CONCEPT RECAPITULATION TEST - III [CONCEPT RECAPITULATION TEST - III]

other 12 Pages
BSP

Contributed by

Bishnu Singh Pau
Loading
  • AITS-CRT-III-PCM(S)-JEE(Main)/14
    FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
    website: www.fiitjee.com
    1
    ANSWERS, HINTS & SOLUTIONS
    CRT –III
    (Main)
    Q. No. PHYSICS CHEMISTRY MATHEMATICS
    1.
    A B A
    2.
    A D D
    3.
    C A B
    4.
    A A C
    5.
    C D A
    6.
    C A C
    7.
    A A D
    8.
    D B C
    9.
    B A B
    10.
    C A B
    11.
    A B C
    12.
    D C B
    13.
    D D A
    14.
    C B B
    15.
    C C C
    16.
    A D B
    17.
    B C B
    18.
    B C A
    19.
    B A C
    20.
    D B A
    21.
    A C C
    22.
    C D C
    23.
    C D D
    24.
    A D A
    25.
    C C B
    26.
    C C C
    27.
    A D C
    28.
    C D D
    29.
    A C B
    30.
    C B B
    ALL INDIA TEST SERIES
    FIITJEE
    JEE (Main), 2014
    From Classroom/Integrated School Programs
    7 in Top 20, 23 in Top 100, 54 in Top 300, 106 in Top 500 All India Ranks & 2314 Students
    from Classroom /Integrated School Programs & 3723 Students from All Programs have been Awarded a Rank in JEE (Advanced), 2013

    Page 1

  • AITS-CRT-III-PCM(S)-JEE(Main)/14
    FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
    website: www.fiitjee.com
    2
    P
    P
    h
    h
    y
    y
    s
    s
    i
    i
    c
    c
    s
    s PART – I
    1. V = Es ;
    E =
    2sm
    qt
    2
    2
    2s m
    V
    qt
    = 11.375 Volt
    .2
    '
    sin30 sin60
    ' 3
    ' 'cos30 vcos30
    cos60 (v ' ') cos60 (2v) v
    = v/ =
    40 3
    400
    0.1 3
    rad/s
    30
    120°
    v
    3. for x<0
    It will oscillate like SHM
    Time period
    1
    m
    T
    k
    For x>0
    It will perform periodic motion under constant force
    2
    2
    2E
    T 2
    mg
    Hence time period
    1 2
    T T T
    2
    m 2E
    T 2
    k
    mg
    4
    2
    0
    0
    mv
    ev B
    R
    0
    mv
    B
    eR
    2 2
    a R b R
    2 2
    b a
    R
    2b
    0
    2 2
    2bmv
    B
    (b a )e

    Page 2

  • AITS-CRT-III-PCM(S)-JEE(Main)/14
    FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
    website: www.fiitjee.com
    3
    5. Here the total time is given T
    Time spent m the region is
    ( 2 )
    Hence total time spent
    2
    T
    2
    2
    v
    2
    v
    6. Torque of Internal force is always zero. All other quantities will vary.
    8. Static friction depends on tendency.
    9.
    2 2
    2
    ML 3ML
    sin 60 2
    12 24
    10. When impulse of external force is zero.
    11. Balance forces.
    N
    R
    mg
    m
    2
    L cos
    N
    (
    )
    12.
    I
    I
    13. y(x, t) =
    2
    a
    x
    t
    b
    e
    v = b
    15. 2L = m (
    1
    /2) =
    m T
    2f 9
    L = n (
    2
    /2) =
    n T
    2f 4
    m = 3n = 3 1 = 3
    16. cos = 1 = 0 L = 1/C
    17. f
    B
    = f f
    0
    =
    0
    0 0
    0
    v v sin30
    f f
    v v sin30
    =
    0 0
    0
    2v f
    2v v

    Page 3

  • AITS-CRT-III-PCM(S)-JEE(Main)/14
    FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
    website: www.fiitjee.com
    4
    18.
    2
    1
    mv h
    2
    19. = D/d, is minimum for violet.
    20.
    PG PC CG
    ˆ ˆ
    V V V 2xj 2i
    Hence, for the paritcle,
    V
    x
    = 2m/s and V
    y
    = 2x = 4t
    a
    x
    = 0 a
    y
    = 4m/s
    2
    x = 2t, y = 2t
    2
    a
    Total
    = 4m/s
    2
    (a) (along Y-axis)
    and trajectory
    2 2
    x x
    y 2.
    4 2
    when
    N
    a a
    , a cos = a sin = 45º
    So, V
    x
    = V
    y
    = 2 V =
    2 2
    2
    C
    o
    N
    V 4 2
    R 2 2
    a 4.cos45
    m =
    4
    m
    2

    Page 4

  • AITS-CRT-III-PCM(S)-JEE(Main)/14
    FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
    website: www.fiitjee.com
    5
    C
    C
    h
    h
    e
    e
    m
    m
    i
    i
    s
    s
    t
    t
    r
    r
    y
    y PART – II
    2.
    O
    O
    H
    
    O
    OH
    O
    OH
    O
    OH
    H
    O
    OH
    Tautomerise
    
    
    O
    OH
    3.
    D
    
    H
    D
    H
    D
    H
    D
    3
    D O
    D
    D
    D
    D
    D
    D
    4. Talc has empirical formula Mg
    2
    (Si
    2
    O
    5
    )
    2
    Mg(OH)
    2
    which
    2n
    2 5
    n
    Si O
    units of two dimensional sheet
    silicates.
    5.
    O
    H
    O
    H
    97
    o
    It has half open book like structure having dihedral angle 97
    o
    6.
    o o 1
    r r 2 4
    H 2 46.1 187.8 2 285.8 H N H kJ / mol
    = - 241.0 kJ/mol
    -1
    o
    2 4
    r
    H N H 50.6 kJ / mol
    Decomposition enthalpy = - 50.6 kJ/mol.
    9. Radius ratio for MgO is = 0.464 which reflect the coordination number 6 and radius ratio for MgS
    is 0.353. Which reflect the coordination number 4.

    Page 5

  • AITS-CRT-III-PCM(S)-JEE(Main)/14
    FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
    website: www.fiitjee.com
    6
    10.
    2 3 4
    H O ClO aq ClO aq 2H aq 2e
    
    3 2 2
    2H aq ClO aq 2e ClO aq H O
    
    3 2 4
    2ClO aq ClO aq ClO aq
    
    o
    cell
    E 0.33 0.36 0.03
    o
    cell
    RT
    E lnK
    nF
    (at equilibrium)
    0.059
    0.03 logK
    2
    logK = - 1
    K = 0.1
    3 4 2
    2ClO ClO ClO
    Initially .1 0 0
    equilibrium .1 2x x x
    
    
    2
    2
    x
    K
    .1 2x
    2
    2
    1 x
    x 0.019 M
    10
    .1 2x
    12. Probability of finding an electron is zero, i.e.
    2
    0
    .
    2
    3/2
    2 /2
    0
    1 1
    1 8 12 e 0
    a
    16 4
    2
    1 8 12 0
    On solving for we get
    = 1, 6, 2 and
    0
    2r
    a
    For = 1;
    0
    0
    a
    2r
    1 r
    a 2
    0
    0
    2r
    2; 2 r a
    a
    0
    0
    2r
    6; 6 r 3a
    a
    13. n-factor of H
    3
    PO
    2
    is 4/3 for the reaction.
    14.
    b b
    T K m i
    2.08 = 0.52 × 1 × i
    i = 4
    This implies that the salt on dissociation gives 4 ions. Thus, the salt that gives four ions is
    K
    3
    [Fe(CN)
    6
    ].

    Page 6

  • AITS-CRT-III-PCM(S)-JEE(Main)/14
    FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
    website: www.fiitjee.com
    7
    15.
    D
    H
    O
    CH
    3
    H
    
    D
    H
    O
    +
    CH
    3
    H
    
    OH
    H
    CH
    3
    D
    -D shift
    OH
    D
    CH
    3
    H
    H
    OH
    D
    CH
    3
    16. O
    2
    SeO
    
    O
    O
    OH
    Bezilic acid rearrangement
    
    OH
    COO
    B
    C
    18. It is a non-degradable pollutant.
    19. Colloidal silver bromide solution is used in photography.
    20. Stability of carbonates increases down the group. Carbonates of alkali metals are more stable
    than those of alkali metals.
    21.
    2px
    and
    2py
    have one nodal plane each. In 1s there is no nodal plane and in case of
    *
    2px
    there
    are two nodal planes.
    22. Le-Chatellier’s principle.
    25.
    COOH
    2
    SOCl
    
    COCl
    3
    AlCl /
    O
    Zn/Hg
    HCl
    26.
    4
    NH Cl Nesseler's reagent Brown colour of iodide
    million's base.
    

    Page 7

  • AITS-CRT-III-PCM(S)-JEE(Main)/14
    FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
    website: www.fiitjee.com
    8
    M
    M
    a
    a
    t
    t
    h
    h
    e
    e
    m
    m
    a
    a
    t
    t
    i
    i
    c
    c
    s
    s PART – III
    1. Using T = S
    1
    take mid–point (h, k) then equation of PQ is
    xh + yk (h
    2
    + k
    2
    ) = 0 ….. (1)
    mx – y + 2m = 0 ….. (2)
    Compare (1) and (2) eliminate m
    2. As f(x) is continuous decreasing function on (0, ), so 0 < f(8x) < f(6x) < f(4x)
    So
    x x x
    f 6x f 4x
    lim 1 lim lim
    f 8x f 8x
     
    By squeeze play theorem
    x
    f 6x
    lim 1
    f 8x
    
    3. The given lines will be parallel to lines ax
    2
    – 6xy + y
    2
    = 0
    So,
    2
    y y
    6 a 0
    x x
    m + m
    2
    = 6 ….. (1)
    Which gives m = –3 or 2
    and mm
    2
    = a ….. (2)
    From (2) a = –27 or 8
    Hence sum of all possible value of a = –19
    4.
    2
    x
    4
    lim
    x
    2
    e lng 2 x lng 2
    0
    form
    0
    x 0
    2
    g' 2
    lim
    1
    2
    g 2
    2
    g' 2 x
    2x
    1
    g 2 x
    e e e
    2x
    e
    5.
    1/ n
    0
    n
    2
    f t dt
    I lim
    1
    n
    
    0
    form
    0
    Using : L’ Hospital and using Leibnitz rule
    3
    2
    n n
    1 1
    n f
    n 1
    n
    n
    I lim lim f
    2 2 n
    
    Put
    1
    n
    h
    hence
    h 0
    1 f 0 h f 0 1
    I lim f ' 0
    2 h 2
    6. For xy = 1
    2
    dy 1
    dx
    x
    Slope of normal to xy = 1 for x
    2
    > 0 (x 0)
    Slope of line (given) =
    2
    2
    log 1 5a a
    0
    5
    So 1 + 5a – a
    2
    > 1

    Page 8

  • AITS-CRT-III-PCM(S)-JEE(Main)/14
    FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
    website: www.fiitjee.com
    9
    a
    2
    – 5a < 0 hence a (0, 5)
    7.
    2
    a
    3
    1 1 2 2 2
    ah
    2 2 3
    3 3
    abc 2 1 1 3 3
    R
    4
    3 4 2 2 2
    A
    1
    1
    M
    a
    B
    C
    2
    3
    1
    3
    1
    3
    8. Area of the parallelogram outside the circle
    = Area of parallelogram – Area of sector OPQ
    =
    1 1
    sin 2 sin f
    2 2 2
    Q
    R
    P
    O
    r = 1
    9. Using
    n n 1 n 2
    r r 1 r 2
    n n n 1
    C C C
    r r r 1
    is equal to
    10. We have |(A
    –1
    ) adj(B
    –1
    ) adj(2A
    –1
    )| =
    2 2
    1 1 64
    A
    B A
    =
    64
    8
    2 1 4
    (By using |A
    –1
    | =
    1
    A
    and |adj B| = |B|
    n–1
    where n is order of square matrix)
    11. This distance between the given parallel lines (h) is
    1
    5
    length of the side of triangle is =
    2h
    3
    Area of triangle =
    2 2
    3 4h h 1
    4 3
    3 5 3
    12. The volume of tetrahedron =
    1 0 0
    1 1 1
    OA OB OC 1 1 0
    6 6 6
    0 1 1
    Area of base =
    1 1
    ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
    i j i j k i i k
    2 2
    =
    1 1
    2
    2
    2
    Hence height =
    3 volume 3 2 1
    Area of base 6
    2
    13. C = 20
    1
    ab
    2
    S – a = 7
    S – b = 13
    a – b = 6 ….. (1)
    a
    2
    + b
    2
    = 400
    C
    b
    A
    7
    D
    13
    B
    a

    Page 9

  • AITS-CRT-III-PCM(S)-JEE(Main)/14
    FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942
    website: www.fiitjee.com
    10
    (a – b)
    2
    + 2ab = 400
    ab = 182
    1
    ab 91
    2
    14. Let
    ˆ ˆ ˆ
    r xi yj zk
    Comparing x = t
    2
    + t – 2 y = t + 2 z = t
    2
    + 6t + 9
    Hence z – x = 5t + 11 z x = 5(y – 2) + 11 x + 5y – z + 1 = 0
    16.
    a b a b 8
    2
    a b 8
    2
    2
    2
    a b sin 8
    2
    2
    8
    b 16
    1
    1
    2
    b 4
    17. If
    2 2
    2 2
    x y
    1
    a b
    , a > b
    So x
    1
    x
    2
    = y
    1
    y
    2
    = b
    2
    3d = 49
    3d 7
    (x
    1
    y
    1
    )
    (x
    2
    y
    2
    )
    y
    1
    y
    2
    18. f(x) = sec x – cosec x
    f(x) = 0 at x
    4
    Also
    f " 0
    4
    so f(x) is minimum at x
    4
    / 4 / 2
    0 / 4
    f sec d cosec d
    4
    =
    2 1
    ln
    2 1
    =
    2ln 2 1
    19.
    a a
    0 0
    cos a x
    ln cota tanx dx ln dx
    sina cosx
    =
    a a a
    0 0 0
    lncos a x dx lnsinadx lncosxdx
    I(a) = – ln (sin a) · a
    I(1) = – ln (sin a) = ln |cosec 1|
    20. Let
    1/ x
    x 1
    x 0
    1
    A lim 2
    2
    (1
    form)
    =
    x
    x 1
    x 0 x 0
    1 1 2 1
    lim 2 1 lim
    x 2 2x
    e e
    =
    1
    ln2
    2
    e 2

    Page 10

Download this file to view remaining 2 pages

logo StudyDocs
StudyDocs is a platform where students and educators can share educational resources such as notes, lecture slides, study guides, and practice exams.

Contacts

Links

Resources

© 2025 StudyDocs. All Rights Reserved.