Chapter 13 Surface Areas And Volumes

Given:

A cuboidal wooden box with the following dimensions:

Length ($l$) = 80 cm

Breadth ($b$) = 40 cm

Height ($h$) = 20 cm

Square sheets of paper with side = 40 cm.

To Find:

The number of square sheets of paper required to cover the entire box.

Solution:

First, we need to find the total surface area of the cuboidal box, as it needs to be completely covered with paper.

The formula for the total surface area of a cuboid is:

Total Surface Area (TSA) = $2(lb + bh + hl)$

Substituting the given dimensions:

TSA = $2((80 \times 40) + (40 \times 20) + (20 \times 80))$

TSA = $2(3200 + 800 + 1600)$

TSA = $2(5600)$

TSA = $11200 \text{ cm}^2$

Next, we find the area of one square sheet of paper.

Area of one sheet = $(\text{side})^2 = (40 \text{ cm})^2 = 1600 \text{ cm}^2$.

To find the number of sheets required, we divide the total surface area of the box by the area of one sheet.

Number of sheets = $\frac{\text{Total Surface Area of Box}}{\text{Area of one sheet}}$

Number of sheets = $\frac{11200}{1600} = \frac{112}{16} = 7$

Mary would require 7 square sheets of paper.

Given:

A cubical water tank with an outer edge length ($a$) = 1.5 m.

The outer surface of the tank, excluding the base, needs to be covered with square tiles.

The side length of each square tile = 25 cm.

The cost of tiles = $\textsf{₹}$ 360 per dozen (12 tiles).

To Find:

The total amount Hameed would spend for the tiles.

Solution:

First, let's find the area of the surface to be tiled. The tank is a cube, which has 6 square faces. The area to be covered includes the four side faces and the top face (lid), but excludes the base. This means 5 faces are to be tiled.

The edge of the cube, $a = 1.5$ m. Since the tile side is in cm, let's convert the edge length to cm:

$a = 1.5 \text{ m} = 1.5 \times 100 = 150$ cm.

The area of one square face of the cube is $a^2 = (150 \text{ cm})^2 = 22500 \text{ cm}^2$.

The total area to be tiled is the area of 5 faces:

Total Area to be Tiled = $5 \times a^2 = 5 \times 22500 = 112500 \text{ cm}^2$.

Next, we find the area of one square tile.

Area of one tile = $(\text{side})^2 = (25 \text{ cm})^2 = 625 \text{ cm}^2$.

Now, we can find the number of tiles required:

Number of tiles = $\frac{\text{Total Area to be Tiled}}{\text{Area of one tile}} = \frac{112500}{625} = 180$ tiles.

The cost is given per dozen. Let's find how many dozens of tiles are needed.

Number of dozens = $\frac{\text{Total number of tiles}}{12} = \frac{180}{12} = 15$ dozens.

The cost of one dozen tiles is $\textsf{₹}$ 360.

Total Cost = Number of dozens $\times$ Cost per dozen

Total Cost = $15 \times 360 = \textsf{₹ } 5400$.

Hameed would spend $\textsf{₹ } 5400$ for the tiles.

Given:

A plastic box, open at the top, with the following dimensions:

Length ($l$) = 1.5 m

Width ($b$) = 1.25 m

Depth (height, $h$) = 65 cm = 0.65 m

Cost of the plastic sheet = $\textsf{₹}$ 20 per m².

To Find:

(i) The area of the sheet required.

(ii) The cost of the sheet.

Solution:

(i) Area of the sheet required

Since the box is open at the top, the plastic sheet is required for the four walls and the base of the box.

Area of the four walls = Lateral Surface Area = $2(l+b)h$.

Area of the base = $l \times b$.

Total area of the sheet required = Area of four walls + Area of the base

Area = $2(l+b)h + lb$

Substituting the given values:

Area = $2(1.5 + 1.25) \times 0.65 + (1.5 \times 1.25)$

Area = $2(2.75) \times 0.65 + 1.875$

Area = $5.5 \times 0.65 + 1.875$

Area = $3.575 + 1.875$

Area = $5.45 \text{ m}^2$.

(ii) Cost of the sheet

The cost of the sheet is given as $\textsf{₹}$ 20 per m².

Total Cost = Total Area $\times$ Rate per m²

Total Cost = $5.45 \times 20$

Total Cost = $\textsf{₹ } 109$.

Final Answer:

(i) The area of the sheet required is 5.45 m².

(ii) The cost of the sheet is $\textsf{₹ } 109$.

Given:

Dimensions of a rectangular room:

Length ($l$) = 5 m

Breadth ($b$) = 4 m

Height ($h$) = 3 m

The rate of whitewashing = $\textsf{₹}$ 7.50 per m².

To Find:

The total cost of whitewashing the walls and the ceiling of the room.

Solution:

First, we need to find the total area to be whitewashed. This includes the area of the four walls and the area of the ceiling. The floor is not whitewashed.

Area of the four walls (Lateral Surface Area) = $2(l+b)h$

Area of the walls = $2(5 + 4) \times 3 = 2 \times 9 \times 3 = 54 \text{ m}^2$.

Area of the ceiling = $l \times b$

Area of the ceiling = $5 \times 4 = 20 \text{ m}^2$.

Total Area to be whitewashed = Area of the walls + Area of the ceiling

Total Area = $54 + 20 = 74 \text{ m}^2$.

Now, we can calculate the total cost of whitewashing.

Total Cost = Total Area $\times$ Rate per m²

Total Cost = $74 \times 7.50$

Total Cost = $\textsf{₹ } 555$.

The cost of whitewashing the walls of the room and the ceiling is $\textsf{₹ } 555$.

Given:

Perimeter of the rectangular floor = 250 m.

Total cost of painting the four walls = $\textsf{₹}$ 15000.

Rate of painting = $\textsf{₹}$ 10 per m².

To Find:

The height of the hall.

Solution:

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

The perimeter of the floor is given by the formula $2(l+b)$.

So, $2(l+b) = 250$ m.

The cost of painting and the rate of painting are given for the four walls. We can use this to find the area of the four walls.

Area of the four walls = $\frac{\text{Total Cost of Painting}}{\text{Rate of Painting}}$

Area of the four walls = $\frac{15000}{10} = 1500 \text{ m}^2$.

The area of the four walls of a rectangular hall is its Lateral Surface Area (LSA), which is given by the formula:

LSA = $2(l+b)h$.

We know that the Area of the four walls is 1500 m² and the perimeter of the floor, $2(l+b)$, is 250 m.

Substituting these values into the formula:

$1500 = (2(l+b)) \times h$

$1500 = 250 \times h$

$h = \frac{1500}{250}$

$h = 6$ m.

The height of the hall is 6 m.

Given:

Total area that can be painted from the container = 9.375 m².

Dimensions of one brick: length ($l$) = 22.5 cm, breadth ($b$) = 10 cm, height ($h$) = 7.5 cm.

To Find:

The number of bricks that can be painted.

Solution:

First, we need to find the total surface area of one brick, which is the area that needs to be painted for each brick.

The formula for the total surface area (TSA) of a cuboid (brick) is:

TSA = $2(lb + bh + hl)$

Substituting the given dimensions in cm:

TSA of one brick = $2((22.5 \times 10) + (10 \times 7.5) + (7.5 \times 22.5))$

TSA = $2(225 + 75 + 168.75)$

TSA = $2(468.75)$

TSA = $937.5 \text{ cm}^2$.

The total area that can be painted is given in m². We need to convert it to cm² to have consistent units.

1 m = 100 cm, so 1 m² = $(100 \text{ cm})^2 = 10000 \text{ cm}^2$.

Total paintable area = $9.375 \text{ m}^2 = 9.375 \times 10000 \text{ cm}^2 = 93750 \text{ cm}^2$.

Now, we can find the number of bricks that can be painted.

Number of bricks = $\frac{\text{Total paintable area}}{\text{Surface area of one brick}}$

Number of bricks = $\frac{93750}{937.5}$

To simplify the division, we can write $937.5$ as $\frac{9375}{10}$.

Number of bricks = $\frac{93750 \times 10}{9375} = 10 \times 10 = 100$.

100 bricks can be painted out of this container.

Given:

A cubical box with edge length ($a$) = 10 cm.

A cuboidal box with length ($l$) = 12.5 cm, width ($b$) = 10 cm, and height ($h$) = 8 cm.

Solution:

(i) Comparison of Lateral Surface Area (LSA)

The LSA of a cube is given by the formula $4a^2$.

LSA of cubical box = $4 \times (10)^2 = 4 \times 100 = 400 \text{ cm}^2$.

The LSA of a cuboid is given by the formula $2(l+b)h$.

LSA of cuboidal box = $2(12.5 + 10) \times 8 = 2(22.5) \times 8 = 45 \times 8 = 360 \text{ cm}^2$.

Comparing the two LSAs, $400 \text{ cm}^2 > 360 \text{ cm}^2$.

The cubical box has the greater lateral surface area.

Difference in LSA = $400 - 360 = 40 \text{ cm}^2$.

(ii) Comparison of Total Surface Area (TSA)

The TSA of a cube is given by the formula $6a^2$.

TSA of cubical box = $6 \times (10)^2 = 6 \times 100 = 600 \text{ cm}^2$.

The TSA of a cuboid is given by the formula $2(lb + bh + hl)$.

TSA of cuboidal box = $2((12.5 \times 10) + (10 \times 8) + (8 \times 12.5))$

TSA = $2(125 + 80 + 100) = 2(305) = 610 \text{ cm}^2$.

Comparing the two TSAs, $600 \text{ cm}^2 < 610 \text{ cm}^2$.

The cubical box has the smaller total surface area.

Difference in TSA = $610 - 600 = 10 \text{ cm}^2$.

Final Answer:

(i) The cubical box has the greater lateral surface area by 40 cm².

(ii) The cubical box has the smaller total surface area by 10 cm².

Given:

A greenhouse (cuboid) with the following dimensions:

Length ($l$) = 30 cm

Width ($b$) = 25 cm

Height ($h$) = 25 cm

Solution:

(i) Area of the glass

The greenhouse is made entirely of glass, including the base. Therefore, the area of the glass is equal to the total surface area (TSA) of the cuboid.

TSA = $2(lb + bh + hl)$

Area of glass = $2((30 \times 25) + (25 \times 25) + (25 \times 30))$

Area of glass = $2(750 + 625 + 750)$

Area of glass = $2(2125)$

Area of glass = $4250 \text{ cm}^2$.

(ii) Tape needed for all 12 edges

The tape is used to join the edges. A cuboid has 12 edges: 4 lengths, 4 breadths, and 4 heights.

Total length of tape = $4l + 4b + 4h = 4(l + b + h)$

Total length of tape = $4(30 + 25 + 25)$

Total length of tape = $4(80)$

Total length of tape = $320 \text{ cm}$.

Final Answer:

(i) The area of the glass is 4250 cm².

(ii) The length of tape needed is 320 cm.

Given:

Bigger box dimensions: $l_1 = 25$ cm, $b_1 = 20$ cm, $h_1 = 5$ cm.

Smaller box dimensions: $l_2 = 15$ cm, $b_2 = 12$ cm, $h_2 = 5$ cm.

Extra cardboard for overlaps = 5% of TSA.

Number of boxes of each kind = 250.

Cost of cardboard = $\textsf{₹}$ 4 per 1000 cm².

To Find:

The total cost of cardboard for 250 boxes of each kind.

Solution:

1. Cardboard for Bigger Boxes:

TSA of one bigger box = $2(l_1b_1 + b_1h_1 + h_1l_1) \ $$ = 2((25 \times 20) + (20 \times 5) + (5 \times 25)) \ $$ = 2(500 + 100 + 125) = 2(725) = 1450 \text{ cm}^2$.

Extra area for overlaps = $5\%$ of $1450 = 0.05 \times 1450 = 72.5 \text{ cm}^2$.

Total cardboard for one bigger box = $1450 + 72.5 = 1522.5 \text{ cm}^2$.

Total cardboard for 250 bigger boxes = $250 \times 1522.5 = 380625 \text{ cm}^2$.

2. Cardboard for Smaller Boxes:

TSA of one smaller box = $2(l_2b_2 + b_2h_2 + h_2l_2) \ $$ = 2((15 \times 12) + (12 \times 5) + (5 \times 15)) \ $$ = 2(180 + 60 + 75) = 2(315) = 630 \text{ cm}^2$.

Extra area for overlaps = $5\%$ of $630 = 0.05 \times 630 = 31.5 \text{ cm}^2$.

Total cardboard for one smaller box = $630 + 31.5 = 661.5 \text{ cm}^2$.

Total cardboard for 250 smaller boxes = $250 \times 661.5 = 165375 \text{ cm}^2$.

3. Total Cost:

Total cardboard area for all boxes = $380625 + 165375 = 546000 \text{ cm}^2$.

The cost is $\textsf{₹}$ 4 for 1000 cm².

Total Cost = $\frac{\text{Total Area}}{1000} \times \text{Cost per 1000 cm}^2$

Total Cost = $\frac{546000}{1000} \times 4$

Total Cost = $546 \times 4 = \textsf{₹ } 2184$.

The cost of cardboard required for supplying 250 boxes of each kind is $\textsf{₹ } 2184$.

Given:

A temporary car shelter (cuboid shape, open at the bottom) with the following dimensions:

Height ($h$) = 2.5 m

Base Length ($l$) = 4 m

Base Breadth ($b$) = 3 m

To Find:

The total area of tarpaulin required.

Solution:

The tarpaulin is required to cover the four sides (walls) and the top (roof) of the shelter. The base is open.

The area of the four walls is the Lateral Surface Area (LSA) of the cuboid.

Area of four walls = $2(l+b)h$

Area of walls = $2(4 + 3) \times 2.5 = 2 \times 7 \times 2.5 = 14 \times 2.5 = 35 \text{ m}^2$.

The area of the top is given by $l \times b$.

Area of top = $4 \times 3 = 12 \text{ m}^2$.

Total tarpaulin required = Area of four walls + Area of top

Total Area = $35 + 12 = 47 \text{ m}^2$.

The amount of tarpaulin required to make the shelter is 47 m².

Given:

A model of a cylindrical kaleidoscope.

Length of the kaleidoscope (which is its height, $h$) = 25 cm.

Radius of the base ($r$) = 3.5 cm.

Value of $\pi$ to be used = $\frac{22}{7}$.

To Find:

The area of chart paper required to make the curved surface of the kaleidoscope.

Solution:

The area of the chart paper required is equal to the Curved Surface Area (CSA) of the cylindrical kaleidoscope.

The formula for the Curved Surface Area of a cylinder is:

CSA = $2\pi rh$

We are given:

$r = 3.5$ cm

$h = 25$ cm

$\pi = \frac{22}{7}$

Substituting these values into the formula:

To simplify the calculation, we can write 3.5 as $\frac{7}{2}$.

Cancel out the common factors:

The area of chart paper required by Savitri would be 550 cm².

Given:

A right circular cylinder with:

Height ($h$) = 14 cm

Curved Surface Area (CSA) = 88 cm²

To Find:

The diameter of the base of the cylinder.

Solution:

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

The formula for the Curved Surface Area of a cylinder is:

CSA = $2\pi rh$

Substitute the given values into the formula:

$88 = 2 \times \frac{22}{7} \times r \times 14$

Now, we solve for the radius $r$.

$88 = 2 \times 22 \times r \times \frac{14}{7}$

$88 = 44 \times r \times 2$

$88 = 88 \times r$

$r = \frac{88}{88} = 1$ cm.

The diameter of the base is twice the radius.

Diameter = $2r = 2 \times 1 = 2$ cm.

The diameter of the base of the cylinder is 2 cm.

Given:

A closed cylindrical tank with:

Height ($h$) = 1 m

Base diameter = 140 cm

To Find:

The area of the metal sheet required in square metres (m²).

Solution:

First, ensure all dimensions are in the same unit (meters).

Height ($h$) = 1 m.

Diameter = 140 cm = $\frac{140}{100}$ m = 1.4 m.

The radius ($r$) is half the diameter.

Radius ($r$) = $\frac{1.4}{2} = 0.7$ m.

Since the tank is closed, the metal sheet required is equal to the Total Surface Area (TSA) of the cylinder.

The formula for the TSA of a closed cylinder is:

TSA = $2\pi r(r+h)$

Substitute the values into the formula:

TSA = $2 \times \frac{22}{7} \times 0.7 \times (0.7 + 1)$

TSA = $2 \times \frac{22}{7} \times 0.7 \times 1.7$

TSA = $2 \times 22 \times 0.1 \times 1.7$

TSA = $4.4 \times 1.7$

TSA = $7.48 \text{ m}^2$.

7.48 m² of the sheet are required to make the tank.

Given:

A hollow cylindrical metal pipe with:

Length (height, $h$) = 77 cm

Inner diameter = 4 cm

Outer diameter = 4.4 cm

Solution:

First, let's determine the radii from the diameters.

Inner radius ($r_1$) = $\frac{4}{2} = 2$ cm.

Outer radius ($r_2$) = $\frac{4.4}{2} = 2.2$ cm.

(i) Inner curved surface area

Inner CSA = $2\pi r_1h$

Inner CSA = $2 \times \frac{22}{7} \times 2 \times 77 = 2 \times 22 \times 2 \times 11 = 968 \text{ cm}^2$.

(ii) Outer curved surface area

Outer CSA = $2\pi r_2h$

Outer CSA = $2 \times \frac{22}{7} \times 2.2 \times 77 = 2 \times 22 \times 2.2 \times 11 = 1064.8 \text{ cm}^2$.

(iii) Total surface area

The total surface area of the pipe is the sum of the inner CSA, the outer CSA, and the area of the two ring-shaped ends.

Area of one ring = $\pi (r_2^2 - r_1^2)$

Area of two rings = $2\pi (r_2^2 - r_1^2) = 2 \times \frac{22}{7} \times ((2.2)^2 - 2^2) \ $$ = \frac{44}{7} \times (4.84 - 4) \ $$ = \frac{44}{7} \times 0.84 = 44 \times 0.12 = 5.28 \text{ cm}^2$.

Total Surface Area = Inner CSA + Outer CSA + Area of two rings

TSA = $968 + 1064.8 + 5.28$

TSA = $2038.08 \text{ cm}^2$.

Final Answer:

(i) Inner curved surface area = 968 cm².

(ii) Outer curved surface area = 1064.8 cm².

(iii) Total surface area = 2038.08 cm².

Given:

A cylindrical roller with:

Diameter = 84 cm

Length (height, $h$) = 120 cm

Number of revolutions to level a playground = 500.

To Find:

The area of the playground in square metres (m²).

Solution:

The area covered by the roller in one revolution is equal to its curved surface area (CSA).

First, let's find the radius and convert dimensions to meters.

Diameter = 84 cm = 0.84 m.

Radius ($r$) = $\frac{0.84}{2} = 0.42$ m.

Height ($h$) = 120 cm = 1.2 m.

Area covered in one revolution = CSA = $2\pi rh$

CSA = $2 \times \frac{22}{7} \times 0.42 \times 1.2$

CSA = $2 \times 22 \times 0.06 \times 1.2$

CSA = $44 \times 0.072$

CSA = $3.168 \text{ m}^2$.

The total area of the playground is the area covered in 500 revolutions.

Area of playground = $500 \times$ Area covered in one revolution

Area of playground = $500 \times 3.168$

Area of playground = $1584 \text{ m}^2$.

The area of the playground is 1584 m².

Given:

A cylindrical pillar with:

Diameter = 50 cm

Height ($h$) = 3.5 m

Rate of painting = $\textsf{₹}$ 12.50 per m².

To Find:

The cost of painting the curved surface of the pillar.

Solution:

First, we need to find the curved surface area (CSA) of the pillar. Let's convert all dimensions to meters.

Diameter = 50 cm = 0.5 m.

Radius ($r$) = $\frac{0.5}{2} = 0.25$ m.

Height ($h$) = 3.5 m.

Area to be painted = CSA = $2\pi rh$

CSA = $2 \times \frac{22}{7} \times 0.25 \times 3.5$

CSA = $2 \times \frac{22}{7} \times \frac{1}{4} \times \frac{7}{2}$

CSA = $\frac{2 \times 22 \times 7}{7 \times 4 \times 2} = \frac{22}{4} = 5.5 \text{ m}^2$.

Now, we can calculate the total cost of painting.

Total Cost = Area to be painted $\times$ Rate of painting

Total Cost = $5.5 \times 12.50$

Total Cost = $\textsf{₹ } 68.75$.

The cost of painting the curved surface of the pillar is $\textsf{₹ } 68.75$.

Given:

A right circular cylinder with:

Curved Surface Area (CSA) = 4.4 m²

Radius of the base ($r$) = 0.7 m

To Find:

The height ($h$) of the cylinder.

Solution:

The formula for the Curved Surface Area of a cylinder is:

CSA = $2\pi rh$

Substitute the given values into the formula:

$4.4 = 2 \times \frac{22}{7} \times 0.7 \times h$

Now, we solve for the height $h$.

$4.4 = 2 \times 22 \times 0.1 \times h$

$4.4 = 44 \times 0.1 \times h$

$4.4 = 4.4 \times h$

$h = \frac{4.4}{4.4} = 1$ m.

The height of the cylinder is 1 m.

Given:

A circular well (cylinder) with:

Inner diameter = 3.5 m

Depth (height, $h$) = 10 m

Rate of plastering = $\textsf{₹}$ 40 per m².

Solution:

First, find the inner radius of the well.

Inner radius ($r$) = $\frac{3.5}{2} = 1.75$ m.

(i) Inner curved surface area

The area to be plastered is the inner curved surface area (CSA).

Inner CSA = $2\pi rh$

Inner CSA = $2 \times \frac{22}{7} \times 1.75 \times 10$

Inner CSA = $2 \times 22 \times 0.25 \times 10$

Inner CSA = $44 \times 2.5$

Inner CSA = $110 \text{ m}^2$.

(ii) Cost of plastering

The total cost is the area to be plastered multiplied by the rate of plastering.

Total Cost = Inner CSA $\times$ Rate

Total Cost = $110 \times 40$

Total Cost = $\textsf{₹ } 4400$.

Final Answer:

(i) The inner curved surface area is 110 m².

(ii) The cost of plastering is $\textsf{₹ } 4400$.

Given:

A cylindrical pipe with:

Length (height, $h$) = 28 m

Diameter = 5 cm

To Find:

The total radiating surface in the system.

Solution:

The total radiating surface is the curved surface area (CSA) of the cylindrical pipe.

First, we need to have consistent units. Let's convert the diameter to meters.

Diameter = 5 cm = 0.05 m.

Radius ($r$) = $\frac{0.05}{2} = 0.025$ m.

Now, calculate the CSA.

Radiating Surface (CSA) = $2\pi rh$

CSA = $2 \times \frac{22}{7} \times 0.025 \times 28$

CSA = $2 \times 22 \times 0.025 \times 4$

CSA = $44 \times 0.1$

CSA = $4.4 \text{ m}^2$.

The total radiating surface in the system is 4.4 m².

Given:

A closed cylindrical petrol storage tank with:

Diameter = 4.2 m

Height ($h$) = 4.5 m

Fraction of steel wasted = $\frac{1}{12}$ of the total steel used.

Solution:

First, find the radius of the tank.

Radius ($r$) = $\frac{4.2}{2} = 2.1$ m.

(i) Lateral or curved surface area (CSA)

CSA = $2\pi rh$

CSA = $2 \times \frac{22}{7} \times 2.1 \times 4.5$

CSA = $2 \times 22 \times 0.3 \times 4.5$

CSA = $13.2 \times 4.5$

CSA = $59.4 \text{ m}^2$.

(ii) Steel actually used

First, we need the total surface area (TSA) of the closed tank, which represents the amount of steel in the final product.

TSA = $2\pi r(r+h)$

TSA = $2 \times \frac{22}{7} \times 2.1 \times (2.1 + 4.5)$

TSA = $2 \times 22 \times 0.3 \times 6.6$

TSA = $13.2 \times 6.6$

TSA = $87.12 \text{ m}^2$.

This TSA represents the steel used after wastage. Let the total steel actually used be $X$ m².

Steel wasted = $\frac{1}{12}X$.

Steel in the tank = Total Steel - Wasted Steel

$87.12 = X - \frac{1}{12}X$

$87.12 = \frac{11}{12}X$

$X = \frac{87.12 \times 12}{11}$

$X = 7.92 \times 12$

$X = 95.04 \text{ m}^2$.

Final Answer:

(i) The lateral surface area is 59.4 m².

(ii) The amount of steel actually used was 95.04 m².

Given:

A cylindrical lampshade frame with:

Base diameter = 20 cm

Height ($h$) = 30 cm

A margin of 2.5 cm is to be given at the top and bottom for folding.

To Find:

The total area of cloth required for covering the lampshade.

Solution:

The cloth required must cover the curved surface of the lampshade, plus the extra margins at the top and bottom.

First, find the radius of the lampshade.

Radius ($r$) = $\frac{\text{Diameter}}{2} = \frac{20}{2} = 10$ cm.

The height of the frame is 30 cm. A margin of 2.5 cm is added to both the top and bottom. So, the total height of the cloth needed will be:

Total Height of Cloth ($H$) = Height of frame + Top margin + Bottom margin

$H = 30 + 2.5 + 2.5 = 35$ cm.

The area of the cloth required is the curved surface area of a cylinder with radius $r=10$ cm and total height $H=35$ cm.

Area of Cloth = $2\pi rH$

Area = $2 \times \frac{22}{7} \times 10 \times 35$

Area = $2 \times 22 \times 10 \times 5$

Area = $44 \times 50$

Area = $2200 \text{ cm}^2$.

2200 cm² of cloth is required for covering the lampshade.

Given:

Cylindrical penholders with a base (open at the top).

Radius ($r$) = 3 cm

Height ($h$) = 10.5 cm

Number of competitors (and penholders) = 35.

To Find:

The total area of cardboard required for the competition.

Solution:

First, we need to find the area of cardboard required for one penholder. Since a penholder is open at the top, the area needed is the sum of the curved surface area and the area of the circular base.

Area for one penholder = CSA + Area of base

Area = $2\pi rh + \pi r^2 = \pi r(2h + r)$

Substitute the values for one penholder:

Area = $\frac{22}{7} \times 3 \times (2 \times 10.5 + 3)$

Area = $\frac{22}{7} \times 3 \times (21 + 3)$

Area = $\frac{22}{7} \times 3 \times 24$

Area = $\frac{1584}{7} \text{ cm}^2$.

This is the area of cardboard for one penholder. There are 35 competitors, so 35 penholders are to be made.

Total Cardboard Required = $35 \times$ Area for one penholder

Total Cardboard = $35 \times \frac{1584}{7}$

Total Cardboard = $5 \times 1584$

Total Cardboard = $7920 \text{ cm}^2$.

7920 cm² of cardboard was required to be bought for the competition.

Given:

A right circular cone with:

Slant height ($l$) = 10 cm

Base radius ($r$) = 7 cm

To Find:

The curved surface area (CSA) of the cone.

Solution:

The formula for the Curved Surface Area of a cone is given by:

CSA = $\pi rl$

We are given $r = 7$ cm and $l = 10$ cm. Using $\pi = \frac{22}{7}$:

CSA = $\frac{22}{7} \times 7 \times 10$

CSA = $22 \times 10$

CSA = $220 \text{ cm}^2$.

The curved surface area of the cone is 220 cm².

Given:

A right circular cone with:

Height ($h$) = 16 cm

Base radius ($r$) = 12 cm

To Find:

1. The curved surface area (CSA) of the cone.

2. The total surface area (TSA) of the cone.

Solution:

First, we need to find the slant height ($l$) of the cone using the Pythagorean theorem, as we have a right-angled triangle formed by the height, radius, and slant height.

$l^2 = r^2 + h^2$

$l = \sqrt{r^2 + h^2}$

$l = \sqrt{(12)^2 + (16)^2} = \sqrt{144 + 256} = \sqrt{400}$

$l = 20$ cm.

1. Curved Surface Area (CSA):

The formula for CSA is $\pi rl$.

CSA = $3.14 \times 12 \times 20$

CSA = $3.14 \times 240$

CSA = $753.6 \text{ cm}^2$.

2. Total Surface Area (TSA):

The formula for TSA is the sum of the curved surface area and the area of the base ($\pi r^2$).

TSA = $\pi rl + \pi r^2 = \pi r(l+r)$

TSA = $3.14 \times 12 \times (20 + 12)$

TSA = $3.14 \times 12 \times 32$

TSA = $3.14 \times 384$

TSA = $1205.76 \text{ cm}^2$.

The curved surface area of the cone is 753.6 cm² and the total surface area is 1205.76 cm².

Given:

A corn cob shaped like a cone with:

Radius of the base ($r$) = 2.1 cm

Length (height, $h$) = 20 cm

Density of grains = 4 grains per cm² of the surface.

To Find:

The total number of grains on the entire cob.

Solution:

The grains are on the curved surface of the cob. So, we first need to find the Curved Surface Area (CSA) of the cone.

To find the CSA, we must first calculate the slant height ($l$) using the formula:

$l = \sqrt{r^2 + h^2}$

$l = \sqrt{(2.1)^2 + (20)^2} = \sqrt{4.41 + 400} = \sqrt{404.41}$

$l \approx 20.11$ cm.

Now, we can calculate the Curved Surface Area.

CSA = $\pi rl$

Using $\pi = \frac{22}{7}$:

CSA = $\frac{22}{7} \times 2.1 \times 20.11$

CSA = $22 \times 0.3 \times 20.11$

CSA = $6.6 \times 20.11$

CSA $\approx 132.726 \text{ cm}^2$.

The number of grains on 1 cm² is 4.

Total number of grains = CSA $\times$ Number of grains per cm²

Total number of grains $\approx 132.726 \times 4$

Total number of grains $\approx 530.9$.

Since the number of grains must be a whole number, we can round it to the nearest integer.

You would find approximately 531 grains on the entire cob.

Given:

A right circular cone with:

Diameter of the base = 10.5 cm

Slant height ($l$) = 10 cm

To Find:

The curved surface area (CSA) of the cone.

Solution:

First, we find the radius of the base of the cone from its diameter.

Radius ($r$) = $\frac{\text{Diameter}}{2} = \frac{10.5}{2} = 5.25$ cm.

The formula for the Curved Surface Area of a cone is:

CSA = $\pi rl$

Substituting the given values ($r = 5.25$ cm, $l = 10$ cm) and using $\pi = \frac{22}{7}$:

CSA = $\frac{22}{7} \times 5.25 \times 10$

CSA = $\frac{22}{7} \times \frac{525}{100} \times 10$

CSA = $\frac{22 \times 75 \times 10}{100}$

CSA = $22 \times 7.5$

CSA = $165 \text{ cm}^2$.

The curved surface area of the cone is 165 cm².

Given:

A cone with:

Slant height ($l$) = 21 m

Diameter of the base = 24 m

To Find:

The total surface area (TSA) of the cone.

Solution:

First, we find the radius of the base of the cone.

Radius ($r$) = $\frac{\text{Diameter}}{2} = \frac{24}{2} = 12$ m.

The formula for the Total Surface Area of a cone is the sum of its curved surface area ($\pi rl$) and the area of its base ($\pi r^2$).

TSA = $\pi r(l+r)$

Substituting the given values ($r = 12$ m, $l = 21$ m) and using $\pi = \frac{22}{7}$:

TSA = $\frac{22}{7} \times 12 \times (21 + 12)$

TSA = $\frac{22}{7} \times 12 \times 33$

TSA = $\frac{8712}{7}$

TSA $\approx 1244.57 \text{ m}^2$.

The total surface area of the cone is $\frac{8712}{7}$ m² (approximately 1244.57 m²).

Given:

A cone with:

Curved Surface Area (CSA) = 308 cm²

Slant height ($l$) = 14 cm

To Find:

(i) Radius of the base ($r$).

(ii) Total surface area (TSA) of the cone.

Solution:

(i) Radius of the base

The formula for the CSA of a cone is $\pi rl$.

We are given CSA = 308 cm² and $l = 14$ cm. Using $\pi = \frac{22}{7}$:

$308 = \frac{22}{7} \times r \times 14$

$308 = 22 \times r \times 2$

$308 = 44r$

$r = \frac{308}{44} = 7$ cm.

The radius of the base is 7 cm.

(ii) Total surface area of the cone

The formula for the TSA of a cone is CSA + Area of Base ($\pi r^2$).

TSA = $308 + \pi (7)^2$

TSA = $308 + \frac{22}{7} \times 49$

TSA = $308 + 22 \times 7$

TSA = $308 + 154 = 462 \text{ cm}^2$.

Final Answer:

(i) The radius of the base is 7 cm.

(ii) The total surface area of the cone is 462 cm².

Given:

A conical tent with:

Height ($h$) = 10 m

Radius of base ($r$) = 24 m

Cost of canvas = $\textsf{₹}$ 70 per m².

To Find:

(i) Slant height ($l$) of the tent.

(ii) Cost of the canvas required.

Solution:

(i) Slant height of the tent

The slant height, height, and radius of a cone form a right-angled triangle. We use the Pythagorean theorem to find the slant height ($l$).

$l = \sqrt{r^2 + h^2}$

$l = \sqrt{(24)^2 + (10)^2} = \sqrt{576 + 100} = \sqrt{676}$

$l = 26$ m.

The slant height of the tent is 26 m.

(ii) Cost of the canvas required

The canvas is required for the curved surface of the tent. The area of the canvas is the Curved Surface Area (CSA) of the cone.

CSA = $\pi rl$

CSA = $\frac{22}{7} \times 24 \times 26 = \frac{13728}{7} \text{ m}^2$.

The cost of the canvas is $\textsf{₹}$ 70 per m².

Total Cost = Area of canvas $\times$ Rate per m²

Total Cost = $\frac{13728}{7} \times 70$

Total Cost = $13728 \times 10 = \textsf{₹ } 137280$.

Final Answer:

(i) The slant height of the tent is 26 m.

(ii) The cost of the canvas required is $\textsf{₹ } 137,280$.

Given:

A conical tent with:

Height ($h$) = 8 m

Base radius ($r$) = 6 m

Width of the tarpaulin cloth = 3 m

Extra length for wastage and stitching = 20 cm = 0.2 m

To Find:

The total length of tarpaulin required.

Solution:

First, we need to find the area of the tarpaulin required, which is the curved surface area (CSA) of the conical tent.

We start by finding the slant height ($l$).

$l = \sqrt{r^2 + h^2} = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ m.

Now, calculate the CSA.

Area of tarpaulin (CSA) = $\pi rl$

Using $\pi = 3.14$:

Area = $3.14 \times 6 \times 10 = 188.4 \text{ m}^2$.

The tarpaulin is a rectangular sheet with a given width of 3 m. The area of this sheet is Length $\times$ Width.

Let the length of the tarpaulin be $L$.

$L \times 3 = 188.4$

$L = \frac{188.4}{3} = 62.8$ m.

This is the length of the material needed for the tent itself. We must also add the extra length for stitching and wastage.

Total Length required = $L$ + Extra length

Total Length = $62.8 + 0.2 = 63$ m.

The total length of tarpaulin required is 63 m.

Given:

A conical tomb with:

Slant height ($l$) = 25 m

Base diameter = 14 m

Rate of whitewashing = $\textsf{₹}$ 210 per 100 m².

To Find:

The cost of whitewashing the curved surface of the tomb.

Solution:

First, we need to find the area to be whitewashed, which is the curved surface area (CSA) of the cone.

Radius ($r$) = $\frac{\text{Diameter}}{2} = \frac{14}{2} = 7$ m.

CSA = $\pi rl$

CSA = $\frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \text{ m}^2$.

The rate of whitewashing is $\textsf{₹}$ 210 for 100 m².

Rate per m² = $\frac{210}{100} = \textsf{₹ } 2.10$ per m².

Total Cost = Area to be whitewashed $\times$ Rate per m²

Total Cost = $550 \times 2.10$

Total Cost = $\textsf{₹ } 1155$.

The cost of whitewashing the curved surface is $\textsf{₹ } 1155$.

Given:

A joker's cap in the shape of a right circular cone with:

Base radius ($r$) = 7 cm

Height ($h$) = 24 cm

Number of caps to make = 10

To Find:

The total area of the sheet required to make 10 caps.

Solution:

The area of the sheet for one cap is its curved surface area (CSA), as a cap is open at the base.

First, we need to find the slant height ($l$).

$l = \sqrt{r^2 + h^2} = \sqrt{(7)^2 + (24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ cm.

Now, calculate the CSA for one cap.

CSA of one cap = $\pi rl$

CSA = $\frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \text{ cm}^2$.

The total area of the sheet required for 10 caps is:

Total Area = $10 \times$ CSA of one cap

Total Area = $10 \times 550 = 5500 \text{ cm}^2$.

The area of the sheet required to make 10 such caps is 5500 cm².

Given:

Number of hollow cones = 50

Dimensions of each cone:

Base diameter = 40 cm

Height ($h$) = 1 m

Cost of painting = $\textsf{₹}$ 12 per m²

Given values: $\pi = 3.14$ and $\sqrt{1.04} = 1.02$.

To Find:

The total cost of painting all 50 cones.

Solution:

The outer side of each cone is to be painted, so we need to calculate the curved surface area (CSA) of one cone and then multiply it by 50.

First, ensure all dimensions are in meters.

Height ($h$) = 1 m.

Diameter = 40 cm = 0.4 m.

Radius ($r$) = $\frac{0.4}{2} = 0.2$ m.

Next, find the slant height ($l$).

$l = \sqrt{r^2 + h^2} = \sqrt{(0.2)^2 + (1)^2} = \sqrt{0.04 + 1} = \sqrt{1.04}$ m.

We are given to use $\sqrt{1.04} = 1.02$ m.

Now, calculate the CSA of one cone.

CSA of one cone = $\pi rl$

CSA = $3.14 \times 0.2 \times 1.02 = 0.64056 \text{ m}^2$.

The total area to be painted is the area of 50 cones.

Total Area = $50 \times$ CSA of one cone

Total Area = $50 \times 0.64056 = 32.028 \text{ m}^2$.

Now, calculate the total cost of painting.

Total Cost = Total Area $\times$ Rate per m²

Total Cost = $32.028 \times 12$

Total Cost = $\textsf{₹ } 384.336$.

Rounding to two decimal places for currency, the cost is $\textsf{₹ } 384.34$.

The cost of painting all these cones will be $\textsf{₹ } 384.34$.

Given:

A sphere with radius ($r$) = 7 cm.

To Find:

The surface area of the sphere.

Solution:

The formula for the surface area of a sphere with radius $r$ is:

Surface Area = $4\pi r^2$

Substitute the given radius and use $\pi = \frac{22}{7}$:

The surface area of the sphere is 616 cm².

Given:

A hemisphere with radius ($r$) = 21 cm.

To Find:

(i) The curved surface area (CSA) of the hemisphere.

(ii) The total surface area (TSA) of the hemisphere.

Solution:

We will use $\pi = \frac{22}{7}$.

(i) Curved surface area (CSA):

The formula for the CSA of a hemisphere is $2\pi r^2$.

(ii) Total surface area (TSA):

The TSA of a hemisphere is the sum of its curved surface area and the area of its circular base ($ \pi r^2 $).

TSA = $2\pi r^2 + \pi r^2 = 3\pi r^2$

Final Answer:

(i) The curved surface area is 2772 cm².

(ii) The total surface area is 4158 cm².

Given:

A hollow sphere with a diameter of 7 m.

To Find:

The area available for the motorcyclist to ride, which is the inner surface area of the sphere.

Solution:

First, we find the radius of the sphere from the given diameter.

The area available for riding is the surface area of the sphere. The formula is:

Surface Area = $4\pi r^2$

Substitute the value of the radius and use $\pi = \frac{22}{7}$:

The area available to the motorcyclist for riding is 154 m².

Given:

A hemispherical dome.

Circumference of the base = 17.6 m.

Cost of painting = $\textsf{₹}$ 5 per 100 cm².

To Find:

The total cost of painting the dome.

Solution:

The area to be painted is the curved surface area (CSA) of the hemisphere. To find the CSA, we first need to determine the radius of the dome from its base circumference.

Circumference of the base, $C = 2\pi r$.

$17.6 = 2 \times \frac{22}{7} \times r$

$r = \frac{17.6 \times 7}{2 \times 22} = \frac{17.6 \times 7}{44} = 0.4 \times 7 = 2.8$ m.

Now, we calculate the curved surface area of the hemisphere.

CSA = $2\pi r^2$

CSA = $2 \times \frac{22}{7} \times (2.8)^2$

CSA = $2 \times \frac{22}{7} \times 2.8 \times 2.8$

CSA = $2 \times 22 \times 0.4 \times 2.8$

CSA = $44 \times 1.12 = 49.28 \text{ m}^2$.

The cost of painting is given as $\textsf{₹}$ 5 per 100 cm². We need to find the cost per m².

1 m² = 10000 cm².

Cost per m² = $\frac{\text{Cost per 100 cm}^2}{100 \text{ cm}^2} \times 10000 \text{ cm}^2 = \frac{5}{100} \times 10000 = \textsf{₹ } 500$ per m².

Now, we can find the total cost of painting.

Total Cost = Total Area $\times$ Rate per m²

Total Cost = $49.28 \times 500$

Total Cost = $\textsf{₹ } 24640$.

The cost of painting the hemispherical dome is $\textsf{₹ } 24,640$.

Solution:

The formula for the surface area of a sphere is $4\pi r^2$.

(i) Radius ($r$) = 10.5 cm

Surface Area = $4 \times \frac{22}{7} \times (10.5)^2$

Surface Area = $4 \times \frac{22}{7} \times 10.5 \times 10.5$

Surface Area = $4 \times 22 \times 1.5 \times 10.5$

Surface Area = $88 \times 15.75 = 1386 \text{ cm}^2$.

(ii) Radius ($r$) = 5.6 cm

Surface Area = $4 \times \frac{22}{7} \times (5.6)^2$

Surface Area = $4 \times \frac{22}{7} \times 5.6 \times 5.6$

Surface Area = $4 \times 22 \times 0.8 \times 5.6$

Surface Area = $88 \times 4.48 = 394.24 \text{ cm}^2$.

(iii) Radius ($r$) = 14 cm

Surface Area = $4 \times \frac{22}{7} \times (14)^2$

Surface Area = $4 \times \frac{22}{7} \times 14 \times 14$

Surface Area = $4 \times 22 \times 2 \times 14$

Surface Area = $88 \times 28 = 2464 \text{ cm}^2$.

Final Answers:

(i) The surface area is 1386 cm².

(ii) The surface area is 394.24 cm².

(iii) The surface area is 2464 cm².

Solution:

The formula for the surface area of a sphere is $4\pi r^2$, where $r = \frac{\text{Diameter}}{2}$.

(i) Diameter = 14 cm

Radius ($r$) = $\frac{14}{2} = 7$ cm.

Surface Area = $4 \times \frac{22}{7} \times (7)^2 = 4 \times 22 \times 7 = 616 \text{ cm}^2$.

(ii) Diameter = 21 cm

Radius ($r$) = $\frac{21}{2} = 10.5$ cm.

Surface Area = $4 \times \frac{22}{7} \times (10.5)^2 = 4 \times \frac{22}{7} \times 10.5 \times 10.5 = 1386 \text{ cm}^2$.

(iii) Diameter = 3.5 m

Radius ($r$) = $\frac{3.5}{2} = 1.75$ m.

Surface Area = $4 \times \frac{22}{7} \times (1.75)^2 = 4 \times \frac{22}{7} \times 1.75 \times 1.75 \ $$ = 4 \times 22 \times 0.25 \times 1.75 \ $$ = 22 \times 1.75 = 38.5 \text{ m}^2$.

Final Answers:

(i) The surface area is 616 cm².

(ii) The surface area is 1386 cm².

(iii) The surface area is 38.5 m².

Given:

A hemisphere with radius ($r$) = 10 cm.

To Find:

The total surface area (TSA) of the hemisphere.

Solution:

The total surface area of a hemisphere is the sum of its curved surface area ($2\pi r^2$) and the area of its circular base ($\pi r^2$).

TSA = $3\pi r^2$

Using the given values, $r = 10$ cm and $\pi = 3.14$:

TSA = $3 \times 3.14 \times (10)^2$

TSA = $3 \times 3.14 \times 100$

TSA = $3 \times 314 = 942 \text{ cm}^2$.

The total surface area of the hemisphere is 942 cm².

Given:

Initial radius of the spherical balloon ($r_1$) = 7 cm.

Final radius of the balloon ($r_2$) = 14 cm.

To Find:

The ratio of the surface areas of the balloon in the two cases.

Solution:

Let SA₁ be the initial surface area and SA₂ be the final surface area.

The formula for the surface area of a sphere is $4\pi r^2$.

SA₁ = $4\pi r_1^2 = 4\pi (7)^2$.

SA₂ = $4\pi r_2^2 = 4\pi (14)^2$.

We need to find the ratio $\frac{\text{SA}_1}{\text{SA}_2}$.

Ratio = $\frac{4\pi (7)^2}{4\pi (14)^2} = \frac{(7)^2}{(14)^2} = \left(\frac{7}{14}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

The ratio of the surface areas is 1:4.

The ratio of the surface areas of the balloon in the two cases is 1:4.

Given:

A hemispherical bowl with an inner diameter of 10.5 cm.

Rate of tin-plating = $\textsf{₹}$ 16 per 100 cm².

To Find:

The cost of tin-plating the inside of the bowl.

Solution:

First, we need to find the area to be tin-plated, which is the inner curved surface area (CSA) of the bowl.

Inner radius ($r$) = $\frac{\text{Inner diameter}}{2} = \frac{10.5}{2} = 5.25$ cm.

Inner CSA = $2\pi r^2$

Inner CSA = $2 \times \frac{22}{7} \times (5.25)^2 = 2 \times \frac{22}{7} \times 5.25 \times 5.25$

Inner CSA = $2 \times 22 \times 0.75 \times 5.25$

Inner CSA = $44 \times 3.9375 = 173.25 \text{ cm}^2$.

Now, we calculate the cost of tin-plating.

The rate is $\textsf{₹}$ 16 for 100 cm².

Cost = $\frac{\text{Area to be plated}}{100} \times \text{Rate per 100 cm}^2$

Cost = $\frac{173.25}{100} \times 16$

Cost = $1.7325 \times 16 = \textsf{₹ } 27.72$.

The cost of tin-plating the inside of the bowl is $\textsf{₹ } 27.72$.

Given:

The surface area of a sphere is 154 cm².

To Find:

The radius ($r$) of the sphere.

Solution:

The formula for the surface area of a sphere is:

Surface Area = $4\pi r^2$

Substitute the given surface area and use $\pi = \frac{22}{7}$:

$154 = 4 \times \frac{22}{7} \times r^2$

$154 = \frac{88}{7} r^2$

$r^2 = \frac{154 \times 7}{88} = \frac{7 \times 7}{4} = \frac{49}{4}$.

$r = \sqrt{\frac{49}{4}} = \frac{7}{2} = 3.5$ cm.

The radius of the sphere is 3.5 cm.

Given:

Diameter of the moon $\approx \frac{1}{4} \times$ Diameter of the earth.

To Find:

The ratio of their surface areas (Moon : Earth).

Solution:

Let $D_m$ and $D_e$ be the diameters of the moon and earth, respectively. Let $r_m$ and $r_e$ be their radii.

We are given $D_m = \frac{1}{4} D_e$.

Since the radius is half the diameter, this also means $r_m = \frac{1}{4} r_e$.

The surface area of a sphere is given by the formula $4\pi r^2$.

The ratio of their surface areas is:

Ratio = $\frac{\text{Surface Area of Moon}}{\text{Surface Area of Earth}} = \frac{4\pi r_m^2}{4\pi r_e^2} = \left(\frac{r_m}{r_e}\right)^2$.

Substitute the relationship between the radii:

Ratio = $\left(\frac{\frac{1}{4} r_e}{r_e}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.

The ratio is 1:16.

The ratio of the surface area of the moon to the surface area of the earth is 1:16.

Given:

A hemispherical bowl with:

Inner radius ($r_{inner}$) = 5 cm

Thickness of the steel = 0.25 cm

To Find:

The outer curved surface area (CSA) of the bowl.

Solution:

To find the outer CSA, we first need to determine the outer radius of the bowl.

Outer radius ($r_{outer}$) = Inner radius + Thickness

$r_{outer} = 5 + 0.25 = 5.25$ cm.

The formula for the curved surface area of a hemisphere is $2\pi r^2$.

Outer CSA = $2\pi r_{outer}^2$

Outer CSA = $2 \times \frac{22}{7} \times (5.25)^2$

Outer CSA = $2 \times \frac{22}{7} \times 5.25 \times 5.25$

Outer CSA = $2 \times 22 \times 0.75 \times 5.25$

Outer CSA = $44 \times 3.9375 = 173.25 \text{ cm}^2$.

The outer curved surface area of the bowl is 173.25 cm².

Given:

A right circular cylinder that perfectly encloses a sphere of radius $r$.

Solution:

When a cylinder just encloses a sphere of radius $r$, the radius of the cylinder's base is also $r$, and the height of the cylinder is equal to the diameter of the sphere, which is $2r$.

(i) Surface area of the sphere

The formula for the surface area of a sphere is:

Surface Area of Sphere = $4\pi r^2$.

(ii) Curved surface area of the cylinder

The formula for the curved surface area (CSA) of a cylinder is $2\pi R H$, where $R$ is the cylinder's radius and $H$ is its height.

In this case, $R = r$ and $H = 2r$.

CSA of Cylinder = $2\pi (r) (2r) = 4\pi r^2$.

(iii) Ratio of the areas obtained in (i) and (ii)

Ratio = $\frac{\text{Surface Area of Sphere}}{\text{CSA of Cylinder}} = \frac{4\pi r^2}{4\pi r^2} = \frac{1}{1}$.

The ratio is 1:1.

Final Answer:

(i) Surface area of the sphere = $4\pi r^2$.

(ii) Curved surface area of the cylinder = $4\pi r^2$.

(iii) The ratio of the areas is 1:1.

Given:

Dimensions of the wall:

Length ($L$) = 10 m

Height ($H$) = 4 m

Thickness ($T$) = 24 cm

Dimensions of one brick:

Length ($l$) = 24 cm

Breadth ($b$) = 12 cm

Height ($h$) = 8 cm

To Find:

The number of bricks required to build the wall.

Solution:

To find the number of bricks required, we need to divide the total volume of the wall by the volume of a single brick. First, we must ensure all dimensions are in the same unit. Let's convert everything to centimeters (cm).

Dimensions of the wall in cm:

$L = 10 \text{ m} = 10 \times 100 = 1000$ cm

$H = 4 \text{ m} = 4 \times 100 = 400$ cm

$T = 24$ cm

Volume of the wall:

Volume of wall = $L \times H \times T$

Volume of wall = $1000 \times 400 \times 24 = 9,600,000 \text{ cm}^3$.

Volume of one brick:

Volume of one brick = $l \times b \times h$

Volume of one brick = $24 \times 12 \times 8 = 2304 \text{ cm}^3$.

Number of bricks required:

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

Number of bricks = $\frac{9,600,000}{2304}$

Number of bricks $\approx 4166.67$.

Since the number of bricks must be a whole number, we would practically need 4167 bricks (rounding up to ensure the wall is completed).

Approximately 4167 bricks would be required.

(Note: The calculation gives approximately 4166.67. This assumes no space is taken by mortar. In a practical scenario, one would buy 4167 bricks.)

Given:

A structure built from identical building blocks shaped like cubes.

The edge of each cube ($a$) = 3 cm.

To Find:

The total volume of the structure built by the child.

Solution:

First, we find the volume of a single cube.

Volume of one cube = $a^3 = (3 \text{ cm})^3 = 27 \text{ cm}^3$.

Next, we count the total number of cubes used to build the structure. By observing the figure, we can count the cubes in each layer or column:

• Bottom layer: 5 cubes
• Second layer: 4 cubes
• Third layer: 3 cubes
• Fourth layer: 2 cubes
• Top layer: 1 cube

Total number of cubes = $5 + 4 + 3 + 2 + 1 = 15$ cubes.

Now, we can find the total volume of the structure.

Volume of the structure = Total number of cubes $\times$ Volume of one cube

Volume of the structure = $15 \times 27 \text{ cm}^3$

Volume of the structure = $405 \text{ cm}^3$.

The volume of the structure built by the child is 405 cm³.

Given:

Dimensions of one matchbox: length ($l$) = 4 cm, breadth ($b$) = 2.5 cm, height ($h$) = 1.5 cm.

A packet contains 12 such matchboxes.

To Find:

The volume of the packet containing 12 matchboxes.

Solution:

First, we calculate the volume of a single matchbox. Since the matchbox is a cuboid, its volume is given by the formula:

Volume of one matchbox = $l \times b \times h$

Volume = $4 \times 2.5 \times 1.5$

Volume = $10 \times 1.5 = 15 \text{ cm}^3$.

The packet contains 12 such boxes. Therefore, the total volume of the packet is the volume of one box multiplied by 12.

Volume of the packet = $12 \times$ Volume of one matchbox

Volume of the packet = $12 \times 15 = 180 \text{ cm}^3$.

The volume of a packet containing 12 such boxes is 180 cm³.

Given:

A cuboidal water tank with dimensions:

Length ($l$) = 6 m

Width ($b$) = 5 m

Depth (height, $h$) = 4.5 m

To Find:

The capacity of the tank in litres.

Solution:

First, we calculate the volume of the tank in cubic meters (m³).

Volume of the tank = $l \times b \times h$

Volume = $6 \times 5 \times 4.5$

Volume = $30 \times 4.5 = 135 \text{ m}^3$.

We are given the conversion factor: 1 m³ = 1000 litres.

To find the capacity in litres, we multiply the volume in cubic meters by 1000.

Capacity in litres = $135 \times 1000 = 135,000$ litres.

The tank can hold 135,000 litres of water.

Given:

A cuboidal vessel with:

Length ($l$) = 10 m

Width ($b$) = 8 m

Volume (Capacity) = 380 m³

To Find:

The required height ($h$) of the vessel.

Solution:

The formula for the volume of a cuboid is:

Volume = $l \times b \times h$

Substitute the given values into the formula:

$380 = 10 \times 8 \times h$

$380 = 80 \times h$

Now, solve for the height $h$.

$h = \frac{380}{80} = \frac{38}{8} = \frac{19}{4} = 4.75$ m.

The vessel must be made 4.75 m high.

Given:

A cuboidal pit with dimensions:

Length ($l$) = 8 m

Breadth ($b$) = 6 m

Depth (height, $h$) = 3 m

Rate of digging = $\textsf{₹}$ 30 per m³.

To Find:

The total cost of digging the pit.

Solution:

First, we need to find the volume of the earth to be dug out, which is the volume of the cuboidal pit.

Volume of the pit = $l \times b \times h$

Volume = $8 \times 6 \times 3 = 144 \text{ m}^3$.

The cost of digging is given at a rate of $\textsf{₹}$ 30 per m³.

Total Cost = Volume of pit $\times$ Rate per m³

Total Cost = $144 \times 30$

Total Cost = $\textsf{₹ } 4320$.

The cost of digging the cuboidal pit is $\textsf{₹ } 4,320$.

Given:

A cuboidal tank with:

Capacity (Volume) = 50,000 litres

Length ($l$) = 2.5 m

Depth (height, $h$) = 10 m

To Find:

The breadth ($b$) of the tank.

Solution:

First, we need to convert the capacity of the tank from litres to cubic meters (m³), since the dimensions are in meters. We know that 1000 litres = 1 m³.

Volume in m³ = $\frac{50000}{1000} = 50 \text{ m}^3$.

The formula for the volume of a cuboid is:

Volume = $l \times b \times h$

Substitute the known values into the formula:

$50 = 2.5 \times b \times 10$

$50 = 25 \times b$

Now, solve for the breadth $b$.

$b = \frac{50}{25} = 2$ m.

The breadth of the tank is 2 m.

Given:

Village population = 4000

Water requirement = 150 litres per person per day

Tank dimensions: $l = 20$ m, $b = 15$ m, $h = 6$ m

To Find:

The number of days the water in the tank will last.

Solution:

First, calculate the total daily water consumption of the village.

Total daily consumption = $4000 \times 150 = 600,000$ litres.

Next, calculate the capacity of the water tank.

Volume of the tank = $l \times b \times h = 20 \times 15 \times 6 = 1800 \text{ m}^3$.

Convert the tank's capacity to litres (1 m³ = 1000 litres).

Capacity of the tank = $1800 \times 1000 = 1,800,000$ litres.

Finally, find the number of days the water will last by dividing the tank's capacity by the daily consumption.

Number of days = $\frac{\text{Capacity of the tank}}{\text{Total daily consumption}}$

Number of days = $\frac{1,800,000}{600,000} = \frac{18}{6} = 3$ days.

The water of this tank will last for 3 days.

Given:

Dimensions of the godown: $L = 40$ m, $B = 25$ m, $H = 15$ m.

Dimensions of one wooden crate: $l = 1.5$ m, $b = 1.25$ m, $h = 0.5$ m.

To Find:

The maximum number of wooden crates that can be stored in the godown.

Solution:

First, calculate the volume of the godown.

Volume of godown = $L \times B \times H = 40 \times 25 \times 15 = 15,000 \text{ m}^3$.

Next, calculate the volume of one wooden crate.

Volume of one crate = $l \times b \times h = 1.5 \times 1.25 \times 0.5 = 0.9375 \text{ m}^3$.

The maximum number of crates that can be stored is the ratio of the volume of the godown to the volume of one crate.

Number of crates = $\frac{\text{Volume of godown}}{\text{Volume of one crate}}$

Number of crates = $\frac{15,000}{0.9375}$

Number of crates = $16,000$.

The maximum number of wooden crates that can be stored in the godown is 16,000.

Given:

A solid cube with side ($a_1$) = 12 cm.

This cube is cut into eight smaller cubes of equal volume.

To Find:

1. The side of the new, smaller cube ($a_2$).

2. The ratio of their surface areas (original cube to new cube).

Solution:

1. Side of the new cube:

First, find the volume of the original cube.

Volume of original cube ($V_1$) = $a_1^3 = (12)^3 = 1728 \text{ cm}^3$.

This volume is divided equally among eight smaller cubes.

Volume of one new cube ($V_2$) = $\frac{V_1}{8} = \frac{1728}{8} = 216 \text{ cm}^3$.

The volume of the new cube is also given by $a_2^3$.

$a_2^3 = 216$

$a_2 = \sqrt[3]{216} = 6$ cm.

The side of the new cube is 6 cm.

2. Ratio of their surface areas:

The surface area of a cube is given by the formula $6a^2$.

Surface area of original cube (SA₁) = $6a_1^2 = 6 \times (12)^2 = 6 \times 144 = 864 \text{ cm}^2$.

Surface area of new cube (SA₂) = $6a_2^2 = 6 \times (6)^2 = 6 \times 36 = 216 \text{ cm}^2$.

The ratio of their surface areas is:

Ratio = $\frac{\text{SA}_1}{\text{SA}_2} = \frac{864}{216} = \frac{4}{1}$.

The ratio is 4:1.

Final Answer:

The side of the new cube is 6 cm.

The ratio between their surface areas is 4:1.

Given:

A river with:

Depth (height, $h$) = 3 m

Width ($b$) = 40 m

Rate of flow (speed) = 2 km/hr

To Find:

The volume of water that falls into the sea in one minute.

Solution:

The volume of water flowing in a certain time can be thought of as the volume of a cuboid where the length is the distance the water travels in that time, and the breadth and height are the width and depth of the river.

First, let's find the distance the water travels in one minute. The speed is given in km/hr, so we convert it to m/min.

Speed = $2 \frac{\text{km}}{\text{hr}} = \frac{2 \times 1000 \text{ m}}{60 \text{ min}} = \frac{2000}{60} \frac{\text{m}}{\text{min}} = \frac{100}{3} \frac{\text{m}}{\text{min}}$.

This means in one minute, the water flows a distance of $\frac{100}{3}$ m. This will be the "length" of our water cuboid.

Volume of water fallen in one minute = Length $\times$ Width $\times$ Depth

Volume = $\frac{100}{3} \times 40 \times 3$

Volume = $100 \times 40 = 4000 \text{ m}^3$.

4000 m³ of water will fall into the sea in a minute.

Given:

Temple pillars are in the shape of cylinders.

Radius of the base of each pillar ($r$) = 20 cm

Height of each pillar ($h$) = 10 m

Number of pillars to be built = 14

To Find:

The total volume of concrete mixture required to build all 14 pillars.

Solution:

First, we need to find the volume of a single cylindrical pillar. It is important to have all dimensions in the same unit. Let's convert the radius to meters.

Radius ($r$) = 20 cm = $\frac{20}{100}$ m = 0.2 m.

Height ($h$) = 10 m.

The formula for the volume of a cylinder is:

Volume = $\pi r^2 h$

Substituting the values for one pillar and using $\pi = \frac{22}{7}$:

Volume of one pillar = $\frac{22}{7} \times (0.2)^2 \times 10$

Volume of one pillar = $\frac{22}{7} \times 0.04 \times 10 = \frac{22}{7} \times 0.4 = \frac{8.8}{7} \text{ m}^3$.

Now, we need to find the total volume of concrete for 14 such pillars.

Total Volume = $14 \times$ Volume of one pillar

Total Volume = $14 \times \frac{8.8}{7}$

Total Volume = $2 \times 8.8 = 17.6 \text{ m}^3$.

17.6 m³ of concrete mixture would be required to build the 14 pillars.

Given:

Large cylindrical vessel: Radius ($R$) = 15 cm, Height ($H$) = 32 cm.

Small cylindrical glass: Radius ($r$) = 3 cm, Height ($h$) = 8 cm.

Selling price of one glass of juice = $\textsf{₹}$ 15.

To Find:

The total amount of money the stall keeper receives by selling all the juice.

Solution:

First, we need to find out how many small glasses can be filled from the large vessel. This can be found by dividing the total volume of juice in the vessel by the volume of one small glass.

The formula for the volume of a cylinder is $V = \pi r^2 h$.

Volume of the large vessel (total juice):

$V_{vessel} = \pi R^2 H = \pi \times (15)^2 \times 32 = \pi \times 225 \times 32 = 7200\pi \text{ cm}^3$.

Volume of one small glass:

$V_{glass} = \pi r^2 h = \pi \times (3)^2 \times 8 = \pi \times 9 \times 8 = 72\pi \text{ cm}^3$.

Number of glasses that can be filled:

Number of glasses = $\frac{\text{Volume of the large vessel}}{\text{Volume of one small glass}}$

Number of glasses = $\frac{7200\pi}{72\pi} = 100$.

So, 100 glasses of juice can be sold.

Total money received:

Total Money = Number of glasses sold $\times$ Price per glass

Total Money = $100 \times 15 = \textsf{₹ } 1500$.

The stall keeper receives $\textsf{₹ } 1500$ by selling the juice completely.

Given:

A cylindrical vessel with:

Circumference of the base = 132 cm

Height ($h$) = 25 cm

To Find:

The capacity of the vessel in litres.

Solution:

First, we need to find the radius ($r$) of the cylindrical vessel from the circumference of its base.

Circumference = $2\pi r$

$132 = 2 \times \frac{22}{7} \times r$

$132 = \frac{44}{7} r$

$r = \frac{132 \times 7}{44} = 3 \times 7 = 21$ cm.

Now, we can find the volume (capacity) of the cylindrical vessel.

Volume = $\pi r^2 h$

Volume = $\frac{22}{7} \times (21)^2 \times 25$

Volume = $\frac{22}{7} \times 21 \times 21 \times 25$

Volume = $22 \times 3 \times 21 \times 25 = 34650 \text{ cm}^3$.

To convert the volume from cm³ to litres, we use the given conversion: 1000 cm³ = 1 litre.

Capacity in litres = $\frac{34650}{1000} = 34.65$ litres.

The cylindrical vessel can hold 34.65 litres of water.

Given:

A cylindrical wooden pipe with:

Inner diameter = 24 cm

Outer diameter = 28 cm

Length (height, $h$) = 35 cm

Density of wood = 0.6 g/cm³

To Find:

The mass of the pipe.

Solution:

First, we find the inner and outer radii of the pipe.

Inner radius ($r$) = $\frac{24}{2} = 12$ cm.

Outer radius ($R$) = $\frac{28}{2} = 14$ cm.

The mass of the pipe is determined by the volume of the wood. The volume of the wood is the volume of the outer cylinder minus the volume of the inner hollow part.

Volume of wood = $\pi R^2 h - \pi r^2 h = \pi h (R^2 - r^2)$

Volume = $\frac{22}{7} \times 35 \times (14^2 - 12^2)$

Using the identity $a^2-b^2 = (a-b)(a+b)$:

Volume = $\frac{22}{7} \times 35 \times (14 - 12)(14 + 12)$

Volume = $22 \times 5 \times (2) \times (26)$

Volume = $110 \times 52 = 5720 \text{ cm}^3$.

Now, we can find the mass of the pipe.

Mass = Volume $\times$ Density

Mass = $5720 \times 0.6 = 3432$ g.

To express the mass in kilograms, we divide by 1000.

Mass = $\frac{3432}{1000} = 3.432$ kg.

The mass of the pipe is 3432 g or 3.432 kg.

Given:

(i) A tin can (cuboid) with $l = 5$ cm, $b = 4$ cm, $h = 15$ cm.

(ii) A plastic cylinder with diameter = 7 cm, height = 10 cm.

To Find:

Which container has a greater capacity and by how much.

Solution:

We need to compare the volumes of the two containers.

(i) Volume of the tin can (cuboid):

Volume = $l \times b \times h$

Volume = $5 \times 4 \times 15 = 300 \text{ cm}^3$.

(ii) Volume of the plastic cylinder:

Radius ($r$) = $\frac{\text{Diameter}}{2} = \frac{7}{2} = 3.5$ cm.

Height ($h$) = 10 cm.

Volume = $\pi r^2 h$

Volume = $\frac{22}{7} \times (3.5)^2 \times 10 = \frac{22}{7} \times 12.25 \times 10 \ $$ = 22 \times 1.75 \times 10 \ $$ = 385 \text{ cm}^3$.

Comparison:

Volume of the cylinder (385 cm³) is greater than the volume of the tin can (300 cm³).

Difference in capacity = $385 - 300 = 85 \text{ cm}^3$.

The plastic cylinder has a greater capacity by 85 cm³.

Given:

A cylinder with:

Lateral Surface Area (LSA or CSA) = 94.2 cm²

Height ($h$) = 5 cm

Use $\pi = 3.14$.

To Find:

(i) Radius of its base ($r$).

(ii) Its volume.

Solution:

(i) Radius of its base:

The formula for the LSA of a cylinder is $2\pi rh$.

$94.2 = 2 \times 3.14 \times r \times 5$

$94.2 = 10 \times 3.14 \times r$

$94.2 = 31.4 \times r$

$r = \frac{94.2}{31.4} = 3$ cm.

The radius of the base is 3 cm.

(ii) Its volume:

The formula for the volume of a cylinder is $\pi r^2 h$.

Volume = $3.14 \times (3)^2 \times 5$

Volume = $3.14 \times 9 \times 5 = 3.14 \times 45$

Volume = $141.3 \text{ cm}^3$.

Final Answer:

(i) The radius of the base is 3 cm.

(ii) The volume is 141.3 cm³.

Given:

A cylindrical vessel with:

Depth (height, $h$) = 10 m

Total cost to paint the inner curved surface = $\textsf{₹}$ 2200

Rate of painting = $\textsf{₹}$ 20 per m²

Solution:

(i) Inner curved surface area (CSA) of the vessel:

Inner CSA = $\frac{\text{Total Cost}}{\text{Rate per m}^2} = \frac{2200}{20} = 110 \text{ m}^2$.

(ii) Radius of the base:

The formula for the CSA of a cylinder is $2\pi rh$.

$110 = 2 \times \frac{22}{7} \times r \times 10$

$110 = \frac{440}{7} r$

$r = \frac{110 \times 7}{440} = \frac{7}{4} = 1.75$ m.

The radius of the base is 1.75 m.

(iii) Capacity of the vessel:

Capacity is the volume of the vessel, given by the formula $\pi r^2 h$.

Volume = $\frac{22}{7} \times (1.75)^2 \times 10$

Volume = $\frac{22}{7} \times (\frac{7}{4})^2 \times 10 = \frac{22}{7} \times \frac{49}{16} \times 10$

Volume = $22 \times \frac{7}{16} \times 10 = \frac{1540}{16} = 96.25 \text{ m}^3$.

The capacity can also be expressed in kilolitres (kl), where 1 m³ = 1 kl.

Capacity = 96.25 kl.

Final Answer:

(i) The inner curved surface area is 110 m².

(ii) The radius of the base is 1.75 m.

(iii) The capacity of the vessel is 96.25 m³ (or 96.25 kl).

Given:

A closed cylindrical vessel with:

Height ($h$) = 1 m

Capacity (Volume) = 15.4 litres

To Find:

The area of metal sheet needed in square metres (m²).

Solution:

First, we convert the capacity from litres to cubic meters (1000 litres = 1 m³).

Volume = $\frac{15.4}{1000} = 0.0154 \text{ m}^3$.

Now, we find the radius of the vessel using the volume formula, $V = \pi r^2 h$.

$0.0154 = \frac{22}{7} \times r^2 \times 1$

$r^2 = \frac{0.0154 \times 7}{22} = 0.0007 \times 7 = 0.0049$.

$r = \sqrt{0.0049} = 0.07$ m.

The area of the metal sheet needed is the Total Surface Area (TSA) of the closed cylinder.

TSA = $2\pi r(r+h)$

TSA = $2 \times \frac{22}{7} \times 0.07 \times (0.07 + 1)$

TSA = $2 \times 22 \times 0.01 \times 1.07$

TSA = $0.44 \times 1.07 = 0.4708 \text{ m}^2$.

0.4708 m² of metal sheet would be needed.

Given:

A cylindrical pencil with:

Diameter of the pencil = 7 mm

Diameter of the graphite = 1 mm

Length (height) of the pencil, $h$ = 14 cm

To Find:

1. Volume of the wood

2. Volume of the graphite

Solution:

First, convert all dimensions to cm.

Outer radius of the pencil ($R$) = $\frac{7 \text{ mm}}{2} = 3.5 \text{ mm} = 0.35$ cm.

Inner radius of the graphite ($r$) = $\frac{1 \text{ mm}}{2} = 0.5 \text{ mm} = 0.05$ cm.

Height ($h$) = 14 cm.

Volume of the Graphite:

Volume = $\pi r^2 h = \frac{22}{7} \times (0.05)^2 \times 14 = 22 \times 0.0025 \times 2 = 0.11 \text{ cm}^3$.

Volume of the Wood:

Volume = Volume of pencil - Volume of graphite

Volume = $\pi R^2 h - \pi r^2 h = \pi h (R^2 - r^2)$

Volume = $\frac{22}{7} \times 14 \times ((0.35)^2 - (0.05)^2)$

Volume = $44 \times (0.1225 - 0.0025) = 44 \times 0.12 = 5.28 \text{ cm}^3$.

The volume of the graphite is 0.11 cm³ and the volume of the wood is 5.28 cm³.

Given:

A cylindrical bowl with:

Diameter = 7 cm

Height of soup ($h$) = 4 cm

Number of patients = 250

To Find:

The total volume of soup to be prepared daily.

Solution:

First, find the volume of soup in one bowl.

Radius of the bowl ($r$) = $\frac{7}{2} = 3.5$ cm.

Volume of soup in one bowl = $\pi r^2 h$

Volume = $\frac{22}{7} \times (3.5)^2 \times 4 = \frac{22}{7} \times 12.25 \times 4 = 22 \times 1.75 \times 4 \ $$ = 154 \text{ cm}^3$.

The total volume of soup required for 250 patients is:

Total Volume = $250 \times$ Volume in one bowl

Total Volume = $250 \times 154 = 38500 \text{ cm}^3$.

To express this in litres (1 litre = 1000 cm³):

Total Volume = $\frac{38500}{1000} = 38.5$ litres.

The hospital has to prepare 38.5 litres of soup daily.

Given:

A right circular cone with:

Height ($h$) = 21 cm

Slant height ($l$) = 28 cm

To Find:

The volume of the cone.

Solution:

To find the volume of the cone, we first need to find the radius ($r$) of its base. We can use the Pythagorean relationship between the height, radius, and slant height of a cone:

$l^2 = r^2 + h^2$

Substituting the given values:

$(28)^2 = r^2 + (21)^2$

$784 = r^2 + 441$

$r^2 = 784 - 441 = 343$.

The formula for the volume of a cone is:

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

Now, we can substitute the values of $r^2$ and $h$ into the volume formula. Using $\pi = \frac{22}{7}$:

$V = \frac{1}{3} \times \frac{22}{7} \times 343 \times 21$

We can simplify the calculation:

$V = \frac{1}{3} \times 22 \times \left(\frac{343}{7}\right) \times 21$

$V = \frac{1}{3} \times 22 \times 49 \times 21$

$V = 22 \times 49 \times \left(\frac{21}{3}\right)$

$V = 22 \times 49 \times 7$

$V = 22 \times 343 = 7546 \text{ cm}^3$.

Therefore, the volume of the cone is 7546 cm³.

Given:

Total area of canvas = 551 m²

Area of canvas wasted = 1 m² (approx.)

Base radius of the conical tent ($r$) = 7 m

To Find:

The volume of the conical tent.

Solution:

First, we determine the actual area of the canvas used to make the tent. This will be the curved surface area (CSA) of the cone.

Area of canvas used (CSA) = Total area - Wastage area

CSA = $551 - 1 = 550 \text{ m}^2$.

The formula for the CSA of a cone is $CSA = \pi r l$, where $l$ is the slant height. We can use this to find $l$.

$550 = \frac{22}{7} \times 7 \times l$

$550 = 22 \times l$

$l = \frac{550}{22} = 25$ m.

Now that we have the slant height ($l=25$ m) and the radius ($r=7$ m), we can find the height ($h$) of the tent using the relation $l^2 = r^2 + h^2$.

$h^2 = l^2 - r^2 = (25)^2 - (7)^2 = 625 - 49 = 576$.

$h = \sqrt{576} = 24$ m.

Finally, we can calculate the volume of the conical tent.

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

$V = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 24$

$V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24$

$V = 1 \times 22 \times 7 \times 8 = 1232 \text{ m}^3$.

The volume of the tent that can be made is 1232 m³.

The formula for the volume of a right circular cone is $V = \frac{1}{3} \pi r^2 h$.

(i) radius 6 cm, height 7 cm

Given: $r = 6$ cm, $h = 7$ cm.

Volume = $\frac{1}{3} \times \frac{22}{7} \times (6)^2 \times 7$

Volume = $\frac{1}{3} \times 22 \times 36$

Volume = $22 \times 12 = 264 \text{ cm}^3$.

(ii) radius 3.5 cm, height 12 cm

Given: $r = 3.5$ cm, $h = 12$ cm.

Volume = $\frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 12$

Volume = $\frac{22}{7} \times 12.25 \times 4$

Volume = $22 \times 1.75 \times 4$

Volume = $22 \times 7 = 154 \text{ cm}^3$.

Final Answers:

(i) The volume of the cone is 264 cm³.

(ii) The volume of the cone is 154 cm³.

The capacity is the volume of the vessel. The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$. We also know that $l^2 = r^2 + h^2$. (1000 cm³ = 1 litre)

(i) radius 7 cm, slant height 25 cm

Given: $r = 7$ cm, $l = 25$ cm.

First, find the height ($h$): $h = \sqrt{l^2 - r^2} = \sqrt{25^2 - 7^2} \ $$ = \sqrt{625 - 49} = \sqrt{576} = 24$ cm.

Volume = $\frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 24 = \frac{1}{3} \times 22 \times 7 \times 24 \ $$ = 22 \times 7 \times 8 \ $$ = 1232 \text{ cm}^3$.

Capacity in litres = $\frac{1232}{1000} = 1.232$ litres.

(ii) height 12 cm, slant height 13 cm

Given: $h = 12$ cm, $l = 13$ cm.

First, find the radius ($r$): $r = \sqrt{l^2 - h^2} = \sqrt{13^2 - 12^2} \ $$ = \sqrt{169 - 144} = \sqrt{25} = 5$ cm.

Volume = $\frac{1}{3} \times \frac{22}{7} \times (5)^2 \times 12 = \frac{22}{7} \times 25 \times 4 = \frac{2200}{7} \text{ cm}^3$.

Capacity in litres = $\frac{2200}{7 \times 1000} = \frac{22}{70} = \frac{11}{35}$ litres.

Final Answers:

(i) The capacity is 1.232 litres.

(ii) The capacity is $\frac{11}{35}$ litres (approx. 0.314 litres).

Given:

A cone with:

Height ($h$) = 15 cm

Volume ($V$) = 1570 cm³

To Find:

The radius of the base ($r$).

Solution:

The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$.

Substitute the given values:

$1570 = \frac{1}{3} \times 3.14 \times r^2 \times 15$

$1570 = 3.14 \times r^2 \times 5$

$1570 = 15.7 \times r^2$

$r^2 = \frac{1570}{15.7} = 100$

$r = \sqrt{100} = 10$ cm.

The radius of the base is 10 cm.

Given:

A right circular cone with:

Height ($h$) = 9 cm

Volume ($V$) = 48$\pi$ cm³

To Find:

The diameter of its base.

Solution:

The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$.

Substitute the given values:

$48\pi = \frac{1}{3} \pi r^2 (9)$

Divide both sides by $\pi$:

$48 = \frac{1}{3} r^2 (9)$

$48 = 3r^2$

$r^2 = \frac{48}{3} = 16$

$r = \sqrt{16} = 4$ cm.

The diameter of the base is twice the radius.

Diameter = $2r = 2 \times 4 = 8$ cm.

The diameter of the base is 8 cm.

Given:

A conical pit with:

Top diameter = 3.5 m

Depth (height, $h$) = 12 m

To Find:

The capacity of the pit in kilolitres (kL).

Solution:

First, find the radius of the pit.

Radius ($r$) = $\frac{\text{Diameter}}{2} = \frac{3.5}{2} = 1.75$ m.

The capacity of the pit is its volume. The formula for the volume of a cone is:

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

Substitute the values:

$V = \frac{1}{3} \times \frac{22}{7} \times (1.75)^2 \times 12$

$V = \frac{22}{7} \times 3.0625 \times 4$

$V = 22 \times 0.4375 \times 4$

$V = 22 \times 1.75 = 38.5 \text{ m}^3$.

We know that 1 m³ is equivalent to 1 kilolitre (kL).

Therefore, the capacity of the pit is 38.5 kL.

The capacity of the conical pit is 38.5 kilolitres.

Given:

A right circular cone with:

Volume ($V$) = 9856 cm³

Diameter of the base = 28 cm

Solution:

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

(i) Height of the cone

Using the volume formula $V = \frac{1}{3} \pi r^2 h$:

$9856 = \frac{1}{3} \times \frac{22}{7} \times (14)^2 \times h$

$9856 = \frac{1}{3} \times \frac{22}{7} \times 196 \times h$

$9856 = \frac{1}{3} \times 22 \times 28 \times h$

$h = \frac{9856 \times 3}{22 \times 28} = \frac{9856 \times 3}{616} = 16 \times 3 = 48$ cm.

The height of the cone is 48 cm.

(ii) Slant height of the cone

Using the formula $l = \sqrt{r^2 + h^2}$:

$l = \sqrt{(14)^2 + (48)^2} = \sqrt{196 + 2304} = \sqrt{2500} = 50$ cm.

The slant height of the cone is 50 cm.

(iii) Curved surface area of the cone

Using the formula CSA = $\pi r l$:

CSA = $\frac{22}{7} \times 14 \times 50 = 22 \times 2 \times 50 = 2200 \text{ cm}^2$.

The curved surface area is 2200 cm².

Final Answer:

(i) Height of the cone = 48 cm.

(ii) Slant height of the cone = 50 cm.

(iii) Curved surface area of the cone = 2200 cm².

Given:

A right triangle with sides 5 cm, 12 cm, and 13 cm.

The triangle is revolved about the side measuring 12 cm.

To Find:

The volume of the solid formed.

Solution:

When a right triangle is revolved about one of its legs, it forms a right circular cone.

The side about which the triangle is revolved becomes the height of the cone, and the other leg becomes the radius of the base.

In this case, the sides are 5 cm, 12 cm, and 13 cm. Since $5^2+12^2 = 169 = 13^2$, the legs are 5 cm and 12 cm.

The triangle is revolved about the 12 cm side, so:

Height of the cone ($h$) = 12 cm

Radius of the base ($r$) = 5 cm

The volume of the cone is given by $V = \frac{1}{3} \pi r^2 h$.

$V = \frac{1}{3} \pi (5)^2 (12)$

$V = \frac{1}{3} \pi \times 25 \times 12 = \pi \times 25 \times 4 = 100\pi \text{ cm}^3$.

The volume of the solid obtained is $100\pi$ cm³.

Given:

A right triangle with sides 5 cm, 12 cm, and 13 cm.

From Question 7, the volume of the solid formed by revolving around the 12 cm side is $V_1 = 100\pi$ cm³.

To Find:

(i) The volume of the solid ($V_2$) formed by revolving the triangle about the 5 cm side.

(ii) The ratio of the volumes ($V_1 : V_2$).

Solution:

(i) Volume of the new solid ($V_2$):

When the triangle is revolved about the 5 cm side, a new cone is formed where:

Height ($h$) = 5 cm

Radius ($r$) = 12 cm

Volume ($V_2$) = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (12)^2 (5) = \frac{1}{3} \pi \times 144 \times 5 = \pi \times 48 \times 5 \ $$ = 240\pi \text{ cm}^3$.

(ii) Ratio of the volumes:

We need to find the ratio of the volume from Question 7 ($V_1$) to the volume from this question ($V_2$).

Ratio = $\frac{V_1}{V_2} = \frac{100\pi}{240\pi} = \frac{100}{240} = \frac{10}{24} = \frac{5}{12}$.

The ratio is 5:12.

Final Answer:

The volume of the solid obtained is $240\pi$ cm³.

The ratio of the volumes is 5:12.

Given:

A conical heap of wheat with:

Diameter = 10.5 m

Height ($h$) = 3 m

To Find:

(i) The volume of the heap.

(ii) The area of the canvas required to cover it.

Solution:

Radius ($r$) = $\frac{10.5}{2} = 5.25$ m.

(i) Volume of the heap:

Volume = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (5.25)^2 \times 3 = \frac{22}{7} \times 27.5625 \ $$ = 86.625 \text{ m}^3$.

(ii) Area of the canvas required:

The canvas covers the curved surface area (CSA) of the cone. First, we need the slant height ($l$).

$l = \sqrt{r^2 + h^2} = \sqrt{(5.25)^2 + (3)^2} = \sqrt{27.5625 + 9} = \sqrt{36.5625} \ $$ \approx 6.0467$ m.

CSA = $\pi r l = \frac{22}{7} \times 5.25 \times 6.0467$

CSA = $22 \times 0.75 \times 6.0467 = 16.5 \times 6.0467 \approx 99.77 \text{ m}^2$.

Final Answer:

The volume of the heap is 86.625 m³.

The area of the canvas required is approximately 99.77 m².

Given:

A sphere with radius ($r$) = 11.2 cm.

To Find:

The volume of the sphere.

Solution:

The formula for the volume of a sphere is given by:

$V = \frac{4}{3} \pi r^3$

Substitute the given radius and use $\pi = \frac{22}{7}$:

$V = \frac{4}{3} \times \frac{22}{7} \times (11.2)^3$

$V = \frac{4}{3} \times \frac{22}{7} \times 11.2 \times 11.2 \times 11.2$

To simplify, we can divide 11.2 by 7:

$V = \frac{4}{3} \times 22 \times 1.6 \times 11.2 \times 11.2$

$V = \frac{88}{3} \times 1.6 \times 125.44$

$V = \frac{140.8 \times 125.44}{3} = \frac{17661.952}{3}$

$V \approx 5887.32 \text{ cm}^3$.

The volume of the sphere is approximately 5887.32 cm³.

Given:

A spherical shot-putt with:

Radius ($r$) = 4.9 cm

Density of the metal ($\rho$) = 7.8 g/cm³

To Find:

The mass of the shot-putt.

Solution:

First, we need to calculate the volume of the spherical shot-putt.

Volume ($V$) = $\frac{4}{3} \pi r^3$

$V = \frac{4}{3} \times \frac{22}{7} \times (4.9)^3$

$V = \frac{4}{3} \times \frac{22}{7} \times 4.9 \times 4.9 \times 4.9$

$V = \frac{4}{3} \times 22 \times 0.7 \times 4.9 \times 4.9$

$V = \frac{88}{3} \times 0.7 \times 24.01 = \frac{1479.016}{3} \approx 493 \text{ cm}^3$.

Now, we can find the mass using the formula: Mass = Density $\times$ Volume.

Mass = $7.8 \times 493.005...$

Mass $\approx 3845.44$ g.

To express the mass in kilograms, divide by 1000:

Mass $\approx 3.845$ kg.

The mass of the shot-putt is approximately 3845.44 g (or about 3.85 kg).

Given:

A hemispherical bowl with a radius ($r$) of 3.5 cm.

To Find:

The volume of water the bowl can contain.

Solution:

The volume of water the bowl can contain is equal to the volume of the hemisphere.

The formula for the volume of a hemisphere is:

$V = \frac{2}{3} \pi r^3$

Substitute the given radius $r = 3.5$ cm (or $\frac{7}{2}$ cm) and use $\pi = \frac{22}{7}$:

$V = \frac{2}{3} \times \frac{22}{7} \times (3.5)^3$

$V = \frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5$

$V = \frac{2}{3} \times 22 \times 0.5 \times 3.5 \times 3.5$

$V = \frac{2}{3} \times 11 \times 12.25 = \frac{269.5}{3} \approx 89.83 \text{ cm}^3$.

The volume of water the bowl would contain is approximately 89.83 cm³.

The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.

(i) radius = 7 cm

Given: $r = 7$ cm.

$V = \frac{4}{3} \times \frac{22}{7} \times (7)^3$

$V = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7$

$V = \frac{4}{3} \times 22 \times 49 = \frac{4312}{3} \approx 1437.33 \text{ cm}^3$.

The volume of the sphere is $\frac{4312}{3}$ cm³ or approximately 1437.33 cm³.

(ii) radius = 0.63 m

Given: $r = 0.63$ m.

$V = \frac{4}{3} \times \frac{22}{7} \times (0.63)^3$

$V = \frac{4}{3} \times \frac{22}{7} \times 0.63 \times 0.63 \times 0.63$

$V = \frac{4}{3} \times 22 \times 0.09 \times (0.63)^2$

$V = 4 \times 22 \times 0.03 \times 0.3969$

$V = 88 \times 0.011907 \approx 1.047816 \text{ m}^3$.

The volume of the sphere is approximately 1.05 m³ (rounded to two decimal places).

The amount of water displaced is equal to the volume of the spherical ball. The volume is given by $V = \frac{4}{3} \pi r^3$.

(i) diameter = 28 cm

Radius, $r = \frac{28}{2} = 14$ cm.

$V = \frac{4}{3} \times \frac{22}{7} \times (14)^3 = \frac{4}{3} \times \frac{22}{7} \times 2744$

$V = \frac{4}{3} \times 22 \times 392 = \frac{34496}{3} \approx 11498.67 \text{ cm}^3$.

The amount of water displaced is $\frac{34496}{3}$ cm³ or approximately 11498.67 cm³.

(ii) diameter = 0.21 m

Radius, $r = \frac{0.21}{2} = 0.105$ m.

$V = \frac{4}{3} \times \frac{22}{7} \times (0.105)^3 = \frac{4}{3} \times \frac{22}{7} \times 0.001157625$

$V = \frac{88}{21} \times 0.001157625 \approx 0.004851 \text{ m}^3$.

The amount of water displaced is 0.004851 m³.

Given:

Diameter of the metallic ball = 4.2 cm

Density of the metal = 8.9 g/cm³

To Find:

The mass of the ball.

Solution:

First, calculate the volume of the spherical ball.

Radius, $r = \frac{4.2}{2} = 2.1$ cm.

Volume, $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (2.1)^3$

$V = \frac{4}{3} \times \frac{22}{7} \times 9.261 = \frac{88}{21} \times 9.261 = 38.808 \text{ cm}^3$.

Now, calculate the mass using the formula: Mass = Volume $\times$ Density.

Mass = $38.808 \times 8.9 = 345.3912$ g.

The mass of the ball is approximately 345.39 g.

Given:

Diameter of Moon ($D_m$) = $\frac{1}{4} \times$ Diameter of Earth ($D_e$).

To Find:

The fraction $\frac{\text{Volume of Moon}}{\text{Volume of Earth}}$.

Solution:

Let $r_m$ and $r_e$ be the radii of the moon and earth, respectively.

From the given relation, we have $2r_m = \frac{1}{4}(2r_e)$, which simplifies to $r_m = \frac{1}{4}r_e$.

The volume of a sphere is $V = \frac{4}{3} \pi r^3$.

The ratio of the volumes is:

$\frac{V_m}{V_e} = \frac{\frac{4}{3} \pi r_m^3}{\frac{4}{3} \pi r_e^3} = \left(\frac{r_m}{r_e}\right)^3$.

Substitute the relationship between the radii:

$\frac{V_m}{V_e} = \left(\frac{\frac{1}{4}r_e}{r_e}\right)^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$.

The volume of the moon is $\frac{1}{64}$ of the volume of the earth.

Given:

A hemispherical bowl with a diameter of 10.5 cm.

To Find:

The capacity of the bowl in litres.

Solution:

First, find the radius of the bowl.

Radius, $r = \frac{10.5}{2} = 5.25$ cm.

The volume (capacity) of a hemisphere is given by the formula $V = \frac{2}{3} \pi r^3$.

$V = \frac{2}{3} \times \frac{22}{7} \times (5.25)^3 = \frac{44}{21} \times 144.703125 \approx 303.1875 \text{ cm}^3$.

To convert the volume to litres, we use the conversion 1 litre = 1000 cm³.

Capacity = $\frac{303.1875}{1000} \approx 0.303$ litres.

The hemispherical bowl can hold approximately 0.303 litres of milk.

Given:

A hemispherical tank with:

Inner radius ($r_{inner}$) = 1 m

Thickness of the iron sheet = 1 cm

To Find:

The volume of the iron used to make the tank.

Solution:

First, ensure all units are consistent. Let's use meters.

Inner radius, $r_{inner} = 1$ m.

Thickness = 1 cm = 0.01 m.

Outer radius, $r_{outer} = r_{inner} + \text{Thickness} = 1 + 0.01 = 1.01$ m.

The volume of the iron used is the difference between the outer volume and the inner volume of the hemisphere.

$V_{iron} = \frac{2}{3} \pi (r_{outer}^3 - r_{inner}^3)$

$V_{iron} = \frac{2}{3} \times \frac{22}{7} \times ((1.01)^3 - (1)^3)$

$V_{iron} = \frac{44}{21} \times (1.030301 - 1) = \frac{44}{21} \times 0.030301$

$V_{iron} \approx 2.095 \times 0.030301 \approx 0.06348 \text{ m}^3$.

The volume of the iron used to make the tank is approximately 0.06348 m³.

Given:

Surface Area of a sphere (SA) = 154 cm².

To Find:

The volume of the sphere.

Solution:

First, we need to find the radius of the sphere from its surface area.

SA = $4 \pi r^2$

$154 = 4 \times \frac{22}{7} \times r^2$

$r^2 = \frac{154 \times 7}{4 \times 22} = \frac{154 \times 7}{88} = \frac{49}{4}$.

$r = \sqrt{\frac{49}{4}} = \frac{7}{2} = 3.5$ cm.

Now, we can find the volume of the sphere.

$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (3.5)^3$

$V = \frac{4}{3} \times \frac{22}{7} \times 42.875 = \frac{539}{3} \approx 179.67 \text{ cm}^3$.

The volume of the sphere is $\frac{539}{3}$ cm³ or approximately 179.67 cm³.

Given:

A hemispherical dome.

Total cost of whitewashing the inside = ₹ 4989.60

Rate of whitewashing = ₹ 20 per m²

Solution:

(i) Inside surface area of the dome

The inside surface area is the curved surface area (CSA) that was painted.

Inside Surface Area = $\frac{\text{Total Cost}}{\text{Rate per m}^2} = \frac{4989.60}{20} = 249.48 \text{ m}^2$.

The inside surface area is 249.48 m².

(ii) Volume of the air inside the dome

First, we find the inner radius ($r$) from the CSA.

CSA = $2 \pi r^2$

$249.48 = 2 \times \frac{22}{7} \times r^2$

$r^2 = \frac{249.48 \times 7}{44} = 39.69$

$r = \sqrt{39.69} = 6.3$ m.

Now, we find the volume of the air inside the dome, which is the volume of the hemisphere.

$V = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (6.3)^3$

$V = \frac{44}{21} \times 250.047 \approx 523.908 \text{ m}^3$.

The volume of the air inside the dome is approximately 523.9 m³.

Given:

27 small solid spheres, each of radius $r$ and surface area $S$.

These are melted to form a new, larger sphere with radius $r'$ and surface area $S'$.

Solution:

(i) Radius $r'$ of the new sphere

The volume of the new sphere is equal to the sum of the volumes of the 27 small spheres.

Volume of new sphere = $27 \times$ Volume of one small sphere

$\frac{4}{3} \pi (r')^3 = 27 \times \frac{4}{3} \pi r^3$

$(r')^3 = 27 r^3$

$r' = \sqrt[3]{27r^3} = 3r$.

The radius of the new sphere is $r' = 3r$.

(ii) Ratio of S and S′

Surface area of a small sphere, $S = 4 \pi r^2$.

Surface area of the new sphere, $S' = 4 \pi (r')^2 = 4 \pi (3r)^2 \ $$ = 4 \pi (9r^2) = 36 \pi r^2$.

The ratio of $S$ to $S'$ is:

$\frac{S}{S'} = \frac{4 \pi r^2}{36 \pi r^2} = \frac{4}{36} = \frac{1}{9}$.

The ratio $S : S'$ is 1 : 9.

Given:

A spherical medicine capsule with a diameter of 3.5 mm.

To Find:

The volume of medicine (in mm³) needed to fill the capsule.

Solution:

The amount of medicine needed is equal to the volume of the sphere.

Radius ($r$) = $\frac{\text{Diameter}}{2} = \frac{3.5}{2} = 1.75$ mm.

Volume of the sphere, $V = \frac{4}{3} \pi r^3$.

$V = \frac{4}{3} \times \frac{22}{7} \times (1.75)^3$

$V = \frac{4}{3} \times \frac{22}{7} \times 5.359375 \approx 22.458 \text{ mm}^3$.

Approximately 22.46 mm³ of medicine is needed to fill the capsule.

Given:

External dimensions of the bookshelf:

• External Height, $H = 110$ cm
• External Depth, $D = 25$ cm
• External Breadth, $B = 85$ cm

Thickness of the plank, $t = 5$ cm.

Rate of polishing = 20 paise/cm² = $\textsf{₹ } 0.20$/cm².

Rate of painting = 10 paise/cm² = $\textsf{₹ } 0.10$/cm².

To Find:

The total expense for polishing and painting the bookshelf.

Solution:

We will calculate the area to be polished and the area to be painted separately.

1. Area to be Polished (External Surfaces)

The external surfaces include the five outer faces (back, top, bottom, and two sides) plus the front frame.

Area of the 5 outer faces = (Area of back) + 2(Area of sides) + (Area of top) + (Area of bottom)

Area = $(85 \times 110) + 2(25 \times 110) + (85 \times 25) + (85 \times 25)$

Area = $9350 + 5500 + 2125 + 2125 = 19100 \text{ cm}^2$.

Area of the front frame = Area of outer front rectangle - Area of inner front opening

Inner breadth = $85 - 2 \times 5 = 75$ cm. Inner height = $110 - 2 \times 5 = 100$ cm.

Area of front frame = $(85 \times 110) - (75 \times 100) = 9350 - 7500 \ $$ = 1850 \text{ cm}^2$.

Total Area to be Polished = $19100 + 1850 = 20950 \text{ cm}^2$.

Cost of Polishing = $20950 \times 0.20 = \textsf{₹ } 4190$.

2. Area to be Painted (Internal Surfaces)

The internal surfaces are those of the three compartments. We can find this by summing the areas of all the inner faces.

Internal dimensions: Breadth ($b_i$) = 75 cm, Depth ($d_i$) = 20 cm, Height ($h_i$) = 100 cm.

Area of inner 5 faces of the box = (Inner Back) + 2(Inner Sides) + (Inner Top) + (Inner Bottom)

Area = $(75 \times 100) + 2(20 \times 100) + (75 \times 20) + (75 \times 20)$

Area = $7500 + 4000 + 1500 + 1500 = 14500 \text{ cm}^2$.

Area of the two shelves (each shelf has a top, bottom, and front edge to be painted):

Area of one shelf = 2(top/bottom) + (front edge) = $2(75 \times 20) + (75 \times 5) = 3000 + 375 = 3375 \text{ cm}^2$.

Area of two shelves = $2 \times 3375 = 6750 \text{ cm}^2$.

Total Area to be Painted = Area of inner 5 faces + Area of two shelves

Total Area to be Painted = $14500 + 6750 = 21250 \text{ cm}^2$.

Cost of Painting = $21250 \times 0.10 = \textsf{₹ } 2125$.

3. Total Expense

Total Expense = Cost of Polishing + Cost of Painting

Total Expense = $\textsf{₹ } 4190 + \textsf{₹ } 2125 = \textsf{₹ } 6315$.

The total expense required is $\textsf{₹} 6315$.

Given:

• Number of spheres = 8, Diameter = 21 cm
• Number of cylindrical supports = 8, Radius ($r$) = 1.5 cm, Height ($h$) = 7 cm
• Silver paint cost = 25 paise/cm² = $\textsf{₹ } 0.25$/cm²
• Black paint cost = 5 paise/cm² = $\textsf{₹ } 0.05$/cm²

To Find:

The total cost of the paint required.

Solution:

1. Cost of Silver Paint (for Spheres):

Radius of one sphere ($R$) = $\frac{21}{2} = 10.5$ cm.

Surface area of one sphere = $4\pi R^2 = 4 \times \frac{22}{7} \times (10.5)^2 = 1386 \text{ cm}^2$.

The area where the sphere rests on the cylinder is not painted. This area is a circle with the radius of the cylinder, $r = 1.5$ cm.

Area of base not painted = $\pi r^2 = \frac{22}{7} \times (1.5)^2 \approx 7.07 \text{ cm}^2$.

Area to be painted silver on one sphere = $1386 - 7.07 = 1378.93 \text{ cm}^2$.

Total area for 8 spheres = $8 \times 1378.93 = 11031.44 \text{ cm}^2$.

Cost of silver paint = $11031.44 \times 0.25 = \textsf{₹ } 2757.86$.

2. Cost of Black Paint (for Cylinders):

The curved surface area of the cylindrical supports is painted black.

CSA of one cylinder = $2\pi rh = 2 \times \frac{22}{7} \times 1.5 \times 7 = 66 \text{ cm}^2$.

Total area for 8 cylinders = $8 \times 66 = 528 \text{ cm}^2$.

Cost of black paint = $528 \times 0.05 = \textsf{₹ } 26.40$.

3. Total Cost:

Total Cost = Cost of Silver Paint + Cost of Black Paint

Total Cost = $\textsf{₹ } 2757.86 + \textsf{₹ } 26.40 = \textsf{₹ } 2784.26$.

The total cost of the paint required is $\textsf{₹} 2784.26$.

Given:

The diameter of a sphere is decreased by 25%.

To Find:

The percentage decrease in its curved surface area.

Solution:

Let the original diameter of the sphere be $D$.

The original radius is $R = \frac{D}{2}$.

The original curved surface area (CSA) is $S = 4\pi R^2 = 4\pi (\frac{D}{2})^2 = \pi D^2$.

The diameter is decreased by 25%. The new diameter, $D'$, is:

$D' = D - (25\% \text{ of } D) = D - 0.25D = 0.75D = \frac{3}{4}D$.

The new radius, $R'$, is $\frac{D'}{2} = \frac{3/4 D}{2} = \frac{3}{8}D$.

The new curved surface area, $S'$, is:

$S' = 4\pi (R')^2 = 4\pi (\frac{3}{8}D)^2 = 4\pi \frac{9}{64}D^2 = \frac{9}{16}\pi D^2$.

The decrease in surface area is $S - S' = \pi D^2 - \frac{9}{16}\pi D^2 = \frac{7}{16}\pi D^2$.

The percentage decrease is calculated as:

Percentage Decrease = $\frac{\text{Decrease in Area}}{\text{Original Area}} \times 100\%$

Percentage Decrease = $\frac{\frac{7}{16}\pi D^2}{\pi D^2} \times 100\% = \frac{7}{16} \times 100\% = 43.75\%$.

The curved surface area of the sphere decreases by 43.75%.