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

Jharkhand Board JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.8 Textbook Exercise Questions and Answers.

JAC Board Class 9th 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.
Solution:
(i) Radius of the sphere (r) = 7 cm
Therefore, Volume of sphere = \(\frac{4}{3}\)πr³
= \(\frac{4312}{3}\) cm³.

(ii) Radius of the sphere (r) = 0.63 m
∴ Volume of the sphere = \(\frac{4}{3}\)πr³
= \(\frac{4}{3} \times \frac{22}{7} \times 0.63 \times 0.63 \times 0.63\) m³
= 1.05 m³

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter :
(i) 28 cm
(ii) 0.21 m.
Solution:
(i) Diameter of spherical ball = 28 cm
Radius (r) = \(\frac{28}{2}\) cm = 14 cm

Amount of water displaced by spherical ball = Volume of spherical ball
= \(\frac{4}{3}\)πr³
= \(\frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14\) cm³
= \(\frac{34496}{3}\) cm³

(ii) Diameter of spherical ball = 0.21 m
∴ Radius (r) = \(\frac{0.21}{2}\) = 0.105 m

Amount of water displaced by spherical ball = Volume of spherical ball
= \(\frac{4}{3}\)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?
Solution:
Radius (r) = \(\frac{4.2}{2}\) = 2.1 cm
Volume of spherical ball = \(\frac{4}{3}\)πr³
= 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

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?
Solution:
Let the diameter of the moon be r.
Radius of the moon = \(\frac{r}{2}\)
Diameter of the earth = 4r
Radius (R) = \(\frac{4r}{2}\) = 2r

Volume of the earth = v = \(\frac{4}{3}\)π\(\frac{r}{2}\)³
8v = \(\frac{4}{3}\)πr³ × \(\frac{1}{83}\) ……(i)

Volume of the earth = V = \(\frac{4}{3}\)π(2r)³
= \(\frac{4}{3}\)π(r)³ × 8
= \(\frac{V}{8}\) = \(\frac{4}{3}\)πr³ ………(ii)

From (i) and (ii), we have
8v = \(\frac{V}{8}\)
v = \(\frac{1}{64}\) V
Thus, the volume of the moon is of \(\frac{1}{64}\) the volume of the earth.

Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold.
Solution:
Diameter of hemispherical bowl = 10.5 cm
Radius (r) = \(\frac{10.5}{2}\) = 5.25 cm

Volume of the bowl = \(\frac{2}{3} \pi r^3\)
= \(\frac{2}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 5.25\) cm³
= 303.1875 cm³

Litres of milk bowl can hold = \(\frac{303.1875}{1000}\)
= 0.3031875 litres (approx.)

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.
Solution:
Inner radius = r = 1 m
External radius = R = (1 + 0.01) m = 1.01 m

Volume of the 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.
Solution:
Let r cm be the radius of sphere.
Surface area of the sphere = 154 cm²
⇒ 4πr² = 154
⇒ 4 × \(\frac{22}{7}\) × r² = 154
⇒ r² = \(\frac{(154×7)}{(4×22)}\) = 12.25
⇒ r = 3.5 cm

Volume = \(\frac{4}{3} \pi r^3\)
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 3.5 × 3.5 × 3.5 cm³
= \(\frac{539}{3}\) 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 ₹ 498.96. If the cost of the white-washing is ₹ 2.00 per square metre, find the :
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Solution:
(i) Inside surface area of the dome = Total cost of white washed/Rate of white washed per square metre
= \(\frac{498.96}{2.00}\) m² = 249.48 m²

(ii) Let r be the radius of the dome.
Surface area = 2πr²
⇒ 2 × \(\frac{22}{7}\) × r² = 249.48
⇒ r² = \(\frac{249.48 \times 7}{22 \times 2}\) = 39.69
⇒ r² = 39.69
⇒ r = 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
= 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) radio of S and S’.
Solution:
Volume of 27 solid spheres of radius
r = 27 × \(\frac{4}{3}\)πr³

Volume of the new sphere of radius
r’ = \(\frac{4}{3} \pi r^{\prime} 3\)
⇒ \(\frac{4}{3} \pi r^{\prime} 3=27 \times \frac{4}{3} \pi r^3\)
⇒ \(r^{\prime 3}=\frac{27 \times \frac{4}{3} \pi r^3}{\frac{4}{3} \pi}\)
⇒ r’³ = 27r³ = (3r)³
⇒ r’= 3r

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

Question 10.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?
Solution:
Diameter of spherical capsule = 3.5 mm
Radius (r)= \(\frac{3.5}{2}\) 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.)