NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

NCERT Solutions Class 9 Maths
Chapter – 13 (Surface Areas and Volumes)

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 13 Surface Areas and Volumes Exercise 13.8 has been provided here to help the students in solving the questions from this exercise.

Chapter 13: Surface Areas and Volumes

Exercise – 13.8

(Assume π = $\mathbf{\frac{22}{7}}$ , unless stated otherwise)

1. Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m

Answer –
(i) Radius of the sphere, r = 7 cm
Volume of sphere = $\frac{4}{3}$ πr³= $\frac{4}{3}$ × $\frac{22}{7}$ × 7³cm³
= 4312/3 cm³
= 1437.33 cm³
Hence, the volume of the sphere is 1437.33 cm³.

(ii) Radius of the sphere, r = 0.63 m
volume of sphere = $\frac{4}{3}$ πr³= $\frac{4}{3}$ × $\frac{22}{7}$ × 0.63³= 1.0478 m³
Hence, the volume of the sphere is 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

Answer –
(i) Diameter = 28 cm
Radius, r = 28/2 cm = 14cm
Volume of the solid spherical ball = $\frac{4}{3}$ πr³Volume of the ball = $\frac{4}{3}$ × $\frac{22}{7}$ × 14³= 34496/3
= 11498.66 cm³
Hence, the volume of the ball is 11498.66 cm³.

(ii) Diameter = 0.21 m
Radius of the ball =0.21/2 m= 0.105 m
Volume of the ball = $\frac{4}{3}$ πr³= $\frac{4}{3}$ × $\frac{22}{7}$ × 0.105³ m³= 0.004851 m³
Hence, the volume of the ball = 0.004851 m³

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

Answer –
Diameter of a metallic ball = 4.2 cm
Radius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cm
Volume formula = $\frac{4}{3}$ πr³= $\frac{4}{3}$ × $\frac{22}{7}$ × 2.1 cm³= 38.808 cm³
Volume of the metallic ball = 38.808 cm³.
Now, Mass = Volume × Density
Mass of the metallic ball = 38.808 cm³ × 8.9g / cm³
= 345.3912 g
= 345.39 g (approx.)
Thus, the mass of the ball is 345.39 g.

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?

Answer –
Let the diameter of the earth = d.
Diameter of the moon = $\frac{1}{4}$ × diameter of the earth
The radius of the moon = $\frac{1}{4}$ × radius of the earth [Since, diameter = 2 × Radius]
r = $\frac{1}{4}$ × R = $\frac{R}{4}$
The volume of the earth = $\frac{4}{3}$ π R³

The volume of the moon = $\frac{4}{3}$ π r³= $\frac{4}{3}$ π $\left ( \frac{R}{4} \right )^3$ [ replacing r with $\frac{R}{4}$ ]

= $\frac{4}{3}$ π × $\frac{1}{64}$ R³The volume of the moon = $\frac{1}{64}$ × Volume of the earth
Hence the volume of the moon is $\frac{1}{64}$ times the volume of the earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22/7)

Answer –
Diameter of the hemispherical ball, d = 10.5 cm
Radius of the hemispherical ball, r = 10.5/2 cm = 5.25 cm
Volume of the hemispherical ball = $\frac{2}{3}$ πr³= $\frac{2}{3}$ × $\frac{22}{7}$ × 5.25 cm × 5.25 cm × 5.25 cm
= 303.1875 cm³= 303.1875 /1000 ( 1000cm³ = 1 L)
= 0.3031875 litres
= 0.303 litres (approx.)
The hemispherical bowl can hold 0.303 litres of milk.

6. A hemi spherical 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.

Answer –
Inner Radius of the tank, (r) = 1 m
Thickness of iron = 1cm = 1/100 m = 0.01 m
Outer radius of the tank, R = 1 m + 0.01m = 1.01 m
Volume of the iron used to make the tank = $\frac{2}{3}$ πR³ – $\frac{2}{3}$ πr³= $\frac{2}{3}$ π (R³ – r³)

= $\frac{2}{3}$ × $\frac{22}{7}$ × [(1.01m)³ – (1m)³]

= $\frac{2}{3}$ × $\frac{22}{7}$ × [1.030301 m³ – 1 m³]

= $\frac{2}{3}$ × $\frac{22}{7}$ × 0.030301 m³= 0.06348 m³ (approx.)
So, the volume of the iron used in the hemispherical tank is 0.06348 m³.

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

Answer –
Let the radius of a sphere = r
Surface area of the sphere = 4πr² = 154cm²
4πr²= 154 cm²
r² = $\frac{154\times 7}{4\times 22}$ = $\frac{7}{2}$

Radius is $\frac{7}{2}$ cm
Therefore, volume of the sphere = $\frac{4}{3}$ πr³= $\frac{4}{3}$ × $\frac{22}{7}$ × $\frac{7}{2}$ cm × $\frac{7}{2}$ cm × $\frac{7}{2}$ cm

= $\frac{539}{3}$ cm³= $179\frac{2}{3}$ cm³.
Therefore, volume of the sphere is $179\frac{2}{3}$ cm³.

8. A dome of a building is in the form of a hemi sphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing is Rs20 per square meter, find the
(i) inside surface area of the dome
(ii) volume of the air inside the dome

Answer –
Cost of white-washing the dome from inside = Rs 4989.60
Cost of white-washing 1 m² area = Rs 20

(i) Inside surface area of the dome = Total cost for whitewashing the dome inside / Rate of whitewashing
= 4989.60/20
= 249.48 m²

(ii) Let the inner radius of the hemispherical dome = r.
Inner surface area of the hemispherical dome = 2πr²2πr²= 249.48 m²⇒ r² = 249.48/2π m²⇒ r² = 249.48 ÷ (2 × $\frac{22}{7}$ )
⇒ r² = 39.69
⇒ r = √39.69
⇒ r = 6.3 m
The volume of the air inside the dome will be the same as the volume of the hemisphere.
Now the volume of the air inside the dome = $\frac{2}{3}$ πr³= $\frac{2}{3}$ × $\frac{22}{7}$ × 6.3 m × 6.3 m × 6.3 m
= 523.9 m³ (approx.)
Therefore, the inner surface area of the dome is 249.48 m² and he volume of the air inside the dome is 523.9 m³.

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 Sand S’.

Answer –
The surface area of a sphere = 4πr²The volume of a sphere = $\frac{4}{3}$ πr³Therefore, The volume of 27 solid spheres with radius r = 27 × $\frac{4}{3}$ πr³ = 36πr³ —————– (1)
Also, the volume of the new sphere with radius r’ = $\frac{4}{3}$ πr’³ —————– (2)

(i) Volume of the new sphere = Volume of 27 solid spheres
$\frac{4}{3}$ πr’³ = 36πr³ [From equation (1) and (2)]
⇒ r’³ = 36πr³ × $\frac{3}{4}$ π
⇒ r’³ = 27r³
⇒ r’ = ∛27r³
r’ = 3r
Radius of the new sphere, r’ = 3r

(ii) Ratio of S and S′
Now, surface area of each iron sphere, S = 4πr²Surface area of the new sphere, S’ = 4πr’² = 4π(3r)² = 36πr²The ratio of the S and S’ = 4πr²/36πr² = 1/9
Hence, the radius of new sphere is 3r and the ratio of S and S’ is 1 : 9.

10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm³) is needed to fill this capsule?

Answer –
Diameter of capsule = 3.5 mm
Radius of capsule, say r = diameter/ 2 = (3.5/2) mm = 1.75 mm
Volume of spherical capsule = $\frac{4}{3}$ πr³= $\frac{4}{3}$ × $\frac{22}{7}$ × (1.75)³
= 22.458
= 22.46mm³ (approx.)
The volume of the spherical capsule is 22.46 mm³.

NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8