Volume And Surface Area Of Solids

Let the length of the side of cube be ‘a’ cm.

Volume of each cube = 27 cm³

Volume of cube = a³

∴ a³ = 27 cm³

⇒ a = (27 cm³)^1/3

⇒ a = 3 cm

Length of a side of cube = 3 cm

Since, two cubes are joined and a cuboid is formed so,

Length of cuboid = l = 2a = 2 × 3 cm = 6 cm

Breadth of cuboid = b = a = 3 cm

Height of cuboid = h = a = 3 cm

Surface area of cuboid = 2 × (l × b + b × h + l × h)

∴ Surface area of resulting cuboid = 2 × (6 × 3 + 3 × 3 + 6 × 3) cm²

= 2 × (18 + 9 + 18) cm²

= 2 × 45 cm²

= 90 cm²

So, surface area of resulting cuboid is 90 cm²

Let the radius of hemisphere be r cm

Volume of hemisphere is given by πr³

Given, volume of hemisphere = 24251/2 cm³

∴ π r³ = 24251/2

⇒ r³ = 4851 × 1/2 × 3/2 × 7/22

⇒ r³ = 1157.625 cm³

⇒ r = (1157.625)^1/3 cm

⇒ r = 10.5 cm

Curved Surface Area of hemisphere = 2πr²

Curved Surface Area of hemisphere = 2 × 22/7 × (10.5)² cm²

= 693 cm²

∴ Curved surface area of hemisphere = 693 cm²

Let the radius of solid sphere be r cm

Total surface area of solid hemisphere = 3πr²

Given, total surface area of solid hemisphere = 462 cm²

∴ 3πr² = 462 cm²

⇒ 3 × 22/7 × r² = 462 cm²

⇒ r² = 462 × 1/3 × 7/22 cm² = 49 cm²

⇒ r = 7 cm

Volume of solid hemisphere = π r³

= 2/3 × 22/7 × 7³ cm³

= 718.67 cm³

∴ Volume of solid hemisphere is 718.67 cm³

Width of cloth used = 5 m

Diameter of conical tent to be made = 14 m

Let the radius of the conical tent be r m

Radius of conical tent = r = diameter ÷ 2 = 14/2 m = 7 m

Height of conical tent = h = 24 m

Let the slant height of conical tent be l

So,

⇒ m²

⇒ l = 25 m

Area of cloth required to make a conical tent = Curved Surface area of conical tent

= πrl

= 22/7 × 7 × 25 m²

= 550 m²

Length of cloth used = Area of cloth required ÷ width of cloth

= 550/5 m

= 110 m

∴ Length of cloth used = 110 m

Cost of cloth used = Rs 25 per meter

Total Cost of cloth required to make a conical tent = 110 × Rs 25

= Rs 2750

∴ Total cost of cloth required to make a conical tent = Rs 2750

Let V₁ be the volume of first cone and V₂ be the volume of second cone.

Then, V₁:V₂ = 1:4

Let d₁ be the diameter of first cone and d₂ be the diameter of second cone.

Then d₁:d₂ = 4:5

Let h₁ be the height of first cone and h₂ be the height of second cone.

We know that volume of cone is given by V = 1/3 × π(d²/4)h

∴

⇒

⇒

⇒

⇒

∴ h₁:h₂ = 25:64

∴ Ratio of height of two cones is 25:64.

Let the radius of base be ‘r’ km and slant height be ‘l’ km

Slant height of conical mountain = 2.5 km

Area of its base = 1.54 km²

Area of base is given by πr²

∴ πr² = 1.54 km²

⇒ 22/7 × r² = 1.54 km²

⇒ r² = 1.54 × 7/22 km² = .49 km²

⇒ r = 0.7 km

Let ‘h’ be the height of the mountain

We know,

l² = r² + h²

Substituting the values of l and r in the above equation

2.5² = 0.7² + h²

h² = 2.5² – 0.7² = 6.25 – 0.49 km²

h² = 5.76 km²

h = 2.4 km

∴ Height of the mountain = 2.4 km

Let the Radius of the solid cylinder be ‘r’ m and its height be ‘h’ m.

Given,

Sum of radius and height of solid cylinder = 37 m

r + h = 37 m

r = 37 – h

Total surface area of solid cylinder = 1628 m²

Total surface area of solid cylinder is given by 2πr (h + r)

∴ 2πr (h + r) =1628 m²

Substituting the value of r + h in the above equation

⇒ 2πr × 37 = 1628 m²

⇒ r = 1628 × 7/22 × 1/2 × 1/37 m

⇒ r = 7 m

Since, r + h = 37 m

h = 37 – r m

h = 37 – 7 m = 30 m

Volume of solid cylinder = πr²h

= 22/7 × 7² × 30 m²

= 4620 m²

Let the radius of sphere be ‘r’ cm

Surface area of sphere = 2464 cm²

Surface area of sphere is given by 4πr²

∴ 4πr² = 2464 cm²

⇒ 4 × 22/7 × r² = 2464 cm²

⇒ r² = 2464 × 1/4 × 7/22 cm² = 196 cm²

⇒ r = 14 cm

Radius of new sphere is double the radius of given sphere.

Let the radius of new sphere be r’ cm

∴ r’ = 2r

r’ = 2 × 14 cm = 28 cm

Surface area of new sphere = 4πr’²

= 4 × 22/7 × 28² cm²

= 9856 cm²

∴ Surface area of new sphere is 9856 cm².

The military tent is made as a combination of right circular cylinder and right circular cone on top.

Total Height of tent = h = 8.25 m

Base diameter of tent = 30 m

Base radius of tent = r = 30/2 m = 15 m

Height of right circular cylinder = 5.5 m

Curved surface area of right circular cylindrical part of tent = 2πrh

Height of conical part = total height of tent – height of cylindrical part

h_cone = 8.25 – 5.5 m = 2.75 m

Base radius of cone = 15 m

Let l be the slant height of cone

Then, l² = h_cone² + r² = 2.75² + 15² m²

l² = 7.5625 + 225 m² = 232.5625

l = 15.25

Curved surface area of conical part of the tent = πrl

Total surface area of the tent = Curved surface area of cylindrical part + curved surface area of conical part

Total surface area of tent = 2πrh + πrl

= πr (2h + l)

= 22/7 × 15 × (2 × 5.5 + 15.25) m²

= 22/7 × 15 × (11 + 15.25) m²

= 22/7 × 15 × 26.25 m²

= 1237.5 m²

Breadth of canvas used = 1.5 m

Length of canvas used = Total surface area of tent ÷ breadth of canvas used

Length of canvas used = 1237.5\1.5 m = 825 m

∴ Length of canvas used is 825 m.

The tent is made as a combination of right circular cylinder and right circular cone on top.

Height of cylindrical part of the tent = h = 3 m

Radius of its base = r = 14 m

Curved surface area of cylindrical part of tent = 2πrh

= 2 × 22/7 × 14 × 3 m²

= 264 m²

Total height of the tent = 13.5 m

Height of conical part of the tent = total height of tent – height of cylindrical part.

Height of conical part of the tent = 13.5 – 3 m =10.5 m

Let the slant height of the conical part be l

l² = h_cone² + r²

l² = 10.5² + 14² = 110.25 + 196 m² = 306.25 m²

l = 17.5 m

Curved surface area of conical part of tent = πrl

= 22/7 × 14 × 17.5 m²

= 770 m²

Total surface area of tent = Curved surface area of cylindrical part of tent + Curved surface area of conical part of tent

Total Surface area of tent = 264 m² + 770 m² = 1034 m²

Cloth required = Total Surface area of tent = 1034 m²

Cost of cloth = Rs 80/m²

Total cost of cloth required = Total surface area of tent × Cost of cloth

= 1034 × Rs 80

= Rs 82720

Cost of cloth required to make the tent is Rs 82720

The Circus tent is made as a combination of cylinder and cone on top.

Height of cylindrical part of tent = h = 3 m

Base radius of tent = r = 52.5 m

Area of canvas required for cylindrical part of tent = 2πrh

= 2 × 22/7 × 52.5 × 3 m²

= 990 m²

Slant height of cone = l = 53 m

Area of canvas required for conical part of the tent = πrl

= 22/7 × 52.5 × 53 m²

= 8745 m²

Area of canvas required to make the tent = Area of canvas required for cylindrical part of tent + Area of canvas required for conical part of tent

Area of canvas required to make the tent = 990 + 8745 m² = 9735 m²

The rocket is in the form of cylinder closed at the bottom and cone on top.

Height of cylindrical part rocket = h = 21 m

Base radius of rocket = r = 2.5 m

Surface Area of cylindrical part of rocket = 2πrh + πr²

= 2 × 22/7 × 2.5 × 21 + 22/7 × 2.5 × 2.5 m²

= 330 + 19.64 m² = 349.64 m²

Slant height of cone = l = 8 m

Surface Area of conical part of the rocket = πrl

= 22/7 × 2.5 × 8 m²

= 62.86 m²

Total surface area of the rocket = Surface Area of cylindrical part of rocket + Surface Area of conical part of rocket

Total surface area of the rocket = 349.64 + 62.86 m² = 412.5 m²

The solid is in the form of a cone surmounted on a hemisphere.

Total height of solid = h = 9.5 m

Radius of Solid = r = 3.5 m

Volume of hemispherical part solid = 2/3 × πr³

= 2/3 × 22/7 × 3.5³ m³

= 89.83 m³

Height of conical part of solid = h_cone = Total height of solid – Radius of solid

Height of conical part of solid = h_cone = 9.5 – 3.5 = 6 m

Volume of conical part of solid = 1/3 × πr²h_cone

= 1/3 × 22/7 × 3.5² × 6 m³

= 77 m³

Volume of solid = Volume of hemispherical part solid + Volume of conical part solid

Volume of solid = 89.83 + 77 m³ = 166.83 m³

The toy is in the form of a cone mounted on a hemisphere.

Total height of toy = h = 31 cm

Radius of toy = r = 7 cm

Surface area of hemispherical part toy = 2πr²

= 2 × 22/7 × 7² cm²

= 308 cm²

Height of conical part of toy = h_cone = Total height of toy – Radius of toy

Height of conical part of toy = h_cone = 31 – 7 = 24 cm

Let the slant height of the conical part be l

l² = h_cone² + r²

l² = 24² + 7² = 576 + 49 cm² = 625 cm²

l = 25 cm

Surface area of conical part of toy = πrl

= 22/7 × 7 × 25 cm²

= 550 cm²

Total surface area of toy = Surface area of hemispherical part of toy + Surface area of conical part of toy

Total Surface area of toy = 308 cm² + 550 cm² = 858 cm²

A toy is in the shape of a cone mounted on a hemisphere of same base radius.

Volume of Toy = 231 cm³

Base Diameter of toy = 7 cm

Base radius of toy = 7/2 cm = 3.5 cm

Volume of hemisphere = 2/3 πr³

= 2/3 × 22/7 × 3.5³ cm³

Volume of cone = Volume of hemisphere – Volume of toy

Volume of cone is given by 1/3 πr²h

Where h is the height of cone

Height of cone = 11 cm

Height of toy = Height of cone + Height of hemisphere

= 11 cm + 3.5 cm = 14.5 cm

[Height of hemisphere = Radius of hemisphere]

∴ Height of toy is 14.5 cm.

Radius of cylindrical container = r = 6 cm

Height of cylindrical container = h = 15 cm

Volume of cylindrical container = πr²h

= 22/7 × 6 × 6 × 15 cm³

= 1697.14 cm³

Whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops.

Let the radius of hemisphere and base of cone be r’

Height of cone = h = 4 times the radius of its base

h’ = 4r’

Volume of Hemisphere = 2/3 π(r’)³

Volume of cone = 1/3 π(r’)²h’ = 1/3 π(r’)² × 4r’

= 2/3 π(r’)³

Volume of ice-cream = Volume of Hemisphere + Volume of cone

= 2/3 π(r’)³ + 4/3 π(r’)³ = 6/3 π(r’)³

Number of ice-creams = 10

∴ total volume of ice-cream = 10 × Volume of ice-cream

= 10 × 6/3 π(r’)³ = 60/3 π(r’)³

Also, total volume of ice-cream = Volume of cylindrical container

⇒ 60/3 π(r’)³ = 1697.14 cm³

⇒ 60/3 × 22/7 × (r’)³ = 1697.14 cm³

⇒ (r’)³ = 1697.14 × 3/60 × 7/22 = 27 cm³

⇒ r = 3 cm

∴ Radius of ice-cream cone = 3 cm

Vessel is in the form of a hemispherical bowl surmounted by a hallow cylinder.

Diameter of hemisphere = 21 cm

Radius of hemisphere = 10.5 cm

Volume of hemisphere = 2/3 πr³ = 2/3 × 22/7 × 10.5³ cm³

= 2425.5 cm³

Total height of vessel = 14.5 cm

Height of cylinder = h = Total height of vessel - Radius of hemisphere

= 14.5 – 10.5 cm = 4 cm

Volume of cylinder = πr²h = 22/7 × 10.5 × 10.5 × 4 cm³

= 1386 cm³

Volume of vessel = Volume of hemisphere + Volume of cylinder

= 2425.5 cm³ + 1386 cm³

= 3811.5 cm³

∴ Capacity of vessel = 3811.5 cm³

Toy is in the form of a cylinder with hemisphere ends

Total length of toy = 90 cm

Diameter of toy = 42 cm

Radius of toy = r = 21 cm

Length of cylinder = l = Total length of toy – 2 × Radius of toy

= 90 – 2 × 21 cm = 48 cm

For cost of painting we need to find out the curved surface area of toy

Curved surface area of cylinder = 2πrl

= 2 × 22/7 × 21 × 48 cm²

= 6336 cm²

Curved surface area of hemispherical ends = 2 × 2πr²

= 2 × 2 × 22/7 × 21 × 21 cm²

= 5544 cm²

Surface area of toy = Curved surface area of cylinder + Curved surface area of hemispherical ends

Surface area of toy = 6336 cm² + 5544 cm² = 11880 cm²

Cost of painting = Rs 0.70/cm²

Total Cost of painting = Surface area of toy × Cost of painting

= 11880 cm² × Rs 0.70/cm² = Rs 8316.00

Total cost of painting the toy = Rs 8316.00

A medicine capsule is in the shape of a cylinder with two hemisphere stuck to each of its ends.

Total length of entire capsule = 14 mm

Diameter of capsule = 5 mm

Radius of capsule = r = Diameter ÷ 2 = 5/2 mm = 2.5 mm

Length of cylindrical part of capsule = l = Total length of entire capsule – 2 × Radius of capsule

= 14 – 2 × 2.5 mm = 14 - 5 mm = 9 mm

Curved surface area of cylindrical part of capsule = 2πrl

= 2 × 3.14 × 9 × 2.5 mm²

= 141.3 mm²

Curved surface area of hemispherical ends = 2 × 2πr²

= 2 × 2 × 3.14 × 2.5 × 2.5 cm²

= 78.5 mm²

Surface area of capsule = Curved surface area of cylindrical part of capsule + Curved surface area of hemispherical ends

Surface area of capsule = 141.3 mm² + 78.5 mm² = 219.8 mm²

The wooden article was made by scooting out a hemisphere from each end of a cylinder

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

Height of cylinder = h = 20 cm

Base diameter of cylinder = 7 cm

Base radius of cylinder = r = diameter ÷ 2 = 7/2 cm = 3.5 cm

Lateral Surface area of cylinder = 2πrh

= 2 × 22/7 × 3.5 × 20 cm²

= 2 × 22/7 × 3.5 × 20 cm²

= 440 cm²

Since, the wooden article was made by scooting out a hemisphere from each end of a cylinder

∴ Two hemispheres are taken out in total

Radius of cylinder = radius of hemisphere

∴ Radius of hemisphere = 3.5 cm

Lateral Surface area of hemisphere = 2πr²

= 2 × 22/7 × 3.5 × 3.5 cm²

= 77 cm²

Total Surface area of two hemispheres = 2 × 77 cm² = 154 cm²

Total surface area of the article when it is ready = Lateral Surface area of cylinder + Lateral Surface area of hemisphere

Total surface area of the article when it is ready = 440 cm² + 154 cm²

= 594 cm²

A solid is in the form of a right circular cone mounted on a hemisphere.

Let r be the radius of hemisphere and cone

Let h be the height of the cone

Radius of hemisphere = r = 2.1 cm

Volume of hemisphere = 2/3 πr³

= 2/3 × 22/7 × 2.1 × 2.1 × 2.1 cm³

= 19.404 cm³

Height of cone = h = 4 cm

Radius of cone = r = 2.1 cm

Volume of cone = 1/3 πr²h

= 1/3 × 22/7 × 2.1 × 2.1 × 4 cm³

= 18.48 cm³

Volume of solid = Volume of hemisphere + Volume of cone

= 19.404 cm³ + 18.48 cm³ = 37.884 cm³

The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water, so, to find the volume of water left in the tub we need to subtract volume of solid from cylindrical tub.

Radius of cylinder = r’ = 5 cm

Height of cylinder = h’ = 9.8 cm

Volume of cylindrical tub = πr’²h’ = 22/7 × 5 × 5 × 9.8 cm³

= 770 cm³

Volume of water left in the tub = Volume of cylindrical tub – Volume of solid

Volume of water left in the tub = 770 cm³ – 37.884 cm³ = 732.116 cm³

∴ Volume of water left in the tub is 732.116 cm³

Height of solid cylinder = h = 8 cm

Radius of solid cylinder = r = 6 cm

Volume of solid cylinder = πr²h

= 3.14 × 6 × 6 × 8 cm³

= 904.32 cm³

Curved Surface area of solid cylinder = 2πrh

Height of conical cavity = h = 8 cm

Radius conical cavity = r = 6 cm

Volume of conical cavity = 1/3 πr²h

= 1/3 × 3.14 × 6 × 6 × 8 cm³

= 301.44 cm³

Let l be the slant height of conical cavity

l² = r² + h²

⇒ l² = (6² + 8²) cm²

⇒ l² = (36 + 64) cm²

⇒ l² = 100 cm²

⇒ l = 10 cm

Curved Surface area of conical cavity = πrl

Since, conical cavity is hollowed out from solid cylinder, so, volume and total surface area of remaining solid will be found out by subtracting volume and total surface area of conical cavity from volume and total surface area of solid cylinder.

Volume of remaining solid = Volume of solid cylinder – Volume of conical cavity

Volume of remaining solid = 904.32 cm³ – 301.44 cm³

= 602.88 cm³

Total surface area of remaining solid = Curved Surface area of solid cylinder + Curved Surface area of conical cavity + Area of circular base

Total surface area of remaining solid = 2πrh + πrl + πr²

= πr × (2h + l + r)

= 3.14 × 6 × (2 × 8 + 10 + 6) cm²

= 3.14 × 6 × 32 cm²

= 602.88 cm²

Height of solid cylinder = h = 2.8 cm

Diameter of solid cylinder = 4.2 cm

Radius of solid cylinder = r = Diameter ÷ 2 = 2.1 cm

Curved Surface area of solid cylinder = 2πrh

= 2 × 22/7 × 2.1 × 2.8 cm²

= 2 × 22/7 × 2.1 × 2.8 cm²

= 36.96 cm²

Height of conical cavity = h = 2.8 cm

Radius conical cavity = r = 2.1 cm

Let l be the slant height of conical cavity

l² = r² + h²

⇒ l² = (2.8² + 2.1²) cm²

⇒ l² = (7.84 + 4.41) cm²

⇒ l² = 12.25 cm²

⇒ l = 3.5 cm

Curved Surface area of conical cavity = πrl

= 22/7 × 2.1 × 3.5

= 23.1 cm²

Total surface area of remaining solid = Curved surface area of solid cylinder + Curved surface area of conical cavity + Area of circular base

Total surface area of remaining solid = (36.96 + 23.1 + 22/7 × 2.1²) cm²

= (36.96 + 23.1 + 13.86) cm²

= 73.92 cm²

Height of solid cylinder = h = 14 cm

Diameter of solid cylinder = 7 cm

Radius of solid cylinder = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm

Volume of solid cylinder = πr²h

= 22/7 × 3.5 × 3.5 × 14 cm³

= 539 cm³

Height of conical cavity = h’ = 4 cm

Radius conical cavity = r’ = 2.1 cm

Volume of conical cavity = 1/3 πr’²h’

= 1/3 × 22/7 × 2.1 × 2.1 × 4 cm³

= 18.48 cm³

Since, there are two conical cavities

∴ Volume of two conical cavities = 2 × 18.48 cm³ = 36.96 cm³

Volume of remaining solid = Volume of solid cylinder – Volume of two conical cavity

Volume of remaining solid = 539 cm³ – 36.96 cm³

= 502.04 cm³

Height of metallic cylinder = h = 5 cm

Radius of metallic cylinder = r = 3 cm

Volume of solid cylinder = πr²h

= π × 3 × 3 × 5 cm³

= 45π cm³

Height of conical hole = h’ = 8/9 cm

Radius conical hole = r’ = 3/2 cm

Volume of conical hole = 1/3 πr’²h’

= 1/3 × π × 3/2 × 3/2 × 8/9 cm³

= 2/3 π cm³

Volume of metal left in cylinder = Volume of metallic cylinder – Volume of conical hole

Volume of metal left in cylinder = 45π - 2/3 π = 133π/3

Ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape = Volume of metal left in cylinder/ Volume of conical hole

Volume of metal left in cylinder : Volume of conical hole = 133π/3 : 2/3 π

Volume of metal left in cylinder: Volume of conical hole = 133: 2

Ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 339:4

Length of cylindrical neck = l = 7 cm

Diameter of cylindrical neck = 4 cm

Radius of cylindrical neck = r = Diameter ÷ 2 = 4/2 cm = 2 cm

Volume of cylindrical neck = πr²l

= 22/7 × 2 × 2 × 7 cm³

= 88 cm³

Diameter of spherical part = 21 cm

Radius of spherical part = r’ = Diameter ÷ 2 = 21/2 cm = 10.5 cm

Volume of spherical part = 4/3 πr’³

= 4/3 × 22/7 × 10.5 × 10.5 × 10.5 cm³

= 4851 cm³

Quantity of water spherical glass vessel with cylindrical neck can hold = Volume of spherical part + Volume of cylindrical neck

Quantity of water spherical glass vessel with cylindrical neck can hold = 4851 cm³ + 88 cm³ = 4939 cm³

Quantity of water spherical glass vessel with cylindrical neck can hold is 4939 cm³.

The solid consisting of a cylinder surmounted by a cone at one end a hemisphere at the other.

Length of cylinder = l = 6.5 cm

Diameter of cylinder = 7 cm

Radius of cylinder = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm

Volume of cylinder = πr²l

= 22/7 × 3.5 × 3.5 × 6.5 cm³

= 250.25 cm³

Length of cone = l’ = 12.8 cm – 6.5 cm = 6.3 cm

Diameter of cone = 7 cm

Radius of cone = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm

Volume of cone = 1/3 πr²l’

= 1/3 × 22/7 × 3.5 × 3.5 × 6.3 cm³

= 80.85 cm³

Diameter of hemisphere = 7 cm

Radius of hemisphere = r = Diameter ÷ 2 = 7/2 cm = 3.5 cm

Volume of hemisphere = 2/3 πr³

= 2/3 × 22/7 × 3.5 × 3.5 × 3.5 cm³

= 89.83 cm³

Volume of the solid = Volume of cylinder + Volume of cone + Volume of hemisphere

Volume of solid = 250.25 cm³ + 80.85 cm³ + 89.83 cm³

= 420.93 cm³

Length of cubical piece of wood = a = 21 cm

Volume of cubical piece of wood = a³

= 21 × 21 × 21 cm³

= 9261 cm³

Surface area of cubical piece of wood = 6a²

= 6 × 21 × 21 cm²

= 2646 cm²

Since, a hemisphere is carved out in such a way that the diameter of the hemisphere is equal to the side of the cubical piece.

So, diameter of hemisphere = length of side of the cubical piece

Diameter of hemisphere = 21 cm

Radius of hemisphere = r = Diameter ÷ 2 = 21/2 cm = 10.5 cm

Volume of hemisphere = 2/3 πr³

= 2/3 × 22/7 × 10.5 × 10.5 × 10.5 cm³

= 2425.5 cm³

Surface area of hemisphere = 2πr²

= 2 × 22/7 × 10.5 × 10.5 cm²

= 693 cm²

A hemisphere is carved out from cubical piece of wood

Volume of remaining solid = Volume of cubical piece of wood – Volume of hemisphere

Volume of remaining solid = 9261cm³ – 2425.5 cm³ = 6835.5 cm³

Surface area remaining piece of solid = surface area of cubical piece of wood – Area of circular base of hemisphere + Curved Surface area of hemisphere

Surface area remaining piece of solid = 6a² – πr² + 2πr²

= (2646 – 22/7 × 10.5² + 693) cm²

= 2992.5 cm²

Length of side of cubical block = a = 10 cm

Since, a cubical block is surmounted by a hemisphere, so, the largest diameter of hemisphere = 10 cm

Since, hemisphere will be touching the sides of cubical block.

Radius of hemisphere = r = Diameter ÷ 2 = 10/2 cm = 5 cm

Surface area of solid = Surface area of cube – Area of circular part of hemisphere + Curved surface area of hemisphere

Total Surface area of solid = 6a² – πr² + 2πr² = 6a² + πr²

= 6 × 10 × 10 cm² + 3.14 × 5 × 5 cm²

= 678.5 cm²

Rate of painting = Rs 5/100 cm²

Cost of painting the total surface area of the solid so formed = Total Surface area of solid × Rate of painting

Cost of painting the total surface area of the solid = Rs 5/100 × 678.5

= Rs 33.925

The toy is in the shape of a right circular cylinder surmounted by a cone at one end a hemisphere at the other.

Total height of toy = 30 cm

Height of cylinder = h = 13 cm

Radius of cylinder = r = 5 cm

Curved surface area of cylinder = 2πrh

= 2 × 22/7 × 5 × 13 cm²

Height of cone = h’ = Total height of toy – Height of cylinder – Radius of hemisphere

Height of cone = h’ = 30 cm – 13 cm – 5 cm = 12 cm

Radius of cone = r = Radius of cylinder

Radius of cone = r = 5 cm

Let the slant height of cone be l

l² = h’² + r²

⇒ l² = 12² + 5² cm² = 144 + 25 cm² = 169 cm²

⇒ l = 13 cm

Curved surface area of cone = πrl

= 22/7 × 5 × 13 cm²

Radius of hemisphere = r = Radius of cylinder

Radius of hemisphere = r = 5 cm

Curved surface area of hemisphere = 2πr²

= 2 × 22/7 × 5 × 5 cm²

Surface area of the toy = Surface area of cylinder + Surface area of cone + Surface area of hemisphere

Surface area of toy = 2πrh + πrl + 2πr²

= πr (2h + l + 2r)

= 22/7 × 5 × (2 × 13 + 13 + 2 × 5) cm²

= 22/7 × 5 × 49 cm²

= 770 cm²

Surface area of toy is 770 cm²

Inner diameter of a glass = 7 cm

Inner radius of glass = r = 7/2 cm = 3.5 cm

Height of glass = h = 16 cm

Apparent capacity of glass = πr²h

= 22/7 × 3.5 × 3.5 × 16 cm³

= 616 cm³

Volume of the hemisphere in the bottom = 2/3 πr³

= 2/3 × 22/7 × 3.5³ cm³

= 89.83 cm³

Actual capacity of the class = Apparent capacity of glass – Volume of the hemisphere

Actual capacity of the class = 616 cm³ – 89.83 cm³ = 526.17 cm³

The wooden toy is in the shape of a cone mounted on a cylinder

Total height of the toy = 26 cm

Height of conical part = H = 6 cm

Height of cylindrical part = Total height of the toy – Height of conical part

h = 26 cm – 6 cm = 20 cm

Diameter of conical part = 5 cm

Radius of conical part = R = Diameter/2 = 5/2 cm = 2.5 cm

Let L be the slant height of the cone

L² = H² + R²

⇒ L² = 6² + 2.5² cm² = 36 + 6.25 cm² = 42.25 cm²

⇒ L = 6.5 cm

Diameter of cylindrical part = 4 cm

Radius of cylindrical part = r = Diameter/2 = 4/2 cm = 2 cm

Area to be painted Red = Curved Surface area of cone + Base area of cone – base area of cylinder

Area to be painted Red = πRL + πR² – πr² = π (RL + R² – r²)

= 22/7 × (2.5 × 6.5 + 2.5 × 2.5 – 2 × 2) cm²

= 22/7 × (16.25 + 6.25 – 4) cm²

= 22/7 × 18.5 cm²

= 58.143 cm²

Area to be painted White = Curved Surface area of cylinder + Base area of cylinder

Area to be painted White = 2πrh + πr² = πr (2h + r)

= 22/7 × 2 × (2 × 20 + 2) cm²

= 22/7 × 2 × (40 + 2) cm²

= 22/7 × 2 × 42 cm² = 264 cm²

∴ Area to be painted red is 58.143 cm² and area to be painted white is 264 cm².

Given,

The dimensions of a metallic cuboid = 100 cm × 80 cm × 64 cm

Let’s find out the volume of the cuboid first;

Volume of the cuboid = l × b × h

⇒ 100 × 80 × 64 = 512000cm³

As cuboid is recast into a cube;

So,

Volume of cube = volume of cuboid

⇒ l³ = 512000 ⇒ l =

⇒ l = 80 cm

Now,

As the length of the side of the cube = 80 cm

The surface area of the cube = 6(l)²

= 6 x 80 x 80

= 38400 cm²

So, the surface area of the cube is 38400 cm²

We have,

The radius of the cone (r) = 5cm and

The height of the cone (h) = 20cm

Let the radius of the sphere be R;

As,

Volume of sphere = Volume of cone

⇒ πR³ = πr²h

⇒ 4R³ = 5 × 5 × 20

⇒ R³ = = 125 cm

⇒ R = 5 cm

Diameter of the sphere = 2R = 2 x 5 = 10 cm

So, the diameter of the sphere is 10 cm

Given,

The radius (r₁) of 1^st sphere = 6 cm

Radius of 2^nd sphere (r₂) = 8 cm and

Radius f third sphere (r₃) = 10 cm

Let the radius of the resulting sphere be R;

So now we have,

Volume of resulting sphere = Volume of three metallic spheres

⇒ πR³ =

⇒πR³ =

⇒ R³ = 216 + 512 + 1000

⇒ R³ = 1728

⇒R =

R = 12 cm

So, the radius of the resulting sphere is 12cm.

Let the number of balls formed = n

Since the metal cone is melted to form spherical balls;

So we have;

Volume of metal cone = Total volume of n spherical balls

Volume of cone = n(Volume of 1 spherical ball)

n = 32

So, 32 spherical balls can be formed.

We have,

The internal base radius of spherical shell, r₁ = 3 cm,

The external base radius of spherical shell, r₂ = 5 cm and

The base radius of solid cylinder, r = 7 cm

Let the height of the cylinder be h

As,

The hollow spherical shell is melted into a solid cylinder;

So,

Volume of solid cylinder = Volume of spherical shell

⇒ πr²h =

⇒ πr²h =

⇒ r²h =

⇒ 49 × h = (125−27)

⇒ h = × 9849

Therefore,

h =

So, the height of the cylinder is .

Given,

Internal diameter of the hemisphere = 6 cm

External diameter of the hemisphere = 10 cm

Diameter of cone = 14 cm

So we have,

Internal radius(r) of the hemisphere = 3 cm

External radius(R) of the hemisphere = 5 cm

Radius of cone = 7 cm

Now,

Volume of the hollow hemisphere = Volume of the cone
Volume of the hollow hemisphere =

Volume of the cone =

So,

= 2 × 98 = 49 × h

= h

So,

h = = 4 cm

Thus, the height of the cone formed is 4 cm.

Given,

Diameter of the copper Rod = 2cm

Length of the Rod = 10 cm

Length of the wire = 10 m = 1000cm

So here we have,

Radius of the copper rod = 1 cm

Let suppose the radius of the wire = r

Volume of the rod = volume of the wire

πr²h = πr²h

1 × 1 × 10 = 1000 × r

10 = 1000r

r = = 0.1 cm

The diameter of the wire = 2r = 2 × 0.1 = 0.2 cm

So, the thickness of the wire is 0.2 cm or 2mm.

Let the required bottles = n

Internal diameter of the hemispherical sphere = 30 cm

Internal Radius of the hemispherical sphere = 15 cm

Diameter of the cylindrical bottle = 5 cm

Radius of the cylindrical bottle = 2.5 cm

Now,

Volume of the hemispherical sphere = πr³

= π × 15 × 15 × 15

= π10 × 15 × 15

= 2,250π cm³

And,

Volume of 1 cylindrical bottle = πr²h = π × 2.5 × 2.5 × 6 = 37.5π cm³

Amount of water in n bottles = Amount of water in bowl

n × 37.5π = 2,250π

n = = 60 bottles

So, 60 numbers of bottles are required to empty the bowl.

Given,

Diameter of the sphere = 21 cm

Diameter of the cone = 3.5 cm = cm

Height of the cone = 3 cm

So here we have,

Radius of the sphere = 10.5 cm

Radius of the cone = 1.75 cm

Volume of the sphere = πr³

=

Volume of the cone =

No of cones =

= 504

So,

504 numbers of cones are formed.

Given,

Diameter of cannon ball = 28cm

So the Radius of cannon ball = 14 cm

Volume of cannon ball = πr³ = π × 14³

Radius of the cone = cm

Let the height of cone be h cm.

Volume of cone = π × h cm³

Here we have,

= π × 14³ = π × h

= h = π × 14 × 14 × 14 ×

= h = 35.84 cm

Hence, the height of the cone = 35.84 cm.

Given,

Radius of the spherical ball = 3 cm

After recasting the ball into three spherical balls;

Radius of the first ball = r₁ = 1.5 cm

Radius of the first ball = r₂ = 2 cm

Let the radius of the third ball be r₃ cm.

Then,

Volume of third ball = Volume of spherical ball – volume of 2 small balls

Volume of the third ball = π × 3³ – π – π × 2³

36π – – 32π/3cm³ = 125π/6cm³

4/3πr³ = 125π/6

r³ = 125π × 3/6 × 4 × π = 125/8

r = = 2.5 cm

Hence, the radius of the third ball is 2.5cm.

External diameter of shell = 24cm and

Internal diameter of shell = 18cm

So,

External radius of shell = 12 cm and

Internal radius = 9 cm

Volume of lead in the shell = π[12³–9³]cm³

Let the radius of the cylinder be r cm

Height of the cylinder = 37 cm

Volume of the cylinder = πr²h = πr²(37)

π[12³–9³] = πr² × 37

π × 999 = πr² × 37

r² = × π × 999 × π = 36cm²

r = = 6cm

Hence, diameter of the base of the cylinder = 12 cm

Volume of hemisphere of radius 9 cm

= × π × 9 × 9 × 9cm³

Volume of circular cone (height = 72 cm)

× π × r² × 72cm

Volume of cone = volume of hemisphere

× π × r² × 72 = π × 9 × 9 × 9

r² = π × 9 × 9 × 9 × π = 20.25

r = = 4.5cm

Hence, radius of the base of the cone = 4.5 cm

Diameter of sphere = 21 cm

Hence, radius of the sphere =

Volume of sphere = π × r³ =

Volume of cube = a³ = 1³

Let the number of cubes formed be n

Volume of sphere = n(volume of cube)

= n × 1

4 × 22 × = n × 1

11 × 21 × 21 = n × 1

4851 = n

So,

Hence, the number of cubes is 4851

Given,

Radius of the sphere = 8 cm

Radius of the ball we made = 1 cm

So,

Volume of sphere (when r = 1 cm) = πr³

= × 1 × 1 × 1 × π cm³

Volume of sphere (when r = 8 cm) = πr³

= × 8 × 8 × 8 × π cm³

Let the number of balls be n;

Then we have;

n × × 1 × 1 × 1 × π = × 8 × 8 × 8 × π

n = = 512

Hence, the number of lead balls can be made is 512.

Given,

Radius of sphere = 3 cm

Volume of sphere = πr³ = × π × 3 × 3 × 3cm³ = 36π cm³

Radius of small sphere = cm = 0.3 cm

Volume of small sphere = × π × 0.3 × 0.3 × 0.3cm³

= cm³

Let the number of small balls be n;

n × = × π × 3 × 3 × 3

n = 1000

Hence, the number of small balls = 1000

Given,

Diameter of sphere = 42 cm

Radius of sphere = 21 cm

Volume of sphere = πr³ = × π × 21 × 21 × 21 cm³

Diameter of cylindrical wire = 2.8 cm

Radius of cylindrical wire = 1.4 cm

Volume of cylindrical wire = πr²h = π × 1.4 × 1.4 × h cm³ = 1.96πh cm³

Volume of cylindrical wire = volume of sphere

1.96πh = × π × 21 × 21 × 21

h = cm

h = 6300

h m = 63 m

Hence, length of the wire = 63 m

Given,

Diameter of sphere = 18 cm

Length of wire = 108 m = 10800 cm

Radius of copper sphere = m = 36 m

Volume of sphere = πr³ = × π × 9 × 9 × 9cm³ = 972π cm³

Let the radius of wire be r cm

= πr²l = πr² × 10800

But the volume of wire = volume of sphere

⇒ πr² × 10800 = 972π

r² = = 0.09 cm²

r = cm = 0.3

Hence, the diameter = 2r = 0.6 cm

Given,

Internal radius of hemispherical bowl (r₁) = 9 cm

Internal radius of cyclical vessel (r₂) = 6 cm

Let the height of water in the cylindrical vessel be h.

So,

Volume of hemispherical bowl = volume of cylindrical vessel

h = 13.5 cm

Hence, height of water in the cylindrical vessel is 13.5 cm

Given,

Diameter of the hemispherical tank = 3 m

Radius of hemispherical tank = m

Volume of tank = m³

= … (1m³ = 1000L)

Half the tank = L

Now,

In 1 sec = L of water is emptied

Required time

=

= 990 sec

So,

To empty half the tank 16.5 min are required.

Length of the roof = 44 m

Breadth of the roof = 20 m

Diameter of the cylindrical tank = 4 m

Radius of the cylindrical tank = 2 m

Height of the cylindrical tank = 3.5 m

Volume of roof = l x b x h
= 44X20 X h
volume of cylinder = x2x2x

Height of roof = height of rainfall
And
Volume of water on roof = Volume of Water in cylindrical tank
44X 20 X h = × 2 x 2 x
44 X 20 X h = 22 × 2 × 2 ×

880 × h = 44
h =
h = .05 m
h = 5 cm

Given,

Length of the roof = 22 m

Breadth of the roof = 20 m

Diameter of the cylindrical vessel = 2 m

Radius of the cylindrical vessel = 1 m

Height of the cylindrical vessel = 3.5 m

Volume of the rainfall = area of the roof × depth of the rainfall

πr²h = area of the roof × d

× 3.14 × 1 × 3.5 = 22 × 20 × d

× 11 = 440d

8.8 = 440d

0.02 = d

d = 0.02 m

So,

The rainfall is 2 cm.

Height of a cone = 60 cm,

Radius of a cone = 30 cm
volume of cone =
Height of cylinder = 180 cm,

Radius of cylinder = 60 cm
Volume of cylinder = πr²h
Volume of the remaining water = volume of cylinder - volume of cone

= π × 60 × 60 × 180 – × π × 30 × 30 × 60

= 648000π – 1800π (Taking 1800 as common)

= 18000(36π-π)

= 18000 x 35 x = 1.98 m³

Given,

Diameter of the cylindrical pipe = 2 cm
Radius of the cylindrical pipe = 1 m
Height of the cylindrical pipe = 0.4m/s = 40 cm/s

Radius of cylindrical tank = 40 cm
Let Height be h,
As we can see the volume of water which passes through the cylindrical pipe is equal to the volume of water present in the cylindrical tank after half an hour.
So,

Volume of water which passes through the cylindrical = volume of water present in the cylindrical tank after half an hour
πr²h = πr²h
(1)² × 72000 = (40)² × h
72000 = 1600 × h
h = 45 cm

Hence,

The rise in level of water in tank in half an hour will be 45 cm.

Given,

Speed of the water flowing through the pipe, H = 6 km/hr

= cm/s

Diameter of the pipe = 14 cm

Radius (R) of pipe = = 7 cm

Length (l) of the rectangular tank = 60 m = 6000 cm,

Breadth (b) of the rectangular tank = 22 m = 2200 cm and

Height (h) or Rise in the level of water in the tank = 7 cm

Now,

Volume of the water in rectangular tank = l × b × h = 6000 x 2200 x 7 = 92400000cm3

Also,

Volume of the water flowing through the pipe in 1 s = πR²H

= cm³

So,

The time taken to rise the water =

cm³

= 92400 x = 3600s = 1 hr

So,

In 1 hour time the level of water in the tank will rise by 7cm.

Given,

Width of the canal = 6 m

Depth of the canal, = 1.5 m

Length of the cuboid = 666.67 m

Water is flowing at a speed of 4 km/hr = 4000m/hr

Thus,

Area irrigated in 10 minutes = hour = x 4000 = 666.67 m

Hence,

Volume of the water flowing in hour = Volume of the canal

⇒ Volume of the water flowing in hour = 666.67 x 6 x 1.5 = 6000.03 = 60000 m³

Let a m² is the area irrigated in hour,

Then,

⇒ a × = 6000

⇒ a = m² = 75000 m²

Given,

Internal diameter of the pipe = 25 cm

Radius (r) of the pipe =

Diameter of the cylindrical tank = 12 m

Radius (R) of the tank = 6 m

Height (h) of the tank = 2.5 m

In 1 hour water comes out of the pipe = 3.6 km = 3600 m

Let the total hrs = n

Volume of the water coming out of the pipe in n hrs = volume of the cylindrical tank

= 6 × 6 × 2.5

3600 n = 36 × 2.5 × 8 × 8

100 n = 8 × 8 × 2.5

1000 n = 8 × 8 × 25

n = = 1.6 hrs

n = 1 hrs 36 minutes

Now calculate the cost;

Cost of the water = Volume of the cylindrical tank × 0.07

= = 19.80 rs

So,

Total time required to fill the tank is 1 hr 36 minutes and the cost is 19.80 rs.

Given,

Diameter of the cylindrical pipe = 7 cm

Radius of the pipe = 3.5 cm

Volumetric flow rate = 192.5 l/min

Flow rate =

Area = pi × r² = 3.14 × 3.5 × 3.5 = 38.5 cm²

Since 1 l = 1dm³ = 1000 cm³

Volumetric flow rate = 192.5 × 1000 cm³/min

So,

Flow rate = = 5000 cm/min

Flow rate in km/h = 5000 cm/min = 5000 × 0.00001 km/( h) = 3 km/h

Given,

Diameter of marble = 1.4 cm

Radius of marble = 0.7 cm

Number of marbles = 150

Diameter of cylinder = 7cm

Radius of cylinder = 3.5cm

150 × Volume of spherical marbles = volume of cylindrical vessel
⇒ 150 x × π × 0.7 × 0.7 × 0.7 = π × 3.5 × 3.5 × h

⇒ 50 x 4 x (0. 7)³ = (3.5)² x h

⇒ 200 x 0.343 = 12.25 x h

⇒ = 5.6 = h

So,

h = 5.6 cm

The rise in the level of water in the vessel (h) = 5.6 cm

Given,

Diameter of marble = 1.4 cm

Diameter of cylinder = 7 cm

Radius of cylinder = 3.5 cm

Cylinder height = 5.6 cm

Radius of marble = 0.7 cm

Volume of 1 marble = πr³

= = 1.4373

So,

Let the number of marbles be n,

Increase in the Volume of cylinder due to n marbles = πr²h

= 3.14 × 3.5 × 3.5 × 5.6 = 215.6 cm³

Hence the total numbers of marble required = = 150 marbles

Given,

Diameter of the well = 10 m

Radius of the well = 5 m
Height of the well = 14 m
Width of the embankment = 5 m
Therefore radius of the embankment = 5 + 5 = 10 m
Let h be the height of the embankment,

Hence,

The volume of the embankment = volume of the well

π(R – r)²h = πr²h
(10² – 5²) x h = 5² x 14
(100 – 25) x h = 25 x 14
h =
Therefore,

The height of the embankment, h = 4.67 m

Given,

Diameter of the well = 14 m

Radius of the well = 7 m

Height of the well = 8 m

Now,

Volume of the earth dug out of the well = πr²h

= × 7 × 7 × 8 = 1,232 m³

Area on which earth dug out is spread = l × b - r²h

= 35 × 22 - × 7 × 7

= 770 - 154 = 616 m

Level of the earth raised = = 2 m

So, the rise in the level of the field is 2 m.

Given,

Diameter of the copper wire = 6 mm = 0.6 cm

Radius of the copper wire = 0.3 cm

Length of the cylinder = 18 cm

Diameter of the cylinder = 49 cm

Radius of the cylinder = cm

Density of the copper = 8.8g

Number of the rotations on the cylinder = = 30 cm

Base circumference of the cylinder = 2πr = = 154 cm

So,

Length of the wire = 154 × 30 = 4620 cm

Volume of the wire = πr²h

= × 0.3 × 0.3 × 4620 = 1306.8

Weight of the wire = volume × density

= 1306.8 × 8.8 = 11,499.84 gm

Let name the triangle, ABC

Given,

Sides of the triangle are 15 and 20 cm,

BD ⟘ AC.

In the given case BD is the radius of the double cone generated by triangle by revolving.

Now by Pythagoras theorem,

AC² = AB² + BC²

AC² = (15)² + (20)²

AC² = 225 + 400

AC² = 625 = (25)²

AC = 25 cm

Let AD be m cm;

∴ CD = (25 – m) cm

By using the Pythagoras theorem in ∆ABD and ∆CBD;

∆ABD,

AD² + BD² = AB²

m² + BD² = 225

BD² = 225 – m² ……….(i)

∆CBD,

BD² + CD² = BC²

BD² + (25 – m)² = (20)²

BD² = (20)² – (25 – m)² …..(ii)

By putting both the equations (i) and (ii) together,

225 – m² = (20)² – (625 – m)²

225 – m² = 400 – (625 + m² – 50m) ……….by (a² – b²)

225 – m² = - 225 – m² + 50m

225 – m² + 225 + m² = 50m

450 = 50m

So,

m = = 9 cm

BD² = (15)² – (9)²

BD² = 225 – 81 = 144 cm

BD = 12 cm

Radius of the generated double cone = 12 cm

Now,

Volume of the cone generated = Volume of the upper cone + Volume of the lower cone

(AD + CD)

(12)²× (25)

= 1200π cm³ = 3,771.42 cm³

Surface area of the double cone formed;

= L.S.A of upper cone + L.S.A of the lower cone

= π (BD) × (AB) + π (BD) × (BC)

= π × 12cm × 15 cm + π × 12 cm × 20 cm

= 420π cm² = 1320 cm²

So, the volume is 1200π cm³ and surface area is 420π cm², of the double cone so formed.

Given: Height of glass = h = 14 cm

Diameter of lower circular end of glass = 12 cm

Diameter of upper circular end of glass = 16 cm

∴ Radius of lower circular end = r = 12/2 = 6 cm

∴ Radius of lower circular end = R = 16/2 = 8 cm

= 44 × 49.33 cm³

2170.52 cm³

∴ Capacity of glass = 2170.52 cm³

Given: Radius of lower circular end = r = 12 cm

Radius of upper circular end = R = 18 cm

Height of frustum = h = 8 cm

Formula: Total surface area of frustum = πr² + πR² + π(R + r)l cm²

Where l = slant height

For slant height we have

∴ l = 10 cm

∴ total surface area of frustum = π × 12² + π × 18² + π × (18 + 12) × 10 cm²

= 3.14 × (144 + 324 + 300) cm²

= 3.14 × 768 cm²

= 2411.52 cm²

∴ total surface area of frustum = 2411.52 cm²

Given: Height of bucket = h = 24 cm

Radius of lower circular end = r = 7 cm

Radius of upper circular end = R = 14 cm

Total surface area of frustum = πr² + πR² + π(R + r)l cm²

Where l = slant height

∴ l = 25 cm

(i) volume of water which will completely fill the bucket = volume of frustum

= 8 × 22 × 49

= 8722 cm³

∴ volume of water which will completely fill the bucket = 8722 cm³

(ii) area of metal sheet used

Since the top is open we need to subtract the area of top/upper circle from total surface area of frustum because we don’t require a metal plate for top.

Radius of top/upper circle = R

Area of upper circle = πR²

∴ area of metal sheet used = (total surface area of frustum)-πR²

∴ Area of metal sheet used = πr² + πR² + π(R + r)l-πR²cm²

= πr² + π(R + r)l cm²

= 22 × 82 cm²

= 1804 cm²

∴ Area of metal sheet used to make bucket = 1804 cm²

Given: height of frustum container = h = 24 cm

Radius of lower circular end = r = 8 cm

Radius of upper circular end = R = 20 cm

Cost of 1 litre milk = 24 Rs

Volume of milk completely fill the container = volume of frustum of cone

= 8 × 3.14 × (400 + 64 + 160) cm³

= 8 × 3.14 × 624 cm³

= 15674.88 cm³

Now, 1 litre is 1000 cm³

∴ 15674.88 cm³ = 15674.88/1000 = 15.67488 litres

∴ Cost of milk which can completely fill the container = 15.67488 × cost of 1 litre milk

= 15.67488 × 24 Rs

= 376.19712 Rs

∴ Cost of milk which can completely fill the container = 376.19712 Rs

Given: height of container frustum = h = 16 cm

Diameter of lower circular end = 16 cm

Diameter of upper circular end = 14 cm

∴ Radius of lower circular end = r = 16/2 = 8 cm

∴ Radius of upper circular end = R = 14/2 = 7 cm

Cost of 100 cm² metal sheet = 10 Rs

∴ Cost of 1 cm² metal sheet = 10/100 = 0.1 Rs

Formula: Total surface area of frustum = πr² + πR² + π(R + r)l cm²

Where l = slant height

∴ l = 16.0312 cm

Since the top is open we need to subtract the area of top/upper circle from total surface area of frustum because we don’t require a metal plate for top.

Radius of top/upper circle = R

Area of upper circle = πR²

∴ area of metal sheet used = (total surface area of frustum)-πR²

= πr² + πR² + π(R + r)l- πR² cm²

= πr² + π(R + r)l cm²

= π × (8² + (7 + 8)16.0312) cm²

= 3.14 × 304.468 cm²

= 956.029 cm²

∴ 956.029 cm² metal sheet is required to make the container.

∴ Cost of 956.029 cm² metal sheet = 956.029 × cost of 1 cm² metal sheet

= 956.029 × 0.1 Rs

= 95.6029 Rs

∴ Cost of metal sheet required to make container = 95.6029 Rs

Given: Radius of lower circular end = r = 27 cm

Radius of upper circular end = R = 33 cm

Slant height = l = 10cm

Total surface area of frustum = πr² + πR² + π(R + r)l cm²

Here h height of frustum is not given and we need h to find the volume of frustum therefore we must first calculate the value of h as follows

using formula for slant height and with the help of given data we get

Squaring both sides

∴ 100 = 36 + h²

∴ h² = 64

∴ h = 8

As length cannot be negative

∴ h = 8 cm

= 22 × 8 × 129

= 22704 cm³

∴ capacity = volume of frustum = 22704 cm³

Total surface area of frustum = πr² + πR² + π(R + r)l cm²

= (22/7) × (1089 + 729 + 600)

= (22/7) × 2418

= 7599.428 cm²

∴ total surface area = 7599.428 cm²

Given: Depth of the bucket = height of frustum = h = 15 cm

Diameter of top of bucket = 56 cm

Diameter of bottom of bucket = 42 cm

∴ Radius of top = R = 56/2 = 28 cm

∴ Radius of bottom = r = 42/2 = 21 cm

Volume of water bucket can hold = volume of bucket which is in form of frustum

= (22/7) × 5 × (784 + 441 + 588) cm³

= (22/7) × 5 × 1813 cm³

= 22 × 5 × 259 cm³

= 28490 cm³

Now 1 litre = 1000 cm³

∴ 28490 cm³ = 28490/1000 litres

= 28.49 litres

∴ bucket can hold 28.49 litres of water

Given: height of container frustum = h = 16 cm

Radius of lower circular end = r = 8 cm

Radius of upper circular end = R = 20 cm

Cost of 100 cm² metal sheet = 15 Rs

∴ Cost of 1 cm² metal sheet = 15/100 = 0.15 Rs

Formula: Total surface area of frustum = πr² + πR² + π(R + r)l cm²

Where l = slant height

∴ l = 20 cm

Since it is given that a bucket is to be made hence the top is open we need to subtract the area of top/upper circle from total surface area of frustum because we don’t require a metal plate for top.

Radius of top/upper circle = R

Area of upper circle = πR²

∴ area of metal sheet used = (total surface area of frustum)-πR²

= πr² + πR² + π(R + r)l- πR² cm²

= πr² + π(R + r)l cm²

= π × (8² + (20 + 8)20) cm²

= 3.14 × 624 cm²

= 1959.36 cm²

∴ 1959.36 cm² metal sheet is required to make the container.

∴ Cost of 1959.36 cm² metal sheet = 1959.36 × cost of 1 cm² metal sheet

= 1959.36 × 0.15 Rs

= 293.904 Rs

∴ Cost of metal sheet required to make container = 293.904 Rs

Given: depth of bucket = height of bucket/frustum = h = 24 cm

Diameter of lower circular end = 10 cm

Diameter of upper circular end = 30 cm

∴ Radius of lower circular end = r = 10/2 = 5 cm

∴ Radius of lower circular end = R = 30/2 = 15 cm

Cost of 100 cm² metal sheet = 10 Rs

∴ Cost of 1 cm² metal sheet = 10/100 = 0.1 Rs

Cost of 1 litre milk = 20 Rs

Total surface area of frustum = πr² + πR² + π(R + r)l cm²

Where l = slant height

∴ l = 26 cm

Since the top is open we need to subtract the area of top/upper circle from total surface area of frustum because we don’t require a metal plate for top.

Radius of top/upper circle = R

Area of upper circle = πR²

∴ area of metal sheet used = (total surface area of frustum)-πR²

= πr² + πR² + π(R + r)l- πR² cm²

= πr² + π(R + r)l cm²

= π × (5² + (15 + 5)26) cm²

= 3.14 × 545 cm²

= 1711.3 cm²

∴ 1711.3 cm² metal sheet is required to make the container.

∴ Cost of 1711.3 cm² metal sheet = 1711.3 × cost of 1 cm² metal sheet

= 1711.3 × 0.1 Rs

= 171.13 Rs

∴ Cost of metal sheet required to make container = 171.13 Rs

Now,

Volume of milk which can completely fill the bucket = volume of frustum

= (1/3) × 3.14 × 26 × 325 cm³

= 26533/3 cm³

= 8844.33 cm³

Now 1 litre = 1000 cm³

∴ 8844.33 cm³ = 8844.33/1000 litres

= 8.84433 litres

∴ Volume of milk which can completely fill the bucket = 8.84433 litres

∴ Cost of milk which can completely fill the bucket = volume of milk which can completely

fill the bucket × cost of 1 litre milk Rs

= 8.84433 × 20 Rs

= 176.8866 Rs

Cost of milk which can completely fill the bucket = 176.8866 Rs

Given: height of container/frustum = h = 14 cm

Diameter of top of container = 35 cm

Diameter of bottom of container = 30 cm

∴ Radius of top = R = 35/2 = 17.5 cm

∴ Radius of bottom = r = 30/2 = 15 cm

1 cm³ of oil = 1.2g of oil

Cost of 1 kg oil = 40 Rs

Volume of oil in container = volume of container which is in form of frustum

= (22/3) × 2 × (306.25 + 225 + 262.5) cm³

= (22/3) × 2 × 793.75 cm³

= 34925/3 cm³

= 11641.667 cm³

∴ Volume of oil in container = 11641.667 cm³

∴ 11641.667 cm³ of oil = 11641.667 × 1.2 g

= 13970.0004 g

We know 1000 g = 1 kg

∴ 13970.0004 g = 13970.0004/1000 kg

= 13.970 kg

∴ Cost of 13.970 kg oil = 13.970 × cost of 1 kg oil Rs

= 13.970 × 40 Rs

= 558.8 Rs

∴ Cost of oil in container = 558.8 Rs

Given: volume of bucket = 28.49 litres

1 litre = 1000 cm³

∴ 28.49 litres = 28.49 × 1000 cm³

∴ Volume of bucket = 28490 cm³

Radius of upper circular end = R = 28 cm

Radius of lower circular end = r = 21 cm

Let ‘h’ be the height of the bucket

Volume of bucket = volume of frustum of cone

∴ 28490 × 21 = h × 22 × (784 + 441 + 588)

∴ h = 598290/39886 cm

∴ h = 15 cm

∴ Height of bucket = h = 15 cm

Given: volume of bucket = 5390 cm³

Radius of upper circular end = R = 14 cm

Radius of lower circular end = r cm & r is less than 14

Height of bucket = h = 15

Volume of bucket = volume of frustum of cone

∴ 5390 × 7 = 22 × 5 × (196 + r² + 14r)

∴ 37730/110 = 196 + r² + 14r

∴ 343 = 196 + r² + 14r

∴ r² + 14r-147 = 0

∴ r² + 21r-7r-147 = 0

∴ r(r + 21)-7(r + 21) = 0

∴ (r-7)(r + 21) = 0

∴ r = 7 or r = -21

Since we require r<14 ∴ r = 7 cm

Given: Radius of lower circular end = r = 27 cm

Radius of upper circular end = R = 33 cm

Slant height = l = 10cm

Formula: Total surface area of frustum = πr² + πR² + π(R + r)l cm²

= 3.14 × (33² + 27² + (33 + 27) × 10) cm²

= 3.14 × (1089 + 729 + 600) cm²

= 3.14 × 2418 cm²

= 7592.52 cm²

∴ total surface area = 7592.52 cm²

Given: Diameter of base of frustum = 20 m

Diameter of top of frustum = 6 m

∴ Radius of base = R = 20/2 = 10 m

∴ Radius of top = r = 6/2 = 3 m

Height of frustum = h_f = 24 m

Height of tent = h_t = 28 m

∴ height of cone = h_c = h_t -h_f = 28-24 = 4 m

Formula: Total surface area of frustum = πr² + πR² + π(R + r)l_f m²

Total surface area of cone = πrl_c

Where l_f = slant height of frustum & l_c = slant height of cone

∴ l_f = 25 m

For slant height of cone consider right angled ΔABC from figure

AB = h_c = 4 m ; BC = r = 3 m ; AC = l_c

By pythagoras theorm

AB² + BC² = AC²

∴ 4² + 3² = l_c²

∴ l_c = 5

Since length cannot be negative l_c = 5 m

Since for tent we don’t require the upper circle of frustum and the lower circle of frustum hence we should subtract their area as we don’t require canvas for that.

Surface area of upper circle = πr²

Surface area of lower circle = πR²

∴ Surface area of frustum for which canvas is required = πr² + πR² + π(R + r)l_f - πr²-πR²cm²

= π(R + r)l_fm²

= (22/7) × (10 + 3) × 25 m²

= (22/7) × 325 m²

= 1021.4285 m²

Surface area of cone = πrl_c m²

= (22/7) × 3 × 5 m²

= (22/7) × 15 m²

= 47.1428 m²

∴ Quantity of canvas required = surface area of frustum + surface area of cone

= 1021.4285 + 47.1428 m²

= 1068.5713 m²

∴ Quantity of canvas required = 1068.5713 m²

Given: Diameter of base of frustum = 26 m

Diameter of top of frustum = 14 m

∴ Radius of base = R = 26/2 = 13 m

∴ Radius of top = r = 14/2 = 7 m

Height of frustum = h_f = 8 m

∴ height of cone = h_c = 12 m

Formula: Total surface area of frustum = πr² + πR² + π(R + r)l_f m²

Total surface area of cone = πrl_c

Where l_f = slant height of frustum & l_c = slant height of cone

∴ l_f = 10 m

Consider right angled ΔABC from figure

AB = h_c = 12 m ; BC = r = 7 m ; AC = l_c

By pythagoras theorm

AB² + BC² = AC²

∴ 12² + 7² = l_c²

∴ l_c = 13.892

Since length cannot be negative l_c = 13.892 m

Since for tent we don’t require the upper circle of frustum and the lower circle of frustum hence we should subtract their area as we don’t require canvas for that.

Surface area of upper circle = πr²

Surface area of lower circle = πR²

∴ Surface area of frustum for which canvas is required = πr² + πR² + π(R + r)l_f - πr²-πR²cm²

= π(R + r)l_fm²

= 3.14 × (13 + 7) × 10 m²

= 3.14 × 200 m²

= 628 m²

Surface area of cone = πrl_c m²

= 3.14 × 7 × 13.892 m²

= 3.14 × 97.244 m²

= 305.346 m²

∴ Quantity of canvas required = surface area of frustum + surface area of cone

= 628 + 305.346 m²

= 933.346 m²

∴ Quantity of canvas required = 933.346 m²

Given: perimeter of upper circle = 36 cm

Perimeter of lower circle = 48 cm

Height of frustum = h = 11 cm

Let r: radius of upper circle & R: radius of lower circle

Total surface area of frustum = πr² + πR² + π(R + r)l cm²

Where l = slant height

Now perimeter of circle = circumference of circle = 2π × radius

∴ Perimeter of upper circle = 2πr

∴ 36 = 2 × 3.14 × r

∴ r = 36/6.28 cm

∴ r = 5.732 cm

∴ Perimeter of lower circle = 2πR

∴ 48 = 2 × 3.14 × R

∴ R = 48/6.28 cm

∴ R = 7.643 cm

∴ l = 11.164 cm

∴ Volume of frustum = (1/3) × 3.14 × 11 × (7.643² + 5.732² + 7.643 × 5.732) cm³

= (1/3) × 34.54 × (58.415 + 32.855 + 43.809) cm³

= 11.513 × 135.079 cm³

= 1555.164 cm³

∴ Volume of frustum = 1555.164 cm³

Now we have asked curved surface area so we should subtract the top and bottom surface areas which are flat circles.

Surface area of top = πr²

Surface area of bottom = πR²

∴ Curved surface area = total surface area - πr² - πR² cm²

= πr² + πR² + π(R + r)l - πr² - πR² cm²

= π(R + r)l cm²

= 3.14 × (7.643 + 5.732) × 11.164 cm²

= 3.14 × 13.375 × 11.164 cm²

= 468.86 cm²

∴ curved surface area = 468.86 cm²

Let ‘H’ be the height of cone ‘R’ be the radius of base of cone.

R = 10 cm

When the cone is cut at midpoint of H by a plane parallel to its base we get a cone of height H/2 and a frustum also of height H/2

Let the radius of the base of the cone which we got after cutting and the radius of upper circle of frustum be ‘r’ as shown in figure

From figure consider ΔABC and ΔADE

∠ABC = ∠ADE = 90˚

∠BAC = ∠DAE …(common angle)

as two angles are equal by AA criteria we can say that

ΔABC∼ΔADE

∴ r = 5 cm

Let V_c be volume of the cone and V_f be the volume of frustum

Volume of cone = (1/3)π(radius)²(height) cm³

∴ V_c = (1/3) × π × r² × (H/2) cm³

= (1/3) × π × 5² × (H/2) cm³

= (1/3) × π × 25 × (H/2) cm³

Volume of frustum = (1/3)πh(R² + r² + Rr) cm³

∴ V_f = (1/3) × π × (H/2)(10² + 5² + 10 × 5) cm³

= (1/3) × π × (H/2) × 175 cm³

∴ The ratio of volumes of two parts after cutting = V_c:V_f = 1:7

Let the cutting plane be passing through points B and C as shown

Height of cone = AD = H = 20 cm

Height of small cone which we get after cutting = AB = h_c

Let ‘r’ be the radius of small cone ∴ we have BC = r

‘R’ be radius of original cone which is to be cut ∴ we have DE = R

From figure consider ΔABC and ΔADE

∠ABC = ∠ADE = 90˚

∠BAC = ∠DAE …(common angle)

as two angles are equal by AA criteria we can say that

ΔABC∼ΔADE

Let V₁ be the volume of cone to be cut

Let V₂ be the volume of small cone which we get after cutting

Volume of cone = (1/3)π(radius)²(height) cm³

∴ V₁ = (1/3) × π × R² × h_c

∴ V₂ = (1/3) × π × r² × 20

Given is that the volume of small cone is (1/8) times the original cone

∴ V₂ = (1/8) V₁

∴ (1/3) × π × r² × 20 = (1/8) × (1/3) × π × R² × h_c

Using equation (i) we get

∴ h_c³ = 20³/8 cm

∴ h_c = 20/2 cm

∴ h_c = 10 cm

But we have to find the height from base i.e. we have to find BD from figure

∴ 20 = BD + h_c

∴ 20 = BD + 10

∴ BD = 10 cm

∴ 10 cm above base the section is made.

Let ‘R’ be the radius of the base of the cone which is also the base of frustum i.e. lower circular end as shown in the figure

DE = R

Let ‘r’ be the radius of the upper circular end of frustum which we get after cutting the cone

BC = r

The height of the cone is 20 cm and we had cut the cone at midpoint therefore height of the frustum so obtained is 10 cm

Vertical angle as shown in the figure is 60˚

Now a wire of diameter 1/12 (i.e. radius 1/24) is made out of the frustum let ‘l’ be the length of the wire

As we are using the full frustum to make wire therefore volumes of both the frustum and the wire must be equal.

∴ Volume of frustum = volume of wire made …(i)

Consider ΔABC

∠BAC = 30˚ ; AB = 10 cm ; BC = r

∴ r = 10/√3 cm

Consider ΔADE

∠DAE = 30˚ ; AD = 20 cm ; DE = R

∴ R = 20/√3 cm

Now using equation (i)

∴ 7000/9 = l/576

∴ 777.778 = l/576

∴ l = 448000 cm

Length of wire = 448000 cm

Given: Radius of lower circular end = R = 10 cm

Radius of upper circular end = r = 4 cm

Slant height = l = 15cm

Formula: Total surface area of frustum = πr² + πR² + π(R + r)l cm²

As the lower circular end is open we need to subtract the area of lower circular end from total surface area since we don’t require material for lower circular end it should be open so that it is wearable.

∴ Total surface area = πr² + πR² + π(R + r)l- πR² cm²

= (22/7) × (4² + (10 + 4) × 15) cm²

= (22/7) × (16 + 210) cm²

= 22 × 226/7 cm²

= 710.285 cm²

∴ total surface area = 710.285 cm²

∴ Area of material used = 710.285 cm²

Divide the funnel into two parts frustum and cylinder as shown in the figure

Parameters of frustum:

Diameter of upper circular end = 18 cm

∴ Radius of upper circular end = r = 18/2 = 9 cm

The radius of cylinder is equal to the radius of lower circular end of frustum

∴ radius of lower circular end = R = 4 cm

Height of frustum = total height – height of cylinder

= 22 – 10

= 12 cm

∴ height of frustum = h = 12 cm

Total surface area of frustum = πr² + πR² + π(R + r)l cm²

Where l = slant height

∴ l = 13 cm

Since for the frustum part of the funnel we don’t require the upper circular end and the lower circular end hence we need to subtract those areas from total surface area.

Area of upper circular end = πr²

Area of lower circular end = πR²

total surface area = πr² + πR² + π(R + r)l- πr²- πR²

= π(R + r)l

= 3.14 × (9 + 4) × 13

= 530.66 cm²

∴ total surface area of frustum for which tin is required = 530.66 cm²

Parameters of cylinder:

Height of cylinder = 10 cm

Radius of cylinder = 4 cm

∴ Area of tin require to make cylinder = 2π × (radius) × (height)

= 2 × 3.14 × 4 × 10

= 251.2 cm²

∴ Area of tin required to make the funnel = area of frustum for which tin is required + area

of tin require to make cylinder

∴ area of tin required to make funnel = 530.66 + 251.2

= 781.86 cm²

∴ Area of tin sheet require to make the funnel = 781.86 cm²

Given,

Depth of the river = 1.5 m

Width of the river = 36 m

Flow rate of river = 3.5 km/hr

Now first change the rate of flow of the water in meter/min

As we know,

1 km = 1000 m

1 hour = 60 minutes

So,

River travel in a minute = 350/6 m

Now,

The amount of water that runs into sea per minute;

350/6 × 1.5 × 36 = 350 × 1.5 × 6 = 3150

So,

The amount of water that runs into the sea per minute is 3150 m³

Given,

Volume of the cube = 729 cm³

Let the edge of the cube = a cm

So,

Volume (v) of the cube = a³
a³ = 729
a³ = (9cm)³
a = 9 cm

Lateral surface area of cube = 4a²
= 4 × 92
= 4 × 81
= 324cm²
Total surface area of the cube = 6a²
= 6 × 9²
= 6 × 81
= 486 cm²

Given,

Edge of the Cubical Box = 1 m

Volume of the Cubical Box = a³

Edge of the cubes = 10 cm

Volume of the cube = a³

Number of Cubes = Volume of box/volume of the cube
= 100 × 100 × 100/10 × 10 × 10
= 1000000/1000
= 1000 cubes

Given,

Edge of the first cube = 6 cm

Volume of the first cube = a³ = (6)³ cm

Edge of the second cube = 8 cm

Volume of the second cube = a³ = (8)³ cm

Edge of the third cube = 10 cm

Volume of the third cube = a³ = (10)³ cm

So,

Volume of the formed cube = Volume of the First + Second + Third Cube

Volume of the formed cube = v1 + v2 + v3
= 6³ + 8³ + 10³
= 216 + 512 + 1000
= 1728
Now volume of new cube = a³= 1728
Edge of new cube = a = ∛1728
a = 12
Therefore surface area of new cube = 6a²
= 6 × 12²
= 6 × 12 × 12
= 864 cm²

Given,

Edge of the given Cube = 5 cm

Now,

Length (l) of the resulting cuboid = Edge × Number of cubes

= 5 x 5 = 25 cm

Breadth (b) of the resulting cuboid = 5 cm

Height (h) of the resulting cuboid = 5 cm

So,

Volume of the resulting cuboid = l × b × h

= 25 x 5 x 5

= 625 cm³

Hence,

The volume of the resulting cube is 625 cm³

Given,

Ratio of two cube = 8:27

Let the edges of the cubes to x and y

As,

…………..(i)

Now,

The ratio of the surface areas of the cubes

⇒ 6x²/6y²

= (x/y)²

= (2/3)² …………………….. [Using (i)]

= 4/9

= 4 : 9

So,

The ratio of the surface areas of the given cubes is 4 : 9.

Given,

Radius of the right circular cylinder = 176/7 cm³

Height of the right circular cylinder = Radius of the right circular cylinder

So,

⇒ h = r

As,

Volume of the right circular Cylinder = 176/7 cm³

⇒ πr²h = 176/7

⇒ 22/7 × h² × h=176/7

⇒ h³ =176 × 7/7 × 22

⇒ h³ = 8

⇒ h = 3√8

Therefore,

h = 2 cm

So,

The height of the right circular cylinder is 2 cm

Given,

Ratio of the base and the height of a cylinder is 2:3

So,

Let the radius of the base = r

And

The height of the cylinder = h

r : h = 2 : 3

That is,

r/h = 2/3

So,

h = 3r/2 – – – – – – – – – (i)

As,

Volume of the cylinder = 12936 cm³

⇒ πr²h=12936

⇒ 22/7 × r² × 3r/2=12936 [Using (i)]

⇒ 33/7 × r³=12936

⇒ r³ = 12936 × 7/33

⇒ r³ = 2744

⇒ r = 3√2744

Therefore,

r = 14 cm

So,

The radius of the base of the cylinder is 14 cm.

Let the radus of the first cylinder = r₁

And the radus of the second cylinder = r₂;

Let the height of first cylinder = h₁

And the height of second cylinder = h₂

Given,

r₁ : r₂ = 2 : 3

r_1/r₂ = 23 ……………….. (i)

And

h₁ : h₂ = 5 : 3

h₁/h₂ = 5/3 ……… (ii)

Now,

The ratio of the volumes of the cylinders =

……… [Using (i) and (ii)]

= 20/27

= 20 : 27

So,

The ratio of the volumes of the given cylinders is 20 : 27

Given,

Volume of the wire = 66 cm³

Diameter of the wire = 1 mm

Radius (r) of wire = 12 = 0.5 mm = 0.05 cm

Let the length of the wire be l

As,

Volume of the wire = 66 cm³

⇒ πr²l = 66

⇒ 227 × 0.05 × 0.05 × l = 66

⇒ l = 66 × 722 × 0.05 × 0.05

Therefore,

l = 8400 cm = 84 m

So,

The length of the wire is 84 m.

Given,

Area of the base of the right circular cone = 3850 cm²

Height of the right circular cone = 84 cm

Let the radius of the cone = r

And the slant height of the cone = l

As,

Area of the base of the cone = 3850 cm²

⇒ πr² = 3850

⇒ 227 × r² = 3850

⇒ r² = 3850 × 722

⇒ r² = 1225

⇒ r = √1225

Therefore,

r = 35 cm

Now,

length = √h² + r²

= √(84)² + (35)²

= √7056 + 1225

= √8281

= 91 cm

So,

The slant height of the given cone is 91 cm.

Given,

Base radius (r) of the cylinder = 8 cm

Height (h) of the cylinder = 2 cm and

Height (H) of the cone = 6 cm

Let the base radius of the cone = R

Now,

As the cylinder is melted to form the cone,

So,

Volume of the cone = Volume of the cylinder

⇒ 1/3πR²H = πr²h

⇒ R² = 3r²hH

⇒ R2=3 × 8 × 8 × 26

⇒ R² = 64

⇒ R = √64

Therefore,

R = 8 cm

So,

The radius of the base of the cone is 8 cm.

Let suppose the radius of the cone = r

And height of the cone = h,

Then,

Radius of the cylindrical vessel = r and

Height of the cylindrical vessel = h

Now,

The required number of cones = Volume of the cylindrical vessel/Volume of a cone

= πr²h/(1/3πr²h)

= 3

So,

The number of the cones that is required to store the water is 3.

The volume of the sphere = 4851 cm³

Let the radius of the sphere = r

As,

Volume of the sphere = 4851 cm³

⇒ 4/3πr³ = 4851

⇒ r³ = 92618

⇒ r = 3√92618

⇒ r = 212cm

Now,

The Curved surface area of the sphere = 4πr²

= 4 × 22/7 × 21/2 × 21/2

= 1386 cm²

So,

The curved surface area of the sphere is 1386 cm²

Given,

Curved surface area of the sphere = 5544 cm³

Let the radius of the sphere = r

As we know,

Curved surface area of the sphere = 4πr²

So,

⇒ 4πr² = 5544

⇒ r² = 441

⇒ r = √441

⇒ r = 21 cm

Now,

Volume of the sphere = 4/3πr³

= 4/3 × 22/7 × 21 × 21 × 21

= 38808 cm³

So,

The volume of the sphere is 38808 cm³.

Let suppose the radius of first spheres = r

And the radius of second spheres = R

As per the question,

Surface area of the first sphere /Surface area of the second sphere = 4/25

⇒ 4πr²/4πR² = 4/25

⇒ (r/R)² = 4/25

⇒ r/R = √4/25

⇒ r/R = 2/5 …………………(i)

Now,

= (r/R)³

= (2/5)³ …………………[Using (i)]

= 8/125

= 8 : 125

So,

The ratio of the volumes of the given spheres will be 8 : 125

Given,

Radius (R) of the solid metallic sphere = 8 cm

Radius (r) of the spherical ball = 2 cm

Now,

The number of spherical balls obtained =

= (R/r)³

= (8/2)³

= 4³

= 64

So,

The number of spherical balls obtained is 64.

Given,

Diameter of the lead shot = 3 mm

Radius (r) of a lead shot = 3/2 = 1.5 mm = 0.15 cm and

Dimensions of the cuboid = 9 cm x 11 cm x 12 cm

Now,

The number of the lead shots =

= 84000

So,

84000 number of lead shots can be made from the cuboid.

Cone:

Radius = 12cm

Height = 24 cm.

Sphere, radius = 2cm

Volume of cone is given as,

⇒

⇒ V = 3620 cm³

Volume of sphere ,

⇒

⇒ V = 33.52 cm³

∴ the number of squares formed will be: 3620/34 = 106

Given,

Radius (R) of the hemisphere = 6 cm

Height (h) of the cone = 75 cm

Let the radius of the base of the cone = r

Now,

Volume of the cone = Volume of the hemisphere

⇒

⇒ r² = 5.76

⇒ r = √5.76

Therefore,

r = 2.4 cm

So,

The radius of the base of the cone is 2.4 cm.

The basic concept required to solve any such question is that the volume of the two figures will be same, so here we will equate the volume of sphere to that of wire which is in shape of a cylinder and subsequently will find out the height of the cylinder.

Given diameter of copper sphere = D = 18 cm

∴ Radius of the sphere = R = d/2=18/2 = 9 cm

As we know the wire is cylindrical in shape so,

Let the height of the cylindrical wire be ‘h’ cm

Given diameter of cylindrical wire = d = 4 mm

⇒ Radius of the cylindrical wire = r = d/2= 4/2= 2 mm= 0.2 cm (∵ 1 mm = 0.1 cm)

Volume of a sphere=(where R=radius of sphere)→eqn1

(putting value of R in eqn 1)

⇒ Volume of sphere = 4π × 243= 972π cm³→eqn2

Volume of cylinder = πr²h

Where r = radius of base of cylinder and h = height of cylinder

⇒ Volume of cylindrical wire = π × (0.2)² × h (putting value of r)

= 0.04π × h cm³→eqn3

Now on equating equation 2 and equation 3, we get,

Volume of sphere = Volume of cylindrical wire

⇒ 972π = 0.04πh

⇒ π(927) = π(0.04h) (taking π common on both sides)

⇒ 927 = 0.04h

∴ h = 24300 cm

The height of cylindrical wire is 24300 cm or 243 m.

Given height of frustum = h =6 cm

Radius of top = r = 6 cm

Radius of bottom = R = 14 cm

Let the slant height of the frustum be ‘ℓ’ cm

We know in frustum

(Slant height)² = (height)² + (R – r)²→eqn1

⇒ ℓ² = 6² + (14 – 6)² (putting values of r, R and h in eqn1)

⇒ ℓ² = 36 + 8²

⇒ ℓ² = 36 + 64

⇒ ℓ² = 100

⇒ ℓ = √100

∴ ℓ = 10 cm

Slant height of the frustum is 10 cm.

Let the radius of the sphere be ‘R’ units

And the cube which will fit inside it be of edge ‘a’ units

Explanation: The longest diagonal of the cube that will fit inside the sphere will be the diameter of the of the sphere.

∴ The longest diagonal of cube = the diameter of the sphere

Consider ΔBCD, ∠BDC = 90°

BD = CD = a units (as they are the edges of cube)

⇒ BC² = a² + a² (putting value of BD and CD)

⇒BC² = 2a²

⇒BC = √(2a²)

∴ BC = a√2 units →eqn1

Now consider ΔABC, ∠ABC = 90°

Here, AB = a units and BC = a√2 units

⇒AC² = a² + (a√2)² (putting values of AB and BC)

⇒ AC² = a² + 2a²

⇒AC² = 3a²

⇒AC = √(3a²)

∴ AC = a√3 units

∴ Diameter of sphere = D = a√3 units

And we know, D = 2 × R

⇒ R = D/2 (put value of D )

Also, Volume of a sphere→eqn2

Put value of R in eqn2

∴ Volume of sphere = πa² cubic units → eqn3

Volume of cube = (edge)³

∴ Volume of cube = a³ cubic units →eqn4

Ratio of volume of cube to that of sphere

(putting values from eqn3 and eqn4)

⇒Ratio of volume of cube to that of sphere

Ratio of volume of cube to that of sphere is a:π

Let the radius of cylinder, cone and sphere be ‘r’ cm

Let the diameter of cylinder, sphere and cone be ‘2r’ units

∴ Height of cylinder, cone, sphere = h = 2r units

Volume of cylinder = π(r²)h

⇒ Volume of cylinder = π × r² × 2r (putting value of h)

∴ Volume of cylinder = 2π × r³→eqn1

Volume of sphere

Volume of cone(put the value of h)

⇒Volume of cone

∴ Volume of cone

Ratio of volume of cylinder to that of a cone to that of a sphere as:

⇒ Volume of cylinder : Volume of Cone : Volume of sphere

(dividing the above relation by (πr³)

(dividing the above ratio by 2)

⇒3 : 1 : 2 (multiplying the above ratio by 3)

Ratio of volume of cylinder to that of cone to that of sphere is 3 : 1 : 2

Given volume of each cube = 125 cm³

Let the edge of each cube be ‘a’ cm

So, Volume of cube = a³

⇒ a³ = 125

⇒ a = ∛(125)

∴ a = 5 cm →eqn1

Now when we join the two cubes then resulting cuboid will be of length twice that of the cube and breadth and height of the resulting cuboid will be same as that of the cube

⇒ length of cuboid = L = 2 × a

⇒ L = 2 × 5 (putting value of a from eqn1)

∴ L = 10 cm

Now, Breadth of cuboid = B = a

⇒ B = 5 cm

Similarly, height of the cuboid = H = 5 cm

Note: Where ever in a question Surface Area is mentioned, it means Total surface area.

Surface area = Total surface area = 2(LB + BH + HL)

⇒ S.A = 2((10 × 5) + (5 × 5) + (10 × 5))

⇒ S.A = 2(50 + 25 + 50)

⇒ S.A = 2 × 125

∴ S.A = 250 cm²

Surface Area of resulting cuboid is 250 cm².

Let the edges of cubes be a₁, a₂ and a₃

So, a₁ = 3 cm, a₂ = 4 cm, and a₃ = 5 cm

Explanation: Here the sum of volumes of all three cubes will be equal to the volume of the resulting larger cube as the resulting cube is formed by melting the three cubes.

Volume of cube with edge a₁ = v₁ = (a₁)³

⇒v₁ = (3)³

∴ v₁ = 27 cm³→eqn1

Similarly

Volume of cube with edge a₂ = v₂ = (a₂)³

⇒v₂ = (4)³

∴ v₂ = 64 cm³→eqn2

Volume of cube with edge a₃ = v₃ = (a₃)³

⇒v₃ = (5)³

∴ v₃ = 125 cm³→eqn3

Now let the volume of resulting cube be ‘V’ cm³

So, V = v₁ + v₂ + v₃

⇒ V = 27 + 64 + 125 (from eqn1, eqn2 and eqn3)

∴ V = 216 cm³→eqn4

Let the edge of resulting cube be ‘a’ cm

So, volume of the resulting cube = V = a³→eqn5

Equate equation 4 and 5,

⇒ a³ = 216

⇒ a = ∛(216)

∴ a = 6 cm

The edge of new cube formed is 6 cm.

Let the diameter of sphere be ‘D’ and Radius of sphere be ‘R’

∴ D = 8 m

Also, we know

R = D/2

⇒ R = 8/2

∴ R = 4 m

Explanation: Here the volume of sphere will be equal to the volume of the resulting cylinder as the resulting cylinder is formed by melting the sphere.

Volume of the sphere, V₁

Let the length/height of the cylinder be ‘H’ and let the radius of the cylinder be ‘r’ and volume of the cylinder be ‘V₂’

∴ H = 12 m

Volume of the cylinder = V₂ = π(r²)H

⇒ V₂ = π(r²) × 12 (putting value of H)

⇒ V₂ = 12π × r² m³→eqn2

Now equate equation 1 and 2,

⇒ V₂ = V₁

∴ r = 2.66 m

Width of cylinder = diameter of cylinder = 2 × radius

⇒ Width of cylinder = 2 × r

⇒ Width of cylinder = 2 × 2.66

∴ Width of cylinder = 5.32 m

Width of the resulting cylinder is 5.32 m

Let the length of the cloth used be ‘L’ cm

Area of cloth used = 5 × L →eqn1

Also, Given Diameter = d = 14 m and height = h = 24 m

∴ Radius = r = D/2

⇒ r = 14/2

∴ r = 7 m

Let the slant height of the cone be ℓ m

So, (Slant height)² = (Height)² + (Radius)²

Put the values in the above relation

⇒ ℓ² = h² + r²

⇒ ℓ² = 24² + 7²

⇒ ℓ² = 576 + 49

⇒ ℓ² = 625

⇒ ℓ = √(625)

∴ ℓ = 25 cm →eqn1

Also, we know Curved Surface Area of cone = πrℓ

Where r = radius of base, ℓ = slant height

C.S.A = π × 7 × 25

⇒ C.S.A = 22 × 25

⇒ C.S.A = 550 m²→eqn2

Now the Curved surface area of conical tent will be equal to the area of the cloth used to make the tent

⇒ C.S.A = Area of cloth

⇒ 550 = 5 × L (from eqn1 and eqn2)

∴ L = 110 m

So, cost of the cloth used = rate of cloth × Length of the cloth

⇒ Cost of cloth used = 25 × 110

⇒ Cost of cloth = Rs.2750

Cost of the cloth used is Rs. 2750

Given height of cylinder = h = 10 cm

Radius of cylinder = r = 3.5 cm

Radius of hemisphere = R = 3.5 cm

Explanation: In this question the volume of wood in toy can be calculated by subtracting the volume of two hemisphere from the volume of cylinder.

So, volume of cylinder = πr²h

Where r = radius of cylinder and h = height of cylinder

⇒ Volume of cylinder = π × (3.5)² × 10 (from given values)

⇒ Volume of cylinder = π × 12.25 × 10

∴ Volume of cylinder = 122.5π cm³→eqn1

Volume of a hemisphere(where R is radius of hemisphere)

∴ Volume of two hemisphere

Volume of wood in toy = eqn1 – eqn2

⇒Volume of wood in toy

(taking π common)

∴ Volume of wood in toy = 205.333 cm³

Volume of wood in toy is 205.333 cm³.

Let the edges of metal cubes be a₁, a₂ and a₃

And it is given that ratio of edges is 3:4:5

So, let a₁ = 3x, a₂ = 4x and a₃ = 5x

Volume of cube with edge a₁ = v₁ = (a₁)³

⇒v₁ = (3x)³

∴ v₁ = 27x³ cm³→eqn1

Similarly

Volume of cube with edge a₂ = v₂ = (a₂)³

⇒v₂ = (4x)³

∴ v₂ = 64x³ cm³→eqn2

Volume of cube with edge a₃ = v₃ = (a₃)³

⇒v₃ = (5x)³

∴ v₃ = 125x³ cm³→eqn3

Now let the volume of resulting cube be ‘V’ cm³

So, V = v₁ + v₂ + v₃

⇒ V = 27x³ + 64x³ + 125x³ (from eqn1, eqn2 and eqn3)

∴ V = 216x³ cm³→eqn4

It is given that the diagonal of the resulting cube is 12√3 cm

Let the edge of resulting cube be ‘a’ cm

Consider ΔBCD, ∠BDC = 90°

BD = CD = a cm (as they are the edges of cube)

⇒ BC² = a² + a² (putting value of BD and CD)

⇒BC² = 2a²

⇒BC = √(2a²)

∴ BC = a√2 cm

Now consider ΔABC, ∠ABC = 90°

Here, AB = a cm and BC = a√2 cm and AC = 12√3 cm

⇒(12√3)² = a² + (a√2)² (putting values of AB, AC and BC)

⇒ 144 × 3 = a² + 2a²

⇒ 144 × 3 = 3a²

⇒ 144 = a²

⇒ a = √144

∴ a = 12 cm

So, volume of the resulting cube = V = a³

⇒ V = 12³ (putting value of a)

∴ V = 1728 cm³ –eqn5

Equate equation 4 and 5

⇒ 216x³ = 1728

⇒ x³ = 1728/216

⇒ x³ = 8

⇒ x = ∛8

∴ x = 2

So a₁ = 3x = 3 × 2

⇒ a₁ = 6 cm

Similarly a₂ = 4x = 4 × 2

∴ a₂ = 8 cm

Similarly a₃ = 5x = 5 × 2

∴ a₃ = 10 cm

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

External diameter of hollow sphere = D = 8 cm

⇒ External radius of hollow sphere = R = D/2

⇒ R = 8/2

⇒ R = 4 cm

Internal diameter of hollow sphere = d = 4 cm

⇒ Internal radius of hollow sphere = r = d/2

⇒ r = 4/2

⇒ r = 2 cm

Let the height of the resulting cone be ‘h’ cm

Let the volume of External sphere be V₁ and that of internal be V₂.

Explanation: Here the volume of hollow sphere will be equal to the volume of the resulting cone as the resulting cone is formed by melting the sphere.

Volume of the External sphere=(put the value of R)

Volume of sphere = V = External volume – Internal volume

Given base radius of the resulting cone = r’ = 8 cm

Le the height be ‘h’ and volume of the resulting cone be V’

(putting value of r' )

Equate equation 3 and 4,

⇒ V = V’

⇒ 224 = 64h

⇒ h = 224/64

∴ h = 3.5 cm

The height of resulting cone is 3.5 cm.

Upper end radius of frustum = d/2= 28/2 = 14 cm

Lower end radius of frustum = D/2 = 42/2 =21 cm

Height of the frustum = 24 cm

And we know,

The amount of milk that bucket can hold = Volume of the bucket

And, Volume of bucket = Volume of Frustum

∴ Amount of milk that bucket can hold = Volume of frustum

⇒Volume of frustum

Where R = Radius of larger or lower end and r = Radius of smaller or upper end and h = height of frustum π = 22/7

⇒Volume of frustum

= 22 × 8 × 133

∴ Volume of frustum = 23408 cm³

Also we know that 1 litre = 1000 cm³

⇒ Volume of frustum in litre = 23408/1000= 23.408 litre

⇒ Amount of milk that the bucket hold = 23.408 litre

⇒ The cost of milk = Rate of milk × amount of milk bucket holds

⇒ The cost of milk = 30 × 23.408

∴ = Rs. 702.24

The cost of milk is Rs.702.24

Explanation: Here in order to find th outer surface area of building we need to simply add the curved surface areas of cone and cylinder.

Given height of cylinder = h = 4 m

Height of cone = h’ = 2.8 m

Diameter of cylinder = diameter of cone = d = 4.2 m

⇒ Radius of cone = Radius of cylinder = d/2 =4.2m/2 = 2.1 m

Outer surface area of building = C.S.A of cylinder + C.S.A of cone

Now, C.S.A of cylinder = 2πrh →eqn1

Where r = radius of base of cylinder, h = height of cylinder

And C.S.A of cone = πrℓ →eqn2

Where r = radius of base of cone, ℓ = Slant height of cone

We know in a cone

(Slant height)² = (height)² + (radius)² (put the given values)

(ℓ)² = (2.8)² + (2.1)²

⇒ (ℓ)² = 7.84 + 4.41

⇒ (ℓ)² = 12.25

⇒ ℓ = √(12.25)

∴ ℓ = 3.5 m

Now, C.S.A of cone = π × 2.1 × 3.5 (putting the values in eqn2)

⇒ C.S.A of cone = 7.35π m²→eqn3

C.S.A of cylinder = 2 × π × 2.1 × 4 (putting the values in eqn1)

⇒ C.S.A of cylinder = 2 × π × 4.41 × 4

∴ C.S.A of cylinder = 16.8π m²→eqn4

Outer surface area of building = eqn3 + eqn4

⇒ Outer surface area = 7.35π + 16.8π

= 24.15π

∴ Outer surface area =75.9 m²

The outer surface area of building is 75.9 m².

Let the Radius of cone be ‘r’ and height of cone be ‘h’

∴ r = 21 cm and h = 84 cm

Explanation: Here the volume of cone will be equal to the volume of the resulting sphere as the resulting sphere is formed by melting the cone.

Volume of cone,

⇒ V₁ = π × 441 × 28

∴ V₁ = 12348π m³→eqn1

Let the Radius of resulting sphere be ‘R’ cm

Volume of the sphere=

Now equate equation 1 and 2,

⇒ V₂ = V₁

⇒ R³ = 3087 × 3

⇒ R³ = 9261

⇒ R =∛(9261)

∴ R = 21 cm

Diameter of Sphere = 2 × radius

⇒ Diameter of Sphere = 2 × R = 42 m

Diameter of the resulting Sphere is 42 m

Explanation: The total surface area of the toy can be calculated by taking the sum of the Curved surface area of cone and that of hemisphere of same radius.

Given total height of toy = H = 15.5 cm

Radius of hemisphere = Radius of cone = r = 3.5 cm

Now the height of cone = h = total height – radius of hemisphere

⇒ Height of cone = h = 15.5 – 3.5

∴ Height of cone = h = 12 cm

Let the slant height of the cone be ‘ℓ’ cm

Also we know that in a cone,

(Slant height)² = (Height)² + (Radius)²

⇒ ℓ² = 12² + 3.5² (putting the values)

⇒ ℓ² = 144 + 12.25

⇒ ℓ² = 156.25

⇒ ℓ = √(156.25)

∴ ℓ = 12.5 cm

C.S.A of cone = πrℓ

⇒ C.S.A of cone = π × 3.5 × 12.5 (putting the given values)

∴ C.S.A of cone = 43.75π m² →eqn1

C.S.A of hemisphere = 2πr²

⇒ C.S.A of hemisphere = 2 × π × 3.5²

= 2 × π × 12.25

= 24.5π m² →eqn2

Now total surface area of toy = eqn1 + eqn2

⇒ Total surface area of toy = 43.75π + 24.5π

= 68.25π

⇒The total surface area of toy (putting π= 22/7)

= 9.75 × 22

= 214.5 m²

The total surface area of the toy is 214.5 m².

Explanation: Here the bucket is in the shape of a frustum. So capacity of bucket will be equal to the volume of the frustum and in order to calculate the total surface area of the bucket we will subtract the top end circular area from the total surface area of the frustum as the bucket is open on top.

Upper end radius of frustum/bucket = R = 28 cm

Lower end radius of frustum/bucket =r = 7 cm

Height of the frustum/bucket = 28 cm

And we know,

The capacity of bucket = Volume of the bucket

And, Volume of bucket = Volume of Frustum

∴ Capacity of bucket = Volume of frustum

⇒Volume of frustum

Where R = Radius of larger or upper end and r = Radius of smaller or lower end and h = height of frustum π = 22/7

⇒Volume of frustum

= 22 × 4 × 343

∴ Volume of frustum = 30184 cm³

∴ Capacity of bucket = 30184 cm³

T.S.A of bucket = T.S.A of frustum – Area of upper circle →eqn1

Let the slant height of the frustum be ‘ℓ’ cm

So, ℓ² = h² + (R – r)²

⇒ ℓ² = 28² + (28 – 7)²

⇒ ℓ² = 784 + (21)²

⇒ ℓ² = 784 + 441

⇒ ℓ² = 1225

⇒ ℓ = √(1225)

∴ ℓ = 35 cm

⇒ T.S.A of frustum = π(R + r)ℓ + πR² + πr²

= π(28 + 7) × 35 + π(28)² + π(7)²

= 35 × 35π + 784π + 49π

= 1225π + 784π + 49π →eqn2

Area of upper circle = πR²

= π(28)²

= 784π →eqn3

T.S.A of bucket = 1225π + 784π + 49π – 784π (from eqn2 and 3)

⇒ T.S.A of bucket = 1274π

⇒T.S.A of bucket=

⇒ T.S.A of bucket = 182 × 22

∴ T.S.A of bucket = 4004 cm²

The capacity and total surface area of the bucket is 30184 cm³ and 4004 cm².

Upper end radius of frustum/bucket = R = 20 cm

Lower end radius of frustum/bucket =r = 12 cm

Height of the frustum/bucket be ‘h’ cm

And we know,

The capacity of bucket = Volume of the bucket

And, Volume of bucket = Volume of Frustum

Volume of frustum/bucket = V = 12308.8 cm³

∴ Capacity of bucket = Volume of frustum

⇒Volume of frustum

Where R = Radius of larger or upper end and r = Radius of smaller or lower end and h = height of frustum π = 3.14

(putting the values)

⇒ 3.14 × h × (144 + 400 + 240) = 12308.8 × 3

⇒ 3.14 × h × 784 = 36926.4

⇒ 2461.76 × h = 36926.4

⇒ h = 36926.4/2461.76

∴ h = 15 cm

The height of the bucket is 15 cm.

Upper end radius of container = R = 20 cm

Lower end radius of container =r = 8 cm

Height of the container be ‘h’ cm

As container is in shape of frustum

Volume of container = Volume of Frustum

Where R = Radius of larger or upper end and r = Radius of smaller or lower end and h = height of frustum π = 22/7

(putting the values)

⇒ 22 × h × 208 = 73216

⇒ 4576 × h = 73216

⇒ h = 73216/4576

∴ h = 16 cm

Let the slant height of the frustum be ‘ℓ’ cm

So, ℓ² = h² + (R – r)²

⇒ ℓ² = 16² + (20 – 8)²

⇒ ℓ² = 256 + (12)²

⇒ ℓ² = 256 + 144

⇒ ℓ² = 400

⇒ ℓ = √(400)

∴ ℓ = 20 cm