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

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

(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

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

(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.).

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

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.

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

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 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 ₹ 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.).

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

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

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