Surface Areas and Volumes: 9 Math
NCERT Exercise 13.2: 9th math
Important Formula
Surface Area of a Right Circular Cylinder or Curved Surface Area of a Cylinder `= 2pi\r`
Total Surface Area of a Cylinder `=2pi\r(r+h)`
NCERT Exercise 13.2: Questions and Solutions
Assume `pi=22/7`, unless stated otherwise.
Surface Areas And Volumes Class nine Math NCERT Exercise 13.2 Question(1) The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.
Solution
Given,
Curved surface area of a cylinder = 88 cm2
Height of cylinder = 14 cm
Thus, Diameter of the base of the cylinder =?
We know that,
Curved surface area of cylinder = 2 π r h
⇒ 88 cm2 `= 2 xx 22/7 xx r xx 14\ cm`
⇒ 88 cm2 = 2 × 22 × 2 cm × r
⇒ 88 cm2 = 48 cm × r
`=>r = (88\ cm^2)/(48 cm)`
⇒ r = 1 cm
Now, we know that, Diameter = 2 r
Therefore diameter of the given cylinder = 2 × 1 cm = 2 cm
Thus, diameter of the base of the cylinder = 2 cm Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.2 Question(2) It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?
Solution
Given, Height of the tank (h) = 1 m
And, Base diameter of the tank (d or 2 r) = 140 cm
= 140/100 m
⇒ Thus, base diameter of the tank ( d or 2 r) = 1.4 m
Therefore, radius of the base of the tank (r) = 1.4/2 = 0.7 m
Thus, metal sheet required make the given closed cylindrical tank = ?
Now we know that, Total surface area cylinder = 2 π r ( r + h)
Therefore, total surface area of given cylindrical tank `= 2xx22/7xx0.7 m(0.7 m+1 m)`
= 2 × 22 × 0.1 m × 1.7 m
= 4.4 m × 1.7 m
= 7.48 m2
Thus, total surface area of given cylindrical tank = 7.48 m2
Now, metal sheet required to make the given tank = total surface area of the given cylindrical tank
Thus, metal sheet required to make the given cylindrical tank = 7.4 m2 Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.2 Question(3) A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm. the outer diameter being 4.4 cm (see Figure). Find its
(i) Inner curved surface area.
(ii) Outer curved surface area.
(iii) Total surface area.
Solution
Given, Height (h) of the metal pipe = 77 cm
Outer diameter of the metal pipe (D or 2R) = 4.4 cm
Therefore, Outer radius (R) = 4.4 cm/2
⇒ Outer radius (R) = 2.2 cm
And, Inner diameter of the metal pipe (d or 2r) = 4 cm
Therefore, Inner radius (r) = 4 cm/2
⇒ Inner radius of the given pipe = 2 cm
Thus, (i) Inner curved surface area.
We know that, Inner curved surface area of a cylinder = 2 π r h
`= 2 xx 22/7 xx 2 cm xx 77 cm`
= 2 × 22 × 2 cm × 11 cm
= 88 × 11 cm2
⇒ Inner curved surface area of the given pipe = 968 cm2
Thus, Inner curved surface area of the given pipe = 968 cm2 Answer
(ii) Outer curved surface area.
We know that, outer curved surface area of a cylinder = 2 π R h
Thus, outer curved surface area of the given cylindrical pipe
`= 2 xx 22/2xx 2.2 cm xx 77 cm`
= 2 × 22 × 2.2 cm × 11 cm
= 44 × 2.2 × 11
= 9.68 cm × 11 cm
= 1064.8 cm2
Thus, outer curved surface area of the given pipe = 1064.8 cm2 Answer
(iii) Total surface area.
Here,
Total surface area of cylinder = Inner surface area + Outer surface area + Area of the thickness of base
= 2 π rh + 2 π RH + [2 π (R2 – r2)]
= 968 cm2 + 1064.8 cm2 + [2 π (R2 – r2)]
[∵ Inner surface area = 968 cm2 and Outer surface area = 1064.8 cm2 as calculated in section (i) and (ii) above]
= 2032.8 cm2 + [2 × `22/7` { (2.2 cm)2 – (2 cm)2}]
= 2032.8 cm2 + [`44/7` × (4.84 cm2 – 4 cm2)]
= 2032.8 cm2 + (`44/7` × 0.84 cm2)
= 2032.8 cm2 + (44 × 0.12 cm2)
= 2032.8 cm2 + 5.28 cm2
= 2038.08 cm2
Thus, total surface area of the given pipe = 2038.08 cm2
Surface Areas And Volumes Class nine Math NCERT Exercise 13.2 Question(4) The diameter of a roller is 84 cm and its length is 120 cm. it takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2
Solution
Given, Diameter of roller (d) = 84 cm
Therefore, radius (r) = d/2
⇒ r = 84/2 = 42 cm
Length of roller i.e. Height of cylinder = 120 cm
Total revolution of roller to move once over playground = 500
Thus, Area of playground = ?
Calculation of curved surface area of roller
We know that, Curved surface area of a cylinder = 2 π r h
Thus, curved surface area of given roller
= 2 × `22/7` × 42 cm × 120 cm
= 2 × 22 × 6 cm × 120 cm
= 44 × 720 cm2
= 31680 cm2
`= 31680/(100xx100)m^2`
= 3.168 m2
Thus, curved surface area of given roller = 3.168 m2
Calculation of Area of Playground
Now, as given the roller completes 500 revolutions to move once to level the play ground
So, area of playground = Curved surface area of roller × number of revolution
= 3.168 cm2 × 500
= 1584 m2
Thus, area of given playground = 1584 m2 Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.2 Question(5) A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar of the rate of 12.50 per m2.
Solution :
Given, Diameter of cylindrical pillar = 50 cm
&therefor; radius of cylindrical pillar (r) = 50/2
⇒ Radius of cylindrical pillar (r) = 25 cm
= 25/100 m
⇒ r = 0.25 m
Height of cylindrical pillar = 3.5 m
And, rate of painting of the pillar Rs 12.50 per m2
Thus, cost of painting of the curved surface area of pillar = ?
Calculation of Curved Surface Area of pillar
Now, we know that, curved surface area of a cylinder = 2 π r h
Thus, curved surface area of given pillar
= 2 × `22/7` × 0.25 m × 3.5 m
= 2 × 22 × 0.25 m × 0.5 m
= 44 × 0.125 m2
= 5.5 m2
Thus, curved surface area of given pillar = 5.5 m2
Calculation of cost of painting of given pillar
The cost of painting of curved surface area of pillar
= Curved surface area of pillar × rate of painting per square meter
Therefore, Cost of painting = 5.5 × 12.50
= Rs 68.75
Thus, cost of painting of the curved surface area of given pillar = Rs 68.75 Answer
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