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

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.8 Textbook Exercise Questions and Answers.

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

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
Answer:
For the given sphere, radius r = 7 cm.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 7 cm³
= \(\frac{4312}{3}\) cm³
= 1437\(\frac{1}{3}\) cm³

(ii) 0.63 m.
Answer:
For the given sphere, radius r = 0.63 m.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.63 × 0.63 × 0.63 m³
= 1.05 m³ (approx.)

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Answer:
Amount of water displaced by a solid spherical ball = Volume of spherical ball

(i) For the given spherical ball, diameter 28
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm
Volume of spherical ball
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 14 × 14 × 14 cm³
= \(\frac{34496}{3}\) cm³
= 11498\(\frac{2}{3}\) cm³
Thus, the amount of water displaced by the given solid spherical ball is = 11498\(\frac{2}{3}\) cm³

(ii) For the given spherical ball, diameter 28
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{0.21}{2}\) m
Volume of spherical ball
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{0.21}{2}\) × \(\frac{0.21}{2}\) × \(\frac{0.21}{2}\) m³
= 11 × 0.01 × 0.21 × 0.21 m³
= 0.004851 m³
Thus, the amount of water displaced by the given solid spherical ball is 0.004851 m³.

Question 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 8.9g per cm³?
Answer:
For the given spherical ball,
radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{4.2}{2}\) cm
= 2.1 cm
= \(\frac{21}{10}\) cm
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{21}{10}\) × \(\frac{21}{10}\) × \(\frac{21}{10}\) cm³
= 38.808 cm³
Now, the density of the metal of the ball is 8.9 g per cm³.
∴ Mass of the ball = Volume × Density
= 38.808 cm³ × 8.9 g/cm³
= 345.39 g (approx.)
Thus, the mass of the metallic ball is 345.39 g (approx.).

Question 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:
As the diameter of the moon is one-fourth of the diameter of the earth, the radius of the moon is also one-fourth of the radius of the earth. In other words, the radius of the earth is four times the radius of the moon. Let, the radius of the moon be r and the radius of the earth be R.
Then, R = 4r
Now, \(=\frac{\text { volume of the moon }}{\text { volume of the earth }}\) = \(\frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R^{3}}\)
= \(\left(\frac{r}{R}\right)^{3}\)
= \(\left(\frac{r}{4 r}\right)^{3}\)
= \(\left(\frac{1}{4}\right)^{3}\)
= \(\frac{1}{64}\)
∴ Volume of the moon
= \(\frac{1}{64}\) × Volume of the earth
Thus, the volume of the moon is \(\frac{1}{64}\) times the volume of the earth.

Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer:
For the hemispherical bowl,
radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{10.5}{2}\) cm
= 5.25 cm
= \(\frac{21}{4}\) cm
Capacity of the hemispherical bowl
= Volume of a hemisphere
= \(\frac{2}{3}\) πr³
= \(\frac{2}{3} \times \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \times \frac{21}{4}\) cm³
= 303.19 cm³ (approx.)
= \(\frac{303.19}{1000}\) liters (approx.)
= 0.303 liters (approx)
Thus, the given hemispherical bowl can hold 0. 303 litres (approx.) of milk.

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.
Answer:
For the hemispherical tank, inner radius r = 1 m and the thickness of the iron sheet = 1 cm = 0.01 m.
∴ For the hemispherical tank, outer radius
R = 1 + 0.01 m = 1.01 m.
Volume of the iron used in the tank
= Volume of outer hemisphere – Volume of Inner hemisphere
= \(\frac{2}{3}\) πR³ – \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) π (R³ – r³)
= \(\frac{2}{3}\) × \(\frac{22}{7}\) (1.01³ – 1³) m³
= \(\frac{44}{21}\) (1.030301 – 1) m³
Thus, the volume of the iron used to make the tank is 0.06349 m³ (approx.).

Question 7.
Find the volume of a sphere whose surface area is 154 cm².
Answer:
For the given sphere, surface area = 154 cm².
Surface area of a sphere = 4πr²
∴ 154 cm² = 4 × \(\frac{22}{7}\) × r²cm²
∴ r² = \(\frac{154 \times 7}{4 \times 22}\) cm²
∴ r² = \(\frac{49}{4}\) cm²
∴ r = \(\frac{7}{2}\)
Thus, the radius of the given sphere is \(\frac{7}{2}\) cm.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}\) cm³
= \(\frac{539}{3}\) cm³
= 179\(\frac{2}{3}\) cm³
Thus, the volume of the given sphere is 179\(\frac{2}{3}\) cm³.

Question 8.
A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ₹ 4989.60. If the s cost of whitewashing is ₹ 20 per square metre, find the
(i) inside surface area of s the dome.
Answer:
(i) Area of the region whitewashed at the
cost of ₹ 20 = 1 m²
∴ Area of the region whitewashed at the cost of ₹ 4989.60 = \(\frac{4989.60}{20}\) m² = 249.48 m²
Hence, the inner surface area of the dome is 249.48 m²

(ii) volume of the air inside the dome.
Answer:
Curved surface area of hemispherical dome = 2πr²
∴ 249.48 m² = 2 × \(\frac{22}{7}\) × r² m²
∴ r² = \(\frac{249.48 \times 7}{2 \times 22}\) m²
∴ r² = 39.69 m²
∴ r² = \(\sqrt{39.69}\) m
∴ r = 6.3 m
Thus. the radius of the hemispherical dome is 6.3m.
Volume of air inside the hemispherical dome = Volume of a hemisphere
= \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 6.3 × 6.3 × 6.3 m³
= 523.908 m³
= 5239 m³ (approx.)
Thus, the volume of the air inside the dome is 523.9 m³ (approx.).

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,
Answer:
(i) 27 solid iron spheres of radius r are melted to form 1 iron sphere of radius r’.
∴ Volume of 1 sphere of radius r’
= Volume of 27 spheres of radius r
∴ \(\frac{4}{3}\) πr’³ = 27 × \(\frac{4}{3}\) πr’³
∴ r’³ = 27r³
∴ r’³ = (3r)³
∴ r’ = 3r

(ii) ratio of S and S’.
Answer:
The surface area of the sphere with radius r is S and the surface area of the sphere with radius r’ is S’.
Then,
\(\frac{\mathrm{s}}{\mathrm{S}^{\prime}}=\frac{4 \pi r^{2}}{4 \pi r^{\prime 2}}=\frac{r^{2}}{r^{\prime 2}}=\frac{r^{2}}{(3 r)^{2}}=\frac{r^{2}}{9 r^{2}}=\frac{1}{9}\) = 1 : 9
Thus, the ratio of S and S’ is 1 : 9.

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 ?
Answer:
For the spherical capsule, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{3.5}{2}\) mm
= 1.75 mm
Capacity of the spherical capsule
= Volume of a sphere
= \(\frac{4}{3}\) πr’³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 1.75 × 1.75 × 1.75 mm³
= 22.46 mm³ (approx.)
Thus, 22.46 mm³ (approx.) medicine is needed to fill the given capsule.