Vidyamandir Classes SOLUTIONS JEE Entrance Examination - Advanced/Paper-1 Code -7

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1 SOLUTIONS 7-JEE Entrance Examination - Advanced/Paper- Code -7 PART-I PHYSICS.(AD) Net external force acting on the system along the x-axis is zero. Along the x-axis Momentum is conserved mv MV From conservation of energy Loss in GPE of particle of mass m = Gain in kinetic energy of both the masses. mgh mv MV m gh v M m m gh V v M M m M along mgh mv mv M M v ve x-axis gh hence option (A) is correct. m M Hence option [B] is incorrect. Since the location of center of mass does not change along the x-axis mr x Mx m x M x m/ G M / G mr x. m M mr xm G m M Hence option (D) is correct. MR xm G R x M m mr Final position of m x m M Hence option (C) is incorrect..(ac) At ~, X L ~ and XC At Current will be nearly zero. 6, X L >> and X C f Hence circuit does not behave like a capacitor. At resonance frequency i.e. when X C X L current will be in phase with voltage and frequency is independent of R. 6 Resonant frequency rad / s LC 3.(ABD/BD) Velocity of the wave depends upon the tension in the rope and mass per unit length of the rope. Velocity is independent of frequency and wavelength VMC/Paper- JEE Entrance Exam-7/Advanced

2 Option (B) is correct Since tension at the mid point is same, therefore speeds at that point will also be same. Option (A) is correct. Since the direction of motion of pulses is opposite, velocities will be equal and opposite. With this point of view A will be incorrect. Since the velocities at position is dependent only on tension at every point (μ = constant) therefore time taken for the wave the reach A from O and O from A will be same. T OA T AO Option (D) is correct Since ν depends on source T Tension decreases as we move from O to A, therefore λ becomes shorter. Option (C) is incorrect. 4.(ACD) Molecules hitting the forward and rear surfaces will bounce back with speeds given above. Let mass of one molecule be m. Then, P forward m ( u v) P Rear m ( u v) Let the rates of collision with front and rear surfaces be R and R respectively So, R ( u v) R ( u v) Force = P. R. So, F R m u v, F R. m u v. So, F F m u v m ( u v) m 4uv uv so, (A) is correct Clearly, the net force due to gas is proportion to v, i.e. it is variable hence acceleration of plate is variable. Finally the plate will start moving with terminal velocity. Hence (C) is correct. Resistive force = P. A V Hence (D) is correct. 5.(A or ACD) Net power radiated a T 4 T 4 above, For small temperature difference, T T T 4 4 P Net A T T T P Net 4AT 3 T T (A) (B) 4 4 T AT T From the above result, P(Net) decreases with A Peak shits to shorter wave lengths for rise in temperature. AT 4 A T 46 watt (C) P(Radiated) 4 (D) 6.(ACD) Angular deviation : i i A Hence (C) is incorrect If P(Net) is taken, then (C) will be correct 4 4T AT T Energy radiated by a body is dependent only on its own temperature, not the temperature of surroundings. Hence (D) is incorrect. If P(Net) is taken, then from the calculation shown (D) will also be correct. VMC/Paper- JEE Entrance Exam-7/Advanced

3 For min deviation : i i so, i A...(i) Also, r r A A i.e. r...(ii) i r (A is correct) By Snell's law, sin i sin r A i.e. sin A sin Hence (B) is incorrect (C) is obviously correct A cos A cos For tangential emergence, i 9. So, sin r cos r Also, r A r sin r sin Acos r cos Asin r sin A cos A By Snell's law on st surface, sin i sin r / sin i sin A cos A i.e. (D) is correct 7.(AB) The angle that area vector makes with B at time t is t. 8.(5) BA cos t BA sin ( t) BA cos( t) BA sin ( t) Due to orientation of loops, the two EMFs will work against each other. So, (net) BA sin( t) So, (A) is correct. A i sin sin A 4 cos cos A We can see that net is maximum when. So, (B) is correct. Obviously, (C) is incorrect. (D) is incorrect as the EMF is proportional to difference in areas. V n So, V V i f n f ni 6.5 Minimum integral value of n f is 5. 9.(8) We know that for the given case, sin constant n f 5.5 n VMC/Paper- 3 JEE Entrance Exam-7/Advanced i

4 So, n m n.6 sin 3 sin 9 Vidyamandir Classes i.e..8 n m n Solving, m = 8.(5) A A e t (6) Here, ln days 8 t hrs day ln A So, A A / exp 8 6 A 6 A 7 If the volume of blood be V ml, then.5 5 A V A V Solving V 5 liters.(6) Frequency received by car Hz Frequency received by source Original frequency f 49 Hz Final frequency f 498 Hz Beat frequency f f 6 Hz 498Hz R r K (Conservation of volume) 3 3 R = radius of bigger drop r = radius of smaller drop 3 3 R r K...(i) U S 4 R i U KS 4 r f 3 KS 4r S 4R 4S Kr R 3 K 3 R R / 3 / (D) For constant velocity, acceleration of particle should be zero, Hence net force should be zero. qvb qe E V (For a = ) B VMC/Paper- 4 JEE Entrance Exam-7/Advanced

5 Electric field and magnetic field should be perpendicular. Option (D) : E V yˆ, E E xˆ, B Bzˆ (for electron) B B F q V B FE E x ˆ E e yˆ B zˆ B E xˆ 4.(B) V, E E yˆ, B B yˆ (Proton) 5.(D) F only force along -y axis is acting due to electric field alone. B E V xˆ ; E Ezˆ ; B B zˆ (Proton) B FB FE along y axis along +z axis So condition for helical path is satisfied 6.(C) (Please note that W is work done on the gas). Process to represents isobaric process W P V V PV PV U Q P V is correct. No other combination is possible. 7.(A) Laplace correction is done is correction in the determination of the speed of sound in an ideal gas, which states that the process assumed was not Isothermal but it is Adiabatic. W P V PV adiabatic Corresponding graph is 8.(B) (A) Incorrect since in Isochoric process W (B) Correct as Wisochoric Graph is also correct. (C) Not correct combination since W PV PV is wrong for Adiabatic process. (D) Incorrect combination of graph and process VMC/Paper- 5 JEE Entrance Exam-7/Advanced

6 PART-II CHEMISTRY 9.(BC).(ABC) * Z represent vapour pressure of pure liquid L and as, the given graph is merging with ideal graph of p L. Hence Option (C) is correct * Since, vapour pressure of liquid L(p L) is higher than the ideal values so L-M interactions are weaker than L-L & M-M interactions. Hence Option (B) is correct choice. In oxoacids acidic strength increases with increase in number of double bonded oxygen atoms because of greater resonance stabilization of conjugate base. Hence HClO 4 is more acidic than HClO. L Central atom in both HClO 4 and HClO is HClO H O ClO H O 3 sp hybridized. This equilibrium remain shifted in forward direction hence ClO4 is weaker base than HO. Because such equilibrium remain shifted towards weaker acid or weaker base side. Cl HO HCl HOCl.(AC) IUPAC name of the given compound is chloro-4-methylbenzene or 4-chlorotoluene.(ABD) NH Cl M H 4 O Cl 6NH 6 3 M NH O 3 Cl3 6HO 6 X (Y) Pink * M NH Cl M NH 3Cl : 3electrolyte Room M HO Cl HCl MCl 6 Temp 4 4HO H3O Excess X Z Pink If M is Co then X : CoH O Cl 6 Y : Co NH 3 Cl 6 3 Z : CoCl 4 Blue x 3.87 due to presence of three unpaired electron in X 7 Co ;3d VMC/Paper- 6 JEE Entrance Exam-7/Advanced

7 3.(CD) HO is weak field ligand hence no pairing Hybridization state is 3 sp d. z 3.87 BM due to presence of three unpaired electron in Z. 7 Co ;3d Cl is weak a field ligand and coordination number is four hence hybridization state is 3 M NH 3 Cl 6 3 3AgNO3 M NH3 6 3NO3 3AgCl(s) Co HO 4Cl CoCl 6 4 6HO : H Pink Blue (As octahedral complex changes to tetrahedral complex with H O ligands replaced by Cl ) 3 sp. At C, equilibrium shifts in reverse direction hence colour of the solution is pink due to CoH O 6 4.(ACD) * Bromination (addition of Br ) proceeds through trans-addition in both the reactions * (M and O) and (N and P) are two pairs of diastereomers State Expansion State p,v,t p,v,t Work done in reversible compression process is smaller than the work done in irreversible compression process. U for reversible isothermal expansion. U for reversible adiabatic expansion. For free expansion q, w and U hence it is simultaneously both isothermal ( U ) as well as adiabatic (q = ). Work done in reversible adiabatic expansion is less than the work done in reversible isothermal expansion as shown in figure. 5.(BC) Electronic configuration of F molecule is Similar electronic configuration for other X molecules. * px and * * * * * s, s, s, s,,, pz px py px, py pz * py are highest energy occupied molecular orbital (HUMO) and * pz is lowest energy unoccupied molecular orbital (LUMO). * * Colour of X molecules of group 7 elements is due to transition of electrons from to. V M C 6.(6) Number of electrons around central atom (N) VMC/Paper- 7 JEE Entrance Exam-7/Advanced

8 7.(6) [TeBr 6 ] : [BrF ] : SNF 3 : [XeF 3] : 6 6 N 7 Number of bp = 6 Number of p = 7 N 4 Number of bp = Number of p = 6 3 N 4 Number of bp = 4 Number of p = 8 3 N 6 Number of bp = 3 Number of p = 3 The sum of the number of lone pairs of electrons on each central atom = = 6 And c K m M HA H A c l R A K c m R l A 7 K 5 c c m 4 m m z m So z 6. 8.(5) Compound having a close loop of (4n ) electrons is aromatic compound VMC/Paper- 8 JEE Entrance Exam-7/Advanced

9 Number of aromatic compounds is five i. e. 3, 5, 7, 8 and 9. 9.(6) H Diamagnetic s * He Paramagnetic s, s * Li Diamagnetic s, s, s * * Be Diamagnetic s, s, s, s * * B Paramagnetic s, s, s, s, px py * * s s s s px py C Diamagnetic,,,, * * s s s s Px py pz * * * * s s s s p z p p p p x y x y * * * * s s s s pz px py px py N Diamagnetic,,,,, O Paramagnetic,,,,,,,, F Diamagnetic,,,,,, The number of diamagnetic species is 6 i.e. H, Li, Be, C, N and F are diamagnetic species 3.() a 4pm d 8g cm 3 w 56g z M d 3 a NA 3 d a NA M z g/mol Number of atoms 6.3 N N N VMC/Paper- 9 JEE Entrance Exam-7/Advanced

10 3-33. The correct matching in column, column and column 3 are Column Column Column 3 (I) (i) (Q) (R) (S) (II) (ii) (P) (Q) (R) (S) (III) (iii) (iv) (S) (IV) (Q) (S) 3.(A) Incorrect combination is (I) (iii) (R) because in the expression of n,,m Zr a must have e. [A] for s-orbital, the exponential part Zr a 3.(D) The correction combination is (I) (i) (s) as n,, m in the column (i) has exponential part as e. 3.6Z 3 3.6Z (4 ) E4 E E 6 E 3.6Z 8 3.6Z 3 (9 ) 36 6 [D] 33.(C) The correct combination for any hydrogen like species is [II] (ii) (P). s-orbital has one radial node and n,, m v/s r plot will start from a finite value and sign changes once from +ve to ve. [C] 34.(B) 35.(A) 36.(B) PART-III MATHEMATICS az b az b 37.(AC) iy z z azz az bz b ( azz bz az b) ( z ) ( z ) ( a b) z z ( a b) iy ( z ) ( z ) iy VMC/Paper- JEE Entrance Exam-7/Advanced

11 ( a b) iy iy ( z ) ( z ) ( z ) ( z ) x y x x x y ( x ) y x y x y x y x y. x f x e f t sin t dt x 38.(CD) (A) f f x f x sin x x f ' x e f x sin x / f x f t sin t dt x (, ) (B) (C) x g x x f t cos t dt / g f t cost dt g f t cost dt (D) g x x 9 f x g f g f 39.(AC) P X / 3...(i) P X Y P X / Y / P 4...(ii) P Y / X / 5 PY X P X...(iii) P X Y.3 5 P X Y 5. 4 (ii) 5 P Y P Y 5. 5 ( ) P X Y P X P Y P X Y VMC/Paper- JEE Entrance Exam-7/Advanced

12 4.(ABD) P X / Y P Y P X Y P Y P X Y P Y cos Lt h h h x Vidyamandir Classes 5 / 5 4 hcos h h cos h h cos h h cos h Lt h cos h hcos h x Lt hcos h x Lt h cos h h cos 3 h h cos h x Lt h cos h h cos 4 h h cos h x 4.(ABD) x y y x x y a 6 c a m b a a 4 7 a. 4.(AB) For (A) and (B) det ( A ), and for (C) 43.(B) For (C): I. y 6x Equation of a chord with a given middle point h, k is T S ky 8 x h k 6h ky 8x 8h k x y p 8 k 8h k p 4 8h 6 p k 4 4 p 8h 6 p h 4 VMC/Paper- JEE Entrance Exam-7/Advanced

13 44.() 45.(5) x! h 3 p / g x f ( t cos ect = g( x) 3 f ( x) cosec( x) x x f lim g x lim 3 3 = x x sin x 9! y C9 C! 46.(6) a d, a, a d a a d a a d a d ad a ad d a 4ad a a 4d 3d 4d 5d A 3 d 4 d 4 6d 4 d 4 d Rejected d a 8 6, 8, 47.() Case I : Touching x-axis When x x y 4 x x y 4y y 4y D (Two real root) p Case II : Touching y-axis Case III : x y x y x 4y 4 y x x 4 D (None real root) Rejected x y x 4y p 48.() (two planes are parallel) (Rejected) VMC/Paper- 3 JEE Entrance Exam-7/Advanced

14 49.(D) y x 8 (two planes are coincident) a a a y 3 x, y mx,, m m m 5.(A) a 5.(C) x y a, ma a y mx a m,, c m x y, x y or y x tangent at a 3 a 4 a 4 x y 4 3x y (D) 53.(A) 54.(B) f x x ln x x ln x n, xln x f ( x) x ln x ln x x x x x ln x x f f x f ( ) f ( e ) f ( x) x x VMC/Paper- 4 JEE Entrance Exam-7/Advanced

JEE ADVANCED 2017 (PAPER-1) (QUESTION WITH ANSWER)

JEE ADVANCED 2017 (PAPER-1) (QUESTION WITH ANSWER) 1 ( P C M ) JEE ADVANCED 017 (PAPER-1) (QUESTION WITH ANSWER) PART I (PHYSICS) SECTION 1 (MAXIMUM MARKS : 8) This section contains SEVEN questions. Each question has FOUR options (A), (B), (C) and (D).