#### NCERT Solutions for Class 9th: Ch 13 Surface Areas and Volumes Maths

Page No: 225**Exercise 13.4**

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 × 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 × 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 × 22/7 × 14 × 14) cm

^{2}

= 2464 cm

^{2}

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

(i) 14 cm (ii) 21 cm (iii) 3.5 m

**Answer**

(i) r = 14/2 cm = 7cm

Surface area = 4πr

^{2}

^{ }= (4 × 22/7 × 7 × 7)cm

^{2}

=616cm

^{2}

(ii) r = 21/2 cm = 10.5 cm

Surface area = 4πr

^{2}

= (4 × 22/7 × 10.5 × 10.5) cm

^{2}

= 1386 cm

^{2}

(iii) r = 3.5/2 m = 1.75 m

Surface area = 4πr

^{2}

= (4 × 22/7 × 1.75 × 1.75) m

^{2}

= 38.5 m

^{2}

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 hemisphere = 3πr

^{2}

= (3 × 3.14 × 10 ×10) cm

^{2}

= 942 cm

^{2}

4. The radius of a spherical balloon increases from 7cm to 14cm 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 balloons.

r = 7cm and R = 14cm

Ratio of the surface area =4πr

^{2}/4πR

^{2}

= r

^{2}/R

^{2}

= (7×7)/(14×14) = 1/4

Thus, the ratio of surface areas = 1 : 4

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 cm

^{2}.

**Answer**

Radius of the bowl (r) = 10.5/2 cm = 5.25 cm

Curved surface area of the hemispherical bowl = 2πr

^{2}

^{ }= (2 × 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 }=

^{ }₹16/100

Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100

= ₹27.72

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 × 22/7 × r

^{2 }= 154

⇒ r

^{2 }= 154/(4 × 22/7)

⇒ r

^{2 }= 49/4

⇒ r = 7/2 = 3.5 cm

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 and that of the moon will be r/4

Radius of the earth = r/2

Radius of the moon = r/8

Ratio of their surface area = 4π(r/8)

^{2}/4π(r/2)

^{2}

^{ }= (1/64)/(1/4)

= 4/64 = 1/16

Thus, the ratio of their surface areas is 1:16

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 = 2πR

^{2}

= (2 × 22/7 × 5.25 × 5.25) cm

^{2}

= 173.25 cm

^{2}

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).

**Answer**

(i) The surface area of the sphere with raius 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

Page No: 228

**Exercise 13.5**

1. A matchbox measures 4cm × 2.5cm × 1.5cm. What will be the volume of a packet containing 12 such boxes?

**Answer**

Dimension of matchbox = 4cm × 2.5cm × 1.5cm

l = 4 cm, b = 2.5 cm and h = 1.5 cm

Volume of one matchbox = (l × b × h)

^{}

= (4 × 2.5 × 1.5) cm

^{3 }= 15 cm

^{3}

Volume of a packet containing 12 such boxes = (12 × 15) cm

^{3 }= 180 cm

^{3}

2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1m

^{3}= 1000 l)

**Answer**

Dimensions of water tank = 6m × 5m × 4.5m

l = 6m , b = 5m and h = 4.5m

Therefore Volume of the tank =ℓbh m

^{3}

=(6×5×4.5)m3=135 m

^{3}

Therefore , the tank can hold = 135 × 1000 litres [Since 1m3=1000litres]

= 135000 litres of water.

3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

**Answer**

Length = 10 m , Breadth = 8 m and Volume = 380 m

^{3}

Volume of cuboid = Length × Breadth × Height

⇒ Height = Volume of cuboid/(Length × Breadth)

= 380/(10×8) m

= 4.75m

4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m

^{3}.

**Answer**

l = 8 m, b = 6 m and h = 3 m

Volume of the pit = lbh m

^{3}

= (8×6×3) m

^{3}

^{ }= 144 m

^{3}

Rate of digging = ₹30 per m

^{3}

Total cost of digging the pit = ₹(144 × 30) = ₹4320

5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

**Answer**

length = 2.5 m, depth = 10 m and volume = 50000 litres

1m

^{3}= 1000 litres

^{}

∴ 50000 litres = 50000/1000 m

^{3}= 50 m

^{3}

Breadth = Volume of cuboid/(Length×Depth)

= 50/(2.5×10) m

= 2 m

6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last?

**Answer**

Dimension of tank = 20m × 15m × 6m

l = 20 m , b = 15 m and h = 6 m

Capacity of the tank = lbh m

^{3}

= (20×15×6) m

^{3}

= 1800 m

^{3}

Water requirement per person per day =150 litres

Water required for 4000 person per day = (4000×150) l

= (4000×150)/1000

= 600 m

^{3}

Number of days the water will last = Capacity of tank Total water required per day

=(1800/600) = 3

The water will last for 3 days.

7. A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates each measuring 1.5m × 1.25m × 0.5m that can be stored in the godown.

**Answer**

Dimension of godown = 40 m × 25 m × 15 m

Volume of the godown = (40 × 25 × 15) m

^{3}= 10000 m

^{3}

Dimension of crates = 1.5m × 1.25m × 0.5m

Volume of 1 crates = (1.5 × 1.25 × 0.5) m

^{3}= 0.9375 m

^{3}

Number of crates that can be stored =Volume of the godown/Volume of 1 crate

= 10000/0.9375 = 10666.66 = 10666

8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

**Answer**

Edge of the cube = 12 cm.

Volume of the cube = (edge)

^{3}cm

^{3}

= (12 × 12 × 12) cm

^{3}

= 1728 cm

^{3}

Number of smaller cube = 8

Volume of the 1 smaller cube =1728/8 cm

^{3}= 216 cm

^{3}

Side of the smaller cube = a

a

^{3 }= 216

⇒ a = 6 cm

Surface area of the cube = 6 (side)

^{2}

Ratio of their surface area = (6 × 12 × 12)/(6 × 6 × 6)

= 4/1 = 4:1

9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

**Answer**

Depth of river (h) = 3 m

Width of river (b) = 40 m

Rate of flow of water (l) = 2 km per hour = (2000/60) m per minute

= 100/3 m per minute

Volume of water flowing into the sea in a minute = lbh m

^{3}

= (100/3 × 40 × 3) m

^{3}

= 4000 m

^{3}

Page No: 230

**Exercise 13.6**

1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm 3 = 1l)

**Answer**

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

Go to Part I (Exercise 3.1 to 3.3)

Go to Part III (Exercise 3.7 to 3.8)

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