Exercise. Question. A plastic box.5 m long,.5 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 m costs ` 0. Solution We have a plastic box of l length 5. m b width.5 m h depth 65 cm 65 m 065. m ( Q m 00 cm) 00 Surface area of the box ( lb + bh + hl) (.5.5 +.5 0.65 + 0.65.5) (.85 + 0.85 + 0.95) (.665).5 m (i) Area of the sheet required for making the box. 5 l b (QBox is opened at the top).5.5.5. 5 85. 5. 45 m (ii) A sheet measuring m costs `0 Sheet measuring 5. 45 m costs `0 5.45 `09 Question. The length, breadth and height of a room are 5 m, 4 m and m, respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of `. 5 0 p e r m. Solution We have a room of l 5m b 4m h m Required area for white washing Area of the four walls + Area of ceiling ( l + b) h + ( l b) 5 ( + 4) + ( 5 4) 9 + 0 54 + 0 4 m White washing m costs `.50 White washing 4 m costs `.50 4 ` 555
Question. The floor of a rectangular hall has a perimeter 50 m. If the cost of painting the four walls at the rate of `0 per m is ` 5000, find the height of the hall. [Hint Area of the four walls Lateral surface area] Solution Let the rectangular hall of length l, breadth b, height h Cost of painting the four walls Now, Area of four walls Cost of painting per m ` 5000 `0 500 m We have, perimeter of the hall ( l + b) 50 m Q Area of four walls 500 m ( l + b) h 500 50 h 500 h 500 50 h 6m Hence, the height of the hall is 6 m. (Given) Question 4. The paint in a certain container is sufficient to paint an area equal to 9.5 m. How many bricks of dimensions. 5 cm 0 cm. 5 cm can be painted out of this container. Solution Given, dimensions of a brick and l.5 cm, b 0 cm h.5 cm Total surface area of bricks ( l b + b h + h l) ( 5. 0+ 0 5. + 5. 5.) ( 5 + 5 + 685. ) 4685. cm 9. 5 cm 9. 5 m 00 00 Number of bricks that painted out of this container Total surface area of container Total surface area of bricks 9. 5 9. 5 00 00 95. 00 00 9. 5 9500 95 00 (Qm 00 cm)
Question 5. A cubical box has each edge 0 cm and another cuboidal box is.5 cm long, 0 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? Solution We have l for cubical box 0 cm For cuboidal box l.5 cm b 0 cm h 8 cm (i) Lateral surface area of cubical box 4 l 4( 0) 4 00 400 cm Lateral surface area of cuboidal box 4( l + b) h ( 5. + 0) 8 (.) 5 8 45 8 60 cm (QLateral surface area of cuboidal box) > (Lateral surface area of cuboidal box) (Q400 > 60) Required area ( 400 60) cm 40 cm (ii) Total surface area of cubical box 6l 6( 0) 6 00 600 cm Total surface area of cuboidal box ( l b + b h + h l) ( 5. 0 + 0 8 + 8 5.) ( 5 + 80 + 00) 05 60 cm (Area of cuboidal box) > (Area of cubical box) (Q60 > 600) Required area ( 60 600) cm 0 cm Question 6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 0 cm long, 5 cm wide and 5 cm high. (i) What is the area of the glass? (ii) How much of tape is needed for all the edges? Solution Dimension for herbarium are l 0 cm, b 5 cm and h 5 cm Area of the glass ( l b + b h + h l) ( 0 5 + 5 5 + 5 0) ( 50 + 65 + 50) ( 5) 450cm Length of the tape 4( l + b + h ) 4( 0 + 5 + 5) [QHerbarium is a shape of cuboid length 4( l + b + h] ) 4 80 0 cm
Question. Shanti Sweets Stalll was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 5 cm 0 cm 5 cm and the smaller of dimensions 5 cm cm 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ` 4 for 000 cm, find the cost of cardboard required for supplying 50 boxes of each kind. Solution Dimension for bigger box, l 5 cm, b 0 cm and h 5 cm Total surface area of the bigger size box ( l b + b h + h l) ( 5 0 + 0 5 + 5 5) ( 500 + 00 + 5) ( 5 ) 450 cm Dimension for smaller box, l 5 cm, b cmand h 5 cm Total surface area of the smaller size box ( 5 + 5 + 5 5) ( 80 + 60 + 5) ( 5 ) 60cm 5 Area for all the overlaps 5% 080 cm 080 cm 04 cm 00 Total surface area of both boxes and area of overlaps ( 080 + 04) cm 84 cm Total surface area for 50 boxes 84 50 cm Cost of the cardboard for 000 cm `4 Costs of the cardboard for cm ` 4 000 4 84 50 Cost of the cardboard for 84 50 cm ` `84 000 Question 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.5 m, with base dimensions 4 m m? Solution Dimension for shetter, l 4m, b mand h. 5 c m Required area of tarpaulin to make the shelter (Area of 4 sides + Area of the top) of the car ( l + b) h + ( l b) ( 4 + ) 5. + ( 4 ) ( 5.) + 5 + 4 m
Exercise. Assume π, unless stated otherwise. Question. The curved surface area of a right circular cylinder of height 4 cm is 88 cm. Find the diameter of the base of the cylinder. Solution We have, height 4 cm Curved surface area of a right circular cylinder 88 cm πrh 88 r 4 88 88 r 4 r cm Diameter Radius cm Question. It is required to make a closed cylindrical tank of height m and base diameter 40 cm from a metal sheet. How many square metres of the sheet are required for the same? Solution Let r be the radius and h be the height of the cylinder. 40 Given, r 0 cm 0. 0 m and h m Metal sheet required to make a closed cylindrical tank Total surface area πr ( h+ r) 0.( + 00. ) 0. 0.. 48m Hence, the sheet required to make a closed cylindrical tank. 48m Question. A metal pipe is 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 We have, h cm Outer diameter ( d ) 44. cm
and inner diameter ( d ) 4 cm Outer radius () r. cm Inner radius ( r ) cm (i) Inner curved surface area πr h 88 968 cm (ii) Outer curved surface area πrh. 44. 064. 8 cm (iii) Total surface area Inner curved surface area + Outer curved surface area + Areas of two bases 968 + 064. 8 + π ( r r ) 968 + 064.8 + [(.) ] [ 0. 8 + (. )] 4 84 4 0.8 + 44 0.84 0.8 + 44 0. 0.8 + 5.8 cm 08.08 cm Question 4. The diameter of a roller is 84 cm and its length is 0 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m. Solution We have, diameter of a roller 84 cm r radius of a roller 4 cm h 0 cm To cover revolution Curved surface area of roller πrh 4 0 44 0 cm 680 cm 680 00 00 m 68. m (Qm 00 cm) Area of the playground Takes 500 complete revolutions 500.68 m 584 m
Question 5. A cylindrical pillar is 50 cm in diameter and.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of `.50 per m. Solution Given, Diameter 50 cm 50 Radius, r 00 m 0. 5 m (Q cm 00 m) and h. 5 m Curved surface area of the pillar πrh 05. 5. Cost of painting per m `.50 Cost of painting 5.5 m `.50 5.5 ` 68.5 0.5 0.5 55. m Question 6. Curved surface area of a right circular cylinder is 4.4 m. If the radius of the base of the cylinder is 0. m, find its height. Solution We have, curved surface area of a right circular cylinder 44. m πrh 44. 0. h 4. 4 h 44 44 h m Hence, the height of the right circular cylinder is m. Question. The inner diameter of a circular well is.5 m. It is 0 m deep. Find (i) its inner curved surface area. (ii) the cost of plastering this curved surface at the rate of `40 per m. Solution We have, inner diameter. 5 m inner radius 5. m and h 0 m 5. (i) Inner curved surface area πrh 0 5 0 m (ii) Cost of plastering per m `40 Cost of plastering 0 m ` 40 0 ` 4400
Question 8. In a hot water heating system, there is a cylindrical pipe of length 8 m and diameter 5 cm. Find the total radiating surface in the system. Solution We have, h 8 m Diameter 5 cm Radius, r 5 5. cm.5 cm m 0.05 m (Q cm 00 00 m) Total radiating surface in the system Curved surface area of the cylindrical pipe πrh 0. 05 8 44. m Question 9. Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4. m in diameter and 4.5 m high. (ii) how much steel was actually used, if of the steel actually used was wasted in making the tank? Solution (i) We have, diameter 4. m Radius, r 4.. m and h 4. 5 m Curved surface area of a closed cylindrical petrol storage tank πrh. 45. 44 0. 45. 594. m (ii) Now, total surface area Curved surface area + πr 59.4 + (.) 59.4 +. 8. m Since, of the actual steel used was wasted, therefore the area of the steel which was actually used for making the tank of x of x x 8. x 8. x 045.44 95.04 m Actually steel used 95.04 m
Question 0. In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 0 cm and height of 0 cm. A margin of.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 Given, r 0 cm 0 cm h 0 cm Since, a margin of.5 cm is used for folding it over the top and bottom so the total height of frame, h 0 + 5. + 5. h 5 cm Cloth required for covering the lampshade Its curved surface area πr h( ) 440 5 440 5 00cm 0 ( 5 ) Question. 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 cm and height 0.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 5 competitors, how much cardboard was required to be bought for the competition? Solution Cardboard required by each competitor Base area + Curved surface area of one penholder πr + πrh [Given, h 0. 5 cm, r cm] () + 05. ( 8. 8 + 98) cm 6. 8 cm For 5 competitors cardboard required 5 6. 8 90 Hence, 90 cm of cardboard was required to be bought for the competition. cm
Exercise. Assume π, unless stated otherwise. Question. Diameter of the base of a cone is 0.5 cm and its slant height is 0 cm. Find its curved surface area. Solution We have, diameter 05. cm. Radius () r 05 5. 5 c m and slant height l 0 cm Curved surface area πrl 5. 5 0 65 cm Question. Find the total surface area of a cone, if its slant height is m and diameter of its base is 4 m. Solution We have, slant height l m and diameter 4 cm r 4 m Total surface area πr( l + r) π ( + ) π 96π 96 8 44 4 m Question. Curved surface area of a cone is 08 cm and its slant height is 4 cm. Find (i) radius of the base and (ii) total surface area of the cone. Solution We have, slant height, l 4 cm Curved surface area of a cone 08 cm πrl 08 r 4 08 08 r (i) Radius cm r cm
(i) Radius cm (ii) Total surface area of a cone πr r( + l ) ( + 4) 46 cm Question 4. A conical tent is 0 m high and the radius of its base is 4 m. Find (i) slant height of the tent. (ii) cost of the canvas required to make the tent, if the cost of m canvas is ` 0. Solution We have, h 0 m r 4m (i) Q l r + h l ( 4 ) + 0 56 + 00 66 6 m Hence, the slant height of the canvas tent is 6 m. (ii) Canvas required to make the tent Curved surface area of tent πrl π 4 6 64 π m Q Cost of m canvas ` 0 Cost of 64 π m canvas ` 0 64 π ` 0 64 ` 0 64 `80 Hence, the cost of the canvas is ` 80. Question 5. What length of tarpaulin m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 0 cm. (Use π. 4 ) Solution respectively. Given, Let r, h and l be the radius, height and slant height of the tent, r 6m, h 8m Q l r + h l 6 + 8 6 + 64 00 0 m Area of the canvas used for the tent Curve surface area of the cone πrl
4 m 4. 6 0 88. 4 m Area of tarpaulin required Length of tarpaulin required Width of tarpaulin 88. 4 (QWidth of tarpaulin, given) 6. 8 The extra material required for stitching margins and cutting 0 cm 0. m Hence, the total length of tarpaulin required 6. 8 + 0. 6 m Question 6. The slant height and base diameter of a conical tomb are 5 m and 4 m respectively. Find the cost of white-washing its curved surface at the rate of `0 per 00 m. Solution We have, slant height, l 5m and diameter 4m Radius, r m Curved surface area of the conical tomb πrl 5 5 550 m Cost of white washing per 00 m ` 0 Cost of white washing per m ` 0 00 Cost of white washing 550 m 0 550 ` 00 4 m 5 m `55 Question. A joker s cap is in the form of a right circular cone of base radius cm and height 4 cm. Find the area of the sheet required to make 0 such caps. Solution We have, r cm, h 4 cm We know that, l h + r l ( 4 ) + 56 + 49 65 l 5 cm Curved surface area of joker's cap πrl 5 5 550 cm QThe sheet required to make cap 550 cm The sheet required to make 0 caps 550 0 5500 cm m
Question 8. A bus stop is barricaded 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 m. If the outer side of each of the cones is to be painted and the cost of painting is ` per m, what will be the cost of painting all these cones? (Use π. 4 and take. 04. 0 ) Solution We have, diameter 40 cm 40 Radius, r 0 cm 0 00 m 0. m Q cm m 00 and height m Q l r + h l ( 0. ) + 0.04 +.04.0 m Curved surface area of a cone πrl 4. 0. 0. 064056. m Cost of painting per m ` Cost of painting 0.64056 m ` 064056. `.686 Cost of painting for cone `.686 Cost of painting 50 cones `.686 50 ` 84.6 ` 84.4
Exercise.4 Question. Find the surface area of a sphere of radius (i) 0.5 cm (ii) 5.6 cm (iii) 4 cm Solution (i) We have, r 05. cm Surface area of a sphere 4 πr 4 05. 05. 88.5 0.5 86 cm (ii) We have, r radius of the sphere 5. 6 c m Surface area 4 πr 4 56. 56. 88 08. 56. 94. 4 cm (iii) We have, r radius of the sphere 4 cm Surface area 4 πr 4 4 4 88 4 464 cm Question. Find the surface area of a sphere of diameter (i) 4 cm (ii) cm (iii).5 m Solution (i) We have, diameter 4 cm Radius, r cm Surface area of a sphere 4 πr 4 4 66 cm (ii) We have, diameter cm r 05. cm Surface area 4 πr 4 05. 05. 88 5. 05. 86 cm (iii) We have, diameter. 5 m r 5. 5. cm Surface area 4πr 4 5. 5. 6950. 8. 5 cm
Question. Find the total surface area of a hemisphere of radius 0 cm. (Use π. 4 ) Solution We have, r 0 cm Total surface area of a hemisphere πr 4. ( 0) 94. 00 94 cm Question 4. The radius of a spherical balloon increases from cm to 4 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. Solution Let initial radius, r cm After increases, r 4 cm Surface area for initial balloon 4 πr 4 88 A 66 cm Surface area for increasing balloon 4 πr 4 4 4 88 8 A 464 cm Required ratio A: A 66 : 464 : 4 Question 5. A hemispherical bowl made of brass has inner diameter 0.5 cm. Find the cost of tin-plating it on the inside at the rate of ` 6 per 00 cm. Solution We have, inner diameter 05. cm Inner radius 05. cm 5. 5 c m Curved surface area of hemispherical bowl of inner side πr (. 5 5) 5. 5 5. 5. 5 cm QCost of tin-plating on inside for 00 cm ` 6 QCost of tin-plating on the inside for.5 cm 6.5 ` `. 00 Question 6. Find the radius of a sphere whose surface area is 54cm. Solution Surface area of a sphere 54 cm 4 πr 54 4 r 54
r 54 4 r. 5 r. 5 c m Hence, the radius of the sphere is.5 cm.. 5 Question. The diameter of the Moon is approximately one-fourth of the diameter of the Earth. Find the ratio of their surface areas. Solution Let diameter of the Earth d Then, diameter of the Moon d 4 d Radius of the Earth, r d d Radius of the Moon, r 4 8 Surface area of the Earth 4 πr d S 4 π πd Surface area of the Moon 4 π d 4 π d 8 64 d S π 6 d d Hence, S: S π : π 6 S: S 6 : Question 8. A hemispherical bowl is made of steel, 0.5 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. Solution Outer radius of the bowl (Inner radius + Thickness) ( 5 + 0. 5) cm 5. 5 c m 5 cm 0.5 cm Outer curved surface area of the bowl πr 5. 5 5. 5. 5 cm
Question 9. A right circular cylinder just encloses a sphere of radius r (see figure). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii). Solution The radius of the sphere r Radius of the cylinder Radius of the sphere r Height of the cylinder Diameter r (i) Surface area of the sphere A 4 πr (ii) Curved surface area of the cylinder πrh A π r r A 4 πr (iii) Required ratio A : A 4πr : 4πr : r r r r
Exercise.5 Question. A matchbox measures 4cm 5. cm. 5cm. What will be the volume of a packet containing such boxes? Solution Volume of a match box 4cm.5 cm 5. cm 5 cm Volume of a packet 5 cm 80 cm Question. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? ( m 000L) Solution Volume of a cuboidal water tank 6m 5m 4.5m 0 4.5 m 5 m 5 000 L 5000 L (Q m 000 L) Question. A cuboidal vessel is 0 m long and 8 m wide. How high must it be made to hold 80 cubic metres of a liquid? Solution Volume of a cuboidal vessel hold liquid l b h 80 m 0 8 h 80 h 80 80 h 4. 5 m Hence, the cuboidal vessel must be made 4.5 m high. Question 4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and m deep at the rate of ` 0 per m. Solution Volume of a cuboidal pit l b h ( 8 6 ) m 44 m QCost of digging per m `0 Cost of digging 44 m ` 0 44 ` 40 Question 5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are.5 m and 0 m, respectively. Solution Given, l.5 m and h 0 m. Let breadth of tank be b. Capacity of a cuboidal tank 50000L 50000 m 50 m QL m 000 000 Q Volume of a cuboidal tank 50 m l b h 50 5. b 0 50 b 50 5 (Given)
b m Hence, breadth of the cuboidal tank is m. Question 6. A village, having a population of 4000, requires 50 litres of water per head per day. It has a tank measuring 0 m 5 m 6 m. For how many days will the water of this tank last? Solution Given, l 0 m,b 5m and h 6m Capacity of the tank Volume of the tank lbh ( 0 5 6) m 800 m QWater requirement per person per day 50 L Water required for 4000 persons per day ( 4000 50) L 4000 50 m (QL 000 000 m) 600 m Capacity of tank Number of days the water will last Total water required per day 800 600 0 Hence, the water will last for 0 days. Question. A godown measures 40 m 5 m 0 m. Find the maximum number of wooden crates each measuring 5. m 5. m 05. m that can be stored in the godown. Solution Dimension for godown, l 40 m, b 5 mand c 0 m Volume of the godown l b h 40 m 5 m 0 m Dimension for wooden gate l.5 m, b.5 m, b.5 m, h 0.5 m Volume of wooden gates l b h.5 m.5 m 0.5m Volume of the godown Number of wooden gates Volume of wooden gates 40m 5m 0m.5 m.5 m 0.5 m 0000 095. 0666
Question 8. A solid cube of side 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. Solution Volume of a solid cube cm cm cm 8 cm The solid cube is cut into eight cubes of equal volume. Hence, the volume of the new cube 8 cm 6 cm 8 (side) 6cm (Taking cube root) side 6cm Hence, side of the new cube 6 cm Surface area of solid cube 6 ( side) 6 ( ) cm S 6 44 864 cm Surface area of new cube 6 (side) 66 ( )cm S 6cm Required ratio S: S 864 : 6 4 : Question 9. A river m deep and 40 m wide is flowing at the rate of km per hour. How much water will fall into the sea in a minute? Solution Given, l km 000 m 000 m b 40 m and h m Since, the water flows at the rate of km h, i.e., the water from km of river flows into the sea in one hour. The volume of water flowing into the sea in one hour Volume of the cuboid l b h (000 40 ) m 40000 m The volume of water flowing into the sea in one minute 40000 m 60 4000 m
Exercise.6 Assume π, unless stated otherwise. Question. The circumference of the base of a cylindrical vessel is cm and its height is 5 cm. How many litres of water can it hold? (000cm L.) Solution We have, circumference of the base cm πr r r r cm Volume of cylinder πr h 5 4650 cm (Given, 000 cm L, cm L) 000 4650 4650 cm L 000 4.65 L Question. The inner diameter of a cylindrical wooden pipe is 4 cm and its outer diameter is 8 cm. The length of the pipe is 5 cm. Find the mass of the pipe, if cm of wood has a mass of 0.6 g. Solution Inner radius () r cm Outer diameter 8 cm Given, inner diameter 4 cm Outer radius ( r ) 4 cm h 5 cm 8 cm 4 cm 5 cm
Volume of cylindrical wooden pipe π( r r ) h ( 4 ) 5 4 + 4 5 ( )( ) 6 5 50 cm Q cm of wood has a mass 0. 6 g 50cm of wood have a mass 06. 50 g 4 g 4 kg (Q 000 000 kg g). 4 kg Question. 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 5 cm. (ii) a plastic cylinder with circular base of diameter cm and height 0 cm. Which container has greater capacity and by how much? Solution (i) We have, h 5 cm l 5 cm, b 4cm, h 5 cm Volume of cuboidical body l b h 5 4 5 00cm (ii) We have, Diameter cm Radius, r cm Height, h 0cm Then, volume of a plastic cylinder πr h 0 5 85 cm (i) (ii) From Eqs. (i) and (ii), we see that volume of a plastic cylinder has greater capacity and its capacity is 85 00 85 cm is more than the tin can. Question 4. If the lateral surface of a cylinder is 94. cm and its height is 5 cm, then find (i) radius of its base, (ii) its volume. (Use π. 4 ) Solution We have, lateral surface of a cylinder 94. cm and h 5 cm πrh 94. 4. r 5 94. r 94.. 4 r cm (i) Hence, radius of base, r cm (ii) Volume of a cylinder πr h 4. () 5 4. 9 5 4. cm
Question 5. It costs ` 00 to paint the inner curved surface of a cylindrical vessel 0 m deep. If the cost of painting is at the rate of `0 per m, find (i) inner curved surface area of the vessel, (ii) radius of the base, (iii) capacity of the vessel. Solution We have, cost to paint the inner curved surface `00 Cost to paint per m `0 Cost to paint the inner curved surface Q Inner curved surface area Cost to paint per m 00 πrh ` 0 m `0 r 0 0 0 r 0 r 4.5 m (i) Inner curved surface area of the vessel 0 m (ii) Hence, radius of the base is.5 m. (iii) Capacity of the vessel Volume of the vessel πr h 0 4 4 8 96.5 m Question 6. The capacity of a closed cylindrical vessel of height m is 5.4 litres. How many square metres of metal sheet would be needed to make it? Solution Capacity of a closed cylindrical vessel 5. 4 L 5. 4 000cm (QL 000 cm ) πr h 5400 cm πr 00 5400 r 54 (Qh m 00cm) r r cm Total surface area of closed cylindrical vessel πr( r+ h) ( + 00) 44 0 408 cm 408 The metal sheet must be required m (Q cm m) 00 00 00 0408. m
Question. 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 mm and the diameter of the graphite is mm. If the length of the pencil is 4 cm, find the volume of the wood and that of the graphite. Solution Diameter of the graphite cylinder mm cm Qmm cm 0 0 Radius, r 0 cm Length of the graphite, h 4 cm Volume of the graphite cylinder πr h 4 cm 0 0 0. cm Also, diameter of the pencil mm cm 0 Radius of the pencil, r 0 cm and length of the pencil, h 4 cm Volume of the pencil πr h 0 0 4 cm 5 9. m Hence, volume of wood Volume of the pencil Volume of the graphite (. 5 9 0. ) cm 5. 8 cm Question 8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 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 50 patients? Solution We have, diameter cm Radius, r cm h 4 cm Capacity of the cylindrical bowl Volume of the cylinder πr h 4 54 cm Volume of soup to be prepared daily to serve 50 patients 54 50 cm 8500 cm 8500 (Q cm 000 000 L) 8. 5 L
Exercise. Assume π, unless stated otherwise. Question. Find the volume of the right circular cone with (i) radius 6 cm, height cm (ii) radius.5 cm, height cm Solution (i) We have, r 6 cm and h cm Then, volume of the right circular cone πr h 6 6 6 64 cm (ii) Given, r. 5 c m and h cm Volume of the cone πr h.5.5 cm ( 05. 5. 4) cm 54 cm Question. Find the capacity in litres of a conical vessel with (i) radius cm, slant height 5 cm (ii) height cm, slant height cm Solution (i) We have, r cm and l 5 cm We know that l r + h h l r 5 65 49 56 4 cm Capacity of conical vessel πr h 4 8 cm 000 L. L Q cm L 000
(ii) We have, We know that, l h + r h cm and l cm r l h 69 44 5 5 cm Capacity of conical vessel πr h 5 5 00 00 cm 000 L 0.4L Q cm L 000 Question. The height of a cone is 5 cm. If its volume is 50 cm, find the radius of the base. (Use π. 4 ) Solution We have, volume of a cone 50 cm πr h 50 4. r 5 50 r 50 4. 5 Hence, radius of the base 0 cm r 0 cm 500 5 Question 4. If the volume of a right circular cone of height 9 cm is 48π cm, find the diameter of its base. Solution We have, volume of a right circular cone 48 π cm r 6 r 4 cm Hence, diameter of the base r 4 8 cm πr h 48 π r 9 48 (Qh 9 cm, given)
Question 5. A conical pit of top diameter.5 m is m deep. What is its capacity in kilolitres? Solution We have, diameter.5 m radius, r.5 m and h m Capacity of a conical pit πr h.5.5.5 0.5.5 8.5 m (m kl) 8.5 kl Question 6. The volume of a right circular cone is 9856 cm. If the diameter of the base is 8 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone Solution We have, d 8 cm r 4 cm Q Volume of a right circular cone 9856 cm πr h 9856 4 h 9856 9856 h 4 4 h (i) Height of the cone 48 cm (ii) We have, h 48 cm and r 4 cm Q l h + r 448 48 cm 8 48 + 4 04 + 96 500 l 50 cm (i) Hence, slant height of the cone is 50 cm.
cm (iii) Curved surface area of the cone πrl 4 50 44 50 00 cm Question. A right triangle ABC with sides 5 cm, cm and cm is revolved about the side cm. Find the volume of the solid so obtained. Solution On revolving the right ABC about the side AB, we get a cone as shown in the adjoining figure. A cm C C 5 cm B Volume of the solid so obtained πr h π 5 5 π 5 5 4 00π cm Hence, the volume of the solid so obtained is 00 π cm. Question 8. If the triangle ABC in the Question 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 and 8. Solution On revolving the right ABC about the side BC( 5 cm, ) wegeta cone as shown in the adjoining figure. A cm B cm 5 cm C Volume of solid so obtainedv M πr h π 5 40 π cm
π In above Question, the volume V obtained by revolving the ABC about the side cm i.e.,v 00πcm. V Hence, 00π 0 5 V 40π 4 Hence, the required ratio 5 : Question 9. A heap of wheat is in the form of a cone whose diameter is 0.5 m and height is m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required. 0.5 Solution We have, d 0.5 m, r 5.5 m and h m Q l r + h (. 5 5) +.565 + 9 6.565 6.046 6.05 m Volume of the heap of cone of wheat πr h 5. 5 5. 5 86.65 m Area of the canvas required curved surface area of the heap πrl 5.5 6.05 m 0.5 6.05 m 99.85 m
Exercise.8 Assume π, unless stated otherwise. Question. (i) cm Solution Find the volume of a sphere whose radius is (ii) 0.6 m (i) We have, r cm Volume of a sphere 4 πr 4 88 49 () 4 4 cm (ii) We have, r 0. 6 m Volume of a sphere 4 4 πr ( 0.6) 4 0.6 0.6 0.6 88 0.9 0. 0.6 0.48 m 0.48 m Question. Find the amount of water displaced by a solid spherical ball of diameter (i) 8 cm (ii) 0. m Solution (i) We have, d 8 cm r 4 cm Water displaced by a solid spherical ball 4 4 πr ( 4) 4 4 4 4 498. 66 cm (ii) Diameter of the spherical ball 0. m Radius 0. m 0.05 m Amount of water displaced by the spherical ball Volume of sphere 4 πr 4 0.05 0.05 0.05 0.00485 m
Question. The diameter of a metallic ball is 4. cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm? Solution We have, d 4. c m r. c m Volume of the metallic ball 4 4 πr (.) 4... 8. 808 cm The density of the metal per cm 89. g The density of the metal 8. 808 cm 8. 808 8. 9 g 45. 9 g Hence, the mass of the ball 45. 9 g Question 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? Solution Let diameter of the Earth is d. We have, diameter of Moon ( d ) diameter of the Earth 4 d d 4 d Hence, the radius of the Moon d 8 The radius of the Earth d 4 d Volume of the Earth ( V) 4 d π π 8 4 d Volume of the Moon ( V) π 8 4 π d 5 Required fraction V V V V V V 5 8 64 4 d π 8 4 d π 5
Question 5. How many litres of milk can a hemispherical bowl of diameter 0.5 cm hold? Solution Diameter 0.5 cm Radius, r 5. 5 c m Volume of a hemisphere πr 5. 5 5. 5 5. 5 085. cm Hemispherical bowl can hold milk 085. L Q cm L 000 000 0.0 L (Approx.) Question 6. A hemispherical tank is made up of an iron sheet cm thick. If the inner radius is m, then find the volume of the iron used to make the tank. Solution Inner radius () r m 00 cm Outer radius ( r ) ( 00 + ) cm (QInner radius + Thickness) 0 cm Outer volume 0 π( ) (QVolume 4 Inner volume 00 π( ) Hence, volume of the iron used to make the tank (Outer volume Inner volume) π[( 0) ( 00) ] ( 0 00) [( 0) + 0 00 + ( 00) ] 00 + 000 + 0000 ( ) 44 ( 00) 648.80 cm 648.80 m 00 00 00 0.06 m (Approx.) πr )
Question. Find the volume of a sphere whose surface area is 54 cm. Solution We have, surface area of a sphere 54 cm 4 πr 54 4 r 54 r 54 4 r cm Hence, volume of a sphere 4 4 π 9.666 cm 9.6 cm Question 8. A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ` 498.96. If the cost of white washing is `.00 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome. Solution Cost of white washed ` 498.96 Cost of white washing per square metre `.00 Cost of white washed Surface area of the dome Cost of white washing per square metre ` 498.96 `.00 49.48 sq m (i) Required surface area 49.48 sq m πr 49. 48 r 49. 48 r 49. 48 44 r 6.69 r 6. m (ii) Volume of the air inside the dome πr (.) 6 5.908 Cum (Approx.)
Question 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 of the new sphere. (ii) ratio of S and S. Solution (i) Volume of solid iron sphere 4 Volume of new sphere 4 πr 6 πr r 4 6 r 4 πr πr r ( r) r r Radius r of the new sphere r (ii) Surface area () S of solid iron sphere 4 πr Surface area ( S ) of new sphere 4 π( r ) 4π( r ) 6πr Required ratio SS : 4 πr : 6πr : 9 Question 0. A capsule of medicine is in the shape of a sphere of diameter.5 mm. How much medicine (in mm ) is needed to fill this capsule? Solution We have, diameter.5 mm Radius, r.5 mm Volume of capsule 4 πr 4.5.5.5 05. 5. 5.. 458.46 mm
Exercise.9 (Optional) Question. A wooden bookshelf has external dimensions as follows : Height 0cm, Depth 5cm, Breadth 85cm (see figure). 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 0 paise per cm and the rate of pointing is 0 paise per cm, find the total expenses required for palishing and painting the surface of the bookshelf. 85 cm 0 cm Solution Here, l 85 cm, b 5 cm and h 0 cm Area of the bookshelf of outer surface lb + bh + hl [ ( 85 5) + ( 0 5) + 85 0] cm ( 450 + 5500 + 950) cm 900 cm 5 cm Cost of polishing of the outer surface of bookshelf 0 900 ` 80 00 Thickness of the plank 5 cm Internal height of bookshelf ( 0 5) 00 cm Internal depth of bookshelf ( 5 5) 0 cm Internal breadth of bookshelf 85 5 5 cm Hence, area of the internal surface of bookshelf ( 5 0) + ( 00 0) + 5 00 000 + 4000 + 500 4500 cm So, cost of painting of internal surface of bookshelf 0 4500 ` 450 00 Hence, total costing of polishing and painting 80 + 450 ` 50
Question. The front compound wall of a house is decorated by wooden spheres of diameter cm, placed on small supports as shown in figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius.5 cm and height cm and is to be painted black. Find the cost of paint required if silver paint costs 5 paise per cm and black paint costs 5 paise per cm. Solution 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 ( 4πR πr ) 8π( 4R r ) (Given, R cm and r.5 cm) 8π 4 (. 5) 44 8π 4. 5 4 8 π( 48.5) cm Therefore, the cost of silver paint at the rate of 5 paise per cm 5 8 48.5 00 905 ` 5. 86 (Approx.) Hence, surface area to be black painted 8 Curved surface area of cylinder 8 πrh 8.5 58 cm Cost of black paint at the rate of 5 paise per cm 5 58 00 ` 6.40 Hence, total costing of painting 5.86 + 6.40 `84.6 (Approx.)
Question. The diameter of a sphere is decreased by 5%. By what per cent does its curved surface area decrease? Solution Let d be the diameter of the sphere. Surface area 4 π d d Qr πd On decreasing its diameter by 5%, the new diameter 5d 5 d d d 00 00 Hence, the new surface area 4 π d 4 π Decrease in surface area πd 4 π 9d πd 9 64 6 d 4 d 4 9 πd 6 6 Percentage decrease in surface area πd 00 % 6 πd 00 6 % 4. 5 %