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IIT JEE Exam 2025 : Main : Prqctice paper

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6001CJA10102125011 - SET B 2025-09-03 JA PHYSICS SECTION-I 1) A long straight wire is parallel to one edge as shown in figure. If current in the long wire is varies in time as , what will be the magnitude of induced emf in the loop ? (A) (B) (C) (D) None of these 2) An aluminium ring B faces an electromagnet A. The current I through A can be altered. Then which of the following statement is correct :- (A) If I decreases A will repel B (B) Whether I increases or decreases, B will not experience any force (C) If I increases, A will repel B (D) If I increases, A will attract B 3) A uniform but time varying magnetic field B = C Kt, where K & C are positive constants and t is time, is applied perpendicular to the plane of a circular loop of radius 'a' and resistance R. The total charge that will pass through any point of the loop by the time B becomes zero is : (A) (B) (C) (D) None of these 4) A coil having an area is placed in a magnetic field which changes from a interval of 2 second. The average e.m.f. induced in the coil will be to in (A) 4 V (B) 3 V (C) 1.5 V (D) 2 V 5) The current in a coil changes from 2 A to 5 A in 0.3 s. The magnitude of emf induced in the coil is 1.0 V. The value of self-inductance of the coil is (A) 1.0 mH (B) 100 mH (C) 0.1 mH (D) 10 mH 6) A metallic rod of length 1 m held along east-west direction is allowed to fall down freely. Given horizontal component of earth's magnetic field BH = 3 10 5 T. The emf induced in the rod at an instant t = 2 s after it is released is [Take g = 10 ms 2] (A) 6 10 4 V (B) 3 10 3 V (C) 3 10 4 V (D) 6 10 3 V 7) Two circular coils A and B are facing each other as shown in figure. The current i through A can be altered (A) There will be repulsion between A and B if i is increased (B) There will be attraction between A and B if i is increased (C) There will be neither attraction nor repulsion when i is changed (D) Attraction or repulsion between A and B depends on the direction of current. It does not depend whether the current is increased or decreased. 8) If magnetic field passing through a coil of area 0.1 m2 is changing according to the equation B = 10 (A) (B) (C) (D) tesla. Find the magnitude of induced emf at t = 0.5s volt volt volt volt 9) A conducting rod PQ of length L = 1.0m is moving with a uniform speed v = 2.0 m s 1 in a uniform magnetic field B = 4.0 T directed into the plane of the paper. A capacitor of capacity C = 10 F is connected as shown in figure, then (A) and (B) and (C) (D) Charge stored in the capacitor increases exponentially with time 10) A conducting ring of radius 'r' with a conducting spoke is in pure rolling on a horizontal surface in a region having a uniform magnetic field 'B' as shown, V being to velocity of the centre of the ring. Then the potential difference VC VA is: (A) (B) (C) (D) 11) A wire is bent into 3 circular segments of radius r = 10 cm as shown in figure. Each segment is a quadrant of a circle, ab is lying in the xy plane, bc is lying in the yz plane and ca is lying in the zx plane. If a magnetic field B points in the positive x direction, what is the magnitude of the emf developed in the wire, when B increases at the rate of 3 mT/s ? (A) 4.8 10 5 V (B) 2.4 10 5 V (C) 1.2 10 5 V (D) 3.6 10 5 V 12) In the given arrangement, the loop is moved with constant velocity v in a uniform magnetic field B in a restricted region of width a. The time for which the emf is induced in the circuit is: (A) (B) (C) (D) 13) A metal disc of radius a rotates with a constant angular velocity about its axis. The potential difference between the center and the rim of the disc is (m = mass of electron, e = charge on electron) (A) (B) (C) (D) 14) The figure shows an isosceles triangle wire frame with apex angle equal to . The frame starts entering into the region of uniform magnetic field B with constant velocity v at t = 0. The longest side of the frame is perpendicular to the direction of velocity. If i is the instantaneous current through the frame then choose the alternative showing the correct variation of i with time. (till the loop enters) (A) (B) (C) (D) 15) Figure shows a uniform magnetic field B confined to a cylindrical volume and is increasing at a constant rate. The instantaneous acceleration experienced by an electron placed at P is (A) zero (B) towards right (C) towards left (D) upwards 16) Two identical conducting rings A & B of radius R are in pure rolling over a horizontal conducting plane with same speed (of center of mass) but in opposite direction. A constant magnetic field B is present pointing inside the plane of paper. Then the potential difference between the highest points of the two rings, is : (A) zero (B) 2 Bvr (C) 4 Bvr (D) None of these 17) An electron moves on a straight line path below a loop as shown in the figure. What will be the direction of current if any, induced in the loop? (A) Clockwise only (B) Anti-clockwise only (C) No current (D) Current reverses its direction as electron goes past 18) A coil carrying a steady current is short-circuited. The current in it becomes 1/ times in time t0. The time constant of the circuit is (A) (B) (C) (D) 19) When a magnet is being moved towards a coil, the induced emf does not depend upon :(A) The number of turns of the coil (B) The motion of the magnet (C) The magnetic moment of the magnet (D) The resistance of the coil 20) A metallic ring is held horizontal and a magnet is allowed to fall vertically through it with N-pole pointing upwards. The acceleration of magnet near the ring is a. Then (A) a = g (B) a < g while approaching but a > g while receding (C) a < g while approaching as well as receding (D) a > g while approaching but a < g while receding. SECTION-II 1) In a coil of resistance 8 the magnetic flux due to an external magnetic field varies with time as . The value of total heat produced in the coil, till the flux becomes zero, will be_______J. 2) In the given figure the magnetic flux through the loop increases according to the relation B(t) = 10t2 + 20t, where B is in milliwebers and t is in seconds. The magnitude of current through R = 2 resistor at t = 5 s is _____ mA. 3) A brass rod of length 3.0m is rotating with angular velocity 200 rad/sec about an axis passing through one of its end in a uniform magnetic field of 0.5 T perpendicular to the plane of rotation of the rod. Find the emf induced across it's ends (in volt). 4) A horizontal straight wire 5 m long extending from east to west falling freely at right angle to horizontal component of earth s magnetic field 0.60 10 4 Wbm 2. The instantaneous value of emf induced in the wire when its velocity is 10 ms 1 is _________ 10 3 V. 5) A part of a complete circuit is shown in the figure. At some instant, the value of current I is 1 A and it is decreasing at a rate of 102A s 1. The value of the potential difference VP VQ, (in volts) at that instant, is. CHEMISTRY SECTION-I 1) The alcohol which show turbidity immediately when reacted with (HCl + ZnCl2): (A) (B) (C) (D) 2) The structure of the compound that gives a tribomo derivative on treatment with bromine water quickly is- (A) (B) (C) (D) 3) The major product U in the following reactions is: (A) (B) (C) (D) 4) The ether when treated with HI produces (A) (B) (C) (D) 5) An organic compound B is formed by the reaction of ethyl magnesium iodide (CH3CH2MgI) with a substance A, followed by treatment with dilute aqueous acid. Compound B does not react with PCC in dichloromethane. Identify A? (A) (B) (C) (D) 6) Which is not the correct combination of names for isomeric alcohols with molecular formula C4H10O is. (A) tert-butylalcohol and 2-methylpropan-2-ol (B) tert-butylalcohol and 1, 1-dimethylethan-1-ol (C) n-butyl alcohol and butan-1-ol (D) Isobutyl alcohol and 2-methylpropan-1-ol 7) Which of the following does not react with aqueous Periodic acid? (A) (B) (C) (D) 8) The major product obtained in the following conversion is:- (A) (B) (C) (D) 9) Statement (I): p Nitrophenol is more acidic than m nitrophenol and o nitrophenol. Statement (II): Ethanol will give immediate turbidity with Lucas reagent. In the light of the above statements, choose the correct answer from the options given below: (A) Both Statement I and Statement II are true (B) Statement I is false but Statement II is true (C) Both Statement I and Statement II are false (D) Statement I is true but Statement II is false 10) The acidic hydrolysis of ether (X) shown below is fastest when: (A) One phenyl group is replaced by a methyl group (B) One phenyl group is replaced by a meta-methoxyphenyl group (C) Two phenyl groups are replaced by two para-methoxyphenyl group (D) No structural change is made to X 11) The major product of the following reaction is (A) (B) (C) (D) 12) The electrophile involved in the above reaction is (A) Trichloromethyl anion (B) Formyl cation (C) Phenoxide ion (D) Dichlorocarbene 13) Product contains: (A) Erythro Racemic (B) Threo Racemic (C) Meso Compound (D) Diastereomers 14) Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives X as the major product. Reaction of X with Hg(OAc)2/H2O followed by NaBH4 , OH gives Y as the major product. Y is: (A) (B) (C) (D) 15) Major Product can be: (A) (B) (C) (D) 16) (A) (B) (C) (D) 17) Dehydration of the alcohols with conc H2SO4 (I) (II) (III) will be in order (IV) (A) IV > III > I > II (B) I > II > III > IV (C) IV > II > III > I (D) II > IV > I > III 18) major product is 'major product' (A) (B) (C) (D) 19) In the Victor-Meyer's test, the colour given by 1 , 2 and 3 alcohols are respectively:(A) Red, blue, colourless (B) Colourless, red, blue (C) Red, blue, violet (D) Red, colourless, blue 20) Find out major product of following reaction: Major product (A) (B) (C) (D) SECTION-II 1) The ratio x/y on completion of the above reaction is ______. 2) An unknown compound (A) of Molecular wt 180 gm is Acetylated to give a compound (B) of molecular wt 390. The No. of hydroxyl groups present in compound (A) 3) Total number of possible alkenes in the below reaction is X. Then find the value of X Alkene 4) (A) Maximum number of moles of Ac2O consumed by reactant (A) is: 5) How many of the following compounds will give immediate turbidity with Lucas reagent [anhydrous ZnCl2/conc.HCl]? (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) MATHEMATICS SECTION-I 1) The integral (where c is a constant of integration) (A) (B) (C) (D) is equal to : 2) is equal to : (where c is a constant of integration) (A) tan x + cot x + c (B) tan x cot x + c (C) cosec x cot x + c (D) sec x cosec x + c 3) is equal to : (where c is a constant of integration) (A) (B) (C) (D) 4) is equal to : (where c is a constant of integration) (A) (B) (C) (D) None 5) If dx = ax + b ln (4ex + 5e x) + c; then: (where c is a constant of integration) (A) a = 1/8, b = 7/8 (B) a = 1/8, b = 7/8 (C) a = 1/8, b = 7/8 (D) a = 1/8, b = 7/8 6) The integral is equal to : (where C is a constant of integration) (A) (B) (C) (D) 7) is equal to : (where c is a constant of integration) (A) (B) (C) (D) 8) is equal to : (where C is constant of integration) (A) (B) (C) (D) 9) is equal to : (where c is a constant of integration) (A) (B) (C) (D) None of these 10) For any natural number m, equals: (Here C is constant of integration) (where x > 0), (A) (B) (C) (D) 11) The value of dx is equal to : (where C is a constant of integration) (A) (B) (C) (D) none of these 12) dx is equal to : (where c is a constant of integration) (A) (B) (C) (D) 13) is equal to : (where c is a constant of integration) (A) 2x + sinx + 2sin2x + c (B) x + 2sinx + 2sin2x + c (C) x + 2sinx + sin2x + c (D) 2x + sinx + sin2x + c 14) The integral is equal to : (Where C is a constant of integration) (A) 3tan 1/3x + C (B) (C) 3cot 1/3x+ C (D) 3tan 1/3x + C 15) (where C is a constant of integration) (A) (B) (C) (D) none of these 16) is equal to : (where C is a constant of integration) (A) (B) (C) (D) cosec ln +C cosec ln +C cosec ln +C cosec ln +C 17) dx is equal to : (where c is a constant of integration) (A) sin 1 + (B) (C) (D) +c + cos 1 + c sec 1 x +c is equal to : 18) Let f(x) = tan x + 2 tan 2x + 4 tan 4x + 8 cot 8x, then primitive of f(x) with respect to x is : (where C is constant of integration) (A) 8 ln(sin 8x) + C (B) ln(sin 8x) + C (C) ln(sin x) + C (D) ln(sec x) + C 19) If, (where C is a constant of integration) . Where , then (0) + 10K is equal to : (A) 0 (B) 10 (C) 20 (D) 21 20) The value of is equal to : (where C is integration constant) (A) x sin2x + C (B) cos5x. sin2x + C (C) xsin3x + C (D) xsin3x.cos5x + C SECTION-II 1) If e being an arbitrary constant, then the value of 20a + b + c + d is ............................ 2) Let and f(x) passes through ( ,0) then the number 3 of solutions of the equation f(x) = x in x [0,2 ] is ............... 3) If (x) = and (0) = 0, then value of ( (1))2 is ................ 4) If and (where C1 and C2 are constants of integration). Let h(x) = f(x) + g(x). If h(1) = 0 then h(e) is equal to ............................ 5) Let and . Also (0) = 0 and g(0) = 0, then the number of solution(s) of equation (x) + g(x) = 0 in [ 4 ,4 ] is ............................. ANSWER KEYS PHYSICS SECTION-I Q. A. 1 B 2 C 3 C 4 B 5 B 6 A 7 A 8 B 9 A 10 D 11 B 12 B 13 B 14 D 15 B 16 C 17 D 18 B 19 D 20 C SECTION-II 21 2 Q. A. 22 60 23 450 24 3 25 33 CHEMISTRY SECTION-I Q. A. 26 B 27 D 28 B 29 A 30 B 31 B 32 C 33 B 34 D 35 C 36 C 37 D 38 B 39 D 40 D 41 B 42 C 43 D 44 A 45 B SECTION-II 46 2 Q. A. 47 5 48 2 49 4 50 7 MATHEMATICS SECTION-I Q. A. 51 B 52 D 53 C 54 A 55 A 56 B 57 C 58 B 59 B 60 C 61 C 62 A 63 C 64 D 65 B 66 A 67 C 68 C 69 C SECTION-II Q. A. 71 37 72 3 73 3 74 1 75 9 70 D SOLUTIONS PHYSICS 1) d = B(adx) 2) According to Lenz' law 3) B = C KT ; At t = 0, B = C initial flux = C a2 , final flux = 0 Total charge flown 4) 5) Induced emf in the coil, where L is the self inductance of the coil and Here, is the rate of change of current in the coil. 6) Given, BH = 3 105 T, t = 2s, l = 1 m Initial velocity, u = 0 Using kinematic equation, v = u + at v = 0 + gt or v = 10 2 = 20 m s 1 e = Bvl e = 3 10 5 20 1 = 6 10 4 V 7) If i is increased than flux through coil B also increases. Than according to Lenz law induced current in coil B is such that it opposes the initial changing magnetic field Hence repulsion occurs. 8) 9) = 80 C = constant Magnetic force on the electron in the conducting rod PQ is towards Q. Therefore, A is positively charged and B is negatively charged. 10) Considering pure rolling of OA, about A, the induced emf across OA will be from lenz's law, O will be negative end, 11) If is in direction then flux changes is yz plane Therefore, E = Used Lenz law from C to B. 12) The emf will induce when flux changes i.e. when the area changes. In the given problem, flux changes from situation (i) to (ii) (in figure). But no flux will change from (ii) to (iii) (as area inside the magnetic field remains unchanged). Again from (iii) to (iv), flux changes. The total time taken from (i) to (iv) is : But during this time, flux does not change from (ii) to (iii) which takes a time: Hence, the time during which flux changes is : 13) Let E be the electric field at a distance r from the centre of the disc. Then or 14) 15) Induced current (i) is in anticlockwise direction. So e moves tangentially right (clockwise). 16) Considering a projected length 2R on the ring in vertical plane. This length will move at a speed v perpendicular to the field. This results in an induced emf : e = Bv(2R) in the ring. In Ring "A" : eA =B(-v)(2R) In Ring "B" : eB = B(v)(2R) Therefore, potential difference between A & B = B(v)(2R) - B(-v)(2R) = 4 BvR. Note : there will be no p.d. across a diameter due to rotation. Alternate Considering rotation of diameter about lowest point : = 2Bvr in A (since pure rotation). and e = 2Bvr in B. Hence (C) 17) Lenz's law 18) Since 19) By theory 20) According to Lenz law 21) t = 3 sec =0 e= Heat produced in 3 sec = 22) at t = 5 = 2J |i| = 60 mA 23) 24) BH = 0.60 10 4 Wb/m2 Induced emf = 0.60 10 4 10 5 = 3 10 3 V 25) VP 5 30 + 2 1 = VQ VP VQ = 33 volt CHEMISTRY 26) 3 - alcohol give turbidity immediate. 27) 28) T = U= 29) 30) Tertiary alcohol does not react with PCC. So, the substance will be: 31) The combination of names for isomeric alcohols with molecular formula C4H10O is/are: Formula Names n-butyl alcohol / butan-1-ol Isobutyl alcohol / 2-methyl propan-1-ol Secondary butyl alcohol / butan-2-ol Tertiary butyl alcohol / 2-metyl propan-2-ol 32) Only vicinal diol / hydroxy carbonyl / dicarbonyol compounds react with HIO4. 33) The major product formed is: 34) p Nitrophenol is more acidic than meta and ortho isomers. The electron withdrawing group, NO2 stabilises the phenoxide ion through R effect at the ortho and para positions. The effective delocalisation of negative charge in phenoxide ion enhances the acidic strength of phenols. The NO2 group at meta position will show I effect only. The NO2 group at ortho position forms hydrogen bond with OH group. Ethanol being primary alcohol does not react with Lucas reagent at room. 35) If 2 Ph groups are substituted by 2 MeO groups then carbocation formed in above sequence will become more stable and rate of above hydrolysis increases. 36) The OH group of Phenal is ortho and Para directing whereas SO3H is meta directing. The 3rd equivalent of Bromine attack at para of OH group. Since, SO3H is a good leaving group, The final product has 3 bromine on ortho and para positon of OH. Reaction: Conclusion: The correct product of the given reaction is: 37) This is Reimer-Tiemann reaction and the electrophile is dichlorocarbene. 38) 39) The product formed is: 40) Threo Racemic mixture 41) P.C.C. (Pyridinium chloro chromate) is mild-oxidising agent. 42) Dehydration of alcohol 43) 44) Red, blue, colourless 45) Stability of first carbocation 46) 47) No. of OH groups present in 'A' 48) 49) No. of alcohol = No. of moles of Ac2O consumed. =4 50) ii, iv, vi, vii, ix, x, xi MATHEMATICS 51) let tan 1 t = u = n |u|+c = 52) = sec x cosec x + c 53) use integration by parts 54) = 55) Let 3ex 5e x = A(4ex + 5e x) + B(4ex 5e x) + c Then 4(A + B) = 3, 5 (A B) = 5, C = 0 A = 1/8, b = 7/8 a = 1/8, b = 7/8 56) 57) 58) 59) = dx = +C= 60) We have: Now put 61) Let 62) Let 2 + 1 = ex 2 d = ex.dx or x = n(42 + 1) Again Option A is correct. +C +C 63) = x + sin2x + 2sinx + c 64) put tanx = t sec2xdx = dt 65) 66) = = cosec = cosec = cosec [log |sin x| log |sin (x + )|] + C = cosec log +C +C 67) I = Put = cos 2 I= = 2 sin 2 d 2 sin 2 d = 68) dx = 2 sin2 + C = cos 1 = = ln (sec x) + ln (sec 2x) + ln(sec 4x) + ln (sin 8x) + C = ln = ln + C = ln = ln(8 sin x) = ln 8 + ln(sin x) + C = ln(sin x) + C 69) = f(x) + K log|x| + C and f( /4) = 1 = tan x + 2log |x| + C f(x) = tanx, K = 2 (by comparing) Now, f(0) + 10K = tan0 + 10 2 = 0 + 20 = 20 70) Let f(x) = sin3x . cos5x f'(x) = sin3x . 5cos4x ( sinx) + cos5x . 3sin2x cosx = = x.sin3x . cos5x 71) Let x = u6, dx = 6u5du = 2u3 3u2 + 6u 6ln(u + 1) + e a = 2, b = 3, c = 6, d = 6 20a + b + c + d = 37 72) (x) = x3sin2x + c ( ) = 0 c = 0 (x) = x3 sin2x Number of solution of (x) = x3 x3sin2x = x3 Total = 3 and sin2x = 1 . 73) (x) = (x) = x2 + x4 + x6 = t2 (x + 2x3 + 3x5)dx = tdt x=0 C = (0) = 0 ( (1))2 = 3 74) = = = h(1) = 0 C = 0 h(x) = n x h(e) = n e = 1 75) Now, (0) = 0 = 0 + c (x) = xesinx Now, g(x) = c =0 Now, g(0) = 0 = c c = 0 g(x) = xecosx Now, (x) + g(x) = 0 esinx = ecosx & x = 0 tanx = 1 & x = 0 9 solution

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