JEE(MAIN) – 2015 TEST PAPER WITH SOLUTION (HELD ON .
CODE-DJEE(MAIN) – 2015 TEST PAPER WITH SOLUTION(HELD ON SATURDAY 04th APRIL, 2015)PART A – PHYSICS1.Distance of the centre of mass of a solid uniformcone from its vertex is z0. If the radius of its baseis R and its height is h then z0 is equal to :3h 2(2)8R5h(1)8(3)3.h24R(4)A pendulum made of a uniform wire of crosssectional area A has time period T. When anadditional mass M is added to its bob, the timeperiod changes to TM. If the Young's modulusof the material of the wire is Y then3h41is equalYto :- (g gravitational acceleration)Ans. (4) T 2 A(1) 1 M T MgSol. T 2 A(2) 1 TM Mgh42 (3) TM 1 A T Mgso z0 h –hfrom base4 for solid cone c.m. ish 3h 44A red LED emits light at 0.1 watt uniformlyaround it. The amplitude of the electric field ofthe light at a distance of 1 m from the diode is:(1) 5.48 V/m(2) 7.75 V/m(3) 1.73 V/m(4) 2.45 V/mAns. (4) 2.Sol. Iav E 1P 0E2C 24 r 22P4 r 2 0 cOn putting value we get 2.45 v/m. T 2 Mg(4) M 1 T AAns. (3)Sol. T 2 LgTM 2 .(1)L Lg.(2)FAY LLFLAYPutting (3) in (2) L .(3)solving the equation we get the value of1asY TM 2 A – 1 T mgJEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.1
JEE(MAIN)-20154.For a simple pendulum, a graph is plottedbetween its kinetic energy (KE) and potentialenergy (PE) against its displacement d. Whichone of the following represents thesecorrectly ? (graphs are schematic and notdrawn to scale)EKE v Sol. f' f v v S d(1)A train is moving on a straight track with speed20 ms –1 . It is blowing its whistle at thefrequency of 1000 Hz. The percentage changein the frequency heard by a person standingnear the track as the train passes him is(speed of sound 320 ms–1) close to :(1) 18% (2) 24%(3) 6%(4) 12%Ans. (4)5. v f" f v v S PEEv vSf" v vf'S(2)KEKEdPE KEdAns. (4)1m 2[A2 – d2], downward parabola21PE m 2d2, upward parabola.2Sol. KE KEvvS 100 12%6.When 5V potential difference is applied acrossa wire of length 0.1 m, the drift speed ofelectrons is 2.5 10 –4 ms–1 . If the electrondensity in the wire is 8 1028 m–3, the resistivityof the material is close to :(2) 1.6 10–5 m(1) 1.6 10–6 m(3) 1.6 10–8 m(4) 1.6 10–7 mAns. (2) Sol. V iR (neAVd ) A V neVd PE21 PE E(4)v vS v v Sf " f ' 10 100 v vSf'2vS v vSE(3) PEVneVd On putting values are got the answer 1.6 10–5 mJEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.2
CODE-D7.6V8. LITwo long current carrying thin wires, both withcurrent I, are held by the insulating threads oflength L and are in equilibrium as shown in thefigure, with threads making an angle ' ' withthe vertical. If wires have mass per unit lengththen the value of I is :(g gravitational acceleration)(2) gL(3) sin µ 0 cos gL(4) 2sin µ 0 cos In the circuit shown, the current in the 1 resistor is :(1) 0.13 A, from Q to P(2) 0.13 A, from P to Q(3) 1.3 A, from P to Q(4) 0AAns. (1)Sol.P6V gLtan µ0 B µ0Iµ0I 2 2 sin 2 rtan fBwhere fB is force per unit length(Bi) gµ0I g tan 2 2 sin I5R0V9VQ96 0 92 35E 3 1 5 5 15 31 1 123 153 1 5E3 0.13R23from ve to –veQ to P9.Assuming human pupil to have a radius of0.25cm and a comfortable viewing distance of25cm, the minimum separation between twoobjects that human eye can resolve at 500 nmwavelength is :(1) 100 µm(2) 300 µm(3) 1 µm(4) 30 µmAns. (4)Sol.i d01.22 d1don solving g I 2 sin µ cos 0R3R Ans. (4)Q 3 gL(1) 2 µ tan 09V1 3 IP 2 d0 1.22 500 10 9 25 10 2 20.5 10d0 1.22 25 10–6d0 30 µmJEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.3
JEE(MAIN)-201510.An inductor (L 0.03 H) and a resistor11.(R 0.15 k ) are connected in series to abattery of 15V EMF in a circuit shown below.An LCR circuit is equivalent to a dampedpendulum. In an LCR circuit the capacitor ischarged to Q0 and then connected to the L andR as shown below :-The key K1 has been kept closed for a longRLtime. Then at t 0, K1 is opened and key K2is closed simultaneously. At t 1ms, the currentin the circuit will be (e5 150) :0.03H0.15k CIf a student plots graphs of the square ofK2maximum charge (Q 2Max ) on the capacitor withtime (t) for two different values L 1 and L 2(L 1 L 2) of L then which of the followingrepresents this graph correctly ? (plots areschematic and not drawn to scale)15V K1(1) 6.7 mA(2) 0.67 mA(3) 100 mA(4) 67 mAL2(1)Ans. (2)Decay of current tRL I I0e10 3 1500.03 15 e 150 L1Q 2MaxtQ2Max Q0 (For both L1 and L2)(2)t1 5e10L12Q MaxL2(3)11500t 6.66 10–4Q 2Max 0.666 10–3(4) 0.67 mAL2L1tAns. (3)JEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.4
CODE-DSol. As damping is happening its amplitude wouldvary asQ0ChargeQ2(3)1µFThe oscillations decay exponentially and willbe proportional to e – t where dependsinversely on L.So as inductance increases decay becomesslower forL13µFCChargeQ2(4)L21µF3µFCAns. (4)Sol.In the given circuit, charge Q 2 on the 2µFcapacitor changes as C is varied from 1µF to3µF. Q2 as a function of 'C' is given properlyby : (figures are drawn schematically and arenot to scale) :- 12.1µF C2µFE ChargeQ2(1)1µF3µF3C3 C2CEQ2 C 3 Ceq CQ Ceq.E3CE C 3 3 2E 1 C 3 1 EQ2/when C 1µF 2E 42 1 Q2/when C 3µF 2E E 2 6E2E 3dQ C 3 1 C 3 2 0 dC6E 2 12Ed 2 023 C 3 C 3 dC 2ChargeQ2(2)1µF3µFCJEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.5
JEE(MAIN)-201513.From a solid sphere of mass M and radius R acube of maximum possible volume is cut.Moment of inertia of cube about an axispassing through its centre and perpendicular toone of its faces is:(1)(3)4MR 2(2)9 3 MR2(4)32 2 3 3 MR 216 2 aOn a hot summer night, the refractive index ofair is smallest near the ground and increaseswith height from the ground. When a light beamis directed horizontally, the Huygens' principleleads us to conclude that as it travels, the lightbeam :(1) bends downwards(2) bends upwards(3) becomes narrower(4) goes horizontally without any deflectionAns. (2) RLet mass and side of cube be M' and aM 3aM' 42R 333 a 2RSol.M 'a 26Normalµ2µ1µ increases Moment of Inertia of cube g 100 2.722% 3%g15.R214. g 1 10 –31 2 90g 20 10 –2100 1004MR 2Ans. (1)Sol. g T 2g T24MR M 3 a 4 3 a 6 4 3 2R 3 The period of oscillation of a simple pendulum L. Measured value of L is 20.0 cmgknown to 1 mm accuracy and time for 100oscillations of the pendulum is found to be 90susing a wrist watch of 1s resolution. Theaccuracy in the determination of g is :(1) 1%(2) 5%(3) 2%(4) 3%Ans. (4)is T 2 Sol. T 2 g 2 1 light bends towards normal light beam bends upwards (as µ with height)16. A single of 5 kHz frequency is amplitudemodulated on a carrier wave of frequency2 MHz. The frequencies of the resultant signalis/are (1) 2005 kHz, 2000 kHz and 1995 kHz(2) 2000 kHz and 1995 kHz(3) 2 MHz only(4) 2005 kHz and 1995 kHzAns. (1)Sol. Frequency present after modulationfc, fc fs 2000 KHz, 2005 KHz and 1995 KHz17.A solid body of constant heat capacity 1 J /C4 2 g T2JEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.6
CODE-Dis being heated by keeping it in contact withreservoirs in two ways (i) Sequentially keeping in contact with 2reservoirs such that each reservoir suppliessame amount of heat.(ii) Sequentially keeping in contact with 8reservoirs such that each reservoir suppliessame amount of heat.In both the cases body is brought from initialtemperature 100 C to final temperature 200 C.Entropy change of the body in the two casesrespectively is (1) ln2, 2ln2(2) 2ln2, 8ln2(3) ln2, 4ln2(4) ln2, ln2Ans. (4) S T1 S1 (1) dT nc T 2001008 12 dT 1 n2 T 240 dT 1 n2 T Consider a spherical shell of radius R attemperature T. The black body radiation insideit can be considered as an ideal gas of photonswith internal energy per unit volume(2)1 U U T 4 and pressure p . If the shell3 V Vnow undergoes an adiabatic expansion therelation between T and R is -(3) S2 200u 1R(3) T e–RAns. (1)Sol.1R3(4) T e–3RP 2401 U 3 V (y2–y1)m812t 812t(s)(y2–y1)mt(s)(y2–y1)m(4)t(s)12Ans. (1)Sol. For particle 14P Tusing PV nRTA1 T3V T t(s)(2) T (1) T 240 10018.Two stones are thrown up simultaneously fromthe edge of a cliff 240 m high with initial speedof 10 m/s and 40 m/s respectively. Which ofthe following graph best represents the timevariation of relative position of the secondstone with respect to the first ?(Assume stones do not rebound after hitting theground and neglect air resistance, take g 10 m/s2)(The figure are schematic and not drawn toscale)(y2–y1)m240 Sol.T219.10m/s240m1RBJEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.7
JEE(MAIN)-2015kQ r 2 5 kQ53 V Vfor .0 2R R2 4 R4RR2 2kQ 3 kQ3 for V V0R3 4 R4 –240 10t – gt225t2 – 10t – 240 0t1 8 secFor particle 240m/sA4R3kQ kQV for V 0R 4 4R 04R4 4R0 .Monochromatic light is incident on a glassprism of angle A. If the refractive index of thematerial of the prism is µ, a ray, incident at anangle , on the face AB would get transmittedthrough the face AC of the prism provided :R3 B 2gt221.5t2 – 40t – 240 0t2 12 secfor 0 t 8 sec arel 0straight line x-t graphfor 8 t 12 sec arel – gdownward parabola –240 40t –for t 12 sec Both particles comes to restBA uniformally charged solid sphere of radiusR has potential V0 (measured with respect to )on its surface. For this sphere the equipotential 20.3V0 5V0 3V0V0,,and2444have radius R1 , R2 , R3 and R4 respectively. Then(1) R1 0 and R2 (R4 – R3)(2) 2R R4(3) R1 0 and R2 (R4 – R3)(4) R1 0 and (R2 – R1) (R4 – R3)Ans. (1 or 2)Sol. surfaces with potentialsV for r R& for r RV kQ r2 3 2R R2 A C 1 1 1 (1) cos µ sin A sin µ 1 (2) cos 1 µ sin A sin 1 µ 1 (3) sin 1 µ sin A sin 1 µ 1 (4) sin 1 µ sin A sin 1 µ Ans. (3)Sol.kQrat r R V0 kQRat r 0 V0 3 kQ 3 V02 R 2 R1 0JEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.8
CODE-DFor all light to come out from face AC anglezof emergence e 90Apply Snell's Law at face ACµ sin r2 1 sin e 1 r2 sin–1 µ r1 A –B(d)(if e 90) 1 µ sin–1 ( r1 r2 A)Apply Snell's law at face AB1 sin µsin(r1)-1 -1 1 sin µ sin(A-sin µ 1 sin -1 µ sin(A-sin -1 µ 22.A rectangular loop of sides 10 cm and 5 cmcarrying a current I of 12 A is place in differentorientations as shown in the figures below :BIyI x23.II(a)zB(b)IIyIIxmagnetic force on the inner solenoid due to the outer one and F2 be the magnetic force on theouter solenoid due to the inner one. Then : (1) F1 is radially inwards and F2 0 (2) F1 is radially outwards and F2 0 (3) F1 F2 0 (4) F1 is radially inwards and F2 is radiallyAns. (3)S1S2BIIIFor unstable equilibriumB magnetic fieldM magnetic momentTwo coaxial solenoids of different radii carry current I in the same direction. Let F1 be theoutwards.z(c)IIf there is a uniform magnetic field of 0.3 T inthe positive z direction, in which orientationsthe loop would be in (i) stable equilibrium and(ii) unstable equilibrium ?(1) (b) and (d), respectively(2) (b) and (c), respectively(3) (a) and (b), respectively(4) (a) and (c), respectivelyAns. (1) Sol. For stable equilibrium B M zxyI for all light transmitted through AC, e 90 IISol.yIxJEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.9
JEE(MAIN)-201524.Field due to solenoid 2 outside the solenoid iszero f1 0 Due to action reaction f2 0or solenoid 2 behave as magnetic dipole Md Bf2 dxdB 0field of 1 is uniform dxf2 0A particle of mass m moving in the x directionwith speed 2 is hit by another particle of mass2m moving in the y direction with speed . If thecollisions perfectly inelastic , the percentage loss25.Consider an ideal gas confined in an isolatedclosed chamber. As the gas undergoes anadiabatic expansion, the average time ofcollision between molecules increases as V q,where V is the volume of the gas. The valueof q is :-Cp Cv 123 5(3)6Ans. (1)in the energy during the collision is close to :(2) 62%(3) 44%(4) 50% Sol.(2)Volumev avg (1) 56 % 123 5(4)6(1) 1Ans. (1)Sol. Before collisonym2v1v2 V1 TFor adiabatic processT V1– V Tcomparing x2m11m(2v)2 2m(v) 222 3 mv226.–1 / 2 V 12 12From a solid sphere of mass M and radius R,q Kinetic energy After collisonApplying momentum conservation for inelasticcollision 2mvjˆ m2viˆ 3m vf8 vvf 9Kf a spherical portion of radiusRis removed, as2shown in the figure. Taking gravitationalpotential V 0 at r , the potential at thecentre of the cavity thus formed is :(G gravitational constant)14mv 2 3m v 2f 23% K K i – K f 5mv 2 / 3 5 56%Ki3mv 29 2GM3R GM(3)2RAns. (4)(1) 2GMR GM(4)R(2)JEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.10
CODE-D28.Sol. By principle of superosition 27.FGM 2 R 2 3G M3R – 2 R2R3 4 82–11 GM GMGM –8R8RRABGiven in the figure are two blocks A and B ofweight 20 N and 100 N, respectively. These arebeing pressed against a wall by a force F asshown. If the coefficient of friction between theblocks is 0.1 and between block B and the wallis 0.15, the frictional force applied by the wallon block B is :(1) 120 N(2) 150 N(3) 100 N(4) 80 NAns. (1)Sol. –– ––– –– (2) –– –––––(3)f2 f1F – – – – ––(1) V –A long cylindrical shell carries positive surfacecharge in the upper half and negative surfacecharge – in the lower half. The electric fieldlines around the cylinder will look like figuregiven in : (figures are schematic and not drawnto scale) –– ––– ––(4)A B20for equllibrrium of Af1 20for equllibrrium of Bf2 f1 100f2 120 Nf1100Ans. (3)It behaves as a dipole.100 –– ––– ––JEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.11
JEE(MAIN)-2015As an electron makes a transition from anexcited state to the ground state of ahydrogen - like atom/ion :(1) kinetic energy decreases, potential energyincreases but total energy remains same(2) kinetic energy and total energy decrease butpotential energy increases(3) its kinetic energy increases but potentialenergy and total energy decreases(4) kinetic energy, potential energy and totalenergy decreaseAns. (3)29.U 27.2z2n2T 13.6z2n2List-Iz2as n decreases k increasesn2 as n decrease U & T lenature of y levelsof atom(C)Davison-GermerExperiment(iiii)Wave natureof electroc(iv)Structure ofatom(1) A-ii, B-i, C-iii(2) A-iv, B-iii, C-ii(3) A-i, B-iv, C-iii(4) A-ii, B-iv, C-iiiAns. (1)Sol. Self Explanatory/Theory(A) Franck-Hertz experiment explains disreteenergy levels of atom(B) Photo-electric experiment explain particlenature of light(C) Davison Germer experiment explain wavenature of electron. K 13.6Match List-I (Fundament Experiment) withList-II (its conclusion) and select the correctoption from the choices given below the list : Sol.30.JEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) forXII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre only.12
JEE(MAIN)-2015 JEE (Main Advanced) Leader Course (Target-2016) for XII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) and JEE (Main) Leader Course (Target-2016) for XII Passed / Appeared students Start on 15th April 2015 (English / Hindi Medium) at Kota Centre
8. JEE (MAIN) 2015 Candidates who wish to appear in JEE (Advanced) 2015 must write Paper-1 of JEE (Main) 2015 which is likely to be held in the month of April 2015. Candidates may visit the website www.jeemain.nic.in for information about JEE (Main) 2015. 9. ELIGIBILITY CRITERIA FOR APPEARING IN JEE (ADVANCED) 2015
Atomic Structure for JEE Main and Advanced (IIT-JEE) m Jupiter (XI) 3 Prepared By: Er. Vineet Loomba (IIT Roorkee) (iv) Some were even scattered in the opposite direction at an angle of 180 [Rutherford was very much surprised by it and remarked that “It was as incredible as if you fired a 15 inch shell at a piece of
iii. Original Admit Card of JEE (Main) 2015 iv. One passport size photograph identical to the one pasted on the JEE (Main)-2015 application form. v. Score Card of JEE (Main)-2015 issued by CBSE (downloaded score card is acceptable). vi. Photo ID proof issued by central Govt./ State Govt./ last attended School/ 12th Admit card vii.
Main Report and Tables A 6 JEE 2011 – Results The Joint Entrance Examination for the year 2011 (JEE 2011) was conducted on Sunday, April 10 . by post to respective zonal JEE offices from where the admit card was issued. The last date for receiving required documents at respective zonal IITs was announced in the Counselling Brochure as 10 .
JEE 2004 MATHEMATICS PAPERS Note: The IIT’s have continued the practice, begun in 2003, of collecting back the JEE question papers from the candidates. At the time of this writing, the JEE 2004 papers were not officially available to the general public, either on a website or on paper. So the text of the questions taken here was based on
JEE (Advanced) 2019 online exam held on 27th may, 2019 and result will be declared on 14th june, 2019. The online exam was held at various cities across india & abroad too. JEE (Advanced) 2019 was conducted in two parts (paper 1 & paper 2). Paper 1 was co
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Secret Wall O2 Pit to Q2 X2 To Level 7 (X3) A1 Portal to L10 (A2) [] Button Q1 From Pit O1 X3 To Level 7 (X1) 0 Pressure Pad Q2 From Pit O2 X4 To Level 5 (X2) Y Nest In the place where you found a lot of Kenkus (bird creatures) is a place called "Nest." After killing both Kenkus, put all ten Kenku eggs on the floor. The wall will disappear, and .
