NCERT Solutions Class 9 Maths Chapter 13 Surface area and Volume Exercise 13.6

Introduction:

In this exercise/article we will learn about Surface Area And Volume. Volume of a cylinder . Just as a cuboid is built up with rectangles of the same size, we have seen that a right circular cylinder can be built up using circles of the same size. So, using the same aargument as for a cuboid, we can see that the volume of a cylinder can be obtained as : base area × height, where r is the base radius and h is the height of the cylinder.

Class 9 Maths Chapter 13 Surface Area And Volume :

Class 9 Maths Exercise 13.6 (Page-230)

Q1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? ( 1000 cm3 = 1 l )

Solution :

According to the question,

Given, Height = 25 cm

Circumference of the base of a cylindrical vessel = 132 cm

So, Circumference of the base of a cylindrical vessel = 2πr

⇒ 132 = 2 × \(\displaystyle \frac{{22}}{7}\) × r

⇒ 132 × 7 = 44r

⇒ 924 = 44r

⇒ \(\displaystyle \frac{{924}}{{44}}\) = r

⇒ 21 = r

Now, volume of a cylindrical vessel = πr2h

= \(\displaystyle \frac{{22}}{7}\) × ( 21 )2 × 25

= \(\displaystyle \frac{{22}}{7}\) × 441 × 25

= \(\displaystyle \frac{{242550}}{7}\)

= 34650 cm3

⇒ Convert into cm3 to litre

= \(\displaystyle \frac{{34650}}{{1000}}\)

= 34.65 litres

∴ 34.65 litres of water can it hold .

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 1 cm3 of wood has a mass of 0.6 g.

Solution :

According to the question,

Given, Inner diameter = 24 cm

R = \(\displaystyle \frac{d}{2}\)

R = \(\displaystyle \frac{{24}}{2}\)

R = 12 cm

Outer diameter = 28 cm

r = \(\displaystyle \frac{d}{2}\)

r = \(\displaystyle \frac{{28}}{2}\)

r = 14 cm

Height/Length = 35 cm

So, volume of a wooden pipe = π ( R2 - r2 ) h

= \(\displaystyle \frac{{22}}{7}\) × ( 122 - 142 ) × 35

= \(\displaystyle \frac{{22}}{7}\) × ( 144 - 196 ) × 35

= \(\displaystyle \frac{{22}}{7}\) × 52 × 35

= \(\displaystyle \frac{{40040}}{7}\)

= 5720 cm3

⇒ Convert into cm3 to gram

= 5720 × 0.6

= 3432 g

⇒ Convert into g to kg

= \(\displaystyle \frac{{3432}}{{1000}}\)

= 3.432 kg

∴ The mass of the pipe = 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?

Solution :

According to the question,

(i) Given, Length = 5 cm

Breadth = 4 cm

Height = 15 cm

So, volume of a cuboid = Length × Breadth × Height

= 5 × 4 × 15

= 300 cm3

(ii) Given, Diameter = 7 cm

Height = 10 cm

R = \(\displaystyle \frac{d}{2}\)

R = \(\displaystyle \frac{{7}}{2}\)

R = 3.5 cm

So, volume of a cylinder = πr2h

= \(\displaystyle \frac{{22}}{7}\) × ( 3.5 )2 × 10

= \(\displaystyle \frac{{22}}{7}\) × 12.52 × 10

= \(\displaystyle \frac{{2695}}{7}\)

= 385 cm3

⇒ Different between capacity = 385 - 300 = 85 cm3

∴ Plastic cylinder container is greater capacity = 85 cm3 .

Q4. If the lateral surface area of a cylinder is 94.2 cm2 and its height is 5 cm, then find

(i) radius of its base (ii) its volume . ( Use π = 3.14 )

Solution :

According to the question,

(i) Given, the lateral surface area of a cylinder = 94.2 cm2

Height = 5 cm

So, Lateral surface area of a cylinder = 2πrh

⇒ 94.2 = 2 × 3.14 × r × 5

⇒ 94.2 = 31.4r

⇒ \(\displaystyle \frac{{94.2}}{{31.4}}\) = r

⇒ 3 cm = r

∴ Radius of its base = 3 cm

(ii) Given, Height = 5 cm

Radius = 3 cm

So, volume of a cylinder = πr2h

= 3.14 × ( 3 )2 × 5

= 3.14 × 9 × 5

= 141.3 cm3

∴ volume of a cylinder = 141.3 cm3 .

Q5. It costs Rs 2200 to paints the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per m2 , find

(i) inner curved surface area of the vessel,

(ii) radius of the base,

(iii) capacity of the vessel .

Solution :

According to the question,

(i) Given, It costs Rs 2200 to paints

Height = 10 m

The rate of Rs 20 per m2

So, inner curved surface area of the vessel = \(\displaystyle \frac{{2200}}{{20}}\)

= 110 m

∴ Inner curved surface area of the vessel = 110 m

(ii) Given, inner curved surface area of the vessel = 110 m

Height = 10 m

So, curved surface area of vessel = 2πrh

⇒ 110 = 2 × \(\displaystyle \frac{{22}}{7}\) × r × 10

⇒ 110 × 7 = 440r

⇒ 770 = 440r

⇒ \(\displaystyle \frac{{770}}{{440}}\) = r

⇒ \(\displaystyle \frac{7}{4}\) = r

⇒ 1.75 m = r

∴ Radius of the base = 1.75 m

(iii) Given, Radius = 1.75 m

Height = 10 m

So, Volume of the vessel = πr2h

= \(\displaystyle \frac{{22}}{7}\) × 1.75 × 1.75 × 10

= \(\displaystyle \frac{{673.75}}{7}\)

= 96.25 m3

∴ Capacity of the vessel = 96.25 m3 or ( 96.25 kl ) .

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

Solution :

According to the question,

Given, Height = 1 m

Capacity of cylindrical vessel = 15.4 litres

Capacity of cylindrical vessel = 0.0154 m3 [ convert litre to m3 ]

So, volume of cylindrical vessel = πr2h = 0.0154

⇒ \(\displaystyle \frac{{22}}{7}\) × r2 × 1 = 0.0154

⇒ 22 r2 = 0.0154 × 7

⇒ 22 r2 = 0.1078

⇒ r2 = \(\displaystyle \frac{{0.1078}}{{22}}\)

⇒ r2 = 0.0049

⇒ r = \(\displaystyle \sqrt{{0.0049}}\)

⇒ 0.07 m

Now, Total surface area of cylindrical vessel = 2πr( r + h )

= 2 × \(\displaystyle \frac{{22}}{7}\) × 0.07 ( 0.07 + 1 )

= 2 × \(\displaystyle \frac{{22}}{7}\) × 0.07 × 1.07

= \(\displaystyle \frac{{3.2956}}{7}\)

= 0.4708 m2

∴ 0.4708 m2 of metal sheet would be needed to make it .

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 .

Solution :

According to the question,

Given, Graphite diameter = 1 mm

Height = 14 cm

Height = 140 mm [ convert cm to mm ]

R = \(\displaystyle \frac{d}{2}\)

R = \(\displaystyle \frac{{1}}{2}\)

So, volume of a graphite = πr2h

= \(\displaystyle \frac{{22}}{7}\) × \(\displaystyle {{\left( {\frac{1}{2}} \right)}^{2}}\) × 140

= \(\displaystyle \frac{{22}}{7}\) × \(\displaystyle \frac{{1}}{4}\) × 140

= \(\displaystyle \frac{{3080}}{{28}}\)

= 110 mm3

⇒ Convert into mm3 to cm3

= \(\displaystyle \frac{{110}}{{1000}}\)

= 0.11 cm3

∴ The volume of the graphite = 0.11 cm3

Now, Given, Wood of pencil diameter = 7

Height = 14 cm

Height = 140 mm [ convert cm to mm ]

r = \(\displaystyle \frac{d}{2}\)

r = \(\displaystyle \frac{{7}}{2}\)

r = 3.5 mm

So, volume of wood used in pencil = π( R2 - r2 )h

= \(\displaystyle \frac{{22}}{7}\) × ( \(\displaystyle {{\frac{1}{2}}^{2}}\) - 3.52 ) × 140

= \(\displaystyle \frac{{22}}{7}\) × ( \(\displaystyle \frac{{1}}{4}\) - 12.25 ) × 140

= \(\displaystyle \frac{{22}}{7}\) × ( \(\displaystyle \frac{{1\,-\,49}}{4}\) ) × 140

= \(\displaystyle \frac{{22}}{7}\) × ( \(\displaystyle \frac{{48}}{4}\) ) × 140

= \(\displaystyle \frac{{22}}{7}\) × 12 × 140

= \(\displaystyle \frac{{36960}}{7}\)

= 5280 mm3

⇒ Convert mm3 to cm3

= \(\displaystyle \frac{{5280}}{{1000}}\)

= 5.28 cm3

∴ The volume of wood of pencil = 5.28 cm3 .

Q8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients ?

Solution:

According to the question,

Given, Diameter = 7 cm

Height = 4 cm

R = \(\displaystyle \frac{d}{2}\)

R = \(\displaystyle \frac{{7}}{2}\)

R = 3.5 cm

So, volume of a cylindrical bowl = πr2h

= \(\displaystyle \frac{{22}}{7}\) × ( 3.5 )2 × 4

= \(\displaystyle \frac{{22}}{7}\) × 12.25 × 4

= \(\displaystyle \frac{{1078}}{7}\)

= 154 cm3

∴ The hospital has to prepare soup daily to serve 250 patients = 250 × 154

= 38500 cm3

∴ 38500 cm3 soup the hospital has to prepare daily to serve 250 patients .

Class 9 Maths Chapter 13 Surface Area And Volume :

NCERT Solutions Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.6

NCERT Solutions Class 9 Maths Chapter 13 Surface area and Volume Exercise 13.6

Introduction:

Class 9 Maths Exercise 13.6 (Page-230)

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