JEE MAIN 016 ONLINE EXAMINATION DATE : 09-04-016 SUBJECT : PHYSICS TEST PAPER WITH SOLUTIONS & ANSWER KEY THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL
1. Two paricles are performing simple harmonic moion in a sraigh line abou he same equilibrium poin. The ampliude and ime period for boh paricles are same and equal o A and T, respecively. A ime = 0 one paricle has displacemen A while he oher one has displacemen A and hey are moving owards each oher.if hey cross each a ime, hen is : (1) T 4 () 5T 6 (3) T 3 (4) T 6 Ans. (4) = 0 = /3 /3 A/ /3 A/ /3 = T 3 6. To find he focal lengh of a convex mirror, a suden records he following daa : Objec pin Convex Convex Mirror Image Pin Lens. cm 3.cm 45.8 cm 71. cm The focal lengh of he convex lens is f 1 and ha of mirror is f. Then aking index correcion o be negligibly small, f 1 and f. are close o : (1) f 1 = 15.6 cm f = 5.4 cm () f 1 = 7.8 cm f = 1.7 cm (3) f 1 = 7.8 cm f = 5.4 cm (4) f 1 = 1.7 cm f = 7.8 cm Ans. () For convex lens O I u = 10 cm v = 39 cm 10 cm 39 cm THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE #
f 1 = uv u v PAPER-1 (B.E./B. TECH.) OF ONLINE JEE (MAIN) CBT 09-04-016 = 390 49 = 7.8 cm For convex mirror O I R = 5.4 cm f = 1.7 cm 5.4 cm 3. Figure shows ellipical pah abcd of a plane around he sun S such ha he area of riangle csa is 1 4 he area of he ellipse. (See figure) Wih db as he semi major axis, and ca as he semi minor axis. If 1 is he ime aken for plane o go over pah abc and for pah aken over cda hen : c d s b a (1) 1 = 3 () 1 = (3) 1 = (4) 1 = 4 Toal area = A Area of sabc = 3A 4 Area of sadc = A 4 3A 4 A 1 3 1 4 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 3
4. A simple pendulum made of a bob of mass m and a meallic wire of negligible mass has ime period s a T = 0 C. If he emperaure of he wire is increased and he corresponding charge in is ime period is ploed agains is emperaure, he resuling graph is a line of slop S. If he coefficien of linear expansion of meal is a hen he value of S is : (1) 1 () (3) (4) Ans. (4) 0 = = 0 g = 0 1 T = (1 + T) 1/ = + T g 0 = T = T S= T 5. The raio of work done by an ideal monaomic gas o he hea supplied o i in an isobaric process is : (1) 3 () 3 (3) 3 5 (4) 5 Ans. (4) w = PV = nrt Q = nc p T = 5 nrt w Q 5 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 4
6. An unknown ransisor needs o be idenified as npn or pnp ype.a mulimeer, wih +ve and ve erminals, is used o measure resisance beween differen erminals ransisor. If erminal is he base of he ransisor hen which of he following is correc for a pnp ransisor? (1) +ve erminal 3, ve erminal, resisance high () +ve erminal, ve erminal 3, resisance low (3) +ve erminal 1, ve erminal, resisance high (4) +ve erminal, ve erminal 1, resisance high Ans. (4) 1 p n p 3 when pn juncion is forward biased resisance is law. When pn juncion reverse biased resisance is high 7. A uniformly apering conical wire is made from a maerial of Young's modulus Y and has a normal, unexended lengh L. The radii, a he upper and lower ends of his conical wire, have values R and 3R, respecively, The upper end of he wire is fixed o a rigid suppor and a mass M is suspended from is lower and. The equilibrium exended lengh, of his wire, would equal (1) 1 Mg L1 3 YR R () Mg L1 3 YR (3) 1 Mg L1 9 YR (4) Mg L1 9 YR x r 3R r R 3R R x r R 1 x L L ; Mg Y dl R dx Mg dx dl R x 1 L L = L Mg dx MgL YR ; L' = L + L = x 3R Y 0 1 L 1 Mg L 1 3 R Y THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 5
8. A cubical block of side 30cm is moving wih velociy ms 1 on a smooh horizonal surface. The surface has a bump aa poin O as shown in figure. The angular velociy (in rad/s) of he block immediaely afer i his he bump, is : a = 30 cm Ans. (3) (1) 9.4 () 6.7 (3) 5.0 (4) 13.3 O Using conservaion of angular momenum a 3v mv ma 5rad / s 3 4a 9. In Young's double sli experimen, he disance beween slis and he screen is 1.0 m and monochromaic ligh of 600 nm is being used. A person sanding near he slis is looking a he fringe paern. When he separaion beween he slis is varied, he inerference paern disappears for a paricular disance d 0 beween he slis. If he angular resoluion of he eye is 1 60 he value of d 0 is close o (1) mm () 1 mm (3) 3mm (4) 4 mm Angular fringe widh = D d 1 d 60 18060 0 d 0 = 18060 = 10 3 m = mm. THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 6
10. Which of he following opion correcly describes he variaion of he speed v and acceleraion 'a' of a poin mass falling verically in a viscous medium ha applies a force F = kv, where 'k' is consan, on he body?( Graphs are schemaic and no drawn o scale) v a (1) a () v v v (3) a (4) a Ans. (4) a = g v dv d = g v v 0 0 dv g v d n g v v V = v 0 (1 e ), a = v dv d 0 0 v e a e a THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 7
11. A series LR circui is conneced o a volage source wih V() = V 0 sin. Afer very large ime curren I() L behaves as 0 R : I() I() (1) () = 0 = 0 I() I() (3) (4) 0 = 0 Ans. (3) Curren will be in he form of I = I 0 sin ( ) Graph will be sinusoidal 1. A car of weigh W is on an inclined road ha rises by 100 m over a disance of 1km and applies a consan fricional force W 0 on he car. While moving uphill on he road a a speed of 10ms 1, he car needs power P. If i needs power P while moving downhill a speed v hen value of v is : (1) 5 ms 1 () 0 ms 1 (3) 10 ms 1 (4) 15 ms 1 Ans. (4) While going up (sin = an = 1 10 ) 10m/s Wsin W/0 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 8
Wsin 10 + W 0 10 = P 3W P While going down v W/0 Wsin W P 3W v 0 4 v = 15 m/s 13. A rocke is fired verically from he earh wih an acceleraion of g, where g is he graviaional acceleraion. On an inclined plane inside he rocke, making an angle. Wih he horizonal, a poin objec of mass m is kep. The minimum coefficien of fricion min beween he mass and he inclined surface such ha he mass does no move is : (1) an () an (3) 3 an (4) an. g ef = 3g bu min = an 14. Two engines pass each oher moving in opposie direcions wih uniform speed of 30m/s. One of hem is blowing a whisle of frequency 540 Hz. Calculae he frequency heard by driver of second engine before pass each oher. Speed sound is 330 m/sec. (1) 540 Hz () 648 Hz (3) 70 Hz (4) 450 Hz Ans. () f=540hz 30m/s 30m/s obs source f 0 = 330 30 540 648Hz 330 30 15. The poenial (in vols) of a charge disribuion is given by V(z) = 30 5z for z 1m V(z) = 35 10 z for z 1m. V(z) does no depend on x and y. If his poenial is generaed by a consan charge per uni volume 0, (in unis of 0 ) which is spread over cerain region,hen choose he correc saemen. (1) 0 = 40 0 in he enire region () 0 = 0 0 in he enire region (3) 0 = 0 0 for z in 1 m and 0 = 0 else where (4) 0 = 10 0 for z in 1 m and 0 = 0 else where Ans. (4) THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 9
PAPER-1 (B.E./B. TECH.) OF ONLINE JEE (MAIN) CBT 09-04-016 10z v / m z 1m E 10 v / m z 1m If should be for a shee lying in x y plane of hickness z = m. For z 1m E = 10 P 10 0 0 0 0 =0 =0 = 10 0 z = 1 z=0 z = +1 16. An audio signal consiss of wo disinc sound : one a human speech signal in he frequency band of 00 Hz o 700 Hz, while he oher is a high frequency music signal in he frequency band of 1000 Hz o 1500 Hz. The raio of he AM signal band widh required o send boh he signals ogeher o he AM signal band widh required o send jus he human speech is : (1) 6 () 5 (3) 3 (4) Band widh for boh signals 1500 Hz 00 Hz = 15000 Hz Band widh for human speech The raio = 15000 6 500 700 Hz 00 Hz = 500 Hz 17. A convex lens, of focal lengh 30cm, a concave lens of focal lengh 10 cm, and a plane mirror are arranged as shown. For an objec kep a a disance of 60 cm from he convex lens, he final image, formed by he combinaion, is a real image, a a disance of : Focal lengh Focal lengh = 30 cm =10 cm 60cm 0cm Ans. 70cm (1) 70 cm from he concave lens () 60 cm from he convex lens (3) 60 cm from he concave lens (4) 70 cm from he convex lens [BONUS] THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 10
PAPER-1 (B.E./B. TECH.) OF ONLINE JEE (MAIN) CBT 09-04-016 f 1 = +30cm f = 10cm O 60cm Even 1 0cm 50cm Refracion from convex lens. f = +30 O 60 60 I 1 1 1 1 v 60 30 v = +60 cm Even Refracion from concave lens 40 I 1 I 1 1 1 v 40 10 v = +60cm Even 3 Refracion from plane mirror I Even 4 I 3 10cm 10cm Refracion from concave lens 30 I 4 40 I 3 1 1 1 v 40 10 v = 30 Even 5 Refracion from convex lens I 75cm 50 I 4 1 1 1 ; v = +75 cm v 50 30 THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 11
18. Three capaciors each of 4F are o be conneced in such a way ha he effecive capaciance is 6F. This can be done by connecing hem : (1) all in series () wo in parallel and one in series (3) wo in series and one in parallel (4) all in parallel Ans. (3) 3c/ where c = 4F 19. To know he resisance G of a galvanomeer by half deflecion mehod, a baery of emf V E and resisance R is used o deflec he galvanomeer by angle. If a shun of resisance S is needed o ga half deflecion hen G, R and S are relaed by he equaion : (1) S = G () G = S (3) S(R+G) = RG (4) S(R+G) = RG Ans. (3) Case -1 Case- R i 1 G R i i 1 / G s i 1 i V E VE i1 R G i1 i1 G i S i 1 (G+S) = i S subsiuing i 1 and i we ge S(R+G) = RG i V E V E GS R G S THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 1
0. In he circui shown, he resisance r is a variable resisance. If for r =fr, he hea generaion in r is maximum hen he value of f is : R R r Ans. (4) (1) 1 () 3 4 (3) 1 4 (4) 1 r r R R E o ge maximum hea generaion from r. r = R eq = R/ R eq = R/ E eq 1. A hydrogen aom makes a ransiion from n = o n=1 and emis a phoon. This phoon srikes a doubly ionized lihium aom (z = 3) in excied sae and compleely removes he orbiing elecron. The leas quanum number for he excied saed of he ion for he process is : (1) 4 () 5 (3) (4) 3 Energy of proon = 13.6 3.4 = 10.eV For removal of elecron z 10.eV > 13.6 n n 9 > 13.6 10. so minimum value of n = 4. 00 g waer is heaed from 40 C o 60 C. Ignoring he sligh expansion of waer, he change in is inernal energy is closed o (Given specific hea of waer = 4184 J/kg/K) : (1) 16.7 kj () 167.4kJ (3) 4. kj (4) 8.4 kj U = mst 10 (4184) (0) = 16736J = 16.7 kj THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 13
3. An experimen is performed o deermine he I - V characerisics of a Zener diode, which has a proecive resisance of R =100, and maximum power of dissipaion raing of 1W. The minimum volage range of he DC source in he circui is : (1) 0 1V () 0 5V (3) 0 4 (4) 0 8V Ans. (3) 100 i i 100i v 100i V P zener = (v 100i)i = 1 vi 100i =1 100i vi + 1 = 0 i mus be real v 4 (100)0 v0 4. Microwave oven acs on he principle of : Ans. () (1) giving roaional energy o waer molecules () giving vibraional energy o waer molecules (3) giving ranslaional energy o waer molecules (4) ransferring elecrons from lower o higher energy levels in waer molecule Energy of microwaves lie in range of vibraion energy of waer molecules. 5. A magneic dipole is aced upon by wo magneic fields which are inclined o each oher a an angle of 75. One of he fields has magniude of 15mT. The dipole aains sable equilibrium a an angle 30 wih his field. The magniude of he oher field (in mt) is close o : (1) 11 () 1060 (3) 36 (4) 1 B =? 45 30 Dipole B 1 =15mT B 1 sin 30 = B sin 45 B = B 1 = 10.60 mt 11mT THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 14
6. A 50 resisance is conneced o a baery of 5V. A galvanomeer of resisance 100 is o be used as ammeer o measure curren hrough he resisance, for his a resisance r s is conneced o he galvanomeer. Which of he following connecions should be employed if he measured curren is wihin 1% of he curren wihou he ammeer in he circui? (1)r s = 0.5 n parallel wih he galvanomeer () r s = 0.5 in series wih he galvanomeer (3) r s = 1 in series wih galvanomeer (4) r s = 1 in parallel wih galvanomeer Is clear, shun should be applied in parallel and leas he shun resisance, beer he ammeer is. 7. When phoons of wavelengh 1, are inciden on an isolaed sphere, he corresponding sopping poenial is found o be V. When phoons of wavelengh, are used, he corresponding sopping poenial was hrice ha of he above value. If ligh of wavelengh 3. is used hen find he sopping poenial for his case hc 1 1 3 (1) e 3 1 hc 1 1 1 (3) e 3 1 hc ev = 0 1 hc 3eV = 0 hc 1 1 1 () e 3 1 hc 1 1 1 (4) e 3 1 hc ev' = 0 3 using hese equaions V' = hc 1 1 3 e 3 1 8. In he following 'I' refers o curren and oher symbols have heir usual meaning choose he opion ha corresponds o he dimensions of elecrical conduciviy : (1) M 1 L 3 T 3 I () M 1 L 3 T 3 I (3) ML 3 T 3 I (4) M 1 L 3 T 3 I J = E = J E = 1 I L 1 1 M L T 1 1 I T = M 1 L 3 T 3 I THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 15
9. Consider a waer jar of radius R ha has waer filled up o heigh H and is kep on a sand of heigh h (see figure). Through a hole of radius r (r<<r) a is boom, he waer leaks ou and he sream of waer coming down owards he ground has a shape like a funnel as shown in he figure. If he radius of he cross-secion of waer sream when i his he ground is x. Then : R r H h x (1) x = r H H h () x = r H H h (3) x = r H H h 1 4 (4) x = r H H h 1 Ans. (4) Using equaion of coninuiy r gh x g(h h) x = r H H h ½ 30. The ruh able given in fig. represens : A B Y 0 0 0 0 1 1 1 0 1 1 1 1 (1) AND - Gae () OR - Gae (3) NOR - Gae (4) NAND Gae Ans. () from ruh able is clear. THIS SOLUTION WAS DOWNLOAD FROM RESONANCE JEE Main 016 SOLUTION PORTAL PAGE # 16