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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.

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.

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?

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?

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?

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.

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.

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.

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’

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?

∴   $Radius(r) = {{3.5} \over 2} = 1.75mm$
Medicine  needed = Volume of capsule $= {4 \over 3}\pi {r^3}$
$= {4 \over 3} \times {{22} \over 7} \times {(1.75)^3}m{m^3}$