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VIEWS: 2 PAGES: 33

PREVIOUS YEAR PAPERS CBSE BOARD EXAM AIEEE BITSAT ISAT VITEEE IIT-JEE STUDY MATERIAL PHYSICS CLASS XI XII SAMPLE PAPERS KEY SOLUTIONS ANSWERS QUESTIONS

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									IIT-JEE 2010
                                Answers by
                                                                                     (Division of Aakash Educational Services Ltd.)

                                                       PAPER - 2 : CODES
     Q.No.            0            1          2             3           4             5            6           7            8            9
      01.            C            D            A            C            D            C            D            C            C            D
      02.            A            A            C            B            C            A            C            B            A            C
      03.            D            C            D            C            B            D             A           D            C            B
      04.            B            C            D            A            D            B            D            D            D            A
      05.            C            B            C            D            A            C            C            A            B            D
      06.            D            D            B            D            C            D             B           C            D            C
      07.            7            2            7            2             7            2            2           3            2             3
      08.            2            6            2            7             2            6            2           7            2             6
      09.            2            3            2            6             3            2            7           6            7             7
      10.            6            2            6            3             2            3            6           2            3             2
      11.            3            7            3            2             6            7            3           2            6             2
      12.            B            B            B            B            B            B             B           B            B            B
      13.            A            C            A            C            A            C             A           C            A            C
      14.            D            B            D            B            D            B            D            B            D            B
      15.            B            B            B            B            B            B             B           B            B            B
      16.            C            A            C            A            C            A            C            A            C            A
      17.            B            D            B            D            B            D             B           D            B            D
      18.         A(r,s)       A(p,s)        A(r,s)      A(p,s)        A(r,s)       A(p,s)       A(r,s)      A(p,s)        A(r,s)       A(p,s)
                    B(t)      B(p,q,r,t)      B(t)      B(p,q,r,t)      B(t)      B(p,q,r,t)      B(t)      B(p,q,r,t)      B(t)      B(p,q,r,t)
                  C(p,q)       C(p,q)       C(p,q)       C(p,q)       C(p,q)       C(p,q)        C(p,q)      C(p,q)       C(p,q)       C(p,q)
                   D(r)         D(p)         D(r)         D(p)          D(r)         D(p)         D(r)        D(p)          D(r)         D(p)
      19.         A(p,s)        A(r,s)      A(p,s)        A(r,s)       A(p,s)       A(r,s)       A(p,s)       A(r,s)      A(p,s)        A(r,s)
                 B(p,q,r,t)      B(t)      B(p,q,r,t)      B(t)      B(p,q,r,t)      B(t)      B(p,q,r,t)      B(t)      B(p,q,r,t)      B(t)
                  C(p,q)       C(p,q)       C(p,q)       C(p,q)       C(p,q)       C(p,q)        C(p,q)      C(p,q)       C(p,q)       C(p,q)
                   D(p)          D(r)        D(p)         D(r)          D(p)         D(r)         D(p)         D(r)        D(p)          D(r)
      20.            A            B            C            B            D            B            D            A            B            B
      21.            C            C            A            D            B            C             A           D            C            B
      22.            B            A            B            A            D            D            C            D            A            D
      23.            D            B            D            C            B            D             B           B            D            C
      24.            B            D            B            D            C            A             B           C            D            D
      25.            D            D            D            B            A            B            D            B            B            A
      26.            0            4            0            4            0            1             4           3            4             3
      27.            1            3            1            0            4            3             1           0            1             3
      28.            4            3            4            3            3            4             0           3            0             0
      29.            3            1            3            3            1            3             3           4            3             4
      30.            3            0            3            1            3            0             3           1            3             1
      31.            C            D            C            D            C            D            C            D            C            D
      32.            A            C            A            C            A            C            A            C            A            C
      33.            B            A            B            A            B            A            B            A            B            A
      34.            D            C            D            C            D            C            D            C            D            C
      35.            C            A            C            A            C            A            C            A            C            A
      36.            A            B            A            B            A            B            A            B            A            B
      37.         A(q,r)         A(t)       A(q,r)         A(t)        A(q,r)        A(t)        A(q,r)        A(t)        A(q,r)        A(t)
                   B(p)        B(p,r)        B(p)        B(p,r)         B(p)        B(p,r)        B(p)        B(p,r)        B(p)        B(p,r)
                 C(p,s,t)       C(q)       C(p,s,t)       C(q)        C(p,s,t)       C(q)       C(p,s,t)      C(q)        C(p,s,t)       C(q)
                D(q,r,s,t)       D(r)      D(q,r,s,t)     D(r)       D(q,r,s,t)      D(r)      D(q,r,s,t)      D(r)      D(q,r,s,t)      D(r)
      38.          A(t)        A(q,r)         A(t)       A(q,r)         A(t)        A(q,r)        A(t)        A(q,r)        A(t)        A(q,r)
                  B(p,r)        B(p)        B(p,r)        B(p)         B(p,r)        B(p)        B(p,r)        B(p)        B(p,r)        B(p)
                   C(q)       C(p,s,t)       C(q)       C(p,s,t)        C(q)       C(p,s,t)       C(q)       C(p,s,t)      C(q)        C(p,s,t)
                   D(r)       D(q,r,s,t)     D(r)       D(q,r,s,t)      D(r)      D(q,r,s,t)      D(r)      D(q,r,s,t)      D(r)      D(q,r,s,t)
      39.            C            D            A            B            B            B            B            B            B            D
      40.            A            A            C            D            B            A            C            D            A            B
      41.            D            C            D            C            D            B            A            C            C            D
      42.            D            B            B            A            D            D            D            B            B            A
      43.            B            D            B            B            A            C            B            A            D            B
      44.            B            B            D            D            C            D            D            D            D            C
      45.            6            8            6            8            6            3             8           2            8             2
      46.            3            4            3            6            8            4             3           6            3             4
      47.            8            2            8            4            2            8             6           4            6             6
      48.            4            3            4            2            3            2             4           8            2             8
      49.            2            6            2            3            4            6             2           3            4             3
      50.            C            D            C            D            C            D            C            D            C            D
      51.            A            B            A            B            A            B            A            B            A            B
      52.            B            C            B            C            B            C            B            C            B            C
      53.            D            C            D            C            D            C            D            C            D            C
      54.            B            A            B            A            B            A            B            A            B            A
      55.            C            B            C            B            C            B            C            B            C            B
      56.         A(p,r)       A(r,s,t)     A(p,r)       A(r,s,t)      A(p,r)      A(r,s,t)      A(p,r)      A(r,s,t)      A(p,r)      A(r,s,t)
                 B(q,s,t)     B(q,r,s,t)   B(q,s,t)     B(q,r,s,t)    B(q,s,t)    B(q,r,s,t)    B(q,s,t)    B(q,r,s,t)    B(q,s,t)    B(q,r,s,t)
                  C(p,r,t)     C(p,q)       C(p,r,t)     C(p,q)       C(p,r,t)     C(p,q)       C(p,r,t)     C(p,q)       C(p,r,t)     C(p,q)
                  D(q,s)      D(q,r,s,t)    D(q,s)      D(q,r,s,t)    D(q,s)      D(q,r,s,t)     D(q,s)     D(q,r,s,t)    D(q,s)      D(q,r,s,t)
      57.         A(r,s,t)     A(p,r)       A(r,s,t)     A(p,r)       A(r,s,t)      A(p,r)      A(r,s,t)      A(p,r)      A(r,s,t)      A(p,r)
                 B(q,r,s,t)   B(q,s,t)     B(q,r,s,t)   B(q,s,t)     B(q,r,s,t)    B(q,s,t)    B(q,r,s,t)    B(q,s,t)    B(q,r,s,t)    B(q,s,t)
                  C(p,q)       C(p,r,t)     C(p,q)       C(p,r,t)     C(p,q)       C(p,r,t)      C(p,q)      C(p,r,t)     C(p,q)       C(p,r,t)
                D(q,r,s,t)     D(q,s)      D(q,r,s,t)    D(q,s)      D(q,r,s,t)    D(q,s)      D(q,r,s,t)    D(q,s)      D(q,r,s,t)     D(q,s)

                 Though every care has been taken to provide the answers correctly
                      but the Institute shall not be responsible for error, if any.




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DATE : 11/04/2010




          Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075
                               Ph.: 011-47623456 Fax : 011-47623472


Time : 3 hrs.
                                    Solutions                                         Max. Marks: 252


                                          to
                                    IIT-JEE 2010
                              PAPER - 2 (Code - 5)


 Instructions :

 1.   The question paper consists of 3 parts (Chemistry, Mathematics and Physics) and each Part
      consists of four Sections.

 2.   For each question in Section I, you will be awarded 5 marks if you have darkened only the bubble
      corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases,
      minus two (–2) mark will be awarded.

 3.   For each question in Section II, you will be awarded 3 marks if you have darkened the bubble
      corresponding to the correct answer and zero mark if no bubbles are darkened. No negative marks
      will be awarded for incorrect answers in this Section.

 4.   For each question in Section III, you will be awarded 3 marks if you have darkened only the bubble
      corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases,
      minus one (–1) mark will be awarded.

 5.   For each question in Section IV, you will be awarded 2 marks for each row in which you have
      darkened the bubble(s) corresponding to the correct answer. Thus, each question in this section
      carries a maximum of 8 marks. There is no negative marks awarded for incorrect answer(s) in this
      Section.




                                     Downloaded from www.icbse.com
                                                                                                                                                IIT-JEE 2010 (Paper-2)


                                                PART–I : CHEMISTRY

                                                                   SECTION - I
                                                    Single Correct Choice Type
This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.


                                                O
                                                           (1) NaOH/Br2
1.   In the reaction H3C                   C                                  O
                                                                                       T , the structure of the product T is
                                                NH2        (2)            C
                                                                              Cl


                                  O
           H3C                C            O
                                                                                                          NH
     (A)                          O    C                                      (B)                                 C                 CH3
                                                                                                        O


                                                                                                                          O
           H3C                NH                                                       H3C                            C                     O
     (C)                                                                      (D)                                         NH       C)
                                  C
                                                                                                                                   td   .
                                                                                                                                sL
                                                                                                                         i ce
                             O

                                                                                                                  lS e rv
                                                                                                          n   a
Answer (C)
                                                                                             u   c a ti o
                                                                                         d
Hints :
                                                                                  s   hE
                                                                            ka
                                                                         Aa
                                                                                        O
                                                                                                                                                O
                                                                       f
                                                                  no
                            O                                                               C


                                                         i si o
                                   NaOH + Br2                                          Cl
     H3C                C                       H3C                           NH2                                                               C
                                                     v
                             NH2                (D i                                                                                NH


                                                                                                              H 3C

2.   Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is
     (A) 1 and diamagnetic                                                    (B) 0 and diamagnetic
     (C) 1 and paramagnetic                                                   (D) 0 and paramagnetic
Answer (A)
Hints :
     Molecular orbital diagram for B2 where Hund’s rule is violated.

                                        π2p2
     σ1s 2 σ * 1s 2   σ2s 2 σ * 2s 2       X

                                        π2p0
                                           Y

     ∴ Bond order = 1
     and diamagnetic.

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 IIT-JEE 2010 (Paper-2)
3.   The species having pyramidal shape is

     (A) SO3                         (B) BrF3                                 (C) SiO2–
                                                                                     3                                                 (D) OSF2

Answer (D)
Hints :


                              Pyramidal
                S
           O            F
                F

4.   The complex showing a spin-only magnetic moment of 2.82 B.M. is
     (A) Ni(CO)4                     (B) [NiCl4]2–                            (C) Ni(PPh3)4                                            (D) [Ni(CN)4]2–
Answer (B)
Hints :    Ni2+ is sp3 hybridized and metal ion is connected with weak ligands.
           ∴ Ni2+ ⇒ d 8




                two unpaired electrons                           sp3

                                                                                                                                      td   .)
                                                                                                                                   sL
     ∴ μs =         2 (2 + 2) B.M.
                                                                                                                            i ce
            =       8 B.M. = 2.83 B.M.
                                                                                                                     lS e rv
                                                                                                             n   a
5.   The compounds P, Q and S
                                                                                                u   c a ti o
                                                                                            d
                                                                                     s   hE
                                                                               Oka
                                                                           f Aa
                        COOH                         OCH3
                                                                 io   no       C

                                                     (D   i vi s                         O

     HO                         H3C
                P                           Q                                        S

     were separately subjected to nitration using HNO3/H2SO4 mixture. The major product formed in each case
     respectively, is

                                                                                                         O
                             COOH                              OCH3
                                                                                                         C
     (A)                                                                                                         O
           HO                             H3C
                      NO2                            NO2                      O2N


                                                                                                O
                             COOH                              OCH3
                                                                                                C
     (B)
                                                                                                         O
           HO                NO2          H3C
                                                     NO2                                            NO2

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                                                                                                                                             IIT-JEE 2010 (Paper-2)

                                                                                                O                             NO2
                          COOH                                     OCH3
                                                                                                C
     (C)                                                                                               O
           HO                             H3C                      NO2
                    NO2

                                                                                                O
                          COOH                                     OCH3
                                                                                                C
     (D)                                                                                               O                      NO2
           HO                             H3C                      NO2
                    NO2

Answer (C)
Hints :

                     COOH                                                    COOH
                                HNO3/H 2SO4


     HO                                               HO
                                                                       NO2

                        OCH3    HNO3/H2SO4                                       OCH3


     H3C                                              H3C                        NO2

                                                                                                                                 td   .)
                O                                                            O                       NO2                      sL
                                                                                                                       i ce
                C                                                            C
                                                                                                                lS e rv
                                         HNO3/H2SO4
                                                                                                        n   a
                                                                                                 a ti o
                    O                                                             O
                                                                                       d   u   c
                                                                                    hE
     The packing efficiency of the two-dimensional squares
                                                       ka unit cell shown below is
6.
                                                                       fA    a
                                                                    no
                                                       v   i si o
                                                  (D i




                                                                                  L

     (A) 39.27%                 (B) 68.02%                                        (C) 74.05%                                               (D) 78.54%
Answer (D)
Hints :

                               Area covered by particle
     Packing efficiency =
                                     Total area

                               2 × πr 2         2 × πr 2               π
                           =                =                      =
                                                (2         )
                                                               2
                                 a   2
                                                      2r               4


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                                                       Downloaded from www.icbse.com
 IIT-JEE 2010 (Paper-2)
                                                               SECTION - II
                                                               Integer Type
This section contains a group of 5 questions. The answer to each of the questions is a Single-digit Integer, ranging from
0 to 9. The correct digit below the equation number in the ORS is to be bubbled.

7.   Among the following, the number of elements showing only one non-zero oxidation state is
     O, Cl, F, N, P, Sn, Tl, Na, Ti
Answer (2)
Hints :   F will only exhibit –1 oxidation state except zero.
          and Na will exhibit +1 oxidation state.

8.   The total number of diprotic acids among the following is
     H3PO4             H2SO4                      H3PO3                             H2CO3                              H2S2O7
     H3BO3             H3PO2                      H2CrO4                            H2SO3
Answer (6)
Hints :   H2SO4, H3PO3, H2CO3, H2CrO4 and H2SO3 and H2S2O7 will behave as dibasic acid.

9.   One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as
     shown in the graph below. If the work done along the solid line path is Ws and that along the dotted line path is
     Wd, then the integer closest to the ratio Wd/Ws is
                                         4.5
                                         4.0     a
                                         3.5
                                                                                                                                 td   .)
                                                                                                                              sL
                                         3.0
                                                                                                                       i ce
                                                                                                                   e rv
                                     P
                                  (atm.) 2.5
                                                                                                        n   a   lS
                                                                                                 a ti o
                                         2.0
                                         1.5                                           d   u   c
                                                                                s   hE
                                                                           ka
                                         1.0
                                         0.5
                                                                  o   f Aa                                            b
                                                             i on
                                                      vi s
                                         0.0

                                               (D i
                                               0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
                                                                                  V
                                                                                (lit.)
Answer (2)
Hints :
     Solid line path work done (Ws) is isothermal beacuse PV is constant at each point & dash line path work done
     (Wd) is isobaric.
                                                                                V2
     Total work done on solid line path (Ws) = 2.303 nRT log
                                                                                V1
                                                                            V2
                                                  = 2.303 PV log
                                                                            V1
                                                                                5.5
                                                  = 2.303 × 2log
                                                                                0.5
                                            ≈ 4.6 l atm
     Total work done on dash line path (Wd) = 4 × 1.5 + 1 × 1 + 0.5 × 2.5
                                            = 8.255 l atm.
      Wd 8.25
         =     ≈ 2(closest integer)
      Ws   4.6

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                                                                         (5)



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                                                                                                                                               IIT-JEE 2010 (Paper-2)
10. Total number of geometrical isomers for the complex [RhCl(CO)(PPh3)(NH3)] is
Answer (3)
Hints :
     Rh Cl(CO)(PPh3)(NH3)] is a square planar complex with four different ligands and hence it will have three
     geometrical isomers
     a                   b      a            b   a                       c

             Rh                     Rh                      Rh

     d                   c      c            d   d                       b


11. Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a surface of
    area 10–12 m2 can be expressed in scientific notation as y × 10x. The value of x is
Answer (7)
Hints :
                                     4 3
     Volume of 1 Ag atom =             πr
                                     3

           4 3           108
     ∴       πr =                   cm3
           3      6.023 × 10 × 10.5
                            23



     ∴ r = 1.6 × 10–8 cm
     r = 1.6 × 10–10 m
     ∴ Number of Ag atoms in 10–12 m2
                                                                                                                                  td   .)
                                                                                                                               sL
     πr2   ×n=    10–12        m2
                                                                                                                        i ce
                         –12
                                                                                                                 lS e rv
     n=
                   10
                                                                                                         n   a
           3.14 × (1.6 × 10–10 )2
                                                                                            u   c a ti o
                                                                                        d
           10 × 10   7
                                                                                 s   hE
     n=                                                                     ka
              8
                                                                       f Aa
                                                             io   no
                                                      i vi s
     n = 1.25 × 107
     ∴ The value of x = 7                        (D
                                                                   SECTION - III
                                                                  Paragraph Type
This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be
answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

                                                 Paragraph for questions 12 to 14
The hydrogen-like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the
ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state
energy of the hydrogen atom.
12. The state S1 is
     (A) 1s                         (B) 2s                                   (C) 2p                                                         (D) 3s
Answer (B)
Hints :
     S1 state is 2s
     In 2s, one radial node is present

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13. Energy of the state S1 in units of the hydrogen atom ground state energy is
     (A) 0.75                   (B) 1.50                                (C) 2.25                                                (D) 4.50
Answer (C)
Hints :
     Energy in S1 state

              –13.6 × 32    9
     E=            2
                         = – 13.6 eV
                 2          4

     E( H1 ) = –13.6 eV [Ground state]
          1



     So energy = 2.25 × energy of e– in ground state in 1H1.

14. The orbital angular momentum quantum number of the state S2 is
     (A) 0                      (B) 1                                   (C) 2                                                   (D) 3
Answer (B)
Hints :
     S2 state is 3p orbital
     Orbital angular momentum of 3p is 1.

                                           Paragraph for questions 15 to 17
Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, which upon treatment
with HCN provides compound S. On acidification and heating, S gives the product shownd.)
                                                                                    t below :                               L
                                                                                                                       es
                                                                      H3C
                                                                                                                  rvic
                                                                                                             Se
                                                                                      OH
                                                                 H3C                                    al
                                                                                                 ti o n
                                                                                      d   u ca
                                                                                   hE
                                                                               O            O
                                                                          ka   s
                                                                       Aa
15. The compounds P and Q respectively are
                                                                     f
                      CH3
                                                           io   no
                     CH                        (D   i vi s
                                  H and H3C                     H
     (A) H3C                C                       C

                            O                       O
                      CH3

                     CH           H and H                       H
     (B) H3C                C                       C

                            O                       O

              H3C           CH2            H     H3C                        H
                     CH             C          and                  C
     (C)
                     CH3            O                               O

              H3C           CH2            H            H                   H
                     CH             C          and                  C
     (D)
                     CH3            O                               O

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16. The compound R is
                          O                                                                                     O
            H3C                                                                        H3C
                          C                                                                                     C
                                H                                                                         C                 H
                   C                                                                  H3C
     (A) H3C                                                                (B)
                  CH2                                                                             CH
                          OH                                                         H3C                        OH
                   CH3          O                                                                 CH3                   O
                  CH            C                                                                CH                     C
           H3C           CH           H                                              H3C                      CH                     H
     (C)                                                                    (D)
                         CH2                                                                                  CH
                                OH                                                            H3C                        OH
17. The compound S is
                   CH3          O                                                                              O
                  CH                                                                  H3C
                                C                                                                              C
           H3C           CH           H                                                           C                     H
     (A)                                                                    (B) H3C
                         CH2                                                                    CH2
                                CN                                                                             CN
                   CH3          CN                                                                             CN
                  CH                                                                  H3C
                                CH                                                                             CH
                                                                                                                            OH .)
           H3C           CH               OH
                                                                            (D) H3C
                                                                                                      C
                                                                                                                           L    td
                                                                                                                      ce s
     (C)
                                                                                                CH2             rvi
                                                                                                             Se
                         CH2
                                 OH
                                                                                                      n   al  OH
                                                                                               ti o
                                                                                          u ca
Hints (15-17) :
15. Answer (B)
                                                                                h    Ed
16. Answer (A)                                                         a   k as
                                                                  fA
17. Answer (D)
                                                          io   no                                               O
                                                   i vi s
                                CH3
                                               (D
                                                                                      H3C
                                                          1. K2CO3                                              C
                                C               H
                                                     2. H                  H                          C                  H
                         H3C          C                                              H3C
                                H                                C
                                                                       (Q)                       CH2
                               (P)    O              3. H   +    O                                              OH
                                                                                                              (R)
                                                                                                                       HCN
                                 OH
                   H3C                                                                                             CN
                                 CH                  O
                           C               C                                              H3C
                  H3C                                                                                              CH
                                                                 Hydrolysis
                                                                                                          C                     OH
                         H2C           HO                                             H3C
                                O     H                                                               CH2
                                           Δ                                                                     OH
                                                                                                                (S)
                                           OH
                               H3C
                                      C                  O
                               H3C
                                     H2C       O

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                                                          SECTION - IV
                                                              Matrix Type
This section contains 2 questions. Each question has four choices (A), (B), (C) and (D) given in Column I and five
statements (p), (q), (r), (s) and (t) in Column II. Any given statement in Column I can have correct matching with
one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the
statements given in (q) and (r), then for that particular question, against statement B, darken the bubbles corresponding
to (q) and (r) in ORS.

18. All the compounds listed in Column I react with water. Match the result of the respective reactions with the
    appropriate options listed in Column II.
     Column I                                                             Column II
     (A) (CH3)2SiCl2                                                (p) Hydrogen halide formation
     (B) XeF4                                                       (q) Redox reaction
     (C) Cl2                                                        (r) Reacts with glass
     (D) VCl5                                                       (s) Polymerization
                                                                    (t) O2 formation
Answer : A(p, s), B(p, q, r, t), C(p, q), D(p)
Hints :

                                                                              CH3
                                    CH3
                                             Polymerise                       Si
     (A) (CH3)2SiCl2 + H2O          Si—OH                         O                      O
                                OH CH3                                        CH3
                                                                                                                           td   .)
                                                                                              n
                                                                                                                        sL
                                                                                                                 i ce
     (B) 6XeF4 + 12H2 O ⎯⎯ 4Xe + 2XeO3 + 24HF + 3O 2
                          →
                                                                                                          lS e rv
                                                                                                  n   a
     (C) Cl2 + H2O ⎯⎯ HCl + HOCl
                     →
                                                                                     u   c a ti o
                                                                                 d
     (D) VCl5 + H2O ⎯⎯ VOCl3 + HCl
                      →                                                   s   hE
                                                                    ka
                                                   o           f Aa
                                                       i on
     Note : Vanadium in (+V) oxidation state from only fluoride. Existence of VCl5 is doubtful.

                                          i vi s
                                              (D
19. Match the reactions in Column I with appropriate options in Column II.
           Column I                                                                                                      Column II
                                            NaOH/H2O
     (A)           N2Cl +              OH                                 N=N                             OH (p)         Racemic mixture
                                              0°C


                                                       O
                OH OH                                                     CH3
                                                        C
                                   H2SO4
     (B) H3C – C — C – CH3                   H3C                C                                                (q)     Addition reaction
                                                                              CH3
                CH3 CH3                                          CH3

                       O                                       OH
                                1. LiAlH4
     (C)           C                   +                  CH                                                     (r)     Substitution reaction
                                 2. H3O
                       CH3                                     CH3

                                  Base
     (D) HS                Cl                      S                                                             (s)     Coupling reaction

                                                                                                                 (t)     Carbocation intermediate
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Answer : A(r, s), B(t), C(p, q), D(r)
Hints :




     (A)
                   N≡ N +                  O—H
                   Cl–
                                      OH


                                N=N                    O
                                      H



                                  N=N                      O




           ∴ Coupling reaction and electrophilic substitution reaction.




                       OH OH                                     OH OH2
                                                                                                                          td   .)
                                                                                                                       sL
                                                                                                                i ce
     (B)                                   +H

                                                                                                            e rv
               CH3 — C — C — CH3                       CH3 — C — C — CH3
                       CH3 CH3                                   CH3 CH3
                                                                                                 n   a   lS
                                                                                    u   c a ti o
                                                                                d
                                                                         s   hE
                                                                    ka
                                                               f Aa
                   O       CH3                              no
                                                            OH
                                                        i—oC — C
                                                         si            CH3
                                                 CHv
           CH3 — C — C — CH3                     (D i 3
                                                                       CH3
                           CH3                             CH3
                                                       Carbocation intermediate




                                                                              O                                                     OH
                            O                                                                             +
                                       LiAlH4    H                                              H3O
     (C)               C          Nucleophilic addition
                                                                              C          H                                          C     H
                            CH3
                                                                              CH3                                                   CH3
                                                                 Racemic mixture




                                                                                         Nucleophilic
               H                                Base       S                             substitution                   S
     (D)
                   S                  Cl                                      Cl
           B


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                                                        PART–II : MATHEMATICS
                                                                            SECTION - I
                                                          (Single Correct Choice Type)
This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.

                                                                                                                                                           x
20. Let f be a real-valued function defined on the interval (–1, 1) such that e f ( x ) = 2 + ∫ t + 1 dt , for all
                                                                               −x                4

                                                                                                                                                           0
     x ∈ (–1, 1), and let f –1 be the inverse function of f. Then (f –1)′ (2) is equal to

                                                    1                                          1                                                      1
     (A) 1                                    (B)                                   (C)                                                      (D)
                                                    3                                          2                                                      e
Answer (B)
Hints :
     We have,

                             x
     e − x f ( x ) = 2 + ∫ t 4 + 1 dt x ∈ (–1, 1)
                             0


     Differentiating w.r.t. x, we get

     e
         −x
              (f ' ( x ) − f ( x )) =   x +1
                                          4
                                                                                                                                            td   .)
                                                                                                                                         sL
                                                                                                                                  i ce
     ⇒ f ' ( x ) = f ( x ) + x 4 + 1e x
                                                                                                                           lS e rv
                                                                                                                   n   a
     ∵ f –1 is the inverse of f                                                                       u   c a ti o
                                                                                                  d
                                                                                           s   hE
     ∴ f –1′(f(x)) = x
                                                                                      ka
                                                                                 f Aa
     ⇒ f –1′(f(x)) f ′(x) = 1
                                                                       io   no
                                                           (D   i vi s
     ⇒ f −1 ' ( f ( x ) ) =
                                    1
                                 f '(x)


     ⇒ f −1 ' ( f ( x ) ) =                    1
                       f ( x ) + x4 + 1 ex
     at x = 0, f(x) = 2

                       1   1
     f ' ( 2) =
      −1
                         =
                      2+1 3

                                                              4        1
21. A signal which can be green or red with probability          and      respectively, is received by station A and then
                                                              5        5
                                                                                                  3
     transmitted to station B. The probability of each station receiving the signal correctly is . If the signal received
                                                                                                  4
     at station B is green, then the probability that the original signal was green is

               3                                    6                                          20                                                      9
     (A)                                      (B)                                   (C)                                                      (D)
               5                                    7                                          23                                                     20
Answer (C)

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                                                                                                                                                                  IIT-JEE 2010 (Paper-2)
Hints :
     From the tree-diagram it follows that
                                                                                                                                                                  S
                                                                                                                                                    4                     1
                          46
                P (BG ) =                                                                                                                           5                     5
                          80
                                                                                                                                                                                    R
                P ( BG | G ) =
                                              10 5
                                                =                                                                                      3G           1                      3            1
                                              16 8                                                                                     4            4                      4            4

                P ( BG ∩ G ) =
                                                  5 4 1                                                                               AG                AR                AR            AG
     ∴                                             × =                                                                           3         1    3             1       1        3             3
                                                  8 5 2                                                                                    4    4             4       4
                                                                                                                                 4                                             4        1    4
                                  1                                                                                                                                                     4
                                  2 = 1 × 80 = 20                                                                              BG              BR                 BG               BR            BG
                P ( G | BG ) =
                               P (BG ) 2 46 23

22. Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to
     (A) 25                                                      (B) 34                                           (C) 42                                     (D) 41
Answer (D)

Hints : Let A ∩ B = φ, A, B ⊂ S

     The number of elements in A : 0                                                         Choices for B : 24
     The number of elements in A : 1                                                         Choices for B : 38
     The number of elements in A : 2                                                         Choices for B : 3
     Total number of possible subsets is 41
                                                                                                                                                  td    .)
                                                                                                                                               sL
                                                                                         i ce
23. For r = 0, 1,..., 10, let Ar , B r and C r denote, respectively, theScoefficient
                                                                                     e rv
                                                                                  al                                                                     of x r in the expansions of
                                                                           ti o n
                                                                      ) ca
                                                  10
    (1 + x)10, (1 + x)20 and (1 + x)30. Then ∑ Ar ( B10Br − C10 Arduis equal to
                                                                    E
                                                 r =1
                                                                 sh
                                                            a ka
                              (B) A10 ( B10 − C10 A10 ) of A (C) 0
                                          2
    (A) B10 – C10                                                                                                                                            (D) C10 – B10
                                                     i on
Answer (D)                                    i vi s
                                            (D10
                            r
Hints : Ar = Coefficient of x in (1 + x)10                                                    =         Cr
     Br = Coefficient of                           xr       in (1 +            x)20    =     20C
                                                                                                r

     Cr = Coefficient of xr in (1 + x)30 =                                                   30C
                                                                                                r
      10                                                         10                              10

     ∑ A (B
     r =1
                r            10
                                  Br − C10 Ar ) = ∑ Ar B10 Br − ∑ Ar C10 Ar
                                                              r =1                               r =1

           10                                               10
     = ∑ Cr                                       Cr − ∑ C r
                    10            20         20                    10          30           10
                                       C10                                          C10 Cr
           r =1                                             r =1

           10                                                      10
     = ∑ C10− r                                        Cr − ∑ C10 − r
                    10                 20         20                      10            30              10
                                            C10                                              C10 Cr
           r =1                                                    r =1

                         10                                                     10
     = C10 ∑ C10 − r . Cr − C10                                                 ∑
          20                      10              20               30                  10               10
                                                                                            C 10− r          Cr
                         r =1                                                   r =1


     = C10
          20
                         (   30
                                              )
                                  C10 − 1 − C10
                                                       30
                                                                   (   20
                                                                            C10 − 1     )
     = C10 − C10 = C10 − B10
          30                  20




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24. If the distance of the point P(1, –2, 1) from the plane x + 2y – 2z = α, where α > 0, is 5, then the foot of the
    perpendicular from P to the plane is

         ⎛8 4 7⎞                      ⎛ 4 4 1⎞                                    ⎛ 1 2 10 ⎞                                          ⎛2 1 5⎞
     (A) ⎜ , , − ⎟                (B) ⎜ , − , ⎟                               (C) ⎜ , ,    ⎟                                      (D) ⎜ , − , ⎟
         ⎝3 3 3⎠                      ⎝3 3 3⎠                                     ⎝3 3 3 ⎠                                            ⎝3 3 2⎠
Answer (A)
Hints : Distance of point P from plane = 5

                1− 4 − 2 − α
     ∴ 5=
                     3                                                                               P(1, –2, 1)

     α = 10

     Foot of perpendicular

      x −1 y + 2 z −1 5
          =     =    =
        1    2    −2   3

           8     4     7
     x=      ,y = ,z=−
           3     3     3
     Thus the foot of the perpendicular is

       ⎛8 4 7⎞
     ∴f⎜ , ,− ⎟
       ⎝3 3 3⎠

                                                                                                                                td   .)
                                                                                                                             sL
25. Two adjacent sides of a parallelogram ABCD are given by
                                                                                                                      i ce
     AB = 2 i + 10 j + 11k and AD = − i + 2 j + 2 k .
                                                                                                               lS e rv
                                                                      ti o                                na
     The side AD is rotated by an acute angle α in the plane of the parallelogram so that AD becomes AD ′. If AD ′
                                                                 d uca α is given by
     makes a right angle with the side AB, then the cosine of the angle
                                                                                         E
                                                                                  sh
            8                                                                 a ka 1
                                                                          of A (C)
                                           17                                                                                                 4 5
     (A)                          (B)                                                                                             (D)
            9                              9
                                                                 io   n              9                                                         9
Answer (B)
                                                     (D   i vi s
Hints :

     AB = 2 i + 10 j + 11k

     AD = − i + 2 j + 2 k                                                                                 D
                                                                                                                                          C
     Angle 'θ' between AB and AD                                                                 α
                                                                                                      θ
                                                                                             A                                B
                           −2 + 20 + 22 8
                       AB . AD
     = cos ( θ ) =       =             =
                   AB AD     (15)( 3 )   9


                        17
     ⇒ sin ( θ ) =
                        9

          α + θ = 90


                                                17
     cos ( α ) = cos ( 90 − θ ) = sin ( θ ) =
                                                9

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                                                                                                                                                IIT-JEE 2010 (Paper-2)
                                                                       SECTION - II
                                                                       (Integer Type)
This Section contains 5 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct
digit below the question no. in the ORS is to be bubbled.

26. Let f be a function defined on R (the set of all real numbers) such that
     f '(x) = 2010 (x – 2009) (x – 2010)2 (x – 2011)3 (x – 2012)4, for all x ∈ R.
     If g is a function defined on R with values in the interval (0, ∞) such that f(x) = ln (g(x)), for all x ∈ R, then the
     number of points in R at which g has a local maximum is
Answer (1)
                      f (x )
Hints : g ( x ) = e            , ∀x ∈R

                          f (x)
     ⇒ g '( x ) = e               .f ' ( x )

     ⇒ f′(x) changes its sign from positive to negative in the neighbourhood of x = 2009
     ⇒ f(x) has local maxima at x = 2009
     So, the number of local maximum is one.


27. Two parallel chords of a circle of radius 2 are at a distance                                       3 + 1 apart. If the chords subtend at the center,
              π       2π
    angles of    and     , where k > 0, then the value of [k] is
              k       k
     [Note : [k] denotes the largest integer less than or equal to k]
                                                                                                                                      td   .)
                                                                                                                                   sL
Answer (3)
                                                                                                                            i ce
                                                                                                                     lS e rv
                     π                                                                                       n   a
Hints : Let θ =
                    2k
                                                                                                u   c a ti o
                                                                                            d
                                                                                     s   hE
                x
                                                                                ka
      cos θ =
                2
                                                                       o   f Aa                     θ x
                                                                   n
                                                            i si o
                                                                                           2
                                                        v
                                                 (D i
                                                                                                         C
                               3 + 1− x
     ⇒ cos2θ =                                                                             2
                                  2                                                             2θ                             3+1–x

                                      3 + 1− x
     ⇒ 2cos2 θ − 1 =
                                         2

        ⎛ x2 ⎞                       3 + 1− x
     ⇒ 2⎜ ⎟ − 1=
        ⎝ 4 ⎠                           2

      ⇒ x2 + x − 3 − 3 = 0

            −1 ± 1 + 12 + 4 3
      x=
                    2

            −1 ± 13 + 4 3
        =
                  2

            −1 + 2 3 + 1
        =                = 3
                  2
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                    3     π
     ∴ cos θ =        ⇒θ=
                   2      6

                               π        π
     ∴ Required angle =          = 2θ =
                               k        3
     ⇒k = 3

28. Let k be a positive real number and let

       ⎡ 2k − 1 2 k           2 k⎤          ⎡ 0           2k − 1         k ⎤
       ⎢                          ⎥         ⎢                              ⎥
     A=⎢2 k      1            −2k ⎥ and B = ⎢1 − 2k             0      2 k ⎥.
       ⎢                          ⎥         ⎢                              ⎥
       ⎢ −2 k 2k
       ⎣                       −1 ⎥
                                  ⎦         ⎢− k
                                            ⎣             −2 k          0 ⎥⎦

     If det (adj A) + det(adj B) = 106, then [k] is equal to
     [Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to
     k]
Answer (4)
Hints :
     |A| = (2k + 1)3, |B| = 0


     ⇒ det (adj A) ≠ det (adj B) = (2k + 1)6 = 106 ⇒ k =
                                                                           9
                                                                                                                     td   .)
                                                                                                                  sL
                                                                             .
                                                                                                           i ce
                                                                           2

                                                                     ⇒ [k] = 4                      lS e rv
                                                                                            n   a
                                                              u                    c a ti o
                                                           Ed
29. Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C
                                                                               h
     respectively. Suppose a = 6, b = 10 and the area of as triangle is 15 3. If ∠ACB is obtuse and if r denotes
                                                              ak the
     the radius of the incircle of the triangle, then r2 of A
                                                         is equal to
                                                              i on
                                                       vi s
                                                  (D i
Answer (3)
Hints :

                 3
     sin C =       and C is given to be obtuse.
                 2

                 2π
     ⇒ C=
                 3
30. Let a1, a2, a3,......, a11 be real numbers satisfying
     a1 = 15, 27 – 2a2 > 0 and ak = 2ak – 1 – ak – 2 for k = 3, 4, ...., 11.

          a1 + a2 + .... + a11
           2    2           2
     If                        = 90,
                  11

                            a1 + a2 + .... + a11
     then the value of                           is equal to
                                    11

Answer (0)

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                                             SECTION-III (Paragraph Type)
This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be
answered. Each of these question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


                                           Paragraph for questions 31 to 33

                                                               x2 y 2
Tangents are drawn from the point P(3, 4) to the ellipse         +    = 1 touching the ellipse at points A and B.
                                                               9   4

31. The coordinates of A and B are

                                                                   ⎛ 8 2 161 ⎞      ⎛ 9 8⎞
     (A) (3, 0) and (0, 2)                                     (B) ⎜ − 5 , 15 ⎟ and ⎜ − 5 , 5 ⎟
                                                                   ⎜          ⎟
                                                                   ⎝          ⎠     ⎝         ⎠


         ⎛ 8 2 161 ⎞                                                          ⎛ 9 8⎞
     (C) ⎜ − 5 , 15 ⎟ and (0, 2)
         ⎜          ⎟                                          (D) (3, 0) and ⎜ − , ⎟
         ⎝          ⎠                                                         ⎝ 5 5⎠

Answer (D)
Hints :
     Figure is self explanatory
                           P
                     D         (3, 4)
               B
                                                                                                                     td   .)
                                                                                                                  sL
                                                                                                           i ce
                    F        A
                                                                                                    lS e rv
                             (3, 0)                                                         n   a
                                                                               u   c a ti o
                                                                           d
                                                                    s   hE
                                                               ka
32. The orthocentre of the triangle PAB is
                                                          f Aa
         ⎛ 8⎞                        ⎛ 7 25 ⎞ isio
                                                     no            ⎛ 11 8 ⎞                                                        ⎛ 8 7⎞
                                            ⎟ iv
                                     ⎝ 5 8 (D
     (A) ⎜ 5, ⎟                  (B) ⎜ ,                       (C) ⎜ , ⎟                                                       (D) ⎜ , ⎟
         ⎝ 7⎠                               ⎠                      ⎝ 5 5⎠                                                          ⎝ 25 5 ⎠
Answer (C)
Hints :
Equation of AB is                                                                                                    P(3, 4)

              8
     y −0 =   5 ( x − 3) = 8 ( x − 3)
             9            −24                                             –9, 8
            − −3                                                           5 5
             5
                                                                             B                                            A
                                                                                                                          (3, 0)
               1
     ⇒    y = − ( x − 3)
               3

     ⇒ x + 3y = 3                ...(i)
     Equation of the straight line perpendicular to AB through P is 3x – y = 5
     Equation of PA is x – 3 = 0

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                                                                 ⎛ −9 8 ⎞    8
     The equation of straight line perpendicular to PA through B ⎜ , ⎟ is y = .
                                                                 ⎝ 5 5⎠      5


     Hence the orthocentre is ⎛ 11, 8 ⎞
                              ⎜ 5 5⎟
                              ⎝       ⎠


33. The equation of the locus of the point whose distances from the point P and the line AB are equal, is
     (A) 9x2 + y2 – 6xy – 54x – 62y + 241 = 0

     (B) x2 + 9y2 + 6xy – 54x + 62y – 241 = 0

     (C) 9x2 + 9y2 – 6xy – 54x – 62y – 241 = 0

     (D) x2 + y2 – 2xy + 27x + 31y – 120 = 0

Answer (A)

Hints :

                               1
     Equation of AB = y − 0 = − ( x − 3)
                               3
     x + 3y – 3 = 0

     |x + 3y – 3|2 = 10[(x – 3)2 + (y – 4)2]

     (Look at coefficient of x2 & y2 in the answers)
                                                                                                     td   .)
                                                                                                  sL
                                                                                           i ce
                                           Paragraph for question 34 to 36
                                                                                    lS e rv
Consider the polynomial f(x) = 1 + 2x + 3x2 + 4x3. Let s be the sum na all distinct real roots of f(x) = 0 and let
                                                                  a ti o
                                                                         of
t = |s|.                                                        c           d   u
                                                                     s   hE
                                                                ka
34. The real number s lies in the interval

                                                       o   f Aa
                               (B) ⎛ −11, − 3 ⎞ sion
     (A) ⎛ − 1 , 0 ⎞                                          (C) ⎛ − 3 , − 1 ⎞
                                                                                                          ⎛ 1⎞
                                            4 ⎟ i vi
                                   ⎜                              ⎜ 4                                 (D) ⎜ 0, ⎟
         ⎜ 4
         ⎝
                   ⎟
                   ⎠               ⎝          D
                                              ⎠                   ⎝         2⎟⎠                           ⎝ 4⎠
                                               (
Answer (C)
Hints :
     f ′(x) = 2(6x2 + 3x + 1) Δ = 9 – 24 < 0
     Hence f(x) = 0 has only one real root.

       ⎛ 1⎞         3 4
     f ⎜ − ⎟ = 1− 1+ − > 0
       ⎝ 2⎠         4 8

       ⎛ 3⎞      6 27 108
     f ⎜ − ⎟ = 1− +  −
       ⎝ 4⎠      4 16 64

                  64 − 96 + 108 − 108
              =                       <0
                          64

                              ⎛ 3 −1 ⎞                               ⎛ −3 −1 ⎞
     f(x) changes its sign in ⎜ − , ⎟ , hence f(x) = 0 has a root in ⎜ , ⎟ .
                              ⎝ 4 2⎠                                 ⎝ 4 2⎠

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35. The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies in the interval

         ⎛3 ⎞                                                          ⎛ 21 11 ⎞
     (A) ⎜ , 3 ⎟                                                   (B) ⎜ ,     ⎟
         ⎝4 ⎠                                                          ⎝ 64 16 ⎠

                                                                       ⎛ 21 ⎞
     (C) (9, 10)                                                   (D) ⎜ 0, ⎟
                                                                       ⎝ 64 ⎠
Answer (A)
Hints :

      1                            3
      2            t               4

     ∫ f ( x )dx < ∫ f ( x )dx < ∫ f ( x )dx                                1
     0             0               0


                                                                                     1 t           3
     ∫ f ( x )dx = x + x       + x3 + x 4
                           2
                                                                                     2             4

      1
      2
                   13      3
     ∫ f ( x )dx = 16 > 4
     0


      3
      4
                   525
      ∫ f ( x )dx = 256 < 3
      0
                                                                                                                  td   .)
                                                                                                               sL
                                                                                                        i ce
36. The function f′(x) is
                                                                                                 lS e rv
                                                                                         n   a
                                                                                c a ti o
     (A) Increasing in ⎜ −t , − ⎟ and decreasing in ⎜ − , t ⎟ Edu
                       ⎛       1⎞                   ⎛ 1 ⎞
                       ⎝       4⎠                   ⎝ 4 sh  ⎠
                                                              a   ka
                                                         of A
                       ⎛       1⎞                 i on ⎛ 1              ⎞
                                           i vi s
     (B) Decreasing in ⎜ −t , − ⎟ and increasing in ⎜ − ,
                       ⎝       4⎠       (D             ⎝ 4
                                                                       t⎟
                                                                        ⎠

     (C) Increasing in (–t, t)

     (D) Decreasing in (–t, t)

Answer (B)

Hints :

     Figure is self explanatory




                               1


            S
      –3          –1                    1      3
       4           2                    2      4



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                                                                     SECTION - IV
                                                                     (Matrix Type)
This Section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five
statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more
statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in
q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.

37. Match the statements in Column-I with the values in Column-II.
          Column I                                                                                     Column II
     (A) A line from the origin meets the lines                                              (p)       –4

          x − 2 y −1 z +1
               =    =     and
            1    −2    1

                8
          x−
                3 = y + 3 = z −1
                                 at P and Q
              2      −1       1
          respectively. If length PQ = d, then d2 is
     (B) The values of x satisfying                                                          (q)       0

                                                ⎛3⎞
          tan−1( x + 3) − tan−1( x − 3) = sin−1 ⎜ ⎟ are
                                                ⎝5⎠

     (C) Non-zero vectors a, b and c satisfy a ⋅ b = 0 ,                                     (r)       4

                                                                                                                                     td   .)
                                                                                                                                  sL
          (b − a) ⋅ (b + c ) = 0 and 2 | b + c | = | b − a | .
                                                                                                                           i ce
          If a = μb + 4c , then the possible values of
                                                                                                                    lS e rv
                                                                                                            n   a
          μ are
                                                                                                   c a ti o
     (D) Let f be the function on [–π, π] given by                                           du
                                                                                            E(s)       5
                                                                                      s   h
                                     ⎛ 9x ⎞     ⎛x⎞                         fA   a ka
          f(0) = 9 and f ( x ) = sin ⎜    ⎟ sin ⎜ 2 ⎟                   o
                                     ⎝ 2 ⎠      ⎝ ⎠                i on
                                                            vi s
                                            π        (D i
                                            ∫ f ( x ) dx is
                                        2
          for x ≠ 0. The value of
                                        π
                                            −π

                                                                                             (t)       6
Answer → A(t), B(p, r), C(q), D(r)
Hints :

                                                                      x y z
     (A) Equation of line passing through origin is                    = =
                                                                      a b c

                  2   1   −1
          ∴       1 −2     1 =0
                a     b    c

          ⇒ a(–1) – b(3) + c(–5) = 0
               –a – 3b – 5c = 0
               a + 3b + 5c = 0                          ... (1)
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               8
                        −3 1
               3
         Also, 2        −1 1 = 0
               a        b c

                     ⎛2⎞     ⎛ 10 ⎞
          a( −2) − b ⎜ ⎟ + c ⎜    ⎟=0
                     ⎝ 3⎠    ⎝ 3 ⎠

                 2b 10c
          2a +     −    =0
                 3   3
         3a + b – 5c = 0                             ... (2)
         Now, a + 3b + 5c = 0
         3a + b – 5c = 0

           a   b   c
             =   =
          −20 20 −8

          a   b   c
            =   =
          5 −5 4
         Equation of line is

          x   y   z
            =   =                                    ... (2)
          5 −5 +4

          x − 2 y −1 z +1                                                                                                 td   .)
               =    =                                                                                                  sL
                                                                                                                i ce
                                                     ... (3)
                 −2
                                                                                                            e rv
            1          1

                       8                                                                         n   a   lS
                                                                                          a ti o
                 x−
                       3 = y + 3 = z −1                                                 c
         Now,                                        ... (4)
                                                                                d   u
                     2      −1       1
                                                                         s   hE
                                                                    ka
         Point on (2) is (5λ, –5λ, + 4λ)
                                                               f Aa
         Point on (3) is (2 + k1, 1 – 2k1, –1 + k1)o
                                                 n
                                                 v   i si o
                         ⎛8                 (D i     ⎞
         Point on (4) is ⎜ + 2k2 , − 3 − k2 , 1 + k2 ⎟
                         ⎝3                          ⎠
         On solving,
         2 + k1 + 1 – 2k1 = 0
         –k1 + 3 = 0
         k1 = 3
         P ≡ (5, –5, 2)
         Again for Q,

          8
            + 2k2 − 3 − k 2 = 0
          3

                 1
          k2 −     =0
                 3

                 1
          k2 =
                 3


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             ⎛ 10 −10 4 ⎞
          Q ≡⎜ ,     ,
             ⎝ 3   3 3⎟ ⎠
                        2       2        2
                   ⎛5⎞ ⎛5⎞ ⎛ 2⎞
         PQ =      ⎜ ⎟ +⎜ ⎟ +⎜ ⎟
                   ⎝3⎠ ⎝3⎠ ⎝3⎠
                   54
              =
                   3
                        54
         PQ2 = d2 =        =6
                         9
               ⎛ x +3−x +3⎞            −1 ⎛ 3 ⎞
     (B) tan−1 ⎜
               ⎜ 1 + ( x 2 − 9) ⎟ = tan ⎝ 4 ⎠
                                ⎟         ⎜ ⎟
               ⎝                ⎠
             6     3
                 =
         x −8 4
           2

         3x2 – 24 = 24
         3x2 = 48
         x = ±4

                    ⎛     a − μb ⎞
     (C) (b − a ) ⋅ ⎜ b +
                    ⎜            ⎟=0
                    ⎝        4 ⎟ ⎠
          (b − a) ⋅ (4b + a − μb ) = 0
         (4 – μ)b2 – a2 = 0                   ... (1)

         Also, 2 b +
                        a − μb
                               = | b − a|                                                                                          td   .)
                                                                                                                                sL
                                                                                                                         i ce
                           4
            (4 − μ)b + a
                                                                                                                  lS e rv
          2
                                                                                                          n   a
                                                                                                   a ti o
                  4
                                                                                         d   u   c
          (4 − μ)2 b 2 a2
                                                                                  s   hE
                      +   = b2 + a 2
                                                                             ka
               4        4
                                                                        f Aa
          3a 2 (4 − μ)2 − 4 2                                 io   no
           4
               =
                      4
                            ⋅b
                                                  (D   i vi s
         3a2 = ((4 – μ)2 – 4)b2               ... (2)
         From equations (1) and (2),
         3(4 – μ) = (4 – μ)2 – 4
         (4 – μ)2 – 3(4 – μ) – 4 = 0
         ⇒ μ = 0, 5
         μ = 5 is not admissible.
     (D) f(0) = 9,

                     ⎛ 9x ⎞
                sin ⎜     ⎟
                     ⎝ 2 ⎠
         f(x) =
                        x
                   sin
                        2
                ⎛           x ⎞⎛            3x ⎞
              = ⎜ 3 − 4 sin2 ⎟ ⎜ 3 − 4 sin2
                ⎝           2 ⎠⎝             2 ⎟
                                               ⎠
                           x           3x          x       3x
              = 9 − 12 sin   − 12 sin2    + 16 sin2 ⋅ sin2
                          2
                           2            2          2        2
              = 9 – 6(1 – cos x) – 6(1 – cos 3x) + 4(1 – cos x)(1 – cos 3x)
              = 1 + 6 cos x + 6 cos 3x – 4 cos x – 4 cos 3x + 4 cos x cos 3x

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                   π        9x
                        sin
                   ∫
              2              2
          I =
              π              x
                   −π   sin
                             2
                   π


                   ∫ 1 + 2 cos x + 2 cos 3x + 2(cos 4x + cos x )
              4
            =
              π
                   0

              4
            =   ×π
              π
            =4
38. Match the statements in Column-I with those in Column-II.
     [Note: Here z takes values in the complex plane and Im z and Re z denote, respectively, the imaginary part and
     the real part of z.]
          Column I                                                          Column II

                                                                                                                        4
     (A) The set of points z satisfying                               (p) An ellipse with eccentricity
                                                                                                                        5
          | z – i | z || = | z + i | z || is contained in
          or equal to
     (B) The set of points z satisfying                               (q) The set of points z satisfying Im z = 0
          | z + 4 | + | z – 4 | = 10 is contained in
                                                                                                              td   .)
          or equal to
                                                                                                      ce   sL
     (C) If | w | = 2, then the set of points                         (r) The set of points vi satisfying | Im z | ≤ 1
                                                                                         er z
                                                                                             a   lS
                 1
                                                                                    a ti o n
          z =w −   is contained in or equal to
                                                                           d   uc
                                                                        hE
                 w
                                                                        s
     (D) If | w | = 1, then the set of points
                                                                fA a ka
                                                                      (s)   The set of points z satisfying | Re z | ≤ 2
                                                            o
                                                     i n
                         is contained in or equal itoo
                                                iv s
                       1
          z =w +
                       w                     (D
                                                                      (t) The set of points z satisfying | z | ≤ 3
Answer → A(q, r), B(p), C(p, s, t), D(q, r, s, t)
Hints :
     (A) z is equidistant from the points i| z | and –i| z |, whose perpendicular bisector is Im (z) = 0.
     (B) Sum of distance of z from (4, 0) and (–4, 0) is a constant 10, hence locus of z is ellipse with semi-major
         axis 5 and focus at (±4, 0).
         ae = 4
                        4
          ∴ e=
                        5

                            1  5
     (C) | z | ≤ |w | +       = <3
                            w  2

                            1
     (D) | z | ≤ |w | +       =2
                            w
          ∴ Re z ≤ | z | ≤ 2


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                                          PART–III : PHYSICS

                                                              SECTION - I
                                             Single Correct Choice Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.

39. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror
    is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is
     (A) Virtual and at a distance of 16 cm from the mirror
     (B) Real and at a distance of 16 cm from the mirror
     (C) Virtual and at a distance of 20 cm from the mirror
     (D) Real and at a distance of 20 cm from the mirror
Answer (B)                                                                                                    16 cm
Hints :                                                                                                      15 cm

     Object is placed at O.                                              O                    B I                                        A
     First refraction through lens

                                                                                                        6 cm
                                                                                                                  10 cm .)
          1   1   1
            −   =
          V −30 +15                                                                    30 cm                          td
                                                                                                                      L         20 cm
                                                                                                       10 cmvi
                                                                                                            r    ce s      30 cm
     ⇒ V = +30 cm                                                                                           e
                                                                                                   n   al S
     ⇒ Real image is formed at A.                                                           ti o             20 cm
                                                                                   d   u ca
                                                                  E
                                                             as h
     This image acts as virtual object for plane mirror. Image is formed 20 cm before mirror (B) This acts as virtual
     object for lens.                                     ak        fA
                                                                o
     Second refraction through lens                        i on
                                                    vi s
          1   1   1
                                             (D i
            −   =
          V +10 +15

     ⇒ V = +6 : Real image
     Distance from mirror is 16 cm.

40. A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It
    is made of two hemispherical shells, held together by pressing them with force F (see figure). F is proportional
    to




                                         F                                                               F




         1 2 2                    1 2                                           1 σ2                                            1 σ2
     (A) ε σ R                (B) ε σ R                                  (C)                                              (D)
          0                        0                                            ε0 R                                            ε0 R 2

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Answer (A)
Hints :
     By dimensional analysis
                 ⎡ 1 q 2 ⎤ ⎡ 1 σ 2 A2 ⎤
          [F ] = ⎢        2 ⎥
                              =⎢   2 ⎥
                 ⎣ 4 πε0 r ⎦ ⎣ ε0 R ⎦

                               ⎡ 1 2 2⎤
                             = ⎢ε σ R ⎥
                               ⎣ 0    ⎦
41. A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in
    its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is
    50 N and the speed of sound is 320 ms–1, the mass of the string is
     (A) 5 grams               (B) 10 grams                            (C) 20 grams                               (D) 40 grams
Answer (B)
Hints :   For hollow pipe, fundamental frequency
                v   320
          f =     =
                4l 4 × 0.8
          For string in 2nd harmonic

                1 T 1 Tl   1 50 × 0.5
          f =      =     =
                l μ l m 0.5     m
          Equating we get,
          m = 0.01 kg = 10 g
                                                                                                        td   .)
                                                                                                     sL
                                                                                  i ce
                                                                              e rv
42. A Vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match
    with 16 main scale divisions. For this Vernier calipers, the least countSis
                                                                          l
                                                                       (C) 0.1 mmio
                                                                                          na
     (A) 0.02 mm               (B) 0.05 mm                                      at  uc
                                                                                                                  (D) 0.2 mm
                                                                                 Ed
                                                                            sh
Answer (D)
                                                                       ka
Hints :
                                                                  f Aa
     20 VSD = 16 MSD                                    io   no
     ⇒ 5 VSD = 4 MSD                        (D   i vi s
          LC = 1 MSD – 1 VSD
                        4     1
                = 1MSD − MSD = MSD = 0.2 mm
                        5     5

43. A block of mass 2 kg is free to move along the x-axis. It is at rest and from t = 0 onwards it is subjected to a
    time-dependent force F(t) in the x direction. The fore F(t) varies with t as shown in the figure. The kinetic energy
    of the block after 4.5 seconds is
                                            (F)t

                                          4N




                                                                                         4.5 s
                                                                                                 t
                                            O                                3s


     (A) 4.50 J                (B) 7.50 J                              (C) 5.06 J                                 (D) 14.06 J

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Answer (C)
Hints :
     Area under the curve of F – t graph provides change in momentum of particle since

          Fdt = d(mV) ⇒          ∫ Fdt = Δ mV
     ⇒ p = 4.5 kg ms–1
                           p2   4.52
     Kinetic energy =         =      = 5.06 J
                           2m 2 × 2

44. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength
    81π
         × 105 Vm −1 . When the field is switched off, the drop is observed to fall with terminal velocity 2 × 10–3 ms–1. Given
      7
    g = 9.8 ms–2, viscosity of the air = 1.8 × 10–5 Ns m–2 and the density of oil = 900 kg m–3, the magnitude of q is
     (A) 1.6 × 10–19 C             (B) 3.2 × 10–19 C                     (C) 4.8 × 10–19 C                                        (D) 8.0 × 10–19 C
Answer (D)
Hints :
     During equilibrium in presence of electric field qE = mg
                 4 3
     ⇒ qE =        πr ρg               …(i)
                 3
     When the drop descends with constant velocity
          mg = 6πηrV
                                                                                                                                 td   .)
                                                                                                                              sL
                                                                                                                       i ce
          4 3
            πr ρg = 6πηrV
                                                                                                                   e rv
     ⇒
                                                                                                                lS
          3
                                                                                                        n   a
                                                                                                 a ti o
     Putting the values, we obtain
                                                                                       d   u   c
                                                                                    hE
                3
          r =     × 10−5 m
                7                                                          ka   s
                                                                      f Aa
     Putting back in equation (i)
                                                            io   no
          q ≈ 8×    10–19    C
                                                (D   i vi s
                                                                 SECTION - II
                                                                 Integer Type
This section contains 5 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The
correct digit below the question number in the ORS is to be bubbled.

45. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to
                   25      50
     move from        m to    m in 30 seconds. What is the speed of the object in km per hour?
                    3      7

Answer (3)
Hints :
     Convex mirror forms virtual image of object behind the mirror.
     By Mirror formula

          1 1 1
           + =
          v u f

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                                                                                                                                      IIT-JEE 2010 (Paper-2)
     For first object

             1     1  1
     ⇒            + =
          + 25 / 3 u +10
                                                                                 O                             I
     ⇒ u = – 50 m
     For second object                                                                      10 cm

             1     1  1
     ⇒            + =
          + 50 / 7 u +10

     ⇒ u = – 25 m

                   25
     Speed =          m/s
                   30

                   25 36
               =     ×   km/h = 3 km/h
                   30 10

                                                                            1
46. A diatomic ideal gas is compressed adiabatically to                       of its initial volume. If the initial temperature of the gas
                                                                           32
     is Ti (in kelvin) and the final temperature is aTi, the value of a is
Answer (4)
Hints :
                                                                                                                            td   .)
                                                                                                                         sL
     TV γ – 1 = constant                                                                                          i ce
                                                                                                           lS e rv
                                                                                                   n   a
                                                                                            a ti o
               7
          γ=
               5
                                                                                  d   u   c
                                                                           s   hE
                            2
                                                                      ka
                      ⎛ V ⎞5                                       Aa
               2
          TiVi = α Ti ⎜ i ⎟                                      f
                                                            no
               5

                      ⎝ 32 ⎠
                                                       io
     ⇒ α=4                                 (D   i vi s

                                                                                         dN (t )                dN (t )
47. To determine the half life of a radioactive element, a student plots a graph of ln           versus t. Here         is
                                                                                           dt                    dt
     the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor
     of p after 4.16 years, the value of p is
                                                    6

                                                    5
                                         dN (t )




                                                    4
                                          dt




                                                    3
                                          n




                                                    2

                                                    1
                                                        2        3     4         5   6                 7        8
                                                                               Years
Answer (8)

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 IIT-JEE 2010 (Paper-2)
Hints :
     Equation for radioactive decay is
                    – λt
          N = N0

          dN                – λt
             = – λ N0
          dt

               dN
          ln      = + ln(λN0 ) – λt
               dt

                                3–4
     ⇒ Slope = – λ =                  (from graph)
                                6–4


                1
     ⇒ λ=         year –1
                2

                   0.693
     Half life =         = 2 × 0.693      1.386
                     λ

     4.16 years is approximately 3 half lives
     ⇒ Nuclei will decay by a factor of 23 = 8

                                                                                           .)
                                                                                      Lt d
48. At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have
                                                                                    s
                                                                                                                     i ce
    no charge initially, at what time (in seconds) does the voltage across them become 4 V?
     [Take : ln 5 = 1.6, ln 3 = 1.1]
                                                                                                              lS e rv
                                                                                                          a
                                                                                                     i on
                                                                                                    t2μF
                                                                                                 ca
                                                  2MΩ
                                                                                         d   u
                                                                                  s   hE
                                                                             ka
                                                                        f Aa
                                      A
                                                              io   no                                                  B

                                                  (D   i vi s
                                                  2MΩ                                               2μF
Answer (2)
Hints :
     Resistance of circuit = 1 MΩ
     Capacitance of circuit = 4 μF
     Time constant = RC = 4 s
     Voltage across capacitor at time 0 = 0 V
     Voltage across capacitor at time ∞ = 10 V
     In general

                           –t
          V =10(1 – α 4 )

     Putting V = 4 V
     We get t = 2 s


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                                                                                                              IIT-JEE 2010 (Paper-2)
49. A large glass slab (μ = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen
    that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?
Answer (6)
Hints :
      Light will appear only from a circular region because rays going out at an angle greater than θc will suffer TIR.
                         1 3
            sin θc =      =
                         μ 5
                                                                                     R
      From figure
                                 R
            sin θc =                                                                                     8 cm
                             R + 82
                                 2


      Equating
                 R               3
                             =
              R +8
                 2       2       5
      Solving
            R = 6 cm

                                                              SECTION - III
                                                            Paragraph Type
This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be
                                                                                         )          .
answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
                                                                                               td
                                                                                            sL
                                                    Paragraph for questions 50 to 52
                                                                                        i ce
                                                                                 lS e rv
      The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when
                                                                             o a
      an electron is revolving around a proton. We will extend this to na general rotational motion to find quantized
                                                                          ti
      rotational energy of a diatomic molecule assuming it to be ca
                                                                  E du rigid. The rule to be applied is Bohr’s quantization
                                                             ashquantization condition its rotational energy in the nth level
      condition.
      A diatomic molecule has moment of inertia l. ByAak
                                                        of
50.                                                        Bohr’s
                                                   i on
      (n = 0 is not allowed) is

            1 ⎛ h ⎞  2
                                  1 ⎛ h (D⎞ i vi s
                                                2
                                                                    ⎛ h ⎞     2
                                                                                                       ⎛ h ⎞    2
                                                                                                                        2

      (A)     ⎜     ⎟               ⎜     ⎟
                                      (B) n                       n⎜
                                                                   (C)     ⎟                        n ⎜ (D) ⎟
            n ⎝ 8π I ⎠
             2       2
                                            ⎝ 8π I ⎠
                                                2
                                                                          ⎝ 8π I ⎠
                                                                              2
                                                                                                                    ⎝ 8π I ⎠
                                                                                                                        2



Answer (D)
Hints :
            nh
      L=
            2π

             L2 n 2 h 2
      KE =     =
             2I 8π2I
51. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is
               4
      close to   × 1011 Hz. Then the moment of inertia of CO molecule about its centre of mass is close to (Take h =
               π
      2π × 10–34 J s)
      (A) 2.76 × 10–46 kg m2                                       (B) 1.87 × 10–46 kg m2
      (C) 4.67 × 10–47 kg m2                                       (D) 1.17 × 10–47 kg m2


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 IIT-JEE 2010 (Paper-2)
Answer (B)
Hints :

          4          (22 – 12 )h 2
     h×     × 1011 =
          π             8π 2I

                 3h
     ⇒ I=                 = 1.87 × 10 –46 kg m2
              32π × 10 22




52. In a CO molecule, the distance between C (mass = 12 a.m.u.) and O (mass = 16 a.m.u.), where
                  5
     1 a.m.u. =     × 10 –27 kg , is close to
                  3

     (A) 2.4 × 10–10 m                                        (B) 1.9 × 10–10 m
     (C) 1.3 × 10–10 m                                        (D) 4.4 × 10–11 m
Answer (C)
Hints :

                                m1m2 2
     I = 1.87 × 10–46 =                r
                               m1 + m2


                       12 × 16 5
     1.87 × 10 –46 =          × × 10–27 r 2
                       12 + 16 3                                                                     td   .)
                                                                                                  sL
                                                                                           i ce
     ⇒ r = 1.3 × 10–10 m
                                                                                    lS e rv
                                                                                a
                                                                      ti n
                                             Paragraph for questions 53oto 55
When liquid medicine of density ρ is to be put in the eye, it isEd
                                                                         u ca
                                                                   sh
                                                                      done with the help of a dropper. As the bulb on the
top of the dropper is pressed, a drop forms at the openingaof the dropper. We wish to estimate the size of the drop.
                                                                 k
                                                           f isa
We first assume that the drop formed at the opening A spherical because that requires a minimum increase in its
                                                        no
                                                 i si o
surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius
                                               v
                                          (D i
of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

53. If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius
    R(assuming r << R) is

                                                                    2πr 2T                                     2πR 2T
     (A) 2 πrT                       (B) 2 πRT               (C)                                      (D)
                                                                      R                                          r

Answer (C)
Hints :
                                                                                                      dF = Tdl
                                                                                                                        r
     Vertical force =      ∫   Tdl sin θ

                                 r                                                                                      R
                       =T×         × 2πr
                                 R

                           2πr 2T
                       =
                             R


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54. If r = 5 × 10–4 m, ρ = 103 kgm–3, g = 10 ms–2, T = 0.11 Nm–1, the radius of the drop when it detaches from the
    dropper is approximately
     (A) 1.4 × 10–3 m            (B) 3.3 × 10–3 m                          (C) 2.0 × 10–3 m                          (D) 4.1 × 10–3 m
Answer (A)
                         2πr 2T 4
Hints :   For equation         = πR 3ρg
                           R    3
                              1/ 4
                    ⎛ 2 r 2T ⎞
          ⇒      R =⎜
                    ⎜ 3 ρg ⎟ ⎟
                    ⎝        ⎠

55. After the drop detaches, its surface energy is
     (A) 1.4 × 10–6 J            (B) 2.7 × 10–6 J                          (C) 5.4 × 10–6 J                          (D) 8.1 × 10–6 J
Answer (B)
Hints :   Surface energy = 4πR2T

                                              SECTION - IV (Matrix Type)
This section contains 2 questions. Each Question has four statements (A, B, C and D) given in Column I and five
statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more
statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in
q are r, then for that particular question, against statement B, darken the bubble corresponding to q and r in the ORS.

56. You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source
    (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as
                                                                                       d. )
    shown in Column II. When a current I (steady state for DC or rms for AC) flows through the circuit, the
    corresponding voltage V1 and V2. (indicated in circuits) are related as shown s Lt
                                                                                  in Column I. Match the two
                                                                                                              i ce
          Column I                                                                Column II
                                                                                                       lS e rv
                                                                                               n   a
                                                                                        a ti o
                                                                                      V1                 V2
                                                                                     uc
                                                                                 h Ed 6 mH               3 μF
                                                                             s
     (A) I ≠ 0, V1 is proportional to I
                                                                   fA   a ka
                                                                           (p)
                                                               o
                                                          i on                                 V

                                            (D   i vi s                               V1                 V2

                                                                                     6 mH               2Ω
     (B) I ≠ 0, V2 > V1                                                    (q)

                                                                                               V
                                                                                      V1                 V2

                                                                                     6 mH               2Ω
     (C) V1 = 0, V2 = V                                                    (r)

                                                                                               V
                                                                                      V1                 V2

                                                                                     6 mH               3 μF
     (D) I ≠ 0, V2 is proportional to I                                    (s)

                                                                                               V
                                                                                      V1                 V2

                                                                                     1 kΩ               3 μF
                                                                           (t)

                                                                                               V
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Answer A(r, s, t), B(q, r, s, t), C(p, q), D(q, r, s, t)
Hints :
     (p) V1 = 0, V2 = V (      XL = 0)
     (q) XL = 0, V1 = 0, V2 = V
     (r) XL = ωL = 2πfL = 1.884 Ω
          V1 = IXL
          V2 = IR
          V1 < V2
     (s) XL = 1.884 Ω

                  1   10
          XC =      =    × 103 Ω
                 ωL 3 π

          XC >> XL
          V2 = IXC, V1 = IXL
     (t) R = 1 kΩ

                 10
          XC =      × 103 Ω
                 3π

          ⇒ XC > R ⇒ V2 > V1
                                                                                                      td   .)
          Also, V2 = IXL, V1 = IR.                                                                 sL
                                                                                rv          i ce
                                                                             Se
57. Two transparent media of refractive indices μ1 and μ3 have a solid lens shaped transparent material of refractive
                                                                          al
                                                                     tion Match them to the ray diagrams shown in
    index μ2 between them as shown in figures in Column II. A ray traversing these media is also shown in figures.
    In Column I different relationships between μ1, μ2 and μ3 are ca
                                                                 u given.
    Column II.                                              h Ed
                                                                          s
          Column I
                                                                    fA aka Column II
                                                          io   no
                                              (D   i vi s
     (A) μ1 < μ2                                                      (p) μ3      μ2   μ1



                                                                                  μ2   μ1
     (B) μ1 > μ2                                                      (q) μ3




     (C) μ2 = μ3                                                      (r)    μ3 μ μ1
                                                                                 2




     (D) μ2 > μ3                                                      (s) μ       μ2 μ1
                                                                           3




                                                                      (t)    μ3   μ2 μ1



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                                                                                                                                     IIT-JEE 2010 (Paper-2)
Answer A(p, r), B(q, s, t), C(p, r, t), D(q, s)
Hints :
     (p) μ2 = μ3, as there is no deviation
          μ1 < μ2, as ray bends towards normal.
     (q) μ1 > μ2, as ray bends away from normal
          μ2 > μ3, for similar reasons
     (r) Similar to (p)
     (s) Similar to (q)
     (t) μ1 > μ2, as ray bends away from normal

          μ2 = μ3, as there is no deviation




                                                                                                                           td   .)
                                                                                                                        sL
                                                                                                                 i ce
                                                                                                          lS e rv
                                                                                                  n   a
                                                                                     u   c a ti o
                                                                                 d
                                                                          s   hE
                                                                     ka
                                                                f Aa
                                                      io   no
                                          (D   i vi s




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