Rd Sharma Solutions for Class 8 Math Chapter 22 Mensuration Iii Surface Area And Volume Of A Right Circular Cylinder are provided here with simple step-by-step explanations. These solutions for Mensuration Iii Surface Area And Volume Of A Right Circular Cylinder are extremely popular among Class 8 students for Math Mensuration Iii Surface Area And Volume Of A Right Circular Cylinder Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Book of Class 8 Math Chapter 22 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Solutions. All Rd Sharma Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 22.10:

Question 1:

Find the curved surface area and total surface area of a cylinder, the diameter of whose base is 7 cm and height is 60 cm.

Answer:

Let r and h be the radius and the height of the cylinder.Given:r=72 cmh=60 cmCurved surface area of the cylinder =2π×r×h                                               =2×227×72×60                                               =22×60=1320 cm2Total surface area of the cylinder=2π×r×(r+h)                                               =2×227×72×(72+60)=22×1272=11×127=1397cm2

Page No 22.10:

Question 2:

The curved surface area of a cylindrical road is 132 cm2. Find its length if the radius is 0.35 cm.

Answer:

Consider h to be the height of the cylindrical rod.Given:Radius, r=0.35 cmCurved surface area=132 cm2We know:Curved surface area=2×π×r×h                       132=2×227×0.35×h                           h=132×72×22×0.35                             h=60Therefore, the length of the cylindrical rod is 60 cm.

Page No 22.10:

Question 3:

The area of the base of a right circular cylinder is 616 cm2 and its height is 2.5 cm. Find the curved surface area of the cylinder.

Answer:

Given:Area of the base of a right circular cylinder= 616 cm2Height= 2.5 cmLet r be the radius of the base of a right circular cylinder.πr2 =616r2=616×722r2=196r=14 cmCurved surface area of the right circular cylinder=2πrh=2×227×14×2.5= 220 cm2

Page No 22.10:

Question 4:

The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area.

Answer:

Given:Height, h=15 cmCircumference of the base of the cylinder=88 cm2Let r be the radius of the cylinder.The circumference of the base of the cylinder=2πr88=2×227×rr=88×72×22=14 cmCurved surface area =2×π×r×h=2×227×14×15=1320 cm2Total surface area =2×π×r×(r+h)=2×227×14×(14+15)=2552 cm2

Page No 22.10:

Question 5:

A rectangular strip 25 cm × 7 cm is rotated about the longer side. Find The total surface area of the solid thus generated.

Answer:

Since the rectangular strip of 25 cm × 7 cm is rotated about the longer side, we have:Height, h=25 cmRadius, r=7cm Total surface area =2πr(r+h) = 2π(7)(25+7)= 14π(32)= 448πcm²=448×227cm²=1408 cm²

Page No 22.10:

Question 6:

A rectangular sheet of paper, 44 cm × 20 cm, is rolled along its length to form a cylinder. Find the total surface area of the cylinder thus generated.

Answer:

The rectangular sheet of paper 44 cm×20 cm is rolled along its length to form a cylinder.  The height of the cylinder is 20 cm and circumference is 44 cm.  We have: Height, h=20 cmCircumference= 2πr=44 cmTotal surface area is S=2πrh=44×20 cm²=880 cm²

Page No 22.10:

Question 7:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface areas.

Answer:

Let the radii of two cylinders be 2r and 3r, respectively, and their heights be 5h and 3h, respectively.Let S1 and S2 be the curved surface areas of the two cylinder.S1= Curved surface area of the cylinder of height 5h and radius 2rS2= Curved surface area of the cylinder of height 3h and radius 3r S1:S2=2×π×r×h : 2×π×r×h=2×π×2r×5h 2×π×3r×3h =10 : 9

Page No 22.10:

Question 8:

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Prove that its height and radius are equal.

Answer:

Let S1 and S2 be the curved surface area and total surface area of the circular cylinder, respectively.Then, S1=2πrh , S2=2πrr+hAccording to the question:                   S1:S2=1:22πrh : 2πrr+h =1 : 2            h : r+h = 1 : 2                    hr+h=12                           2h=r+h                              h=rTherefore, the height and the radius are equal.



Page No 22.11:

Question 9:

The curved surface area of a cylinder is 1320 cm2 and its base has diameter 21 cm. Find the height of the cylinder.

Answer:

Let h be the height of the cylinder.Given:Curved surface area, S=1320 cm2 Diameter, d=21 cmRadius, r=10.5     S=2πrh1320=2π×10.5×h     h=13202π×10.5     h=20 cm

Page No 22.11:

Question 10:

The height of a right circular cylinder is 10.5 cm. If three times the sum of the areas of its two circular faces is twice the area of the curved surface area. Find the radius of its base.

Answer:

Let r be the radius of the circular cylinder.Height, h= 10.5 cmArea of the curved surface, S1=2πrhSum of the areas of its two circular faces, S2=2πr2According to question:       3S2=2S13×2πr2=2×2πrh            6r=4h             3r=2h                  r=23×10.5 cm                    =7 cm

Page No 22.11:

Question 11:

Find the cost of plastering the inner surface of a well at Rs 9.50 per m2, if it is 21 m deep and diameter of its top is 6 m.

Answer:

Given:Height, h=21 mDiameter, d= 6 mRadius, r= 3 mArea of the inner surface of the well, S=2πrh=2π×3×21 m2= 2×227×3×21 m2=396 m2According to question, the cost per m2 is Rs 9.50.  Inner surface cost is Rs 396×9.50=Rs 3762

Page No 22.11:

Question 12:

A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of tin-plating it on the inside at the rate of 50 paise per hundred square centimetre.

Answer:

Given:Diameter, d=20 cmRadius, r= 10 cmHeight, h=14 cmArea inside the cylindrical vessel that is to be tin-plated=SS=2πrh+πr2=2π×10×14+π×102=280π+100π=380×227 cm2=83607 cm2According to question:Cost per 100 cm2 = 50 paiseCost per cm2 = Rs 0.005Cost of tin-plating the area inside the cylindrical vessel= Rs 0.005×83607=Rs 41.87=Rs 5.97

Page No 22.11:

Question 13:

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curved surface at Rs 4 per square metre.

Answer:

Given:Inner diameter of the circular well =3.5 m Inner radius of the circular well, r=1.75 mDepth of the circular well, h=10 mInner curved surface area, S=2πrhS=2π×1.75×10 m2=2×227×1.75×10 m2   =110 m2Cost of plastering 1m2 area =Rs 4Cost of plastering 110 m2 area =Rs 110×4=Rs 440

Page No 22.11:

Question 14:

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions moving once over to level a playground. What is the area of the playground?

Answer:

Given:Diameter of the roller =84 cm Radius, r=Diameter2=42 cmIn 1 revolution, it covers the distance of its lateral surface area.Roller is a cylinder of height,  h= 120 cm  Radius = 42 cmLateral surface area of the cylinder=2πrh=2×227×42×120=31680 cm2It takes 500 complete revolutions to level a playground. Area of the field =31680×500=15840000 cm2                       1 cm2=110000m2 15840000cm2=1584 m2.Thus, the area of the field in m2 is 1584 m2.

Page No 22.11:

Question 15:

Twenty one cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m, what will be the cost of cleaning them at the rate of Rs 2.50 per square metre?

Answer:

Given:Diameter of the pillars=0.5 mRadius of the pillars, r=0.25 mHeight of the pillars, h=4 mNumber of pillars =21Rate of cleaning = Rs 2.50 per square metreCurved surface area of one pillar=2πrh=2×227×0.25×4=2×227=447 m2 Curved surface area of one pillar=447 m2Cost of cleaning 21 pillars at the rate of Rs 2.50 per m2= Rs 2.5×21×447 =7.5×44 Cost of cleaning 21 pillars at the rate of Rs 2.50  per m2= Rs 330

Page No 22.11:

Question 16:

The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, area of base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.

Answer:

Given:Total surface area of the cylinder=4620 cm2Area of the base ring= 115.5 cm2Height, h=7 cmLet R be the radius of the outer ring and r be the radius of the inner ring.Area of the base ring =πR2-πr2115.5=πR2- r2R2- r2=115.5×722     (R+r)(R-r)=36.75      ...........   (i)Total surface area = Inner curved surface area + Outer curved surface area + Area of bottom and top rings4620=2πrh+2πRh+2×115.52πh(R+r)=4620-231R+r=4389×72×22×7R+r=3994        ...........   (ii)   Substituting the value of R+r from the equation (ii) in (i):3994(R-r)=36.75(R-r)=36.75×4399=0.368 cm Thickness of the cylinder= (R-r)=0.368 cm

Page No 22.11:

Question 17:

The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m2, find the circumference of its base.

Answer:

Let r and h be the radius and height of the solid cylinder.Given:r+h=37 mTotal surface area, S=2πrr+h1628 =2π×r×37        r=16282π×37          =1628232.477          =7 mCircumference of its base, S1=2πr    =2×227×7  m    =44 m

Page No 22.11:

Question 18:

Find the ratio between the total surface area of a cylinder to its curved surface area, given that its height and radius are 7.5 cm and 3.5 cm.

Answer:

Let S1 and S2 be the total surface area and curved surface area, respectively.Given:Height, h=7.5 cmRadius, r= 3.5 cmS1=2πrr+hS2=2πrhAccording to the question:S1S2=2πrr+h2πrhS1S2=r+hhS1S2=3.5+7.57.5S1S2=117.5=11075=2215Therefore, the ratio is 22:15.

Page No 22.11:

Question 19:

A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs 3.50 per 1000 cm2.

Answer:

Let r cm and h cm be the radius of the cylindrical vessel.Given: Radius, r= 70 cmHeight, h=1.4 m=140 cmRate of tin-plating = Rs 3.50 per 1000 square centimetreCost of tin-plating the cylindrical vessel on both the surfaces (inner and outer):Total suface area of a vessel= Area of the inner and the outer side of the base + Area of the inner and the outer curved surface=2πr2+2πrh=2πrr+2h=2×227×70×70+2×140=44×10×350=154000 cm2Cost of painting at the rate of Rs 3.50 per 1000 cm2=154000×3.501000=Rs 539Therefore, cost of painting is Rs 539.



Page No 22.25:

Question 1:

Find the volume of a cylinder whose
(i) r = 3.5 cm, h = 40 cm
(ii) r = 2.8 m, h = 15 m

Answer:

(i) Given :r=3.5 cm, h=40 cmVolume of cylinder, V=πr2h  =227×3.52×40  =1540 cm3(ii) Given:r=2.8 m, h=15 mVolume of cylinder, V=πr2h  =227×2.82×15  =369.6 m3

Page No 22.25:

Question 2:

Find the volume of a cylinder, if the diameter (d) of its base and its altitude (h) are:
(i) d = 21 cm, h = 10 cm
(ii) d = 7 m, h = 24 m

Answer:

(i) Given:d=21 cm, radius, r=d2=10.5 cmheight, h= 10 cmVolume of the cylinder, V=πr2h=227×10.52×10=3465 cm3(ii) Given: d=7 m, radius, r=d2=3.5 mheight h= 24 mVolume of the cylinder, V=πr2h=227×3.52×24=924 m3

Page No 22.25:

Question 3:

The area of the base of a right circular cylinder is 616 cm2 and its height is 25 cm. Find the volume of the cylinder.

Answer:

Let the area of the base of a right circular cylinder be S cm2.Given:S=616 cm2Height, h=25 cmLet the radius of a right circular cylinder be r cm.    S=πr2616=227×r2 r   2=616×722  r2 =196     r=14 cmVolume of the cylinder, V=πr2h  =π×142×25  =15400 cm3

Page No 22.25:

Question 4:

The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find the volume of the cylinder.

Answer:

Let r cm be the radius of a cylinder.Circumference of the cylinder, S=2πrGiven:Height, h=15 cmCircumference, S=88 cm S=2πr88=2×227×r  r=88×744  r=14 cmVolume of cylinder, V=πr2h   =227×142×15   =9240 cm3

Page No 22.25:

Question 5:

A hollow cylindrical pipe is 21 dm long. Its outer and inner diameters are 10 cm and 6 cm respectively. Find the volume of the copper used in making the pipe.

Answer:

Let the length of the cylinder pipe be h=21, dm=210 cm.Let the outer and the inner radius of the pipe be R cm and r cm, respectively. 2R=10  and 2r=6R=5 cm and r= 3cmVolume of the copper used in making the pipe, V=πR2-r2h   =227×52-32×210   =22×25-9×30   =22×16×30   =10560 cm3

Page No 22.25:

Question 6:

Find the (i) curved surface area (ii) total surface area and (iii) volume of a right circular cylinder whose height is 15 cm and the radius of the base is 7 cm.

Answer:

Given:Height, h= 15 cmRadius, r= 7 cm(i) Curved surface area, S1=2πrh   =2×227×7×15   =660 cm2(ii) Total surface area, S2=2πrr+h   =2×227×7×7+15   =44×22   =968 cm2(iii) Volume of the right circular cylinder, V=πr2h  =227×72×15  =2310 cm3

Page No 22.25:

Question 7:

The diameter of the base of a right circular cylinder is 42 cm and its height is 10 cm. Find the volume of the cylinder.

Answer:

Given:Diameter, d= 42 cmRadius, r=d2=21 cmHeight, h=10 cmVolume of the cylinder, V=πr2h  =227×212×10  =13860 cm3

Page No 22.25:

Question 8:

Find the volume of a cylinder, the diameter of whose base is 7 cm and height being 60 cm. Also, find the capacity of the cylinder in litres.

Answer:

Given:Diameter, d=7 cmRadius, r=3.5 cmHeight, h= 60 cmVolume of the cylinder, V=πr2h  =227×3.52×60  =2310 cm3Capacity of the cylinder in litres =23101000            1litre= 1000 cubic cm=2.31 L

Page No 22.25:

Question 9:

A rectangular strip 25 cm × 7 cm is rotated about the longer side. Find the volume of the solid, thus generated.

Answer:

Given:Rectangular strip has radius, r=7 cmHeight, h=25 cmVolume of the solid, V=πr2h  =227×72×25  =3850 cm3

Page No 22.25:

Question 10:

A rectangular sheet of paper, 44 cm × 20 cm, is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.

Answer:

The length (l) and breadth (b) of the rectangular sheet are 44 cm and 20 cmNow, the sheet is rolled along the length to form a cylinder.Let the radius of the cylinder be r cm.Height, h=b=20 cmCircumference, S=44 cm             2πr=44 cm2×227×r=44 cm                  r=7 cmVolume of the cylinder, V=πr2h   =227×72×20 cm3   =3080 cm3

Page No 22.25:

Question 11:

The volume and the curved surface area of a cylinder are 1650 cm3 and 660 cm2 respectively. Find the radius and height of the cylinder.

Answer:

Curved surface area of the cylinder = 2πrh  =660 cm2     ... (1)
Volume of the cylinder = πr2h   =1650 cm3                       ... (2)

From (1) and (2), we can calculate the radius (r) and the height of cylinder (h).
We know the volume of the cylinder, i.e. 1650 cm3
∴  1650 = πr2h
     
   h = 1650πr2

Substituting h into (1):
660 = 2πrh

660 = 2πr × 1650πr2
660r = 2(1650)
r = 5 cm

h = 1650πr2     = 1650227×52 = 21 cm.

Hence, the radius and the height of the cylinder are 5 cm and 21 cm, respectively.

Page No 22.25:

Question 12:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes.

Answer:

Here, r1 = Radius of cylinder 1
         h1 = Height of cylinder 1
         r2 = Radius of cylinder 2
         h2 = Height of cylinder 2
         V1 = Volume of cylinder 1
         V2 = Volume of cylinder 2
Ratio of the radii of two cylinders = 2:3
Ratio of the heights of two cylinders = 5:3

Volume of the cylinder =  πr2h
V1/V2 = (πr12h1)/(πr22h2) = (π(2r)25h)/(π(3r)23h
V1/V2 = (π4r25h)/(π9r23h) = 20 / 27
Hence, the ratio of their volumes is 20:27

Page No 22.25:

Question 13:

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. Find the volume of the cylinder, if its total surface area is 616 cm2.

Answer:

Let r cm be the radius and h cm be the length of the cylinder. The curved surface area and the total surface area is 1:2.
The total surface area is 616 cm2.
The curved surface area is the half of 616 cm2, i.e. 308 cm2​.

Curved area = 2πrh

       So, h = 3082πr
Total surface area = Curved surface area + Top and bottom area
Top and bottom area = 616 - 308 = 308 cm2 = 2πr2

r2 = 3082×227
r = 7 cm

h = 3082×π×7 = 7cm

Then, the volume of the cylinder can be calculated as follows:

V = 227× 72 × 7 = 1078
Hence, it is obtained that the volume of the cylinder is 1078 cm3.

Page No 22.25:

Question 14:

The curved surface area of a cylinder is 1320 cm2 and its base has diameter 21 cm. Find the volume of the cylinder.

Answer:

r cm = Radius of the cylinder
h cm = Height of the cylinder

Diameter of the cylinder is 21 cm. Thus, the radius is 10.5 cm.
Since the curved surface area has been known, we can calculate h by the equation given below:

The curved surface area of the cylinder = 2πrh
1320 cm2= 2πrh
1320 cm2= 2 x 22 x (10.5 cm) x h
​                       7
h = 20 cm

∴ Volume of the cylinder (V) = πr2h
V = 22(10.5 cm)2(20 cm)
        7
V = 6930 cm3

Page No 22.25:

Question 15:

The ratio between the radius of the base and the height of a cylinder is 2 : 3. Find the total surface area of the cylinder, if its volume is 1617 cm3.

Answer:

Let r cm be the radius and h cm be the height of the cylinder. It is given that the ratio of r and h is 2:3, so h = 1.5r
The volume of the cylinder (V) is 1617 cm3.

So, we can find the radius and the height of the cylinder from the equation given below:
V= πr2h
1617 = πr2h
1617 = πr2(1.5r)
 r3 =343
r = 7 cm and h = 10.5 cm

Total surface area = 2πr2+2πrh
​                                =2×227×72+2×227×7×10.5=770 cm2
                              
Hence, the total surface area of the cylinder is 770 cm2.

Page No 22.25:

Question 16:

The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the diameter and the height of the pillar.

Answer:

Here, m= radius of the cylinder
         h m= height of the cylinder

Curved surface area of the cylinder = 2πrh     ... (1)
Volume of the cylinder = πr2h                        ... (2)
            924 = πr2h
            h=924πr2

Then, substitute h into equation (1):
264 = 2πrh
264=2πr924πr2
264r = 2(924)
      r=2×924264
r = 7 m, so d = 14 m

h=924πr2

 h=924227×72=6 m
Hence, the diameter and the height of the cylinder are 14 m and 6 m, respectively.

Page No 22.25:

Question 17:

Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find the ratio of their radii.

Answer:

Here, V1 = Volume of cylinder 1
         V2 = Volume of cylinder 2
          r1 = Radius of cylinder 1
          r2 = Radius of cylinder 2
          h1 = Height of cylinder 1
          h2 = Height of cylinder 2

Volumes of cylinder 1 and 2 are equal.
Height of cylinder 1 is half the height of cylinder 2.
V1 = V2
r12h1) = (πr22h2
r12h) = (πr222h
r12r22=21
r1r2=21

Thus, the ratio of their radii is 2 : 1.

Page No 22.25:

Question 18:

The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder.

Answer:

It is known that three times the sum of the areas of the two circular faces, of the right circular cylinder, is twice the area of the curved surface.
Hence, it can be written using the following formula:
3 (2πr2) = 2(2πrh)
r2 = 2πrh
3r = 2h
It is known that the height of the cylinder (h) is 10.5 m.
Substituting this number in the equation:
3r = 2(10.5)
r = 7 m

Volume of the cylinder = πr2h
                             = 22 (72) (10.5)
                                7
                             = 1617 m3
Thus, the volume of the cylinder is 1617 m3.

Page No 22.25:

Question 19:

How many cubic metres of earth must be dug-out to sink a well 21 m deep and 6 m diameter?

Answer:

The volume of the earth that must be dug out is similar to the volume of the cylinder which is equal to πr2h.
Height of the well =21 m
Diameter of the well= 6 m
∴ Volume of the earth that must be dug out = (π (32) (21)) m3= 594 m3



Page No 22.26:

Question 20:

The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m, find the volume of the timber that can be obtained from the trunk.

Answer:

Circumference of the tree = 176 cm = 2πr
Length of the trunk, h= 3 m =300 cm
So, the radius (r) can be calculated by:
r=1762×227=28 cm
Thus, the volume (V) of the timber can be calculated using the following formula:
V = πr2(h) =22 (28 )2 (300) cm3 = 739200 cm3 = 0.74 m3
                         

Page No 22.26:

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?

Answer:

Height of the well = h m = 20 m
Diameter of the well = d m =7 m
Radius of the well = r m = 3.5 m
Volume of the well = πr2h = 22(3.5)2(20 ) m3= 770 m3
​                                                  7
Volume of the well = Volume of the rectangular plot
Length of the rectangular plot = 22 m
Breadth of the rectangular plot =14 m
Volume of the rectangular plot = 770 m3 = (Length× Breadth ×Height) of the rectangular plot
Height =77022×14=2.5 m

Thus, the height of the platform is 2.5 m.

Page No 22.26:

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.

Answer:

Diameter of the well = d m = 14 m
Height of the well = h m = 8 m
Radius of the well = r m = 7 m
Volume of the well = πr2 h = π(7 m)2(8 m) = 1232 m3
Volume of the well = Volume of the embankment 

An embankment is a hollow cylinder with thickness. Its inner radius would be equal to the radius of the well, i.e. r = 7 m, and its outer radius is R = 7 + 21 = 28 m.

Volume of the embankment = πh(R2-r2)


To find the height (h), we use the fact that the volume of the embankment is equal to the volume of the well.
1232= πh ((28)2-(7)2)
h=1232227×[(28)2-(7)2]=0.533 m

Hence, the height of the embankment is 0.533 m or 53.3 cm.

Page No 22.26:

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.

Answer:

Diameter of the cylindrical container = d cm = 56 cm
Radius of the cylindrical container = r cm = 28 cm
Volume of cylindrical container = Volume of the rectangular solid
Length of the rectangular solid = 32 cm
Breadth of the rectangular solid = 22 cm
Height of the rectangular solid = 14 cm
Volume of the rectangular solid = Length x Breadth x Height = 32 cm x 22 cm x 14 cm = 9856 cm3
Volume of the cylindrical container = 9856 cm3 = πr2h
                                                   9856 cm3 = 22(28 cm)2h
​                                                                       7
                                                                h = 4 cm
Thus, when the solid is completely submerged, the water will rise up to 4 cm.

Page No 22.26:

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.

Answer:

Case 1:
Height = h cm = 30 cm
Diameter = d cm = 18 cm
Radius = r cm = 9 cm
∴ Volume = (πr2h) or (π(9 )2(30 )) = 2430π cm3

Case 2:
Height = h cm = 18 cm
Diameter = d cm = 30 cm,
Radius = r cm = 15 cm
∴ Volume = (πr2h) or (π(15 )2(18 )) = 4050π cm3

Hence, the ratio of the volumes of the two cylinders formed is 3:5.

Page No 22.26:

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?

Answer:

Length of the water on a roof = 18 m
Breadth of the water on a roof = 16.5 m
Height of the water on a roof = 10 cm=0.1 m
Volume of the water on a roof = Length × Breadth  × Height = 18 m × 16.5 m × 0.1 m = 29.7 m3
Since water is to be stored in the cylindrical tank, the volume of water on a roof is equal to the volume of a cylindrical tank.
Volume of cylindrical tank = πr2h = 29.7 m3
h=29.7227×(4)2=0.5906 m=59.06 cm
Thus, the rise of water level in the tank is 59.06 cm.

Page No 22.26:

Question 26:

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

Answer:

Diameter of the ductile metal = 1 cm
Radius of the ductile metal = 0.5 cm
Volume of the ductile metal = πr2(length) = π(0.5 cm)2(5 cm) = 1.25π cm3
Ductile metal is drawn into a wire of diameter 1 mm.
Radius of the wire = 0.5 mm = 0.05 cm
Length of wire=1.25π cm3π0.05 cm2=500 cm =5 m
Thus, the length of wire is 5 m.

Page No 22.26:

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.

Answer:

Density of copper = Weight/Volume = 8.4 gram/1 cm3 = 8.4 gram/cm3
Volume = Weight/Density = 13.2 kg × 1000 gram/kg/8.4 gram/cm3 = 1571.43 cm3



Thus, length of 13.2 kg of copper is 125 m.

Page No 22.26:

Question 28:

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.

Answer:

Diameter of the cylindrical wire = 0.25 cm
Radius of the cylindrical wire = 0.125 cm
 Volume of the brass = 2.2 dm3 = 2200 cm3

Volume of the brass = Volume of the cylindrical wire
Length of wire =2200 cm3227×0.125 cm2=44800 cm=448 m
Thus, length of the wire is 448 m.

Page No 22.26:

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.

Answer:

r = Inner radii of the tube
R = Outer radii of the tube
h = Length of the tube

h(R-r) = 88                          ... (1)      
πh(R2-r2) = 176                      ... (2)

Substituting h = 14 cm in equation (1) and (2):

π​(R-r) = 88/28                        ... (1)                                       
π(R-r)(R+r)= 176/14​             ... (2)

Simplifying the second equation by substituting it with the first equation:
R+r=4 cm or R=(4-r) cm
Re-substituting R=4-r into equation (1):
22 (4-r-r) = 88
7                 28
4-2r = 1
r = 1.5 cm
R = 4-1.5 = 2.5 cm

Hence, the inner and the outer radii of the tube are 1.5 and 2.5 cm, respectively.

Page No 22.26:

Question 30:

Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres 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?

Answer:

Radius of the circular pipe= 0.01 mLength of the water column in 1 sec= 6 mVolume of the water flowing in 1 s =πr2 h=π(0.01)2(6) m3Volume of the water flowing in 30 mins=π(0.01)2(6)×30×60  m3Let h m be the rise in the level of water in the cylindrical tank.Volume of the cylindrical tank in which water is being flown=π(0.6)2×hVolume of water flowing in 30 mins=Volume of the cylindrical tank in which water is being flownπ(0.01)2(6)×30×60 =π(0.6)2×hh=6(0.01)2 ×30×600.6×0.6h=3 m

Page No 22.26:

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.

Answer:

Here, r = Inner radius = 5.2 cm
         R = Outer radius
         t = Thickness = 0.8 cm
        h = Length = 25 cm

R = r + t = 5.2 cm + 0.8 cm = 6 cm
Volume of the metal = π h (R2 - r2)
                          = 22 × (25) × ((6 )2 - (5.2 )2)
                             7
                          = 704 cm3
Thus, the volume of the metal is 704 cm3.

Page No 22.26:

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 litres of water delivered by the pipe in one hour.

Answer:

Radius of the water tap = 0.75 cm =0.0075 mLength of the water flowing in 1 s=7 m=700 cmVolume of water flowing in 1s =π(0.0075)2×700Volume of the water flowing in 1 hour =π(0.0075)2×700×60×60Volume of the water flowing in 1 hour=227×(0.0075)2×7×60×60=4.455 m3=4455 l                     (1000 l=1 m3)

Page No 22.26:

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 metre per second. In how much time the tank will be filled?

Answer:

Radius of the cylindrical tank = 0.7 mHeight of the cylindrical tank = 2.1 mVolume of the cylindrical tank= π(0.7)2(2.1) m3Length of the water column flown from the pipe in 1 s= 2 mLet the time taken to completely fill the water tank be x sec.Length of the water column flown from the pipe in x sec= 2 x mRadius of the pipe = 1.75 cm= 0.0175 m Volume of the water column flown from the pipe in x sec=π(0.0175)2 (2 x) m3Volume of the cylindrical tank=Volume of the water column flown from the pipeπ(0.7)2(2.1)=π(0.0175)2 (2 x)x=0.7)2(2.1)0.0175)2 (2 )=1680 sec= 28 min

Thus, the time required to fill the water tank is 28 min.

Page No 22.26:

Question 34:

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.

Answer:

Let h cm be the length of the paper and r cm be the radius of the paper.

We know that the rectangular sheet of paper 30 cm x 18 cm can be transformed into two types of cylinder.

Type 1:
Length = 30 cm 
Diameter = 18 cm
Volume = (
πr2h) = (π(9cm)(30cm)) = 2430π cm3

Type 2:
Length = 18 cm 
Diameter = 30 cm
Volume = (
πr2h) or (π(15cm)(18cm)) = 4050π cm3

Hence, the ratio of the volumes of the two cylinders formed is 3:5.

Page No 22.26:

Question 35:

How many litres of water flow out of a pipe having an area of cross-section of 5 cm2 in one minute, if the speed of water in the pipe is 30 cm/sec?

Answer:

We know:
Area of cross section = 5 cm2
Rate = 30 cm/s and
Time =1 min
So, the volume of water flow is:
Volume = Volumetric rate × Time = (30 cm/s)(5 cm2)(60 s/min) = 9000 cm3 = 9 litres

Thus, 9 litres of water flows out of the pipe.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                          

Page No 22.26:

Question 36:

A solid cylinder has a total surface area of 231 cm2. Its curved surface area is 23 of the total surface area. Find the volume of the cylinder.

Answer:

We know that the total surface area of the cylinder is 231 cm2 and the curved surface area is 2/3 of the total surface area.
So, the curved surface area is:
2/3 × (231 cm2) = 154 cm2

Then, the radius of the cylinder can be calculated in the following manner:
Curved surface area = 2πrh
​154 cm2 = 2πrh         ... (1)
Here, r  cm is the radius of the cylinder and h cm is the length of the cylinder.
2πr2 = (231-154) cm2 = 77 cm2
77 cm2 = 2π​r2
From here, the radius (r) can be calculated in the following manner:

r = 772×227
r = 3.5 cm

Substituting this result into equation (1):
154 cm2 = 2π(3.5 cm)h
h= 154 cm2 / (2x 22 x (3.5cm))
                             7
h = 7 cm

V = π​r2h227 ​x (3.5 cm)2 x (7 cm) = 269.5 cm3

Hence, the volume of the cylinder is 269.5 cm3.



Page No 22.27:

Question 37:

Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of Rs 3.60 per cubic metre. Find also the cost of cementing its inner curved surface at Rs 2.50 per square metre.

Answer:

Cost of sinking a tube well = Volume of the tube well × Cost of sinking a tube well per cubic metre
                                              = 22 x (1.52) x (280) m3 x Rs 3.6/m3 = Rs 7128.
                                                  7
Cost of cementing = Inner surface area of the tube well × Cost of cementing per square metre
                                  = ((2 x 22 x 1.5x 280) m2) x Rs 2.5/m2 = Rs 6600
                                             7                      

Page No 22.27:

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.

Answer:

Since we know the weight and the volume of copper,  we can calculate its density.
density of copper=weightvolume=8.4 gram1 cm3=8.4gramcm3
If the weight of copper wire is 13.2 kg and the density of copper is 8.4 g/cm3, then:
Volume = Weight / Density = 13.2 kg x 1000 gram/kg / 8.4 gram/
cm3 = 1571.43 cm3

The radius of copper wire is 2 mm or 0.2 cm. So, the length of the wire can be determined in the following way:
L=Vπr2=1571.43 m3π0.2 cm2=125050.01 cm=125 m

Thus, the length of 13.2 kg of copper is 125 m.

Page No 22.27:

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.

Answer:

Let r cm be the radius of the wire and h cm be the length of the wire.
Volume of brass = Volume of the wire
We know that the volume of brass = 2.2 dm3= 2200 cm3 
Volume of the wire=  πr2h = (0.125 cm)2 (h)
h=2200 cm3π0.125 cm2=44800 cm=448 m
Thus, length of the wire is 448 m.

Page No 22.27:

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.

Answer:

Let r m be the radius and d m be the depth of the well that is dug.
Volume of the well = πr2d = π(5 m)2(8.4 m) = 660 m3

An embankment has the shape of hollow cylinder with thickness. Its inner radii is equal to the well's radii, i.e. r = 5 m, and its outer radii is R = (5 + 7.5 )= 12.5 cm.
Then, the volume of the embankment = πh(
R − r2)

Volume of the well = Volume of the embankment
659.73 m3 = πh((12.5 m)2
(5 m)2)​
h=660 cm3227×12.5 m2-5 m2=1.6 m
Hence, the height of the embankment is 1.6 m.

Page No 22.27:

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.

Answer:

Here, R = Outer radius
          r = Inner radius
          t = Thickness = 4 cm
        w = Width = 63 cm

Girth = 440 cm = 2πR
R=4402×227=70cm

r = R − t = 70 cm − 4 cm = 66 cm

Volume of the iron = π (R2r2) w227 − (702 − 662) − (63) = 107712 cm3
​                                                        
Hence, volume of the iron is 107712 cm3.

Page No 22.27:

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?

Answer:

Here,= Internal radius
        R = External radius = 10 cm
        h = Length of the cylinder
       = Thickness = 0.25 cm

Volume of the hollow cylinder = πh(R2 - r2) = π (16) (102 - (10-0.25)2) = 79 π cm3

Volume of the solid cylinder = Volume of the hollow cylinder
We know that the radius of the solid cylinder is 1 cm.
∴ π(12)h = 79 π​
h = 79 cm
Hence, length of the solid cylinder that gives the same volume as the hollow cylinder is 79 cm.

Page No 22.27:

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.

Answer:

Let r  m be the radius and h m be the depth of the well that is dug.
Volume of the well = π​r2h = 22 × (3.5 m)2 × (10 m) = 385 m3
                                                 7
Volume of the well = Volume of the rectangular field
Volume of the rectangular field = 385 m3 = 30 m × 20 m × height
Height=385 cm230m×20m=0.641 m=64.1 cm

Hence, the height through which the level of the field is raised is 64.1 cm.



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