Surface Areas and Volumes – Exercise 13.8 – (MATHEMATICS) – 9th Class

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Exercise 13.8

Assume π ={{22} \over 7}, unless stated otherwiese.

1. Find the volume of a sphere whose radius is
(i) 7 cm

(ii) 0.63 m.

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Sol. (i) Radius (r) = 7 cm

Volume =   {4 \over 3}\pi {r^3} = {4 \over 3} \times {{22} \over 7} \times {(7)^3}c{m^3}
= 1437 {1 \over 3} cm³

(ii) Volume = {4 \over 3} \times {{22} \over 7} \times {(0.63)^3}c{m^3}

= 1.05 m³ (approx.).


2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm

(ii) 0.21 m.

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Sol. Volume of water displaced = Volume of solid spherical

(i) Diameter = 28 cm     ⇒   Radius = 14 cm.
∴ Volume of water displaced = {4 \over 3} \times {{22} \over 7} \times {(14)^3}c{m^3}

= 11498  = {2 \over 3}c{m^3}

(ii) Diameter = 0.21 m      ⇒     Radius = 0.105 m.

∴ Volume of water displaced = {4 \over 3} \times {{22} \over 7} \times {(0.105)^3}c{m^3}
= 0.004851 m³.


3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

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Sol. Diameter of the ball = 4.2 cm.
⇒  Radius of the ball = 2.1 cm.

Volume of the ball = {4 \over 3} \times {{22} \over 7} \times {(2.1)^3}c{m^3} = 38.808c{m^3}

Mass of the ball = density × volume
= 8.9 × 38.808 g
= 345.39 g (approx.)


4. The diameter of the 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?

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Sol. Let diameter of the earth = d units.

⇒      Radius of the earth = {d \over 2} units.

Diameter of the moon =  = {d \over 4}units

⇒    Radius of the moon =  = {d \over 8}units

Volume of the earth /Volume of the moon =  = {{{4 \over 3}\pi {{\left( {{d \over 2}} \right)}^3}} \over {{4 \over 3}\pi {{\left( {{d \over 8}} \right)}^3}}} = 64

Volume of the moon = {1 \over {64}} volume of the earth.

Hence, the volume of the moon is {1 \over {64}} of the volume of the earth.


5. How many liters milk can a hemispherical bowl of diameter 10.5 cm hold?

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Sol. Volume of the hemisphere = {2 \over 3} \times {{22} \over 7} \times {(5.25)^3}c{m^3}
= 303.19 cm³

So, the capacity of the bowl   {{303.19} \over {1000}}l
= 0.303 l (approx.)


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.

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Sol. Inner radius = 1 m, thickness

= 1 cm = 0.01 m

Outer radius = 1 + 0.01 m
= 1.01 m

Volume of iron used

 = {2 \over 3} \times {{22} \over 7} \times \left\{ {{{(1.01)}^3} - {{(1)}^3}} \right\}{m^3}

 = {{44} \over {21}} \times \left\{ {1.0303 - 1} \right\}{m^3} = {{44} \over {21}} \times 0.0303{m^3}

= 0.06348 m(approx)


7. Find the volume of a sphere whose surface area is 154 cm².

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Sol. Surface area = 154 cm²

 \Rightarrow 4 \times {{22} \over 7} \times {r^2} = 154 \Rightarrow {r^2} = {{154 \times 7} \over {88}} = 12.25

 \Rightarrow r = 3.5cm

Volume of the sphere = {4 \over 3} \times {{22} \over 7} \times {(3.5)^3}c{m^3} = 179{2 \over 3}c{m^3}


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 metre, find the

(i) inside surface area of the dome,
(ii) volume of the air inside the dome. Sol. Cost of white washing = 498.96. Rate of white-washing = 2 per sq. m.

Sol. Cost of white washing = 498.96. Rate of white-washing = 2 per sq. m.

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Sol. Cost of white-washing = Rs.498.96.

Rate of white-washing = Rs.2 per sq. m.

(i) Inside surface area = {{498.96} \over 2}{m^2} = 249.48{m^2}

(ii) We have 2 \times {{22} \over 7} \times {r^2} = 249.48

 \Rightarrow {r^2} \times {{249.48 \times 7} \over {44}} \times 39.69 \Rightarrow r = 6.3m

∴  Volume of the dome = {2 \over 3}\pi {r^3}

 = {2 \over 3} \times {{22} \over 7} \times {(6.3)^3}{m^3}

= 523.m3(approx.)


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’

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Sol. Total volume of 27 sphere =  = 27 \times {4 \over 3}\pi {r^3} = 36\pi {r^3}

(i) Volume of a new sphere =  = {4 \over 3}\pi {r^{              …..(i)

∴      {4 \over 3}\pi r{

⇒             r{

(ii) Surface area of each of 27 spheres (S) = 4πr²                                     …(a)
Surface area of a new sphere (S’) = 4π(3r)² = 36π²                …(b)
From (a) and (la), we get

{S \over {S

Hence, S : S’ = 1 : 9.


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?

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Sol.    Diameter (d) = 3.5 mm

∴   Radius(r) = {{3.5} \over 2} = 1.75mm

Medicine  needed = Volume of capsule  = {4 \over 3}\pi {r^3}

 = {4 \over 3} \times {{22} \over 7} \times {(1.75)^3}m{m^3}

= 22.46 mm3  (approx)


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