Q.1. The circumference of the base of cylindrical vessel is 132cm and its height is 25cm. How many litres of water can it hold? (1000 cm^{3} = 1L) (Assume π = 22 / 7)
Solution: Circumference of the base of cylindrical vessel = 132 cm
Height of vessel, h = 25 cm
Let r be the radius of the cylindrical vessel.
Step 1: Find the radius of vessel
We know that, circumference of base = 2πr, so
2πr = 132 (given)
r = (132 / (2 π))
r = 66 × 7 / 22 = 21
Radius is 21 cm
Step 2: Find the volume of vessel
Formula: Volume of cylindrical vessel = πr^{2}h
= (22 / 7) × 21^{2} × 25
= 34650
Therefore, volume is 34650 cm^{3}
Since, 1000 cm^{3} = 1L
So, Volume = 34650 / 1000 L= 34.65L
Therefore, vessel can hold 34.65 litres of water.
Q.2. The inner diameter of a cylindrical wooden pipe is 24cm and its outer diameter is 28 cm. The length of the pipe is 35cm. Find the mass of the pipe, if 1cm^{3} of wood has a mass of 0.6g. (Assume π = 22 / 7)
Solution: Inner radius of cylindrical pipe, say r_{1} = diameter_{1} / 2 = 24 / 2 cm = 12cm
Outer radius of cylindrical pipe, say r_{2} = diameter_{2} / 2 = 28 / 2 cm = 14 cm
Height of pipe, h = Length of pipe = 35cm
Now, the Volume of pipe = π(r_{2}^{2} - r_{1}^{2})h cm^{3}
Substitute the values.
Volume of pipe = 110 × 52 cm^{3} = 5720 cm^{3}
Since, Mass of 1 cm^{3} wood = 0.6 g
Mass of 5720 cm^{3} wood = (5720 × 0.6) g = 3432 g or 3.432 kg.
Q.3. A soft drink is available in two packs –
(i) a tin can with a rectangular base of length 5cm and width 4cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7cm and height 10cm. Which container has greater capacity and by how much? (Assume π = 22 / 7)
Solution: (i) tin can will be cuboidal in shape
Dimensions of tin can are
Length, l = 5 cm
Breadth, b = 4 cm
Height, h = 15 cm
Capacity of tin can = l × b × h = (5 × 4 × 15) cm^{3 }= 300 cm^{3}
(ii) Plastic cylinder will be cylindrical in shape.
Dimensions of plastic can are:
Radius of circular end of plastic cylinder, r = 3.5cm
Height, H = 10 cm
Capacity of plastic cylinder = πr^{2}H
Capacity of plastic cylinder = (22 / 7) × (3.5)^{2} × 10 = 385
Capacity of plastic cylinder is 385 cm^{3}
From results of (i) and (ii), plastic cylinder has more capacity.
Difference in capacity = (385 - 300) cm^{3} = 85cm^{3}
Q.4. If the lateral surface of a cylinder is 94.2cm^{2} and its height is 5cm, then find
(i) radius of its base
(ii) its volume. [Use π = 3.14]
Solution: CSA of cylinder = 94.2 cm^{2}
Height of cylinder, h = 5cm
(i) Let radius of cylinder be r.
Using CSA of cylinder, we get
2πrh = 94.2
2 × 3.14 × r × 5 = 94.2
r = 3
radius is 3 cm
(ii) Volume of cylinder
Formula for volume of cylinder = πr^{2}h
Now, πr^{2}h = (3.14 × (3)^{2} × 5) (using value of r from (i))
= 141.3
Volume is 141.3 cm^{3}.
Q.5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10m deep. If the cost of painting is at the rate of Rs 20 per m^{2}, find
(i) inner curved surface area of the vessel
(ii) radius of the base
(iii) capacity of the vessel
(Assume π = 22 / 7)
Solution: (i) Rs 20 is the cost of painting 1 m^{2} area.
Rs 1 is the cost to paint 1 / 20 m^{2} area
So, Rs 2200 is the cost of painting = (1 / 20 × 2200) m^{2}
= 110 m^{2} area
The inner surface area of the vessel is 110m^{2}.
(ii) Radius of the base of the vessel, let us say r.
Height (h) = 10 m and
Surface area formula = 2πrh
Using result of (i)
2πrh = 110 m^{2}
2 × 22 / 7 × r × 10 = 110
r = 1.75
Radius is 1.75 m.
(iii) Volume of vessel formula = πr^{2}h
Here r = 1.75 and h = 10
Volume = (22 / 7) × (1.75)^{2 }× 10 = 96.25
Volume of vessel is 96.25 m^{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 1m is 15.4 liters. How many square meters of metal sheet would be needed to make it? (Assume π = 22 / 7)
Solution: Height of cylindrical vessel, h = 1 m
Capacity of cylindrical vessel = 15.4 litres = 0.0154 m^{3}
Let r be the radius of the circular end.
Now,
Capacity of cylindrical vessel = (22 / 7) × r^{2} × 1 = 0.0154
After simplifying, we get, r = 0.07 m
Again, total surface area of vessel = 2πr(r + h)
= 2 × 22 / 7 × 0.07(0.07 + 1)
= 0.44 × 1.07 = 0.4708
Total surface area of vessel is 0.4708 m^{2}
Therefore, 0.4708 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 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. (Assume π = 22 / 7)
Solution:
Radius of pencil, r_{1} = 7 / 2 mm = 0.7 / 2 cm = 0.35 cm
Radius of graphite, r_{2} = 1 / 2 mm = 0.1 / 2 cm = 0.05 cm
Height of pencil, h = 14 cm
Formula to find, volume of wood in pencil = (r_{1}^{2} - r_{2}^{2})h cubic units
Substitute the values, we have
= [(22 / 7) × (0.352 - 0.052) × 14]
= 44 × 0.12 = 5.28
This implies, volume of wood in pencil = 5.28 cm^{3}
Again,
Volume of graphite = r_{2}^{2}h cubic units
Substitute the values, we have
= (22 / 7) × 0.052 × 14
= 44 × 0.0025
= 0.11
So, the volume of graphite is 0.11 cm^{3}.
Q.8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients? (Assume π = 22 / 7)
Solution: Diameter of cylindrical bowl = 7 cm
Radius of cylindrical bowl, r = 7 / 2 cm = 3.5 cm
Bowl is filled with soup to a height of 4cm, so h = 4 cm
Volume of soup in one bowl = πr^{2}h
(22 / 7) × 3.52 × 4 = 154
Volume of soup in one bowl is 154 cm^{3}
Therefore, Volume of soup given to 250 patients = (250 × 154) cm^{3 }= 38500 cm^{3}
= 38.5 litres.