0
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 = 4πr2
                    = (4 × 22/7 × 10.5 × 10.5) cm2
                    = 1386 cm2

(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr2
                     = (4 × 22/7 × 5.6 × 5.6) cm2
                     = 394.24 cm2

(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr2
                     = (4 × 22/7 × 14 × 14) cm2
                     = 2464 cm2

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 = 4πr2
                         = (4 × 22/7 × 7 × 7)cm2
                                                =616cm2

(ii) r = 21/2 cm = 10.5 cm
Surface area = 4πr2
                     = (4 × 22/7 × 10.5 × 10.5) cm2
                     = 1386 cm2

(iii) r = 3.5/2 m = 1.75 m
Surface area = 4πr2 
                    = (4 × 22/7 × 1.75 × 1.75) m2
                    = 38.5 m2

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 = 3πr2
                                                     = (3 × 3.14 × 10 ×10) cm2
                                                     = 942 cm2

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 =4πr2/4πR2
                                        = r2/R2
                                        = (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 cm2.
Answer

Radius of the bowl (r) = 10.5/2 cm = 5.25 cm
Curved surface area of the hemispherical bowl = 2πr2
                                                                        = (2 × 22/7 × 5.25 × 5.25) cm2
                                                           = 173.25 cm2
Rate of tin - plating is = ₹16 per 100 cm2
Therefor, cost of 1 cm= ₹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 cm2.Answer

Let r be the radius of the sphere.
Surface area = 154 cm2
⇒ 4πr= 154
⇒ 4 × 22/7 × r= 154
⇒ r= 154/(4 × 22/7)
⇒ r= 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 = 2πR2
                                  = (2 × 22/7 × 5.25 × 5.25) cm2
                                  = 173.25 cm2

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 = 4πr2

(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
                                               = 4πr2
(iii) Ratio of the areas = 4πr2:4πr2 = 1:1

Exercise 13.51. 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= 15 cm3
Volume of a packet containing 12 such boxes = (12 × 15)  cm= 180 cm3

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? (1m3 = 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 m3
                                                   =(6×5×4.5)m3=135 m3
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 m3
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 m3.
Answer

l = 8 m, b = 6 m and h = 3 m
Volume of the pit = lbh m3
                             = (8×6×3) m3
                                    = 144 m3
Rate of digging = ₹30 per m3
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
1m3 = 1000 litres

∴ 50000 litres = 50000/1000 m3 = 50 m3
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 m3
                                 = (20×15×6) m3
                                 = 1800 m3
Water requirement per person per day =150 litres
Water required for 4000 person per day = (4000×150) l
                                                                = (4000×150)/1000
                                                                = 600 m3
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) m3 = 10000 m3
Dimension of crates = 1.5m × 1.25m × 0.5m 
Volume of 1 crates = (1.5 × 1.25 × 0.5) m3 = 0.9375 m3
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 cm3
                                = (12 × 12 × 12) cm3
                                = 1728 cm3 
Number of smaller cube = 8
Volume of the 1 smaller cube =1728/8 cm3 = 216 cm3
Side of the smaller cube = a
a= 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 m3
                                                                               = (100/3 × 40 × 3) m3
                                                                               = 4000 m3

Exercise 13.61. 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

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 = √l- r2
⇒ h = √25- 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 = √lh2
⇒ r = √13- 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 × r× 15 = 1570
⇒ r= 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 × r× 9 = 48
⇒ 3r= 48
⇒ r= 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 m= 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 = h+ r2
⇒ l2 = 48+ 142
⇒ l2 = 2304+196
⇒ l= 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 = h+ r2
⇒ l2 = 3+ (5.25)2
⇒ l2 = 9 + 27.5625
⇒ l= 36.5625
⇒ l = √36.5625 = 6.05 m
Area of canvas = Curve surface area
                         = πrl = (22/7 × 5.25 × 6.05) m2
                         = 99.825 m(approx)

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 πr× 1/8
⇒ 8v = 4/3 πr--- (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 πR- 2/3 πr3
                                  = 2/3 π(R- 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
⇒ 4πr= 154
⇒ 4 × 22/7 × r= 154
⇒ r= (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) m= 249.48 m

 (ii) Let r be the radius of the dome.
Surface area = 2πr2
⇒ 2 × 22/7 × r= 249.48
⇒ r= (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 m(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'= 27r3
⇒ r'3 = (3r)3
⇒ r′ = 3r
(ii) Required ratio = S/S′ = 4 πr2/4πr′= 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 mm(approx.)

Post a Comment

 
Top