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BIG Brands Research Excellence Service Award - 2011LSOFT TECHNOLOGIES, PUNEPresentsXI, XII, JEE-MAINS, NEET/MH-CETTEST SERIESWITH CLASS NAME AND LOGOISO 9001 : 2008 CertifiedRLSOFT TECHNOLOGIES, PUNE303, Nilgiri Apts., Opp. S.N.D.T. College, Karve Road, Erandwane,Pune - 411 004 Ph: 020-25455771, 9370949449, 8421808284, 9423456101Web : www.Lsoft.co Email : info@Lsoft.coMore than 500 educational organizations are getting benefitted from our software and Services.
To,The concerned authority,Respected Sir/Madam,The world is moving ahead very fast and technology is accelerating this speed. If we do notadopt new technology, we may lag behind the world. The students, especially from the rural areaare not introduced to new technology at the right time. That gives a sense of inferiority complex tothem. I, being from the rural area, can better understand this fact and I decided to modernize thiseducational sector with the help of technology and equip students with modern techniques ofstudy. Hence I introduce educational software to the rural & urban area, in year 2002. That was thebeginning of Lsoft Technologies.We also noticed that the time, efforts and energy spent by the teachers of the educationalorganization in preparing question papers, checking the answer sheets, declaring the result etc.deviate their focus from their core activity i.e. teaching. So we have started offering total solutionfor competitive examinations. That includes question paper and OMR answer sheets designing,printing, computerized checking of the answer sheets, declaring results, performance analysis ofindividual student etc. The educational organizations to which we are offering our services savestime, effort and energy as well as cost.Honesty and sincerity are the two virtues we have been following since the inception of LsoftTechnologies and our achievements speak for the rewards we got by these virtues.We assure you that we will keep on doing the quality work in the future as well and hope toget the same kind of support from you all.Thanking you,Amol Bijwe,Director,Lsoft Technologies, PuneEmail: admin@Lsoft.co
ABOUT LSOFT TECHNOLOGIESABOUT LSOFT TECHNOLOGIESVISIONTo revolutionize the education sector using the latest Information Technology.MISSIONTo deliver high quality educational products and services.GOALTo be the leading solution provider in the education sector.ABOUT USLsoft Technologies is an ISO 9001:2008 certified company functional in educational arenafor last 12 years. It was incorporated in 2002 with a mission to create state of the art softwaresolutions for domestic as well as international markets using the latest Information Technology torevolutionize the education sector.CORPORATE PROFILELsoft Technologies is run by young and dynamic Engineers, Management Graduates andAcademicians. It combines strong management and infrastructure resources with state of the arttechnology and a highly skilled and versatile workforce of dedicated software professionals toconceptualize, design, develop and market software solutions for demanding and qualityconscious customers.ACHIEVEMENTSWithin a short period of 12 years, Lsoft Technologies has carved a niche for itself in thefiercely competitive market. It has become the leading educational solutions provider inMaharashtra and neighbouring states. Lsoft has developed a range of quality products for theeducational institutes and students and created strong brand image through efficient service andcustomer satisfaction. In 2011, Lsoft was honored by BIG Brands Research Excellence ServiceAward for its quality services in education sector. Lsoft has designed products both for domesticas well as international markets with its presence in Maharashtra, Goa, Gujarat, M. P., A. P., WestBengal etc. This proves its capabilities and long term commitment to Education Sector.
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SAMPLE QUESTION PAPERCLASS NAME & LOGO XII-JEE (MAINS)-YEARTopic Names: Circular motionTest Number000001Test Booklet No.110001 Write/Check this Code on yourAnswer SheetWrite this number on yourAnswer Sheet: IMPORTANT INSTRUCTIONS y fill in the particulars on this page of the Test Booklet with Blue/Black Ball point Pen. Use of pencil is strictlyprohibited.The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fillin the particulars carefully.The test is of 60 Min. duration.The Test Booklet consists of 30 questions. The maximum marks are 120. All the Ques. consists of FOUR (4) marks each.PHYSICS- 30 Ques. (120 marks).Candidates will be awarded marks as stated above in Instruction No.5 for correct response of each question. ¼ (one-fourth) markswill be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response isindicated for an item in the Answer Sheet.Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet. Use ofpencil is strictly prohibited.No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device,etc., except the Admit Card inside the examination hall/room.Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of eachpage of the booklet.On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, thecandidates are allowed to take away this Test Booklet with them.The CODE for this Booklet is A. Make Sure that the CODE printed on Side-2 of the Answer Sheet is the same as that on this booklet.In case of discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the Test Bookletand the Answer Sheet.Do not fold or make any stray marks on the Answer Sheet.No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.Name of the Candidate:Roll Number : In figures :In words :Examination Centre Number :Name of Examination Centre (in Capital letters) :Candidate’s Signature :Invigilator’s Signature :
1. The angle turned by a body undergoingcircular motion depends on time asθ θ0 θ1t θ2t2 . Then the angularacceleration of the body is(2) θ2(1) θ1(3) 2θ1(4) 2θ22. The maximum and minimum tension in thestring whirling in a circle of radius 2.5 m withconstant velocity are in the ratio 5 : 3 then itsvelocity is(1) 7 m/s(2) 98 m/s(3)4.9(4)490 m/s3. A bucket full of water is revolved in verticalcircle of radius 2m. What should be themaximum time-period of revolution so thatthe water doesn't fall off the bucket(1) 1 sec(2) 2 sec(3) 3 sec(4) 6 sec4. The coordinates of a moving particle at anytime ‘t’ are given by x αt3 and y βt3. Thespeed of the particle at time ‘t’ is given by(1)α 2 β2(3) 3 t2 α2 β2(2) t2 α2 β2(4) 3 t α2 β25. A simple pendulum oscillates in a verticalplane. When it passes through the meanposition, the tension in the string is 3 timesthe weight of the pendulum bob. What is themaximum displacement of the pendulum ofthe string with respect to the vertical(2) 45 o(1) 10o(3) 60 o(4) 90 o6. A stone of mass m is tied to a string and ismoved in a vertical circle of radius r making nrevolutions per minute. The total tension inthe string when the stone is at its lowest pointis(1) mg(2) m(g π n r)(3) m(g π n r 2 )(4) m{g (π2 n2 r)/ 900}7. A stone tied with a string, is rotated in avertical circle. The minimum speed withwhich the string has to be rotated(1) Is independent of the length of the string(2) Is independent of the mass of the stone(3) Decreases with increasing in length of thestring(4) Decreases with increasing mass of thestone8. A fan is making 600 revolutions perminute. If after some time it makes 1200revolutions per minute, then increase in itsangular velocity is(1) 10 π rad /sec(2) 20 π rad /sec(3) 40 π rad /sec(4) 50 π rad /sec9. The tension in the string revolving in avertical circle with a mass m at the endwhich is at the lowest position(1) mgmv2 mgrmv2 mg(3)rmv 2(4)r(2)10. If the equation for the displacement of aparticle moving on a circular path is given by(θ) 2t3 0.5 , where θ is in radians and t inseconds, then the angular velocity of theparticle after 2 sec from its start is(1) 8 rad/sec(2) 12 rad/sec(3) 24 rad/sec(4) 48 rad/sec11. A weightless thread can bear tension upto3.7 kg wt. A stone of mass 500 gms is tied toit and revolved in a circular path of radius 4m in a vertical plane. If g 10 ms 2 , then themaximum angular velocity of the stone will be(1) 2 radians/sec(2) 21 radians/sec(3) 16 radians/sec(4) 4 radians/sec12. A weightless thread can support tensionupto 30 N. A stone of mass 0.5 kg is tied to itand is revolved in a circular path of radius 2m in a vertical plane. If g 10m /s2 , then themaximum angular velocity of the stone will be(1) 10 rad /s(3)30 rad/s(2)60 rad/s(4) 5 rad /s13. A 2 kg stone at the end of a string 1 mlong is whirled in a vertical circle at aconstant speed. The speed of the stone is 4m/sec. The tension in the string will be 52 N,when the stone is(1) Halfway down(2) At the bottom of the circle(3) At the top of the circle(4) None of the above
14. A wheel is subjected to uniform angularacceleration about its axis. Initially its angularvelocity is zero. In the first 2 sec, it rotatesthrough an angle θ1 . In the next 2 sec, itrotates through an additional angle θ2 . Theratio of θ2 / θ1 is(1) 1(3) 3(2) 2(4) 415. In a circus stuntman rides a motorbike ina circular track of radius R in the verticalplane. The minimum speed at highest point oftrack will be(1) 2gR(2) 2gR(3)gR(4)3gR16. If the length of the second's hand in a stopclock is 3 cm the angular velocity and linearvelocity of the tip is(1) 0.2547 rad/sec., 0.314 m/sec(2) 0.2047 rad/sec., 0.0314 m/sec(3) 0.1047 rad/sec., 0.00314 m/sec(4) 0.1472 rad/sec., 0.06314 m/sec17. In 1.0 s, a particle goes from point A topoint B, moving in a semicircle of radius 1.0m (see figure). The magnitude of the averagevelocity is(1) Zero(3) 2.0 m/s(2) 1.0 m/s(4) 3.14 m /s20. A cyclist goes round a circular path ofcircumference 34.3 m in 22 sec. the anglemade by him, with the vertical, will be(1) 40 o(2) 45oo(3) 48(4) 42 o21. If a particle covers half the circle of radiusR with constant speed then(1) Change in K.E. is zero(2) Change in K.E. is mv2(3) Change in K.E. is 1/2 mv2(4) Momentum change is mvr22. A wheel completes 2000 revolutions tocover the 9.5 km. distance. then the diameterof the wheel is(1) 1.5 cm(2) 1.5 m(3) 7.5 m(4) 7.5 cm23. If a particle of mass m is moving in ahorizontal circle of radius r with acentripetal force ( k / r 2 ) , the total energy iskk(2) (1) r2r4k2k(3) (4) rr24. Radius of the curved road on nationalhighway is R . Width of the road is b . Theouter edge of the road is raised by h withrespect to inner edge so that a car withvelocity v can pass safe over it. The value ofh isv 2bv(2)(1)RgRgb(3)v 2bR(4)v2Rg18. A person with his hands in his pockets isskating on ice at the velocity of 10 m/s anddescribes a circle of radius 50 m. What is hisinclination with vertical 1 (2) tan 1(1)(1) tan 1 5 3 1 (3) tan 1 (4) tan 1 5 10 25. A particle of mass m is executing uniformcircular motion on a path of radius r . If p isthe magnitude of its linear momentum. Theradial force acting on the particle isrm(1)(2) pmrp19. A bob of mass 10 kg is attached to wire0.3 m long. Its breaking stress is 4.8 107N/m2. The area of cross section of the wire is10-6 m2. The maximum angular velocity withwhich it can be rotated in a horizontal circle(1) 1 rad/sec(2) 2 rad/sec(3) 4 rad/sec(4) 8 rad/sec26. The length of second's hand in a watch is1 cm. The change in velocity of its tip in 15seconds isππ 2(2)(1)cm /seccm / sec3030π(3)(4) Zerocm /sec30 2(3)p2rm(4)mp2r
27. A car is moving on a circular path andtakes a turn. If R1 and R2 be the reactionson the inner and outer wheels respectively,then(1) R1 R 2(2) R1 R 2(3) R1 R 2(4) R1 R 228. Two bodies of mass 10 kg and 5 kgmoving in concentric orbits of radii R and rsuch that their periods are the same. Then theratio between their centripetal acceleration is(1) r / R(2) R / r(3) r 2 /R 2(4) R 2 / r 229. A particle P is moving in a circle of radius'a ' with a uniform speed v . C is the centre ofthe circle and AB is a diameter. Whenpassing through B the angular velocity of Pabout A and C are in the ratio(1) 1 : 1(2) 1 : 2(3) 2 : 1(4) 4 : 130. Two racing cars of masses m1 and m 2 aremoving in circles of radii r1 and r2respectively. Their speeds are such that eachmakes a complete circle in the same durationof time t . The ratio of the angular speed of thefirst to the second car is(2) m1 : m2(1) r1 : r2(3) m1r1 : m2r2(4) 1 : 1
SAMPLE SOLUTIONSCLASS NAME & LOGOTOPIC NAME: Circular motion1. Ans: (4) 2θ2Sol: Angular acceleration d2 θ 2θ2dt22. Ans: (2) 98 m /sSol: In this problem it is assumed that particlealthough moving in a vertical loop but itsspeed remain constant.mv 2Tension at lowest point Tmax mgrmv 2Tension at highest point Tmin mgrmv 2 mgTmax5 r2 mvTmin3 mgrby solving we get, v 4gr 4 9.8 2.5 98 m/s3. Ans: (3) 3 secSol: Minimum angular velocity ωmin g / R Tmax 22πR 2π 2 2 3s 2π10gωmin4. Ans: (3) 3 t 2 α 2 β2Sol: x α t 3 and y βt 3 (given)dxdyvx 3αt 2 and v y 3βt 2dtdtResultant velocity v v 2x v 2y 3t 2 α 2 β25. Ans: (4) 90 oSol: Tension at mean position,mv 2mg 3mgr (i)v 2gland if the body displaces by angle θ with thevertical then v 2gl(1 cos θ) (ii)Comparing (i) and (ii), cos θ 0 θ 90 6. Ans: (4) m{g (π2 n2 r)/900}Sol: T mg mω2r m{g 4π2n2r}2 π2n2r n m g 4π2 rmg 60 900 Test No. : 0000017. Ans: (2) Is independent of the mass of thestoneSol: Is independent of the mass of the stone8. Ans: (2) 20 π rad /secSol: Increment in angular velocityω 2π(n2 n1 )ω 2π(1200 600) 20 πrad 2π 600 rad ŵŝŶ60sradsmv 2 mgrSol: Tension Centrifugal force weightmv 2 mgr9. Ans: (2)10. Ans: (3) 24 rad/secdθ dSol: ω (2t 3 0.5) 6t 2 dt dtat t 2 s, ω 6 (2)2 24rad/s11. Ans: (4) 4 radians/secSol: Max. tension that string can bear 3.7kgwt 37NTension at lowest point of verticalloop mg mω2r 0.5 10 0.5 ω2 4 5 2ω237 5 2w2 ω 4 rad/s.12. Ans: (4) 5 rad /sSol: Tmax mω2max r mg Tmax ω2r gm30 10 ω2max r 0.55050ωmax 5 rad /sr2 13. Ans: (2) At the bottom of the circlemv 2 2 (4)2Sol: mg 20N and 32Nr1It is clear that 52 N tension will be at thebottom of the circle. Because we know thatmv 2TBottom mg r
14. Ans: (3) 31Sol: Using relation θ ω0 t at221 (i) (As ω0 0, t 2 sec )θ1 (α )(2)2 2α2Now using same equation for t 4 sec, ω0 01 (ii)θ1 θ2 α(4)2 8α2θFrom (i) and (ii), θ1 2α and θ2 6α 2 3θ115. Ans: (3) gRSol: Minimum speed at the highest point ofvertical circular path v gR16. Ans: (3) 0.1047 rad/sec., 0.00314 m/sec2π 2πSol: ω 0.1047 rad/sT60and v ωr 0.1047 3 10 2 0.00314 m/s21. Ans: (1) Change in K.E. is zeroSol: As momentum is vector quantitychange in momentum P 2mv sin(θ/ 2) 2mv sin(90) 2mvBut kinetic energy remains always constantso change in kinetic energy is zero.22. Ans: (2) 1.5 mSol: Distance covered in ‘n’ revolution n 2πr nπ D 2000 π D 9500 [As n 2000, distance 95009500 m] D 1.5 m2000 π23. Ans: (2) k2rmv 2kk1kK.E. mv 2 2 mv 2 rrr22rkkP.E. F dr 2 dr rrk kkTotal energy K.E. P.E. 2r r2rSol:17. Ans: (3) 2.0 m /sSol: AverageTotal displacement 2m 2ms 1velocity time1s 1 18. Ans: (1) tan 1 5 Sol: The inclination of person from vertical isgiven by,v2(10)21θ tan 1(1/5)tan θ rg 50 10 524. Ans: (2)19. Ans: (3) 4 rad/secSol: Centripetal force breaking force mω2 r breaking stress cross sectionalarea mω2r p A 25. Ans: (3)p A4.8 107 10 6 mr10 0.3ω 4 rad /secω Sol: We know that tan θ Hencehv2and tan θ Rgbh v2v 2b h b RgRgp2rm2Sol: Radial force mv 2 m p p2[As p rr m mrmv]π 2cm / sec30Sol: In 15 second's hand rotate through 90 .26. Ans: (1) Change in velocity v 2v sin(θ /2)20. Ans: (2) 45oSol: 2πr 34.3 r v2bRg34.32πr2πrand v T2π22 v2 Angle of binding θ tan 1 45 rg 2(rω)sin(90 / 2) 2 1 4π60 2 π 2 cm30 sec2π 1 T2[As T 60 sec]
27. Ans: (3) R1 R 2Sol: Reaction on inner wheel1 v 2h R1 M g 2 ra Reaction on outer wheel R 2 1 v2h M g 2 ra where, r radius of circular path, 2a distance between two wheels and h height ofcentre of gravity of car.28. Ans: (2) R / rSol:a R ωR2 R Tr2 R R 2 2 arωr rTR rr[As Tr TR]29. Ans: (2) 1 : 2Sol: Angular velocity of particle P about pointA,vv rAB 2rAngular velocity of particle P about point C,vvωC rBC rωv/2r 1 .Ratio A v/r2ωCωA 30. Ans: (4) 1 : 1Sol: As time periods are equal therefore ratio2πof angular speeds will be same. ω T
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XI-STD TEST SERIES SCHEDULE �ϭϭϬϮ ĐĂůĂƌƐ ĂŶĚ ŝůĞ �ϮϬϱϰϭϭϭϬϱ&ƌŝĐƚŝŽŶ ŝŶ ƐŽůŝĚƐ ĂŶĚ ůŝƋƵŝĚƐϯϬϭϮϬϲϰϭϭϭϬϲ ŽƵŶĚ tĂǀĞƐϯϬϭϮϬϳϰϭϭϭϬϳdŚĞƌŵĂů ƉƌŽƉĞƌƚŝĞƐ ŽĨ �ŽŶ ŽĨ ŝŐŚƚϯϬϭϮϬϵϰϭϭϭϬϵZĂLJ ƐƚĂƚŝĐƐϯϬϭϮϬϭϭϰϭϭϭϭϭ ƵƌƌĞŶƚ �ŐŶĞƚŝĐ ĞĨĨĞĐƚ ŽĨ ĞůĞĐƚƌŝĐ �ŶĞƚŝĐ ǁĂǀĞƐϯϬϭϮϬ ! ϭϰϭϭϮϬϭ ŽŵĞ ĂƐŝĐ ŽŶĐĞƉƚƐ ŽĨ ŚĞŵŝƐƚƌLJϯϬϭϮϬϮϰϭϭϮϬϮ ƚĂƚĞƐ ŽĨ ŵĂƚƚĞƌ ͗ 'ĂƐĞƐ ĂŶĚ ůŝƋƵŝĚƐϯϬϭϮϬϯϰϭϭϮϬϯ ƚƌƵĐƚƵƌĞ ŽĨ ĂƚŽŵϯϬϭϮϬϰϰϭϭϮϬϰWĞƌŝŽĚŝĐ ƚĂďůĞϯϬϭϮϬϱϰϭϭϮϬϱZĞĚŽdž ƌĞĂĐƚŝŽŶϯϬϭϮϬϲϰϭϭϮϬϲ ŚĞŵŝĐĂů ĞƋƵŝůŝďƌŝƵŵϯϬϭϮϬϳϰϭϭϮϬϳ ƵƌĨĂĐĞ �Ğ ŽĨ ĐŚĞŵŝĐĂů �ϭϮϬϭϬϰϭϭϮϭϬƐͲ ůŽĐŬ ĞůĞŵĞŶƚƐ ;ŐƌŽƵƉ ϭ Θ ϮͿϯϬϭϮϬϭϭϰϭϭϮϭϭƉͲ ůŽĐŬ ĞůĞŵĞŶƚƐ ;ŐƌƉƵƉ ϭϯ ΘϭϰͿϯϬϭϮϬϭϮϰϭϭϮϭϮ ĂƐŝĐ ƉƌŝŶĐŝƉůĞƐ ĂŶĚ ƚĞĐŚŶŝƋƵĞƐ ŝŶ ŽƌŐĂŶŝĐ ĐŚĞŵŝƐƚƌLJϯϬϭϮϬϭϯϰϭϭϮϭϯ ůŬĂŶĞƐϯϬϭϮϬϭϰϰϭϭϮϭϰ ůŬĞŶĞƐϯϬϭϮϬϭϱϰϭϭϮϭϱ ůŬLJŶĞƐϯϬϭϮϬϭϲϰϭϭϮϭϲ ƌŽŵĂƚŝĐ ƌŽŶŵĞŶƚĂů ĐŚĞŵŝƐƚƌLJϯϬϭϮϬ
* .-., ϭϰϭϭϯϬϭϮ ŝǀĞƌƐŝƚLJ ŝŶ ŽƌŐĂŶŝƐŵƐϲϬϮϰϬϰϭϭϯϬϮ ŝŶŐĚŽŵ WůĂŶƚĂĞϲϬϮϰϬϯϰϭϭϯϬϯ ŝŽĐŚĞŵŝƐƚƌLJ ŽĨ ĐĞůůϲϬϮϰϬϰϰϭϭϯϬϰ Ğůů �ŽŐLJ ŽĨ WůĂŶƚƐϲϬϮϰϬϲϰϭϭϯϬϲWĂŶƚ tĂƚĞƌ ZĞůĂƚŝŽŶƐ ĂŶĚ DŝŶĞƌĂů EƵƚƌŝƚŝŽŶϲϬϮϰϬϳϰϭϭϯϬϳWůĂŶƚ 'ƌŽǁƚŚ ĂŶĚ ĞǀĞůŽƉŵĞŶƚϲϬϮϰϬϴϰϭϭϯϬϴ ŝŶŐĚŽŵ �ĂƚŝŽŶ ŽĨ ĞůůϲϬϮϰϬϭϬϰϭϭϯϭϬ ƚƵĚLJ ŽĨ ŶŝŵĂů dŝƐƐƵĞƐϲϬϮϰϬϭϭϰϭϭϯϭϭ ƚƵĚLJ ŽĨ ŶŝŵĂů dLJƉĞϲϬϮϰϬϭϮϰϭϭϯϭϮ,ƵŵĂŶ EƵ
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