CG PAT / PVPT 2021 Question Paper Science

Question Booklet No.
SET – A


A Subject Code : 22103/BE – SC/ENT – M

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Number of Pages in this Question Booklet : 88 Number of Questions in this Question Booklet : 200

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A
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11. Bg àíZ-nwpñVH$m _| VrZ ^mJ hmoJ| : 11. This Question Paper consists of Three Parts namely :
(i) àW_ ^mJ :- ^m¡{VH$ emñÌ - à.g§. 1 – 60$ (i) First Part : – Physics – Q. No. 1 – 60
(ii) {ÛVr` ^mJ :- agm`Z emñÌ - à.g§. 61 – 120 (ii) Second Part : – Chemistry – Q. No. 61 – 120
(iii) V¥Vr` ^mJ :- (A) J{UV - à.g§. 121 – 200 (iii) Third Part : – (A) Mathematics – Q. No. 121 – 200
(~) Ord {dkmZ - à.g§. 121 – 200 (B) Biology – Q. No. 121 – 200
Candidates are directed to choose only one subject from
narjm{W©`m| H$mo {ZX}e {X`m OmVm h¡ {H$ do V¥Vr` ^mJ (A) Ed§ (~) _| go Third Part (A) and (B).
Ho$db EH$ hr {df` MwZ| &
12. In case of any ambiguity in English version the Hindi version shall
12. `{X A§JOo« r ^mfm _| H$moB© g§Xho h¡, Vmo {hÝXr ^mfm H$mo hr àm_m{UH$ _mZm Om`oJm Ÿ& be considered authentic.

-1-

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NO
DO

-2- Set-A

PART – I
Physics ^m¡{VH$ emñÌ
1. Two simple harmonic Oscillations 1. Xmo gab AmdV© J{V`m| H$mo {ZåZ g_rH$aU Ûmam
(SHM) are represented by the equation àX{e©V {H$`m J`m h¡
 π
y1 = 0.1sin  100πt +  and
 3 y1 = 0.1sin  100πt + π  Am¡a
 3
y2 = 0.1 cos(100πt)
y2 = 0.1 cos(100πt)
The phase difference of velocity of
particle 2 with respect to the velocity H$U 2 Ho$ doJ H$m H$bmÝVa H$U 1 Ho$ doJ Ho$ gmnoj
of particle 1 is hmoJm
π π
(A) (A)
6 6
π π
(B) − (B) −
6 6
π π
(C) (C)
3 3
π π
(D) − (D) −
3 3
2. The spring constant for the combination 2. {XE JE {MÌ Ho$ AZwgma pñà¨Jm| Ho$ g_m`moOZ Ho$
of springs as shown in figure is {bE pñà¨J {Z`Vm§H$ hmoJm


(A) K (A) K
(B) 2K (B) 2K
(C) 4K (C) 4K
5K 5K
(D) (D)
2 2

-3- Set-A

3. A source of frequency n gives 5 beats/s, 3. EH$ òmoV {OgH$s Amd¥{Îm n h¡, H$mo O~ 200/goH$ÝS>
when sounded with a source of H$s Amd¥{Îm dmbo òmoV Ho$ gmW ~Om`m OmVm h¡,
frequency 200/sec. The second Vmo à{V goH$ÝS> 5 {dñn§X CËnÝZ H$aVm h¡ & {ÛVr`
harmonics (2n) gives 10 beats/s, when
sounded with a source of frequency
g§ZmXr (2n) H$mo O~ 420/goH$ÝS> H$s Amd¥{Îm
420/sec, then n is equal to dmbo òmoV Ho$ gmW ~Om`m OmVm h¡, Vmo à{V goH$ÝS>
10 {dñn§X CËnÝZ H$aVm h¡, Vmo n H$m _mZ h¡
(A) 200/sec (B) 195/sec (A) 200/goH$ÝS> (B) 195/goH$ÝS>
(C) 210/sec (D) 205/sec (C) 210/goH$ÝS> (D) 205/goH$ÝS>

4. Two wavelength of light λ1 and λ2 4. àH$me H$s Xmo Va§JX¡¿`© λ1 Ed§ λ2 H$mo `§J {Û-pñbQ>
are sent through young’s double slit CnH$aU Ho$ Ûmam EH$ gmW ^oOm OmVm h¡ & λ1 Ed§ λ2
apparatus simultaneously. What must Ho$ ~mao _| Š`m ghr h¡, `{X λ1 H$m V¥Vr` H«$_ H$m
be true about λ1 and λ2, if the third
order bright fringe of λ1 coincides with
M_H$sbm q\«$O λ2 Ho$ n§M_ H«$_ Ho$ H$mbr q\«$O Ho$
fifth order dark fringe of λ2 ? gmW g§nmVr hmo ?
(A) 3λ1 = 2λ2 (B) 2λ1 = 3λ2 (A) 3λ1 = 2λ2 (B) 2λ1 = 3λ2
(C) 3λ1 = 5λ2 (D) 5λ1 = 3λ2 (C) 3λ1 = 5λ2 (D) 5λ1 = 3λ2

5. A bar magnet of magnetic moment M 5. M Mwå~H$s` AmKyU© H$m EH$ XÊS> Mwå~H$ Mwå~H$s`
is placed at right angles to a magnetic joÌ B Ho$ bå~dV aIm J`m h¡ & `{X XÊS> Mwå~H$
induction B. If a force F is experienced Ho$ àË`oH$ Y«wd na bJZo dmbo ~b H$m _mZ F h¡, Vmo
by each pole of the bar magnet, the XÊS> Mwå~H$ H$s bå~mB© hmoJr
length of the bar magnet will be
(A) MB (B) BF (A) MB (B) BF
F M F M
(C) MF (D) F (C) MF (D) F
B MB B MB
6. A magnetic needle has magnetic 6. EH$ Mwå~H$s` gwB© H$m Mwå~H$s` AmKyU©
moment of 6.7 × 10–2 Am2 and moment 6.7 × 10–2 Am2 VWm O‹S>Ëd AmKyU©
of inertia 7.5 × 10–6 kg m2. It performs 7.5 × 10–6 kg m2 h¡ & `{X Mwå~H$s` gwB©
10 complete oscillations in 6.70 s. 10 XmobZ H$aZo _| 6.70 s boVm h¡, Vmo
What is the magnitude of the magnetic Mwå~H$s` joÌ H$s n[a_mU Š`m hmoJr ?
field ?
(A) 0.04 T (B) 0.03 T (A) 0.04 T (B) 0.03 T
(C) 0.01 T (D) 0.02 T (C) 0.01 T (D) 0.02 T

-4- Set-A

7. A square of side ‘a’ has charge ‘Q’ at 7. ‘a’ bå~mB© dmbr ^wOm Ho$ EH$ dJ© Ho$ Ho$ÝÐ na
its centre and charge ‘q’ at one of its Amdoe ‘Q’ aIm J`m h¡ Ed§ BgHo$ EH$ H$moZo na
corners. The work required in moving Amdoe ‘q’ pñWV h¡ & Amdoe ‘q’ H$mo EH$ H$moZo go
the charge ‘q’ from one corner to the {dnarV {gao ({dH$U© Ho$ Xÿgao {gao) VH$ MbmZo _|
diagonally opposite corner is {H$`m J`m H$m`© hmoJm
Qq Qq 2 Qq Qq 2
(A) 4π ∈0 Ω (B) (A) 4π ∈0 Ω (B)
4π ∈0 a 4π ∈0 a
Qq Qq
(C) Zero (D) (C) eyÝ` (D)
2π ∈0 a
2π ∈0 a


8. The magnitude of electric field intensity 8. {ÛY«wd AmKyU© p = i + j dmbo EH$ {dÚwV {ÛY«wd Ho$
at point (2, 0, 0) due to a electric dipole H$maU, {~ÝXþ (2, 0, 0) na {dÚwV joÌ H$s Vrd«Vm

of dipole moment p = i + j kept at origin H$m n[a_mU h¡ ( K = 1 )
1 4π ∈0
(where K = )
4π ∈0
7 5 7 5
(A) K (B) K (A) K (B) K
8 8 8 8
5 7 5 7
(C) K (D) K (C) K (D) K
4 4 4 4

9. A nucleus of mass M + ∆m is at rest and 9. EH$ M + ∆m Ðì`_mZ H$m Zm{^H$ {dam_mdñWm
decays into two daughter nuclei of equal go Xmo g§V{V Zm{^H$m| {OZ_| go àË`oH$ H$m Ðì`_mZ
M M
mass each. The speed of daughter h¡, _| j` hmo OmVm h¡ & g§V{V Zm{^H$ H$s
2 2
nuclei is (speed of light is cm/s) Mmb h¡ (àH$me H$m Mmb go_r./go. h¡)
 ∆m  2∆m  ∆m  2∆m
(A) c   (B) c (A) c   (B) c
 M + ∆m  M  M + ∆m  M

∆m ∆m ∆m ∆m
(C) c (D) c (C) c (D) c
M M + ∆m M M + ∆m

-5- Set-A

10. A parallel plate is made of two dielectric 10. EH$ g_mZm§Va ßboQ> Xmo namd¡ÚwV ãbm°H$ H$mo loUr
blocks in series. one of the block has H«$_ _| Omo‹S>H$a ~Zm`m h¡ & EH$ namd¡ÚwV ãbm°H$ H$s
thickness d1 and dielectric constant
_moQ>mB© d1 VWm namd¡ÚwVm§H$ k1 VWm Xÿgao namd¡ÚwV
k1 and other has thickness d2 and
dielectric constant k2 as shown in figure. ãbm°H$ H$s _moQ>mB© d2 Ed§ namd¡ÚwVm§H$ k2 h¡ O¡gm
This arrangement can be thought as a {H$ {MÌ _| {XIm`m J`m h¡ & `{X Bg ì`dñWm H$mo
dielectric slab of thickness, d (=d1 + d2) EH$ namd¡ÚwV n{Å>H$m H$s Vah _mZ| {OgH$s _moQ>mB©
and effective dielectric constant k, then d (=d1 + d2) h¡, Vmo BgH$m à^mdr namd¡ÚwVm§H$
k hmoJm



k1d1 + k 2 d2 k1d1 + k 2 d2
(A) k = (A) k =
d1 + d2 d1 + d2
k1d1 + k 2 d2 k1d1 + k 2 d2
(B) k = (B) k =
k1 + k 2 k1 + k 2
k1k 2 ( d1 + d2 ) k1k 2 ( d1 + d2 )
(C) k = (C) k =
k1d1 + k 2 d2 k1d1 + k 2 d2
2 k1k 2 2 k1k 2
(D) k = (D) k =
k1 + k 2 k1 + k 2

11. Two cities are 150 km apart. Electric 11. Xmo eha EH$ Xÿgao go 150 km H$s Xÿar na h¢ & {dÚwV
power is sent from one city to another epŠV EH$ eha go Xÿgao eha Vm§~o Ho$ Vma Ûmam ^oOr
city through copper wires. The fall of OmVr h¡ & `{X à{V {H$bmo_rQ>a {d^d nVZ 8 Volts
potential per km is 8 Volts and the VWm Vma H$m à{V {H$bmo_rQ>a Am¡gV à{VamoYH$Vm
average resistance per km is 0.5 Ω. 0.5 Ω h¡, Vmo Vma _| Hw$b epŠV j` (power
The power loss in the wire is loss) hmoJm
(A) 19.2 W (B) 19.2 kW (A) 19.2 W (B) 19.2 kW
(C) 19.2 J (D) 12.2 kW (C) 19.2 J (D) 12.2 kW

-6- Set-A

12. A wire bent in the form of right angled 12. EH$ Vma H$mo EH$ g_H$moU {Ì^wO PQR Ho$ AmH$ma
triangle PQR, carries a current 1A. _| _mo‹S>m OmVm h¡& {Ì^wO PQR go 1A H$s Ymam
It is placed in a region of a uniform àdm{hV hmoVr h¡ & Bgo EH$ g_mZ Mwå~H$s` joÌ
magnetic field B = 0.2 T. If PR = 1 m, B = 0.2 T _| aIm OmVm h¡ & `{X PR = 1 m
the net force on the wire is hmo, Vmo Vma na Hw$b ~b hmoJm



(A) 0 (B) 1.73 N (A) 0 (B) 1.73 N
(C) 3.46 N (D) 2.732 N (C) 3.46 N (D) 2.732 N

13. The velocity v of a particle at time t is 13. {H$gr H$U H$m g_` t na doJ v H$m _mZ
b b
given by v = at + , where a, b and c v = at + Ûmam {X`m OmVm h¡, Ohm± a, b d
t+c t+c
are constants. The dimensions of a, b c {Z`Vm§H$ h¢ & a, b d c H$s {d_mE± H«$_e: hmoJr
and c are respectively
(A) [LT–1], [L] and [T] (A) [LT–1], [L] d [T]
(B) [LT–2], [L] and [T] (B) [LT–2], [L] d [T]
(C) [LT–2], [LT–1] and [T] (C) [LT–2], [LT–1] d [T]
(D) [LT2], [LT] and [T] (D) [LT2], [LT] d [T]

14. If the energy E, velocity v and force F 14. `{X D$Om© E, doJ v d ~b F H$mo _yb^yV am{e
be taken as fundamental quantities, {b`m Om`, Vmo Ðì`_mZ H$s {d_m hmoJr
then the dimension of mass will be
(A) F v –2 (B) F v –1 (A) F v –2 (B) F v –1
(C) E v –2
­ (D) E v2 (C) E v –2
­ (D) E v2

-7- Set-A

15. 15.

As shown in above figure, a heavy D$na {XImE JE {MÌmZwgma M Ðì`_mZ dmbo EH$
block of mass M kept on a frictionless ^mar ãbm°H$ H$mo Kf©Ua{hV gVh na Xmo g_mZ
surface and being pulled by two ropes Ðì`_mZ m H$s apñg`m| go ItMm OmVm h¡ & g_`
of equal mass m. At time t = 0, the
t = 0 na ~m`t añgr go Amamo{nV ~b H$mo hQ>m {X`m
force on the left rope is withdrawn but
the force on the right end continues to OmVm h¡, O~{H$ Xm`t añgr go ~b Amamo{nV hmoVm
act. Let F1 and F2 be the magnitude of ahVm h¡ & `{X F1 d F2 H«$_e: Xm{hZr añgr d ~m`t
the forces by the right rope and the left añgr Ûmam ãbm°H$ na Amamo{nV ~b hmo, Vmo
rope on the block respectively, then
(A) F1 = F2 = F for t < 0 (A) F1 = F2 = F O~ t < 0
(B) F1 = F2 = F + mg for t < 0 (B) F1 = F2 = F + mg O~ t < 0
(C) F1 = F, F2 = F for t > 0 (C) F1 = F, F2 = F O~ t > 0
(D) F1 < F, F2 = F for t > 0 (D) F1 < F, F2 = F O~ t > 0

16. 16.

A string of length L is fixed at one end A L bå~mB© H$s EH$ S>moar H$m EH$ {gam A ~m±YH$a Xÿgao
and carries a mass M at the other {gao na M Ðì`_mZ bQ>H$m`m J`m h¡ & pñWa {gao
end. The string makes 2 revolutions go D$Üdm©Ya Aj Ho$ n[aV: S>moar 2 π MŠH$a à{V
π
per second around the vertical axis
goH|$S> _| {MÌmZwgma Ky_ ahr h¡, Vmo S>moar _| VZmd T
through the fixed end as shown in
above figure, then the tension T in the hmoJm
string is
(A) ML (A) ML
(B) 2ML (B) 2ML
(C) 4ML (C) 4ML
(D) 16ML (D) 16ML

-8- Set-A

17. A circuit consists of a battery, a resistor R 17. {MÌ _| Xem©`o J`o n[anW _| EH$ ~¡Q>ar, à{VamoY R
and two light bulbs A and B as shown Am¡a Xmo àH$me ~ë~m| A Am¡a B Ho$ gmW OwS>r h¡ &
in figure. If the filament of light bulb A
`{X ~ë~ A H$m {\$bm_oÊQ> Ob OmVm h¡, V~ {ZåZ
burns out, then which of the following
is true for light bulb B ? _| go H$m¡Z-gm H$WZ ~ë~ B Ho$ {bE gË` hmoJm ?

(A) It is turned off (A) `h ~wP Om`oJm
(B) Its brightness does not change (B) BgH$s XrpßV _| H$moB© n[adV©Z Zht hmoJm
(C) It gets dimmer (C) `h Yw§Ybm àH$me XoJm
(D) It gets brighter (D) `h A{YH$ XrßV hmo Om`oJm

18. A resistance of 4 Ω and a wire of length 18. 1 Ω AmÝV[aH$ à{VamoY Am¡a 10 dmoëQ> {dÚwV dmhH$
5 m and resistance 5 Ω are joined in ~b H$s gob Ho$ gmW loUr H«$_ _| EH$ à{VamoY
series and connected to a cell of emf 4 Ω Am¡a EH$ 5 _rQ>a bå~m Vma {OgH$m à{VamoY
10 V and internal resistance 1 Ω. A 5 Ω h¡ OwS>m h¡ & BgHo$ gmW Xmo g_mZ gobm| H$m
parallel combination of two identical g§`moOZ Vma H$s 3 _rQ>a bå~mB© na A{djon H$s
cells is balanced across 3 m of the pñW{V XoVm h¡, V~ àË`oH$ gob H$m {dÚwV dmhH$
wire. The emf E of each cell is ~b E H$m _mZ h¡
1Ω 1Ω

4Ω 10 V 4Ω 10 V
3m 3m
X X
5 Ω, 5 Ω,
E E
5m 5m
E G E G

(A) 1.5 V (B) 3.0 V (A) 1.5 dmoëQ> (B) 3.0 dmoëQ>
(C) 0.67 V (D) 1.33 V (C) 0.67 dmoëQ> (D) 1.33 dmoëQ>

-9- Set-A

19. n identical cells are joined in series 19. n EH$ g_mZ gob loUrH«$_ _| Xmo gobm|, gobm| A Am¡a
with two cells A and B with reverse B Ho$ gmW Ow‹S>m h¡, A Am¡a B gobm| H$s Y«wdUVm
polarities. EMF of each cell is E and H$mo {dnarV aIm J`m h¡ & à`wŠV àË`oH$ gob H$m
internal resistance is Ω. Potential
across cell A or B is (n > 4)
{dÚwV dmhH$ ~b E h¡ VWm AmÝV[aH$ à{VamoY Ω
h¡ & gob A `m B Ho$ {gam| na {d^d H$m _mZ hmoJm
(n > 4)

2E 2E
(A) (A)
n n
4E 4E
(B) (B)
n n
 1  1
(C) 2E  1 −  (C) 2E  1 − 
 n  n

 2  2
(D) 2E  1 −  (D) 2E  1 − 
 n  n

20. A conductor is bent in the form of 20. EH$ MmbH$ Vma H$mo {MÌ Ho$ AZwgma g§Ho$ÝÐr AÕ©
concentric semicircles as shown in the d¥Îmm| _| _mo‹S>m J`m h¡ & Bg g§`moOZ Ho$ Ho$ÝÐ O na
figure. The magnetic field at the point
O is
Mwå~H$s` joÌ H$m _mZ h¡
I I

I I

I I
I I

upto ∞ AZ§V VH$
O° O°
a 2a 4a 8a a 2a 4a 8a

µ 0i µ 0i
(A) Zero (B) (A) eyÝ` (B)
a a
µ 0i µ 0i µ 0i µ 0i
(C) (D) (C) (D)
4a 6a 4a 6a

-10- Set-A

21.
The binding energy per nucleon for a 21. S>çyQ>oam°Z Am¡a α-H$U H$s à{V Ý`ypŠbAm°Z ~ÝYZ
deuteron and α-particle are x1 and x2 D$Om© H«$_e: x1 Am¡a x2 h¢ & A{^{H«$`m
respectively. What will be the energy Q
released in the reaction 1
H2 + 1H2 → 2He4 + Q _| {ZH$br hþB© D$Om© Q
1H2 + 1H2 → 2He4 + Q ? {H$VZr hmoJr ?
(A) 4(x2 – x1) (B) 4(x1 + x2) (A) 4(x2 – x1) (B) 4(x1 + x2)
(C) 2(x1 + x2) (D) 2(x2 – x1) (C) 2(x1 + x2) (D) 2(x2 – x1)

22. Three equal weights A, B, C of mass 22. Xmo {H$bmoJ«m_ Ho$ ~am~a Ðì`_mZ Ho$ VrZ ~mQ>>
2 kg each are hanging on a string A, B, C EH$ pñWa Kf©Ua{hV {KaZr Ho$ D$na go
passing over a fixed frictionless pulley JwOaZo dmbr Vma go bQ>H$s h¡, O¡gm {H$ {MÌ _|
as shown in figure. The tension in the {XIm`m J`m h¡ & Ðì`_mZ B d C H$s ~rM Ow‹S>o
string connecting weight B and C is Vma _| VZmd hmoJm


(A) Zero (B) 13 Newton (A) eyÝ` (B) 13 Ý`yQ>Z
(C) 3.3 Newton (D) 19.6 Newton (C) 3.3 Ý`yQ>Z (D) 19.6 Ý`yQ>Z

23. A magnetic field given by 23. 1.8 go_r {ÌÁ`m Am¡a 25 Koam| dmbr d¥ÎmmH$ma
B(t) = 0.2 t – 0.05 t2 tesla (where t denotes Hw$ÊS>br {OgH$m Hw$b à{VamoY 1.5 Ω h¡, Bg
the time), is directed perpendicular to Hw$ÊS>br Ho$ Vb Ho$ bå~dV {Xem _| EH$ Mwå~H$s`
the plane of a circular coil containing joÌ B(t) = 0.2 t – 0.05 t2 Q>ogbm (t g_`
25 turns of radius 1.8 cm and whose H$mo B§{JV H$aVm h¡) Amamo{nV {H$`m J`m h¡ & g_`
total resistance is 1.5 Ω. The power t = 3 goH$ÊS> na Hw$b epŠV ì`` H$m _mZ h¡
dissipation at t = 3 s is approximately
bJ^J
(A) Zero (B) 1.37 µW (A) eyÝ` (B) 1.37 µW
(C) 7 µW (D) 4 µW (C) 7 µW (D) 4 µW

-11- Set-A

24. Three rings each of mass P and radius Q 24. VrZ db`, àË`oH$ H$m Ðì`_mZ P Ed§ {ÌÁ`m Q
are arranged as shown in figure. The h¡, {MÌ Ho$ AZwgma ì`dpñWV h¡ & yy′ Ho$ n[aV:
moment of inertia of the arrangement Bg ì`dñWm _| OS>Ëd AmKyU© hmoJm
about yy′ will be



7 5 7 5
(A) PQ 2 (B) PQ 2 (A) PQ 2 (B) PQ 2
2 2 2 2
2 2 2 2
(C) PQ 2 (D) PQ 2 (C) PQ 2 (D) PQ 2
7 5 7 5

25. 25.

A particle originally at rest at the EH$ {MH$Zo D$Üdm©Ya d¥Îm na EH$ H$U pñWa AdñWm
highest point of a smooth verticle circle go CÀMV_ {~ÝXþ go WmoS‹ >m {dñWm{nV {H$`m OmVm h¡ &
is slightly displaced. It will leave the D$Üdm©Ya d¥Îm H$s gVh N>mo‹S>Vo dŠV CÀMV_ {~ÝXþ
circle at a vertical distance h below the go ZrMo D$Üdm©Ya D±$MmB© h H$m _mZ hmoJm
highest point such that (R is the radius (d¥Îm H$s {ÌÁ`m R h¡)
of circle)
R R
(A) h = R (B) h = (A) h = R (B) h =
3 3
R 2R R 2R
(C) h = (D) h = (C) h = (D) h =
2 3 2 3

-12- Set-A

26. 26.

For frictionless surface, the acceleration Kf©Ua{hV gVh Ho$ {bE 500 g ãbm°H$$ H$m ËdaU
of the 500 g block is hmoJm
8g 8g
(A) upward (A) D$na H$s Amoa
13 13
13 13
(B)
8g
upward (B)
8g
D$na H$s Amoa
13 13
(C)
8g
downward (C)
8g
ZrMo H$s Amoa
8g 8g
(D) downward (D) ZrMo H$s Amoa
13 13

27. A projectile is projected with velocity 27. EH$ àjoß` H$mo gVh go AÝV[aj _| nve doJ go
nve in vertically upward direction from D$Üdm©Ya D$na H$s Amoa àjo{nV {H$`m OmVm h¡,
the ground into the space (ve is escape (Ohm± ve nbm`Z doJ d n < 1 h¡) & dm`w à{VamoY
velocity and n < 1). If air resistance is
considered to be negligible, then the
H$mo ZJÊ` _mZVo hþE n¥Ïdr Ho$ Ho$ÝÐ go àjoß` Ûmam
maximum height from the centre of àmßV A{YH$V_ D±$MmB© hmoJr (Ohm± R n¥Ïdr H$s
earth to which it can go, will be (R is {ÌÁ`m h¡)
the radius of earth)
Rn2 Rn2
(A) (A)
1 − n2 1 − n2
R R
(B) 2 (B) 2
n −1 n −1
R R
(C) (C)
1 − n2 1 − n2
R R
(D) (D)
n+1 n+1

-13- Set-A

28. The correct graph representing the 28. EH$ CnJ«h H$s n¥Ïdr Ho$ Ho$ÝÐ go Xÿar (r) Ho$ gmW Hw$b
variation of total energy (E), kinetic D$Om© (E), J{VO D$Om© (EK) d pñW{VO D$Om© (U)
energy (EK) and potential energy (U) H$m n[adV©Z {ZåZ _| go H$m¡Z-go ghr J«m\$$ Ûmam
of a satellite with distance (r) from the
centre of earth is
àX{e©V h¡ ?

(A) (A)

(B) (B)

(C) (C)

(D) (D)

-14- Set-A

29. One end of the metal rod is kept 29. YmVw H$s EH$ N>‹S> H$m EH$ {gam ^Å>r _| aIm OmVm
in a furnace. In steady state, the h¡ & gmå`mdñWm _| N>‹S> H$m Vmn
temperature of the rod
(A) Increases (A) ~‹T>oJm
(B) Decreases (B) KQ>oJm
(C) Remains constant (C) {Z`V ahoJm
(D) is non-uniform (D) Ag_mZ hmoJm
 dv 
− 
30. Which of the following graph correctly 30. {Z`V Vmn na AmXe© J¡g Ho$ {bE β =  dp 
V
 dv 
−  H$m Xm~ P Ho$ gmW n[adV©Z {ZåZ _| go {H$g J«m\$
 dp 
represents the variation of β = Ho$ Ûmam ghr àX{e©V hmoJm ?
V
with P for an ideal gas at constant
temperature ?

(A) (A)

(B) (B)

(C) (C)

(D) (D)

-15- Set-A

31. A cyclic process ABCD is shown in the 31. EH$ MH«$s` àH«$_ ABCD H$mo PV AmaoI _| àX{e©V
PV diagram. Which of the following PT {H$`m J`m h¡ & {ZåZ _| go H$m¡Z-gm PT dH«$ BgHo$
curves represent the same process ? g_mZ àH«$_ H$mo àX{e©V H$aoJm ?



(A) (A)

(B) (B)

(C) (C)

(D) (D)

-16- Set-A

32. A hot liquid is kept in a big room. Its 32. EH$ J_© Ðd H$mo EH$ ~‹S> o H$_ao _| aIm OmVm h¡ &
temperature is plotted as a function CgHo$ Vmn H$mo g_` Ho$ \$bZ Ho$ ê$n _| Ambo{IV
of time. Which of the following curves {H$`m J`m h¡ & {ZåZ _| go H$m¡Z-gm dH«$ Cg AmboI
may represents the plot ?
H$mo àX{e©V H$a gH$Vm h¡ ?


(A) (a) (B) (b) (A) (a) (B) (b)
(C) (c) (D) None of the above (C) (c) (D) Cn`w©ŠV _| go H$moB© Zht

33. 33.

From above T-P graph, what conclusion Cn`w©ŠV T-P J«m\$ go Š`m {ZîH$f© {ZH$mbm Om
can be drawn ? gH$Vm h¡ ?
(A) V2 > V1 (A) V2 > V1
(B) V2 < V1 (B) V2 < V1
(C) V2 = V1 (C) V2 = V1
(D) Nothing can be predicted (D) Hw$N> H$hm Zht Om gH$Vm

34. The root mean square speed of the 34. EH$ {Ûna_mpÊdH$ J¡g Ho$ AUwAm| H$s _yb dJ©
molecules of a diatomic gas is v. _mÜ` Mmb v h¡ & Vmn_mZ XþJwZm H$aZo na AUw Xmo
When the temperature is doubled, the na_mUwAm| _| {d^ŠV hmo OmVm h¡, Vmo na_mUw H$s
molecules dissociate into two atoms.
The new root mean square speed of
ZB© _yb dJ© _mÜ` Mmb hmoJr
the atom is
(A) v (B) v (A) v (B) v
(C) 2v (D) 4v (C) 2v (D) 4v

-17- Set-A

35. There is a horizontal film of soap 35. gm~wZ Ho$ Kmob H$s EH$ j¡{VO {\$ë_ h¡, {Og na
solution. On it, a thread is placed in the EH$ YmJm byn Ho$ ê$n _| aIm OmVm h¡ & {\$ë_ H$mo
form of a loop. The film is pierced inside byn Ho$ AÝXa {N>{ÐV H$a {X`m OmVm h¡ Am¡a YmJm R
the loop and the thread becomes a
circular loop of radius R. If the surface
{ÌÁ`m H$m d¥ÎmmH$ma byn ~Zm boVm h¡ & `{X byn H$m
tension of the loop is T, then what will n¥îR> VZmd T hmo, Vmo YmJo _| VZmd H$m _mZ {H$VZm
be the tension in the thread ? hmoJm ?
πR2 πR2
(A) (B) πR2T (A) (B) πR2T
T T
(C) 2πRT (D) 2RT (C) 2πRT (D) 2RT

36. The adiabatic bulk modulus of an ideal 36. Xm~ P na AmXe© J¡g Ho$ {bE éÕmoî_ ~ëH$ JwUm§H$
 Cp   Cp 
gas at pressure P is  γ = hmoVm h¡  γ =
 Cv   C 
v

(A) P (B) γP (A) P (B) γ P
P P P P
(C) (D) (C) (D)
2 γ 2 γ

37. A Satellite of mass m revolves around 37. EH$ CnJ«h {OgH$m Ðì`_mZ m h¡, R {ÌÁ`m Ho$
the earth of radius R at a height x from n¥Ïdr Ho$ gVh go x D±$MmB© na n¥Ïdr H$m n[a^«_U H$a
its surface. If g is the acceleration due ahm h¡ & `{X g n¥Ïdr H$s gVh na JwéËdr` ËdaU
to gravity on the surface of the earth, h¡, Vmo CnJ«h H$m H$jr` Mmb hmoJm
then orbital speed of the satellite is
(A) gx (A) gx

gR gR
(B) (B)
(R − x ) (R − x )

gR2 gR2
(C) (C)
(R − x ) (R − x )
gR2 gR2
(D) (D)
(R + x ) (R + x )

-18- Set-A

38. A metallic sphere cools from 50°C 38. EH$ YmVw Ho$ Jmobm H$mo 50°C go 40°C VH$ R>§S>m
to 40°C in 300 seconds. If the hmoZo _| 300 goH$ÊS> bJVm h¡& `{X H$_ao H$m Vmn
room temperature is 20°C, then its 20°C h¡, Vmo AJbo 5 {_ZQ> _| CgH$m Vmn hmoJm
temperature in next 5 minutes will be
(A) 30°C (A) 30°C
(B) 33.3°C (B) 33.3°C
(C) 36.3°C (C) 36.3°C
(D) 38°C (D) 38°C

39. A convex lens is made up of three 39. {MÌ _| {XImE AZwgma EH$ CÎmb b|g VrZ {^Þ
different materials as shown in the nXmWm] go {_bH$a ~Zm h¡& EH$ {~ÝXþ AmH$ma H$s dñVw
figure. for a point object placed on b|g Ho$ Aj na aIr OmVr h¡, Vmo b|g go ~ZZo dmbo
its axis, the number of images formed à{V{~å~m| H$s g§»`m hmoJr
are


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

40. The critical wavelength for producing 40. Q>J§ ñQ>Z YmVw go àH$me {dÚwV à^md CËnÞ H$aZo Ho$ {bE
the photoelectric effect in tungsten H«$mpÝVH$ Va§JX¡¿`© 2600 Å h¡& Q>§JñQ>Z go 2200 Å
metal is 2600 Å. What wavelength
would be necessary to produce
go CËnÝZ J{VO D$Om© go XþJwZr J{VO D$Om© Ho$ {dÚwV
photoelectrons from tungsten having BboŠQ´>m°Z CËg{O©V H$amZo Ho$ {bE {H$g Va§JX¡¿`© H$m
twice the kinetic energy of those àH$me Amdí`H$ hmoJm ?
produced at 2200 Å ?
(A) 1800 Å (A) 1800 Å
(B) 1907 Å (B) 1907 Å
(C) 1926 Å (C) 1926 Å
(D) 2015 Å (D) 2015 Å

-19- Set-A

41. What focal length should be reading 41. EH$ ì`pŠV Ho$ arqS>J Mí_o H$s \$moH$g Xÿar {H$VZr
spectacles have for a person whose hmoJr, `{X CZH$m ZOXrH$ H$m q~Xþ 50 go_r h¡ ?
near point is 50 cm ?
(A) 25 cm (A) 25 go_r
(B) 30 cm (B) 30 go_r
(C) 40 cm (C) 40 go_r
(D) 50 cm (D) 50 go_r

42. The ratio of resolving power of an 42. Xmo Va§JX¡¿`m] λ1 = 4000 Å VWm λ2 = 6000 Å Ho$
optical microscope for two wavelengths {b`o {H$gr àH$mer` gyú_Xeu Ho$ {d^oXZ j_VmAm|
λ1 = 4000 Å and λ2 = 6000 Å is H$m AZwnmV hmoJm
(A) 9 : 4 (B) 4 : 9 (A) 9 : 4 (B) 4 : 9
(C) 3 : 2 (D) 2 : 3 (C) 3 : 2 (D) 2 : 3

43. The momentum of a photon is 43. EH$ \$moQ>mZ H$m g§dJo 3.3 × 10–29 kg-m/sec h¡ &
3.3 × 10–29 kg-m/sec. Its frequency BgH$s Amd¥{Îm hmoJr
will be
(A) 3 × 103 Hz (A) 3 × 103 Hz
(B) 6 × 103 Hz (B) 6 × 103 Hz
(C) 7.5 × 1012 Hz (C) 7.5 × 1012 Hz
(D) 1.5 × 1013 Hz (D) 1.5 × 1013 Hz

44. The Bohr model of atom 44. na_mUw H$m ~moa _m°S>b
(A) Assumes that the angular (A) Bg VÏ` na AmYm[aV h¡ {H$ BboŠQ´>m°Z H$m
momentum of electron is H$moUr` g§doJ Šdm§Q>rH¥$V h¡
quantized
(B) Uses Einstein’s photoelectric (B) AmB§ñQ>rZ Ho$ àH$me {dÚwV à^md H$m Cn`moJ
effect H$aVm h¡
(C) Predicts continuous emission (C) na_mUw Ho$ gVV CËgO©Z dU©H«$_ Ho$ ~mao _|
spectra for atoms OmZH$mar XoVm h¡
(D) Predicts the same emission (D) `h ~VmVm h¡ {H$ g^r na_mUwAm| H$m EH$ hr
spectra for all types of atoms Vah H$m CËgO©Z dU©H«$_ hmoVm h¡

-20- Set-A

45. Two coherent monochromatic light 45. Xmo H$bm g§~Õ EH$dUu àH$me {H$aU {OgH$s
beams of intensities I and 4I are Vrd«VmE± I Ed§ 4I h¡, H$mo AÜ`mamo{nV {H$`m OmVm
superposed. The maximum and h¡& n[aUm_r {H$aU H$m A{YH$V_ Ed§ Ý`yZV_ g§^d
minimum possible intensities in the
resulting beam are respectively
Vrd«VmE± H«$_e: hmoJr
(A) 5I and I (A) 5I Ed§ I
(B) 9I and I (B) 9I Ed§ I
(C) 5I and 3I (C) 5I Ed§ 3I
(D) 9I and 3I (D) 9I Ed§ 3I

46. An object is placed between two plane 46. Xmo g_Vb Xn©U EH$ Xÿgao go 60° H$moU _| PwHo$ hþE
mirrors inclined at 60° to each other. h¡, Ho$ ~rM _| EH$ dñVw aIm OmVm h¡, Vmo Amn {H$VZo
How many images do you expect to à{V{~å~ XoI gH$Vo h¢ ?
see ?
(A) 5 (A) 5
(B) 6 (B) 6
(C) 7 (C) 7
(D) 9 (D) 9

47. The focal length of a plane convex 47. EH$ g_VbmoÎmb b|g H$s \$moH$g Xÿar Š`m hmoJr,
lens, when radius of curved surface is `{X dH«$s` gVh H$s {ÌÁ`m 15 go_r Ed§
15 cm and µ = 1.5 µ = 1.5 h¡ ?
(A) 15 cm (A) 15 go_r
(B) 20 cm (B) 20 go_r
(C) 30 cm (C) 30 go_r
(D) 45 cm (D) 45 go_r

48. If the refractive index of a material of 48. `{X g_^wO {àÁ_ Ho$ nXmW© H$m AndV©Zm§H$ 3
equilateral prism is 3 , then the angle h¡, Vmo {àÁ_ H$m Ý`yZV_ {dMbZ H$moU h¡
of minimum deviation of prism is
(A) 75° (A) 75°
(B) 60° (B) 60°
(C) 45° (C) 45°
(D) 30° (D) 30°

-21- Set-A

49. A parallel plate capacitor is connected 49. EH$ g_mÝVa ßboQ> g§Ym[aÌ H$mo EH$ ~¡Q>ar go Omo‹S>m
to a battery. The plates are pulled J`m h¡ & Bg g§Ym[aÌ H$s ßboQ>m| H$mo EH$ g_mZ Mmb
apart with a uniform speed. If x is go ~mha H$s Amoa ItMm OmVm h¡ & `{X BgH$s ßboQ>m|
the separation between the plates, Ho$ ~rM H$s Xÿar x h¡, V~ Bg g§Ym[aÌ H$s d¡ÚwV
then the time rate of change of the
ñW¡{VH$ D$Om© _| n[adV©Z H$s g_` Xa AZwH$« _mZwnmVr
electrostatic energy of the capacitor
is proportional to
hmoVr h¡
(A) x2 (A) x2 Ho$
(B) x (B) x Ho$
1 1
(C) (C) Ho$
x2 x2
1 1
(D) (D) Ho$
x x

50. In the given circuit, a charge of + 80 µc 50. {MÌ _| {X`o J`o n[anW _| 4 µF Ym[aVm Ho$ g§Ym[aÌ
is given to the upper plate of the H$s D$nar ßboQ> H$mo + 80 µC H$m Amdoe {X`m OmVm
4 µF capacitor. Then in the steady h¡ & ñWm`r AdñWm _| 3 µF Ym[aVm dmbo g§Ym[aÌ
state, the charge on the upper plate H$s D$nar ßboQ> na Amdoe H$m _mZ h¡
of the 3 µF capacitor is

(A) + 32 µC (B) + 40 µC (A) + 32 µC (B) + 40 µC
(C) + 80 µC (D) + 48 µC (C) + 80 µC (D) + 48 µC

-22- Set-A

51. A parallel plate condenser contains a 51. {H$gr g_mÝVa nÅ> g§Km[aÌ _| EH$ naV A^«H$ H$s
mica sheet (thickness 10–3 m) and a (_moQ>mB© 10–3 _r) Am¡a EH$ naV {H$gr AÝ` \$mB~a
sheet of a fibre (thickness 0.5 × 10–3 m). nXmW© (_moQ>mB© 0.5 × 10–3 _r) H$s bJr h¡ & A^«H$
The dielectric constant of mica is 8 and VWm \$mB~a Ho$ namd¡ÚVw m§H$ H«$_e: 8 VWm 2.5 h¢ & `h
that of fibre is 2.5. Assuming that the _mZH$a {H$ \$mB~a 6.4 × 106 dmoëQ>/_r H$m {dÚwV
fibre breaks down when subjected to joÌ bJZo na ^§J hmo OmVm h¡ & V~ Bg g§Km[aÌ na
an electric field of 6.4 × 106 volt/meter. bJZo dmbr gwa{jV A{YH$V_ dmoëQ>Vm H$m _mZ hmo
Then the maximum safe voltage that
gH$Vm h¡
can be applied to condenser is

(A) 5200 volts (A) 5200 dmoëQ>
(B) 2600 volts (B) 2600 dmoëQ
(C) 6800 volts (C) 6800 dmoëQ
(D) 6400 volts (D) 6400 dmoëQ

52. Figure shows a network of eight 52. {MÌ _| Xem©`m J`m n[anW 8 g_mZ à{VamoYm| (àË`oH$
resistor, each equal to 2 Ω, connected 2 Ω) go {_bH$a ~Zm EH$ ZoQ>dH©$ ZJÊ` AmÝV[aH$
to a 3 V battery of negligible internal à{VamoY H$s 3 V ~¡Q>ar go OwS‹ >m h¡ & n[anW _| Ymam I
resistance. The current I in the circuit is H$m _mZ h¡

(A) 0.25 A (A) 0.25 Epån`a
(B) 0.50 A (B) 0.50 Epån`a
(C) 0.75 A (C) 0.75 Epån`a
(D) 1.0 A (D) 1.0 Epån`a

-23- Set-A

53.
Two long current carrying thin wires, 53. Xmo Ymamdmhr Vma {Og_| Ymam I ~h ahr h¡ & BZ
both with current I, are held by Ymamdmhr Vmam| H$mo L bå~mB© Ho$ Hw$MmbH$ YmJm|
insulating threads of length L and are Ûmam {MÌ _| Xem©B© JB© pñW{V Ho$ AZwgma gÝVw{bV
in equilibrium as shown in the figure,
with threads making an angle θ with
AdñWm _| aIm J`m h¡ & gÝVwbZ H$s pñW{V _| Ymam,
the vertical. If wires have mass λ per D$Üdm©Ya Ho$ gmW θ H$moU ~ZmVm h¡ & `{X Vma H$m
unit length, then the value of I is à{V BH$mB© bå~mB© H$m Ðì`_mZ λ hmo, V~ Ymam I
(g is gravitational acceleration) H$m _mZ h¡ (g JwéËdr` ËdaU h¡)

πgL tan θ πgL tan θ
(A) 2 (A) 2
µ0 µ0

πλgL tan θ πλgL tan θ
(B) (B)
µ0 µ0
πλgL πλgL
(C) sinθ (C) sinθ
µ 0 cos θ µ 0 cos θ
πλgL πλgL
(D) 2 sinθ (D) 2 sinθ
µ 0 cos θ µ 0 cos θ

54. A solenoid of 0.4 m length with 500 54. 0.4 _rQ>a bå~mB© VWm 500 Koam| dmbr n[aZm{bH$m
turns carries a current of 3 A. A coil of go 3 Epån`a H$s Ymam ~h ahr h¡ & EH$ 0.01 _rQ>a
10 turns and of radius 0.01 m carries a {ÌÁ`m VWm 10 Koam| dmbr Hw$ÊS>br go 0.4 Epån`a
current of 0.4 A. The torque required to H$s Ymam ~h ahr h¡ & Bg Hw$ÊS>br H$mo n[aZm{bH$m
hold the coil with its axis at right angle
to that of solenoid in the middle point Ho$ _Ü` {~ÝXþ na Aj Ho$ bå~dV² pñW{V _| ~Zm`o
of it is aIZo Ho$ {bE Amdí`H$ ~b AmKyU© H$m _mZ h¡
(A) 6π2 ×10–7 Nm (A) 6π2 ×10–7 Ý`yQ>Z _rQ>a
(B) 3π2 ×10–7 Nm (B) 3π2 ×10–7 Ý`yQ>Z _rQ>a
(C) 9π2 ×10–7 Nm (C) 9π2 ×10–7 Ý`yQ>Z _rQ>a
(D) 12π2 ×10–7 Nm (D) 12π2 ×10–7 Ý`yQ>Z _rQ>a

-24- Set-A


55. A dipole of magnetic moment

55. Mwå~H$s` AmKyU© m = 30j Epån`a 
_r2 dmbo
m = 30j Am2 is placed along the y-axis EH$ {ÛY«wd H$mo EH$ Mwå~H$s` joÌ B = 2i + 5j
in a uniform magnetic field B = 2i + 5j . _| y-Aj Ho$ AZw{Xe aIm J`m h¡ & Bg {ÛY«wd na
The torque acting on it is bJZo dmbo ~b AmKyU© H$m _mZ h¡
(A) – 60 k Nm (A) – 60 k Ý`yQ>Z _rQ>a
(B) – 50 k Nm (B) – 50 k Ý`yQ>Z _rQ>a
(C) – 70 k Nm (C) – 70 k Ý`yQ>Z _rQ>a
(D) 50 k Nm (D) 50 k Ý`yQ>Z _rQ>a

56. An ammeter has resistance R0 and 56. EH$ A_rQ>a H$m à{VamoY R0 VWm namg I Epån`a
range IA. What resistance should be h¡ & Bg A_rQ>a Ho$ g_mÝVa H«$_ _| {H$VZm à{VamoY
connected in parallel with it to increase Omo‹S>Zm n‹S>oJm {Oggo BgH$s namg nI hmo Om`o ?
its range to nI ?
R0 R0
(A) (A)
n n
R0 R0
(B) (n − 1) (B) (n − 1)

R0 R0
(C) (C)
(n + 1) (n + 1)

(D) None of these (D) BZ_| go H$moB© Zht

57. Solar energy is mainly caused due to 57. gm¡a D$Om© H$m _w»` H$maU h¡
(A) Burning of hydrogen in oxygen (A) hmBS´>moOZ H$m Am°ŠgrOZ _| ObZo Ho$ H$maU
(B) Fission of uranium present in the (B) gy`© _| CnpñWV `yao{Z`_ Ho$ {dIÊS>Z Ho$
sun H$maU
(C) Fusion of protons during synthesis (C) àmoQ>mZ H$m g§b`Z H$a ^mar Ad`d ~ZmZo Ho$
of heavier elements H$maU
(D) Gravitational contraction (D) JwéËd Ho$ H$maU {gHw$‹S>Zo na

-25- Set-A

58. A sample of radioactive element has 58. {H$gr jU t = 0 na ao{S>`moEpŠQ>d nXmW© H$m Ðì`_mZ
a mass of 5 g at an instant t = 0. The 5 g h¡ & Xmo Am¡gV AÕ© H$mb Am`w Ho$ níMmV² Cg
approximate mass of this element in nXmW© H$m Ðì`_mZ hmoJm
sample after two mean half life period is
(A) 3.15 gm (A) 3.15 gm
(B) 1.85 gm (B) 1.85 gm
(C) 1.25 gm (C) 1.25 gm
(D) 0.67 gm (D) 0.67 gm

59. In an n-type silicon, which of the 59. n-àH$ma Ho$ {g{bH$m°Z Ho$ {b`o H$m¡Z-gm H$WZ
following statement is true ? ghr h¡ ?
(A) Electrons are majority carriers and (A) BboŠQ´>m°Z ~hþg§»`H$ AmdoedmhH$ h¡ VWm
trivalent atoms are the dopant {Ìg§`moOr nXmW© H$m An{_lU {H$`m OmVm h¡
(B) Electrons are minority carriers (B) BboŠQ´>mZ° Aëng§»`H$ AmdoedmhH$ h¡ VWm n§M
and pentavalent atoms are the g§`moOr nXmW© H$m An{_lU {H$`m OmVm h¡
dopants
(C) Holes are minority carriers (C) hmob Aëng§»`H$ AmdoedmhH$ h¡ VWm n§M
and pentavalent atoms are the g§`moOr nXmW© H$m An{_lU {H$`m OmVm h¡
dopants
(D) Holes are the majority carriers and hmob ~hþg»§ `H$ AmdoedmhH$ h¡ VWm
(D)
the trivalent atoms are dopants {Ìg§`moOr nXmW© H$m An{_lU {H$`m OmVm h¡

60. The barrier potential of a p-n junction 60. p-n g§{Y S>m`moS> H$m amo{YH$m {d^d {Z^©a H$aVm h¡
depends on
1. Types of semiconductor material AÕ© MmbH$ nXmW© na
1.
2. Amount of doping An{_lU H$s _mÌm na
2.
3. Temperature Vmn na
3.
Which one of the following is correct ? {ZåZ _| go H$m¡Z-gm ghr h¡ ?
(A) 1 and 2 only (A) Ho$db 1 Am¡a 2
(B) 2 only (B) Ho$db 2
(C) 2 and 3 only (C) Ho$db 2 Am¡a 3
(D) 1, 2 and 3 (D) 1, 2 Am¡a 3

-26- Set-A

PART – II
Chemistry agm`Z emñÌ
61. The major product of the following 61. {ZåZ A{^{H«$`m H$m _w»` CËnmX h¡
reaction :
(i) Hg(OAc )2 / THF − H2O (i) Hg(OAc ) / THF − H O
C H CH = CH − CH  2 2
C6H5CH = CH − CH3  → 6 5 3 →
(ii) NaBH4 (ii) NaBH4

(A) (A)

(B) (B)

(C) (C)

(D) (D)

62. Phenol reacts with PCl5 to give 62. \$sZm°b PCl5 Ho$ gmW A{^{H«$`m H$m _w»` ê$n go
mainly XoVm h¡
(A) Triphenyl phosphate (A) Q´>mB{\$ZmBb \$m°ñ\o$Q>
(B) Chlorobenzene (B) Šbmoamo~oÝOrZ
(C) P-chlorophenol (C) P-Šbmoamo\$sZm°b
(D) O- and P-chlorophenol (D) O- Am¡a P-Šbmoamo\$sZm°b

63. 63.

Here is `hm±
h¡
(A) Glycolic acid (A) ½bmBH$mo{bH$ Aåb
(B) Succinic acid (B) gpŠg{ZH$ Aåb
(C) Malonic acid (C) _obmo{ZH$ Aåb
(D) α-hydroxypropionic acid (D) α-hmBS´>mŠgràmonrAmo{ZH$ Aåb

-27- Set-A

64. 64.



The compound is `m¡{JH$ h¡

(A) (B) R – C ≡ N (A) (B) R – C ≡ N
(C) RCH2OH (D) RCHO (C) RCH2OH (D) RCHO

65. What is the major product (A) ? 65. _w»` CËnmX (A) Š`m h¡ ?

CH3 − CH2 − C
|
= CH − CH3 CH3 − CH2 − C
|
= CH − CH3

C2H5 C2H5

(i) Hg(OAc ) / H O (i) Hg(OAc ) / H O

2 2
→ A 
2 2
→ A
(ii) NaBH4 (ii) NaBH4

O O
|| ||
(A) CH3 − CH2 − CH − C − CH3 (A) CH3 − CH2 − CH − C − CH3
| |
C2H5 C2H5

(B) CH3 − CH2 − CH − CH − CH3 (B) CH3 − CH2 − CH − CH − CH3
| | | |
OH C2H5 OH C2H5

OH OH
| |
(C) CH3 − CH2 − CH − CH − CH3 (C) CH3 − CH2 − CH − CH − CH3
| |
C2H5 C2H5
(D) (C2H5)3COH (D) (C2H5)3COH

-28- Set-A

66. In the presence of Ni(CN)2, acetylene 66. Ni(CN)2> H$s CnpñW{V _|, Egr{Q>brZ XoVm h¡
gives
(A) cyclooctatetraene (A) gmBŠbmoAmŠQ>mQ>oQ´>mBZ
(B) cyclohexatriene (B) gmBŠbmohoŠgmQ´>mBZ
(C) benzene (C) ~oÝOrZ
(D) cyclobutadiene (D) gmBŠbmoã`yQ>mS>mBZ
67. The insecticide gammexane is 67. H$sQ>mUwZmeH$ J¡_mŠgoZ h¡
(A) Hexachlorocyclobenzene (A) hoŠgmŠbmoamogmBŠbmo~|OrZ
(B) Y-BHC (B) Y-BHC
(C) Lindane (C) {bÊS>oZ
(D) All of the above (D) CnamoŠV g^r

68. Consider the reaction : 68. {ZåZ A{^{H«$`m na {dMma H$s{O`o :

Ortho and Para-isomers can be AmWm} Ed§ n¡am-g_md`dr H$mo n¥WH$ {H$`m Om
separated by gH$Vm h¡
(A) Fractional distillation (A) à^mJr AmgdZ Ûmam
(B) Fractional crystallization (B) à^mJr {H«$ñQ>brH$aU Ûmam
(C) Steam distillation (C) ^mn AmgdZ
(D) Sublimation (D) CÜd©nmVZ

69. Polythene is 69. nm°{bWrZ h¡
(A) Cross linked copolymer (A) H«$mg qbŠS> H$monm°{b_a
(B) Alternate copolymer (B) AëQ>aZoQ> H$monm°{b_a
(C) Homo polymer (C) hmo_mo nm°{b_a
(D) Random copolymer (D) a¡ÊS>_ H$monm°{b_a

-29- Set-A

70. In which of the following cationic 70. Aåb Ho$ Ûmam àmaå^ {H$`m J`m YZm`{ZV `moJmË_H$
addition polymerization reaction ~hþbrH$aU A{^{H«$`m {ZåZ{b{IV _| go {H$g_| g~go
initiated by an acid is maximum ? A{YH$ hmoJm ?
CH3 CH3
| |
(A) CH3 − C
|
− CH3 (A) CH3 − C − CH3
|
CH3 CH3
CH3 CH3
| |
(B) CH3 − C = CH2 (B) CH3 − C = CH2
(C) CH3CH = CH2 (C) CH3CH = CH2
(D) CH2 = CH2 (D) CH2 = CH2

71. Artificial sweetener which is stable 71. H¥${Ì_ ñdrQ>Za Omo Ho$db R>§S>r pñW{V _| ñWm`r
under cold conditions only is hmoVm h¡
(A) Saccharin (B) Sucralose (A) g¡Ho$[aZ (B) gwH«$mbmog
(C) Aspartame (D) Alitame (C) EgnmaQ>o_ (D) E{bQ>o_

72. Which of the following best represents 72. jmar` _mÜ`_$ (pH = 11) _| EH$ A_rZmo Aåb H$s
the structure of an amino acid in basic g§aMZm {ZåZ _| go H$m¡Z-gm gdm©{YH$ ghr àX{e©V
medium (pH = 11) ? H$aVm h¡ ?
COOH COOH
| ⊕ | ⊕
(A) R − CH − NH3 (A) R − CH − NH3

(B) (B)

(C) (C)

COOH COOH
| |
(D) R − CH − NH2 (D) R − CH − NH2

-30- Set-A

73. Ferric ion forms a prussian blue ppt./ 73. ________ Ho$ JR>Z Ho$ H$maU \o$[aH$ Am`Z EH$
colouration due to the formation of à{e`m Zrbm nrnrQ>r/a§J ~ZmVm h¡ &
(A) K4[Fe(CN)6] (B) Fe4[Fe(CN)6] (A) K4[Fe(CN)6] (B) Fe4[Fe(CN)6]
(C) KMnO4 (D) Fe(OH)3 (C) KMnO4 (D) Fe(OH)3

74. Which of the following acids does not 74. {ZåZ _| go H$m¡Z-gm Aåb {Ì{d_$ g_md`dVm
exhibit stereoisomerism ? àX{e©V Zht H$aVm ?
(A) Fumaric acid (A) â`y_o[aH$ Aåb
(B) Lactic acid (B) bopŠQ>H$ Aåb
(C) Citric acid (C) gmB{Q´>H$ Aåb
(D) Tartaric acid (D) Q>mQ>©[aH$ Aåb

75. IUPAC name of Allyl carbinol is 75. EbmBb H$m{~©Zmb H$m AmB© `y nr E gr Zm_ h¡
(A) prop-2-en-1-al (A) prop-2-en-1-al
(B) but-3-en-1-al (B) but-3-en-1-al
(C) but-3-en-1-ol (C) but-3-en-1-ol
(D) prop-2-en-1-ol (D) prop-2-en-1-ol

76. Which of the following fractions of 76. {ZåZ _| go {H$g noQ´>mo{b`_ à^mJ H$m ŠdWZm§H$
petroleum has the lowest boiling Ý`yZV_ h¡ ?
point ?
(A) Kerosene (B) Heavy oil (A) {H$amogrZ (B) ^mar Vob
(C) Gasoline (D) Diesel oil (C) J¡gmobrZ (D) S>rOb Vob

77. The edge length of the unit cell of LiCl 77. LiCl H$s EH$H$ gob Ho$ {H$Zmao H$s b§~mB© 0.514 nm
is 0.514 nm. Assuming that the lithium h¡ & `h _mZVo hþE {H$ {b{W`_ Am`Z BVZm N>mQo >m h¡
ion is small enough so that chloride {H$ ŠbmoamBS> Am`Z g§nH©$ _| hmoVo h¢, ŠbmoamBS> Am`Z
ions are in contact, the ionic radius for H$s Am`{ZH$ {ÌÁ`m {H$VZr hmoJr ?
the chloride ion is
(A) 0.257 nm (A) 0.257 nm
(B) 1.33 Å (B) 1.33 Å
(C) 0.182 nm (C) 0.182 nm
(D) None of the above (D) Cn`w©ŠV _| go H$moB© Zht

-31- Set-A

78. Consider the following statements : 78. {ZåZ H$WZm| na {dMma H$s{O`o :
I. The coordination number of Cs+ I. CsCl _| Cs+ Am`Z H$s g_Ýd` g§»`m
ion in CsCl is 8. 8 h¡ &
II. A metal crystallizes in b.c.c. structure II. b.c.c. g§aMZm _| {H«$ñQ>brH¥$V EH$ YmVw H$s
has coordination number 12. g_Ýd` g§»`m 12 hmoVm h¡ &
III. A unit cell of an ionic crystal III. Am`{ZH$ {H«$ñQ>b Ho$ EH$H$ gob Ho$ Am`Z Xÿgao
shares some of its ions with ions EH$H$ gob Ho$ Am`Zm| Ho$ gmW ^mJ boVo h¢ &
of other unit cell.
IV. The radius ratio in octahedral AîQ>\$bH$s` ~ÝX {Z{d‹S> _| {ÌÁ`mZwnmV
IV.
close packing is 0.225. 0.225 hmoVm h¡ &
Correct statements are : ghr H$WZ h¢ :
(A) I and III (B) I and IV (A) I Ed§ III (B) I Ed§ IV
(C) II and III (D) III and IV (C) II Ed§ III (D) III Ed§ IV

79. Among the following nuclear reactions 79. {ZåZ Zm{^H$s` {H«$`mAm| Ho$ _Ü` H$m¡Z EH$ H¥${Ì_
which is not a nuclear reaction of VËdm§VaU Zm{^H$s` {H«$`m Zht h¡ ?
artificial transmutation ?
14 4 17 1 14 4 17 1
(A) 7 N + 2He → 8 O + 1H (A) 7 N + 2He → 8 O + 1H

9 4 12 1 9 4 12 1
(B) 4 Be + 2He → 6 C + 0n (B) 4 Be + 2He → 6 C + 0n

7 1 7 1 7 1 7 1
(C) 3 Li + 1H → 4Be + 0n
(C) 3 Li + 1H → 4Be + 0n

27 1 28 27 1 28
(D) 13 Al + 0n → 13 Al + y (D) 13 Al + 0n → 13 Al + y

80. The decay time t for a radioactive 80. EH$ ao{S>`moEpŠQ>d VËd H$m j` g_` t, 4 AY© Am`w
element proceeds to 4 half lives. The VH$ hmoVm h¡ & j` g_` Ed§ Am¡gV Am`w Ta Ho$ _Ü`
decay time and average life T a is gå~ÝY h¡
related by
(A) t = 2Talog2 (B) t = 4Talog2 (A) t = 2Talog2 (B) t = 4Talog2
1 1
(C) t = 2 Ta4 log 2 (D) t = log 2 (C) t = 2 Ta4 log 2 (D) t = log 2
Ta2 Ta2

-32- Set-A

81. The first (IE1) and the second (IE2) 81. Mma VËdm| H$m àW_ (IE1) Ed§ {ÛVr` (IE2) Am`ZZ
ionisation energy of four elements are D$Om© H$m _mZ {ZåZmZwgma h¡ :
shown below :
Element IE1 IE2 VËd IE1 IE2
a 2372 5251 a 2372 5251
b 520 7300 b 520 7300
c 900 1760 c 900 1760
d 1680 3380 d 1680 3380
Which of the above elements likely to be CnamoŠV VËdm| H$s g§^mdZmE§ Š`m h¢ ?
(A) a – alkali metal, b – alkaline (A) a – jmar` YmVw, b – jmar` _¥Xm YmVw,
earth metal, c – reactive c – {H«$`merb AYmVw, d – A{H«$` J¡g
non-metal, d – noble gas
(B) a – noble gas, b – alkali metal, (B) a – A{H«$` J¡g, b – jmar` YmVw,
c – reactive non-metal, c – {H«$`merb AYmVw, d – jmar` _¥Xm YmVw
d – alkaline earth metal
(C) a – noble gas, b – alkali metal, (C) a – A{H«$` J¡g, b – jmar` YmVw,
c – alkaline earth metal, c – jmar` _¥Xm YmVw, d – {H«$`merb
d – reactive non-metal AYmVw
(D) a – alkaline earth metal, b – alkali (D) a – jmar` _¥Xm YmVw, b – jmar` YmVw,
metal, c – reactive non-metal, c – {H«$`merb AYmVw, d – A{H«$` J¡g
d – noble gas

82. Correct order of electronegativity 82. {dÚwVF$UmË_H$Vm Ho$ _mZ H$m ghr H«$_ h¢
values are
(A) SP-hybridised C > Cl > F > (A) SP-hmB{~«S>rH¥$V C > Cl > F >
SP-hybridised N SP-hmB{~«S>rH¥$V N
(B) SP-hybridised N > F > Cl > (B) SP-hmB{~«S>rH¥$V N > F > Cl >
SP-hybridised C SP-hmB{~«S>rH¥$V C
(C) SP-hybridised N > F > (C) SP-hmB{~«S>rH¥$V N > F >
SP-hybridised C > Cl SP-hmB{~«S>rH¥$V C > Cl
(D) F > Cl > SP-hybridised C > (D) F > Cl > SP-hmB{~«S>rH¥$V C >
SP-hybridised N SP-hmB{~«S>rH¥$V N

-33- Set-A

−
83. The hybridisation and geometry of the 83. XeO46 Am`Z H$m hmB~«rS>rH$aU Ed§ Á`m{_Vr h¡
−
XeO46 ion is
(A) sp3, tetrahedral (A) sp3, MVwî\$bH$
(B) sp3d2, octahedral (B) sp3d2, AîQ>\$bH$
(C) sp3, pyramidal (C) sp3, {nam{_S>r`
(D) sp3d3, pentagonal bipyramidal (D) sp3d3, n§M^wOr` {Û{nam{_S>r`

84. Which one of the following does 84. {ZåZ _| go H$m¡Z Aåbr` KMnO4 {db`Z H$mo
not decolourise an acidified KMnO4 a§JhrZ Zht H$aVm ?
solution ?
(A) H2O2 (B) FeSO4 (A) H2O2 (B) FeSO4
(C) SO2 (D) FeCl3 (C) SO2 (D) FeCl3

85. The correct decreasing order of rate 85. H2SO4 Ûmam {ZO©brH$aU H$s Xa H$m ghr KQ>Vm
of dehydration with H2SO4 is H«$_ h¡
OH OH
| |
a. CH3 − CH2 − CH − CH2 − CH2 − CH3 a. CH3 − CH2 − CH − CH2 − CH2 − CH3

OH OH
| |
b. (CH3 )2 − C − CH2 − CH2 − CH3 b. (CH3 )2 − C − CH2 − CH2 − CH3

OH OH
| |
c. (CH3 )2 − C − CH (CH3 )2 c. (CH3 )2 − C − CH (CH3 )2

d. CH3 − CH2 − CH − CH (CH3 )2 d. CH3 − CH2 − CH − CH (CH3 )2
| |
OH OH

e. CH3-CH2-CH2-CH2-CH2-CH2-OH e. CH3-CH2-CH2-CH2-CH2-CH2-OH
(A) c>b>d>a>e (A) c>b>d>a>e
(B) e>a>d>b>c (B) e>a>d>b>c
(C) b>c>d>e>a (C) b>c>d>e>a
(D) a>b>c>d>e (D) a>b>c>d>e

-34- Set-A

86. In the following reaction sequence Y is 86. {ZåZ{b{IV A{^{H«$`m H«$_ _| Y h¡
Cl2 Cl
CH3CHO → X CH3CHO 
2
→X
Ca(OH)2 Ca(OH)2
C H NH C H NH

6 5 2
→y 
6 5 2
→y
alcoholic EëH$mohmo{bH$
KOH KOH
(A) CH3CH = NHC6H5 (A) CH3CH = NHC6H5

(B) C6H5NHCH3 (B) C6H5NHCH3

(C) C6H5NC (C) C6H5NC

(D) C6H5NCO (D) C6H5NCO

87. An optically active compound X has 87. EH$ àH$me g{H«$` `m¡{JH$ X H$m AUwgyÌ C4H8O3
molecular formula C4H8O3. It evolves h¡ & Obr` NaHCO3 Ho$ gmW `h CO2 XoVm h¡ &
CO2 with aqueous NaHCO3. X reacts X, LiAlH4 go {H«$`m H$a EH$ EH$mB©ab `m¡{JH$ XoVm
with LiAlH4 to give an achiral compound. h¡ & X h¡
X is

(A) CH3 − CH2 − CHCOOH (A) CH3 − CH2 − CHCOOH
| |
OH OH

(B) CH3 − CHCOOH (B) CH3 − CHCOOH
| |
OCH3 OCH3

(C) CH3 − CHCOOH (C) CH3 − CHCOOH
| |
CH2OH CH2OH

(D) CH3 − CH − CH2COOH (D) CH3 − CH − CH2COOH
| |
OH OH

-35- Set-A

88. End product of the following sequence 88. {ZåZ{b{IV A{^{H«$`m Ho$ H«$_ _| ApÝV_
of reaction is CËnmX h¡
NBS NBS

→ 
→

Mg / ether i) CH CN Mg / B©Wa i) CH CN

→ 
+
3
→ → 
+
3
→
ii) H3 O ii) H3 O


(A) (A)

(B) (B)

(C) (C)

(D) (D)

89. Balmer lines occur between wavelength 89. ~m_a aoImE± Va§JX¡¿`© namg Ho$ _Ü` nm`m OmVm h¡
range
(A) 2000Å to 3500Å (A) 2000Å go 3500Å
(B) 3647Å to 6563Å (B) 3647Å go 6563Å
(C) 6653Å to 7600Å (C) 6653Å go 7600Å
(D) 1650Å to 3267Å (D) 1650Å go 3267Å

-36- Set-A

90. The total energy of an electron of 90. H-na_mUw Ho$ EH$ BboŠQ´>mZ H$s {ÛVr` ¹$mÊQ>_
H-atom in the second quantum state AdñWm _| Hw$b D$Om© (–E2) h¡, Vmo He+ na_mUw
is (–E2), the total energy of He+ atom H$m V¥Vr` ¹$mÊQ>_ AdñWm _| Hw$b D$Om© hmoJr
in third quantum state will be
3 3
(A) −   E2 (A) −   E2
2 2
2 2
(B) −   E2 (B) −   E2
3 3
4 4
(C) −   E2 (C) −   E2
9 9
 16   16 
(D) −   E2 (D) −   E2
 9   9 

91. Consider the following iodides : 91. {ZåZ Am`moS>mB©S>m| na {dMma H$s{O`o :
PI3 AsI3 SbI3 PI3 AsI3 SbI3
bond angle : 102° 100.2° 99° Am~ÝY H$moU : 102° 100.2° 99°
The bond angle is maximum in PI3 due to PI3 H$m Am~ÝY H$moU gdm©{YH$ hmoZo H$m H$maU h¡
(A) Small size of phosphorous atom (A) \$m°ñ\$moag na_mUw H$m N>moQ>m AmH$ma
(B) More bP-bP repulsion in PI3 (B) PI3_| bP-bP {dH$f©U A{YH$ hmoZm
(C) Less electronegativity of (C) \$m°ñ\$moag H$m {dÚwVF$UmË_H$Vm {ZåZ hmoZm
phosphorous
(D) None of these (D) BZ_| go H$moB© Zht

92.
A solution of 10 g of a non-ionic solute EH$ J¡a-Am`{ZH$ {dbo` Ho$ 10 g H$m Kmob
92.
in 100 g of benzene freezes at 4.2°C. 100 g ~|OrZ _| 4.2°C na O_ OmVm h¡ & {dbo`
The molecular weight of the solute is H$m AmU{dH$ ^ma {H$VZm h¡ ?
(Given that Kf of benzene is 5.12°C/m) ({X`m J`m h¡ {H$ ~|OrZ H$m Kf = 5.12° C/m h¡)
(A) 400 g/mol (A) 400 g/mol
(B) 180 g/mol (B) 180 g/mol
(C) 200 g/mol (C) 200 g/mol
(D) 512 g/mol (D) 512 g/mol

-37- Set-A

93. Among the following which are 93. {ZåZ Ho$ _Ü` H$m¡Z gw_o{bV Zht h¢ ?
mismatched ?
I. Milk – O/W emulsion I. XÿY – O/W nm`g
II. Gelatin – Protective colloid II. {Obo{Q>Z – ajr H$mobm`S>
III. Chemisorption – Multilayer in III. amgm`{ZH$ A{YemofU – ~hþnaVr` àH¥${V
nature
IV. Colour of colloidal particles – IV. H$mobm`S>b H$Um| Ho$ a§J – H$Um| Ho$ AmH$ma
Shape and size of particle Ed§ AmH¥${V
V. Adsorption – ∆G increases V. A{YemofU – ∆G _| d¥{Õ
(A) I and II (B) II and III (A) I Ed§ II (B) II Ed§ III
(C) III and V (D) I and IV (C) III Ed§ V (D) I Ed§ IV

94. Consider following reactions involved 94. Q>mBQ>{o Z`_ Ho$ n[aemoYZ _| em{_b {ZåZ A{^{H«$`mAm|
in purification of titanium : na {dMma H$s{O`o :
523 K 1700 K 523 K 1700 K
Ti + 2i2 
→ Tii4 → Ti + 2i2 Ti + 2i3 
→ Tii4 → Ti + 2i2

The method of purification is n[aemoYZ H$s {d{Y h¡
(A) Cupellation (A) Š`ynobrH$aU
(B) Poling (B) XÊS>rH$aU
(C) Zone refining (C) µOmoZ n[aîH$aU
(D) Van Arkel method (D) dm°Z AmH}$b {d{Y

95. 95.

chemical formulae of [P], [Q] and [R] [P], [Q] Ed§ [R] Ho$ amgm`{ZH$ gyÌ h¢
are
[P] [Q] [R] [P] [Q] [R]
(A) Ca(OH)2, NaOH, CaCO3 (A) Ca(OH)2, NaOH, CaCO3
(B) NaOH, Ca(OH)2, CaCO3 (B) NaOH, Ca(OH)2, CaCO3
(C) NaOH, CaO, CaCO3 (C) NaOH, CaO, CaCO3
(D) CaO, Ca(OH)2, NaOH (D) CaO, Ca(OH)2, NaOH

-38- Set-A

96. When PbO2 is treated with conc. HNO3, 96. O~ PbO2 H$s {H«$`m gmÝÐ HNO3 go H$amB© OmVr
the gas evolved is h¡, Vmo _wŠV J¡g h¡
(A) NO2 (A) NO2
(B) N2O (B) N2O
(C) NO (C) NO
(D) O2 (D) O2

97. Aldoses and ketoses are 97. EëS>moO Am¡a H$sQ>moO _| {d^oX {H$`m Om gH$Vm h¡
differentiated by
(A) Tollen’s reagent (A) Q>mboÝg A{^H$_©H$ Ûmam
(B) Fehling solution (B) \o$hqbJ Kmob Ûmam
(C) dil.NaOH (C) VZw NaOH Ûmam
(D) Br2 water (D) Br2 Ob Ûmam

For the reaction, A + B → C,
98. 98. à{V{H«$`m A + B → C Ho$ {bE,
∆H°= + 30 kJ; ∆S° = + 50 J/K. ∆H°= + 30 kJ; ∆S° = + 50 J/K, Bg{bE
Therefore the reaction is à{V{H«$`m h¡
(A) Spontaneous at temperatures less (A) 600 K go H$_ Vmn_mZ na ghO
than 600 K
(B) Spontaneous at temperatures (B) 600 K go A{YH$ Vmn_mZ na ghO
greater than 600 K
(C) Spontaneous at all temperatures (C) g^r Vmn_mZm| na ghO
(D) Spontaneous only at 27°C (D) Ho$db 27°C na ghO

99. According to electrochemical theory of 99. {dÚwVamgm`{ZH$ {gÕm§V Ho$ AZwgma g§jmaU _|
corrosion involves em{_b h¡
(A) Cathodic deposition of metal (A) YmVw H$m H¡$Wmo{S>H$ {ZjonU
(B) Anodic deposition of metal (B) YmVw H$m EZmo{S>H$ {ZjonU
(C) Cathodic dissolution of metal (C) YmVw H$m H¡$Wmo{SH$> {dKQ>Z
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

-39- Set-A

100. A dispersion of a liquid in a gas, a 100. J¡g _| Ðd H$m n[ajonU, Ðd _| J¡g H$m n[ajonU Am¡a
dispersion of a gas in a liquid and a Ðd _| R>mgo Ho$ n[ajonU H$mo Š`m H$hm OmVm h¡ ?
dispersion of a solid in a liquid are
known as
(A) Aerosol, foam, sol (A) Eamogmob, \$mo_, gmob
(B) Sol, foam, aerosol (B) gmob, \$mo_, Eamogmob
(C) Aerosol, emulsion, foam (C) Eamogmob, B_ëeZ, \$mo_
(D) Emulsion, aerosol, sol (D) B_ëeZ, Eamogmob, gmob

101. In the dichromate anion 101. S>mBH«$mo_oQ> F$Um`Zm| _|
(A) 4 Cr – O bonds are equivalent (A) 4 Cr – O ~m°ÊS>> g_Vwë` h¢
(B) 6 Cr – O bonds are equivalent (B) 6 Cr – O ~m°ÊS>> g_Vwë` h¢
(C) All Cr – O bonds are equivalent (C) g^r Cr – O ~m°ÊS> g_Vwë` h¢
(D) All Cr – O bonds are non-equivalent (D) g^r Cr – O ~m°ÊS> J¡a-g_Vwë` h¢

102. The IUPAC name for 102. K2 [Cr(CN)2O2(O2) NH3] H$m IUPAC Zm_ h¡
K2 [Cr(CN)2O2(O2) NH3] is
(A) Potassium aminedicyano- (A) nmoQ>o{e`_ E{_ZoS>mBgm`Zmo-
dioxoperoxochromate (VI) S>mBAm°ŠgmonoamoŠgmoH«$mo_oQ> (VI)
(B) Potassium aminocyano- (B) nmoQ>{o e`_ E{_Zmogm`ZmoS>mBnoamoŠgmoH$« mo_Qo > (VI)
diperoxochromate (VI)
(C) Potassium aminedicyano- (C) nmoQ>o{e`_ E{_ZoS>mBgm`ZmoH«$mo_oQ> (IV)
chromate (IV)
(D) None of the above (D) Cn`w©ŠV _| go H$moB© Zht

103. Among the following metal carbonyls, 103. {ZåZ{b{IV YmVw H$m~m}{Zëg _| go, C – O ~m°ÊS>
the C – O bond order is lowest in Am°S>©a __________ _| g~go H$_ h¡ &
(A) [Mn(CO)6]+ (A) [Mn(CO)6]+
(B) [Fe(CO)5] (B) [Fe(CO)5]
(C) [Cr(CO)6] (C) [Cr(CO)6]
(D) [V(CO)6]– (D) [V(CO)6]–

-40- Set-A

104. In nitroprusside ion, the iron and NO 104. ZmBQ´>moàwgmBS> Am`Z _| FeIII Am¡a NO Ho$ ~Om`
exist as FeII and NO+ rather than FeII Am¡a NO+ Ho$ ê$n _| Am`aZ Am¡a NO
FeIII and No. These forms can be _m¡OyX hmoVo h¢ & `o ê$n ________ Ûmam {d^o{XV
differentiated by hmo gH$Vo h¢ &
(A) Estimating the concentration of (A) Am`aZ H$s gm§ÐVm H$m AmH$bZ
iron
(B) Measuring the concentration (B) CN– H$s gm§ÐVm H$mo _mnZm
of CN–
(C) Measuring the solid state magnetic (C) gmo{bS> ñQ>oQ> _¡½Zo{Q>H$ _mo_|Q> H$mo _mnZm
moment
(D) Thermally decomposing the (D) `m¡{JH$ H$mo D$î_r` ê$n go {dK{Q>V H$aZm
compound

105. Among the following which are 105. {ZåZ _| go H$m¡Z-gm gw_o{bV Zht h¡ ?
mismatched ?
I. XeF4 – Tetrahedral I. XeF4 – MVwî\$bH$s`
II. XeO4 – Tetrahedral II. XeO4 – MVwî\$bH$s`
III. XeF2 – Trigonal bipyramidal III. XeF2 – {Ì^wOr` {Û{nam{_{S>`
IV. xeOF4 – Square pyramidal IV. xeOF4 – dJ© {nam{_{S>`
V. XeO3 – Trigonal pyramidal V. XeO3 – {Ì^wOr` {nam{_{S>`
(A) I and III (A) I Ed§ III
(B) II and III (B) II Ed§ III
(C) IV and V (C) IV Ed§ V
(D) II and V (D) II Ed§ V

106. Which one of the following reactions of 106. µOoZm°Z `m¡{JH$ Ho$ {ZåZ{b{IV _| go H$m¡Z-gr
xenon compounds is not feasible ? A{^{H«$`m g§^d Zht h¡ ?
(A) XeO3 + 6HF → XeF6 + 3H2O (A) XeO3 + 6HF → XeF6 + 3H2O
(B) 3XeF4 + 6H2O → 2Xe + XeO3 (B) 3XeF4 + 6H2O → 2Xe + XeO3
+ 12 HF + 1.5 O2 + 12 HF + 1.5 O2
(C) 2XeF2 + 2H2O → 2Xe + 4HF + O2 (C) 2XeF2 + 2H2O → 2Xe + 4HF + O2
(D) XeF6 + RbF → Rb [XeF7] (D) XeF6 + RbF → Rb [XeF7]

-41- Set-A

107. The pair of the compounds in which 107. `m¡{JH$m| H$s Omo‹S>r {Og_| XmoZm| YmVwE± g§^m{dV
both the metals are in the highest CÀMV_ Am°ŠgrH$aU AdñWm _| hmoVo h¢
possible oxidation state is
(A) [Fe(CN)6]3–, [Co(CN)6]3– (A) [Fe(CN)6]3–, [Co(CN)6]3–
(B) CrO2Cl2 , MnO4− (B) CrO2Cl2 , MnO4−
(C) TiO2, MnO2 (C) TiO2, MnO2
(D) [Co(CN)6]3–, MnO2 (D) [Co(CN)6]3–, MnO2

108. In an alkaline condition KMnO4 reacts 108. EH$ jmar` AdñWm _| KMnO4 {ZåZmZwgma
as follows : à{V{H«$`m H$aVm h¡ :
2KMnO4 + 2KOH → 2K2MnO4 + H2O + [O] 2KMnO4 + 2KOH → 2K2MnO4 + H2O + [O]
its equivalent weight is BgH$m Vwë` ^ma h¡
(A) 31.6 (B) 52.7 (A) 31.6 (B) 52.7
(C) 49 (D) 158 (C) 49 (D) 158
ka ka
109. For the reaction A  → products, 109. A  → CËnmX, A{^{H«$`m Ho$ {bE EH$sH¥$V
the integrated rate law is as follows : [A]
[A] Xa {Z`_ Bg àH$ma h¡ : in = −k a t ; A H$s
in = −k a t ; the concentration of A [ A ]0
[ A ]0 gm§ÐVm [A]0 h¡ Am¡a ~mX _| g_` t _| `h [A] h¡ &
is [A]0 and at a later time t it is [A].
Pick the correct statements from the {ZåZ{b{IV _| go ghr H$WZm| H$m M`Z H$s{OE :
following :
I. [A]0 decreases exponentially with I. [A]0 g_` Ho$ gmW KmVr` ê$n go KQVm h¡ &
time.
II. A plot of In[A]0/[A] versus t gives II. In[A]0/[A] ~Zm_ t H$m AmboI ka H$m _mZ
the value of ka. XoVm h¡ &
1 1
III. A plot of
[A]
versus t gives the III.
[A]
~Zm_ t H$m AmboI ka H$m _mZ XoVm h¡ &
value of ka.
IV. A plot of In[A] versus t gives a IV. In[A] ~Zm_ t H$m AmboI grYr aoIm XoVm h¡ &
straight line.
(A) I (B) I and III (A) I (B) I Ed§ III
(C) III (D) II and IV (C) III (D) II Ed§ IV

-42- Set-A

110. The rate constants for two reactions 110. Xmo A{^{H«$`mAm| Ho$ {bE Xa pñWam§H$ {OZH$m g_mZ
that have the same A value but have A _mZ h¡ bo{H$Z Ea _mZ 1 kcal/mol go {^Þ h¡ &
Ea values differ by 1 kcal/mol. The
room temperature ratio will be H$_ao Ho$ Vmn_mZ H$m AZwnmV hmoJm
(R = 1.987 cal mol–1 K–1, T = 298 K) (R = 1.987 cal mol–1 K–1, T = 298 K)
(A) 2.0 (B) 2.7 (A) 2.0 (B) 2.7
(C) 5.4 (D) 2 × 107 (C) 5.4 (D) 2 × 107

111. Consider the following statements : {ZåZ H$WZm| na {dMma H$s{O`o :
111.
I. CuSO4(aq) cannot be stored in zinc I. CuSO4(aq) H$mo qOH$ Ho$ nmÌ _| aIm Zht
vessel. Om gH$Vm &
II. Electrolysis of molten CaH 2 II. CaH2 (J{bV) Ho$ {dÚwV{dÀN>oXZ go H2
evolves H2 at anode. J¡g EZmoS> na _wŠV hmoVr h¡ &
III. Electrolysis of a solution containing III. Cu2+(aq) Ed§ Fe2+(aq) _wŠV {db`Z H$m
Cu2+(aq) and Fe2+(aq) ions deposits {dÚwV{dÀN>oXZ H$aZo na Fe2+(aq) Am`Z
iron first. nhbo {Zjo{nV hmoVm h¡ &
IV. Electrolysis of NaCl(aq) gives a IV. NaCl(aq) {db`Z Ho$ {dÚwV{dÀN>oXZ
solution of pH < 7. {db`Z H$m pH < 7 XoVm h¡ &
V. In recharging lead storage battery V. bo‹S> ñQ>moaoO ~¡Q>ar Ho$ [aMm{OªJ _| H2SO4
H2SO4 is produced. CËnÞ hmoVm h¡ &
Correct statements are : ghr H$WZ h¢ :
(A) I, II and III (A) I, II Ed§ III
(B) II, III and IV (B) II, III Ed§ IV
(C) III, IV and V (C) III, IV Ed§ V
(D) I, II and V (D) I, II Ed§ V

112.
How much charge is required to produce 112. STP na J{bV NaCl Ho$ {dÚwV{dÀN>oXZ H$aZo
H2 gas at the rate of 1 ml sec–1 by na 1 ml sec–1 H$s Xa go _wŠV H2 J¡g Ho$ {b`o
electrolysis of molten NaCl at STP ? Amdí`H$ {dÚwV H$s _mÌm h¡
(A) 8.6 C (B) 18.4 C (A) 8.6 C (B) 18.4 C
(C) 11.2 C (D) 1.4 C (C) 11.2 C (D) 1.4 C

-43- Set-A

113.
The correct statement among the 113. {ZåZ{b{IV _| go ghr H$WZ H$m¡Z-gm h¡ ?
following is
(A) The Henderson-Hasselbalch (A) h|S>agZ-hmgob~ëM g_rH$aU H$mo {ZåZ
equation can be expressed as ê$n _| ì`ŠV {H$`m Om gH$Vm h¡
[HA ] pH = pKa + log
[HA ] &
pH = pKa + log − .
 A   A − 
(B) Conjugate acid, the species (B) àmoQ>m°Z ñdrH$ma H$aZo Ho$ ~mX jma go ~Zr
formed after a base accepts a àOm{V`m±, g§`w½_ Aåb, Am_Vm¡a na EH$
proton, typically a strong acid. à~b Aåb hmoVo h¢ &
(C) The pKa of H3O+ is –1.74. (C) H3O+ H$m pKa –1.74 h¡ &
(D) For a given equilibrium, a reaction (D) {H$gr {XE JE g§Vbw Z Ho$ {bE, _m¡OXy gm_mÝ`
with a common ion present has Am`Z Ho$ gmW A{^{H«$`m _| ~¥hÎma Ksp Am¡a
a greater Ksp and the reaction
Am`Z Ho$ {~Zm A{^{H«$`m _| H$_ Ksp hmoVm h¡ &
without the ion has a lower Ksp.

114. Consider the following statements : 114. {ZåZ H$WZm| na {dMma H$s{O`o :
I. All natural processes are reversible I. g^r àmH¥${VH$ àH«$_ CËH«$_Ur` hmoVo h¢ &
in nature.
II. Enthalpy of neutralization of weak II. Xþ~©b Aåb Ed§ Xþ~©b jma Ho$ {b`o
acid and weak base is always less CXmgrZrH$aU BÝW°ënr h_oem 57.1 KJ
than 57.1 KJ. go H$_ hmoVm h¡ &
III. Resonance energy = calculated III. AZwZmX D$Om© = n[aH${bV {Z_m©U BÝW°ënr
enthalpy of formation – observed – ào{jV {Z_m©U BÝW°ënr &
enthalpy of formation.
IV. For exothermic reactions IV. Cî_jonr A{^{H«$`m Ho$ {b`o
H product < Hreactant. HCËnmX < HA{^H$maH$ &
V. The inversion temperature of a EH$ J¡g Ho$ {b`o à{Vbmo_ VmnH«$_ CgHo$
V.
gas depends upon vander Waals’ dmÊS>admb pñWam§H$ ‘a’ Ed§ ‘b’ na {Z^©a
constant ‘a’ and ‘b’. H$aVm h¡ &
Wrong statements are : JbV H$WZ h¢ :
(A) II and III (B) IV and V (A) II Ed§ III (B) IV Ed§ V
(C) I and II (D) I and III (C) I Ed§ II (D) I Ed§ III

-44- Set-A

115. The equilibrium constant Kp for the 115. 1000 K na A{^{H«$`m
reaction 2SO 2(g ) + O 2(g )  2SO 3(g ) at 2SO 2(g ) + O 2(g )  2SO 3(g ) Ho$ {b`o gmå`
1000 K is 3.5. The partial pressure pñWam§H$ Kp 3.5 h¡ &
of oxygen gas to give equal moles of SO2(g) Ed§ SO3(g) Ho$ g_mZ _mobm| Ho$ {b`o
SO2(g) and SO3(g) is
Am°ŠgrOZ J¡g Ho$ Am°§{eH$ Xm~ h¡
(A) 0.29 atm (A) 0.29 atm
(B) 3.5 atm (B) 3.5 atm
(C) 0.53 atm (C) 0.53 atm
(D) 1.87 atm (D) 1.87 atm

116. The volume of a perfect gas is doubled 116. pñWa Vmn_mZ na {dnarV ê$n go EH$ AmXe© J¡g H$m
reversibly at constant temperature. Am`VZ XmoJZw m hmo OmVm h¡ & àUmbr Ho$ _moba EÝQ´>mn° r
The total change in molar entropy of _| Hw$b n[adV©Z (R = 8.3145 JK–1 mol–1) h¡
the system is (R = 8.3145 JK–1 mol–1)
(A) 0 (A) 0
(B) 5.76 JK–1 mol–1 (B) 5.76 JK–1 mol–1
(C) –5.76 JK–1 mol–1 (C) –5.76 JK–1 mol–1

(D) 16.62 JK–1 mol–1 (D) 16.62 JK–1 mol–1

117. Which set of elements have nearly the 117. H$m¡Z-go VËdm| Ho$ goQ> H$s na_mUw {ÌÁ`m bJ^J
same atomic radii ? g_mZ h¡ ?
(A) F, Cl, Br, I (A) F, Cl, Br, I
(B) Na, K, Rb, Cs (B) Na, K, Rb, Cs
(C) Li, Be, B, C (C) Li, Be, B, C
(D) Fe, Co, Ni, Cu (D) Fe, Co, Ni, Cu

-45- Set-A

118.
The correct decreasing order of atomic 118. {ZåZ ñnogrO Ho$ _Ü` na_mUw AmH$ma H$m ghr KQ>Vm
size among following species is H«$_ h¡
Ar, K+, Cl–, S2–, Ca2+ Ar, K+, Cl–, S2–, Ca2+
(A) Ca2+ > K+ > Ar > Cl– > S2– (A) Ca2+ > K+ > Ar > Cl– > S2–
(B) K+ > Ca2+ > Cl– > Ar > S2– (B) K+ > Ca2+ > Cl– > Ar > S2–
(C) S2– > Cl– > Ar > K+ > Ca2+ (C) S2– > Cl– > Ar > K+ > Ca2+
(D) S2– > Ar > Cl– > Ca2+ > K+ (D) S2– > Ar > Cl– > Ca2+ > K+

119. Consider following statements : 119. {ZåZ H$WZm| na {dMma H$s{O`o :
I. The second ionization enthalpy of I. na_mUw Ho$ àW_ Am`ZZ BÝW¡ënr go {ÛVr`
atom is always less than the first Am`ZZ BÝW¡ënr h_oem H$_ hmoVr h¡ &
ionization enthalpy.
II. The size of cation is always less II. YZm`Z H$m AmH$ma h_oem _yb na_mUw go H$_
than the parent atom. hmoVm h¡ &
iii. The first electron gain enthalpy of iii. ZmBQ´>moOZ Ho$ àW_ BboŠQ´>mZ àmpßV BÝW¡ënr
nitrogen is +31 kJ mol–1. H$m _mZ +31 kJ mol–1 hmoVm h¡ &
IV. Electronegativity of carbon IV. H$m~©Z H$s {dÚwV F$UmË_H$Vm H$m ghr KQ>Vm
decreases from sp3 > sp2 > sp. H«$_ sp3 > sp2 > sp h¡ &
v. Higher the screening effect v. n[aajU à^md H$m _mZ {OVZm CÀM hmoJm,
lesser is the value of ionization Am`ZZ BÝW¡ënr H$m _mZ CVZm H$_ hmoJm &
enthalpy.
Incorrect statements are JbV H$WZ h¢
(A) II and III (A) II Ed§ III
(B) II and V (B) II Ed§ V
(C) I and IV (C) I Ed§ IV
(D) III and V (D) III Ed§ V

120. Peroxide [– O – O – ] bond is present in 120. naAm°ŠgmBS> [– O – O – ] Am~ÝY CnpñWV h¡
(A) S2O2−
7 (A) S2O2− 7 _|
2− 2−
(B) S2O8 (B) S2O8 _|
2− 2−
(C) S4O6 (C) S4O6 _|
(D) SO2−
3 (D) SO2− 3
_|

-46- Set-A

PART – III (A)
Mathematics J{UV
π π
2
a tan x 2
a tan x
121. ∫ atan x + a cot x
dx is equal to 121. ∫ atan x + acot x dx H$m _mZ hmoJm
0 0
(A) 0 (A) 0
π π
(B) (B)
2 2
π π
(C) (C)
4 4
(D) π (D) π

π π
2
sin(2n + 1)θ 2
sin(2n + 1)θ
122. If Sn = ∫ dθ and 122. `{X Sn = ∫ dθ VWm
0
sin θ sin θ
0
π
π 2 2
2 2  sin nθ 
Vn = ∫
 sin nθ  Vn = ∫  sin θ  dθθ, V~ Vn+1 – Vn =
 sin θ  dθ , then Vn+1 – Vn =  
0   0

(A) 2Sn (B) Sn (A) 2Sn (B) Sn
(C) Sn + 1 (D) 2Sn + 1 (C) Sn + 1 (D) 2Sn + 1

123. Let f be a function continuous on R, 123. _mZ br{OE R _| f EH$ g§VV \$bZ h¡, Vmo
π π
2 2
then ∫ (x + sin x) {f(x) + f(−x)} dx is ∫ (x + sin x) {f(x) + f(−x)} dx H$m _mZ
−π −π
2 2
equal to hmoJm
(A) π + 2 (A) π + 2
(B) 0 (B) 0
π2 π2
(C) +2 (C) +2
4 4
π2 + 2 π2 + 2
(D) (D)
4 4

-47- Set-A

124. The area bounded by the curve 124. dH«$ y = logex, x-Aj VWm gab aoIm x = e
y = logex, the x-axis and the straight Ûmam n[a~Õ joÌ\$b hmoJm
line x = e is
(A) e sq. units (A) e dJ© BH$mB©
(B) 1 sq. units (B) 1 dJ© BH$mB©
 1
(C)  1 −  dJ© BH$mB©
1
(C)  1 −  sq. units
 e  e
 1  1
(D)  1 +  sq. units (D)  1 +  dJ© BH$mB©
 e  e

125. If the given planes ax + by + cz + d = 0 125. `{X {X`o J`o g_Vb ax + by + cz + d = 0 VWm
and a′x + b′y + c′z + d′ = 0 be mutually a′x + b′y + c′z + d′ = 0 nañna b§~dV h¡, Vmo
perpendicular, then
a b c a b c
(A) = = (A) = =
a′ b′ c′ a′ b′ c′
a b c a b c
(B) + + (B) + +
a′ b′ c′ a′ b′ c′
(C) aa′ + bb′ + cc′ + dd′ = 0 (C) aa′ + bb′ + cc′ + dd′ = 0
(D) aa′ + bb′ + cc′ = 0 (D) aa′ + bb′ + cc′ = 0

126. The distance of the point (1, 8, 4) from 126. {~ÝXþ (1, 8, 4) go g_Vb 12x – 4y + 3z = 3
the plane 12x – 4y + 3z = 3 is H$s Xÿar h¡
14 14
(A) − (A) −
13 13
7 7
(B) (B)
13 13
5 5
(C) (C)
13 13
(D) None of these (D) BZ_| go H$moB© Zht

-48- Set-A

127. If u = i × (a × i) + j × (a × j) + k × (a × k), 127. `{X u = i × (a × i) + j × (a × j) + k × (a × k),
then u = Vmo u =
(A) 0 (B) i + j + k (A) 0 (B) i + j + k
(C) 2a (D) a (C) 2a (D) a

128. If the vectors i – 3j + 2k, –i + 2j represents 128. `{X g{Xe i – 3j + 2k, –i + 2j g_mÝVa MVw^w©O
the diagonals of a parallelogram, then Ho$ {dH$Um] H$mo àX{e©V H$aVo h¢, Vmo CgH$m joÌ\$b
its area will be hmoJm
21 21
(A) 21 (B) (A) 21 (B)
2 2
21 21
(C) 2 21 (D) (C) 2 21 (D)
4 4

129. A bag contains 10 black and 5 white 129. EH$ W¡bo _| 10 H$mbr Am¡a 5 g\o$X J|Xo h¢ & EH$ Ho$
balls. Two balls are drawn one after the ~mX EH$ H$aHo$ {~Zm {dñWmnZ Ho$ Xmo J|Xo {ZH$mbr
other without replacement. Probability OmVr h¢ & XmoZm| {ZH$mbr J|Xm| Ho$ H$mbr hmoZo H$s
that both drawn balls are black
àm{`H$Vm hmoJr
2 2
(A) (A)
15 15
2 2
(B) (B)
3 3
1 1
(C) (C)
5 5
3 3
(D) (D)
7 7

130. The smallest positive root of x3 – 5x +3 = 0 130. x3 – 5x +3 = 0 H$m g~go N>moQ>m YZmË_H$ _yb
lies between BgHo$ ~rM hmoVo h¢
(A) 1 and 2 (A) 1 Am¡a 2
(B) 2 and 3 (B) 2 Am¡a 3
(C) 0 and 1 (C) 0 Am¡a 1
(D) None of these (D) BZ_| go H$moB© Zht

-49- Set-A

131. Inequations 3x – y > 3 and 4x – y > 4 131. Ag_{`H$mAm| 3x – y > 3 Am¡a 4x – y > 4 Ho$
have solution hb àmßV hm|Jo
(A) for positive x and y (A) YZmË_H$ x Am¡a y Ho$ {bE
(B) no solution for positive x and y (B) YZmË_H$ x Am¡a y Ho$ {bE hb àmßV Zht
hm|Jo
(C) for all x (C) g^r x Ho$ {bE
(D) for all y (D) g^r y Ho$ {bE

132. In which quadrant, the bounded region 132. x + y < 1 Am¡a x – y < 1 go {Kam hþAm n[a~Õ
for x + y < 1 and x – y < 1 is situated ? joÌ {H$g MVwWmªe _| hmoJm ?
(A) I, II (B) I, III (A) I, II (B) I, III
(C) only I (D) All the four (C) Ho$db I (D) g^r Mmam| _|

133. The sum of two positive numbers is 133. Xmo YZmË_H$ g§»`mAm| H$m `moJ 2n Ho$ ~am~a h¡ &
equal to 2n. The probability that their 3
3 CZH$m JwUZ\$b CZHo$ CÀMV_ JwUZ\$b go 4 JwZm
product is not less than times their H$_ Zht hmoZo H$s àm{`H$Vm hmoJr
4
greatest product is
3 1 3 1
(A) (B) (A) (B)
4 2 4 2
1 3 1 3
(C) (D) (C) (D)
4 2 4 2

134. If P(E1) = p1 and P(E2) = p2 and E1 and 134. `{X P(E1) = p1 VWm P(E2) = p2 Am¡a E1 VWm
E2 are independent, then P (neither E1 E2 ñdVÝÌ h¡, V~ P (Z E1 Am¡a Z hr E2) ~am~a
nor E2) is equal to hmoJm
(A) (1 – p1) (1 – p2) (A) (1 – p1) (1 – p2)
(B) 1 – (p1 + p2) (B) 1 – (p1 + p2)

(C) p1 + p2 – p1p2 (C) p1 + p2 – p1p2
1 1 1 1 1 1
(D) + − (D) + −
p1 p 2 p1 p2 p1 p 2 p1 p2

-50- Set-A

135. If line of regression of y on x is 135. `{X y H$s x na g_ml`U aoIm y = m1x + c1
y = m1 x + c1 and line of regression of x on y VWm x H$s y na g_ml`U aoIm y = m2x + c2
is y = m2x + c2, then which is correct ? hmo, Vmo {ZåZ _| go gË` hmoJm
(A) m1 m2 < 1 (A) m1 m2 < 1

(B) 0 ≤ m1 m2 ≤ 1 (B) 0 ≤ m1 m2 ≤ 1

m1 m1
(C) −1 ≤ (C) −1 ≤
m2 m2

m2 m2
(D) −1 ≤ (D) −1 ≤
m1 m1

136. If X follows a binomial distribution with 136. `{X X {ÛnX ~§Q>Z H$m nmbZ àmMbm| n = 100 VWm
1 1
parameters n = 100 and p = , then p= Ho$ {bE H$aVm h¡, V~ p(x = r) A{YH$V_
3 3
p(x = r) is maximum when r is equal to h¡, O~ r ~am~a h¡
(A) 16 (A) 16
(B) 32 (B) 32
(C) 33 (C) 33
(D) None of these (D) BZ_| go H$moB© Zht

137. The new co-ordinates of a point (4, 5), 137. EH$ {~ÝXþ (4, 5) Ho$ Z`o {ZX}em§H$ O~{H$ _yb {~ÝXþ
when the origin is shifted to the point H$mo {~ÝXþ (1, –2) _| ñWmZm§V[aV {H$`m OmVm h¡
(1, –2) are
(A) (5, 3) (A) (5, 3)
(B) (3, 5) (B) (3, 5)
(C) (3, 7) (C) (3, 7)
(D) None of these (D) BZ_| go H$moB© Zht

-51- Set-A

138. If the straight lines y = 3x + 1 and 138. `{X gab aoImE± y = 3x + 1 VWm 2y = x + 3 H$m
2y = x + 3 are equally inclined to the gab aoIm y = mx + 4 go ~am~a PwH$md hmo, Vmo
straight line y = mx + 4, then m = m=
1+ 3 2 1+ 3 2
(A) (A)
7 7
1− 3 2 1− 3 2
(B) (B)
7 7
1± 3 2 1± 3 2
(C) (C)
7 7
1± 5 2 1± 5 2
(D) (D)
7 7
139. Length of the common chord of the 139. d¥Îmm| x2 + y2 + 5x + 7y + 9 = 0 VWm
circles x2 + y2 + 5x + 7y + 9 = 0 and x2 + y2 + 7x + 5y + 9 = 0 Ho C^`{ZîQ>
x2 + y2 + 7x + 5y + 9 = 0 is Ordm H$s bå~mB© h¡$
(A) 9 (B) 8 (A) 9 (B) 8
(C) 7 (D) 6 (C) 7 (D) 6

140. The direction cosine of the straight line 140. gab aoIm x = y = z H$s {XH²$ H$moÁ`m`| h¢
x = y = z are
1 1 1 1 1 1
(A) , , (A) , ,
3 3 3 3 3 3
1 1 1 1 1 1
(B) , , (B) , ,
3 3 3 3 3 3
(C) 1, 1, 1 (C) 1, 1, 1
(D) None of these (D) BZ_| go H$moB© Zht

 π π
141. If xr = cis  r  , then 141. `{X xr = cis  r  , Vmo
2  2 

Lt x1x 2 x 3 ....xn = Lt x1x 2 x 3 ....xn =
n→∞ n → ∞
(A) 0 (B) 1 (A) 0 (B) 1
(C) – 1 (D) ∞ (C) – 1 (D) ∞

-52- Set-A

142. If the cube roots of unity are 1, w, ω2, `{X 1, w, ω2 BH$mB© Ho$ KZ_yb h¢, Vmo g_rH$aU
142.
then the roots of the equation (x – 1)3 + 8 = 0 Ho$ _yb h¢
(x – 1)3 + 8 = 0, are
(A) – 1, – 1 + 2w, – 1 – 2w2 (A) – 1, – 1 + 2w, – 1 – 2w2
(B) – 1, – 1, – 1 (B) – 1, – 1, – 1
(C) – 1, 1 + 2w, 1 + 2w2 (C) – 1, 1 + 2w, 1 + 2w2
(D) – 1, 1 – 2w, 1 – 2w2 (D) – 1, 1 – 2w, 1 – 2w2

143. If the roots of the cubic equation 143. `{X {ÌKmV g_rH$aU ax3 + bx2 + cx + d = 0
ax3 + bx2 + cx + d = 0 are in G.P., Ho$ _yb JwUmoÎma lo‹T>r _| h¡, Vmo
then
(A) c3a = b3d (B) ca3 = bd3 (A) c3a = b3d (B) ca3 = bd3
(C) a3b = c3d (D) ab3 = cd3 (C) a3b = c3d (D) ab3 = cd3

144. If log 2, log (2x – 1) and log (2x + 3) are 144. `{X log 2, log (2x – 1) Am¡a log (2x + 3)
in A.P., then x is equal to g_m§Va lo‹T> r _| h¢, Vmo x ~am~a h¡
(A) 5/2 (B) log25 (A) 5/2 (B) log25
(C) log32 (D) 3/2 (C) log32 (D) 3/2

145. The lines a1x + b1y + c1 = 0 and 145. aoIm`| a1x + b1y + c1 = 0 Am¡a
a2x + b2y + c2 = 0 are perpendicular a2x + b2y + c2 = 0 nañna b§~dV h¡, `{X
to each other, if
(A) a1b2 – b1a2 = 0 (A) a1b2 – b1a2 = 0
(B) a1a2 + b1b2 = 0 (B) a1a2 + b1b2 = 0
2 2 2 2
(C) a1 b 2 + b1 a 2 = 0 (C) a1 b 2 + b1 a 2 = 0
(D) a1b1 + a2b2 = 0 (D) a1b1 + a2b2 = 0

146. If two circles (x – 1)2 + (y – 3)2 = r2 and 146. `{X Xmo d¥Îm (x – 1)2 + (y – 3)2 = r2 Am¡a
x2 + y2 – 8x + 2y + 8 = 0 intersect in x2 + y2 – 8x + 2y + 8 = 0 Xmo {^ÝZ q~XþAm|
two distinct points, then
na à{VÀN>oX H$aVo h¢, Vmo
(A) 2<r<8 (A) 2<r<8
(B) r<2 (B) r<2
(C) r=2 (C) r=2
(D) r>2 (D) r>2

-53- Set-A

147. The slope of the line touching both the 147. Xmo nadb`m| y2 = 4x Am¡a x2 = – 32y H$mo ñne©
parabolas y2 = 4x and x2 = – 32y is H$aZo dmbr aoIm H$s T>mb h¡
(A) 1/8 (A) 1/8
(B) 2/3 (B) 2/3
(C) 1/2 (C) 1/2
(D) 3/2 (D) 3/2

148. The normal to the curve, 148. q~Xþ (1, 1) na dH«$ x2 + 2xy – 3y2 = 0
x2 + 2xy – 3y2 = 0 at a point (1, 1) H$m A{^b§~
(A) meets the curve again in the (A) nwZ: dH«$ go {ÛVr` MVwWmªe _| {_bVm h¡
second quadrant
(B) meets the curve again in the third (B) nwZ: dH«$ go V¥Vr` MVwWmªe _| {_bVm h¡
quadrant
(C) meets the curve again in the fourth (C) nwZ: dH«$ go MVwW© MVwWmªe _| {_bVm h¡
quadrant
(D) does not meet the curve again (D) dH«$ go nwZ: Zht {_bVm
149. If R(x) = 13x2 + 26x + 15 represents the 149. `{X R(x) = 13x2 + 26x + 15 go {H$gr CËnmX
revenue received from selling x units of Ho$ x BH$mB© ~oMZo na àmßV amOñd H$mo àX{e©V H$aVm
a product, then the marginal revenue h¡, Vmo x = 7 BH$mB© ~oMZo na gr_m§V amOñd hmoJm
of selling x = 7 units will be
(A) 834 (B) 90 (A) 834 (B) 90
(C) 208 (D) 126 (C) 208 (D) 126

 x   x 
150. The selling price of x balls is  5 − 150. x Jo§Xm| H$m {dH«$` _yë`  5 − à{V J|X Am¡a
 100   100 
each and the cost price of x balls is CËnmXZ bmJV x Jo§Xm| Ho$ {bE  x + 500  h¡,
 x  5 
 5 + 500  , the profit will be maximum
  Vmo bm^ A{YH$V_ hmoJm `{X {H$VZr JoX§ | ~oMr OmE ?
if number of ball sold will be
(A) 1080 (A) 1080
(B) 500 (B) 500
(C) 250 (C) 250
(D) 240 (D) 240

-54- Set-A

2 2
x dx x dx
151. ∫1 (x + 1) (x + 2) = ? 151. ∫ (x + 1) (x + 2) = ?
1

3 32 3 32
(A) 4 log (B) 4 log (A) 4 log (B) 4 log
2 3 2 3
32 27 32 27
(C) log (D) log (C) log (D) log
27 32 27 32

b b
152. If f (a + b − y ) = f(y ) , then ∫ y f(y ) dy 152. `{X f (a + b − y) = f(y) , Vmo ∫ y f(y) dy
a a
is equal to H$m _mZ hmoJm
 a+b  b  a+b  b
(A)   ∫ f(x ) dx (A)   ∫ f(x ) dx
 2  a  2  a
a+b b a+b b
2 ∫a 2 ∫a
(B) f(b − x ) dx (B) f(b − x ) dx

a+b b a+b b
2 ∫a 2 ∫a
(C) f(a − x ) dx (C) f(a − x ) dx

a−b b a−b b
2 ∫a 2 ∫a
(D) f(x ) dx (D) f(x ) dx

153. If A is the singular matrix, then Adj A is 153. `{X A Aì`wËH«$_Ur` Amì`yh h¡, Vmo Adj A hmoJm
(A) Singular (A) Aì`wËH«$_Ur`
(B) Non-singular (B) ì`wËH«$_Ur`
(C) Symmetric (C) g_{_V
(D) Not defined (D) n[a^m{fV Zht {H$`m Om gH$Vm

154. The number of diagonals in a polygon 154. m ^wOm dmbr ~hþ^wO _| {dH$Um] H$s g§»`m hmoJr
of m sides is
1 1 1 1
(A) m (m − 5) (B) m (m − 1) (A) m (m − 5) (B) m (m − 1)
2! 2! 2! 2!
1 1 1 1
(C) m (m − 3) (D) m (m − 2) (C) m (m − 3) (D) m (m − 2)
2! 2! 2! 2!

-55- Set-A

1 1 1 1
155. If sin A = and sin B = , where 155. `{X sin A = VWm sin B = , Ohm±
10 5 10 5
A and B are positive acute angle, then A VWm B YZmË_H$ Ý`yZH$moU h¡, Vmo A + B =
A+B=
π π
(A) p (B) (A) p (B)
2 2
π π
(C) (D) None of the above (C) (D) Cn`w©ŠV _| go H$moB© Zht
3 3

 2a   2b   2a   2b 
If sin−1 
156. 2 
+ sin−1  2 
156. `{X sin−1  2 
+ sin−1  2 
 1+ a   1+ b   1+ a   1+ b 
= 2tan–1x, then x = = 2tan–1x, Vmo x =
a−b b a−b b
(A) (B) (A) (B)
1 + ab 1+ ab 1 + ab 1+ ab
b a+b b a+b
(C) (D) (C) (D)
1− ab 1 − ab 1− ab 1 − ab

157. The shaded region is displayed by 157. N>m`m§{H$V joÌ {ZåZ Ûmam àX{e©V hmoVm h¡


(A) 4x – 2y ≤ 3 (A) 4x – 2y ≤ 3
(B) 4x – 2y ≤ – 3 (B) 4x – 2y ≤ – 3
(C) 4x – 2y ≥ 3 (C) 4x – 2y ≥ 3
(D) 4x – 2y ≥ –3 (D) 4x – 2y ≥ –3

-56- Set-A

158. For objective function z = x1 + x2, CÔoí` \$bZ z = x1 + x2 Ho$ ì`damoYm|
158.
subject to the constraints : x1 + x2 ≤ 1, 3x1 + x2 ≥ 3 VWm x1, x2 ≥ 0 Ho$
x1 + x2 ≤ 1, 3x1 + x2 ≥ 3 and x1, x2 ≥ 0 {bE hm|Jo
have
(A) Two feasible space (A) Xmo gå^mì` joÌ
(B) Infinite feasible space (B) AZÝV gå^mì` joÌ
(C) No feasible space (C) H$moB© gå^mì` joÌ Zht
(D) None of these (D) BZ_| go H$moB© Zht

159. The value of i is 159. i H$m _mZ hmoJm
1+ i 1+ i
(A) (A)
2 2
1− i 1− i
(B) ± (B) ±
2 2
1+ i 1+ i
(C) ± (C) ±
2 2
(D) None of these (D) BZ_| go H$moB© Zht

160. The numbers 2 + 1, 1, 2 − 1 will be 160. g§»`mE± 2 + 1, 1, 2 − 1 hm|Jo
in
(A) A.P. (A) A.P. _|
(B) G.P. (B) G.P. _|
(C) H.P. (C) H.P. _|
(D) None of these (D) BZ_| go H$moB© Zht

ex + ax, x < 0 ex + ax, x < 0
161. If f(x ) =  161. `{X f(x) = 
2 2
b(x − 1) , x ≥ 0 b(x − 1) , x ≥ 0
is differentiable at x = 0, then the x = 0 na AdH$bZr` h¡, V~ (a, b) H$m _mZ h¡
value of (a, b) is
(A) (–3, –1) (B) (–3, 1) (A) (–3, –1) (B) (–3, 1)
(C) (3, 1) (D) (3, –1) (C) (3, 1) (D) (3, –1)

-57- Set-A

162. Let f(x) = |x – 1|, then 162. `{X f(x) = |x – 1|, V~
(A) f(x2) = (f(x))2 (A) f(x2) = (f(x))2
(B) f(x + y) = f(x) + f(y) (B) f(x + y) = f(x) + f(y)
(C) f(|x|) = |f(x)| (C) f(|x|) = |f(x)|
(D) f(x) is not differentiable at x = 1 (D) x = 1 na f(x) AdH$bZr` Zht h¡

163. If the function f(x) = x3 – 6x2 + ax + b 163. `{X [1, 3] na \$bZ f(x) = x3 – 6x2 + ax + b
defined on [1, 3], satisfies the Rolle’s 2 3 +1
n[a^m{fV h¡ VWm c = na amobo à_o` H$mo
2 3 +1 3
theorem for c = , then g§VwîQ> H$aVm h¡, Vmo
3
(A) a = 11, b ∈ R (A) a = 11, b ∈ R
(B) a = –11, b = 6 (B) a = –11, b = 6
(C) a = 11, b = 6 (C) a = 11, b = 6
(D) None of these (D) BZ_| go H$moB© Zht

164. The minimum value of (A 2 + A + 1) (B2 + B + 1) (C2 + C + 1) (D2 + D + 1)
164.
ABCD
(A 2 + A + 1) (B2 + B + 1) (C2 + C + 1) (D2 + D + 1)
ABCD H$m {ZpåZîR> _mZ hmoJm
given by
(A) 34 (B) 3–4 (A) 34 (B) 3–4
(C) 24 (D) 2–4 (C) 24 (D) 2–4

165. The number of solutions of the equation 165. g_rH$aU x2 + 3|x| + 2 = 0 Ho$ hbm| H$s
x2 + 3|x| + 2 = 0 is g§»`m h¡
(A) 0 (A) 0
(B) 1 (B) 1
(C) 2 (C) 2
(D) 4 (D) 4

∞
(loge x )n ∞
(loge x )n
166. ∑ n! is equal to 166. ∑ n! ~am~a h¡
n=0 n=0

(A) logex (B) x (A) logex (B) x
(C) logxe (D) none of these (C) logxe (D) BZ_| go H$moB© Zht

-58- Set-A

167. In the binomial expansion of (a – b)n, 167. (a – b)n, n ≥ 5 Ho$ {ÛnX àgma _| nm±Mdo Am¡a N>Ro
n ≥ 5, the sum of 5th and 6th terms is a
a nXm| H$m `moJ eyÝ` h¡, Vmo ~am~a h¡
b
zero, then equals to
b
5 6 5 6
(A) (B) (A) (B)
n−4 n−5 n−4 n−5
n−5 n−4 n−5 n−4
(C) (D) (C) (D)
6 5 6 5
b+c a −b a
168.
Value of the determinant 168. gma{UH$ c+a b−c b H$m _mZ h¡
b+c a −b a a+b c −a c
is
c+a b−c b
a+b c −a c
(A) 3abc – a3 – b3 – c3 (A) 3abc – a3 – b3 – c3
(B) 3abc + a3 + b3 + c3 (B) 3abc + a3 + b3 + c3
(C) 1 (C) 1
(D) 0 (D) 0

169.
The distance between point (–1, –5, –10) 169. q~Xþ (–1, –5, –10) H$s aoIm
and the point of intersection of line x − 2 y +1 z − 2
x − 2 y +1 z − 2 = = Am¡a g_Vb
= = and the plane 3 4 12
3 4 12 x – y + z = 5 Ho$ à{VÀN>oX q~Xþ Ho$ ~rM Xÿar h¡
x – y + z = 5 is
(A) 10 (B) 8 (A) 10 (B) 8
(C) 21 (D) 13 (C) 21 (D) 13

170. The centroid of a triangle ABC is 170. EH$ {Ì^wO ABC H$m H|$ÐH$ (1, 1, 1) h¡ & `{X
(1, 1, 1). If co-ordinates of A and B are A Am¡a B Ho$ {ZX}em§H$ H«$_e: (3, –5, 7) d
(3, –5, 7) and (–1, 7, –6) respectively,
(–1, 7, –6) h¢ & q~Xþ C Ho$ {ZX}em§H$ kmV H$s{OE &
then co-ordinates of point C are
(A) (1, 1, 2) (B) (2, 1, 1) (A) (1, 1, 2) (B) (2, 1, 1)
(C) (1, 2, 1) (D) (2, 2, 1) (C) (1, 2, 1) (D) (2, 2, 1)

-59- Set-A

171. The x-coordinate of a point on the 171. q~Xþ P(2, 2, 1) Am¡a Q(5, 1,–2) H$mo Omo‹S>Zo
line joining the points P(2, 2, 1) and dmbr aoIm na EH$ q~Xþ H$m x-{ZX}em§H$ 4 h¡ & CgH$m
Q(5, 1,–2) is 4. Find its z-coordinate. z-{ZX}em§H$ kmV H$s{O`o &
(A) 1 (B) –1 (A) 1 (B) –1
(C) 2 (D) –2 (C) 2 (D) –2

172. The equation of the plane passing 172. g_Vbm| 2x – 3y + z – 4 = 0 Am¡a x – y + z
through the intersection of the planes + 1 = 0 Ho$ à{VÀN>oX go JwOaZo dmbr Am¡a g_Vb
2x – 3y + z – 4 = 0 and x – y + z + 1 = 0 x + 2y – 3z + 6 = 0 Ho$ b§~dV² g_Vb H$m
and perpendicular to the plane g_rH$aU h¡
x + 2y – 3z + 6 = 0 is
(A) x – 5y + 3z – 23 = 0 (A) x – 5y + 3z – 23 = 0
(B) x + 5y – 3z + 23 = 0 (B) x + 5y – 3z + 23 = 0
(C) x – 5y – 3z – 23 = 0 (C) x – 5y – 3z – 23 = 0
(D) x – 5y + 3z + 23 = 0 (D) x – 5y + 3z + 23 = 0

173. A particle moves in a straight line with 173. `{X y Mbr JB© Xÿar hmo, Vmo EH$ H$U Ûmam {H$gr
dy
a velocity = y + 1 in time t, where g_` t na gab aoIm _| doJ dy = y + 1 hmo, Vmo
dt dt
y is the distance travelled, then the EH$ H$U Ûmam 99 _rQ>a H$s Xÿar V` H$aZo na bJm
time taken by a particle to traverse a g_` hmoJm
distance of 99 meters is
(A) log10 e (B) 2 loge 10 (A) log10 e (B) 2 loge 10
1 1
(C) 2 log10 e (D) loge 10 (C) 2 log10 e (D) loge 10
2 2
174. A family has two children. What is the 174. EH$ n[adma _| Xmo ~ÀMo h¢ & XmoZm| ~ÀMo b‹S>H$s hmoZo
probability that both the children are H$s àm{`H$Vm Š`m hmoJr, {X`m h¡ {H$ CZ_| H$_ go
girls, given that at least one of them is H$_ EH$ b‹S>H$s h¡ ?
a girl ?
3 1 3 1
(A) (B) (A) (B)
4 2 4 2
1 1
(C) 3 (D) (C) 3 (D)
2 3 2 3

-60- Set-A

175.
An inverted cone of height ‘h’ and 175. EH$ D§$MmB© ‘h’ Am¡a AmYma H$s {ÌÁ`m ‘r’ dmbm e§Hw$
radius ‘r’ is pointed at bottom. It is erf© H$s ghm`Vm go I‹S>m h¡ Am¡a EH$ C‹S>Zerb Ðd
filled with a volatile liquid completely. go nyam ^am h¡ & `{X dmînrH$aU H$s Xa {H$gr g_`
If the rate of evaporation is directly
proportional to the surface area of the
Cg Ðd H$s gVh Ho$ joÌ\$b go grYo AZwnm{VH$
liquid in contact with air, then the time hmo, Vmo dh g_`, O~ nyam Ðd dmpînV hmo OmEJm,
in which whole liquid evaporated hmoJm (k AZwnm{VH$ pñWam§H$ h¡)
(k is proportion constant)
r πr 2 r πr 2
(A) (B) (A) (B)
k k k k
h h
(C) (D) khr (C) (D) khr
k k

176. A multiple choice examination has 176. EH$ ~hþ{dH$ënr` narjm _| 5 àíZ h¢ & àË`oH$ àíZ _|
5 questions. Each question has {XE J`o 3 CÎmam| _| go Ho$db EH$ ghr h¡ & àm{`H$Vm
3 alternative answers of exactly
one is correct. The probability that
Š`m hmoJr O~{H$ H$moB© {dÚmWu 4 `m A{YH$ ghr
a student will get 4 or more correct CÎma XoVm h¡ ?
answer is
17 13 17 13
(A) (B) (A) (B)
3 5
35 35 35
11 10 11 10
(C) (D) (C) (D)
3 5
35 35 35

π π π π
177. cos2 − sin2 = 177. cos2 − sin2 =
4 12 4 12
(A) 3 2 (B) 4 3 (A) 3 2 (B) 4 3
(C) 3 4 (D) 2 3 (C) 3 4 (D) 2 3

178. If in any ∆ABC, ∠C =105°, ∠B= 45°, 178. `{X {H$gr ∆ABC _|, ∠C =105°, ∠B= 45°,
a = 2, then the value of b is a = 2, Vmo b H$m _mZ h¡
(A) 1 2 (B) 2 2 (A) 1 2 (B) 2 2
(C) 3 2 (D) 2 3 (C) 3 2 (D) 2 3

-61- Set-A

 4  2   4  2 
179. The value of tan cos −1   + tan−1    179. tan cos −1   + tan−1    H$m _mZ hmoJm
will be  5  3   5  3 
(A) 6/11 (B) 6/17 (A) 6/11 (B) 6/17
(C) 11/6 (D) 17/6 (C) 11/6 (D) 17/6

180. If 3 cos θ + sin θ = 2 , then general 180. `{X 3 cos θ + sin θ = 2 , Vmo θ H$m
value of θ is ì`mnH$ _mZ h¡
n π n π
(A) nπ + (−1) (A) nπ + (−1)
4 4
n π π n π π
(B) (−1) − (B) (−1) −
4 3 4 3
π π π π
(C) nπ + − (C) nπ + −
4 3 4 3
π π π π
(D) nπ + (−1)n − (D) nπ + (−1)n −
4 3 4 3
     
181. If a = 5, b = 4 and c = 3, then what 181. `{X a = 5, b = 4 Am¡a c = 3 , Vmo
           
will be the value of a ⋅ b + b ⋅ c + c ⋅ a ? a ⋅ b + b ⋅ c + c ⋅ a H$m _mZ Š`m hmoJm ?
(A) –25 (B) 25 (A) –25 (B) 25
(C) –50 (D) 50 (C) –50 (D) 50
   
182. Let a = i − k , b = xi + j + (1 − x ) k and 182. `{X a = i − k , b = xi + j + (1 − x ) k Am¡a
       
c = y i + x j + (1 + x − y ) k , then [a, b, c ] c = y i + x j + (1 + x − y ) k h¡, Vmo [a, b, c ]
depends on {Z^©a H$aVm h¡
(A) Only y (B) Only x (A) Ho$db y na (B) Ho$db x na
(C) Both x and y (D) Neither x nor y (C) x Am¡a y XmoZm| na (D) Zm x na Zm y na
   
183. Let a = j − k and c = i − j − k , then 183. `{X a = j − k Am¡a c = i − j − k , Vmo
          
vector b satisfying a × b + c = 0 and a × b + c = 0 Am¡a a ⋅ b = 3 H$mo g§VwîQ> H$aZo
  
a ⋅ b = 3 is dmbm g{Xe b h¡
(A) 2i − j + 2k (B) i − j − 2k (A) 2i − j + 2k (B) i − j − 2k
(C) i + j − 2k (D) − i + j − 2k (C) i + j − 2k (D) −i + j − 2k

-62- Set-A

         
If a + b + c = 0 and a = b = 1,
184. 184. `{X a + b + c = 0 Am¡a a = b = 1,
    
c = 3 , then angle between a c = 3 , Vmo a Am¡a b Ho$ ~rM H$m H$moU h¡

and b is
π π
(A) 6 (B) π 3 (A) 6 (B) π 3

(C) 2 π (D) π 2 (C) 2 π (D) π 2
3 3
θ θ
d dx d dx
185. The function f(θ) = ∫
dθ 0 1 − cos θ cos x
185. \$bZ f(θ) = ∫
dθ 0 1 − cos θ cos x
{H$g
satisfies the differential equation AdH$b g_rH$aU H$mo g§VwîQ> H$aVm h¡ ?
df df
(A) + 2f(θ)cot θ = 0 (A) + 2f(θ)cot θ = 0
dθ dθ
df df
(B) dθ − 2f(θ)cot θ = 0 (B) dθ − 2f(θ)cot θ = 0
df df
(C) + 2f(θ) = 0 (C) + 2f(θ) = 0
dθ dθ
df df
(D) dθ − 2f(θ) = 0 (D) dθ − 2f(θ) = 0

dy
186. The solution of the differential equation 186. AdH$b g_rH$aU x = y(log y − log x + 1)
dy dx
x = y(log y − log x + 1) is H$m hb h¡
dx
y2 y2
(A) y = xe cx
(B) = cx (A) y = xe cx
(B) = cx
x x
(C) y2 = cxlogx (D) logy = cx (C) y2 = cxlogx (D) logy = cx

187. The slope of the tangent to the curve 187. dH«$ Ho$ {H$gr {~ÝXþ na ItMr JB© ñne© aoIm H$s àdUVm
at any point is the reciprocal of twice Cg {~ÝXþ Ho$ H$mo{Q> Ho$ XwJZw o Ho$ ì`wËH«$_ h¡, VWm dH«$
the ordinate of that point. The curve
passes through point (4, 3), then the {~ÝXþ (4, 3) go hmoH$a OmVm h¡, Vmo dH«$ h¡
curve is a
(A) Circle (B) Parabola (A) d¥Îm (B) nadb`
(C) Ellipse (D) Hyperbola (C) XrK©d¥Îm (D) A{Vnadb`

-63- Set-A

188. A particle starting from origin and move 188. EH$ H$U _yb{~ÝXþ go àmaå^ H$aHo$$ x-Aj Ho$
along to x-axis in such a way that AZw{Xe Bg àH$ma J_Z H$aVm h¡ {H$ {~ÝXþ (x, 0)
the formula of the motion at the point dy
dy
na BgH$s J{V gyÌ = cos2 π x Ûmam Xr JB© h¡,
(x, 0) is = cos2 π x , then particle dt
dt V~ H$U {H$g {~ÝXþ na H$^r Zht nhþM
± Jo m ?
never reach a point
1 1
(A) x = (A) x =
4 4
3 3
(B) x = (B) x =
4 4
1 1
(C) x = (C) x =
2 2
(D) x = 1 (D) x = 1

189. If a and a + h are the two consecutive 189. `{X a VWm a + h Ý`yQ>Z Ho$ {Z`_ Ûmam ImoOo JE,
and approximate roots of the equation g_rH$aU f(x) = 0 Ho$ Xmo H«${_H$ d g{ÞH$Q> _yb
f(x) = 0 as obtained by Newton’s hmo, Vmo h H$m _mZ {ZåZ hmoJm
method, then h is equal to
(A) f(a) / f ′(a) (B) f ′(a) / f(a) (A) f(a) / f ′(a) (B) f ′(a) / f(a)
(C) − f ′(a) / f(a) (D) − f (a) / f ′(a) (C) − f ′(a) / f(a) (D) − f (a) / f ′(a)

1 2
190. The value of f(x) given only at 190. `{X f(x) H$m _mZ Ho$db x = 0, , , 1 na
1 2 1 2 3
x = 0, , , 1, which of the following kmV hmo, Vmo ∫ f(x) dx H$m g{ÞH$Q> _mZ {ZH$mbZo
2 3 1
0
can be used to evaluate ∫ f(x ) dx Ho$ {bE {ZåZ _| go H$m¡Z-gm à`moJ {H$`m Om
0
approximately ? gH$Vm h¡ ?
(A) Trapezoidal rule (A) g_bå~r {Z`_
(B) Simpson’s rule (B) {gångZ {Z`_
(C) Trapezoidal as well as Simpson’s (C) g_bå~r VWm {gångZ XmoZm| {Z`_
rule
(D) None of the above (D) Cn`w©ŠV _| go H$moB© Zht

-64- Set-A

191. The Newton-Raphson method 191. Ý`yQ>Z-am\$gZ {d{Y H$s A{^g[aVm Vrd« hmoJr, `{X
converges fast, if f ′(α) is (α is the f ′(α) hmoJm (α gQ>rH$ _yb h¡)
exact value of root)
(A) small (B) large (A) N>moQ>m (B) ~‹S>m
(C) equal to 0 (D) none of these (C) eyÝ` Ho$ ~am~a (D) BZ_| go H$moB© Zht
192. Which of the following terms is not used 192. EH$ aoIr` àH«$_Z g_ñ`m _| {ZåZ _| go H$m¡Z-gm
in a linear programming problem ? eãX à`wŠV Zht hmoVm h¡ ?
(A) Slack variables (A) ñb¡H$ Ma
(B) Objective function (B) CÔoí` \$bZ
(C) Concave region (C) AdVb joÌ
(D) feasible region (D) gwg§JV hb

 1+ x   2x   1+ x   2x 
193. If f(x ) = log   , then f  2 
is 193. `{X f(x) = log  , Vmo f 
equal to  1 − x   1+ x   1 − x  2 
 1+ x 
~am~a hmoJm
(A) [f(x)]2 (B) [f(x)]3 (A) [f(x)]2 (B) [f(x)]3
(C) 2f(x) (D) 3f(x) (C) 2f(x) (D) 3f(x)

194. If f(x) = xa ⋅ logx with f(0) = 0, f(x) satisfies 194. `{X f(x) = xa ⋅ logx, f(0) = 0 Bg àH$ma hmo {H$
Rolle’s theorem is [0, 1], then value of f(x), amobr Ho$ à_o` H$mo [0, 1] _| g§VwîQ>> H$aVm h¡,
a is Vmo a H$m _mZ hmoJm
(A) –2 (B) –1 (A) –2 (B) –1
1 1
(C) 0 (D) 2 (C) 0 (D) 2

195. A triangular park is enclosed on two 195. EH$ {Ì^wOr` nmH©$ Xmo Va\$ go g_mZ b§~mB© x Ho$
sides by a fence and third side by Vma go Am¡a Vrgar Amoa ZXr Ho$ grYo VQ> go {Kam
straight river bank. The two sides fence hþAm h¡ & nmH©$ Ûmam {Kam A{YH$V_ joÌ\$b hmoJm
are of same length x. The maximum
area enclosed by the park is
x3 1 2 x3 1 2
(A) (B) x (A) (B) x
8 2 8 2
3 2 3 2
(C) πx (C) πx
2 2
(D) x (D) x
2 2

-65- Set-A

x2 − 1 x2 − 1
196. f( x ) = , value of f(1) will be 196. f( x ) = , Vmo f(1) H$m _mZ hmoJm
x3 − 1 x3 − 1
0 2 0 2
(A) (B) (A) (B)
0 3 0 3
2 2
(C) ∞ (D) − (C) ∞ (D) −
3 3
197. If f : R → R satisfies f(x + y) = f(x) 197. `{X f : R → R Bg àH$ma h¡ {H$
+ f(y), for all x, y ∈ R and f(1) = 7, f(x + y) = f(x) + f(y), g^r x, y ∈ R Am¡a
n n
then ∑ f(r ) is f(1) = 7, Vmo ∑ f(r ) h¡
r =1 r =1
7n(n + 1) 7(n + 1) 7n(n + 1) 7(n + 1)
(A) (B) (A) (B)
2 2 2 2
7n 7n
(C) 7n (n + 1) (D) (C) 7n (n + 1) (D)
2 2
−1 −1
198. The value of 4[cot 3 + cosec 5 ] is 198. 4[cot −1 3 + cosec −1 5 ] H$m _mZ ~am~a
equal to hmoJm
π π
(A) π (B) (A) π (B)
2 2
π 3π π 3π
(C) (D) (C) (D)
4 4 4 4

1 1
199. lim (cos x + a sin bx ) x is equal to 199. lim (cos x + a sin bx ) x H$m _mZ hmoJm
x →0 x →0
a b a b
(A) e b (B) e a (A) e b (B) e a
(C) eab (D) None of these (C) eab (D) BZ_| go H$moB© Zht

d2 y d2 y
200. If ey + xy = e, then the value of 2
for 200. `{X ey + xy = e h¡, Vmo 2 H$m x = 0 na _mZ
dx dx
x = 0 is hmoJm
1 1 1 1
(A) (B) 2 (A) (B)
e e e e2
1 1
(C) (D) None of these (C) (D) BZ_| go H$moB© Zht
e3 e3

-66- Set-A

PART – III (B)
Biology Ord {dkmZ
121. Which of the following Mendelian law 121. ghb½ZVm Ho$ Ûmam _oÊS>b
o Ho$ {H$g {Z`_ H$m Cëb§KZ
will be violated by linkage ? {H$`m OmVm h¡ ?
(A) Law of dominance (A) à^m{dVm H$m {Z`_
(B) Law of segregation of gametes (B) Jo_oQ>²g n¥WŠH$aU H$m {Z`_
(C) Law of independent assortment (C) ñdV§Ì Anì`yhZ H$m {Z`_
(D) All the above (D) CnamoŠV g^r

122. DNA polymerase III holoenzyme 122. DNA nm°br_oaoO III hmobmoEÝOmB_ _| hmoVr h¡
possesses
(A) 3′→ 5′ endonuclease and (A) 3′ → 5′ EÊS>moÝ`ypŠbEO Ed§ nm°br_oaoO
polymerase activity {H«$`m
(B) 3 → 5 endonuclease activity only (B) Ho$db 3 → 5 EÊS>moÝ`ypŠbEO {H«$`m
(C) Polymerase activity only (C) Ho$db nm°br_oaoO {H«$`m
(D) None of the above (D) Cn`w©ŠV _| go H$moB© Zht

123. Elongation of peptide chain involves 123. noßQ>mBS> ûm¥§Ibm Ho$ XrKuH$aU _| {H$go N>mo‹S>H$a ~mH$s
all except g^r ^mJ boVo h¢ ?
(A) EE-Tu, EE-Ts and EF-G factor (A) EE-Tu, EE-Ts Ed§ EF-G H$maH$
(B) GTP and peptidyl transferase (B) GTP Ed§ noßQ>r{S>b Q´>m§g\o$aoO
(C) mRNA (C) mRNA
(D) Formyl met tRNA (D) \$m{_©b _oQ> tRNA

124. T2 - phage is a 124. T2- \o$O h¡
(A) ss DNA phage (A) ss DNA \o$O
(B) ds DNA phage (B) ds DNA \o$O
(C) ss RNA phage (C) ss RNA \o$O
(D) ds RNA phage (D) ds RNA \o$O

125. In Spirogyra, meiosis takes place in 125.ñnmBamoJm`am _| AY©gÌy r {d^mOZ {H$g_| hmoVm h¡ ?
(A) Developing zygospore (A) {dH${gV hmoVo hþE Om`Jmoñnmoa _|
(B) Conjugation process (B) g§`w½_Z à{H«$`m
(C) During gamete formation (C) `w½_H$m| Ho$ {Z_m©U Ho$ Xm¡amZ
(D) During vegetative reproduction (D) dYu àOZZ Ho$ Xm¡amZ

-67- Set-A

126. Protonema forms leafy gametophyte in 126. {H$g_| n{U©b `w½_H$moX²{^X H$m {Z_m©U àmoQ>moZr_m go
hmoVm h¡ ?
(A) Riccia (A) [apŠg`m
(B) Funaria (B) â`yZo[a`m
(C) Anthoceros (C) EÝWmo{gamog
(D) Marchantia (D) _mH}$pÝe`m

127. Which of the following is known as 127. {ZåZ _o§ go {H$go Imgr-nm_ Ho$ Zm_ go OmZm
Khasi-palm ? OmVm h¡ ?
(A) Pinus insularis (A) nmBZg BÝgwbo[ag
(B) Cycas revoluta (B) gm`H$g [admoë`yQ>m
(C) Pinus wallichiana (C) nmBZg d¡br{M`mZm
(D) Araucaria (D) AamD$Ho$[a`m

128. Male gametophyte of angiosperm is 128. Amd¥Îm~r{O`m| _| Za `w½_H$moX²{^X {H$gHo$ ê$n _|
shed as {JaVm h¡ ?
(A) Four celled pollen grain (A) Mma H$mo{eH$s` namJH$U
(B) Three shed pollen grain (B) VrZ H$mo{eH$s` namJH$U
(C) Microspore mother cell (C) bKw~rOmUw _mV¥ H$mo{eH$m
(D) Anther (D) namJH$moe

129. Which of the following is a botanical 129. _gya H$m dmZñn{VH$ Zm_ {ZåZ _| go Š`m h¡ ?
name of Masoor ?
(A) Lens eculenta (A) boÝg EñŠ`wboÝQ>m
(B) Cajanus cajan (B) H¡$OoZg H$OmZ
(C) Phaseolus aureus (C) \o${gAmobg Amo[a`g
(D) Cicer arietinum (D) gmBga E[a{Q>Z_

130. Drugs are obtained from which part of 130. ~«÷r nmXn Ho$ {H$g ^mJ go Am¡fYr àmßV H$s
Brahmi plant ? OmVr h¡ ?
(A) Root (B) Bark (A) O‹S> (B) N>mb
(C) Stem (D) Leaf (C) VZm (D) nÎmr

-68- Set-A

131. Crotalaria juncea is a name of which 131. H«$moQ>obo[a`m OZ{g`m {ZåZ _| go {H$gH$m Zm_ h¡ ?
of the following ?
(A) Jute (B) Flax (A) nQ>gZ (B) âboŠg
(C) Sunn Hemp (D) Hemp (C) gZB© (D) hoån
132. Which of the following acid is used in 132. {ZåZ _| go {H$g Aåb H$m Cn`moJ ^moÁ` nXmWm] Ho$
preservation of food materials ? n[aajU _| {H$`m OmVm h¡ ?
(A) Acetic acid (B) Lactic acid (A) E{gQ>rH$ Aåb (B) bopŠQ>H$ Aåb
(C) Benzoic acid (D) All the above (C) ~oݵOmoBH$ Aåb (D) CnamoŠV g^r

133. Deficiency of which of the following 133. {ZåZ _| go {H$g I{ZO VËd H$s H$_r go nwZ:J«hU
mineral element causes Reclamation amoJ hmoVm h¡ ?
disease ?
(A) Copper (B) Zinc (A) H$m°na (B) qOH$
(C) Boron (D) Molybdenum (C) ~moamZ (D) _m{bãS>oZ_
134. Which of the following is a site of 134. nm`é{dH$ Aåb go E{gQ>mBb H$moEÝOmB_ H$m {Z_m©U
synthesis of Acetyl CoA from Pyruvic ñWb {ZåZ _| go H$m¡Z-gm h¡ ?
acid ?
(A) Cytoplasm (B) Chloroplast (A) H$mo{eH$m Ðì` (B) ŠbmoamoßbmñQ>
(C) Mitochondria (D) Nucleus (C) _mBQ>moH$m°pÊS´>`m (D) Ho$ÝÐH$
135. Phosphoglycolate is synthesised in 135. \$mñ\$mo½bm`H$moboQ> H$m {Z_m©U {ZåZ _| go {H$g
which of the following process ? à{H«$`m _| hmoVm h¡ ?
(A) Calvin cycle (A) Ho$pëdZ MH«$
(B) Hatch and Slack pathway (B) hoM Ed§ ñboH$ MH«$
(C) Photorespiration (C) àH$mer`ídgZ
(D) C.A.M. cycle (D) gr.E.E_. MH«$

136. Soyabean is an example of which of 136. gmo`m~rZ {ZåZ _| go {H$gH$m CXmhaU h¡ ?
the following ?
(A) Short day plant (A) Aën àH$mer` nmXn
(B) Long day plant (B) XrK© àH$mer` nmXn
(C) Short night plant (C) Aën {Zem nmXn
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

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137. ‘Jaya’ and ‘Ratna’ are varieties of 137. "O`m' Ed§ "aËZm' {ZåZ _| go {H$gH$s doamBQ>r h¢ ?
which of the following ?
(A) Rice (B) Wheat (A) Mmdb (B) J|hÿ
(C) Maize (D) Sugar-cane (C) _ŠH$m (D) JÝZm
138. Bt cotton provides security against 138. ~rQ>r H$m°Q>Z {ZåZ _| go {H$gHo$ à{V gwajm àXmZ
which of the following ? H$aVm h¡ ?
(A) Fungi (B) Bacteria (A) H$dH$ (B) OrdmUw
(C) Insects (D) All the above (C) H$sQ> (D) CnamoŠV g^r

139. R.N.A. interference gives security 139. Ama.EZ.E. B§Q>a\o$aÝg H$m Cn`moJ {ZåZ _| go {H$gHo$
against which of the following? à{V gwajm àXmZ H$aVm h¡ ?
(A) Nematodes (B) Mycoplasma (A) Zr_oQ>moS> (B) _mBH$moßbmÁ_m
(C) Bacteria (D) Virus (C) OrdmUw (D) {dfmUw

140. Lactobacillus bulgaricus is mainly 140. boŠQ>mo~ogrbg ~ëJoarH$g H$m _w»` Cn`moJ {ZåZ _|
used for the production of which of the go {H$gHo$ {Z_m©U Ho$ {bE {H$`m OmVm h¡ ?
following ?
(A) Yoghurt (B) Cheese (A) `moJQ>© (B) nZra
(C) Alcohol (D) Antibiotics (C) AëH$mohb (D) à{VO¡{dH$

141. Gradual increase of D.D.T. in aquatic 141. Obr` ImÚ ûm§¥Ibm _| S>r.S>r.Q>r. H$s _mÌm H$m
food chain is known as H«$_e: ~‹T> Zm H$hbmVm h¡
(A) Eutrophication (A) gwnmofU
(B) Algal bloom (B) e¡dmb àñ\y$Q>Z
(C) Bio-magnification (C) O¡d-AmdY©Z
(D) Biodegradation (D) O¡d-{ZåZrH$aU

142. Which of the following is C.N.G. ? 142. gr.EZ.Or. {ZåZ _| go Š`m h¡ ?
(A) Petrol (A) noQ´>mob
(B) Petrol without lead (B) grgm a{hV noQ´>mob
(C) Diesel (C) S>rµOb
(D) Natural gas (D) àmH¥${VH$ J¡g

-70- Set-A

143. Which of the following is absorbed from 143. n¥Ïdr go J«rZ hmCg J¡g {ZåZ _| go {H$gH$m
earth by green house gas ? AdemofU H$aVr h¡ ?
(A) Ultra violet radiation (A) nam~¢JZr {d{H$aU
(B) Infra-red radiation (B) XrK© Va§J {d{H$aU
(C) Ozone (C) AmoµOmoZ
(D) Carbon dioxide (D) H$m~©Z S>mBAm°ŠgmBS>

144. Which of the following is a non-drying 144. {ZåZ _| go H$m¡Z EH$ Z gyIZo dmbm Vob h¡ ?
oil ?
(A) Linseed (A) Abgr
(B) Soyabean (B) gmo`m~rZ
(C) Margosa (C) Zr_
(D) Ground nut (D) _y±J\$br

145. Which of the following statement is true 145. `yHo$[a`m°Q²>g Ho$ g§X^© _| H$m¡Z-gm H$WZ ghr h¡ ?
about eukaryotes ?
a. They are cells with a nucleus. a. `o Zm{^H$ `wŠV H$mo{eH$m dmbo hmoVo h¡§ &
b. They are found in human and b. `o _Zwî` _| nm`o OmVo h¢ VWm `o ~hþH$mo{eH$s`
multicellular organisms. Ord hmoVo h¢ &
c. Endoplasmic reticulum is present c. `yHo$[a`m°Q²>g _| AÝV:àX«ì`r Om{bH$m nmB©
in eukaryotes. OmVm h¡ &
d. They have chemically complexed d. BZ_| amgm`{ZH$ ê$n go O{Q>b H$mo{eH$m {^{Îm
cell wall. hmoVr h¡ &
(A) a, b and c (B) a, c and d (A) a, b Ed§ c (B) a, c Ed§ d
(C) a, b and d (D) All are correct (C) a, b Ed§ d (D) CnamoŠV g^r

146. Important site for formation of 146. ½bmBH$moàmoQ>rZ Ed§ ½bmBH$mo{b{nS²>g H$m _hËdnyU©
glycoproteins and glycolipids is {Z_m©U ñWb h¡
(A) Endoplasmic reticulum (A) AÝV:àÐì`r Om{bH$m
(B) Golgi apparatus (B) Jm°ëJr H$m`
(C) Lysosome (C) bm`gmogmo_
(D) All the above (D) CnamoŠV g^r

-71- Set-A

147. Cell organelle is present only in plants is 147. Ho$db nm¡Ym| _| nm`m OmZo dmbm H$mo{eH$m§J h¡
(A) Glyoxysome (A) ½bmBAm°Šgrgmo_
(B) Ribosome (B) amB~mogmo_
(C) Lysosome (C) bm`gmogmo_
(D) Peroxysome (D) nam°Šgrgmo_

148. Phagosomes and pinosomes are 148. \¡$Jmogmo_ Ed§ {nZmogmo_ H$mo gm_y{hH$ ê$n go H$hVo h¢
collectively called
(A) Digestive vacuoles (A) nmMH$ [apŠVH$m
(B) Residual bodies (B) Ad{eîQ> H$m`
(C) Endosomes (C) EÊS>mogmo_
(D) Autophagic bodies (D) ñd`§^jH$ H$m`
149. Which of the following pair have 149. {ZåZ _| go {H$g OmoS‹ >r _| AJw{UV g§aMZmE± hmoVr h¡§ ?
haploid structures ?
(A) Nucellus and antipodal cells (A) Ý`ygobg Ed§ EÊQ>rnmoS>b H$mo{eH$mE±
(B) Antipodal cells and egg cell (B) EÊQ>rnmoS>b H$mo{eH$mE± Ed§ AÊS> H$mo{eH$m
(C) Antipodal cells and megaspore (C) EÊQ>rnmoS>b H$mo{eH$mE± Ed§ Jwé~rOmUw _mV¥
mother cells H$mo{eH$mE±
(D) Nucellus and primary endosperm (D) ~rOmÊS>H$m` Ed§ àmW{_H$ ^«yUnmof Zm{^H$
nucleus
150. Filiform apparatus is found in 150. {\$br\$m_© EnoaoQ>g nm`m OmVm h¡
(A) Synergids (A) {gZ{O©S²>g _|
(B) Antipodals (B) EÊQ>rnmoS>ëg _|
(C) Polar nuclei (C) nmoba Zm{^H$m| _|
(D) Zygote (D) Om`JmoQ> _|

151. In angiosperms all the four microspores Amd¥Îm~r{O`m| _| Q>oQ´>mS> _| CnpñWV g^r Mma
151.
of tetrad are covered by a layer which bKw~rOmUw EH$ AmdaU `m ñVa Ho$ Ûmam T>Ho$ hmoVo
is formed by h¢ Omo {H$ ~Zm hmoVm h¡
(A) Callose (A) Ho$bmoO H$m
(B) Cellulose (B) goë`wbmoO H$m
(C) Sporopollenin (C) ñnmoamonmobo{ZZ H$m
(D) Pectolactose (D) noŠQ>mob¡ŠQ>moO H$m

-72- Set-A

152. Phloem fibres are made up of which 152. âbmoE_ V§Vw {ZåZ _| go {H$g D$VH$ Ho$ ~Zo hmoVo h¢ ?
of the following tissue ?
(A) Parenchyma (A) _¥XÿVH$
(B) Collenchyma (B) ñWybH$moU D$VH$
(C) Sclerenchyma (C) Ñ‹T> moVH$
(D) All the above (D) CnamoŠV g^r

153. Pfr stage of phytochrome stimulates 153. \$mBQ>moH«$mo_ H$s Pfr AdñWm {ZåZ _| go {H$Z nmXnm|
flowering in which of the following _| nwî>nZ H$m CÔrnZ H$aVr h¡ ?
plants ?
(A) Short day plants (A) Aën àH$mer` nmXn
(B) Long day plants (B) XrK© àH$mer` nmXn
(C) Long night plants (C) XrK© {Zem nmXn
(D) All the above (D) CnamoŠV g^r

154. Which of the following hormone is a 154. {ZåZ _| go H$m¡Z d¥{ÕamoYH$ hm_m}Z h¡ ?
growth inhibitor ?
(A) Auxin (A) AmpŠgZ
(B) Gibberellin (B) {O~ao{bZ
(C) Cytokinin (C) gm`Q>moH$m`{ZZ
(D) A.B.A. (D) E.~r.E.

155. AH2 → A + 2H 155. AH2 → A + 2H
(A = substrate) (A = nXmW©)
Which of the following is a class of Bg A{^{H«$`m H$mo CËào[aV H$aZo dmbo {dH$a H$m dJ©
enzyme catalysing this reaction ? {ZåZ _| go H$m¡Z-gm h¡ ?
(A) Oxido-reductase (A) AmpŠgS>mo-[aS>ŠQ>og
(B) Transferase (B) Q´>mÝg\$aoµO
(C) Hydrolases (C) hmBS´>moboOog
(D) Isomerases (D) AmBgmo_aoµO

-73- Set-A

156. Copepods are known as which of the 156. Vmbm~ Ho$ nm[apñW{VH$ V§Ì _| H$mononmoS>²g H$mo {ZåZ
following in Pond Ecosystem ? _| go Š`m H$hm OmVm h¡ ?
(A) Producer (A) CËnmXH$
(B) Primary consumer (B) àmW{_H$ Cn^moŠVm
(C) Secondary consumer (C) {ÛVr`H$ Cn^moŠVm
(D) Decomposer (D) AnKQ>H$

157. Which of the following organism is 157. {ZåZ _| go H$m¡Z-gm OrdYmar namgaUr` ê$n go
osmotically inactive ? A{H«$` hmoVm h¡ ?
(A) Mycoplasma (B) Bacteria (A) _mBH$moßbmÁ_m (B) OrdmUw
(C) Nostoc (D) All the above (C) Zm°ñQ>m°H$ (D) CnamoŠV g^r

158. The amino acid found only in bacteria 158. Ho$db ~¡ŠQ>r[a`m Ed§ Zrbr har e¡dmbm| _| nm`m OmZo
and blue green algae is dmbm A_rZm| Aåb h¡
(A) Aspartic acid (A) Eñnm{Q>©H$ Aåb
(B) Glumatic acid (B) ½byQ>o{_H$ Aåb
(C) Diamino-pimetic acid (C) S>mB©E_rZmo-nm_o{Q>H$ Aåb
(D) Methionine (D) {_{W`mo{ZZ

159. Muramic acid is found in the cell walls of 159. å`yao{_H$ Aåb {H$gH$s H$mo{eH$m {^{Îm`m| _| nm`m
OmVm h¡ ?
(A) Blue-green algae (B) Bacteria (A) Zrbr-har e¡dmb (B) OrdmUw
(C) Green algae (D) Fungi (C) har e¡dmb (D) H$dH$

160. Which of the following is true about cell 160. J«m_-YZmË_H$ OrdmUwAm| H$s H$mo{eH$m {^{Îm Ho$
wall of gram-positive bacteria ? g§X^© _| {ZåZ _| go Š`m ghr h¡ ?
(A) Cell wall comprises of many (A) H$mo{eH$m {^{Îm H$B© ñVam| H$s ~Zr hmoVr h¡
layers
(B) Cell wall is thicker than the (B) H$mo{eH$m {^{Îm g§~pÝYV J«m_ F$UmË_H$
associated gram negative bacteria OrdmUwAm| H$s Anojm _moQ>r hmoVr h¡
(C) Cell wall comprises teichoic acid (C) H$mo{eH$m {^{Îm {Q>H$moBH$ Aåb H$s ~Zr
hmoVr h¡
(D) All the above (D) CnamoŠV g^r

-74- Set-A

161. Which of the following hormone is a 161. H$m¡Z-gm hm_m}Z A_rZmo Aåb H$m _m°{S>\$mBS>
modified amino acid ? ê$n h¡ ?
(A) Prostaglandin (A) àmoñQ>o½bopÝS>Z
(B) Estrogen (B) EñQ´>moOoZ
(C) Epinephrine (C) BnrZo{\«$Z
(D) Progesterone (D) àmoOoñQ´>m°Z

162. Human placenta is 162. _Zwî` H$m ßbogoÝQ>m h¡
(A) Haemochorial (A) hr_moH$m°[a`b
(B) Haemo-endothelial (B) hr_mo-EÝS>moWr{b`b
(C) Syndesmo-chorial (C) {gÝS>oñ_mo-H$m°[a`b
(D) Epitheliochorial (D) EnrWr{b`moH$m°[a`b

163. Which one of the following statement 163. _mZd ewH«$mUw Ho$ {df` _| {ZåZ{b{IV _| go
about human sperm is correct ? H$m¡Z-gm EH$ H$WZ ghr h¡ ?
(A) Acrosome serve no particular (A) EH«$mogmo_ H$m H$moB© {deof H$m`© Zht hmoVm
function
(B) Acrosome has a conical pointed (B) EH«$mogmo_ _| EH$ e§ŠdmH$ma ZwH$sbr g§aMZm
structure used for piercing and hmoVr h¡ {OgH$m Cn`moJ AÊS>o H$mo ^oYZo _|
penetrating the egg, resulting in Am¡a CgHo$ ^rVa àdoe H$aZo Ho$ {bE {H$`m
fertilization OmVm h¡ {Oggo {ZfoMZ hmoVm h¡
(C) The sperm lysins in the acrosome (C) EH«$mogmo_ _| ewH«$mUw bmB{gZ AÊS>o Ho$ AmdaU
dissolve the egg envelope H$mo Kmob XoVo h¢ {Oggo {ZfoMZ Ho$ {bE gw{dYm
facilitating fertilization hmo OmVr h¡
(D) Acrosome serves as a sensory (D) EH«$mogmo_ EH$ g§doXr g§aMZm H$m H$m`© H$aVm h¡
structure leading the sperm {Oggo ewH«$mUw AÊS>mUw H$s Amoa AmJo-AmJo
towards to ovum ~‹T>Vm OmVm h¡

-75- Set-A

164. The eggs of some mammals have 164. Hw$N> ñVZYm[a`m| Ho$ EJ _| Á`mXm `moH$ hmoVm h¡ &
more yolk. They are do h¢
(A) Metatherian (A) _oQ>mWr[a`Z
(B) Prototherians (B) àmoQ>moWr[a`Z
(C) Eutherians (C) `yWr[a`Z
(D) Aquatic mammals (D) Obr` ñVZYmar

165. Match the following and choose the {ZåZ H$mo gw_o{bV H$s{OE Am¡a ghr {dH$ën
165.
correct option. Mw{ZE &
Kingdom Proposed by qH$JS>_ Ho$ Ûmam à{Vnm{XV
1. Four kingdom a. R.H. Whittaker 1. Mma qH$JS>_ a. Ama.EM. pìhQ>Ho $a
2. Five kingdom b. E. Haeckle 2. nm±M qH$JS>_ b. B©. hHo$b
3. Three kingdom c. C. Linnaeus 3. VrZ qH$JS>_ c. gr. brZr`g
4. Two kingdom d. Copeland 4. Xmo qH$JS>_ d. H$monrb¢S>
(A) 1 – d, 2 – a, 3 – b, 4 – c (A) 1 – d, 2 – a, 3 – b, 4 – c
(B) 1 – b, 2 – a, 3 – c, 4 – d (B) 1 – b, 2 – a, 3 – c, 4 – d
(C) 1 – d, 2 – c, 3 – b, 4 – a (C) 1 – d, 2 – c, 3 – b, 4 – a
(D) 1 – a, 2 – c, 3 – b, 4 – d (D) 1 – a, 2 – c, 3 – b, 4 – d

166. Match the following and choose the 166. {ZåZ H$mo gw_o{bV H$s{OE Am¡a ghr {dH$ën
correct option. Mw{ZE &
List – I List – II gyMr> - I gyMr> - II
a. Daphnia I. Insecta a. S>¡\$Zr`m I. BÝgoŠQ>m
b. Lepisma II. Arachnida b. bonrÁ_m II. Ea¡ŠZrS>m
c. Palamnacus III. Merostomata c. n¡bo_¡Zg III. _ramoñQ>mo_mQ>m
d. Limulus IV. Crustacea d. brå`ybg IV. H«$ñQ>oer`m
(A) a – I, b – II, c – III, d – IV (A) a – I, b – II, c – III, d – IV
(B) a – IV, b – I, c – II, d – III (B) a – IV, b – I, c – II, d – III
(C) a – IV, b – II, c – III, d – I (C) a – IV, b – II, c – III, d – I
(D) a – II, b – III, c – IV, d – I (D) a – II, b – III, c – IV, d – I

-76- Set-A

167. The characteristic feature of mammal 167. ñVZYmar H$m {deofVmnyU© bjU S>m`\«$m_ H$s
is the presence of diaphragm which is CnpñW{V h¡, Omo hmoVm h¡
(A) Cartilaginous (A) H$mQ>uboOrZg (H$mQ>uboO H$m ~Zm)
(B) Bony (B) ~moZr (ApñW H$m ~Zm)
(C) Membranous (C) _oå~«oZg ({Pëbr Zw_m)
(D) Muscular (D) _ñHw$ba (_m§gnoer `wŠV)

168. Due to discovery of which of the 168. 1980 _| {H$gHo$ Am{dîH$ma Ho$ H$maU {dH$mg H$mo
following in 1980, the evolution was RNA H$s Xþ{Z`m H$hm J`m ?
termed as RNA world ?
(A) mRNA, t-RNA, r-RNA synthesis (A) mRNA, t-RNA, r-RNA go àmoQ>rZ
to protein g§íbofU
(B) in some virus, RNA is genetic (B) Hw$N> {dfmUw _| RNA AmZwd§{eH$s nXmW©
materials hmoVm h¡
(C) RNA have enzymatic property (C) RNA _| EÝOmB_ H$m JwU hmoVm h¡
(D) RNA is not found in all cells (D) RNA g^r H$mo{eH$mAm| _| Zht nm`m OmVm h¡

169. The synthetic drugs structurally similar 169. ES´>oZo{bZ Ho$ g§aMZm go g_Vwë` g§íbo{fV S´>J
to adrenaline are hmoVm h¡
(A) amphetamines (A) Eå\o$Q>m_mBÝg
(B) barbiturates (B) ~m~uQ²>`yaoQ>
(C) hallucinogens (C) h¡bwgrZmoOZ
(D) nicotinic derivatives (D) {ZH$mo{Q>ZrH$ ì`wËn{Îm

170. ‘MERLIN’ is new environmental body 170. "MERLIN' EH$ ZdrZ dmVmdaUr` g§aMZm h¡,
which determines Omo {ZYm©[aV H$aVm h¡
(A) photosynthetic rate (A) àH$me g§íbofU H$s Xa
(B) humidity in atmosphere (B) dmVmdaU _| AmЩVm
(C) water pollution (C) Obr` àXyfU
(D) amount of ozone (D) AmoµOmoZ H$s _mÌm

-77- Set-A

171. The pollutant that directly affects the 171. àXyfH$ Omo grYm Ho$ÝÐr` V§{ÌH$m V§Ì H$mo à^m{dV
nervous system is H$aVm h¡
(A) Organophosphates (A) AmaJoZmo\$mñ\o$Q>
(B) DDT (B) S>r.S>r.Q>r.
(C) Aldrin (C) EëS´>rZ
(D) Lindane (D) {bÝS>oZ

172.
Match the following and choose the {ZåZ{b{IV H$mo gw_o{bV H$s{OE Am¡a ghr {dH$ën
172.
correct option. MwZ| &
I II I II
a. DDT 1. CO, CO2 a. S>r.S>r.Q>r. 1. CO, CO2
b. PAN 2. Smog b. PAN 2. ñ_m°J
c. Acid rain 3. Biological c. Aåb dfm© 3. O¡{dH$ AmdY©Z
magnification
d. Global 4. SO2 d. d¡pídH$ Vmn_mZ 4. SO2
warming
(A) a – 4, b – 3, c – 2, d – 1 (A) a – 4, b – 3, c – 2, d – 1
(B) a – 1, b – 3, c – 2, d – 4 (B) a – 1, b – 3, c – 2, d – 4
(C) a – 2, b – 3, c – 4, d – 1 (C) a – 2, b – 3, c – 4, d – 1
(D) a – 3, b – 2, c – 4, d – 1 (D) a – 3, b – 2, c – 4, d – 1

173. Mammals can regenerate 173. ñVZYmar [aOZaoQ> H$a gH$Vo h¢
(A) Liver (A) `H¥$V
(B) Brain (B) _pñVîH$
(C) Lung (C) \o$\$‹S>m
(D) Kidney (D) d¥ŠH$

174. Which pair is a chromosomal aberration ? 174. H$m¡Z-gr Omo‹S>r H«$mo_mogmo_b E~oaoeZ h¡ ?
(A) Duplication and translocation (A) Sy>ßbrHo$eZ d Q´>m§gbmoHo$eZ
(B) Duplication and transduction (B) Sy>ßbrHo$eZ d Q´>m§gS>ŠeZ
(C) Duplication and transversion (C) Sy>ßbrHo$eZ d Q´>m§gdg©Z
(D) All of these (D) `o g^r

-78- Set-A

175. A person who is trisomic for twenty first 175. BŠH$sgdt Omo‹S>r Ho$ JwUgyÌmo§ Ho$ {bE {Ìg_gyÌr
pair of chromosome is (trisomic) ì`pŠV hmoVm h¡
(A) Klinfelter’s syndrome (A) pŠbZ\o$ëQ>a {gÝS´>mo_
(B) Down’s syndrome (B) S>mCÝg {gÝS´>mo_
(C) Turner’s syndrome (C) Q>Z©a {gÝS´>mo_
(D) None of these (D) BZ_| go H$moB© Zht

176. Sex linked characters are usually 176. gm_mÝ`V: goŠg qbŠS> JwU hmoVo h¢
(A) Dominant (A) à^mdr
(B) Found only in male (B) Ho$db Za _| nm`o OmVo h¢
(C) Found only in female (C) Ho$db _mXm _| nm`o OmVo h¢
(D) Recessive (D) aogo{gd

177. Which evidence suggest that man is 177. {H$g gmú` go ñnîQ> hmoVm h¡ {H$ _Zwî` AÝ`
more closely related with chimpanzee hmo_moZm°`S> Eßg H$s VwbZm _| {MånmÝOr Ho$ µÁ`mXm
than with other homonoid apes ? ZOXrH$ h¡ ?
(A) Evidence from DNA of sex (A) Ho$db goŠg JwUgyÌ go àmßV DNA Ho$
chromosome only gmú` go
(B) Comparison of chromosome (B) Ho$db JwUgyÌ Ho$ AmH$m[aH$s H$s$ VwbZm go
morphology only
(C) Evidence from fossil remains and (C) Ordmí_ VWm Ordmí_ _mBQ>moH$mBÝS´>r`m Ho$
the fossil mitochondrial DNA DNA Ho$ gmú` go
(D) Evidence from DNA extracted (D) goŠg JwUgyÌ, Am°Q>mg
o mo_ VWm _mBQ>mHo $mBÝS´>r`m
from sex chromosome, autosome go {ZH$mbo JE DNA Ho$ gmú` go
and mitochondria
178. Secretion of silk gland comes through 178. {gëH$ J«§{W H$m òmd EH$ AË`§V N>moQ>o {N>Ð go ~mha
a small pore. This pore is situated on AmVm h¡ & `h {N>Ð pñWV hmoVm h¡
(A) Exopodite of 2nd maxilla (A) {ÛVr` _¡pŠgbm Ho$ ~mønmXm§e na
(B) Endopodite of 1st maxilla (B) àW_ _¡pŠgbm Ho$ A§V:nmXm§e na
(C) Prostheca of mandible (C) _¡pÝS>~b Ho$ àm°pñWH$m na
(D) Anterior part of hypopharynx (D) AYmoJ«gZr Ho$ AJ« ^mJ na

-79- Set-A

179. In malaria infection, Iysolecithin, that 179. _bo[a`m Ho$ g§H«$_U _| RBC H$mo ZîQ> H$aZo dmbo
destroys the RBC, is secreted by bmBgmobogrWrZ H$m òmdU H$aVm h¡
(A) Spleen (A) ßbrhm
(B) WBC (B) g\o$X aŠV H$U (WBC)
(C) Liver (C) `H¥$V
(D) Plasmodium (D) ßbmµÁ_mo{S>`_

180. Malarial parasite is best obtained from 180. _arµO go g~go AÀN>m _bo[a`m naOrdr àmßV {H$`m
a patient Om gH$Vm h¡
(A) an hour before rise of temperature (A) Vmn_mZ d¥{Õ go EH$ K§Q>m nyd©
(B) after temperature become normal (B) O~ Áda gm_mÝ` hmo OmVm h¡
(C) while temperature is rising sharply (C) O~ Áda Vrd«Vm go ~‹T>Vm ahVm h¡
(D) after 24 hours when temperature (D) Áda gm_mÝ` hmoZo Ho$ 24 K§Q>o níMmV²
is normal
181. Match List – I and List – II and select gyMr> – I VWm gyMr – II H$mo gw_o{bV H$a| VWm ghr
181.
correct answer. CÎma MwZ| &
List – I List – II gyMr – I gyMr – II
a. Crypt of p. Loop of a. H«$sßQ> Am°\$ p. S²>`yS>oZ_ H$m byn
Leiberkuhn duodenum br~aIwZ
b. Pancreas q. Stomach b. A½Zme` q. Am_me`
c. Adrenal gland r. Intestine c. ES>arZb J«§Wr r. Am§V
d. Gastric gland s. Kidney d. OR>a J«§Wr s. d¥ŠH$
(A) a – q, b – r, c – s, d – p (A) a – q, b – r, c – s, d – p
(B) a – r, b – p, c – s, d – q (B) a – r, b – p, c – s, d – q
(C) a – r, b – p, c – q, d – s (C) a – r, b – p, c – q, d – s
(D) a – p, b – s, c – r, d – q (D) a – p, b – s, c – r, d – q

182. Antibodies in our body are complex of 182. h_mao eara H$s E§Q>r~m°S>r {H$gHo$ H$måßboŠg hmoVo h¡ ?
(A) Glycoproteins (A) ½bm`H$moàmoQ>rZ
(B) Lipoproteins (B) bmBnmoàmoQ>rZ
(C) Steroids (C) ñQ>ram°`S>
(D) Prostaglandins (D) àmñQ>o½boÝS>rZ

-80- Set-A

183. Substance which is not absorbed in 183. _yÌ _| H$m¡Z-gm nXmW© Ademo{fV Zht hmoVm h¡ ?
urine is
(A) Carbohydrate (B) Fat (A) H$m~m}hm`S´>oQ> (B) dgm
(C) Vitamins (D) Proteins (C) {dQ>m{_Z (D) àmoQ>rZ
184. Fabellae are the sesamoid bones 184. \¡${~br grg_m°BS> hS²>S>r g§~§{YV hmoVr h¡
associated with
(A) Elbow joint (B) Knee joint (A) Eë~mo g§{Y (B) Zr g§{Y
(C) Wrist joint (D) Neck joint (C) H$bmB© g§{Y (D) ZoH$ g§{Y

185. Cancer cells show unlimited growth 185. H¢$ga H$mo{eH$mE± A{Z`§{ÌV d¥{Õ Xem©Vr h¢,
because of Š`m|{H$
(A) difference in surface proteins, (A) gVh àmoQ>rZ gmBQ>moH$mBZ _| \$H©$ Ho$ H$maU
cytokines
(B) difference in cholesterol (B) H$moboñQ>oamb _| \$H©$ Ho$ H$maU
(C) abnormal cytoplasm (C) Agm_mÝ` OrdÐd Ho$ H$maU
(D) abnormal nucleus (D) Agm_mÝ` Ho$ÝÐH$ Ho$ H$maU

186.
Read the following statement having {ZåZ{b{IV H$WZ H$mo n{‹T>E {Og_| Xmo [aŠVVmE±
186.
two blanks (a and b) : a VWm b h¡ :
‘‘A drug used for ___(a)
_____ patients ""EH$ Am¡fYr H$m ___(a) _____ amo{J`m| Ho$ {bE
is obtained from a species of the
Cn`moJ {H$`m OmVm h¡, Omo ___(b)_____ àmUr
organism ___(b)
_____ .”
The one correct option for two
Ho$ ñnoerµO go àmßV {H$`m OmVm h¡ &''
blanks is Xmo [aŠVVmE± hoVw EH$ ghr CÎma h¡
(A) a – Swine flue (A) a – ñdmBZ âby
b – Monascus b – _moZ¡ñH$g
(B) a – AIDS (B) a – ES²>g
b – Pseudomonas b – ñ`yS>mo_moZmg
(C) a – Heart (C) a – öX`
b – Penicillium b – no{Zgr{b`_
(D) a – Organ transplant (D) a – A§J àË`mamonU
b – Trichoderma b – Q´>mBH$moS>_m©

-81- Set-A

187. An allergic reaction is initiated by 187. EbOu H$s à{V{H«$`m H$mo {H$g EÊQ>r~m°S>r Ûmam Ama§^
antibodies of {H$`m OmVm h¡ ?
(A) IgG group (A) IgG g_yh
(B) IgM group (B) IgM g_yh
(C) IgA group (C) IgA g_yh
(D) IgE group (D) IgE g_yh

188. Read both Assertion (A) and Reason (R) H$WZ (A) VWm H$maU (R) XmoZm| n‹T>| Am¡a VËníMmV²
188.
and then choose right answer. ghr CÎma MwZ| &
Assertion (A) : Alcohol acts as H$WZ (A) : AëH$mohb S>m``yaoQ>rH$ (A{V_yÌ
diuretic. òmd) H$s Vah H$m`© H$aVm h¡ &
Reason (R) : Sweating is enhanced H$maU (R) : AëH$mohb Ûmam ngrZm òmd ~‹T>m
by alcohol and so it is {X`m OmVm h¡, AV: `h AÝV:Vmnr`
endothermic. hmoVm h¡ &
(A) Both assertion and reason are true (A) H$WZ VWm H$maU XmoZm| ghr h¢ VWm H$maU,
and reason is correct explanation H$WZ H$s ghr ì`m»`m h¡
of assertion
(B) Assertion is true but reason is (B) H$WZ ghr h¡, naÝVw H$maU JbV h¡
false
(C) Both assertion and reason are (C) H$WZ VWm H$maU XmoZm| JbV h¢
false
(D) Assertion is false but reason is true (D) H$WZ JbV h¡ naÝVw H$maU ghr h¡

189. Vertebro arterial canal is found in 189. dQu~«mo AmaQ>o[a`b H°$Z¡b nm`m OmVm h¡
(A) Lumber vertebrae (A) bå~a H$eoéH$
(B) Cervical vertebrae (B) J«rdm H$eoéH$
(C) Thoracic vertebrae (C) dj H$eoéH$
(D) Vertebral column (D) d{Q>©~«b H$m°b_

190. Axis is different from atlas in 190. EŠgrg, EQ>bmg go {^Þ hmoVm h¡
(A) Presence of odontoid process (A) AmoS>moÝQ>m`S> àmogog H$s CnpñW{V go
(B) Presence of central cavity (B) Ho$ÝÐr` Jwhm H$s CnpñW{V go
(C) Absence of central cavity (C) Ho$ÝÐr` Jwhm H$s AZwnpñW{V go
(D) Presence of neural cavity (D) V§{ÌH$m Jwhm H$s CnpñW{V go

-82- Set-A

191. A steroid hormone which regulates 191. ½byH$mog _oQ>m~m{bÁ_ H$mo {Z`§{ÌV H$aZo dmbm
glucose metabolism is ñQ>ram`S> hm_m}Z h¡
(A) Cortisone (A) H$m{Q>©gmoZ
(B) Cortisol (B) H$m{Q>©gmob
(C) Corticosterone (C) H$m{Q>©H$moñQ>ramZ
(D) 11-deoxycorticosterone (D) 11-S>rAm°ŠgrH$m{Q>©H$moñQ>ramZ

192.
Match the following diseases with 192. {ZåZ{b{IV amoJm| H$mo Cggo g§~§{YV J«§{W`m| go
their associated glands and select the gw_o{bV H$s{OE Am¡a ghr {dH$ën Mw{ZE &
correct options.
Diseases Glands amoJ JǤWr
a. Addison’s 1. Pituitary a. E{S>eZ amoJ 1. nr`yf J«§Wr
disease
b. Tetany 2. Thyroid b. Q>rQ>oZr 2. Wm`am°`S>
c. Acromegaly 3. Adrenal cortex c. EH«$mo{_Jobr 3. ES´>rZb H$maQ>oŠg
d. Myxoedema 4. Parathyroid d. {_ŠgmoB©{S>_m 4. nmamWm`am°`S>
(A) a – 3, b – 4, c – 1, d – 2 (A) a – 3, b – 4, c – 1, d – 2
(B) a – 1, b – 3, c – 2, d – 4 (B) a – 1, b – 3, c – 2, d – 4
(C) a – 3, b – 4, c – 2, d – 1 (C) a – 3, b – 4, c – 2, d – 1
(D) a – 3, b – 2, c – 1, d – 4 (D) a – 3, b – 2, c – 1, d – 4

193. A woman is married for second time. 193. EH$ Am¡aV Zo Xygar ~ma emXr H$s & CgHo$ nhbo n{V
Her first husband was ABO blood H$m ABO aŠV àH$ma A Wm Am¡a Cg emXr go Omo
type A and her child by that marriage ~ÀMm Wm CgH$m aŠV àH$ma O Wm & CgH$m Z`m n{V
was type O. Her new husband is AB àH$ma H$m Wm, V~ Am¡aV H$m ABO OrZmoQ>mBn
type AB. What is the woman’s ABO Am¡a aŠV àH$ma Š`m h¡ ?
genotype and blood type ?
(A) IAIO, blood type A (A) IAIO, aŠV àH$ma A
(B) IAIB, blood type AB (B) IAIB, aŠV àH$ma AB
(C) IBIO, blood type B (C) IBIO, aŠV àH$ma B
(D) IOIO, blood type O (D) IOIO, aŠV àH$ma O

-83- Set-A

194. An example of gene therapy is 194. OrZ W¡aonr H$m CXmhaU h¡
(A) Production of injectable (A) à{dîQ>`mo½` {hnoQ>mB{Q>g B doŠgrZ H$m
Hepatitis B vaccine CËnmXZ
(B) Production of vaccine in food (B) ^moÁ` \$gbm| (O¡go-Amby {OÝh| Im`m Om
crops like potatoes which can be gH$Vm h¡) _| doŠgrZ H$m CËnmXZ
eaten
(C) Introduction of gene for (C) gr{d`a H$å~mBÝS> Bå`yZmo {S>{\${g`oÝgr
adenosine deaminase in person (SCID) go J«{gV _Zwî`m| _| ES>rZmo{gZ
suffering from Severe Combined S>rE_rZoO Ho$ {bE OrZ H$m àdoe
Immuno Deficiency (SCID)
(D) Production of test tube babies (D) {Zfo{MV AÊS>m| Ho$ àË`mamonU Am¡a H¥${Ì_
by artificial insemination and BZgo_rZoeZ Ho$ Ûmam Q>oñQ> Q²>`y~ ~o~rµO H$m
implantation of fertilized eggs CËnmXZ

195. Which is correct about white tiger ? 195. {ZåZ _| go H$m¡Z-gm g\o$X Q>mBJa Ho$ {bE gË` h¡ ?
(A) White tiger is recessive mutant (A) g\o$X Q>mBJa Aà^mdr å`yQ>oÝQ> h¡
(B) It is part of subspecies of Indian (B) `h ^maVr` Q>mBJa H$s CnOm{V H$m EH$ ^mJ h¡
tiger
(C) It is a type of albino (C) `h Epë~Zmo H$m EH$ àH$ma h¡
(D) It is a transgenic animal (D) `h EH$ Q´>m§gOo{ZH$ OÝVw h¡

196. Classification that considers several 196. dJuH$aU Omo nyd©O H$m AZoH$ bjU H$mo ñdrH$ma
characters of ancestors is H$aVm h¡
(A) Phylogenetic (A) d§emdbr
(B) Artificial (B) H¥${Ì_
(C) Natural (C) àmH¥${VH$
(D) Phyllotaxy (D) nU© {dÝ`mg

-84- Set-A

197. Bowman’s glands are located in 197. ~mo_oÝg J«§Wr AdpñWV hmoVm h¡
(A) Olfactory epithelium of nose (A) ZmH$ Ho$ K«mU E{nWo{b`_ _|
(B) Anterior pituitary gland (B) AJ« nr`yf J«§Wr _|
(C) Female reproductive system of (C) {VbMÅ>m Ho$ _mXm àOZZ V§Ì _|
cockroach
(D) Proximal end of uriniferous (D) `y[aZo\o$ag Z{bH$m| Ho$ AJ« ^mJ _|
tubule

198. The intestine and stomach in mammals 198. ñVZYm[a`m| Ho$ Am_me` d B§Q>ogQ>mBZ H$s bmBqZJ
is lined by hmoVr h¡
(A) Stratified epithelium (A) pñQ´>`oQ>oS> BnrWr{b`_
(B) Cuboidal epithelium (B) Š`y~m`S>b BnrWr{b`_
(C) Columnar epithelium (C) H$më`yåZa BnrWr{b`_
(D) Squamous epithelium (D) ñŠdo_g BnrWr{b`_

199. Immediately after ovulation the 199. AÊS>moËgO© (ovulation) Ho$ VËH$mbníMmV
mammalian egg is covered by a ñVZYmar H$m AÊS>mUw EH$ {Pëbr go T>H$m hmoVm h¡
membrane known as {Ogo H$hVo h¢
(A) Corona radiata (A) H$moamoZm ao{S>`mQ>m
(B) Vitelline membrane (B) {dQ>obmBZ {Pëbr
(C) Zona pellucida (C) OmoZm noë`wgrS>m
(D) Chorion (D) H$m°[aAmoZ

200. Which of the following vitamins can be 200. {ZåZ _| go H$m¡Z-gm {dQ>m{_Z JQ> Ho$ A§Xa ~¡ŠQ>r[a`m
synthesized by bacteria inside the gut ? Ûmam {Z{_©V hmoVm h¡ ?
(A) B (B) E (A) B (B) E
(C) D (D) K (C) D (D) K

-85- Set-A

-86- Set-A

-87- Set-A

SET – A

CÎma A§{H$V H$aZo H$m g_` : 3 K§Q>o A{YH$V_ A§H$ : 200
Time for marking answers : 3 Hours Maximum Marks : 200

ZmoQ> :
Bg àíZ nwpñVH$m _| VrZ ^mJ - àW_ ^mJ - ^m¡{VH$ emñÌ, {ÛVr` ^mJ - agm`Z emñÌ, V¥Vr` ^mJ - (A) J{UV
1.
Ed§ (~) Ord {dkmZ h¡ - BZ ^mJm| _| A§H$m| H$m {ddaU {ZåZmZwgma h¡ -
(i) àW_ ^mJ :- ^m¡{VH$ emñÌ 1 – 60 60 A§H$
(ii) {ÛVr` ^mJ :- agm`Z emñÌ 61 – 120 60 A§H$
(iii) V¥Vr` ^mJ :- (A) J{UV 121 – 200 80 A§H$
(~) Ord {dkmZ 121 – 200 80 A§H$
àW_ Ed§ {ÛVr` ^mJ A{Zdm`© h¡ & V¥Vr` ^mJ Ho$ Cn^mJ (A) AWdm (~) H$m CÎma X| &
2. àíZm| Ho$ CÎma, Xr JB© OMR CÎma-erQ> (Answer-Sheet>) na A§{H$V H$s{OE Ÿ&
3. F$UmË_H$ _yë`m§H$Z Zht {H$`m OmdoJm Ÿ&
4. {H$gr ^r Vah Ho$ H¡$bHw$boQ>a `m bm°J Q>o~b Ed§ _mo~mBb \$moZ H$m à`moJ d{O©V h¡ Ÿ&
5. OMR CÎma-erQ> (Answer-Sheet>) H$m à`moJ H$aVo g_` Eogr H$moB© AgmdYmZr Z H$a| ~aV| {Oggo `h \$Q>
Om`o `m Cg_| _mo‹S> `m {gbdQ> Am{X n‹S> Om`o {OgHo$ \$bñdê$n dh Iam~ hmo Om`o Ÿ&
Note :

1. This Question Booklet contains Three Parts – First Part – Physics, Second Part –
Chemistry and Third Part – (A) Mathematics and (B) Biology. The detailed account of
marks of these parts are as follows –
(i) First Part : – Physics 1 – 60 60 Marks
(ii) Second Part : – Chemistry 61 – 120 60 Marks
(iii) Third Part : – (A) Mathematics 121 – 200 80 Marks
(B) Biology 121 – 200 80 Marks
First and Second Part are Compulsory. In the Third Part Answer Sub Part (A) or
Sub Part (B).
2. Indicate your answers on the OMR Answer-Sheet provided.
3. No negative marking will be done. a\$ H$m`© Ho$ {bE ñWmZ
4. Use of any type of calculator orSpace for Rough
log table Workphone is prohibited.
and mobile
5. While using OMR Answer-Sheet care should be taken so that the Answer-Sheet does
not get torn or spoiled due to folds and wrinkles.
-88- Set-A