EXERCISE - 13.1
1. A
plastic box 1.5 m long, 1.25 m wide and 65 cm deep, is to be made. It is to be
open at the top. Ignoring the thickness of the plastic sheet, determine:
(i)
The area of the sheet required for making the box.
(ii)
The cost of sheet for it, if a sheet measuring 1 m2 costs Rs
20.
Answer:
It is given that, length (l) of box =
1.5 m
Breadth (b) of box = 1.25 m
Depth (h) of box = 0.65 m
since the box is open at the top, so area = 2lh + 2bh + lb
= 2 × 1.5 × 0.65 + 2 × 1.25 × 0.65 + 1.5 ×
1.25 m2
= (1.95 + 1.625 + 1.875) m2 =
5.45 m2
(ii) Cost of 1m2 of
sheet = Rs. 20
Cost of sheet of 5.45 m2 area
= Rs (5.45 × 20)
= Rs 109
2. The length, breadth and height of a room are 5
m, 4 m and 3 m respectively. Find the cost of white washing the walls of the
room and the ceiling at the rate of Rs 7.50 per m2.
Answer:
It
is given that
Length
(l) = 5 m
Breadth
(b) = 4 m
Height
(h) = 3 m
∴ Area for white washing = [Lateal surface
area] + [Area of the ceiling]
=
[2h(1 + b)] + [1 × b]
=
[2x3(5 + 4)] + [5 × 4]
=
54m2 + 20m2 = 74m2
Cost
of white washing:
Cost
of white washing for 1 m2 = Rs. 7.50
∴
Cost of white washing for 74 m2 = Rs. 7.50 × 74
The
required cost of white washing is Rs 555.
3.The floor of a rectangular hall has a perimeter
250 m. If the cost of panting the four walls at the rate of Rs.10 per m2 is
Rs.15000, find the height of the hall.
[Hint: Area
of the four walls = Lateral surface area.]
Answer:
Cost
of painting the four walls = Rs 15000
Rate
of painting the four walls = Rs 10 / m2
Area
of four walls = 1500/10m2
= 1500 m2
Area
of four walls = 2(I + b) h (or perimeter
of base x h)
According
to the question,
∴ 250 x h = 1500
∴
h= 1500/25 =
6
∴height
of the hall = 6m
4.The paint in a certain container is sufficient
to paint an area equal to 9.375 m2. How many bricks of dimensions
22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Answer:
∴
Total surface area of a brick = 2[lb + bh + hl]
=
2[(22.5 × 10) + (10 × 7.5) + (7.5 × 22.5)] cm2
=
2[(225) + (75) + (168.75)] cm2
= 2x 468.75 = 937.5 cm2
9.375
m2 = 93750 cm2
Number
of bricks that can be painted out of the
given container = 
=
100 bricks.
5. A cubical box has each edge 10 cm and another
cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface
area and by how much?
(ii) Which box has the smaller total surface area
and by how much?
Answer:
(i)
For the cubical box
∵Edge
of the cubical box = 10 cm
∴
Lateral surface area = 4a2
=
4 × 102 cm2
=
4 × 100 cm2
=
400 cm2
For
the cuboidal box, l = 12.5 cm, b = 10 cm, h = 8 cm
∴
Lateral surface area = 2h(l + b)
=
2x 8 (12.5 + 10) cm2
=
16 x 22.5 cm2
=
360 cm2
Clearly,
the lateral surface area of the cubical box is greater than the lateral surface
area of the cuboidal box.
Lateral
surface area of cubical box − Lateral surface area of cuboidal box = 400 cm2 −
360 cm2 = 40 cm2
(ii) Total surface area = 6a2
=
6 × 102 = 6 × 100 cm2
=
600 cm2
Total
surface area = 2[lb + bh + hl]
=
2[(12.5 × 10) + (10 × 8) + (8 × 12.5)] cm2
=
2[125 + 80 + 100] cm2
=
2[305] cm2 = 610 cm2
Clearly,
the total surface area of the cubical box is smaller than that of the cuboidal
box.
Total
surface area of cuboidal box − Total surface area of cubical box = 610 cm2 −
600 cm2 = 10 cm2.
6.A small indoor greenhouse (herbarium) is made
entirely of glass panes (including base) held together with tape. It is 30 cm
long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12
edges?
Answer:
The
herbarium is like a cuboid
Here,
l = 30 cm, b = 25 cm, h = 25 cm
(i)
∵ Area of a cuboid = 2[lb + bh + hl]
∴
Surface area of the herbarium (glass) = 2[(30 × 25) + (25 × 25) + (25 × 30)] cm2
=
2[750 + 625 + 750] cm2
=
2[2125] cm2
=
4250 cm2
Thus,
the required area of glass is 4250 cm2.
(ii)
Total length of 12 edges = 4l + 4b + 4h
=
4(1 + b + h) = 4(30 + 25 + 25) cm
=
4 × 80 cm = 320 cm
Thus,
length of tape needed = 320 cm
7.Shanti Sweets Stall was placing an order for
making cardboard boxes for packing their sweets. Two sizes of boxes were
required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of
dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface
area is required extra. If the cost of the cardboard is Rs 4 for 1000 cm2,
find the cost of cardboard required for supplying 250 boxes of each kind.
Answer:
For bigger box:
Length
(1) = 25 cm, Breadth (b) = 20 cm, Height (h) = 5 cm
∵
The box is like a cuboid and total surface area of a cuboid = 2(lb + bh + hl)
Area
of a box = 2([25 × 20) + (20 × 5) + (5 × 25)] cm2
=
2[500 + 100 + 125] cm2
=
2[725] cm2 = 1450 cm2
Total
surface area of 250 boxes = 250 × 1450 cm2 = 362500 cm2
For
smaller box:
l
= 15 cm, b = 12 cm, h = 5 cm
Total
surface area of a box = 2[lb + bh + hl]
=
2[(15 × 12) + (12 × 5) + (5 × 15)] cm2
=
2[180 + 60 + 75] cm2
=
2[315] cm2 = 630 cm2
⇒
Total surface area of 250 boxes = 250 × 630 cm2
=
157500 cm2
Now,
total surface area of both kinds of boxes
=
362500 cm2 + 157500 cm2
=
5,20,000 cm2
Area
for overlaps = 5% of [total surface area]
= 
x
520000
= 26000 cm2
∴
Total area of the cardboard required = [Total area of 250 boxes] + [5% of total
surface area]
=
520000 cm2 + 26000 cm2
=
546000 cm2
Cost
of cardboard:
∵
Cost of 1000 cm2 = Rs. 4
∴ cost of 1 cm2 = 
∴ Cost of 546000 cm2 = x
546000
= 4 x 546
= 2184
Therefore
the cost of cardboard = Rs 2184
8. Parveen wanted to make a temporary shelter for
her car, by making a box-like structure with tarpaulin that covers all the four
sides and the top of the car (with the front face as a flap which can be rolled
up). Assuming that the stitching margins are very small, and therefore
negligible, how much tarpaulin would be required to make the shelter of height
2.5 m, with base dimensions 4 m × 3 m?
Answer:
l=4
m , b = 3 m , h = 2.5 m
Therefore Tarpaulin
required =
2(lh + bh) + l b
=
[2(4 × 2.5 + 3 × 2.5) + 4 × 3] m2
=
[2(10 + 7.5) + 12] m2
=
47 m2
Therefore,
47 m2 tarpaulin required.
EXERCISE 13.2
1. The curved surface area of a right
circular cylinder of height 14 cm is 88 cm2. Find the diameter of
the base of the cylinder.
Answer:
Let r be the radius of the base and h = 14 cm be the
height of the cylinder.
Then, curved surface area of cylinder =2 πrh
According to the question, 2Ï€rh =
88
Therefore Diameter of the base = 2r =
2 x 1 = 2 cm
2. It
is required to make a closed cylindrical tank of height 1 m and base diameter
140 cm from a metal sheet. How many square meters of the sheet are required for
the same?
Answer:
Here
h = 1 m
Diameter
= 140 cm, so radius = 70 cm = 0.7 m
TSA
= 2 π r(r + h)
= 2 x 22 x 0.1 x 1.7
= 7.48 m2
Required
area of the sheet =7.48 m2
3. A metal pipe is 77 cm long. The inner diameter
of a cross section is 4 cm, the outer diameter being 4.4 cm.
(i) Inner curved surface area,
(ii) Outer curved surface area,
(iii) Total surface area.
Answer:
(i)Inner diameter = 4 cm, inner radius = 2 cm
Height = 77 cm
So inner curved surface
area = 2Ï€rh
=2 x
x 2 x 77
= 4 x 22 x 11 = 968 cm2
(ii we have outer
diameter D= 4.4 cm, so outer radius R =
2.2 cm
Outer curved surface area
= 2 πRh
= 2 x
x 2.2 x 77
= 4.4 x 22 x 11 =
1064.8 cm2
4. The diameter of a
roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to
move once over to level a playground. Find the area of the playground in m2?
Answer:
Height of cylindrical roller, h = 120 cm = 1.2 m
Diameter = 84 cm, Radius r = 42cm= 0.42
CSA of roller = 2Ï€rh
= 2
x 22 x0.0 6 x 1.20
= 3.168 m2
Area of field = 500 × CSA of roller
= (500 × 3.168) m2
= 1584m2
The area of the playground = 1584m2
5. A cylindrical pillar is 50 cm in diameter and
3.5 m in height. Find the cost of painting the curved surface of the pillar at
the rate of Rs.12.50 per m2.
Answer:
Height ,h = 3.5 m
Radius,r =
= 25cm =
0.25 m
CSA of pillar = 2Ï€rh
= 2 x 22 x 0.25 x
0.5
= 5.5 m2
Cost of painting 1 m2 area
= Rs 12.50
Cost of painting 5.5 m2 area
= Rs (5.5 × 12.50)
= Rs 68.75
6. Curved surface area of a right circular
cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7
m, find its height.
Answer:
Let the height of the circular cylinder be h.
Radius (r) of the base of cylinder = 0.7 m
CSA of cylinder = 4.4 m2
2Ï€rh = 4.4 m2
2 x 22/7 x
0.7x h = 4.4
h = 1 m
Therefore, the height of the cylinder is 1 m.
7. The inner diameter of a circular well is 3.5
m. It is 10 m deep. Find
(i) Its inner curved surface area,
(ii) The cost of plastering this curved surface
at the rate of Rs 40 per m2.
Answer:
Inner radius of circular well,
r =3.5/2
Height of circular
well, h = 10 m
Inner curved surface area = 2Ï€rh
= (22 × 0.5 × 10) m2
= 110 m2
Therefore, the inner curved surface area of the circular
well is 110 m2.
Cost of plastering 1 m2 area = Rs 40
Cost of plastering 110 m2 area = Rs (110 ×
40)
= Rs 4400
Therefore, the cost of plastering = Rs 4400.
8.In a hot water heating system, there is a
cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating
surface in the system.
Answer:
Height of cylinder = 28 m
Radius, r = 2.5 cm = 0.025 m
CSA of cylindrical pipe = 2Ï€rh
= 4.4 m2
The area of the radiating surface of the system is 4.4 m2.
9. Find
(i) The lateral or curved surface area of a
closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m
high.
(ii) How much steel was actually used, if 
of the steel actually used
was wasted in making the tank.
Answer:
Height h = 4.5 m
Radius , r = 2.1 m
(i) Lateral or curved surface area
of tank = 2Ï€rh
= 2 x 
x 2.1 x 4.5
= 44 × 0.3 × 4.5
= 59.4 m2
Therefore, CSA of tank is 59.4 m2.
(ii) Total surface area of tank = 2Ï€r (r
+ h)
= 2 x x 2.1
x (2.1 + 4.5)
= 44 × 0.3 × 6.6 m2
=
87.12 m2
let x be the
actual area of the steel used.
since 
of the actual sheet was wasted, so 
of the total sheet used to make the tank.
x = 95.04
Therefore the total area of the sheet that was actually used 95.04 m2
10. In the given figure, you see the frame of a
lampshade. It is to be covered with a decorative cloth. The frame has a base
diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for
folding it over the top and bottom of the frame. Find how much cloth is
required for covering the lampshade.
Answer:
The lampshade is in the form of a cylinder, where
radius
= 10 cm
∵A
margin of 2.5 cm is to be added to top and bottom.
∴
Total height of the cylinder
h
= 30 cm + 2.5 cm + 2.5 cm
=
35 cm
Now,
curved surface area = 2Ï€rh
2 x x 35 x 10
=
2 × 22 × 10 × 5 cm2 =2200
cm2
Thus,
the required area of the cloth = 2200 cm2
11.The students of a Vidyalaya were asked to
participate in a competition for making and decorating penholders in the shape
of a cylinder with a base, using cardboard. Each penholder was to be of radius
3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with
cardboard. If there were 35 competitors, how much cardboard was required to be
bought for the competition?
Answer:
Radius, r = 3 cm
Height, h = 10.5 cm
Surface area of 1 penholder = CSA of
penholder + Area of base of penholder
= 2πrh + πr2
= 2 x x 3 x 10.5 + x 3 x 3
= 44 x 3 x 1.5 +
= 198 +
=
cm2
Area of cardboard sheet used by 35
competitors
x 35 = 7920 cm2
EXERCISE 13.3
1. Diameter of the base of a cone is 10.5 cm and
its slant height is 10 cm. Find its curved surface area.
Answer:
Radius (r) of the base of
cone = 
= 5.25 cm
Slant height (l) of cone = 10
cm
CSA of cone = πrl
= 22/7 
x 5.25 x 10 = 165
Therefore, the curved surface area
of the cone is 165 cm2.
2.Find the total surface area of a cone, if its
slant height is 21 m and diameter of its base is 24 m.
Answer:
Radius (r) of the base of
cone = 12 m
Slant height (l) of cone = 21
m
Total surface area of cone = πr(r + l)
= x 12(12 + 21)
= x 12 x 33
= 1244.57 m3
3. Curved surface area of a cone is 308 cm2 and
its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface
area of the cone.
Answer:
(i) Slant height (l) of cone
= 14 cm
Let the radius of the circular end
of the cone be r.
Therefore πrℓ=308
x 14 x r = 308
r = 
= 7 cm
(ii) Total surface area of cone = CSA of
cone + Area of base
= πrl + πr2
= πr(l + r)
= x 7 (14 + 7)
=22 x 21 = 462 sq cm.
Therefore the total area of the cone = 462 sqcm.
4.A conical tent is 10 m high and the radius of its
base is 24 m. Find
(i) slant height of the tent
(ii) cost of the canvas required to make the
tent, if the cost of 1 m2 canvas is Rs 70.
Answer:
Height (h) of conical tent =
10 m
Radius (r) of conical tent =
24 m
Let the slant height of the tent be l
l2 = h2 + r2
= (10 m)2 + (24 m)2
= 676 m2
∴ l = 26 m
Therefore, the slant height of the
tent is 26 m.
(ii) CSA of tent = πrl
(ii) CSA of tent = πrl
Therefore, the cost of the canvas required to
make such a tent is Rs 137280.
5.What length of tarpaulin 3 m wide will be
required to make conical tent of height 8 m and base radius 6 m? Assume that
the extra length of material that will be required for stitching margins and
wastage in cutting is approximately 20 cm. [Use π = 3.14]
Answer:
Height (h) of conical tent =
8 m
Radius (r) of base of tent =
6 m
Slant height (l) of tent 
CSA of conical tent = πrl
= (3.14 × 6 × 10) m2
= 188.4 m2
width of the tarpaulin cloth = 3 m
Area of the tarpaulin sheet = CSA of the conical tent
length x breadth = 188.4
length of the tarpaulin = 188.4 / 3 = 62.8 m
The extra material required for stitching margins and cutting is 20 cm = 0.2 m
Total length of tarpaulin that will be required = (62.8 + 0.2) = 63 m
6. The slant height and base diameter of a conical tomb are 25 m
and 14 m respectively. Find the cost of white-washing its curved surface at the
rate of Rs 210 per 100 m2.
Answer:
Slant height, l = 25 m
Base radius, r = 7 m
CSA of conical tomb = πrl
Cost of white-washing 100 m2 area
= Rs 210
Cost of white-washing 550 m2 area
=
=
Rs 1155
Cost white-washing such a conical
tomb = Rs1155
7.A joker’s cap is in the form of right circular
cone of base radius 7 cm and height 24 cm. Find the area of the sheet required
to make 10 such caps.
Answer:
Radius, r = 7 cm
Height, h = 24 cm
CSA of 10 such conical caps = (10 ×
550) cm2 = 5500 cm2
Therefore, 5500 cm2 sheet
will be required.
8. A bus stop is barricaded from the remaining
part of the road, by using 50 hollow cones made of recycled cardboard. Each
cone has a base diameter of 40 cm and height 1 m. If the outer side of each of
the cones is to be painted and the cost of painting is Rs 12 per m2,
what will be the cost of painting all these cones? (Use π = 3.14 and take= 
= 1.02).
Answer:
Radius, r =20 cm = 0.2 m
Height, h = 1 m
CSA of each cone = πrl
= (3.14 × 0.2 × 1.02) m2 =
0.64056 m2
CSA of 50 such cones = (50 ×
0.64056) m2
= 32.028 m2
Cost of painting 1 m2 area
= Rs 12
Cost of painting 32.028 m2 area
= Rs (32.028 × 12)
= Rs 384.336
= Rs 384.34 (approximately)
Therefore, it will cost Rs 384.34 in
painting 50 such hollow cones.
EXERCISE 13.3
1.Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Answer:
(i) Radius (r) of sphere =
10.5 cm
Surface area of sphere = 4Ï€r2
Therefore, the surface area is 1386
cm2.
ii) Radius(r) of sphere = 5.6
cm
Surface area of sphere = 4Ï€r2
Therefore, the surface area is
394.24 cm2.
(iii) Radius (r) of sphere =
14 cm
Surface area of sphere = 4Ï€r2
Therefore, the surface area is 2464
cm2.
2.Find the surface area of a sphere of diameter:
(i) 14 cm (ii) 21 cm (iii) 3.5 m
Answer:
(i) Diameter = 14 cm, so radius = 7
cm
Surface area of sphere = 4Ï€r2
Therefore, the surface area is 616 cm2.
(ii) Diameter = 21 cm, so radius =
10.5 cm
Surface area of sphere = 4Ï€r2
Therefore, the surface area is 1386
cm2.
3.The radius of a spherical balloon increases
from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface
areas of the balloon in the two cases.
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
Rs 16 per 100 cm2.
6. Find the radius of a
sphere whose surface area is 154 cm2.
7. The diameter of the moon is approximately
one-fourth of the diameter of the earth. Find the ratio of their surface area.
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.
 |
| EXERCISE 13.4 QUESTION 9 |
1. A matchbox measures 4 cm × 2.5 cm × 1.5 cm.
What will be the volume of a packet containing 12 such boxes?
Answer:
Length of the matchbox = 4 cm
Breadth of the matchbox = 2.5 cm
Height of the matchbox = 1.5 cm
Volume of the match box = l × b × h
=
4 × 2.5 × 1.5 = 15 cm3
Volume of 12 such matchboxes = 12 x
15 = 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? (1 m3 =
1000l)
Answer:
Length
of the tank = 6 m
Breadth
of the tank = 5 m
Height
of the tank = 4.5 m
Volume
of the tank = l × b × h
= 6 x 5 x
4.5 = 135 m3
1
m3 = 1000
Volume
of the tank = 135 x 1000 = 135000 litres.
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
of a cuboidal vessel = 10 m
Breadth
of a cuboidal vessel = 8 m
Let
the height of a vessel be h m
Volume of vessel = l × b × h
∴ l × b × h = 380 m3
10 x
8 x h = 380 m3
h=
= 4.75
Therefore the height must be 4.75 m
4. Find the cost of digging a cuboidal pit 8 m
long, 6 m broad and 3 m deep at the rate of Rs 30 per m3.
Answer:
Length
of a pit, l = 8 m
Breadth
of a pit, b = 6 m
Depth
of a pit, h = 3 m
Volume
of a pit = l x b x h
=
8 x 6 x 3
= 144 m3
Rate
of digging 1 m3 = Rs 30
So cost
of digging the pit = 144 × 30 = Rs 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 of a tank, l = 2.5m
Depth of a tank, h = 10 m
breadth of a tank = ?
Volume = l x b x b
Capacity of a tank(volume) = 50000 litres
2.5 x 10 x b = 50
⇒ b = 
= 2 m
Therefore, the breadth of the tank is 2 m.
7. A village, having a population of 4000,
requires 150 litres of water per head per day. It has a tank measuring 20 m ×
15 m × 6 m. For how many days will the water of this tank last?
Answer:
Length of a tank, l = 20 m
Breadth of a tank,
b = 10 m
Height of a tank,
h = 6 m
Volume = l x b x b
= 20
x 10 x 6
=
1800 m3
Number of peoples of a village = 4000
Water consumption per head = 150 litres
∴ total consumption of water per
day = 4000 x 150
=
6,00,000 litres
Number of days the water of this tank last = 
= 3days
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:
Side
of a larger cube = 12 cm
Volume
of a larger cube = (a)3 = (12 cm)3 = 1728 cm3
Number
of smaller cube = 8
∴ volume of one smaller cube =
Side3
= 216 cm3
Side
of smaller cube, a =
= 6 cm
Surface
area of a smaller cube = 6a2
= 6
x 62 = 216 cm2
Surface
area of a smaller cube = 6a2
= 6
x 122 = 864 cm2
Ratio between surface areas of cubes =
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:
Rate of water flow in 1 hour = 2 km =
2000 m
Rate
of water flow in 1 minute = 
Length of river flow in one minute,l =
m
Height of a river, h = 3 m
Breadth of a river, b = 40 m
Volume of water fall into the sea in
a minute = l x b x h
= 3 x 40 x 
= 4000 m3
EXERCISE 13.61. The circumference of the base of cylindrical
vessel is 132 cm and its height is 25 cm. How many litres of water can it hold?
(1000 cm3 = 1l) Answer: Height h of vessel, h = 25 cm Circumference of vessel = 132 cm 2Ï€r = 132 cm
2 x  x r = 132 
Volume of cylindrical vessel = πr2h = x 21
x 252
=
34650 cm3 Volume of water in litres =  = 34.65 litres 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 cm3 of wood has a mass of 0.6
g. Answer: Inner
diameter = 24 cm Inner
radius, r = 12 cm Outer
diameter = 28 cm Outer
radius, R = 14 cm Height
of pipe, h = Length of pipe = 35 cm Volume
of pipe = πh(R2
– r2)
= s x 35(142 - 122) = 22 x 5 (14 + 2)(14 – 2) = 110 x 26 x 2 = 5720 cm3 Mass
of 1 cm3 = 0.6 g Mass
of 5720 cm3 wood = (5720 × 0.6) g =
3432 g
=
3.432 kg 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? Answer: Tin is in the shape of a cuboid: l = 5 cm, b = 4 cm, h =15 Capacity of tin can = l × b × h =
(5 × 4 × 15) cm3
=
300 cm3 For a plastic can having circular
base
Diameter = 7 cmRadius =  Height = 10 cm
Volume of plastic cylinder = πr2h  Therefore, plastic cylinder has the
greater capacity by 385 − 300 cm3 = 85 cm3 4. If the lateral surface of a cylinder is 94.2
cm2 and its height is 5 cm, then find (i) radius of its base
(ii) its volume. [Use Ï€ = 3.14] (i) Height of a cylinder = 5 cm Let radius of cylinder be r. LSA of cylinder = 94.2 cm2 2Ï€rh = 94.2 cm2 2 × 3.14 × r × 5 = 94.2 cm2 
r =
3 cm (ii) Volume of cylinder = Ï€r2h = 3.14 × (3)2 × 5 cm3
= 141.3 cm3 5. It costs Rs 2200 to paint the inner curved
surface of a cylindrical vessel 10 m deep. If the cost of painting is at the
rate of Rs 20 per m2, find
(i) Inner curved surface area of the vessel
(ii) Radius of the base
(iii) Capacity of the vessel Answer: (i) Rate of painting the CSA = 20
per m2 cost of painting = Rs 2200 ∴ Inner curved surface area = 
= 110 m2 (ii) Inner Curved Surface area of a cylinder = 2Ï€rh ∴2Ï€rh = 110 m2 2 x  x r x
10 = 110 
(iii) volume of the
vessel = πr2h 
Therefore, the capacity of the
vessel is 96.25 m3 or 96250 litres. 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? 
7. A lead pencil consists of a cylinder of wood
with 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. 
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? Answer: Diameter of a base of cylindrical bowl
= 7 cm Radius of a base = 
cm Height of a bowl, h = 4 cm Volume of soup in 1 bowl = πr2h =
 x x x 4 =
154 cm3 Volume of soup given to 250 patients
= 250 × 154 = 38500 cm3
= 38.5l. EXERCISE - 13.71.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 of a cone = 6 cm Height of a cone = 7 cm Volume of cone =  π r2h =x  x
x 6 x6 x 7 = 264 cm3 (ii) Radius of a cone = 3.5 cm Height of a cone = 12 cm Volume of cone = πr2h = x
x 3.5 x3.5
x 12 = 154
∴ the
volume of the cone is 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 of cone, r = 7 cm
Slant height of cone, l = 25
cm
3. The height of a cone is 15 cm. If its
volume is 1570 cm3, find the diameter of its base.(Use π = 3.14)
Answer:
Height of a cone, h = 15 cm
Volume of cone = 1570 cm3
4. If the volume of a right circular cone of
height 9 cm is 48Ï€ cm3, find the diameter of its base.
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.
EXERCISE 13.8
1. Find the volume of a sphere whose radius is
(i) 7 cm (ii) 0.63 m
Answer:
(i) Radius of sphere = 7 cm
2. Find the amount of water displaced by a
solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m
Answer:
(i) Radius of a ball = 14 cm
No comments:
Post a Comment