WBJEE 2023 Question Paper Physics Chemistry

Subject : PHYSICS & CHEMISTRY
(Booklet Number)
Duration : 2 Hours Full Marks : 100

INSTRUCTIONS
1. All questions are of objective type having four answer options for each.
2. Category-1: Carries 1 mark each and only one option is correct. In case of incorrect answer or
any combination of more than one answer, ¼ mark will be deducted.
3. Category-2: Carries 2 marks each and only one option is correct. In case of incorrect answer or
any combination of more than one answer, ½ mark will be deducted.
4. Category-3: Carries 2 marks each and one or more option(s) is/are correct. If all correct
answers are not marked and no incorrect answer is marked, then score = 2 × number of correct
answers marked ÷ actual number of correct answers. If any wrong option is marked or if any
combination including a wrong option is marked, the answer will be considered wrong, but
there is no negative marking for the same and zero mark will be awarded.
5. Questions must be answered on OMR sheet by darkening the appropriate bubble marked A, B, C, or D.
6. Use only Black/Blue ink ball point pen to mark the answer by filling up of the respective
bubbles completely.
7. Write Question Booklet number and your roll number carefully in the specified locations of the
OMR Sheet. Also fill appropriate bubbles.
8. Write your name (in block letter), name of the examination center and put your signature (as is
appeared in Admit Card) in appropriate boxes in the OMR Sheet.
9. The OMR Sheet is liable to become invalid if there is any mistake in filling the correct bubbles
for Question Booklet number/roll number or if there is any discrepancy in the name/ signature
of the candidate, name of the examination center. The OMR Sheet 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 the sole
responsibility of candidate.
10. Candidates are not allowed to carry any written or printed material, calculator, pen, log-table,
wristwatch, any communication device like mobile phones, bluetooth device etc. inside the
examination hall. Any candidate found with such prohibited items will be reported against
and his/her candidature will be summarily cancelled.
11.
Question work must be done on the Question Booklet itself. Additional blank pages are given in
Rough
Bookletthe Question Booklet for rough work. PC-2023
Series
12. Hand over the OMR Sheet to the invigilator before leaving the Examination Hall.
13. This Booklet containsSubject : PHYSICS
questions & and
in both English CHEMISTRY
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.
14. Candidates are allowed to take the Question Booklet after examination is over.

Signature of the Candidate : ______________________________
(as in Admit Card)

Signature of the Invigilator : ______________________________
A 1 P.T.O.
PC-2023 A

SPACE FOR ROUGH WORK

A 2

PHYSICS
Category-1 (Q. 1 to 30)
(Carry 1 mark each. Only one option is correct. Negative marks : – ¼)

1. In a simple harmonic motion, let f be the acceleration and T be the time period. If x denotes
the displacement, then |f T| vs. x graph will look like,
HL¢V plm ­c¡mN¢al ­r­œ f q'm aÆlZ J T q'm fkÑ¡uL¡m z k¢c x qu plZ a­h |f T| he¡j x
­mM¢Qœ¢V q­h,
|f T|
|f T|

(A) (B)

x x

|f T| |f T|

(C) (D)

x x

2. The displacement of a plane progressive wave in a medium, travelling towards positive
x-axis with velocity 4 m/s at t = 0 is given by y = 3 sin 2    . Then the expression for
x
 3
the displacement at a later time t = 4 sec will be

HL¢V j¡dÉ­j de¡aÈL x-Ar A¢ij¤­M 4 m/s ­h­N N¢an£m HL¢V pjam Qmal­‰l t = 0 pj­u
pl­Zl pj£LlZ y = 3 sin 2   x  z a¡q­m t = 4 ­p­L™ pj­u pl­Zl l¡¢n¢V q­h,
 3

x  16    x  16 
(A) y = 3 sin 2    (B) y = 3 sin 2  
 3   3 
  x 1   x 1
(C) y = 3 sin 2   (D) y = 3 sin 2  
 3   3 

A 3 P.T.O.

3. As shown in the figure, a liquid is at same levels in two
arms of a U-tube of uniform cross-section when at rest. If
the U-tube moves with an acceleration ‘f’ towards right,
the difference between liquid heights between two arms
of the U-tube will be, (acceleration due to gravity = g)
¢Q­œ fËc¢nÑa ¢ÙÛl AhÙÛ¡u pjfËÙÛ­µRck¤š² U-e­ml c¤¢V h¡ý­a
HL¢V alm pj¡e EµQa¡u b¡­L z k¢c U-em¢V ‘f’ aÆl­Z
X¡e¢c­L N¢an£m qu a­h U-e­ml c¤C h¡ýl al­ml EµQa¡l
f¡bÑLÉ q­h, (A¢iLoÑS aÆlZ = g)

f g
(A) a (B) a (C) a (D) 0
g f

4. Six molecules of an ideal gas have velocities 1, 3, 5, 5, 6 and 5 m/s respectively. At any
given temperature, if V and Vrms represent average and rms speed of the molecules, then
HL¢V BcnÑ NÉ¡­pl Ru¢V AZ¤l ­hN kb¡œ²­j 1, 3, 5, 5, 6 J 5 m/s z k¢c V J Vrms ­L¡e ¢e¢cÑø
a¡fj¡œ¡u AZ¤…¢ml Ns­hN J rms ­hN ­h¡T¡u a­h,
2
(A) V = 5 m/s (B) Vrms > V (C) Vrms  V2 (D) Vrms = V

5.

As shown in the figure, a pump is designed as horizontal cylinder with a piston having area
A and an outlet orifice having an area ‘a’. The piston moves with a constant velocity under
the action of force F. If the density of the liquid is ρ, then the speed of the liquid emerging
from the orifice is, (assume A >> a)
¢Q­œ fËc¢nÑa HL¢V f¡Çf Ae¤i¨¢jL ­Q¡­Pl BL¡­l N¢Wa k¡l ¢fø­el fËÙÛ­µR­cl ­rœgm A J
¢eNÑje j¤­Ml fËÙÛ­µR­cl ­rœgm ‘a’ z HL¢V ¢ÙÛl hm F-Hl fË­u¡­N ¢føe¢V ¢ÙÛl ­h­N ANËpl qu z
al­ml OeaÆ k¢c ρ qu a­h f¡­Çfl j¤M ­b­L ¢eNÑa al­ml ­hN q­h (d­l e¡J A >> a),

F a F 2F A 2F
(A) (B) (C) (D)
ρA A ρA ρA a ρA

A 4

6. Two substances A and B of same mass are heated at constant rate. The variation of
temperature θ of the substances with time t is shown in the figure. Choose the correct
statement.
θ
(A) Specific heat of A is greater than that of B.
B
(B) Specific heat of B is greater than that of A.

(C) Both have same specific heat. A

(D) None of the above is true. t
0

HLC i­ll c¤¢V fc¡bÑ A J B ­L HLC q¡­l Ešç Ll¡ q­µR z pju (t)-Hl p¡­b a¡fj¡œ¡ (θ)-Hl
f¢lhaÑe ¢Q­œ ­cM¡­e¡ q­u­R z ­p­r­œ p¢WL E¢š²¢V ¢ehÑ¡Qe Ll z
(A) B-Hl ­Q­u A-Hl B­f¢rL a¡f ­h¢n z

(B) A-Hl ­Q­u B-Hl B­f¢rL a¡f ­h¢n z

(C) c¤¢V fc¡­bÑlC B­f¢rL a¡f pj¡e z
(D) Ef­ll ­L¡­e¡ E¢š²C p¢WL eu z

7. A given quantity of gas is taken from A to C in two ways; p
a) directly from A → C along a straight line and b) in two A
steps, from A → B and then from B → C. Work done and heat
absorbed along the direct path A→C is 200 J and 280 J
B C
respectively.
V
If the work done along A→B→C is 80 J, then heat absorbed
along this path is,

HL¢V ¢e¢cÑø f¢lj¡Z NÉ¡p­L A ­b­L C AhÙÛ¡­e c¤C i¡­h ¢e­u k¡Ju¡ q'm, a) A → C plm ­lM¡
hl¡hl J b) A → B J B → C HC c¤C d¡­f z A→C f­b L«a L¡kÑÉ J ­n¡¢oa a¡f kb¡œ²­j 200 J
J 280 J z A→B→C f­b L«a L¡kÑÉ k¢c 80 J qu a­h I f­b ­n¡¢oa a¡f q­h,
(A) 80 J (B) 0 (C) 160 J (D) 120 J

A 5 P.T.O.

8. A thin glass rod is bent in a semicircle of radius R. A charge is y
non-uniformly distributed along the rod with a linear charge
density    0 sin  (0 is a positive constant). The electric field R
θ
at the centre P of the semicircle is, x

HL¢V pl¦ Ly¡­Ql c™­L R hÉ¡p¡­dÑl AdÑhš
ª ¡L¡­l hy¡L¡­e¡ q'm z c™¢V­a
Apj ¢hÙ¹«a Bd¡e k¤š² B­R k¡l °l¢ML OeaÆ q'm    0 sin  (0 HL¢V de¡aÈL dËh¤ L) z ­p­r­œ
AdÑhª­šl ­L¾cÊ P-­a a¢sv ­r­œl j¡e q­h,

o o o o
(A) – ^j (B) ^j (C) ^i (D) – ^i
8oR 8oR 8oR 8oR

9. Consider a positively charged infinite cylinder with uniform volume charge
Q
density ρ > 0. An electric dipole consisting of + Q and ˗ Q charges attached to
opposite ends of a massless rod is oriented as shown in the figure. At the
˗Q
instant as shown in the figure, the dipole will experience,

(A) a force to the left and no torque.

(B) a force to the right and a clockwise torque.

(C) a force to the right and a counter clockwise torque.

(D) no force but only a clockwise torque.

j­e Ll de¡aÈL Bd¡ek¤š² HL¢V A¢a c£OÑ ­Q¡­Pl Bua¢eL Bd¡e OeaÆ ρ > 0 z Bh¡l HL¢V ilq£e
c­™l c¤C fË¡­¿¹ + Q J ˗ Q Bd¡e k¤š² HL¢V a¢sv ¢à­jl¦ ¢Q­œ fËc¢nÑa A¢ij¤­M l¡M¡ B­R z ¢Q­œ
fËc¢nÑa j¤q¨­aÑ a¢sv ¢à­jl¦¢Vl Efl ¢œ²u¡ Ll­h,

(A) h¡j A¢ij¤­M HL¢V hm ¢L¿¹¥ ­L¡­e¡ ï¡jL eu z

(B) X¡e A¢ij¤­M HL¢V hm J O¢sl Ly¡V¡l A¢ij¤­M HL¢V ï¡jL z

(C) X¡e A¢ij¤­M HL¢V hm J O¢sl Ly¡V¡l ¢hfl£a A¢ij¤­M HL¢V ï¡jL z

(D) ­L¡e hm eu, öd¤ O¢sl Ly¡V¡l A¢ij¤­M HL¢V ï¡jL z

A 6

10. 12 μC and 6 μC charges are given to the two conducting plates
12 μC 6 μC
having same cross-sectional area and placed face to face close
to each other as shown in the figure. The resulting charge
distribution in μC on surfaces A, B, C and D are respectively,

¢Q­œ fËc¢nÑa HLC fËÙÛ­µRc k¤š² c¤¢V f¢lh¡q£ f¡a­L kb¡œ²­j
A
12 μC J 6 μC Bd¡e ­cJu¡ q'm Hhw M¤h L¡R¡L¡¢R j¤­M¡j¤¢M l¡M¡ B C D

q'm z ­p­r­œ A, B, C J D fª­ù h¢¸Va Bd¡­el j¡e (μC HL­L)
q­h kb¡œ²­j,
(A) 9, 3, – 3, 9 (B) 3, 9, – 9, 3 (C) 6, 6, – 6, 12 (D) 6, 6, 3, 3

A wire carrying a steady current I is kept in the x-y plane along the curve y  A sin  x  .
2π
11.
 λ 
A magnetic field B exists in the z-direction. The magnitude of the magnetic force in the
portion of the wire between x = 0 and x = λ is

¢ÙÛl a¢svfËh¡q I k¤š² HL¢V pl¦ a¡l­L x-y a­m y  A sin  2π x  hœ²­lM¡ hl¡hl l¡M¡ q'm z I
 λ 
ÙÛ¡­e z-Ar A¢ij¤­M HL¢V ­Q±ðL ­rœ B ¢œ²u¡ L­l z a¡q­m a¡l¢Vl x = 0 J x = λ -Hl jdÉha£Ñ
Aw­nl Efl ­k ­Q±ðL hm ¢œ²u¡ L­l a¡l j¡e q'm,
(A) 0 (B) 2IB (C) IB (D) IB/2

12. The figure represents two equipotential lines in x-y y(cm) 2 Volt
plane for an electric field. The x-component Ex of the 3
4 Volt
electric field in space between these equipotential lines 2
is,
1
x-y a­m HL¢V a¢sa­r­œl j­dÉ c¤¢V pj¢hih ­lM¡ ¢Q­œ
0 x(cm)
­cM¡­e¡ q­u­R z I c¤¢V ­lM¡l jdÉhaÑ£ ÙÛ¡­e a¢sv­r­œl 2 4 6 8

x-Ef¡wn Ex-Hl j¡e q'm,

(A) 100 V/m (B) –100 V/m (C) 200 V/m (D) –200 V/m

A 7 P.T.O.


13. An electric dipole of dipole moment p is placed at the origin of the co-ordinate system
along the z-axis. The amount of work required to move a charge ‘q’ from the point (a, 0,
0) to the point (0, 0, a) is,

p ¢à­jl¦ ï¡j­Ll HL¢V a¢sv-¢à­jl¦ L¡­VÑS£u ¢e­cÑna­¿»l j§m¢h¾c¤­a l¡M¡ B­R z ¢à­jl¦¢Vl A¢ij¤M
z-Ar hl¡hl z a¡q­m HL¢V ‘q’ Bd¡e­L (a, 0, 0) ¢h¾c¤ ­b­L (0, 0, a) ¢h¾c¤­a ¢e­u ­k­a fË­u¡Se£u
L«aL¡kÑÉ q'm,
pq  pq pq
(A) (B) 0 (C) (D)
4 ε 0 a 4ε 0a 2 4ε 0a 2

14. The electric field of a plane electromagnetic wave of wave number k and angular frequency

ω is given by E = E0(^i + ^j) sin(kz – t). Which of the following gives the direction of the

associated magnetic field B ?
al‰ pwMÉ¡ k J ω ®L±¢ZL LÇf¡­ˆl HL¢V pjam a¢sv ®Q±ðL£u al­‰l a¢s­vrœ
 
E = E0(^i + ^j) sin(kz – t) z a¡q­m ¢e­Ql ­L¡e¢V Be¤p¢‰L ­Q±ðL­rœ B-Hl ¢cL ¢e­cÑn L­l ?

(A) k^ (B) – ^i + ^j (C) – ^i – ^j (D) ^i – k^


15. A charged particle in a uniform magnetic field B = B0k^ starts moving from the origin with

velocity v = 3^i + 4k^ m/s. The trajectory of the particle and the time t at which it reaches
2 m above x-y plane are,
(A) Circular path, ½ sec. (B) Helical path, ½ sec.

(C) Circular path, 2/3 sec. (D) Helical path, 2/3 sec.

p¤¤oj a¢sv­rœ B = B0k^ -Hl j­dÉ HL¢V B¢qa LZ¡ j§m¢h¾c¤ ­b­L v = 3^i + 4k^ m/s ­h­N Qm­a
öl¦ L­l z LZ¡¢Vl Njefb J ­k pju t-­a LZ¡¢V x-y-a­ml 2 m EµQa¡u ­f±yR­h a¡ q'm,
(A) hªš¡L¡l fb, ½ ­p­L™ (B) L¥™m¡¢ua fb, ½ ­p­L™

(C) hªš¡L¡l fb, 2/3 ­p­L™ (D) L¥™m¡¢ua fb, 2/3 ­p­L™

A 8

16. In an experiment on a circuit as shown in the figure, the voltmeter shows 8 V reading. The
resistance of the voltmeter is,

¢Q­œ fËc¢nÑa haÑe£l fl£r¡u ­i¡ÒV¢jV¡l 8 V f¡W ­cu z a¡q­m ­i¡ÒV¢jV¡­ll ­l¡d La ?
(A) 20 Ω (B) 320 Ω (C) 160 Ω (D) 1.44 k Ω

17. An interference pattern is obtained with two coherent sources of intensity ratio n:1. The
I I
ratio Max Min will be maximum if
I Max  I Min
fË¡h­mÉl Ae¤f¡a n:1 ¢h¢nø c¤¢V p¤¤pwNa Evp à¡l¡ HL¢V hÉ¡¢aQ¡l f¢V N¢Wa q'm z ­p­r­œ
I Max  I Min
Ae¤f¡a¢V p­hÑ¡µQ q­h kMe,
I Max  I Min
(A) n = 1 (B) n=2 (C) n=3 (D) n = 4

18. A circular coil is placed near a current carrying conductor, both Up
lying on the plane of the paper. The current is flowing through the
conductor in such a way that the induced current in the loop is
clockwise as shown in the figure. The current in the wire is,
(A) time dependent and downward.
(B) steady and upward. I
(C) time dependent and upward.
(D) An alternating current. Down

HL¢V hªš¡L¡l L¥™m£­L HL¢V a¢svh¡q£ f¢lh¡q£l L¡­R l¡M¡ q'm; Ei­uC L¡N­Sl a­m Ah¢ÙÛa z
f¢lh¡q£l jdÉ ¢c­u a¢sv Hje i¡­h fËh¡¢qa q­µR ­k L¥™m£­a B¢hø a¢sv fËh¡­ql A¢ij¤M O¢sl
Ly¡V¡l ¢c­L (¢Q­œ fËc¢nÑa) z ­p­r­œ f¢lh¡q£l j­dÉ a¢svfËh¡q
(A) pj­ul p¡­b f¢lhaÑen£m J ¢ejj À ¤M£
(B) FdÑÄj¤M£ J ¢ÙÛlfËh¡q
(C) pj­ul p¡­b f¢lhaÑen£m J FdÑÄj¤M£
(D) HL¢V f¢lhaÑ£ a¢svfËh¡q

A 9 P.T.O.

19. Three identical convex lenses each of focal
length f are placed in a straight line separated
by a distance f from each other. An object is
located at f/2 in front of the leftmost lens. Object
Then,
(A) Final image will be at f/2 behind the rightmost lens and its magnification will be −1.
(B) Final image will be at f/2 behind the rightmost lens and its magnification will be +1.
(C) Final image will be at f behind the rightmost lens and its magnification will be −1.
(D) Final image will be at f behind the rightmost lens and its magnification will be +1.
f ­g¡L¡p °c­OÑÉl ¢ae¢V A¢iæ Ešm ­m¾p­L HL¢V plm ­lM¡u H­L Afl ­b­L f c§l­aÄ ÙÛ¡fe Ll¡
q'm z hy¡ ¢c­Ll fËbj ­m¾p¢Vl p¡j­e f/2 c§l­aÄ HL¢V hÙ¹¥ l¡M¡ q'm z ­p­r­œ,
(A) A¢¿¹j fË¢a¢hð¢V X¡e¢c­Ll ­no ­m­¾pl ¢fR­e f/2 c§l­aÄ pª¢ø q­h Hhw ¢hhdÑe q­h −1
(B) A¢¿¹j fË¢a¢hð¢V X¡e¢c­Ll ­no ­m­¾pl ¢fR­e f/2 c§l­aÄ pª¢ø q­h Hhw ¢hhdÑe q­h +1
(C) A¢¿¹j fË¢a¢hð¢V X¡e¢c­Ll ­no ­m­¾pl ¢fR­e f c§l­aÄ pª¢ø q­h Hhw ¢hhdÑe q­h −1
(D) A¢¿¹j fË¢a¢hð¢V X¡e¢c­Ll ­no ­m­¾pl ¢fR­e f c§l­aÄ pª¢ø q­h Hhw ¢hhdÑe q­h +1

20. A ray of monochromatic light is incident on the plane
surface of separation between two media X and Y with sin r
angle of incidence ‘i’ in medium X and angle of refraction
‘r’ in medium Y. The given graph shows the relation
between sin i and sin r. If VX and VY are the velocities of
the ray in media X and Y respectively, then which of the
following is true ? 30O
1 0 sin i
(A) VX = VY
3
(B) VX = 3 VY
(C) Total internal reflection can happen when the light is incident in medium X.
(D)  X  3 Y , where νX and νY are frequencies of the light in medium X and Y
respectively.
HL¢V HLhZÑ£ B­m¡l l¢nÈ X J Y c¤¢V j¡dÉ­jl pjam ¢h­µRc a­ml Efl Bf¢aa q'm, ­kM¡­e X
j¡dÉ­j Bfae ­L¡Z ‘i’ J Y j¡dÉ­j fË¢aplZ ­L¡Z ‘r’ z sin i J sin r-Hl j­dÉ pÇfLÑ ¢Q­œ ­cM¡­e¡
q­u­R z k¢c VX J VY kb¡œ²­j X J Y j¡dÉ­j l¢nÈ¢Vl ­hN qu a­h ¢e­Ql ­L¡e E¢š²¢V p¢WL ?
1
(A) VX = VY
3
(B) VX = 3 VY
(C) l¢nÈ¢V X j¡dÉ­j Bf¢aa q'­m BiÉ¿¹l£Z f§ZÑ fË¢agme q­a f¡­l z
(D)  X  3 Y , ­kM¡­e νX J νY q'm kb¡œ²­j X J Y j¡dÉ­j B­m¡­Ll LÇf¡ˆ z

A 10

21. If the potential energy of a hydrogen atom in the first excited state is assumed to be zero,
then the total energy of n = ∞ state is,
q¡C­XÊ¡­Se flj¡Z¤l fËbj EŸ£¢fa Ù¹­ll ¢ÙÛ¢an¢š² n§eÉ dl­m n = ∞ Ù¹­l ­j¡V n¢š² q­h
(A) 3.4 eV (B) 6.8 eV (C) 0 (D) ∞

22.

In the given circuit, find the voltage drop VL in the load resistance RL.
fËcš haÑe£­a ­m¡X RL-H ¢hih fË­ic VL-Hl j¡e La ?
(A) 5 V (B) 3V (C) 9V (D) 6 V

23.

Consider the logic circuit with inputs A, B, C and output Y. How many combinations of
A, B and C gives the output Y = 0 ?
Cef¤V A, B, C J BEVf¤V Y-Hl fËcš haÑe£¢V ¢h­hQe¡ Ll z A, B J C-Hl La lL­jl ­r­œ
Y = 0 q­h ?
(A) 8 (B) 5 (C) 7 (D) 1

24. X-rays of wavelength λ gets reflected from parallel
planes of atoms in a crystal with spacing d between θ
two planes as shown in the figure. If the two d
reflected beams interfere constructively, then the
condition for maxima will be, (n is the order of
interference fringe)
λ al‰°c­OÉÑl X-l¢nÈ HL¢V ­Lm¡­pl pj¡¿¹l¡m c¤¢V am (k¡­cl j­dÉ c§laÄ q'm d) ­b­L fË¢ag¢ma
q'm z fË¢ag¢ma l¢nÈc¤¢Vl NWej§mL hÉ¢aQ¡­ll naÑ q­h, (n q'm hÉ¢aQ¡l f¢Vl œ²j)
(A) d tan  = n (B) d sin  = n (C) 2d cos  = n (D) 2d sin  = n

A 11 P.T.O.

25. A particle of mass m is projected at a velocity u, making an angle θ with the horizontal (x-
axis). If the angle of projection θ is varied keeping all other parameters same, then
magnitude of angular momentum (L) at its maximum height about the point of projection
varies with θ as,

m i­ll HL¢V hÙ¹¥LZ¡­L u ­h­N Ae¤i¨¢jL (x-A­rl) p­‰ θ ­L¡­Z fË­rf Ll¡ q'm z AeÉ pjÙ¹
¢LR¥ HLC ­l­M k¢c öd¤ fË­rf­L¡Z θ f¢lhaÑe Ll¡ qu a­h I hÙ¹¥LZ¡l p­hÑ¡µQ AhÙÛ¡­e fË­rf
¢h¾c¤l p¡­f­r ­L±¢ZL il­h­Nl (L) f¢lhaÑe ¢e­Ql ­L¡e ­mM¢Qœ¢V à¡l¡ ¢e­cÑn Ll¡ k¡u ?

(A) (B)

(C) (D)

26. A body of mass 2 kg moves in a horizontal circular path of radius 5 m. At an instant, its
speed is 2√5 m/s and is increasing at the rate of 3 m/s2. The magnitude of force acting on
the body at that instant is,
2 kg i­ll HL¢V hÙ¹¥ 5 m hÉ¡p¡­dÑl Ae¤i¨¢jL hªš¡L¡l f­b O¤l­R z ­L¡­e¡ HL j¤q¨­aÑ hÙ¹¥¢Vl ⤢a
2√5 m/s Hhw I ⤢a 3 m/s2 q¡­l hª¢Ü f¡­µR z I j¤q¨­aÑ hÙ¹¥¢Vl Efl ¢œ²u¡la h­ml j¡e q'm,
(A) 6 N (B) 8N (C) 14 N (D) 10 N

27. In an experiment, the length of an object is measured to be 6.50 cm. This measured value
can be written as 0.0650 m. The number of significant figures on 0.0650 m is

HL¢V fl£r¡u HL¢V hÙ¹¥l °cOÑÉ ­j­f f¡Ju¡ ­Nm 6.50 cm z f¢lj¡­fl j¡e¢Vl HLL f¢lhaÑe L­l
­mM¡ k¡u 0.0650 m z 0.0650 m-H a¡vfkÑf§ZÑ AˆpwMÉ¡ q'm
(A) 3 (B) 4 (C) 2 (D) 5

A 12

28. A mouse of mass m jumps on the outside edge of a rotating ceiling fan of moment of inertia
I and radius R. The fractional loss of angular velocity of the fan as a result is,
m i­ll HL¢V Cyc¤l R hÉ¡p¡dÑ J I S¡XÉï¡jL ¢h¢nø HL¢V O¤ZÑ¡uj¡e ¢p¢mw gÉ¡­el h¢qxfË¡­¿¹ m¡g
¢c­u EWm z a¡q­m gÉ¡e¢Vl ­L±¢ZL ­h­Nl Bw¢nL qÊÊ¡p La q­h ?

mR 2 I I  mR 2 I  mR 2
(A) (B) (C) (D)
I  mR 2 I  mR 2 I I  mR 2

29. Acceleration due to gravity at a height H from the surface of a planet is the same as that at
a depth of H below the surface. If R be the radius of the planet, then H vs. R graph for
different planets will be,
HL¢V NË­ql fªù ­b­L H EµQa¡u J HLC Ni£la¡u A¢iLoÑS aÄl­Zl j¡e pj¡e z NËq¢Vl hÉ¡p¡dÑ
k¢c R qu a­h ¢h¢iæ NË­ql SeÉ H he¡j R ­mM¢Qœ¢V q­h,

(A) (B)

(C) (D)

30. A uniform rope of length 4 m and mass 0.4 kg is held on a frictionless table in such a way
that 0.6 m of the rope is hanging over the edge. The work done to pull the hanging part of
the rope on to the table is, (Assume g = 10 m/s2)
0.4 kg i­ll J 4 m °c­OÉÑl HL¢V c¢s­L HL¢V OoÑZ ¢hq£e ­V¢h­ml Efl Hje i¡­h l¡M¡ B­R ­k
c¢s¢Vl 0.6 m Awn ­V¢h­ml h¡C­l T¥m­R z c¢s¢V­L ­V¢h­ml Efl ­V­e a¥m­a La L¡kÑÉ Ll­a
q­h ? (d­l e¡J g = 10 m/s2)
(A) 0.36 J (B) 0.24 J (C) 0.12 J (D) 0.18 J

A 13 P.T.O.

Category-2 (Q. 31 to 35)
(Carry 2 marks each. Only one option is correct. Negative marks : – ½)

31. There are n elastic balls placed on a smooth horizontal plane. The masses of the balls are
m m m
m, , 2 ,.... n 1 respectively. If the first ball hits the second ball with velocity v0, then
2 2 2
the velocity of the nth ball will be,

HL¢V jpªZ Ae¤i¨¢jL a­ml Efl n pwMÉL ¢ÙÛ¢aÙÛ¡fL hm l¡M¡ q'm z hm…¢ml il kb¡œ²­j
m m m
m, , 2 ,.... n 1 z fËbj hm¢V k¢c ¢àa£u hm­L v0 ­h­N A¡O¡a L­l a¡q­m n-aj h­ml ­hN q­h,
2 2 2
n n 1
4 4 4
(A) v0 (B)   v0 (C)   v0 (D) v0
3 3 3

32. An earth’s satellite near the surface of the earth takes about 90 min per revolution. A
satellite orbiting the moon also takes about 90 min per revolution. Then which of the
following is true ?
(A) ρm < ρe
(B) ρ m > ρe
(C) ρm = ρ e
(D) No conclusion can be made about the densities.
[where ρm is density of the moon and ρe is density of the earth.]

fª¢bh£ fª­ùl L¡R¡L¡¢R HL¢V EfNÊq 90 min-H HLh¡l fª¢bh£­L fËc¢rZ L­l z Q¾cÊfª­ùl L¡R¡L¡¢R
HL¢V EfNÊqJ 90 min-H HLh¡l Q¾cÊ­L fËc¢rZ L­l z ­p­r­œ ¢e­Ql ­L¡e¢V p¢WL ?
(A) ρm < ρe
(B) ρm > ρe
(C) ρm = ρe

(D) OeaÄ pÇf­LÑ ­L¡­e¡ Efpwq¡l Ll¡ pñh eu z

[­kM¡­e ρm q'm Q­¾cÊl OeaÄ J ρe q'm fª¢bh£l OeaÄ z]

A 14

33. A bar magnet falls from rest under gravity through the centre of a
horizontal ring of conducting wire as shown in figure. Which of the
following graph best represents the speed (v) vs. time (t) graph of the bar
magnet ?
¢Q­œ fËc¢nÑa f¢lh¡q£ a¡­ll HL¢V Ae¤i¨¢jL ¢lw-Hl ­L¾cÊ hl¡hl HL¢V
c™Q¥ðL­L A¢iL­oÑl fËi¡­h Eõð i¡­h ¢e­Q ­gm¡ q'm z a¡q­m ¢e­Ql ­L¡e¢V
ph­Q­u p¢WL i¡­h c™Q¥ðL¢Vl ­hN (v) he¡j pju (t) ­mM¢Qœ¢V ¢e­cÑn L­l ?

(A) (B)

(C) (D)

34. An amount of charge Q passes through a coil of resistance R. If the current in the coil
decreases to zero at a uniform rate during time T, then the amount of heat generated in the
coil will be,
R ­l¡­dl HL¢V L¥äm£l jdÉ ¢c­u Q Bd¡e Nje L­l z k¢c T pj­u L¥äm£­a fËh¡q p¤¤oj q¡­l L­j
n§eÉ qu a¡q­m L¥äm£­a Evfæ a¡­fl f¢lj¡Z q­h,
4Q2R 2QR Q 2T
(A) (B) (C) (D) Q2RT
3T 3T 4R
GM A
35. A modified gravitational potential is given by V    2 . If the constant A is
r r
expressed in terms of gravitational constant (G), mass (M) and velocity of light (c), then
from dimensional analysis, A is,
GM A
j­e Ll HL¢V f¢lh¢aÑa jq¡LoÑ£u ¢hi­hl j¡e ­cJu¡ B­R V    2 z dˤhL A-­L k¢c
r r
jq¡LoÑ£u dˤhL (G) il (M) J A¡­m¡­Ll ­hN (c)-Hl p¡q¡­kÉ ¢eZÑu Ll¡ k¡u a¡q­m A q'm
G 2M 2 GM
(A) (B)
c 2
c2
1
(C) (D) Dimensionless (j¡œ¡¢hq£e)
c2

A 15 P.T.O.

Category-3 (Q. 36 to 40)
(Carry 2 marks each. One or more options are correct. No negative marks)
36. p T 2
3
3
1 2 1

v S
A cyclic process is shown in p-v diagram and T-S diagram. Which of the following
statement(s) is/are true ?
(A) 1→2: Isobaric, 2→3: Isothermal.
(B) 3→1: Isochoric, 2→3: adiabatic.
(C) Work done by the system in the complete cyclic process is non-zero.
(D) The heat absorbed by the system in the complete cyclic process is non-zero.
HL¢V Qœ²£u fË¢œ²u¡­L p-v ­mM¢Qœ J T-S ­mM¢Q­œl p¡q¡­kÉ ­cM¡­e¡ q­u­R z ¢e­Ql ­L¡e
E¢š²(…¢m) p¢WL ?
(A) 1→2: pjQ¡f£, 2→3: p­j¡o·
(B) 3→1: pj¡uae£, 2→3: l©Üa¡f
(C) pÇf¨ZÑ Qœ²£u fË¢œ²u¡¢V­a L«a L¡kÑÉ n§eÉ eu z
(D) pÇf¨ZÑ Qœ²£u fË¢œ²u¡­a a¿»¢V à¡l¡ ­n¡¢oa a¡f n§eÉ eu z
37. The figure shows two identical parallel plate capacitors A and B K
of capacitances C connected to a battery. The key K is initially
closed. The switch is now opened and the free spaces between C C
V
the plates of the capacitors are filled with a dielectric constant 3. A B
Then which of the following statement(s) is/are true ?
(A) When the switch is closed, total energy stored in the two capacitors is CV 2.
(B) When the switch is opened, no charge is stored in the capacitor B.
3
(C) When the switch is opened, energy stored in capacitor B is CV2.
2
5
(D) When the switch is opened, total energy stored in two capacitors is CV2.
3
¢Q­œ fËc¢nÑa C d¡lLaÄ ¢h¢nø A J B c¤¢V pj¡¿¹l¡m f¡a d¡lL­L HL¢V hÉ¡V¡l£l c¤C fË¡­¿¹ k¤š² Ll¡
q­u­R z K Q¡¢h¢V fËb­j hå Ll¡ B­R z Hh¡l Q¡¢h¢V M¤­m ¢c­u d¡lL c¤¢Vl f¡a c¤¢Vl j­dÉ 3
fl¡°hc¤É¢aL ¢ÙÛl¡ˆ ¢h¢nø j¡dÉj ¢c­u f§ZÑ Ll¡ q'm z a¡q­m ¢e­Ql ­L¡eÚ E¢š²(…¢m) p¢WL ?
(A) Q¡¢h¢V hå b¡L¡l pj­u d¡lLc¤¢V­a p¢’a ­j¡V n¢š² q'm CV2
(B) Q¡¢h¢V kMe ­M¡m¡ qu aMe B d¡l­L ­L¡­e¡ Bd¡e p¢’a b¡­L e¡ z
3
(C) Q¡¢h¢V kMe ­M¡m¡ qu aMe B d¡l­L p¢’a n¢š² q'm CV2
2
5
(D) Q¡¢h¢V kMe ­M¡m¡ qu aMe d¡lL c¤¢V­a p¢’a ­j¡V n¢š² q'm CV2
3

A 16

38. A charged particle of charge q and mass m is placed at a distance 2R from the centre of a


vertical cylindrical region of radius R where magnetic field varies as B = (4t2 – 2t + 6)k^ ,

where t is time. Then which of the following statement(s) is/are true ?

(A) Induced electric field lines form closed loops.

(B) Electric field varies linearly with r if r < R, where r is the radial distance from the

centerline of the cylinder.

(C) The charged particle will move in clockwise direction when viewed from top.

7q
(D) Acceleration of the charged particle is when t = 2 sec.
2m


R hÉ¡p¡dÑ ¢h¢nø HL¢V Eõð ­Q¡P¡L«¢a ÙÛ¡­e ­Q±ðL ­rœ B = (4t2 – 2t + 6)k^ Ae¤k¡u£ f¢lh¢aÑa qu,

­kM¡­e t q'm pju z ­pM¡­e q Bd¡e J m il ¢h¢nø HL¢V hÙ¹¥LZ¡­L ­Q¡­Pl Ar­lM¡ ­b­L 2R

c§l­aÄ l¡M¡ q'm z a¡q­m ¢e­Ql ­L¡e E¢š²(…¢m) p¢WL ?

(A) B¢hø a¢svhm­lM¡…¢m hÜ m¤­fl BL¡­l qu z

(B) a¢sv­r­œl j¡e r-Hl p­‰ plm°l¢ML pÇf­LÑ f¢lh¢aÑa qu k¢c r < R qu, ­kM¡­e r q'm

­Q¡­Pl Ar­lM¡ ­b­L hÉ¡p hl¡hl c§laÄ z

(C) Efl ­b­L ­cM­m LZ¡¢V O¢sl Ly¡V¡l ¢c­L O¤l­h z

7q
(D) t = 2 sec pj­u LZ¡¢Vl aÄlZ q'm z
2m

A 17 P.T.O.

39. A uniform magnetic field B exists in a region. An electron of charge q and mass m moving
with velocity v enters the region in a direction perpendicular to the magnetic field.
Considering Bohr angular momentum quantization, which of the following statement(s)
is/are true ?
(A) The radius of nth orbit rn  n .
qB 
(B) The minimum velocity of the electron is .
m
(C) Energy of the nth level En n.
(D) Transition frequency ω between two successive levels is independent of n.
­L¡­e¡ ÙÛ¡­e HL¢V p¤¤oj ­Q±ðL ­rœ B l­u­R z I ÙÛ¡­e HL¢V m il J q Bd¡epÇfæ C­mƒÊe ­Q±ðL
­r­œl Eõð ¢cL hl¡hl v ­h­N fË­hn Llm z ­h¡­ll ­L±¢ZL il­h­Nl ­L¡u¡¾V¡C­Sne naÑ Ae¤k¡u£
¢e­Ql ­L¡eÚ E¢š²(…¢m) p¢WL ?
(A) n-aj L­rl hÉ¡p¡dÑ rn  n .
qB 
(B) C­mƒÊ­el eɨeaj ­hN
m
(C) n-aj L­rl n¢š² En n.
(D) flØfl c¤¢V d¡­fl j­dÉ ÙÛ¡e¡¿¹l LÇf¡ˆ ω, n-Hl Efl ¢eiÑln£m eu z

40. A train is moving along the tracks at a constant speed u. A girl on the train throws a ball
of mass m straight ahead along the direction of motion of the train with speed v with respect
to herself. Then
(A) Kinetic energy of the ball as measured by the girl on the train is mv 2/2.
(B) Work done by the girl in throwing the ball is mv 2/2.
(C) Work done by the train is mvu.
(D) The gain in kinetic energy of the ball as measured by a person standing by the rail
track is mv2/2.
HL¢V ­VÊe u ¢ÙÛl ­h­N Qm­R z ­VÊ­el k¡œ£ HL¢V h¡¢mL¡ m i­ll HL¢V hm­L ­VÊ­el N¢al A¢ij¤­M
pl¡p¢l p¡j­el ¢c­L v ­h­N Ry¥­s ¢cm z a¡q­m
(A) ­VÊ­el k¡œ£ h¡¢mL¡l p¡­f­r hm¢Vl N¢an¢š² q­h mv2/2 z
(B) hm¢V ­Ry¡s¡l SeÉ h¡¢mL¡ à¡l¡ L«a L¡kÑÉ q'm mv2/2 z
(C) ­VÊe¢V à¡l¡ L«a L¡kÑÉ q'm mvu z
(D) ­lmm¡C­el d¡­l cy¡s¡­e¡ ­L¡­e¡ hÉ¢š²l p¡­f­r hm¢Vl N¢an¢š² q'm mv2/2 z

A 18

CHEMISTRY
Category-1 (Q 41 to 70)
(Carry 1 mark each. Only one option is correct. Negative marks : – ¼)

41. The correct order of boiling points of N-ethylethanamine (I), ethoxyethane (II) and
butan-2-ol (III) is
N-Cb¡CmCb¡eÉ¡¢je (I), Cb¢„C­be (II) Hhw ¢hEV¡e-2-Am (III)-Hl Øg¥Ve¡­ˆl p¢WL œ²j qm
(A) III < II < I (B) II < III < I
(C) II < I < III (D) III < I < II

42. H2, Lindlar Catalyst Ozonolysis
M N O+P
(CH3CO)2O
Ammoniacal AgNO3 CH3COONa
White precipitate
Cinnamic acid
Structure of M is,

M-Hl NWe q'm

(A) Ph – C  CH (B) Ph – C  C – CH3

(C) H3C – C  CH (D) H3C – C  C – CH3

OH OH OH OH
43.

CH3 NMe2 CH2
CH = CH
CH NO2
2
(I) (II) (III) (IV)

The correct order of acidity of above compounds is
Ef¢l¢õ¢Ma ­k±N…¢ml B¢ÇmLa¡l p¢WL œ²j qm
(A) II > IV > I > III (B) III > IV > II > I
(C) IV > II > III > I (D) IV > III > I > II

A 19 P.T.O.

44. O
Br2/AcOH Br2/NaOH
X Y+Z
CH3 CH3

The correct option for the above reaction is
Ef­l¡š² ¢h¢œ²u¡u p¢WL ¢hLÒf¢V q'm
O
(A) Y= CHBr3 Z= CH3CO2Na
X=
CH3 CH2Br

O
O

(B) Y= CHBr3 Z= CH3CO2Na
X=
CH3 CBr3

O
(C) Y= CHBr3 Z= CH2CO2Na
X=
CH3 CH2Br
Br
O

(D) Y= CHBr3 Z= CH2CO2Na
X=
CH2 CH2
Br

BrBr Br

45. H3CH2C
aqueous H3C SO2Cl
NaOH NaI
H Cl C4H10O C11H16SO3 C 4 H9 I
pyridine acetone
H3 C F
E
If all the nucleophilic substitution reactions at saturated carbon atoms in the above
sequence of reactions follow SN2 mechanism, then E and F will be respectively,
k¢c Ef­l h¢ZÑa ¢h¢œ²u¡œ²­j pÇf«š² L¡hÑe flj¡Z¤­a O¢Va phL'¢V ¢eJ¢LÓJ¢g¢mL fË¢aÙÛ¡fe ¢h¢œ²u¡
SN2 ¢œ²u¡­L±nm Ae¤plZ L­l, a­h E Hhw F q­h kb¡œ²­j

H3CH2C H3CH2C CH2CH3 H3C

OH and I HO H and I
H H H
H3 C CH3 H3CH2C
H3 C
(A) (B) H3CH2C
CH2CH3 CH2CH3 CH2CH3

and I
HO H and I H HO H H
H3C
CH3 CH3 CH3
(C) (D)

A 20

46. Two base balls (masses: m1 = 100 g, and m2 = 50 g) are thrown. Both of them move with
uniform velocity, but the velocity of m2 is 1.5 times that of m1 . The ratio of de Broglie
wavelengths   m 1  :   m 2  is given by

c¤C¢V ­hphm (il m1 = 100 NË¡j, Hhw m2 = 50 NË¡j) ­Ry¡s¡ qm z fË­aÉL¢VC ¢ÙÛl ­h­N N¢an£m z
¢L¿¹¥ m2 -Hl N¢a­hN m1 -Hl ­b­L 1.5 …Z ­h¢n z ­hphm c¤C¢Vl p­‰ k¤š² ¢X hËN¢m al‰°cOÑÉ-Hl
Ae¤f¡a   m1  :   m 2  qm
(A) 4 : 3 (B) 3:4 (C) 2:1 (D) 1 : 2

47. What is the edge length of the unit cell of a body centred cubic crystal of an element whose
atomic radius is 75 pm ?
HL¢V ­cq­L¢¾cÊL OeL¡L¡l ­Lm¡­pl HL¢V ­j±­ml flj¡Z¤l hÉ¡p¡dÑ 75 pm z I ­Lm¡­pl HLL
­L¡­ol fË¡¢¿¹L£l °cOÑÉ La ?
(A) 170 pm (B) 175 pm (C) 178 pm (D) 173.2 pm

48. The root mean square (rms) speed of X2 gas is x m/s at a given temperature. When the
temperature is doubled, the X2 molecules dissociated completely into atoms. The root
mean square speed of the sample of gas then becomes (in m/s)
HL¢V ¢e¢cÑø Eo·a¡u X2 -Hl Ns hNÑ⤢al hNÑj§m qm x m/s z Eo·a¡ ¢à…Z Ll­m X2 -Hl AZ¤…¢m
flj¡Z¤­a ¢h­u¡¢Sa qu z HC AhÙÛ¡u HC NÉ¡­pl Ns hNÑ⤢al hNÑj§m (m/s HL­L) q­h
(A) x/2 (B) x (C) 2x (D) 4x

49. Arrange the following in order of increasing mass
I. 1 mole of N2 II. 0.5 mole of O3

III. 3.011  1023 molecules of O2 IV. 0.5 gram atom of O2

i­ll EÜÑœ²j¡e¤p¡­l p¡S¡J
I. 1 ­j¡m N2 II. 0.5 ­j¡m O3
III. O2 -Hl 3.011×1023 pwMÉL AZ¤ IV. 0.5 NË¡j flj¡Z¤ O2

(A) IV < III < II < I (B) IV < I < III < II
(C) III < II < IV < I (D) I < III < II < IV

A 21 P.T.O.

50. Which of the following would give a linear plot ?
¢e­Ql ­L¡e fÔV¢V (­lM¡¢Qœ¢V) plm°l¢ML q­h ?
(A) k vs T (B) k vs 1/T (C) ln k vs T (D) ln k vs 1/T
(k is the rate constant of an elementary reaction and T is temp. in absolute scale)
(k HL d¡f ¢h¢œ²u¡l N¢adËh¤ L, T flj a¡fj¡œ¡)

51. The equivalent conductance of NaCl, HCl and CH3COONa at infinite dilution are 126.45,
426.16 and 91 ohm–1cm2eq–1 respectively at 25C. The equivalent conductance of acetic
acid (at infinite dilution) would be
(A) 461.61 ohm–1 cm2 eq–1 (B) 390.71 ohm–1 cm2 eq–1
(C) cannot be determined from the given data. (D) 208.71 ohm–1 cm2 eq–1
Ap£j mO¤a¡u NaCl, HCl Hhw CH3COONa -Hl a¥mÉ¡wL f¢lh¡¢qa¡ kb¡œ²­j 126.45, 426.16
Hhw 91 ohm–1cm2eq–1 (25℃ -H) z HLC AhÙÛ¡u AÉ¡­p¢VL AÉ¡¢p­Xl a¥mÉ¡wL f¢lh¡¢qa¡
(A) 461.61 ohm–1 cm2 eq–1 (B) 390.71 ohm–1 cm2 eq–1
(C) k¡ abÉ ­cJu¡ q­u­R a¡ ­b­L ­hl Ll¡ k¡­h e¡ z(D) 208.71 ohm–1 cm2 eq–1

52. For the reaction A + B → C, we have the following data :
Initial concentration of A Initial concentration of B Rate (initial)
(in molarity) (in molarity) (Relevant unit)
1 10 100
1 1 1
10 1 10
The order of the reaction with respect to A and B are
(A) Not possible to tell with the given data.
(B) First order with respect to both A and B.
(C) First order with respect to A and second order with respect to B.
(D) Second order with respect to A and first order with respect to B.
A J B Hl j­dÉ A + B → C ¢h¢œ²u¡¢Vl N¢a­hN pÇf­LÑ ¢e­Ql abÉ…¢m S¡e¡ B­R z
A -Hl fË¡l¢ñL N¡taÄ B -Hl fË¡l¢ñL N¡taÄ ¢h¢œ²u¡l fË¡l¢ñL N¢a­hN
(­j¡m¡¢l¢V­a) (­j¡m¡¢l¢V­a) (Efk¤š² HL­L)
1 10 100
1 1 1
10 1 10
­p­r­œ A J B -Hl p¡­f­r ¢h¢œ²u¡¢Vl œ²j
(A) hm¡l ja k­bø abÉ ­cJu¡ ­eC z
(B) A J B c¤C­ul p¡­f­rC œ²j HL z
(C) A-Hl p¡­f­r HL Hhw B-Hl p¡­f­r c¤C z
(D) A-Hl p¡­f­r c¤C Hhw B-Hl p¡­f­r HL z

A 22

53. The equivalent weight of KIO3 in the given reaction is (M = molecular mass):
2Cr(OH)3 + 4OH – + KIO3  2CrO42– + 5H2O + KI
fËcš ¢h¢œ²u¡u 2Cr(OH)3 + 4OH – + KIO3  2CrO42– + 5H2O + KI,
KIO3-Hl a¥mÉ¡ˆ il (M = Be¢hL il)
(A) M (B) M/2 (C) M/6 (D) M/8
54. At STP, the dissociation reaction of water is H2O H+ (aq.) + OH– (aq.) , and the pH
of water is 7.0. The change of standard free energy (∆G°) for the above dissociation process
is given by
STP n­aÑ, S­ml ¢h­u¡Se ¢h¢œ²u¡¢V qm H2O H+ (aq.) + OH– (aq.), Hhw S­ml pH qm
7.0 z Ef¢lš² ¢h­u¡Se ¢h¢œ²u¡u BcnÑ j¤š²n¢š²l f¢lhaÑe (∆G°) qm
(A) 20301 cal/mol (B) 19091cal/mol (C) 20096 cal/mol (D) 21301 cal/mol

55. Na2CO3 is prepared by Solvay process but K2CO3 cannot be prepared by the same because
(A) K2CO3 is highly soluble in H2O (B) KHCO3 is sparingly soluble
(C) KHCO3is appreciably soluble (D) KHCO3 decomposes
Na 2 CO3 fËÙ¹¥¢a­a pmÚ­i fÜ¢a hýmhÉhq©a q­mJ K 2 CO3 fËÙ¹¥¢a­a hÉhq¡l Ll¡ k¡u e¡ L¡lZ
(A) K2CO3 S­m A¢a j¡œ¡u â¡hÉ (B) KHCO3üÒfâ¡hÉ
(C) KHCO3 k­bø j¡œ¡u â¡hÉ (D) KHCO3 ¢h­u¡¢Sa q­u k¡u

56. If in case of a radio isotope the value of half-life (T1/2) and decay constant (λ) are identical
in magnitude, then their value should be
k¢c HL¢V ­aS¢ûu BC­p¡­V¡­fl AdÑ¡u¤×L¡m (T1/2) Hhw ru¡ˆ (λ)-Hl j¡e f¢lj¡ZNai¡­h pj¡e
qu, a¡q­m ­pC j¡e¢V qJu¡ E¢Qv
(A) 0.693/2 (B) (0.693)1/2 (C) (0.693)2 (D) 0.693

57. Suppose a gaseous mixture of He, Ne, Ar and Kr is treated with photons of the frequency
appropriate to ionize Ar. What ion(s) will be present in the mixture ?
HL¢V h¡uh£u ¢jnË­Zl j­dÉ He, Ne, Ar Hhw Kr haÑj¡e z HC ¢jnË­Zl j­dÉ Ar-­L Bu¢ea Ll­a
f¡­l HlLj LÇf¡­ˆl ­g¡Ve f¡W¡­e¡ qm z HMe ¢jnË­Zl j­dÉ ­k Bue…¢m b¡L­h a¡l¡ qm
(A) Ar+ (B) Ar+ + Kr+
(C) Ar+ + He+ + Ne+ (D) He+ + Ar+ + Kr+
58. A solution containing 4g of polymer in 4.0 litre solution at 27C shows an osmotic pressure
of 3.0×10 – 4 atm. The molar mass of the polymer in g/mol is
27℃ Eo·a¡u 4.0 L âh­Z HL¢V f¢mj¡­ll 4.0 g âh£i¨a b¡L­m 3.0×10 – 4 atm A¢ipËhe Q¡­fl
pª¢ø qu z f¢mj¡l¢Vl Be¢hL il (NË¡j/­j¡m HL­L) q­h
(A) 820000 (B) 82000 (C) 8200 (D) 820

A 23 P.T.O.

59. The molecular shapes of SF4, CF4 and XeF4 are
(A) the same with 2, 0 and 1 lone pairs of electrons on the central atoms, respectively.
(B) the same with 1, 1 and 1 lone pairs of electrons on the central atoms, respectively.
(C) different with 0, 1 and 2 lone pairs of electrons on the central atoms, respectively.
(D) different with 1, 0 and 2 lone pairs of electrons on the central atoms, respectively.
SF4, CF4 Hhw XeF4 Hl Be¢hL BL«¢a qm

(A) HLC Hhw ­L¾cÊ£u flj¡Z¤l ¢exp‰ C­mLVÊe ­S¡s kb¡œ²­j 2, 0 Hhw 1 z
(B) HLC Hhw ­L¾cÊ£u flj¡Z¤l ¢exp‰ C­mLVÊe ­S¡s kb¡œ²­j 1, 1 Hhw 1 z
(C) Bm¡c¡ Hhw ­L¾cÊ£u flj¡Z¤l ¢exp‰ C­mLVÊe ­S¡s kb¡œ²­j 0, 1 Hhw 2 z
(D) Bm¡c¡ Hhw ­L¾cÊ£u flj¡Z¤l ¢exp‰ C­mLVÊe ­S¡s kb¡œ²­j 1, 0 Hhw 2 z

60. The species in which nitrogen atom is in a state of sp hybridisation is
e£­Ql ­L¡eÚ ­k±­N e¡C­VÊ¡­Se flj¡Z¤¢V sp pwLl¡¢ua (hybridised)
(A) NO3– (B) NO2 (C) NO2+ (D) NO2–

3
The correct statement about the magnetic properties of  Fe  CN 6  and  FeF6 
3
61. is

(A) Both are paramagnetic
(B) Both are diamagnetic
3
 Fe  CN 6  is diamagnetic,  FeF6  is paramagnetic
3
(C)
3
 Fe  CN 6  is paramagnetic,  FeF6 
3
(D) is diamagnetic
3
 Fe  CN 6  Hhw  FeF6 3 Hl ­Q±ðL djÑ pÇf¢LÑa p¢WL hš²hÉ¢V qm
(A) c¤¢V ­k±NC EfQ¥ðL£u
(B) c¤¢V ­k±NC AfQ¥ðL£u
3
 Fe  CN 6  AfQ¥ðL£u,  FeF6 
3
(C) EfQ¥ðL£u
3
(D)  Fe  CN 6  EfQ¥ðL£u,  FeF6 3 AfQ¥ðL£u

A 24

3
The calculated spin-only magnetic moment values in BM for  FeCl 4  and  Fe  CN 6 

62.
are
3
 FeCl 4  Hhw  Fe  CN 6  Hl NZe¡L«a ¢Øfe Q¥ðL£u ï¡jL (spin magnetic moment) (BM)


qm
(A) 5.9 BM, 1.732 BM (B) 4.89 BM, 1.732 BM
(C) 3.87 BM, 1.732 BM (D) 1.732 BM, 2.82 BM

63. BrF3 self ionises as following
BrF3 ¢e­jÀ¡š²i¡­h üaxBu¢ea qu
(A) 2BrF3 BrF+ + BrF5– (B) 2BrF3 BrF2+ + BrF4–
(C) 2BrF3 BrF4+ + BrF2– (D) 2BrF3 BrF3+ + BrF3–

64. 4f2 electronic configuration is found in
­L¡eÚ¢Vl ­r­œ 4f2 C­mLVÊe ¢heÉ¡p ­cM¡ k¡u ?
(A) Pr (B) Pr3+ (C) Nd3+ (D) Pm3+

65. Which of the following statements is incorrect ?
 VF6 
3
(A)A is paramagnetic with 2 unpaired electrons.

(B)B  CuCl4 
2
is paramagnetic with 1 unpaired electron.
3
(C)C  Co  NH 3 6  is diamagnetic

(D)D  CoF6 
3
is paramagnetic with 2 unpaired electrons.

¢ej¢À m¢Ma ­L¡e ¢hhª¢a¢V p¢WL eu ?
 VF6  EfQ¥ðL£u, 2¢V ­S¡sq£e C­mLVÊe p­ja
3
(A)A

(B)B  CuCl4  EfQ¥ðL£u, 1¢V ­S¡sq£e C­mLVÊe p­ja
2

3
(C)C  Co  NH 3 6  AfQ¥ðL£u

(D)D  CoF6 
3
EfQ¥ðL£u, 2¢V ­S¡sq£e C­mLVÊe p­ja

A 25 P.T.O.

O O O

66.
O O O
I II III

The correct order of C  O bond length in ethyl propanoate (I), ethyl propenoate (II) and
ethenyl propanoate (III) is

Cb¡Cm ­fË¡f¡­e¡­uV (I), Cb¡Cm ­fË¡¢f­e¡­uV (II) Hhw C¢be¡Cm ­fË¡f¡­e¡­uV (III) -H C  O håe
°c­OÑÉl p¢WL œ²j qm
(A) I > II > III (B) III > II > I (C) I > III > II (D) II > I > III

67. Select the molecule in which all the atoms may lie on a single plane is

(A) 4-Nitrobenzaldehyde (B) 4-Methoxybenzaldehyde

(C) 4-Methylnitrobenzene (D) 4-Nitroacetophenone

¢e­j¡À š² ­k AZ¤­a pLm flj¡Z¤…¢m HL¢V j¡œ a­m AhÙÛ¡e L­l ­p¢V qm
(A) 4-e¡C­VÊ¡­h”É¡m¢Xq¡CX (B) 4-¢jb¢„­h”É¡m¢Xq¡CX

(C) 4-¢jb¡Cme¡C­VÊ¡­h¢”e (D) 4-e¡C­VÊ¡AÉ¡¢p­V¡­g­e¡e

68. The IUPAC name of CH 3CH= C– CH2–CH3 is :
|
CHO

(A) 3-Formyl-2-pentene (B) 2-Ethylbut-2-enal

(C) 3-Ethylbut-3-enal (D) 2-Ethylcrotonaldehyde

CH3CH= C– CH2–CH3 ­k±N¢Vl e¡j q'm:
|
CHO
(A) 3-glj¡Cm-2-­f¢¾Ve (B) 2-Cb¡Cm¢hEV-2-CeÉ¡m

(C) 3-Cb¡Cm¢hEV-3-CeÉ¡m (D) 2-Cb¡Cm­œ²¡­V¡eÉ¡m¢Xq¡CX

A 26

69. The correct stability order of the following carbocations is
¢ejÀ¢m¢Ma L¡­hÑ¡LÉ¡V¡ue…¢ml ÙÛ¡u£­aÄl p¢WL œ²j qm
 
H2C – CH = CH – CH3 CH2 – CH = CH – BMe2
I II
 
H2C – CH = CH – NMe2 H2C – CH = CH – OMe
III IV

(A) II > I > III > IV (B) III > I > II > IV
(C) III > IV > I > II (D) IV > III > II > I

HO
70. H
H OH
1
and
And

CO2H HO2C

CH3 C2H5

2 Br C2H5 and
And Br H

H CH3

3

And

The relationship between the pair of compounds shown above are respectively,
(A) enantiomer, diastereomer, diastereomer
(B) enantiomer, enantiomer, diastereomer
(C) enantiomer, homomer (identical), diastereomer
(D) homomer (identical), diastereomer, geometrical isomer
Ef­l¡š² ­k±N­S¡s…¢ml j­dÉ pÇfLÑ qm kb¡œ²­j
(A) He¡e¢nJj¡l (fË¢a¢hð pj¡huh), X¡u¡¢ØV¢lJj¡l, X¡u¡¢ØV¢lJj¡l
(B) He¡e¢nJj¡l (fË¢a¢hð pj¡huh), He¡e¢nJj¡l (fË¢a¢hð pj¡huh), X¡u¡¢ØV¢lJj¡l
(C) He¡e¢nJj¡l (fË¢a¢hð pj¡huh), ­q¡­j¡j¡l, X¡u¡¢ØV¢lJj¡l
(D) ­q¡­j¡j¡l, X¡u¡¢ØV¢lJj¡l, SÉ¡¢j¢aL pj¡huh

A 27 P.T.O.

Category-2 (Q 71 to 75)
(Carry 2 marks each. Only one option is correct. Negative marks : – ½)

71.
(i) OH/H2O,
C6H12O2 H + I
(ii) H3O+
G
CrO3/H+

‘G’ in the above sequence of reactions is
Ef­l¡š² ¢h¢œ²u¡œ²­j ‘G’ q'm
(A) (CH3)2CHCOOCH2CH3 (B) CH3CH2CH2COOCH2CH3
(C) CH3CH2COOCH2CH2CH3 (D) CH3CH2COOCH(CH3)2

72. Case – 1: An ideal gas of molecular weight M at temperature T.
Case – 2: Another ideal gas of molecular weight 2M at temperature T/2.
Identify the correct statement in context of above two cases.
(A) Average kinetic energy and average speed will be the same in the two cases.
(B) Both the averages are halved.
(C) Both the averages are doubled.
(D) Only average speed is halved in the second case.
fËbj ­rœ : M Be¢hL i­ll HL¢V BcnÑ NÉ¡p T Eo·a¡u z
¢àa£u ­rœ : 2M Be¢hL i­ll HL¢V BcnÑ NÉ¡p T/2 Eo·a¡u z
Hl©f f¢l¢ÙÛ¢a­a p¢WL hš²hÉ¢V ¢el©fZ Ll:
(A) Ns N¢an¢š² J Ns ⤢a Eiu­r­œ HLC b¡L­h z
(B) Eiu l¡¢nl f¢lj¡f A­dÑL q­h z
(C) Eiu l¡¢nl f¢lj¡f ¢à…Z q­h z
(D) ­Lhm Ns ⤢al f¢lj¡e ¢àa£u ­r­œ A­dÑL q­h z

73. 63 g of a compound (Mol. Wt. = 126) was dissolved in 500 g distilled water. The density
of the resultant solution as 1.126 g/ml. The molarity of the solution is
HL¢V ­k±­Nl (Be¢hL il =126) 63 NË¡j ¢e­u 500 NË¡j f¡¢aa S­m âh£i¨a Ll¡ q­m
âh­Zl OeaÄ qm 1.126 NË¡j/¢j¢m z I âh­Zl ­j¡m¡¢l¢V qm
(A) 1.25 M (B) 1.0 M (C) 0.75 M (D) 1.1 M

A 28

74. Nickel combines with a uninegative monodentate ligand ( X  ) to form a paramagnetic
complex  NiX 4  . The hybridisation involved and number of unpaired electrons present
2

in the complex are respectively
(A) sp3, two (B) dsp2, zero (C) dsp2, one (D) sp3, one
¢e­Lm HL¢V HLL GZ¡aÈL Bd¡ek¤š² HL­k¡S£ ¢mNÉ¡­äl ( X  ) p­‰ k¤š² q­u HL¢V EfQ¥ðL£u
­k±N  NiX 4 2  NWe L­l z ­k±N¢V­a ¢e­L­ml pwLl¡uZ (hybridisation) Hhw Ak¤NÈ C­mLVÊ­el
pwMÉ¡ qm
(A) sp3, c¤¢V (B) dsp2, HL¢VJ eu (C) dsp2, HL¢V (D) sp3, HL¢V

(i) PhMgBr CrO3/H Ph3 P = CH2
75. L   M  N  Ph2 C = CH2
═ (ii) H3O ═ ═
‘L’ in the above sequence of reaction is/are (where L ≠ M ≠ N)
(A) Benzaldehyde (B) Methyl benzoate
(C) Benzoyl chloride (D) Benzonitrile
Ef­l¡š² ¢h¢œ²u¡œ²­j ‘L’ q'm (HM¡­e L ≠ M ≠ N)
(A) ­h”É¡m¢Xq¡CX (B) ¢jb¡Cm ­h­”¡­uV
(C) ­h­”¡¢um ­LÓ¡l¡CX (D) ­h­”¡e¡CVÊ¡Cm
Category-3 (Q 76 to 80)
(Carry 2 marks each. One or more options are correct. No negative marks)
76. The correct set(s) of reactions to synthesize benzoic acid starting from benzene is/are
 A   i  Br2 / Fe  ii  Mg / dry ether  iii  CO2  iv  (iv)
H3OH3O
 B  i  Br2 / Fe  ii  NH3 , 25o C  iii  NaNO2 , dil. HCl, 0o to 5o C
(iv) CuCN / KCN (v) dil.HCl, 
 
 C   i  CH3Cl, Anhydrous AlCl3  ii  KMnO
(ii) KMn O4 4 O H,  (iii) H 3 O

H 3 O H3O
(D) (i) CH 3COCl, Anhydrous AlCl3 (ii) Br2 , NaOH (iii) (iii)

­h¢”e ­b­L öl¦ L­l ­h­”¢uL AÉ¡¢pX pw­nÔo­Zl SeÉ p¢WL ¢h¢œ²u¡ œ²j¢V/œ²j…¢m q'm
ö×L Cb¡l (iv) H3O
mO¤
mO¤ HCl, 
Ae¡âÑ AlCl3 (ii) KMnO4

Ae¡âÑ AlCl3 (iii) H3O

A 29 P.T.O.

77. Which statement(s) is/are applicable above critical temperature ?
(A) A gas cannot be liquified.
(B) Surface tension of a liquid is very high.
(C) A liq. phase cannot be distinguished from a gas phase.
(D) Density changes continuously with P or V.
pwLV a¡fj¡œ¡l Jf­l ­L¡eÚ ¢hhª¢a¢V/¢hhª¢a…¢m fË­k¡SÉ ?
(A) ­L¡­e¡ NÉ¡p­L alm Ll¡ k¡u e¡ z
(B) al­ml fªùV¡e M¤h ­h¢n qu z
(C) NÉ¡p h¡ alm­L Bm¡c¡ i¡­h ­Qe¡ k¡u e¡ z
(D) OeaÄ, P Abh¡ V-Hl p­‰ d¡l¡h¡¢qLi¡­h f¢lhaÑe qu z
78. Which of the following mixtures act(s) as buffer solution ?
¢ej¢À m¢Ma ¢jnËZ…¢ml j­dÉ ­L¡e¢V/­L¡e…¢m h¡g¡l âhZ ¢qp¡­h L¡S Ll­h ?
(A) NaOH + CH3COOH (1 : 1 mole ratio)
(B) NH4OH + HCl (2 : 1 mole ratio)
(C) CH3COOH + NaOH (2 : 1 mole ratio)
(D) CH3COOH + NaOH (1 : 2 mole ratio)

79. An electron in the 5d orbital can be represented by the following (n, l, m ℓ,) values

5d EfL­rl (orbital) HL¢V C­mLVÊe­L ¢ejÀ¢m¢Ma (n, l, mℓ,) Hl j¡e…¢m à¡l¡ hm¡ k¡u
(A) (5, 2, 1) (B) (5, 1, –1) (C) (5, 0, 1) (D) (5, 2, –1)

80. The conversion(s) that can be carried out by bromine in carbon tetrachloride solvent is/are
L¡hÑe ­VVÊ¡­LÓ¡l¡CX â¡h­L ­hË¡¢je ­k l©f¡¿¹l¢V­L/l©f¡¿¹l…¢m­L pÇfæ Ll­a f¡­l ­p…¢m q'm
(A) PhCH = CHCH3  PhCHBrCHBrCH3

COOH COOH

(B)
Br

(C) CH3CH2COOH  CH3CHBrCOOH

Br

COOAg

(D)

_____________

A 30

SPACE FOR ROUGH WORK

A 31 P.T.O.

Subject : PHYSICS & CHEMISTRY

pju: 2 O¾V¡ f§ZÑj¡e : 100

¢e­cÑn¡hm£
1. HC fËnÀf­œ pjÙ¹ fËnC À Ah­S¢ƒi fËnÀ Hhw fË¢a¢V fË­nÀl Q¡l¢V pñ¡hÉ Ešl ­cJu¡ B­R z
2. Category-1 : HL¢V Ešl p¢WL z p¢WL Ešl ¢c­m 1 eðl f¡­h z i¥m Ešl ¢c­m Abh¡ ­k ­L¡e
HL¡¢dL Ešl ¢c­m ¼ eðl L¡V¡ k¡­h z
3. Category-2 : HL¢V Ešl p¢WL z p¢WL Ešl ¢c­m 2 eðl f¡­h z i¥m Ešl ¢c­m Abh¡ ­k ­L¡e
HL¡¢dL Ešl ¢c­m ½ eðl L¡V¡ k¡­h z
4. Category-3: HL h¡ HL¡¢dL Ešl p¢WL z ph L¢V p¢WL Ešl ¢c­m 2 eðl f¡­h z k¢c ­L¡e i¥m
Ešl e¡ b¡­L Hhw p¢WL EšlJ ph L¢V e¡ b¡­L a¡q­m f¡­h 2  ­k L¢V p¢WL Ešl ­cJu¡ q­u­R
a¡l pwMÉ¡  Bp­m ­k L¢V Ešl p¢WL a¡l pwMÉ¡ z k¢c ­L¡­e¡ i¥m Ešl ­cJu¡ qu h¡ HL¡¢dL Eš­ll
j­dÉ HL¢VJ i¥m b¡­L a¡q­m Ešl¢V i¥m d­l ­eJu¡ q­h z ¢L¿¹¥ ­p­r­œ ­L¡­e¡ eðl L¡V¡ k¡­h e¡,
AbÑ¡v n§eÉ eðl f¡­h z
5. OMR f­œ A, B, C, D ¢Q¢q²a p¢WL Ol¢V il¡V L­l Ešl ¢c­a q­h z
6. OMR f­œ Ešl ¢c­a öd¤j¡œ L¡­m¡ h¡ e£m L¡¢ml hm f­u¾V ­fe hÉhq¡l Ll­h z
7. OMR f­œ ¢e¢cÑø ÙÛ¡e R¡s¡ AeÉ ­L¡b¡J ­L¡e c¡N ­c­h e¡ z
8. OMR f­œ ¢e¢cÑø ÙÛ¡­e fËnÀf­œl eðl Hhw ¢e­Sl ­l¡m eðl A¢a p¡hd¡ea¡l p¡­b ¢mM­a q­h Hhw
fË­u¡Se£u Ol…¢m f§lZ Ll­a q­h z
9. OMR f­œ ¢e¢cÑø ÙÛ¡­e ¢e­Sl e¡j J fl£r¡ ­L­¾cÊl e¡j ¢mM­a q­h Hhw ¢e­Sl (Admit Card-H
E­õ¢Ma) pÄ¡rl Ll­a q­h z
10. fËnÀf­œl eðl h¡ ­l¡m eðl i¥m ¢mM­m Abh¡ i¥m Ol il¡V Ll­m, fl£r¡bÑ£l e¡j, fl£r¡ ­L­¾cÊl e¡j
h¡ pÄ¡r­l ­L¡e i¥m b¡L­m Ešl fœ h¡¢am q­u ­k­a f¡­l z OMR fœ¢V i¡yS q­m h¡ a¡­a Ae¡hnÉL
c¡N fs­mJ h¡¢am q­u ­k­a f¡­l z fl£r¡bÑ£l HC dl­el i¥m h¡ ApaÑLa¡l SeÉ Ešlfœ h¡¢am
q­m HLj¡œ fl£r¡bÑ£ ¢e­SC a¡l SeÉ c¡u£ b¡L­h z
11. ­j¡h¡Cm­g¡e, LÉ¡mL¥­mVl, pÔ¡CXl¦m, mN­Vhm, q¡aO¢s, ­lM¡¢Qœ, NË¡g h¡ ­L¡e dl­Zl a¡¢mL¡ fl£r¡
L­r Be¡ k¡­h e¡ z Be­m ­p¢V h¡­Su¡ç q­h Hhw fl£r¡bÑ£l JC fl£r¡ h¡¢am Ll¡ q­h z
12. fËnf À ­œ l¡g L¡S Ll¡l SeÉ gy¡L¡ S¡uN¡ ­cJu¡ B­R z AeÉ ­L¡e L¡NS HC L¡­S hÉhq¡l Ll¡ k¡­h
e¡ z
13. fl£r¡ Lr R¡s¡l B­N OMR fœ AhnÉ C f¢lcnÑL­L ¢c­u k¡­h z
14. HC fËnf À ­œ Cwl¡S£ J h¡wm¡ Eiu i¡o¡­aC fËnÀ ­cJu¡ B­R z h¡wm¡ j¡dÉ­j fËnÀ °al£l pju fË­u¡Se£u
p¡hd¡ea¡ J paLÑa¡ Ahmðe Ll¡ q­u­R z a¡ p­šÄJ k¢c ­L¡e Ap‰¢a mrÉ Ll¡ k¡u, ­p­r­œ Cwl¡S£
j¡dÉ­j ­cJu¡ fËnÀ ¢WL J Q¨s¡¿¹ h­m ¢h­h¢Qa q­h z
15. fl£r¡ ®n­o fl£r¡b£Ñl¡ fËnÀfœ¢V ¢e­u k¡­h z

A 32
A