# NCERT Solutions for Class 9th: Ch 13 Surface Areas and Volumes Maths (Part-3)

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

Page No: 233**Exercise 13.7**

1. Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm

**Answer**

(i) Radius (r) = 6 cm

Height (h) = 7 cm

Volume of the cone = 1/3 πr

^{2}h

= (1/3 × 22/7 × 6 × 6 × 7) cm

^{3}

= 264 cm

^{3}

(ii) Radius (r) = 3.5 cm

Height (h) = 12 cm

Volume of the cone = 1/3 πr

^{2}h

= (1/3 × 22/7 × 3.5 × 3.5 × 12) cm

^{3}

= 154 cm

^{3}

2. Find the capacity in litres of a conical vessel with

(i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm

**Answer**

(i) Radius (r) = 7 cm

Slant height (l) = 25 cm

Let h be the height of the conical vessel.

∴ h = √l

^{2 }- r

^{2}

⇒ h = √25

^{2 }- 7

^{2}

⇒ h = √576

⇒ h = 24 cm

Volume of the cone = 1/3 πr

^{2}h

= (1/3 × 22/7 × 7 × 7 × 24) cm

^{3}

= 1232 cm

^{3}

Capacity of the vessel = (1232/1000) l = 1.232 l

(i) Height (h) = 12 cm

Slant height (l) = 13 cm

Let r be the radius of the conical vessel.

∴ r = √l

^{2 }-

^{h}

^{2}

⇒ r = √13

^{2 }- 12

^{2}

⇒ r = √25

⇒ r = 5 cm

Volume of the cone = 1/3 πr

^{2}h

= (1/3 × 22/7 × 5 × 5 × 12) cm

^{3}

= (2200/7) cm

^{3}

Capacity of the vessel = (2200/7000) l = 11/35 l

3. The height of a cone is 15 cm. If its volume is 1570 cm

^{3}, find the radius of the base. (Use π = 3.14)

**Answer**

Height (h) = 15 cm

Volume = 1570 cm

^{3}

Let the radius of the base of cone be r cm

∴ Volume = 1570 cm

^{3}

⇒ 1/3 πr

^{2}h = 1570

⇒ 13 × 3.14 × r

^{2 }× 15 = 1570

⇒ r

^{2 }= 1570/(3.14×5) = 100

⇒ r = 10

4. If the volume of a right circular cone of height 9 cm is 48π cm

^{3}, find the diameter of its base.

**Answer**

Height (h) = 9 cm

Volume = 48π cm

^{3}

Let the radius of the base of the cone be r cm

∴ Volume = 48π cm

^{3}

⇒ 1/3 πr

^{2}h = 48π

⇒ 13 × r

^{2 }× 9 = 48

⇒ 3r

^{2 }= 48

⇒ r

^{2 }= 48/3 = 16

⇒ r = 4

5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

**Answer**

Diameter of the top of the conical pit = 3.5 m

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

Depth of the pit (h) = 12 m

Volume = 1/3 πr

^{2}h

= (13 × 22/7 × 1.75 × 1.75 × 12) m

^{3}

= 38.5 m

^{3}

1 m

^{3 }= 1 kilolitre

^{}

Capacity of pit = 38.5 kilolitres.

6. The volume of a right circular cone is 9856 cm

^{3}. If the diameter of the base is 28 cm, find

(i) height of the cone (ii) slant height of the cone

(iii) curved surface area of the cone

**Answer**

(i) Diameter of the base of the cone = 28 cm

Radius (r) = 28/2 cm = 14 cm

Let the height of the cone be h cm

Volume of the cone = 13πr

^{2}h = 9856 cm

^{3}

⇒ 1/3 πr

^{2}h = 9856

⇒ 1/3 × 22/7 × 14 × 14 × h = 9856

⇒ h = (9856×3)/(22/7 × 14 × 14)

⇒ h = 48 cm

(ii) Radius (r) = 14 m

Height (h) = 48 cm

Let l be the slant height of the cone

l

^{2}= h

^{2 }+ r

^{2}

⇒ l

^{2}= 48

^{2 }+ 14

^{2}

⇒ l

^{2}= 2304+196

⇒ l

^{2 }= 2500

⇒ ℓ = √2500 = 50 cm

(iii) Radius (r) = 14 m

Slant height (l) = 50 cm

Curved surface area = πrl

= (22/7 × 14 × 50) cm

^{2}

= 2200 cm

^{2}

7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

**Answer**

On revolving the ⃤ ABC along the side 12 cm, a right circular cone of height(h) 12 cm, radius(r) 5 cm and slant height(l) 13 cm will be formed.

Volume of solid so obtained = 1/3 πr

^{2}h

= (1/3 × π × 5 × 5 × 12) cm

^{3}

= 100π cm

^{3}

8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

**Answer**

On revolving the ⃤ ABC along the side 12 cm, a cone of radius(r) 12 cm, height(h) 5 cm, and slant height(l) 13 cm will be formed.

Volume of solid so obtained =1/3 πr

^{2}h

= (1/3 × π × 12 × 12 × 5) cm

^{3}

= 240π cm

^{3}

Ratio of the volumes = 100π/240π = 5/12 = 5:12

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

**Answer**

Diameter of the base of the cone = 10.5 m

Radius (r) = 10.5/2 m = 5.25 m

Height of the cone = 3 m

Volume of the heap = 1/3 πr

^{2}h

= (1/3 × 22/7 × 5.25 × 5.25 × 3) m

^{3}

= 86.625 m

^{3}

Also,

l

^{2}= h

^{2 }+ r

^{2}

⇒ l

^{2}= 3

^{2 }+ (5.25)

^{2}

⇒ l

^{2}= 9 + 27.5625

⇒ l

^{2 }= 36.5625

⇒ l = √36.5625 = 6.05 m

Area of canvas = Curve surface area

= πrl = (22/7 × 5.25 × 6.05) m

^{2}

= 99.825 m

^{2 }(approx)

Page No: 236

**Exercise 13.8**

1. Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m

**Answer**

(i) Radius of the sphere(r) = 7 cm

Therefore, Volume of the sphere = 4/3 πr

^{3}

= (4/3 × 22/7 × 7 × 7 × 7) cm

^{3}

= 4312/3 cm

^{3}

(ii) Radius of the sphere(r) = 0.63 m

Volume of the sphere = 4/3 πr

^{3}

= (4/3 × 22/7 × 0.63 × 0.63 × 0.63) m

^{3}

= 1.05 m

^{3}

2. Find the amount of water displaced by a solid spherical ball of diameter.

(i) 28 cm (ii) 0.21 m

**Answer**

(i) Diameter of the spherical ball = 28 cm

Radius = 28/2 cm = 14 cm

Amount of water displaced by the spherical ball = Volume

= 4/3 πr

^{3}

= (4/3 × 22/7 × 14 × 14 × 14) cm

^{3}

= 34496/3 cm

^{3}

(ii) Diameter of the spherical ball = 0.21 m

Radius (r) = 0.21/2 m = 0.105 m

Amount of water displaced by the spherical ball = Volume

= 4/3 πr

^{3}

= (43×227×0.105×0.105×0.105) m

^{3}

= 0.004851 m

^{3}

3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm

^{3}?

**Answer**

Diameter of the ball = 4.2 cm

Radius = (4.2/2) cm = 2.1 cm

Volume of the ball = 4/3 πr

^{3}

= (4/3 × 22/7 × 2.1 × 2.1 × 2.1) cm

^{3}

= 38.808 cm

^{3}

Density of the metal is 8.9g per cm3

Mass of the ball = (38.808 × 8.9) g = 345.3912 g

4. The diameter of the moon is approximately one-fourth of the diameter of the earth.What fraction of the volume of the earth is the volume of the moon?

**Answer**

Let the diameter of the moon be r

Radius of the moon = r/2

A/q,

Diameter of the earth = 4r

Radius(r) = 4r/2 = 2r

Volume of the moon = v = 4/3 π(r/2)

^{3}

= 4/3 πr

^{3 }× 1/8

⇒ 8v = 4/3 πr

^{3 }--- (i)

Volume of the earth = r

^{3}= 4/3 π(2r)

^{3}

= 4/3 πr

^{3}× 8

⇒ V/8 = 4/3 πr

^{3}--- (ii)

From (i) and (ii), we have

8v = V/8

⇒ v = 1/64 V

Thus, the volume of the moon is 1/64 of the volume of the earth.

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

**Answer**

Diameter of a hemispherical bowl = 10.5 cm

Radius(r) = (10.5/2) cm = 5.25cm

Volume of the bowl = 2/3 πr

^{3}

= (2/3 × 22/7 × 5.25 × 5.25 × 5.25) cm

^{3}

= 303.1875 cm

^{3}

Litres of milk bowl can hold = (303.1875/1000) litres

= 0.3031875 litres (approx.)

6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

**Answer**

Internal radius = r = 1m

External radius = R = (1 + 0.1) cm = 1.01 cm

Volume of iron used = External volume – Internal volume

= 2/3 πR

^{3 }- 2/3 πr

^{3}

= 2/3 π(R

^{3 }- r

^{3})

= 2/3 × 22/7 × [(1.01)

^{3}−(1)

^{3}] m

^{3}

= 44/21 × (1.030301 - 1) m

^{3}

= (44/21 × 0.030301) m

^{3}

= 0.06348 m

^{3}(approx)

7. Find the volume of a sphere whose surface area is 154 cm

^{2}.

**Answer**

Let r cm be the radius of the sphere

So, surface area = 154cm

^{2}

⇒ 4πr

^{2 }= 154

⇒ 4 × 22/7 × r

^{2 }= 154

⇒ r

^{2 }= (154×7)/(4×22) = 12.25

⇒ r = 3.5 cm

Volume = 4/3 πr

^{3}

= (4/3 × 22/7 × 3.5 × 3.5 × 3.5) cm

^{3}

= 539/3 cm

^{3}

8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹498.96. If the cost of white-washing is ₹2.00 per square metre, find the

(i) inside surface area of the dome, (ii) volume of the air inside the dome.

**Answer**

(i) Inside surface area of the dome =Total cost of white washing/Rate of white washing

= (498.96/2.00) m

^{2 }= 249.48 m

^{2 }

(ii) Let r be the radius of the dome.

Surface area = 2πr

^{2}

^{}

⇒ 2 × 22/7 × r

^{2 }= 249.48

⇒ r

^{2 }= (249.48×7)/(2×22) = 39.69

⇒ r

^{2}= 39.69

⇒ r = 6.3m

Volume of the air inside the dome = Volume of the dome

= 2/3 πr

^{3}

^{ }= (2/3 × 22/7 × 6.3 × 6.3 × 6.3) m

^{3}

= 523.9 m

^{3 }(approx.)

9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the

(i) radius r′ of the new sphere, (ii) ratio of S and S′.

**Answer**

(i) Volume of 27 solid sphere of radius r = 27 × 4/3 πr

^{3}--- (i)

Volume of the new sphere of radius r′ = 4/3 πr'

^{3}--- (ii)

A/q,

4/3 πr'

^{3}= 27 × 4/3 πr

^{3}

⇒ r'

^{3 }= 27r

^{3}

⇒ r'

^{3 }= (3r)

^{3}

⇒ r′ = 3r

(ii) Required ratio = S/S′ = 4 πr

^{2}/4πr′

^{2 }= r

^{2}/(3r)

^{2}

= r

^{2}/9r

^{2}= 1/9 = 1:9

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm

^{3}) is needed to fill this capsule?

**Answer**

Diameter of the spherical capsule = 3.5 mm

Radius(r) = 3.52mm

= 1.75mm

Medicine needed for its filling = Volume of spherical capsule

= 4/3 πr

^{3}

= (4/3 × 22/7 × 1.75 × 1.75 × 1.75) mm

^{3}

= 22.46 mm

^{3 }(approx.)

Go to Part I (Exercise 3.1 to 3.3)

Go to Part II (Exercise 3.4 to 3.6)

**Go To Chapters**