JEE Main 2017 Answers & Explanations

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1 Test Booklet Code A JEE Main 7 Answers & Eplanations Physics Chemistry M athematics CL Educate Limited Registered Office / Corporate Office A -, Lower Ground Floor, Espire Building, Mohan Co-operative Industrial Area Main Mathura Road, New Delhi, Contact No. -88 / I

2 PART A PHYSICS. Weight Density Volume Stress Area Area If Dimensions increase by a factor of 9 the stress becomes 9 times. Velocity decreases linearly. mdv kv dt Now R L = constant dr RL R dl or dr R...() dl L using () & () MR R ML L 6 R L 6. v v dv k dt v m k t v m at t = s m k kg m v t dv m 6t dt t m dv 6 tdt mv = t v = t m Work = Change in K.E. m v v =.5 J or L R 6. = I l Ml mg sin 7. Inside: g sin l d g = g g R R Outside: GM g = 8. Using concept at calorimetry. (T 75 ) =. (75 ) + 7 (75 ) Solving we get T = 885 C 5. MR ML I di MR dr ML...() dl dl 6 9. P K V V V P V K......

3 Coefficient of volume epansion = P T, K P T K. C P C V = R for hidrogen gas and C P C V = R for nitrogen gas 8 Because if is the angle with -ais, 9 will be the angle with y-ais. Since T T PE sin= ten = = 6. PEcos 5. W e can connect capacitors, with capacitors in series & 8 such series in parallel. Here, C P and C V are defined as specific heat per unit mass Thus, the answer will be (). Note: If C P and C V are taken as molar specific heat then C P C V for the both the gases will be R.. Using PV = nrt PV PV n i,n f R 9 R 6. E i r r There will be no current through the branch of capacitor. Er V ir V r c r r CEr qc CVC r r n = n f n i = PV R 9 The number of molecules = n Avogadro number = n 6. Solving we get = K.E is ma at t =, Then it decreases to zero at t = T.. For light the doppler effect formulas is c c c v n n n n c v c c. T = PE sin T P E sin 9 7. The p.d. across each resistor is zero. 8. I T MB M Magnetic moment B Magnetic field I Moment of Inertia T = Time for oscillations = 6.67 s 9. = 5 (5 + R) = 5 + R R = 985 =

4 . From the graph: i = t. = R.5 idt tdt.5.5 t t = (.5.5) = (5.5) = 5 Wb. D D. n n for coinciding of bright d d fringes. or n = n n 65 = n 5 5n = n n = and n = 5 D Minimum distance n d. hc min ev log(hc) log( min ) = log e + log V. hc log( min ) = log log V e This is a straight line with positive y-intercept and slope negative = = = 7.8 mm. Before Collision A(m) B(m/). 5 cm After Collision V A B V Because P sys = constt m mv = mv + v. v = v v...() f = 5 cm f = cm Image by diverging lens is formed at 5 cm, to the left of the lens. This image serves as object for the converging lens. u = cm; f = + cm. v u f v v v = + cm Image is Real and at a distance of cm from convergent lens. Further e = v v v v v v Solving the equations we get, v v ; v mv pa ; pb A v. h A h p mv B mv h B h ; p mv A B...()......

5 5. hc E E E hc E E E r. 6. A B N A = N e t N B = N ( e t ) A t hc E hc E N B ( e ) t e. (given) t N e e t =. t = log. Also; log T Tlog. t log 7. conceptual. 8. conceptual. 9. conceptual. or log. t PART B CHEMISTRY. 89. = [H r (CH ) + ] [( 9.5) + ( ( 85.8)] 89. = H r (CH ) H r (CH ) = = 7.8 KJ mol. M CO + HCl MCl + H O + CO M g of M CO gives mole of CO g of M CO gives = moles M So,.86 M M.86 M 8.. U = q + w for adiabatic q = U = w. Two conditions should be satisfied regarding tyndall effect: (i) The diameter of the dispersed particles should be smaller but not much than the wavelength of light used. (ii) The refractive indices of dispersion medium and the dispersed phase must vary in magnitude to a large scale.. dhg T d diameter h rise of water g acceleration due to gravity T d h g T d h g...5% For F.C.C. lattice a r closest approach in r = a 6. Strongest reducing agent will be Cr as E Cr / Cr.7V

6 7. T f =.5 i wa Tf ikf M A w B.5 i i ph = [Pk w Pk a Pk b ] ph = [..] 6.9. Lithium and Magnesium, both form basic carbonates.. CO = = s, s, s, s, y z p p, p It has no unpaired electrons so diamagnetic or =.8 = r A Z.59 r.98a.a Ea K Ae RT For R, K = For R, K = Ea Ae RT Ea Ae RT. XeF OF XeF6 O Oidation no. of Xe changes from + to +6 Oidation no. of O changes from + to. As both oidation and reduction takes place so a redo reaction.. Permissible value is ppm for Flouride ions. 5. O = electrons Na + = electrons F = electrons Mg + = electrons 6. NaOH Cl NaCl NaOCl HO (cold and dilute) 7. ZnONaO NaZnO (acid) (base) lnk = Ea ln A RT... (i) ZnO CO ZnCO (base) (acid) Ea lnk = ln A... (ii) RT (ii) (i) gives, K ln (Ea Ea ) K RT K ln.9 K H + 8. NaCO H C O Conc HSO HCO CaC O CO + CO + H O CaCl (White ppt.)

7 9. % of Hydrogen is = The weight of person is = 75 kg Weight of Hydrogen in that is =7.5 kg If H is replaced by H Weight gain = 7.5 kg 5. Number of milli moles of CoCl.6H O =. = Number of milli moles of ions. = = 6 [Co(H O) 5 Cl]Cl.H O + AgNO Co(H O) Cl 5 AgCl NO H O 5. O has least resonating structures. 55. Due to stability of carbocations CH + OCH (+) > CH CH CH CH (+) > CH CH CH NH NH Br H 5. Conc HNO HSO 56. C 6 H 5 C 6 H 5 tbuok C 6 H 5 CH = CH C 6 H 5 NH NO 5% NH 57. As on C, OCOCH will get converted to OH group so due to this free group it will act as reducing sugar. + + NO 58. CH CH = C CH CH HBr Peroide NO Br CH 7% % CH CH CH CH CH CH 5. Br Elimination reaction (Two chiral center in the product.) Number of stereoisomers = not possible. 5. Nylon

8 O O PART C MATHEMATICS 59. CHO H / CHOH CH MgBr [Ag(NH ) ] OH HO O COOCH CH COO 6. y y + y = y + y = D y y (y + ) (y ) y Range = Co-domain therefore function is dy surjective ( ) Function is increasing & decreasing therefore function is not injective HO CH CH 6. ( + ) + ( + ) ( + ) + ( n ) ( n) n 6. O O DIBALH n n n (n ) (n )(n) n n n n(n ) n n n OH COOH CHO n n n n CHO n n, Let roots are k & k + k + k + = n n k(k ) n n n

9 n n n = n n = n = = z z 65. a a b a a b a a C C C C C C i a 6. i i, 7 = ( ) = ( ) = i = z = k k z A C C C C A A Adj b b No value of b 66. Let X invites a ladies and b men Y invites c ladies and d men so a + c =, b + d =, a + b =, c + d = X (a ladies, b men) (, ) (, ) (, ) (, ) Y (c ladies, d men) (, ) (, ) (, ) (, ) ( C C ) + ( C C ) + ( C C ) + ( C C ) = = ( C C C ) ( C C C ) ( ) ( )

10 68. 5a + 9b + 5c 7cac 5ab 5bc = Multiply two both sides (5a b) + (b 5c) + (5c 5a) = 5a b = b 5c = 5c 5a = b = 5a c = a b, c, a are in AP. 5a, a, a 5a a a. 69. f( + y) + f() + f(y) + y.u R a( + y) + b( + y) + c = (a + b + c) + (ay + by + c) + c + y 5 a,c so b f() = 5 5 n n. n cot cos lim 7. ( ) Put h tanh sinh lim h (h) ( cosh)sinh cosh lim lim h h (h) 8h h sin lim. h h tan 9 tan /, d / 9 tan ( ) Curve intersects y-ais at : =, y( )( ) = + 6 y = Differentiating dy ( )( ) y[ 5] dy dy at, y is Normal at (, ) in y = ( ) + y =. r Given r + r = Area = r = r r( r) d(area) ( r) dr r = 5 Maimum area = 5 ( ) 5 sq. m. n n I tan I n + I n = n (tan ) C n l n tan (sec ) n tan sec......

11 If n = 6 I 6 + I = 5 (tan ) 5 a, b. 5 I cos C cos( ) cos cos I cosec cot cot cot = I =. = y + y = y = + Area enclosed ( ) ( ) dy + sin dy + y cos + cos = Integrating on both sides y + sin y +sin = c =, y = c = 9t,y. 78. A(, 6) B(5, ) Area = F E D C(, ) k 5 k k k k k 56 k,, 5 BE : y = Ortho centre :,

12 79. r y (, ) r () in () a a ( a ) a = a a a 9a + 8 = a = (or) a = 8. a 8 a + b = a = tangent at (, ) is 8. r = r r = a e, a e b = a ( e ) = Equation of elipse is y Equation of normal is y y = y Option satisfies. 8. Normal vector is y ˆ ˆ ˆ i j k = ˆ i(5) ˆ j( 7) k() ˆ = 5i ˆ 7j ˆ k. ˆ Equation of a plane is 5( ) + 7(y + ) + z+ = distance from (,, 7) to plane is d 8. distance = Let Hyerperbola a y b Passes through P(, )... () a b foci is (, ) ae = b = a (e ) b = a e a a + b =... () P(,, ) d = 6/ 9 + y z + = Q......

13 6 cos d PQ cos 8. a b c sin a b c 6 i j k a b i j k a b c = c a c a a.e 9 a.c... () 87. n {,, 8} Any n + {, 5, 9} n + {, 6, } Any n + {, 7} Probability = 88. Let cos = t 89. C C C t t 5 t 9 t 5( t t ) = t + t + 8 9t + t + = t = cos = B 7, cos P(g) = 5, P(y) = 5 Its a Binomial distribution whose variance is npq = P(AB) = P(BC) = P(AC) = P(A B C) 6 P(A B C) [P(A B) P(A C) P(B C)] P(A B C) C A tan = tan(( + ) ) tan( ) tan tan( ) tan p q pq ~p ~p q (~pq)q s T T T F T T T T F F F T F T F T T T T T T F F T T F T T P......