JEE (ADVANCED) Test Code-3

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1 JEE (ADVANCED)-013 Test Code-3

2 JEE (Advanced)-13 Solution Paper-I PART-I [PHYSICS] SECTION - 1 (Straight Objective Type) This section contains 10 multiple choice questions. Each question has 4 choice (A), (B), (C) and (D) out of which ONLY ONE is correct. 1. A particle of mass m is projected from the ground with an initial speed u 0 at an angle a with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u 0. The angle that the composite system makes with the horizontal immediately after the collision is (A) Sol. [A] 4 (B) (C) (D) 4 4 u 0 u.cos mu0cos + gh 0 m u x = u y = mu cos 0 m u sin u g. g u cos 0 = u 1 sin. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is times the wavelength in 3 free space. The radius of the curved surface of the lens is (A) 1m (B) m (C) 3m (D) 6m Sol. [C] M = 3 3 m m = (3 / ) u (u 1) f R1 R (u 1) v u R1 R 3. The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 division equivalent to.45 cm. The 4th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is (A) 5.11 cm (B) 5.14 cm (C) cm (D) cm Sol. [B] L.C. = cm, Patel Nagar, Bikaner. Ph: , 4035, Web. :

3 VSR = 4 MSR = 5.10 cm Total reading = MSR + (VSR) (LC) = (0.001) = 5.14 cm JEE (Advanced)-13 Solution x 4. The work done on a particle of mass m by a force, K ˆ y ˆ i j 3/ 3/ (K being a constant (x y ) (x y ) of appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is : (A) K a Sol. [D] (B) K a (C) K a (D) 0 rsin (x,y) r rcos xi ˆ yj ˆ F = K 3/ 3/ (x y ) (x y ) Let x = rcos y = rsin r cos r sin F = K ˆ i ˆ j r cos 3 3 r r F = K ˆ ˆ [cos i sin j] r Workdone along path = O ( force is acting radially) 5. One end of a horizontal thick copper wire of length L and radius R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is (A) 0.5 (B) 0.50 (C).00 (D) 4.00 Sol. [C] R L FL FL L = Ay r y L R r. 1 1 r1 = R R , Patel Nagar, Bikaner. Ph: , 4035, Web. : 3

4 JEE (Advanced)-13 Solution 6. A ray of light travelling in the direction 1 (i ˆ 3 ˆj) is incident on a plane mirror. After reflection, it travels along the direction 1 (i ˆ 3 ˆj). The angle of incidence is : (A) 30º (B) 45º (C) 60º (D) 75º Sol. [A] 1 (i ^ + 3 ^ j) tan = y 1 x 3 1 = 30º = 30º 7. Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity k and the other k. The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuration II is Configuration-I K K Configuration-II (A).0 s (B) 3.0 s (C) 4.5 s (D) 6.0 s Sol. [A] K K x K K R 1 = KA KA R 1 = 3 KA K K 1 KA KA R 1 3KA R R 3KA R1 9 R H = T R t R t R t R, Patel Nagar, Bikaner. Ph: , 4035, Web. : 4

5 JEE (Advanced)-13 Solution 8. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mv and the speed of light is 3 10 m/s. The final momentum of the object is (A) kg ms 1 (B) kg ms 1 (C) kg ms 1 (D) kg ms 1 Sol. [B] P = hc c P t P C P = E C hc E 9. In the Young s double slit experiment using a monochromatic light of wavelength, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is (A) (n 1) Sol. [B] I = I 1 + I + I1 I cos I 0 = I 0 + I 0 + I 0 cos cos = 0 = x x = 4 Hence, (B) (n 1) (C) (n 1) 4 8 x =. n (D) (n 1) Two non-reactive monoatomic ideal gases have their atomic masses in the ratio : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is (A) 1 : 4 (B) 1 : (C) 6 : 9 (D) 8 : 9 Sol. [D] MO 1 M 3 O P1 4 P 3 (Constant temperature) P = = RT M 0 PM 0 RT P M 4 8 M M O 1. O SECTION - (One or more option correct type) This section contains 5 multiple choices questions. Each question has 4 choice (A), (B), (C) and (D) out of which ONLY ONE or MORE is correct. 11. Two non-conducting solid spheres of radii R and R, having uniform volume charge densities 1 and respectively, touch each other. The net electric field at a distance R from the centre of the smaller sphere, along the line joining the centres of the spheres is zero. The ratio 1 / can be (A) 4 (B) 3 5 (C), Patel Nagar, Bikaner. Ph: , 4035, Web. : (D) 4

6 Sol. [B] E 1 + E = 0 Kq 4R Kq 5R 1 0 JEE (Advanced)-13 Solution R Q R R R R P p 1, q1 p q R 8R A horizontal stretched string fixed at two ends, is vibrating in its fifth harmonic according to the equation y(x, t) = 0.01m sin [(6.8m 1 )x] cos[(68s 1 )t]. Assuming = 3.14, the correct statement(s) is (are) (A) The number of nodes is 5. (B) The length of the string is 0.5 m. (C) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01m. (D) The fundamental frequency is 100 Hz. Sol. [B,C] N 1 N N 3 N 4 N 5 N 6 A 1 A A 3 A 4 A 5 Y = 0.01 sin [6.8x] cos [68t] y = A sin Kxcos t No. of nodes 6 No. of antinodes 5 K = X = L = 5 = m A = 0.01 = max. amplitude. 13. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S 1 is pressed first to fully charge the capacitor C 1 and then released. The switch S is then pressed to charge the capacitor C. After some time, S is released and then S 3 is pressed. After some time, S 1 S S 3 V 0 C 1 C V 0 (A) the charge on the upper plate of C 1 is CV 0. (B) the charge on the upper plate of C 1 is CV 0. (C) the charge on the upper plate of C is 0. (D) The charge on the upper plate of C is CV 0. S 1 S S 3 Sol. [B,D] V + 0 C 1 C V 0, Patel Nagar, Bikaner. Ph: , 4035, Web. : 6

7 Step 1 Q 1 = C 1 V = C 1 (V 0 ) = CV 0 Step Q' 1 Q' CV0 Step 3 V 0 + C V 0 C = CV 0 + Q Q = CV 0 CV0 Q 0 JEE (Advanced)-13 Solution C QCV0 +QCV0 + V A particle of mass M and positive charge Q, moving with a constant velocity 1 u1 4i ˆ ms, enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity 1 u ( 3 ˆi ˆ j)ms. The correct statements is (are) : (A) The direction of the magnetic field is z direction (B) The direction of the magnetic field is +z direction (C) The magnitude of the magnetic field is 50M 3Q units (D) The magnitude of the magnetic field is 100 M units 3Q Sol. [A,C] y 4i ^ j^ x = L 3 x î tan = y x t = sec = 1 = 30º 3 3 = t qb m qb.t t, 6 m qb m B = m rm 6q 3q = qb m 15. A solid sphere of radius R and density is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3. The complete arrangement is placed in a liquid of density and is allowed to reach equilibrium. The correct statement(s) is (are) (A) The net elongation of the spring is 3 4R g 3k 3 8R g (B) The net elongation of the spring is 3k (C) The light sphere is partially submerged (D) The light sphere is completely submerged Sol. [A,D], Patel Nagar, Bikaner. Ph: , 4035, Web. : 7

8 JEE (Advanced)-13 Solution F 1 R 3 R F F 1 = F 1 = R g R g R 3 g F = R g 3 = kx SECTION - 3 (Integer value correct type) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive) 16. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad/s 1 about its own axis, which is vertical. Two uniform circular rings, each of mass 6.5 kg and radius 0. m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s-1) of the system is. Sol. [8] 1 =10Rad/s r 1 I 1 = mr 1 = 10 Rad. mr R' I = m' I 1 1 = I 17. The work functions of Silver and Sodium are 4.6 and.3 ev, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is. Sol. [1] Ag = 4.6 Na =.3 h = 0 + ev 0 v = e v0 h 0 h, Patel Nagar, Bikaner. Ph: , 4035, Web. : 8

9 JEE (Advanced)-13 Solution e cons tant h m 1 1 m (Equation of straight line) 18. A bob of mass m, suspended by a string of length 1 is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio Sol. [5] 1 is. 5g 5g1 5g 5g 1 g A particle of mass 0. kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed (in m/s) of the particle is zero, the speed (in m/s) after 5 s is : Sol. [5] m = 0. kg P = 0.5 watt u = 0, v =? Pt = 1 mv P t v = = 5 m/s m 0. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 10 3 disintegrations per second. Given that ln = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is Sol. [4] A 0 = 10 3 decay/sec ln = T 1/ = 1386 sec A = A 0 e t R = R = A A A 0 0 (1 e t ) A A 0 0 t (1 e ) 100% = (1 0.96) 100% = = 4 (e =.7), Patel Nagar, Bikaner. Ph: , 4035, Web. : 9

10 JEE-13 (Advanced) Solution Paper-I PART-II [CHEMISTRY] SECTIO N - 1 (Straight Objective Type) This section contains 10 multiple choice questions. Each question has 4 choice (A), (B), (C) and (D) out of which ONLY ONE is correct. 1. The standard enthalpies of formation of CO (g), H O(l) and glucose(s) at 5 C are 400 kj/mol, 300 kj/ mol and 1300 kj/mol, respectively. The standard enthalpy of combustion per gram of glucose at 5 C is: (A) +900 kj (B) 900 kj (C) kj (D) kj Sol. [C] For combustion of glucose : C 6 H 1 O 6 (s) + 6O (g) 6CO (g) + 6H O(l) ( c H) reaction = (H f ) pdt (H f ) reactant = {6H f (CO ) + 6H f H O (l)} {6 (H f (O )) + H f (C 6 H 1 O 6 (s)} = {6 ( )} { } = 900 kj/mol Enthalpy of combustion per gram = 900 = kj/gm KI in acetone, undergoes S N reaction with each P, Q, R and S. The rates of the reaction vary as H C Cl 3 (P) Cl (Q) (A) P > Q > R > S (B) S > P > R > Q (C) P > R > Q > S (D) R > P > S > Q Sol. [B] (R) Cl (S) O Cl O Cl > CH Cl > > Cl 3 Cl Excellent substrate for SN. 3. The compound that does NOT liberate CO, on treatment with aqueous sodium bicarbonate solution, is : (A) Benzoic acid (B) Benzenesulphonic acid (C) Salicylic acid (D) Carbolic acid (phenol) Sol. [D] Carbolic acid means phenol Phenol does not give NaHCO 3 test because it is weaker acid than H CO Consider the following complex ions, P, Q and R. P = [FeF 6 ] 3, Q = [V(H O) 6 ] + and R = [Fe(H O) 6 ] +. the correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is : (A) R < Q < P (B) Q < R < P (C) R < P < Q (D) Q < P < R Sol. [B] P [fef 6 ] 3 : Fe +3 : 3d 5 4s 0 F : WFL 5(5 ) B.M. = 35 B.M. Q [V(H O) 6 ] + V + : 3d 3 4s = 3(3 ) R [Fe(H O) 6 ] + : Fe + : 3d 6 4s = 15 B.M., Patel Nagar, Bikaner. Ph: , 4035, Web. : 10

11 H O : WFL JEE-13 (Advanced) Solution = 4(4 ) = 4 B.M. Order of magnetic moment () P > R > Q. 5. In the reaction, P + Q R + S the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentrationof Q varies with reaction time as shown in the figure. The overall order of the reaction is : [Q] 0 [Q] (A) (B) 3 (C) 0 (D) 1 Sol. [D] From given graph of concentration [Q] and time t : Time Q Q 0 Time Kt = A 0 A t K t = Q 0 Q It represent zero order reaction (ZOR) w.r.t. Q. from data : (t ½ ) = (t 3/4 ) for P For first order reaction (FOR) : K t 3/4 = ln (A/4) l n() t 3/4 = t1/ K Data represent order of reaction wrt P is 1. For reaction : P + Q R + S Rate = K[P] 1 [Q] 0 Overall OOR = = 1 6. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of (A) NO (B) NO (C) N O (D) N O 4 Sol. [B] HNO 3 H O + NO + 3 O NO gas is brown yellow due to odd electron. 7. The arrangement of X ions around A + ion in solid AX is given inthe figure (not drawn to scale). If the radius of X is 50 pm, the radius of A + is: X A + (A) 104 pm (B) 15 pm (C) 183 pm (D) 57 pm, Patel Nagar, Bikaner. Ph: , 4035, Web. : 11

12 Sol. [A] For ionic solid (AB type) given arrangement represents octahedral void r < r JEE-13 (Advanced) Solution Case A : r r Limiting value r + = = pm Case B : r r r + < r + < 183 pm To exact fit in O v r + = 104 pm 8. Upon treatement with ammoniacal H S, the metal ion that precipitates as a sulphide is : (A) Fe(III) (B) Al(III) (C) Mg(II) (D) Zn(II) Sol. [D] As H S + NH 4 OH is reagent for group IV only Zn + ppt in form of ZnS having white colour. 9. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 5 C. for this process, the correct statement is : (A) The adsorption requires activation at 5 C (B) The adsorption is accompanied by a decreases in enthalpy (C) The adsorption increases with increase of temperature (D) The adsorption is irreversible Sol. [B] Methylene blue on charcol at room temperature (5 C) is physical adosrption and it is accompanied by decrease in enthalpy always. 30. Sulphide ores are common for the metals : (A) Ag, Cu and Pb (B) Ag, Cu and Sn (C) Ag, Mg and Pb (D) Al, Cu and Pb Sol. [A] Ores of Ag :- Ag S (argentite)- sulphide ore Cu :- CuFeS (chalcopyrite) sulphide ore Pb :- PbS(galena) sulphide ore SECTIO N - (One or more option correct type) This section contains 5 multiple choices questions. Each question has 4 choice (A), (B), (C) and (D) out of which ONLY ONE or MORE is correct. 31. The benzene and napthalene from an ideal solution at room temperature. For this process, the true statement(s) is (are) (A) G is positive (B) S system is positive (C) S surroundings = 0 (D) H = 0 Sol. [BCD] G < 0 ; S system > 0 (for solution formation) Dissolution process is spontaneous and randomness increases. H = 0 (for ideal solution) S surrounding = 0 (No heat exchanged with system (solution) and surrounding. 3. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are) (A) [Cr(NH 3 ) 5 Cl]Cl and [Cr(NH 3 ) 4 Cl ]Cl (B) [Co(NH 3 ) 4 Cl ] + and [Pt(NH 3 ) (H O)Cl] + (C) [CoBr Cl ] and [PtBr Cl ] (D) [Pt(NH 3 ) 3 (NO 3 )]Cl and [Pt(NH 3 ) 3 Cl]Br Sol. [BD] A : [Cr(NH 3 ) 5 Cl]Cl Cr +3 : 3d 3 (d sp 3 ) [Cr(NH 3 ) 4 Cl ]Cl Cr +3 : 3d 3 (d sp 3 ) Both not showing geometrical/sterio isomerism. B : [Co(NH 3 ) Cl ] + (d sp 3 ) [Pt(NH 3 ) (H O) (Cl)] + (dsp ) Both are geometrically active. C : [Co(Br) (Cl )] : Co + 3d 5 (sp 3 ) [PtBr Cl ] : Pt + (dsp ), Patel Nagar, Bikaner. Ph: , 4035, Web. : 1

13 JEE-13 (Advanced) Solution can not shows same kind of isomerism. D : [Pt(NH 3 ) 3 (NO 3 )]Cl [Pt(NH 3 ) 3 Cl] Br Due to formation of AgCl (white) and AgBr (Pale yellow) can shows ionization (structural isomerism). 33. The initial rate of hydrolysis of methyl acetate (1 M) by a weak acid (HA, 1M) is 1/100th of that of a strong acid (HX, 1M), at 5 C. The Ka of HA is (A) (B) (C) (D) Sol. [A] O H + CH 3 C OCH3 + H O CH3COOH + CH3OH Rate = K[ester] 1 [H 3 O + ] 1 K [ester] (XS) where K 1 = K[H 3 O + ] 1 (Psuedounimolecular) OOR = 1(wrt conc. of H + ) (Rate)HA (H )HA (Rate)HX (H )HX 1 (H )HA (H ) HA = 10 M HA H + + A 1 x x x x = 0.01 {(1%) < 5% so can be neglected} K a = (H )(A ) (0.01)(0.01) 1 10 (HA) The hyperconjugative stabilities of tert-butyl cation and -butene, respectively, are due to (A) p (empty) and * electron delocalisations. (B) * and electron delocalisations. (C) p (filled) and * electron delocalisations. (D) p(filled)*and *electron delocalisations. Sol. [A] 35. Among P, Q, R and S, the aromatic compound(s) is/are Cl (A) P (C) R Sol. [ABCD ] AlCl 3 P, NaH Q, (B) Q (D) S O O (NH 4) CO C R, O HCl S, Patel Nagar, Bikaner. Ph: , 4035, Web. : 13

14 JEE-13 (Advanced) Solution Paper-I SECTIO N - 3 (Integer value correct type) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive) 36. The total number of lone-pairs of electrons in melamine is Sol. [6] melanine 37. The total number of carboxylic acid groups in the product P is Sol. [] 38. The atomic masses of He and Ne are 4 and 0 a.m.u., respectively. The value of the de Broglie wavelength of He gas at 73 C is M times that of the de Broglie wavelength of Ne at 77 C. M is Sol. [5] From debroglie hypothesis : = h (K.E.) m KE Temp. ( (K.E.) (m)) He ((KE) (m)) Ne and K.E. = 3 RT (for 1 mole) He Ne (K.E. m) He (K.E. m) 4 00 Ne Ne He EDTA 4 is ethylenediaminetetraacetate ion. The total number of N Co O bond angles in [Co(EDTA)] 1 complex ion is Sol. [8], Patel Nagar, Bikaner. Ph: , 4035, Web. : 14

15 JEE-13 (Advanced) Solution 40. A tetrapeptide has COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with NH group attached to a chiral center is Sol. [4], Patel Nagar, Bikaner. Ph: , 4035, Web. : 15

16 JEE (Advanced)-013 Solution Paper-I PART-III [MATHS] SECTION - I (Straight Objective Type) This section contains 10 questions. Each question has 4 choice (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. 41. (A) (B) (C) (D) 3 1 cot cot (1 ( n)) n 1 4. (A) 5 (B) 0 (C) 10 (D) 30 = PR SQ PR SQ PT Applying R 1 R 1 + R and then R R R 1, Patel Nagar, Bikaner. Ph: , 4035, Web. : 16

17 JEE (Advanced)-013 Solution = = 43., Patel Nagar, Bikaner. Ph: , 4035, Web. : 17

18 44. JEE (Advanced)-013 Solution Solving the equations ax + by + c = 0 and bx + ay + c = Sol. (D) 46. Sol. (A) P(A B C D) =, Patel Nagar, Bikaner. Ph: , 4035, Web. : 18

19 JEE (Advanced)-013 Solution 47. Sol. (B) sin(x / 4) 0 4 cos(x / 4) Sol. (A), Patel Nagar, Bikaner. Ph: , 4035, Web. : 19

20 JEE (Advanced)-013 Solution 49. Sol. (D) 50. Sol. (C) Let f(x) = x x sinx cosx f (x) = x x cosx sinx + sinx = x( cosx) f(x) is increasing for x > 0 and decreasing for x < 0 and f(0) = 1 Graph of f(x) can be drawn as 51. Clear f(x) = 0 has solutions (0, 1) SECTION - II (Multiple Correct Answer Type) This section contains 5 questions. Each question has 4 choice (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. Sol. (A,C), Patel Nagar, Bikaner. Ph: , 4035, Web. : 0

21 JEE (Advanced)-013 Solution, Patel Nagar, Bikaner. Ph: , 4035, Web. : 1

22 JEE (Advanced)-013 Solution 5. Sol. (A,D) 53. Sol. (B,D) 1 is given by r 3i ˆˆj 4k t (i ˆj ˆk) ˆ and is given by r (3i ˆ3j ˆkˆ s(i ˆj ˆk). x 4 Hence, equation of is k1 3, Patel Nagar, Bikaner. Ph: , 4035, Web. :

23 JEE (Advanced)-013 Solution Let A is intersection point of and 1 A ( 1, 3 1, 1 ) (3 +, 1 +, 4 + ) 1 = = = 4 + Solving, we get 1 = 1 A (, 3, ) 54. Sol. (B, C) 55. Sol. (C, D), Patel Nagar, Bikaner. Ph: , 4035, Web. : 3

24 JEE (Advanced)-013 Solution SECTION - III (Integer value correct Type) This section contains 5 questions. The answer to each question is a single-digit integer, ranging from 0 to 9 (both inclusive). 56. Sol. (9), Patel Nagar, Bikaner. Ph: , 4035, Web. : 4

25 57. Sol. (6) JEE (Advanced)-013 Solution 58. Sol. (5) 59. Sol. (6) 60. Sol. (5) n + n 4k = 450 for n = 50 (least value) = 4k k = 5 k 0 = 5, Patel Nagar, Bikaner. Ph: , 4035, Web. : 5

JEE ADVANCE PREVIOUS YEAR PAPER Paper-I. Name of the Student:.. Father s Name:. Signature:. Branch Name: Contact No:.

JEE Advance 016 Time:3Hr JEE ADVANCE PREVIOUS YEAR PAPER-013 Paper-I Topics Covered: Maximum Marks: 40 Physics Chemistry Mathematics : Full Syllabus : Full Syllabus : Full Syllabus Important Instruction: