EXERCISE 13.6
Q 1, Ex. 13.6, Page No 230, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Page No 230:
Question 1:
The circumference of the base of cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm3 = 1l) $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Let the radius of the cylindrical vessel be r.Height (h) of vessel = 25 cm
Circumference of vessel = 132 cm
2πr = 132 cm
$r=\left(\frac{132 \times 7}{2 \times 22}\right) \mathrm{cm}$
=21cm
Volume of cylindrical vessel = πr2h
$=\left[\frac{22}{7} \times(21)^{2} \times 25\right] \mathrm{cm}^{3}$
= 34650 cm3
$=\left(\frac{34650}{1000}\right)$ litres $\left[\because 1 \text { litre }=1000 \mathrm{cm}^{3}\right]$
= 34.65 litres
Therefore, such vessel can hold 34.65 litres of water.
Q 2, Ex. 13.6, Page No 230, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Inner radius (r1) of cylindrical pipe $=\left(\frac{24}{2}\right) \mathrm{cm}$=12cm
Outer radius (r2) of cylindrical pipe $=\left(\frac{28}{2}\right) \mathrm{cm}$
=14cm
Height (h) of pipe = Length of pipe = 35 cm
Volume of pipe $=\pi\left(r_{2}^{2}-r_{1}^{2}\right) h$
$=\left[\frac{22}{7} \times\left(14^{2}-12^{2}\right) \times 35\right] \mathrm{cm}^{3}$
$=110 \times 52 \mathrm{cm}^{3}$
=5720 cm3
Mass of 1 cm3 wood = 0.6 g
Mass of 5720 cm3 wood = (5720 × 0.6) g
= 3432 g
= 3.432 kg
Q 3, Ex. 13.6, Page No 230, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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? $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
The tin can will be cuboidal in shape while the plastic cylinder will be cylindrical in shape.
Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) of tin can = 15 cm
Capacity of tin can = l × b × h
= (5 × 4 × 15) cm3
= 300 cm3
Radius (r) of circular end of plastic cylinder $=\left(\frac{7}{2}\right) \mathrm{cm}$
=3.5cm
Height (H) of plastic cylinder = 10 cm
Capacity of plastic cylinder = πr2H
$=\left[\frac{22}{7} \times(3.5)^{2} \times 10\right] \mathrm{cm}^{3}$
$=(11 \times 35) \mathrm{cm}^{3}$
=385 cm3
Therefore, plastic cylinder has the greater capacity.
Difference in capacity = (385 − 300) cm3 = 85 cm3
Q 4, Ex. 13.6, Page No 230, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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]Answer:
(i) Height (h) of cylinder = 5 cmLet radius of cylinder be r.
CSA of cylinder = 94.2 cm2
2πrh = 94.2 cm2
(2 × 3.14 × r × 5) cm = 94.2 cm2
r = 3 cm
(ii) Volume of cylinder = πr2h
= (3.14 × (3)2 × 5) cm3
= 141.3 cm3
Q 5, Ex. 13.6, Page No 231, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Page No 231:
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
$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Rs 20 is the cost of painting 1 m2 area.Rs 2200 is the cost of painting $=\left(\frac{1}{20} \times 2200\right) \mathrm{m}^{2}$ area
= 110 m2 area
Therefore, the inner surface area of the vessel is 110 m2.
(ii) Let the radius of the base of the vessel be r.
Height (h) of vessel = 10 m
Surface area = 2πrh = 110 m2
$\Rightarrow\left(2 \times \frac{22}{7} \times r \times 10\right) \mathrm{m}=110 \mathrm{m}^{2}$
$\Rightarrow r=\left(\frac{7}{4}\right) \mathrm{m}=1.75 \mathrm{m}$
(iii) Volume of vessel = πr2h
$=\left[\frac{22}{7} \times(1.75)^{2} \times 10\right] \mathrm{m}^{3}$
= 96.25 m3
Therefore, the capacity of the vessel is 96.25 m3 or 96250 litres.
Q 6, Ex. 13.6, Page No 231, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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? $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Let the radius of the circular end be r.Height (h) of cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m3
$\pi r^{2} h=0.0154 \mathrm{m}^{3}$
$\left(\frac{22}{7} \times r^{2} \times 1\right) \mathrm{m}=0.0154 \mathrm{m}^{3}$
⇒ r = 0.07 m
Total surface area of vessel $=2 \pi r(r+h)$
$=\left[2 \times \frac{22}{7} \times 0.07(0.07+1)\right] \mathrm{m}^{2}$
$=0.44 \times 1.07 \mathrm{m}^{2}$
$=0.4708 \mathrm{m}^{2}$
Therefore, 0.4708 m2 of the metal sheet would be required to make the cylindrical vessel.
Q 7, Ex. 13.6, Page No 231, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 7:
A lead pencil consists of a cylinder of wood with 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. $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r1) of pencil $=\left(\frac{7}{2}\right) \mathrm{mm}=\left(\frac{0.7}{2}\right) \mathrm{cm}$= 0.35 cm
Radius (r2) of graphite $=\left(\frac{1}{2}\right) \mathrm{mm}=\left(\frac{0.1}{2}\right) \mathrm{cm}$ = 0.05 cm
Height (h) of pencil = 14 cm
Volume of wood in pencil $=\pi\left(r_{1}^{2}-r_{2}^{2}\right) h$
$=\left[\frac{22}{7}\left\{(0.35)^{2}-(0.05)^{2} \times 14\right\}\right] \mathrm{cm}^{3}$
$=\left[\frac{22}{7}(0.1225-0.0025) \times 14\right] \mathrm{cm}^{3}$
$=(44 \times 0.12) \mathrm{cm}^{3}$
$=5.28 \mathrm{cm}^{3}$
Volume of graphite $=\pi r_{2}^{2} h=\left[\frac{22}{7} \times(0.05)^{2} \times 14\right] \mathrm{cm}^{3}$
$=(44 \times 0.0025) \mathrm{cm}^{3}$
$=0.11 \mathrm{cm}^{3}$
Q 8, Ex. 13.6, Page No 231, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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? $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r) of cylindrical bowl $=\left(\frac{7}{2}\right) \mathrm{cm}=3.5 \mathrm{cm}$
Height (h) of bowl, up to which bowl is filled with soup = 4 cm
Volume of soup in 1 bowl = πr2h
$=\left(\frac{22}{7} \times(3.5)^{2} \times 4\right) \mathrm{cm}^{3}$
= (11 × 3.5 × 4) cm3
= 154 cm3
Volume of soup given to 250 patients = (250 × 154) cm3
= 38500 cm3
= 38.5 litres.
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