NCERT Solutions for Exercise 13.6 Class 9 Maths Chapter 13 - Surface Area and Volumes

Q2 The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if $\small 1\hspace{1mm}cm^3$ of wood has a mass of $\small 0.6\hspace{1mm}g$ .

Answer:

Given, A hollow cylinder made of wood.

The inner diameter of the cylindrical wooden pipe = $r_1 = 24\ cm$
Outer diameter = $r_2 = 28\ cm$

Length of the pipe = $h= 35\ cm$

$\therefore$ The volume of the wooden cylinder = $\pi r^2 h = \pi (r_2^2 - r_1^2)h$

$\\ = \frac{22}{7}\times (14^2-12^2) \times35 \\ = 22\times (14-12)(14+12) \times5 \\ = 22\times (2)(26) \times5 \\ = 5720\ cm^3$

Mass of $\small 1\hspace{1mm}cm^3$ wood = $\small 0.6\hspace{1mm}g$
Mass of $5720\ cm^3$ wood = $5720\times0.6\ g$
$\\ = 3432\ g \\ = 3.432\ kg$

Therefore, the mass of the wooden hollow cylindrical pipe is $3.432\ kg$

Q3 A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Answer:

Given,

(i) The dimension of the rectangular base of the tin can = $5\ cm\times4\ cm$

Height of the can = $h = 15\ cm$

$\therefore$ The volume of the tin can = $Rectangular\ area\times height$

$(5\times4\times15)\ cm^3 = 300\ cm^3$

(ii) The radius of the circular base of the plastic cylinder = $r = \frac{7}{2}= 3.5\ cm$

Height of the cylinder = $h = 10\ cm$

$\therefore$ The volume of the plastic cylinder = $\pi r^2 h$

$\\ = \frac{22}{7}\times (3.5)^2 \times10 \\ = 22\times0.5\times3.5\times10 \\ = 11\times35 \\ = 385\ cm^3$

Clearly, the plastic cylinder has more capacity than the rectangular tin can.

The difference in capacity = $(385-300)\ cm^3 = 85\ cm^3$

Q4 (i) If the lateral surface of a cylinder is $\small 94.2\hspace{1mm}cm^2$ and its height is 5 cm, then find radius of its base (Use $\small \pi =3.14$ )

Answer:

Given,

The lateral surface area of the cylinder = $\small 94.2\hspace{1mm}cm^2$

Height of the cylinder = $h = 5\ cm$

(i) Let the radius of the base be $r\ cm$

We know,

The lateral surface area of a cylinder = $2\pi r h$

$\\ \therefore 2\pi r h = 94.2 \\ \Rightarrow 2\times3.14\times r\times5 = 94.2 \\ \\ \Rightarrow r = \frac{94.2}{31.4} = 3\ cm$

Therefore, the radius of the base is $3\ cm$

Q4 (ii) If the lateral surface of a cylinder is $\small 94.2\hspace{1mm}cm^2$ and its height is 5 cm, then find it's volume. (Use $\small \pi =3.14$ )

Answer:

Given,

The lateral surface area of the cylinder = $\small 94.2\hspace{1mm}cm^2$

Height of the cylinder = $h = 5\ cm$

The radius of the base is $3\ cm$

(ii) We know,

The volume of a cylinder = $\pi r^2 h$

$\\ = 3.14\times3^2\times5 \\ = 3.14 \times 9 \times 5= 3.14 \times 45 \\ = 141.3\ cm^3$

Therefore, the volume of the cylinder is $141.3\ cm^3$ .

Q5 (i) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per $\small m^2$ , find inner curved surface area of the vessel.

Answer:

(i) Given,

Rs 20 is the cost of painting $1\small m^2$ area of the inner curved surface of the cylinder.

$\therefore$ Rs 2200 is the cost of painting $\frac{1}{20}\times2200 = 110\ m^2$ area of the inner curved surface of the cylinder.

$\therefore$ The inner curved surface area of the cylindrical vessel = $110\ m^2$

Q5 (ii) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per $\small m^2$ , find radius of the base.

Answer:

The inner curved surface area of the cylindrical vessel = $110\ m^2$

Height of the cylinder = $h = 10\ m$

Let the radius of the circular base be $r \ m$

$\therefore$ The inner curved surface area of the cylindrical vessel = $2\pi r h$

$\\ \Rightarrow 2\times\frac{22}{7}\times r \times10 = 110 \\ \Rightarrow 44\times r = 11\times7 \\ \Rightarrow 4\times r = 7 \\ \Rightarrow r = \frac{7}{4} = 1.75\ m$

Therefore, the radius of the base of the vessel is $1.75 \ m$

Q5 (iii) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per $\small m^2$ , find capacity of the vessel.

Answer:

(iii) Height of the cylinder = $h = 10\ m$

Radius of the base of the vessel = $r =1.75 \ m$

$\therefore$ Volume of the cylindrical vessel = $\pi r^2 h$

$\\ = \frac{22}{7}\times (1.75)^2 \times10 \\ = 22\times2.5\times1.75\times10 \\ = 55\times17.5 \\ = 96.25 \ m^3$

Therefore, the capacity of the cylindrical vessel is $96.25 \ m^3$

Q6 The capacity of a closed cylindrical vessel of height 1 m is $\small 15.4$ litres. How many square metres of metal sheet would be needed to make it?

Answer:

(Using capacity(volume), we will find the radius and then find the surface area)

The capacity of the vessel = Volume of the vessel = $\small 15.4$ litres

Height of the cylindrical vessel = $h = 1\ m$

Let the radius of the circular base be $r \ m$

$\therefore$ The volume of the cylindrical vessel = $\pi r^2 h$

$\\ \Rightarrow \frac{22}{7}\times r^2 \times1 = 15.4\ litres= 0.0154\ m^3 \\ \Rightarrow r^2 = \frac{0.0014\times11\times7}{22} \\ \Rightarrow r^2 = \frac{7\times7}{10000} \\ \Rightarrow r = \frac{7}{100}= 0.07\ m$

Therefore, the total surface area of the vessel = $2\pi r h+ 2\pi r^2 = 2\pi r(r+h)$

$\\ = 2\times\frac{22}{7}\times0.07\times(0.07+1) \\ = 0.44 \times 1.07 \\ = 0.4708\ m^2$

Therefore, square metres of metal sheet needed to make it is $0.4708\ m^2$

Q7 A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Answer:

Given,

Length of the cylindrical pencil = $h = 14\ cm$

The radius of the graphite (Inner solid cylinder) = $r_1 = \frac{1}{2}\ mm = 0.05\ cm$

Radius of the pencil (Inner solid graphite cylinder + Hollow wooden cylinder) =

= $r_2 = \frac{7}{2}\ mm = 0.35\ cm$

We know, Volume of a cylinder= $\pi r^2 h$

$\therefore$ The volume of graphite = $\pi r_1^2 h$

$\\ = \frac{22}{7}\times 0.05^2 \times14 \\ = 44\times0.0025 \\ = 0.11\ cm^3$

And, Volume of wood = $\pi (r_2^2- r_1^2) h$

$\\ = \frac{22}{7}\times (0.35^2-0.05^2) \times14 \\ = 44\times(0.35-0.05)(0.35+0.05) \\ = 44\times(0.30)(0.40) \\ = 5.28 \ cm^3$

Therefore, the volume of wood is $5.28 \ cm^3$ and the volume of graphite is $0.11 \ cm^3$