9th Maths 13.6
Chapter 13
Surface Areas and Volumes
NCERT Class 9th solution of Exercise 13.1
NCERT Class 9th solution of Exercise 13.2
NCERT Class 9th solution of Exercise 13.3
NCERT Class 9th solution of Exercise 13.4
NCERT Class 9th solution of Exercise 13.5
NCERT Class 9th solution of Exercise 13.7
NCERT Class 9th solution of Exercise 13.8
Exercise 13.6
Assume `pi=(22)/7`, unless stated otherwise.
Q1. The circumference of the base of a cylindrical vessel is `132cm` and its height is `25cm`. How many litres of water can it hold?(`1000cm^3=1l`)
Sol. :
Given:
Circumference of Base `=132cm`
Height of Cylinder `h=25cm`
To find:
Amount of water vessel can hold.
Solve:
Circumference of Base `=2pir`
`132=2 times (22)/7 times r`
`r = (132 times7)/(22 times 2)`
`r=21 cm`
Volume of Cylinder `=pir^2h`
`=(22)/7(21)^2 times 25`
`=34650 cm^3`
Amount of water vessel can hold
`=((34650)/1000)` [`1000cm^3 = 1l`]
`=34.65` litres
Answer:
The vessel can hold `34.65` litres of water.
Q2. The inner diameter of a cylindrical wooden pipe is `24cm` and its outer diameter is `28cm`. The length of the pipe is `35cm`. Find the mass of pipe, if `1cm^3` of wood has a mass of `0.6g`.
Sol. :
Given:
Inner Radius of Cylinder `r=(24)/2=12 cm`
Outer Radius of Cylinder `R=(28)/2=14 cm`
Length of Cylinder `h=35 cm`
Rate `0.6g` per `1cm^3`
To Find:
The mass of pipe.
Solve:
Volume of the wooden pipe
= Volume of the external Cylinder - Volume of the Internal Cylinder
`= piR^2h - pir^2h`
`=pih(R^2-r^2)`
`=(22)/7 times 35((14)^2 - (12)^2)`
`=22 times 5(196-144)`
`=110 times 52`
`= 5720 cm^3`
Mass of the pipe = Volume`times`Rate
`= 5720times 0.6g`
`= ((343.2)/1000)kg` [`1kg = 1/(1000)g`]
`=3.432 kg`
Answer:
The Mass of pipe `3.432 kg`.
Q3. A soft drink is available in two packs - i) a tin can with a rectangular base of length `5cm` and width `4cm`, having a height of `15cm` and ii) a plastic cylinder with circular base of diameter `7cm` and height `10cm`. Which container has greater capacity and by how much?
Sol. :
Given:
Length of tin can `l=5cm`
Breadth of tin can `b=4cm`
Height of tin can `h=15cm`
Radius of plastic cylinder `r=7/2cm`
Height of plastic cylinder `h=10cm`
To Find:
Capacity of tin can and plastic Cylinder.
Solve:
Capacity of tin can `=l times b times h`
`=5 times 4 times 15`
`=300 cm^2`
Capacity of plastic Cylinder `=pir^2h`
`=(22)/7(7/2)^2times10`
`=385cm^3`
Differnce = Capacity of plastic cylinder - Capacity of tin can
`=385-300=85cm^3`
Answer:
The plastic cylinder has greater capacity by `85cm^3`.
Q4. If the lateral surface of a cylinder is `94.2cm^2` and its height is `5cm`, then find
i) radius of its base ii) its volume. (Use `pi=3.14`)
Sol, :
Given:
Lateral Surface of a Cylinder `94.2cm^2`
Height of a Cylinder `5cm`
To Find:
i) radius of its base
ii) its volume
Solve:
i) Lateral Surface of a Cylinder `=2pirh`
`94.2=2times3.14times r times 5`
`r = (94.2 )/(2 times 3.14 times 5)`
`r = 3cm`
Answer:
Radius of Cylinder `3cm`.
ii) Volume of Cylinder `=pir^2h`
`=3.14(3)^2times 5`
`=141.3cm^3`
Answer:
Volume of Cylinder `141.3cm^3`.
Q5. It costs `₹2200` to paint the inner curved surface of a cylindrical vessel `10m` deep. If the cost of painting is at the rate of `₹20` per `m^2`, find
i) inner curved surface area of the vessel,
ii) radius of the base,
iii) capacity of the vessel.
Sol. :
Given:
Total Cost of Paint `₹2200`
Height of Cylinder `h=10m`
Rate of Paint `₹20` per`m^2`
To Find:
i) inner curved surface area of the vessel,
ii) radius of the base,
iii) capacity of the vessel.
Solve:
i) Inner Curved Surface Area of the vessel = (Total cost of paint)/(Rate of paint)
`= ((2200)/20)=110m^2`
Answer:
Inner Curved surface Area of the vessel `110m^2`.
ii) Curved Surface Area of Cylinder `=2pirh`
`110 = 2(22)/7times r times10`
`r = (110times 7)/(2times22times10)`
`r = 7/4m`
Answer:
Radius of base `7/4m`
iii) Capacity of vessel `=pir^2h`
`=(22)/7times7/4times7/4times10`
`=96.25m^3`
Answer:
Capacity of vessel `96.25m^3`.
Q6. The capacity of a closed cylindrical vessel of height `1m` is `15.4litres`. How many square meters of metres of metal sheet would be needed to make it?
Sol. :
Given:
Capacity of closed Cylinder `15.4 litres=((15.4)/1000)=0.0154m^3`
Height of Cylinder `h=1m`
To Find:
Area of metal sheet.
Solve:
Volume of closed Cylinder `=pir^2h`
`0.0154=(22)/7timesr^2times 1`
`r^2=(0.0154times7)/22=0.0049`
`r=sqrt0.0049`
`r=0.07`
Area of Metal sheet `=2pirh+2pir^2`
`=2pir(h+r)`
`=2(22)/7times0.07(1+0.07)`
`=0.4708m^2`
Answer:
Area of metal sheet needed to make the vessel `0.4708m^2`.
Q7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencils `7mm` and the diameter of the graphite is `1mm`. If the length of the pencil is `14cm`, find the volume of the wood and that of the graphite.
Sol. :
Given:
Radius of graphit cylinder `1/2mm=1/20cm`
Radius of pencil `R=7/2mm=7/20cm`
Height of the pencil `h=14cm`
To Find:
Volume of the wood and graphit.
Solve:
Volume of graphit cylinder `=pir^2h`
`=((22)/7times(1/20)^2times14)`
`=0.11cm^3`
Volume of the pencil`=piR^2h`
`=((22)/7times(7/20)^2times14)`
`=5.39cm^3`
Volume of wood = Volume of pencil - Volume of graphit
`=5.39 - 0.11`
`=5.28cm^3`.
Answer:
Volume of wood `5.39cm^3`.
Q8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter `7cm`. If the bowl is filled with soup to a height of `4cm`, how much soup the hospital has to prepare daily to serve `250` patients?
Sol. :
Given:
Radius of Bowl `r=7/2cm`
Height of Bowl `h=4cm`
To Find:
Volume of soup.
Solve:
Volume of Bowl `=pir^2h`
`=((22)/7times(7/2)^2times4)`
`=1.54cm^3`
Volume of soup `= Volume of a Bowl`times 250`
`=1.54times250`
`=38500cm^3=38.5litres`
Answer:
Volume of Soup `38.5litres.`
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