HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.4

Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm.
Solution:
(i) We have,
Radius of sphere (r) = 10.5 cm
∴ Surface area of a sphere = 4πr²
= 4 × \(\frac{22}{7}\) × (10.5)²
= \(\frac{9702}{7}\) cm² Ans.
Hence,surface area of a sphere = 1386 cm².

(ii) We have,
Radius of sphere (r) = 5.6 cm²
∴ Surface area of a sphere = 4πr²
= 4 × \(\frac{22}{7}\) × (5.6)²
= \(\frac{2759.68}{7}\)
= 394.24 cm²
Hence,surface area of a sphere
= 394.24 cm².

(iii) We have,
Radius of sphere (r) = 14 cm
∴ Surface area of a sphere = 4πr²
= 4 × \(\frac{22}{7}\) × (14)²
\(\frac{17248}{7}\) = 2464 cm²
Hence, surface area of a sphere = 2464 cm².

Question 2.
Find the surface area of a sphere of diameter :
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m.
Solution:
(i) We have,
Diameter of a sphere (d) = 14 cm
∴ Radius of a sphere (r) = \(\frac{14}{2}\) = 7 cm
∴ Surface area of a sphere = 4πr²
= 4 × \(\frac{22}{7}\) × (7)²
= \(\frac{4312}{7}\)
= 616 cm²
Hence,surface area of a sphere = 616 cm².

(ii) We have,
Diameter of a sphere (d) = 21 cm
∴ Radius of a sphere (r) = \(\frac{21}{2}\) = 10.5 cm
∴ Surface area of a sphere = 4πr²
= 4 × \(\frac{22}{7}\) × (10.5)²
= \(\frac{9702}{7}\) cm² = 1386 m²
Hence,surface area of a sphere = 1386 cm².

(iii) We have,
Diameter of a sphere (d) = 3.5 m
∴ Radius of a sphere (r) = \(\frac{3.5}{2}\) = 1.75 m
∴ Surface area of a sphere = 4πr²
= 4 × \(\frac{22}{7}\) × (1.75)²
= \(\frac{269.5}{7}\)
= 38.5 m²
Hence,surface area of a sphere= 38.5 m².

Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Solution:
We have, Radius of a hemisphere (r) = 10 cm
∴ Total surface area of a hemisphere= 3πr²
= 3 × 3.14 × (10)²
= 942 cm²
Hence, total surface area of a hemisphere = 942 cm².

Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
We have,
Radius of a spherical balloon (r₁) = 7 cm
∴ Surface area of a spherical balloon (S₁) = 4π(r₁)² = 4 × \(\frac{22}{7}\) × (7)²
= 616 cm²
and increased radius of a spherical ballon (r₂)= 14 cm
∴ Surface area of a spherical ballon in this case (S₂) = 4π(r₁)²
= 4 × \(\frac{22}{7}\) × (14)²
= 2464 cm²
Now,
\(\frac{S_1}{S_2}=\frac{616}{2464}\)
⇒ \(\frac{S_1}{S_2}=\frac{1}{4}\)
⇒ S₁ : S₂ = 1 : 4
Hence, required ratio of the balloon in two cases = 1 : 4.

Question 5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm².
Solution:
We have,
Diameter of hemispherical bowl
(d) = 10.5 cm
∴ Radius of the hemispherical bowl (r) = \(\frac{10.5}{2}\) = 5.25 cm
∴ Inner curved surface area of the bowl = 2πr²
= 2 × \(\frac{22}{7}\) × (5.25)²
= \(\frac{44 \times 27.5625}{7}\)
= 173.25 cm²
Rate of tin-plating = Rs. 16 per 100 cm²
∴ Cost of tin-plating inside of bowl = \(\frac{173.25 \times 16}{100}\)
= Rs. 27.72
Hence, cost of tin-plating the inside of a bowl = Rs. 27.72.

Question 6.
Find the radius of a sphere whose surface area is 154 cm².
Solution:
Let the radius of a sphere be r cm.
Surface area of a sphere = 154 cm², (given)
⇒ 4πr² = 154
⇒ 4 × \(\frac{22}{7}\) × r² = 154
⇒ \(\frac{88}{7}\) × r² = 154
⇒ r² = \(\frac{154 \times 7}{88}\)
⇒ r² = \(\frac{7 \times 7}{4}\)
⇒ r = \(\sqrt{\frac{7 \times 7}{2 \times 2}}\)
⇒ r = \(\frac{7}{2}\) = 3.5 cm
Hence, radius of the sphere = 3.5 cm.

Question 7.
The diameter of the Moon is approximately one fourth of the diameter of the Earth. Find the ratio of their surface areas.
Solution:
Let diameter of the Earth be 2x.
∴ Radius of the Earth (R) = x
∴ Surface area of the Earth (S₁) = 4πR² = 4πx²
According to question,
Diameter of the Moon = \(\frac{1}{4}\) × diameter of the Earth
= \(\frac{1}{4}\) × 2x = \(\frac{x}{4}\)
∴ Radius of the Moon (r) = \(\frac{\frac{x}{2}}{2}=\frac{x}{4}\)
∴ Surface area of the Moon (S₂) = 4πr²

Hence, required ratio of the their surface areas = 1 : 16.

Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
We have,
Inner radius of the bowl = 5 cm
and thickness of the bowl = 0.25 cm

Fig. 13.34
∴ Outer radius of the bowl (r) = 5 + 0.25
= 5.25 cm
∴ Outer curved surface area of the bowl = 2πr²
= 2 × \(\frac{22}{7}\) × (5.25)²
= \(\frac{44 \times 27.5625}{7}\)
= 44 × 3.9375
= 173.25 cm²
Hence, outer curved surface area of the bowl = 173.25 cm².

Question 9.
A right circular cylinder just encloses a sphere of radius r (see Fig. 13.35). Find :
(i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).

Fig. 13.35
Solution:
We have,
(i) Radius of sphere = r
∴ Surface area of the sphere (S₁) = 4πr².
(ii) Radius of the cylinder = radius of the sphere = r.
Height of the cylinder (h) = 2 × r = 2r
∴ Curved surface area of the cylinder
(S₂) = 2πrh = 2π × r × 2r
= 4πr²

Hence, (i) surface area of sphere = 4πr²
(ii) curved surface area of the cylinder = 4πr²
(iii) required ratio of the areas obtained in (i) and (ii) = 1 : 1.