Surface Area And Volume Of A Right Circular Cylinder

Given,

Curved surface area of a right circular cylinder = 4.4m²

Radius of base of cylinder = 0.7 m

=2πrh = 4.4

= h =

∴ Height of cylinder = 1m.

Given,

Length of cylindrical pipe = 28 m =2800 cm

Diameter of pipe = 5cm

So, radius of pipe =

Total radiating surface area = 2πrh

=

Given,

Diameter of cylindrical pillar = 50 cm

So, radius of pillar =

Height of pillar = 3.5 m

So, curved surface area of pillar = 2πrh =

Cost of painting 1 m² area = 12.50 rs.

∴ cost of painting 55 m² =

Given,

Base diameter of cylindrical tank = 140 cm

So, radius of tank = = .70m

Height of tank = 1m

So, area of sheet required for making the tank =

=

Given,

Total surface area of cylinder = 462 cm²

Curved surface area = (total surface area)

=

= 2πrh = 154

= h =

And,

=

=

=

= 2r² = 98

= r = 7

So, radius of cylinder = 7 cm

∴ height =

2πrh = 154

=

Given,

Total surface area of hollow cylinder = 4620 cm²

Area of base ring = 115.5 cm²

Height of cylinder = 7 cm

Let outer radius of hollow cylinder = R cm

Let inner radius be = r cm

= πR² –πr² = 115.5 ………. (i)

And,

2πrh +2πrh+ 2(πR² – πr²) = 4620 ……………….(ii)

(putting value of (πR² – πr²) from (i) to (ii) )

=2πh(R+r) = 4620 -231 = 4389

= π(R+r) =………………..(iii)

From equation (i)

=π(R² – r²) = 115.5

= π[(R+r)(R-r)] = 115.5

(putting value of π(R+r) from equation (iii))

=

Given,

Height of cylinder = 7.5 cm

Radius of cylinder = 3.5 cm

So, curved surface area of cylinder = 2πrh = cm²

Total surface area = 2πr(h+r) = cm²

So,

Given,

Total surface area of hollow cylinder = 338π cm²

Height = 10 cm

External radius R = 8 cm

Let internal radius = r cm

∴ 2πRh+2πrh+2π(R²-r²) = 338π

=

=80+10r+(8+r)(8-r) =169

= 10r+64-8r+8r-r² = 89

=r²-10r+25 = 0

= r(r-5) -5(r-5) =0

= r = 5 cm

∴ Thickness of hollow cylinder = (R-r) = 8-5 = 3 cm

Given,

Radius of base = 70 cm

Height = 1.4 m = 140 cm

Because tin is coated on both side of vessel so surface area of side should be calculated,

= 2(2πrh) = cm²

Area of circular bases = 2πr² =cm²

So, total area to paint = 61600+30800 = 154000 cm²

Cost of coating 1000 cm² area = Rs. 3.50

∴cost of coating 154000 cm² =

Given,

Inner diameter of circular well = 3.5 m

Inner radius =

Depth of well = 10 m

i) Inner curved surface area of well = 2πrh = m²

ii) Cost of plastering 1m² area = Rs.40

Cost of plastering 110 m² = 40×110 = 4400 rs.

Given,

Diameter of cylindrical tank = 4.3 m

Radius of tank =

Height of tank = 4.5 m

Lateral curved surface area of tank = 2πrh =m²

Let total steel used in making of tank = X m²

Wasted steel = m²

Actual steel used = X-²

∴

=

= x =

Given,

Radius of each cylindrical pen holder = 3cm

Height = 10.5 cm

So, area of one cardboard = 2πrh+πr²

= πr(2h+r) =

Number of competitors = 35

∴ total area of cardboard needed = 35×226.28 = 7920 cm²

Given,

Diameter of roller = 84 cm

So radius of roller =

Length of roller = 1.5 m

Area covered by roller in one revolution = 2πrh =²

Area covered by it in 100 revolution = 100×3.96 = 396 m²

Cost of levelling 1m² area = Rs. .50

∴cost of levelling 396m² = .50×396 = Rs. 198

Given,

Diameter of one pillar = .50m

Radius of pillar =

Height of each pillar = 4m

Curved surface area of one pillar = 2πrh = ²

Area of 20 pillars = 20×6.28 = 125.60 m²

Cost of cleaning 1 m² area = Rs. 2.50

∴cost of cleaning 125.60 m² = 125.60×2.50 = Rs 314.28.

Given,

i) Length of rectangular can = 5cm

Breadth = 4cm

Height = 15 cm

So, volume of rectangular can = length×breadth×height

= 5×4×15 = 300cm³

ii) Diameter of plastic cylindrical box = 7 cm

So radius =

Height = 10 cm

Volume of cylindrical box = πr²h =

So,

V₂-V₁ = 85 cm . (plastic cylinder has more capacity).

Given,

Radius of base of cylindrical pillar = 20 cm = .20 m

Height = 10 m

So, concrete require to make a pillar = πr²h =

=

Concrete required for 14 such pillars =
= 0.00176 m³
= 1.76 litres

Given,

Inner diameter of cylindrical pipe = 24 cm

So, inner radius =

Outer diameter of that pipe = 28 cm

So, outer radius =

Length of pipe = 35 cm

Volume of pipe = π(R² – r²) =

= 22×26×2×5 = 5720 m³

1 cm³ wwod has mass = 0.6 gm

∴ 5720 cm³ wood has mass = 5720 ×0.6 = 3432 gm = 3.432 kg.

Given,

Lateral surface area of cylinder = 9.42 cm²

Height of cylinder = 5 cm

i) Radius of cylinder =

2πrh = 9.42

R =

ii) Volume of cylinder = πr²h =

Given,

Volume of cylindrical vessel = 15.4 litre = 15400 cm³

Height of vessel = 1 m = 100 cm

= πr²h = 15400

= r² =

= r =

Area of metal sheet required = 2πr(h+r)

=

Given,

Diameter of cylindrical bowl = 7 cm

So radius of bowl =

Height of soup in bowl = 4 cm

Volume of soup in 1 bowl = πr²h =

For 250 patients volume of soup prepared = 250 ×154 = 38500 cm³

=

Given,

Width of hollow cylinder = 63 cm

Girth (perimeter) = 440 cm

Let external radius of cylinder = R cm

So, 2πr = 440

= R =

= inner radius = outer radius – thickness = 70 – 4 = 66 cm

∴ volume of iron = π( R²-r²)h =

Given,

Total surface area of cylinder = 231 cm²

Curved surface area =

= 2πrh + 2πr² = 231

= 154 +2πr² = 231

= 2πr² = 77 = r² =

= r =

And , 2πrh = 154

(put value of r )

=

= height = h = 7 cm

So, volume of cylinder = πr²h =

Given,

Height of tank = 6 × radius of tank

Let radius of tank = r decimeter

So, height of tank = 6r dm

Total surface area of tank =

= 2πr(h+r) = 198×2

= 2πr(7r) = 396

= r² =

= r = 3 dm

Height = 6r = 6×3 = 18 dm

So, volume of tank = πr²h =

Given,

Ratio of radii of cylinders = r₁:r₂ =

Ratio oh heights of cylinders = h₁:h₂ =

So, ratio of their volumes = ₁²h₁ : πr₂²h₂ =

Ratio of their curved surface areas = 2πr₁h₁ : 2πr₂h₂ =

Given,

= = and, total surface area of cylinder = 616 cm²

=

So, 2πr(h+r) = 616

= 2πr(2r) = 616

= r² =

= r =

Hence, volume of cylinder = πr²h =

Given,

Curved surface area of cylinder = 1320 cm²

Diameter of base = 21 cm

So, radius of base =

= 2πrh = 1320

=

=

Hence volume of cylinder = πr²h =

Given,

=

Volume of cylinder = 1617 cm³

Let radius = 2x cm and height = 3x cm

So, volume of cylinder = πr²h =

=

= x =

Radius of cylinder = 2x = 2×3.5 = 7 cm

Height of cylinder = 3x = 3×3.5 = 10.5 cm

Hence , total surface area of cylinder = 2πr(h+r) =

Given,

Dimension of rectangular sheet = 44cm×20cm

(sheet is rolled along length so length become perimeter and breadth become its height)

=2πr = 44 = r =

Height = 20 cm

So, volume of cylinder = πr²h =

Given,

Curved surface area of cylinder = 264 m²

Volume of cylinder = 924 m³

= 2πrh = 264

= h =

And, πr²h = 924

(putting value of h from equation (I)

= πr²×

=

So, diameter of cylinder = 2r = 2×7 = 14 m

From equation (i)

=

Given,

Volume of cylinder 1 = volume of cylinder 2

=

= πr₁²h₁ = r₂²h₂

= r₁²/ r₂² = h₂/h₁ =

= ratio of their radii = r₁:r₂ =

Given,

Height of right circular cylinder = 10.5 m

And 3(area of both circular ends) = 2(curved surface area of cylinder)

= 3×πr² = 2×2πrh

= 6πr² = 4πrh

=

So, volume of cylinder = πr²h =

Given,

Depth of well = 21m

Diameter of well = 6m

So, radius of well =

Volume of earth that can be dug out = πr²h =

Inner curved surface area of cylindrical well = 2πrh =

Cost of plastering 1 m² area = Rs. 9.50

∴cost of plastering 396 m² area = 396×9.50 = Rs. 3762

Given,

Circumference of cylindrical tank = 176 cm

Length of tank = 3m = 300 cm

= 2πr = 176 = r =

∴ volume of timber = πr²h=

=

Given,

Diameter of well = 14 m

So, radius of well =

Depth of well = 8 m

Width of embankment = 21 m

So outer radius R = 21+7 = 28 m

Let height of embankment = H m

So, volume of earth in embankment = volume of earth dug out

= π(R²-r²) ×H = πr²h

= π (28² – 7²) ×H = π ×49×8

= H =

Given,

Difference between inner and outer surface area of cylindrical tube = 88 cm²

Length of tube = 14 cm

Volume of tube = 176 cm³

= 2πRh – 2πrh = 88

= 2πh(R-r) = 88

= R-r = ………(i)

And, 2π(R²-r²)h = 176

= (R+r) (R-r) =

= R+r = 4 cm ………..(ii)

From equation (i) and (ii)

= 2R = 5 = R = 2.5cm

= r = 4-2.5 = 1.5 cm

Given,

Internal diameter of circular pipe = 2 cm

So, radius of pipe =

Rate of flow of water through pipe = 6 m/s

Radius of base of cylindrical tank = 60 cm

So, volume of water flows through pipe in 1 second = πr²h =

Volume of flows in 30 minute (30×60) second =

Let rise of water level in cylindrical tank = h m

So, volume of water collected in tank in 30 minute =

=

= h =

Given,

Diameter of base of cylindrical tank = 56 cm

So, radius of tank =

Dimension of rectangular solid = 32cm×22cm×14 cm

Let height of cylinder = h cm

So,

= h =

Given,

Internal diameter of tube = 10.4 cm

So radius of tube =

Length of tube = 25 cm

Thickness of metal of tube = 8 mm =

Hence , external radius of tube = 5.2+.8 = 6 cm

So, volume of metal = π(R² – r²)h =

Given,

Inner radius of tap = 0.75 cm

Rate of flow of water through tap = 7 m/s = 700cm/s

Volume of water delivered in 1 second = πr²h =

Volume of water delivered in 1 hour (60×60 sec) =

=

=

Given,

Diameter of cylindrical tank = 1.4 m

So, radius of tank =

Height of tank = 2.1 m

Diameter of pipe = 3.5 cm

So radius of pipe =

Rate of flow of water through pipe = 2 m/s

Suppose the tank is filled in = x minute

So, volume of water flows through pipe in x minute= volume of tank

=

=

=

Given,

Dimension of rectangular sheet of paper = 30cm×18cm

Case (i)

When paper is rolled along its length ,

Then, 2πr = 30 = r =

= h = 18 cm

So, volume of cylinder V₁ = πr²h =

Case (ii)

When paper is rolled along its width,

Then 2πr = 18 = r=

= h = 30 cm

So, volume of cylinder thus form =

Rato of volumes V₁ : V₂ =

Given,

Area of cross section of pipe = 5 cm²

Speed of water flows through pipe = 30 cm/s

So, volume of water flows through pipe in 1 second = area of cross section × rate of flow

= 5×30 = 150 cm³

Hence , volume of water flows through pipe in 1 minute(60second) =

= 60×150 = 9000 cm³ =

Given,

Sum of base radius and height of cylinder = 37 m

Total surface area of cylinder = 1628 m²

= 2πr(h+r) = 1628

= 2πr ×37 = 1628 = r =

∵ h+r = 37

∴ h = 37- 7 = 30 m

Hence, volume of cylinder = πr²h =

Given,

Depth of tubewell = 280 m

Diameter of tubewell = 3 m

So radius =

Volume of tubewell = πr²h =

Curved surface area of tubewell = 2πrh =

∴ cost of sinking at rate Rs 3.60/m³ = 3.60 ×1980 = Rs. 7128.

∴ cost of cementing at rate Rs. 2.50/m² = 2.50×2640 = Rs. 6600.

Given,

Weight of copper wire = 13.2 kg = (13.2×1000) gm

Diameter of wire = 4mm

So, radius of wire =

Weight of 1 cm³ wire = 8.4 gm

Let length of wire = h cm

So, volume ×8.4 = 13.2×1000

= πr²h×8.4 = 13.2×1000

=

= h =

Given,

Inside diameter of well = 10 m

So, inside radius of well =

Depth of well = 8.4 m

Earth spread out at height up to = 7.5m

So, outer radius R = 7.5+5= 12.5 m

Let height of embankment = H meter

So ,

volume of earth in embankment = volume of earth dug out

= π(R² –r²) = πr²h

= π(12.5² – 5²) ×H = π×5²×8.4

= 17.5×7.5×H = 25×8.4

=

Let Radius of cylinder = r

Let height of cylinder = h

So, curved surface area of cylinder = 2πrh

Now radius become = 2r

Height becomes =

So curved surface area =

Hence curved surface area is SAME.

There are 3 surfaces in a cylinder. (top, bottom, side)

Given,

Ratio of diameters of cylinders = 3:1

So, ratio of their radii = r₁:r₂ = 3:1

Ratio of their heights = h₁:h₂ = 1:3

So, ratio of their volumes = π(r₁)²h₁ : π(r₂)²h₂ =

Given,

Volume of cylinder = 1617 cm³

Let radius of cylinder = 2x cm

Let height of cylinder = 3x cm

= πr²h = 1617

=

= x³ =

= x = 3.5 cm

So radius = 2x = 2 ×3.5 = 7 cm

Height = 3x = 3×3.5 = 10.5 cm

∴ total surface area of cylinder = 2πr(h+r)

= ²

The number of surfaces in a right cylinder = 3 (top, bottom, side)

Given,

Ratio of radii = ₁:r₂ = 2:3

Ratio of heights = h₁:h₂ = 5:3

Ratio of volumes of cylinders = ₁²h₁) : (πr₂²h₂) =

Vertical cross section of a cylinder is a RECTANGLE

Radius of cylinder = r

Height of cylinder = h

So, volume of cylinder = πr²h

Number of surfaces in a hollow cylinder = 4

Let radius of cylinder = r

Let height of cylinder = h

So, volume of cylinder = πr²h

Now radius becomes = 2r

And radius = h

So ,

volume = π×4r²×h = 4πr²h

It becomes FOUR times.

Let radius of cylinder = r

Let height of cylinder = h

So, volume of cylinder = πr²h

New radius = r

New height = 2h

So,

volume becomes = πr²× 2h = 2πr²h

It becomes DOUBLED.

Let radius of cylinder = r

Let height of cylinder = h

So, volume of cylinder = πr²h

Now radius become =

Now height becomes = 2h

So, volume =

It become HALF .

Let radius of cylinder = r

Let height of cylinder = h

So, volume = πr²h

Now height is doubled = 2h

So , volume = πr²×2h = 2πr²h.

To gain a volume same as original volume we have to multiply r by

New r =

Height = 2h

Volume = πr²h =

Given

Radius of cylinder = r

Side length of square base =x

Height of rectangular box = height of cylinder

= πr²h =

= r^{2 =}

= r =

= r =

Given,

Height of cylinder = h

= h = 2πr

= r =

Volume of cylinder = πr²h =

Given,

Radius of cylinder = r

Height of cylinder = h

Total surface area = area of curved surface + area of circular bases

= 2πrh + 2πr²

= 2πr(h+r)

Given,

1 cubic feet =12×12×12 = 1728 inch³

Culinder of volume decrease = πr²×3 inch

= πr²×3 inches³ = 1728 inches³

= πr² =

= =

Diameter of cylinder = 2r = 2×

Given,

Diameter of cylinder = height of cylinder

2r = h or

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

= hπ (h+) =

Given,

Diameter of cylindrical tunnel = 2 m

So radius =

Length of tunnel = 40 m

Area of iron sheet required = 2πrh

= 2π×1×40 = 80π m²

In case (i)

2πr= a₁

= r = a₁/2π

= h₁ = a₂

So, v₁ = πr²h = π×(a₁/2π)²h = a₁²a₂/4π…………..(1)

In case (ii)

2πr = a₂

= r = a₂/2π and h = a₁

So, v₂ = πr²h = π× (a₂/4π)²a₁ = a₂²a₁/4π……………….(2)

From equations 1 and 2 …

= a₂v₁ = a₁v₂

= v_1/a₁ = v₂/a₂ ( so both options (B) and (D) are correct.)

Given,

Ratio of heights of cylinder = h₁:h₂ = 1:2

V₁=V₂

So, πr₁²h₁= πr₂²h₂

=

Given,

R₂ = ₁

V₁=V₂

= πR₁²h₁ = πR₂²h₂

= R₁²h₁ = R₁²h₂

= h₂= 9h₁

Given,

Let Initial height of cylinder = h

Initial curved surface area = 2πrh

Now altitude become = 6h

And base area =

In first case radius = r

In second case it become =

So , area =

Factor by which lateral surface of cylinder increase =