# 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.)