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www.eenadupratibha.net www.eenadupratibha.net EAMCET ENGINEERING MODEL GRAND TEST No. of Questions: 160 Marks: 160 Time: 3 Hrs. MATHEMATICS 1. If f (x) = cos [π] x + cos [πx] where [ . ] denotes the greatest integer function. Then f ( π 2 ) = 1) cos 4 2) cos 3 3) 0 4) None of these 2. f (2x + 3y, 2x 7y) = 20x. Then f (x, y) = 1) 7x 3y 2) 7x + 3y 3) 3x 7y 4) x y 3. The domain of f(x) = is v°æü˨¡ç 4 x + 8 2 3 (x 2) 13 2 2(x 1) 1) [1, ) 2) [0, ) 3) [2, ) 4) does not exist ´u´-Æœnûªç é¬ü¿’ a b c 4. For a ABC, if b c a = 0, then sin 1/3 A + sin 1/3 B + sin 1/3 C = c a b a b c vA¶µº’ïç ABC b c a = 0, Å®·ûË sin 1/3 A + sin 1/3 B + sin 1/3 C = c a b 1) 3 ( 1 2 ) 1/3 2) 3 ( 3 2 ) 1/3 3) ( 1 2 ) 1/3 4) ( 3 2 ) 1/3 0 a + 1 b 2 5. If A = [ 2a 1 0 c 2 ] is a skew symmetric then a + b + c = 2b + 1 2 + c 0 0 a + 1 b 2 A= [ 2a 1 0 c 2 ] Å≤˘-≠æ d´ ´÷vAéπ Å®·ûË a + b + c = 2b + 1 2 + c 0 1) 3 2) 3 3) 7 3 4) 1 3 6. The number of solutions of the system x +y + z = 0, 2x + 5y + 7z = 0, 2x 5y + 3z = 0 is x + y + z = 0, 2x + 5y + 7z = 0, 2x 5y + 3z = 0 ´u´Ææn ≤ƒüµ¿-†© Ææçêu www.eenadupratibha.net www.eenadupratibha.net REG-14

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    EAMCET ENGINEERING MODEL GRAND TEST

    No. of Questions: 160 Marks: 160 Time: 3 Hrs.

    MATHEMATICS1. If f (x) = cos [π] x + cos [πx] where [ . ] denotes the greatest integer function.

    Then f ( π2 ) = 1) cos 4 2) cos 3 3) 0 4) None of these

    2. f (2x + 3y, 2x − 7y) = 20x. Then f (x, y) =

    1) 7x − 3y 2) 7x + 3y 3) 3x − 7y 4) x − y

    3. The domain of f(x) = √

    is v°æü˨¡ç4x + 8

    2−3

    (x − 2)− 13 − 2

    2(x − 1)

    1) [1, ∞) 2) [0, ∞)

    3) [2, ∞) 4) does not exist ´u -́Æœnûªç é¬ü¿’a b c

    4. For a ∆ ABC, if b c a = 0, then sin1/3 A + sin1/3 B + sin1/3 C = c a ba b c

    vA¶µº’ïç ABC ™ b c a = 0, Å®·ûË sin1/3 A + sin1/3 B + sin1/3 C = c a b1) 3 (12)

    1/3

    2) 3 (√3

    2 )

    1/3

    3) (12)1/3

    4) (√3

    2 )

    1/3

    0 a + 1 b − 2

    5. If A = [ 2a − 1 0 c − 2 ] is a skew symmetric then a + b + c =2b + 1 2 + c 0

    0 a + 1 b − 2

    A = [ 2a − 1 0 c − 2 ] Å≤˘-≠æd´ ´÷vAéπ Å®·ûË a + b + c =2b + 1 2 + c 0

    1) 3 2) −3 3) 73

    4) 13

    6. The number of solutions of the system x +y + z = 0, 2x + 5y + 7z = 0, 2x−5y + 3z = 0 is

    x + y + z = 0, 2x + 5y + 7z = 0, 2x − 5y + 3z = 0 ´u´Ææn ≤ƒüµ¿-†© Ææçêu

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    1) 1 2) 2 3) 3 4) Infinite (ņçûªç)

    1 x x + 1

    7. If f(x) = 2x x(x − 1) (x + 1)x 3x(x − 1) x(x−1)(x−2) x(x−1)(x + 1)

    then f(50) + f(51) + ... + f(99) =

    1 x x + 1

    If f(x) = 2x x(x − 1) (x + 1)x 3x(x − 1) x(x−1)(x−2) x(x−1)(x + 1)

    -Å®·-ûË f(50) + f(51) + ... + f(99) =

    1) 3725 2) 0 3) 1275 4) None àD-é¬ü¿’

    8. The number of different nine digit numbers can be formed from the number223355888 by rearranging its digits. So that the odd digits occupy even position is

    223355888 ™E ÅçÈé-©†’ Ö°æ-ßÁ÷-TÆæ÷h à®Ωp-JîË 9 ÅçÈé© Ææçêu™x ¶‰Æœ Ææçêu©’ ÆæJ-≤ƒnØ√™xÖçúË Ææçêu© Ææçêu

    1) 16 2) 36 3) 180 4) 60

    17 17 1 7(1 + 17) [1 + ] [1 + ]........... [1 + ]2 3 19

    9. The value of =19 19 19

    (1 + 19) [1 + ] [1 + ]........... [1 + ]2 3 171) 1 2) 36C17 3)

    36C19 4)219

    10. The greatest number of points of intersection of 8 lines and 4 circles is

    8 Ææ®Ω-∞¡-Í®-ê©’, 4 ´%û√h©’ í∫J-≠æeçí¬ êçúÕç--èπ◊ØË êçúø† Gçü¿’-´¤© Ææçêu

    1) 64 2) 92 3) 104 4) 96

    11. If f(x) is a periodic function having period 7 and g(x) is a periodic functionhaving period 11, then the period of

    f(x) ÅØËC 7 Ç´-®Ωh-†çí¬ Ö†o Ç´-®Ωh† v°æ¢Ë’ߪ’ç, g (x) ÅØËC 11 Ç´-®Ωh-†çí¬ Ö†o Ç´®Ωh†v°æ¢Ë’ߪ’ç Å®·ûË

    xf(x) f ( )3

    D(x) = x is Ç´-®Ωh†çg(x) g( )51) 77 2) 231 3) 385 4) 1155

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    12. 15C2 + 2.15C3 + 3.15C4 + ... + 14.15C15 =

    1) 14.215 + 1 2) 13.214 + 1 3) 14.215 − 1 4) 13.214 − 1

    13. The last two digits of the number 3400 are

    3400 ™E *´J È®çúø’ Ææçêu©’

    1) 39 2) 29 3) 01 4) 43

    14. If α, β are roots of x2 − p (x + 1) − c = 0 then

    α2 + 2α + 1 β2 + 2β + 1the value of + =

    α2 + 2α + c β2 + 2β + c

    α2 + 2α + 1 β2 + 2β + 1x2 − p (x + 1) − c = 0 ´‚™«©’ α, β Å®·ûË + =

    α2 + 2α + c β2 + 2β + c

    1) 1 2) −1 3) 2 4) 0

    β2 + γ215. If α, β, γ are the roots of x3 + px2 + qx + r = 0, then ∑ =

    βγ

    β2 + γ2x2 + px2 + qx + r = 0 ´‚™«©’ α, β, γ Å®·ûË ∑ =

    βγ

    pq pq pq pq1) − 1 2) − 2 3) − 3 4) − 4

    r r r r

    16. The geometric mean of 3, 32, .......................... 3

    nis

    3, 32, .......................... 3

    ní∫’ù ´’üµ¿u´’ç

    1) 3n/2

    2) 3

    n( n+12

    )3) 4

    n/24) 3

    n+12

    17. Variance of the data 2, 4, 6, 8, 10 is

    2, 4, 6, 8, 10 © NÆæh %A

    1) 6 2) 7 3) 8 4) 9

    x2 + 5x + 1 A B C18. If = + + then B =

    (x + 1)(x + 2)(x + 3) x +1 (x +1) (x + 2) (x + 1) (x + 2) (x + 3)

    x2 + 5x + 1 A B C = + + Å®·ûË B =(x + 1)(x + 2)(x + 3) x +1 (x +1) (x + 2) (x + 1) (x + 2) (x + 3)

    1) 1 2) −5 3) 4 4) 0

    19. A determinant is chosen at random from the set of all determinants of order2 × 2 with elements 0 or 1 only. The probability that the determinant is positiveis

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    0 ™‰ü∆ 1 ´‚©-é¬-©ûÓ ´÷vûªç à®ΩpúË 2 × 2 ûª®Ω-í∫A E®√l¥-®Ω-鬩 †’ç* ߪ÷ü∑¿%-*a ¥-éπçí¬ äéπE®√l¥®Ωé¬Eo ᆒo-èπ◊çõ‰ ü∆E E®√l¥-®Ωéπç N©’´ üµ¿Ø√-ûªtéπç é¬ -́ú≈-EéÀ Ææ綵«-́ uûª

    3 3 5 71) 2) 3) 4)

    16 8 8 8

    20. The probability that in a random arrangement of the letters of the word COLLEGE,the two E's and two L's do not come together.

    COLLEGE ÅØË °æü¿ç-™E Åéπ~-®√-©†’ ߪ÷ü∑¿%-*a ¥-éπçí¬ äéπ ´®Ω’-Ææ™ Å´’-JaûË 2 E ©’, 2 L ©’äÍé-îÓô Öçúø-éπ-§Ú-́ -ú≈-EéÀ Ææ綵«´uûª

    18 20 191) 2) 3) 4) None

    21 21 21

    21. If α, β are the roots of the equation x2 − 15x + 1 = 0 then the value of

    1 1( − 15)−2 + ( − 15)−2 isα β1 1

    x2 − 15x + 1 = 0 ´‚™«©’ α, β Å®·ûË ( − 15)−2 + ( − 15)−2 =α β1) 225 2) 900 3) 223 4) 0

    22. Match the following: éÀçC¢√-öÀE ïûª-°æ-®Ω-îªçúÕ:

    List - I List - II

    A) Number of distinct terms 1) 212

    in the expansion of (x + y − z)16

    (x + y − z)16 ™E NGµ†o °æü∆© Ææçêu

    B) Number of terms in the expansion 2) 97

    (x + √x2 −1)6 + (x − √x2 −1)6

    (x + √x2 −1)6 + (x − √x2 −1)6 ™E °æü∆© ÆæçêuC) The number of irrational terms in ( 8√5 + 6√2 )

    1003) 4

    ( 8√5 + 6√2 )100

    ™E Åéπ-®Ω-ùÃߪ’ °æü∆© Ææçêu

    D) The sum of numerical 4) 153

    coefficients in (1 + x3 + 2y3 )12

    (1 + x3 + 2y3 )12

    í∫’ù-鬩 ¢Á·ûªhç 5) 1

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    A B C D

    1) 2 1 3 4

    2) 1 2 3 4

    3) 4 3 2 1

    4) 1 4 3 2

    23. A letter is known to have come from "INDORE or BANGLORE". On the postmark only three consecutive letters ORE are visible. The probability that the

    letter came from BANGLORE is

    INDORE †’ç* ™‰ü∆ BANGLORE †’ç* äéπ Öûªh®Ωç ´*açC ÅE ûÁL-ÆœçC. Ç Öûªh®ΩçO’C §ÚÆæd™¸ ´·vü¿™ ORE ÅØË ´‚úø’ ´®Ω’Ææ Åéπ~-®√©’ ´÷vûª¢Ë’ éπE°œÆæ÷h Öçõ‰ ÅCBANGLORE †’ç* ´*a Öçúø-ö«-EéÀ Ææ綵«-́ uûª

    1) 35

    2) 15

    3) 25

    4)45

    24. If the variance of the random variable X is 9, then the S.D. of the randomvariable −4x + 8 is

    äéπ ߪ÷ü∑¿%-*a ¥éπ -®√P X NÆæh %A 9 Å®·ûË −4x + 8 ÅØË îª©-®√P véπ´’ Nîª-©†ç

    1) 144 2) 27 3) 12 4) 16

    25. If x is a random poisson variate such that 2P(x = 0) + P(x = 2) = 2P(x = 1) thenE(x) =

    2P(x = 0) + P(x = 2) = 2P(x = 1) ÅßË’u™« x ߪ÷ü∑¿%-*a ¥é𠧃®·-ñ«Ø˛ -®√P Å®·ûËE(x) =

    1) 4 2) 3 3) 2 4) 1

    26. If tan π9

    , x, tan 5π18

    are in A.P. and tan π9

    , y, tan 7π18

    are in A.P. Then

    tan π9

    , x, tan 5π18

    ©’ A.P.™ -Ö-Ø√o®· -´’-J-ߪ· tan π9

    , y, tan 7π18

    ©’ A.P.™ Öçõ‰

    1) y = 2x 2) x = 2y 3) x = y 4) None

    27. The number of solutions of 16sin2x + 16cos

    2x = 10, 0 ≤ x ≤ 2π

    0 ≤ x ≤ 2π Å®·ûË 16sin2x + 16cos

    2x = 10 ≤ƒüµ¿†© Ææçêu

    1) 4 2) 6 3) 8 4) None

    28. If A = cot−1 √tanθ − tan−1 √

    tanθ then tan ( π4 − A2 ) =

    A = cot−1 √

    tanθ − tan−1 √

    tanθ Å®·ûË tan ( π4 − A2 ) =

    1) √cot θ 2) tanθ 3) cotθ 4) √

    tan θ

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    a cos A + b cos B + c cos C29. In a ∆ABC, =

    a + b + c

    a cos A + b cos B + c cos C∆ ABC ™ =

    a + b + c

    1) Rr

    2) R2r

    3) rR

    4)2rR

    30. In a ∆ABC, ∠B = π3

    and ∠C = π4

    and D divides BC internally in the

    sin ∠BADratio 1 : 3, then =

    sin ∠CAD

    ∆ ABC ™, ∠B = π3

    , ∠C = π4

    Å´¤ûª÷ BCE D Åçûª-®Ωçí¬ N¶µº->çîË

    sin ∠BADE≠æpAh 1 : 3 Å®·ûË =

    sin ∠CAD

    1 1 1 √21) 2) 3) 4)

    3 √3 √

    6 3

    (k + 1) + √k2+1

    31. If sec.h−1 (2/3) = log [ ], then k =k1) 2 2) 3 3) 5 4) 6

    32. A spherical balloon of radius 5 subtends an angle 60° at the eye of the observerwhile the angle of elevation of its centre is 30°. The height of the centre of theballoon is

    ¢√u≤ƒ®Ωl¥ç 5 Ö†o äéπ íÓ∞«-é¬-®Ω°æ¤ ¶„©÷Ø˛ °æJ-Q-©-èπ◊úÕ éπçöÀ ´ü¿l îËÊÆ éÓùç 60°, ¶„©÷Ø˛ Íéçvü¿çÜ®Ωl¥ y-éÓùç 30° Å®·ûË Íéçvü¿ç ´ü¿l ¶„©÷Ø˛ áûª’h

    1) 5 Units 2) 10 Units 3) 15 Units 4) 20 Units

    α α33. If fr (α) = (cos + i sin )r2 r2

    2α 2α α α(cos + i sin )................. (cos + i sin ) then limn→∞ fn(π) =r2 r2 r r 1) −1 2) 1 3) −i 4) i

    2n

    (1 + i)2n34. If n ∈ I, then + =

    (1 + i)2n 2n

    1) 0 2) 2 3) (1 + (−1)n).in 4) None of these

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    35. The maximum value of 3z + 9 − 7i if z + 2 − i = 5 is

    z + 2 − i = 5 Å®·ûË 3z + 9 − 7i í∫J≠æe N©’´

    1) 10 2) 15 3) 20 4) 5

    36. If r−

    = 3i−

    + 2 j−

    − 5 k−

    , a− = 2i−

    − j−

    + k−

    , b−

    = i−

    + 3 j−

    − 2k−

    and c−

    = −2i−

    + j−

    3 k−

    such that r−

    = λ a− + µ b−

    + ν c−

    then

    1) µ, λ−2

    , ν are in A.P. 2) λ, µ, ν are in A.P.

    3) λ, µ, ν are in H.P. 4) µ, λ, ν are in G.P.

    37. Let V−

    = 2 i−

    + j−

    − k−

    and W−

    = i−

    + 3 k−

    , if U−

    is a unit vector, then the

    maximum value of [ U−

    V−

    W−] is

    U−

    ߪ‚Eö¸ ÆæC¨¡ Å´¤ûª÷ V−

    = 2 i−

    + j−

    − k−

    , W−

    = i−

    + 3 k−

    Å®·ûË [ U−

    V−

    W−] í∫J≠æe

    N©’´

    1) −1 2) √10 + √

    6 3) √

    59 4) √

    60

    38. The vectors AB−

    = 3i−

    + 4k−

    and AC−

    = 5i−

    − 2j−

    + 4k−

    are the sides of a triangle

    ABC. The length of the median through A is

    AB−

    = 3i−

    + 4k−

    , AC−

    = 5i−

    − 2j−

    + 4 k−

    ©’ vA¶µº’ïç ABC ™ ¶µº’ñ«©’ Å®·ûË A O’ü¿’í¬

    ¢Á∞Ïx ´’üµ¿u-í∫-ûª-Í®ê §Òúø´¤

    1) √72 2) √

    33 3)√

    288 4) √

    18

    −r

    39. If a−

    = i−

    + j−

    , b−

    = 2j−

    − k−

    and r−

    × a−

    = b−

    × a−

    , r−

    × b−

    = a−

    × b−

    then =r−

    1 11) ( i

    −+ 3 j

    −− k

    −) 2) ( i

    −− 3 j

    −+ k

    −)

    √11 √

    11

    13) ( i

    −− j−

    + k−

    ) 4) None√

    3

    40. If (b−

    × c−

    ) × (c−

    × a−

    ) = 3c−

    then [ b−

    × c−

    c−

    × a−

    a−

    × b−

    ] =

    1) 2 2) 7 3) 9 4) 11

    1n

    41. If ak = for k = 1, 2, 3, ...........n then ( Σ ak)2

    =k (k + 1) k = 1

    1 n

    k = 1, 2, 3, ... ...........n èπ◊ ak = Å®·ûË ( Σ ak)2

    =k(k + 1) k = 1

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    n n2 n4 n61) 2) 3) 4)

    n + 1 (n + 1)2 (n + 1)4 (n + 1)6

    cosθ sinθ a b42. If = then + =

    a b sec 2θ cosec 2θa

    1) a 2) b 3) 4) a + bb

    43. The locus of the point (r cos α cos β, r cos α sinβ, r sinα) is

    (r cos α cos β, r cos α sinβ, r sinα) Gçü¿’´¤ Gçü¿’-°æü∑¿ç

    1) x2 − y2 − z2 = r2 2) x2 + y2 + z2 = r2

    3) x2 + y2 − z2 = r2 4) x2 − y2 + z2 = r2

    44. When the angle of rotation of axes is tan−12, then transformed equation of4xy − 3x2 = a2 is

    Åé~¬-©†’ tan−1 2 éÓùç™ v¶µº´’ù °æJ-́ -®Ωh† îËÊÆh 4xy − 3x2 = a2 †÷ûª† ÆæO’-éπ-®Ωùç

    1) X2 − 4Y2 = a2 2) 2X2 − Y2 = a2

    3) 2X2 − 3Y2 = a2 4) 3X2 − 2Y2 = a2

    45. The coordinates of a point on x + y + 3 = 0 whose distance from x + 2y + 2 = 0is √

    5 are

    x + 2y + 2 = 0 †’ç* √5 ü¿÷®Ωç™ Öçô÷ x + y + 3 = 0 O’ü¿ Ö†o Gçü¿’´¤

    1) (9, 6) 2) (−9, 6) 3) (6, −9) 4) (−9, −6)

    46. If a ≠ b ≠ c and if ax + by + c = 0, bx + cy + a = 0, cx + ay + b = 0

    are concurrent then 2a2b−1c−1

    . 2b2c−1a−1

    . 2c2a−1b−1

    =

    a ≠ b ≠ c- Å®·u ax + by + c = 0, bx + cy + a = 0, cx + ay + b = 0 -Å-†’-≠æé¬h-™„j-ûË

    2a2b−1c−1

    . 2b2c−1a−1

    . 2c2a−1b−1

    =

    1) 8 2) 0 3) 2 4) None

    47. If (−4, 5) is one vertex and 7x − y + 8 = 0 is one diagonal of a square then theperpendicular distance from (−3, 4) to the other diagonal is

    îªûª’®ΩvÆæç™E äéπ Q®Ω{ç (–4, 5), äéπ éπ®Ωgç 7x − y + 8 = 0 Å®·ûË (–3, 4) Gçü¿’´¤ †’ç*È®çúÓ éπ®√g-EéÀ Ö†o ©ç•-ü¿÷®Ωç

    4√2 2 √

    2 √

    2 3√

    2

    1) 2) 3) 4) 5 5 5 5

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    48. The distance between the parallel lines given by

    (x + 7y)2 + 4√2(x + 7y) − 42 = 0

    (x + 7y)2 + 4√2(x + 7y) − 42 = 0 Ææ÷*çîË Ææ´÷ç-ûª®Ω Í®ê© ´’üµ¿u ü¿÷®Ωç

    41) 2) 4 √

    2 3) 2 4) 10 √

    2

    5 49. The reflection of the point p(1, 0, 0) in the line

    x − 1 y + 1 z + 10 = = is

    2 −3 8x − 1 y + 1 z + 10 = = Í®ê°j p (1, 0, 0) Gçü¿’´¤ v°æA-Gç•ç

    2 −3 8

    1) (3, −4, −2) 2) (5, −8, −4) 3) (1, −1, −10) 4) (2, −3, 8)

    50. A piece of ice in the form of a cube melts so that the percentage error in the sideis 0.7, then the percentage error in the volume is

    äéπ °∂æ’Ø√-é¬-®Ω°æ¤ ´’ç-í∫úøf éπJ-Íí-ô-°æ¤púø’ v°æA¶µº’ïç™ 0.7% ûª®Ω’-í∫’-ü¿© éπE-°œÊÆh, Ç °∂æ’†ç°∂æ’†°æJ-´÷-ùç-™E Ææ’´÷®Ω’ üÓ≠æ-¨»ûªç

    1) 0.7 2) 1.4 3) 2.1 4) None

    51. The value of

    1 2 n − 1 n n nlim 1 e e e

    n→∞ [ + + + ...... + ]isn n n n

    1) 1 2) 0 3) e + 1 4) e − 1

    [(a − n)nx − tan x ] sin nx52. If lim = 0, n ≠ 0 then a =

    x→0 x2

    1 11) 0 2) n + 3) 1 + 4) n

    n n

    x2 − x d f −1(x)53. If f(x) = then =

    x2 + 2x dx

    3 −3 3 −31) 2) 3) 4)

    1− x2 1 − x2 (1 − x)2 (1 − x)2

    1 + x 1 dy54. If y = log(

    1

    )4 − tan−1 x then =1 − x 2 dxx2 2x2

    1) 2) 1 − x4 1 − x4

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    x23) 4) None of these

    2(1 − x4)

    55. Minimum value of f(x) = (sin−1 x)3 + (cos−1 x)3 (−1 < x < 1)

    7π3 3π3 π3 3π31) 2) 3) 4)

    8 8 32 16

    56. The acute angle between the curves y = x2 − 1 and y = x2 − 3 at their pointof intersection when x > 0 is

    x > 0 Å®·-†-°æ¤púø’ y = x2 − 1, y = x2 − 3 © êçúø† Gçü¿’´¤ ´ü¿l ´v鬩 Å©p-éÓùç

    4 4√2

    1) tan−1 () 2) tan−1()7√2 74

    3) tan ( ) 4) None7dx

    57. ∫ = x(xn + 1)

    1 xn 1 xn + 11) log ( ) + c 2) log ( ) + cn xn + 1 n xn

    xn3) log ( ) + c 4) Nonexn + 1

    (x2 − 1)dx 58. ∫ =

    (x4 + 3x2 + 1) tan−1 x2 + 1( )x

    1 11) log tan (x + ) + C 2) log tan−1(x + ) + Cx x

    13) log sec−1(x + ) + C 4) None of thesex3π−2

    2x59. ∫

    0sin [ ] dx where [ . ] denotes the greatest integerπ

    1) π−2

    (sin 1 + cos 1) 2) π−2

    (sin 1 − cos 2)

    3) π−2

    (sin 1 − cos 1) 4) π−2

    (sin 1 + cos 2)

    cos3x60. If ∫ dx = log sin x+ A sin x + c then A =

    sin2x + sin x

    1) 0 2) 1 3) −1 4) 2

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    61. If the area lying between the curves y = ax2 and x = ay2 is 1 square unit. Then a =

    y = ax2, x = ay2 ´v鬩 ´’üµ¿u v°æü˨¡ ¢Áj¨»©uç 1 îª.ߪ‚-Eö¸ Å®·ûË a =

    1) 1−√

    3

    2) 1−2

    3) 1−3

    4) √3

    62. For x > 0 the integrating factor of x2dydx

    + 3xy − 1 = 0

    x > 0 Å®·ûË x2dydx

    + 3xy − 1 = 0 Ææ´÷-éπ-©† 鬮Ω-ù«çéπç

    1) x 2) x2 3) x3 4)1x3

    63. The general solution of eydydx

    + 2 ey

    x

    = sin x

    1) xey = −x cos x + sin x + c

    2) x2ey = −x2 cos x − 2x sin x − cos x + c

    3) xey = x cos x − sin x + c

    2 sin x cos x c4) ey = −cos x + + +

    x x2 x2

    64. Equation of circle whose radius is 5 and which touch the circle x2 + y2 − 2x− 4y −20 = 0 at (5, 5) is

    ´%ûªh ¢√u≤ƒ®Ωl¥ç 5 Ö†o x2 + y2 −2x −4y − 20 = 0 ´%û√hEo (5, 5) Gçü¿’´¤ ´ü¿l Ææp %PçîË´%ûªh ÆæO’éπ®Ωùç

    1) (x − 9)2 + (y − 8)2 = 5 2) (x − 9)2 + (y + 8)2 = 25

    3) (x − 1)2 + (y − 2)2 = 25 4) (x − 9)2 + (y − 8)2 = 25

    65. Equations of the tangents of

    x2 + y2 + 4x + 6y − 12 = 0 perpendicular to 4x + 3y + 1 = 0 are

    4x + 3y + 1 = 0 Í®êèπ◊ ©ç•çí¬ ÖçúË x2 + y2 + 4x + 6y − 12 = 0 Ææp®Ωz-Í®-ê©’

    1) 3x − 4y − 31 = 0, 3x − 4y + 19 = 0

    2) 3x − 4y + 31 = 0, 3x − 4y − 19 = 0

    3) 3x − 4y − 21 = 0, 3x − 4y + 20 = 0

    4) 3x − 4y − 41 = 0, 3x − 4y − 20 = 0

    66. If (a, −3), ( 8−15 , 2) are conjugate points w.r.to circle 2x2 + 2y2 + 7x + 14y + 8 = 0 the value of a is

    2x2 + 2y2 + 7x + 14y + 8 = 0 ´%ûªhç ü¿%≥ƒd u (a, −3), ( 8−15 , 2) Ææçߪ·í∫t Gçü¿’-´¤©’Å®·ûË a N©’´

    1) −2 2) −3 3) 2 4) 3

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    67. The circles whose equations are x2 + y2 + 10x − 2y + 22 = 0 andx2 + y2 + 2x − 8y + 8 = 0 touch each other, the circle which touch both circlesat the point of contact and passes through (0, 0) is

    x2 + y2 + 10x − 2y + 22 = 0, x2 + y2 + 2x − 8y + 8 = 0 ÅØË ´%û√h©†’ Ææp®Ωz Gçü¿’´¤´ü¿l Ææp %PÆæ÷h (0, 0) O’ü¿’í¬ ¢Á∞Ïx ´%ûªh ÆæO’-éπ-®Ωùç

    1) 9(x2 + y2) − 15x − 20y = 0 2) 5(x2 + y2) − 18x − 80y = 0

    3) 7(x2 + y2) − 18x − 80y = 0 4) x2 + y2 − 9x − 40y = 0

    68. The radical axis of two circles x2 + y2 + 4x + 2y − 4 = 0,x2 + y2 − 2x − 4y − 20 = 0 divides the line segment joining the centres of circlein the ratio is

    x2 + y2 + 4x + 2y − 4 = 0, x2 + y2 − 2x − 4y − 20 = 0 ÅØË ´%û√h© ´‚™«éπ~ç´%ûªhÍéçvü∆-©†’ éπLÊ° Í®ë«-êç-ú≈Eo N¶µº>çîË E≠æpAh

    1) 1 : 4 2) 4 : 1 3) 1 : 17 4) 17 : 1

    69. If the ends of a focal chord of the parabola y2 = 4ax are (x1, y1) and (x2, y2) thenx1 x2 + y1y2 =

    y2 = 4ax °æ®√-́ -©ßª’ç Ø√Gµ ñ«u * -́®Ω©’ (x1, y1), (x2, y2) Å®·ûË x1 x2 + y1y2 =

    1) a2 2) −3a2 3) 5a2 4) −5a2

    70. Match the eccentricities of the following.

    éÀçC¢√öÀ ÖûË\ç-vü¿-ûª-©†’ ïûª-°æ-®Ω-îªçúÕ.

    List - I List - II

    x2 y2 7A) + = 1 1) √

    9 16 10

    x2 y2 √3

    B) + = 1 2) 25 9 2

    C) 3(x − 1)2 + 12(y − 1)2 = 36 3) 45

    √7

    D) 3x3 + 10y2 = 30 4) 4

    1) A−3, B−1, C−2, D−4

    2) A−3, B−2, C−1, D−4

    3) A−4, B−1, C−3, D−2

    4) A−4, B−3, C−2, D−1

    71. If e1 and e2 are the eccentricities of the hyperbola xy = c2, x2 − y2 = c2

    then e12+ e

    22 =

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    xy = c2, x2 − y2 = c2 ÅA °æ®√-´-©-ߪ÷© ÖûË\ç-vü¿-ûª©’ e1, e2 Å®·ûË e12 + e

    22 =

    1) 1 2) 4 3) 6 4) 8

    72. The tangent at any point on the curve x = acos3θ, y = asin3θ meets the axes inP and Q. The locus of the midpoint of PQ is

    x = acos3θ, y = asin3θ ´vé¬-©èπ◊ àüÁjØ√ Gçü¿’´¤ †’ç* UÆœ† Ææp®Ωz-Í®ê Åé~¬-©†’ P, Q © ´ü¿lÆæp %PÊÆh PQ ´’üµ¿u Gçü¿’´¤ Gçü¿’-°æü∑¿ç

    1) x3−2 + y

    3−2 = a

    3−2 2) x

    2−3 + y

    2−3 = a

    2−3

    3) 4(x + y) = a 4) 4 (x2 + y2) = a2

    73. If β is the acute angle between the lines px + qy = p + q and p(x − y) + q(x + y) = 2q. Then the value of sin β

    px + qy = p + q, p(x − y) + q(x + y) = 2q ÅØË Í®ê© ´’üµ¿u Å©p-éÓùç β Å®·ûË sin β N©’´

    √3 1

    1) 2) 34

    3) 12

    4) 2 √

    2x − 3 y − 8 z − 3

    74. The shortest distance between lines = = and3 −1 1

    x + 3 y + 7 z − 6 = = is−3 2 4

    x − 3 y − 8 z − 3 x + 3 y + 7 z − 6 = = , = = ÅØË Í®ê© ´’üµ¿u éπE≠æe ü¿÷®Ωç

    3 −1 1 −3 2 4

    1) 3 √30 2) √

    30 3) 2 √

    30 4) 4√

    30

    75. 72n + 3n − 1. 23n − 3 is divisible by

    72n + 3n − 1. 23n − 3 †’ ¶µ«TçîË Ææçêu

    1) 24 2) 25 3) 9 4) 13

    76. f(x) = kx3 − 9x2 + 9x + 3 is increasing for all x then

    x ÅEo N©’-´-©èπ◊ f(x) = kx3 − 9x2 + 9x + 3 -Ǯӣæ«ùç Å®·ûË

    1) k < 3 2) k < −1 3) k > 3 4) k < −2

    77. On the interval [0, 1] the function x25(1 − x)75 takes maximum value at

    [0, 1] Åçûª-®Ωç™ x25(1 − x)75 v°æ¢Ë’ߪ’ç í∫J≠æeç 鬢√-©çõ‰

    1) x = 0 2) x = 12

    3) x = 1 4) x = 14

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    78. If α, β, γ, δ are four solutions of the equation tan (θ + π4) = 3 tan 3θ, no two of which have equal tangents then the value of tan α + tan β + tan γ + tan δ is

    à È®çúø’ Ææp®Ωz-Í®-ê©’ Ææ´÷†ç é¬E tan (θ + π4) = 3 tan 3θ ÆæO’-éπ-®Ω-ù«-EéÀ 4 ´‚™«©’α, β, γ, δ Å®·ûË tan α + tan β + tan γ + tan δ =

    1) 1 2) 0 3) −2 4) None

    9 √x

    79. ∫ dx =4 √

    x − 1

    1) 7 + 2 log 2 2) 5 + 2 log 2 3) 2 + log 2 4) None

    80. f is function such that f(x + y) = f(x) + f(y) ∀ x, y and f(1) = 2

    If φ (x) =x

    0∫ ƒ (2t) dt then φ'(3) =

    xy ÅEo N©’-´-©èπ◊ f(x + y) = f(x) + f(y), f(1) = 2 ÅßË’u™« f äéπ v°æ¢Ë’ߪ’ç Å´¤ûª÷

    φ (x) = x

    0∫ ƒ (2t) dt Å®·ûË φ'(3) =

    1) 3 2) 4 3) 6 4) 12

    PHYSICS1000V

    81. The current, voltage relation of a diode is given by i = (e T - 1) mA, wherethe applied voltage 'V' is in volts and the temperature 'T' is in kelvin. If a student

    makes an error measuring ± 0.01 V while measuring the current of 5 mA at300 K. The error in measuring the current value in mA will be

    1000V

    äéπ úøßÁ÷ú˛ Nü¿’uû˝, ¢Ó™‰d>© ´’üµ¿u Ææç•ç-üµ∆Eo Ææ÷*çîË ÆæO’-éπ-®Ωùç i = (e T - 1) mA,v°æßÁ÷-Tç-*† ¢Ó™‰d> 'V' 㙸d©™, Ö≥Úg-ví∫ûª 'T' ÈéLy-Ø˛™ ÖØ√o®·. äéπ Nü∆uJn 300 KÖ≥Úg-ví∫ûª ´ü¿l 5 mA Nü¿’uû˝ v°æ¢√-£æ…Eo éÌL-îË-ô-°æ¤púø’ îËÆœ† üÓ≠æç ± 0.01VÅ®·ûË, Nü¿’uû˝v°æ¢√-£æ…Eo éÌ©-´-ôç™ üÓ≠æç N’Mx Çç°œ-ߪ’-®Ωx™

    1) 0.5 2) 0.2 3) 0.05 4) 0.02

    82. The special number π is defined as the ratio of circumference of a circle to itsdiameter. The number of significant digit in it

    v°æûËuéπ Ææçêu π E äéπ ´%ûªhç °æJ-CµéÀ, ü∆E ¢√u≤ƒ-EéÀ Ö†o E≠æpAhí¬ E®Ωy-*ÊÆh, Åçü¿’-™E ≤ƒ®Ωnéπ

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    Ææçêu©’

    1) 1 2) 2 3) 3 4) infinity (ņçûªç)

    83. Six coplanar forces having magnitudes 1 N, 2 N, 3 N, 4 N, 5 N and 6 N areacting simultaneously on a particle. If their directions are represented by the six

    sides of a regular hexagon taken in order, the magnitude of their resultant is

    1 N, 2 N, 3 N, 4 N, 5 N, 6 N °æJ-´÷-ù«©’†o Ç®Ω’ Ææûª-Mߪ’ •™«©’ àéπ-é¬-©ç™ äéπ éπùçO’ü¿ °æE-îËÆæ’hØ√o®·. ¢√öÀ C¨¡-©†’ äéπ véπ´’ç™ BÆæ’èπ◊çõ‰, äéπ véπ´’-≠æ-úø’s¥> Ç®Ω’ ¶µº’ñ«©’Ææ÷*ÊÆh, ¢√öÀ °∂æLûª •© °æJ-´÷ùç

    1) zero (Ææ’Ø√o) 2) 3√3 N 3) 6 N 4) 9 N

    84. The position of an object moving along X - axis is given by x = a + bt2 wherea = 8.5 m, b = 2.5 ms−2 and 't' is in seconds. The average velocity of the objectbetween t = 2.0 s and t = 4.0 s is

    X- Åéπ~ç ¢Áç•úÕ v°æߪ÷-ùÀçîË äéπ ´Ææ’h´¤ ≤ƒnØ√Eo Ææ÷*çîË ÆæO’-éπ-®Ωùç x = a + bt2 ™a = 8.5 m, b = 2.5 ms−2, 't' ÂÆéπ-†x™ ÖØ√o®·. t = 2.0 s, t = 4.0 s ´’üµ¿u ´Ææ’h´¤ Ææí∫ô’¢Ëí∫ç

    1) 0 ms−1 2) 10 ms−1 3) 1.5 ms−1 4) 15 ms−1

    85. A body moves with a speed t − 2 ms−1, the distance travelled by it in first fourseconds is

    äéπ ´Ææ’h´¤ ´úÕ t − 2 ms−1 Å®·ûË ¢Á·ü¿öÀ Ø√©’í∫’ ÂÆéπ-†x™ ÅC v°æߪ÷-ùÀçîË ü¿÷®Ωç

    1) 4 m 2) 8 m 3) 2 m 4) 0 m

    86. For a particle thrown slantwise from a point on the ground, the time of flight is

    T. At time t = T4

    and t = T2

    , it is found to be at heights 'h' and 'H' above the

    ground respectively. The ratio hH is¶µº÷N’ O’ü¿ Ö†o Gçü¿’´¤ †’ç* àô-¢√©’í¬ v°æéÀ~°æhç îËÆœ† äéπ éπù«-EéÀ °æ™«-ߪ’† 鬩ç T.

    鬩ç t = T4 , t = T2

    ´ü¿l ÅC ¶µº÷N’ †’ç*´®Ω’-Ææí¬ 'h', 'H' áûª’h© ´ü¿l Öçõ‰ hH

    E≠æpAh N©’´

    4 3 2 11) 2) 3) 4)

    5 4 3 2

    87. A force 'F' acting on a body depends on its displacement 'S' as F ∝ (S)−1/3. Thepower delivered by 'F' will depend on displacement as

    äéπ ´Ææ’h-´¤Â°j °æE-îËÊÆ •©ç 'F' ü∆E ≤ƒn†-v¶µºç¨¡ç 'S' O’ü¿ F ∝ (S)−1/3 ÅßË’u Nüµ¿çí¬Çüµ∆®Ω°æúø’-ûª’çC. •©ç 'F' ´©x éπLÍí ≤ƒ´’®Ωn uç, ≤ƒn†-v¶µºç-̈ ¡ç°j éÀçC Nüµ¿çí¬ Çüµ∆-®Ω-°æ-úø’-ûª’çC

    1) S1/2

    2) S0 3) S2/3

    4) S−5/3

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    88. When a ball is released from a height 'h' above a horizontal sand surface, it

    penetrates into the sand to a depth of h−8

    . If 'g' is acceleration due to gravity the

    initial vertical speed that must be given to the ball from that point, so that it again

    rises to the initial point of release is

    éÀ~Aï Ææ´÷ç-ûª®Ω ÉÆæ’éπ ûª©ç †’ç* 'h' áûª’h† Ö†o äéπ Gçü¿’´¤ †’ç* äéπ •çAE éÀçCéÀñ«®Ω-N-úÕ-*-†-°æ¤púø’, •çA ÉÆæ’-éπ-™éÀ h−

    8™ûª’èπ◊ îÌa-éÌ-E-§Ú-ûª’çC. •çA ¢Á·ü¿ô ñ«®Ω-N-úÕ-*†

    Gçü¿’-´¤èπ◊ îË®√-©çõ‰, ÉÆæ’-éπ-™E Ç Gçü¿’´¤ †’ç* •çAE Eôd-E-©’´¤í¬ °jéÀ NÆæ-®√-Lq† ûÌL´úÕ N©’´ (g í∫’®Ω’ûªy ûªy®Ωùç)

    9gh1) √

    2) √4gh 3) √

    9gh 4) √

    8gh

    2

    89. A particle of mass 1 kg and carrying a charge 0.01 C is at rest on an inclined

    490plane of angle 30° with horizontal when an electric field of NC

    −1is applied

    √3

    parallel to horizontal. The coefficient of friction between the particle and inclinedplane is

    490éÀ~Aï Ææ´÷ç-ûª®Ω C¨¡™ NC−1 °æJ-´÷ùç Ö†o äéπ Nü¿’uû˝ Íé~vû√Eo v°æßÁ÷-Tç-*-†-°æ¤púø’

    √3

    1 Kg vü¿´u-®√P, 0.01 C Nü¿’u-ü∆-¢Ë¨¡çûÓ Ö†o äéπ éπùç éÀ~Aï Ææ´÷ç-ûª-®ΩçûÓ 30° éÓùç îËÆæ’h†o¢√©’-ûª©ç°j E¨¡a© ÆœnA™ ÖçC. éπùç, ¢√©’-ûª©ç ´’üµ¿u °∂æ’®Ω{ù í∫’ùéπç

    √3 1 1 √

    3

    1) 2) 3) 4) 2 2 √

    3 7

    90. An inclined plane makes an angle of 30° with the horizontal. The upper half ofthe inclined plane is perfectly smooth and the lower half is rough. Starting from

    rest a small block begins to slide down the inclined plane from the top and comes

    to rest at the bottom. The coefficient of friction in the lower half is

    äéπ ¢√©’-ûª©ç éÀ~Aï Ææ´÷ç-ûª-®ΩçûÓ 30° éÓùç îË≤ÚhçC. ¢√©’-ûª©ç áí∫’´ Ææí∫ç †’†oí¬, Cí∫’´Ææí∫ç í∫®Ω’-èπ◊í¬ ÖØ√o®·. ¢√©’-ûª©ç °j¶µ«-í¬† E¨¡a© ÆœnA™ Ö†o äéπ *†o C¢Á’t ¢√©’-ûª©ç¢Áç•úÕ éÀçCéÀ ñ«®Ω’ûª÷ ´*a ¢√©’-ûª©ç Åúø’-í∫’-¶µ«-í¬† E¨¡a© ÆœnAéÀ ´*açC. ¢√©’-ûª©ç Cí∫’´Ææí¬-EéÀ °∂æ’®Ω{ù í∫’ùéπç

    1 2 4 11) 2) 3) 4)

    2√3 √

    3 √

    3 √

    3

    91. A neutron of mass 'm' is approaching a stationary deuterium nucleus of mass'2 m'. The ratio of the de Broglie wavelengths of the neutron and deuterium

    nucleus in the frame of reference of the centre of mass of the system is

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    m vü¿´u-®√P Ö†o äéπ †÷uvö«Ø˛, E¨¡a© ÆœnA™ Ö†o 2 m vü¿´u-®√P Ö†o äéπ úø’uöÃ-Jߪ’çÍéçvü¿-é¬Eo ÆæO’-°œ-≤ÚhçC. ´u´Ææn vü¿´u-®√P Íéçvü¿ EÍ®l¨¡ îªvôç ü¿%≥ƒd u †÷uvö«Ø˛, úø’uöÃ-Jߪ’çÍéçvü¿-鬩 úŒ v¶-Ux ûª®Ωç-í∫-üÁj-®√` u© E≠æpAh

    1) 1 : 2 2) 2 : 1 3) 1 : 1 4) √2 : 1

    92. Two particles moving in the opposite direction along the same horizontal withvelocities v1 and v2 collide with each other. If 'e' is the coefficient of restitution,

    then their distance of separation at time 't' after collision is

    È®çúø’ éπù«©’ äéπ éÀ~Aï Ææ´÷ç-ûª®Ω Í®ê ¢Áç•úÕ v1, v2 ¢Ëí¬-©ûÓ °æ®Ω-Ææp®Ωç ´uA-Í®éπ C¨¡™xv°æߪ÷-ùÀÆæ÷h ÅGµ-°∂æ÷ûªç îÁçü∆®·. v°æû√u-́ -≤ƒn† í∫’ùéπç 'e' Å®·ûË, ÅGµ-°∂æ÷ûªç ïJ-T† 't' 鬩çûª®√yûª ¢√öÀ ´’üµ¿u ü¿÷®Ωç

    1) te (v1 + v2) 2) te (v1 − v2) 3) te(v2 − v1) 4) te ( v1v2)93. A uniform solid sphere of radius 0.2 m and mass 5 kg rotates about its diameter.

    Angular velocity of the sphere changes with time 't' according to the equation

    ω = (3 + 5t) rad s−1. The tangential force applied to the sphere is

    0.2 m ¢√u≤ƒ®Ωl¥ç, 5 kg vü¿´u-®√P Ö†o äéπ àéπ-KA °∂æ’†-íÓ∞¡ç ü∆E ¢√uÆæç ü¿%≥ƒd u v¶µº´’-ù«©’îË≤ÚhçC. íÓ∞¡ç éÓùÃߪ’¢Ëí∫ç 鬩ç 't' ü¿%≥ƒd u ω = (3 + 5t) rad s−1 ÆæO’-éπ-®Ωùç ü¿%≥ƒd u´÷®Ω’-ûÓçC. íÓ∞¡ç°j °æE-îËÊÆ Ææp®Ωz-Í®-&ߪ’ •©ç

    1) 4 N 2) 2 N 3) 3 N 4) 1 N

    94. Two blocks of masses 3 kg and 6 kg are kept on a smooth horizontal floor. Theblocks are connected together by a light spring of force constant 20 N/m. Steady

    forces of 5 N and 5 N are applied on the blocks to stretch the spring. If the

    applied forces are suddenly removed, the minimum time after which the spring

    will have its minimum length (π2 = 10)

    3 kg, 6 kg vü¿´u-®√¨¡Ÿ©’†o È®çúø’ C¢Á’t©†’ éÀ~Aï Ææ´÷ç-ûª®Ω †’†’°j† ûª©ç°j Öçî√®Ω’.È®çúø’ C¢Á’t©†’ 20 N/m •©-Æœn-®√çéπç Ö†o äéπ ûËL-Èéj† vÆœpçí∫’ Ææ£æ…-ߪ’çûÓ éπL§ƒ®Ω’. vÆœpçí∫’≤ƒÍíNüµ¿çí¬ 5 N, 5 N Æœn®Ω •™«©†’ C¢Á’t-©Â°j v°æßÁ÷-Tç-î√®Ω’. v°æßÁ÷-Tç-*† •™«-©†’Åéπ-≤ƒtûª’hí¬ ûÌ©-TÊÆh, vÆœpçí∫’ éπE≠æe §Òúø-´¤†’ §Òçü¿-ú≈-EéÀ °æõ‰d éπFÆæ Æ洒ߪ’ç (π2 = 10)

    1) 2 S 2) 1 S 3) 14

    S 4) 12

    S

    95. If the minimum and maximum distances of an orbiting satellite from the Earth's

    centre are 2 R and 4 R respectively (R = Radius of Earth), then the ratio of

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    smooth floor

    5 N5 N

    3 kg 6 kgK = 20 N/m

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    minimum to maximum speed of the satellite is

    ¶µº÷N’ ô÷d °æJ-v¶µº-N’-Ææ’h†o äéπ Ö°æ-ví∫-£æ…-EéÀ ¶µº÷N’ Íéçvü¿ç †’ç* Ö†o éπE≠æe, í∫J≠æe ü¿÷®√©’´®Ω’-Ææí¬ 2 R, 4 R (R = ¶µº÷N’ ¢√u≤ƒ®Ωl¥ç) Å®·ûË, Ö°æ-ví∫£æ«ç éπE≠æe ´úÕéÀ, í∫J≠æe ´úÕéÀ Ö†oE≠æpAh

    1) 1 : 3 2) 2 : 3 3) 1 : 2 4) 3 : 5

    96. A uniform metal wire elongates by 2 mm when a load of weight W is suspendedfrom its free end. If this wire is passed over a smooth, light pulley and weights

    each of W are hung at two free ends of the wire, then the elongation of the wire is

    äéπ àéπ-KA ™£æ«°æ¤ Bí∫ ÊÆyî√a ¥ é̆ †’ç* W ¶µ«®Ωç Ö†o äéπ C¢Á’t†’ ¢Ë™«úø-D-Æœ-†-°æ¤púø’Bí∫-™E ≤ƒí∫’-ü¿© 2 mm. Ñ Bí∫†’ äéπ ûËL-Èéj†, °∂æ’®Ω{-ù-®Ω-£œ«ûª éπ°‘p O’ü¿’í¬ §ÚE*a, Bí∫ È®çúø’ÊÆyî√a ¥ é̆© †’ç* äéÌ\-éπ\öÀ W ¶µ«®Ωç Ö†o È®çúø’ C¢Á’t-©†’ ¢Ë™«úø-DÊÆh Bí∫-™E ≤ƒí∫’-ü¿©

    1) 12

    mm 2) 4 mm 3) 1 mm 4) 2 mm

    97. When a soap bubble is blown, the work done per unit volume of the bubble is (P

    is excess pressure inside the soap bubble)

    äéπ Ææ•’s •’úø-í∫†’ ÜC-†-°æ¤púø’, •’úøí∫ àé¬çéπ °∂æ’†-°æ-J-´÷-ù«-EéÀ ïJT† °æE (Ææ•’s •’úø-í∫-™EÅü¿-†°æ¤ °‘úø†ç P)

    3P 3P 2P P1) 2) 3) 4)

    2 4 3 2

    98. From the given graph between centigrade (°C) and Fahrenheit (°F) temperature of a body, the angle

    made by the graph line with Fahrenheit axis is

    äéπ ´Ææ’h´¤ ÂÆçöÃ-vÍíú˛ Ö≥Úg-ví∫ûª (°C), §∂ƒÈ®-Ø˛-£‘«ö¸ Ö≥Úg-ví∫ûª (°F) ©èπ◊ UÆœ† ví¬°∂ˇ †’ç* ví¬°∂ˇ Í®ê

    §∂ƒÈ®-Ø˛-£‘«ö¸ Åéπ~çûÓ îËÊÆ éÓùç

    5 5 5 51) tan−1 () 2) cos−1() 3) sin−1() 4) cos−1 ()106 106 √ 106 √ 106

    99. An insulated container contains 4 moles of an ideal diatomic gas at atemperature T. Heat Q is supplied to this gas, due to which 2 moles of the gas are

    dissociated into atoms, but temperature of the gas remains constant. Then

    (R = universal gas constant)

    äéπ Ö≠æg-•ç-üµ¿é𠧃vûª™ T Ö≥Úg-ví∫ûª ´ü¿l 4 ¢Á÷™¸© äéπ Çü¿®Ωz Cy°æ-®Ω-´÷-ù’éπ ¢√ߪ·´¤ ÖçC.Ñ ¢√ߪ·-´¤èπ◊ Q Ö≠æg-¨¡-éÀhE Åçü¿ñ‰ßª’úøç ´©x 2 ¢Á÷™¸© ¢√ߪ·´¤ °æ®Ω-´÷-ù’-´¤-©’í¬NúÕ§Úûª’çC. R ≤ƒ®Ωy-vAéπ ¢√ߪ· Æœn®√çéπç Å®·ûË

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    °C

    α°F

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    1) Q = 2RT 2) Q = RT 3) Q = 4RT 4) Q = 3RT

    100. Nitrogen though a diatomic gas exhibits vibrations at higher temperatures. The ratioof the specific heat at constant pressure to the specific heat at constant volume is

    ØÁjvö-ïØ˛ Cy°æ-®Ω-´÷-ù’éπ ¢√ߪ·´¤ Å®·-†-°æp-öÀéÃ, ÅCµéπ Ö≥Úg-ví∫-ûª© ´ü¿l éπç°æ-Ø√-©†’ v°æü¿-Jz-Ææ’hçC.Æœn®Ω-°‘-úø†ç ´ü¿l ¢√ߪ· NP-≥Úd-≥ƒg-EéÀ, Æœn®Ω °∂æ’†-°æ-J-´÷ùç ´ü¿l ¢√ߪ· NP-≥Úd--≥ƒg-EéÀ Ö†o E≠æpAh

    1) 9−7

    2) 5−3

    3) 7−5

    4) 8−6

    101. A transverse wave of amplitude 10 cm is generated at one end (x = 0) of a long

    stretched string by a tuning fork of frequency 500 Hz. At a certain instant of time,

    the displacement of a particle at 'A' (at x = 100 cm) is −5 cm and of a particle B

    (at x = 200 cm) is +5 cm. The speed of the wave is

    500 Hz §˘†”-°æ¤†uç Ö†o äéπ ¨¡%A ü¿çúøç Ææ£æ…-ߪ’çûÓ äéπ §Òúø-¢Áj† ≤ƒí∫-D-Æœ† Bí∫ äéπ é̆ (x = 0) ´ü¿l 10 cm éπç°æ† °æJ-N’A Ö†o äéπ A®Ωué˙ ûª®Ωç-í¬Eo à®Ωp-J-î√®Ω’. äé¬-ØÌéπ 鬩ç'A' ´ü¿l Ö†o äéπ éπùç (x = 100 cm ´ü¿l) ≤ƒn†-v¶µºç¨¡ç −5 cm, B ´ü¿l Ö†o ´’®Ó éπùç(x = 200) ≤ƒn†-v¶µºç¨¡ç +5 cm -Å®·-ûË ûª®Ωçí∫ ¢Ëí∫ç

    1) 500 ms−1 2) 1000 ms−1 3) 750 ms−1 4) 250 ms−1

    102. In a closed organ pipe of length l air particles are vibrating with a maximumamplitude A in the third overtone. Then the amplitude of vibration for the air

    lparticles at a distance of from the closed end is

    14

    l §Òúǿ ¤†o äéπ ´‚Æœ† íÌôdç™E í¬LÆæh綵ºç ´‚úÓ ÅAÆæy®√Eo ¢Á©’-́ -Jç-*-†-°æ¤púø’ Åç-ü¿’™E

    lí¬L éπù«©’ A í∫J≠æe éπç°æ† °æJ-N’-AûÓ éπç°æ-Ø√©’ îË≤ÚhçC. íÌôdç ´‚Æœ† é̆ †’ç*

    14

    ü¿÷®Ωç™ Ö†o í¬L éπù«© éπç°æ† °æJ-N’A

    A A1) A 2) 2 A 3) 4)

    2 √2

    103. A point object is placed at 8 cm infront of a glass slab of thickness 6 cm whose

    back face is silvered. If the final image is formed 10 cm behind the silvered face,

    then the refractive index of the glass is

    6 cm ´’çü¿ç Ö†o äéπ í¬V C¢Á’t ´·çü¿’ äéπ Gçü¿’-®Ω÷°æ ´Ææ’h-´¤†’ 8 cm ü¿÷®Ωç™

    Öçî√®Ω’. í¬V C¢Á’t ¢Á†’éπ ûª™«-EéÀ éπ∞«®· °æ‹¨»®Ω’. ûª’C v°æA-Gç•ç éπ∞«®· °æ‹Æœ† ûª™«-EéÀ

    ¢Á†’éπ 10 cm ü¿÷®Ωç™ à®Ωp-úÕûË, í¬V ´véÃ-¶µº-́ † í∫’ùéπç N©’´

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    2 5 √3

    1) 1.5 2) 3) 4) 3 4 2

    104. A rectangular block of glass is placed on a printed page lying on a horizontalsurface. The minimum value of refractive index of glass for which the letters on

    the page are not visible from any of the vertical faces of the block will be

    äéπ éÀ~Aï Ææ´÷ç-ûª®Ω ûª©ç°j Åéπ~-®√©’ ´·vCç-*† é¬T-û√Eo °öÀd, é¬T-ûªç°j äéπ D®Ω`-îª-ûª’-®Ω-v≤ƒé¬®Ωí¬V-C-¢Á’t†’ Öçî√®Ω’. C¢Á’t à E©’´¤ ûª©ç ü∆y®√ èπÿú≈ é¬T-ûªç°j Ö†o Åéπ~-®√©’éπE°œçîªèπ◊çú≈ Öçú≈©çõ‰ í¬V éπE≠æe ´véÃ-¶µº-́ † í∫’ùéπç N©’´

    1) 1.732 2) 1.414 3) 1.5 4) 1.707

    105. White light may be considered to be a mixture of waves of wavelength rangingbetween 3000 A° and 7800 A°. An oil film of thickness 10,000 A° is examinednormally by the reflected light. If refractive index of oil is 1.4, then the film

    appears bright for the wavelengths

    ûÁ©’-°æ¤-®Ωçí∫’ é¬çAE 3000 A°, 7800 A° ûª®Ωçí∫ üÁj®Ω` u Å -́Cµ-™E é¬çA ûª®Ωç-í¬© Ææ¢Ë’t-∞¡-†çí¬¶µ«Nç-îª-́ -a. 10,000 A° ´’çü¿´·†o äéπ ûÁj©°æ¤ §Ò®Ω†’ ÅGµ©ç• °æ®√-́ -®Ωh† é¬çAûÓ°æKéÀ~≤ƒh®Ω’. ûÁj©ç ´véÃ-¶µº-́ † í∫’ùéπç 1.4, Å®·ûË ûÁj©°æ¤ §Ò®Ω Ñ ûª®Ωçí∫ üÁj®√` u-©èπ◊ é¬çA-´ç-ûªçí¬ éπE-°œ-Ææ’hçC.

    1) 4308 A°, 5091 A°, 6222 A° 2) 4000 A°, 5091 A°, 5600 A°

    3) 4667 A°, 6222 A°, 7000 A° 4) 4000 A°, 4667 A°, 5600 A°, 7000 A°

    106. Three identical charges each of + 0.1 C are placed at three vertices of anequilateral triangle of side 1 m. Energy is supplied to the system at a rate of 1

    kilowatt and the charge at any one of the three vertices is taken to top the mid

    point of the line joining the other two charges. The number of days required to do

    so is nearly

    äéÌ\-éπ\öÀ + 0.1 C Ö†o ´‚úø’ Ææ®Ωy-Ææ-́ ÷† Nü¿’u-ü∆-¢Ë-̈ »-©†’ 1 m ¶µº’ïç Ö†o äéπ Æǽ ’-¶«£æ›vA¶µº’ïç ´‚úø’ Q®√{© ´ü¿l Öçî√®Ω’. ´u -́Æænèπ◊ 1 éÀ™-¢√ö¸ Í®ô’™ ¨¡éÀhE ÅçCç*, ´‚úø’ Q®√{©´ü¿l Ö†o Nü¿’u-ü∆-¢Ë̈ »™x àüÁjØ√ äéπü∆Eo N’T-L† È®çúø’ Ç¢Ë-̈ »-©†’ éπLÊ° Ææ®Ω-∞¡-Í®ê ´’üµ¿u Gçü¿’´¤´ü¿lèπ◊ BÆæ’èπ◊ ´î√a®Ω’. Å™« îËߪ’ú≈-EéÀ °æõ‰d ®ÓV© Ææçêu Ææ’´÷-®Ω’í¬

    1) 20 2) 28 3) 2 4) 4

    107. A parallel plate condenser consists of two circular plates each of radius 2 cmseparated by a distance of 0.1 mm. A time varying potential difference of

    5 × 1013 VS−1 is applied across the plates of the condenser. The displacementcurrent is

    äéπ Ææ´÷ç-ûª®Ω °æ©-éπ© È駃-Æœô®˝ È®çúø’ °æ©-éπ© ´’üµ¿u ü¿÷®Ωç 0.1 mm, äéÌ\-éπ\öÀ 2 cm

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    ¢√u≤ƒ®Ωl¥ç Ö†o È®çúø’ ´%û√h-鬮Ω °æ©-éπ©’. È®çúø’ °æ©-éπ© ´’üµ¿u 鬩ç-ûÓ-§ƒô’5 × 1013 VS−1 Í®ô’™ ´÷®Ω’-ûª’†o äéπ §Òõ„-E{-ߪ’™¸ --¶µ‰-ü∆Eo éπ©-í∫-ñ‰-¨»®Ω’. ≤ƒn†-v¶µºç¨¡ Nü¿’uû˝v°æ¢√£æ«ç

    1) 5.50 A 2) 2.28 × 104 A

    3) 5.56 × 102 A 4) 5.56 × 103

    108. In the circuit shown, the cell is ideal with an emf of 30 V, each resistance is3 ohm. The electric charges stored on the

    capacitor of capacity 3 µF is

    °æôç™ îª÷°œ† ´©-ߪ’ç™ Nü¿’uû˝ °∂æ’ôç Çü¿-®Ωz-°∂æ’ôç,

    ü∆E Nî√• 30 V, v°æA-E-®Óüµ¿ç -N©’´ 3 㢒̨©’.

    3 µF È駃-Æœô®˝ O’ü¿ E©y Ö†o Nü¿’u-ü∆-¢Ë¨¡ç N©’´

    1) Zero 2) 72 µC

    3) 18 µC 4) 36 µC

    109. In an oscillating LC circuit the maximum charge on the capacitor is Q. Thecharge on the capacitor when the energy is stored equally between the electric

    and magnetic field is

    äéπ éπç°æ-Ø√-ûªtéπ LC ´©ßª’ç™E È駃-Æœô®˝ O’ü¿ í∫J≠æe Nü¿’u-ü∆-¢Ë¨¡ç Q. Nü¿’uû˝, Åߪ’-≤ƒ\çûªÍé~vû√© ´’üµ¿u ¨¡éÀh Ææ´÷-†çí¬ E©y Ö†o-°æ¤púø’ È駃-Æœô®˝ O’ü¿ Nü¿’u-ü∆-¢Ë¨¡ç

    Q Q Q1) 2) Q 3) 4)

    √3 2 √

    2

    110. Each cell has emf 2 V and internal resistance 1 Ω. Minimum number of such cellsrequired to supply a maximum current of 2 A in a load resistance of 6 Ω is

    äéÌ\éπ\ Nü¿’uû˝ °∂æ’ôç Nî√• 2 V, Åçûª-Jo-®Óüµ¿ç 1 Ω . 6 Ω ¶µ«®Ω E®Óüµ¿ç ü∆y®√ 2 A í∫J≠æeNü¿’u-û˝†’ v°æ´-£œ«ç-°æ---ñ‰-ߪ’-ú≈-EéÀ Å´-Ææ-®Ω-¢Á’i† Å™«ç-öÀ Nü¿’uû˝ °∂æ’ö«© éπE≠æe Ææçêu

    1) 24 2) 48 3) 12 4) 6

    111. An electron moves with a constant speed V along a circle of circumference 2 πr.Its gyromagnetic ratio is (m is mass of electron, 'e' is magnitude of charge on

    electron)

    2 πr °æJCµ Ö†o äéπ ´%û√h-鬮Ω ´÷®Ω_ç™ äéπ á©-é¬ZØ˛ V Æœn®Ώ úÕûÓ °æJ-v¶µº-N’-≤ÚhçC. ü∆E v¶µº´’ùÅߪ’-≤ƒ\çûª E≠æpAh N©’´ (m á©-é¬ZØ˛ vü¿́ u-®√P, 'e' á©-é¬ZØ˛ O’-C Nü¿’u-ü∆-¢Ë̈ ¡ °æJ-́ ÷ùç)

    e m e 2e1) − 2) − 3) − 4) − m e 2m m

    112. Two conducting circular loops of radii R1 and R2 are placed in the same plane withtheir centres coinciding. Assuming R2

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    R1, R2 ¢√u≤ƒ-®√l¥©’ Ö†o È®çúø’ ´%û√h-鬮Ω ¢√£æ«-éπ°æ¤ Bí∫--ôd-©†’ ¢√öÀ Íéçvü∆©’àéöµºNçîËôô’xí¬ äÍé ûª©ç™ Öçî√®Ω’. R2 éπçõ‰ π

    4 Í®úÕߪ’†’x -´·ç-ü¿’ ÖçC. ï†éπ §˘†”-°æ¤-Ø√uEo ƒ2 éÀ °ç*-†-°æ¤púø’

    Nü¿’uû˝ v°æ¢√£æ« 㙉dñ¸ éπçõ‰ π4

    Í®úÕߪ’-Ø˛©’ ¢Á†’-éπ-•úÕ ÖçC. ´©ßª’ç ņ’--Ø√ü¿ §˘†”-°æ¤†uçN©’´

    ƒ1ƒ2 2ƒ1ƒ2 ƒ1 + ƒ21) 2) 3) √

    ƒ1ƒ2 4) ƒ1 + ƒ2 ƒ1 + ƒ2 2

    115. If a magnet is suspended at an angle 30° to the magnetic meridian, the dipneedle makes an angle of 45° with the horizontal. The real dip is

    äéπ Åߪ’-≤ƒ\ç-û√Eo Åߪ’-≤ƒ\çûª ߪ÷¢Á÷u-ûªh-®ΩçûÓ 30° éÓùç îËÊÆ-™« ÊÆyîªa ¥í¬ ¢Ë™«-úø-D-Æœ-†-°æ¤púø’,

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    úÕ°ˇ Ææ÷*éπ éÀ~Aï Ææ´÷ç-ûª-®ΩçûÓ 45° éÓùç îËÆæ’hçõ‰, Eï úÕ°ˇ N©’´

    √2 3 3

    1) tan−1 ( ) 2) tan−1 ( ) 3) tan−1(√3 ) 4) tan−1 ( )√3 2 √2116. Two radioactive nucleii x and y initially contain equal number of atoms. Their

    half lives are 1 hour and 2 hours respectively. The ratio of their rates of

    disintegration after 2 hours from the start is

    x, y ÅØË È®çúø’ Í®úÕ-ßÁ÷-üµ∆-Jtéπ Íéçvü¿-鬩’ ûÌ©’ûª Ææ´÷-†-Ææç-êu™ °æ®Ω-´÷-ù’-´¤-©†’ éπLTÖØ√o®·. ¢√öÀ Å®Ωl¥@Nûª 鬙«©’ ´®Ω’-Ææí¬ 1 í∫çô, 2 í∫çô©’. v§ƒ®Ω綵ºç †’ç* 2 í∫çô©ûª®√yûª ¢√öÀ N°∂æ’-ô† Í®ôx E≠æpAh

    1) 1 : 1 2) 1 : 3 3) 1 : 2 4) 2 : 3

    117. The de Broglie wavelength of a particle moving with a velocity2.25 × 108 ms−1 is equal to the wavelength of photon. The ratio of kineticenergy of the particle to the energy of the photon is (Speed of light in vacuum =

    3 × 108 ms−1)

    2.25 × 108 ms−1 ¢Ëí∫çûÓ îªL-Ææ’h†o äéπ éπùç úŒ v¶--Ux ûª®Ωçí∫üÁj®Ω` uç, §∂Úö«Ø˛ ûª®Ωçí∫üÁj®√` u-EéÀÆæ´÷†ç. éπùç í∫A-ï-̈ ¡-éÀhéÀ, §∂Úö«Ø˛ ¨¡éÀhéÀ Ö†o E≠æpAh

    (¨¡⁄†uç™ é¬çA ´úÕ = 3 × 108 ms−1).

    5 3 7 11) 2) 3) 4)

    8 8 8 8

    118. The ionization potential of Hydrogen is 13.6 eV. When it is excited from groundstate by a monochromatic radiation of wavelength 970.6 A°. The number ofemission lines according to Bohr's theory will be

    £j«vúÓ-ïØ˛ °æ®Ω-´÷-ù’´¤ Åߪ’-F-éπ-®Ωù §Òõ„-E{-ߪ’™¸ N©’´ 13.6 eVÖç-C. 970.6 A° ûª®Ωç-í∫-üÁj®Ω` uçÖ†o äéπ àéπ-́ ®Ωg NéÀ-®Ω-ùçûÓ ü∆Eo ¶µº÷≤ƒn®· †’ç* ÖûËhïç îÁçCÊÆh, ¶®˝ Æœü∆l¥ç-û√-Eo ņ’-Ææ-Jç* Öü∆_-®Ω-Í®-ê© Ææçêu

    1) 4 2) 6 3) 3 4) 10

    119. A transmitting antenna has a height of 40 m and the receiving antenna has heightof 60 m. The maximum distance between them for satisfactory communication is

    nearly

    äéπ v°æ≤ƒ®Ω ߪ÷çõ„Ø√o áûª’h 40 m, ÊÆéπ-JùÀ ߪ÷çõ„Ø√o áûª’h 60 m. Ææçûª%-°œh-éπ®Ω v°æ≤ƒ-®√-EéÀ¢√öÀ ´’üµ¿u Öçú≈Lq† í∫J≠æe ü¿÷®Ωç Ææ’´÷-®Ω’í¬

    1) 50 km 2) 25 km 3) 22.5 km 4) 27.5 km

    1 1

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    120. For the transistor, x = and y = where α and β are current gains in common α β

    base and common emitter configuration of the transistor. Then

    1 1äéπ vö«Eq-Ææd-®˝èπ◊ x = , y = , Ééπ\úø α, β ©’ vö«EqÆæd®˝ Ö´’túÕ Çüµ∆®Ω NØ√uÆæç, Ö´’túÕ

    α β

    Öü∆_®Ω NØ√u≤ƒ© Nü¿’uû˝ v°æ¢√£æ« ´%Cl¥ Å®·ûË

    1) x + y = 0 2) x + y = 1 3) 2x + y = 1 4) x − y = 1

    CHEMISTRY

    121. The first ionization enthalpy values of the elements Na, Mg and Si are 496, 737 and786 KJ.mol−1 respectively. Then the first ionization enthalpy of 'Al' will be (in

    KJ.mol−1)

    Na, Mg, Si © v°æü∑¿´’ ÅßÁ·-ØÁj-ñ‰-≠æØ˛ áçü∑∆-Lp N©’-´©’ ´®Ω’-Ææí¬ 496, 737, 78 KJ.mol −1

    Å®·ûË 'Al' v°æü∑¿´’ ÅßÁ·-ØÁj-ñ‰-≠æØ˛ áçü∑∆-Lp N©’´.

    1) 419 2) 575 3) 762 4) 1061

    122. According to MO theory, Lithium molecule (Li2) is

    Åù’ -ÇJs-ö«™¸ Æœü∆l¥çûªç v°æ鬮Ωç LC∑ߪ’ç Åù’´¤ (Li2)

    1) stable and paramagnetic

    Æœn®Ω-¢Á’içC, §ƒ®√ Åߪ’-≤ƒ\çûª üµ¿®Ωtç Ö†oC.

    2) unstable and paramagnetic

    ÅÆœn-®Ω-¢Á’içC, §ƒ®√ Åߪ’-≤ƒ\çûª üµ¿®Ωtç Ö†oC.

    3) stable and diamagnetic

    Æœn®Ω-¢Á’içC, v°æûªu-ߪ’-≤ƒ\çûª üµ¿®ΩtçûÓ Öçô’çC.

    4) unstable and does not exist

    ÅÆœn-®Ω-¢Á’içC, Åù’´¤ à®ΩpúË Å´-鬨¡ç Öçúøü¿’.

    123. 360 cm3 of methane gas diffused through a porous membrane in 15 minutes.Under similar conditions, 120 cm3 of another gas 'X' is diffused in 10 minutes.

    The molar mass of the gas 'X' is

    360 cm3 -O’ü∑ËØ˛ ¢√ߪ·´¤ äéπ Ææ*avü¿ §ƒvûª †’ç* 15 EN’-≥ƒ™x ¢√u°æ†ç îÁçCçC. ÅüË°æJ-Æœnûª’™x 120 cm3 äéπ ¢√ߪ·´¤ 10 EN’-≥ƒ™x ¢√u°æ†ç îÁçCûË Ç ¢√ߪ·´¤ ¢Á÷™«®˝ vü¿´u-®√P

    1) 16 2) 32 3) 64 4) 72

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    124. The energy associated with the first orbit of He+ is

    He+ ¢Á·ü¿öÀ éπéπ~u ¨¡éÀh

    1) −8.72 × 10−18 J 2) −4.36 × 10−18 J

    3) −5.45 × 10−19 J 4) −10.9 × 10−19 J

    125. The species that does not show disproportionation reaction among the following is

    éÀçC-¢√-öÀ™ ņ-†’-§ƒûª Ωu ï®Ω°æ-EC

    1) ClO− 2) ClO2− 3) ClO3

    − 4) ClO4−

    126. The enthalpy of combustion of benzene at 298 K and 1 atm. is −3261 KJ. Standardenthalpies of formation of CO2 and H2O are −393 KJ and −285 KJ respectively.

    Then the standard enthalpy of formation (∆f H0) of benzene is

    298 K, 1 Åö«t -°‘-úø†ç ´ü¿l äéπ ¢Á÷™¸ ¶„ç[email protected]Ø˛ ü¿£æ«†ç îÁçC −3261 KJ Ö≥ƒgEo Núø’-ü¿©îËÆæ’hçC. CO2, H2O © v°æ´÷ù Ææç°∂æ’-ô† áçü∑∆Lp N©’-´©’ ´®Ω’-Ææí¬ −393 KJ, −285 KJÅ®·ûË ¶„ç[email protected]Ø˛ v°æ´÷ù Ææç°∂æ’-ô† áçü∑∆-Lp N©’´

    1) −48 KJ 2) +48 KJ 3) +24.83 KJ 4) −24.83 KJ

    127. The value of Kc for the reaction, 2A B + C is 2 × 10−3. At a given time, the

    composition of reaction mixture is [A] = [B] = [C] = 3 × 10−4 M. In which

    direction the reaction will proceed?

    2A B + C Ωuèπ◊ Kc N©’´ 2 × 10−3. äéπ EÍ®l-Pûª é¬©ç ´ü¿l √u- N’-v¨¡-´’ç™

    [A] = [B] = [C] = 3 × 10−4 M. Å®·ûË à C¨¡™ Ωu °æ¤®Ó-í∫-N’-Ææ’hçC?

    1) Will proceed in reverse direction.

    Ωu A®Ó-í¬N’ C¨¡í¬ ï®Ω’-í∫’-ûª’çC.

    2) Will proceed in forward direction.

    Ωu °æ¤®Ó-í¬N’ C¨¡í¬ ï®Ω’-í∫’-ûª’çC.

    3) Equilibrium is established.

    Ωu™ Ææ´’-û√-ÆœnA à®Ωp-úø’-ûª’çC.

    4) Cannot be predicted.

    Ωu í∫´’† C¨¡†’ ûÁ©’-Ææ’-éÓ™‰ç.

    128. A solution of HCl has [H+] = 10−3 M. If 1 ml of it is diluted to 1 litre, then thepH of the resulting solution will be

    HCl vü∆ -́ùç™ [H+] = 10−3 M. Ñ vü∆´ùç †’ç* 1 ml vü∆´-ù«Eo 1 Mô-®Ω’èπ◊ NM†çîËÊÆh °∂æLûª vü∆´-ù pH N©’´

    1) 3 2) 6 3) 8 4) 11

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    129. Electron-rich hydride among the following is

    éÀçC-¢√-öÀ™ á©-é¬Z-Ø˛©’ ÅCµ-éπçí¬ Ö†o £j«v-úÁjú˛

    1) MgH2 2) CH4 3) NH3 4) B2H6130. The alkali metal that imparts crimson red colour to an oxidizing flame is

    ÇéÃq-éπ-®Ωù ñ«y©™ Èéç°æ¤- ®Ωçí∫’-†’ ÉîËa é~¬®Ω ™£æ«ç

    1) Li 2) Na 3) K 4) Rb

    131. 'Plaster of paris' is

    '§ƒxÆæd®˝ Ç°∂ˇ §ƒJÆˇ— ÅØËC

    1) CaSO4.2H2O 2) 2 CaSO4.H2O

    3) CaSO4.H2O 4) (CaSO4.H2O)2132. Incorrect statement about orthoboric acid is

    --Ç®Ón-¶-Jé˙ -Ç´’xç í∫’Jç* ÆæJ-é¬-E -¢√u-êu

    1) It is a protonic acid and a strong acid

    ÉC äéπ v§Úö«Ø˛ ü∆ûª Ç´’xç, •©¢Á’içC.

    2) It is a white crystalline solid with soapy touch.

    ÉC Ææ•’s™« ´’%ü¿’-́ ¤í¬ ÖçúË, ûÁ©xE Ææp¥öÀéπ °æü∆®Ωnç

    3) It has a layer structure in which planar BO3 units are joined by hydrogen

    bonds.

    DEéÀ §Ò®Ω© E®√tùç Öçô’çC. Ñ E®√t-ùç™ Ææ´’-ûª© BO3 ߪ‚Eô’x £j«vúÓ-ïØ˛ •çüµ¿çûÓéπL°œ Öçö«®·.

    4) It acts as a Lewis acid.

    ÉC ©÷®·Æˇ Ç´’xçí¬ °æE-îË-Ææ’hçC.

    133. The basic structural unit of silicates is

    'ÆœL-Íé-ö¸—-©-™E v§ƒü∑¿-N’éπ E®√t-ù«-ûªtéπ ߪ‚Eö¸

    1) SiO2 2) SiO32− 3) SiO4

    4− 4) [− R2SiO−]

    134. The maximum limit of nitrate in drinking water is 50 ppm. Excess nitrate indrinking water can cause the disease known as

    û√Íí FöÀ™ ØÁjvõ‰ö¸ í∫J≠æe í¬úµøûª 50 ppm í¬ ÖçúÌa. ØÁjvõ‰ö¸ í¬úµøûª ÅCµéπ °æJ-́ ÷-ùç™ Öçõ‰--́ -îËa ¢√uCµ

    1) cataract é¬ô-®√é˙d 2) skin cancer Ωt- é¬u-†q®˝

    3) fluorosis §∂Úx®Ó-ÆœÆˇ 4) 'blue baby' syndrome FL -Gúøf ÆœçvúÓ¢’̨

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    135. An aqueous solution of sodium acetate on electrolysis gives

    ≤ÚúÕߪ’ç áÆœ-õ‰ö¸ ï©-vü∆-´-ù«Eo Nü¿’uû˝ N¨Ïx-≠æù îËÊÆh à®ΩpúËC

    1) methane -O’ü∑ËØ˛ 2) ethane Ñü∑ËØ˛

    3) ethylene áC∑-LØ˛ 4) acetylene áÆœ-öÀ-LØ˛

    136. Glycerol can be separated from spent-lye in soap industry by using the technique

    Ææ•’s© °æJ-v¨¡-´’™ Ö°æ-ßÁ÷TçîË í¬úµø é~¬®Ω vü∆´ùç †’ç* Tx-ï-®√-™¸†’ ¢Ë®Ω’îËÊÆ °æü¿l¥A

    1) steam distillation ï©--¶«≠æp ÊÆyü¿†ç

    2) fractional distillation -ÅçPéπ ÊÆyü¿†ç

    3) differential extraction -¶µ‰-ü∆ûªtéπ E≠æ \-®Ω{ù

    4) distillation under reduced pressure ûªèπ◊\´ °‘úø-†çûÓ ÊÆyü¿†ç

    137. Which of the following compounds will show cis-trans isomerism?

    éÀçC-¢√-öÀ™ 'ÆœÆˇ – -vö«Ø˛q— ≤ƒü¿%-¨»uEo v°æü¿-Jzç-îËN

    A) (CH3)2 C = CH − C2H5 B) CH2 = CBr2C) C6H5CH = CH − CH3 D) CH3 − CH = CClCH31) A and B 2) C and D 3) A only 4) D only

    138. Ortho and para directing group among the following is

    éÀçC-¢√-öÀ™ Ç®Ón, §ƒ®√ ≤ƒn† EÍ®l-¨¡éπ Ææ´‚£æ«ç

    1) −OH 2) −NO2 3) −CHO 4) −SO3H

    139. 20 ml of 0.1 M NH4OH is added to 20 ml of 1 M NH4Cl solution. The pH of theresulting solution is 8.2. The pKb of NH4OH is

    20 ml 0.1 M NH4OH vü∆´-ù«Eo 20 ml 1 M NH4Cl vü∆´-ù«-EéÀ éπL-§ƒ®Ω’. °∂æLûª vü∆´-ù°æ¤pH N©’´ 8.2 Å®·ûË NH4OH pKb N©’´ ---

    1) 8.2 2) 4.8 3) 5.8 4) 3.8

    140. The correct set of quantum numbers for the valence shell electron of Rb (Z = 37)is

    ®Ω’H-úÕߪ’ç °æ®Ω-´÷--ù’-´¤ (Z = 37) ¢Ë©Fq ≤ƒn®·-™E á©-é¬Z-Ø˛èπ◊ Ö†o Ø√©’í∫’ é¬yçôç Ææçêu©ÆæN’A

    1) 5, 0, 0, +12

    2) 5, 1, 0, +12

    3) 5, 1, 1, ---−12

    4) 6, 0, 0, −12

    141. The sulphur modification that shows paramagnetic behaviour is

    §ƒ®√ Åߪ’-≤ƒ\çûª üµ¿®Ωtç Ö†o Ææ©p¥®˝ ®Ω÷§ƒç-ûª®Ωç

    1) S2 2) S4 3) S6 4) S8

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    142. The chemical compound formed in the brown ring test of nitrate ions is

    ØÁjvõ‰ö¸ Åߪ÷Ø˛© ñ‰í∫’®Ω’ ®Ωçí∫’ ´©ßª’ °æK-éπ~™ à®ΩpúË ®Ω≤ƒ-ߪ’† Ææ¢Ë’t-∞¡†ç

    1) Fe[NO3]2 2) Fe2[SO4]3.NO

    3) [Fe(H2O)6]3+ 4) [Fe(H2O)5NO]

    2+

    143. The time required for the completion of 90% of a first order reaction is60 minutes. Then the time required for the completion of 99% of the same

    reaction is

    äéπ v°æü∑¿´’ véπ´÷çéπ Ωu 90% °æ‹Jh 鬴-ú≈-EéÀ 60 EN’-≥ƒ©’ °æúø’-ûª’çC. ÅüË îª®Ωu 99%°æ‹Jh 鬴-ú≈-EéÀ °æõ‰d 鬩ç

    1) 60 min 2) 66 min 3) 90 min 4) 120 min

    144. With hot and concentrated alkalies, chlorine gives

    éÓxJØ˛ ¢ËúÕ í¬úµø, é~¬®Ω vü∆´-ù«-©ûÓ îª®Ωu ïJ-°œ-†-°æ¤púø’ à®ΩpúËN

    1) a mixture of chloride and hypochlorite

    éÓxÈ®jú˛, £j«§ÚéÓxÈ®j-ö¸© N’v¨¡´’ç

    2) a mixture of chloride and chlorate

    éÓxÈ®jú˛, éÓxÍ®-ö¸© N’v¨¡´’ç

    3) a mixture of chloride and perchlorate

    éÓxÈ®jú˛, °æ®˝-éÓx-Í®-ö¸© N’v¨¡´’ç

    4) a mixture of hypochlorite and perchlorate

    £j«§ÚéÓxÈ®j-ö¸, °æ®˝-éÓx-Í®-ö¸© N’v¨¡´’ç

    145. XeOF4 is a colourless, volatile liquid and has a

    XeOF4 ®Ωçí∫’-™‰E -¶«≠æpQ© vü¿´ç, DE Åù’ E®√tùç

    1) distorted octahedral structure N®Ω÷-°æù Å≠æd-´·-&ߪ’ç

    2) square pyramidal structure îªûª’-®ΩvÆæ °œ®Ω-N’-úø™¸

    3) square planar structure Ææ´’-ûª© îªûª’-®ΩvÆæç

    4) distorted tetrahedral structure N®Ω÷-°æù îªûª’®Ω’t-&ߪ’ç

    146. 'Argyrol' is a silver sol, used

    'ÇJb-®√™¸— ÅØË Æœ©y®˝ ≤ƒ™¸ Ö°æ-ßÁ÷í∫ç

    1) as an eye lotion

    éπçöÀ ™≠æ-Ø˛í¬ ¢√úø-û√®Ω’

    2) in curing kala azar

    鬙« Åñ«®˝ ÅØË ¢√uCµE ûªT_ç-îª-ú≈-EéÀ ¢√úø-û√®Ω’

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    3) as an intramuscular injection

    éπçúø-®√ç-ûª®Ω Éçñ-éπ~-Ø˛í¬ ¢√úø-û√®Ω’

    4) for stomach disorders

    Öü¿®Ω ÅÆæy-Æænûª-©èπ◊ ¢√úø-û√®Ω’

    147. Carboxylic acids having an α - hydrogen are halogenated at the α - position ontreatment with chlorine or bromine in the presence of small amount of red

    phosphorus to give α - halocarboxylic acids. The reaction is known as

    α - £j«vúÓ-ï-Ø˛-©’†o 鬮√s-éÀq-Lé˙ Ç´÷x™x α - ≤ƒn†ç™ £æ…™-ï-Ø˛†’ v°æA-Íé~-°œç-îª-´îª’a. ÑÇ´÷x©†’ éÓxJØ˛ ™‰ü∆ v¶N’Ø˛ûÓ áv®Ω §∂ƒÆæp¥-®ΩÆˇ Ææ´’-éπ~ç™ îª®Ωu ïJ-°œûË α - £æ…™-é¬-®√s-éÀq-Lé˙Ç´÷x-©†’ à®Ωp-®Ω’-≤ƒh®·. Ñ îª®Ωu

    1) Gattermann - Koch reaction í¬ô-®˝-´’-Ø˛ - -éÓî˝ îª®Ωu

    2) Etard reaction Éö«® f̋ Ωu

    3) Hell - Volhard - Zelinsky reaction £«™¸ - -¢Ó-™«-®˝f - -ñ„-©-Ø˛Æœ\ Ωu

    4) Stephen reaction Æ‘d°∂Ø˛ Ωu

    dil. NaOH ∆148. 2CH3 − CHO A → But - 2 - enal −H2O

    The compound 'A' in the reaction is

    dil.NaOH ∆2CH3 − CHO A → •’uö¸– 2 – -É-Ø√™¸.−H2O

    °j Ωu™ 'A' ÅØËC

    1) aldol Ç™«f™¸ 2) ketol éÃ-ö™¸

    3) mesityloxide ¢Á’iÆœ-õ„j™¸ ÇÈéj qú˛ 4) crotonaldehyde véÓô-Ø√-Lf-£j«ú˛

    149. The compound with highest boiling point among the following is

    éÀçC-¢√-öÀ™ Åûªu-Cµé𠶫≠‘p-¶µº-́ † ≤ƒn†ç Ö†oC

    1) pentan - 1 - ol °çô-Ø˛ – 1 – -㙸 2) n - butane n - -•÷uõ‰Ø˛

    3) pentanal °çô-Ø√™¸ 4) ethoxyethane Éü∑∆-éÃq-Ñ-ü∑ËØ˛

    150. The amine produced by the Hoffmann degradation of benzamide is

    £æ…°∂ˇ-´’Ø˛ EO’o-éπ-®Ωù Ωu™ ¶„çï-¢Á’iú˛ †’ç* à®ΩpúË áO’Ø˛

    1) phenyl ethanamine °∂œØÁj™¸ Éü∑¿-†-¢Á’iØ˛

    2) benzenamine ¶„ç@-†-O’Ø˛

    3) N - N - dimethyl aniline N - N - úÁ--O’-ü∑Áj™¸ áE-LØ˛

    4) hexan - 1, 6 - diamine £«é¬qØ˛ – 1, 6 – úÁjá-O’Ø˛

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    151. Match the following:

    éÀçC-¢√-öÀE ïûª°æ®Ω-îªçúÕ:

    Compound Ææ¢Ë’t-∞¡†ç Geometry ñ«uN’AI) [Ni(CO)4] A) Octahedral

    Çé¬d-£«-vúø™¸

    II) [PtCl4]2− B) Trigonal bipyramidal

    wõ„jíÓ-†-™¸ -¶„j-°œ-®Ω-N’-úø™¸

    III) [Fe(CO)5] C) Square planar

    Ææ´’-ûª© îªûª’-®ΩvÆæç

    IV) [Co(NH3)6]3+ D) Tetrahedral

    õ„vö«-£«-vúø™¸

    1) I-C, II-D, III-B, IV-A 2) I-B, II-A, III-D, IV-C

    3) I-A, II-B, III-C, IV-D 4) I-D, II-C, III-B, IV-A

    152. 1.00 g of a non electrolyte solute dissolved in 50 g of benzene lowered the

    freezing point by 0.40 K. The freezing point depression constant of benzene is

    5.12 K kg mol −1. The molar mass of solute is

    50 ví¬. ¶„ç[email protected]Ø˛™ 1.00 ví¬. ÅN-ü¿’uû˝ N¨Ïx--≠æuéπ vü∆N-û√Eo éπJ-Tç-*-†-°æ¤púø’, °∂æ’F-¶µº-́ † Ö≥Úg-ví∫ûª0.40 K ûªT_çC. ¶„ç[email protected]Ø˛ ¢Á÷™«®˝ E´’oA Æœn®√çéπç 5.12 K kg mol−1. Å®·ûË vü∆Nûª°æ¤¢Á÷™«®˝ ¶µ«®Ωç

    1) 41 2) 392 3) 180 4) 256

    153. Λ0m for NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol−1.

    Λ0 for CH3COOH is (in S cm2 mol−1)

    NaCl, HCl, CH3COONa © Λ0m N©’-´©’ ´®Ω’-Ææí¬ 126.4, 425.9, 91.0 S cm

    2

    mol −1 Å®·ûË CH3COOH ¢Á÷™«®˝ ¢√£æ«-éπûª

    1) 643.3 2) 228.5 3) 451.3 4) 390.5

    154. The electronic configuration of cerium is

    'Æ‘Jߪ’ç— ´‚©éπç á©-é¬ZØ˛ NØ√uÆæç

    1) [Xe]4f05d16s2 2) [Xe]4f

    15d16s2

    3) [Xe]4f25d06s2 4) 2 and 3

    155. RNA does not contain

    RNA ™ ÖçúøEC

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    1) Thymine ü∑Áj-O’Ø˛ 2) Uracil ߪ·®√-Æœ™¸

    3) Cytosine ÂÆjö-ÆœØ˛ 4) Adenine áúÕ-ØÁjØ˛

    156. Nylon − 6, 6 is obtained by condensation polymerization of

    éÀçC Åù’-´¤©’ Ææç°∂æ’†† §ƒL´’-K-éπ-®Ωùç îÁçCûË

    ØÁj™«Ø˛ − 6, 6 §ƒL-´’®˝ à®Ωp-úø’-ûª’çC

    1) Adipic acid and hexamethylene diamine

    áúÕ-°œé˙ Ç´’xç, £«é¬q-N’-C∑-MØ˛ úÁjá-O’Ø˛

    2) Phenol and formaldehyde °∂‘Ø√™¸, §∂ƒ®√t-Lf-£j«ú˛

    3) Terephthalic acid and ethylene glycol

    õ„J-ü∑∆-Lé˙ Ç´’xç, áC∑-M--Ø˛ Èíkx-鬙¸

    4) Caprolactam and formaldehyde

    é¬v§Ò-™«éπdç, §∂ƒ®√t--Lf-£j«ú˛

    157. The antibiotic that is supposed to be toxic towards certain strains of cancer cellsis

    éÌEo ®Ω鬩 NÆæh %ûª ®Ω÷°æ é¬u†q®˝ éπù«-©èπ◊ N≠æ-°æü∆-®Ωnçí¬ °æE-îËÊÆ ßª÷ç-öà -•--ߪ÷-öÀé˙

    1) Vancomycin ¢√çéÓ-¢Á’i-ÆœØ˛

    2) Ofloxacin ã§∂Úx-é¬q-ÆœØ˛

    3) Dysidazirine úÕÆœ-ú≈->-JØ˛

    4) Prontosil v§ƒØ -̨ö-Æœ™¸

    158. If the ore alone but not the gangue is soluble in some suitable solvent, then oreis concentrated by

    êEï ´÷L†uç é¬èπ◊çú≈ ´·úÕ-ê-Eïç ´÷vûª¢Ë’ àüÁjØ√ äéπ vü∆ -́ùÀ™ éπJ-T-ûË, Ç êE-ñ«Eo ¨¡ŸCl¥îËÊÆ°æü¿l¥A

    1) Levigation FöÀûÓ é~¬∞¡†ç îËߪ’úøç

    2) Leaching Eé~¬-∞¡†ç

    3) Froth floatation method °æx´† v°ævéÀߪ’ °æü¿l¥A

    4) Liquation í∫©-Eéπ °æ%-ü∑¿-éπ\-®Ωùç

    159. The incorrect statement among the following is éÀçC-¢√-öÀ™ Ææ-J é¬E -¢√u-êu

    1) Under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO.

    éÌEo °æJ-Æœnûª’™x ¢Á’Uo-≠œßª’ç, Al2O3 †’ éπ~ߪ’-éπ-®Ωùç îËߪ’í∫-©ü¿’, Å©÷u-N’-Eߪ’ç, MgO †’éπ~ߪ’-éπ-®Ωùç îËߪ’-í∫-©ü¿’.

    2) The reaction, Cr2O3 + 2 Al → Al2O3 + 2 Cr is thermodynamically feasible.

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    Cr2O3 + 2 Al → Al2O3 + 2 Cr, Ñ îª®Ωu Ö≠æg-í∫-A-éπçí¬ ïJÍí Å´-é¬-¨¡-´·çC.

    3) Ellingham diagram normally consists of plots of ∆G0 Vs T for formation of

    oxides of elements

    äéπ áLxç-í -̊£æ…¢’̨ °æôç™, ≤ƒüµ∆-®Ωù ™£æ…© ÇéÃq-éπ-®Ω-ù«-EéÀ Öç-úË ∆G0, Ö≥Úg-ví∫ûª (T)©ví¬°∂椩’çö«®·.

    4) Ellingham diagram explains the kinetics of the reduction process.

    áLxç-í˚-£æ…¢’̨ °æôç éπ~ߪ’-éπ-®Ωù v°ævéÀߪ’ í∫A-ï-¨»ÆæYç í∫’Jç* ûÁL-ߪ’-ñ‰-Ææ’hçC.

    160. The crystal system of a compound with unit cell dimensions a = 0.387, b = 0.387and c = 0.504 nm and α = β = 90° and γ = 120° is

    äéπ Ææ¢Ë’t-∞¡-†ç™E ߪ‚Eö¸ ÂÆ™¸™ a = 0.387, b = 0.387, c = 0.504 nm, α = β = 90°, γ = 120° Å®·ûË ü∆E Ææp¥öÀéπ E®√tùç

    1) cubic °∂æ’†ç 2) hexagonal ≠æö\-ùÃߪ’

    3) orthorhombic Ææ´’-îª-ûª’-®Ω’s¥ï 4) rhombohedral Ææ´÷ç-ûª®Ω ≠æôp¥-©-éÃߪ’

    KEY1-1; 2-2 ; 3-3; 4-2; 5-4 ; 6-1 ; 7-2 ; 8-4 ; 9-1 ; 10-4; 11-4; 12-2; 13-3; 14-1; 15-3;

    16-4; 17-3; 18-4; 19-1; 20-3; 21-3; 22-3; 23-3; 24-3; 25-3; 26-1; 27-3; 28-4; 29-3;

    30-3; 31-1; 32-1; 33-4; 34-3; 35-3; 36-1; 37-3; 38-2; 39-1; 40-3; 41-2; 42-1; 43-2;

    44-1; 45-2; 46-1; 47-4; 48-3; 49-2; 50-3; 51-4; 52-2; 53-3; 54-1; 55-3; 56-2; 57-1;

    58-2; 59-4; 60-3; 61-1; 62-3; 63-4; 64-4; 65-1; 66-3; 67-3; 68-3; 69-2; 70-4; 71-2;

    72-4; 73-4; 74-1; 75-2; 76-3; 77-4; 78-2; 79-1; 80-4; 81-2; 82-4; 83-3; 84-4; 85-1;

    86-2; 87-2; 88-1; 89-4; 90-2; 91-3; 92-1; 93-2; 94-2; 95-3; 96-4; 97-1; 98-3; 99-2;

    100-1; 101-2; 102-4; 103-1; 104-2; 105-1; 106-3; 107-4; 108-2; 109-4; 110-1; 111-3;

    112-1; 113-4; 114-3; 115-2; 116-1; 117-2; 118-2; 119-1; 120-4; 121-2; 122-3;

    123-3; 124-1; 125-4; 126-2; 127-1; 128-2; 129-3; 130-1; 131-2; 132-1; 133-3; 134-4;

    135-2; 136-4; 137-2; 138-1; 139-2; 140-1; 141-1; 142-4; 143-4; 144-2; 145-2; 146-1;

    147-3; 148-1; 149-1; 150-2; 151-4; 152-4; 153-4; 154-2; 155-1; 156-1; 157-3; 158-2;

    159-4; 160-2.

    (-Ñ -†-´‚-Ø√ -v°æ-¨¡o°æ-vû√-Eo ®Ω÷-§Òç-Cç-*-†-¢√®Ω’: -öÀ. éÀ---≥Ú®˝, ->.-N. -îªç-vü¿-¨Ï-ê®˝, --â. -ÅçéÓ[email protected]®√--´¤ – -Ø√®√ߪ’-ù -N-ü∆uÆæçÆæn-©’, -ØÁ-©÷x®Ω’)

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