NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6
NCERT Solutions for Class 9 Maths Chapter 13 Surface
Areas and Volumes Ex 13.6 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 13 Surface
Areas and Volumes Ex 13.6.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6
Ex 13.6 Class 9 Maths 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 = 1 L)
Solution:
Let the base radius of the
cylindrical vessel be r cm.
∴ Circumference of the base = 2Ï€r
⇒ 2Ï€r = 132 [Circumference = 132 cm,
(given)]
= 2 × 22/7 × r = 132
⇒ r = 132 × 7/44 cm = 21
cm
Since, height of the vessel (h) = 25 cm
Volume of a vessel (h) = πr2h
= 22/7 × (21)2 ×
25 cm3
= 22/7 × 21 ×
21 × 25 cm3
= 22 × 3 × 21 × 25 cm3
= 34650 cm3
∵ Capacity of the vessel = Volume of
the vessel
∴ Capacity of cylindrical vessel =
34650 cm3
Since, 1000 cm3 = 1 litre
⇒ 34650 cm3 = 34650/1000 litres = 34.65 litres
Thus, the vessel can hold 34.65 litres of water.
Ex 13.6 Class 9 Maths 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:
Inner diameter of the cylindrical
pipe = 24 cm
⇒ Inner radius of the pipe (r)
= 24/2 cm = 12 cm
Outer diameter of the pipe = 28 cm
Outer radius of the pipe (R) = 14 cm
Length of the pipe (h) = 35 cm
∴ Amount of wood in the pipe = Outer
volume – Inner volume
= Ï€R2h – Ï€r2h
= Ï€h (R + r) (R – r) [∵ a2 – b2 = (a + b)(a –
b)]
= 22/7 × 35 ×
(14 + 12) × (14 – 12) cm3
= 22 × 5 × 26 × 2 cm3
Mass of wood in the pipe = (Mass of wood in 1 cm3 of
wood) x (Volume of wood in the pipe)
= (0.6 g) × (22 × 5 × 26 × 2) cm3
= 6/10 × 22 ×
10 × 26 g = 3432 g
= 3432/1000 kg
= 3.432 kg [∵ 1000 g = 1 kg]
Thus, the required mass of the pipe is 3.432 kg.
Ex 13.6 Class 9 Maths 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.
(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) For rectangular pack,
Length (l) = 5 cm,
Breadth (b) = 4 cm
Height (h) = 15 cm
Volume = l × b × h = 5 × 4 × 15 cm3 = 300 cm3
∴ Capacity of the rectangular pack = 300 cm3
(ii)
For cylindrical pack,
Base diameter = 7 cm
∴ Radius of the base (r) = 7/2 cm
Height (h) = 10 cm
∴ Volume = Ï€r2h = 22/7 × (7/2)2 ×
10 cm
= 22/7 × 7/2 × 7/2 × 10 cm
= 11 × 7 ×
5 cm3 = 385 cm3
∴ Capacity of the cylindrical pack = 385 cm3
So, the cylindrical pack has greater capacity by (385 – 300) cm3 =
85 cm3
Ex 13.6 Class 9 Maths 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:
Height of the cylinder (h) = 5 cm
Let the base radius of the cylinder be ‘r’.
(i) Since lateral surface area of the cylinder =
2Ï€rh
But the lateral surface of the cylinder is 94.2 cm2 [given]
Therefore,
Thus, the radius of the cylinder = 3 cm
(ii)
Volume of a cylinder = πr2h
⇒ Volume of the given cylinder
= 3.14 (3)2 × 5 cm3
= 314/100 × 3 × 3 × 5 cm3
= 1413/10 cm3 =
141.3 cm3
Thus, the required volume = 141.3 cm3
Ex 13.6 Class 9 Maths 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:
(i) Total cost of painting = ₹ 2200
Cost of painting of area 1 m2 = ₹ 20
∴ Inner curved surface area of the vessel = 110 m2
(ii)
Let r be the base radius of the cylindrical vessel.
Curved surface area of a cylinder = 2Ï€rh
Ex 13.6 Class 9 Maths 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:
Capacity of the cylindrical vessel
= 15.4 litres = 15.4 × 1000 cm3 [1 litre = 1000 cm3]
Now, total surface area of the cylindrical vessel
Thus, the required metal sheet = 0.4708 m2.
Ex
13.6 Class 9 Maths 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:
Since, 10 mm = 1 cm
∴ 1 mm = 1/10 cm
For graphite cylinder,
∴ Radius of the pencil (R) = 7/20 cm
Height of the pencil (h) = 14 cm
Volume of the pencil = πR2h
= 5.39 cm3 – 0.11 cm3
= 5.28 cm3
Thus, the required volume of the wood is 5.28 cm3.
Ex 13.6 Class 9 Maths 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:
∵ The bowl is cylindrical, where diameter of the base = 7 cm
Thus,
the hospital needs to prepare 38.5 litres of soup daily for 250 patients.
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