## Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 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:

Surface area of the sphere = \(4 \pi r^{2}\)

= 22 × 3 × 21

= 1386 cm^{2}.

(ii) r = 5.6 cm. = \(\frac{56}{10}\)cm.

Surface area of the sphere = \(4 \pi r^{2}\)

= 394.24 cm^{2}.

(iii) r = 14 cm.

Surface area of the sphere = \(4 \pi r^{2}\)

= 88 × 28

= 2464 cm^{2}.

Question 2.

Find the surface area of a sphere of diameter :

(i) 14 cm

(ii) 21 cm

(iii) 3.5 m.

Solution:

(i) Diameter, d = 14 cm.

Surface area of a sphere = \(4 \pi r^{2}\)

= 616 cm^{2}.

(ii) Diameter, d = 21 cm.

Surface area of a sphere = \(4 \pi r^{2}\)

= 1386 cm^{2}.

(iii) Diameter, d = 3.5 m. = \(\frac{7}{2}\)

Surface area of a sphere = \(4 \pi r^{2}\)

= 38.5 m^{2}.

Question 3.

Find the total surface area of a hemisphere of radius 10 cm.

(Use π = 3.14 )

Solution:

r = 10 cm.

Total surface area of a hemisphere = \(3 \pi r^{2}\)

= 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.

Solution:

Radius of spherical balloons be r_{1} and r_{2}, r_{1} = 7 cm. and r_{2} = 14 cm.

Ratio of surface area

∴ Ratio of surface areas of the balloons = 1 : 4.

Question 5.

A hemispherical bowl made of brass has an inner diameter of 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm^{2}.

Solution:

Inner diameter of hemispherical bowl = 10.5 cm.

Curved surface area of the bowl = \(2 \pi r^{2}\)

= 173.25 cm^{2}.

Cost of tin-plating for 100 cm^{2} is Rs. 16.

Cost of tin-plating for 173.25 sq.cm. …?…

= Rs. 27.72.

Question 6.

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

Solution:

Total surface area of a sphere= 154 cm^{2}

Radius, r =?

∴ r = 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 the diameter of earth be d’ unit

∴ Diameter of Moon = \(\frac{d}{4}\)

radius of the earth = \(\frac{\mathrm{d}}{2}\)

radius of the moon = \(\frac{1}{2} \times \frac{\mathrm{d}}{4}=\frac{\mathrm{d}}{8}\)

Ratio of surface area,

= 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:

Thickness of the hemispherical bowl = 0.25 cm.

inner radius, r = 5 cm.

∴ Outer Surfce area= 5 + 0.25 = 5.25 cm.

Outer Curved Surface area = 2πr^{2}

= 173.25 cm^{2}.

Question 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) & (ii).

Solution:

(i) If radius of a sphere is ‘r’ cm, its Surface area = 4πr^{2}

(ii) height of cylinder, h = diameter of sphere

h = r + r

∴ h = 2r

∴ Curved surface area of cylinder = 2πrh

= 2πr × h

= 2πr × 2r

= 4πr^{2}

= 1 : 1