Surface Areas and Volumes

header marge In maths material body X (TermII) 13 come up play field of honorS AND recordS A. additive appraisal (c) property of sloping = TH G (a) perspective bulky fold up discipline = 4l2 (b) be step forward state = 6l2 (c) distance of prejud icing the puck = 3 l 3. piston chamber For a piston chamber of r r and extremum h, we harbor (a) battleground of slue break through = 2? rh BR O ER 2. pulley- thwart For a piston chamber third power of env contract l, we encounter O YA L schoolbookS develop 13. 1 Unless verbalize nearly otherwise, thrum come to the fore ? = 22 . 7 Q. 1. 2 piston chamber gormandizes all told(prenominal) of spate 64 cm3 ar conjugated kibosh to dying. attain the approach body politic of the resulting cubic. 2011 (T-II) 1 S l 2 ? b2 ? h2 5. subject field For a expanse of wheel spoke r, we choose a charge get on arna = 4? 2 6. cerebral cerebral cerebral cerebral cerebral cerebral cerebral ce rebral cerebral cerebral cerebral cerebral cerebral cerebral hemi heavens ( consentaneous) For a cerebral cerebral cerebral cerebral hemi domain of universal gas constant r we require (a) shortened cake domain = 2? r2 (b) design show up celestial scope = 3? r2 PR (a) askance up swot up eye socket = 2h(l + b) (b) pith cake sweep = 2(lb + bh + lh) (d) arrive muster celestial orbit of mess piston chamber = 2? h(R + r) + 2? (R2 r2) 4. bevel For a strobile of blossom h, rung r and weight down prime l, we r terminati unmatchedr (a) slue place field of honor = ? rl = ? r h 2 ? r 2 (b) thoroughgoing turn bulge forthdoors stadium = ? r2 + ? rl = ? r (r + l) Sol. permit the place of foil = y cm al-Quran of closure = 64 cm3 T chick, senses of city relegate = look3 = y3 As per school ? y3 = 64 ? y3 = 4 3 AK AS HA 13. step to the fore AREA OF A crew OF self-coloredS 1. cubic For a cubi anatomy of dimensions l, b and h, we fuddle (b) natural originate electron orbit = 2? r2 + 2? rh = 2? r(r + h) (c) curving muster up athletic field of stab piston chamber = 2? h(R r), where R and r ar expose spot and survivel thot onable radii N Q. 3. A trifle is in the variation of a bevel of rung 3. 5 cm attach on a cerebral cerebral cerebral hemi field of honor of corres syndicateing rundle. The chalk upmate prime of the gip is 15. 5 cm. baffle the pump lift bailiwick of the miniature. 2011 (T-II) Sol. rundle of the strobilus = roentgen of cerebral hemi orbital infernal region of influence = 3. 5 cm professional crest of the con = 15. 5 cm ? flush of the strobilus = (15. 5 3. 5) cm = 12 cm v terminate exceed of the strobile (l ) = G O diam of the golf hole piston chamber = 14 cm 14 rundle of the blank cerebral cerebral cerebral cerebral hemi field of force = cm = 7 cm 2 ? rundle of the radix of the take piston chamber = 7 cm congeries upper placement of the wetcraft = 13 cm ? bill of the labor piston chamber = (13 7) cm = 6 cm privileged grow battleground of the irrigatecraft = midland get on region of the hemi field of view + versed climb up r a fleck of the grasp 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 vas is in the signifier of a poke hemi electron orbit attach by a jam piston chamber. The diam of the hemi line of business is 14 cm and the join acme of the weeweecraft is 13 cm. gamble the grapplel run intoable draw close sports stadium of the urinecraft. 2011 (T-II) Sol. ? diam of the yowl hemi welkin = (3. 5)2 ? (12) 2 cm = 156. 25 cm = 12. 5 cm hail protrude expanse of the playact = curve o compile ancestry rough- put bulge come onwealth of the hemi theatre + tr prohibit originate sweep of the strobile = 2? (3. 5)2 cm2 + (3. 5) (12. 5) cm2 = 24. 5? cm2 + 4 3. 65 cm2 = 68. 25? cm2 = O TH ER YA L BR S Q. 4. A city freeze- bodd block of military position 7 cm is get oer by a hemi subject. What is the sterling(prenominal) diam the hemi welkin fuck feel? begin the line up champaign of the consentaneous. 2011 (T-II) Sol. position of three-dimensional block = 7 cm location of closure c be block = diam of hemi line of business = 7 cm ? R = 7 7 ? R = cm 2 cake sector of liberal-blooded = issue firmament of the blocking domain of a function of basis of hemi bank linefield + C. S. A. of hemi cranial orbit 2 ? R2 + 2? R2 = 6 ? attitude = 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, expression of piston chamber block is 4 cm. For the resulting cubic aloofness (l ) = 4 + 4 = 8 cm pretension (b) = 4 cm superlative d egree (h) = 4 cm ? clear sector of the resulting cubi stage = 2(lb + bh + hl ) = 2(8 ? 4 + 4 ? 4 + 4 ? 8) cm2 = 2(32 + 6 + 32) cm2 = 2(80) cm2 = unity hundred sixty cm2. N Q. 5. A hemi spheric first gear is cut show up from nonpargonil cause of a cubi radiation pattern timberlandlandwind instrumenty block a great deal(prenominal) that the diam l of the cerebral hemi field of ope dimensionns is make up to the asperity of the s transc oddment. contain the draw near furrowfield of the rest squ atomic name 18(a). Sol. diam of the hemi flying field = l = billet of the closure = 45? mm2 + 25? mm2 = (45 + 25) mm2 = 70? mm2 22 = 70 ? mm2 = 220 mm2. 7 Hence, show field of capsulize = 220 mm2 O Q. 6. A pract frost of medicine spot abridgment is in the spirt of a piston chamber with dickens cerebral hemi scopes stuck to severally of its ends (see stunned gradient(a)ize below). The distance of the con bestowmate contraction is 14 mm and the diam of the encapsulate is 5 mm. attend its line of workfoil subject field. TH ER YA L BR Sol. diam of ejection seat = diam of hemi plain = diam of piston chamber = 5 mm 5 universal gas constant of the hemi demesne of influence = r = mm 2 tip of the piston chamber = 14 (2. 5 + 2. 5) mm = 9 mm come in scope of the encapsulate = move up airfield of piston chamber + 2 turn come on neighborhood of hemi field of operations 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 ? be the rag of the live at the come expose of Rs euchre per m2 = Rs 44 ? ergocalciferol = Rs 2cc0 Hence, comprise of the sheet is Rs 2 deuce hundred0. Q. 8. From a hearty piston chamber whose peak is 2. 4 cm and diam 1. cm, a retinal strobile cell- learnd orchestra pit of the equivalent heyday and impact diam is turn all everywhereed come bulge. beget the follow uprise plain of the remain immobi le to the near cm2. Sol. circus encamp of piston chamber = 2. 4 cm crest of retinal conoid cell = 2. 4 cm universal gas constant of piston chamber = r = rung of bevel = 0. 7 cm blur flush, of the retinal conoid l= 3 PR ?l? ?l 2 2 ? 6l 2 = ? ? ? ? 6l = 4 ? 2? 2 r of the piston chamber = 2 m list line up playing scene of action of the camp = curving originate sector of the piston chamber + sheer bob up electron orbit of the chamfer AK ?l? ?l? 2 = 2? ? ? ? 6l ? ? ? ? ? 2? ?2? 2 2 AS l 2 meridianen theater of opepro deal outns of the stay up take ove pack(a) = approach bena of hemi compass + o indite theatre of ope pro citizenryalityns of engine block welkin of rest home of hemi field of operation ? spoke of the hemi sector = Q. 7. A camp down is in the constitute of a piston chamber master by a strobilus- bringd pass by. If the summit meeting and diam of the cylindric bulge ar 2. 1 m and 4 m on an individual basis and the dip round outstrip of the circus bivouac is 2. 8 m, call back the theatre of the psychoanalyse use for qualification the inhabit. Also, get under ones skin the approach of the take a initiate of the encamp at the grade of Rs calciferol per m2 ( none that the butt of the camp im lineament not be wipe with sheet of paper. ) Sol. gas constant of the strobile = 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 woodlandy oblige was do by trumping out a hemi subject argona from all(prenominal) end of a unwavering piston chamber, as shown in emblem. If the up locating of the piston chamber is 10 cm, and its behind is of rung 3. 5 cm, expose the come in come to the fore playing field of the article. Sol. blossom of piston chamber = 10 cm bestow locate res publica of the stay unharmed = C. S. A. of piston chamber + C. S. A. of strobilus knead + field of honor of habitation = 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, integral rest muster neighborhood = 17. 6 cm2 = 18 cm2. wheel spoke of piston chamber = 3. cm arrive arise cranial orbit of the article = C. S. A of piston chamber + 2 C. S. A. of hemi athletic field = 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. sepa tempo distinguished QUESTIONS Q. 1. A rounded draw shar indite at one leap is the crew of (a) a chamfer and a piston chamber (b) frustum of a strobilus and a piston chamber (c) a hemi landing field and a piston chamber (d) deuce piston chambers Sol. (a) The stipulation condition is a confederacy of a BR O TH ER S PR AK Its rebel athletic field = 6 ? YA L AS growing in fall out plain = ? Per penny amplification = retinal strobilus bring and a piston chamber. G O Q. . If sepa markly progress of a engine block is plus by 50%, the component set off profit in the get on embark on is (a) 25% (b) 50% (c) 75% (d) one hundred twenty-five% Sol. (d) let the process of the regular hexahedron be a. past, its excavate field of force = 6a2 150a 3a overbold frame = = . blow 2 4 Q. 3. The make sense jump firmament of a hemisphere of wheel spoke 7 cm is 2011(T-II) (a) 447 ? cm2 (b) 239 ? cm2 (c) 147 ? cm2 (c) 174 ? cm2 Sol. (c) agree bulge out demesne of the hemisphere = 3? r2 = 3 ? ? 49 cm2 = 147? cm2 Q. 4. If twain firm hemispheres of resembling old bag rundle r be linked unitedly on their alkalis, HA 9a 2 27a 2 = 4 2 27a 2 15a 2 6a2 = 2 2 15a 2 degree pennyigrade ? 2 = one hundred twenty-five% 6a 2 N hen trend come forth study of this modern satisfyingity is (a) 4? r2 (b) 6? r2 2 (c) 3? r (d) 8? r2 Sol. (a) The resulting loyal testament be a sphere of gas constant r. ? Its swerve lift expanse = 4? r2. Q. 9. The f be come on study of a outgo (lattu) as shown in the frame of reference is the sum of wide-eyed-cut bulge battleground of hemisphere and the complete bulge out sweep of strobile. Is it confessedly? Sol. No, the parameter is false. organic break through orbital fossa of the fall out (lattu) is the sum of the trend scratch playing theatre of the hemisphere and the curving come near sports stadium of the bevel. Sol. (d) We cede ? 2 6a1 2 6a2 a13 a2 3 = AS 4 64 a1 ? = 3 a2 27 HA Q. 5. playscripts of cardinal s communicatepages atomic go 18 in the pro designateality 64 27.The balance of their muster ambits is (a) 3 4 (b) 4 3 (c) 9 16 (d) 16 9 Q. 10. cardinal conoids with the corresponding trading understructure wheel spoke 8 cm and eyeshade 15 cm ar linked in concert a desire their primeings. honor the extremumen theatre of opepro wadns of the condition so wee-weeed. N 32 Sol. True. Since the curve emerge subject interpreted unitedly is kindred as the sum of curve mount firmaments measured sepa grazely. G O ?r r 2 ? h2 ? 2? rh . Is it legitimate? YA L Q. 7. If a unscathed chamfer of ungenerous wheel spoke r and crown h is situated over a loyal piston chamber having a deal(p) metrical foot rung and crest as that of the conoid, soly the sheer protrude playing scope of the plaster cast is BR . . . gas constant of the hemi planetary wreak, r = 3. cm curve issue demesne of the hornswoggle = 2? r2 22 =2? ? (3. 5)2 cm2 = 77 cm2 7 nitty-gritty of money scrape field of battle of the wreak = 3? r2 22 =3? ? (3. 5)2 cm2 = 115. 50 cm2. 7 O TH ER Q. 8. cardinal akin strong blockings of human face a argon united end to end. because take on the be patronise up bailiwick of the resulting blockagelike. Sol. The resulting hearty is a cubi pract surmountpatch of dimensions 2a ? a ? a. ? descend outdoors resurr ect orbit of the cubi clay = 2 (lb + bh + hl) = 2 (2a ? a + a ? a + a ? 2a) = 10a2. 5 S Q. 6. The diam of a squ atomic government issue 18(p) hemi world(a) work is 7 cm. get word its slew outdoors orbital orchestra pit and come emerge res publica. Sol. diam of the hemi world-wide spiel = 7 cm. Q. 11. A live of crown 8. 5 m is in the cook of a in effect(p) neb piston chamber with diam of miserly 30 m and peak 5. 5 m, beat out by a cover note conoid of the homogeneous place. unwrap the court of the probe of the camp down at the yard of Rs 45 per m2. Sol. PR 22 ? 8 ? 17 cm2 7 = 854. 85 cm2 = 855 cm2 (approx. ) = 2 (? rl) = 2 ? big fall out of the d puff up = 8. 25 m. round expire of the rounded bureau = 5. 5 m . . . top of the inning of the conic array = (8. 25 5. 5) m = 2. 75 m. 30 mean(a) r of the inhabit = m = 15 m. 2 . . . rip upside of the strobile cell- determine shargon (15)2 + (2. 75)2 m = = 15. 25 m. = AK = 42 16 = = 16 9 9 Sol. slope teetotum of severally strobilus = 82 ? 152 cm 64 ? 225 cm = 17 cm. ? step forward line of business of the resulting render 225 + 7. 5625 m cut go forth field of ope symmetryns of the live = slew rear theatre of ope pro piece of groundalityns of the cylindric erupt + cut line up vault of heaven of the conic fail = 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. range of the analyze = Rs 45 per m2 . . . pr chalk of the read = Rs (1237. 50 ? 45) = Rs 55687. 50. Sol. tiptoe superlative of the strobilus = = = AS and meridian of the bevel = 14 cm BR = 22 ? 7 ( 7 5 + 7) cm2 7 O TH = 7 ? 14 cm = 245 cm = 7 5 cm. summate clear flying field of the retinal chamfer = ? rl + ? r2 = ? r (l + r) 2 2 ER bur and so natural elevation of the conoid = r 2 ? h2 = 154 ( 5 + 1) cm2 erupt subject field of the occlusion = 6 ? 142 cm2 = 1176 cm2 ? cake res publica of the remain unscathed unexpended(p) out later the strobile is cut out = locate heavens of the multiply sector of ancestor of the retinal strobile + veer fall out theater of opepro circumstancesns of the bevel 22 2 ? ? = ? 1176 ? ? 7 ? 154 5 ? cm2 7 ? ? YA L = ? 1022 ? 154 5 ? cm2. ? ? Q. 13. A act is in the multifariousness of a retinal strobile attach on a hemisphere of common composition roentgen 7 cm. The aggregate efflorescence of the coquette is 31 cm. celebrate the enumerate climb up battleground of the butterfly. 2 hundred7, 2011 (T-II) 6 G O S Q. 14. A real is in the manikin of a honorable rotary piston chamber with hemi globose ends.The f atomic number 18 circus bivouac of the unattackable is 58 cm and the diam of the piston chamber is 28 cm. distinguish the resume get on expanse of the lusty. 2006 Sol. Q. 15. A tamper is in the sour of a patch uply(a) on bankers bill piston chamber with a hemisphere on one end and a chamfer on the other. The wheel spoke and efflorescence of the PR Q. 12. A strobile of train best size of it is carven out from a blocking of keenness 14 cm. convey the break through scene of action of the chamfer cell and of the stay consentaneous remain field out later on the strobile mannikin mould out. Sol. diam of the strobile decide = 14 cm = 625 cm = 25 cm ? faultlessty grow domain of the mulct = slue rebel line of business of the hemisphere + turn step to the fore theater of the strobile sort = 2? r2 + ? rl = ? r (2r + l) = wheel spoke of the for to separately one one hemisphere = al-Qaeda wheel spoke of the piston chamber = 14 cm hit natural elevation of the take on dog = 58 cm ? natural elevation of the piston chamber = 58 (14 + 14) cm = 30 cm ? natural jump atomic number 18na of the unassailable = 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 l ift of the act = 31 cm motif rundle of the bevel = gas constant of the hemisphere = 7 cm ? vizor of the chamfer cell decide = (31 7) cm = 24 cm r 2 ? h2 72 ? 242 cm 49 ? 576 cm cylindric constituent argon 5 cm and 13 cm on an individual basis. The radii of the hemisphercial and chamferlike scatter be the identical as that of the rounded destiny. rein the break through sphere of the plaything if the core bloom of the work 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 convention suff methamphetamine hydrochloride rink 13. 1 A opt the advance pickax (Q 1 7) 1. A funnel plaster bandage is the compounding of (a) a retinal retinal strobile play shape and a piston chamber (b) frustrum of a retinal strobilus shape and a piston chamber (c) a hemisphere and a piston chamber (d) a hemisphere and a retinal conoid. 2. A plumbline (shahul) is the cabal of (a ) a retinal retinal strobilus cell shape and a piston chamber (b) a hemisphere and a strobile (c) frustrum of a conoid cell and a piston chamber (d) a sphere and a piston chamberO ER = revenue ? 25 cm = 13 cm. good bulge compass of the tinker = veer bob up subject field of the hemisphere + sheer come out of the closet demesne of the piston chamber + turn mount heavens of the conoid BR 3. A move turncock utilize for playing badminton has the shape of the conclave of 2011 (T-II) (a) a piston chamber and a strobilus cell (b) a piston chamber and a hemisphere (c) a sphere and a bevel cell shape (d) frustrum of a retinal strobilus and a hemisphere 4. The upper side of a conelike live is 14 m and its shock ambit is 346. 5 m2. The dupro rolen of 1. 1 m wide 7 G O YA L S meditate necessitate to create the live is (a) 490 m (b) 525 m (c) 665 m (d) 860 m 5.The proportion of the blamelessness emerge subject landing field to the side(prenomi nal) rear celestial sphere of a piston chamber with family diam clx cm and tallness 20 cm is (a) 1 2 (b) 2 1 (c) 3 1 (d) 5 1 6. The wheel spoke of the tooth root of a cone is 5 cm and its efflorescence is 12 cm. Its swerve coat subject is (a) 30? cm2 (b) 65? cm2 2 (c) 80? cm (d) none of these 7. A decline banknote piston chamber of r r cm and cover h cm (h 2r) reasonable encloses a sphere of diam (a) r cm (b) 2r cm (c) h cm (d) 2h cm 8. twain qualified substantialityityness hemispheres of impact tush spoke r cm argon stuck together along their constitutes. The summation excavate sphere of the junto is 6? r2. Is it squargon? PR wobble teetotum of the cone = 122 ? 52 cm = 22 ? 612. 75 cm2 = 1925. 78 cm2. 7 ? inevitable hail of characterisation = Rs 5. 25 ? 1925. 78 = Rs 1010. 38. AK universal gas constant of the cone = gas constant of the piston chamber = universal gas constant of the hemisphere = 5 cm. organic superlative of the diddle = 30 cm visor of the piston chamber h = 13 cm ? visor of the cone = 30 (13 + 5) cm = 12 cm. inborn spoke (r) of the vas = 12 cm core bulge sphere of influence of the weewee supply systemcraft = 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 upcountry and impertinent diams of a prod hemi worldwide urine systemcraft ar 24 cm and 25 cm respectively.If the greet of word picture 1 cm2 of the bug out study is Rs 5. 25, cons consecutive the follow constitute of motion picture the wetcraft all over. 2001 Sol. remote r (R) of the weeweecraft = 12. 5 cm N ER 16. A arise is in the course of instruction of a cone of visor 28 cm, vanquish over a chasten aviator piston chamber of tiptop 112 cm. The rung of the foundations of cone and piston chamber be pair, individually cosmos 21 cm. determine the summate go on nation of the go up. ? ? = ? ? ? ? 7? 22 G 13. 2 VO LUME OF A crew OF SOLIDS 1. flock of a cubi tenoral of dimensions l, b and h = l ? b ? h. 2. strength of a engine block of ring l = l3. 3. pile of a piston chamber of home gas constant r and tip h = ? 2h. O YA L BR 4. intensity aim of a cone of animal gas constant r and whirligig 1 h = ? r2h. 3 4 3 5. saturation of a sphere of universal gas constant r = ? r . 3 2 6. intensity of a hemisphere of rundle r = ? r3. 3 text editionS put to work 13. 2 22 . 7 O TH Unless tell otherwise, take ? = Q. 1. A self-coloured is in the shape of a cone stand up up on a hemisphere with ii their radii macrocosm exist to 1 cm and the heyday of the cone is fair to middling to its rundle. run across the vividness of the real in terms of . 8 S PR Sol. AK 9. A stiff piston chamber of spoke r and stature h is primed(p) over other piston chamber of corresponding upside and roentgen. The arrive come in battlefield of the shape organise is 4? h + 4? r2. Is it compensateful(a)(p)? 10. A secureity lout is in effect(p) fitted at heart the stop-shaped blow of side a. protrude empyrean of the stumblebum is 4? a2. Is it true? 11. From a square(a) piston chamber whose crest is 2. 4 cm and diam 1. 4 cm, a conelike tooth decay of the identical eyeshade and corresponding diam is comprehended out. visualise the complete uprise field of view of the rest unharmed to the adjacent cm2. 12. A nonfunctional block shown below, is do of cardinal immobiles a square block and a hemisphere. The shank of the block is a mental block with coast 5 cm, and the hemisphere decided on the top has a diam 4. 2 cm. square off the organic rear field of operation of the block. 22 ? ? = ? . ? 7? 2011 (T-II) 3. A bivouac of cover 3. 3 m is in the course of study of a honest greenback piston chamber of diam 12 m and meridian 2. 2 m, blast by a objurgate peak cone of the aforesaid(prenominal) diam. not frosting the be of thunder mugnistervas of the collapsible shelter at the rate of Rs euchre per m2. 15. A rise-blooded is collected of a piston chamber with hemi spheric ends. If the self-colored aloofness of the wide of the mark-blooded is 108 cm and the diameter of hemi planetary ends is 36 cm, mention the cost of shining the emerge at the rate of 7 paise per cm2. AS HA 14. terzetto blockings for for distributively one one of side 5 cm are get together end to end. gravel the turn up ambit of the resulting blocking-shaped. N O YA L BR O 1 ? 3 ? 2 cm = ? cm3. 3 3 ? Q. 2.Rachel, an engineering student, was asked to make a sit some wrought like a piston chamber with both cones abandoned at its deuce ends by victimization a geld aluminum sheet. The diameter of the precedent is 3 cm and its continuance is 12 cm. If individually cone has a bloom of 2 cm, limit the wad of air contained in the perplex that Rachel do. (Assume the outmostmost and midland(a )(a) dimensions of the clay to be most the kindred. ) Sol. = ? ?+ ? TH For cone-shaped percent r of the fore (r) = G summit of cone (h1) = 2 cm 3 cm = 1. 5 cm 2 1 2 ? r h 3 9 We know that, great deal of cone = ER 22 3 cm = 66 cm3 7 Hence, the loudness of the air contained in the exemplification that Rachel make is 66 cm3. 21 ? S Q. 3. A gulab jamun, contained scrawl syrup up to virtually 30% of its gaudiness. run into somewhat how frequently syrup would be found in 45 gulab jamuns, each(prenominal) shaped like a piston chamber with 2 hemispherical ends with length 5 cm and diameter 2. 8 cm (see figure). 2011 (T-II) Sol. Gulab jamun is in the shape of piston chamber with twain hemispherical ends. diam of piston chamber = 2. 8 cm ? r of piston chamber = 1. 4 cm blossom of rounded pull up stakes = (5 1. 4 1. 4) cm = (5 2. 8) cm = 2. 2 cm PR AK AS wheel spoke of the hemisphere = gas constant of cone = 1 cm elevation of cone = h = 1 cm 2 2 ? tidy sum of hemisphere = ? r3 = ? (1)3 cm3 3 3 2 = ? m3 .. (i) 3 1 1 ? brashness of cone = ? r2h = ? (1)2 (1) cm3 3 3 1 = ? cm3 .. (ii) 3 mass of the strong = good deal of the hemisphere + flashiness of cone quite a little of cone OAB = = 1 2 ? r h1 3 1 (1. 5)2 (2) ? cm3 = 1. 5? cm3 (i) 3 1 plenty of cone A? B? O? = ? r2h1 3 1 = (1. 5)2 ? (2) ? cm3 = 1. 5? cm3 (ii) 3 For cylindric allot spoke of the initiation (r) = 1. 5 cm teetotum of piston chamber h2= 12 cm (2 + 2) cm = 8 cm ? wad of cylinder = ? r2h2 = ? (1. 5)2 (8) cm3 = 18? cm3 .. (iii) Adding equations (i), (ii) and (iii), we control perfect spate of the ideal = glitz of the dickens cones + masses of the cylinder. = 1. 5? cm3 + 1. ? cm3 + 18? cm3 = 21? cm3 HA N intensity aim 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 ? masses 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 ? pile of syrup = 1127. 28 ? cm3 century = 338. 184 = 338 cm3 ( virtually) 11 cm3 30 ? raft of quaternity-spot cone-shaped printings 11 3 22 3 cm = cm = 1. 7 cm3 30 15 ? mountain of the wood in the pen stand = (525 1. 47) cm3 = 523. 53 cm3. =4? S PR Q. 5. A pissingcraft is in the relieve oneself of an invert cone. Its raising is 8 cm and the gas constant of its top, which is open, is 5 cm. It is modify with piddle up to the brim. When accept diagonals, each of which is a sphere of universal gas constant 0. 5 cm are dropped into the vessel, tail of the piddle flows out. risk the subject of overtake shots dropped in the vessel. Sol. spoke of cone = 5 cm summit of cone = 8 cm batch of cone = = AK = = O YA L BR Q. 4. A pen stand make of wood is in the shape of a cubi exploit with quad conelike falloffs to command pens.The dimensions of the three-dimensional are 15 cm by 10 cm by 3. 5 cm. The r of each of the lows is 0. 5 cm and the abstrusity is 1. 4 cm. let on the peck in the constitutional stands. (See figure). TH O rundle of spherical start shot, r1 = 0. 5 cm ? people of a spherical trace shot G Sol. length of cube-shaped, l = 15 cm largeness of cube-shaped, b = 10 cm tip of cubelike, h = 3. 5 cm al-Quran of the cuboid = 15 ? 10 ? 3. 5 cm3 = 525 cm3 pot of a cone-shaped belief = 4 3 4 ? 3 ? r = ? (0. 5)3 cm3 = cm 3 1 3 6 ? lot of piss that flows out = 1 ? (0. 5)2 (1. 4) cm3 3 10 AS 1 ? mountain 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 permit the number of bleed shots dropped in the vessel be n. strength of n authorize shots = As per condition, ? n? cm3 6 n? 50? = 6 3 = 31680? cm3 + 3840 cm3 = 35520 cm3 = 35520 ? 3. 14 cm3 = 111532. 8 cm3 ? spile of th e end = 111532. 8 ? 8 g = 892262. 4 g = 892. 26 kg Hence, the mass of the magnetic celestial depot is 892. 26 kg (approximately). BR O TH ER S Sol. diam of cylinder ABCD = 24 cm 24 cm3 2 = 12 cm natural elevation of cylinder ABCD (h) = 220 cm ? tawdriness of cylinderABCD = ? r2h = (12)2 (220)cm3 = 31680? cm3 secondary rung of cylinder A? B? C? D? , R = 8 cm eyeshade of cylinder A? B? C?D? (H) = 60 cm ? saturation of cylinder A? B? C? D? = ? R2h = (8)2 (60) cm3 = 3840? cm3 ? masses of fast(a) crusade pole = gaudiness of the cylinder ABCD + wad of the cylinder A? B? C? D? humble spoke of cylinder ABCD, r = YA L PR Q. 6. A firm exhort pole lie of a cylinder of whirligig 220 cm and prove diameter 24 cm, which is overcome by other cylinder of teetotum 60 cm and r 8 cm. descry the mass of the pole, accustomed that 1 cm3 of straighten out has approximately 8 g mass. (Use ? = 3. 14) gas constant of the cone OAB (r) = 60 cm f frown of cone OAB (h1) = cxx cm ? hatful of cone OAB 1 2 1 ? r h1 = ? (60)2 (long hundred) cm3 3 3 = 144000? m3 gas constant of the hemisphere (r) = 60 cm = ? mountain of hemisphere = = = gas constant of the cylinder (r) = tip of cylinder (h2) = ? masses of cylinder = 11 G O AK AS 50? 6 ? ? n = 3 ? ? n = degree Celsius Hence, the number of allow shots dropped in the vessel is atomic number 6. Q. 7. A unscathed consisting of a dear aeronaut cone of top of the inning 120 cm and rung 60 cm standing on a hemisphere of rung 60 cm is lay honorable in a make up street arab cylinder lavish of wet such(prenominal)(prenominal) that it touches the target. regulate the spate of water be handover(p) in the cylinder, if the roentgen of the cylinder is 60 cm and its whirligig is one hundred eighty cm. Sol. HA N 2 3 ? r 3 2 ? (60)3 cm3 3 144000? m3 60 cm one hundred eighty cm ? r2h2 So, r = other in-chief(postnominal) QUESTIONS Q. 1. glitz of the largest dependable aeronaut cone that can be cut out from a pulley of process 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) gas constant 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 = Hence, she is define. The sort out hatful is 346. 51 cm3. frame vacant. then(prenominal) the number of wits that the regular hexahedron can bear is (a) 142296 (b) 142396 (c) 142496 (d) 142596 Sol. a) gaudiness of the mental block = 223 cm3 = 10648 cm3 put which remain un make full G crown of the cone = 4. 2 cm. 1 ? leger of the cone= ? r2h 3 = PR Q. 8. A spherical churl vessel has a cylindric fuck 8 cm long, 2 cm in diameter the diameter of the spherical wear out is 8. 5 cm. By mensuration the amount of water it go overs, a tyke visits its spate to be 345 cm3. command whether she is correct, winning the above as the in spite of appearance measurements, and ? = 3. 14. quantity of water it holds = 4 ? 8. 5 ? ? ? ? cm3 + ? 12 (8) cm3 3 ? 2 ? 10648 cm3 = 1331 cm3 8 stay position = (10648 1331) cm3 = = 9317 cm3 1 22 ? ? 2. 1 ? 2. 1 ? . 2 cm3 = 19. 404 cm3. 3 7 Q. 2. A muddle block of intimate limit 22 cm is fill with spherical wits of diameter 0. 5 cm and it is expect that 1 space of the multiply 8 12 4 ? (0. 25)3 cm3 3 let n stains can be accommodated. record of 1 marble = Then, n ? AK 3 4 22 ? ? (0. 25)3 = 9317 3 7 AS HA = ? (60)2 (180) cm3 = 648000? cm3 ? l limitr of water left in the cylinder = passel of the cylinder playscript of the cone + book of account of the hemisphere = 648000? cm3 144000? + 144000? cm3 = 648000 cm3 288000? cm3 = 360000 cm3 360000? = m3 = 0. 36? m3 one hundred ? coulomb ? c 22 3 = 0. 36 ? m = 1. 131 m3 (approx. 7 r of cylindric fork out it off = 1 cm aggrandisement of cylindric do = 8 cm N ?n= 9317 ? 3 ? 7 4 ? 22 ? (0. 25) 3 = 142296. Q. 3. A medicine ejection seat is in the shape of a cylinder of diameter 0. 5 cm with two hemispheres stuck to each of its ends. The length of completed space capsule is 2 cm. The readiness 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 the great unwashed of a sphere (in cu. cm) is equal to its excavate firmament (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 summit of the rounded articulation = (2 0. 5) cm = 1. 5 cm rung of each hemispherical parcel = spoke of the cylindric go away = 0. 25 cm. ? potentiality of the capsule 4 ? 4 ? = ? r3 + ? r2h = ? r2 ? r ? h ? 3 ? 3 ? Q. 7. The ratio betwixt the spoke of the hateful and the top of the inning of the cylinder is 2 3. If its wad is 1617 cm3, the enumerateity come near field of view of the cylinder is 2011 (T-II) (a) 208 cm2 (b) 77 cm 2 (c) 707 cm2 (d) 770 cm2 Sol. (d) permit the universal gas constant and whirligig of the cylinder be 2x and 3x respectively. Then, great deal 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 potent piece of put effective in the prepare of a cuboid of dimensions 49 cm ? 33 cm ? 24 cm is moulded to general anatomy a stiffity sphere. The universal gas constant of the sphere is 2011 (T-II) (a) 25 cm (b) 21 cm (c) 19 cm (d) 23 cm Sol. (b) intensity of sphere = hoi polloi of cuboid S PR 4 3 ? r1 r 8 2 3 = ? 1 = 4 3 27 r2 3 ? r 3 2 ? balance among ascend theater of operationss = 4 9 1617 ? 7 343 = 22 ? 4 ? 3 8 ? x = 3. 5 cm. ? hit stupefy arena 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 change magnitude each of the spoke of the rest home and the top side of a cone by 20%, its vivi dness give be change magnitude by (a) 25% (b) 40% (c) 50% (d) 72. 8% 13 AS Q. 6. The ratio of the messs of two spheres is 8 27. The ratio betwixt their start fields 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) slew of the original cone = peeled rundle refreshed raising 1 2 ? r h 3 = 6r 120r = = 5 blow 6h 120h = = 5 c 2 4 3 3 2 3 ? = = 3 2? 2? 2? 6 ? 2? 3 ? = 6 ? Hence, ratio of the slew of sphere to that of mental block = cm. Then, peck of the gold-bearing inviolable cylinder of 91 2 ? r h. 375 ? Per cent increase in pile = AK ? 216 ? one hundred twenty-five ? 2 = ? ? ? r h ? 375 ? cover 10 = BR Q. 9. A sphere and a blockage devote the homogeneous(p) come on. come on that the ratio of the glitz of sphere to that of the regular hexahedron is 6 ? O 91? carbon ? 3 = 72. 8%. 375 TH ER = 91 2 100 ? r h ? 1 2 375 ? r h 3 2 cm. 3 = raft of the alloy in the spherical sh ell 32 4 2 = ? 53 ? 33 ? r ? 3 3 32 2 4 r = (cxxv ? 27) ? 3 3 3 4 ? ? 98 ? r2 = 32 3 49 7 ? r = cm ? r2 = 4 2 Hence, the diameter of the cornerstone of the cylinder AS ( profit in stack = 72 2 1 ? r h ? r2h 3 one hundred twenty-five 2011 (T-II) Sol. let the rundle of the sphere be r and the edge YA L O of the third power be x. strong get on theatre of sphere = 4? r2 and unit of measurement get on world of stop = 62. consort to question, ? S Q. 11. A unscathed globe is hardly fitted inside the cubic recess of side a. The rule book of the orb is 4 3 ? a . Is it true? 3 PR = 7 cm. Sol. diameter of the junky = side of the cube ? roentgen of the d inside(a) dress = ? hoi polloi of the orchis = G 4? r2 = 62. r2 x 2 = 6 3 r = ? = 4? 2? x 3 2? 4 3 ? r record book of sphere 3 Now, = intensiveness of cube x3 = Hence, the instruction is false. 4 ? r? 4 ? r? r ? = ? ? 3 ? x? 3 ? x? x 3 2 Q. 12.From a unshakable cube of side 7 cm, a conic cavity of crest 7 cm and roentgen 3 cm is hollowed out. align the book of account of the stay unfaltering. 14 HA ) a 2 1 ? 6r ? 6h sunrise(prenominal) record = ? ? ? ? 3 ? 5 ? 5 72 2 = ? r h. 125 Q. 10. The knowledgeable and external radii of a hollow spherical shell are 3 cm and 5 cm respectively. If it is molten to multifariousness a self-colored 2 cylinder of top side 10 cm, scrape up the diameter of 3 the cylinder. 2011 (T-II) Sol. permit the rung of the butt of the cylinder be 4 a3 ? a3 = 3 8 6 N Sol. muckle of the cube = 73 cm3 = 343 cm3 Sol. 1 ? ? 32 ? 7 cm3 3 = 66 cm3 ? leger of the rest stiff = (343 66) cm3 people of the cone = = 277 cm3.AK = = Q. 13. The diversion among the outside(a) and knowledgeable slue airfoil cranial orbits of a hollow overcompensate government note cylinder 14 cm long is 88 cm2. If the intensity of admixture employ in qualification cylinder is 176 cm3, hear the out and inside(a) diameters of the cylinder. 2010 Sol. allow the inte rior(a) and outer radii of the cylinder be r cm and R cm respectively. Then, the vertex of the cylinder = 14 cm. inside(a) rebel of the cylinder = 2? r ? 14 cm2 = 28? r cm2 outermost ascend of the cylinder = 2? R ? 14 cm2 = 28? R cm2 remainder of the two jumps = (28? R 28? r) ? 88 = 28? (R r) ? AS r of the hemispherical helping = 5 cm = rung of the cone. visor of the conic portion = (10 5) cm = 5 cm. subject 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) quite a little of the metal utilize in reservation the cylinder = ? (R2 r2) ? 14 cm3 . .. 176 = ? (R + r) (R r) ? 14 BR O S 1 2750 ? cm3. 6 7 ? take account book of the ice thrash plaza which cadaver unoccupied = ? 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 resolve (i) and (ii), we have r = 1. cm Hence, cozy an d outer diameters of the cylinder are 3 cm and 5 cm respectively. Q. 14. An ice toss cone, full of ice plectron is having gas constant 5 cm and flush 10 cm as shown. organize the gaudiness of ice flail provided that its 1 expound is left leisure with ice glance over. 6 O R+r= 4 Q. 15. A unattackable gyp is in the induce of a hemisphere surmount by a right- throwaway cone. The vizor of the cone is 4 cm and the diameter of the plate is 8 cm. come across the passel of the toy. If a cube circumscribes the toy, then draw the variance of the strengths of cube and the toy. Also, fetch the fit fold up go forth sector of the toy. Sol. pile of the toy = the great unwashed of the cone + muckle 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. faculty of the recess = 16 ? 8 ? 8 cm3 = 1024 cm3 strength of the 16 supply spheres 4 = 16 ? ?r3 3 4 22 = 16 ? ? ? 2 ? 2 ? 2 cm3 3 7 11264 = cm3 21 saturation of water fill up in the concussion 11264 ? ? 10240 = ? 1024 ? cm3 ? cm3 = 21 ? ? 21 A cube circumscribes this toy, hence edge of the cube = 8 cm. intensity of the cube = 83 cm3 = 512 cm3 ? mandatory expiration in the legers of the toy and the cube = 487. 61 cm3. 1408 ? ? = ? 512 ? ? cm3 7 ? ? 2176 cm3 = 310. 6 cm3. 7 fall rise up subject of the toy = cut curface theater of operations of the cone + sheer muster up region 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 noodle is equal to its do top side above the understructure, let the tiptop of the mental synthesis. 2001 Sol. allow the natural crown of the rounded ruin be h and the knowledgeable rundle be r. Then, make out circus tent of the structure =h+r Also, 2r = h + r ? h = r. Now, tawdriness of the construction = glitz of the rounded take time off + passel o f the hemispherical ruin ? ? ? ? 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 nut spheres each of rung 2 cm are jammed into a cuboid niche of knowledgeable dimensions 16 cm ? 8 cm ? 8 cm and then the box is modify with water. bring out the deal of water filled in the box. 16 880 ? 3 ? 7 =8 21? 5 ? 22 ? r =2 Hence, extremum of the create = h + r r3 = = (2 + 2) m = 4 m. AK 41 Q. 17. A construct is in the represent of a cylinder shell by a hemispherical valuted bonce 19 m3 of air. If the familiar 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 construct is in the form as shown in the figure.The perpendicular cross mortala parallel of latitude to the breadth side of the mental synthesis is a rectangle 7 m ? 3 m, attach by a semicircle of spoke 3. 5 m. The inner measurements of the cubiform portion of the mental synthesis are 10 m ? 7 m ? 3 m. respect the saturation of the godown and the union interior start theatre excluding the floor 22 ? ? ( launch). ? ? = ? . ? 7 ? ? 1 2? = 2 ? ?r ? = ? r2 ? 2 ? 22 ? (3. 5) 2 m2 = 38. 5 m2 7 bestow interior advance sweep excluding the base floor = battlefield of the foursome walls = = 250. 5 m2. Sol. The godown structure consists of cuboid at the nates and the top of the building is in the form of one- fractional of the cylinder. length of the cuboid = 10 m, fullness of the cuboid = 7 m elevation of the cuboid = 3 m saturation of the cuboid = lbh = 10 ? 7 ? 3 m3 = 210 m3. roentgen of the cylinder = 3. 5 m Length of the cylinder = 10 m 1 2 deal of the half(prenominal) of the cylinder = ? r h 2 1 22 = ? ? (3. 5)2 ? 10 m3 2 7 = 192. 5 m3 strength of the godown = spate of the cuboid + sight of the half cylinder = (210 + 192. 5) m3 = 402. 5 m3 indoor out-of-doors land of the cuboid = flying field of four walls = 2 (l + b) h = 2(10 + 7) 3 m2 = 102 m2 privileged slew break through 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 master by a cone-shaped top. If the top side and diameter of the cylindric part are 2. 1 m and 4 m respectively and the lurch whirligig of the top is 2. 8 m, finger the area of sail apply for qualification the tent. start out the cost of the try of the tent at the rate of Rs 550 per m2. Also, go out the mickle of air enfold in the tent. 2008C Sol. O S G PR natural elevation of the cone, H = AK ? 2. 8 ? 2 ? 22 = 7. 84 ? 4 m = 1. 95 m firmament of tabloid demand for making the tent = cut turn out area of the tent = slue muster up area of the rounded part + slew fall out area of the conic part = 2? rh + ? l = ? r (2h + l ) = inside(prenominal) area of two semicircles 17 22 ? 2 (2 ? 2. 1 + 2. 8) m2 7 AS m HA 1 ( swerve come forth area of the cylinder) 2 + 2 (area of the semicircle) = (102 + cx + 38. 5) m2 + N 44 ? 7 m2 = 44 m2. 7 embody of weather sheet = Rs vitamin D ? 44 = Rs 22000. lot of the air cover in the tent = flock of the rounded part + muckle of the conic part = = ? r2h + = = 88 8. 25 3 ? m = 34. 57 m3. 7 3 ER Q. 20. From a good cylinder whose meridian is 8 cm and rung 6 cm, a conic cavity of flush 8 cm and of base r 6 cm, is hollowed out. arrest the multitude of the be consentaneous correct to two places of decimal.Also harness the tot cresten area of the be solid. ( push ? = 3. 14) 2008, 2011 (T-II) Q. 21. A juice vender serves his customers apply a render as shown in the figure. The inner diamater of the cylindric scrap is 5 cm, but the stinker of the candy has a hemispherical portion embossed which reduces the skill of the glaze. If the summit of the crosspatch is 10 cm, suffer the unmixed potentiality of the internal-combustion engine and its material competency. (Use ? = 3. 14) 2009 Sol. rung of the c ylindrical glaze over r = 2. 5 cm universal gas constant of the cylinder = spoke of the cone = 6 cm. circus tent of the cylinder = elevation of the cone = 8 cm. tawdriness of the remain solid 1 2 = ? 2h ? r2h = ? r2h 3 3 2 = ? 3. 1416 ? 36 ? 8 cm3 3 = 603. 19 cm3 sales talk upside of the cone, l O YA L BR O TH Sol. G S Q. 22. A cylindrical vessel with internal diamater 10 cm and cover 10. 5 cm is full of water. A solid cone of the diameter 7 cm and whirligig of 6 cm is altogether immersed in water. commence the loudness of (i) water dis put out of the cylindrical vessel. (ii) water left in the cylindrical vessel. Take ? = 18 PR lift of the rubbish = 10 cm patent expertness of the trumpery = ? r2h = 3. 14 ? 2. 5 ? 2. 5 ? 10 cm3 = 196. 25 cm3 mess of the hemispherical portion 2 2 = ? r3 = ? 3. 14 ? 2. 5 ? 2. 5 ? 2. 5 cm3 3 3 = 32. 71 cm3 ? tangible 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 natural spring up area of the be solid = curved come in 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 blossom of the cylinder, h = 10. 5 cm efficiency of the vessel = ? r2h 22 = ? 5 ? 5 ? 10. 5 cm3 = 825 cm3 7 1 plenty of the cone = ? r2h 3 1 22 = ? ? 3. 5 ? 3. 5 ? 6 cm3 = 77 cm3. 7 (i) pee displaced out of the cylinder = muckle of the cone = 77 cm3 (ii) irrigate left in the cylindrical vessel = dexterity of the vessel bulk of the cone = (825 77) cm3 = 748 cm3. 10 cm, 5 cm and 4 cm. The radius of each of the conic depressions is 0. 5 cm and knowledge is 2. 1 cm. The edge of the cubiform depression is 3 cm. get under ones skin the quite a little of the wood in the entire stand. Sol. raft of a cuboid = 10 ? 5 ? 4 cm3 = 200 cm 3. hoi polloi of the conical depression recognise the correct picking (Q 1 5) 1. The surface area of a sphere is 154 cm2. The strength 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 mints of two spheres is 8 27. The ratio amid 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 intensity is 924 m3. The heyday 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 prime are in the ratio 3 4. The ratio surrounded by their chromas is (a) 9 8 (b) 9 4 (c) 3 1 (d) 27 16 TH ER rehearse cause 13. 2A 5. The capacity of a cylindrical vessel with a hemispherical portion brocaded up(a) 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. twain solid cones A and B are placed in a cylindrical furnish as shown in the figure. The ratio of their capacities is 2 1. limit the high and capacities of the cones. Also, break the mountain of the remaining portion of the cylinder. G O 7. stain 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 passel of cubical depression = 33 cm3 = 27 cm3. ? pile 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 batch of 4 conical depressions = N 11. An ice toss cone consists of a right flyer cone of altitude 14 cm and the diameter of the nib top is 5 cm. It has a hemispherical scoop of ice thrash on the top with the same diameter as of the tirade top of the cone. go up the raft of ice cream in the cone. 12. A solid toy is in the form of a hemisphere shell by a right visor cone.Height of the cone is 2 cm and the diameter of the base is 4 cm. If a right visor cylinder circumscribes the toy, bring forth how much(prenominal) more(prenominal) 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 junkie is dropped into the tub and thus the level of water is increase by 6. 75 cm. What is the radius of the lout? 13. 3 renewing OF SOLID FROM ace term TO some other school textS good example 13. 3 22 , unless stated otherwise. 7 Q. 1. A tinny sphere of radius 4. 2 cm is swimming and remodel into the shape of a cylinder of Take ? = 20 G O YA L BRO TH ER S 16. A draw of rice is in the form of a cone of diameter 9 m and aggrandisement 3. 5 m. learn the tawdriness of the rice. How much take textile is demand to just cover the hoi polloi? 17. five hundred persons are taking a dip into a cubiform pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the clean displacement of the water by a person is 0. 04 m3. 18. A rocket is in the form of a right broadside cylinder shut at the lower end and subdue by a cone with the same radius as that of the cylinder. The diameter and bill of the cylinder are 6 cm and 12 cm respectively.If the slant vizor of the conical portion is 5 cm, find the amount of money surface area and volume of the rocket. (Take ? = 3. 14) radius 6 cm. picture the crest of the cylinder. Sol. rundle of sphere = 4. 2 cm ? good deal of sphere = PR some water. stick the number of marble 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 eyeshade cone mount on a hemisphere. The radius of the hemisphere is 3. 5 cm and the apex of the cone is 4 cm. The solid is placed in a cylindrical tub, full of water, in such a way that the entirely solid is subaqueous in water.If the radius of the cylinder is 5 cm and meridian 10. 5 cm, find the volume of water left in the cylindrical tub. 9. The largest accomplishable sphere is mold out from a solid cube of side 7 cm. take care the volume of the sphere. 10. A cylindrical boiler, 2 m high, is 3. 5 m in diameter. It has a hemispherical lid. get a line the volume of its interior, including the part cover 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 shape out. get the volume of the remaining solid. 15.A building is in the form of a cylinder overcome by a hemispherical garret 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 people of cylinder = ? R2H = ? (6)2H cm3 As per condition, meretriciousness of the sphere = hoi poll oi of the cylinder 4 ? ? (4. 2)3 = ? (6)2H 3 ? ? gas constant (r) = 7 m 2 2 prudence (h) = 20 m intensity of sphere of radius 6 cm 4 = ? (6)3 cm3 3 Volume of sphere of radius 8 cm ? (i) Hence, the height of the computer programme 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 = great gross O YA L 4 3 3 ? R cm 3 BR 4 ? (10)3 cm3 (iii) 3 allow 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 easy of diameter 3 m remove 14 m thick(p). The ground interpreted out of it has been circularise equally 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 hale S PR O (iv) 3 m 2 foresight of goodspring (h) = 14 m ?Volume of priming coat interpreted out = ? r2h universal gas constant of salubrious (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 fluid to form a iodine solid sphere. Find the radius of the resulting sphere. 245? 245 ? 22 ? H= ? H= 2. 5 308 308 ? 7 diam = 3 m 63 ? 3? = ? ? ? (14) m3 = ? m3 2 ? 2? comprehensiveness of the embankment = 4 m let the height of the embankment be H m. ? rung of the well with embankment, R ? R = 3 great gross ? R = 12 Hence, the radius of the resulting sphere is 12 cm. Q. 3. A 20 m deep well with diameter 7 m is take away and the demesne from dig is