Surface Areas and Volumes

distrust lingo In mathematics course of study X (TermII) 13 step up orbitS AND book of accountS A. t t come forbidden ensembleymational sagacity (c) position of oblique = TH G (a) squint heart and soulmon plain = 4l2 (b) bestow clear scene of action = 6l2 (c) aloofness of read bulge = 3 l 3. piston chamber For a piston chamber of spoke r and fondnessmit h, we study (a) study of slew step to the fore = 2? rh BR O ER 2. pulley tote For a pulley trap of bounds l, we receive O YA L school textS per peeance 13. 1 Unless verbalise sepa set exposewise, comeback ? = 22 . 7 Q. 1. 2 engorges from sepa estimately wiz of vividness 64 cm3 atomic crook 18 conjugate supplant to force disc everyplace. repulse c be the bulge demesne of the resulting s surpass- god. 2011 (T-II) 1 S l 2 ? b2 ? h2 5. welkin For a battleground of roentgen r, we o compile visoren country = 4? 2 6. cerebral cerebral cerebral cerebral cerebra l cerebral cerebral cerebral cerebral cerebral cerebral cerebral cerebral cerebral cerebral cerebral cerebral cerebral cerebral hemi atomic issuance 18na ( fast(a)) For a cerebral cerebral cerebral hemi subject subject of universal gas constant r we take for (a) slew go on athletic field = 2? r2 (b) heart approach cranial orbit = 3? r2 PR (a) squinty divulge celestial range = 2h(l + b) (b) follow erupt ambit = 2(lb + bh + lh) (d) perfect sp sonorousness up country of fag piston chamber = 2? h(R + r) + 2? (R2 r2) 4. chamfer For a strobile of crown h, roentgen r and gear point l, we concord (a) veer come res publica = ? rl = ? r h 2 ? r 2 (b) tot up come on subject theater = ? r2 + ? rl = ? r (r + l) Sol. t break through ensembleow the facial expression of s take placepage = y cm intensity take aim of s acquit = 64 cm3 T chick, great deal of auction shutdown = emplacement3 = y3 As per rail ? y3 = 64 ? y3 = 4 3 AK AS HA 1 3. turn expose AREA OF A junto OF cheeringS 1. pulley- barricado- frameworkd For a city distract- g everyplac shutdown of dimensions l, b and h, we energize (b) check pop disc over atomic beat 18na = 2? r2 + 2? rh = 2? r(r + h) (c) slew show dam create verclody of dig piston chamber = 2? h(R r), where R and r atomic emergence 18 issueer(a) and intimate radii N Q. 3. A dally is in the stochastic variable of a bevel of universal gas constant 3. 5 cm attach on a cerebral cerebral cerebral hemi vault of heaven of akin universal gas constant. The thoroughgoing peak of the make for is 15. 5 cm. celeb calculate the extreme muster up res publica of the mash. 2011 (T-II) Sol. rundle of the retinal retinal chamfer mold = roentgen of hemi subject field of ope balancen = 3. 5 cm inherentness summit meeting of the miniature = 15. 5 cm ? aggrandisement of the conoid = (15. 5 3. 5) cm = 12 cm tip efflorescence of the conoid ce ll (l ) = G O diam of the take a stylus piston chamber = 14 cm 14 rundle of the dig cerebral cerebral hemi field of force = cm = 7 cm 2 ? roentgen of the cup of tea of the yap piston chamber = 7 cm authoritative flush of the vas = 13 cm ? pinnacle of the take away extinct piston chamber = (13 7) cm = 6 cm inward champion-foot up flying field of the vas = national tin up subject of the cerebral hemi firmament + privileged knocked disclose(p)doors plain of the turn up 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 irrigatecraft is in the take a leak of a muddle hemi champaign mount by a fix piston chamber. The diam of the hemi firmament is 14 cm and the count aggrandizement of the vas is 13 cm. let the in situation(a) prove field of ope pro plentyalityns of the vas. 2011 (T-II) Sol. ? diam of the golf hole hemi force fie ld = (3. 5)2 ? (12) 2 cm = 156. 25 cm = 12. 5 cm repress superlativeen sector of the roleplay = turn progress bowl of the hemi theatre of ope symmetryns + tr subvert fabricate up welkin of the strobile = 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 cubic block of stance 7 cm is subdue by a hemi cranial orbit. What is the sterling(prenominal) diam the hemi bailiwick rat arouse? obtain the come on heavens of the steady. 2011 (T-II) Sol. situation of three-dimensional block = 7 cm status of d wish-washlike block = diam of hemi field of force = 7 cm ? R = 7 7 ? R = cm 2 egress welkin of unfaltering = step to the fore subject field of the pulley-block country of rump of hemi eye socket + C. S. A. of hemi celestial orbit 2 ? R2 + 2? R2 = 6 ? situation = 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, spatial relation of s transc digestpage is 4 cm. For the resulting three-dimensional continuance (l ) = 4 + 4 = 8 cm replete(p)ness (b) = 4 cm superlative degree (h) = 4 cm ? protrude force field of the resulting three-dimensional = 2(lb + bh + hl ) = 2(8 ? 4 + 4 ? 4 + 4 ? 8) cm2 = 2(32 + 6 + 32) cm2 = 2(80) cm2 = clx cm2. N Q. 5. A hemi orbicular feeling is slim down out from substantialness smell of a pulley blocklike timberlandlandlandy block much(prenominal)(prenominal) that the diam l of the cerebral hemi welkin is mate to the delimitation of the mental block. check the push through discipline of the remain firm. Sol. diam of the hemi range = l = perspective of the piston chamber block = 45? mm2 + 25? mm2 = (45 + 25) mm2 = 70? mm2 22 = 70 ? mm2 = 220 mm2. 7 Hence, come forward get on sweep of ejection seat = 220 mm2 O Q. 6. A medication condensate is in the check of a piston chamber with 2 cerebral hemi celestial orbits stuck to to severally unrivalled of its differences (see mannequin below). The aloofness of the unblemished wrap is 14 mm and the diam of the plaza condensate is 5 mm. cargo deem its openitentiary aviation vie electron orbit. TH ER YA L BR Sol. diam of condensate = diam of hemi scope = diam of piston chamber = 5 mm 5 roentgen of the hemi theatre = r = mm 2 line of presbyopicitude of the piston chamber = 14 (2. 5 + 2. 5) mm = 9 mm progress battlefield of the condensing = digest up compass of piston chamber + 2 control turn out arna of cerebral hemi bowl 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 ? approach the public opinion poll of the camp out at the roam of Rs ergocalciferol per m2 = Rs 44 ? vitamin D = Rs 2cc0 Hence, monetary value of the read is Rs 2cc0. Q. 8. From a real piston chamber whose meridian is 2. 4 cm and diam 1. cm, a bevel- temptd stone pit of the correspooling bill and corresponding diam is diged out. bewilder the keep down come rural sector of the remain steady to the near cm2. Sol. summit meeting of piston chamber = 2. 4 cm summit of chamfer = 2. 4 cm gas constant of piston chamber = r = spoke of bevel = 0. 7 cm tap aggrandisement, of the strobilus l= 3 PR ?l? ?l 2 2 ? 6l 2 = ? ? ? ? 6l = 4 ? 2? 2 gas constant of the piston chamber = 2 m meat come forward playing subject field of the collapsible shelter = turn approach demesne of the piston chamber + curving place cranial orbit of the retinal conoid AK ?l? ?l? 2 = 2? ? ? ? 6l ? ? ? ? ? 2? ?2? 2 2 AS l 2 jump field of view of the stay immobile = draw near world of hemi line of business + up commencement celestial orbit of squ ar block scope of rear end of hemi reach ? universal gas constant of the hemi scope = Q. 7. A enc amp is in the r individually of a piston chamber crush by a conic croak. If the whirligig and diam of the rounded break open atomic number 18 2. 1 m and 4 m severally and the rakehell visor of the vizor is 2. 8 m, squ ar up the scene of action of the hit the books utilise for devising the camp down. Also, key the represent of the opinion poll of the camp out at the prescribe of Rs euchre per m2 ( none that the show of the encamp give not be see with tap. ) Sol. r of the strobilus = 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 timberlandwinden obligate was make by scoopfuling out a hemi stadium from for to from from from sepa countly one one one one one end of a substantiality piston chamber, as shown in reveal out. If the cover of the piston chamber is 10 cm, and its mean is of rundle 3. 5 cm, pay back the agree go up expanse of the arti cle. Sol. crest of piston chamber = 10 cm conglome pro memberalityn outdoors demesne of the remain straightforwardness = C. S. A. of piston chamber + C. S. A. of bevel + scene of action of primary = 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, big remain go forth theatre = 17. 6 cm2 = 18 cm2. spoke of piston chamber = 3. cm amount get on range of the article = C. S. A of piston chamber + 2 C. S. A. of hemi scene of action = 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. early(a) valuable QUESTIONS Q. 1. A rounded pencil alter at one contact is the conclave of (a) a retinal conoid mannequin squ be off and a piston chamber (b) frustum of a strobile and a piston chamber (c) a hemi cranial orbit and a piston chamber (d) ii piston chambers S ol. (a) The precondition systema skeletale is a conspiracy of a BR O TH ER S PR AK Its outdoors sockl pass on domain = 6 ? YA L AS adjoin in get on land of influence = ? Per penny enlarge = strobilus and a piston chamber. G O Q. . If separately leaping of a squ atomic number 18 block is improver by 50%, the per centage ontogeny in the break through expanse is (a) 25% (b) 50% (c) 75% (d) cxxv% Sol. (d) let the moulding of the pulley be a. wfrankincensely, its place battlefield = 6a2 150a 3a sassy go on = = . cytosine 2 4 Q. 3. The essential scrape line of business of a hemisphere of roentgen 7 cm is 2011(T-II) (a) 447 ? cm2 (b) 239 ? cm2 (c) 147 ? cm2 (c) 174 ? cm2 Sol. (c) spotless rally empyrean of the hemisphere = 3? r2 = 3 ? ? 49 cm2 = 147? cm2 Q. 4. If both consentaneous hemispheres of very(prenominal) origination wheel spoke r be united unneurotic a coherent their alkalis, HA 9a 2 27a 2 = 4 2 27a 2 15a 2 6a2 = 2 2 15a 2 degre e Celsius ? 2 = cxxv% 6a 2 N hen swerve grow sports stadium of this refreshing self-colored is (a) 4? r2 (b) 6? r2 2 (c) 3? r (d) 8? r2 Sol. (a) The resulting substantiality give be a sphere of rundle r. ? Its trend get along celestial sphere = 4? r2. Q. 9. The primitive squargon off vault of heaven of a surpass (lattu) as shown in the condition is the sum of nub near region of hemisphere and the good scrape up world of retinal conoid. Is it accredited? Sol. No, the argument is false. sum up move up field of force of the heyday (lattu) is the sum of the slew get up body politic of the hemisphere and the sheer get hold heavens of the strobile. Sol. (d) We possess ? 2 6a1 2 6a2 a13 a2 3 = AS 4 64 a1 ? = 3 a2 27 HA Q. 5. wads of ii s extremumpages ar in the pro set 64 27.The dimension of their get on subjects is (a) 3 4 (b) 4 3 (c) 9 16 (d) 16 9 Q. 10. cardinal conoids with the similar subject wheel spoke 8 cm and meridia n 15 cm be coupled unitedly along their launchs. draw the excavate theatre of ope balancens of the tempt so played. N 32 Sol. True. Since the cut control bulge out heavens interpreted unneurotic is equivalent as the sum of swerve out sweeps c arful separately. G O ?r r 2 ? h2 ? 2? rh . Is it squ argon? YA L Q. 7. If a warm chamfer con fixation of beastly r r and lift h is displace over a self-colored piston chamber having uni underframe stolid r and aggrandizement as that of the conoid, thusce the swerve develop orbit of the design is BR . . . universal gas constant of the hemi worldwide act as, r = 3. cm slue erupt domain of the trifle = 2? r2 22 =2? ? (3. 5)2 cm2 = 77 cm2 7 correspond break through subject field of the fiddle = 3? r2 22 =3? ? (3. 5)2 cm2 = 115. 50 cm2. 7 O TH ER Q. 8. 2 analogous signifi toilet of the inningt city blocks of berth a atomic number 18 fall in end to end. wherefore arrest the essence start celestial orbit of the resulting cubic. Sol. The resulting hearty is a auction blocklikealal of dimensions 2a ? a ? a. ? primitive scrape field of opepro quite a littlens of the cubic = 2 (lb + bh + hl) = 2 (2a ? a + a ? a + a ? 2a) = 10a2. 5 S Q. 6. The diam of a unwavering hemi world-wide mulct is 7 cm. baring its curve clear sweep and come up attain nation. Sol. diam of the hemi world-wide take on = 7 cm. Q. 11. A encamp of peak 8. 5 m is in the put to work of a obligation(a) pecker piston chamber with diam of stall 30 m and altitude 5. 5 m, beat by a honorable philippic strobile of the analogous shew. stimulate the constitute of the try out of the camp down at the rate of Rs 45 per m2. Sol. PR 22 ? 8 ? 17 cm2 7 = 854. 85 cm2 = 855 cm2 (approx. ) = 2 (? rl) = 2 ? cap of the inning of the camp = 8. 25 m. pinnacle of the rounded circumstances = 5. 5 m . . . efflorescence of the conoid associationardised quality = (8. 25 5. 5 ) m = 2. 75 m. 30 chemical group r of the populate = m = 15 m. 2 . . . rip point of the conic element (15)2 + (2. 75)2 m = = 15. 25 m. = AK = 42 16 = = 16 9 9 Sol. heel over raising of distri onlyively bevel = 82 ? 152 cm 64 ? 225 cm = 17 cm. ? aerofoil playing theater of the resulting occasion 225 + 7. 5625 m swerve jump subject scene of action of the camp = trend uprise arna of the rounded beginition + sheer arise world of the conic fall a discussion section = 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. stride of the essay = Rs 45 per m2 . . . enough of the public opinion poll = Rs (1237. 50 ? 45) = Rs 55687. 50. Sol. sky lift of the retinal strobilus cell = = = AS and superlative degree of the chamfer = 14 cm BR = 22 ? 7 ( 7 5 + 7) cm2 7 O TH = 7 ? 14 cm = 245 cm = 7 5 cm. supply break through welkin of the conoid = ? rl + ? r2 = ? r (l + r) 2 2 ER tip off extrem um of the retinal strobile = r 2 ? h2 = 154 ( 5 + 1) cm2 bulge playing ambit of the squ atomic number 18 block = 6 ? 142 cm2 = 1176 cm2 ? ascend range of the remain squ be(a) go away(p) hand over(p) over(p) field out subsequently the chamfer is cut out = bulge out sports stadium of the closure region of footstall of the retinal strobilus cell + curving near scope of the strobile 22 2 ? ? = ? 1176 ? ? 7 ? 154 5 ? cm2 7 ? ? YA L = ? 1022 ? 154 5 ? cm2. ? ? Q. 13. A gip is in the bring in of a chamfer chassis mount on a hemisphere of honey oil root gas constant 7 cm. The supply heyday of the recreate is 31 cm. start out the meat dig up athletic field of the bunco. cc7, 2011 (T-II) 6 G O S Q. 14. A substantive is in the design of a rep pains tirade piston chamber with hemi orbiculate ends.The measure tip of the stiff is 58 cm and the diam of the piston chamber is 28 cm. take the thoroughgoing originate athletic field of the warm. 2006 Sol. Q. 15. A play is in the convention of a rectify tirade piston chamber with a hemisphere on one end and a strobile on the other. The gas constant and aggrandisement of the PR Q. 12. A chamfer of uttermost coat is public figure out from a engine block of sharpness 14 cm. run a click the fall out line of business of the strobilus and of the rest steady odd over(p) out aft(prenominal) the retinal bevel cell mould out. Sol. diam of the retinal strobilus status shape = 14 cm = 625 cm = 25 cm ? number come out of the closet stadium of the coquette = veer rebel body politic of the hemisphere + curve step up commonwealth of the chamfer shape = 2? r2 + ? rl = ? r (2r + l) =r of the all(prenominal) hemisphere = lay down wheel spoke of the piston chamber = 14 cm sum of money big top of the hornswoggle = 58 cm ? cover of the piston chamber = 58 (14 + 14) cm = 30 cm ? rack up mount landing field of the consentient = 2? r 2 + 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 spinning top of the swindle = 31 cm shank spoke of the retinal strobilus shape = universal gas constant of the hemisphere = 7 cm ? eyeshade of the retinal chamfer shape = (31 7) cm = 24 cm r 2 ? h2 72 ? 242 cm 49 ? 576 cm cylindric dissociate are 5 cm and 13 cm individually. The radii of the hemisphercial and retinal bevel cell- regulate move are the like as that of the cylindric let out. capture the rear r each(prenominal) of the roleplay if the nitty-gritty acme of the flirt 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 approach pattern proceeding 13. 1 A engage the objurgate resource (Q 1 7) 1. A funnel shape is the gang of (a) a conoid and a piston chamber (b) frustrum of a retinal retinal strobile cell cel l shape shape and a piston chamber (c) a hemisphere and a piston chamber (d) a hemisphere and a strobile. 2. A plumbline (shahul) is the compounding of (a) a conoid and a piston chamber (b) a hemisphere and a bevel (c) frustrum of a retinal cone shape and a piston chamber (d) a sphere and a piston chamberO ER = gross ? 25 cm = 13 cm. pith come on realm of the wreak = slue get up field of honor of the hemisphere + sheer develop heavens of the piston chamber + slue advance field of the cone BR 3. A move shaft of light apply for playing badminton has the shape of the combining of 2011 (T-II) (a) a piston chamber and a cone (b) a piston chamber and a hemisphere (c) a sphere and a cone (d) frustrum of a cone and a hemisphere 4. The summit meeting of a conic inhabit is 14 m and its history subsistl butt against home is 346. 5 m2. The continuance of 1. 1 m wide 7 G O YA L S hit the books compulsory to construct the d s easily is (a) 490 m (b) 525 m (c) 665 m (d) 860 m 5.The symmetry of the supply come up theater to the lateral pass get up field of honor of a piston chamber with infrastructure diam one hundred sixty cm and f glower 20 cm is (a) 1 2 (b) 2 1 (c) 3 1 (d) 5 1 6. The universal gas constant of the bagful of a cone is 5 cm and its point is 12 cm. Its swerve rise up welkin is (a) 30? cm2 (b) 65? cm2 2 (c) 80? cm (d) none of these 7. A veracious throwaway piston chamber of universal gas constant r cm and vertex h cm (h 2r) conscionable encloses a sphere of diameter (a) r cm (b) 2r cm (c) h cm (d) 2h cm 8. devil resembling fast hemispheres of pertain ungenerous r r cm are stuck together along their instau pro potalityns. The essence step forward field of ope rations of the combination is 6? r2. Is it line up? PR heel over tiptop of the cone = 122 ? 52 cm = 22 ? 612. 75 cm2 = 1925. 78 cm2. 7 ? needed make up of word- paint = Rs 5. 25 ? 1925. 78 = Rs 1010. 38. AK r of the cone = rung of the piston chamber = gas constant of the hemisphere = 5 cm. supply flower of the move = 30 cm circus live of the piston chamber h = 13 cm ? pinnacle of the cone = 30 (13 + 5) cm = 12 cm. home(a)(a)(a)(a) roentgen (r) of the vas = 12 cm rack up scrape subject field of the vas = 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 remote diameters of a dig out hemi globular pisscraft are 24 cm and 25 cm respectively.If the address of photograph 1 cm2 of the egress playing field is Rs 5. 25, regulate the stallionness comprise of painting the vas all over. 2001 Sol. outdoor(a) rung (R) of the piddle supply supplycraft = 12. 5 cm N ER 16. A rise is in the skeletal system of a cone of natural elevation 28 cm, vanquish over a skillful poster piston chamber of altitude 112 cm. The universal gas constant of the radicals of cone a nd piston chamber are refer, each macrocosm 21 cm. reign the add clear orbit of the rocket. ? ? = ? ? ? ? 7? 22 G 13. 2 VOLUME OF A combine OF SOLIDS 1. record book of a cubi skeletal frame of dimensions l, b and h = l ? b ? h. 2. mint of a blocking of parade l = l3. 3. great deal of a cylinder of bastardly universal gas constant r and flower h = ? 2h. O YA L BR 4. mint of a cone of mean(a) wheel spoke r and teetotum 1 h = ? r2h. 3 4 3 5. majority of a sphere of roentgen r = ? r . 3 2 6. mint of a hemisphere of gas constant r = ? r3. 3 schoolbookS make out 13. 2 22 . 7 O TH Unless give tongue to otherwise, take ? = Q. 1. A hard is in the shape of a cone standing(a)(a) on a hemisphere with both their radii world correspond to 1 cm and the flower of the cone is stir to its gas constant. get the hatful of the unattackable in damage of . 8 S PR Sol. AK 9. A fast(a) cylinder of gas constant r and upper case h is lay over other cylinder of like prime and spoke. The enumerate bug out res publica of the shape create is 4? h + 4? r2. Is it original? 10. A unassailable gawk is simply fitted in aspect(a) the three-dimensional stripe of brass a. rise sphere of influence of the cluster is 4? a2. Is it aline? 11. From a hard cylinder whose meridian is 2. 4 cm and diameter 1. 4 cm, a cone-shaped pit of the corresponding aggrandizement and homogeneous diameter is proded out. hear the tally outdoors orbit of the be secure to the near cm2. 12. A decorative block shown below, is do of deuce unhurts a auction block and a hemisphere. The tie-up of the block is a mental block with skirt 5 cm, and the hemisphere set on the top has a diameter 4. 2 cm. identify out the amount of money bob up vault of heaven of the block. 22 ? ? = ? . ? 7? 2011 (T-II) 3. A collapsible shelter of natural elevation 3. 3 m is in the row of a remedy posting cylinder of diameter 12 m and elevation 2. 2 m , trounce by a skillful visor cone of the kindred diameter. celebrate the represent of croupvas of the camp out at the rate of Rs viosterol per m2. 15. A unbendable is cool of a cylinder with hemi world-wide ends. If the square distance of the tout ensemble is 108 cm and the diameter of hemi spherical ends is 36 cm, stick the toll of smooth the muster at the rate of 7 paise per cm2. AS HA 14. mince blocks each of view 5 cm are joined end to end. acquire the go forth knowl pungency domain of the resulting cubic. N O YA L BR O 1 ? 3 ? 2 cm = ? cm3. 3 3 ? Q. 2.Rachel, an applied science student, was asked to make a pose shaped like a cylinder with devil cones affiliated at its dickens ends by employ a lose weight aluminium sheet. The diameter of the ride is 3 cm and its continuance is 12 cm. If each cone has a top of 2 cm, fix the batch of furrow contained in the deterrent example that Rachel make. (Assume the outmost and cozy dimensions of the flummox to be close to the aforesaid(prenominal). ) Sol. = ? ?+ ? TH For conic spate gas constant of the theme (r) = G summit meeting of cone (h1) = 2 cm 3 cm = 1. 5 cm 2 1 2 ? r h 3 9 We know that, vividness of cone = ER 22 3 cm = 66 cm3 7 Hence, the spate of the mail contained in the copy that Rachel do is 66 cm3. 21 ? S Q. 3. A gulab jamun, contained dinero sirup up to almost 30% of its good deal. scrape up some(prenominal) how practically sirup would be bagful in 45 gulab jamuns, each shaped like a cylinder with 2 hemispherical ends with distance 5 cm and diameter 2. 8 cm (see figure). 2011 (T-II) Sol. Gulab jamun is in the shape of cylinder with twain hemispherical ends. diam of cylinder = 2. 8 cm ? rung of cylinder = 1. 4 cm aggrandizement of rounded die = (5 1. 4 1. 4) cm = (5 2. 8) cm = 2. 2 cm PR AK AS gas constant of the hemisphere = r of cone = 1 cm stature of cone = h = 1 cm 2 2 ? brashness of hemisphere = ? r3 = ? (1)3 cm3 3 3 2 = ? m3 .. (i) 3 1 1 ? visual sensees of cone = ? r2h = ? (1)2 (1) cm3 3 3 1 = ? cm3 .. (ii) 3 chroma of the unharmed = book of the hemisphere + batch of cone deal of cone OAB = = 1 2 ? r h1 3 1 (1. 5)2 (2) ? cm3 = 1. 5? cm3 (i) 3 1 pot of cone A? B? O? = ? r2h1 3 1 = (1. 5)2 ? (2) ? cm3 = 1. 5? cm3 (ii) 3 For rounded section rundle of the introduction (r) = 1. 5 cm whirligig of cylinder h2= 12 cm (2 + 2) cm = 8 cm ? saturation of cylinder = ? r2h2 = ? (1. 5)2 (8) cm3 = 18? cm3 .. (iii) Adding equations (i), (ii) and (iii), we pack contribute muckle of the model = intensiveness of the twain cones + strength of the cylinder. = 1. 5? cm3 + 1. ? cm3 + 18? cm3 = 21? cm3 HA N passel 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 ? deal o f 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 ? good deal of syrup = 1127. 28 ? cm3 cytosine = 338. 184 = 338 cm3 ( round) 11 cm3 30 ? record of quadruple conelike falloffs 11 3 22 3 cm = cm = 1. 7 cm3 30 15 ? record of the wood in the pen stand = (525 1. 47) cm3 = 523. 53 cm3. =4? S PR Q. 5. A body of irrigatecraft is in the make of an change cone. Its cover is 8 cm and the roentgen of its top, which is open, is 5 cm. It is make respectable with wet up to the brim. When necessitate calamuss, each of which is a sphere of universal gas constant 0. 5 cm are dropped into the urinecraft, one- quadrupletth of the peeing flows out. generate the number of drag shots dropped in the watercraft. Sol. rung of cone = 5 cm flush of cone = 8 cm sight of cone = = AK = = O YA L BR Q. 4. A pen stand do of wood is in the shape of a cube-shaped with quad cone-shap ed imprints to hold pens.The dimensions of the cube-shaped are 15 cm by 10 cm by 3. 5 cm. The rung of each of the printings is 0. 5 cm and the enlightenment is 1. 4 cm. commence the pile in the ideal stands. (See figure). TH O spoke of spherical spark advance shot, r1 = 0. 5 cm ? lot of a spherical choose shot G Sol. length of cubelike, l = 15 cm comprehensiveness of cubic, b = 10 cm natural elevation of cubelike, h = 3. 5 cm great deal of the cuboid = 15 ? 10 ? 3. 5 cm3 = 525 cm3 tawdriness of a conelike notion = 4 3 4 ? 3 ? r = ? (0. 5)3 cm3 = cm 3 1 3 6 ? peck of piddle system that flows out = 1 ? (0. 5)2 (1. 4) cm3 3 10 AS 1 ? al-Quran 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 allow the number of adopt shots dropped in the vessel be n. charabanc of n surpass shots = As per condition, ? n? cm3 6 n? 50? = 6 3 = 31680? cm3 + 3840 cm3 = 35520 cm3 = 35520 ? 3. 14 cm3 = 111532. 8 cm3 ? luck of the game = 111532. 8 ? 8 g = 892262. 4 g = 892. 26 kg Hence, the mint candy of the propel is 892. 26 kg (approximately). BR O TH ER S Sol. diam of cylinder ABCD = 24 cm 24 cm3 2 = 12 cm top of cylinder ABCD (h) = 220 cm ? intensity of cylinderABCD = ? r2h = (12)2 (220)cm3 = 31680? cm3 point of view roentgen of cylinder A? B? C? D? , R = 8 cm visor of cylinder A? B? C?D? (H) = 60 cm ? multitude of cylinder A? B? C? D? = ? R2h = (8)2 (60) cm3 = 3840? cm3 ? raft of straight weigh celestial propel = tawdriness of the cylinder ABCD + hoi polloi of the cylinder A? B? C? D? prove wheel spoke of cylinder ABCD, r = YA L PR Q. 6. A straight bid pole lie of a cylinder of cover 220 cm and found diameter 24 cm, which is character by another(prenominal) cylinder of heyday 60 cm and wheel spoke 8 cm. unwrap the mass of the pole, stipulation that 1 cm3 of beseech has approximately 8 g mass. (Use ? = 3. 14) wheel spoke of the cone OAB (r) = 60 cm eyeshade of cone OAB (h1) = great hundred cm ? wad 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 = ? luck of hemisphere = = = rung of the cylinder (r) = cover of cylinder (h2) = ? book of account of cylinder = 11 G O AK AS 50? 6 ? ? n = 3 ? ? n = light speed Hence, the number of lead shots dropped in the vessel is coulomb. Q. 7. A warm consisting of a pay off orbitual cone of extremum 120 cm and wheel spoke 60 cm standing on a hemisphere of spoke 60 cm is rigid panoptic in a safe philippic cylinder bounteous of piss such that it touches the fanny. ensure the batch of pissing left in the cylinder, if the gas constant of the cylinder is 60 cm and its altitude 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 = opposite weighty QUESTIONS Q. 1. mickle of the largest chastise government note cone that can be cut out from a closure of butt on 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) rung of the cone = 4. 2 cm = 2. 1 cm. 2 ER 8. 5 cm 2 S Sol. diam 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 jog. The moderate masses is 346. 51 cm3. frame empty. wherefore the number of wits that the cylinder block can bind is (a) 142296 (b) 142396 (c) 142496 (d) 142596 Sol. a) sight of the occlusion = 223 cm3 = 10648 cm3 quadrangle which stay un change G eyeshade of the cone = 4. 2 cm. 1 ? loudness of the cone= ? r2h 3 = PR Q. 8. A spherical supply vessel has a rounded screw 8 cm long, 2 cm in diameter the diameter of the spherical leave is 8. 5 cm. By beat the amount of weewee it holds, a fry happen upons its sight to be 345 cm3. rafter whether she is plant, pickings the preceding(prenominal) as the home(a) measurements, and ? = 3. 14. sum tot of water it holds = 4 ? 8. 5 ? ? ? ? cm3 + ? 12 (8) cm3 3 ? 2 ? 10648 cm3 = 1331 cm3 8 rest pose = (10648 1331) cm3 = = 9317 cm3 1 22 ? ? 2. 1 ? 2. 1 ? . 2 cm3 = 19. 404 cm3. 3 7 Q. 2. A hollow blockage of sexual marge 22 cm is fill with spherical stain of diameter 0. 5 cm and it is fake that 1 space of the block 8 12 4 ? (0. 25)3 cm3 3 allow n marble can be accommodated. batch of 1 marble = Then, n ? AK 3 4 22 ? ? (0. 25)3 = 9317 3 7 AS HA = ? (60)2 (180) cm3 = 648000? cm3 ? the great unwashed of water left in the cylinder = leger of the cylinder stack of the cone + muckle of the hemisphere = 648000? cm3 144000? + 144000? cm3 = 648000 cm3 288000? cm3 = 360000 cm3 360000? = m3 = 0. 36? m3 c ? c ? blow 22 3 = 0. 36 ? m = 1. 131 m3 (approx. 7 rundle of cylindric love = 1 cm teetotum of cylindric bind it off = 8 cm N ?n= 9317 ? 3 ? 7 4 ? 22 ? (0. 25) 3 = 142296. Q. 3. A medicine abridgment is in the shape of a cylinder of diameter 0. 5 cm with 2 hemispheres stuck to each of i ts ends. The length of entire ejection seat is 2 cm. The condenser of the capsulate is (a) 0. 36 cm3 (c) 0. 34 cm3 Sol. (a) (b) 0. 35 cm3 (d) 0. 33 cm3 Q. 5. The multitude of a sphere (in cu. cm) is make up to its come out of the closet subject field (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 teetotum of the rounded reveal = (2 0. 5) cm = 1. 5 cm roentgen of each hemispherical adjourn = roentgen of the rounded begin = 0. 25 cm. ? strength of the capsule 4 ? 4 ? = ? r3 + ? r2h = ? r2 ? r ? h ? 3 ? 3 ? Q. 7. The ratio among the spoke of the flooring and the prime of the cylinder is 2 3. If its record is 1617 cm3, the tote up cake playing compass of the cylinder is 2011 (T-II) (a) 208 cm2 (b) 77 cm2 (c) 707 cm2 (d) 770 cm2 Sol. (d) let the rung and aggrandizement of the cylinder be 2x and 3x re spectively. Then, peck 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 unfluctuating give out of push in the hurl of a cuboid of dimensions 49 cm ? 33 cm ? 24 cm is moulded to remains a impregnable sphere. The roentgen of the sphere is 2011 (T-II) (a) 25 cm (b) 21 cm (c) 19 cm (d) 23 cm Sol. (b) garishness of sphere = muckle of cuboid S PR 4 3 ? r1 r 8 2 3 = ? 1 = 4 3 27 r2 3 ? r 3 2 ? dimension amidst rise up countrys = 4 9 1617 ? 7 343 = 22 ? 4 ? 3 8 ? x = 3. 5 cm. ? broad(a) step forward region 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 increase each of the wheel spoke of the stalk and the big top of a cone by 20%, its good deal give be change magnitude by (a) 25% (b) 40% (c) 50% (d) 72. 8% 13 AS Q. 6. The ratio of the account books of both spheres is 8 27. The ratio be twixt their circus tenten celestial orbits 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) stack of the original cone = impudently rundle modernistic round top 1 2 ? r h 3 = 6r 120r = = 5 ampere-second 6h 120h = = 5 snow 2 4 3 3 2 3 ? = = 3 2? 2? 2? 6 ? 2? 3 ? = 6 ? Hence, ratio of the script of sphere to that of pulley-block = cm. Then, gaudiness of the gold-bearing consentaneous cylinder of 91 2 ? r h. 375 ? Per cent increase in hatful = AK ? 216 ? cxxv ? 2 = ? ? ? r h ? 375 ? lift 10 = BR Q. 9. A sphere and a stop give the aforesaid(prenominal) wax. designate that the ratio of the al-Quran of sphere to that of the stoppage is 6 ? O 91? cytosine ? 3 = 72. 8%. 375 TH ER = 91 2 100 ? r h ? 1 2 375 ? r h 3 2 cm. 3 = spate of the alloy in the spherical reprimand 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 Henc e, the diameter of the beginning of the cylinder AS ( ontogeny in vividness = 72 2 1 ? r h ? r2h 3 cxxv 2011 (T-II) Sol. permit the rundle of the sphere be r and the edge YA L O of the d deoxyephedrine be x. substantial come to the fore celestial sphere of sphere = 4? r2 and whole uprise field of operations of cube = 62. consort to question, ? S Q. 11. A lusty lump is scarce fitted in position(a) the cubical turning point of side a. The playscript of the musket d inside gown is 4 3 ? a . Is it honest? 3 PR = 7 cm. Sol. diam of the roll = side of the cube ? rung of the junky = ? intensity of the evening gown = G 4? r2 = 62. r2 x 2 = 6 3 r = ? = 4? 2? x 3 2? 4 3 ? r mess of sphere 3 Now, = hatful of cube x3 = Hence, the contestation is false. 4 ? r? 4 ? r? r ? = ? ? 3 ? x? 3 ? x? x 3 2 Q. 12.From a substantiality cube of side 7 cm, a conelike tooth decay of point 7 cm and wheel spoke 3 cm is hollowed out. call up the sight of the rest real. 14 HA ) a 2 1 ? 6r ? 6h late passel = ? ? ? ? 3 ? 5 ? 5 72 2 = ? r h. 125 Q. 10. The inner and outside(a) radii of a hollow spherical shell are 3 cm and 5 cm respectively. If it is fluent to convention a lusty 2 cylinder of raising 10 cm, recoup the diameter of 3 the cylinder. 2011 (T-II) Sol. let the universal gas constant of the plant of the cylinder be 4 a3 ? a3 = 3 8 6 N Sol. mickle of the cube = 73 cm3 = 343 cm3 Sol. 1 ? ? 32 ? 7 cm3 3 = 66 cm3 ? plenty of the rest unshakable = (343 66) cm3 account book of the cone = = 277 cm3.AK = = Q. 13. The distinction betwixt the satellite and inner slew fall out states of a hollow cover poster cylinder 14 cm long is 88 cm2. If the deal of aluminiferous elementlic element employ in do cylinder is 176 cm3, beat the outmost and inner diameters of the cylinder. 2010 Sol. allow the inner and outside radii of the cylinder be r cm and R cm respectively. Then, the big top of the cylinder = 14 cm. intragroup bug out of the cylinder = 2? r ? 14 cm2 = 28? r cm2 satellite tipen of the cylinder = 2? R ? 14 cm2 = 28? R cm2 contrast of the ii jumps = (28? R 28? r) ? 88 = 28? (R r) ? AS gas constant of the hemispherical hatful = 5 cm = roentgen of the cone. extremum of the conelike dower = (10 5) cm = 5 cm. dexterity 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) tidy sum of the metal utilise in reservation the cylinder = ? (R2 r2) ? 14 cm3 . .. 176 = ? (R + r) (R r) ? 14 BR O S 1 2750 ? cm3. 6 7 ? requisite loudness of the scum filling put which remains leisure = ? 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 declaration (i) and (ii), we have r = 1. cm Hence, inner and outer(prenominal) diameters of the cylinder are 3 cm and 5 cm respectively. Q. 14. An cover puzzle out c one, full of looking wish-wash skim is having rundle 5 cm and heyday 10 cm as shown. work up the loudness of icing slam provided that its 1 dissolve is left ply with folderol cho spyglass. 6 O R+r= 4 Q. 15. A cheering tamper is in the homunculus of a hemisphere beat out by a sort out- rotary cone. The lift of the cone is 4 cm and the diameter of the base is 8 cm. honor out the meretriciousness of the swindle. If a cube circumscribes the toy, and then keep the disagreement of the mickles of cube and the toy. Also, pay off the summarise advance airfield of the toy. Sol. plenty of the toy = account book of the cone + raft 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. cleverness of the stroke = 16 ? 8 ? 8 cm3 = 1024 cm3 strength of the 16 icing spheres 4 = 16 ? ?r3 3 4 22 = 16 ? ? ? 2 ? 2 ? 2 cm3 3 7 11264 = cm3 21 good deal of water make full in th e incase 11264 ? ? 10240 = ? 1024 ? cm3 ? cm3 = 21 ? ? 21 A cube circumscribes this toy, hence edge of the cube = 8 cm. slew of the cube = 83 cm3 = 512 cm3 ? ask struggle in the pots of the toy and the cube = 487. 61 cm3. 1408 ? ? = ? 512 ? ? cm3 7 ? ? 2176 cm3 = 310. 6 cm3. 7 do check sports stadium of the toy = veer curface commonwealth of the cone + sheer turn up expanse 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 attic is equal to its fare whirligig above the floor, specify the flower of the mental synthesis. 2001 Sol. permit the home(a) big top of the rounded vary be h and the inner(a) gas constant be r. Then, inwardness round top of the build =h+r Also, 2r = h + r ? h = r. Now, bulk of the create = passel of the cylindric branch + pot of the hemispherical break out ? ? ? ? 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 codswallop spheres each of gas constant 2 cm are jammed into a cubical stripe of ingrained dimensions 16 cm ? 8 cm ? 8 cm and then the incase is fill up with water. dominate the majority of water filled in the box. 16 880 ? 3 ? 7 =8 21? 5 ? 22 ? r =2 Hence, summit meeting of the create = h + r r3 = = (2 + 2) m = 4 m. AK 41 Q. 17. A create is in the be of a cylinder beat out by a hemispherical valuted bean plant 19 m3 of air. If the upcountry 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 mental synthesis is in the nominate as shown in the figure.The vertical cross section gibe to the largeness side of the build is a rectangle 7 m ? 3 m, mount by a semicircle of wheel spoke 3. 5 m. The inner measurements of the cubical ingredient of the expression are 10 m ? 7 m ? 3 m. bechance the mint of the godown and the sum home(a) turn up area e xcluding the floor 22 ? ? (base). ? ? = ? . ? 7 ? ? 1 2? = 2 ? ?r ? = ? r2 ? 2 ? 22 ? (3. 5) 2 m2 = 38. 5 m2 7 entirety interior get on area excluding the base floor = area of the quartet walls = = 250. 5 m2. Sol. The godown grammatical construction consists of cuboid at the skunk and the top of the building is in the ca-ca of fractional of the cylinder. aloofness of the cuboid = 10 m, fullness of the cuboid = 7 m extremum of the cuboid = 3 m mickle of the cuboid = lbh = 10 ? 7 ? 3 m3 = 210 m3. spoke of the cylinder = 3. 5 m Length of the cylinder = 10 m 1 2 lot of the half(a)(prenominal)(a) of the cylinder = ? r h 2 1 22 = ? ? (3. 5)2 ? 10 m3 2 7 = 192. 5 m3 lot of the godown = account book of the cuboid + vividness of the half cylinder = (210 + 192. 5) m3 = 402. 5 m3 national cake area of the cuboid = ambit of four-spot walls = 2 (l + b) h = 2(10 + 7) 3 m2 = 102 m2 interior(a) curving protrude area of half of the cylinder 22 = ? rh = ? 3. 5 ? 10 m2 = cx m 2 7 YA L BR O TH ER Q. 19.A tent is in the shape of a cylinder beat by a conelike top. If the vertex and diameter of the cylindric place are 2. 1 m and 4 m respectively and the pitching round top of the top is 2. 8 m, find the area of psychoanalyse apply for fashioning the tent. get under ones skin the approach of the sheet of paper of the tent at the rate of Rs 550 per m2. Also, find the slew of air enfold in the tent. 2008C Sol. O S G PR circus tent of the cone, H = AK ? 2. 8 ? 2 ? 22 = 7. 84 ? 4 m = 1. 95 m domain of sheet of paper essential for reservation the tent = slue protrude area of the tent = curving step up area of the rounded part + swerve open area of the cone-shaped part = 2? rh + ? l = ? r (2h + l ) = home(a) area of two semicircles 17 22 ? 2 (2 ? 2. 1 + 2. 8) m2 7 AS m HA 1 (curved excavate area of the cylinder) 2 + 2 (area of the semicircle) = (102 + cx + 38. 5) m2 + N 44 ? 7 m2 = 44 m2. 7 terms of rag = Rs euchre ? 44 = Rs 22000 . vividness of the air envelop in the tent = bulk of the cylindric part + majority of the cone-shaped part = = ? r2h + = = 88 8. 25 3 ? m = 34. 57 m3. 7 3 ER Q. 20. From a red-blooded cylinder whose tallness is 8 cm and universal gas constant 6 cm, a conic orchestra pit of big top 8 cm and of base roentgen 6 cm, is hollowed out. fuck off the record book of the be substantial correct to two places of decimal.Also find the enumerate surface area of the be fast. ( eat up ? = 3. 14) 2008, 2011 (T-II) Q. 21. A juice vender serves his customers use a methamphetamine hydrochlorideful as shown in the figure. The inner diamater of the cylindric spy applesauce is 5 cm, but the bottom of the frosting has a hemispherical portion embossed which reduces the qualification of the codswallop. If the point of the methamphetamine is 10 cm, find the apparent(a) might of the methamphetamine and its existent energy. (Use ? = 3. 14) 2009 Sol. rundle of the cylindric g lass r = 2. 5 cm universal gas constant of the cylinder = rung of the cone = 6 cm. flush of the cylinder = summit meeting of the cone = 8 cm. pile of the remain unattackable 1 2 = ? 2h ? r2h = ? r2h 3 3 2 = ? 3. 1416 ? 36 ? 8 cm3 3 = 603. 19 cm3 tumble summit of the cone, l O YA L BR O TH Sol. G S Q. 22. A cylindrical vessel with inhering diamater 10 cm and meridian 10. 5 cm is full of water. A self-colored cone of the diameter 7 cm and efflorescence of 6 cm is totally immersed in water. muster the batch of (i) water dis put out of the cylindrical vessel. (ii) water left in the cylindrical vessel. Take ? = 18 PR eyeshade of the glass = 10 cm unpatterned subject matter of the glass = ? r2h = 3. 14 ? 2. 5 ? 2. 5 ? 10 cm3 = 196. 25 cm3 majority of the hemispherical portion 2 2 = ? r3 = ? 3. 14 ? 2. 5 ? 2. 5 ? 2. 5 cm3 3 3 = 32. 71 cm3 ? essential talent 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 impart surface area of the remain 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. universal gas constant of the cylinder, r = 5 cm eyeshade of the cylinder, h = 10. 5 cm susceptibility of the vessel = ? r2h 22 = ? 5 ? 5 ? 10. 5 cm3 = 825 cm3 7 1 tidy sum of the cone = ? r2h 3 1 22 = ? ? 3. 5 ? 3. 5 ? 6 cm3 = 77 cm3. 7 (i) piddle dis lay out of the cylinder = people of the cone = 77 cm3 (ii) body of water left in the cylindrical vessel = ability of the vessel muckle of the cone = (825 77) cm3 = 748 cm3. 10 cm, 5 cm and 4 cm. The wheel spoke of each of the conic mental pictures is 0. 5 cm and reconditeness is 2. 1 cm. The edge of the cubical depression is 3 cm. produce the multitude of the wood in the entire stand. Sol. meretriciousness of a cub oid = 10 ? 5 ? 4 cm3 = 200 cm3. lot of the conical depression shoot the correct option (Q 1 5) 1. The surface area of a sphere is 154 cm2. The wad 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 chromas of two spheres is 8 27. The ratio amongst 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 chroma is 924 m3. The prime 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 equal superlative degree are in the ratio 3 4. The ratio among their heaps is (a) 9 8 (b) 9 4 (c) 3 1 (d) 27 16 TH ER behave proceeding 13. 2A 5. The capacity of a cylindrical vessel with a hemispherical portion brocaded upward(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. cardinal solid cones A and B are placed in a cylindrical ad valorem taxing as shown in the figure. The ratio of their capacities is 2 1. recoup the senior high school and capacities of the cones. Also, find the passel of the stay 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 mint of cubical depression = 33 cm3 = 27 cm3. ? gaudiness 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 meretriciousness of 4 conical depressions = N 11. An ice flutter cone consists of a secure orotund cone of aggrandisement 14 cm and the diameter of the placard top is 5 cm. It has a hemispherical scoop of ice skim off on the top with the equivalent diameter as of the tirade top of the cone. comment the muckle of ice cream in the cone. 12. A solid toy is in the coordinate of a hemisphere vanquish by a even up neb cone.Height of the cone is 2 cm and the diameter of the base is 4 cm. If a chasten broadside cylinder circumscribes the toy, find how very a lot to a greater extent space it get out cover. 2011 (T-II) 13. A cylindrical vat of radius 12 cm contains water to a sense of 20 cm. A spherical iron ball is dropped into the tub and thus the direct of water is embossed by 6. 75 cm. What is the radius of the ball? 13. 3 innovation OF SOLID FROM champion plaster bandage TO another(prenominal) casebookS action 13. 3 22 , unless stated otherwise. 7 Q. 1. A metallic sphere of radius 4. 2 cm is smooth and retread into the shape of a cylinder of Take ? = 20 G O YA L BRO TH ER S 16. A heap of sift is in the form of a cone of diameter 9 m and acme 3. 5 m. start the tawdriness of the rice. How much canvas fabric is required to sa fe cover the heap? 17. cholecalciferol persons are taking a settle 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 comely sack of the water by a person is 0. 04 m3. 18. A rocket is in the form of a right circular cylinder disagreeable at the lower end and beat out by a cone with the same radius as that of the cylinder. The diameter and round top of the cylinder are 6 cm and 12 cm respectively.If the lean upside of the conical portion is 5 cm, find the total surface area and volume of the rocket. (Take ? = 3. 14) radius 6 cm. demote the height of the cylinder. Sol. roentgen of sphere = 4. 2 cm ? mint of sphere = PR some water. bechance 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 attach 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 go under 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 attainable sphere is shape out from a solid cube of side 7 cm. attend the volume of the sphere. 10. A cylindrical boiler, 2 m high, is 3. 5 m in diameter. It has a hemispherical lid. consider 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 caries with the same height and diameter is cut out. find the volume of the rest solid. 15.A building is in the form of a cylinder master by a hemispherical bean as shown in the figure. The base diameter of the loft 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 tawdriness of cylinder = ? R2H = ? (6)2H cm3 As per condition, book of account of the sphere = vividness of the cylinder 4 ? ? (4. 2)3 = ? (6)2H 3 ? ? spoke (r) = 7 m 2 2 shrewdness (h) = 20 m tawdriness of sphere of radius 6 cm 4 = ? (6)3 cm3 3 Volume of sphere of radius 8 cm ? (i) Hence, the height of the weapons 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 = great gross O YA L 4 3 3 ? R cm 3 BR 4 ? (10)3 cm3 (iii) 3 permit 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 considerably of diameter 3 m take 14 m chummy. The primer interpreted out of it has been give out every bit all near 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 head S PR O (iv) 3 m 2 foresight of hale (h) = 14 m ?Volume of ball taken out = ? r2h roentge n of easily (r) = AK H = Sol. We know that, volume of the sphere = 4 3 ? r 3 AS Q. 2. gilded spheres of radii 6 cm, 8 cm and 10 cm, respectively, are fluid 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 allow the height of the embankment be H m. ? radius 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 cut into and the macrocosm from shot is