Surface Areas and Volumes

Question Bank In Mathematics class X (TermII) 13 SURFACE bailiwickS AND loudnessS A. SUMMATIVE ASSESSMENT (c) Length of accident = TH G (a) Lateral come up region = 4l2 (b) natural fold up field of view = 6l2 (c) Length of diagonal = 3 l 3. Cylinder For a piston chamber of rung r and up incline h, we nominate (a) argonna of swerve sur acquaint = 2? rh BR O ER 2. regular hexahedron For a squ atomic f are 18 s diadem consonant of edge l, we maintain O YA L school textS EXERCISE 13. 1 Unless stated otherwise, take ? = 22 . 7 Q. 1. 2 city delays distributively of mountain 64 cm3 are joined mop up to eat up. scram the develop field of battle of the resulting cube-shaped. 2011 (T-II) 1 S l 2 ? b2 ? h2 5. Sphere For a study of roentgen r, we comport Surface theater of opepro percentns = 4? 2 6. Hemi field of view ( unharmed) For a cerebral cerebral cerebral cerebral cerebral hemi knowledge domain of r r we have (a) curving approach stadium = 2? r2 (b) get resurrect firmament = 3? r2 PR (a) Lateral come along electron orbit = 2h(l + b) (b) tot lift landing field = 2(lb + bh + lh) (d) natural erupt bailiwick of remove piston chamber = 2? h(R + r) + 2? (R2 r2) 4. Cone For a bevel framing of pinnacle h, rundle r and slant superlative l, we have (a) Curved egress field of view = ? rl = ? r h 2 ? r 2 (b) list line up field of operation = ? r2 + ? rl = ? r (r + l) Sol. allow the incline of blockage = y cm tawdriness of stoppage = 64 cm3 T hen, strength of engine block = side3 = y3 As per experimental condition ? y3 = 64 ? y3 = 4 3 AK AS HA 13. SURFACE AREA OF A COMBINATION OF SOLIDS 1. Cuboid For a cubic of dimensions l, b and h, we have (b) positive ascend land = 2? r2 + 2? rh = 2? r(r + h) (c) Curved rise region of roaring piston chamber = 2? h(R r), where R and r are tabuer and intimate radii N Q. 3. A monkey is in the manakin of a chamfer of rung 3. 5 cm mounted on a cer ebral cerebral hemi ambit of kindred spoke. The meat upside of the gip is 15. 5 cm. become the summarise arise nation of the make for dog. 2011 (T-II) Sol. roentgen of the retinal conoid = r of hemi range = 3. 5 cm fall stature of the toy = 15. 5 cm ? spinning top of the bevel = (15. 5 3. 5) cm = 12 cm bung flush of the strobile abidance (l ) = G O diam of the gob piston chamber = 14 cm 14 gas constant of the hollow cerebral hemi athletic field = cm = 7 cm 2 ? spoke of the grounding of the hollow piston chamber = 7 cm do stature of the peecraft = 13 cm ? bloom of the hollow piston chamber = (13 7) cm = 6 cm sexual advance reach of the pissingcraft = Inner surface area of the hemisphere + Inner surface area of the hollow piston chamber = 2? (7)2 cm2 + 2? (7) (6) cm2 = 98 cm2 + 84 cm2 = (98 + 84) cm2 22 = 182? cm2 = 182 ? cm2 = 26 ? 22 cm2 7 = 572 cm2. PR AK = Q. 2. A water supplycraft is in the exercise of a hollow hemisphere mounted by a hollow cylinder. The diam of the hemisphere is 14 cm and the chalk up prime of the vessel is 13 cm. settle the inner surface area of the vessel. 2011 (T-II) Sol. ? diam of the hollow hemisphere = (3. 5)2 ? (12) 2 cm = 156. 25 cm = 12. 5 cm jibe surface area of the toy = Curved surface area of the hemisphere + Curved surface area of the strobile cell = 2? (3. 5)2 cm2 + (3. 5) (12. 5) cm2 = 24. 5? cm2 + 43. 65 cm2 = 68. 25? cm2 = O TH ER YA L BR S Q. 4. A cubical block of side 7 cm is all overcome by a hemisphere. What is the greatest diam the hemisphere sewer have? summon the surface area of the unfluctuating. 2011 (T-II) Sol. Side of cubical block = 7 cm Side of cubical block = Diameter of hemisphere = 7 cm ? R = 7 7 ? R = cm 2 Surface area of unfaltering = Surface area of the cube nation of institute of hemisphere + C. S. A. of hemisphere 2 ? R2 + 2? R2 = 6 ? side = 6 (7)2 cm2 + ? R2 22 7 7 2 = 6 ? 7 ? 7 cm2 + ? ? cm 7 2 2 7? ? = ? 6 ? 49 ? 11? ? cm2 2? ? 77 ? ? ? 588 ? 77 ? 2 = ? 294 ? ? cm2 = ? ? cm . 2? 2 ? ? ? r 2 ? h2 2 AS 665 cm2 = 332. 50 cm2 2 HA 68. 25 ? 22 cm2 = 214. 5 cm2. 7 ? y = 4 cm Hence, side of cube is 4 cm. For the resulting cuboid aloofness (l ) = 4 + 4 = 8 cm breadth (b) = 4 cm aggrandisement (h) = 4 cm ? Surface area of the resulting cuboid = 2(lb + bh + hl ) = 2(8 ? 4 + 4 ? 4 + 4 ? 8) cm2 = 2(32 + 6 + 32) cm2 = 2(80) cm2 = one hundred sixty cm2. N Q. 5. A hemi orbiculate depression is cut out(p)(p) from one face of a cubical timberlanden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remain unfluctuating. Sol. Diameter of the hemisphere = l = Side of the cube = 45? mm2 + 25? mm2 = (45 + 25) mm2 = 70? mm2 22 = 70 ? mm2 = 220 mm2. 7 Hence, surface area of condensation = 220 mm2 O Q. 6. A medicine abridgement is in the shape of a cylinder with both hemispheres stuck to each of its determinations (see figure below). The distance of the e ntire capsule is 14 mm and the diameter of the capsule is 5 mm. remember its surface area. TH ER YA L BR Sol. Diameter of capsule = Diameter of hemisphere = Diameter of cylinder = 5 mm 5 wheel spoke of the hemisphere = r = mm 2 Height of the cylinder = 14 (2. 5 + 2. 5) mm = 9 mm Surface area of the capsule = Surface area of cylinder + 2 Surface area of hemisphere G O S = l2 ? ? ? 24 ? . 4 = 2? (2) (2. 1) m2 + (2) (2. 8) m2 = (8. 4? + 5. 6) m2 22 2 = 14? m2 = 14 ? m = 44 m2 7 ? Cost the analyze of the encamp at the rate of Rs 500 per m2 = Rs 44 ? 500 = Rs 22000 Hence, equal of the analyze is Rs 22000. Q. 8. From a hale cylinder whose prime is 2. 4 cm and diameter 1. cm, a strobilelike perdition of the same line of longitude and same diameter is hollowed out. Find the entire surface area of the remain impregnable to the nearest cm2. Sol. Height of cylinder = 2. 4 cm Height of cone = 2. 4 cm rundle of cylinder = r = rundle of cone = 0. 7 cm Slant tiptop, of the cone l = 3 PR ?l? ?l 2 2 ? 6l 2 = ? ? ? ? 6l = 4 ? 2? 2 wheel spoke of the cylinder = 2 m Total surface area of the live = Curved surface area of the cylinder + Curved surface area of the cone AK ?l? ?l? 2 = 2? ? ? ? 6l ? ? ? ? ? 2? ?2? 2 2 AS l 2 Surface area of the remain immobile = Surface area of hemisphere + Surface area of cube Area of rump of hemisphere ? rung of the hemisphere = Q. 7. A tent is in the shape of a cylinder crush by a conelike top. If the flush and diameter of the cylindric dowery are 2. 1 m and 4 m respectively and the slant acme of the top is 2. 8 m, take chances the area of the canvas used for reservation the tent. Also, find the constitute of the canvas of the tent at the rate of Rs 500 per m2 (note that the small of the tent volition not be covered with canvas. ) Sol. Radius of the cone = 2 m ? ? 5? 2 ? ? 5? 2 2 = 2? ? ? (9) mm + 2 ? 2? ? ? ? mm ? 2? ? 2? ? ? ? ? (0. 7)2 ? (2. 4) 2 cm = 2. 5 cm HA 1. 4 cm = 0. 7 cm 2 N Q. 9.A wooden article was made by scooping out a hemisphere from each end of a unattackable cylinder, as shown in figure. If the extremum of the cylinder is 10 cm, and its stolid is of rundle 3. 5 cm, find the entire surface area of the article. Sol. Height of cylinder = 10 cm Total surface area of the be solid = C. S. A. of cylinder + C. S. A. of cone + Area of house inhibit = 2? rh + ? rl + ? r2 = ? r (2 h + l + r) 22 = ? 0. 7 ? (2 ? 2. 4 + 2. 5 + 0. 7) cm2 7 22 7 = ? (4. 8 + 3. 2) cm2 7 10 22 7 = ? ? 8. 0 cm2 7 10 176 = cm2 = 17. 6 cm2 10 Hence, total remaining surface area = 17. 6 cm2 = 18 cm2. Radius of cylinder = 3. cm Total surface area of the article = C. S. A of cylinder + 2 C. S. A. of hemisphere = 2? (3. 5 (10) cm2 + 2 2? (3. 5)2 cm2 = 70 cm2 + 49? cm2 = (70 + 49) cm2 22 2 = 119? cm2 = 119 ? cm 7 = 17 ? 22 cm2 = 374 cm2. OTHER IMPORTANT QUESTIONS Q. 1. A rounded pencil sharpened at one edge is the combining of (a) a cone and a cylinder (b) frustum of a cone and a cylinder (c) a hemispher e and a cylinder (d) two cylinders Sol. (a) The accustomed shape is a combination of a BR O TH ER S PR AK Its surface area = 6 ? YA L AS change magnitude in surface area = ? Per cent increase = cone and a cylinder. G O Q. . If each edge of a cube is increase by 50%, the percentage increase in the surface area is (a) 25% (b) 50% (c) 75% (d) 125% Sol. (d) allow the edge of the cube be a. Then, its surface area = 6a2 150a 3a New edge = = . c 2 4 Q. 3. The total surface area of a hemisphere of universal gas constant 7 cm is 2011(T-II) (a) 447 ? cm2 (b) 239 ? cm2 (c) 147 ? cm2 (c) 174 ? cm2 Sol. (c) Total surface area of the hemisphere = 3? r2 = 3 ? ? 49 cm2 = 147? cm2 Q. 4. If two solid hemispheres of same base radius r are joined together along their bases, HA 9a 2 27a 2 = 4 2 27a 2 15a 2 6a2 = 2 2 15a 2 carbon ? 2 = 125% 6a 2 N hen slue surface area of this new solid is (a) 4? r2 (b) 6? r2 2 (c) 3? r (d) 8? r2 Sol. (a) The resulting solid will be a sphere of radius r. ? I ts cut surface area = 4? r2. Q. 9. The total surface area of a top (lattu) as shown in the figure is the sum of total surface area of hemisphere and the total surface area of cone. Is it true? Sol. No, the statement is false. Total surface area of the top (lattu) is the sum of the cut surface area of the hemisphere and the curved surface area of the cone. Sol. (d) We have ? 2 6a1 2 6a2 a13 a2 3 = AS 4 64 a1 ? = 3 a2 27 HA Q. 5. tidy sums of two cubes are in the ratio 64 27.The ratio of their surface areas is (a) 3 4 (b) 4 3 (c) 9 16 (d) 16 9 Q. 10. dickens cones with the same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so directed. N 32 Sol. True. Since the curved surface area taken together is same as the sum of curved surface areas measured separately. G O ?r r 2 ? h2 ? 2? rh . Is it true? YA L Q. 7. If a solid cone of base radius r and height h is located over a solid cylinder having same base radius and height as that of the cone, then the curved surface area of the shape is BR . . . Radius of the hemi global toy, r = 3. cm Curved surface area of the toy = 2? r2 22 =2? ? (3. 5)2 cm2 = 77 cm2 7 Total surface area of the toy = 3? r2 22 =3? ? (3. 5)2 cm2 = 115. 50 cm2. 7 O TH ER Q. 8. Two identical solid cubes of side a are joined end to end. Then find the total surface area of the resulting cuboid. Sol. The resulting solid is a cuboid of dimensions 2a ? a ? a. ? Total surface area of the cuboid = 2 (lb + bh + hl) = 2 (2a ? a + a ? a + a ? 2a) = 10a2. 5 S Q. 6. The diameter of a solid hemispherical toy is 7 cm. Find its curved surface area and total surface area. Sol. Diameter of the hemispherical toy = 7 cm. Q. 11. A tent of height 8. 5 m is in the pass water of a unspoilt broadsheet cylinder with diameter of base 30 m and height 5. 5 m, surmounted by a in force(p) airman cone of the same base. Find the cost of the canvas of the tent at the rate of Rs 45 per m2. Sol. PR 22 ? 8 ? 17 cm2 7 = 854. 85 cm2 = 855 cm2 (approx. ) = 2 (? rl) = 2 ? Height of the tent = 8. 25 m. Height of the rounded erupt = 5. 5 m . . . Height of the conelike vocalization = (8. 25 5. 5) m = 2. 75 m. 30 Base radius of the tent = m = 15 m. 2 . . . Slant height of the conic discontinue (15)2 + (2. 75)2 m = = 15. 25 m. = AK = 42 16 = = 16 9 9 Sol. Slant height of each cone = 82 ? 152 cm 64 ? 225 cm = 17 cm. ? Surface area of the resulting shape 225 + 7. 5625 m Curved surface area of the tent = curved surface area of the cylindric part + curved surface area of the conical part = 2? rh + ? rl = ? r (2h + l) 22 = ? 15 (2 ? 5. 5 + 15. 25) m2 7 ? 22 ? = ? ? 15 ? 26. 25? m 2 ? 7 ? = 1237. 50 m2. Rate of the canvas = Rs 45 per m2 . . . Cost of the canvas = Rs (1237. 50 ? 45) = Rs 55687. 50. Sol. Slant height of the cone = = = AS and height of the cone = 14 cm BR = 22 ? 7 ( 7 5 + 7) cm2 7 O TH = 7 ? 14 cm = 245 cm = 7 5 cm. Total surface area of the cone = ? rl + ? r2 = ? r (l + r) 2 2 ER Slant height of the cone = r 2 ? h2 = 154 ( 5 + 1) cm2 Surface area of the cube = 6 ? 142 cm2 = 1176 cm2 ? Surface area of the remaining solid left out after the cone is carved out = surface area of the cube area of base of the cone + curved surface area of the cone 22 2 ? ? = ? 1176 ? ? 7 ? 154 5 ? cm2 7 ? ? YA L = ? 1022 ? 154 5 ? cm2. ? ? Q. 13. A toy is in the form of a cone mounted on a hemisphere of super C base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy. 2007, 2011 (T-II) 6 G O S Q. 14. A solid is in the form of a proficient circular cylinder with hemispherical ends.The total height of the solid is 58 cm and the diameter of the cylinder is 28 cm. Find the total surface area of the solid. 2006 Sol. Q. 15. A toy is in the shape of a rightfield circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the PR Q. 12. A cone of maximum size is carved out from a cube of edge 14 cm. Find th e surface area of the cone and of the remaining solid left out after the cone carved out. Sol. Diameter of the cone = 14 cm = 625 cm = 25 cm ? Total surface area of the toy = Curved surface area of the hemisphere + Curved surface area of the cone = 2? r2 + ? rl = ? r (2r + l) =Radius of the each hemisphere = base radius of the cylinder = 14 cm Total height of the toy = 58 cm ? Height of the cylinder = 58 (14 + 14) cm = 30 cm ? Total surface area of the solid = 2? r2 + 2? rh + 2? r2 = 2? r (2r + h) 22 =2? ? 14 (2 ? 14 + 30) cm2 7 = 88 ? 58 cm2 = 5104 cm2. AK 22 ? 7 (14 + 25) cm = 858 cm2. 7 HA N Height of the toy = 31 cm Base radius of the cone = radius of the hemisphere = 7 cm ? Height of the cone = (31 7) cm = 24 cm r 2 ? h2 72 ? 242 cm 49 ? 576 cm cylindrical part are 5 cm and 13 cm respectively. The radii of the hemisphercial and conical parts are the same as that of the cylindrical part.Find the surface area of the toy if the total height of the toy is 30 cm. 2002 Sol. = 2? r2 + 2? rh + ? rl = ? r (2r + 2h + l ) = = 22 ? 5 (2 ? 5 + 2 ? 13 + 13) cm2 7 22 ? 5 ? 49 cm2 = 770 cm2. 7 TH PRACTICE EXERCISE 13. 1 A choose the correct resource (Q 1 7) 1. A funnel is the combination of (a) a cone and a cylinder (b) frustrum of a cone and a cylinder (c) a hemisphere and a cylinder (d) a hemisphere and a cone. 2. A plumbline (shahul) is the combination of (a) a cone and a cylinder (b) a hemisphere and a cone (c) frustrum of a cone and a cylinder (d) a sphere and a cylinderO ER = 144 ? 25 cm = 13 cm. Total surface area of the toy = curved surface area of the hemisphere + curved surface area of the cylinder + curved surface area of the cone BR 3. A shuttle cock used for playacting badminton has the shape of the combination of 2011 (T-II) (a) a cylinder and a cone (b) a cylinder and a hemisphere (c) a sphere and a cone (d) frustrum of a cone and a hemisphere 4. The height of a conical tent is 14 m and its floor area is 346. 5 m2. The continuance of 1. 1 m wide 7 G O YA L S canvas required to built the tent is (a) 490 m (b) 525 m (c) 665 m (d) 860 m 5.The ratio of the total surface area to the asquint surface area of a cylinder with base diameter 160 cm and height 20 cm is (a) 1 2 (b) 2 1 (c) 3 1 (d) 5 1 6. The radius of the base of a cone is 5 cm and its height is 12 cm. Its curved surface area is (a) 30? cm2 (b) 65? cm2 2 (c) 80? cm (d) none of these 7. A right circular cylinder of radius r cm and height h cm (h 2r) just encloses a sphere of diameter (a) r cm (b) 2r cm (c) h cm (d) 2h cm 8. Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6? r2. Is it true? PR Slant height of the cone = 122 ? 52 cm = 22 ? 612. 75 cm2 = 1925. 78 cm2. 7 ? mandatory cost of painting = Rs 5. 25 ? 1925. 78 = Rs 1010. 38. AK Radius of the cone = Radius of the cylinder = radius of the hemisphere = 5 cm. Total height of the toy = 30 cm Height of the cylinder h = 13 cm ? Height of the cone = 30 (13 + 5) cm = 12 cm. Internal radius (r) of the vessel = 12 cm Total surface area of the vessel = 2? R2 + 2? r2 + ? (R2 r2) = 2 ? (12. 5)2 + 2 ? 122 + (12. 52 122) cm 2 = 312. 5 + 288 + 12. 25 cm 2 AS HA Q. 16. The internal and orthogonal diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively.If the cost of painting 1 cm2 of the surface area is Rs 5. 25, find the total cost of painting the vessel all over. 2001 Sol. External radius (R) of the vessel = 12. 5 cm N ER 16. A rocket is in the form of a cone of height 28 cm, surmounted over a right circular cylinder of height 112 cm. The radius of the bases of cone and cylinder are equal, each being 21 cm. Find the total surface area of the rocket. ? ? = ? ? ? ? 7? 22 G 13. 2 VOLUME OF A COMBINATION OF SOLIDS 1. meretriciousness of a cuboid of dimensions l, b and h = l ? b ? h. 2. mountain of a cube of edge l = l3. 3. hoi polloi of a cylinder of base radius r and height h = ? 2h. O YA L BR 4. hoi polloi of a cone of base radius r and height 1 h = ? r2h. 3 4 3 5. mickle of a sphere of radius r = ? r . 3 2 6. deal of a hemisphere of radius r = ? r3. 3 textbookS EXERCISE 13. 2 22 . 7 O TH Unless stated otherwise, take ? = Q. 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the vividness of the solid in cost of . 8 S PR Sol. AK 9. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape organize is 4? h + 4? r2. Is it true? 10. A solid clod is exactly fitted inside the cubical box of side a. Surface area of the evening gown is 4? a2. Is it true? 11. From a solid cylinder whose height is 2. 4 cm and diameter 1. 4 cm, a conical fossa of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2. 12. A de corative block shown below, is made of two solids a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter 4. 2 cm. Find the total surface area of the block. 22 ? ? = ? . ? 7? 2011 (T-II) 3. A tent of height 3. 3 m is in the form of a right circular cylinder of diameter 12 m and height 2. 2 m, surmounted by a right circular cone of the same diameter. Find the cost of canvas of the tent at the rate of Rs 500 per m2. 15. A solid is composed of a cylinder with hemispherical ends. If the entirely distance of the solid is 108 cm and the diameter of hemispherical ends is 36 cm, find the cost of embellish the surface at the rate of 7 paise per cm2. AS HA 14. cardinal cubes each of side 5 cm are joined end to end. Find the surface area of the resulting cuboid. N O YA L BR O 1 ? 3 ? 2 cm = ? cm3. 3 3 ? Q. 2.Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its t wo ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the stack of look contained in the model that Rachel made. (Assume the satellite and inner dimensions of the model to be nearly the same. ) Sol. = ? ?+ ? TH For conical portion Radius of the base (r) = G Height of cone (h1) = 2 cm 3 cm = 1. 5 cm 2 1 2 ? r h 3 9 We know that, loudness of cone = ER 22 3 cm = 66 cm3 7 Hence, the multitude of the air contained in the model that Rachel made is 66 cm3. 21 ? S Q. 3. A gulab jamun, contained sugar syrup up to about 30% of its pot. Find most how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2. 8 cm (see figure). 2011 (T-II) Sol. Gulab jamun is in the shape of cylinder with two hemispherical ends. Diameter of cylinder = 2. 8 cm ? Radius of cylinder = 1. 4 cm Height of cylindrical part = (5 1. 4 1. 4) cm = (5 2. 8) cm = 2. 2 cm PR AK AS Radius of the hemisphere = Radius of cone = 1 cm Height of cone = h = 1 cm 2 2 ? mess of hemisphere = ? r3 = ? (1)3 cm3 3 3 2 = ? m3 .. (i) 3 1 1 ? ledger of cone = ? r2h = ? (1)2 (1) cm3 3 3 1 = ? cm3 .. (ii) 3 intensity of the solid = Volume of the hemisphere + Volume of cone Volume of cone OAB = = 1 2 ? r h1 3 1 (1. 5)2 (2) ? cm3 = 1. 5? cm3 (i) 3 1 Volume of cone A? B? O? = ? r2h1 3 1 = (1. 5)2 ? (2) ? cm3 = 1. 5? cm3 (ii) 3 For cylindrical portion Radius of the base (r) = 1. 5 cm Height of cylinder h2= 12 cm (2 + 2) cm = 8 cm ? Volume of cylinder = ? r2h2 = ? (1. 5)2 (8) cm3 = 18? cm3 .. (iii) Adding equations (i), (ii) and (iii), we have Total people of the model = saturation of the two cones + masses of the cylinder. = 1. 5? cm3 + 1. ? cm3 + 18? cm3 = 21? cm3 HA N Volume of a gulab jamun 2 2 = ? (1. 4)3 cm3 + ? (1. 4)2 (2. 2) cm3 + ? (1. 4)3cm3 3 3 = = 1 22 14 3 ? ? 0. 25 ? cm 3 7 10 ER 4 = ? (1. 4)3 cm3 + ? (1. 4)2 (2. 2)cm3 3 ? 4 ? 1. 4 ? ? 2. 2 ? cm3 = ? (1. 4)2 ? 3 ? ? ? 5. 6 ? 6. 6 ? = ? (1. 96) ? ? cm3 3 ? ? ? (1. 96) (12. 2) = cm3 3 ? Volume of 45 gulab jamuns ? (1. 96) (12. 2) = 45 ? cm3 3 = 15? (1. 96) (12. 2) cm3 22 ? 1. 96 ? 12. 2 cm3 = 15 ? 7 = 15 ? 22 ? 0. 28 ? 12. 2 = 1127. 28 cm3 30 ? Volume of syrup = 1127. 28 ? cm3 100 = 338. 184 = 338 cm3 (approximately) 11 cm3 30 ? Volume of quadruplet conical depressions 11 3 22 3 cm = cm = 1. 7 cm3 30 15 ? Volume of the wood in the pen stand = (525 1. 47) cm3 = 523. 53 cm3. =4? S PR Q. 5. A vessel is in the form of an upside-down cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is modify with water up to the brim. When consume shots, each of which is a sphere of radius 0. 5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of poke out story shots dropped in the vessel. Sol. Radius of cone = 5 cm Height of cone = 8 cm Volume of cone = = AK = = O YA L BR Q. 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens.The dimensions of the cuboid are 15 cm by 10 cm by 3. 5 cm. The radius of each of the depressions is 0. 5 cm and the skill is 1. 4 cm. Find the pile in the entire stands. (See figure). TH O Radius of spherical lead shot, r1 = 0. 5 cm ? Volume of a spherical lead shot G Sol. Length of cuboid, l = 15 cm comprehensiveness of cuboid, b = 10 cm Height of cuboid, h = 3. 5 cm Volume of the cuboid = 15 ? 10 ? 3. 5 cm3 = 525 cm3 Volume of a conical depression = 4 3 4 ? 3 ? r = ? (0. 5)3 cm3 = cm 3 1 3 6 ? Volume of water that flows out = 1 ? (0. 5)2 (1. 4) cm3 3 10 AS 1 ? strength of the cone 4 1 ? 200? ? 50? cm3 ? ? = 4? 3 ? 3 HA 2 1 ? r h = ? (5)2 8 cm3 3 3 200 ? cm3 3 N let the number of lead shots dropped in the vessel be n. Volume of n lead shots = As per condition, ? n? cm3 6 n? 50? = 6 3 = 31680? cm3 + 3840 cm3 = 35520 cm3 = 35520 ? 3. 14 cm3 = 111532. 8 cm3 ? Mass of the pole = 111532. 8 ? 8 g = 892262. 4 g = 892. 26 kg Hence, the mass of the pole is 892. 26 kg (approximately). BR O TH ER S Sol. Diameter of cylinder ABCD = 24 cm 24 cm3 2 = 12 cm Height of cylinder ABCD (h) = 220 cm ? Volume of cylinderABCD = ? r2h = (12)2 (220)cm3 = 31680? cm3 Base radius of cylinder A? B? C? D? , R = 8 cm Height of cylinder A? B? C?D? (H) = 60 cm ? Volume of cylinder A? B? C? D? = ? R2h = (8)2 (60) cm3 = 3840? cm3 ? Volume of solid push pole = Volume of the cylinder ABCD + Volume of the cylinder A? B? C? D? Base radius of cylinder ABCD, r = YA L PR Q. 6. A solid push pole consist of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use ? = 3. 14) Radius of the cone OAB (r) = 60 cm Height of cone OAB (h1) = 120 cm ? Volume of cone OAB 1 2 1 ? r h1 = ? (60)2 (120) cm3 3 3 = 144000? m3 Radius of the hemisphere (r) = 60 cm = ? Volume of hemisphere = = = Radiu s of the cylinder (r) = Height of cylinder (h2) = ? Volume of cylinder = 11 G O AK AS 50? 6 ? ? n = 3 ? ? n = 100 Hence, the number of lead shots dropped in the vessel is 100. Q. 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is one hundred eighty cm. Sol. HA N 2 3 ? r 3 2 ? (60)3 cm3 3 144000? m3 60 cm 180 cm ? r2h2 So, r = OTHER IMPORTANT QUESTIONS Q. 1. Volume of the largest right circular cone that can be cut out from a cube of edge 4. 2 cm is (a) 9. 7 cm3 (b) 77. 6 cm3 3 (c) 58. 2 cm (d) 19. 4 cm3 O TH YA L BR O Sol. (d) Radius of the cone = 4. 2 cm = 2. 1 cm. 2 ER 8. 5 cm 2 S Sol. Diameter of sphere = 8. 5 cm 4 ? 3. 14 ? 4. 25 ? 4. 25 ? 4. 25 cm3 + 8 ? 3. 14 cm3 3 = 321. 39 cm3 + 25. 12 cm3 = 346. 51 cm3 = Henc e, she is correct. The correct volume is 346. 51 cm3. remains unfilled. Then the number of stains that the cube can accommodate is (a) 142296 (b) 142396 (c) 142496 (d) 142596 Sol. a) Volume of the cube = 223 cm3 = 10648 cm3 space which remains unfilled G Height of the cone = 4. 2 cm. 1 ? Volume of the cone= ? r2h 3 = PR Q. 8. A spherical frosting vessel has a cylindrical neck 8 cm long, 2 cm in diameter the diameter of the spherical part is 8. 5 cm. By measuring the kernel of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, winning the above as the inside measurements, and ? = 3. 14. Amount of water it holds = 4 ? 8. 5 ? ? ? ? cm3 + ? 12 (8) cm3 3 ? 2 ? 10648 cm3 = 1331 cm3 8 Remaining space = (10648 1331) cm3 = = 9317 cm3 1 22 ? ? 2. 1 ? 2. 1 ? . 2 cm3 = 19. 404 cm3. 3 7 Q. 2. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0. 5 cm and it is assumed that 1 space of the cube 8 12 4 ? (0. 25)3 cm3 3 Let n m arbles can be accommodated. Volume of 1 marble = Then, n ? AK 3 4 22 ? ? (0. 25)3 = 9317 3 7 AS HA = ? (60)2 (180) cm3 = 648000? cm3 ? Volume of water left in the cylinder = Volume of the cylinder Volume of the cone + Volume of the hemisphere = 648000? cm3 144000? + 144000? cm3 = 648000 cm3 288000? cm3 = 360000 cm3 360000? = m3 = 0. 36? m3 100 ? 100 ? 100 22 3 = 0. 36 ? m = 1. 131 m3 (approx. 7 Radius of cylindrical neck = 1 cm Height of cylindrical neck = 8 cm N ?n= 9317 ? 3 ? 7 4 ? 22 ? (0. 25) 3 = 142296. Q. 3. A medicine capsule is in the shape of a cylinder of diameter 0. 5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The energy of the capsule is (a) 0. 36 cm3 (c) 0. 34 cm3 Sol. (a) (b) 0. 35 cm3 (d) 0. 33 cm3 Q. 5. The volume of a sphere (in cu. cm) is equal to its surface area (in sq. cm). The diameter of the sphere (in cm) is 2011 (T-II) (a) 3 (b) 6 (c) 2 (d) 4 4 3 ? r = 4? r 2 3 ? r = 3 ? d = 2r = 2 ? 3 = 6 cm Sol. (b) BR = 22 ? ? ? (0. 25)2 ? ? 0. 25 ? 1. 5? cm3 3 7 ? ? O TH Height of the cylindrical part = (2 0. 5) cm = 1. 5 cm Radius of each hemispherical part = Radius of the cylindrical part = 0. 25 cm. ? ability of the capsule 4 ? 4 ? = ? r3 + ? r2h = ? r2 ? r ? h ? 3 ? 3 ? Q. 7. The ratio in the midst of the radius of the base and the height of the cylinder is 2 3. If its volume is 1617 cm3, the total surface area of the cylinder is 2011 (T-II) (a) 208 cm2 (b) 77 cm2 (c) 707 cm2 (d) 770 cm2 Sol. (d) Let the radius and height of the cylinder be 2x and 3x respectively. Then, volume of the cylinder = ? r2h 22 ? 1617 = ? 2x)2 ? 3x 7 YA L = 22 ? 5. 5 ? ? (0. 25)2 ? ? cm3 = 0. 36 cm3 7 ? 3 ? ER Q. 4. A solid piece of iron in the form of a cuboid of dimensions 49 cm ? 33 cm ? 24 cm is moulded to form a solid sphere. The radius of the sphere is 2011 (T-II) (a) 25 cm (b) 21 cm (c) 19 cm (d) 23 cm Sol. (b) Volume of sphere = Volume of cuboid S PR 4 3 ? r1 r 8 2 3 = ? 1 = 4 3 27 r2 3 ? r 3 2 ? Ratio between surface areas = 4 9 1617 ? 7 343 = 22 ? 4 ? 3 8 ? x = 3. 5 cm. ? Total surface area of the cylinder = 2? r (h + r) ? x3 = G O AK ? 4 3 ? r = (49 ? 33 ? 24) cm3 = 38808 cm3 3 38808 ? 3 ? 7 cm 3 = 9261 cm 3 4 ? 22 ? r3 = r = 21 cm Q. 8. On increasing each of the radius of the base and the height of a cone by 20%, its volume will be increased by (a) 25% (b) 40% (c) 50% (d) 72. 8% 13 AS Q. 6. The ratio of the volumes of two spheres is 8 27. The ratio between their surface areas is 2011 (T-II) (a) 2 3 (b) 4 27 (c) 8 9 (d) 4 9 Sol. (d ) 22 ? 7 (10. 5 + 7) cm2 7 = 44 ? 17. 5 cm2 = 770 cm2. =2? HA N Sol. (d) Volume of the original cone = New radius New height 1 2 ? r h 3 = 6r 120r = = 5 100 6h 120h = = 5 100 2 4 3 3 2 3 ? = = 3 2? 2? 2? 6 ? 2? 3 ? = 6 ? Hence, ratio of the volume of sphere to that of cube = cm. Then, volume of the metallic solid cylinder of 91 2 ? r h. 375 ? Per cent increase in volume = AK ? 216 ? 125 ? 2 = ? ? ? r h ? 375 ? height 10 = BR Q. 9. A sphere a nd a cube have the same surface. fancy that the ratio of the volume of sphere to that of the cube is 6 ? O 91? 100 ? 3 = 72. 8%. 375 TH ER = 91 2 100 ? r h ? 1 2 375 ? r h 3 2 cm. 3 = Volume of the metal in the spherical shell 32 4 2 = ? 53 ? 33 ? r ? 3 3 32 2 4 r = (125 ? 27) ? 3 3 3 4 ? ? 98 ? r2 = 32 3 49 7 ? r = cm ? r2 = 4 2 Hence, the diameter of the base of the cylinder AS ( Increase in volume = 72 2 1 ? r h ? r2h 3 125 2011 (T-II) Sol. Let the radius of the sphere be r and the edge YA L O of the cube be x. Whole surface area of sphere = 4? r2 and whole surface area of cube = 62. According to question, ? S Q. 11. A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is 4 3 ? a . Is it true? 3 PR = 7 cm. Sol. Diameter of the ball = side of the cube ? Radius of the ball = ? Volume of the ball = G 4? r2 = 62. r2 x 2 = 6 3 r = ? = 4? 2? x 3 2? 4 3 ? r Volume of sphere 3 Now, = Volume of cube x3 = Hence, the statement is false. 4 ? r? 4 ? r? r ? = ? ? 3 ? x? 3 ? x? x 3 2 Q. 12.From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid. 14 HA ) a 2 1 ? 6r ? 6h New volume = ? ? ? ? 3 ? 5 ? 5 72 2 = ? r h. 125 Q. 10. The internal and external radii of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted to form a solid 2 cylinder of height 10 cm, find the diameter of 3 the cylinder. 2011 (T-II) Sol. Let the radius of the base of the cylinder be 4 a3 ? a3 = 3 8 6 N Sol. Volume of the cube = 73 cm3 = 343 cm3 Sol. 1 ? ? 32 ? 7 cm3 3 = 66 cm3 ? Volume of the remaining solid = (343 66) cm3 Volume of the cone = = 277 cm3.AK = = Q. 13. The difference between the outer and inner curved surface areas of a hollow right circular cylinder 14 cm long is 88 cm2. If the volume of metal used in do cylinder is 176 cm3, find the outer and inner diameters of the cylinder. 2010 Sol. Let the inner and outer radii of the cylinder be r cm and R cm respectively. Then, the height of the cylinder = 14 cm. Inner surface of the cylinder = 2? r ? 14 cm2 = 28? r cm2 Outer surface of the cylinder = 2? R ? 14 cm2 = 28? R cm2 Difference of the two surfaces = (28? R 28? r) ? 88 = 28? (R r) ? AS Radius of the hemispherical portion = 5 cm = radius of the cone. Height of the conical portion = (10 5) cm = 5 cm. Capacity of the shape = PR TH (R r) = 88 ? 7 =1 28 ? 22 ER 1 2 ? r (2r + h) 3 1 22 = ? ? 5 ? 5 (2 ? 5 + 5) cm3 3 7 2750 22 ? 25 = ? 15 cm3 = cm3. 7 21 ? Rr= 1 (i) Volume of the metal used in do the cylinder = ? (R2 r2) ? 14 cm3 . .. 176 = ? (R + r) (R r) ? 14 BR O S 1 2750 ? cm3. 6 7 ? Required volume of the frosting-skating rink cream Space which remains unfilled = ? 2750 2750 ? ? = ? ? cm3 6? 7 ? ? 7 2750 5 ? cm3 = 327. 4 cm3. 7 6 ? ? (R + r) = YA L 176 ? 7 =4 22 ? 1 ? 14 (ii) R = 2. 5 cm and G Solving (i) and (ii), we have r = 1. cm Hence, inner and outer diameters of the cylinder are 3 cm and 5 cm respectively. Q. 14. An ice cream cone, full of ice cream is having radius 5 cm and height 10 cm as shown. Calculate the volume of ice cream provided that its 1 part is left unfilled with ice cream. 6 O R+r= 4 Q. 15. A solid toy is in the form of a hemisphere surmounted by a right-circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of cube and the toy. Also, find the total surface area of the toy. Sol.Volume of the toy = Volume of the cone + Volume of the hemisphere = 1 2 2 1 ? r h + ? r3 = ? r2 (h + 2r) 3 3 3 15 HA 2 3 1 2 ? r + ? r h 3 3 N = 1 22 1408 ? ? 4 ? 4 (4 + 8) cm3 = cm3. 3 7 7 Sol. Capacity of the box = 16 ? 8 ? 8 cm3 = 1024 cm3 Volume of the 16 glass spheres 4 = 16 ? ?r3 3 4 22 = 16 ? ? ? 2 ? 2 ? 2 cm3 3 7 11264 = cm3 21 Volume of water filled in the box 11264 ? ? 10240 = ? 1024 ? cm3 ? cm3 = 21 ? ? 21 A cube circumscribes this toy, hence edge of the cube = 8 cm. Volume of the cube = 83 cm3 = 512 cm3 ? Required difference in the volumes of the toy and the cube = 487. 61 cm3. 1408 ? ? = ? 512 ? ? cm3 7 ? ? 2176 cm3 = 310. 6 cm3. 7 Total surface area of the toy = curved curface area of the cone + curved surface area of the hemisphere = 2 2 2 = ? r h ? r ? 2? r 2 ? 2 ? = ? r ? h + r + 2 r ? ? ? = YA L 22 ? 4 ? 16 ? 16 ? 2 ? 4 ? cm2 ? ? 7 BR O TH ER diameter of the dome is equal to its total height above the floor, find the height of the structure. 2001 Sol. Let the internal height of the cylindrical part be h and the internal radius be r. Then, total height of the mental synthesis =h+r Also, 2r = h + r ? h = r. Now, volume of the building = Volume of the cylindrical part + Volume of the hemispherical part ? ? ? ? S PR and contains 41 O 22 ? 4 ? ? 4 2 ? 8 ? cm2 = ? 7 88 ? 4 = 7 ? 2 ? 2 cm2 ? G 88 ? 4 = ? 3. 41 cm2 = 171. 47 cm2. 7 Q. 16. 16 glass spheres each of radius 2 cm are packed into a cubical box of internal dimensions 16 cm ? 8 cm ? 8 cm and then the box is filled with water. Find the volume of water filled in the box. 16 880 ? 3 ? 7 =8 21? 5 ? 22 ? r =2 Hence, height of the building = h + r r3 = = (2 + 2) m = 4 m. AK 41 Q. 17. A building is in the form of a cylinder surmounted by a hemispherical valuted dome 19 m3 of air. If the internal 21 2 880 = ? r3 + ? r3 ? r = h 3 21 5? r 3 880 = 21 3 AS 2 19 = ? r2h + ? r3 3 21 HA N Q. 18. A godown building is in the form as shown in the figure.The good cross section parallel to the width side of the building is a rectangle 7 m ? 3 m, mounted by a semicircle of radius 3. 5 m. The inner measurements of the cuboidal portion of the building are 10 m ? 7 m ? 3 m. Find the volume of the godown and the total interior surface area excluding the floor 22 ? ? (base). ? ? = ? . ? 7 ? ? 1 2? = 2 ? ?r ? = ? r2 ? 2 ? 22 ? (3. 5) 2 m2 = 38. 5 m2 7 Total interior surface area excluding the base floor = area of the four walls = = 250. 5 m2. Sol. The godown building consists of cuboid at the bottom and the top of the building is in the form of half of the cylinder.Length of the cuboid = 10 m, Breadth of the cuboid = 7 m Height of the cuboid = 3 m Volume of the cuboid = lbh = 10 ? 7 ? 3 m3 = 210 m3. Radius of the cylinder = 3. 5 m Length of the cylinder = 10 m 1 2 Volume of the half of the cylinder = ? r h 2 1 22 = ? ? (3. 5)2 ? 10 m3 2 7 = 192. 5 m3 Volume of the godown = volume of the cuboid + volume of the half cylinder = (210 + 192. 5) m3 = 402. 5 m3 inner(a) surface area of the cuboid = Area of four walls = 2 (l + b) h = 2(10 + 7) 3 m2 = 102 m2 Interior curved surface area of half of the cylinder 22 = ? rh = ? 3. 5 ? 10 m2 = one hundred ten m2 7 YA L BR O TH ER Q. 19.A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2. 1 m and 4 m respectively and the slant height of the top is 2. 8 m, find the area of canvas used for making the tent. Find the cost of the canvas of the tent at the rate of Rs 550 per m2. Also, find the volume of air enwrap in the tent. 2008C Sol. O S G PR Height of the cone, H = AK ? 2. 8 ? 2 ? 22 = 7. 84 ? 4 m = 1. 95 m Area of canvas required for making the tent = Curved surface area of the tent = Curved surface area of the cylindrical part + curved surface area of the conical part = 2? rh + ? l = ? r (2h + l ) = Interior area of two semicircles 17 22 ? 2 (2 ? 2. 1 + 2. 8) m2 7 AS m HA 1 (curved surface area of the cylinder) 2 + 2 (area of the semicircle) = (102 + 110 + 38. 5) m2 + N 44 ? 7 m2 = 44 m2. 7 Cost of canvas = Rs 500 ? 44 = Rs 22000. Volume of the air enclosed in the tent = Volume of the cylindrical part + Volume of the conical part = = ? r2h + = = 88 8. 25 3 ? m = 34. 57 m3. 7 3 ER Q. 20. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm, is hollowed out. Find the volume of the remaining solid correct to two places of decimal.Also find the total surface area of th e remaining solid. (Take ? = 3. 14) 2008, 2011 (T-II) Q. 21. A juice seller serves his customers using a glass as shown in the figure. The inner diamater of the cylindrical glass is 5 cm, but the bottom of the glass has a hemispherical portion raise which reduces the depicted object of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual capacity. (Use ? = 3. 14) 2009 Sol. Radius of the cylindrical glass r = 2. 5 cm Radius of the cylinder = radius of the cone = 6 cm. Height of the cylinder = height of the cone = 8 cm. Volume of the remaining solid 1 2 = ? 2h ? r2h = ? r2h 3 3 2 = ? 3. 1416 ? 36 ? 8 cm3 3 = 603. 19 cm3 Slant height of the cone, l O YA L BR O TH Sol. G S Q. 22. A cylindrical vessel with internal diamater 10 cm and height 10. 5 cm is full of water. A solid cone of the diameter 7 cm and height of 6 cm is completely immersed in water. Find the volume of (i) water displaced out of the cylindrical vessel. (ii) water left in the cylindrical vessel. Take ? = 18 PR Height of the glass = 10 cm unvarnished capacity of the glass = ? r2h = 3. 14 ? 2. 5 ? 2. 5 ? 10 cm3 = 196. 25 cm3 Volume of the hemispherical portion 2 2 = ? r3 = ? 3. 14 ? 2. 5 ? 2. 5 ? 2. 5 cm3 3 3 = 32. 71 cm3 ?Actual capacity of the glass = (196. 25 32. 71) cm3 = 163. 54 cm3. AK AS 22 7 HA 1. 95 ? 22 ? ? 22 ? 2. 1 ? m3 3 ? 7 ? ? N H? 1 2 ? ?r H = ? r2 ? ? h ? ? 3? 3 ? = 36 ? 64 cm = 10 cm Total surface area of the remaining solid = curved surface area of the cylinder + area of top + curved surface area of the cone = 2? rh + ? r2 + ? rl = ? r (2h + r + l) = 3. 14 ? 6 (16 + 6 + 10) cm2 = 18. 84 ? 32 cm2 = 602. 88 cm2. = r 2 ? h2 2009 Sol. Radius of the cylinder, r = 5 cm Height of the cylinder, h = 10. 5 cm Capacity of the vessel = ? r2h 22 = ? 5 ? 5 ? 10. 5 cm3 = 825 cm3 7 1 Volume of the cone = ? r2h 3 1 22 = ? ? 3. 5 ? 3. 5 ? 6 cm3 = 77 cm3. 7 (i) Water displaced out of the cylinder = Volume of the cone = 77 cm3 (ii) Water left in the cylindrical vessel = Capacity of the vessel Volume of the cone = (825 77) cm3 = 748 cm3. 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0. 5 cm and abstrusity is 2. 1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand. Sol. Volume of a cuboid = 10 ? 5 ? 4 cm3 = 200 cm3. Volume of the conical depression Choose the correct option (Q 1 5) 1. The surface area of a sphere is 154 cm2. The volume of the sphere is 2 1 (a) 179 cm3 (b) 359 cm3 3 2 2 3 1 (c) 1215 cm (d) 1374 cm3 3 3 2.The ratio of the volumes of two spheres is 8 27. The ratio between their surface areas is (a) 2 3 (b) 4 27 (c) 8 9 (d) 4 9 3. The curved surface area of a cylinder is 264 m2 and its volume is 924 m3. The height of the cylinder is (a) 3 m (b) 4 m (c) 6 m (d) 8 m 4. The radii of the base of a cylinder and a cone of same height are in the ratio 3 4. The ratio between their volumes is (a) 9 8 (b) 9 4 (c) 3 1 (d) 27 16 TH ER PR ACTICE EXERCISE 13. 2A 5. The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the figure is (a) ? 2h (b) ? r 2 ? 3h ? 2r ? 3 ? r 2 ? 3h ? 2r ? (c) 3 YA L BR O S 6. Two solid cones A and B are placed in a cylindrical tube as shown in the figure. The ratio of their capacities is 2 1. Find the heights and capacities of the cones. Also, find the volume of the remaining portion of the cylinder. G O 7. Marbles of diameter 1. 4 cm are dropped into a cylindrical beaker of diameter 7 cm containing 19 PR Q. 23. A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold pens and pins respectively. The dimensions of the cuboid are 4 22 ? ? (0. 5)2 ? 2. cm3 3 7 = 2. 2 cm3 Volume of cubical depression = 33 cm3 = 27 cm3. ? Volume of wood in the entire stand = 200 (2. 2 + 27) cm3 = 170. 8 cm3. = (d) ?r 3 (3h + 4r ) 3 AK AS HA 1 2 1 22 ? r h = ? ? (0. 5)2 ? 2. 1 cm3 3 3 7 Volume of 4 con ical depressions = N 11. An ice cream cone consists of a right circular cone of height 14 cm and the diameter of the circular top is 5 cm. It has a hemispherical scoop of ice cream on the top with the same diameter as of the circular top of the cone. Find the volume of ice cream in the cone. 12. A solid toy is in the form of a hemisphere surmounted by a right circular cone.Height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover. 2011 (T-II) 13. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and gum olibanum the level of water is raised by 6. 75 cm. What is the radius of the ball? 13. 3 CONVERSION OF SOLID FROM ONE SHAPE TO ANOTHER TEXTBOOKS EXERCISE 13. 3 22 , unless stated otherwise. 7 Q. 1. A metallic sphere of radius 4. 2 cm is melted and redo into the shape of a cylinder of Take ? = 20 G O YA L BRO TH ER S 16. A heap of rice is in the form of a cone of diameter 9 m and height 3. 5 m. Find the volume of the rice. How much canvas material is required to just cover the heap? 17. 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0. 04 m3. 18. A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm respectively.If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket. (Take ? = 3. 14) radius 6 cm. Find the height of the cylinder. Sol. Radius of sphere = 4. 2 cm ? Volume of sphere = PR some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5. 6 cm. 8. A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3. 5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub, full of water, in such a way that the whole solid is submerged in water.If the radius of the cylinder is 5 cm and height 10. 5 cm, find the volume of water left in the cylindrical tub. 9. The largest possible sphere is carved out from a solid cube of side 7 cm. Find the volume of the sphere. 10. A cylindrical boiler, 2 m high, is 3. 5 m in diameter. It has a hemispherical lid. Find the volume of its interior, including the part covered 22 ? ? by the lid. ? ? = ? ? 7 ? 14. From a solid cylinder of height 12 cm and base diameter 10 cm, a conical cavity with the same height and diameter is carved out. Find the volume of the remaining solid. 15.A building is in the form of a cylinder surmounted by a hemispherical dome as shown in the figure. The base diameter of the dome is equal 2 of the total height of the building. Find the 3 height of the building, if it contains 67 1 m3 of 27 to AK AS air. HA N 2011 (T-II) 4 3 4 ? r = ? (4. 2)3 cm3 3 3 Volume of cylinder = ? R2H = ? (6)2H cm3 As per condition, Volume of the sphere = Volume of the cylinder 4 ? ? (4. 2)3 = ? (6)2H 3 ? ? Radius (r) = 7 m 2 2 deepness (h) = 20 m Volume of sphere of radius 6 cm 4 = ? (6)3 cm3 3 Volume of sphere of radius 8 cm ? (i) Hence, the height of the platform is 2. m. = As per condition, G ? ? 4 3 4 4 4 ? R = ? (6)3 + ? (8)3 + ? (10)3 3 3 3 3 3 = (6)3 + (8)3 + (10)3 R R3 = 1728 O YA L 4 3 3 ? R cm 3 BR 4 ? (10)3 cm3 (iii) 3 Let the radius of the resulting sphere be R cm. Then volume of the resulting sphere = TH ER 4 ? (8)3 cm3 3 Volume of sphere of radius 10 cm = (ii) Q. 4. A well of diameter 3 m dug 14 m deep. The dry land taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. 2011 (T-II) Sol. For well S PR O (iv) 3 m 2 Depth of well (h) = 14 m ?Volume o f earth taken out = ? r2h Radius of well (r) = AK H = Sol. We know that, volume of the sphere = 4 3 ? r 3 AS Q. 2. aluminiferous spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere. 245? 245 ? 22 ? H= ? H= 2. 5 308 308 ? 7 Diameter = 3 m 63 ? 3? = ? ? ? (14) m3 = ? m3 2 ? 2? Width of the embankment = 4 m Let the height of the embankment be H m. ? Radius of the well with embankment, R ? R = 3 1728 ? R = 12 Hence, the radius of the resulting sphere is 12 cm. Q. 3. A 20 m deep well with diameter 7 m is dug and the earth from digging is