Rd Sharma 2019 2020 for Class 8 Maths Chapter 22 - Mensuration Iii Surface Area And Volume Of A Right Circular Cylinder

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:

$Let r and h be the radius and the height of the cylinder . Given : r = \frac{7}{2} cm h = 60 cm Curved surface area of the cylinder = 2 π \times r \times h = 2 \times \frac{22}{7} \times \frac{7}{2} \times 60 = 22 \times 60 = 1320 {cm}^{2} Total surface area of the cylinder = 2 π \times r \times (r + h) = 2 \times \frac{22}{7} \times \frac{7}{2} \times (\frac{7}{2} + 60) = 22 \times \frac{127}{2} = 11 \times 127 = 1397 {cm}^{2}$

Question 2:

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

Answer:

$Consider h to be the height of the cylindrical rod . Given : Radius, r = 0.35 cm Curved surface area = 132 {cm}^{2} We know : Curved surface area = 2 \times π \times r \times h 132 = 2 \times \frac{22}{7} \times 0.35 \times h h = \frac{132 \times 7}{2 \times 22 \times 0.35} h = 60 Therefore, the length of the cylindrical rod is 60 cm .$

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:

$Given : Area of the base of a right circular cylinder = 616 {cm}^{2} Height = 2 .5 cm Let r be the radius of the base of a right circular cylinder . {πr}^{2} = 616 \Rightarrow r^{2} = 616 \times \frac{7}{22} \Rightarrow r^{2} = 196 \Rightarrow r = 14 cm Curved surface area of the right circular cylinder = 2 πrh = 2 \times \frac{22}{7} \times 14 \times 2.5 = 220 {cm}^{2}$

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:

$Given : Height, h = 15 c m Circumference of the base of the cylinder = 88 {cm}^{2} Let r be the radius of the cylinder . The circumference of the base of the cylinder = 2 πr 88 = 2 \times \frac{22}{7} \times r r = \frac{88 \times 7}{2 \times 22} = 14 cm Curved surface area = 2 \times π \times r \times h = 2 \times \frac{22}{7} \times 14 \times 15 = 1320 {cm}^{2} Total surface area = 2 \times π \times r \times (r + h) = 2 \times \frac{22}{7} \times 14 \times (14 + 15) = 2552 c m^{2}$

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:

$Since the rectangular strip of 25 cm \times 7 cm is rotated about the longer side, we have : Height, h = 25 cm Radius, r = 7 cm ∴ Total surface area = 2πr(r+h) = 2π(7)(25+7) = 14π(32) = 448π cm ² = 448 \times \frac{22}{7} cm ² = 1408 cm ²$

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:

$The rectangular sheet of paper 44 cm \times 20 cm is rolled along its length to form a cylinder . The height of the cylinder is 20 cm and circumference is 44 cm . We have : Height, h = 20 cm Circumference = 2 π r = 44 cm ∴ Total surface area is S = 2 π r h = 44 \times 20 cm ² = 880 cm ²$

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:

$Let the radii of two cylinders be 2 r and 3 r, respectively, and their heights b e 5 h and 3 h, respectively . Let S_{1} and S_{2} be the curved surface areas of the two cylinder . S_{1} = Curved surface area of the cylinder of height 5 h and radius 2 r S_{2} = Curved surface area of the cylinder of height 3 h and radius 3 r ∴ S_{1} : S_{2} = 2 \times π \times r \times h : 2 \times π \times r \times h = \frac{2 \times π \times 2 r \times 5 h}{2 \times π \times 3 r \times 3 h} = 10 : 9$

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:

$Let S_{1} and S_{2} be the curved surface area and total surface area of the circular cylinder, respectively . Then, S_{1} = 2 π r h, S_{2} = 2 π r (r + h) According to the question : S_{1} : S_{2} = 1 : 2 2 π r h : 2 π r (r + h) = 1 : 2 h : (r + h) = 1 : 2 \frac{h}{r + h} = \frac{1}{2} 2 h = r + h h = r Therefore, the height and the radiu s are equal.$

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:

$Let h be the height of the cylinder . Given : Curved surface area, S = 1320 {cm}^{2} Diameter, d = 21 cm Radius, r = 10.5 S = 2 π r h 1320 = 2 π \times 10.5 \times h h = \frac{1320}{2 π \times 10.5} h = 20 cm$

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:

$Let r be the radius of the circular cylinder . Height, h = 10.5 cm Area of the curved surface, S_{1} = 2 π r h S um of the areas of its two circular faces, S_{2} = 2 π r^{2} According to question : 3 S_{2} = 2 S_{1} 3 \times 2 {πr}^{2} = 2 \times 2 π r h 6 r = 4 h 3 r = 2 h r = \frac{2}{3} \times 10.5 cm = 7 cm$

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:

$Given : Height, h = 21 m Diameter, d = 6 m Radius, r = 3 m Area of the inner surface of the well, S = 2 π r h = 2 π \times 3 \times 21 m^{2} = 2 \times \frac{22}{7} \times 3 \times 21 m^{2} = 396 m^{2} According to question, the cost per m^{2} is Rs 9.50 . ∴ Inner surface cost is Rs 396 \times 9.50 = Rs 3762$

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:

$Given : Diameter, d = 20 cm Radius, r = 10 cm Height, h = 14 cm Area inside the cylindrical vessel that is to be tin - plated = S S = 2 π r h + π r^{2} = 2 π \times 10 \times 14 + π \times 10^{2} = 280 π + 100 π = 380 \times \frac{22}{7} {cm}^{2} = \frac{8360}{7} {cm}^{2} According to question : Cost per 100 {cm}^{2} = 50 paise Cost per {cm}^{2} = Rs 0.005 Cost of tin - plating the area inside the cylindrical vessel = R s 0.005 \times \frac{8360}{7} = Rs \frac{41.8}{7} = Rs 5.97$

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:

$Given : Inner diameter of the circular well = 3.5 m ∴ Inner radius of the circular well, r = 1.75 m Depth of the circular well, h = 10 m Inner curved surface area, S = 2 π r h S = 2 π \times 1.75 \times 10 m^{2} = 2 \times \frac{22}{7} \times 1.75 \times 10 m^{2} = 110 m^{2} Cost of plastering 1 m^{2} area = Rs 4 Cost of plastering 110 m^{2} area = Rs (110 \times 4) = Rs 440$

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:

$Given : Diameter of the roller = 84 cm ∴ Radius, r = \frac{Diameter}{2} = 42 cm In 1 revolution, it covers the distance of its lateral surface area . Roller is a cylinder of height, h = 120 cm Radius = 42 cm Lateral surface area of the cylinder = 2 π r h = 2 \times \frac{22}{7} \times 42 \times 120 = 31680 {cm}^{2} It takes 500 complete revolutions to level a playground . ∴ Area of the field = 31680 \times 500 = 15840000 {cm}^{2} ({1 cm}^{2} = \frac{1}{10000} m^{2}) ∴ {15840000cm}^{2} =1584 m^{2} . Thus, the area of the field in m^{2} is 1584 m^{2} .$

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:

$Given : Diameter of the pillars = 0.5 m Radius of the pillars, r = 0.25 m Height of the pillars, h = 4 m Number of pillars = 21 Rate of cleaning = Rs 2.50 per square metre Curved surface area of one pillar = 2 π r h = 2 \times \frac{22}{7} \times 0.25 \times 4 = 2 \times \frac{22}{7} = \frac{44}{7} m^{2} ∴ Curved surface area of one pillar = \frac{44}{7} m^{2} Cost of cleaning 21 pillars at the rate of Rs 2.50 per m^{2} = Rs 2.5 \times 21 \times \frac{44}{7} = 7.5 \times 44 ∴ Cost of cleaning 21 pillars at the rate of Rs 2.50 per m^{2} = Rs 330$

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:

$Given : Total surface area of the cylinder = 4620 {cm}^{2} Area of the base ring = 115.5 {cm}^{2} Height, h = 7 cm Let R be the radius of the outer ring and r be the radius of the inner ring . Area of the base ring = π R^{2} - π r^{2} 115.5 = π (R^{2} - r^{2}) R^{2} - r^{2} = 115.5 \times \frac{7}{22} (R + r) (R - r) = 36.75 . . . . . . . . . . . (i) Total surface area = Inner curved surface area + Outer curved surface area + Area of bottom and top rings 4620 = 2 π r h + 2 π R h + 2 \times 115.5 2 π h (R + r) = 4620 - 231 R + r = \frac{4389 \times 7}{2 \times 22 \times 7} R + r = \frac{399}{4} . . . . . . . . . . . (ii) Substituting the value of R + r from the equation (ii) in (i) : \frac{399}{4} (R - r) = 36.75 (R - r) = 36.75 \times \frac{4}{399} = 0.368 cm ∴ Thickness of the cylinder = (R - r) = 0.368 cm$

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:

$Let r and h be the radius and height of the solid cylinder . Given : r + h = 37 m Total surface area, S = 2 π r (r + h) 1628 = 2 π \times r \times 37 r = \frac{1628}{2 π \times 37} = \frac{1628}{232.477} = 7 m Circumference of its base, S_{1} = 2 πr = (2 \times \frac{22}{7} \times 7) m = 44 m$

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:

$Let S_{1} and S_{2} be the total surface area and curved surface area, respectively . Given : Height, h = 7.5 cm Radius, r = 3.5 cm S_{1} = 2 π r (r + h) S_{2} = 2 π r h According to the question : \frac{S_{1}}{S_{2}} = \frac{2 π r (r + h)}{2 π r h} \frac{S_{1}}{S_{2}} = \frac{r + h}{h} \frac{S_{1}}{S_{2}} = \frac{3.5 + 7.5}{7.5} \frac{S_{1}}{S_{2}} = \frac{11}{7.5} = \frac{110}{75} = \frac{22}{15} Therefore, the ratio is 22 : 15 .$

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:

$Let r cm and h cm be the radius of the cylindrical vessel . Given : Radius, r = 70 cm Height, h = 1.4 m = 140 cm Rate of tin - plating = Rs 3.50 per 1000 square centimetre Cost of tin - plating the cylindrical vessel on both the surfaces (inner and outer) : Total suface area of a vessel = Area of the inner and the outer side of the base + Area of the inner and the outer curved surface = 2 (π r^{2} + 2 π r h) = 2 π r (r + 2 h) = 2 \times \frac{22}{7} \times 70 \times (70 + 2 \times 140) = 44 \times 10 \times 350 = 154000 {cm}^{2} Cost of painting at the rate of Rs 3.50 per 1000 {cm}^{2} = 154000 \times \frac{3.50}{1000} = Rs 539 Therefore, cost of painting is Rs 539 .$

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:

$(i) Given : r = 3.5 cm, h = 40 cm Volume of cylinder, V = π r^{2} h = \frac{22}{7} \times {(3.5)}^{2} \times 40 = 1540 {cm}^{3} (ii) Given : r = 2.8 m, h = 15 m Volume of cylinder, V = π r^{2} h = \frac{22}{7} \times {(2.8)}^{2} \times 15 = 369.6 m^{3}$

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:

$(i) Given : d = 21 cm, radius, r = \frac{d}{2} = 10.5 cm height, h = 10 cm Volume of the cylinder, V = π r^{2} h = \frac{22}{7} \times {(10.5)}^{2} \times 10 = 3465 {cm}^{3} (ii) Given : d = 7 m, radius, r = \frac{d}{2} = 3.5 m height h = 24 m Volume of the cylinder, V = π r^{2} h = \frac{22}{7} \times {(3.5)}^{2} \times 24 = 924 m^{3}$

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:

$Let the area of the base of a right circular cylinder be S {cm}^{2} . Given : S = 616 {cm}^{2} Height, h = 25 cm Let the radius of a right circular cylinder be r cm . S = π r^{2} 616 = \frac{22}{7} \times r^{2} r^{2} = \frac{616 \times 7}{22} r^{2} = 196 r = 14 cm Volume of the cylinder, V = π r^{2} h = π \times {(14)}^{2} \times 25 = 15400 {cm}^{3}$

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:

$Let r cm be the radius of a cylinder . Circumference of the cylinder, S = 2 π r Given : Height, h = 15 cm Circumference, S = 88 cm S = 2 πr 88 = 2 \times \frac{22}{7} \times r r = \frac{88 \times 7}{44} r = 14 cm Volume of cylinder, V = π r^{2} h = \frac{22}{7} \times 14^{2} \times 15 = 9240 {cm}^{3}$

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:

$Let the length of the cylinder pipe be h = 21, dm = 210 cm . Let the outer and the inner radius of the pipe b e R cm and r cm, re s pectively . ∴ 2 R = 10 and 2 r = 6 R = 5 cm and r = 3 cm Volume of the copper used in making the pipe, V = π (R^{2} - r^{2}) h = \frac{22}{7} \times (5^{2} - 3^{2}) \times 210 = 22 \times (25 - 9) \times 30 = 22 \times 16 \times 30 = 10560 {cm}^{3}$

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:

$Given : Height, h = 15 cm Radius, r = 7 cm (i) Curved surface area, S_{1} = 2 π r h = 2 \times \frac{22}{7} \times 7 \times 15 = 660 {cm}^{2} (ii) Total surface area, S_{2} = 2 π r (r + h) = 2 \times \frac{22}{7} \times 7 \times (7 + 15) = 44 \times 22 = 968 {cm}^{2} (iii) Volume of the right circular cylinder, V = π r^{2} h = \frac{22}{7} \times 7^{2} \times 15 = 2310 {cm}^{3}$

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:

$Given : Diameter, d = 42 cm Radius, r = \frac{d}{2} = 21 cm Height, h = 10 cm Volume of the cylinder, V = π r^{2} h = \frac{22}{7} \times 21^{2} \times 10 = 13860 {cm}^{3}$

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:

$Given : Diameter, d = 7 cm Radius, r = 3.5 cm Height, h = 60 cm Volume of the cylinder, V = π r^{2} h = \frac{22}{7} \times 3 . 5^{2} \times 60 = 2310 {cm}^{3} Capacity of the cylinder in litres = \frac{2310}{1000} (1 litre = 1000 cubic cm) = 2.31 L$

Question 9:

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

Answer:

$Given : Rectangular strip has radius, r = 7 cm Height, h = 25 cm Volume of the solid, V = π r^{2} h = \frac{22}{7} \times 7^{2} \times 25 = 3850 {cm}^{3}$

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:

$The length (l) and breadth (b) of the rectangular sheet are 44 cm and 20 cm Now, the sheet is rolled along the length to form a cylinder . Let the radius of the cylinder be r cm . Height, h = b = 20 cm Circumference, S = 44 cm 2 π r = 44 cm 2 \times \frac{22}{7} \times r = 44 cm r = 7 cm Volume of the cylinder, V = π r^{2} h = \frac{22}{7} \times 7^{2} \times 20 {cm}^{3} = 3080 {cm}^{3}$

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

   $h = \frac{1650}{{πr}^{2}}$

Substituting h into (1):
660 = 2πrh

$660 = 2 πr \times \frac{1650}{{πr}^{2}}$
660r = 2(1650)
r = 5 cm

$h = \frac{1650}{{πr}^{2}} = \frac{1650}{\frac{22}{7} \times 5^{2}} = 21 c m .$

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

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

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, $h = \frac{308}{2 πr}$
Total surface area = Curved surface area + Top and bottom area
Top and bottom area = 616 - 308 = 308 cm² = 2πr²

$r^{2} = \frac{308}{2 \times \frac{22}{7}}$
r = 7 cm

$h = \frac{308}{2 \times π \times 7}$ = 7cm

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

$V = \frac{22}{7} \times 7^{2} \times 7 = 1078$
Hence, it is obtained that the volume of the cylinder is 1078 cm³.

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

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
= $2 \times \frac{22}{7} \times 7^{2} + 2 \times \frac{22}{7} \times 7 \times 10.5 = 770 {cm}^{2}$

Hence, the total surface area of the cylinder is 770 cm².

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, r 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
$h = \frac{924}{π r^{2}}$

Then, substitute h into equation (1):
264 = 2πrh
$264 = 2 π r (\frac{924}{π r^{2}})$
264r = 2(924)
$r = \frac{2 \times 924}{264}$
r = 7 m, so d = 14 m

$h = \frac{924}{π r^{2}}$

h= $\frac{924}{\frac{22}{7} \times 7^{2}} = 6 m$
Hence, the diameter and the height of the cylinder are 14 m and 6 m, respectively.

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
(πr₁²h₁) = (πr₂²h₂)
(πr₁²h) = (πr₂²2h)
$\frac{{r_{1}}^{2}}{{r_{2}}^{2}} = \frac{2}{1}$
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{2}{1}}$

Thus, the ratio of their radii is $\sqrt{2}$ : 1.

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)
3πr² = 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.

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

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:
$r = \frac{176}{2 \times \frac{22}{7}} = 28 cm$
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

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 $\times$ Breadth $\times$ Height) of the rectangular plot
$Height = \frac{770}{22 \times 14} = 2.5 m$

Thus, the height of the platform is 2.5 m.

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 = πr²h = 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)
$h = \frac{1232}{\frac{22}{7} \times [(28)^{2} - (7)^{2}]} = 0.533 m$
Hence, the height of the embankment is 0.533 m or 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.

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.

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.

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 $\times$ Breadth $\times$ Height = 18 m $\times$ 16.5 m $\times$ 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
$h = \frac{29.7}{\frac{22}{7} \times (4)^{2}} = 0.5906 m = 59.06 cm$
Thus, the rise of water level in the tank is 59.06 cm.

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
$Length of wire = \frac{1.25 π {cm}^{3}}{π {(0.05 cm)}^{2}} = 500 cm = 5 m$
Thus, the length of wire is 5 m.

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 $\times$ 1000 gram/kg/8.4 gram/cm³ = 1571.43 cm³

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

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
$Length of wire = \frac{2200 {cm}^{3}}{\frac{22}{7} \times {(0.125 cm)}^{2}} = 44800 cm = 448 m$
Thus, length of the wire is 448 m.

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

2πh(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.

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:

$Radius of the circular pipe = 0.01 m Length of the water column in 1 s e c = 6 m Volume of the water flowing in 1 s = π r^{2} h = π (0.01)^{2} (6) m^{3} Volume of the water flowing in 30 mins = π (0.01)^{2} (6) \times 30 \times 60 m^{3} Let h m be the rise in the level of water in the cylindrical tank . Volume of the cylindrical tank in which water is being flown = π (0.6)^{2} \times h Volume of water flowing in 30 mins = Volume of the cylindrical tank in which water is being flown π (0.01)^{2} (6) \times 30 \times 60 = π (0.6)^{2} \times h h = \frac{6 (0.01)^{2} \times 30 \times 60}{0.6 \times 0.6} h = 3 m$

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 $\times$ (25) $\times$ ((6 )² - (5.2 )²)
7
= 704 cm³
Thus, the volume of the metal is 704 cm³.

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:

$Radius of the water tap = 0.75 cm = 0.0075 m Length of the water flowing in 1 s = 7 m = 700 cm Volume of water flowing in 1 s = π (0.0075)^{2} \times 700 Volume of the water flowing in 1 hour = π (0.0075)^{2} \times 700 \times 60 \times 60 Volume of the water flowing in 1 hour = \frac{22}{7} \times (0.0075)^{2} \times 7 \times 60 \times 60 = 4.455 m^{3} = 4455 l (1000 l = 1 m^{3})$

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:

$Radius of the cylindrical tank = 0.7 m Height of the cylindrical tank = 2.1 m Volume of the cylindrical tank = π (0.7)^{2} (2.1) m^{3} Length of the water column flown from the pipe in 1 s = 2 m Let the time taken to completely fill the water tank be x \sec . Length of the water column flown from the pipe in x \sec = 2 x m Radius of the pipe = 1.75 cm = 0.0175 m Volume of the water column flown from the pipe in x \sec = π (0.0175)^{2} (2 x) m^{3} V o l u m e o f t h e c y l i n d r i c a l \tan k = Volume of the water column flown from the pipe π (0.7)^{2} (2.1) = π (0.0175)^{2} (2 x) x = \frac{0.7)^{2} (2.1)}{0.0175)^{2} (2)} = 1680 \sec = 28 \min$

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

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)²(30cm)) = 2430π cm³

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

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

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 $\times$ Time = (30 cm/s)(5 cm²)(60 s/min) = 9000 cm³ = 9 litres

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

Question 36:

A solid cylinder has a total surface area of 231 cm². Its curved surface area is $\frac{2}{3}$ 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 $\times$ (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 = 2πr²
From here, the radius (r) can be calculated in the following manner:

$r = \sqrt{\frac{77}{2 \times \frac{22}{7}}}$
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 = πr²h = $\frac{22}{7}$ x (3.5 cm)² x (7 cm) = 269.5 cm³

Hence, the volume of the cylinder is 269.5 cm³.

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 $\times$ Cost of sinking a tube well per cubic metre
= 22 x (1.5²) x (280) m³x Rs 3.6/m³ = Rs 7128.
7
Cost of cementing = Inner surface area of the tube well $\times$ Cost of cementing per square metre
= ((2 x 22 x 1.5x 280) m²) x Rs 2.5/m² = Rs 6600
7

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.
$density of copper = \frac{weight}{volume} = \frac{8.4 gram}{1 {cm}^{3}} = 8.4 \frac{gram}{{cm}^{3}}$
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:
$L = \frac{V}{π r^{2}} = \frac{1571.43 m^{3}}{π {(0.2 cm)}^{2}} = 125050.01 cm = 125 m$
Thus, the length of 13.2 kg of copper is 125 m.

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)
$h = \frac{2200 {cm}^{3}}{π {(0.125 cm)}^{2}} = 44800 cm = 448 m$
Thus, length of the wire is 448 m.

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)
$h = \frac{660 {cm}^{3}}{\frac{22}{7} \times [{(12.5 m)}^{2} - {(5 m)}^{2}]} = 1.6 m$
Hence, the height of the embankment is 1.6 m.

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
$R = \frac{440}{2 \times \frac{22}{7}} = 70 cm$
r = R − t = 70 cm − 4 cm = 66 cm

Volume of the iron = π (R² − r²) w = $\frac{22}{7}$ − (70² − 66²) − (63) = 107712 cm³

Hence, volume of the iron is 107712 cm³.

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, r = Internal radius
R = External radius = 10 cm
h = Length of the cylinder
t = 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.

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
$Height = \frac{385 {cm}^{2}}{30 m \times 20 m} = 0.641 m = 64.1 cm$

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

Rd Sharma 2019 2020 for Class 8 Maths Chapter 22 - Mensuration Iii Surface Area And Volume Of A Right Circular Cylinder

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