Class IX Math NCERT Solution for Surface Areas and Volumes 13.2

EXERCISE 13.2 (Page 216)
Q1.   The curved surface area of a right circular cylinder of height 14 cm is the diameter of the base of the cylinder.
Sol: Let 'r' be the radius of the cylinder.
           Here, height (h) = 14 cm
           and curved surface area = 88 cm2
           Curved surface area of a cylinder = 2πrh
           ∴ 2πrh = 88
           
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?
Sol: Here, height (h) = 1 m
           ∵ Diameter of the base = 140 cm = 1.40 m
           
           
           Hence, the required sheet is 7.48 m2
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.
Sol: Length of the metal pipe = 77 cm
           ∵ It is in the form of a cylinder.
           ∴ Height (h) of the cylinder = 77 cm
           Inner diameter = 4 cm
           
           (iii) Total surface area
                 = [Inner curved surface] + [outer curved surface area] + [Two base circular lamina]
                 = [27πrh] + [2πRh] + [2π(R2 – r2)]
                 
           
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.
Sol: The roller is in the form of a cylinder diameter of the roller = 84 cm
           
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.
Sol: Diameter of the pillar = 50 cm
           
           
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.
Sol: Radius (r) = 0.7 m
           Let height of the cylinder be 'h' metres.
           ∴ Curved surface area = 2πrh
           
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 40 per W.
Sol: Inner diameter of the well = 3.5 m
           
           Depth of the well (= height of the cylinder) h = 10 m
           (i) Inner curved surface area = 2πrh
                 
           (ii) Cost of plastering
                 ∵ Rate of plastering = Rs. 40 per m2
                 ∴ Total cost of plastering 110 m2 = Rs.110 × 40 = Rs. 4400
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.
Sol: Length of the cylindrical pipe = 28 m
           ⇒                                                                                        h = 28m
           Diameter of the pipe = 5 cm
           
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 12 of the steel actually used was wasted in making the tank.
Sol: The storage tank is in the form of a cylinder, and diameter of the tank = 4.2 m
           
           
           Thus, the required area of the steel that was actually used is 95.04 m2
Q10.   In the figure you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has 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.
Sol: The lampshade is in the form of a cylinder, where
                 
           ∵A margin of 2.5 cm is to be added to top and bottom.
           ∴ Total height of the cylinder
                 h = 30 cm + 2.5 cm + 2.5 cm
                 = 35 cm
           Now, curved surface area = 2πrh
                 
                 = 2 × 22 × 10 × 5 cm2 =2200 cm2
           Thus, the required area of the cloth = 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 Vdyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Sol: Here, the penholders are in the form of cylinders
           Radius of a cylinder (r) = 3 cm
           Height of a cylinder (h) = 10.5 cm
           Since, a penholder must be open from the top.
           ∴ Surface area of a penholder (cylinder) = [Lateral surface area] + [Base area]
                 

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