iit jee model paper

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keys and soluton of iit jee model test 2

Transcript of iit jee model paper

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    Max Marks:360KEY & HINTS

    PHYSICS1) 1 2) 2 3) 2 4) 1 5) 1 6) 27) 2 8) 2 9) 4 10) 2 11) 4 12) 113) 1 14) 3 15) 3 16) 3 17) 4 18) 319) 2 20) 3 21) 2 22) 3 23) 1 24) 325) 2 26) 1 27) 4 28) 4 29) 3 30) 2

    CHEMISTRY31) 1 32) 1 33) 1 34) 1 35) 3 36) 437) 2 38) 1 39) 3 40) 1 41) 3 42) 143) 3 44) 3 45) 4 46) 2 47) 1 48) 349) 1 50) 4 51) 3 52) 1 53) 4 54) 455) 4 56) 3 57) 1 58) 3 59) 3 60) 3

    MATHEMATICS61) 2 62) 1 63) 3 64) 2 65) 1 66) 467) 4 68) 3 69) 4 70) 3 71) 1 72) 373) 2 74) 3 75) 3 76) 4 77) 3 78) 379) 4 80) 3 81) 3 82) 1 83) 3 84) 485) 2 86) 1 87) 3 88) 1 89) 1 90) 4

    HINTSPHYSICS

    1. VS=4, MS=2.7= MS+VS x LC=2.7 + 0.04=2.74 ]

    3. Spring force does not change instantaneouslyThus for 1 1 0;m a aFor 2m 2 2pSF m a .(i) instantaneously after 2F is withdrawn

    Initially 2 2 0pSF F m a

    2a

    pSF

    2 2 0Fsp F m a ii

    From (i) and (ii) 22 02

    Fa am

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    4. WD k

    21 21 .6 02 6m mmgl v vm m

    1 273v v gl

    5. sin60tan30 tan60o

    oo

    eV

    2tan30 tan 30tan60o

    ooe

    6.1 2 22cme eV gh 22 1 2

    max 2 4cm e e hVh g

    8.1

    2 rad/hr1

    22 rad/hr8

    2

    1 12 3

    T RT R

    2

    1

    R 4R 4

    2R 4x10 km41

    12 RV 2 x10 km/hr1h

    422 2 RV x10 km/hr8h

    At closest separation4

    rel4

    V toline joining x10 km/hr rad/hr.lengthof line journing 3x10 km 3

    ]

    9.When source is at origin, the observer receives the sound emitted by the source,when it was at P.Such that 50t 1cos 200t 4

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    0s

    v (v) 90x200v 50v v cos 200 4

    V=96 Hz

    10. 3 3L2 2L

    0y A sinkxsin t 2 3

    02 L 5A sin x sin t2L 2 6

    0

    5A sin t 6

    11. Comment for discussion. In the ideal case that we normally consider each

    collision transfers twice the magnitude of its normal momentum. On the faceEFGH, it transfers only half of that.

    13. Let x mole of the gas dissociate at 1000 KNo. of mole of diatomic gas molecule =1-xNo. of moles of monatomic gas molecules=2 x XEnergy of diatomic molecules = energy of monatomic molecule 5 31 x RT 2x x RT2 2 x 5/11Now new no.of moles = (1-x)+2x=1+x=(16/11)P= nRTVPressure initially at 300K= i 300RP V

    Pressure finally at 1000K = f 1 x R x1000 16 RP 100V 11 V ]

    15. 13 3 27 1x x 3x+39=-x+27x=-3

    a b27SoV V (x 13)17

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    16. iE A 2

    2kr iE R r kiE r

    17. 4 4 1I' I4 G 4 G 5 (given) where G=164 3I" I4 163

    1I" I13

    or 1 5I" 51' I '13 13 i 4xG 4x i G 16W5 5

    In second case (i-i) x 16= 4 i '3 eq2 4 4R 2 4 3

    i '4i 4i ' 3 13i ' 124i i ' i3 13

    1 1i i ' i x0.65mA 0.05mA13 13 ]19. Introducing two equal and opposite current 1I and also 2I between A & C.

    Force on ABCA closed loop zeroForce on ADCA closed loop zeroForce on extra 1I & 2I

    1 2F I I lB=IlB20. 2 21 1Li C2 2

    128

    3C 9x10 12x3i 12 10L 2.5x10 5

    4i 7.2x10 A ]

    22. maxV V 2 2WA

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    24. 2eV

    1

    hc 8eV 2 1T 2TIf 1l is the wavelength corresponding to maximum intensity at 1 2 2T & T at T ;Then 2 1 /2 (by weins displacement Law)

    2 1

    hc 2hc 16ev 2ev max hcK.E. 14ev

    26. longest minEfor Balmer longest is for n 2 3 for Lyman longest is for n 1 2 longest Balmer> longest Lyman

    CHEMISTRY:31. 3 2 2 22 2 2AgClO Cl AgCl ClO O 50. The nitrogen has 7 electrons while the oxygen has 8 electron. The MO electronic

    configuration of NO molecule is 2 * 2 2 * 2 2 2 2 * 11 . 1 . 2 . 2 . 2 . 2 . 2 . 2z x y ys s s s p P P P . Bond order is 2.5. When NO molecule is converted in to NO ion, the electron in

    * MO is removed. Therefore, bond order increases, 2.5 to 3, bond lengthdecreases and becomes diamagnetic. The N-O bond length in NO is 106 pm.

    MATHEMATICS61) Conceptual

    62)

    63)

    64) Since x, y, z are in GP,

    Hence,

    \

    are in AP.

    are in HP.

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    65)

    66) Conceptual

    67)

    68) =

    69)0 1 1 0 0 0

    | | 0 1 00 1

    A a b c d c d cdab cd a b ab

    70)

    71) Reflexive but not Symmetric

    72)

    No. of non negative inetegral solutions

    Total no. of solutions = 16 x 21 = 336

    73) a = p(A getting 6)

    b = p(B getting 7)

    74)

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    75) Adjust the series and apply the formula

    76)

    The tangent at

    ----- (1)

    The normal at

    ----- (2)

    (1), (2) are identical ;

    Eliminating

    77)

    78)

    79)

    80) Given that

    Differentiable with respective x,

    81)

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    82)

    83) For continuity,

    This is possible only when

    For to exist ; f(x) will not be differentiable if ;

    84) Equation of pair of tangents,

    angle between them,

    85) Passing through

    86) Common tangents to and are

    87) Slope of tangent = 1/2 and slope of PS = 4/3

    89) Let the three given points lie on the line , where l, m and n are constants, them

    ; ; For

    i.e are roots ; Them

    90) ; ;

    ; ;

    Required line passing through .

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