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 = \(\frac{4}{3}\)πr^{3}

= \(\frac{4312}{3}\) cm^{3}.

(ii) Radius of the sphere (r) = 0.63 m

∴ Volume of the sphere = \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3} \times \frac{22}{7} \times 0.63 \times 0.63 \times 0.63\) m^{3}

= 1.05 m^{3}

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

= \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14\) cm^{3}

= \(\frac{34496}{3}\) cm^{3}

(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

= \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3} \times \frac{22}{7}\) × 0.105 × 0.105 × 0.105 m^{3}

= 0.004851 m^{3}

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 cm^{3}?

Solution:

Radius (r) = \(\frac{4.2}{2}\) = 2.1 cm

Volume of spherical ball = \(\frac{4}{3}\)πr^{3}

= 38.808 cm^{3}

Density of the metal is 8.9 g per cm^{3}

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 = \(\frac{4}{3}\)πr^{3} × \(\frac{1}{83}\) ……(i)

Volume of the earth = V = \(\frac{4}{3}\)π(2r)^{3}

= \(\frac{4}{3}\)π(r)^{3} × 8

= \(\frac{V}{8}\) = \(\frac{4}{3}\)πr^{3} ………(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 = \(\frac{2}{3} \pi r^3\)

= \(\frac{2}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 5.25\) cm^{3}

= 303.1875 cm^{3}

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

= \(\frac{2}{3}\)πR^{3} – \(\frac{2}{3}\)πr^{3}

= \(\frac{2}{3}\)π[R^{3} – r^{3}]

= \(\frac{2}{3} \times \frac{22}{7}\)[(1.01)^{3} – (1)^{3}] m^{3}

= \(\frac{44}{21}\)(1.030301 – 1) m^{3}

= \(\frac{44}{21}\) × 0.030301 m^{3}

= 0·06348 m^{3} (approx.)

Question 7.

Find the volume of a sphere whose surface area is 154 cm^{2}.

Solution:

Let r cm be the radius of sphere.

Surface area of the sphere = 154 cm^{2}

⇒ 4πr^{2} = 154

⇒ 4 × \(\frac{22}{7}\) × r^{2} = 154

⇒ r^{2} = \(\frac{(154×7)}{(4×22)}\) = 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 cm^{3}

= \(\frac{539}{3}\) cm^{3}.

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

= \(\frac{498.96}{2.00}\) m^{2} = 249.48 m^{2}

(ii) Let r be the radius of the dome.

Surface area = 2πr^{2}

⇒ 2 × \(\frac{22}{7}\) × r^{2} = 249.48

⇒ r^{2} = \(\frac{249.48 \times 7}{22 \times 2}\) = 39.69

⇒ r^{2} = 39.69

⇒ r = 6.3 m.

Volume of the air inside the dome = Volume of the dome

= \(\frac{2}{3}\)πr^{3}

= \(\frac{2}{3} \times \frac{22}{7}\) × 6.3 × 6.3 × 6.3

= 523.9 m^{3} (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 × \(\frac{4}{3}\)πr^{3}

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} = 27r^{3} = (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 mm^{3}) 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 = \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3} \times \frac{22}{7}\) × 1.75 × 1.75 × 1.75 mm^{3}

= 22.46 mm^{3} (approx.)