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The volume of water flows in the canal in one hour = width of the canal × depth
of the canal × speed of the canal water = 3 × 1.2 × 20 × 1000 m
3
= 72000 m
3
In 20 minutes the volume of water = (72000 × 20)/ 60 = 24000 m
3
Area irrigated in 20 minutes, if 8 cm, i.e., 0.08 m standing water is required
=24000/0.08
= 300000 m
2
= 30 hectares
Q.2: Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What is the ratio of their volumes?
Solution:
Given,
Ratio of heights of two cones = 1 : 3
Ratio of radii = 3 : 1
Let h and 3h be the height of two cones.
Also, 3r and r be the corresponding radii of cones.
So, r
1
= 3r, h
1
= h, r
2
= r, h
2
= 3h.
Ratio of volumes = [(1/3)πr
1
2
h
1
]/ [(1/3)πr
2
2
h
2
]
= [(3r)
2
h]/[r
2
(3h)]
= (9r
2
h)/(3r
2
h)
= 3/1
Hence, the ratio of volumes = 3 : 1
Q.3: A cubical ice-cream brick of edge 22 cm is to be distributed among some children by filling ice-cream cones of radius 2 cm and height 7 cm up to its brim. How many children will get the ice cream cones?
Solution:
Let n be the number of ice-cream cones.
Volume of cubical ice-cream brick = 22 cm × 22 cm × 22 cm
Radius of cone = r = 2 cm
Height of cone = h = 7 cm
Volume of cone = (1/3)πr
2
h = (1/3) × (22/7) × 2 × 2 × 7
So,
n × Volume of one cone = Volume of cubical ice-cream brick
n × (1/3) × (22/7) × 2 × 2 × 7 = 22 × 22 × 22
n × (1/3) × 4 = 22 × 22
n = (22 × 22 × 3)/4
n = 363
Therefore, 363 children will get the ice cream cones.
Q.4: Three cubes of a metal whose edges are in the ratio 3:4:5 are melted and converted into a single cube whose diagonal is 12√3 cm. Find the edges of the three cubes.
Solution:
Let the edges of three cubes (in cm) be 3x, 4x and 5x, respectively.
Volume of the cubes after melting is = (3x)
3
+ (4x)
3
+ (5x)
3
= 216x
3
cm
3
Let a be the side of a new cube so formed after melting.
Therefore, a
3
= 216x
3
So, a = 6x
Given that, diagonal of a single cube = 12√3 cm
i.e. √(a
2
+ a
2
+ a
2
) = 12√3
a√3 = 12√3
Therefore, a = 12
Thus, 12 = 6x
x = 2
Now, 3x = 3 × 2 = 6
4x = 4 × 2 = 8
5x = 5 × 2 = 10
Therefore, the edges of the three cubes are 6 cm, 8 cm and 10 cm, respectively.
Q.5: Find the number of solid spheres each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm.
Solution:
Given,
Diameter of solid sphere = 6 cm
Diameter of cylinder = 4 cm
Height of cylinder = h = 45 cm
Radius of sphere = r
1
= 6/2 = 3 cm
Radius of cylinder = r
2
= 4/2 = 2 cm
Let n be the number of spheres.
n × Volume of one sphere = Volume of cylinder
n × (4/3)πr
1
3
= πr
2
2
h
n × (4/3) × (22/7) × 3 × 3 × 3 = (22/7) × 2 × 2 × 45
n × 9 = 45
n = 45/9
n = 5
TSA of the toy = CSA of hemisphere + CSA of cone
Curved surface area of the hemisphere = 1/ 2 (4πr
2
) = 2π r
2
= 2(22/7) × (3.5/2) × (3.5/2) = 19.25 cm
2
Height of the cone = Height of the top – Radius of the hemispherical part
= (5 – 3.5/2) cm = 3.25 cm
Let h be the height of the cylinder and r be the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath = CSA of cylinder + CSA of the hemisphere
= 2πrh + 2πr
2
= 2π r(h + r)
= 2 (22/7) × 30 × (145 + 30) cm
2
= 33000 cm
2
= 3.3 m
2
Given,
The Volume (V) of each cube is = 64 cm
3
This implies that a
3
= 64 cm
3
∴ The side of the cube, i.e. a = 4 cm
Also, the breadth and length of the resulting cuboid will be 4 cm each while its height will be 8 cm.
So, the surface area of the cuboid (TSA) = 2(lb + bh + lh)
Now, by putting the values, we get,
= 2(8×4 + 4×4 + 4×8) cm
2
= (2 × 80) cm
2
Hence, TSA of the cuboid = 160 cm
2
It is known that a tent is a combination of a cone and a cylinder, as shown below.
From the question, we know that
The diameter = D = 4 m
l = 2.8 m (slant height)
The radius of the cylinder is equal to the radius of the cylinder
So, r = 4/2 = 2 m
Also, we know the height of the cylinder (h) is 2.1 m
So, the required surface area of the given tent = surface area of the cone (the top) + surface area of the cylinder (the base)
= πrl + 2πrh
= πr (l+2h)
Now, substituting the values and solving it we get the value as 44 m
2
∴ The cost of the canvas at the rate of Rs. 500 per m
2
for the tent will be
= Surface area × cost/ m
2
= 44 × 500
So, Rs. 22000 will be the total cost of the canvas.
Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere as shown in the above figure.
The radius BO of the hemisphere (as well as of the cone) =( ½) × 4 cm = 2 cm.
So, volume of the toy = (⅔) πr
3
+ (⅓) πr
2
h
= (⅔) × 3.14 × 2
3
+ (⅓)× 3.14 × 2
2
× 2
= 25.12 cm
3
Now, let the right circular cylinder EFGH circumscribe the given solid.
The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is
EH = AO + OP = (2 + 2) cm = 4 cm
So, the volume required = volume of the right circular cylinder – the volume of the toy
= (3.14 × 2
2
× 4 – 25.12) cm
3
= 25.12 cm
3
Hence, the required difference between the two volumes = 25.12 cm
3
