9th Maths 13.8
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 13.6
NCERT Class 9th solution of Exercise 13.7
NCERT Class 9th solution of Exercise 14.1
Exercise 13.8
Assume `pi=(22)/7`, unless stated otherwise.
Q1. Find the volume of a sphere whose radius, is
i) `7cm` ii) `0.63m`
Sol. :
i)
Volume of a sphere `=4/3pir^3`
`V=4/3times22/7times(7)^3`
`V=4/3times22/7times7times7times7`
`V=(4times22times49)/3`
`V=(4312)/3`
`V=1437.33 cm^3` (approx)
Answer:
`V=1437.33 cm^3`.
ii)
Volume of a sphere `=4/3pir^3`
`V=4/3times22/7times(0.63)^3`
`V=4/3times22times0.63times0.63times0.09`
`V=1.05 m^3`
Answer:
`V=1.05 m^3`.
Q2. Find the amount of water displaced by a solid spherical ball of diameter
i) `28cm` ii) `0.21m`
Sol. :
i)
Given:
Radius `r=28/2=14cm`,
Solve:
Amount of water displaced = Volume of the spherical ball
`=4/3pir^3`
`=4/3times22/7times(14)^3`
`=4/3times22times14times14times2`
`=34496/3`
`=11498.67 cm^3` (approx)
Answer:
Amount of water displaced `11498.67 cm^3`
ii)
Given:
Radius `r=0.21/2cm`
Solve:
Amount of water displaced = Volume of the spherical ball
`=4/3pir^3`
`=4/3times22/7times(0.21/2)^3`
`=4/3times22/7times0.21/2times0.21/2times0.21/2`
`=11times0.03times0.07times0.21`
`=0.004851 m^3`
Answer:
Amount of water displaced `0.004851 m^3`.
Q3. The diameter of metallic ball is `4.2cm`. What is the mass of the ball, if the density of the metal is `8.9g` per `cm^3`?
Sol. :
Given:
Radius of metallic ball `r=4.2/2=2.1cm`
Density of metal is `8.9g` per `cm^3`
To find:
Mass of the metal ball.
Solve:
Volume of the metal ball `=4/3pir^3`
`=4/3times22/7times(2.1)^3`
`=38.808cm^3`
Mass of the metal ball
= Volume of the metal ball`times`Density of metal ball
`=38.808times8.9`
`=345.3912 gm`
Answer:
Mass of the metal ball `345.3912 gm`.
Q4. The diameter of moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Sol. :
Given:
Radius of the earth `r_e=D/2`
Radius of the moon `r_m=D/(2times4)=D/8`
To Find:
Fraction of the Volume of the earth is the volume of the moon.
Solve:
`=(V_(mo on))/(V_(earth))`
`=(4/3pi(r_m)^3)/(4/3pi(r_e)^3)`
`=(r_m)^3/(r_e)^3`
`=((D/2)/(D/8))^3`
`=(2times2times2)/(8times8times8)`
`=1/64`
Answer:
Fraction of the Volume of the earth is the Volume of the moon `1/64`.
Q5. How many litres of milk can a hemispherical bowl of diameter `10.5cm` hold?
Sol. :
Given:
Radius of a bowl `r = 10.5/2cm`
To find:
The volume of milk in litre.
Solve:
Volume of bowl `=2/3pir^3`
`=2/3times22/7times(10.5/2)^3`
`=303.1875 cm^3`
`=303.2cm^3`
`=0.3032 litres` [`1cm^3 = 1/1000litres`]
Answer:
The volume of milk is ` 0.3032 litres`.
Q6. A hemispherical tank is made up of an iron sheet `1cm` thick. If the inner radius is `1m`, then find the volume of the iron used to make the tank.
Sol. :
Given:
The thickness of the iron sheet `=1 cm = 0.01 m`
Inner Radius of the iron tank `r_2=1 m`
Outer Radius of the iron tank `r_1=1+0.01=1.01 m`
To find:
The volume of the iron used to make the tank.
Solve:
Volume of hemisphere `=2/3pi(r_1^3-r_2^3)`
`=2/3times22/7[(1.01)^3-(1)^3]`
`=44/21(1.03-1)`
`=44/21times0.03`
`=0.44/7`
`=0.06286 m^3` (approx)
Answer:
The volume of the iron used to make the tank `0.06286 m^3` (approx)`
Q7. Find the volume of a sphere whose surface area is `154cm^2.`
Sol. :
Given:
The surface area of the sphere is `154 m^3`
To Find:
The volume of the sphere.
Solve:
The surface area of the sphere `=4pir^2`
`154 = 4times22/7times r^2`
`r^2 = (154times7)/(4times22)`
`r^2 = (7times7)/4`
`r^2 =(7/2)^2`
`r = 7/2 `
The volume of the sphere `=4/3pir^3`
`=4/3times22/7times(7/2)^3`
`=(11times49)/3`
`=539/3`
`=179.67 cm^3` (approx)
Answer:
The volume of the sphere is `179.67 cm^3` (approx).
Q8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of `₹498.96.` If the cost of white-washing is `₹2.00` per square metre., find the
i) inside surface area of the dome, ii) volume of the air inside the dome.
Sol. :
Given:
The total cost of white-wash `₹ 498.96`
The rate of the white-wash `₹ 2.00`.
To Find:
i) inside surface area of the dome
ii) volume of the air inside the dome.
Solve:
i) inside surface area of the dome = (total cost)/rate
`=498.96/2=249.48 m^2`
Answer:
The inside surface area of the dome `249.48 m^2`.
ii) The inside surface area of the dome `=2pir^2`
`249.48 = 2times22/7times r^2`
`r^2 =(249.48times7)/44`
`r^2 =39.69`
`r^2 =(6.3)^2`
`r = 6.3`
The volume of the air inside the dome`=2/3pir^3`
`=2/3times22/7times(6.3)^3`
`=44times6.3times6.3times0.3`
`=523.9 m^3` (approx)
Answer:
The volume of the air inside the dome `523.9 m^3`.
Q9. Twenty seven solid iron spheres, each of radius `r` and surface area `S` are melted to form a sphere with surface area `S'`. Find the
i) radius `r'` of the new sphere, ii) ratio of `S` and `S'`.
Sol. :
Given:
Twenty-seven spheres each Radius `r` and
Surface area `S`,
new Surface area `S'`
To Find:
i) radius `r'` of the new sphere,
ii) ratio of `S` and `S'`
Solve:
i) The volume of new sphere `= 27times`Volume of sphere
`4/3pi(r')^3=27times4/3pi(r)^3`
`(r')^3=(3)^3times(r)^3`
`(r')^3=(3r)^3`
`r'=3r`
Answer:
`r'=3r`
ii) Ratio `=S:S'=4pir^2:4pir'^2`
`S: S'=r^2:r'^2`
`S: S'=r^2:(3r)^2 = r^2:9r^2=1:9`
Answer:
The ratio `S:S`=1:9`.
Q10. A capsule of medicine is in the shape of a sphere of diameter `3.5mm`. How much medicine (in `mm`) is needed to fill this capsule.
Sol. :
Given:
The radius of capsule `3.5/2mm`
To Find:
Volume of capsule
Solve:
The volume of the sphere capsule `=4/3pir^3`
`=4/3times22/7times(3.5/2)^3`
`=11/3times3.5times3.5times3.5`
`=22.46mm^3` (approx)
Answer:
The volume of the sphere capsule `22.46mm^3`.
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