AP State Syllabus AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.4 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.4

Question 1.

The radius of a sphere is 3.5 cm. Find its surface area and volume.

Solution:

Radius of the sphere, r = 3.5 cm

Question 2.

The surface area of a sphere is 1018\(\frac{2}{7}\) cm^{2} . What is its volume ?

Solution:

Surface area of sphere = 4πr^{2}

= 1018\(\frac{2}{7}\) cm^{2}

= 3054.857cm^{3}

≅ 3054.86cm^{3}

Question 3.

The length of equator of the globe is 44 cm. Find its surface area.

Solution:

Length of the equator of the globe 2πr = 44 cm.

2 × \(\frac{22}{7}\) × r = 44

∴ r = \(\frac{44 \times 7}{2 \times 22}\) = 7cm

∴ surface area = 4πr^{2}

= 4 × \(\frac{22}{7}\) × 7 × 7

= 4 × 22 × 7

= 616cm^{2}

Question 4.

The diameter of a spherical ball is 21 cm. How much leather is required to prepare 5 such balls?

Solution:

Diameter of the spherical ball d’ = 21 cm

Thus, its radius r = \(\frac{d}{2}=\frac{21}{2}\) = 10.5 cm

Surface area of one ball = 4πr^{2}

= 4 × \(\frac{22}{7}\) × 10.5 × 10.5

= 88 × 1.5 × 10.5 = 1386 cm^{2}

∴ Leather required for 5 such balls

= 5 × 1386 = 6930 cm^{2}

Question 5.

The ratio of radii of two spheres is 2 : 3. Find the ratio of their surface areas and volumes.

Solution:

Ratio of radii r_{1} : r_{2} = 2 : 3

Ratio of surface area

= 4πr_{1}^{2} : 4πr_{2}^{2}

= 2^{2}: 3^{2} = 4 : 9

Ratio of volumes

= 4/3 πr_{1}^{3} : 4/3 πr_{2}^{3}

= 2^{3} : 3^{3} = 8 : 27

Question 6.

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

Solution:

Radius of the hemisphere = 10 cm

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

= 3 × 3.14 × 10 × 10

= 9.42 × 100

= 942 cm^{2}

Question 7.

The diameter of a spherical balloon increases from 14 cm. to 28 cm. as air is being pumped into it. Find the ratio of surface areas of the balloons in the

two cases.

Solution:

The diameter of the balloon, d = 14 cm

Thus, its radius, r = \(\frac{d}{2}=\frac{14}{2}\) = 7 cm

∴ Surface area = 4πr^{2} = 4 × \(\frac{22}{7}\) × 7 × 7

= 88 × 7 = 616cm^{2}

When air is pumped, the diameter = 28 cm

thus its radius = \(\frac{d}{2}=\frac{28}{2}\) = 14 cm

Its surface area = 4πr^{2}

= 4 × \(\frac{22}{7}\) × 14 × 14

= 88 × 28 = 2464 cm^{2}

Ratio of areas = 616 : 2464

= 1 : 4

(OR)

Original radius = \(\frac{14}{2}\) = 7 cm

Increased radius = \(\frac{28}{2}\) = 14cm

Ratio of areas = r_{1}^{2} : r_{2}^{2}

= 7^{2} : 14^{2}

= 7 × 7 : 14 × 14

= 1:4

Question 8.

A hemispherical bowl is made of brass, 0.25 cm thickness. The inner radius of the bowl is 5 cm. Find the ratio of outer surface area to inner surface area.

Solution:

Inner radius of the hemisphere ‘r’ = 5 cm

Outer radius of the hemisphere ‘R’

= inner radius + thickness

= (5 + 0.25) cm = 5.25 cm

Ratio of areas = 3πR^{2}: 3πr^{2}

= R^{2} : r^{2}

= (5.25)^{2}: 5^{2}

= 27.5625 : 25

= 1.1025:1

= 11025 : 10000

= 441 : 400

[Note : If we read “radius as diameter” then we get the T.B. answer]

Question 9.

The diameter of a lead ball is 2.1 cm. The density of the lead used is 11.34 g/c^{3}. What is the, weight of the ball ?

Solution:

The diameter of the ball = 2.1 cm

Thus, its radius, r = \(\frac{d}{2}=\frac{2.1}{2}\) = 1.05 cm

Volume of the ball V’ = \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3} \times \frac{22}{7}\) x 1.05^{3} = \(\frac{101.87}{21}\)

∴Weight of the ball = Volume × density

= 4.851 × 1.34

= 55.010

Question 10.

A metallic cylinder of diameter 5 cm 1 and height 3 \(\frac{1}{3}\) cm is melted and cast into a sphere. What is its diameter ?

Solution:

Diameter of the cylinder’d’ = 5 cm

Thus, its radius, r = \(\frac{d}{2}=\frac{5}{2}\) = 2.5 cm

Height of the cylinder,

Volume of the cylinder

Given that cylinder melted to form sphere

∴ Volume of the sphere = Volume of the cylinder

(Where r is the radius of the sphere)

r^{3} = \(\frac{3}{4}\) × 2.5 × 2.5 × \(\frac{10}{3}\)

r^{3} = 2.5^{3}

∴ r = 2.5 cm

Hence its diameter, d = 2r

= 2 × 2.5 = 5 cm

Question 11.

How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold ?

Solution:

Diameter of the hemispherical bowl = 10.5 cm

Thus its radius = \(\frac{d}{2}=\frac{10.5}{2}\) = 5.25cm

Quantity of milk, the bowl can hold = Volume of the bowl = \(\frac{2}{3}\)πr^{3}

= \(\frac{2}{3} \times \frac{22}{7}\) × 5.25 × 5.25 × 5.25

= 303.1875 cm^{3}

= \(\frac{303.1875}{1000}\) lit = 0.303 lit.

Question 12.

A hemispherical bowl has diameter 9 cm. The liquid is poured into cylindrical bottles of diameter 3 cm and height 3 cm. If a full bowl of liquid is Riled in the bottles, find how many

bottles are required ?

Solution:

Diameter of the hemispherical bowl ‘d’ = 9 cm

Its radius, r = \(\frac{d}{2}=\frac{9}{2}\) = 4.5cm

Volume of its liquid = Volume of the bowl = \(\frac{2}{3}\) πr^{3}

= \(\frac{2}{3} \times \frac{22}{7}\) × 4.5 × 45 × 4.5

Diameter of the cylindrical bottle, d = 3 cm

Its radius, r = \(\frac{d}{2}\)

= \(\frac{3.0}{2}\)

= 1.5cm

Height of the bottle, h = 3 cm

Let the number of bottles required = n

Then total volumes of these n bottles = n πr^{2}h

But this is equal to volume of the bowl

Hence n. \(\frac{22}{7}\) × 1.5 × 1.5 × 3

= \(\frac{2}{3} \times \frac{22}{7}\) × 4.5 × 4.5 × 4.5

∴ n = \(\frac{2}{3} \times \frac{20.25}{1.5}\) = 9

∴ Number of bottles required = 9