PUBDET 2023 Question Paper Chemistry Physics Geology

Paper - VII
Subject : Physics, Chemistry, Mathematics
for admission in
Chemistry/Physics/Geology
(Booklet Number)
Duration : 90 Minutes No. of Questions : 50 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 mark 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 ink ball point pen to mark the answer by complete filling up of the
respective bubbles.
4. Mark the answers only in the space provided. 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 Sheet. Also fill appropriate bubbles.
6. Write your name (in block letter), name of the examination centre and put your signature (as
is appeared in Admit Card) in appropriate boxes in the OMR Sheet.
7. 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 centre. 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.
8. Candidates are not allowed to carry any written or printed material, calculator, pen, docu-
pen, log table, wristwatch, any communication device like mobile phones, bluetooth etc.
inside the examination hall. Any candidate found with such prohibited items will be
reported against and his/her candidature will be summarily cancelled.
9. Rough work must be done on the question booklet itself. Additional blank pages are given in
the question booklet for rough work.
10. Hand over the OMR Sheet to the invigilator before leaving the Examination Hall.
11. This booklet 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.
12. Candidates are allowed to take the Question Booklet after Examination is over.

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

Signature of the Invigilator : ______________________________

Che./Phy./Geo. 

SPACE FOR ROUGH WORK / l¡g L¡­Sl SeÉ S¡uN¡

Che.+Phy.+Geo. 2 

MATHEMATICS

1. If for the number x > 0, log1/7 x – 2+ 3 = 5 – log1/7 x,then
(A) the smallest integral value of x is 1
(B) there is no smallest value of x
1
(C) the largest value of x is
49
 1 
(D) x lies in the interval  ,  
 49 
log1/7 x – 2+ 3 = 5 – log1/7 x, x > 0 , q­m
(A) x-Hl r¥âaj f§ZÑpwMÉ¡l j¡e q­h 1 (B) x-Hl ­L¡e r¥âaj j¡e ­eC
1  1 
(C) x-Hl p­îÑ¡µQ j¡e q­h (D) x,  ,   A¿¹l¡­m Ah¢ÙÛa
49  49 

2. In complex plane, a point z moves in such a way that z  z0  z  z1  k , where z0 and z1
are two fixed points and k is a real constant. Then the locus of z is a conic section whose
eccentricity is
S¢Vm a­m HL¢V ¢h¾c¤ z Hje i¡­h N¢an£m ­k z  z0  z  z1  k (­kM¡­e z0 J z1 c¤¢V ¢e¢cÑø ¢h¾c¤
Hhw k qm h¡Ù¹h dˤhL)z ­p­r­œ z -Hl p’¡lfb qm HL¢V L¢eL ­pLpe k¡l Ev­L¾cÊa¡ qm
z1  z 0 z1  z 0 z1  z 0 z1  z0
(A) (B) (C) (D)
2k 2k k k

3. The roots of the equation  x  a  x  b    x  b  x  c    x  c  x  a   0 are
(A) negative (B) positive (C) real (D) imaginary
 x  a  x  b    x  b  x  c   x  c x  a   0 pj£Ll­Zl h£Sàu
(A) GZ¡aÈL (B) de¡aÈL (C) h¡Ù¹h (D) L¡Òf¢eL

4. Consider the expression f(n) = 72n + 16n – 1, n 
(A) f(n) is divisible by 26. (B) f(n) is not divisible by 26.
(C) f(n) is divisible by 5. (D) f(n) is divisible by 7.
f(n) = 72n + 16n – 1, n  l¡¢n¢V­L ¢h­hQe¡ Ll z ­p­r­œ
(A) f(n), 26 à¡l¡ ¢hi¡SÉ (B) f(n), 26 à¡l¡ ¢hi¡SÉ eu
(C) f(n), 5 à¡l¡ ¢hi¡SÉ (D) f(n), 7 à¡l¡ ¢hi¡SÉ

Che.+Phy.+Geo. 3 

cos  sin  cos 
5. Let r      sin  cos  sin  then range of r(θ) is
 cos   sin  cos 
cos  sin  cos 
k¢c r      sin cos  sin  , a­h r(θ)-Hl ¢hÙ¹¡l (range)q­h
 cos   sin  cos 
(A) [0, 1] (B) [0, 2 2 ] (C) [–2, 2] (D) [–2 + 5,– 5 + 2]

 2 0 5
 
6. Let A   1 2 3  . The system of linear equations AX=Y has a solution
 1 5 1 
 
x 0
   
(A) only for Y   0  , x  (B) only for Y   y  , y 
0 0
   
0
 
(C) only for Y   y  , y, z  (D) for all Y  3
z
 
 2 0 5
j­e Ll, A   1 2 3  z °l¢ML pj£LlZ fËZ¡m£ AX=Y -Hl pj¡d¡e B­R
 1 5 1 
 
x
 
(A) ­Lhmj¡œ k¢c Y   0  , x 
0
 
0
(B) ­Lhmj¡œ Y   y  , y  -Hl ­r­œ
0
 
0
(C) ­Lhmj¡œ Y   y  , y, z  -Hl ­r­œ
z
 
(D) pLm Y  3 Hl ­r­œ

Che.+Phy.+Geo. 4 

a  a  a  a 
x x 2 x x 2
1

7. The value of the determinant  b  b   b  b 
x x 2 x x 2
1 is

c  c  c  c 
x x 2 x x 2
1

a  a  a  a 
x x 2 x x 2
1

¢eZÑ¡uL  b  b   b  b 
x x 2 x x 2
1 -Hl j¡e qm

c  c  c  c 
x x 2 x x 2
1

1
(A) 0 (B) 2abc (C) abc (D)
abc

8. In ABC , A   a,0 , B   a, 0 and the difference between A and B is α. Then the
locus of C is
ABC ¢œi¥S A   a,0 , B   a,0 Hl©f ­k A J B -Hl A¿¹l qm α z ­p­r­œ C ¢h¾c¤l
p’¡lfb q­h
(A) x2 – 2xy cot  – y2 = a2 (B) x2 + 2xy tan  – y2 = a2
(C) x2 – 2xy tan  – y2 = a2 (D) x2 + 2xy cot  – y2 = a2

x
9. Let f :  be continuous at x=0 and such that f  x   f    0 , x  . Then
2
(A) value of f(x) cannot be ascertained on the given conditions
(B) f(x)=0 x 
(C) f(x) is negative valued function
(D) f(x) is positive valued function for all x

A­frL¢V x=0 ¢h¾c¤­a p¿¹a Hhw pLm x  -Hl SeÉ f  x   f    0 pÇfLÑ j¡¢eu¡
x
f: 
2
Q­m z ­p­r­œ
(A) fËcš naÑ¡hm£l Ad£­e f(x) -Hl j¡e ¢el©fZ pñh eu
(B) pLm x  -Hl SeÉ f(x)=0 q­h
(C) f(x) GZ¡aÈL j¡e ¢h¢nø A­frL
(D) pLm x-Hl j¡­el SeÉ f(x) de¡aÈL j¡e ¢h¢nø A­frL

Che.+Phy.+Geo. 5 

10. Consider the function f(x) = (x – 3)logex.. Then the equation x loge x = 3 – x
(A) has at least one root in (1, 3) (B) has no root in (1, 3)
(C) is not at all solvable (D) has infinitely many roots in (–2, 1)
A­frL f(x) = (x – 3)logex ¢h­hQe¡ Ll z ­p­r­œ x loge x = 3 – x pj£LlZ¢Vl
(A) (1, 3)-­a A¿¹a HL¢V h£S B­R (B) (1, 3)-­a ­L¡e h£S ­eC
(C) B­c± pj¡d¡e­k¡NÉ eu (D) (–2, 1)-­a Ap£j pwMÉL h£S B­R

2x 1
 x 1 
11. lim  
x  x  2
 
(A) does not exist (B) is 1
(C) is e6 (D) is e13
2x 1
 x 1 
lim  
x  x  2
 
(A) -Hl A¢Ù¹aÆ ­eC (B) q­h 1
(C) q­h e6 (D) q­h e13

  d2 y
, then the value of 1  x  2  x
dy
n
12. If y  x  1  x 2 2
is
dx dx

 
k¢c y  x  1  x 2 qu, a­h 1  x 2 
d2 y dy
n

2
x q­h
dx dx
(A) n2y (B) –n2y (C) ny (D) –ny

cos 4x  1
13. If  dx  k cos 2 2x  c , then k =
cot x  tan x

cos 4x  1
k¢c  dx  k cos 2 2x  c qu, a¡q­m k =
cot x  tan x
1 1 1 1
(A) (B) (C)  (D) 
4 8 4 8

Che.+Phy.+Geo. 6 

14. If 
x tan 1 x
1 x 2 
dx  1  x 2 f  x   k log x  x 2  1  c , then 

k¢c 
x tan 1 x
1 x 2  
dx  1  x 2 f  x   k log x  x 2  1  c qu, a¡q­m

(A) f(x) = tan–1 x, k = –1 (B) f(x) = tan–1 x, k = 1

(C) f(x) = 2tan–1 x, k = –1 (D) f(x) = 2tan–1 x, k = 1

15. Consider the curve :

x   t 2  2  sin t  2t cos t
y   2  t 2  cos t  2t sin t

 2 2
 dx   dy 
then       dt is equal to
0  dt   dt 

x   t 2  2  sin t  2t cos t
y   2  t 2  cos t  2t sin t

 2 2
 dx   dy 
à¡l¡ ¢eZÑ£a hœ²­lM¡¢V ¢h­hQe¡ Ll z ­p­r­œ       dt -Hl j¡e q­h
0  dt   dt 

3 2
(A) (B)
3 6
2
(C) (D) 2 + 1
3

Che.+Phy.+Geo. 7 

dd22xy dx
16. Let the characteristic equation for the ODE 22
 p0  q 0 x  0 have distinct real roots
dx
dt dt
1 and  2 . Then

e1t  e 2 t
(A) is a solution of the ODE.
1   2

e1t  e 2 t
(B) is not a solution of the ODE.
1   2

e 1 t  e  2 t
(C) is not a solution of the ODE.
1   2

e1t  e 2 t
(D) + A sin t is a solution of the ODE where A is arbitrary constant.
1   2

dd22xy dx
p¡d¡lZ A¿¹lLm pj£Ll­Zl 22  p 0  q 0 x  0 °h¢nøÉ pj£Ll­Zl ¢iæ ¢iæ h¡Ù¹h h£S qm
dx
dt dt
1 J  2 z ­p­r­œ

e1t  e 2 t
(A) pj£LlZ¢Vl HL¢V pj¡d¡e
1   2

e1t  e 2 t
(B) pj£LlZ¢Vl pj¡d¡e eu
1   2

e 1 t  e  2 t
(C) pj£LlZ¢Vl pj¡d¡e eu
1   2
e1t  e 2 t
(D) + A sin t pj£LlZ¢Vl HL¢V pj¡d¡e q­h, ­kM¡­e A kcªµR dˤhL
1   2

17. The family of curves y = ea cos x, where ‘a’ is arbitrary constant, is represented by the
differential equation
hœ²­lM¡ f¢lh¡l y = ea cos x ­kM¡­e ‘a’ -HL¢V kcªµR dˤhL, ¢ejÀ AhLm pj£Ll­Zl à¡l¡ p§¢Qa qu
dy dy
(A) y log y + cot x =0 (B) y log y + cot x
dx dx
dy dy
(C) log y + cot x =0 (D) log y + cot x
dx dx

Che.+Phy.+Geo. 8 

18. Consider the curve x2 – y2 = a2. Then the locus of point of intersection of tangent at any
point on the curve and perpendicular to it, passing through (0,0) is
hœ²­lM¡ x2 – y2 = a2 ¢h­hQe¡ Ll z I hœ²­lM¡l Ef¢lÙÛ ­L¡e ¢h¾c¤­a A¢ˆa ØfnÑL Hhw (0,0)
¢h¾c¤N¡j£ mð­lM¡l ­Rc¢h¾c¤l p’¡lfb q­h
(A) x2 + y2 = a2 (B) (x2 + y2)2 = a2 (x2 – y2)
(C) x2 – y2 = a2 (D) y2 = 4ax

2 x  1,  1  x  0

19. Let f :  1,1  be defined by f  x   2 x , x  0
2 x  1, 0  x  1

Then
(A) f(x) attains its maximum value 2
(B) f(x) attains its minimum value 2
(C) f(x) does not attain its maximum and minimum value
(D) f(x) is a continuous function
f :  1,1  ¢eji
À ¡­h pw‘¡a B­R z
2 x  1,  1  x  0

f  x   2 x , x  0
2 x  1, 0  x  1

(A) f(x) p­îÑ¡µQ j¡e 2 f¢lNËq L­l
(B) f(x) phÑ¢ejÀ j¡e 2 f¢lNËq L­l
(C) f(x) p­îÑ¡µQ J phÑ¢ejÀ j¡e ­L¡e¢VC f¢lNËq L­l e¡
(D) f(x) p¿¹a A­frL


The vector of magnitude 6 and perpendicular to both  = 2^i+ 2^j+ k^ and  = ^i– 2^j+ 2k^ is

20.

 = 2^i+ 2^j + k^ ,  = ^i– 2^j+ 2k^ ­iƒlà­ul Efl mð J 6 j¡e ¢h¢nø ­iƒl¢V q­h


(A)  2(2^i– ^j – 2k)
^ (B) 2(2^i– ^j– k)
^

(C) 2^i– ^j – k^ (D)  (2^i– ^j– k)
^

Che.+Phy.+Geo. 9 

PHYSICS
21. The velocity (v) of a particle moving in a straight line varies with distance (x) as shown in
figure
plm°l¢ML f­b Qmj¡e HL¢V LZ¡l ­hN (v) he¡j c§l­aÆl (x) ­mM¢Qœ ¢e­Ql R¢h­a ­cM¡­e¡ q­u­R z

The variation of its acceleration (a) with distance (x) is like
LZ¡¢Vl aÆlZ (a) he¡j c§l­aÆl (x) ­mM¢Qœ q­h

(A) (B)

(C) (D)

22.

A vessel filled with water is moving horizontally with constant acceleration (a).
AB is the liquid surface when the vessel is at rest. The shape of the liquid surface (AB)
when it is accelerated will be as
HL¢V Smf§ZÑ f¡œ Ae¤i¨¢jL i¡­h pj aÆl­Z (a) Qm­R z ¢ÙÛl¡hÙÛ¡u f¡­œl al­ml Ef¢lfªù am AB z
aÆlZpq Qm¡L¡m£e al­ml Ef¢lfªùa­ml (AB) BL«¢a q­h

(A) (B)

(C) (D)

Che.+Phy.+Geo. 10 

23. The angle between the velocity and the acceleration of a particle at any instant is θ (≠0).
Then

(A) the particle moves in a curved path with uniform speed

(B) the particle moves in a straight line with uniform acceleration

(C) the particle moves in a curved path with non - uniform speed

(D) the particle moves in a straight line with non - uniform acceleration

­L¡­e¡ HL¢V j¤q¨­aÑ HL¢V Qmj¡e LZ¡l ­hN Hhw aÆl­Zl j­dÉL¡l ­L¡Z θ (≠0) z a¡q­m

(A) LZ¡¢V pj­h­N hœ²¡L¡l f­b N¢an£m z

(B) LZ¡¢V pjaÆl­Z plm°l¢ML f­b N¢an£m z

(C) LZ¡¢V Apj­h­N hœ²¡L¡l f­b N¢an£m z

(D) LZ¡¢V ApjaÆl­Z plm°l¢ML f­b N¢an£m z

24. A particle of mass 2 m moves to the east, and a particle of mass m moves to the west both
with speed V0. If they collide elastically and if it is observed that the mass 2 m ends up
moving northward, then its speed is

2m i­ll HL¢V LZ¡ f§hÑ¢c­L Hhw m i­ll HL¢V LZ¡ f¢ÕQj¢c­L, EiuC V0 ­h­N N¢an£m z k¢c

H­cl j­dÉ ¢ÙÛ¢aÙÛ¡fL pwOoÑ O­V Hhw k¢c pwO­oÑl fl 2m i­ll LZ¡¢V Ešl¢cL hl¡hl N¢an£m
qu, a­h a¡l ­hN q­h

V0
(A) V0 (B)
2

V0 V0
(C) (D)
2 3

Che.+Phy.+Geo. 11 

25. The figure shows three points on a steadily rotating wheel. If the angular velocities at three

points 1, 2 and 3 are 1 , 2 , 3 and linear velocities are v1 , v 2 and v 3 respectively, then

(A) 3 > 2 > 1; v 3 > v2 > v1

(B) 1 = 2 > 3; v 1 = v2 < v3

(C) 1 = 2 = 3; v 1 = v2 = v3

(D) 1 = 2 = 3; v 1 = v2 < v3

¢Q­œ HL¢V p¤¤oji¡­h O§ZÑ¡uj¡e Q¡L¢al Efl ¢ae¢V ¢h¾c¤ 1, 2 Hhw 3 ­cM¡­e¡ q­u­R z k¢c HC

¢h¾c¤…¢m­a ­L±¢ZL ­hN kb¡œ²­j 1 , 2 , 3 Hhw °l¢ML ­hN kb¡œ²­j v1 , v 2 Hhw v3 qu, a­h

(A) 3 > 2 > 1; v 3 > v2 > v1

(B) 1 = 2 > 3; v 1 = v2 < v3

(C) 1 = 2 = 3; v 1 = v2 = v3

(D) 1 = 2 = 3; v 1 = v2 < v3

Che.+Phy.+Geo. 12 

26. Two vibrations (1) and (2) are shown in figure.

c¤¢V al‰ (1) J (2) ¢Q­œ ­cM¡­e¡ q­u­Rz

Which of the following statements is true ?

(A) Frequencies of two vibrations are same, but the vibration (2) is leading in phase by

.
2

(B) Frequencies of two vibrations are same, but the vibration (2) lags behind (1) in phase

by .
2

(C) Amplitudes of two vibrations are different, however both are in same phase.

(D) Two vibrations have same frequency, different amplitudes but vibration (2) lags
behind vibration (1) in phase by π.

¢e­Ql ­L¡eÚ E¢š²¢V paÉ ?


(A) c¤¢V al­‰l LÇf¡ˆ pj¡e, ¢L¿¹¥ (2) ew al‰ cn¡u H¢N­u b¡­L z
2


(B) c¤¢V al­‰l LÇf¡ˆ pj¡e, ¢L¿¹¥ (2) ew al‰ cn¡u ¢f¢R­u b¡­L z
2

(C) c¤¢V al­‰l ¢hÙ¹¡l Bm¡c¡ Hhw Ei­uC pjcn¡ pÇfæ z

(D) c¤¢V al­‰l LÇf¡ˆ pj¡e, ¢hÙ¹¡l Bm¡c¡ ¢L¿¹¥ (2) ew al‰ (1) ew al­‰l ­b­L π cn¡u

¢f¢R­u b¡­L z

Che.+Phy.+Geo. 13 

27. The position (x) vs time (t) graph of two particles each of mass m are performing SHM,
with A2 = 2A1, that is, the amplitude of 2 is equal to 2 times of (1) is shown in figure.

E2
The ratio of their kinetic energies is
E1
plm ­c¡mN¢a pÇfæ c¤¢V m i­ll LZ¡l plZ-pju ­mM¢Qœ R¢h­a ­cM¡­e¡ q­u­R z c¤¢V ­c¡mN¢al
E2
¢hÙ¹¡l A1 , A2 Hhw A2 = 2A1 q­m H­cl N¢an¢š²l Ae¤f¡a q­h
E1
1 1
(A) (B) (C) 1 (D) 2
4 2

28. The Figure shows a cyclic process in P-T diagram

The same process in P-V diagram will be
R¢h­a HL¢V BhaÑ fË¢œ²u¡l P-T ­mM¢Qœ ­cM¡­e¡ q­u­R, fË¢œ²u¡¢Vl P-V ­mM¢Qœ¢V q­h

(A) (B)

(C) (D)

Che.+Phy.+Geo. 14 

29. An ideal gas at pressure p1 and volume v1 expands to volume v2 in two ways

(a) adiabatically and (b) isothermally. The final pressures in adiabatic and isothermal

processes are pad and piso respectively. Wad represents the work done in adiabatic

expansion and Wiso represents the work done in isothermal expansion. Then which of the

following statements is true ?

(A) pad < piso but Wad > Wiso (B) pad > piso and Wad > Wiso

(C) pad < piso and Wad < Wiso (D) pad > piso but Wad < Wiso

p1 Q¡­f HL¢V BcnÑ NÉ¡­pl Buae v1 c¤¢V fË¢œ²u¡u ­h­s v2 qu, (a) l©Üa¡f fË¢œ²u¡u Hhw (b)

p­j¡o· fË¢œ²u¡u z l©Üa¡f Hhw p­j¡o· fË¢œ²u¡u A¢¿¹j Q¡f kb¡œ²­j pad Hhw piso z Wad l©Üa¡f

fË¢œ²u¡u L«aL¡kÑ Hhw Wiso p­j¡o· fË¢œ²u¡u L«aL¡kÑ q­m ¢e­Ql ­L¡eÚ E¢š²¢V p¢WL ?

(A) pad < piso ¢L¿¹¥ Wad > Wiso (B) pad > piso Hhw Wad > Wiso

(C) pad < piso Hhw Wad < Wiso (D) pad > piso ¢L¿¹¥ Wad < Wiso

30. In the given circuit, if the points a and b are at the same potential, then

fËcš haÑe£­a a Hhw b ¢h¾c¤­a ¢hih pj¡e q­m

C1 R1 C1 R 2
(A)  (B) 
C2 R 2 C 2 R1

C1 R 2 C1  C 2
(C)  (D) 1
R1 C 2 R1  R 2

Che.+Phy.+Geo. 15 

q q
31. Along the x-axis, three charges ,  2q and are placed at x = 0, x = a and x = 2a
2 2
respectively. The resultant electric potential V(r) at a point P located at a distance r (r >> a)
1
from the charge – 2q is V  n , where n equals to
r
q q
x -Ar hl¡hl x = 0, x = a Hhw x = 2a ­a kb¡œ²­j ¢ae¢V Bd¡e ,  2q Hhw l¡M¡ B­R z – 2q
2 2
1
Bd¡e ­b­L r (r >> a) c§l­aÆ P ¢h¾c¤­a mì a¢sv¢hih V  n , ­kM¡­e n-Hl j¡e
r
(A) 2 (B) 3 (C) 0 (D) 1

32. OABC is a current carrying square loop. A positive charge is projected from the centre of
the loop along its diagonal AC as shown. Unit vector in the direction of initial acceleration
will be
OABC HL¢V a¢sv f¢lh¡q£ hNÑ¡L¡l m¤f z HL¢V de¡aÈL Bd¡e­L m¤­fl ­L¾cÊ ­b­L AC LZÑ hl¡hl
­R¡ys¡ qm z fË¡l¢ñL aÆl­Zl ¢cL hl¡hl HLL ­iƒl¢V q­h

ˆ k  ˆi ˆjˆi ˆj   ˆi ˆjˆi ˆj 
(A) k̂ A kA (B)B B    (C) C– k̂Ckˆ k D(D)D   
 2  2   2 2
       

33. In the given circuit, potential difference across 100 Ω resistance, that is, between the points
A and B is Vi. However, if one measures the potential difference between the same points
V  Vf
A and B by a voltmeter of 900 Ω resistance, the result is Vf. Then i is
Vi
fËcš haÑe£­a 100 Ω ­l¡­dl c¤¢c­L A Hhw B ¢h¾c¤l j­dÉ ¢hih fË­ic Vi z k¢c HLC A Hhw B
V V
¢h¾c¤l j­dÉ 900 Ω ­l¡­dl ­i¡ÒV¢jV¡l ¢c­u ¢hihfË­ic j¡f¡ qu a­h a¡ qu Vf z a¡q­m i f
Vi
qm

1 1
(A) (B) 1 (C) (D) 10
10 100

Che.+Phy.+Geo. 16 

34. Two identical coherent sources of light separated by a distance d produce an interference
pattern on a screen. If the wavelength of the monochromatic light emitted by the source is
, then the maximum number of fringes that can be observed on the screen is

d c§l­aÆ Ah¢ÙÛa c¤¢V A¢iæ p¤¤pwNa B­m¡L Evp fcÑ¡u hÉ¡¢aQ¡l f¢V °a¢l L­l z k¢c Evp ­b­L

¢eNÑa HLhZÑ£ B­m¡­Ll al‰°cOÉÑ  qu a­h fcÑ¡u ­k p­h¡ÑµQ pwMÉ¡l T¡ml ­cM¡ k¡­h a¡ qm
d d d
(A) (B)  (C) (D) 2
λ 2λ λ

35. In hydrogen atom, if electron moves from a lower energy level to a higher energy level,
then

(A) its kinetic energy increases while potential energy decreases.

(B) its kinetic energy decreases, but potential energy increases.

(C) both the kinetic energy and potential energy increases.

(D) kinetic energy remains the same, only the potential energy and hence the total energy
increases.

HL¢V q¡C­XÊ¡­Se flj¡Z¤­a k¢c C­mLVÊe HL¢V ¢e­Ql n¢š²Ù¹l ­b­L Ef­ll n¢š²Ù¹­l Nje L­l,
a­h C­mLVÊ­el

(A) N¢an¢š² hª¢Ü f¡u ¢L¿¹¥ ¢ÙÛ¢an¢š² qÊÊ¡p f¡u z

(B) N¢an¢š² qÊÊ¡p f¡u ¢L¿¹¥ ¢ÙÛ¢an¢š² hª¢Ü f¡u z

(C) N¢an¢š² J ¢ÙÛ¢an¢š² EiuC hª¢Ü f¡u z

(D) N¢an¢š² HLC b¡­L, öd¤j¡œ ¢ÙÛ¢an¢š² Hhw gmül©f ­j¡V n¢š² hª¢Ü f¡u z

Che.+Phy.+Geo. 17 

CHEMISTRY

36. How many cyclic compounds are possible for the molecular formula C 3H6O ?

C3H6O BZ¢hL pw­La pÇfæ ­k±­Nl La…¢m Qœ²¡L«¢a ­k±N pñh ?

(A) 1 (B) 2 (C) 3 (D) 4

37. Hybridisation state of all the carbon atoms (from left to right) of the following carbocation
are :
¢ejÀ¢m¢Ma L¡­h¡ÑLÉ¡V¡ue¢Vl L¡hÑe flj¡Z¤…¢ml (h¡j¢cL ­b­L X¡e¢c­L) pwLl¡uZ AhÙÛ¡ qm

H2C – CH = C = CH2

(A) sp2, sp2, sp, sp2 (B) sp, sp2, sp, sp2

(C) sp3, sp2, sp2, sp2 (D) sp2, sp2, sp2, sp2

38. Two components (A and B) of same molecular formula, C 2H5NO, on treatment with
P2O5 gives X. Both A and X on treatment with alkali evolve NH 3. A and B are respectively

(A) CH3CH = N – OH and CH3CONH2

(B) HCONHCH3 and CH3CONH2

(C) CH3CONH2 and CH3CH = N – OH

(D) CH3CONH2 and HCONHCH3

C2H5NO BZ¢hL pw­La pÇfæ c¤¢V ­k±N (A Hhw B) P2O5Hl p¡­b ¢h¢œ²u¡u X ­cu z A Hhw X

Eiu ­k±Nc¤¢V r¡­ll p¡­b ¢h¢œ²u¡u NH3 Evfæ L­l z A Hhw B ­k±Nc¤¢V kb¡œ²­j
(A) CH3CH = N – OH Hhw CH3CONH2

(B) HCONHCH3 Hhw CH3CONH2

(C) CH3CONH2 Hhw CH3CH = N – OH

(D) CH3CONH2 Hhw HCONHCH3

Che.+Phy.+Geo. 18 

39. The product of the following reaction sequence is :
1. acetic anhydride
2. Br2 in HOAC
Aniline  Product
3. dil. HCl
4. NaNO2 and dil. HCl
5. CuBr, HBr

(A) p – bromoaniline (B) 1, 4 –dibromobenzene
(C) p – bromoacetanilide (D) bromobenzene

e£­Ql ¢h¢œ²u¡œ²­j Evf¡¢ca ­k±N¢V qm
1. AÉ¡­p¢VL AÉ¡eq¡CXÊ¡CX
2. Br2 /HOAC
AÉ¡¢e¢me Evf¡¢ca ­k±N
3. mO¤ HCl
4. NaNO2 Hhw mO¤ HCl
5. CuBr, HBr
(A) p – ­hË¡­j¡AÉ¡¢e¢me (B) 1, 4 - X¡C­hË¡­j¡­h¢”e
(C) p - ­hË¡­j¡AÉ¡¢pV¡¢em¡CX (D) ­hË¡­j¡­h¢”e

40. The gas that is evolved on treatment of an ethereal solution of methylacetylene with a
solution of vinyl magnesium bromide in ether is
(A) methane (B) ethylene (C) acetylene (D) isoprene
¢jb¡Cm AÉ¡¢p¢V¢me J ¢ie¡Cm jÉ¡N­e¢pu¡j ­hË¡j¡C­Xl Cb¡l âh­Zl j­dÉ ¢h¢œ²u¡u ­k NÉ¡p Eá¥a
qu, ­p¢V qm
(A) ¢j­be (B) C¢b¢me (C) AÉ¡¢p¢V¢me (D) BC­p¡¢fËe

41. The IUPAC nomenclature of Na  Ag  CN 2  is
(A) Sodium dicyanoargentate (II)
(B) Sodium dicyanidoargentate (II)
(C) Sodium dicyanidoargentate (I)
(D) Sodium dicyanoargentate (III)
Na  Ag  CN 2  Hl IUPAC e¡jLlZ qm
(A) ­p¡¢Xu¡j X¡Cp¡u¡­e¡B­SÑ­¾VV (II)
(B) ­p¡¢Xu¡j X¡Cp¡u¡e¡C­X¡B­SÑ­¾VV (II)
(C) ­p¡¢Xu¡j X¡Cp¡u¡e¡C­X¡B­SÑ­¾VV (I)
(D) ­p¡¢Xu¡j X¡Cp¡u¡­e¡B­SÑ­¾VV (III)

Che.+Phy.+Geo. 19 

42. Identify the polar molecule – pairs from the following molecules :

XeF4, SF4, O3, B2H6

e£­Ql AZ¤…¢ml ­b­L dËh¤ £u AZ¤k¤Nm ¢Q¢q²a Ll
XeF4, SF4, O3, B2H6

(A) SF4, XeF4 (B) O3, B2H6

(C) SF4, O3 (D) XeF4, B2H6

43. Aqueous solution of silver nitrate causes precipitation reactions separately with aqueous
solutions of H3PO3, HI and H2S respectively. The formula of the respective precipitates are

Sm£u ¢pmi¡l e¡C­VÊV âh­Zl p¢qa H3PO3, HI J H2S Hl Sm£u âh­Zl fªbL ¢h¢œ²u¡u
Adx­rfZ O­V z Evfæ Adx­rf…¢ml pw­La kb¡œ²­j
(A) Ag2HPO3, AgI, Ag2S (B) Ag3P, Ag, S

(C) Ag, AgI, Ag2S (D) Ag3PO4, AgI, (Ag + S)

44. Which of the following orders are wrong ?

(A) Electron affinity : N < O < F < Cl

(B) First ionisation energy : Be < B < N < O

(C) Basic property : BeO < MgO < CaO < BaO

(D) Electronegativity : F > O > N > C

e£­Ql ­L¡eÚ œ²j¢V i¥m ?
(A) C­mLVÊe Bp¢š² : N < O < F < Cl

(B) fËbj Bue£ihe n¢š² : Be < B < N < O
(C) r¡lL£u djÑ : BeO < MgO < CaO < BaO
(D) a¢svGZ¡aÈLa¡ : F > O > N > C

Che.+Phy.+Geo. 20 

45. Identify the reagent that produces colourless product on reaction with aqueous H 2O2.
(A) Acidic Fe(II) Sulphate Solution (B) Cold acidic K2Cr2O7 Solution
(C) Acidic KMnO4 Solution (D) Acidic KI Solution

Sm£u H2O2 -Hl p¢qa ¢h¢œ²u¡u ­k ¢hL¡lL¢V hZÑq£e âhZ Evfæ L­l a¡q¡­L ¢Q¢q²a Ll
(A) B¢ÇmL Fe(II) p¡m­gV âhZ (B) n£am B¢ÇmL K2Cr2O7 âhZ
(C) B¢ÇmL KMnO4 âhZ (D) B¢ÇmL KI âhZ

46. X2(g) + Y2(g) 2XY (g)
At 400 K the value of Kp is 64, for the above equilibrium. If an equimolar mixture of X 2
and Y2 is heated in a closed container at 400 K, then the mole fraction of Y 2 at equilibrium
will be
X2(NÉ¡p) + Y2(NÉ¡p) 2XY (NÉ¡p)
400 K a¡fj¡œ¡u Ef­l¡š² p¡jÉ¢Vl p¡jÉdˤhL Kp Hl j¡e 64 z pj¡e pj¡e ­j¡m pwMÉ¡u HL¢V hÜ
f¡­œ X2 Hhw Y2 ­L ¢j¢n­u 400 K a¡fj¡œ¡u Ešç Ll¡ qm z p¡jÉ¡hÙÛ¡u Y2-Hl ­j¡m iNÀ¡wn
(mole fraction) q­h,
1 4
(A) (B)
5 5
1 1
(C) (D)
8 10

47. Pressure (P) of a given mass of gaseous oxygen (obeying van der Waals equation) is given
by
RT a
P=  , where V and T are the volume and temperature (K). The mass of the
2V  b 4V 2
gas is
HL¢V ¢e¢cÑø JS­el NÉ¡p£u A¢„­S­el Q¡f (P) van der Waals pj£LlZ Ae¤k¡u£ ­mM¡ k¡u :
RT a
P=  , ­kM¡­e V Hhw T kb¡œ²­j Buae Hhw a¡fj¡œ¡ (K) z NÉ¡p¢Vl il qm
2V  b 4V 2
(A) 32 g (B) 64 g (C) 16 g (D) 4 g

Che.+Phy.+Geo. 21 

1

48. Adiabatic expansion of a gas (ideal) is found to obey the relation T  V 3 . The value of

 (= Cp, m / Cv, m) of the gas is
1

HL¢V BcnÑ NÉ¡­pl l©Üa¡f pÇfËp¡l­Zl ­r­œ T  V 3 pÇfLÑ¢V l¢ra qu z NÉ¡p¢Vl

 (= Cp, m / Cv, m) Hl j¡e q­h

4 2
(A) (B)
3 3

5
(C) (D) 1
2

49. The compressibility factor for a real gas at high pressure is

EµQ Q¡­f h¡Ù¹h NÉ¡­pl pwejÉa¡ …ZL

RT Pb
(A) 1 + (B) 1+
Pb RT

Pb
(C) 1– (D) 1
RT

50. A buffer is prepared by mixing 10 mL of 0.4(N) CH 3CO2Na solution and 10 mL of

0.2(N) CH3CO2H solution. [Given: Ka of CH3CO2H is 1.8  10–5 and log(1.8) = 0.2553].

The pH of the resulting solution is given by
10 ¢j¢m 0.4(N) CH3CO2Na âhZ Hhw 10 ¢j¢m 0.2(N) CH3CO2H âhZ ¢j¢nËa L­l HL¢V h¡g¡l

âhZ °a¢l Ll¡ q­u­R z [fËcš: CH3CO2H Hl Ka = 1.8  10–5 Hhw log(1.8) = 0.2553] h¡g¡l
âh­Zl pH qm
(A) 3.04 (B) 4.04
(C) 6.06 (D) 5.04
_______________

Che.+Phy.+Geo. 22 

SPACE FOR ROUGH WORK / l¡g L¡­Sl SeÉ S¡uN¡

Che.+Phy.+Geo. 23 

Paper - VII
Subject : Physics, Chemistry, Mathematics
for admission in
Chemistry/Physics/Geology

pju: 90 ¢j¢eV ®j¡V fËnÀ : 50 ¢V f§ ZÑj¡e : 100

1. HC fËnÀf­œl ph fËnÀC Ah­S¢ƒi fËnÀ Hhw fË¢a¢V fË­nÀl Q¡l¢V pñ¡hÉ Ešl ­cJu¡ B­R
k¡l HL¢V j¡œ p¢WL z p¢WL Ešl ¢Q¢q²a Ll­m 2 eðl f¡­h z i¥m Ešl ¢Q¢q²a Ll­m
Abh¡ HL¡¢dL Ešl ¢Q¢q²a Ll­m ½ eðl L¡V¡ k¡­h z
2. OMR f­œ A, B, C, D ¢Q¢q²a p¢WL Ol¢V il¡V L­l Ešl ¢c­a q­h z
3. OMR f­œ Ešl ¢c­a öd¤j¡œ L¡­m¡ h¡ e£m L¡¢ml hm f­u¾V ­fe hÉhq¡l Ll­h z
4. OMR f­œ ¢e¢cÑø ÙÛ¡e R¡s¡ AeÉ ­L¡b¡J ­L¡­e¡ c¡N ­c­h e¡ z
5. 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
6. OMR f­œ ¢e¢cÑø ÙÛ¡­e ¢e­Sl e¡j J fl£r¡­L­¾cÊl e¡j ¢mM­a q­h Hhw ¢e­Sl (Admit
Card H E­õ¢Ma) ü¡rl Ll­a q­h z
7. 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¡ ü¡r­l ­L¡­e¡ i¥m b¡L­m Ešlfœ 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
8. ­j¡h¡Cm ­g¡e h¡ ®k ®L¡e dl®el C®mLVÌ¢eL NÉ¡®SV, LÉ¡mL¥­mVl, pÔ¡CXl¦m, mN­Vhm,
q¡aO¢s, ­lM¡¢Qœ, NË¡g h¡ ­L¡­e¡ dl­el a¡¢mL¡ , Lmj CaÉ¡¢c 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
9. fËnÀf­œ l¡g L¡S Ll¡l SeÉ gy¡L¡ S¡uN¡ ­cJu¡ B­R z AeÉ ­L¡­e¡ L¡NS HC L¡­S
hÉhq¡l Ll­h e¡ z
10. fl£r¡Lr R¡s¡l B­N OMR fœ AhnÉC f¢lcnÑL­L ¢c­u k¡­h z
11. HC fËnÀf­œ 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¡ p­šJ Ä 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
12. fl£r¡­n­o fl£r¡b£Ñl¡ fËnÀfœ¢V ¢e­u k¡­h z

Che.+Phy.+Geo. 24 
