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Class 9 NCERT Solutions- Chapter 13 Surface Areas And Volumes – Exercise 13.6

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  • Difficulty Level : Medium
  • Last Updated : 10 Mar, 2021
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Question 1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many liters of water can it hold? (1000 cm3 = 1l)

Solution:

Given values,

Circumference of the base of a cylindrical = 132 cm

Height of cylinder (h)= 25 cm

Base of cylinder is of circle shape, having circumference = 2πr (r is radius)

Hence, 2πr = 132 cm

r = \frac{132}{2 \times \frac{22}{7}}                  (taking π=\frac{22}{7}  )

r =\frac{132 \times 7}{2 \times 22}

r = 21 cm

So, volume of cylinder = πr2h

= 22/7 × 21 × 21 × 25                                 (taking π=\frac{22}{7}  )

= 34650 cm3

As, 1000 cm3 = 1 litre

34650 cm3\frac{1}{1000}   × 34650

= \frac{34650}{1000}

= 34.650 litres

Question 2. 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:

Given values,

Inner radius of cylinder (r1)= \frac{24}{2}   = 12 cm

Outer radius of cylinder (r2)= \frac{28}{2}   = 14 cm

Height of cylinder (h)= 35 cm

So, volume used to make wood = volume of outer cylinder – volume of outer cylinder

= π(r22)h – π(r12)h

= π(r22 – r12)h

\frac{22}{7}   × (142 – 122) × 35                     (taking π=\frac{22}{7}  )

\frac{22}{7}   × (52) × 35

= 5720 cm3

As, 1 cm3 = 0.6 g

5720 cm3 = 0.6 × 5720 g

= 3432 grams

= 3.432 kg

Question 3. 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:

Let’s see each case,

(i) The shape of can is cuboid here, as having rectangular base 

Given values,

Length of can (l) = 5 cm

Width of can (b) = 4 cm

Height of can (h) = 15 cm

So, The amount of soft drink it can hold = volume of cuboid

= (l × b × h)

= 5 × 4 × 15 cm3

= 300 cm3

(ii)The shape of can is cylinder here, as having circular base

Given values,

Radius of can (r) = \frac{7}{2}   cm

Height of can (h) = 10 cm

So, The amount of soft drink it can hold = volume of Cylinder

= (πr2h)

\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}   × 10 cm3                                                  (taking π=\frac{22}{7}  )

= 385 cm3

Hence, we can see the can having circular base can contain (385 – 300 = 85 cm3) more amount of soft drinks than first can.

Question 4. If the lateral surface 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:

Given values,

Lateral surface of cylinder = 94.2 cm2

Height of cylinder (h) = 5 cm

Let’s see each case,

(i) So, the lateral surface is of rectangle shape whose

length = (circumference of base circle of cylinder) and width = height of cylinder 

Let the base radius = r

Lateral surface = length × width

94.2 cm2 = (2πr) × h                         (circumference of circle = 2πr)

94.2 cm2 = (2 × 3.14 × r) × 5               (taking π = 3.14)

r = \frac{94.2}{2 \times 3.14 \times 5}

r = 3 cm

(ii) Given values,

Radius of cylinder (r)= 3 cm

So, the volume of cylinder = (πr2h)

= π × 3 × 3 × 5 cm3

= 3.14 × 3 × 3 × 5 cm3                    (taking π = 3.14)

= 141.3 cm3

Question 5. It costs ₹2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹20 per m2, find

(i) Inner curved surface area of the vessel,

(ii) Radius of the base,

(iii) Capacity of the vessel

Solution:

Given values,

Height of cylinder (h) = 10 m

Cost of painting rate = ₹20 per m2

Let’s see each case,

(i) For 1 m2 = ₹20

For lateral surface = ₹2200

So the lateral surface = \frac{2200}{20}

= 110 m2

(ii) Let the base radius = r

So as, Lateral surface = (circumference of base circle of cylinder) × height

110 m2= (2πr) × h                      

110 = (2 × \frac{22}{7}   × r) × 10               (taking π=\frac{22}{7}  )

r = \frac{110 \times 7}{2 \times 22 \times 10}   cm

r = \frac{7}{4}   cm

r = 1.75 cm

(iii) Volume of cylinder = (πr2h)

\frac{22}{7} \times \frac{7}{4} \times \frac{7}{4}   × 10 cm3                  (taking π=\frac{22}{7}  )

= 96.25 cm3

Question 6. The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of metal sheet would be needed to make it? 

Solution:

Given values,

Height of cylinder (h) = 1 m = 100 cm

Volume of cylinder (V) = 15.4 liters

As 1 liter = 1000 cm3

15.4 liters = 15.4 × 1000 cm3

V = 15,400 cm3

Volume of cylinder = (πr2h)

15,400 = \frac{22}{7}   × r2 × 100                                       (taking π=\frac{22}{7}  )

r2\frac{15400 \times 7}{22 \times 100}

r2 = 49

r = √49

r = 7 cm

Surface area of a closed cylinder = (curve surface area + top and bottom circle) = 2πrh + (2 × πr2)

= 2πr (r+h)

= 2 × \frac{22}{7}   × 7 × (7 + 100) cm2                                                   (taking π=\frac{22}{7}  )

= 2 × 22 × 107

= 4708 cm2

= 0.4708 m2

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

Question 7. 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:

So here pencil = (cylinder of wood + cylinder of graphite)

Given values,

Height of wood (and graphite) cylinder (h) = 14 cm = 140 mm

Radius of pencil (R)= \frac{7}{2}   mm

Radius of graphite (r)= \frac{1}{2}   mm

Volume of Graphite = (πr2h)

\frac{22}{7} \times \frac{1}{2} \times \frac{1}{2}   × 140 mm3                                              (taking π=\frac{22}{7}  )

= 110 mm3

= 0.11 cm3

Volume of wood = Volume of pencil – Volume of graphite

= (πR2h) – (πr2h) = π(R2 – r2)h

\frac{22}{7}   × ((\frac{7}{2}  )2 – (\frac{1}{2}  )2) × 140 mm3                                      (taking π=\frac{22}{7}  )

= 22 × 20 × (\frac{49}{4}   – \frac{1}{4}  ) mm3

= 22 × 20 × 12 mm3

= 5280 mm3

= 52.80cm3

Question 8. 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:

So here Volume of soup for each patient = Volume of cylinder.

Given values,

Height of cylinder (h) = 4 cm

Radius of cylinder (r)= \frac{7}{2}   cm

Volume of Cylinder = (πr2h)

\frac{22}{7}   × \frac{7}{2} \times \frac{7}{2}   × 4 cm3                                             (taking π=\frac{22}{7}  )

= 154 cm

Volume of soup for 250 patient = 250 × Volume of cylinder.

= 250 × 154

= 38,500 cm3

Hence, 38,500cm3 soup is needed daily to serve 250 patients.


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