NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes (Ex 13.6 - Ex 13.8)

NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes (Ex 13.6 - Ex 13.8)

Page No: 230

Exercise 13.6

Q.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 $c{m}^3$ = 1l )

Ans : Let the radius Of the cylindrical vessel be r. 
Height (h) of vessel = 25 cm 
Circumference of vessel = 132 cm 
 $2\pi r$= 132 cm 
$r=\left(\frac{132\times 7}{2\times 22}\right)cm=21cm$
Volume of cylindrical vessel=${pir}^{2}h$
$=[\frac{22}{7}\times (21{)}^2\times 25]{cm}^3$
=34650${cm}^3$
$=\left(\frac{34650}{1000}\right)$liter
=34.65 liter
Therefore, such vessel can hold 34.65 litres Of water. 

Q.2: The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 $c{m}^3$ of wood has a mass of 0.6 g.

Ans : Inner radius of cylindrical pipe =$\left(\frac{28}{2}\right)cm=14cm$
Outer radius of cylindrical pipe =$\left(\frac{28}{2}\right)cm=14cm$

Height (h) of pipe = Length of pipe = 35 cm 
Volume of pipe =$\pi (r_2^2-r_1^2)h$
$=[\frac{22}{7}\times ({14}^2-{12}^2)\times 35]{cm}^3$
=110x52 ${cm}^3$
=5720  ${cm}^3$
Mass of 1  ${cm}^3$ wood= 0.6 g
Mass of 5720  ${cm}^3$ wood = (5720 x0.6) g
=3432 g
=3.432 kg 

Q.3: A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Ans : The tin can will be cuboidal in shape while the plastic cylinder Hill be cylindrical in shape. 


Length (l) of tin can = 5 cm 
Breadth (b) of tin can = 4 cm 
Height (h) Of tin can = 15 cm 
Capacity of tin can l x b x h 
(5 x 4 x 15) ${cm}^3$
=300 ${cm}^3$

Radius (r) of circular end of plastic cylinder $=\left(\frac{7}{2}\right)cm=3.5cm$
Height (H) of plastic cylinder = 10 cm 
Capacity of plastic cylinder=$\pi r^{2}H$
$[\frac{22}{7}\times (3.5{)}^2\times 10]$ ${cm}^3$
=(11x35) ${cm}^3$
Therefore, plastic cylinder has the greater capacity. 
Difference in capacity = (385 — 300)  ${cm}^3$
= 85  ${cm}^3$ 

Q.4: If the lateral surface of a cylinder is 94.2 $c{m}^2$ and its height is 5 cm, then find (i) radius of its base (ii) its volume. (Use π = 3.14)

Ans : (i) Height (h) of cylinder =5 cm 
Let radius of cylinder be r. 
CSA of cylinder = 94.2 ${cm}^2$
$2\pi rh$= 94.2 ${cm}^2$
(2 x 3.14 x r x 5) cm =94.2 ${cm}^2$
r = 3 cm 
(ii) Volume Of cylinder =$\pi r^{2}H$
 =$=(3.14\times (3{)}^2\times 5){cm}^3$
= 141.3 ${cm}^3$ 

Q.5: It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per $m^2$ , find 
(i) inner curved surface area of the vessel, 
(ii) radius of the base, 
(iii) capacity of the vessel.

Ans : (i) Rs 20 is the cost Of painting 1$m^2$: area. 
RS 2200 is the cost Of painting =$(\frac{1}{20}\times 2200)m^2$ 
= 110$m^2$ area 
Therefore, the inner surface area of the vessel is 110$m^2$
(ii) Let the radius of the base of the vessel be r. 
Height (h) Of vessel = 10 m 
Surface area $2\pi rh$ = 110 $m^2$
$\Rightarrow (2\times \frac{22}{7}\times r\times 10)m=110m^2$
$\Rightarrow r=\left(\frac{7}{4}\right)m=1.75m$
(iii) volume of vessel =$\pi r^{2}H$
$=[\frac{22}{7}\times (1.75{)}^2\times 10]m^3$
$=96.25m^3$
Therefore, the capacity of the vessel is 96.25 $m^3$ or 96250 litres 

Q.6: The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Ans : Let the radius of the circular end be r. 
Height (h) Of cylindrical vessel =1 m 
Volume of cylindrical vessel = 15.4 litres = 0.0154 $m^3$
$\pi r^{2}h=0.0154m^3$
r=0.07m
Total surface area of vessel = $2\pi r(r+h)$
$=[2\times \frac{22}{7}\times 0.07(0.07+1\left)\right]m^2$
$=0.44\times 1.07m^2$
$=0.4708m^2$
Therefore, 0.4708 (\mathrm{m}^{2}\) of the metal sheet would be required to make the cylindrical vessel. 

Q.7: A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Ans : 
Radius (r) of pencil =$\left(\frac{7}{2}\right)mm=\left(\frac{0.7}{2}\right)cm$=0.35cm
Radius of graphite = $\left(\frac{1}{2}\right)mm=\left(\frac{0.1}{2}\right)cm$=0.05
Height (h) of pencil = 14cm
Volume of wood in pencil =$\pi (r_1^2-r_2^2)h$v
$=\left[\frac{22}{7}\left\{\right(0.35{)}^2-(0.05{)}^2\times 14\}\right]{cm}^3$
$=\left[\frac{22}{7}\right(0.1225-0.0025)\times 14]{cm}^3$
$=(44\times 0.12){cm}^3$
$=5.28{cm}^3$
$=\pi r_2^{2}h=[\frac{22}{7}\times (0.05{)}^2\times 14]{cm}^3$
$=(44\times 0.0025)c{m}^3$
$=0.11{cm}^3$ 

Q.8: A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Ans : 
Radius (r) of cylindrical bowl =$\left(\frac{7}{2}\right)cm=3.5cm$
Height (h) of bowl, up to which bowl is filled with soup = 4 cm 
Volume Of soup in I bowl = $\pi r^{2}H$
$=(\frac{22}{7}\times (3.5{)}^2\times 4){cm}^3$
$=(11\times 3.5\times 4){cm}^3$
$=154{cm}^3$
Volume of soup given to 250 patients =(250x154)${cm}^3$
$=38500{cm}^3$
$=38.5$

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 πr²h
                                = (1/3 × 22/7 × 6 × 6 × 7) cm³
                                = 264 cm³
(ii) Radius (r) = 3.5 cm
Height (h) = 12 cm
Volume of the cone = 1/3 πr²h
                                = (1/3 × 22/7 × 3.5 × 3.5 × 12) cm³
                                = 154 cm³

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 = √l² - r² 
⇒ h = √25² - 7² 
⇒ h = √576 
⇒ h = 24 cm
Volume of the cone = 1/3 πr²h
                                = (1/3 × 22/7 × 7 × 7 × 24) cm³
                                = 1232 cm³ 
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 = √l² - ^h2 
⇒ r = √13² - 12² 
⇒ r = √25 
⇒ r = 5 cm
Volume of the cone = 1/3 πr²h
                                = (1/3 × 22/7 × 5 × 5 × 12) cm³
                                = (2200/7) cm³ 
Capacity of the vessel = (2200/7000) l = 11/35 l

3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Answer

Height (h) = 15 cm
Volume = 1570 cm³
Let the radius of the base of cone be r cm
∴ Volume = 1570 cm³
⇒ 1/3 πr²h = 1570
⇒ 13 × 3.14 × r² × 15 = 1570
⇒ r² = 1570/(3.14×5) = 100
⇒ r = 10

4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

Answer

Height (h) = 9 cm
Volume = 48π cm³
Let the radius of the base of the cone be r cm
∴ Volume = 48π cm³
⇒ 1/3 πr²h = 48π
⇒ 13 × r² × 9 = 48
⇒ 3r² = 48
⇒ r² = 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 πr²h
             = (13 × 22/7 × 1.75 × 1.75 × 12) m³
             = 38.5 m³
1 m³ = 1 kilolitre

Capacity of pit = 38.5 kilolitres.

6. The volume of a right circular cone is 9856 cm³. 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 = 13πr²h = 9856 cm³ 
⇒ 1/3 πr²h = 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
l² = h² + r²
⇒ l² = 48² + 14²
⇒ l² = 2304+196
⇒ l² = 2500
⇒ ℓ = √2500 = 50 cm

(iii) Radius (r) = 14 m
Slant height (l) = 50 cm
Curved surface area = πrl
                                 = (22/7 × 14 × 50) cm²

                                 = 2200 cm²

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 πr²h
                                             = (1/3 × π × 5 × 5 × 12) cm³
                                             = 100π cm³

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 πr²h
                                              = (1/3 × π × 12 × 12 × 5) cm³
                                              = 240π cm³
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 πr²h
                                = (1/3 × 22/7 × 5.25 × 5.25 × 3) m³
                                = 86.625 m³
Also,
l² = h² + r²
⇒ l² = 3² + (5.25)²
⇒ l² = 9 + 27.5625
⇒ l² = 36.5625
⇒ l = √36.5625 = 6.05 m
Area of canvas = Curve surface area
                         = πrl = (22/7 × 5.25 × 6.05) m²
                         = 99.825 m² (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 πr³
                                                     = (4/3 × 22/7 × 7 × 7 × 7) cm³
                                                     = 4312/3 cm³

(ii) Radius of the sphere(r) = 0.63 m
Volume of the sphere = 4/3 πr³
                                   = (4/3 × 22/7 × 0.63 × 0.63 × 0.63) m³
                                   = 1.05 m³

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 πr³
                              = (4/3 × 22/7 × 14 × 14 × 14) cm³
                              = 34496/3 cm³

(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 πr³
                              = (43×227×0.105×0.105×0.105) m³
                              = 0.004851 m³

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 cm³?

Answer

Diameter of the ball = 4.2 cm
Radius = (4.2/2) cm = 2.1 cm
Volume of the ball = 4/3 πr³
                               = (4/3 × 22/7 × 2.1 × 2.1 × 2.1) cm³
                               = 38.808 cm³
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 πr³ × 1/8
⇒ 8v = 4/3 πr³ --- (i)
Volume of the earth = r³ = 4/3 π(2r)³
                                = 4/3 πr³× 8
⇒ V/8 = 4/3 πr³ --- (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 πr³
                                = (2/3 × 22/7 × 5.25 × 5.25 × 5.25) cm³
                                = 303.1875 cm³
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 πR³ - 2/3 πr³
                                  = 2/3 π(R³ - r³)
                                  = 2/3 × 22/7 × [(1.01)³−(1)³] m³
                                  = 44/21 × (1.030301 - 1) m³
                                  = (44/21 × 0.030301) m³
                                  = 0.06348 m³(approx)

7. Find the volume of a sphere whose surface area is 154 cm².

Answer

Let r cm be the radius of the sphere
So, surface area = 154cm²
⇒ 4πr² = 154
⇒ 4 × 22/7 × r² = 154
⇒ r² = (154×7)/(4×22) = 12.25
⇒ r = 3.5 cm
Volume = 4/3 πr³
             = (4/3 × 22/7 × 3.5 × 3.5 × 3.5) cm³
             = 539/3 cm³

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) m² = 249.48 m² 

 (ii) Let r be the radius of the dome.
Surface area = 2πr²
⇒ 2 × 22/7 × r² = 249.48
⇒ r² = (249.48×7)/(2×22) = 39.69
⇒ r²=  39.69
⇒ r = 6.3m
Volume of the air inside the dome = Volume of the dome
                                                       = 2/3 πr³
                                                                    = (2/3 × 22/7 × 6.3 × 6.3 × 6.3) m³
                                                       = 523.9 m³ (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 πr³ --- (i)
Volume of the new sphere of radius r′ = 4/3 πr'³ --- (ii)
A/q,
4/3 πr'³= 27 × 4/3 πr³
⇒ r'³ = 27r³
⇒ r'³ = (3r)³
⇒ r′ = 3r
(ii) Required ratio = S/S′ = 4 πr²/4πr′² = r²/(3r)²
                              = r²/9r² = 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 mm³) 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 πr³
                                                  = (4/3 × 22/7 × 1.75 × 1.75 × 1.75) mm³
                                                  = 22.46 mm³ (approx.)