SURFACE AREA AND VOLUME
EXERCISE 13.2
EXERCISE 13.2
Solution: 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
Solution: Here, height (h) = 1 m
∵ Diameter of the base = 140 cm = 1.40 m
Hence, the required sheet is 7.48 m2
Q 3. 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.
Solution: 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)]
Q 4. 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.
Solution: The roller is in the form of a cylinder diameter of the roller = 84 cm
Q 5. 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.
Solution: Diameter of the pillar = 50 cm
Q 6. 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.
Solution: Radius (r) = 0.7 m
Let height of the cylinder be ‘h’ metres.
∴ Curved surface area = 2πrh
Q 7. 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.
Solution: 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
Q 8. 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.
Solution: Length of the cylindrical pipe = 28 m
h = 28m
Diameter of the pipe = 5 cm
Q 9. 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.
Solution: 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
Q 10. 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.
Solution: 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
Q 11. 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?
Solution: 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]
