# Class 8 RD Sharma Solutions- Chapter 22 Mensuration III (Surface Area And Volume Of Right Circular Cylinder) – Exercise 22.2 | Set 2

### Chapter 22 Mensuration III (Surface Area And Volume Of Right Circular Cylinder) – Exercise 22.2 | Set 1

**Question 21. A well is dug 20 m deep and it has a diameter of 7 m. The earth which is so dug out is spread out on a rectangular plot 22 m long and 14 m broad. What is the height of the platform so formed?**

**Solution: **

Given that,

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Depth of well = 20m

Diameter of well = 7 m

Radius of well = d/2 = 7/2 m

Dimension of rectangular field = 22m × 14m

As we know that Volume of earth dug out from well = πr

^{2}h= 22/7 × 7/2 × 7/2 × 20 = 770 m

^{3}When this earth will spread on rectangular field,

then the Height of platform formed on Rectangular Field = Volume of Earth dug out / Area of Field

= 770/ (22×14) =

2.5m

**Question 22. A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an** **embankment. Find the height of the embankment.**

**Solution: **

Given that,

Diameter of well = 14 m,

Radius of well = d/2 = 14/2 = 7m,

Depth of well = 8 m.

We know that Volume of Earth dug out from well = πr

^{2}h= 22/7 × 7 × 7 × 8 = 1232 m

^{3}.This earth spread out on width of 21 m.

(Given)Area × h = 1232

π(R

^{2}– r^{2}) h = 1232π(28

^{2}– 7^{2}) h = 123222/7 (735) h = 1232

h = 1232×7 / 22×735 = 8624/16170 = 0.533 m =

53.3 cm

Hence, the Height of embankment is 53.3 cm.

**Question 23. A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm×22 cm×14 cm. Find the rise in the level of the water when the solid is completely submerged.**

**Solution: **

Given that,

Diameter of base of cylindrical vessel = 56 cm

Radius of base = d/2 = 56/2 = 28cm

Dimensions of rectangular Solid Vessel = 32cm × 22cm × 14cm

Volume of rectangular Solid Vessel = 32 × 22 × 14 = 9856 cm

^{3}Let us assume that the rise of water level is ‘h’ cm.

Volume of cylindrical container = Volume of rectangular solid vessel

(Given)πr

^{2}h = 985622/7 × 28 × 28 × h = 9856

h = 9856×7 / 22×28×28 =

4cm

Hence, the Rise in Water Level is 4cm.

**Question 24. A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.**

**Solution:**

Given that,

Dimensions of rectangular sheet = 30 cm × 18 cm

Case 1.When the paper is rolled along its length,2 πr = 30

r = 30 / 2π cm

Height = 18 cm

We know that Volume of Cylinder V1 = πr

^{2}h= π × (30/2π)

^{2}× 18 cm^{3}

Case 2.When paper is rolled along its breadth2πr = 18

r = 18/2π cm

Height = 30 cm

We know that Volume of Cylinder V2 = πr

^{2}h= π × (18/2π)

^{2}× 30 cm^{3}Hence Volume of Cylinder V1 / Volume of Cylinder V2 = (π × (30/2π)

^{2}× 18} / {π × (18/2π)^{2}× 30)= (π × (30/2π)

^{2}× 18}×1/{π × (2π/18)^{2}× 30)= 30

^{2}× 18 / 18^{2}× 30 =5/3

Hence the ratio of two volumes is 5:3

**Question 25. The rain which falls on a roof 18 m long and 16.5 m wide is allowed to be stored in a cylindrical tank 8 m in diameter. If it rains 10 cm on a day, what** **is the rise of water level in the tank due to it?**

**Solution: **

Given that,

Dimensions of roof = 18 m × 16.5 m,

Diameter of cylindrical tank = 8 m,

Radius of tank = d/2 = 8/2 = 4m,

It rains 10 cm a day

Let us assume that the rise in level of tank be ‘h’

Volume of tank = Volume of Roof

πr

^{2}h = lbh22/7 × 4 × 4 × h = 18 × 16.5 × 0.1

h = (18 × 16.5 × 0.1 × 7) / 22×4×4

= 207.9/352 = 0.5906m =

59.06cm

Hence, Rise in water level is 59.06cm

**Question 26. A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 5 cm. It is drawn-out into a wire of diameter 1 mm. What will be the length of the wire so formed?**

**Solution:**

Given that,

Diameter of metallic cylinder = 1 cm

Radius of metallic cylinder = d/2 = 1/2 = 0.5 cm

Length of cylinder = 5 cm

Diameter of wire drawn from it = 1 mm = 0.1 cm

Radius of wire = 0.5mm = 0.05cm

Let us assume that the length of wire be ‘h’ cm

Length of wire drawn from metal = Volume of Metal / Volume of Wire

(Given)= πr

^{2}h / πr^{2}= (½)

^{2}× 5 / (0.05)^{2}= 1.25/0.0025 =500 cm or 5m

Hence, Length of the wire is 5m.

**Question 27. Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.**

**Solution: **

Given that,

Weight of copper wire = 13.2 kg = 13200 gm,

Diameter of wire = 4 mm,

Radius of wire = d/2 = 4/2 = 2mm = 0.2cm,

Let us assume the length of wire is ‘h’ cm

Weight of 1 cubic cm wire = 8.4 gm

As we know that Volume = Weight / Density

Volume × Density = Weight

πr

^{2}h × 8.4 = 1320022/7 × 0.2 × 0.2 × h × 8.4 = 13200

h = 13200×7 / 22×0.2×0.2×8.4 = 12500 cm =

125m

Hence, the Length of 13.2kg of copper wire is 125 m.

**Question 28. 2.2 cubic dm of brass is to be drawn into cylindrical wire 0.25 cm in diameter. Find the length of the wire.**

**Solution:**

Given that,

Diameter of cylindrical wire = 0.25 cm,

Radius of wire = d/2 = 0.25/2 = 0.125cm,

Volume of brass wire = 2.2 dm

^{3}= 2200 cm^{3}.Let us assume that the length of wire is ‘h’ cm

πr

^{2}h = 220022/7 × 0.125 × 0.125 × h = 2200

h = 2200×7 / 22×0.125×0.125 =

44800 cm = 448m

Hence, the Length of the wire is 448m.

**Question 29. The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube.**

**Solution: **

Given that,

Length of Cylindrical tube = 14 cm

Let us assume that the outer radius of tube is = R cm

Let us assume that the inner radius of tube is = r cm

Difference between inside and outside surface of tube = 88 cm

^{2}2π (R-r) h = 88

——— (i)Volume of Cylinder = 176 cm

^{3}(Given)π (R2 – r2) h = 176

——— (ii)Dividing equation (i) by equation (ii),

2π (R-r) h / π (R

^{2}– r^{2}) h = 88/1762 / R + r = 1/2

R + r = 4

——— (iii)From equation (ii) we get,

π (R

^{2}– r^{2}) h = 176π (R+r) (R-r) h = 176

22/7 × 4 × (R-r) × 14 = 176

R-r = 176×7 / 22×4×14 = 1232/1232

R-r = 1

———— (iv)By adding equation

(iii)and(iv)we will get,R+r = 4

R-r = 1

2R = 5

R = 5/2 = 2.5cmR-r = 1

r = 2.5 – 1 = 1.5cm

Hence, the Inner and outer radii are 2.5cm and 1.5cm.

**Question 30. Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 meters per second into a cylindrical tank, the radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?**

**Solution: **

Given that,

Internal diameter of pipe = 2 cm

Internal radius of pipe = d/2 = 2/2 = 1cm

Rate of flow of water = 6 m/s = 600 cm/s

Radius of base of cylindrical tank = 60 cm

Rise in height in Cylindrical tank = Rate of flow of water × Total time × Volume of pipe / Volume of cylindrical tank

= (600 × 30 × 60 × π × 1 × 1) / (π × 60 × 60) =

300 cm = 3m

Hence, Rise in water level is 3m.

**Question 31. A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the** **metal is 8 mm everywhere. Calculate the volume of the metal.**

**Solution:**

Given that,

Internal diameter of cylindrical tube = 10.4 cm

Internal radius of tube = d/2 = 10.4/2 = 5.2cm

Length of tube = 25 cm

Thickness of metal = 8 mm = 0.8 cm

Outer radius of tube = R = 5.2+0.8 = 6 cm

We know that, Volume of metal = π(R

^{2}– r^{2}) × l= 22/7 × (6

^{2}– 5.2^{2}) × 25= 22/7 × (36 – 27.04) × 25

= 704 cm^{3}

**Question 32. From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in liters of water delivered by the pipe in one hour.**

**Solution: **

Given that,

Inner radius of tap = 0.75 cm

Length of water flowing in 1s = 7m = 700 cm

Volume of water per second derived from tap = πr ‑ 2l

= 22/7 × 0.75 × 0.75 × 700 =

1237.5 cm^{3}

Hence, Volume of water derived in 1 hour (3600 sec) = (1237.5 × 3600)/1000 = 4455 liters.

**Question 33. A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 meter per second. In how much time the tank will be filled?**

**Solution:**

Given that,

Diameter of cylindrical tank = 1.4 m

Radius of tank = d/2 = 1.4/2 = 0.7m

Height of tank = 2.1 m

Diameter of pipe flowing water in tank = 3.5 cm

Radius of pipe = d/2 = 3.5/2 cm

Rate of flow of water = 2 m/s

Time taken to fill the tank = Volume of tank / Volume of pipe × Rate of flow

= πr

^{2}h/(πr^{2}× 2)= (π × 0.7 × 0.7 × 2.1) / (π × 3.5/2 × 3.5/2 × 2) = 1.029 / 6.125 = 0.168=

1680 seconds = 28 minutes.

Hence, Time taken to fill the tank is 28 minutes.

**Question 34. A rectangular sheet of paper 30 cm × 18 cm be transformed into the curved surface of a right circular cylinder in two ways i.e., either by rolling the** **paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.**

**Solution: **

Given that,

Dimensions of rectangular sheet = 30 cm × 18 cm

Case 1.When paper is rolled along its length2 πr = 30

r = 30 / 2π cm

Height = 18 cm

Volume of Cylinder V1 = πr

^{2}h= π × (30/2π)

^{2 }× 18 cm^{3}

Case 2.When paper is rolled along its breadth2πr = 18

r = 18/2π cm

Height = 30 cm

Volume of cylinder V2 = πr

^{2}h= π × (18/2π)

^{2 }× 30 cm^{3}Hence, the Volume of Cylinder V1 / Volume of Cylinder V2 = (π × (30/2π)

^{2}× 18) / (π × (18/2π)^{2}× 30)= (π × (30/2π)

^{2}× 18) × 1/(π × (2π/18)^{2}× 30) = 30^{2}× 18 / 18^{2}× 30 =5/3

Hence, the ratio of two volumes is 5:3

**Question 35. How many liters of water flows out of a pipe having an area of cross-section of 5 cm**^{2} in one minute, if the speed of water in the pipe is 30 cm/sec?

^{2}in one minute, if the speed of water in the pipe is 30 cm/sec?

**Solution: **

Given that,

Cross-section area of pipe = 5 cm

^{2},Speed of water = 30 cm/s,

Time = 1 minute = 60 sec.

As we know that Volume of water flows through pipe = Area of cross-section × speed of flow × time

= 5 × 30 × 60 = 9000 cm

^{3}= 9000/1000 =9 liters

Hence, 9 liters of water flows out of pipe.

**Question 36. A solid cylinder has a total surface area of 231 cm**^{2}. It curved surface area is 2/3 of the total surface area. Find the volume of the cylinder.

^{2}. It curved surface area is 2/3 of the total surface area. Find the volume of the cylinder.

**Solution: **

Given that,

Total surface area of cylinder = 231 cm

^{2}Curved surface area = 2/3 total surface area = 2/3 × 231 = 154 cm

^{2}(Given)2πrh = 2/3 2πr(r + h)

3h = 2(r + h)

3h = 2h + 2r

h = 2r

———– (i)2πr(h + r) = 231

(Given)2 × 22/7 × r × (2r+r) =231

2 × 22/7 × r × 3r = 231

3r

^{2}= 231×7 / 2×22 = 36.75r

^{2}= 36.75 / 3 = 12.25r = √12.25 = 3.5 cm

Since, h = 2r = 2×3.5 = 7cm

We know that Volume of cylinder = πr

^{2}h= 22/7 × 3.5 × 3.5 × 7 =

269.5 cm3

**Question 37. Find the cost of sinking a tube well 280 m deep, having diameter 3 m at the rate of Rs 3.60 per cubic meter. Find also the cost of cementing its inner** **curved surface at Rs 2.50 per square meter.**

**Solution: **

Given that,

Depth of tube well = 280 m

Diameter of tube well = 3 m

Radius of well = d/2 = 3/2 = 1.5 m

As we know that Volume of Cylinder= πr

^{2}h= 22/7 × 1.5 × 1.5 × 280 = 1980 m

^{3}Cost of Sinking tube well at rate of Rs 3.60/m

^{3 }= 1980 × 3.60 = Rs 7128(Given)We know that Curved Surface Area = 2πrh

= 2 × 22/7 × 1.5 × 280 =

2640 m^{2}

Hence, Cost of cementing its inner curved surface at rate Rs 2.50/m^{2 }= 2.50 × 2640 = Rs 6600

**Question 38. Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.**

**Solution:**

Given that,

Weight of copper wire = 13.2 kg = 13200 gm

Diameter of wire = 4 mm

Radius of wire = d/2 = 4/2 = 2mm = 0.2cm

Let us assume that length of wire is ‘h’ cm.

Weight of 1 cubic cm wire = 8.4 gm

(Given)As we know that the Volume = Weight / Density

Volume × Density = Weight

πr

^{2}h × 8.4 = 1320022/7 × 0.2 × 0.2 × h × 8.4 = 13200

h = 13200×7 / 22×0.2×0.2×8.4 =

12500 cm = 125m

Hence, the Length of 13.2kg of copper wire is 125m.

**Question 39. 2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.**

**Solution: **

Given that,

Diameter of cylindrical wire = 0.25 cm

Radius of wire = d/2 = 0.25/2 = 0.125cm

Let us assume that length of wire is ‘h’ cm

Volume of brass wire = 2.2 dm3 = 2200 cm

^{3}(Given)πr

^{2}h = 220022/7 × 0.125 × 0.125 × h = 2200

h = 2200×7 / 22×0.125×0.125 =

44800 cm = 448m

Hence, the Length of the wire is 448m.

**Question 40. A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.**

**Solution:**

Given that,

Diameter of well = 10 m

Radius of well = d/2 = 10/2 = 5m

Depth of well = 8.4 m

Volume of earth dug out from well = πr

^{2}h= 22/7 × 5 × 5 × 8.4 = 660 m

^{3}This earth is spread out on width of 7.5 m.

Inner radii r = 5m and Outer radii R = (5+7.5) = 12.5cm

Area × h = 660

π(R

^{2}– r^{2}) h = 660π(12.5

^{2 }– 5^{2}) h = 66022/7 (131.25) h = 660

h = 660×7 / 22×131.25 = 4620/2887.5 =

1.6 m

Hence, the Height of embankment is 1.6m.

**Question 41. A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.**

**Solution:**

Given that,

Width of roller = 63 cm

Thickness of roller = 4 cm

Girth (Perimeter) = 440 cm

As we know that Perimeter = 2πR

2πR = 440

2 × 22/7 × R = 440

R = 440×7 / 2×22 = 70cm

Inner radius = R – Thickness = 70 – 4 = 66 cm

Volume of Cylindrical Iron = π(R

^{2}– r^{2}) l= 22/7 × (70

^{2}– 66^{2}) × 63 =107712 cm^{3}

Hence, the Volume of Iron is 107712cm^{3}.

**Question 42. What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm?**

**Solution:**

Given that,

Length of Solid Cylinder = L

Diameter of Cylinder = 2 cm

Radius of Cylinder = d/2 = 2/2 = 1cm

Volume of Cylinder = πr

^{2}L——— (i)Length of Hollow Cylinder = 16 cm

External Diameter = 20 cm

External Radius = 20/2 = 10cm

Thickness = 2.5 mm = 0.25 cm

Inner Radius = 10 – 0.25 = 9.75 cm

Volume = π (R

^{2}– r^{2}) l————– (ii)From equation (i) and (ii)

πr

^{2}L = π (R^{2}– r^{2}) lπ × 1 × 1 × L = π × (10

^{2}– 9.75^{2}) × 16L =

79cm

Hence, the length of the solid cylinder should be 79cm.

**Question 43. In the middle of a rectangular field measuring 30m × 20m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread** **over the remaining part of the field. Find the height through which the level of the field is raised.**

**Solution:**

Given that,

Diameter of well = 7 m

Radius of well = d/2 = 7/2 = 3.5m

Depth of well = 10 m

As we know that Volume of well = πr

^{2}h= 22/7 × 3.5 × 3.5 × 10 = 385 m

^{3}Area of Embankment field = 30 × 20 – 22/7 × 7/2 × 7/2

= 600 – 38.5 = 561.5 m

^{2}Volume of well = Area of Embankment field × Height of Embankment

385 = 561.5 × h

h = 385/561.5 =

0.6856m = 68.56 cm

Hence, the Height of Embankment is 68.56 cm.