Surface Areas and Volumes (Cylinder)
EXERCISE 13.6
Q1: 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 cm³ = 1l)
Answer: Given, h = 25 cm, 2πr = 132 cm.
⇒ 2πr = 132
⇒ 2 × (22/7) × r = 132
⇒ r = (132 × 7)/ (2 × 22) = 21cm
Volume = πr²h = (22/7) × 21 × 21 × 25
= 34650 cm³
= 34650/1000
= 34.65 litres (Answer)
Q2: 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: Here, inner radius (r) 24/2 = 12cm
Outer radius (R) = 28/2 = 14cm
h = 35 cm
Volume of the wood used in the pipe = π(R² – r²) h
= (22/7) × (14² – 12²) × 35
= (22/7) × 26 × 2 × 35
= 5720 cm³
Mass of 1 cm³ of wood = 0.6 g
∴ Mass of 5720 cm³ of wood = 0.6 × 5720 g = 3432 g = 3.432 kg (Answer)
Q3: 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:
For tin can with rectangular base (cuboid).
Given,
l = 5 cm, b = 4 cm, h = 15 cm
Volume of the tin can = lbh = 5 × 4 × 15 = 300 cm³
For plastic cylinder with circular base.
Given,
r = 7/2= 3.5cm,
h = 10 cm
Volume of the plastic cylinder = πr²h
= (22/7) × 3.5 × 3.5 × 10 = 385 cm³
Difference in the capacities of the two containers = (385 – 300) cm3 = 85 cm³
Thus, the plastic cylinder with circular base has greater capacity by 85 cm³ (Answer)
Q4: 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 π = 3.14)
Answer: Given LSA = 2πrh = 94.2 cm²
h = 5 cm,
(i) 2πrh = 94.2 cm²
⇒ 2 × 3.14 × r × 5 = 94.2
⇒ r = 94.2 / (2 × 3.14 × 5) = 3cm (Answer)
(ii) Volume of the cylinder = πr²h
= 3.14 × 3 × 3 × 5 = 141.3 cm³ (Answer)
Q5: 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 m², find
(i) Inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.
Answer: Given, h = 10 m
(i) Inner curved surface area =
CSA = | Total cost |
Cost of painting per m² |
= 2200/20 = 110m² (Answer)
(ii) Since, 2πrh = 110
⇒ 2 × (22/7) × r × 10 = 110
⇒ r = (110 × 7) / (2 × 22 × 10)
= 1.75 m (Answer)
(iii) Capacity of the vessel = πr²h
= (22/7) × 1.75 × 1.75 × 10 = 96.25 m³
= 96.25 kl (Answer) [1 m³ = 1 kl]
Q6: 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?
Answer: Given, h = 1 m, volume = 15.4 litres
= 15.4 / 1000 = 0.0154 m³
Vol. of Cylinder = πr²h = 0.0154 m³
⇒ r² = (0.0154 × 7) / 22 = 0.0049
⇒ r = 0.07m
∴ Total surface area of the cylinder = 2πr (h + r)
= 2 × (22/7) × 0.07 (1 + 0.07) m²
= 44 × 0.01 × 1.07 m2 = 0.4708 m² (Answer)
Q7: A lead pencil consists of a cylinder of wood with a 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.
Answer: Given, h = 14 cm.
Radius of the pencil (R) = 7/2 = 3.5mm = 0.35 cm.
Radius of the graphite (r) = ½ mm = 0.05 cm.
Volume of the the graphite = πr²h
= (22/7) × 0.05 × 0.05 × 14 = 0.11 cm³
Volume of the the wood = π (R² – r²)h
= (22/7) × [(0.35)² – (0.05)²] × 14 cm³
= (22/7) × 0.4 × 0.3 × 14 cm³ = 5.28 cm³
Thus, volume of the wood = 5.28 cm³ and volume of the graphite = 0.11 cm³ (Answer).
Q8: 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:
Given, r = 7/2 = 3.5 cm, h = 4 cm
Capacity of 1 cylindrical bowl = πr²h
= (22/7) × 3.5 × 3.5 × 4 = 154 cm³
Thus, soup consumed by 250 patients per day = 250 × 154 cm³ = 38500 cm³ (Answer)
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