Surface Areas and Volumes

142 Lifeskills' Complete NCERT Solutions Class-IX Mathematics 13 Surface Areas and Volumes Chapter In the Chapter O In this chapter, you will study the following points: Surface area of a cuboid = 2 (lb + bh + hl) Surface area of a cube = 6a2 Curved surface area of a cylinder = 2πrh Total surface area of a cylinder = 2πr(r + h) Curved surface area of a cone = πrl Total surface area of a right circular cone = πrl + πr2, i.e., πr (l + r) Surface area of a sphere of radius r = 4 π r2 Curved surface area of a hemisphere = 2πr2 Total surface area of a hemisphere = 3πr2 Volume of a cuboid = l × b × h Volume of a cube = a3 Volume of a cylinder = πr2h O Volume of a cone = O Volume of a sphere of radius r = O Volume of a hemisphere = O O O O O O O O O O O 1 2 πr h 3 4 3 πr 3 2 3 πr 3 [Here, letters l, b, h, a, r, etc. have been used in their usual meaning, depending on the context.] TEXT BOOK QUESTIONS (SOLVED) EXERCISE 13.1 Q.1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box. (ii) The cost of sheet for it, if a sheet measuring 1m2 costs Rs 20. Ans. For plastic box : Length = 1.5 m Breadth = 1.25 m Height (depth) = 0.65 m (i) The box is open at top, therefore area of sheet required = 2 [lb + bh + hl] – lb = 2 [1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5] – 1.5 × 1.25 = 2[1.875 + 0.8125 + 0.975] – 1.875 = 2 × 3.6625 – 1.875 = 7.325 – 1.875 = 5.45 m2 (ii) Cost of the sheet = Rs. (5.45 × 20) = Rs. 109. Ans. Q.2. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs.7.50 per m2. Ans. For room : length = 5 m breadth = 4 m height = 3 m Area of four walls = 2(l + b) × h = 2( 5 + 4) × 3 = 54m2 Area of ceiling = l × b = 5 × 4 = 20m2 Total area to be white washed = 54 + 20 = 74m2 Rate of white-washing = Rs. 7.50 per m2 Cost of white-washing = Rs. (74 × 7.50) = Rs. 555. Q.3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m2 is Rs. 15000, find the height of the hall. [Hint : Area of the four walls = Lateral surface area.] Ans. length = l breadth = b height = h Given, Perimeter = 250 m ∴ 2 (l + b) = 250 Cost of painting the four walls = Rs. 15000 Rate of painting the four walls = Rs. 10 per m2 15000 2 m = 1500m2 10 Also area of four walls= 2 (l + b) × h ∴ 2 (l + b) × h = 1500 or 250 × h = 1500 Area of four walls = 1500 m=6m 250 Q.4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container? Ans. Area that can be painted by the container = 9.375 m2 = 9.375 × 100 × 100 or h = = 93750 cm2 Area of the brick = 2(22.5 × 10 + 10 × 7.5 + 75 × 22.5) = 2(225 + 75 + 168.75) = 2 × 468.75 = 937.5 cm2 93750 = 100 bricks. 937.5 Q.5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high. (i) Which box has the greater lateral surface area and by how much? (ii) Which box has the smaller total surface area and by how much? Ans. We have l1 for cubical box = 10 cm For cuboidal box l = 12.5 cm b = 10 cm h = 8 cm (i) Lateral surface area of cubical box = 4t2 = 4(10)2 = 4 × 100 = 400 cm2 Lateral surface area of cuboidal box = 4 (l + b) ×h = 2 (12.5 + 10) × 8 = 2 (22.5) × 8 = 45 × 8 = 360 cm2 (Lateral surface area of cuboidal box) > (Lateral surface area of cuboidal box) (400 > 360) ∴ Required Area = (400 – 360)cm2 = 40 cm2 (ii) Total surface area of cubical box = 2 ( l × b + b × h + h × l) = 2 (12.5 × 10 + 10 × 8 + 8 × 12.5) = 2(125 + 80 + 100) = 2 × 305 = 610 cm2 ∴ (area of cuboidal box) > (Area of cubical box) (610 > 600) Required area = (610 – 600) cm2 = 10 cm2 Q.6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. (i) What is the area of the glass? (ii) How much of tape is needed for all the 12 edges? Ans. Dimension for herbarium are l = 30 cm, b = 25 cm and h = 25 cm Number of bricks = Area of the glass = 2 (l × b + b × h + h × l) = 2 (30 × 25 + 25 × 25 + 25 × 30) = 2 (750 + 625 + 750) = 2 (2125) = 4250cm2 ∴ Length of the taps = 4(l + b + h) = 4(30 + 25 + 25) [Herbarium is a shape of cuboid length = 2 (l + b + h)) = 4 × 80 = 320 cm. Q.7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind. Ans. For bigger box length = 25 cm breadth = 20 cm height = 5 cm Surface area of the box = 2 (lb + bh + hl) = 2(25 × 20 + 20 × 5 + 5 × 25) = 2(500 + 100 + 125) = 2 × 725 = 1450 cm2 For smaller box length = 15 cm breadth = 12 cm height = 5 cm Surface area of the box = 2 (lb + bh + hl) = 2 (15 × 12 + 12 × 5 + 15 × 5) = 2(180 + 60 + 72) = 2 × 315 = 630 cm2 Total surface area of the two boxes = 1450 + 630 = 2080 cm2 Surface area of cardboard required for overlapping. 5 × 2080 = 104 cm2 100 Total surface required of one box of each kind = 2080 + 104 = 2184 cm2 Total surface area of 250 boxes of each kind = 250 × 2184 cm2 Rate of 1000 cm2 of cardboard = Rs. 4 Cost of total cardboard = 250 × 2184 × 4 = Rs. 2184. 1000 Q.8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m? Ans. Length = 4 m Breadth = 3 m Height = 2.5 m Surface area of the shelter = 2(lb + bh + hl) – lb = 2(4 × 3 + 4 × 2,5 + 2,5 × 4) – 4 × 3 = 2(12 + 7.5 + 10) – 12 = 2 × 29.5 – 12 = 59 – 12 = 47 m2 = EXERCISE 13.2 Q.1. 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. Ans. We have, height = 14 cm Curved surface area of a right circular cylinder = 88 cm2 2πrh = 88 ⇒ ⇒ 2× 22 × r × 14 7 = r = 88 r = 1 cm Diameter = 2 × Radius = 2 × 1 = 2 cm Q.2. 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? Ans. For cylindrical tank height = 1 m Base diameter = 140 cm 140 cm = 70 cm = 0.7 m 2 Total surface area of the tank radius r = 88 × 7 2 × 22 × 14 = 2πrh + 2πr2 = 2πr(h + r) 22 × 0.7 (1+ 0.7) 7 = 44 × 0.1 × 1.7 m2 = 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 Fig.). Find its (i) inner curved surface area, (ii) outer curved surface area, (iii) total surface area. Ans. Length = 77 cm Inner diameter = 4 cm Outer diameter = 4.4 cm Inner radius r = 2 cm Outer radius R = 2.2 cm (i) Inner curved surface area =2× = 2πrh = 2 × 22 × 2 × 77 7 = 968 cm2 (ii) Outer curved surface area = 2πRh = 2 × 22 × 2.2 × 77 7 = 1064.80 cm2 (iii) Total surface area = Inner curved surface area + outer curved surface area + area of two bases = 968 + 1064.80 + 2π(R2 – r2) = 968 + 1064.80 + 2 × 22 (2.22 – 22) 7 = 968 + 1064.80 + 2 × 22 × 4.4 × 0.4 7 = 2032.80 + 5.28 = 20.38.08 cm2 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. Ans. Diameter of the roller = 84 cm Radius of the roller = 42 cm = 0.42 m Length of the roller h = 120 cm = 1.2 m Curved surface area = 2πrh =2× 22 × 0.42 × 1.2 7 Area covered by the roller in 1 revolution = 3.168 m2 Area covered in 500 revolutions = 500 × 3.168 = 1584 m2 Hence area of playground = 1584 m2 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. Ans. For cylindrical pillar : Diameter= 50 cm = 0.5 m Radius = 0.25 m Height = 3.5 m Area of curved surface = 2πrh 22 × 0.25 × 3.5 7 = 5.5 m2 Rate of painting = Rs. 12.50 per m2 Hence Cost of painting = Rs. 12.50 × 5.5 = Rs. 68.75. 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. Ans. We have, curved surface area of a right circular cylinder = 4.4m2 ∴ 2πrh = 4.4 =2× ⇒ 2× 22 × 0.7 × h = 7 4.4 44 44 ⇒ h =1m Hence, the height of the right circular cylinder is ⇒ h = 1 m. 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 Rs. 40 per m2. Ans. We have, inner diameter = 3.5 m 3 .5 m 2 and h = 10 m (i) Inner curved surface area ∴ inner radius = = 2πrh = 2× 22 3.5 × × 10 = 22 × 5 = 110m2 7 2 (ii) Cost of plastering per m2 = Rs. 40 Cost of plastering 110m2 = Rs. 40 × 110 × 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. Ans. Height (length) of the cylindrical pipe = 28 m Diameter of the pipe = 5 cm 5 cm = 2.5 cm 2 = 0.025 m Total radiating surface area = 2πrh Radius of the pipe (r) = 22 × 0.025 × 28 7 = 176 × 0.025 = 4.4 m2 =2× 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. 1 of 12 the steel actually used was wasted in making the tank. Ans. Diameter = 4.2 m Radius = 2.1 m Height = 4.5 m (i) Lateral or curved surface area = 2πrh ∴ 11 of x = 87.12 m2 12 87.12 ×12 = 64.8m2 11 Q.10. In Fig. , 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. ⇒ x= 20 cm = 10 cm 2 h = 30 cm Since, a margin of 2.5 cm is used for folding it over the top and bottom so the total height of frame. h1 = 30 + 2.5 + 2.5 h1 = 35 cm ∴ Cloth required for covering the lampshade = its curved surface area Ans. Given : r = = 2πr (h1) = 2 × (ii) how much steel was actually used, if 22 × 2.1 × 4.5 7 = 59.4 m2 (ii) Total surface area of the tank = 2πrh (h + r) =2× 22 × 2.1 (4.5 + 2.1) 7 = 44 × 0.3 × 6.6 = 87.12 m2 Let the actual area of steel used be x m2. Since =2× 1 of the actual steel used was wasted, the area of 12 steel which has gone into the tank. 1⎞ ⎛ 11 = ⎜1 − ⎟ of x = of x 12 ⎝ ⎠ 12 22 × 10(35) 7 440 × 35 = 440 × 5 7 = 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 Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? Ans. For one penholder. r = 3 cm h = 10.5 cm Surface area of a penholder = 2πrh + πr2 = πr (2h + r) = = 22 × 3 × (2 × 10.5 + 3) 7 = 22 × 3 × (24) cm2 7 Area of cardboard required for 35 boxes = 35 × = 5 × 22 × 72 = 110 × 72 = 7920 cm2 22 × 3 × 24 7 EXERCISE 13.3 Q.1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area. Ans. We have, diameter = 10.5 cm 10.5 = 5.25 cm 2 slant height l = 10 cm Radius (r) = and 22 Curved surface area = πrl = × 5.25 × 10 7 = 165cm2 Q.2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. Ans. Slant height l = 21 m Diameter of base = 24 m Radius of base = 12 m Total surface area of the cone = πr ( l + r) = 22 × 12 (21 + 12) 7 22 × 12 × 33 7 = 1244.57 m2. Q.3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone. Ans. Curved surface area of cone = 308 cm2 Slant height l = 14 cm (i) Curved surface area = πrl ∴ πrl = 308 = or 22 × r × 14 = 308 7 44r = 308 or r or = 308 = 7 cm 44 Hence radius of base = 7 cm (ii) Total surface area of cone = πr (l + r) 22 × 7 (14 + 7) 7 = 22 × 21 = 462 cm2 Q.4. A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent. (ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs. 70. Ans. We have, h = 10 m r = 24 m (i) l= r 2 + h2 ∴ l= (24) 2 + 10 2 = 576 + 100 = 676 = 26 m Hence, the slant height of the canvas tent is 26m (ii) Canvas required to make the tent = Curved surface area of tent = πrl = π × 24 × 26 = 624π m2 2 Cost of 1m canvas = Rs. 70 Cost of 624πm2 canvas = Rs. 70 × 624π 22 7 = Rs. 10 × 624 × 22 = Rs. 137280 Hence, the cost of the canvas is Rs. 137280. Q.5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14). Ans. For conical tent, height = 8 m and Cadius of base = 6m = Rs. 70 × 624 × Now, l = r 2 + h2 l = 62 + 82 l = 36 + 64 = 100 = 10m = Area of tarpaulin = πrl = 3.14 × 6 × 10m2 = 188.4 m2 Given, width of tarpaulin = 3 m l = 25 cm Area of the sheet required for 10 caps = 10 × curved surface area of cone = 10 × πrl 188.4m = 62.8 m 3 The extra material required for stitching margins and cutting is 20 cm = 0.2m So, the total length of tarpaulin bought = (62.8 + 0.2) m = 63 m. Q.6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2. Ans. We have, slant height, l = 25 m and diameter = 14 ∴ Radius, r = 7 m Curved surface area of the conical tomb Length of tarpaulin = 22 × 7 × 25 7 = 5500 cm2 Q.8. A bus stop is barricated from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02) Ans. For one cone, Diameter = 40 cm Radius, r = 20 cm = 0.2 m Height, h =1m = 10 × Slant height, l 22 = πrl = × 7 × 25 7 = 22 × 25 = 550m2 Cost of white washing per 100m2 = Rs. 210 Cost of white washing per 1m2 = Rs. 210 × 550 100 = Rs. 1155 Q.7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. Ans. Base radius = 7 cm Height = 24 cm Let l be the slant height, then in ∆OAB, l 2 = 72 + 242 l 2 = 49 + 576 l = 625 h2 + r 2 = 12 + (0.2) 2 Curved surface 210 100 Cost of white washing 550m2 = Rs. = = 1.04 = 1.02m = πrl = 22 × 0.2 × 1.02m2 7 22 × 0.204m2 7 Curved surface area of such 50 hollow cones = = 50 × 22 × 0.204 m2 7 224.4 2 m = 32.057m2 7 Rate of painting the cardboard = Rs. 12 per m2 Hence, cost of painting the outer surface of all the cones = Rs. (12 × 32.057) = Rs. 384.68 = EXERCISE 13.4 Q.1. Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm Ans. (i) For the given sphere r = 10.5 cm Surface Area = 4 πr2 22 × 10.5 × 10.5 7 = 1386 cm2 Ans. = 4× (ii) r = Surface Area = = = (iii) r = Surface Area = 5.6 cm 4πr2 22 × 5.6 × 5.6 7 394.24 cm2 14 cm 4πr2 4× 22 × 14 × 14 7 = 2464 cm2 Q.2. Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m Ans. (i) We have, diameter = 14 cm ∴ Radius, r = 7 cm = 4× 22 × 72 7 = 4 × 22 × 7 = 616 cm2 (ii) We have, diameter = 21 cm Surface area of a sphere = 4πr2 = 4 × r= 21 = 105 cm 2 22 × 105 × 105 7 =88 × 1.5 × 105 = 1386 cm2 (iii) We have, diameter 3.5 m ∴ Surface area = 4πr2 = 4 × ∴ r= 3 .5 = 1.75 cm 2 ∴ Surface area = 4πr2 = 4 × = 22 × 1.75 × 1.75 7 26950 = 38.5 cm2 7 Q.3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14) Ans. Here r = 10 cm Total surface area of hemisphere = 3πr2 = 3 × 3.14 × 10 × 10 cm2 = 942 cm2 Q.4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. Ans. Let initial radius, r1 = 7 cm Area increases, r2 = 14 cm Surface area of initial balloon = 4πr12 = 4 × 22 × 7 × 7 = 88 × 7 7 A1 = 616 cm2 Surface area for increasing balloon 22 × 14 × 14 = 88 × 28 7 A2 = 2464 cm2 ∴ Required ratio = A1 : A2 = 616 : 2464 = 1 : 4 Q.5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm2. Ans. For bowl inner diameter = 10.5 cm = 4πr22 = 4 × 10.5 cm 2 Inner Surface area of the bowl = 2πr2 Inner radius = 22 10.5 10.5 × × cm2 7 2 2 = 173.25 cm2 Rate of tin-plating = Rs. 16 per 100 cm2 Hence cost of tin plating =2× 16 × 173.25 100 = Rs. 27.72 Ans. Q.6. Find the radius of a sphere whose surface area is 154 cm2. Ans. Surface area of a sphere = 154 cm2 Let its radius = r cm = ∴ 4× or 22 2 × r = 154 7 r2 = 154 × 7 49 = 4 × 22 4 49 7 = = 3.5 cm 4 2 Hence radius of the sphere = 3.5 cm. Q.7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. Ans. Let diameter of moon = 2r or r= and ∴ and diameter of earth Radius of moon radius of earth Surface area of moon S1 and surface area of earth S2 Now = = = = = = 4 × 2r = 8r r 4r 4πr2 4π(4r)2 64πr2 S1 4πr 2 = S 2 64πr 2 4 1 = or 1 : 16 Ans. 64 16 Q.8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. Ans. For hemispherical bowl Inner radius r = 5 cm Outer radius R = inner radius + thickness = (5 + 0.25) = 5.25 cm. Outer curved surface area = 2πR2 = Q.9. A right circular cylinder just encloses a sphere of radius r (see Fig. ). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii). Ans. (i) For sphere, radius = r Surface area of sphere, S1 = 4pr2 (ii) For cylinder, Radius = r Height = 2r [Diameter of sphere =height of the cylinder] Curved surface area, S2 = 2prh (iii) Requried ratio, S1 4πr 2 = S 2 4πr 2 = 22 × 5.25 × 5.25 7 = 173.25 cm2 1 or 1 : 1. 1 =2× EXERCISE 13.5 Q.1. A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes? Ans. Volume of a match box = 4 cm × 2.5 cm × 1.5 cm = 15 cm2 Volume of a packet = 12 × 15 cm3 = 180 cm3 Q.2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m3 = 1000 l) Ans. Volume of a cuboidal water tank = 6 cm × 5 cm × 4.5 = 30 × 4.5 m3 = 135m3 = 135 × 1000 L = 135000L (1m3 = 1000L) Q.3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid? Ans. Volume of cuboidal vessel = 380 cm3 Length = 10 m Breadth = 8 m Volume Now Height = l × b 380 19 = 10 × 8 = 4 = 4.75 m. Q.4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m3. Ans. For tank : Length = 8 m Breadth = 6 m Height (depth) = 3 m Volume = l × b × h = 8 × 6 × 3 m3 = 144 m3 Rate of digging = Rs. 30 per m3 Hence total cost of digging = Rs. 30 × 144 = Rs. 4320. Q.5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m. Ans. Capacity of a cuboidal tank = 50,000 litres Volume of the tank = 50,000 3 m = 50 m3 1000 Length = 2.5 m Height (depth) = 10 m Breadth = Volume l×b 50 = 2.5 × 10 = 2 m. Q.6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last? Ans. Population = 4000 Consumption of water = 150 litres per head per day Total consumption of water per day = 4000 × 150 litres 4000 × 150 3 m 1000 = 600 m3 Volume of tank = 20 × 15 × 6 = 1800 m3 Number of days the water of the tank will last = 1800 = 3 days. 600 Q.7. A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown. Ans. Volume of godown = 40 × 25 × 10 = 10000 m3 Volume of one wooden crate = 1.5 × 1.25 × 0.5 = 0.9375 m3 = that can be stored in the godown is 10666. Q.8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas. Ans. Volume of bigger cube = 12 × 12 × 12 cm3 = 1728 cm3 Number of smaller cubes = 8 Volume of one smaller cube = 1728 cm3 8 ⇒ a3 = 216 cm3 Side of smaller cube (a) 3 216 cm3 = 6 cm Surface area of bigger cube = 6a2 = 6 × (12)2 = 6 × 144 = 864 cm2 Surface of 8 small cubes = 6 × (6)2 = 216 cm2 = 864 1 = or 1 : 4. 216 4 Q.9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute? Ans. Height (depth) = 3 m Breadth = 40 m Rate of water flown = 2 km per hour = 2000 m per hour Radio = = 2000 m per minutes 60 Hence volume of water 10,000 Number of crates = 0.9375 = 10.666.66 crates. Hence, the maximum number of wooden crates = 3 × 40× = 4000 m3. EXERCISE 13.6 Q.1. 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) Ans. For cylindrical vessel, Height = 25 cm Circumferecne of base = 132 cm ∴ 2πr = 132 or 2 × 22 × r = 132 7 132 × 7 r = 2 × 22 = 21 cm 2000 3 m 60 Volume of the cylindricxal vessel = πr2h 22 = (21)2 × 25 7 22 = × 21 × 21 × 25 7 = 22 × 63 × 25 = 34650 cm3 Volume of water in litres 34650 litres 1000 = 34.65 litres Q.2. 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. Ans. For cylindrical vessel, inner diameter = 24 cm and outer diameter = 28 cm = ∴ Inner radius r = 24 = 12 cm 2 28 = 14 cm 2 Height (length) of pipe = 35 cm. Volume of the wood = πR2h – πr2h = πh (R2 – r2) 22 = × 35 (142 – 122) 7 22 = × 35 (14 + 12) (14 – 12) 7 22 = × 35 × 26 × 2 = 5720 cm3 7 Mass of 1 cm3 = 0.6 g ∴ Required mass = 0.6 × 5720 gm = 3432 g = 3.432 kg. Q.3. 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? Ans. (i) We have, h = 15 cm l = 5 cm, b = 4 cm, h = 15 cm Volume of cuboidical body = l × b × h = 5× 4 × 15 = 300 cm3 ...(i) (ii) We have, Diameter = 7 cm Outer radius R = 7 cm 2 Height, h = 10 cm Then, volume of a plastic cylinder Radius , r = 22 7 7 × × ×10 7 2 2 = 77 × 5 = 385 cm3 ∴ From Eqs. (i) and (ii), we see that volume of a plastic cylinder has greater capacity and its capacity is 385 – 300 = 85 cm3 is more than the tin can. Q.4. 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) Ans. We have, lateral surface of a cylinder = 94.2 cm2 and h = 5 cm ∴ 2πrh = 94.2 ⇒ 2 × 3.14 × r × 5 = 94.2 = πr2h = 94 .2 31 .4 ⇒ r = 3 cm (i) Hence, radius of base, r = 3 cm (ii) Volume of a cylinder = πr2h = 3.14(3)2 × 5 = 3.14 × 9 × 5 = 141.3 cm3 Q.5. It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m 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. Ans. We have, cost to paint the inner curved surface = Rs. 2200 Cost to paint per m2 = Rs. 20 Inner curved surface area ⇒ r = Cost to paint the inner curved surface Cost to paint per m 2 2πrh = ⇒ = 2× Rs.2200 = 110m2 Rs.20 22 × r × 10 = 110 7 110 × 7 r = 2 × 220 r= 7 = 1.75 m 4 (i) Inner curved surface area of the vessel = 110m2 (ii) Hence, radius of the base is 1.75 m (iii) Capacity of the vessel = Volume of the vessel = πr2h 22 7 7 77 × × × 10 = = 7 4 4 8 = 96.25 m3 Q.6. 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? Ans. Capacity of a closed cylindrical vessel = 15.4 L = 15.4 × 1000 cm3 (1L = 1000 cm2) ∴ πr2 h = 15400 cm3 πr2 × 100 = 15400 r2 = 154 × 7 (h = 1m = 100 cm) 22 ⇒ r2 = 7 × 7 ⇒ r = 7 cm Total surface area of closed cylindrcial vessel = 2πr( r + h) 22 =2× × 7(7 + 100) 7 = 44 × 107 = 4708 cm2 4708 The metal sheet must be required = 100 ×100 m2 (1 cm = 0 .1 cm = 0.05 cm 2 Length h = 14 cm Volume of graphite = πr2h 22 = × (0.05)2 × 14 7 22 = × 0.0025 × 14 7 = 22 × 0.0025 × 2 cm2 = 0.11 cm3 Volume of wood = πR2h – πr2h Radius of graphite r = 22 × h (R2 – r2) 7 = 22 × 14 [(0.35)2 – (0.05)2] cm3 7 22 × 14 (0.4 × 0.3) cm3 7 = 44 × 0.12 cm3 = 5.28 cm3 Q.8. 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? Ans. Diameter of Base= 7 cm = 7 cm 2 Height, h = 4 cm Volume = pr2h 1 m) 100 = 0.4708 m2 Q.7. 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. Ans. Diameter of pencil = 7 mm = 0.7 cm Diameter of graphite = 1 mm 0.1 cm 0 .7 Radius of pencil R = cm = 0.35 cm 2 = Radius of Base, r = 2 22 ⎛ 7 ⎞ ⎜ ⎟ × 4 cm3 = 7 ⎝2⎠ = 154 cm2 Soup needed for 250 patients = (250 × 154) cm3 = 38500 cm3 38500 = litres 1000 = 38.5 litres. EXERCISE 13.7 Q.1. Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm Ans. For cone : (i) Radius = 6 cm, Height = 7 cm Volume of the cone = 1 2 πr h 3 = 1 22 × × (6)2 × 7 3 7 = 1 22 × × 36 × 7 3 7 = 264 cm3 Ans. (ii) Radius = 3.5 cm, Height = 12 cm = Volume of the cone = 1 2 πr h 3 1 ⎛ ⎞ 3 L⎟ ⎜1cm = 1000 ⎠ ⎝ Q.3. The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π = 3.14) Ans. We have, volume of a cone = 1570 cm3 1 22 = × × (3.5)2 × 12cm3 3 7 ⇒ 1 22 × × 12.25 × 12 cm3 3 7 = 22 × 1.75 × 4 cm3 = 154 cm3 Q.2. Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm Ans. (i) We have, r = 7 cm and l = 25 cm We know that l 2 = r 2 + h2 ⇒ = ⇒ h= t 2 − r 2 = 25 2 − 7 2 = 625 − 49 = 576 = 24 cm Capacity of conical vessel = 1 2 πr h 3 1 22 × × 7 × 7 × 24 3 7 = 22 × 7 × 8 = 1232 cm3 = 1232 L = 1.232 L. 1000 h = 12 cm and l = 13 cm l 2 = h 2 + r2 ⇒ r= 1 2 πr h = 1570 3 1 × 3.14 × r2 × 15 = 1570 3 1570 15700 = 3.14 × 5 157 r = 10 cm Hence, radius of the base = 10 cm. Q.4. If the volume of a right circular cone of πcm3, find the diameter of its base. height 9 cm is 48π Ans. We have, volume of a right circular cone = 48π cm3 ∴ r2 = 1 2 1 πr h = 48p ⇒ r2 × 9 = 48 3 3 (h = 9 cm, given) ⇒ r2 = 16 – r = 4 cm Hence, diameter of the base = 2r = 2 × 4 = 8 cm. Q.5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? Ans. We have, diameter = 3.5 m ∴ 3 .5 m 2 h = 12 m ∴ radius, r = = (ii) We have, we know that, 22 × 100 L = 0.314L 7 × 1000 and Capacity of a conical pit = l 2 − h 2 = 132 − 12 2 1 2 πr h 3 2 = 169 − 144 = 25 = 5 cm = 1 2 πr h 3 = Capacity of conical vessel = = 1 22 × × 5 × 5 × 12 3 7 = 22 × 100 cm3 7 1 22 ⎛ 3.5 ⎞ × ×⎜ ⎟ × 12 3 7 ⎝ 2 ⎠ 1 22 3.5 3.5 × × × × 12 3 7 2 2 = 22 × 0.5 × 3.5 = 38.5 m3 (1 m3 = 1kL) = 38.5 kL Q.6. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone Ans. For right circular cone : Volume = 9856 cm3 Diameter of the base = 28 cm Radius of the base r = 14 cm (i) ∴ 1 2 πr h = 3 1 22 × (14)2 h 3 7 h = = 9856 = 9856 × 7 × 3 22 × 14 × 14 cm 48 cm. = (ii) For slant height (l) l2 = l2 = l 9856 Ans. If we revolve the right ∆ABC about the side BC (=5 cm), we get a cone as shown in the figure: Height of the cone, h = 5 cm Radius of the cone, r = 12 cm ∴ Volume of the solid cone so obtained = 1 π (12)2 × 5 cm3 3 = 240π cm3 Now, ratio of the volumes obtained in Q.7 and = Q.8. h2 + r 2 (48)2 + (14)2 2304 + 196 cm = 2500 cm = 50 cm Hence, l = 50 cm (iii) Curved surface area of the cone = πrl 22 × 14 × 50 cm2 7 = 2200 cm2 Q.7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained. Ans. If we revolve the right ΔABC about the side AB ( = 12 cm), we get a cone as shown in the figure : Height of the cone h = 12 cm Radius of the cone r = 5 cm ∴ Volume of the solid cone so obtained 1 = πr2h 3 1 = π (5)2 × 12 cm3 3 = 100 p cm3 Q.8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8. = 1 2 πr h 3 100π 5 = = 5 :12 240π 12 Q.9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required. = Ans. We have, d = 10.5 m, r = 10.5 = 5.25m and 2 h = 3m l = r 2 + h2 = (5.25) 2 + 32 = 27.5625 + 9 = 36.5625 = 6.046 = 6.05 m Volume of the heap of cone of wheat 1 2 = πr h 3 1 22 = × × 5.25 × 5.25 × 3 3 7 = 86.625 m2 Area of the canvas required = curved surface area of the heap = πrl 22 = × 5.25 × 6.05m2 7 = 99.825 m2 EXERCISE 13.8 Q.1. Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m Ans. (i) r = 7 cm Volume of sphere = 4 3 4 22 πr = × ×7×7×7 3 3 7 = 88 × 49 4312 × 3 3 1 cm3 3 = 1437.3 cm3 r = 0.63 m = 1437 (ii) Volume of sphere = 4 × pr3 3 4 22 63 63 63 3 × × × × m 3 7 100 100 100 = 1.047816m3 = 1.05m3 (Approx.) Q.2. Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m Ans. (i) For sphere, Diameter= 28 cm Radius = 14 cm = 1 × 8x = 2x 4 = 4x = x and diameter of moon = Now and radius of earth radius of moon 4 3 πr 3 Volume of moon V1 = 4 π ( x) 3 3 = 4 22 × × 14 × 14 × 14 3 7 Volume of earth V2 = 4 π (4 x) 3 3 = 88 × 392 34496 = 3 3 Volume = 2 3 cm 3 = 11498.67 cm3 Diameter= 0.21 m 0.21 Radius r = m = 0.105 m 2 4 3 Volume = πr 3 4 22 × = × (0.105)3 m3 3 7 = 0.004851 m3 = 11498 (ii) Q.3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3? Ans. Diameter= 4.2 cm 4.2 Radius r = cm = 2.1 cm 2 4 3 Volume = πr 3 4 22 21 21 21 3 × × × × cm = 3 7 10 10 10 38808 = = 38.808 cm3 1000 Mass of 1 cm3 = 8.9 gm Hence total mass = 38.808 × 8.9 gm = 345.3912 gm = 345.39 g. (Approx) Q.4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? Ans. Let diameter of earth = 8x 4 π ( x) 3 V1 3 = Now required ratio = V2 4 π (4 x) 3 3 1 or 1 : 64. 64 Q.5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold? Ans. Diameter = 10.5 cm ⇒ Radius, r = 5.25 cm = Volume of hemisphere = = 2 3 πr 3 2 22 × × 5.25 × 5.25 × 5.25 3 7 = 303.1875 cm3 Hemispherical bowl can hold milk = 303.1875 L 1000 4 22 ⎛ 7 ⎞ 4 Now volume = πr3 = × × ⎜ ⎟ 3 7 ⎝2⎠ 3 4 22 7 × 7 × 7 = 3 × 7 × 2× 2× 2 1 ⎛ ⎞ 3 L⎟ ⎜1cm = 1000 ⎠ ⎝ = 0.303 L (Approx.) Q.6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. Ans. For hemispherical tank Inner radius r = 1 m - 100 cm Outer radius R = (100 + 1) = 101 cm Volume of the iron-sheet = 2 2 2 π R3 − π r 3 = π (R3 − r 3 ) 3 3 3 = 2 22 × [(101) 3 − (100) 3 ] 3 7 = = = 44 [1030301 – 1000000] 21 44 1333244 3 [30301] = cm 21 21 63487.809 cm3 63487.809 Volume = 100 × 100 ×100 m3 1 1 1 3⎤ ⎡ 3 ⎢1 cm = 100 × 100 × 100 m ⎥ ⎣ ⎦ 3 = 0.063487809 m = 0.06348 m3 (Approx.) Q.7. Find the volume of a sphere whose surface area is 154 cm2. Ans. Surface area = 154 cm2 ∴ 4πr2 = 154 cm2 4× 22 2 ×r 7 = 154 cm2 r2 = 154 × 7 49 = 4 × 22 4 r = 49 7 = cm 4 2 3 = 11× 49 539 = 3 3 2 3 cm 3 Q.8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 498.96. If the cost of white-washing is Rs. 2.00 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome. Ans. (i) Rate of white-washing = Rs. 2 per sq. m Cost of white-washing = Rs. 498.96 Inside surface area of the hemisphere = 179 498.96 = 249.48m2. 2 Now inside surface area of hemisphere = 2πr2. (ii) ∴ 2πr2 = 249.48 = 2× 22 2 ×r = 7 249.48 r2 = 249.48 × 7 2 × 22 = 5.67 × 7 r2 = 39.69 r = 39.69 = 6.3 cm Volume of air inside the dome 2 2 22 π r 3 = × × (6.3) 3 3 3 7 = 523.9 m3 Q. 9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S' . Find the (i) radius r 2 of the new sphere, (ii) ratio of S and S' . Ans. (i) For smallest sphere, radius = r ∴ For one sphere S = 4πr2 = Volume of one smaller sphere = 4 4πr3 3 Now volume of 27 smaller spheres 4 = 27× 4πr2 = 36 πr3 3 ∴ Volume of one big sphere = 36 πr3 Given : Radius of one big sphere = r ∴ Its volume = ∴ ⇒ 4 3 πr 3 Radius r = 4 3 πr = 36 πr3 3 r'3 = 27r3 (r')3 = (3r)3 r' = 3r ⇒ For new sphere, Radius r'= (ii) Now S and S' 3r = = = 3 .5 mm = 1.75 mm 2 175 7 mm = mm 100 4 4 3 4 22 ⎛ 7 ⎞ Volume = π r = × × ⎜ ⎟ 3 3 7 ⎝4⎠ r 1 = r' 3 3 2 = 2 4πr 4πr2 2 S 4π r ⎛ r ⎞ ⎛1⎞ × =⎜ ⎟ =⎜ ⎟ 2 S ' 4π r ' ⎝ r ' ⎠ ⎝ 3 ⎠ 2 ∴ S 1 = or 1 : 9 S' 9 Q.10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule? Ans. For spherical capsule : Diameter= 3.5 mm 4 22 7 ⎛ 7 ⎞ 22 49 539 × × ⎜ ⎟ = × = 3 7 4⎝ 4⎠ 3 16 24 = 22.46 mm3 (Approx.) 2 EXERCISE 13.9 (Optional) Q.1. A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see Fig.). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf. Ans. External surface area to be polished = [(110 × 85) + 2(110 × 25) + 2(85 × 25) + 2(110 ×5) + 4(75 ×5)] = (9350 + 5500 + 4250 + 110 + 1500)cm2 = 21700 cm2 ∴ Expenses requried for polishing @ 20 paise per cm2 = 21700 × 20 paise 21700× 20 = Rs. 4240 100 Internal surface area to be painted = [2(20 × 90) + 6(75 × 20) + (75 × 90)] = (3600 + 9000 + 6750) cm2 = 19350 cm2 ∴ Expenses required for painting @ 10 paise per = Rs. cm2 = 19350 × 10 paise 19350 × 10 = Rs. 1935 100 ∴ Total expenses required for polishing and painting the surface of the bookshelf = Rs. 4340 + Rs. 1935 = Rs. 6275. Ans. Q.2. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is = Rs. to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2. Ans. It is obvious, we have to subtract the cost of the sphere that is resting on the supports while calculating the cost of silver paint. Surface area to be silver paint = 8 (Curved surface area of the sphere – Area of circle on which sphere is resting) = 8 (4πR2 – πr2) = 8π(4R2 – r2)(Given, R = 21 cm and r = 1.5 cm) 2 5 ⎞ ⎛ ⎟ = Rs. 26.40 = ⎜ 528 × 100 ⎠ ⎝ Hence, total costing of painting = 2757.86 + 26.40 = Rs. 2784.26 (Approx.) Q.3. The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease? Ans. Let diameter = 2r ∴ Radius = r ∴ Now diameter = 2r – ⎛ ⎛ 21 ⎞ 2 ⎞ ⎜ 4 × ⎜ ⎟ – (15) 2 ⎟ = 8π ⎜ ⎟ ⎝ ⎝ 2⎠ ⎠ 25 ⎞ ⎛ ⎛ 75 ⎞ ⎟ = 2r ⎜ ⎟ = 2 r ⎜1 – 100 ⎝ ⎠ ⎝ 100 ⎠ ⎛ 441 ⎞ – 2.25 ⎟ = 8π ⎜ 4 × 4 ⎝ ⎠ 2 = 8π (438.75) cm Therefore, the cost of silver paint at the rate of 25 paise per cm2 25 ⎞ ⎛ 22 ⎟ = ⎜ 8 × × 438.75 × 7 100 ⎠ ⎝ 3 3r = 4 2 = 2r × 3r 3r + 2= 2 4 Original curved surface area = 4π(r)2 = 4πr2 New curved surface area ∴ Now radius = 2 19305 = Rs. 2757 (Approx) 7 Hence, surface area to be black painted = 8 × Curved surface area of cylinder = 8 × 2πrh = 22 × 1.5 × 7 7 =8×2× 25 × 2r 100 ⎛ 3r ⎞ 9π r 2 = 4π ⎜ ⎟ = ⎝ 4⎠ 4 Decrease in curved surface area 9π r 2 7π r 2 = 4 4 Hence percentage decrease = 4πr2 – 7π r 2 = 4 × 100% = 7 ×100% 16 4πr 2 2 = 528 cm ∴ Cost of black paint at the rate of 5 paise per cm2 = 175 % = 43.75% 4 Additional Questions Q.1. A storage tank is in the form of a cube. When it is full of water, the volume of water is 15.625 m3. If the present depth of water is 1.3 m, find the volume of water already used from the tank. Ans. Volume of cubical tank = 15.625 m2 depth = 1.3 m Let, side of cube = x cm x3 ⇒ x3 x Now, height of water Height of water used Volume of water used = 15.625 m2 = (2.5)3 = 2.5 cm = 1.3 m = 2.5 – 1.5 = 1.2 m = 1.2 × 2.5 × 2.5 = 7.5 m3 Q.2. Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 cm. When 16 spheres as packed the box is filled with preservative liquid. Find the volume of this liquid. Give your answer to the nearest integer. [Use π = 3.14] Ans. Volume of cuboidal box = l × b × h = 16 × 8 × 8 = 1024 cm3 Volume of sphere = 4 3 4 4 πr = π(2)3 = π × 8 3 3 3 32π = 33.49 cm3 3 Volume of 16 sphere = 535.84 cm3 Volume of liquid = volume of box – volume of 16 spheres ∴ Volume of liquid = 1024 – 535.84 = 488.16 m3 or = 488 cm3 Q.3. A shopkeeper has one spherical laddo of radius 5 cm. With the same amount of material, how many laddoos of radius 2.5 cm can be made ? Ans. Radius of big spherical laddoo R = 5 cm = 4 Volume of laddo = πR3 3 4 3 500π = ×5 π = cm3 3 3 Volume of each small laddoo = 4 π × (2.5)3 3 = 4 62.5π π × 15.625 = cm3 3 3 500π 3 500 Number of laddoos = 62.5π = =8 62.5 3 Number of laddoo = 8. Q.4. A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. If the glass is filled with milk upto an height of 12 cm, find how many litres of milk is needed to serve 1600 students. Ans. Diameter of glass = 7 cm Radius of glass = 7 cm 2 Milk contained = πr2h 22 7 7 × × × 12 7 2 2 = 462 cm3 Milk required for 1600 students = 462 × 1600 cm2 = 739.2 litres Q.5. A cylindrical roller 2.5 m in length, 1.75m in radius when rolled on a road was found to cover the area of 5500 m2. How many revolutions did it make ? Ans. Length of roller = height of roller (h) = 2.5 m radius (r) = 1.75 m ∴ Curved surface area of cylindrical roller = 2πrh = ⎛ 22 ⎞ = ⎜ 2 × × 1.75 × 2.5 ⎟ m2 7 ⎝ ⎠ = (2 × 22 × 0.25 × 2.5)m2 = 44 × 0.625)m2 = 27.5 m2 ∴ Area covered in 1 revolution = 27.5 m2 Thus, No. of revolutions = Total area Area covered in 1 revolution 5500 = 200 27.5 Hence, the road roller will make 200 complete revolutions. Q.6. A cube of side 4 cm contains a sphere touching its sides. Find the volume of the gap in between. Ans. Volume of gap in between = Volume of cube volume of sphere = = a3 – Volume of gap ⎞ 4 3 ⎛ a πr ⎜ r = Given ⎟ 2 3 ⎝ ⎠ = (4)3 – 4 22 3 × ×2 3 7 4 22 × ×8 3 7 = 64 – 33.52 = 30.48 cm3 Q.7. 30 circular plates, each of radius 14 cm and thickness 3 cm are placed one above the another to form a cylindrical solid. Find : = 64 – (i) the total surface area (ii) volume of the cylinder so formed Ans. Radius of each circular plate (r) = 14 m Thickness of each plate = 3 cm ∴ Height of cylinder so formed h = 30 × 3 = 90 cm (i) Total surface area = 2πr(h + r) 22 × 14 × (90 + 14) 7 = 44 × 2 × 104 = 9152 cm2 (ii) Volume of cylinder = pr2h = 2× 22 × 14 ×14 × 90 7 = 22 × 2 × 14 × 90 = (44 × 14 × 90) cm3 = 55440 cm3 Ans. Q.8. A hemispherical bowl is made of steel 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bow. (use π = = 2× 22 ) 7 Ans. Thickness of the bowl 25 1 = cm 100 4 The inner radius of the bowl = 5 cm ∴ Outer radius = Inner radius + Thickness = 0.25 cm = 1⎞ 1 21 ⎛ = ⎜ 5 + ⎟ = 5 cm – cm 4⎠ 4 4 ⎝ Curved surface area= 2πr2 ⎛ 22 21 21 ⎞ = ⎜ 2 × × × ⎟ cm2 7 4 4⎠ ⎝ 693 2 cm = 173.25 cm2 4 Q.9. If slant height of a cone is 21 m and diameter of base is 24 m, then find its total surface area. Ans. We have slant height (l) = 21 m = 24 m = 12 m 2 ∴ Total surface area of the cone = πr(l + r) and radius (r) = = 22 × 12(21 + 12) m2 7 22 × 12 × 33 m3 7 = 12.44.57 m2 Q.10. Surface area of a sphere is 154 cm2 Find its radius. Ans. Let the radius of the sphere be r cm. Then, surface area of the sphere = 4πr2 So, 4πr2 = 154 = ⇒ r2 154 154 × 7 49 = 4π = 4 × 22 = 4 ⇒ r = 7 cm. 2 Multiple Choice Questions Q.1. Three solid cubes of side 4 cm each are joined to form a cuboid. Then its surface area is: (a) 288 cm2 (b) 200cm2 (c) 192cm2 (d) 224cm2 Ans. (d) 224cm2 Q.2. The number of planks of dimensions 4 m × 5 m × 2 m that can be stored in a pit which is 40 m long, 12 m wide and 160 m deep is: (a) 1900 (b) 1920 (c) 1800 (d) 1840 Ans. (b) 1920 Q.3. In a cylinder, radius is doubled and height is halved, curved surface area will be: (a) halved (b) doubled (c) same (d) four times Ans.(c) same Q.4. The volume of a cube whose surface area is 54 cm2 is: (a) 9 cm3 (b) 27 cm3 (c) 3 cm3 (d)82 cm3 Ans. (b) 27 cm3 Q.5. The radii of cylinders of the same height are in the ratio 4:5, the ratio of their volumes is: (a) 16:25 (b) 4:5 (c) 64:125 (d) 25:16 Ans. (a) 16:25 Q.6. The surface area of sphere of diameter 14 cm is: (a) 616cm2 (b) 632cm2 (c) 742cm2 (d) 246cm2 Ans. (a) 616cm2 Q.7. If the volume of a sphere is numerically equal to its surface area, then diameter of the sphere is: (a) 6cm (b) 8cm (c) 10cm (d) 4cm Ans. (a) 6cm Q.8. The radius of largest sphere that is carved out of a cube of side 7 cm is: (a) 7 cm (b) 12 cm (c) 13 cm (d) 3.5 cm Ans. (d) 3.5 cm Q.9. The sheet required to make closed cylindrical vessel of hight 1 m and diameter 140 cm is: (a) 7.48 m2 (b) 7.43 m2 (c) 5.48 m2 (d)6.49 m2 Ans. (a) 7.48 m2 Q.10.If the areas of three adjacent faces of a cuboid are 8 cm2, 18 cm2 and 25 cm2, then volume of the cuboid is: (a) 70 cm3 (b) 60 cm3 (c) 80 cm3 (d) 90 cm3 Ans. (b) 60 cm3