Surface Areas and Volumes: 9 Math

NCERT Exercise 13.8 Q6-10: 9th math

Surface Areas And Volumes Class nine Math NCERT Exercise 13.8  Question (6) A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Solution

Given, Thickness of the iron sheet = 1 cm = 0.01 m

And, Inner radius of the hemispherical tank (r) = 1 m

Therefore, Outer radius of the given tank (R) = inner radius + thickness

⇒ R = 1 m + 0.01 m = 1.01 m

Now, volume of the iron sheet used to make the given tank = ?

Now, we know that, We know that, Volume of a hemisphere `=2/3\ pi\ r^3`

And, volume of the metal sheet used in making of given tank

= Outer volume of the tank – Inner volume of the tank

Thus, volume of the iron used to make the given tank = 0.06348 m³ Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.8  Question (7) Find the volume of a sphere whose surface area is 154 cm²

Solution

Given, Surface Area of a Sphere = 154 cm²

Therefore, Volume of the given sphere = ?

Now, we know that, Surface Area of a Sphere = 4 ℼ r²

Thus, radius of the given sphere = 3.5 cm

Now, we know that Volume of a sphere = 4/3 × ℼ r³

Thus, volume of the given sphere

= `4/3xx22/7` × 3.5 cm × 3.5 cm × 3.5 cm

= 179.66 cm³

Thus, volume of the given sphere = 179.66 cm³ Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.8  Question (8) A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of Rs 498.96. If the cost of white washing is Rs 2.00 per square meter, find the

(i) Inside surface area of the dome,

(ii) Volume of the air inside the dome.

Solution

Given, Cost of white washing of the given hemispherical dome from inside = Rs 498.96

And, Rate of white washing = Rs 2 per square meter

Therefore, inside surface are of the dome and volume of the air inside the dome = ?

(i) Inside surface area of the dome

Here inside surface are of the dome = cost of white washing/rate

= 498.96/2

= 249.48 m²

Thus, inside surface area of the given dome = 249.48 m² Answer

(ii) Volume of the air inside the dome.

Here as calculated in section (i) inner surface area of the given hemispherical dome = 249.48 m²

Now we know that, Curved surface area of a hemisphere `= 2\ pi\ r^2`

Thus, inner surface area of the given hemispherical dome,

⇒ r = 6.2939 m

⇒ r ≃ 6.3 m

Thus, inner radius of the given hemisphere = 6.3 m

Now, we know that, Volume of a hemisphere `=2/3\ pi\ r^3`

Thus, volume of the given hemispherical dome

= `2/3xx22/7` × 6.3 m × 6.3 m × 6.3 m

= 2 × 22 × 2.1 × 0.9 × 6.3 m³

= 44 × 11.907 m²

= 523.908 m³

Now, volume of air inside the given dome = volume of the given hemispherical dome

= 523.908 m³

Thus, volume of air inside the given dome = 523.908 m³ Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.8  Question (9) 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'

Solution

Given, Number of small spheres which are melted to form one big sphere = 27

And, Radius of given small sphere = r

And, Surface area of small sphere = S

Radius of big sphere = r'

And, Surface area of big sphere = S'

Thus, radius (r') of new sphere and ratio of S and S' = ?

(i) Radius r' of the new sphere

We know that Volume of a Sphere `=4/3 ℼ r³

And as per question,

Volume of 27 small sphere = volume of big sphere

⇒ 27 × `4/3` ℼ r ³ = `4/3` ℼ r'³

⇒ 27 r³ = r'³

⇒ (3 r)³ = r'³

⇒ 3 r = r'

Thus, radius of new big sphere = 3 r Answer

(ii) Ratio of S and S'

Now, we know that Surface of a Sphere = 4 ℼ r²

Thus, Surface area of big sphere (S') = 4 ℼ r'²

⇒ S' = 4 ℼ (3r)²

[Because r' = 3 r]

⇒ S' = 4 ℼ 9 r² - - - - (i)

Thus, Surface area of small sphere (S) = 4 ℼ r²

⇒ S = 4 ℼ (r)² - - - - (ii)

Now, ratio of surface area of small sphere and new (big) sphere

= S : S'

= 4 ℼ (r)² : 4 ℼ 9 r²

= 1 : 9

Thus, ratio of S and S' = 1 : 9 Answer

Alternate Method (Shortcut Method)

Given, Number of small spheres which are melted to form one big sphere = 27

And, Radius of given small sphere = r

And, Surface area of small sphere = S

Radius of big sphere = r'

And, Surface area of big sphere = S'

Thus, radius (r') of new sphere and ratio of S and S' = ?

(i) Radius r' of the new sphere

Here given, volumes are in the ratio of 27 : 1

Now, we know that, radii of spheres are in triplicate ratio of volumes of sphere or circle.

Thus, ratio of radii of given sphere = (3)³ : (1)³

⇒ Ratio of radii of given sphere = 3 : 1

Now, since, radius of small sphere = r

And, here r = 1

And, r'

Therefore, radius (r') of big (new) sphere = 3 r Answer

(ii) Ratio of S and S'

Now, we know that, ratio of surface area of given spheres = r² : r'²

Thus, S : S' = (r)² : (3r)²

= r² : 9 r²

= 1 : 9

Thus, ratio of surface areas of small and new sphere = 1 : 9 Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.8  Question (10) A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Solution

Given, Diameter of the spherical capsule = 3.5 mm

Therefore, Radius (r) = 3.5/2 = 1.75 mm

Thus, amount of medicine in mm³ needed to fill the capsule = ?

Here, volume of medicine needed to fill the given capsule = Volume of given capsule

Now we know that, Volume of a Sphere = 4/3 ℼ r³

Thus, volume of the given spherical capsule

= `4/3xx22/7` × (1.75 mm)³

= `4/3xx22/7` × 1.75 × 1.75 × 1.75 mm³

= `4/3` × 22 × 0.25 × 1.75 × 1.75 mm³

`=4/3xx16.84375` mm³

= 67.375/3 mm³

= 22.4583 mm³

Thus, medicine needed to fill the given capsule = 22.4583 mm³ Answer

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