NCERT Solution For Class 9, Maths, Chapter 13, Surface Areas And Volumes, Exercise 13.4

NCERT Solution For Class 9, Maths, Chapter 13, Surface Areas And Volumes, Exercise 13.4 is all about study of curved surface area of sphere and hemisphere which are well explained. Exercise 13.4 class 9, maths solutions contains total nine questions which are given below.

Table of Contents

Class 9, Maths, Chapter 13, Exercise 13.4 Solutions (Page no. 225)

Q.1. Find the surface area of a sphere of radius:

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Ans: (i) We have: r = radius of the sphere = 10.5 cm
Surface area = 4πr²

$=4\times \frac{22}{7}\times 10.5\times 10.5\,\,c{{m}^{2}}$

=1386 cm²

(ii) We have: r = radius of the sphere = 5.6 cm

Surface area = 4πr²

$=4\times \frac{22}{7}\times 5.6\times 5.6\,c{{m}^{2}}$

= 394.24 cm²

(iii) We have: r = radius of the sphere = 14 cm

Surface area = 4πr²

$=4\times \frac{22}{7}\times 14\times 14\,c{{m}^{2}}$

=2464 cm²

Q.2. Find the surface area of a sphere of diameter:

(i) 14 cm

(ii) 21 cm

(iii) 3.5 m

Ans: (i) Here r =$\frac{14}{2}$cm = 7 cm

Surface area = 4πr²

$=4\times \frac{22}{7}\times 7\times 7\,\,c{{m}^{2}}$

= 616 cm²

(ii) Here r = $\frac{21}{2}$cm = 10.5 cm

Surface area = 4πr²

$=4\times \frac{22}{7}\times 10.5\times 10.5\,\,c{{m}^{2}}$

= 1386 cm²

(iii) Here r = $\frac{3.5}{2}$cm = 1.75 cm

Surface area = 4πr²

$=4\times \frac{22}{7}\times 1.75\times 1.75\,\,c{{m}^{2}}$

= 38.5 cm²

Q.3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Ans: Here r = 10 cm,

Total surface area of hemisphere

= 3πr² = (3 × 3.14 × 10 ×10) cm²

= 942 cm²

Q.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.

Ans: Let r₁ and r₂ be the radius of balloons in the two cases.

Here r₁ = 7 cm and r₂ = 14 cm

Ratio of their surface areas = $\frac{4\pi r_{1}^{2}}{4\pi r_{2}^{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{7\times 7}{14\times 14}=\frac{1}{4}$

Thus, the required ratio of their surface areas = 1 : 4

Q.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: Here $r=\frac{10.5}{2}$cm = 5.25 cm

Curved surface area of the hemisphere

$=2\pi {{r}^{2}}=\left( 2\times \frac{22}{7}\times 5.25\times 5.25 \right)c{{m}^{2}}$

= 173.25 cm²

Rate of tin-plating is Rs 16 per 100 cm²

Cost of tin-plating the hemisphere

= ₹ $\left( 173.25\times \frac{16}{100} \right)$ = ₹ 27.72

Q.6. Find the radius of a sphere whose surface area is 154 cm².

Ans: Let r be the radius of the sphere.

Surface area = 154 cm²

⇒ 4πr² = 154

⇒ $4\times \frac{22}{7}\times \,\,{{r}^{2}}=154$

⇒ $\,{{r}^{2}}=\frac{154\times 7}{4\times 22}=12.25$

⇒ $r=\sqrt{12.25}=3.5$ cm

Thus, the radius of the sphere is 3.5 cm.

Q.7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Ans: Let the diameter of earth be R and that of the moon R will be $\frac{R}{4}$

The radii of moon and earth are $\frac{R}{8}$ and $\frac{R}{2}$respectively.

Ratio of their surface area = $\frac{4\pi {{\left( \frac{R}{8} \right)}^{2}}}{4\pi {{\left( \frac{R}{2} \right)}^{2}}}=\frac{\frac{1}{64}}{\frac{1}{4}}$

⟹ $\frac{1}{64}\times \frac{4}{1}=\frac{1}{16}\,\,i.e.,\,1\,\,\,:\,\,16$

Q.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.

Ans: Inner radius, r = 5 cm

Thickness of steel = 0.25 cm

Outer radius, R = (r + 0.25) cm = (5 + 0.25) cm = 5.25

Now, outer curved surface = 2πR²

$=2\times \frac{22}{7}\times 5.25\times 5.25\,\,c{{m}^{2}}$

= 173.25 cm²

Q.9. A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in (i) and (ii).