# JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Jharkhand Board JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

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Question 1.
Find the surface area of a sphere of radius:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
(i) Radius of the sphere (r) = 10.5 cm
Surface area
= 4πr2 = (4 × $$\frac{22}{7}$$ × 10.5 × 10.5) cm2
= 1386 cm2

(ii) Radius of the sphere (r) = 5.6 cm
Surface area
= 4πr2 = (4 × $$\frac{22}{7}$$ × 5.6 × 5.6) cm2
= 394.24 cm2

(iii) Radius of the sphere (r) = 14 cm
Surface area
= 4πr2 = (4 × $$\frac{22}{7}$$ × 14 × 14) cm2
= 2464 cm2

Question 2.
Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21cm
(iii) 3.5 m
(i) Radius of a sphere (r) = $$\frac{14}{2}$$ = 7 cm
Surface area = 4πr2
= (4 × $$\frac{22}{7}$$ × 7 × 7) cm2
= 616 cm2

(ii) Radius of a sphere (r) = $$\frac{21}{2}$$ = 10.5 cm
Surface area = 4πr2
= (4 × $$\frac{22}{7}$$ × 10.5 × 10.5) cm2
= 1386 m2

(iii) Radius of a sphere (r) = $$\frac{3.5}{2}$$ = 1.75 m
Surface area = 4πr2
= (4 × $$\frac{22}{7}$$ × 1.75 × 1.75) cm2
= 38.5 m2 Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
r = 10 cm
∴ Total surface area of a hemisphere= 3πr2
= 3 × 3.14 × 10 × 10
= 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.
Let r be the initial radius and R be the increased radius of balloon.
r = 7 cm and R = 14 cm
Ratio of the surface area
= $$\frac{4 \pi r^2}{4 \pi \mathrm{R}^2}$$ = $$\frac{r^{2}}{R^{2}}$$ =
Thus, the ratio of surface areas = 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 ₹ 16 per 100 cm2.

Ans. Radius of the bowl (r) =$$\frac{10.5}{2}$$ = 5.25 cm

Curved surface area of the hemispherical bowl = 2πr2
= (2 × $$\frac{22}{7}$$ × 5.25 × 5.25) cm2
= 173.25 cm2

Rate of tin-plating is ₹ 16 per 100 cm2
Therefor, cost of 1 cm2 = ₹ $$\frac{16}{100}$$
Total cost of tin-plating the hemisphere bowl
= 173.25 × $$\frac{16}{100}$$ = ₹ 121.72

Question 6.
Find the radius of a sphere whose surface area is 154 cm2.
Let r be the radius of the sphere.
Surface area =154 cm2
⇒ 4πr2 = 154
⇒ 4 × $$\frac{22}{7}$$ × r2 = 154
⇒ r2 = $$\frac{154}{\left(4 \times \frac{22}{7}\right)}$$
⇒ r2 = $$\frac{49}{4}$$
⇒ r = $$\frac{7}{2}$$ = 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.
Let the diameter of earth be r then the diameter of the moon = r/4
Radius of the earth = $$\frac{r}{2}$$
Radius of the moon = $$\frac{r}{8}$$
Ratio of their surface areas = $$\frac{4 \pi\left(\frac{r}{8}\right)^2}{4 \pi\left(\frac{r}{2}\right)^2}$$
= $$\frac{\left(\frac{1}{64}\right)}{\left(\frac{1}{4}\right)}$$
= $$\frac{4}{64}$$ = $$\frac{1}{16}$$

Thus, the ratio of their surface areas is 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.
Inner radius of the bowl (r) = 5 cm
Thickness of the steel = 0.25 cm
= (r + 0.25) cm
= (5 + 0.25) cm
= 5.25 cm

Outer curved surface area = 2πR2
= 2 × $$\frac{22}{7}$$ × 5.25 × 5.25
= 173.25 cm2

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