

Let $a$ be the number of bricks.

Number of bricks = $\frac{\text{Volume of wall}}{\text{Volume of one brick}}$

$$\begin{aligned}\Rightarrow a = \frac{12960000}{2160}\\[1em] \Rightarrow a = 6000\end{aligned}$$

Hence, 6000 bricks will be needed to construct the wall.

Given:

Length of the place = 15 m = 1500 cm

Width of the place = 4 m = 400 cm

Depth of the place = 60 cm

Volume of the place = length × width × depth

= 1500 × 400 × 60 cm³

= 36000000 cm³

Length of the plank = 5 m = 500 cm

Width of the plank = 24 cm

Height of the plank = 10 cm

Volume of one plank = length × width × height

= 500 × 24 × 10 cm³

= 120000 cm³

Here, it is assumed that the planks can be packed in the place without leaving any unused space.

Let $a$ be the number of planks.

Number of planks = $\frac{\text{Volume of the place}}{\text{Volume of one plank}}$

$$\begin{aligned}\Rightarrow a = \frac{36000000}{120000}\\[1em] \Rightarrow a = 300\end{aligned}$$

Hence, 300 planks can be stored in the place.

Given:

Length of the classroom (l) = 10 m

Breadth of the classroom (b) = 6.4 m

Height of the classroom (h) = 5 m

Area of the floor of the classroom = l × b

= 10 × 6.4 m²

= 64 m²

Floor area given to each student = 1.6 m²

Number of students = $\frac{\text{Area of the floor}}{\text{Floor area for each student}} = \frac{64}{1.6} = 40$

Volume of the room = l × b × h

= 10 × 6.4 × 5 m³

= 320 m³

Air available to each student = $\frac{\text{Volume of the room}}{\text{Number of students}}$

= $\frac{320}{40}$ m³

= 8 m³

Hence, 40 students can be accommodated in the room and each student gets 8 m³ of air.

Given:

Length of the room ($l$) = 12 m

Breadth of the room ($b$) = 8 m

Height of the room ($h$) = 9 m

Length of the longest pole = Length of the diagonal of the room

The diagonal of the floor, the height of the room and the diagonal of the room form a right-angled triangle.

By Pythagoras' theorem,

(Diagonal of the floor)² = $l^2 + b^2$

(Diagonal of the room)² = (Diagonal of the floor)² + $h^2 = l^2 + b^2 + h^2$

Length of the diagonal of the room ($d$) = $\sqrt{l^2 + b^2 + h^2}$

$$\begin{aligned}= \sqrt{12^2 + 8^2 + 9^2}\\[1em] = \sqrt{144 + 64 + 81}\\[1em] = \sqrt{289}\\[1em] = 17 \text{ m}\end{aligned}$$

Hence, the length of the longest pole that can be placed in the room is 17 m.

Given:

Volume of the cuboid = 972 m³

Length of the cuboid ($l$) = 16 m

Breadth of the cuboid ($b$) = 13.5 m

Let $h$ be the height of the cuboid.

Volume of a cuboid = $l \times b \times h$

$$\begin{aligned}\Rightarrow 16 \times 13.5 \times h = 972\\[1em] \Rightarrow 216 \times h = 972\\[1em] \Rightarrow h = \frac{972}{216}\\[1em] \Rightarrow h = 4.5 \text{ m}\end{aligned}$$

Hence, the height of the cuboid is 4.5 m.

Given:

Volume of the cuboid = 1296 m³

Length of the cuboid ($l$) = 24 m

The breadth and height of the cuboid are in the ratio 3 : 2.

Let breadth of the cuboid ($b$) = $3a$ m and height of the cuboid ($h$) = $2a$ m, where $a \gt 0$.

Volume of a cuboid = $l \times b \times h$

$$\begin{aligned}\Rightarrow 24 \times 3a \times 2a = 1296\\[1em] \Rightarrow 144a^2 = 1296\\[1em] \Rightarrow a^2 = \frac{1296}{144}\\[1em] \Rightarrow a^2 = 9\end{aligned}$$

Since lengths cannot be negative, we take the positive square root.

$$\begin{aligned}\Rightarrow a = \sqrt{9}\\[1em] \Rightarrow a = 3\end{aligned}$$

Thus, breadth of the cuboid = $3a$ = 3 × 3 m = 9 m

Height of the cuboid = $2a$ = 2 × 3 m = 6 m

Hence, the breadth of the cuboid is 9 m and the height is 6 m.

Given:

Surface area of the cuboid = 468 cm²

Length of the cuboid ($l$) = 12 cm

Breadth of the cuboid ($b$) = 9 cm

Let $h$ be the height of the cuboid.

As we know, the total surface area of the cuboid = $2(l \times b + b \times h + h \times l)$

$$\begin{aligned}\Rightarrow 2(12 \times 9 + 9 \times h + h \times 12) = 468\\[1em] \Rightarrow 12 \times 9 + 9 \times h + h \times 12 = \frac{468}{2}\\[1em] \Rightarrow 108 + 9h + 12h = 234\\[1em] \Rightarrow 21h = 234 - 108\\[1em] \Rightarrow 21h = 126\\[1em] \Rightarrow h = \frac{126}{21}\\[1em] \Rightarrow h = 6 \text{ cm}\end{aligned}$$

Hence, the height of the cuboid is 6 cm.

(i) Given:

Length of the room (l) = 8 m

Breadth of the room (b) = 6.5 m

Height of the room (h) = 3.5 m

As we know, the area of 4 walls of the room = 2(l + b) × h

= 2(8 + 6.5) × 3.5 m²

= 2 × 14.5 × 3.5 m²

= 101.5 m²

Hence, the area of 4 walls of the room is 101.5 m².

(ii) Area of the floor of the room = l × b

= 8 × 6.5 m²

= 52 m²

Hence, the area of the floor of the room is 52 m².

(i) Given:

Length of the room (l) = 9 m

Breadth of the room (b) = 6 m

Height of the room (h) = 3.6 m

As we know, the area of 4 walls of the room = 2(l + b) × h

= 2(9 + 6) × 3.6 m²

= 2 × 15 × 3.6 m²

= 108 m²

Area of one door = 1.4 × 2 m² = 2.8 m²

Area of two windows = 2 × (1.6 × 0.75) m²

= 2 × 1.2 m²

= 2.4 m²

Area not to be white-washed = (Area of 1 door) + (Area of 2 windows)

= 2.8 + 2.4 m²

= 5.2 m²

Area of 4 walls, excluding doors and windows = 108 - 5.2 m²

= 102.8 m²

Hence, the area of 4 walls, excluding doors and windows, is 102.8 m².

(ii) Area to be white-washed = 102.8 m²

Rate of white-washing = ₹ 22.50 per m²

Cost of white-washing = Area to be white-washed × Rate of white-washing

= 102.8 × ₹ 22.50

= ₹ 2,313

Hence, the cost of white-washing the walls from inside is ₹ 2,313.

(iii) Area of the ceiling = l × b

= 9 × 6 m²

= 54 m²

Rate of painting = ₹ 25 per m²

Cost of painting the ceiling = Area of the ceiling × Rate of painting

= 54 × ₹ 25

= ₹ 1,350

Hence, the cost of painting the ceiling is ₹ 1,350.

(i) Given:

Length of the assembly hall (l) = 45 m

Breadth of the assembly hall (b) = 30 m

Height of the assembly hall (h) = 16 m

As we know, the area of 4 walls of the hall = 2(l + b) × h

= 2(45 + 30) × 16 m²

= 2 × 75 × 16 m²

= 2400 m²

Area of five doors = 5 × (4 × 3.5) m²

= 5 × 14 m²

= 70 m²

Area of four windows = 4 × (2.5 × 1.6) m²

= 4 × 4 m²

= 16 m²

Area not to be papered = (Area of 5 doors) + (Area of 4 windows)

= 70 + 16 m²

= 86 m²

Area to be papered = 2400 - 86 m²

= 2314 m²

Rate of papering = ₹ 35 per m²

Cost of papering the walls = Area to be papered × Rate of papering

= 2314 × ₹ 35

= ₹ 80,990

Hence, the cost of papering the walls of the assembly hall is ₹ 80,990.

(ii) Area of the floor of the hall = l × b

= 45 × 30 m²

= 1350 m²

Rate of carpeting = ₹ 154 per m²

Cost of carpeting the floor = Area of the floor × Rate of carpeting

= 1350 × ₹ 154

= ₹ 2,07,900

Hence, the cost of carpeting the floor of the assembly hall is ₹ 2,07,900.

Given:

The length, breadth and height of a cuboid are in the ratio 7 : 6 : 5.

Surface area of the cuboid = 1926 cm²

Let the length, breadth and height of the cuboid be $7a, 6a$ and $5a$.

As we know, the total surface area of the cuboid = 2(l × b + b × h + h × l)

$$\begin{aligned}\Rightarrow 2(7a \times 6a + 6a \times 5a + 5a \times 7a) = 1926\\[1em] \Rightarrow 2(42a^2 + 30a^2 + 35a^2) = 1926\\[1em] \Rightarrow 2 \times 107a^2 = 1926\\[1em] \Rightarrow 214a^2 = 1926\\[1em] \Rightarrow a^2 = \frac{1926}{214}\\[1em] \Rightarrow a^2 = 9\\[1em] \Rightarrow a = \sqrt{9}\\[1em] \Rightarrow a = 3\end{aligned}$$

Thus, length of the cuboid (l) = $7a$ = 7 × 3 cm = 21 cm

Breadth of the cuboid (b) = $6a$ = 6 × 3 cm = 18 cm

Height of the cuboid (h) = $5a$ = 5 × 3 cm = 15 cm

Volume of a cuboid = l × b × h

= 21 × 18 × 15 cm³

= 5670 cm³

Hence, the dimensions of the cuboid are 21 cm, 18 cm and 15 cm and the volume is 5670 cm³.

Let $l, b$ and $h$ be the length, breadth and height of the cuboidal box.

Then, $l \times b$ = 120 cm², $b \times h$ = 72 cm² and $l \times h$ = 60 cm².

Multiplying the three areas, we get :

$$\begin{aligned}\Rightarrow (l \times b) \times (b \times h) \times (l \times h) = 120 \times 72 \times 60\\[1em] \Rightarrow l^2 \times b^2 \times h^2 = 518400\\[1em] \Rightarrow (lbh)^2 = 518400\\[1em] \Rightarrow lbh = \sqrt{518400}\\[1em] \Rightarrow lbh = 720\end{aligned}$$

Volume of a cuboid = $l \times b \times h$

= 720 cm³

Hence, the volume of the box is 720 cm³.

Given:

Depth of the river = 2 m

Width of the river = 40 m

Rate of flow of the river = 4.5 km per hour = 4.5 × 1000 m per hour = 4500 m per hour

The water that runs into the sea in 1 hour forms a cuboid whose length is 4500 m, breadth is 40 m and height is 2 m.

Volume of water flowing in 1 hour = length × width × depth

= 4500 × 40 × 2 m³

= 360000 m³

As we know, 1 hour = 60 minutes

Volume of water flowing in 1 minute = $\frac{360000}{60}$ m³

= 6000 m³

Hence, 6000 m³ of water runs into the sea per minute.

(i) Given:

External length of the box (L) = 80 cm

External width of the box (B) = 65 cm

External height of the box (H) = 45 cm

Thickness of wood = 2.5 cm

As the box is closed, the thickness is to be subtracted from both the ends of each dimension.

Internal length of the box (l) = (80 - 2 × 2.5) cm = (80 - 5) cm = 75 cm

Internal width of the box (b) = (65 - 2 × 2.5) cm = (65 - 5) cm = 60 cm

Internal height of the box (h) = (45 - 2 × 2.5) cm = (45 - 5) cm = 40 cm

Capacity of the box = Internal volume of the box = l × b × h

= 75 × 60 × 40 cm³

= 180000 cm³

Hence, the capacity of the box is 180000 cm³.

(ii) External volume of the box = L × B × H

= 80 × 65 × 45 cm³

= 234000 cm³

Volume of wood = (External volume) - (Internal volume)

= 234000 - 180000 cm³

= 54000 cm³

Weight of 100 cm³ of wood = 8 grams

Weight of the box = $\frac{54000}{100}$ × 8 grams

= 540 × 8 grams

= 4320 grams

= $\frac{4320}{1000}$ kg

= 4.32 kg

Hence, the weight of the box is 4.32 kg.

(i) Given:

External length of the box (L) = 54 cm

External width of the box (B) = 30 cm

External height of the box (H) = 16 cm

Thickness of wood = 2 cm

Since the box is open at the top, the thickness is to be subtracted from both the ends of the length and the width, but only from the bottom of the height.

Internal length of the box (l) = (54 - 2 × 2) cm = (54 - 4) cm = 50 cm

Internal width of the box (b) = (30 - 2 × 2) cm = (30 - 4) cm = 26 cm

Internal height of the box (h) = (16 - 2) cm = 14 cm

Capacity of the box = Internal volume of the box = l × b × h

= 50 × 26 × 14 cm³

= 18200 cm³

Hence, the capacity of the box is 18200 cm³.

(ii) External volume of the box = L × B × H

= 54 × 30 × 16 cm³

= 25920 cm³

Volume of wood = (External volume) - (Internal volume)

= 25920 - 18200 cm³

= 7720 cm³

Hence, the volume of wood is 7720 cm³.

Given:

Internal length of the box (l) = 24 cm

Internal width of the box (b) = 18 cm

Internal height of the box (h) = 12 cm

Thickness of iron = 1 cm

Internal volume of the box = l × b × h

= 24 × 18 × 12 cm³

= 5184 cm³

As the box is closed, the thickness is to be added to both the ends of each dimension.

External length of the box (L) = (24 + 2 × 1) cm = (24 + 2) cm = 26 cm

External width of the box (B) = (18 + 2 × 1) cm = (18 + 2) cm = 20 cm

External height of the box (H) = (12 + 2 × 1) cm = (12 + 2) cm = 14 cm

External volume of the box = L × B × H

= 26 × 20 × 14 cm³

= 7280 cm³

Volume of iron = (External volume) - (Internal volume)

= 7280 - 5184 cm³

= 2096 cm³

Hence, the volume of iron in the box is 2096 cm³.

Given:

Edge of the cube (a) = 8 m

Volume of a cube = a³

= 8³ m³

= 512 m³

As we know, the total surface area of the cube = 6a²

= 6 × 8² m²

= 6 × 64 m²

= 384 m²

As we know, the lateral surface area of the cube = 4a²

= 4 × 8² m²

= 4 × 64 m²

= 256 m²

As we know, the diagonal of the cube = a$\sqrt{3}$

= 8 × 1.73 m

= 13.84 m

Hence, the volume is 512 m³, the total surface area is 384 m², the lateral surface area is 256 m² and the diagonal is 13.84 m.

Given:

Edge of the cube (a) = 6.5 cm

Volume of a cube = a³

= (6.5)³ cm³

= 274.625 cm³

As we know, the total surface area of the cube = 6a²

= 6 × (6.5)² cm²

= 6 × 42.25 cm²

= 253.5 cm²

As we know, the lateral surface area of the cube = 4a²

= 4 × (6.5)² cm²

= 4 × 42.25 cm²

= 169 cm²

As we know, the diagonal of the cube = a$\sqrt{3}$

= 6.5 × 1.73 cm

= 11.245 cm

Hence, the volume is 274.625 cm³, the total surface area is 253.5 cm², the lateral surface area is 169 cm² and the diagonal is 11.245 cm.

Given:

Edge of the cube (a) = 2 cm 6 mm = 2.6 cm

Volume of a cube = a³

= (2.6)³ cm³

= 17.576 cm³

As we know, the total surface area of the cube = 6a²

= 6 × (2.6)² cm²

= 6 × 6.76 cm²

= 40.56 cm²

As we know, the lateral surface area of the cube = 4a²

= 4 × (2.6)² cm²

= 4 × 6.76 cm²

= 27.04 cm²

As we know, the diagonal of the cube = a$\sqrt{3}$

= 2.6 × 1.73 cm

= 4.498 cm

Hence, the volume is 17.576 cm³, the total surface area is 40.56 cm², the lateral surface area is 27.04 cm² and the diagonal is 4.498 cm.

Given:

Surface area of the cube = 1176 cm²

Let $s$ be the side of the cube.

Surface area of a cube = $6s^2$

$$\begin{aligned}\Rightarrow 6 \times s^2 = 1176\\[1em] \Rightarrow s^2 = \frac{1176}{6}\\[1em] \Rightarrow s^2 = 196\\[1em] \Rightarrow s = \sqrt{196}\\[1em] \Rightarrow s = 14 \text{ cm}\end{aligned}$$

Volume of a cube = $s^3$

= 14³ cm³

= 2744 cm³

Hence, the volume of the cube is 2744 cm³.

Given:

Volume of the cube = 216 cm³

Let $s$ be the side of the cube.

Volume of a cube = $s^3$

$$\begin{aligned}\Rightarrow s^3 = 216\\[1em] \Rightarrow s = \sqrt[3]{216}\\[1em] \Rightarrow s = 6 \text{ cm}\end{aligned}$$

Surface area of a cube = $6s^2$

= 6 × 6² cm²

= 6 × 36 cm²

= 216 cm²

Hence, the surface area of the cube is 216 cm².

Given:

Volume of the cube = 343 cm³

Let $s$ be the side of the cube.

Volume of a cube = $s^3$

$$\begin{aligned}\Rightarrow s^3 = 343\\[1em] \Rightarrow s = \sqrt[3]{343}\\[1em] \Rightarrow s = 7 \text{ cm}\end{aligned}$$

Surface area of a cube = $6s^2$

= 6 × 7² cm²

= 6 × 49 cm²

= 294 cm²

Hence, the surface area of the cube is 294 cm².

Given:

The dimensions of the solid cuboid are 24 cm × 18 cm × 4 cm.

The cuboid is melted and recast into a cube, i.e., the volume of both the solids remains the same.

Let $s$ be the edge of the cube.

Volume of the cube = Volume of the cuboid

$$\begin{aligned}\Rightarrow s^3 = 24 \times 18 \times 4 \text{ cm}^3\\[1em] \Rightarrow s^3 = 1728 \text{ cm}^3\\[1em] \Rightarrow s = \sqrt[3]{1728} \text{ cm}\\[1em] \Rightarrow s = 12 \text{ cm}\end{aligned}$$

Hence, the length of each edge of the cube is 12 cm.

Given:

The edges of the three cubes are 5 cm, 4 cm and 3 cm.

The three cubes are melted to form a single cube, i.e., the sum of the volumes of the three cubes is equal to the volume of the new cube.

Let $x$ be the edge of the new cube.

Sum of the volumes of the three cubes = Volume of the new cube

$$\begin{aligned}\Rightarrow 5^3 + 4^3 + 3^3 = x^3\\[1em] \Rightarrow 125 + 64 + 27 = x^3\\[1em] \Rightarrow x^3 = 216\\[1em] \Rightarrow x = \sqrt[3]{216}\\[1em] \Rightarrow x = 6 \text{ cm}\end{aligned}$$

Lateral surface area of the new cube = $4x^2$

= 4 × 6² cm²

= 4 × 36 cm²

= 144 cm²

Hence, the lateral surface area of the new cube formed is 144 cm².

Given:

Radius of the cylinder (r) = 10 cm

Height of the cylinder (h) = 42 cm

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 10^2 \times 42\\[1em] = \frac{22}{7} \times 100 \times 42\\[1em] = \frac{92400}{7}\\[1em] = 13200 \text{ cm}^3\end{aligned}$$

Hence, the volume of the cylinder is 13200 cm³.

Given:

Diameter of the cylinder = 70 cm

Radius of the cylinder (r) = $\frac{70}{2}$ cm = 35 cm = 0.35 m

Height of the cylinder (h) = 20 m

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times (0.35)^2 \times 20\\[1em] = \frac{22}{7} \times 0.1225 \times 20\\[1em] = \frac{53.9}{7}\\[1em] = 7.7 \text{ m}^3\end{aligned}$$

Hence, the volume of the cylinder is 7.7 m³.

Given:

Radius of the cylinder (r) = 14 cm

Height of the cylinder (h) = 30 cm

Curved surface area of a cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 14 \times 30\\[1em] = \frac{18480}{7}\\[1em] = 2640 \text{ cm}^2\end{aligned}$$

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

$$\begin{aligned}= 2 \times \frac{22}{7} \times 14 \times (14 + 30)\\[1em] = 2 \times \frac{22}{7} \times 14 \times 44\\[1em] = \frac{27104}{7}\\[1em] = 3872 \text{ cm}^2\end{aligned}$$

Hence, the curved surface area is 2640 cm² and the total surface area is 3872 cm².

Given:

Radius of the cylinder (r) = 10.5 cm

Height of the cylinder (h) = 32 cm

Curved surface area of a cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 10.5 \times 32\\[1em] = \frac{14784}{7}\\[1em] = 2112 \text{ cm}^2\end{aligned}$$

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

$$\begin{aligned}= 2 \times \frac{22}{7} \times 10.5 \times (10.5 + 32)\\[1em] = 2 \times \frac{22}{7} \times 10.5 \times 42.5\\[1em] = \frac{19635}{7}\\[1em] = 2805 \text{ cm}^2\end{aligned}$$

Hence, the curved surface area is 2112 cm² and the total surface area is 2805 cm².

Given:

Radius of the cylinder (r) = 21 cm

Height of the cylinder (h) = 35 cm

Curved surface area of a cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 21 \times 35\\[1em] = \frac{32340}{7}\\[1em] = 4620 \text{ cm}^2\end{aligned}$$

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

$$\begin{aligned}= 2 \times \frac{22}{7} \times 21 \times (21 + 35)\\[1em] = 2 \times \frac{22}{7} \times 21 \times 56\\[1em] = \frac{51744}{7}\\[1em] = 7392 \text{ cm}^2\end{aligned}$$

Hence, the curved surface area is 4620 cm² and the total surface area is 7392 cm².

Given:

Radius of the cylinder (r) = 35 cm

Height of the cylinder (h) = 5 m = 500 cm

Curved surface area of a cylinder = 2πrh

$$\begin{aligned}= 2 \times \frac{22}{7} \times 35 \times 500\\[1em] = \frac{770000}{7}\\[1em] = 110000 \text{ cm}^2\end{aligned}$$

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

$$\begin{aligned}= 2 \times \frac{22}{7} \times 35 \times (35 + 500)\\[1em] = 2 \times \frac{22}{7} \times 35 \times 535\\[1em] = \frac{823900}{7}\\[1em] = 117700 \text{ cm}^2\end{aligned}$$

Hence, the curved surface area is 110000 cm² and the total surface area is 117700 cm².

Given:

Radius of the cylinder (r) = 10.5 cm

Height of the cylinder (h) = 60 cm

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times (10.5)^2 \times 60\\[1em] = \frac{22}{7} \times 110.25 \times 60\\[1em] = \frac{145530}{7}\\[1em] = 20790\end{aligned}$$

Volume of the cylinder = 20790 cm³

Weight of 1 cm³ of the material = 5 grams

Weight of the cylinder = Volume of the cylinder × Weight of 1 cm³ of the material

= 20790 × 5 grams

= 103950 grams

= $\frac{103950}{1000}$ kg

= 103.95 kg

Hence, the weight of the solid cylinder is 103.95 kg.

Given:

Capacity of the cylindrical tank = 6160 m³

Radius of the cylindrical tank ($r$) = 14 m

Let $h$ be the depth of the tank.

Volume of a cylinder = $\pi r^2h$

$$\begin{aligned}\Rightarrow \frac{22}{7} \times 14^2 \times h = 6160\\[1em] \Rightarrow \frac{22}{7} \times 196 \times h = 6160\\[1em] \Rightarrow 616 \times h = 6160\\[1em] \Rightarrow h = \frac{6160}{616}\\[1em] \Rightarrow h = 10\end{aligned}$$

Depth of the tank ($h$) = 10 m

Curved surface area of a cylinder = $2\pi rh$

$$\begin{aligned}= 2 \times \frac{22}{7} \times 14 \times 10\\[1em] = \frac{6160}{7}\\[1em] = 880\end{aligned}$$

Curved surface area of the tank = 880 m²

Rate of painting = ₹ 30 per m²

Cost of painting = Curved surface area × Rate of painting

= 880 × ₹ 30

= ₹ 26,400

Hence, the depth of the tank is 10 m and the cost of painting its curved surface is ₹ 26,400.

Given:

Curved surface area of the cylinder = 4400 cm²

Circumference of its base = 110 cm

Let $r$ be the radius and $h$ be the height of the cylinder.

Circumference of the base = $2\pi r$

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r = 110\\[1em] \Rightarrow \frac{44}{7} \times r = 110\\[1em] \Rightarrow r = \frac{110 \times 7}{44}\\[1em] \Rightarrow r = \frac{770}{44}\\[1em] \Rightarrow r = 17.5\end{aligned}$$

Radius of the cylinder ($r$) = 17.5 cm

Curved surface area of a cylinder = $2\pi rh$ = (Circumference of the base) × $h$

$$\begin{aligned}\Rightarrow 110 \times h = 4400\\[1em] \Rightarrow h = \frac{4400}{110}\\[1em] \Rightarrow h = 40\end{aligned}$$

Height of the cylinder ($h$) = 40 cm

Volume of a cylinder = $\pi r^2h$

$$\begin{aligned}= \frac{22}{7} \times (17.5)^2 \times 40\\[1em] = \frac{22}{7} \times 306.25 \times 40\\[1em] = \frac{269500}{7}\\[1em] = 38500 \text{ cm}^3\end{aligned}$$

Hence, the height of the cylinder is 40 cm and its volume is 38500 cm³.

Given:

Total surface area of the solid cylinder = 462 cm²

Curved surface area of the cylinder = $\frac{1}{3}$ × Total surface area

Let $r$ be the radius and $h$ be the height of the cylinder.

Curved surface area of the cylinder = $\frac{462}{3}$ cm² = 154 cm²

As we know, the total surface area of cylinder = $2\pi rh + 2\pi r^2$

$$\begin{aligned}\Rightarrow 154 + 2\pi r^2 = 462\\[1em] \Rightarrow 2\pi r^2 = 462 - 154\\[1em] \Rightarrow 2\pi r^2 = 308 \\[1em] \Rightarrow 2 \times \frac{22}{7} \times r^2 = 308\\[1em] \Rightarrow \frac{44}{7} \times r^2 = 308\\[1em] \Rightarrow r^2 = \frac{308 \times 7}{44}\\[1em] \Rightarrow r^2 = 49\\[1em] \Rightarrow r = \sqrt{49}\\[1em] \Rightarrow r = 7\end{aligned}$$

Radius of the cylinder ($r$) = 7 cm

Curved surface area of a cylinder = $2\pi rh$

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times 7 \times h = 154\\[1em] \Rightarrow \frac{44}{\cancel{7}} \times \cancel{7} \times h = 154\\[1em] \Rightarrow 44 \times h = 154\\[1em] \Rightarrow h = \frac{154}{44}\\[1em] \Rightarrow h = 3.5\end{aligned}$$

Height of the cylinder ($h$) = 3.5 cm

Volume of a cylinder = $\pi r^2h$

$$\begin{aligned}= \frac{22}{7} \times 7^2 \times 3.5\\[1em] = \frac{22}{7} \times 49 \times 3.5\\[1em] = \frac{3773}{7}\\[1em] = 539 \text{ cm}^3\end{aligned}$$

Hence, the volume of the cylinder is 539 cm³.

Given:

Sum of the radius of the base and the height of the cylinder = 37 m

Total surface area of the cylinder = 1628 m²

Let $r$ be the radius and $h$ be the height of the cylinder.

So, $r + h$ = 37

Total surface area of a cylinder = $2\pi r(r + h)$

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r \times 37 = 1628\\[1em] \Rightarrow \frac{1628}{7} \times r = 1628\\[1em] \Rightarrow r = \frac{1628 \times 7}{1628}\\[1em] \Rightarrow r = \frac{\cancel{1628} \times 7}{\cancel{1628}}\\[1em] \Rightarrow r = 7\end{aligned}$$

Radius of the cylinder ($r$) = 7 m

It is given that $r + h$ = 37

$$\begin{aligned}\Rightarrow 7 + h = 37\\[1em] \Rightarrow h = 37 - 7\\[1em] \Rightarrow h = 30\end{aligned}$$

Height of the cylinder ($h$) = 30 m

Volume of a cylinder = $\pi r^2h$

$$\begin{aligned}= \frac{22}{7} \times 7^2 \times 30\\[1em] = \frac{22}{7} \times 49 \times 30\\[1em] = \frac{32340}{7}\\[1em] = 4620 \text{ m}^3\end{aligned}$$

Hence, the volume of the cylinder is 4620 m³.

Given:

Total surface area of the circular cylinder = 1320 cm²

Radius of the cylinder ($r$) = 10 cm

Let $h$ be the height of the cylinder.

Total surface area of a cylinder = $2\pi r(r + h)$

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times 10 \times (10 + h) = 1320\\[1em] \Rightarrow \frac{440}{7} \times (10 + h) = 1320\\[1em] \Rightarrow (10 + h) = \frac{1320 \times 7}{440}\\[1em] \Rightarrow (10 + h) = 21\\[1em] \Rightarrow h = 21 - 10\\[1em] \Rightarrow h = 11 \text{ cm}\end{aligned}$$

Hence, the height of the cylinder is 11 cm.

Given:

Length of the swimming pool (l) = 70 m

Breadth of the swimming pool (b) = 44 m

Depth of the swimming pool (h) = 3 m

Diameter of the pipe = 35 cm = 0.35 m

Rate of flow of water = 6 m per second

Volume of the pool = l × b × h

= 70 × 44 × 3 m³

= 9240 m³

Radius of the pipe (r) = $\frac{0.35}{2}$ m = 0.175 m

The water issuing from the pipe in 1 second forms a cylinder of radius 0.175 m whose length is 6 m (the distance the water flows in 1 second).

Volume of water released in 1 second = π × (0.175)² × 6 m³

$$\begin{aligned}= \frac{22}{7} \times (0.175)^2 \times 6\\[1em] = \frac{22}{7} \times 0.030625 \times 6\\[1em] = \frac{4.0425}{7}\\[1em] = 0.5775\end{aligned}$$

Volume of water released in 1 second = 0.5775 m³

Time taken to fill the pool = $\frac{\text{Volume of the pool}}{\text{Volume of water released in 1 second}}$

= $\frac{9240}{0.5775}$ seconds

= 16000 seconds

As we know, 1 hour = 3600 seconds

Time taken to fill the pool = $\frac{16000}{3600}$ hours

= $\frac{40}{9}$ hours

= $4\frac{4}{9}$ hours

Hence, the pool is filled in $\mathbf{4\frac{4}{9}}$ hours.

Given:

Depth of the well (d) = 20 m

Diameter of the well = 7 m

Radius of the well (r) = $\frac{7}{2}$ m = 3.5 m

Dimensions of the platform = 22 m × 14 m

The well is in the shape of a cylinder.

Volume of earth dug out = Volume of the well = πr²d

$$\begin{aligned}= \frac{22}{7} \times (3.5)^2 \times 20\\[1em] = \frac{22}{7} \times 12.25 \times 20\\[1em] = \frac{5390}{7}\\[1em] = 770\end{aligned}$$

Volume of earth dug out = 770 m³

The earth dug out is spread evenly to form a cuboidal platform.

Let $H$ be the height of the platform.

Volume of the platform = Volume of earth dug out

$$\begin{aligned}\Rightarrow 22 \times 14 \times H = 770\\[1em] \Rightarrow 308 \times H = 770\\[1em] \Rightarrow H = \frac{770}{308}\\[1em] \Rightarrow H = 2.5 \text{ m}\end{aligned}$$

Hence, the height of the platform is 2.5 m.

Given:

Inside diameter of the well = 10 m

Inside radius of the well (r) = $\frac{10}{2}$ m = 5 m

Depth of the well (h) = 8.4 m

Width of the embankment = 7.5 m

Volume of earth dug out = Volume of the well = πr²h

$$\begin{aligned}= \frac{22}{7} \times 5 \times 5 \times 8.4\\[1em] = \frac{22}{7} \times 25 \times 8.4\\[1em] = \frac{4620}{7}\\[1em] = 660\end{aligned}$$

Volume of earth dug out = 660 m³

The embankment is in the shape of a hollow cylinder whose inner radius is 5 m.

Outer radius of the embankment (R) = (5 + 7.5) m = 12.5 m

Area of the embankment = π(R² - r²)

$$\begin{aligned}= \frac{22}{7} \times \lbrace(12.5)^2 - 5^2\rbrace\\[1em] = \frac{22}{7} \times (156.25 - 25)\\[1em] = \frac{22}{7} \times 131.25\\[1em] = \frac{2887.5}{7}\\[1em] = 412.5\end{aligned}$$

Area of the embankment = 412.5 m²

Height of the embankment = $\frac{\text{Volume of earth dug out}}{\text{Area of the embankment}}$

= $\frac{660}{412.5}$ m

= 1.6 m

Hence, the height of the embankment is 1.6 m.

Given:

Length of the cuboid (l) = 2 m

Breadth of the cuboid (b) = 2 m

Height of the cuboid (h) = 1 m

As we know, the total surface area of the cuboid = 2(l × b + b × h + h × l)

= 2(2 × 2 + 2 × 1 + 1 × 2) m²

= 2(4 + 2 + 2) m²

= 2 × 8 m²

= 16 m²

Hence, option 3 is the correct option.

Given:

The length, breadth and height of the cuboid are in the ratio 6 : 5 : 4.

Total surface area of the cuboid = 33300 cm²

Let the length, breadth and height of the cuboid be $6a, 5a$ and $4a$.

As we know, the total surface area of the cuboid = 2(l × b + b × h + h × l)

$$\begin{aligned}\Rightarrow 2(6a \times 5a + 5a \times 4a + 4a \times 6a) = 33300\\[1em] \Rightarrow 2(30a^2 + 20a^2 + 24a^2) = 33300\\[1em] \Rightarrow 2 \times 74a^2 = 33300\\[1em] \Rightarrow 148a^2 = 33300\\[1em] \Rightarrow a^2 = \frac{33300}{148}\\[1em] \Rightarrow a^2 = 225\\[1em] \Rightarrow a = \sqrt{225}\\[1em] \Rightarrow a = 15\end{aligned}$$

Breadth of the cuboid = $5a$ = 5 × 15 cm = 75 cm

Hence, option 2 is the correct option.

Given:

Width of the sand box (b) = 5 m

Length of the sand box (l) = 2 m

Depth upto which it is to be filled (h) = 10 cm = 0.1 m

Volume of sand needed = l × b × h

= 2 × 5 × 0.1 m³

= 1 m³

Hence, option 1 is the correct option.

Given:

Length of the beam (l) = 9 m

Width of the beam (b) = 40 cm = 0.4 m

Depth of the beam (h) = 20 cm = 0.2 m

Volume of a cuboid = l × b × h

= 9 × 0.4 × 0.2 m³

= 0.72 m³

Weight of 1 m³ of iron = 50 kg

Weight of the beam = Volume of the beam × Weight of 1 m³ of iron

= 0.72 × 50 kg

= 36 kg

Hence, option 2 is the correct option.

Given:

Length of the room ($l$) = 30 m

Breadth of the room ($b$) = 24 m

Height of the room ($h$) = 18 m

Length of the longest rod = Length of the diagonal of the room

The diagonal of the floor, the height of the room and the diagonal of the room form a right-angled triangle.

By Pythagoras' theorem,

(Diagonal of the floor)² = $l^2 + b^2$

(Diagonal of the room)² = (Diagonal of the floor)² + $h^2 = l^2 + b^2 + h^2$

Length of the diagonal of the room (d) = $\sqrt{l^2 + b^2 + h^2}$

$$\begin{aligned}= \sqrt{30^2 + 24^2 + 18^2}\\[1em] = \sqrt{900 + 576 + 324}\\[1em] = \sqrt{1800}\\[1em] = \sqrt{900 \times 2}\\[1em] = 30\sqrt{2} \text{ m}\end{aligned}$$

Hence, option 3 is the correct option.

Given:

Length of the wall = 15 m = 1500 cm

Height of the wall = 1.8 m = 180 cm

Thickness of the wall = 37.5 cm

Volume of the wall = length × thickness × height

= 1500 × 37.5 × 180 cm³

= 10125000 cm³

Volume of one brick = length × width × height

= 25 × 12.5 × 7.5 cm³

= 2343.75 cm³

Number of bricks = $\frac{\text{Volume of wall}}{\text{Volume of one brick}}$

= $\frac{10125000}{2343.75}$

= 4320

Hence, option 1 is the correct option.

Given:

Dimensions of the wooden box = 8 m × 7 m × 6 m = 800 cm × 700 cm × 600 cm

Dimensions of each rectangular box = 8 cm × 7 cm × 6 cm

Volume of the wooden box = length × breadth × height

= 800 × 700 × 600 cm³

= 336000000 cm³

Volume of one rectangular box = length × breadth × height

= 8 × 7 × 6 cm³

= 336 cm³

Maximum number of boxes = $\frac{\text{Volume of the wooden box}}{\text{Volume of one rectangular box}}$

= $\frac{336000000}{336}$

= 1000000

The boxes fit exactly, without any unused space, because

Number of boxes along the length = $\frac{800}{8}$ = 100

Number of boxes along the breadth = $\frac{700}{7}$ = 100

Number of boxes along the height = $\frac{600}{6}$ = 100

Total number of boxes = 100 × 100 × 100 = 1000000

Hence, option 1 is the correct option.

Given:

Length of the wall = 6 m = 600 cm

Height of the wall = 5 m = 500 cm

Thickness of the wall = 0.5 m = 50 cm

Volume of the wall = length × thickness × height

= 600 × 50 × 500 cm³

= 15000000 cm³

Volume occupied by the mortar = 5% of 15000000 cm³

= $\frac{5}{100}$ × 15000000 cm³

= 750000 cm³

Volume occupied by the bricks = 15000000 - 750000 cm³

= 14250000 cm³

Volume of one brick = length × width × height

= 25 × 12.5 × 7.5 cm³

= 2343.75 cm³

Number of bricks = $\frac{\text{Volume occupied by the bricks}}{\text{Volume of one brick}}$

= $\frac{14250000}{2343.75}$

= 6080

Hence, option 3 is the correct option.

Given:

Side of the cube = 27 cm

Surface area of a cube = 6 × (side)²

= 6 × (27)² cm²

= 6 × 729 cm²

= 4374 cm²

Hence, option 3 is the correct option.

Given:

Perimeter of one face of the cube = 40 cm

Each face of a cube is a square.

Let $s$ be the side of the cube.

Perimeter of one face = 4 × side

$$\begin{aligned}\Rightarrow 4 \times s = 40\\[1em] \Rightarrow s = \frac{40}{4}\\[1em] \Rightarrow s = 10 \text{ cm}\end{aligned}$$

Volume of a cube = (side)³

= 10³ cm³

= 1000 cm³

Hence, option 3 is the correct option.

Given:

Surface area of the cube = 384 cm²

Let $s$ be the side of the cube.

Surface area of a cube = 6 × (side)²

$$\begin{aligned}\Rightarrow 6 \times s^2 = 384\\[1em] \Rightarrow s^2 = \frac{384}{6}\\[1em] \Rightarrow s^2 = 64\\[1em] \Rightarrow s = \sqrt{64}\\[1em] \Rightarrow s = 8 \text{ cm}\end{aligned}$$

Volume of a cube = (side)³

= 8³ cm³

= 512 cm³

Hence, option 4 is the correct option.

Given:

Volume of the cube = 2744 cm³

Let $s$ be the side of the cube.

Volume of a cube = (side)³

$$\begin{aligned}\Rightarrow s^3 = 2744\\[1em] \Rightarrow s = \sqrt[3]{2744}\\[1em] \Rightarrow s = 14 \text{ cm}\end{aligned}$$

Surface area of a cube = 6 × (side)²

= 6 × (14)² cm²

= 6 × 196 cm²

= 1176 cm²

Hence, option 4 is the correct option.

Given:

Length of the diagonal of the cube = $4\sqrt{3}$ cm

Let $s$ be the edge of the cube.

Diagonal of a cube = $\sqrt{3}$ × side

$$\begin{aligned}\Rightarrow \sqrt{3} \times s = 4\sqrt{3}\\[1em] \Rightarrow \cancel{\sqrt{3}} \times s = 4 \times \cancel{\sqrt{3}}\\[1em] \Rightarrow s = 4 \text{ cm}\end{aligned}$$

Hence, option 2 is the correct option.

Given:

Length of the longest pole that can be placed on the floor of the room = 10 m

Length of the longest pole that can be placed in the room = $10\sqrt{2}$ m

Let $l, b$ and $h$ be the length, breadth and height of the room.

The longest pole that can be placed on the floor is the diagonal of the floor.

By Pythagoras' theorem, diagonal of the floor = $\sqrt{(l^2 + b^2)}$

$$\begin{aligned}\Rightarrow \sqrt{l^2 + b^2} = 10\\[1em] \Rightarrow l^2 + b^2 = 10^2\\[1em] \Rightarrow l^2 + b^2 = 100 \text{ ..........(1)}\end{aligned}$$

The longest pole that can be placed in the room is the diagonal of the room.

The diagonal of the floor, the height of the room and the diagonal of the room form a right-angled triangle. By Pythagoras' theorem, diagonal of the room = $\sqrt{(l^2 + b^2) + h^2} = \sqrt{l^2 + b^2 + h^2}$

$$\begin{aligned}\Rightarrow \sqrt{l^2 + b^2 + h^2} = 10\sqrt{2}\\[1em] \Rightarrow l^2 + b^2 + h^2 = \left(10\sqrt{2}\right)^2\\[1em] \Rightarrow l^2 + b^2 + h^2 = 200\end{aligned}$$

Putting the value of $l^2 + b^2$ from equation (1), we get

$$\begin{aligned}\Rightarrow 100 + h^2 = 200\\[1em] \Rightarrow h^2 = 200 - 100\\[1em] \Rightarrow h^2 = 100\\[1em] \Rightarrow h = \sqrt{100}\\[1em] \Rightarrow h = 10 \text{ m}\end{aligned}$$

Hence, option 4 is the correct option.

Given:

Volume of the wall = 12.8 m³

Let $b$ be the breadth of the wall.

The wall is 5 times as high as it is broad.

⇒ Height of the wall $(h) = 5b$

The wall is 8 times as long as it is high.

⇒ Length of the wall ($l$) = 8 × $5b = 40b$

Volume of a cuboid = $l \times b \times h$

$$\begin{aligned}\Rightarrow 40b \times b \times 5b = 12.8\\[1em] \Rightarrow 200b^3 = 12.8\\[1em] \Rightarrow b^3 = \frac{12.8}{200}\\[1em] \Rightarrow b^3 = 0.064\\[1em] \Rightarrow b = \sqrt[3]{0.064}\\[1em] \Rightarrow b = 0.4 \text{ m}\\[1em] \Rightarrow b = 0.4 \times 100 \text{ cm} = 40 \text{ cm}\end{aligned}$$

Hence, option 4 is the correct option.

Given:

Edge of each cube = 8 cm

When two cubes of edge 8 cm each are joined end to end, a cuboid is formed whose

Length (l) = (8 + 8) cm = 16 cm

Breadth (b) = 8 cm

Height (h) = 8 cm

As we know, the total surface area of the cuboid = 2(l × b + b × h + h × l)

= 2(16 × 8 + 8 × 8 + 8 × 16) cm²

= 2(128 + 64 + 128) cm²

= 2 × 320 cm²

= 640 cm²

Hence, option 2 is the correct option.

Given:

The edges of the three cubes are 6 cm, 8 cm and 10 cm.

The three cubes are melted to make a single cube, i.e., the sum of the volumes of the three cubes is equal to the volume of the new cube.

Let $x$ be the edge of the new cube.

Sum of the volumes of the three cubes = Volume of the new cube

$$\begin{aligned}\Rightarrow 6^3 + 8^3 + 10^3 = x^3\\[1em] \Rightarrow 216 + 512 + 1000 = x^3\\[1em] \Rightarrow x^3 = 1728\\[1em] \Rightarrow x = \sqrt[3]{1728}\\[1em] \Rightarrow x = 12 \text{ cm}\end{aligned}$$

Hence, option 3 is the correct option.

Given:

Side of the bigger cube = 6 cm

Side of each smaller cube = 2 cm

Volume of a cube = (side)³

Volume of the bigger cube = 6³ cm³ = 216 cm³

Volume of each smaller cube = 2³ cm³ = 8 cm³

Number of cubes formed = $\frac{\text{Volume of the bigger cube}}{\text{Volume of each smaller cube}}$

= $\frac{216}{8}$

= 27

Hence, option 4 is the correct option.

Given:

Side of the bigger cube = 4 cm

Side of each smaller cube = 1 cm

Volume of a cube = (side)³

Volume of the bigger cube = 4³ cm³ = 64 cm³

Volume of each smaller cube = 1³ cm³ = 1 cm³

Number of smaller cubes = $\frac{64}{1}$ = 64

Surface area of a cube = 6 × (side)²

Surface area of each smaller cube = 6 × 1² cm² = 6 cm²

Total surface area of all the small cubes = 64 × 6 cm²

= 384 cm²

Hence, option 4 is the correct option.

Let $s$ be the side of the cube.

So, side of the new cube = $2s$

Volume of a cube = (side)³

Original volume = $s^3$

New volume = $(2s)^3 = (8s)^3$

$\therefore \frac{\text{New volume}}{\text{Original volume}} = \frac{8s^3}{s^3} = 8$

So, the volume becomes 8 times the original volume.

Hence, option 4 is the correct option.

Given:

The volumes of the two cubes are in the ratio 1 : 27.

Let $a$ and $b$ be the sides of the two cubes.

Volume of a cube = (side)³

$$\begin{aligned}\Rightarrow \frac{a^3}{b^3} = \frac{1}{27}\\[1em] \Rightarrow \left(\frac{a}{b}\right)^3 = \left(\frac{1}{3}\right)^3\\[1em] \Rightarrow \frac{a}{b} = \frac{1}{3}\end{aligned}$$

Surface area of a cube = 6 × (side)²

$$\begin{aligned}\Rightarrow \frac{\text{Surface area of the first cube}}{\text{Surface area of the second cube}} = \frac{6a^2}{6b^2}\\[1em] = \frac{\cancel{6}a^2}{\cancel{6}b^2}\\[1em] = \left(\frac{a}{b}\right)^2\\[1em] = \left(\frac{1}{3}\right)^2\\[1em] = \frac{1}{9}\end{aligned}$$

Hence, option 3 is the correct option.

Given:

Height of the cylinder (h) = 14 cm

Diameter of the base = 4 cm

Radius of the base (r) = $\frac{4}{2}$ cm = 2 cm

Volume of a cylinder = πr²h

$$\begin{aligned}= \frac{22}{7} \times 2^2 \times 14\\[1em] = \frac{22}{7} \times 4 \times 14\\[1em] = \frac{1232}{7}\\[1em] = 176 \text{ cm}^3\end{aligned}$$

Hence, option 1 is the correct option.

Given:

Diameter of the cylinder = 28 cm

Radius of the cylinder (r) = $\frac{28}{2}$ cm = 14 cm

Height of the cylinder (h) = 20 cm

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

$$\begin{aligned}= 2 \times \frac{22}{7} \times 14 \times (14 + 20)\\[1em] = 2 \times \frac{22}{7} \times 14 \times 34\\[1em] = \frac{20944}{7}\\[1em] = 2992 \text{ cm}^2\end{aligned}$$

Hence, option 3 is the correct option.

Given:

Curved surface area of the cylinder = 1760 cm²

Base radius of the cylinder ($r$) = 14 cm

Let $h$ be the height of the cylinder.

Curved surface area of a cylinder = $2\pi rh$

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times 14 \times h = 1760\\[1em] \Rightarrow \frac{616}{7} \times h = 1760\\[1em] \Rightarrow 88 \times h = 1760\\[1em] \Rightarrow h = \frac{1760}{88}\\[1em] \Rightarrow h = 20 \text{ cm}\end{aligned}$$

Hence, option 3 is the correct option.

Given:

Height of the cylinder ($h$) = 14 cm

Curved surface area of the cylinder = 264 cm²

Let $r$ be the radius of the cylinder.

Curved surface area of a cylinder = $2\pi rh$

$$\begin{aligned}\Rightarrow 2 \times \frac{22}{7} \times r \times 14 = 264\\[1em] \Rightarrow \frac{616}{7} \times r = 264\\[1em] \Rightarrow 88 \times r = 264\\[1em] \Rightarrow r = \frac{264}{88}\\[1em] \Rightarrow r = 3\end{aligned}$$

Radius of the cylinder ($r$) = 3 cm

Volume of a cylinder = $\pi r^2h$

$$\begin{aligned}= \frac{22}{7} \times 3^2 \times 14\\[1em] = \frac{22}{7} \times 9 \times 14\\[1em] = \frac{2772}{7}\\[1em] = 396 \text{ cm}^3\end{aligned}$$

Hence, option 2 is the correct option.

Given:

Base radius of the cylinder ($r$) = 80 cm

Height of the cylinder ($h$) = 20 cm

As we know, the total surface area of cylinder = $2\pi r(r + h)$ and the lateral surface area of cylinder = $2\pi rh$

$$\begin{aligned}\Rightarrow \frac{\text{Total surface area of the cylinder}}{\text{Lateral surface area of the cylinder}} = \frac{2\pi r(r + h)}{2\pi rh}\\[1em] = \frac{\cancel{2\pi r}(r + h)}{\cancel{2\pi r}h}\\[1em] = \frac{r + h}{h}\\[1em] = \frac{80 + 20}{20}\\[1em] = \frac{100}{20}\\[1em] = \frac{5}{1}\end{aligned}$$

Hence, option 4 is the correct option.

Given:

The ratio between the radius of the base and the height of the cylinder is 2 : 3.

Volume of the cylinder = 1617 cm³

Let the radius of the base and the height of the cylinder be $2a$ and $3a$.

Volume of a cylinder = $\pi r^2h$

$$\begin{aligned}\Rightarrow \frac{22}{7} \times (2a)^2 \times 3a = 1617\\[1em] \Rightarrow \frac{22}{7} \times 4a^2 \times 3a = 1617\\[1em] \Rightarrow \frac{264a^3}{7} = 1617\\[1em] \Rightarrow a^3 = \frac{1617 \times 7}{264}\\[1em] \Rightarrow a^3 = \frac{343}{8}\\[1em] \Rightarrow a = \sqrt[3]{\frac{343}{8}}\\[1em] \Rightarrow a = \frac{7}{2}\\[1em] \Rightarrow a = 3.5\end{aligned}$$

Radius of the base ($r$) = $2a$ = 2 × 3.5 cm = 7 cm

Height of the cylinder ($h$) = $3a$ = 3 × 3.5 cm = 10.5 cm

Total surface area of a cylinder = $2\pi r(r + h)$

$$\begin{aligned}= 2 \times \frac{22}{7} \times 7 \times (7 + 10.5)\\[1em] = \frac{44}{\cancel{7}} \times \cancel{7} \times 17.5\\[1em] = 44 \times 17.5\\[1em] = 770 \text{ cm}^2\end{aligned}$$

Hence, option 4 is the correct option.

Given:

Diameter of each coin = 1.5 cm

Radius of each coin, r = $\frac{1.5}{2}$ cm = 0.75 cm

Thickness of each coin, h = 0.2 cm

Diameter of the cylinder = 4.5 cm

Radius of the cylinder, R = $\frac{4.5}{2}$ cm = 2.25 cm

Height of the cylinder, H = 10 cm

Each coin is in the shape of a cylinder.

Volume of each coin = πr²h

$$\begin{aligned}= \frac{22}{7} \times (0.75)^2 \times 0.2\\[1em] = \frac{22}{7} \times 0.5625 \times 0.2\\[1em] = \frac{2.475}{7}\end{aligned}$$

Volume of the required cylinder = πR²H

$$\begin{aligned}= \frac{22}{7} \times (2.25)^2 \times 10\\[1em] = \frac{22}{7} \times 5.0625 \times 10\\[1em] = \frac{1113.75}{7}\end{aligned}$$

Let $n$ be the required number of coins.

Number of coins = $\frac{\text{Volume of the required cylinder}}{\text{Volume of each coin}}$

$$\begin{aligned}\Rightarrow n = \frac{\frac{1113.75}{7}}{\frac{2.475}{7}}\\[1em] \Rightarrow n = \frac{1113.75}{2.475}\\[1em] \Rightarrow n = 450\end{aligned}$$

Hence, option 2 is the correct option.

(i) As we know, 1 m = 100 cm

⇒ 1 m³ = (100 × 100 × 100) cm³ = 1000000 cm³

Hence, 1 m³ = 1000000 cm³.

(ii) As we know, 1 dm = 10 cm

⇒ 1 dm³ = (10 × 10 × 10) cm³ = 1000 cm³

⇒ 1 cm³ = $\frac{1}{1000}$ dm³ = 0.001 dm³

Hence, 1 cm³ = 0.001 dm³.

(iii) Let $a$ be the edge of the cube.

Surface area of the cube = $6a^2$ and volume of the cube = $a^3$

It is given that the surface area of the cube is numerically equal to its volume.

$$\begin{aligned}\Rightarrow a^3 = 6a^2\\[1em] \Rightarrow \frac{a^3}{a^2} = 6\\[1em] \Rightarrow a = 6\end{aligned}$$

Hence, the edge of the cube is 6 units.

(iv) A road roller is cylindrical in shape. The area of the road paved by it in one revolution is equal to the curved surface area of the roller.

Hence, to calculate the area of the road paved by a road roller, we need to find the curved surface area of the roller.

(v) When a rectangle is rolled along its length or its breadth, the side along which it is rolled forms the circumference of the circular base and the other side gives the height of the cylinder.

Hence, a rectangle when rolled along its length or breadth forms a cylinder.

(vi) Let $r$ be the radius and $h$ be the height of the cylinder.

Volume of the cylinder = $\pi r^2h$ and curved surface area of the cylinder = $2\pi rh$

It is given that the volume of the cylinder is numerically equal to its curved surface area.

$$\begin{aligned}\Rightarrow \pi r^2 h = 2\pi rh\\[1em] \Rightarrow \frac{\pi r^2 h}{\pi rh} = 2\\[1em] \Rightarrow r = 2\end{aligned}$$

Hence, the radius of its base is 2 units.

The volume of a cylinder with radius r and height h is πr²h.

So, Reason (R) is true.

Two cylinders can have the same volume with different radii and heights.

For example, consider a cylinder with r = 2 units and h = 9 units.

Volume = π × 2² × 9 = 36π cubic units

Total surface area = 2π × 2 × (2 + 9) = 44π square units

Now, consider another cylinder with r = 3 units and h = 4 units.

Volume = π × 3² × 4 = 36π cubic units

Total surface area = 2π × 3 × (3 + 4) = 42π square units

Both the cylinders have equal volumes but their surface areas are not equal.

So, Assertion (A) is false.

∴ Assertion (A) is false but Reason (R) is true.

Hence, option 4 is the correct option.

The lateral surface area of a cuboid is the sum of the areas of its four side faces, excluding the top and the bottom faces.

Lateral surface area of a cuboid = 2h(l + b)

So, Reason (R) is true.

A room is a cuboid whose four walls are its four side faces, while the floor and the ceiling are its bottom and top faces.

Area of 4 walls of a room = 2h(l + b) = Lateral surface area of the cuboid

So, Assertion (A) is true and Reason (R) correctly explains assertion (A).

Hence, option 1 is the correct option.

Given:

Sum of the length, breadth and depth of the cuboid = 19 cm

Diagonal of the cuboid = $5\sqrt{5}$ cm

Let $l, b$ and $h$ be the length, breadth and depth of the cuboid.

So, $l + b + h$ = 19

By Pythagoras' theorem (applied first to the diagonal of the base and then to the diagonal of the cuboid), the diagonal of the cuboid = $\sqrt{l^2 + b^2 + h^2}$

$$\begin{aligned}\Rightarrow \sqrt{l^2 + b^2 + h^2} = 5\sqrt{5}\\[1em] \Rightarrow l^2 + b^2 + h^2 = \left(5\sqrt{5}\right)^2\\[1em] \Rightarrow l^2 + b^2 + h^2 = 125\end{aligned}$$

Squaring both the sides of $l + b + h$ = 19, we get

$$\begin{aligned}\Rightarrow (l + b + h)^2 = 19^2\\[1em] \Rightarrow l^2 + b^2 + h^2 + 2(lb + bh + hl) = 361\\[1em] \Rightarrow 125 + 2(lb + bh + hl) = 361\\[1em] \Rightarrow 2(lb + bh + hl) = 361 - 125\\[1em] \Rightarrow 2(lb + bh + hl) = 236\end{aligned}$$

As we know, the surface area of the cuboid = $2(lb + bh + hl)$

= 236 cm²

Hence, option 2 is the correct option.

A wire is cylindrical in shape.

Let $r$ be the radius and $l$ be the length of the wire.

Volume of a cylinder = $\pi r^2h$

Volume of the wire = $\pi r^2l$

Radius of the new wire = $\frac{r}{3}$

Let $L$ be the length of the new wire.

It is given that the volume remains the same.

$$\begin{aligned}\Rightarrow \pi \times \left(\frac{r}{3}\right)^2 \times L = \pi r^2l\\[1em] \Rightarrow \pi \times \frac{r^2}{9} \times L = \pi r^2l\\[1em] \Rightarrow \frac{\cancel{\pi r^2}}{9} \times L = \cancel{\pi r^2}l\\[1em] \Rightarrow L = 9l\end{aligned}$$

So, the new length of the wire is 9 times the original length (i.e., the length increases by 8 times the original length).

Hence, option 4 is the correct option.

Let $r$ be the radius and $h$ be the height of the cylinder.

Volume of the cylinder = $\pi r^2h$

Curved surface area of the cylinder = $2\pi rh$

It is given that the volume and the curved surface area are numerically same.

$$\begin{aligned}\Rightarrow \pi r^2 h = 2\pi rh\\[1em] \Rightarrow \frac{\pi r^2 h}{\pi rh} = 2\\[1em] \Rightarrow r = 2\end{aligned}$$

The height $h$ gets cancelled, so the equality holds for any height of the cylinder provided its radius is 2 units.

Hence, option 2 is the correct option.

Given:

Length of the swimming pool (l) = 200 m

Breadth of the swimming pool (b) = 50 m

Drop in the water level (h) = 2 cm = 0.02 m

The water used forms a cuboid of length 200 m, breadth 50 m and height equal to the drop in the water level.

Volume of water used = l × b × h

= 200 × 50 × 0.02 m³

= 200 m³

Hence, option 1 is the correct option.

Given:

Inner radius of the pipe (r) = 0.75 cm

Rate of flow of water = 7 m per second = 700 cm per second

The water delivered by the pipe in 1 second forms a cylinder of radius r = 0.75 cm and height h = 700 cm.

Volume of water delivered in 1 second = πr²h

$$\begin{aligned}= \frac{22}{7} \times (0.75)^2 \times 700\\[1em] = \frac{22}{7} \times 0.5625 \times 700\\[1em] = \frac{8662.5}{7}\\[1em] = 1237.5\end{aligned}$$

Volume of water delivered in 1 second = 1237.5 cm³

As we know, 1 hour = 3600 seconds

Volume of water delivered in 1 hour = 1237.5 × 3600 cm³

= 4455000 cm³

As we know, 1 litre = 1000 cm³

Volume of water delivered in 1 hour = $\frac{4455000}{1000}$ litres

= 4455 litres

Hence, option 4 is the correct option.

The sheet of paper is folded along its length $l$.

So, the length of the sheet forms the circumference of the base of the cylinder and the breadth of the sheet becomes the height of the cylinder.

Let $r$ be the radius of the base of the cylinder.

⇒ $2\pi r = l$

Height of the cylinder, $h = b$

Lateral surface area of a cylinder = $2\pi rh$

$$\begin{aligned}= (2\pi r) \times h\\[1em] = l \times b\\[1em] = lb\end{aligned}$$

Hence, option 3 is the correct option.

Given:

Side of the original cube = $a$

Surface area of a cube = 6 × (side)²

Surface area of the original cube = $6a^2$

Half of the surface area of the original cube = $\frac{6a^2}{2} = 3a^2$

When the cube is cut into two equal halves, each half is a cuboid whose

Length (l) = $a$, Breadth (b) = $a$ and Height (h) = $\frac{a}{2}$

As we know, the total surface area of the cuboid = 2(l × b + b × h + h × l)

$$\begin{aligned}= 2\left(a \times a + a \times \frac{a}{2} + \frac{a}{2} \times a\right)\\[1em] = 2\left(a^2 + \frac{a^2}{2} + \frac{a^2}{2}\right)\\[1em] = 2\left(a^2 + a^2\right)\\[1em] = 2 \times 2a^2\\[1em] = 4a^2\end{aligned}$$

As $4a^2 \gt 3a^2$, the surface area of each part is more than half of the surface area of the original cube.

This happens because cutting the cube creates two new faces, each of area $a^2$.

Hence, option 3 is the correct option.

Let l, b and h be the length, breadth and height of the cuboid.

ABCD is the base of the cuboid.

Area of the rectangle ABCD = l × b = $x$ square units

As we know, the total surface area of the cuboid = 2(l × b + b × h + h × l)

= 2($x$ + bh + hl) square units

The height h of the cuboid is not known, so the terms bh and hl cannot be found in terms of $x$.

Therefore, the surface area of the cuboid cannot be determined from the area of the rectangle ABCD alone.

Hence, option 4 is the correct option.