Page 1 (1) IIT - JEE 2014 (Advanced) CODE : 1 Page 2 (1) IIT - JEE 2014 (Advanced) CODE : 1 (2) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution (2) Page 3 (1) IIT - JEE 2014 (Advanced) CODE : 1 (2) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution (2) IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (3) (3) PART I : PHYSICS SECTION - 1 : (One or More Than One Options Correct Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE THAN ONE are correct. 1. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t) = I 0 cos (?t), with I 0 = 1A and ? = 500 rad s -1 starts flowing in it with the initial direction shown in the figure. At t = 7 6 p ? , the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20µF, R = 10 O and the battery is ideal with emf of 50V, identify the correct statement (s). (A) Magnitude of the maximum charge on the capacitor before t = 7 6 p ? is 1 × 10 -3 C. (B) The current in the left part of the circuit just before t = 7 6 p ? is clockwise. (C) Immediately after A is connected to D, the current in R is 10A. (D) Q = 2 × 10 -3 C. 1. (C), (D) If q represents the charge on capacitor’s upper plate: I (t) = I 0 cos(?t) = dq dt ? q(t) = 0 I ? sin(?t) Max charge = 1 1A 500 rad s - = 2 × 10 -3 C Charge on upper plate at t = 7 6 p ? = 0 I ? sin 7 6 p ?? ?? ?? = 0 I 2 - ? When capacitor is fully charged; charge on upper plate = 50 V × 20µF = 1 × 10 -3 C ? Q = 1 × 10 -2 C - 0 I 2 ?? - ?? ? ?? = 2 × 10 -3 C Voltage across capacitor when A and D are connected = 3 6 110 C 20 10 F - - × × µ = 50V Total voltage across resistor = 100V ? current = 100v 10O = 10A Current is anti clockwise. Page 4 (1) IIT - JEE 2014 (Advanced) CODE : 1 (2) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution (2) IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (3) (3) PART I : PHYSICS SECTION - 1 : (One or More Than One Options Correct Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE THAN ONE are correct. 1. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t) = I 0 cos (?t), with I 0 = 1A and ? = 500 rad s -1 starts flowing in it with the initial direction shown in the figure. At t = 7 6 p ? , the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20µF, R = 10 O and the battery is ideal with emf of 50V, identify the correct statement (s). (A) Magnitude of the maximum charge on the capacitor before t = 7 6 p ? is 1 × 10 -3 C. (B) The current in the left part of the circuit just before t = 7 6 p ? is clockwise. (C) Immediately after A is connected to D, the current in R is 10A. (D) Q = 2 × 10 -3 C. 1. (C), (D) If q represents the charge on capacitor’s upper plate: I (t) = I 0 cos(?t) = dq dt ? q(t) = 0 I ? sin(?t) Max charge = 1 1A 500 rad s - = 2 × 10 -3 C Charge on upper plate at t = 7 6 p ? = 0 I ? sin 7 6 p ?? ?? ?? = 0 I 2 - ? When capacitor is fully charged; charge on upper plate = 50 V × 20µF = 1 × 10 -3 C ? Q = 1 × 10 -2 C - 0 I 2 ?? - ?? ? ?? = 2 × 10 -3 C Voltage across capacitor when A and D are connected = 3 6 110 C 20 10 F - - × × µ = 50V Total voltage across resistor = 100V ? current = 100v 10O = 10A Current is anti clockwise. (4) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution (4) 2. A light source, which emits two wavelengths ? 1 = 400 nm and ? 2 = 600 nm, is used in a Young's double slit experiment. If recorded fringe widths for ? 1 and ? 2 are ß 1 and ß 2 and the number of fringes for them within a distance y on one side of the central maximum are m 1 and m 2 , respectively, then (A) ß 2 > ß 1 (B) m 1 > m 2 (C) From the central maximum, 3 rd maximum of ? 2 overlaps with 5 th minimum of ? 1 (D) The angular separation of fringes for ? 1 is greater than ? 2 2. (A), (B), (C) ß = ? D d 1 22 1 nm 2 ? ?= n 2 600 = 1 400 m 2 3n 2 = m 1 For n 2 = 3 m 1 = 9 which is 5 th minima. 3. One end of a taut string of length 3m along the x axis is fixed at x = 0. The speed of the waves in the string is 100 ms -1 . The other end of the string is vibrating in the y direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are) (A) y(t) = x50t Asin cos 63 pp (B) y(t) = x100t Asin cos 33 pp (C) y(t) = 5 x 250 t Asin cos 63 pp (D) y(t) = 5x Asin cos t 2 p 250 p 3. (A), (C), (D) ? = k ? 1 AB C D 100ms - ? = ? = ? = ? = x = 3 is an antinode. This eliminates (B) 4. A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is C 1 . When the capacitor is charged, the plate area covered by the dielectric gets charge Q 1 and the rest of the area gets charge Q 2 . The electric field in the dielectric is E 1 and that in the other portion is E 2 . Choose the correct option/options, ignoring edge effects. (A) 1 2 E E = 1 (B) 1 2 E E = 1 K (C) 1 2 Q Q = 3 K (D) 1 C C = 2 K + ? Page 5 (1) IIT - JEE 2014 (Advanced) CODE : 1 (2) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution (2) IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (3) (3) PART I : PHYSICS SECTION - 1 : (One or More Than One Options Correct Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE THAN ONE are correct. 1. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t) = I 0 cos (?t), with I 0 = 1A and ? = 500 rad s -1 starts flowing in it with the initial direction shown in the figure. At t = 7 6 p ? , the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20µF, R = 10 O and the battery is ideal with emf of 50V, identify the correct statement (s). (A) Magnitude of the maximum charge on the capacitor before t = 7 6 p ? is 1 × 10 -3 C. (B) The current in the left part of the circuit just before t = 7 6 p ? is clockwise. (C) Immediately after A is connected to D, the current in R is 10A. (D) Q = 2 × 10 -3 C. 1. (C), (D) If q represents the charge on capacitor’s upper plate: I (t) = I 0 cos(?t) = dq dt ? q(t) = 0 I ? sin(?t) Max charge = 1 1A 500 rad s - = 2 × 10 -3 C Charge on upper plate at t = 7 6 p ? = 0 I ? sin 7 6 p ?? ?? ?? = 0 I 2 - ? When capacitor is fully charged; charge on upper plate = 50 V × 20µF = 1 × 10 -3 C ? Q = 1 × 10 -2 C - 0 I 2 ?? - ?? ? ?? = 2 × 10 -3 C Voltage across capacitor when A and D are connected = 3 6 110 C 20 10 F - - × × µ = 50V Total voltage across resistor = 100V ? current = 100v 10O = 10A Current is anti clockwise. (4) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution (4) 2. A light source, which emits two wavelengths ? 1 = 400 nm and ? 2 = 600 nm, is used in a Young's double slit experiment. If recorded fringe widths for ? 1 and ? 2 are ß 1 and ß 2 and the number of fringes for them within a distance y on one side of the central maximum are m 1 and m 2 , respectively, then (A) ß 2 > ß 1 (B) m 1 > m 2 (C) From the central maximum, 3 rd maximum of ? 2 overlaps with 5 th minimum of ? 1 (D) The angular separation of fringes for ? 1 is greater than ? 2 2. (A), (B), (C) ß = ? D d 1 22 1 nm 2 ? ?= n 2 600 = 1 400 m 2 3n 2 = m 1 For n 2 = 3 m 1 = 9 which is 5 th minima. 3. One end of a taut string of length 3m along the x axis is fixed at x = 0. The speed of the waves in the string is 100 ms -1 . The other end of the string is vibrating in the y direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are) (A) y(t) = x50t Asin cos 63 pp (B) y(t) = x100t Asin cos 33 pp (C) y(t) = 5 x 250 t Asin cos 63 pp (D) y(t) = 5x Asin cos t 2 p 250 p 3. (A), (C), (D) ? = k ? 1 AB C D 100ms - ? = ? = ? = ? = x = 3 is an antinode. This eliminates (B) 4. A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is C 1 . When the capacitor is charged, the plate area covered by the dielectric gets charge Q 1 and the rest of the area gets charge Q 2 . The electric field in the dielectric is E 1 and that in the other portion is E 2 . Choose the correct option/options, ignoring edge effects. (A) 1 2 E E = 1 (B) 1 2 E E = 1 K (C) 1 2 Q Q = 3 K (D) 1 C C = 2 K + ? IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (5) (5) 4. (A), (D) If c = o A 3d e then c 1 = kc c 2 = 2c c 1 + c 2 = C ? (k + 2) c = C ? c = c k2 + ? c 1 = kc k2 + c 2 = 2C kz + If charging voltage is V: charges will be in the ratio of capacities and as potential difference is same. Electric field should be equal. 5. Let E 1 (r), E 2 (r) and E 3 (r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density ?, and an infinite plane with uniform surface charge density s. If E 1 (r 0 ) = E 2 (r 0 ) = E 3 (r 0 ) at a given distance r 0 , then (A) Q = 2 0 4r sp (B) r 0 = 2 ? ps (C) E 1 (r 0 /2) = 2E 2 (r 0 /2) (D) E 2 (r 0 /2) = 4E 3 (r 0 /2) 5. (C) 2 0 2 00 000 0 2 0 0 2 00 0 1q 2r q 2r 2 r 42r r 1q q2r r 42 r ? ? = ? ? = ? p s = ? ? pe pe ? ? ? s ? s = ? = ? = p s p ? pe e ? 10 E(r /2) 4 = 0 E(r / 2) 2 6. A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s -1 . He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350 ± 0.005) m, the gas in the tube is (Useful information : 167RT = 640 j 1/2 mole -1/2 ; 140RT = 590 J 1/2 mole -1/2 . THe molar masses M in grams are given in the options. Take the values of 10 M for each gas as given there) (A) Neon 10 7 M20, 20 10 ?? = = ?? ?? (B) Nitrogen 10 3 M28, 28 5 ?? = = ?? ?? (C) Oxygen 10 9 M32, 32 16 ?? = = ?? ?? (D) Argon 10 17 M36, 36 32 ?? = = ?? ?? 6. (D) f = 244 Hz 4 ? = 0.356 ± 0.005 ? ? = 1.400 ± 0.020 ? ? = (341.6 ± 4.88) m s ? = ?f = ? = rRT M = 100 rRT 100M = 3 100RrT 100 M 10 kg/mol - × ×Read More

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