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

Page-225

Question 1.

Find the surface area of a sphere of radius:

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Answer:

(i) Radius of the sphere (r) = 10.5 cm

Surface area

= 4πr^{2} = (4 × \(\frac{22}{7}\) × 10.5 × 10.5) cm^{2}

= 1386 cm^{2}

(ii) Radius of the sphere (r) = 5.6 cm

Surface area

= 4πr^{2} = (4 × \(\frac{22}{7}\) × 5.6 × 5.6) cm^{2}

= 394.24 cm^{2}

(iii) Radius of the sphere (r) = 14 cm

Surface area

= 4πr^{2} = (4 × \(\frac{22}{7}\) × 14 × 14) cm^{2}

= 2464 cm^{2}

Question 2.

Find the surface area of a sphere of diameter:

(i) 14 cm

(ii) 21cm

(iii) 3.5 m

Answer:

(i) Radius of a sphere (r) = \(\frac{14}{2}\) = 7 cm

Surface area = 4πr^{2}

= (4 × \(\frac{22}{7}\) × 7 × 7) cm^{2}

= 616 cm^{2}

(ii) Radius of a sphere (r) = \(\frac{21}{2}\) = 10.5 cm

Surface area = 4πr^{2}

= (4 × \(\frac{22}{7}\) × 10.5 × 10.5) cm^{2}

= 1386 m^{2}

(iii) Radius of a sphere (r) = \(\frac{3.5}{2}\) = 1.75 m

Surface area = 4πr^{2}

= (4 × \(\frac{22}{7}\) × 1.75 × 1.75) cm^{2}

= 38.5 m^{2}

Question 3.

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

Answer:

r = 10 cm

∴ Total surface area of a hemisphere= 3πr^{2}

= 3 × 3.14 × 10 × 10

= 942 cm^{2}

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.

Answer:

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 = 2πr^{2}

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

= 173.25 cm^{2}

Rate of tin-plating is ₹ 16 per 100 cm^{2}

Therefor, cost of 1 cm^{2} = ₹ \(\frac{16}{100}\)

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 cm^{2}.

Answer:

Let r be the radius of the sphere.

Surface area =154 cm^{2}

⇒ 4πr^{2} = 154

⇒ 4 × \(\frac{22}{7}\) × r^{2} = 154

⇒ r^{2} = \(\frac{154}{\left(4 \times \frac{22}{7}\right)}\)

⇒ r^{2} = \(\frac{49}{4}\)

⇒ 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.

Answer:

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.

Answer:

Inner radius of the bowl (r) = 5 cm

Thickness of the steel = 0.25 cm

∴ Outer radius (R)

= (r + 0.25) cm

= (5 + 0.25) cm

= 5.25 cm

Outer curved surface area = 2πR^{2}

= 2 × \(\frac{22}{7}\) × 5.25 × 5.25

= 173.25 cm^{2}

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)

Answer:

(i) The surface area of the sphere with radius r = 4πr^{2}

(ii) The right circular cylinder just encloses a sphere of radius r.

∴ The radius of the cylinder = r and its height = 2r

∴ Curved surface of cylinder = 2πrh = 2π × r × 2r = 4πr^{2}

(iii) Ratio of the areas = 4πr^{2} : 4πr^{2} = 1 : 1.