CBSE - Surface Areas and Volumes

Answer

1. Length of plastic box = 1.5 m
Width of plastic box = 1.25 m
Depth of plastic box = 1.25 m
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.
Surface area of the box = Lateral surface area + Area of the base
= 2(l+b)×h + (l×b)
= 2[(1.5 + 1.25)×1.25] + (1.5 × 1.25) m² 
                                              = (3.575 + 1.875) m² 
= 5.45 m² 
The sheet required required to make the box is 5.45 m² 
(ii) Cost of 1 m² of sheet = Rs 20
∴ Cost of 5.45 m² of sheet = Rs (20 × 5.45) = Rs 109

2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of rs 7.50 per m².

Answer

length of the room = 5m
breadth of the room = 4m
height of the room = 3m
Area of four walls including the ceiling = 2(l+b)×h + (l×b)
= 2(5+4)×3 + (5×4) m² 
 = (54 + 20) m² 
= 74 m² Cost of white washing = ���7.50 per m²
Total cost = rs (74×7.50) = rs 555

Answer

Perimeter of rectangular hall = 2(l + b) = 250 m
Total cost of painting = ���15000
Rate per m² = rs10
Area of four walls = 2(l + b) h m² = (250×h) m²
A/q,
(250×h)×10 = rs15000
⇒ 2500×h = rs15000
⇒ h = 15000/2500 m
⇒ h = 6 m
Thus the height of the hall is 6 m.

Answer

Volume of paint = 9.375 m² = 93750 cm²
Dimension of brick = 22.5 cm×10 cm×7.5 cm
Total surface area of a brick = 2(lb + bh + lh) cm²
= 2(22.5×10 + 10×7.5 + 22.5×7.5) cm²
= 2(225 + 75 + 168.75) cm²
= 2×468.75 cm² = 937.5 cm²
Number of bricks can be painted = 93750/937.5 = 100

(i) Lateral surface area of cubical box of edge 10cm = 4×10² cm² = 400 cm²
Lateral surface area of cuboid box = 2(l+b)×h
= 2×(12.5+10)×8 cm²
= 2×22.5×8 cm² = 360 cm²
Thus, lateral surface area of the cubical box is greater by (400 – 360) cm² = 40 cm²

(ii) Total surface area of cubical box of edge 10 cm =6×102cm²=600cm²
Total surface area of cuboidal box = 2(lb + bh + lh)
= 2(12.5×10 + 10×8 + 8×12.5)cm²
= 2(125+80+100) cm²
= (2×305) cm² = 610 cm²
Thus, total surface area of cubical box is smaller by 10 cm²

(i) Dimensions of greenhouse:
l = 30 cm, b = 25 cm, h = 25 cm
Total surface area of green house = 2(lb + bh + lh)
= 2(30×25 + 25×25 + 25×30) cm²
= 2(750 + 625 + 750) cm²
= 4250 cm²
(ii) Length of the tape needed = 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4×80 cm = 320 cm

Dimension of bigger box = 25 cm × 20 cm × 5 cm
Total surface area of bigger box = 2(lb + bh + lh)
= 2(25×20 + 20×5 + 25×5) cm²
= 2(500 + 100 + 125) cm²
= 1450 cm²
Dimension of smaller box = 15 cm × 12 cm × 5 cm
Total surface area of smaller box = 2(lb + bh + lh)
= 2(15×12 + 12×5 + 15×5) cm²
= 2(180 + 60 + 75) cm²
= 630 cm²
Total surface area of 250 boxes of each type = 250(1450 + 630)cm²
= 250×2080 cm² = 520000 cm²
Extra area required = 5/100(1450 + 630) × 250 cm² = 26000 cm²

Total Cardboard required = 520000 + 26000 cm² = 546000 cm²
Total cost of cardboard sheet = rs (546000 × 4)/1000 = rs 2184

Dimensions of the box- like structure = 4m × 3m × 2.5
Tarpaulin only required for all the four sides and top.
Thus, Tarpaulin required = 2(l+b)×h + lb
= [2(4+3)×2.5 + 4×3] m²
= (35×12) m²
= 47 m²

Let r be the radius of the base and h = 14 cm be the height of the cylinder.
Curved surface area of cylinder = 2πrh = 88 cm²
⇒ 2 × 22/7 × r × 14 = 88
⇒ r = 88/ (2 × 22/7 × 14)
⇒ r = 1 cm
Thus, the diameter of the base = 2r = 2×1 = 2cm

Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 cm and Height (h) = 1m
Radius of base (r) = 140/2 = 70 cm = 0.7 m
Metal sheet required to make a closed cylindrical tank = 2πr(h + r)
= (2 × 22/7 × 0.7) (1 + 0.7) m²
= (2 × 22 × 0.1 × 1.7) m²
=7.48 m²

Let R be external radius and r be the internal radius h be the length of the pipe.

R  = 4.4/2 cm = 2.2 cm

r = 4/2 cm = 2 cm
h = 77 cm
(i) Inner curved surface = 2πrh cm²
= 2 × 22/7 × 2 × 77cm²
= 968 cm²

(ii) Outer curved surface = 2πRh cm²

= 2 × 22/7 × 2.2 × 77 cm²
= 1064.8 cm²

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases

= 2πrh + 2πRh + 2π(R^{2 –} r²)
= [968 + 1064.8 + (2 × 22/7) (4.84 – 4)] cm²
= (2032.8 + 44/7 × 0.84) cm²
= (2032.8 + 5.28) cm² = 2038.08 cm²

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

Answer

Length of the roller (h) = 120 cm = 1.2 m
Radius of the cylinder = 84/2 cm = 42 cm = 0.42 m
Total no. of revolutions = 500
Distance covered by roller in one revolution = Curved surface area = 2πrh
= (2 × 22/7 × 0.42 × 1.2) m²
= 3.168 m²
Area of the playground = (500 × 3.168) m² = 1584 m²

Radius of the pillar (r) = 50/2 cm = 25 cm = 0.25 m
Height of the pillar (h) = 3.5 m.
Rate of painting = rs12.50 per m²
Curved surface = 2πrh
= (2 × 22/7 × 0.25 × 3.5) m²
                              =5.5 m²
Total cost of painting = (5.5 × 12.5) = rs 68.75

Let r be the radius of the base and h be the height of the cylinder.
Curved surface area = 2πrh = 4.4 m²
⇒ 2 × 22/7 × 0.7 × h = 4.4
⇒ h = 4.4/(2 × 22/7 × 0.7) = 1m
⇒ h = 1m

Radius of circular well (r) = 3.5/2 m = 1.75 m
Depth of the well (h) = 10 m
Rate of plastering = rs 40 per m²
(i) Curved surface = 2πrh
= (2 × 22/7 × 1.75 × 10) m²
= 110 m²

(ii) Cost of plastering = ���(110 × 40) = rs4400

Radius of the pipe (r) = 5/2 cm = 2.5 cm = 0.025 m
Length of the pipe (h) = 28/2 m = 14 m
Total radiating surface = Curved surface area of the pipe = 2πrh
= (2 × 22/7 × 0.025 × 28) m²
                                             = 4.4 m²

(i) Radius of the tank (r) = 4.2/2 m = 2.1 m
Height of the tank (h) = 4.5 m
Curved surface area = 2πrh m²
= (2 × 22/7 × 2.1 × 4.5) m²
= 59.4 m²
(ii) Total surface area of the tank = 2πr(r + h) m²
= [2 × 22/7 × 2.1 (2.1 + 4.5)] m²
= 87.12 m²

Let x be the actual steel used in making tank.
∴ (1 – 1/12) × x = 87.12
⇒ x = 87.12 × 12/11
⇒ x = 95.04 m²

Radius of the frame (r) = 20/2 cm = 10 cm
Height of the frame (h) = 30 cm + 2×2.5 cm = 35 cm
2.5 cm of margin will be added both side in the height.
Cloth required for covering the lampshade = curved surface area = 2πrh
= (2 × 22/7 × 10 × 35) cm²
= 2200 cm²

Radius of the penholder (r) = 3cm
Height of the penholder (h) = 10.5cm
Cardboard required by 1 competitor = CSA of one penholder + area of the base
= 2πrh + πr²
= [(2 × 22/7 × 3 × 10.5) + 22/7 × 3²] cm²
= (198 + 198/7) cm²
= 1584/7 cm²
Cardboard required for 35 competitors = (35 × 1584/7) cm²
= 7920 cm²

Radius (r) = 10.5/2 cm = 5.25 cm

Slant height (l) = 10 cm

Curved surface area of the cone = (πrl) cm²

=(22/7 × 5.25 × 10) cm²

=165 cm²

Radius (r) = 24/2 m = 12 m
Slant height (l) = 21 m
Total surface area of the cone = πr (l + r) m²
= 22/7 × 12 × (21 + 12) m²
= (22/7 × 12 × 33) m²
= 1244.57 m²

(i) Curved surface of a cone = 308 cm²
Slant height (l) = 14cm
Let r be the radius of the base
∴ πr��� = 308
⇒ 22/7 × r × 14 = 308
⇒ r =308/(22/7 × 14) = 7 cm

(ii) TSA of the cone = πr(l + r) cm²
= 22/7 × 7 ×(14 + 7) cm²
= (22 × 21) cm²
= 462 cm²

Radius of the base (r) = 24 m
Height of the conical tent (h) = 10 m
Let l be the slant height of the cone.
∴ l² = h² + r²
⇒ l = √h² + r²
⇒ l = √10² + 24²
⇒ l = √100 + 576
⇒ l = 26 m
(ii) Canvas required to make the conical tent = Curved surface of the cone
Cost of 1 m² canvas = rs70
∴ πrl = (22/7 × 24 × 26) m² = 13728/7 m²
∴ Cost of canvas = rs 13728/7 × 70 = rs 137280

Radius of the base (r) = 6 m
Height of the conical tent (h) = 8 m
Let l be the slant height of the cone.
∴ l = √h² + r²
⇒ l = √8² + 6²
⇒ l = √100
⇒ l = 10 m
CSA of conical tent = πrl
= (3.14 × 6 × 10) m² = 188.4 m²
Breadth of tarpaulin = 3 m
Let length of tarpaulin sheet required be x.
20 cm will be wasted in cutting.
So, the length will be (x – 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(x – 0.2 m) × 3] m = 188.4 m²
⇒ x – 0.2 m = 62.8 m
⇒ x = 63 m

Radius (r) = 14/2 m = 7 m
Slant height tomb (l) = 25 m
Curved surface area = πrl m²
=(227×25×7) m²
=550  m²
Rate of white- washing = rs 210 per 100 m²
Total cost of white-washing the tomb = rs (550 × 210/100) = rs1155

Radius of the cone (r) = 7 cm
Height of the cone (h) = 24 cm
Let l be the slant height
∴ l = √h² + r²
⇒ l = √24² + 7²
⇒ l = √625
⇒ l = 25 m
Sheet required for one cap = Curved surface of the cone
= πrl cm²
= (22/7 × 7 × 25) cm²
= 550 cm²
Sheet required for 10 caps = 550 × 10 cm² = 5500 cm²

Radius of the cone (r) = 40/2 cm = 20 cm = 0.2 m
Height of the cone (h) = 1 m
Let l  be the slant height of a cone.
∴ l = √h² + r²
⇒ l = √1² + 0.2²
⇒ l = √1.04
⇒ l = 1.02 m
Rate of painting = rs12 per m²

Curved surface of 1 cone = πrl m²
= (3.14 × 0.2 × 1.02) m²
= 0.64056 m²
Curved surface of such 50 cones = (50 × 0.64056) m²
                                                                 = 32.028 m²
Cost of painting all these cones = rs(32.028 × 12)
= rs 384.34

(i) Radius of the sphere (r) = 10.5 cm
Surface area = 4πr²
                    = (4 × 22/7 × 10.5 × 10.5) cm²
                    = 1386 cm²

(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr²
                     = (4 × 22/7 × 5.6 × 5.6) cm²
                     = 394.24 cm²

(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr²
                     = (4 × 22/7 × 14 × 14) cm²
                     = 2464 cm²

(i) Radius of the sphere (r) = 10.5 cm
Surface area = 4πr²
                    = (4 × 22/7 × 10.5 × 10.5) cm²
                    = 1386 cm²

(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr²
                     = (4 × 22/7 × 5.6 × 5.6) cm²
                     = 394.24 cm²

(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr²
                     = (4 × 22/7 × 14 × 14) cm²
                     = 2464 cm²

r = 10 cm  
Total surface area of hemisphere = 3πr²
                                                     = (3 × 3.14 × 10 ×10) cm²
                                                     = 942 cm²

Let r be the initial radius and R be the increased radius of balloons.
r = 7cm and R = 14cm
Ratio of the surface area =4πr²/4πR²
                                        = r²/R²
                                        = (7×7)/(14×14) = 1/4
Thus, the ratio of surface areas = 1 : 4

Radius of the bowl (r) = 10.5/2 cm = 5.25 cm
Curved surface area of the hemispherical bowl = 2πr²
                                                                        = (2 × 22/7 × 5.25 × 5.25) cm²
                                                           = 173.25 cm²
Rate of tin - plating is = rs16 per 100 cm²
Therefor, cost of 1 cm² = rs16/100 
Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100
                                                                         = rs 27.72

Let r be the radius of the sphere.
Surface area = 154 cm²
⇒ 4πr² = 154
⇒ 4 × 22/7 × r² = 154
⇒ r² = 154/(4 × 22/7)
⇒ r² = 49/4
⇒ r = 7/2 = 3.5 cm

Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = r/2
Radius of the moon = r/8
Ratio of their surface area = 4π(r/8)²/4π(r/2)²
                                                    = (1/64)/(1/4)
                                          = 4/64 = 1/16
Thus, the ratio of their surface areas is 1:16

Inner radius of the bowl (r) = 5 cm
Thickness of the steel = 0.25 cm
∴ outer radius (R) = (r + 0.25) cm
                              = (5 + 0.25) cm  = 5.25 cm
Outer curved surface = 2πR²
                                  = (2 × 22/7 × 5.25 × 5.25) cm²
                                  = 173.25 cm²

(i) The surface area of the sphere with raius r = 4πr²

(ii) The right circular cylinder just encloses a sphere of radius r.
∴ the radius of the cylinder = r and its height = 2r
∴ Curved surface of cylinder =2πrh
                                               = 2π × r × 2r
                                               = 4πr²
(iii) Ratio of the areas = 4πr²:4πr² = 1:1

Dimension of matchbox = 4cm × 2.5cm × 1.5cm
l = 4 cm, b = 2.5 cm and h = 1.5 cm
Volume of one matchbox = (l × b × h)
                                         = (4 × 2.5 × 1.5)  cm³ = 15 cm³
Volume of a packet containing 12 such boxes = (12 × 15)  cm³ = 180 cm³

Dimensions of water tank = 6m x  5m x 4.5m
l = 6m , b = 5m and h = 4.5m
Therefore Volume of the tank =lbh m³
 =(6 x 5 x 4.5)m3=135 m³
Therefore , the tank can hold = 135 x 1000 litres[Since 1m3=1000litres]
= 135000 litres of water.

Length = 10 m , Breadth = 8 m and Volume = 380 m³
Volume of cuboid = Length × Breadth × Height
⇒ Height = Volume of cuboid/(Length × Breadth) 
= 380/(10×8) m
= 4.75m

l = 8 m, b = 6 m and h = 3 m
Volume of the pit = lbh m³
 = (8×6×3) m³
 = 144 m³
Rate of digging = rs 30 per m³
Total cost of digging the pit =  rs(144 × 30) =  rs 4320

length = 2.5 m, depth = 10 m and volume = 50000 litres
1m³ = 1000 litres

∴ 50000 litres = 50000/1000 m³ = 50 m³
Breadth = Volume of cuboid/(Length x Depth)
= 50/(2.5 x 10) m
= 2 m

Dimension of tank = 20m x 15m x 6m
l = 20 m , b = 15 m and h = 6 m
Capacity of the tank = lbh m³
= (20 x 15 x 6) m³
= 1800 m³
Water requirement per person per day =150 litres
Water required for 4000 person per day = (4000 x 150) l
                                                                = (4000 x 150) / 1000
                                                                = 600 m³
Number of days the water will last = Capacity of tank Total water required per day
                                                        =(1800/600) = 3
The water will last for 3 days.

Dimension of godown = 40 m x 25 m x 15 m
Volume of the godown = (40 x 25 x 15) m³ = 10000 m³
Dimension of crates = 1.5m x 1.25m x 0.5m 
Volume of 1 crates = (1.5 x 1.25 x 0.5) m³ = 0.9375 m³
Number of crates that can be stored =Volume of the godown/Volume of 1 crate
= 10000/0.9375 = 10666.66 = 10666

Edge of the cube = 12 cm. 
Volume of the cube = (edge)³ cm³
= (12 x 12 x 12) cm³
= 1728 cm³ 
Number of smaller cube = 8
Volume of the 1 smaller cube =1728/8 cm³ = 216 cm³
Side of the smaller cube = a
a³ = 216
⇒ a = 6 cm
Surface area of the cube = 6 (side)²
Ratio of their surface area = (6 x 12 x 12)/(6 x 6 x 6)
                                           = 4/1 = 4:1

Depth of river (h) = 3 m
Width of river (b) = 40 m
Rate of flow of water (l) = 2 km per hour = (2000/60) m per minute
= 100/3 m per minute 
Volume of water flowing into the sea in a minute = lbh m³
= (100/3 x  40 x 3) m³
= 4000 m³

Given; circumference = 132 cm, h = 25 cm
Circumference = 2 π r
Or, 132 = 2 πr
Or, r = (132 X7)/(2 X 22) = 21 cm

Volume of cylinder = π r²h
= (22/7) X 21² X 25
= 34650 cubic cm = 34650/1000 litre
= 34.65 litre

Given; R = 28 cm, r = 24 cm, h = 35 cm
Volume of pipe = π R²h – π r²h
= π h (R² - r²)
= (22/7) X 35 (28² - 24²)
= 110 (28 + 24)(28 – 24)
= 110 X 52 X 4 = 22880 cubic cm
Mass = 22880 X 0.6 g = 13728 g

Rectangular can: l = 5 cm, b = 4 cm, h = 15 cm
Volume of cubcoid = l X b X h
= 5 X 4 X 15 = 300 cubic cm

Cylindrical can: r = 35 cm, h = 10 cm
Volume of cylinder = π r² h
= (22/7) X 35² X 10
= 385 cubic cm
Difference = 385 – 300 = 85 cubic cm

Given; curved surface area of cylinder = 94.2 sq cm, h = 5 cm
CSA of cylinder = 2 π rh
Or, 94.2 = 2 X (3.14) X r X 5
Or, r = 94.2/(2 X 3.14 X 5) = 3 cm

Now, volume of cylinder = π r² h
= 3.14 X 3² X 5 = 141.30 cubic cm

Cost = Rs. 2200, rate = Rs. 20 per sq m, h = 10 m
Curved surface area = Cost/Rate
= 2200/20 = 110 sq cm

CSA of cylinder = 2 π rh
Or, 110 = 2 X (22/7) X r X 10
Or, r = (110 X 7)/(2 X 22 X 10) = 1.75 cm

Now, volume of cylinder = π r² h
= (22/7) X (1.75)² X 10 = 96.25 cubic m

Capacity = 15.4 litre = 15400 cubic cm, h = 1 m = 100 cm
Volume of cylinder = π r² h
Or, 15400 = (22/7) X r² X 100
Or, r² = (15400 X 7)/(22 X 100) = 49
Or, r = 7 cm

Now, total surface area of cylinder = 2 π r(r + h)
= 2 X (22/7) X 7 (7 + 10)
= 2 X 22 X 17 = 748 sq cm

Radius of pencil = 3.5 mm, h = 14 cm = 140 mm
Volume of pencil = π r²h
= (22/7) X (3.5)² X 140 = 5390 cubic mm

Radius of lead = 0.5 mm, h = 140 mm
Volume of lead = π r² h
= (22/7) X (0.5)² X 140 = 110 cubic mm
Hence, volume of wood = 5390 – 110 = 5280 cubic mm

Given; r = 3.5 cm, h = 4 cm
Volume of cylinder = π r² h
= (22/7) X (3.5)² X 4 = 154 cubic cm
Volume of 250 bowls = 250 X 154 = 38500 cubic cm = 38.5 litre

(i) radius 6 cm, height 7 cm

Given; r = 6 cm, h = 7 cm
Volume of cone = 1/3 π r²h
= (1/3) X (22/7) X 6² X 7 = 264 cubic cm

(ii) radius 3.5 cm, height 12 cm

 Given; r = 3.5 cm, h = 12 cm
Volume of cone = 1/3 π r² h
= (1/3) X (22/7) X (3.5)² X 12 = 154 cubic cm

(i) radius 7 cm, slant height 25 cm

Answer: Given; r = 7 cm, l = 25 cm
Here; h² = l² - r²
= 25² - 7²
= 625 – 49 = 576
Or, h = 24 cm

Volume of cone = 1/3 π r² h
= (1/3) X (22/7) X 7² X 24 = 1232 cubic cm
= 1.232 litre

(ii) height 12 cm, slant height 13 cm

Answer: Given; h = 12 cm, l = 13 cm
Here, r² = l² - h²
= 13² - 12²
= 169 – 144 = 25
Or, r = 5 cm

Volume of cone = 1/3 π r² h
= (1/3) X (3.14) X 5² X 12 = 314 cubic cm
= 0.314 litre

Given; volume = 1570 cubic cm, h = 15 cm, r = ?
Volume of cone = 1/3 π r²h
Or, 1570 = (1/3) X (3.14) X r² X 15
Or, r² = (1570 X 3)/(3.14 X 15) = 100
Or, r = 10 cm

Given; volume = 48 π cubic cm, h = 9 cm, r = ?
Volume of cone = 1/3 π r² h
Or, 48 π = 1/3 π r² X 9
Or, r² = 48/3 = 16
Or, r = 4 cm

Given; r = 3.5 m, h = 12 m
Volume of cone = 1/3 π r² h
= (1/3) X (22/7) X (3.5)² X 12 = 154 cubic cm
= 154 kilo litre

Given; volume = 9856 cubic cm, d = 28 cm so, r = 14 cm
Volume of cone = 1/3 π r² h
Or, 9856 = (1/3) X (22/7) X 14² X h
Or, h = (9856 X 3 X 7)/(22 X 196) = 48 cm

Slant Height:
Here; l² = h² + r²
= 48² + 14² = 2500
Or, l = 50 cm

Curved surface area of cone = π rl
= (22/7) X 14 X 50 = 2200 sq cm

Given; h = 12 cm, r = 5 cm
Volume of cone = 1/3 π r² h
= (1/3) π X 5² X 12 = 100 π cubic cm

Given; h = 5 cm, r = 12 cm
Volume of cone = 1/3 π r² h
= (1/3) π X 12² X 5 = 240 π cubic cm

Ratio = 100/240 = 5 : 12

Given; r = 5.25 m, h = 3 m
Volume of cone = 1/3 π r²h
= (1/3) X (22/7) X (5.25) X 3

= 50.625 cubic m

(i) 7 cm

Solution: Volume of sphere = 4/3 π r²
= (4/3) X (22/7) X 7³ = 1437.33 cubic cm

(ii) 0.63 m

Solution: Volume of sphere = 4/3 π r³
= (4/3) X (22/7) X (0.63)³ = 1.047 cubic m

(i) 28 cm

Solution: Volume of sphere = 4/3 π r³
= (4/3) X (22/7) X 28³ = 91989.33 cubic cm

(ii) 0.21 m

Solution: Volume of sphere = 4/3 π r³
= (4/3) X (22/7) X (0.21)³ = 0.38808 cubic m

Solution: 

Volume of sphere = 4/3 π r³
= (4/3) X (22/7) X (4.2)³ = 310.464 cubic cm

Mass = volume X density
= 310.464 X 8.9
= 2763.1296 gm = 2.76 kg

Solution: 

Volume of two similar shapes are in triplicate ratio of their dimensions. For example; if radii are R and r then ratio of volumes = R³ : r³
Hence, volume of earth/volume of moon
= 4³ : 1³
= 64 : 1

Solution: 

Volume of hemisphere = 2/3 π r³
= (2/3) X (22/7) X (5.25)³ = 303.1875 cubic cm = 0.303 litre

Solution: 

Inner radius r = 1 m, outer radius r = 1.01 m
Volume of metal = 4/3 π (R³ - r³)
= (4/3) X (22/7) [(1.01)³ - 1³]
= (4/3) X (22/7) X 0.030301 = 0.06348 cubic m

Solution: 

Surface area of sphere = 4 π r²
Or, 154 = 4 X (22/7) X r²
Or, r² = (154 X 7)/(22 X 4) = 49/4
Or, r = 7/2 = 3.5 cm

Volume of sphere = 4/3 π r³
= (4/3) X (22/7) X (3.5)³ = 179.67 cubic cm

Solution: 

Curved surface area of hemisphere = cost/rate
= 498.96/2 = 249.48 sq m
Or, 2 π r² = 249.48
Or, r² = (249.48 X 7)/(2 X 22)
Or, r = 6.3 m

Volume of hemisphere = 2/3 π r³
= (2/3) X (22/7) X (6.3)³ = 523.908 cubic m

Solution: 

Here; ratio of volumes = 27 : 1
Radii shall be in sub-triplicate ratio, i.e. 3 : 1
Because 3³ : 1³ = 27 : 1
Now, surface areas shall be in duplicate ratio of radii
Hence, ratio of surface areas = 3² : 1² = 9 : 1

Solution: 

Volume of sphere = 4/3 π r³
= (4/3) X (22/7) X (1.75)³
= 22.46 cubic mm

Solution: 

External height (l) of book self = 85 cm

External breadth (b) of book self = 25 cm

External height (h) of book self = 110 cm

External surface area of shelf while leaving out the front face of the shelf

= lh + 2 (lb + bh)

= [85 x 110 + 2 (85 x 25 + 25 x 110)] cm²

= (9350 + 9750) cm²

= 19100 cm²

Area of front face = [85 x 110 - 75 x 100 + 2 (75 x 5)] cm²

= 1850 + 750 cm²

= 2600 cm²

Area to be polished = (19100 + 2600) cm² = 21700 cm²

Cost of polishing 1 cm² area = Rs 0.20

Cost of polishing 21700 cm² area Rs (21700 x 0.20) = Rs 4340

 

It can be observed that length (l), breadth (b), and height (h) of each row of the book shelf is 75 cm, 20 cm, and 30 cm respectively.

Area to be painted in 1 row = 2 (l + h) b + lh

= [2 (75 + 30) x 20 + 75 x 30] cm²

= (4200 + 2250) cm²

= 6450 cm²

Area to be painted in 3 rows = (3 x 6450) cm² = 19350 cm²

Cost of painting 1 cm² area = Rs 0.10

Cost of painting 19350 cm² area = Rs (19350 x 0.1)

= Rs 1935

Total expense required for polishing and painting = Rs (4340 + 1935)

= Rs 6275

Therefore, it will cost Rs 6275 for polishing and painting the surface of the bookshelf.

Solution: 

Radius (r) of wooden sphere = 21/2 cm = 10.5 cm
Surface area of wooden sphere = 4πr² 

Radius (r₁) of the circular end of cylindrical support = 1.5 cm

Height (h) of cylindrical support = 7 cm

CSA of cylindrical support = 2πrh

= 7.07 cm²

Area to be painted silver = [8 × (1386 - 7.07)] cm²

= (8 × 1378.93) cm² = 11031.44 cm²

Cost for painting with silver colour = Rs (11031.44 × 0.25) = Rs 2757.86

Area to be painted black = (8 × 66) cm² = 528 cm²

Cost for painting with black colour = Rs (528 × 0.05) = Rs 26.40

Total cost in painting = Rs (2757.86 + 26.40)

= Rs 2784.26

Therefore, it will cost Rs 2784.26 in painting in such a way.

Let the diameter of the sphere be d.