Monday, January 23, 2017

CBSE Class 9 Maths - Surface Areas and Volumes - NCERT Exercise 13.2 (#cbseclass9Notes)

Surface Areas and Volumes

CBSE Class 9 Maths - Surface Areas and Volumes - NCERT Exercise 13.2 (#cbseclass9Notes)

NCERT Exercise 13.2

Q1: 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: Given, h = 14 cm,
curved, surface area (CSA) = 88 cm², r = ?
CSA  = 2πrh
88 = 2 × (22/7) × r × 14
88 = 44 × 2 × r
r = 88/88 = 1
diameter = 2r = 2 cm


Q2: 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 metres of the sheet are required for the same?.



Answer: Given, h = 1m
r = 140 / 2 = 70cm = 0.7m
TSA (cylinder) = 2πr (h + r)
= 2 × (22/7) × 0.7(1 + 0.7)
= 44 × 0.1 × 1.7
= 7.48 m²  (Answer)


Q3: A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its.
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.


Answer: Given, h = 77 cm,
Outer radius (R) = 4.4/2 = 2.2 cm,
Inner radius (r) = 4/2 = 2 cm

(i) Inner curved surface area (CSA₁)) of the pipe
= 2πrh
= 2 × (22/7) × 2 × 77
= 2 × 22 × 22 = 968 cm² (Answer)

(ii) Outer curved surface area (CSA₂) of the pipe = 2πRh
= 2 ×(22/7) × 2.2 × 77
= 44 × 24.2
= 1064.80 cm² (Answer)

(iii) Total surface area of the pipe = CSA₁ + CSA₂ + areas of the two base rings.
= 968 + 1064.80 + 2π (R² – r²)
= 2032.80 + 2(22/7)(2.2² - 2²)
= 2032.80 + 5.28 
= 2038.08 cm² (Answer)


Q4: 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: Radius of roller (r) = 84/2 = 42 cm
Length of the roller (h) = 120 cm
CSA = 2πrh
= 2 × (22/7) × 42 × 120
= 44 × 720
= 31680 cm²
∴ area covered by the roller in 1 revolution = 31680 cm²
area covered by the roller in 500 revolutions = 31680 × 500 = 15840000 cm²

Hence, area of of the playground = 15840000 / 10,000 = 1584 m² (Answer)


Q5: 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: Given,
r = 50/2 = 25 cm = 0.25 m, h = 3.5 m
CSA of pillar = 2πrh
= 2 × (22/7) × 0.25 × 3.5
= 5.5 m²

Cost of painting 1 m² = ₹ 12.50
∴ Total cost of painting the curved surface of the pillar
= ₹ 12.50 × 5.5 = ₹ 68.75 (Answer)



Q6: 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: Given,
CSA of the cylinder = 4.4 m²,
r = 0.7 m,
h = ?
CSA = 2πrh
4.4 = 2 × (22/7) × 0.7 × h
h = 4.4/4.4 = 1m (Answer)



Q7: 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: Given, r = 3.5/2 = 1.75 m,

h = 10m


(i) Inner curved surface area of the well

= 2πrh = 2 × (22/7) × 1.75 × 10

= 22 × 5

= 110 m² (Answer)


(ii) Cost of plastering 1 m² = ₹ 40

∴ Cost of plastering the curved surface area of the well

= ₹ 110 × 40 = ₹ 4400 (Answer)

Q8: 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: Given, r = 5/2 = 2.5 cm = 0.025m, h = 28m
Total radiating surface in the system = total surface area of the cylinder
= 2π r(h + r)
= 2 ×  (22/7) ×  (0.025) × (0.025 + 28)
= (44 × 0.025 ×  28.025)/7 = 4.403 m² (Answer)


Q9: 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 1/12th of the steel actually used was wasted in making the tank.


Answer: Here, r = 4.2/2 = 2.1 m, h = 4.5 m
(i) CSA of the storage tank = 2πrh
= 2 × (22/7) × 2.1 × 4.5 m²
= 59.4 m² (Answer)


(ii) Total surface area of the tank = 2πr (h + r)
= 2 × (22/7) × 2.1 (4.5 + 2.1)
= 44 × 0.3 × 6.6 m2 = 87.12 m²
Let the actual area of steel used be x m².
Area of steel wasted = (1/12) of x = (x/12) m².
∴ area of the steel used in the tank = x - (x/12) = (11x/12)
⇒ 87.12 = (11x/12)
⇒ x = 87.12 × 12/11 =  95.04 m²

Thus, 95.04 m² of steel was actually used.


Q10: In the 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.
CBSE Class 9 Maths - Surface Areas and Volumes - NCERT Exercise 13.2 (#cbseclass9Notes)


Answer: Given, r =20/2 = 10cm, h = 30 cm
Circumference of the frame = 2πr
= 2π × 10 cm = 20π cm
Height of the frame = 30 cm
Height of the cloth needed to cover the frame (including margin) = 30 + 2.5 + 2.5 = 35cm
Also, breadth of the cloth = circumference of the base of the frame.
∴ Area of the cloth required for covering the lampshade = length × breadth
= 35 × 20π
= 35 × 20 × (22/7)
= 2200 cm²


Q11: 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:
Given,  r = 3 cm, h = 10.5 cm
The penholders have only one base i.e., these are open at one end.
Total surface area (TSA) of 1 penholder = 2πrh + πr²
= πr (2h + + r)
= (22/7) × 3 (2 × 10.5 + 3)
= (22/7) × 3 × 24
TSA for 35 pen holders = (22/7) × 3 × 24 × 35

= 7920 cm² (Answer)

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