PUBDET 2018 Question Paper Geology

PUBDET-2018 81220001
Subject: Geology (Booklet Number)

Duration: 90 minutes Full Marks: 100

Instructions
1. All questions are of objective type having four answer options for each. Only one option is
correct. Correct answer will carry full marks 2. In case of incorrect answer or any combination
of more than one answer, ½ marks will be deducted.
2. Questions must be answered on OMR sheet by darkening the appropriate bubble marked A, B,
C, or D.
3. Use only Black/Blue ball point pen to mark the answer by complete filling up of the respective
bubbles.
4. Do not make any stray mark on the OMR.
5. Write question booklet number and your roll number carefully in the specified locations of the
OMR. Also fill appropriate bubbles.
6. Write your name (in block letter), name of the examination centre and put your full signature in
appropriate boxes in the OMR.
7. The OMRs will be processed by electronic means. Hence it is liable to become invalid if there
is any mistake in the question booklet number or roll number entered or if there is any mistake
in filling corresponding bubbles. Also it may become invalid if there is any discrepancy in the
name of the candidate, name of the examination centre or signature of the candidate vis-a-vis
what is given in the candidate’s admit card. The OMR may also become invalid due to folding
or putting stray marks on it or any damage to it. The consequence of such invalidation due to
incorrect marking or careless handling by the candidate will be sole responsibility of candidate.
8. Candidates are not allowed to carry any written or printed material, calculator, pen, docu-pen,
log table, any communication device like mobile phones etc. inside the examination hall. Any
candidate found with such items will be reported against & his/her candidature will be
summarily cancelled.
9. Rough work must be done on the question paper itself. Additional blank pages are given in the
question paper for rough work.
10. Hand over the OMR to the invigilator before leaving the Examination Hall.
11. This paper contains questions in both English and Bengali. Necessary care and precaution were
taken while framing the Bengali version. However, if any discrepancy(ies) is /are found
between the two versions, the information provided in the English version will stand and will
be treated as final

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Mathematics

1. If a x = b y = c z and a, b, c are in G.P., then x, y, z are
(A) in A.P. (B) in G.P. (C) in H.P. (D) Consecutive numbers

k¢c a x = b y = cz Hhw a, b, c …−Z¡šl fËN¢a−a b¡−Lz a¡q−m x, y, z
(A) pj¡¿¹l fËN¢a−a b¡L−h (B) …−Z¡šl fËN¢a−a b¡L−h
(C) ¢hfl£a fËN¢a−a b¡L−h (D) flØfl œ²¢jL q−h

2. 1 1
Given +2= . then x is
log x 10 log 0.5 10

1 1
−cJu¡ B−R +2= z a¡q−m x q−h
log x 10 log 0.5 10

1 1
(A) (B) 0.0025 (C) (D) 0.125
2 4

3. 3 − 4ix
A real value of x will satisfy the equation = α − iβ (α, β ∈ ℝ ) if
3 + 4ix
3 − 4ix
x -Hl h¡Ù¹h j¡e = α − iβ (α, β ∈ ℝ ) pj£LlZ¢V−L ¢pÜ Ll−h, k¢c
3 + 4ix
(A) α2 − β2 = −1 (B) α2 − β2 = 1
(C) α2 + β2 = 1 (D) α2 + β2 = −1

4. The number of ways that the letters of the word “TREES” can be arranged such that each word
starts with a consonant and end with a vowel is
“TREES” n−ël Arl…¢m Hjei¡−h p¡S¡−e¡ qm k¡−a n−ël fËbj Arl¢V consonant Hhw
−n−ol¢V vowel q−h a¡q−m p¡S¡−e¡l pwMÉ¡¢V q−h

(A) 9 (B) 18 (C) 24 (D) 27

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5.  −1 2 −2 
1 
Let A =  −2 1 2  Then
3 
 2 2 1
(A) A is symmetric matrix (B) A is diagonal matrix
(C) A is orthogonal matrix (D) none of the above

 −1 2 −2 
j−e L¢l A =  −2 1 2 
1
a¡q−m
3 
 2 2 1
(A) A pcªnÉ jÉ¡¢VÊ„ (B) A LZÑ jÉ¡¢VÊ„
(C) A mð jÉ¡¢VÊ„ (D) I…¢ml −L¡e¢VC euz

6. A card is drawn from a pack of cards numbered 2 to 53. The probability that the number on the
card is a prime number less than 20 is
2 −b−L 53 ¢m¢Ma HL…µR L¡XÑ −b−L HL¢V L¡XÑ V¡e¡ qmz L¡−XÑl Efl ¢m¢Ma pwMÉ¡¢V 20 Hl
Lj −j±¢mL pwMÉ¡ qJu¡l pñhe¡ q−h
2 4 5 8
(A) (B) (C) (D)
13 13 13 13

7. The straight lines x + y = 0, 5x + y = 4 and x + 5y = 4 form
(A) an isosceles triangle (B) an equilateral angle
(C) a scalene triangle (D) a right angled triangle

x + y = 0, 5x + y = 4 Hhw x + 5y = 4 plm−lM¡…¢ml à¡l¡ BhÜ −rœ¢V qm

(A) pj¢àh¡ý ¢œi¥S (B) pjh¡ý ¢œi¥S
(C) ¢hojh¡ý ¢œi¥S (D) pj−L¡Z£ ¢œi¥S

8. x
Let f (x) = . Then domain of differentiability of ‘f’ is
1+ x

x
−cJu¡ B−R f (x) = z ‘f’ A−frL¢Vl AhLme−k¡NÉ A’m q−h
1+ x

(A) [ 0, ∞ ) (B) ( −∞, 0] (C) ( −∞, ∞ ) (D) ( 0, ∞ )

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9. dy
If y = log x , x ≠ 0, then the value of is
dx
dy
y = log x , x ≠ 0, q−m, -Hl j¡e q−h
dx
1 1 1 1
(A) (B) ± (C) − (D)
x x x x

10. sin ( π sin 2 x )
lim =
x →0 x2

(A) π2 (B) 3π (C) 2π (D) π

11. dx
The value of the integration ∫ is
(1 + x ) 1 − x 2
dx
∫ (1 + x ) 1 − x -pj¡Lme¢Vl j¡e qm
2

1−x 1−x
(A) +c (B) − +c
1+ x 1+ x
1+ x 1+ x
(C) +c (D) − +c
1− x 1− x

1
12. 2
1+ x
The value of ∫ cos x. log 1 − x dx
−1
2

π 1 π 1
(A) − (B) + (C) 0 (D) none of the above
2 4 4 2
1
2
1+ x
∫ cos x. log 1 − x dx -Hl j¡e
−1
2

π 1 π 1
(A) − (B) + (C) 0 (D) I…¢ml −L¡e¢VC eu
2 4 4 2

13. dy
The integrating factor of the differential equation ( x log x ) + y = 2 log x is
dx
dy
( x log x ) + y = 2 log x A¿¹lLm pj£Ll−Zl pj¡Lm …ZL q−h
dx
(A) e x (B) log x (C) log ( log x ) (D) x

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14. The expression ax 2 + bx + c has the same sign as ‘a’ if
(A) b 2 − 4 ac > 0 (B) b 2 − 4ac ≤ 0
(C) a + b + c > 0 (D) b and c have the same sign as ‘a’.

k¢c ax 2 + bx + c Hl j¡e phÑc¡ ‘a’-l -¢Qq²k¤š² qu a−h
(A) b 2 − 4 ac > 0 (B) b2 − 4ac ≤ 0

(C) a + b + c > 0 (D) b Hhw c -Hl ¢Qq² ‘a’ Hl ja q−h

15. The probability of getting ‘4’ at least once in two tosses of a fair die is
HL¢V m¤−X¡l Oy¥¢V 2 h¡l −gm¡ q−m A¿¹a HLh¡l ‘4’ qh¡l pñ¡he¡
11 1 1 1
(A) (B) (C) (D)
36 18 3 4

16. If the line 2 x − y + λ = 0 is a diameter of the circle x 2 + y 2 + 6x − 6y + 5 = 0 then the value of λ
will be
λ -Hl −k j¡−el SeÉ 2 x − y + λ = 0 plm−lM¡¢V x 2 + y 2 + 6x − 6y + 5 = 0 hª−šl hÉ¡p q−h, a¡ qm

(A) 6 (B) 9 (C) 3 (D) 12

17. The inequality z − 6 < z − 2 represents the region given by

z − 6 < z − 2 Apja¡¢V −k A’m p§¢Qa L−l, a¡ qm

(A) Re ( z ) < 0 (B) Re ( z ) < 1 (C) Re ( z ) > 1 (D) Re ( z ) > 4

n

∑ f ( x − μ ) is equal to ( f i & x i have their usual
18.
If μ is the mean of a distribution, then i i
i =1

meaning)
(A) Mean deviation (B) Standard deviation (C) 0 (D) Median
n
HL¢V h¾V−el Ns μ q−m ∑ f i ( x i − μ ) q−h, (−kM¡−e f i J x i fËQ¢ma AbÑhq)
i =1

(A) Ns ¢hQ¥É¢a (B) pjL ¢hQ¥É¢a (C) 0 (D) jdÉj¡

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19. If α and β be the roots of k 2 x 2 + ( kx + 1)( x + k ) + 1 = 0, ( k ≠ 0, −1) , then the value of
α2β2 + ( αβ + 1 )( α + β ) + 1 is

k 2 x 2 + ( kx + 1)( x + k ) + 1 = 0, ( k ≠ 0, −1) -Hl h£Sàu α, β q−m α2β2 + ( αβ + 1 )( α + β ) + 1 -Hl
j¡e q−h
(A) 0 (B) 1 (C) - 1 (D) 2

If the coefficient of x2 and x3 in the expansion of ( 3 + ax ) be same, then ‘a’ is
9
20.

( 3 + ax ) Hl ¢àfc pÇfËp¡l−Z x2 J x3 Hl pqN k¢c pj¡e qu a−h Hl ’a’ j¡e
9

3 7 7 9
(A) (B) (C) (D)
7 3 9 7

Physics

21. Which of the following is not a unit of pressure?
¢ejÀ¢m¢Ma l¡¢n…¢ml −L¡e¢V Q¡−fl HLL eu?
(A) millibar (B) atmosphere (C) gm/cm2 (D) N-m2

22. A concave mirror of focal length ‘f’ produces an erect image when the object distance is
(A) less than f (B) equal to f (C) between f and 2f (D) greater than 2f

‘f’ −g¡L¡p °cOÑ ¢h¢nø HL¢V Aham cfÑ−el hÙ¹¥ c¤laÆ L£ qC−m pjn£oÑ fË¢a¢hð f¡Ju¡ k¡−h?
(A) f Hl Lj (B) f Hl pj¡e (C) f J 2f Hl j−dÉ (D) 2f Hl −hn£

23. The half life of At215 is 100μs.The time taken for the radioactivity of a sample of At215 to decay to
1/16 th of its initial value is
At215 −j±−ml AdÑ¡u¤ 100μsz −aS¢ûu HC −j±m¢V rufË¡ç q−u fË¡b¢jL i−ll 1/16 i¡N q−a pju
m¡N−h
(A) 40 μs (B) 400 μs (C) 6.3 μs (D) 300 μs

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24. With increasing quantum number, the energy difference between adjacent energy levels in a
hydrogen atom
(A) increases (B) remains the same
(C) decreases (D) sometimes increases, sometimes decreases

q¡C−XÊ¡−Se flj¡Z¤−a n¢š²Ù¹l…¢ml −L¡u¡¾V¡j pwMÉ¡ hª¢Ü f¡Ju¡l p¡−b p¡−b c¤¢V f¡n¡f¡¢n
n¢š²Ù¹l…¢ml n¢š²l ag¡v
(A) hª¢Ü f¡u (B) HLC b¡−L

(C) qÊÊ¡p qu (D) LMeJ hª¢Ü qu, LMeJ qÊÊ¡p qu

25. The slope of the graph of frequency of the incident light versus the stopping potential for a given
metallic surface is [e is charge of an electron, h is Plank’s constant ]
−L¡e fËcš d¡a¥fª−ùl ¢ehª¢š ¢hi−hl p¡−f−r Bf¢aa B−m¡l LÇf¡ˆ −mM¢Q−œ Aˆe Ll−m
−mM¢Q−œl e¢al j¡e q−h (e-C−mƒÊ−el Bd¡e, h-fÔ¡ˆ dË¥hL)
e h
(A) (B) (C) h (D) eh
h e

26. A 0-10 mA galvanometer with a coil resistance of 20Ω is converted to 0-10 A ammeter by using a
shunt resistor of value
HL¢V 0-10 mA j¡f¡l 20Ω BiÉ¿¹l£Z −l¡−dl NÉ¡mi¡−e¡¢jV¡−l−L 0-10A AÉ¡j¢jV¡−l l©f¡¿¹¢la
Ll−a La −l¡−dl n¡¾V hÉhq¡l Ll−a q−h?
(A) 0.02 Ω (B) 0.2 Ω (C) 9.99 Ω (D) 0.99 Ω

27. Two capacitors each having capacitance C and breakdown voltage V are joined in series. The
capacitance and the breakdown voltage of the combination will be
c¤¢V d¡lL−L k¡−cl fË¢a¢Vl d¡lLaÆ C J i”L −i¡−ÒVS V −nËZ£p‹¡u k¤š² Ll¡ qmz HC
−S¡s¢Vl −j¡V d¡lLaÆ J i”L −i¡−ÒV−Sl j¡e
(A) 2C and 2V (B) C/2 and V/2 (C) 2C and V/2 (D) C/2 and 2V

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28. A point charge +q is rotated along a circle around another point charge +Q. The work done by q in
going through a complete circle is
(A) zero (B) positive
(C) negative (D) zero if Q is at the centre of the circle , non zero otherwise.

HL¢V °hc¤É¢aL Bd¡e +q Bl HL¢V °hc¤É¢aL Bd¡e +Q -Hl Q¡¢l¢c−L hªš¡L¡−l O¤l−Rz HLh¡l
pÇf¨ZÑ hªš O¤−l H−m +q Bd¡e¢V à¡l¡ L«a L¡−kÑÉl f¢lj¡e q−h,

(A) n§eÉ (B) de¡aÆL
(C) GZ¡aÆL (D) Q k¢c hª−šl −L¾cÊ¢h¾c¥−a b¡−L a−h n§eÉ, AeÉb¡u n§eÉ eu

29. If the length of a heater wire is reduced by 10% then the power of the heater will
(A) increase by about 10% (B) increase by about 9%
(C) increase by about 11% (D) decrease by about 19%

HL¢V ¢qV¡−ll a¡−ll °cOÑÉ 10% Lj−m, ¢qV¡−ll rja¡
(A) ~ 10% h¡s−h (B) ~ 9% h¡s−h

(C) ~ 11% h¡s−h (D) ~ 19% Lj−h

30. A particle of mass m is moving with a velocity v. If the velocity of the particle is increased by 2v
then its kinetic energy would increase by
m i−ll HL¢V LZ¡ v −h−N N¢an£mz k¢c LZ¡¢Vl N¢a−hN 2v f¢lj¡Z hª¢Ü f¡u a¡q−m a¡l
N¢an¢š² hª¢Ü f¡−h
(A) 200% (C) 300% (C) 800% (D) 400%

31. The angle between the two vectors î and 2ˆi + ˆj is

î Hhw 2ˆi + ˆj HC c¤¢V −iƒ−ll j−dÉ −L¡Z

2 2 2 2
(A) cos −1 (B) cos −1 (C) cos −1 (D) cos −1
3 3 5 5

32. R
Accelerations due to gravity at a height from the earth’s surface is [ g = 10 m / s2, R is radius of
2
the earth]

fª¢bh£fªù −b−L R EµQa¡u A¢iLoÑS aÆl−Zl j¡e q−h (g = 10 m / s2, R qm fª¢bh£l hÉ¡p¡dÑ)
2
(A) 0 (B) 5 m/s2 (C) 0.44 m/s2 (D) 4.4 m/s2

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33. The rms speed of oxygen at room temperature is 500 m/s. The rms speed of hydrogen at the same
temperature is
ü¡i¡¢hL a¡fj¡œ¡u A¢„−S−el h¢NÑa −h−Nl N−sl j§m 500 m /s. HC a¡fj¡œ¡u q¡C−XÊ¡−S−el
h¢NÑa −h−Nl N−sl j§m La?
(A) 125 m /s (B) 2000 m / s (C) 8000 m / s (D) 16000 m / s

34. What is the work done by an ideal gas in expanding adiabatically from a state (p1,v1) to (p2,v2)?
HL¢V BcnÑ NÉ¡p l¦Üa¡f fË¢œ²u¡u (p1, v1) AhÙÛ¡ −b−L (p2, v2) AhÙÛ¡u fËp¡¢la qmz NÉ¡p¢Vl
c
L«a L¡−kÑÉl f¢lj¡Z ( γ = p )
cv
p 1 v 2 − p 2 v1 p1 v1 − p 2 v2
(A) (B)
γ γ −1
1  p1 p 2  (p1 − p2 ) (v1 − v2 )
(C)  −  (D)
γ  v1 v 2  γ −1

35. The electric field of an electric dipole of moment p at a large distance r will be proportional to
p ï¡jL¢h¢nø HL¢V a¢sv ¢à−jl¦ −b−L r c§l−aÆ °hc¤É¢aL −rœ k¡l pj¡e¤f¡¢aL,

1 1 1 1 1 1
(A) and 2 (B) p and (C) and 3 (D) p and
p r r2 p 2
r r3

Chemistry

36. A solution contains Fe2+ , Fe3+ and I – ions. This solution is treated with iodine at 250C. The possible
redox reaction will be : [E 0 ( Fe3+ / Fe 2+ ) = 0.770 V and E 0 (I 2 / 2I− ) = 0.536 V]

(A) I2 will be reduced to I − (B) There will be no redox reaction
(C) I − will be oxidized to I2 (D) Fe2+ will be oxidized to Fe3+

HL¢V âh−Z Fe2+ , Fe3+ Hhw I – Bue …¢m B−Rz 250C a¡fj¡œ¡u âhZ¢V−L B−u¡¢X−el p¡−b
¢h¢œ²u¡ Ll¡e qmz pñ¡hÉ S¡le-¢hS¡le ¢h¢œ²u¡¢V q−h
[E 0 ( Fe3+ / Fe 2+ ) = 0.770 V and E 0 (I 2 / 2I− ) = 0.536 V]
−
(A) I2 ¢hS¡¢la q−u I q−h (B) −L¡eJ S¡le-¢hS¡le ¢h¢œ²u¡ q−h e¡
−
(C) I S¡¢la q−u I2 q−h (D) Fe2+ S¡¢la q−u Fe3+ q−h

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37. Which of the following expressions is correct for the rate of the reaction given below
fËcš ¢h¢œ²u¡¢Vl SeÉ q¡−ll −L¡e pj£Lle¢V p¢WL
5Br − (aq) + Br O 3 − (aq) + 6H + (aq) → 3Br2 (aq) + 3H 2 O(ℓ)

(A) d[Br − ] / dt = 5 × d[H + ] / dt (B) d[Br − ] / dt = 6( 5 ) × d[H ] / dt
+

( 6 ) × d[H ] / dt
(C) d[Br − ] / dt = 5 +
(D) d[Br − ] / dt = 6 × d[H + ] / dt

38. In a buffer solution containing equal concentrations of B − and HB, the Kb for B − is 10 −10 . The pH
of the buffer solution is
HL¢V h¡g¡l âh−Z pj j¡œ¡u B− Hhw HB B−R, B− Hl Kb qm 10−10 z h¡g¡l âhZ¢Vl pH
(A) 4 (B) 10 (C) 7 (D) 6

39. The most effective ion in KCl , ZnCl2 and AlCl3 for coagulation of As2S3 sol in aqueous medium is
Sm£u j¡dÉ−j As2S3 p−ml a’−el SeÉ KCl , ZnCl2 J AlCl3 Hl j−dÉ phÑ¡−fr¡ p¢œ²u Bue
(A) K+ ion (B) Al3+ ion (C) Zn2+ ion (D) Cl - ion

40. At what temperature will the root mean square (r. m. s) velocity of oxygen gas at 1 bar pressure be
twice of its value at S.T.P? [Consider that the gas behaves ideally]
−L¡e Eo·a¡u A¢„−Se NÉ¡−pl Ns hNÑ−h−Nl hNÑjm
§ , 1 bar Q¡−f S.T.P −a j¡−el ¢à…e q−h? [d−l
e¡J NÉ¡p¢V BcÑn NÉ¡−pl eÉ¡u BQlZ L−l ]
(A) 546.3 K (B) 1092.6 K (C) 819.45 K (D) 2185.2 K

41. A mixture of solid KClO3 and KCl is strongly heated. The maximum volume of evolved O2 gas is
67.2 litre at S.T.P and the residual mass needs 5 moles of AgNO3 for complete precipitation of
AgCl . The ratio of moles of KClO3 and KCl respectively in the solid mixture is
L¢We KClO3 J KCl Hl ¢jnËZ a£hËi¡−h Ešç Ll¡ qmz Hl g−m Evf¡¢ca O2 NÉ¡−pl phÑ¡¢dL
Buae S.T.P −a 67.2 litre Hhw Ah¢nø il J 5 moles AgNO3 Hl ¢h¢œ²u¡u AgCl pÇf¨ZÑl©−f
Adx¢rç quz j§m L¢We ¢jnË−Z KClO3 J KCl Hl −j¡m Ae¤f¡a kb¡œ²−j
(A) 2 : 3 (B) 3 : 2 (C) 1 : 3 (D) 5 : 2

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42. The oxidation number of S in Na2S4O6
(A) +2.5 for all the four S atoms
(B) +2 for two S and +3 for two other S atoms
(C) +2 for three S and +1 for the rest one S atom
(D) 0 for two S and +5 for two other S atoms

Na2S4O6 H S Hl S¡le pwMÉ¡
(A) Q¡l¢V S Hl SeÉC +2.5
(B) c¤¢V S Hl SeÉ +2, h¡¢L c¤C¢V S Hl SeÉ +3
(C) ¢ae¢V S Hl SeÉ +2, h¡¢L HL¢Vl SeÉ +1
(D) c¤C¢V S Hl SeÉ n§eÉ, h¡¢L c¤C¢Vl SeÉ +5

43. ClO3 is mixed anhydride of
(A) HClO and HClO2 (B) HClO2 and HClO3
(C) HClO3 and HClO4 (D) HClO and HClO4

¢ejÀ¢m¢Ma −L¡e¢Vl ¢jnË AÉ¡eq¡CXÊ¡CX qm ClO3?
(A) HClO Hhw HClO2 (B) HClO2 Hhw HClO3
(C) HClO3 Hhw HClO4 (D) HClO Hhw HClO4

44. Fluorine does not show disproportionation reaction because
(A) fluorine is the strongest oxidizing agent
(B) fluorine is gaseous
(C) fluorine is a moderate reducing agent
(D) fluorine is unstable

−gÓ¡¢le Apj”pÉ ¢h¢œ²u¡ L−l e¡ L¡le
(A) −gÓ¡¢le ph−Q−u n¢š²n¡m£ S¡lL âhÉ
(B) −gÓ¡¢le HL¢V NÉ¡p
(C) −gÓ¡¢le HL¢V j¡T¡¢l j¡−el ¢hS¡lL âhÉ
(D) −gÓ¡¢le AÙÛ¡u£

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45. The number of moles of KMnO4 that can be reduced by one mole of KI in alkaline medium is
r¡l£u j¡dÉ−j HL −j¡m KI La −j¡m KMnO4 −L ¢hS¡¢la Ll−a f¡−l?
(A) 1 (B) 2 (C) 5 (D) 1
5

46. Which of the following properties is not due to H- bonding ?
(A) High boiling point of water (B) High viscosity of glycerol.
(C) Solubility of ammonia in water (D) Polar nature of halogen acids

−L¡e djÑ¢V q¡C−XÊ¡−Se hå−el SeÉ eu?
(A) S−ml EµQ Øg¥Ve¡ˆ (B) ¢NÔp¡l−ml EµQ p¡¾cÊa¡
(C) AÉ¡−j¡¢eu¡l S−m âhÉa¡ (D) qÉ¡−m¡−Se AÉ¡¢pX…¢ml dËh
¥ £u fËL«¢a

47. The IUPAC name of the given compound is :
CH2COOH

HO C COOH

CH2COOH

(A) 3 – hydroxy – 3 - carboxy - 1, 5- pentanedioic acid
(B) 3 – carboxy-3-hydroxy-1, 5-pentanedioic acid
(C) 2 - hydroxy propane – 1, 2, 3 – tricarboxylic acid
(D) 3 - hydroxy pentane – 1,3,5 – trioic acid

fËcš −k±N¢Vl IUPAC e¡j qm
CH2COOH

HO C COOH

CH2COOH

(A) 3 – q¡CXÊ¢„ – 3 - L¡hÑ¢„ - 1, 5- −f−¾VeX¡CJ¢uL AÉ¡¢pX
(B) 3 – L¡hÑ¢„ - 3 - q¡CXÊ¢„ -1, 5-−f−¾VeX¡CJ¢uL AÉ¡¢pX
(C) 2 - q¡CXÊ¢„ −fË¡−fe – 1, 2, 3 – VÊ¡CL¡hÑ¢„¢mL AÉ¡¢pX
(D) 3 - q¡CXÊ¢„ −f−¾Ve – 1,3,5 – VÊ¡CJ¢uL AÉ¡¢pX

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48. The major product of the reaction of HOBr with CCl3 CH = CH2 is

HOBr J CCl3 CH = CH2 ¢h¢œ²u¡u Evfæ j¤MÉ −k±N¢V qm -

(A) CCl3 CH (OH)CH2 Br (B) CCl3 CH (Br)CH2 OH
(C) CCl2 (Br)CH (Cl) CH2 OH (D) CCl2 (OH)CH (Cl) CH2 Br

49. Among the following molecules, which one will not show keto – enol tantomerism
fËcš −k±N…¢ml j−dÉ −L¡e¢V ¢L−V¡-Hem V−V¡−j¢lpj −cM¡−h e¡?

O O
O O O
O

(A) (C) (D)
(B)

50. The major product in the following reaction will be
Cl
Cl I. Conc. HNO / Cone. H SO
3 2 4
?
II. NaOMe, MeOH
Cl Cl OMe Cl
Cl MeO Cl Cl (D) OMe
(A) (B) (C)

OMe NO2 NO2 NO2

¢ejÀ¢m¢Ma ¢h¢œ²u¡¢V−a Evf¡¢ca j¤MÉ −k±N¢V qm
Cl
Cl I. N¡t HNO / N¡t H SO
3 2 4
?
II. NaOMe, MeOH
Cl Cl OMe Cl
Cl MeO Cl Cl (D) OMe
(A) (B) (C)

OMe NO2 NO2 NO2

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PUBDET-2018
Subject: Geology

pju: 90 ¢j¢eV phÑ¡¢dL eðl: 100

¢e−cÑn¡hm£
1. HC fËnÀf−œl ph fËnÀC Ah−S¢ƒi fËnÀ Hhw fË¢a¢V fË−nÀl Q¡l¢V pñ¡hÉ Ešl −cJu¡
B−R k¡l HL¢V j¡œ p¢WLz p¢WL Ešl ¢c−m 2 eðl f¡−hz i¥m Ešl ¢c−m Abh¡
HL¡¢dL Ešl ¢c−m ½ eðl L¡V¡ k¡−hz
2. OMR f−œ A,B,C,D ¢Q¢q²a p¢WL Ol¢V il¡V L−l Ešl ¢c−a q−hz
3. OMR f−œ Ešl ¢c−a öd¤j¡œ L¡−m¡ h¡ e£m hm f−u¾V −fe hÉ¡hq¡l Ll−hz
4. OMR f−œ ¢e¢cÑø ÙÛ¡e R¡s¡ AeÉ −L¡b¡J −L¡e c¡N −c−h e¡z
5. OMR f−œ ¢e¢cÑø ÙÛ¡−e fËnÀf−œl eðl Hhw ¢e−Sl −l¡m eðl A¢a p¡hd¡ea¡l p¡−b
¢mM−a q−h Hhw fË−u¡Se£u Ol…¢m f§lZ Ll−a q−hz
6. OMR f−œ ¢e¢cÑø ÙÛ¡−e ¢e−Sl e¡j J fl£r¡ −L−¾cÊl e¡j ¢mM−a q−h Hhw ¢e−Sl pÇf¨ZÑ
p¡rl ¢c−a q−hz
7. OMR Ešlfœ¢V C−mLVÊ¢eL k−¿»l p¡q¡−kÉ fs¡ q−hz p¤a
¤ l¡w fËnÀ−fœl eðl h¡ −l¡m
eðl i¥m ¢mM−m Abh¡ i¥m Ol il¡V Ll−m Ešlfœ¢V A¢eh¡kÑ L¡l−Z h¡¢am q−a
f¡−lz HR¡s¡ fl£r¡bÑ£l e¡j, fl£r¡ −L−¾cÊl e¡j h¡ p¡r−l −L¡e i¥m b¡L−mJ Ešl fœ
h¡¢am q−u −k−a f¡−lz OMR Ešlfœ¢V i¡yS q−m h¡ a¡−a Ae¡hnÉL c¡N fs−mJ
h¡¢am q−u −k−a f¡−lz fl£r¡bÑ£l HC dl−el i¥m h¡ ApaÑLa¡l SeÉ Ešlfœ h¡¢am
q−m HLj¡œ fl£r¡bÑ£ ¢e−SC a¡l SeÉ c¡u£ b¡L−hz
8. −j¡h¡Cm−g¡e, LÉ¡mL¥−mVl, pÔ¡CXl¦m, mN−Vhm, −lM¡¢Qœ, NË¡g h¡ −L¡e dl−Zl a¡¢mL¡
fl£r¡ L−r Be¡ k¡−h e¡z Be−m −p¢V h¡−Su¡ç q−h Hhw fl£r¡bÑ£l JC fl£r¡
h¡¢am Ll¡ q−hz
9. fËnÀf−œl −n−o l¡g L¡S Ll¡l SeÉ gy¡L¡ S¡uN¡ −cJu¡ B−Rz AeÉ −L¡e L¡NS HC
L¡−S hÉhq¡l Ll−h e¡z
10. fl£r¡ Lr R¡s¡l B−N OMR fœ AhnÉ C f¢lcnÑL−L ¢c−u k¡−hz
11. HC fËnÀf−œ Cwl¡S£ J h¡wm¡ Eiu i¡o¡−aC fËnÀ −cJu¡ B−Rz h¡wm¡ j¡dÉ−j fËnÀ °al£l
pju fË−u¡Se£u p¡hd¡ea¡ J paLÑa¡ Ahmðe Ll¡ q−u−Rz a¡ p−šÄJ k¢c −L¡e Ap‰¢a
mr Ll¡ k¡u, −p−r−œ Cwl¡S£ j¡dÉ−j −cJu¡ fËnÀ ¢WL J Q¨s¡¿¹ h−m ¢h−h¢Qa q−hz

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