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TOPICS
Exercise - 13.3

Question-1 :-  A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. (Take π = 22/7)

Solution :-
```    Given that:-
Radius(r1) of the metallic sphere = 4.2 cm
Radius(r2) of the cylinder = 6 cm

According to question, metalic sphere melted and recast into the shape of cylinder.
So, Volume of sphere = Volume of cylinder
4/3 x πr13 = πr22h
4/3 x r13 = r22h
4/3 x 4.2 x 4.2 x 4.2 = 6 x 6 x h
98.784 = 36 x h
h = 98.784/36
h = 2.744 cm
Hence, height(h) of the cylinder = 2.744 cm
```

Question-2 :-  Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere. (Take π = 22/7)

Solution :-
```    Given that:-
Radius(r1) of metallic sphere = 6 cm
Radius(r2) of metallic sphere = 8 cm
Radius(r3) of metallic sphere = 10 cm
Let R be the radius of resulting sphere.

According to question, Three metallic sphere are melted to form a single solid sphere.
So, Volume of resulting sphere = Volume of 1st metallic sphere + Volume of 2nd metallic sphere + Volume of 3rd metallic sphere
4/3 x πR3 = 4/3 x πr13 + 4/3 x πr23 + 4/3 x πr33
R3 = r13 + r23 + r33
R3 = (6 x 6 x 6) + (8 x 8 x 8) + (10 x 10 x 10)
R3 = 216 + 512 + 1000
R3 = 1728
R3 = (12)3
R = 12 cm
Therefore, the radius of the sphere so formed will be 12 cm.
```

Question-3 :-  A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform. (Take π = 22/7)

Solution :-
``` Given that:-
The shape of the well will be cylindrical.
Diameter of the well = 7 m
Radius(r) of the well = 7/2 m
Depth(h) of the well = 20 m
Area of platform = Length × Breadth = L x B = 22 × 14 m2
Let H be the height of the platform.

Therefore, Volume of soil dug from the well will be equal to the volume of soil scattered on the platform.
So, Volume of soil from well = Volume of soil used to make such platform
πr2h = L x B x H
22/7 x 7/2 x 7/2 x 20 = 22 x 14 x H
H = 22/7 x 49/4 x 20 x 1/22 x 1/14
H = 5/2 m
H = 2.5 m
Therefore, the height of such platform will be 2.5 m.
```

Question-4 :-  A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. (Take π = 22/7)

Solution :-
``` Give that:-
The shape of the well will be cylindrical.
Diameter of the well = 3 m
Radius(r) of the well = 3/2 m
Depth(h) of the well = 14 m
Width of embankment = 4 m
From the figure, it can be observed that our embankment will be in a cylindrical shape having
outer radius(R) = 4 + 3/2 = 11/2 m and
Let H be the height of the embankment.
Volume of soil dug from well = Volume of earth used to form embankment
πr2h = π(R2 - r2)H
3/2 x 3/2 x 14 = [(11/2 x 11/2) - (3/2 x 3/2)] x H
63/2 = [121/4 - 9/4] x H
63/2 = 112/4 x H
63/2 = 28 x H
H = 63/56
H = 9/8
H = 1.125 m
```

Question-5 :-  A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream. (Take π = 22/7)

Solution :-
```
Give that:-
Diameter of circular cylinder = 12 cm
Radius(r1) of the circular cylinder = 12/2 = 6 cm
Height(h1) of the cicular cylinder = 15 cm
Height(h2) of the cone = 12 cm
Diameter of the cone = 6 cm
Radius(r2) of the cone = 6/2 = 3 cm

Let n ice-cream cones be filled with ice-cream of the container.
Therefore, Volume of ice-cream in cylinder = n × (Volume of 1 ice-cream cone + Volume of hemispherical shape on the top)
πr12h1 = n x [(1/3 x πr22h2) + (2/3 x πr23)]
π(r12h1) = nπ x [(1/3 x r22h2) + (2/3 x r23)]
(r12h1) = n x [(1/3 x r22h2) + (2/3 x r23)]
6 x 6 x 15 = n x [(1/3 x 3 x 3 x 12) + (2/3 x 3 x 3 x 3)]
540 = n x [36 + 18]
540 = n x 54
n = 540/54
n = 10
Therefore, 10 ice-cream cones can be filled with the ice-cream in the container.
```

Question-6 :- How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm? (Use π = 22/7)

Solution :-
``` Given that:-
Coins are cylindrical in shape.
Diameter of coins = 1.75 cm
Radius(r) of coins = 1.75/2 = 175/200 = 7/8 cm
Thickness(h) of coins = 2 mm = 0.2 cm

Let n coins be melted to form the required cuboids.
Cuboid dimensions = 5.5 cm x 10 cm x 3.5 cm
Therefore, Volume of n coins = Volume of cuboids
n x πr2h = l x b x h
n x 22/7 x 7/8 x 7/8 x 0.2 = 5.5 x 10 x 3.5
n x 0.48125 = 192.5
n = 192.5/0.48125
n = 400
Therefore, the number of coins melted to form such a cuboid is 400.
```

Question-7 :-  A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap. (Take π = 22/7)

Solution :-
``` Given that:-
Height(h1) of the cylinderical bucket = 32 cm
Radius(r1) of the cylinderical bucket = 18 cm
Height(h2) of the conical heap = 24 cm

Let the radius of the circular end of conical heap be r2.
The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.
Therefore, Volume of sand in the cylindrical bucket = Volume of sand in conical heap
πr12h1 = 1/3 x πr22h2
r12h1 = 1/3 x r22h2
18 x 18 x 32 = 1/3 x r22 x 24
18 x 18 x 32 = r22 x 8
18 x 18 x 4 = r22
r22 = 1296
r22 = (36)2
r2 = 36 cm

Slant height(l) of the conical heap = √r2² + h2²
= √362 + 242
= √1296 + 576
= √1872
= 12√13 cm
Therefore, the radius and slant height of the conical heap are 36 cm and 12√13 cm respectively.
```

Question-8 :- Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed? (Take π = 22/7)

Solution :-
``` Given that:-
Consider an area of cross-section of canal as ABCD.
Area of cross-section = 6 × 1.5 = 9 m2
Speed of water = 10 km/h = 10000/60 = m/min
Volume of water that flows in 1 minute from canal = 9 x 10000/60 = 15000 m3
Volume of water that flows in 30 minutes from canal = 30 × 1500 = 45000 m3 Let the irrigated area be A.
Volume of water irrigating the required area will be equal to the volume of water that flowed in 30 minutes from the canal.
Therefore, Vol. of water flowing in 30 minutes from canal = Vol. of water irrigating the required area
45000 = (A x 8)/100
A = 4500000/8
A = 562500 m2
Therefore, area irrigated in 30 minutes is 562500 m2.
```

Question-9 :- A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled? (Take π = 22/7)

Solution :-
``` Consider an area of cross-section of pipe as shown in the figure.
Radius(r1) of circular end of pipe = 20/200 = 0.1 m
Area of cross-section = πr12 = π x 0.1 x 0.1 = 0.01π m2
Speed of water = 3 km/h = 3000/60 = 50 m/min
Volume of water that flows in 1 minute from pipe = 50 × 0.01π = 0.5π m3
Volume of water that flows in t minutes from pipe = t × 0.5π m3 Radius(r2) of circular end of cylindrical tank = 10/2 = 5 m
Depth(h2) of cylindrical tank = 2 m
Let the tank be filled completely in t minutes.
Volume of water filled in tank in t minutes is equal to the volume of water flowed in t minutes from the pipe.
Therefore, Volume of water that flows in t minutes from pipe = Volume of water in tank
Volume of water that flows in t minutes from pipe = Volume of water in tank
t × 0.5π = π(r2)2h2
t × 0.5 = 52 × 2
t = 100
Therefore, the cylindrical tank will be filled in 100 minutes.
```
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