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

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Textbook Exercise Questions and Answers.

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

Question 1.
Find the volume of a sphere whose radius is :
(i) 7 cm
(ii) 0.63 m
Answer:
(i) We have : r = radius of the sphere = 7 cm.
Volume of the sphere = \(\frac{4}{3}\)πr³ = (\(\frac{4}{3} \times \frac{22}{7}\) × 7 × 7 × 7) cm³
= \(\frac{4312}{3}\)πr² = 1437 g cm³

(ii) We have : r = radius of the sphere = 0.63 m
∴ Volume of the sphere = \(\frac{4}{3}\)πr³ = (\(\frac{4}{3} \times \frac{22}{7}\) × 0.63 × 0.63 × 0.63) cm³
= 1.05 m³ (approx.).

Question 2.
Find the amount of water displaced by a solid spherical hall of diameter :
(i) 28 cm
(ii) 0.21 m
Answer:
(i) Diameter of the spherical ball = 28 cm
∴ Radius = \(\left(\frac{28}{2}\right)\) cm = 14 cm
Amount of water displaced by the spherical ball
= Its volume = \(\frac{4}{3}\)πr³
= (\(\frac{4}{3} \times \frac{22}{7}\) × 14 × 14 × 14) cm³
= \(\frac{34496}{3}\)cm³ = 11498g cm³

(ii) Diameter of the spherical ball = 0.21 m
Radius = \(\left(\frac{0.21}{2}\right)\) m = 0.105 m

Amount of water displaced by the spherical ball
= Its volume = \(\frac{4}{3}\)πr³
= (\(\frac{4}{3} \times \frac{22}{7}\) × 0.105 × 0.105 × 0.105) m³
= 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 8.9 g per cm³?
Answer:
Diameter of the ball = 4.2 cm
Radius = \(\left(\frac{4.2}{2}\right)\) cm = 2.1 cm
Volume of the ball = \(\frac{4}{3}\)πr³ = (\(\frac{4}{3} \times \frac{22}{7}\) × 2.1 × 2.1 × 2.1) cm³ = 38.808 cm³
Density of the metal is 8.9 g per cm³.
Mass of the ball = (38.808 × 8.9) g = 345.3912 g = 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:
Let the diameter of the moon be r. Then, the radius of the moon = \(\frac{r}{2}\)
According to the question, diameter of the earth is 4r, so its radius = \(\frac{4r}{2}\) = 2r
V₁ = The volume of the moon = \(\frac{4}{3} \pi\left(\frac{r}{2}\right)^{3}=\frac{4}{3}\)πr³ × \(\frac{1}{8}\)
or 8V₁ = \(\frac{4}{3}\)πr³ .............(1)
V₂ = The volume of the earth = \(\frac{4}{3}\)π(2r)³ = \(\frac{4}{3}\)πr³ × 8
or \(\frac{V_{2}}{8}=\frac{4}{3}\)πr³ ..........(2)

From (1) and (2), we have :
8V₁ = \(\frac{\mathrm{V}_{2}}{8}\) or V₁ = \(\frac{1}{16}\)V₂
Hence, the volume of the moon is \(\frac{1}{16}\) of the volume of the earth.

Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer:
Diameter of a hemispherical bowl =10.5 cm.
Its radius = \(\left(\frac{10.5}{2}\right)\)cm = 5.25 cm
Volume of the bowl = \(\frac{2}{3}\)πr³ = (\(\frac{2}{3} \times \frac{22}{7}\) × 5.25 × 5.25 × 5.25) cm³
= 303.1875 cm³
Hence, the hemispherical bowl can hold 303 mL i.e. 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:
Let R and r be respectively the external and internal radii of the hemispherical vessel. Then,
R = 1.01 m (As thickness = 1 cm = 0.01 m)
and r = 1 m.
Volume of iron used = External volume - Internal volume
= \(\frac{2}{3}\)πr³ - \(\frac{2}{3}\)πr³ = \(\frac{2}{3}\)π(R³ - r³)
= \(\frac{2}{3} \times \frac{22}{7}\) × [(1.01)3 - (1)3] m³
= \(\frac{44}{21}\) × (1.030301 - 1) m³
= (\(\frac{44}{21}\) × 0.030301)m³ = 0.06348 m³ (approx).

Question 7.
Find the volume of a sphere whose surface area is 154 cm².
Answer:
Let r cm be the radius of the sphere.
Also, surface area = 154 cm²
So, 4πr² = 154
or 4 × \(\frac{22}{7}\) × r² = 154
or r² = \(\frac{154 \times 7}{4 \times 22}=\frac{49}{4}\)
or r = g cm = 3.5 cm.
Now, volume = \(\frac{4}{3}\)πr³ = (\(\frac{4}{3} \times \frac{22}{7}\) × 3.5 × 3.5 × 3.5) cm³
= \(\frac{539}{3}\)cm³ = 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 cost of white-washing is ₹ 20.00 per square metre, find the :
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Answer:
(i) Inside surface area of the dome = \(\frac{\text { Total cost of whitewashing }}{\text { Rate of whitewashing }}\)
= \(\left(\frac{498.96}{2.00}\right)\)m² = 249.49 m².

(ii) Let r be the radius of the dome.
∴ Surface area = 2πr²
So, 2 × \(\frac{22}{7}\) × r² = 249.48
or r² = \(\frac{249.48 \times 7}{2 \times 22}\) = 39.69
or r = \(\sqrt{39.69}\) m = 6.3 m

Volume of the air inside the dome = Volume of the dome
= \(\frac{2}{3}\)πr³ = \(\frac{2}{3} \times \frac{22}{7}\) × 6.3 × 6.3 × 6.3 m³
= 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,
(ii) ratio of S and S'.
Answer:
(i) Volume of 27 solid spheres of radius r = 27 × \(\frac{4}{3}\)πr³ .............(1)
Volume of the new sphere of radius r' = \(\frac{4}{3}\)πr³ ......(2)

According to the problem, we have
\(\frac{4}{3}\)πr'³ = 27 × \(\frac{4}{3}\)πr³
r'³ = 27 r³ = (3r)³
r' = 3r.

(ii) Required ratio = \(\frac{\mathrm{S}}{\mathrm{S}^{\prime}}=\frac{4 \pi r^{2}}{4 \pi r^{\prime 2}}=\frac{r^{2}}{(3 r)^{2}}=\frac{r^{2}}{9 r^{2}}=\frac{1}{9}\) = 1 : 9

Question 10.
A capsule of medicine is in the shape of a sphere of diameter 3.5 much medicine (in mm³) is needed to fill this capsule?
Answer:
Diameter of the spherical capsule = 3.5 mm
Radius = \(\frac{3.5}{2}\)mm = 1.75 mm.
∴ Medicine needed for its filling = Volume of spherical capsule
= \(\frac{4}{3}\)πr³ = (\(\frac{4}{3} \times \frac{22}{7}\) × 1.75 × 1.75 × 1.75) mm³
= 22.46 mm³ (approx.).