HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.6

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.6

Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
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 = 11)
Solution:
We have,
Height of cylindrical vessel (h) = 25 cm
Circumference of the base of the cylindrical vessel = 132 cm …..(1)
Let the radius of the base of the cylindrical vessel be r сm, then
Circumference of the base of the cylindrical vessel = 2πr
⇒ 132 = 2 × \(\frac{22}{7}\) × r [Using (1)]
⇒ \(\frac{132 \times 7}{2 \times 22}=r\)
⇒ r = 21 cm
∴ volume of the water in the vessel = πr2h
\(\frac{22}{7}\) × (21)² × 25
= 34650 cm³
= \(\frac{34650}{1000}\)
= 34.65 litres
Hence, volume of the water in the vessel = 34.65 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:
We have,
Length of the cylindrical pipe (h) = 35 cm
Inner diameter of the cylindrical pipe = 24 cm
∴ Inner radius of the cylindrical pipe (r) = \(\frac{24}{2}\) = 12 cm
and outer diameter of the cylindrical pipe = 28 cm
∴ Outer radius of the cylindrical pipe (R) = \(\frac{28}{2}\) = 14 cm
∴ Volume of wooden used in making the pipe
= π(R² – r²) × h
= \(\frac{22}{7}\)(14² – 12²) × 35
= \(\frac{22}{7}\)(14 + 12)(14 – 12) × 35
= 110 × 26 × 2 = 5720 cm³
∵ Weight of 1 cm³ wood = 0·6 gram
∴ Weight of 5720 cm³ wood = 0·6 × 5720 gram
= 3432 gram
= \(\frac{3432}{1000}\) kg = 3.432
Hence,mass of the wooden pipe = 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:
(i) We have,
Length of rectangular tin (l) = 5 cm
Breadth of rectangular tin (b) = 4 cm
Height of rectangular tin (h) = 15 cm
∴ Volume of the rectangular tin = l × b × h
= 5 × 4 × 15 = 300 cm³

(ii) Diameter of the base of the plastic cylinder (d) = 7 cm
∴ Radius of the base of the plastic cylinder (r) = \(\frac{7}{2}\) = 3.5 cm
and height of the plastic cylinder (h) = 10 cm
∴ Volume of the plastic cylinder = πr²h
= \(\frac{22}{7}\) × (3.5)² × 10
= 385 cm³
Hence, plastic cylinder has greater capacity and it is equal (385 – 300) cm³ = 85 cm³.

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:
We have,
Height of the cylinder (h) = 5 cm
Let radius of the cylinder be r сm.
and lateral surface of the cylinder = 94.2 cm²
⇒ 2πrh = 94.2
⇒ 2 × 3.14 × r × 5 = 94.2
⇒ 31.4 × r = 94.2
⇒ r = \(\frac{94.2}{31.4}\)
⇒ r = 3 cm
(ii) Volume of the cylinder = πr²h
= 3.14 × (3)² × 5
= 141.3 cm³
Hence, (i)Radius of the cylinder = 3 cm
(ii) Volume of the cylinder = 141.3 cm³.

Question 5.
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 m², find :
(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.
Solution:
We have,
Depth of the cylindrical vessel (h) = 10 m
cost of painting = Rs. 20 per m²
Total cost of painting = Rs. 2200
(i) Inner curved surface area of the vessel = \(\frac{\text { Total cost }}{\text { Cost of painting }}=\frac{2200}{20}\) = 110 m²
(ii) Let the radius of the base of vessel ber m.
Curved surface area of the vessel = 110 m²,
[As solved in (i)]
⇒ 2πrh = 110
⇒ 2 × \(\frac{22}{7}\) × r × 10 = 110
⇒ \(\frac{440}{7}\) × r = 110
⇒ r = \(\frac{110 \times 7}{440}\)
⇒ r = 1.75 m.
(iii) Capacity of the vessel = πr²h
= \(\frac{22}{7}\) × (175)² × 10
= 96.25 m³ = 96.25 kl
Hence, (i) Inner curved surface area of the vessel = 110 m²
(ii) Radius of the base = 1.75 m
(iii) Capacity of the vessel = 96.25 m³ or 96.25 kl.

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?
Solution:
We have,
Height of the cylindrical vessel (h) = 1 m
and volume of the cylindrical vessel = 15.4 litres …..(1)
Let the radius of the cylindrical vessel be r m, then
Volume of the cylindrical vessel = πr²h
⇒ 15.4 litres = \(\frac{22}{7}\) × r² × 1[Using (1)]
⇒ \(\frac{15.4}{1000} \mathrm{~m}^3=\frac{22}{7} r^2\)
⇒ \(\frac{15.4 \times 7}{1000 \times 22}=r^2\)
⇒ 0.0049 = r²
⇒ r = \(\sqrt{0.0049}\)
⇒ r = \(\sqrt{\frac{49}{10000}}\)
r = \(\frac{7}{100}\)
Required area of the metal sheet = Total surface area of the cylindrical vessel
= 2πrh + 2πr²
= 2 \(\frac{22}{7}\) × 1 + 2 × \(\frac{22}{7} \times\left(\frac{7}{100}\right)^2\)
= \(\frac{44}{100}\) + 0.0308
= 0.44 + 0.0308
= 0.4708 m²
Hence, area of the metal sheet required to make the vessel = 0.4708 m².

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:
We have,
Length of the cylindrical pencil (h)= 14 cm
Diameter of the cylindrical pencil = 7 mm
= \(\frac{7}{10}\) cm
∴ Radius of the cylindrical pencil (R) = \(\frac{7}{20}\) cm
and diameter of the cylindrical graphite
= 1 mm = \(\frac{1}{10}\) cm
∴ Radius of the cylindrical graphite (r) = \(\frac{1}{10}\)
∴ Volume of the cylindrical pencil = πR²h
= \(\frac{22}{7} \times\left(\frac{7}{20}\right)^2 \times 14\)
= 5.39 cm³
and volume of the cylindrical graphite = πr²h
= \(\frac{22}{7} \times\left(\frac{1}{20}\right)^2 \times 14\)
= 0·11 cm³
Volume of the wood = Volume of the pencil – Volume of the graphite
= 5.39 – 0·11 = 5.28 cm³
Hence, volume of the wood = 5.28 cm³ and volume of the graphite = 0·11 cm³.

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:
We have,
Height of the cylidrical bowl filled with soup (h) = 4 cm
Diameter of the cylindrical bowl (d) = 7 cm
∴ Radius of the cylindrical bowl (r) = \(\frac{7}{2}\) cm
Volume of the soup for 1 patient = πr²h
= \(\frac{22}{7} \times\left(\frac{7}{2}\right)^2 \times 4\)
= 154 cm³
Volume of the soup for 250 patients
= 154 × 250 = 38500 cm³
= \(\frac{38500}{1000}\) litres
= 38.5 litres
Hence, volume of the soup for 250 patients = 38500 cm³ or 38.5 litres.