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# Jee main 2014 qpaper key solutions

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• Hello people.. Hope all of u maxed ur IIT JEE...
But still no time to relax... Gear up for all the other exams.. Found this article giving information about SRMEE...
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### Jee main 2014 qpaper key solutions

1. 1. Sri Chaitanya IIT Academy., India Corporate Office: Plot No:304,Kasetty Heights, Ayyappa Society, Madhapur - Hyd Email:iconcohyd@srichaitanyacollege.net, Web: www.srichaitanya.net JEE-MAIN
3. 3. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 3 MATHEMATICS 1. The image of the line x 1 y 3 z 4 3 1 5       in the plane 2x – y + z + 3 = 0 is the line: 1) x 3 y 5 z 2 3 1 5       2) x 3 y 5 z 2 3 1 5        3) x 3 y 5 z 2 3 1 5       4) x 3 y 5 z 2 3 1 5        Key: 1 Sol : x 1 y 3 z 4 is parallel to 2x y z 3 0 3 1 5           Image h 1 k 3 l 4 2 2 1 1            h, k, l 3, 5, 2   Required equation x 3 y 5 z 2 3 1 5       2. If the coefficients of 3 4 x and x in the expansion of    182 1 ax bx 1 2x   in powers of ‘x’ are both zero, then (a, b) is equal to: 1) 251 16, 3      2) 251 14, 3      3) 272 14, 3      4) 272 16, 3     
4. 4. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 4 Key : 4 Sol : Equating coefficients of 3 4 x and x is zero 32 17 17a b 0 3      1615 – 32 a + 3b = 0 Solving (a, b) = 272 16, 3      3. If a R and the equation       2 2 3 x x 2 x x a 0      (where [x] denotes the greatest integer x ) has no integral solution, then all possible values of a lie in the interval : 1)    1, 0 0, 1  2) (1, 2) 3) (-2, -1) 4)    , 2 2,    Key : 1 Sol : 2 2 3f 2f a 0   where   x x 2 1 1 3a f 1 3     1 a 1; a 0       a 1, 0 0, 1    4. If 2 a b b c c a a b c                      then  is equal to: 1) 2 2) 3 3) 0 4) 1
5. 5. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 5 Key : 4 Sol : 2 a b b c c a a b c                     5. The variance of first 50 even natural numbers is : 1) 833 4 2) 833 3) 437 4) 437 4 Key : 2 Sol : Variance of first ‘n’ even natural numbers is equal to 2 n 1 3  2 50 1 833 3    6. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point ‘O’ on the ground is 0 45 . It flies off horizontally straight away from the point ‘O’. After one second, the elevation of the bird from ‘O’ is reduced to 0 30 . Then the speed (in m/s) of the bird is: 1)  40 2 1 2)  40 3 2 3) 20 2 4)  20 3 1
6. 6. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 6 Key : 4 Sol : P Q A B C 450 300 0 A 45 AB PB 20 mts and CQ 20      0 20 Tan30 BC 20 3 1 PQ 20 BC        distance speed 20 3 1 mts time    7. The integral 2 0 x x 1 4sin 4sin dx 2 2    equals: 1) 4 2) 2 4 4 3 3    3) 4 3 4 4) 4 3 4 3    Key : 4 Sol : 3 0 0 3 x x x 1 2sin dx 1 2sin dx 2sin 1 dx 2 2 2                        3 0 3 x x x 4cos 4cos x 2 2                      4 3 4 3    
7. 7. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 7 8. The statement  ~ p ~ q is : 1) equivalent to p q 2) equivalent to ~ p q 3) a tautology 4) a fallacy Key : 1 Sol : The statement  ~ p ~ q is P q ~q p ~ q  ~ p ~ q p q T T F F T T T F T T F F F T F T F F F F T F T T  ~ p ~ q  is equivalent to p q 9. If A is an 3 3 non-singular matrix such that 1 AA' A'A and B A A', then BB'   equals : 1) I + B 2) I 3) 1 B 4)  1 B Key : 2 Sol :    11 1 1 1 BB' A A A A     11 1 1 A AA A I. I I    
8. 8. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 8 10. The integral 1 x x 1 1 x e dx x        is equal to : 1)   1 x x x 1 e c    2) 1 x x xe c   3)   1 x x x 1 e c    4) 1 x x xe c    Key : 2 Sol : 1 x x 1 1 x e dx x        1 x x 2 1 1 x 1 e dx x            1 1 x x x x 2 1 e dx x 1 e dx x          By parts = 1 x x xe c   11. If ‘z’ is a complex number such that | z | 2, then the minimum value of 1 z 2  : 1) is equal to 5 2 2) lies in the interval (1, 2) 3) is strictly greater than 5 3 4) is strictly greater than 3 2 but less than 5 2
9. 9. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 9 Key: 2 Sol : z rcis then | z | r 2    2 2 2 21 1 z rcos r sin 2 2          2 1 r rcos 4     17 9 25 2cos which lies in , 4 4 4               1 3 5 z , 2 2 2         Minimum value 3 2 lies in (1, 2) 12. If ‘g’ is the inverse of a function ‘f’ and    5 1 f ' x , then g' x 1 x   is equal to: 1) 5 1 x 2) 4 5x 3)    5 1 1 g x 4)    5 1 g x Key : 4 Sol : If ‘g’ is the inverse of a function ‘f’, then   x f g x     f ' g x g' x 1         51 g' x 1 g x f ' g x        
10. 10. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 10 13. If                   n n 3 1 f 1 1 f 2 , 0, and f n and 1 f 1 1 f 2 1 f 3 1 f 2 1 f 3 1 f 4               =       2 2 2 K 1 1 , then K   is equal to : 1)  2) 1  3) 1 4) -1 Key : 3 Sol : Given determinant = 2 2 2 2 1 1 1 1 1 1 1 . 1 1 1         K 1  14. Let    k k k 1 f x sin x cos x k   where x R and k 1  . Then    4 6f x f x equals: 1) 1 6 2) 1 3 3) 1 4 4) 1 12 Key : 4 Sol :    4 6f x f x = 2 2 2 21 1 [1 2sin xcos x] 1 3sin xcos x 4 6      
11. 11. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 11 15. Let and  be the roots of equation 2 px qx r 0, p 0    . If p, q, r are in A.P and 1 1 4,    then the value of | |  is : 1) 61 9 2) 2 17 9 3) 34 9 4) 2 13 9 Key : 4 Sol : q p    r . p    2q p q  By given condition 1 1 4    q 4 r     2 4       16. Let A and B be two events such that       1 1 1 P A B , P A B and P A , where A 6 4 4      stands for the complement of the event A. Then the events A and B are : 1) mutually exclusive and independent 2) equally likely but not independent 3) independent but not equally likely 4) independent and equally likely
12. 12. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 12 Key : 3 Sol :   1 5 P A B 1 6 6         1 5 P A P B 4 6      3 1 5 P B 4 4 6      5 1 5 3 1 P B 6 2 6 3            3 1 1 P A .P B P A B 4 3 4      A, B are independent but not equally likely     P A P B 17. If ‘f’ and ‘g’ are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then for some c ]0, 1[ : 1)    2f ' c g' c 2)    2f ' c 3g' c 3)    f ' c g' c 4)    f ' c 2g' c Key : 4 Sol :               f ' c f 1 f 0 for c 0, 1 g' c g 1 g 0     18. Let the population of rabbits surviving at a time ‘t’ be governed by the differentiable equation     dp t 1 p t 200 dt 2   . If p(0) = 100, then p(t) equals : 1) t/2 400 300 e 2) t/2 300 200 e  3) t/2 600 500 e 4) t/2 400 300 e 
13. 13. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 13 Key : 1 Sol :   dp 1 p t 200 dt 2   This is linear d.e in ‘p’ 1 1 t t 2 2 p.e 400e c      By given condition p(0) = 100, c = -300 1 t 2 p 400 300e   19. Let ‘C’ be the circle with centre at (1, 1) and radius = 1. If ‘T’ is the circle centred at (0, y), passing through origin and touching the circle ‘C’ externally, then the radius of ‘T’ is equal to : 1) 3 2 2) 3 2 3) 1 2 4) 1 4 Key : 4 Sol :    1 2c 1, 1 c 0, y  1 2r 1, r y  1 2 1 2c c r r      2 2 1 y 1 1 y    1= 4y
14. 14. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 14 20. The area of the region described by   2 2 2 A x, y :x y 1 and y 1 x     is 1) 4 2 3   2) 4 2 3   3) 2 2 3   4) 2 2 3   Key : 1 Sol. Required area = 1 0 4 2 1 x dx 2 2 3       21. Let a, b, c and d be non-zero numbers. If the point of intersection of the line 4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then : 1) 2bc – 3ad = 0 2) 2bc + 3ad = 0 3) 3bc – 2ad = 0 4) 3bc + 2ad = 0 Key : 3 Sol. Point of intersection 2ad 2bc 5bc 4ad , 2ab 2ab         Equidistance from the both axis in fourth quadrant 3bc 2ad 0 
15. 15. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 15 22. Let PS be the median of the triangle with vertices P(2, 2), Q(6, -1) and R(7, 3). The equation of the line passing through (1, -1) and parallel to PS is 1) 4x 7y 11 0   2) 2x 9y 7 0   3) 4x 7y 3 0   4) 2x 9y 11 0   Key : 2 Sol : Mid point 13 ,1 2      Slope PS 2 9   23.  2 2x 0 sin cos x lim x  is equal to : 1) 2  2) 1 3)  4)  Key : 4 Sol :  2 2x 0 sin sin x lim x   2 2 2 2x 0 sin sin x sin x lim 1 1 sin x x         24. If  n X 4 3n 1: n N    and   Y 9 n 1 :n N ,   where N is the set of natural numbers, then X Y is equal to: 1) N 2) Y – X 3) X 4) Y
16. 16. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 16 Key : 4 Sol : X and Y are multiplies of 9 and X Y X Y Y   25. The locus of the foot of perpendicular drawn from the centre of the ellipse 2 2 x 3y 6  on any tangent to it is : 1)   22 2 2 2 x y 6x 2y   2)   22 2 2 2 x y 6x 2y   3)   22 2 2 2 x y 6x 2y   4)   22 2 2 2 x y 6x 2y   Key : 3 Sol. Given that ellipse 2 2 x y 1 6 2   C P A B Slope CP = 1 1 y x Slope of AB = 1 1 x y  = m Equation of the tangent is 2 y mx 6m 2   By above equations we get   22 2 2 2 x y 6x 2y  
17. 17. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 17 26. Three positive numbers from an increasing G.P. If the middle term in this G.P is doubled, the new numbers are in A.P. Then the common ratio of the G.P is : 1) 2 3 2) 3 2 3) 2 3 4) 2 3 Key : 4 Sol : 2 a,ar,ar G.P . 2 a ar 2ar 2   2 4ar a ar  2 r 4r 1 0   4 16 4 r 2    r 2 3  r 2 3   27. If               9 1 8 2 7 9 9 10 2 11 10 3 11 10 ..... 10 11 k 10 ,     then ‘k’ is equal to 1) 121 10 2) 441 100 3) 100 4) 110
18. 18. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 18 Key : 3 Sol :             9 1 8 2 7 9 S 10 2 11 10 3 11 10 .....10 11     ……… (1)           2 3 6 108 711 S 11.10 2 11 10 3 11 10 .....9 11 11 10      ……… (2) (2) ... (1)   29 8 7 9 101 S 10 11.10 11 10 .... 11 11 10         9 S 100 10  K 100  28. The angle between the lines whose direction cosines satisfy the equations 2 2 2 m n 0 and m nl l     is : 1) 3  2) 4  3) 6  4) 2  Key : 1 Sol : 2 2 2 l m n 0    l m n     2 2 2 m n m n 0    2mn = 0 m = 0 (or) n = 0 If m = 0 l = -(0 + n) l = -n l : m : n = -n : 0 : n = -1 : 0 : 1 similarly if n = 0 then l : m : n = -1, : 1 : 0 1 0 0 cos 2 2     3   
19. 19. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 19 29. The slope of the line touching both the parabolas 2 2 y 4x and x 32y  is : 1) 1 2 2) 3 2 3) 1 8 4) 2 3 Key : 1 Sol : Common tangent of   2/32 2 1/3 1/3 y 4ax and x 4by is a x b y ab 0     Here 4a = 4 and 4b = -32 Slope of the common tangent = 1/2 30. If x = -1 and x = 2 are extreme points of   2 f x log | x | x x    then: 1) 1 6, 2     2) 1 6, 2     3) 1 2, 2     4) 1 2, 2     Key : 3 Sol :    f ' 1 0 and f ' 2 0   Solving above equations we get 1 2, 2    
20. 20. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 20 PHYSICS 31. When a rubber-band is stretched by a distance x , it exerts a restoring force of magnitude 2 F ax bx  where a and b are constants. The work done in stretching the unstretched rubber-band by L is: 1) 2 3 1 2 2 3 aL bL      2) 2 3 aL bL 3)  2 31 2 aL bL 4) 2 3 2 3 aL bL  Key: 4 Sol:   2 3 2 0 2 3 L aL bL w Fdx ax bx dx      32. The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 1 3 10 Am  . The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is: 1) 6A 2) 30 mA 3) 60 mA 4) 3 A Key: 4 Sol: 3100 I=3 10 0.1 nI H    3I A
21. 21. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 21 33. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 Wand 1heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be: 1) 14A 2) 8A 3) 10A 4) 12A Key: 4 Sol:   1 40 15 100 5 80 5 1000 220 P IV I         2500 11.4 12A 220 A   34. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46cm. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg) 1) 6 cm 2) 16 cm 3) 22 cm 4) 38 cm Key: 2 Sol:  76 8 22x x   By trail and error method 16x 
22. 22. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 22 35. A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed  rad/s about the vertical. About the point of suspension: 1) angular momentum changes both in direction and magnitude 2) angular momentum is conversed 3) angular momentum changes in magnitude but not in direction 4) angular momentum changes in direction but not in magnitude Key: 4 Sol: Angular Momentum Changes in directions 36. The current voltage relation of diode is given by  1000 / 1 ,V T I e mA  where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring 0.01V while measuring the current of 5mA at 300 K, what will be the error in the value of current in mA? 1) 0.05 Ma 2) 0.2 mA 3) 0.02 mA 4) 0.5 mA Key: 2 Sol: I+1 = 1000 /V T e   1000V ln I 1 T   1000 V I+1 T I   31000 I= 6 10 0.01 3      = 0.2 mA
23. 23. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 23 37. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is: 1)   2 2gH n u  2) 2 2 2gH n u 3)   2 2 2gH n u  4)  2 2 2gH nu n  Key: 4 Sol: H u Let 1t is the time taken to reach max height 1 u t g  Let 2t is the time taken to reach ground from tower equation 2 2 2 1 2 H ut gt   2 2 2 4 8 2 u u gH t g    2 2 2u u gH g g    Given that 2 1t nt 2 2 2u u gH nu g g g    ,   2 2 2 2 2 1 u gH u n g g      22 2 1 1gH u n      2 2 2 2gH u n n     2 2 2gH u n n 
24. 24. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 24 38. A thin convex lens made from crown glass 3 2        has focal length f . When it is measured in two different liquids having refractive indices 4 3 and 5 3 , it has the focal lengths 1f and 2f respectively. The correct relation between the focal lengths is: 1) 1f and 2f both become negative 2) 1 2f f f  3) 1f f and 2f becomes negative 4) 2f f and 1f becomes negative Key: 3 Sol: From   1 2 1 f R         1 1 f R   3 2/  f R For medium 4 3/  3 1 2 12 1 4 3 4f / R R             1 4f R for medium 2 1 3 2 2 5 3 1 5 3 / M / f / R          1 5R   2 5f R 
25. 25. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 25 39. A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 4 3 10 /V m , the charge density of the positive plate will be close to: 1) 4 2 6 10 /C m 2) 7 2 6 10 /C m  3) 7 2 3 10 /C m  4) 4 2 3 10 /C m Key: 2 Sol: 4 0 3 10 K     12 4 2.2 8.8 10 3 10       7 6 10    40. In the circuit shown here, the point ‘C’ is kept connected to point ‘A’ till the current flowing through the circuit becomes constant. Afterward, suddenly, point ‘C’ is disconnected from point ‘A’ and connected to point ‘B’at time t=0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to: L R B A C 1) 1 e e  2) 1 e e 3) 1 4) 1 Key: 4 Sol: 0R LV V  always for 0t 
26. 26. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 26 41. Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30makes the two beams appear equally bright. If the initial intensities of two beams are AI and BI respectively, then A B I I equals: 1) 1 3 2) 3 3) 3 2 4) 1 Key: 1 Sol: 1 1 2 2 30 60A B A BI I I cos I cos   2 2 cos 60 1 cos 30 3 A B I I    42. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities 1d and 2d are filled in the tube. Each liquid subtends 90 angle at centre. Radius joining their interface makes an angle  with vertical. Ratio 1 2 d d is:  1d 2d 1) 1 sin 1 cos     2) 1 sin 1 sin     3) 1 cos 1 cos     4) 1 tan 1 tan    
27. 27. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 27 Key: 4 Sol: A BP P 1 1 2 2h d g h d g     1 2cos sin cos sinR d R d       1 2 cos sin 1 tan cos sin 1 tan d d             A B R 2h 2d 1d 1h R  43. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100 C is: (For steel Young’s modulus is 11 2 2 10 N m  and coefficient of thermal expansion is 5 1 1.1 10 K   ) 1) 6 2.2 10 Pa 2) 8 2.2 10 Pa 3) 9 2.2 10 Pa 4) 7 2.2 10 Pa Key: 2 Sol: Pressure 11 5 8 α 2 10 1.1 10 100 2.2 10y           
28. 28. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 28 44. A block of mass m is placed on a surface with a vertical cross section given by 3 6 x y  . If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is: 1) 1 2 m 2) 1 6 m 3) 2 3 m 4) 1 3 m Key: 2 Sol: At the instant of just sliding  H=? tan  0.5 dy dx  2 1 2 x x    1 ' ' 1 6 H y at x m  
29. 29. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 29 45. Three rods of Copper, Brass and Steel are welded together to form a Y – shaped structure. Area of cross – section of each rod 2 4cm . End of copper rod is maintained at 100 C where as ends of brass and steel are kept at 0 C . Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivites of copper, brass and steel are 0.92,0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is: 1) 6.0 cal/s 2) 1.2 cal/s 3) 2.4 cal/s 4) 4.8 cal/s Key: 4 Sol: Let T is the junction temperature then cu Brass steelH H H     100cu B S cu B S K A T K A T K AT l l l    ,   0 96 0 26 0 12 100 46 13 12 . . . T T T   200 5T 40T C  Heat flow per sec through copper  100KA T l    0 96 4 60 46 .   4 8. cal/sec 0 C0 C 100 C Brass Copper Steell=13 cm k=0.26 l=12 cm k=0.12 l=46 k=0.96 T
30. 30. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 30 46. A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release? m R m 1) g 2) 2 3 g 3) 2 g 4) 5 6 g Key: 3 Sol: mg T ma  2 2 . a TR mR mR T ma R     2ma mg  or / 2a g
31. 31. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 31 47. Match List-I (Electromagnetic wave type) with List – II (Its association/application) and select the correct option from the choices given below the lists: List-I List-II (a) Infrared waves (i) To treat muscular strain (b) Radio waves (ii) For broadcasting (c) X-rays (iii) To detect fracture of bones (d) Ultra violet rays (iv) Absorbed by the ozone layer of the atmosphere (a) (b) (c) (d) 1) (i) (ii) (iii) (iv) 2) (iv) (iii) (ii) (i) 3) (i) (ii) (iv) (iii) 4) (iii) (ii) (i) (iv) Key: 1 Sol: Conceptual 48. The radiation corresponding to 3 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 4 3 10 T  . If the radius of largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to: 1) 1.6 eV 2) 1.8 eV 3) 1.1 eV 4) 0.8 eV
32. 32. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 32 Key: 3 Sol: In magnetic field ,  2 KE m r Bq    28 19 42 2 2 31 19 9 10 1.6 10 10 1 2 2 9 10 1.6 10 B q r KE m                0.8ev 1 1 13.6 1.89 4 9 E eV         0 1.89 0.8 1.1W E KE eV     49. During the propagation of electromagnetic waves in a medium: 1) Both electric and magnetic energy densities are zero 2) Electric energy density is double of the magnetic energy density 3) Electric energy density is half of the magnetic energy density 4) Electric energy density is equal to the magnetic energy density Key: 4 Sol: Conceptual 50. A green light is incident from the water the air – water interface at the critical angle   . Select the correct statement. 1) The entire spectrum of visible light will come out of the water at various angles to the normal 2) The entire spectrum of visible light will come out of the water at an angle of 90 to the normal 3) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium 4) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium
33. 33. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 33 Key: 3 Sol: With increase in frequency, critical angle c decreases i.e., present angle of incidence in water will be more than respective critical angles of high frequencies radiation. Hence high frequency radiations will get TIR and low frequencies radiations will come out in to air 51. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is: 1)  1 1 2 2 2 GM R  2) GM R 3) 2 2 GM R 4)  1 2 2 GM R  Key: 1 Sol: V R R 2 2 2 2 2 mv 1 1 1 2 2 42 Gm R R R           2 1 v 2 2 1 4 Gm R    1 v 2 2 1 2 Gm R  
34. 34. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 34 52. A particle moves with simple harmonic motion in a straight line. In first s , after starting from rest it travels a distance a, and in next s it travels 2a, in same direction, then: 1) time period of oscillations is 6 2) amplitude of motion is 3a 3) time period of oscillations is 8 4) amplitude of motion is 4a Key: 1 Sol: a a a  0 2 6 T   53. A conductor lies along the z – axis at 1.5 1.5z m   and carries a fixed current of 10.0 A in ˆza direction (see figure). For a field 4 0.2 ˆ3.0 10 x yB e a T     , find the power required to move the conductor at constant speed to 2.0x m , 0y m in 3 5 10 s  . Assume parallel motion along the x  axis. x yB I 2.0 1.5 1.5 1) 29.7 W 2) 1.57 W 3) 2.97 W 4) 14.85 W
35. 35. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 35 Key: 3 Sol: 4 0.2 Instant 3 2 3 10 3 5 10 x P B V e          1 0.212 10 5 x P e    2.97range pdx P W dx     54. The forward biased diode connection is: 1) 2V 2V 2) 2V2V 3) 3V3V 4) 2V 4V Key: 2 Sol: Conceptual 55. Hydrogen 1 1 H , Deuterium 2 1 H , singly ionized Helium  4 2 He  and doubly ionized lithium  6 3 Li  all have one electron around the nucleus. Consider an electron transition from 2n  to 1n  . If the wavelengths of emitted radiation are 1 2 3, ,   and 4 respectively then approximately which one of the following is correct? 1) 1 2 3 42 3 4      2) 1 2 3 44 2 2      3) 1 2 3 42 2      4) 1 2 3 44 9     
36. 36. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 36 Key: 4 Sol: 2 2 1 1 1 1 4 Rz z           1 2  as 1z  for both 1 1H and 2 1 H 2 31 1 32 3 1 4 4 z z         2 1 4 1 42 4 1 9 9 z z          1 2 3 44 9      56. On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If ,r R and the surface tension of water is T, value of r just before bubbles detach is: (density of water is w ) R 2r 1) 2 3 wg R T  2) 2 3 wg R T  3) 2 6 wg R T  4) 2 wg R T 
37. 37. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 37 Key: 3(This is an ambigious Question, so expected key is 3) Sol: sinTdl   buoyancy force on bubble R  r r  2 r T r R   34 3 wR g  4 2 2 3 wR g r T   2 2 3 wg r R T    57. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s. 1) 4 2) 12 3) 8 4) 6 Key: 4 Sol: From 2 340 100 4 4 85 10 V n HZ l       Possible frequencies in closed pipe is odd harmonics only So 100 300 500 700 900HZ, HZ, HZ, HZ, HZ and 1100 1300 1500HZ, HZ, HZ..... etc below 1250HZ , 6 natural oscillations are possible
38. 38. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 38 58. Assume that an electric field 2ˆ30E x i  exists in space. Then the potential difference A OV V , where OV is the potential at the origin and AV the potential at 2x m is: 1) 80 J 2) 120 J 3) – 120 J 4)– 80 J Key: 4 Sol: 0 2 2 0 0 0 30 80 AV A A V dv Edx V V x dx J          59. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? 1) A screw guage having 50 divisions in the circular scale and pitch as 1mm 2) A meter scale 3) A vernier calipers where the 10 divisions in vernier scale matches with 9 division in main scale has 10 divisions in 1cm 4) A screw guage having 100 divisions in the circular scale and pitch as 1mm
39. 39. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 39 Key: 3 Sol: Least count of vernier 1 10 mm 60. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement: P V A 400K C B 800 K 600 K 1) The change in internal energy in the process BC is 500R 2) The change in internal energy in whole cyclic process is 250 R 3) The change in internal energy in the process CA is 700 R 4) The change in internal energy in the process AB is 350R Key: 1 Sol: 1) du for isochoric process VnC dT   3 1 400 600 2 R R   2) for Process BC    1 2 200 500 1 1 7 / 5 nR T T R du R r         3) for whole process 0du 
40. 40. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 40 CHEMSITRY 61. Which one is classified as a condensation polymer? 1) Acrylonitrile 2) Dacron 3) Neoprene 4) Teflon Key: 2 Sol: Dacron is the condensation polymer n Neoprene Add polymers Teflon    62. Which one of the following properties is not shown by NO? 1) It’s bond order is 2.5 2) It is diamagnetic in gaseous state 3) It is a neutral oxide 4) It combines with oxygen to form nitrogen dioxide Key: 2 Sol: NO is paramagnetic
41. 41. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 41 63. Sodium phenoxide when heated with 2CO under pressure at 0 125 C yields a product which on acetylation produces C. 2CO 0 125 5Atm  2 H AC O B C   ONa The major product C would be: 1) 3OCOCH COOH 2) 3OCOCH COOH 3) OH 3COCH 3COCH 4) OH 3COOCH Key: 2
42. 42. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 42 Sol: ONa 0125 C 5atm2 CO  OH COOH H AC O2   COOH 3 OCOCH (B) (C) Aspirin 64. Given below are the half-cell reactions: 2 0 2 ; 1.18Mn e Mn E V       3 2 0 2 ; 1.51Mn e Mn E V       The 0 E for 2 3 3 2Mn Mn Mn    will be: 1) - 0.33 V; the reaction will occur 2) – 2.69 V; the reaction will not occur 3) – 2.69 V; the reaction will occur 4) – 0.33 V; the reaction will not occur Key: 2 Sol: 0 1.18 1.51E    = –2.69 V
43. 43. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 43 65. For complete combustion of ethanol.        2 5 2 2 23 2 3C H OH l O g CO g H O l   , the amount of heat produced as measured in bomb calorimeter, is 1364.47 1 kJmol at 0 25 C . Assuming ideally the Enthalpy of combustion, cH , for the reaction will be:  1 8.314R kJmol  1) 1 1350.50kJmol  2) 1 1366.95kJmol  3) 1 1361.95kJmol  4) 1 1460.50kJmol  Key: 2 Sol: H U nRT      3 1364 1 8.314 298 10      66. For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of 10 M sulphuric acid. The unreacted acid required 20 mL of 10 M sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is: 1) 5% 2) 6% 3) 10% 4) 3%
44. 44. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 44 Key: 3 Sol: Meq of 2 4 1 H SO 60 2 12 10     Meq of base used for back titration 1 20 2 10    = Meq of acid left Meq of acid reacted with liberated Ammonia = 12-2 = 10 meq = meq of 3 NH Millimoles of 3 NH liberated = 10 [ n-factor = 1] Moles of 3 3 NH 10 10   Moles “N” 3 10 10   Wt of “N” 3 10 10 14    % of “N” in the organic compound 3 10 10 14 100 10% 1.4       67. The major organic compound formed by the reaction of 1, 1, 1 – trichloroethane with silver powder is: 1) 2 – Butene 2) Acetylene 3) Ethene 4) 2 – Butyne Key: 4
45. 45. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 45 Sol: 3 2CH C Cl  Cl Cl 3 3 6Ag CH C C CH     2-Butyne + 6 AgCl 68. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is: 1) 3 : 16 2) 1 : 4 3) 7 : 32 4) 1 : 8 Key: 3 Sol: Ratio of the molecules = 1 4 : 32 28 = 7: 32 69. The metal that cannot be obtained by electrolysis of an aqueous solution of its salt is: 1) Cr 2) Ag 3) Ca 4) Cu Key: 3 Sol: Ca cannot be obtained by the electrolysis of aqueous solution of its salt.
46. 46. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 46 70. The equivalent conductance of NaCl at concentration C and at infinite dilution are C and , respectively. The correct relationship between C and is given as (where the constant B is positive) 1)  C B C   2)  C B C   3)  C B C   4)  C B C   Key: 4 Sol:  C B C   71. The correct set of four quantum numbers for the valence electrons of rubidium atom (Z=37) is: 1) 1 5,0,1, 2  2) 1 5,0,0, 2  3) 1 5,1,0, 2  4) 1 5,1,1, 2  Key: 2 Sol: 1 5s 72. Consider separate solutions of 0.500 M  2 5 ,C H OH aq    3 4 2 0.100 ,M Mg PO aq  0.250M KBr aq and  3 40.125M Na PO aq at 0 25 C . Which statement is true about these solutions, assuming all salts to be strong electrolytes? 1)  2 50.500M C H OH aq has the highest osmotic pressure 2) They all have the same osmotic pressure 3)    3 4 2 0.100M Mg PO aq has the highest osmotic pressure 4)  3 40.125M Na PO aq has the highest osmotic pressure
47. 47. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 47 Key: 2 Sol: iCST  73. The most suitable reagent for the conversion of 2R CH OH R CHO    is: 1) PCC (Pyridinium Chlorochromate) 2) 4KMnO 3) 2 2 7K Cr O 4) 3CrO Key: 1 Sol: Most suitable reagent for the conversion of 0 1 alcohol to aldehyde is “PCC” 74. CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct? 1) 3Cs Cl r r a   2) 3Cs Cl r r a   3) 3 2Cs Cl a r r   4) 3 2Cs Cl r r a   Key: 4 Sol: c a3a 2(r r )  c a 3a r r 2  
48. 48. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 48 75. In which of the following reactions 2 2H O acts as reducing agent? (a) 2 2 22 2 2H O H e H O     (b) 2 2 22 2H O e O H     (c) 2 2 2 2H O e OH    (d) 2 2 2 22 2 2H O OH e O H O      1) (b), (d) 2) (a), (b) 3) (c), (d) 4) (a), (c) Key: 1 Sol: While acting as reducing agent it gives up electrons and release oxygen gas. 76. For which of the following molecule significant 0  ? a) Cl Cl b) CN CN c) OH OH d) SH SH 1) (c) and (d) 2) Only (a) 3) (a) and (b) 4) Only (c) Key: 1
49. 49. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 49 Sol: Cl Cl 0  CN CN 0 ; 0  0 ; O H O H S H S H But dipole moment of "SH" SH is less than dipolemoment of OH OH , but is significant 77. On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is: 1) an alkyl isocyanide 2) an alkanol 3) an alkanediol 4) an alkyl cyanide Key: 1 Sol: 2 3 R NH CHCl KOH R NC     alkylisocyanide
50. 50. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 50 78. In 2NS reactions, the correct order of reactivity for the following compounds:  3 3 2 3 2 , ,CH Cl CH CH Cl CH CHCl and  3 3 CH CCl is 1)    3 3 2 3 32 3 CH CHCl CH CH Cl CH Cl CH CCl   2)    3 3 3 2 32 3 CH Cl CH CHCl CH CH Cl CH CCl   3)    3 3 2 3 32 3 CH Cl CH CH Cl CH CHCl CH CCl   4)    3 2 3 3 32 3 CH CH Cl CH Cl CH CHCl CH CCl   Key: 3 Sol:    3 3 2 3 32 3 CH Cl CH CH Cl CH CHCl CH CCl   79. The octahedral complex of a metal ion 3 M  with four monodentate ligands 1 2 3, ,L L L and 4L absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is: 1) 1 2 4 3L L L L   2) 4 3 2 1L L L L   3) 1 3 2 4L L L L   4) 3 2 4 1L L L L   Key: 3 Sol: Order of wavelength of colours Red >Yellow > Green > Blue
51. 51. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 51 80. The equation which is balanced and represents the correct product(s) is: 1)  4 2 2 44 4CuSO KCN K Cu CN K SO     2) 2 22 2Li O KCl LiCl K O   3)   2 3 45 5 5CoCl NH H Co NH Cl             4)       2 24 2 26 6excess NaOH Mg H O EDTA Mg EDTA H O            Key: 3 Sol:   2 3 45 CoCl NH 5H Co 5NH Cl             81. In the reaction, 54 . 3 ,PClLiAlH Alc KOH CH COOH A B C   the product C is: 1) Acetyl chloride 2) Acetaldehyde 3) Acetylene 4) Ethylene Key: 4 Sol: 54 3 3 2 3 2 2 2 PClLiAlH Alc.KOH CH C OH CH CH OH CH CH Cl CH CH          O 82. The correct statement for the molecule, 3CsI , is: 1) it contains ,Cs I  and lattice 2I molecule 2) it is a covalent molecule 3) it contains Cs and 3I ions 4) it contains 3 Cs  and I  ions
52. 52. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 52 Key: 3 Sol: 3 CsI contains Cs and 3 I ions 83. For the reaction      2 2 3 1 2g g g SO O SO  , if   x p cK K RT where the symbols have usual meaning then the value of x is: (assuming ideality) 1) 1 2) 1 3) 1 2  4) 1 2 Key: 3 Sol: n P CK K (RT)  3 1 n 1 2 2     84. For the non-stoichiometre reaction 2 ,A B C D   the following kinetic data were obtained in three separate experiments, all at 298 K. Initial Concentration (A) Initial Concentration (B) Initial rate of formation of C  mol L S  0.1M 0.1 M 0.2 M 0.1 M 0.2 M 0.1 M 3 1.2 10  3 1.2 10  3 2.4 10  The rate law for the formation of ‘C’ is? 1)   dc k A dt  2)    dc k A B dt  3)     2dc k A B dt  4)     2dc k A B dt 
53. 53. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 53 Key: 1 Sol: m m r K(A) (B) 85. Resistance of 0.2 M solution of an electrolyte is 50. The specific conductance of the solution is 1 1.4 S m . The resistance of 0.5 M solution of the same electrolyte is 280. The molar conductivity of 0.5 M solution of the electrolyte in 2 1 S m mol is: 1) 2 5 10 2) 4 5 10  3) 3 5 10  4) 3 5 10 Key: 2 Sol: l R a   1 l 50 1.4 a  70 a l  280 (70)  4  K = 0.25 3 410 0.25 5 10 0.5       86. Among the following oxoacids, the correct decreasing order of acid strength is: 1) 2 4 3HClO HClO HClO HOCl   2) 2 3 4HOCl HClO HClO HClO   3) 4 2 3HClO HOCl HClO HClO   4) 4 3 2HClO HClO HClO HOCl   Key: 4 Sol: With increase in unprotonated oxygen atoms acidic strength increases
54. 54. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 54 87. Which one of the following bases is not present in DNA? 1) Thymine 2) Quinoline 3) Adenine 4) Cytosine Key: 2 Sol: Qunoline is the base which is not present in DNA 88. Considering the basic strength of amines in aqueous solution, which one has the smallest bpK value? 1) 6 5 2C H NH 2)  3 2 CH NH 3) 3 2CH NH 4)  3 3 CH N Key: 2 Sol:  3 2 CH NH Stronger base in aqueous medium than 3 Me N . So its pKb value is smallest 89. If Z is a compressibility factor, vander Waals equation at low pressure can be written as: 1) 1 Pb Z RT   2) 1 RT Z Pb   3) 1 a Z VRT   4) 1 Pb Z RT  
55. 55. 2014 JEE-MAIN Q. PAPER WITH SOLUTIONS #CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD EMAIL: ICONCOHYD@SRICHAITANYACOLLEGE.NET, WEB: WWW.SRICHAITANYA.NET 55 Key: 3 Sol: 2 a P V RT V        a PV RT V   a Z 1 RTV   90. Which series of reactions correctly represents chemical relations related to iron and its compound? 1) 0 0 2 , ,600 ,700 3 4 O heat CO C CO C Fe Fe O FeO Fe   2)  2 4 2 4 2dil , 4 2 4 3 H SO H SO O heat Fe FeSO Fe SO Fe   3) 2 2 4, dil 4 O heat H SO heat Fe FeO FeSO Fe   4) 2 , , 3 2 Cl heat heat air Zn Fe FeCl FeCl Fe   Key: 1 Sol: O heat CO CO2 600 7003 4 Fe Fe O FeO Fe  