RD Sharma solution class 8 chapter 22 Mensuration III(Volume and Surface Area of a Right Circular Cylinder) Exercise 22.2

Exercise 22.2

Page-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 1:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Given} : \\ r=3.5 \mathrm{cm}, h=40 \mathrm{cm} \\ \mathrm{Volume} \mathrm{of} \mathrm{cylinder}, \mathrm{V}=\mathrm{\pi }{r}^{2}h \\ =\frac{22}{7}\times {\left(3.5\right)}^2\times 40 \\ =1540 {\mathrm{cm}}^3 \\ \left(\mathrm{ii}\right) \mathrm{Given}: \\ r=2.8 \mathrm{m}, h=15 \mathrm{m} \\ \mathrm{Volume} \mathrm{of} \mathrm{cylinder}, \mathrm{V}=\mathrm{\pi }{r}^{\mathit{2}}h \\ =\frac{22}{7}\times {\left(2.8\right)}^2\times 15 \\ =369.6 {\mathrm{m}}^3 \end{gathered}$$

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 2:

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

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 3:

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

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 4:

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

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 5:

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

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 6:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Height}, h= 15 \mathrm{cm} \\ \mathrm{Radius}, r= 7 \mathrm{cm} \\ \left(\mathrm{i}\right) \mathrm{Curved} \mathrm{surface} \mathrm{area}, S_1=2\mathrm{\pi }rh \\ =2\times \frac{22}{7}\times 7\times 15 \\ =660 {\mathrm{cm}}^2 \\ \left(\mathrm{ii}\right) \mathrm{Total} \mathrm{surface} \mathrm{area}, S_2=2\mathrm{\pi }r\left(r+h\right) \\ =2\times \frac{22}{7}\times 7\times \left(7+15\right) \\ =44\times 22 \\ =968 {\mathrm{cm}}^2 \\ \left(\mathrm{iii}\right) \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{right} \mathrm{circular} \mathrm{cylinder}, V=\mathrm{\pi }{r}^{2}h \\ =\frac{22}{7}\times 7^2\times 15 \\ =2310 {\mathrm{cm}}^3 \end{gathered}$$

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 7:

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

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 8:

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

Question 9:

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

Answer 9:

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

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 10:

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

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 11:

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
     
   $h = \frac{1650}{{\mathrm{\pi r}}^2}$

Substituting h into (1):
660 = 2πrh

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

$$\begin{gathered} h = \frac{1650}{{\mathrm{\pi r}}^2} \\ = \frac{1650}{{\displaystyle \frac{22}{7}}\times 5^2} = 21 cm. \end{gathered}$$

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 12:

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 13:

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

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

$h = \frac{308}{2\times \mathrm{\pi }\times 7}$ = 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 14:

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 15:

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
​                                =$2\times \frac{22}{7}\times 7^2+2\times \frac{22}{7}\times 7\times 10.5=770 {\mathrm{cm}}^2$
                              
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 16:

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
            $h=\frac{924}{\mathrm{\pi }{r}^2}$

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

$h=\frac{924}{\mathrm{\pi }{r}^2}$

 h=$\frac{924}{{\displaystyle \frac{22}{7}}\times 7^2}=6 \mathrm{m}$
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 17:

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 18:

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 19:

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-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 20:

Circumference of the tree = 176 cm = 2πr
Length of the trunk, h= 3 m =300 cm
So, the radius (r) can be calculated by:
$r=\frac{176}{2\times {\displaystyle \frac{22}{7}}}=28 \mathrm{cm}$
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 21:

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
$\mathrm{Height} =\frac{770}{22\times 14}=2.5 \mathrm{m}$

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 22:

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)
$h=\frac{1232}{{\displaystyle \frac{22}{7}}\times \left[\right(28{)}^2-\left(7{)}^2\right]}=0.533 \mathrm{m}$
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 23:

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 24:

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 25:

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
$h=\frac{29.7}{{\displaystyle \frac{22}{7}}\times (4{)}^2}=0.5906 \mathrm{m}=59.06 \mathrm{cm}$
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 26:

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
$\mathrm{Length} \mathrm{of} \mathrm{wire}=\frac{1.25\mathrm{\pi } {\mathrm{cm}}^3}{\mathrm{\pi }{\left(0.05 \mathrm{cm}\right)}^2}=500 \mathrm{cm} =5 \mathrm{m}$
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 27:

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 28:

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
$\mathrm{Length} \mathrm{of} \mathrm{wire} =\frac{2200 {\mathrm{cm}}^3}{{\displaystyle \frac{22}{7}}\times {\left(0.125 \mathrm{cm}\right)}^2}=44800 \mathrm{cm}=448 \mathrm{m}$
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 29:

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 30:

$$\begin{gathered} \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{circular} \mathrm{pipe}= 0.01 \mathrm{m} \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{water} \mathrm{column} \mathrm{in} 1 sec= 6 \mathrm{m} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{water} \mathrm{flowing} \mathrm{in} 1 s =\mathrm{\pi }{r}^{2 }h=\mathrm{\pi }(0.01{)}^2(6) {\mathrm{m}}^3 \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{water} \mathrm{flowing} \mathrm{in} 30 \mathrm{mins}=\mathrm{\pi }(0.01{)}^2\left(6\right)\times 30\times 60 {\mathrm{m}}^3 \\ \mathrm{Let}\mathit{ }h \mathrm{m} \mathrm{be} \mathrm{the} \mathrm{rise} \mathrm{in} \mathrm{the} \mathrm{level} \mathrm{of} \mathrm{water} \mathrm{in} \mathrm{the} \mathrm{cylindrical} \mathrm{tank}. \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylindrical} \mathrm{tank} \mathrm{in} \mathrm{which} \mathrm{water} \mathrm{is} \mathrm{being} \mathrm{flown}=\mathrm{\pi }(0.6{)}^2\times h \\ \mathrm{Volume} \mathrm{of} \mathrm{water} \mathrm{flowing} \mathrm{in} 30 \mathrm{mins}=\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylindrical} \mathrm{tank} \mathrm{in} \mathrm{which} \mathrm{water} \mathrm{is} \mathrm{being} \mathrm{flown} \\ \mathrm{\pi }(0.01{)}^2\left(6\right)\times 30\times 60 =\mathrm{\pi }(0.6{)}^2\times h \\ h\mathit{=}\frac{6(0.01{)}^2 \times 30\times 60}{0.6\times 0.6} \\ h\mathit{=}3 \mathrm{m} \end{gathered}$$

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 31:

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 32:

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

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 33:

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

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 34:

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 35:

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 36:

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 = 2π​r²
From here, the radius (r) can be calculated in the following manner:

$r = \sqrt{\frac{77}{2\times {\displaystyle \frac{22}{7}}}}$
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 = π​r²h = $\frac{22}{7}$ ​x (3.5 cm)² x (7 cm) = 269.5 cm³

Hence, the volume of the cylinder is 269.5 cm³.

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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 37:

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) m³ x 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 38:

Since we know the weight and the volume of copper,  we can calculate its density.
$\mathrm{density} \mathrm{of} \mathrm{copper}=\frac{\mathrm{weight}}{\mathrm{volume}}=\frac{8.4 \mathrm{gram}}{1 {\mathrm{cm}}^3}=8.4\frac{\mathrm{gram}}{{\mathrm{cm}}^3}$
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:
$\mathrm{L}=\frac{\mathrm{V}}{\pi r^2}=\frac{1571.43 {\mathrm{m}}^3}{\pi {\left(0.2 \mathrm{cm}\right)}^2}=125050.01 \mathrm{cm}=125 \mathrm{m}$
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 39:

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)
$h=\frac{2200 {\mathrm{cm}}^3}{\mathrm{\pi }{\left(0.125 \mathrm{cm}\right)}^2}=44800 \mathrm{cm}=448 \mathrm{m}$
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 40:

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)​
$\mathrm{h}=\frac{660 {\mathrm{cm}}^3}{{\displaystyle \frac{22}{7}}\times \left[{\left(12.5 \mathrm{m}\right)}^2-{\left(5 \mathrm{m}\right)}^2\right]}=1.6 \mathrm{m}$
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 41:

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

Girth = 440 cm = 2πR
$\mathrm{R}=\frac{440}{2\times {\displaystyle \frac{22}{7}}}=70\mathrm{cm}$
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 42:

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 43:

Let r  m be the radius and h m be the depth of the well that is dug.
Volume of the well = $\mathrm{\pi }$​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
$\mathrm{Height}=\frac{385 {\mathrm{cm}}^2}{30\mathrm{m}\times 20\mathrm{m}}=0.641 \mathrm{m}=64.1 \mathrm{cm}$

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