CG PPHT 2021 Question Paper

Question Booklet No.
SET – A


A Subject Code : 12203/ UE – PHT/ENT – E

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Number of Pages in this Question Booklet : 64 Number of Questions in this Question Booklet : 150

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11. Bg àíZ nwpñVH$m _| VrZ ^mJ hmo§Jo :- 11. This Question Paper consists of Three Parts namely :
(i) àW_ ^mJ :- ^m¡{VH$ emñÌ - à.g§. 1-50 (i) First Part : – Physics – Q. No. 1-50
(ii) {ÛVr` ^mJ :- agm`Z emñÌ - à.g§. 51-100 (ii) Second Part : – Chemistry – Q. No. 51-100
(iii) V¥Vr` ^mJ :- (A) J{UV - à.g§. 101-150 (iii) Third Part : – (A) Mathematics – Q. No. 101-150
(~) Ord {dkmZ - à.g§. 101-150 (B) Biology – Q. No. 101-150
^mJ àW_ Ed§ {ÛVr` A{Zdm`© h¢& Aä`Wu ^mJ V¥Vr` (A) VWm (~) _| Part First & Second are compulsory. Candidates should attempt
ANY ONE PART from part Third (A) and (B).
go {H$gr EH$ ^mJ H$m hr M`Z H$a|&
12. In case of any ambiguity in English version the Hindi version shall
12. `{X A§JO
o« 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. A soap bubble of radius r1 and another 1. r1 {ÌÁ`m Ho$ gm~wZ Ho$ EH$ ~wb~wbo H$mo EH$ AÝ`
soap bubble of radius r2 (r2 > r1) are r2 {ÌÁ`m Ho$ gm~wZ Ho$ ~wb~wbo Ho$ (r2 > r1)$ BVZo nmg
brought together so that they have a
bm`m OmVm h¡ {H$ CZH$s C^`{ZîR n[agr_m EH$ hmo
common interface. The radius of the
interface is Om`o V~ Bg C^`{ZîR> n[agr_m H$s {ÌÁ`m h¡
(A) (r2 + r1) (B) (r2 – r1) (A) (r2 + r1) (B) (r2 – r1)
r1 r2 r1 r2
(C) 2(r2 – r1) (D) (r − r ) (C) 2(r2 – r1) (D) (r − r )
2 1
2 1

2. To decrease the volume of a gas by 5% 2. {Z`V Vmn na {H$gr J¡g H$m Am`VZ 5% KQ>mZo
at constant temperature, the pressure Ho$ {bE Xm~
should be
(A) decreased by 5.26% (A) 5.26% KQ>mZm Mm{hE
(B) increased by 5.26% (B) 5.26% ~‹T>mZm Mm{hE
(C) decreased by 10% (C) 10% KQ>m>Zm Mm{hE
(D) increased by 10% (D) 10% ~‹T>mZm Mm{hE

3. An ideal gas is compressed to half its initial 3. EH$ AmXe© J¡g H$mo {d{^Þ àH«$_m| Ho$ Ûmam CgHo$
volume by means of several process. àmapå^H$ Am`VZ Ho$ AmYo Am`VZ hmoZo VH$ gånr{S>V
Which of the process results in the {H$`m OmVm h¡& {ZåZ{b{IV _| go {H$g àH«$_ _| J¡g
maximum work done on the gas ? na {H$`m J`m H$m`© A{YH$V_ hmoJm ?
(A) Adiabatic (A) éÕmoî_
(B) Isobaric (B) g_Xm~r
(C) Isochoric (C) g_Am`V{ZH$
(D) Isothermal (D) g_Vmnr

4. A sound source with a frequency of 790 Hz 4. EH$ Üd{Z òmoV {OgH$s Amd¥{Îm 790 Hz h¡, EH$
moves away from a stationary observer pñWa lmoVm go 15m/s H$s J{V go Xÿa Om ahm h¡&
at a rate of 15m/s. What frequency does pñWa lmoVm Ho$ {b`o Üd{Z H$s Amd¥{Îm {H$VZr hmoJr?
the stationary observer hear ? (The
speed of sound is 340m/s.)
(Üd{Z H$s Mmb 340m/s h¡ &)
(A) 775 Hz (B) 757 Hz (A) 775 Hz (B) 757 Hz
(C) 826 Hz (D) 655 Hz (C) 826 Hz (D) 655 Hz

-3- Set-A

5. Interference was observed in interference 5. ì`{VH$aU àH$moîR> _| dm`w H$s CnpñW{V _| ì`{VH$aU
chamber where air was present, now the à{Vê$n àmßV {H$`m OmVm h¡& `{X ì`{VH$aU àH$moîR>
chamber is evacuated, and if the same _| {Zdm©V CËnÞ H$a {X`m Om`, Vmo àmßV ì`{VH$aU
light is used, a careful observer will see
à{Vê$n _| Š`m n[adV©Z hmoJm ?
(A) Fringe width will increase in the (A) ì`{VH$aU à{Vê$n _| q\«$O AÝVamb ~‹T>
interference pattern Om`oJm
(B) Interference with decreased fringe (B) ì`{VH$aU à{Vê$n _| q\«$O AÝVamb KQ>
width Om`oJm
(C) No interference pattern (C) H$moB© ì`{VH$aU à{Vê$n Zht ~ZoJm
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

6. A rocket is going towards moon with 6. EH$ amHo$Q> v doJ go Mm§X H$s Amoa Om ahr h¡& A§V[aj
a speed v. The astronaut in the rocket `mÌr amHo$Q> go Mm§X H$s Amoa EH$ {g½Zb ^oOVm h¡,
sends signals of frequency ν towards {OgH$s Amd¥{Îm ν h¡& CŠV {g½Zb Mm§X go namd{V©V
the moon and receives them back on hmoH$a nwZ: A§V[aj `mÌr Ho$ nmg AmVr h¡& namd{V©V
reflection from the moon. What will be hmoH$a dmng AmZo dmbr {g½Zb H$s Amd¥{Îm Š`m
the frequency of signal received by the hmoJr ?
astronaut ?
C C
(A) ν (A) ν
C−v C−v

C C
(B) ν (B) ν
C − 2v C − 2v

2v 2v
(C) ν (C) ν
C C
2C 2C
(D) ν (D) ν
v v
7. The flux linked with a coil at any instant 7. {H$gr jU t na Hw§$S>br go g§~§{YV âbŠg
t is given by d = 10t2 – 50t + 250. The d = 10t2 – 50t + 250 Ho$ Ûmam {X`m J`m h¡& Vmo
induced e.m.f. at t = 3 second is ào[aV {d.dm. ~b H$m _mZ t = 3 goH$ÊS> _| hmoJm
(A) – 190 V (B) 190 V (A) – 190 V (B) 190 V
(C) –10 V (D) 10 V (C) –10 V (D) 10 V

-4- Set-A

8. In an A.C. circuit, a resistance of 8. EH$ E.gr. n[anW _| EH$ à{VamoY R Amo_ EH$
R ohm is connected in series with an àoaH$Ëd L Ho$ gmW loUrH«$_ _| Ow‹S>m hþAm h¡& `{X
inductance L. If phase angle between {d^d d Ymam Ho$ ~rM H$bm H$m H$moU 45° h¡, Vmo
voltage and current be 45°, the value àoaH$Ëd à{VKmV H$m _mZ hmoJm
of inductive reactance will be
R R
(A) (A)
4 4
R R
(B) (B)
2 2

(C) R (C) R

(D) Cannot be found with given data (D) {XE JE S>mQ>m go Zht {ZH$mbm Om gH$Vm h¡
A A

9. ~ C
RL
D 9. ~ C
RL
D

B B
In above circuit output waveform across D$na {XImE n[anW _| bmoS> à{VamoY RL Ho$ {gam|
load Resistance RL is given by na {ZJ©V dod\$m_© hmoJm

(A) (A)

(B) (B)

(C) (C)

(D) (D)

-5- Set-A

10. The curve between charge density and 10. PN g§{Y Ho$ {ZH$Q> Amdoe KZËd d Xÿar Ho$ ~rM
distance near PN junction will be dH«$ hmoJm
Charge density Amdoe KZËd
N N
P P
(A) (A)
Distance Xÿar

Charge density Amdoe KZËd
P P
N N
(B) (B)
Distance Xÿar

Charge density Amdoe KZËd
P N P N
(C) (C)
Distance Xÿar

Charge density Amdoe KZËd
P N P N

(D) (D)
Distance Xÿar

11. Vibration magnetometer is used for 11. H$ånZ Mwå~H$Ëd_mnr H$m Cn`moJ VwbZm H$aZo _|
comparing {H$`m OmVm h¡
(A) Magnetic field (A) Mwå~H$s` joÌ H$s
(B) Earth’s field (B) n¥Ïdr Ho$ joÌ H$s
(C) Magnetic moments (C) Mwå~H$s` AmKyU© H$s
(D) All of the above (D) CnamoŠV g^r H$s

-6- Set-A

12. a particle of energy 400 kev are 12. 400 kev D$Om© dmbm a H$U 82pb Ho$ Zm{^H$
bombarded on nucleus of 82Pb. Its na ~_df©H$ Ho$ ê$n _| Amamo{nV {H$`m OmVm h¡&
scattering of a particles, its minimum a H$U Ho$ àH$sU©U Ho$ {bE Zm{^H$ go Ý`yZV_ Xÿar
distance from nucleus will be hmoJr
(A) 5.9 × 10–10 m (A) 5.9 × 10–10 _rQ>a

(B) .59 × 10–10 m (B) .59 × 10–10 _rQ>a

(C) 5.9 × 10–13 m (C) 5.9 × 10–13 _rQ>a

(D) .59 × 10–13 m (D) .59 × 10–13 _rQ>a

Y Y
13. 13.
(1) (1)

(4) (4)
activity (3) g{H«$`Vm (3)

(2) (2)
X X
time g_`
In above graph time and activity of a D$na {XImE J«m\$ _| g_` Ed§ ao{S>`moEpŠQ>d nXmW©
radioactive sample are taken along H$s g{H«$`Vm H$mo H«$_e: X d Y Aj Ho$ AZw{Xe
X and Y axis respectively. Then the {H$`m J`m h¡& nXmW© H$s g{H«$`Vm H$m g_` Ho$ gmW
activity of sample varies with time n[adV©Z {H$E dH«$ Ho$ AZwgma hmoJm
according to the curve

(A) (1) (A) (1)

(B) (2) (B) (2)

(C) (3) (C) (3)

(D) (4) (D) (4)

-7- Set-A

14. From a newly formed radioactive 14. EH$ ZE ~Zo ao{S>`moEpŠQ>d nXmW© (AY© Am`w 2 K§Q>o)
substance (half life 2 hours), the intensity go {d{H$aU (ao{S>EeZ) H$s Vrd«Vm, AZw_V gwa{jV
of radiation is 64 times the permissible bodb$ Ho$ 64 JwZm h¡& Bg òmoV go gwa{jV ê$n _|
safe level. The minimum time after H$m`© H$aZo Ho$ {bE bJZo dmbm Ý`yZV_ g_` h¡
which work can be done safely from this
source is
(A) 6 hours (A) 6 K§Q>o

(B) 12 hours (B) 12 K§Q>o

(C) 24 hours (C) 24 K§Q>o

(D) 8 hours (D) 8 K§Q>o

15. A cell can be balanced against 110 cm 15. EH$ gob H«$_e: 110 go_r Am¡a 100 go_r Ho$
and 100 cm of potentiometer wire {d^d_mnr Vma Ho$ {déÕ g§Vw{bV H$a gH$Vm h¡,
respectively. When in open circuit O~ n[anW Iwbm Ed§ O~ n[anW 10Ω à{VamoY Ho$
and when short circuited through Ûmam emQ>© g{H©$Q>> hmo& gob H$m Am§V[aH$ à{VamoY h¡
a resistance of 10Ω. The internal
resistance of the cell is
(A) 0.5Ω (A) 0.5Ω
(B) 0.75Ω (B) 0.75Ω
(C) 1Ω (C) 1Ω
(D) 1.5Ω (D) 1.5Ω

16. A 500 W heating device is designed to 16. EH$ 500 W H$m Cî_r` CnH$aU (hrqQ>J {S>dmBg)
operate on a 220 V line. If the voltage H$mo 220 V bmBZ _| H$m`© Ho$ {bE ~Zm`m J`m h¡&
drop to 200 V the percentage drop in `{X dmoëQ>oO 200 V VH$ {Ja OmVm h¡, Vmo {ZJ©V
heat output is Cî_m _| {JamdQ> à{VeV _| hmoJm
(A) 30.26% (A) 30.26%
(B) 21.36% (B) 21.36%
(C) 17.36% (C) 17.36%
(D) None of the above
(D) CnamoŠV _| go H$moB© Zht

-8- Set-A

17. Two long parallel wires P and Q are 17. Xmo bå~o g_mZmÝVa Vma P Am¡a Q H$mo EH$ Xÿgao go
held perpendicular to the plane of 5 _r. H$s Xÿar na H$mJO Ho$ Vb Ho$ bå~dV aIm
paper with distance 5 m between them. J`m h¡& `{X P Am¡a Q na H«$_e: 2.5 A Am¡a
If P and Q carry current of 2.5 A and 5 A H$s Ymam EH$ hr {Xem _| àdm{hV hmo ahr h¡, Vmo
5 A respectively in the same direction, XmoZmo§ Vmam| Ho$ _Ü` q~Xþ na Mwå~H$s` joÌ h¡
then the magnetic field at a point half
way between the wires is
(A) m 0 (A) m 0
π π
m0 m0
(B) (B)
2π 2π
3m 0 3m 0
(C) (C)
2π 2π
3m 0 3m 0
(D) (D)
π π
18. Heat is supplied to a diatomic gas at {Z`V Xm~ na EH$ {Ûna_mUwH J¡g H$mo Cî_m àXm`
18.
constant pressure, with the usual notation H$s OmVr h¡& gm_mÝ` g§Ho$VH$m| _|, AZwnmV
the ratio ∆Q : ∆U : ∆W is ∆Q : ∆U : ∆W H$m _mZ h¡
(A) 5 : 2 : 2 (B) 5 : 2 : 3 (A) 5 : 2 : 2 (B) 5 : 2 : 3
(C) 7 : 5 : 2 (D) 7 : 2 : 5 (C) 7 : 5 : 2 (D) 7 : 2 : 5
19. The efficiency of a Carnot engine which 19. {H$gr H$mZm} B§OZ H$s XjVm Omo {H$ Vmn T1 Ed§ T2
is working between temperatureT1 and T2 Ho$ ~rM H$m`© H$a ahm h¡, hmoJr
is given by
T1 T1
(A) 1 − (A) 1 −
T2 T2
T2 − T1 T2 − T1
(B) (B) T1
T1
T1 − T2 T1 − T2
(C) (C)
T1 T1
T1 T1
(D) (D)
T1 − T2 T1 − T2

-9- Set-A

20. The heat transfer by conduction through 20. MmbZ {d{Y go EH$ _moQ>o Jmobo Ûmam Cî_m ñWmZm§VaU
a thick sphere is given by Ho$ {b`o ghr g§~§Y H$m MwZmd H$a| &
(A) Q = 2πkr1r2 ( T1 − T2 ) (A) Q = 2πkr1r2 ( T1 − T2 )
(r2 − r1) (r2 − r1)
(B) Q = 4πkr1r2 ( T1 − T2 ) (B) Q = 4πkr1r2 ( T1 − T2 )
(r2 − r1) (r2 − r1)
(C) Q = 6πkr1r2 ( T1 − T2 ) (C) Q = 6πkr1r2 ( T1 − T2 )
(r2 − r1) (r2 − r1)
(D) Q = 8πkr1r2 ( T1 − T2 ) (D) Q = 8πkr1r2 ( T1 − T2 )
(r2 − r1) (r2 − r1)
21. The magnifying power of a simple 21. {H$gr gab gyú_Xeu H$s àdY©Z j_Vm {ZåZ _| go
microscope can be increased, if we {H$g àH$ma Ho$ Zo{ÌH$m b|g Ho$ à`moJ go ~‹T>m`r Om
use eye piece of gH$Vr h¡ ?
(A) Higher focal length (A) CƒVa \$moH$g Xÿar dmbo
(B) Smaller focal length (B) {ZåZVa \$moH$g Xÿar dmbo
(C) Higher diameter (C) CƒVa ì`mg dmbo
(D) Smaller diameter (D) {ZåZVa ì`mg dmbo
22. The diameter of the objective of a 22. EH$ XÿaXeu Ho$ A{^Ñí`H$ H$m ì`mg a h¡ VWm
telescope is a, its magnifying power CgH$s AmdY©Z j_Vm m d àH$me H$s Va§JX¡¿`© λ h¡&
is m and wavelength of light is λ. The Q>o{bñH$mon H$s {d^oXZ j_Vm h¡
resolving power of the telescope is
1.22a 1.22a
(A) 1.22λ (B) (A) 1.22λ (B)
λ a λ
a
λm a λm a
(C) (D) (C) (D)
1.22a 1.22λ 1.22a 1.22λ
23.
A photon of energy 7 eV is incident on 23. 7 eV D$Om© dmbm EH$ \$moQ>mZ YmVw gVh na Amn{VV
metal surface of threshold frequency hmoVm h¡ {OgH$s Xohbr Amd¥{Îm 1.6 × 1015 hQ>©µO
1.6 × 1015 Hz. The maximum kinetic h¡& CËg{O©V \$moQ>moBboŠQ´>mZ H$s A{YH$V_ J{VO
energy of the photoelectron emitted
(in eV) is D$Om© (eV _|) h¡
(h = 6 × 10–34 Js) (h = 6 × 10–34 Oyb goH|$S>>)
(A) 1.6 (B) 6 (A) 1.6 (B) 6
(C) 2 (D) 1 (C) 2 (D) 1

-10- Set-A

3 µF 10 µF 15 µF 3 µF 10 µF 15 µF

24. 24.

100 V 100 V
In above circuit the charge on 15 mF is D$na {XImE n[anW _| 15 mF na Amdoe hmoJm
(A) 50 mC (A) 50 mC

(B) 100 mC (B) 100 mC

(C) 200 mC (C) 200 mC

(D) 280 mC (D) 280 mC

a a
25. 25.
k1 k1
k2 d k2 d
k3 k3

b b
The plate area of above combination D$na {XImE JE g§`moOZ _| ßboQ> H$m joÌ\$b A
is A and the separation between the d ßboQ>m§o Ho$ ~rM H$s Xÿar d h¡& a d b Ho$ _Ü`
plates is d. The equivalent capacity Vwë`Ym[aVm hmoJr
between a and b is
3 ∈0 A k1k 2k 3 3 ∈0 A k1k 2k 3
(A) (A)
d(k1k 2 + k 2k 3 + k 3k1) d(k1k 2 + k 2k 3 + k 3k1)

(B) 3 ∈0 A (k1 + k 2 + k 3 ) (B) 3 ∈0 A (k1 + k 2 + k 3 )
d d
∈0 A k1k 2k 3 ∈0 A k1k 2k 3
(C) (C)
d(k1k 2 + k 2k 3 + k 3k1) d(k1k 2 + k 2k 3 + k 3k1)

∈0 A ∈0 A
(D) (k1 + k 2 + k 3 ) (D) (k1 + k 2 + k 3 )
d d

-11- Set-A

26. An electric dipole is placed at an angle 26. EH$ {dÚwV {ÛY«wd H$mo EH$ Ag_mZ {dÚwV joÌ Ho$
of 30° to a non-uniform electric field. gmW 30° H$moU _| aIm J`m h¡& {dÚwV {ÛY«wd d
The electric dipole will experience AZw^d H$aoJm
(A) a torque as well as translational (A) ~b AmKyU© Ho$ gmW-gmW ñWmZmÝVar` ~b
force
(B) a torque only (B) Ho$db ~b AmKyU©
(C) a translational force only along the (C) joÌ Ho$ {Xem _| Ho$db ñWmZmÝVar` ~b
field
(D) a translational force only along (D) joÌ Ho$ bå~dV {Xem _| Ho$db ñWmZm§Var`
normal to field ~b
27. A particle of mass m is moving in 27. m Ðì`_mZ H$m EH$ H$U r {ÌÁ`m Ho$ EH$ j¡{VO
a horizontal circle of radius r under k
k d¥Îm _| A{^Ho$ÝÐr ~b − 2 Ho$ AÝVJ©V J{V H$a
centripetal force equal to − 2 , where r
r ahm h¡, Ohm± k EH$ {Z`Vm§H$ h¡ & Bg H$U H$s Hw$b
k is a constant. The total energy of the
particle is D$Om© H$m _mZ h¡
k k k k
(A) − r (B) − 2r (A) − r (B) − 2r
k k k k
(C) (D) (C) (D)
r 2r r 2r

28. In the given spring – block arrangement, 28. {MÌ _| Xem©`o J`o pñà¨J-{nÊS> g_m`moOZ _| pñà§J
find maximum elongation in the spring. _| CËnÞ A{YH$V_ àgma (elongation) H$m _mZ h¡
K K
m m

2m 2m

2mg 3mg 2mg 3mg
(A) (B) (A) k (B) k
k k
4mg 5mg 4mg 5mg
(C) (D) (C) (D)
k k k k

-12- Set-A

29. A solid cylinder of mass M and radius R 29. M Ðì`_mZ Am¡a R {ÌÁ`m H$m EH$ R>mog ~obZ
rolls from rest down a plane inclined {H$gr AmZV Vb Omo j¡{VO go θ Ho$ H$moU na h¡, go
at an angle θ to the horizontal. The {dam_mdñWm go ZrMo H$s Amoa bw‹T>H$Vm h¡ & Bg ~obZ
velocity of the centre of mass of the Ho$ X«ì`_mZ Ho$ÝÐ Ho$ doJ H$m _mZ hmoJm O~{H$ `h
cylinder after it has rolled down a Xÿar d bw‹T>H$ (rolled) MwH$m hmo
distance d is
2 2
(A) gd tanθ (B) gd tanθ (A) gd tanθ (B) gd tanθ
3 3

3 4 3 4
(C) gd sinθ (D) gd sinθ (C) gdsinθ (D) gd sinθ
4 3 4 3
30. A curve between magnetic moment (M) 30. Mwå~H$s` AmKyU© (M) Ed§ Vmn (T) Ho$ _Ü` ItMm
and temperature (T) of magnet is J`m dH«$ hmoJm
M M

(A) (A)

T T

M M

(B) (B)
T T

M M

(C) (C)
T T

M M

(D) (D)

T T

-13- Set-A

31. The electric charge in uniform motion 31. EH$ g_mZ J{V H$aVm {dÚwV Amdoe CËnÝZ H$aVm h¡
produces

(A) An electric field only (A) Ho$db {dÚwV joÌ
(B) A magnetic field only (B) Ho$db Mwå~H$s` joÌ
(C) Both electric and magnetic field (C) {dÚwV Ed§ Mwå~H$s` joÌ XmoZm|
(D) Neither electric nor magnetic (D) Z Vmo {dÚwV joÌ Z hr Mwå~H$s` joÌ
field

32. A charge Q is placed at the centre of 32. EH$ Amdoe Q EH$ H$mën{ZH$ AY©Jmobo Ho$ Ho$ÝÐ na
an imaginary hemispherical surface. aIm h¡ & Bg Amdoe Ho$ H$maU AY©Jmobo H$s gVh go
The flux of the electric field due to {ZH$bZo dmbm {dÚwV âbŠg hmoJm
this charge through the surface of the
hemisphere

Q Q
R R


Q Q
(A) (A)
∈0 ∈0
Q Q
(B) (B)
2 ∈0 2 ∈0
(C) Zero (C) eyÝ`
Q Q
(D) (D)
4πR2 ∈0 4πR2 ∈0
33. An infinite number of point masses 33. g_mZ Ðì`_mZ m Ho$ {~ÝXþ Ðì`_mZ, {OZH$s g§»`m
each equal to m are placed at x = 1, AZÝV h¡, x = 1, x = 2, x = 4, x = 8, . . . .,
x = 2, x = 4, x = 8, . . . ., what is the total pñW{V`m| na aIo h¢& x = 0 na Hw$b JwéËdr` {d^d
gravitational potential at x = 0 ? H$m _mZ Š`m h¡ ?
(A) – Gm (B) – 2 Gm (A) – Gm (B) – 2 Gm
(C) – 4 Gm (D) – 8 Gm (C) – 4 Gm (D) – 8 Gm

-14- Set-A

34. A satellite is orbiting around the earth 34. EH$ CnJ«h J{VO D$Om© k Ho$ gmW n¥Ïdr H$s n[aH«$_m
with kinetic energy k. What will happen H$a ahm h¡& `{X CnJ«h H$s J{VO D$Om© 2k H$a Xr
if the satellite’s kinetic energy is made
2k ? Om`o V~
(A) Radius of the orbit of the satellite (A) CnJ«h H$s H$j H$s {ÌÁ`m XþJwZr hmo Om`oJr
is doubled
(B) Radius of the orbit of the satellite (B) CnJ«h H$s H$j H$s {ÌÁ`m AmYr hmo Om`oJr
is halved
(C) Period of revolution of the satellite (C) CnJ«h H$m n[a^«_U H$mb XþJwZm hmo Om`oJm
is doubled
(D) Satellite escapes away (D) CnJ«h nbm`Z H$a Om`oJm
35. One end of a horizontal thick copper 35. 2L bå~mB© VWm 2R {ÌÁ`m Ho$ EH$ _moQ>o j¡{VO
wire of length 2L and radius 2R is Vm±~o Ho$ Vma Ho$ EH$ {gao H$mo EH$ AÝ` L bå~mB© VWm
welded to an end of another horizontal R {ÌÁ`m Ho$ nVbo Vm±~o Ho$ Vma Ho$ {gao go Omo‹S>
thin copper wire of length L and radius
R when the arrangement is stretched by (doëS>) {X`m J`m h¡ & Bg g_m`moOZ Ho$ XmoZm| {gam|
applying forces at two ends, the ratio of na ~b bJm H$a BÝh| ItMm OmVm h¡& V~ nVbo Vma>
the elongation in the thin wire to that in VWm _moQ>o Vma H$s bå~mB©`m§o _| d¥{Õ H$m AZwnmV h¡
the thick wire
(A) 0.25 (B) 0.50 (A) 0.25 (B) 0.50
(C) 2.00 (D) 4.00 (C) 2.00 (D) 4.00

36.
The current required to deposit 0.972 gm 36. 0.972 J«m_ H$m H«$mo{_`_ VrZ K§Q>o _| O_m hmoZo
of chromium in 3 hours is Ho$ {bE Amdí`H$ Ymam h¡
(E.C.E. of chromium = 0.00018 g/coul.) (H«$mo{_`_ H$m E.C.E. = 0.00018 J«m_/Hy$bå~)
(A) 0.5 A (B) 1.0 A (A) 0.5 A (B) 1.0 A
(C) 1.5 A (D) 2.0 A (C) 1.5 A (D) 2.0 A

37. A wire of length 2 m carries a current 37. EH$ 2 _r. bå~mB© Ho$ Vma {Og_| 1 Eånr`a H$s Ymam
of 1 ampere is bend to form a circle. àdm{hV hmo ahr h¡, Cgo d¥ÎmmH$ma _| _mo‹S>m OmVm h¡&
The magnetic moment of the coil is Hw§$S>br H$m Mwå~H$s` AmKyU© h¡
(A) 2π (B) π 2 (A) 2π (B) π 2
1 1
(C) π 4 (D) (C) π 4 (D)
π π

-15- Set-A

38. An ammeter gives full scale deflection 38. EH$ A_rQ>a 1 Eånr`a Ymam Ho$ gmW nyU© ñHo$b
with a current of 1 ampere. It is {djonU XoVm h¡& Bgo 10 Eånr`>a namg dmbo EH$
converted into an ammeter of range A_rQ>a _| ~Xbm J`m h¡& A_rQ>a Ho$ à{VamoY Ho$ gmW
10 ampere, the ratio of the resistance Cn`moJ {H$E JE e§Q> à{VamoY H$m AZwnmV hmoJm
of ammeter to the shunt resistance
used
(A) 10 : 9 (B) 9 : 10 (A) 10 : 9 (B) 9 : 10
(C) 11 : 10 (D) 10 : 11 (C) 11 : 10 (D) 10 : 11

39.
When a wave traverses a medium, the {H$gr _mÜ`_ go JwOaVr {H$gr Va§J _§o EH$ H$U H$m
39.
displacement of a particle located at x pñW{V$ x VWm g_` t na {dñWmnZ {ZåZmZwgma
at a time t is given by ì`ŠV hmoVm h¡
y = a sin (bt – cx), where a, b and c y = a sin (bt – cx), Ohm± a, b Am¡a c {Z`Vm§H$ h¡&
are constants of the wave, which of the
following is a quantity with dimensions ?
V~ {ZåZ{b{IV {H$g am{e H$s {d_m`| hmoVr h¡ ?
y y
(A) a (A) a
(B) bt (B) bt
(C) cx (C) cx
b b
(D) (D)
c c
40.
There are two values of time for which g_` Ho$ Xmo _mZm| Ho$ {bE H$moB© àjoß` g_mZ D±$MmB©
40.
a projectile is at the same height. The na hmoVm h¡& BZ XmoZm| g_`m| H$m `moJ ~am~a hmoVm h¡
sum of these two times is equal to
(T = time of flight of the projectile)
(T = àjoß` H$m CS²>S>`Z H$mb)
3T 3T
(A) (A)
2 2
4T 4T
(B) (B)
3 3
3T 3T
(C) (C)
4 4
(D) T (D) T

-16- Set-A

41. In the three cases, as shown in 41. VrZ pñW{V`m| _| O¡gm {H$ {MÌ _| Xem©`m J`m h¡,
figure, blocks are moving with constant ãbmH$ {Z`V doJ go J{V_mZ h¡, Vmo (a), (b) VWm
velocity, the friction acting in (a), (b) and (c) _| H$m`©H$mar Kf©U fa, fb d fc h¡¡, V~
(c) is fa, fb and fc, then
F F F F
θ θ θ θ
m F m m m F m m

µ µ µ µ µ µ
(a) (b) (c) (a) (b) (c)

(A) fa > fb > fc (A) fa > fb > fc
(B) fc > fa > fb (B) fc > fa > fb
(C) fa = fb = fc (C) fa = fb = fc
(D) none of the above (D) CnamoŠV _| go H$moB© Zht
42. The resistance between A and B in 42. {XE J`o n[anW _| A d B Ho$ ~rM à{VamoY h¡
given circuit is

B B
A C D A C D

R 2R R 2R
(A) (B) (A) (B)
3 3 3 3
3R 3R
(C) (D) 3R (C) (D) 3R
2 2
43. A wire of resistance 5Ω is drawn out so 43. EH$ 5Ω à{VamoY Ho$ Vma H$mo Bg àH$ma Ir§Mm OmVm
that its length is increased to twice its h¡ {H$ CgH$s bå~mB© àma§{^H$ bå~mB© H$s XmoJwZr hmo
original length. Its new resistance is OmVr h¡& BgH$m Z`m à{VamoY h¡
(A) 10 Ω (A) 10 Ω
(B) 20 Ω (B) 20 Ω
(C) 30 Ω (C) 30 Ω
(D) 40 Ω (D) 40 Ω

-17- Set-A

44. Resistance of wire at 20°C is 20Ω and 44. EH$ Vma H$m à{VamoY 20°C na 20Ω h¡ Am¡a
at 500°C is 60Ω. At what temperature 500°C na 60Ω h¡& {H$g Vmn na CgH$m à{VamoY
its resistance is 25Ω ? 25Ω hmoJm ?
(A) 250°C (B) 160°C (A) 250°C (B) 160°C
(C) 100°C (D) 80°C (C) 100°C (D) 80°C
45. Light enters at an angle of incidence 45. EH$ nmaXeu N>‹S> {OgH$m AndV©Zm§H$ n h¡ & Bg_|
in a transparent rod of refractive index n, àH$me {H$aU Amn{VV H$moU go àdoe H$aVr h¡&
as shown in figure. ({MÌ _| XoI|)

r r
α n α n

For what value of refractive index of nmaXeu nXmW© H$m AndV©Zm§H$ Š`m hmoZm Mm{h`o {H$
the material of the rod of light, entered Cg_| àdoe H$aZo dmbr àH$me CgHo$ n¥îR> ^mJ go
into it will not leave it through its lateral
face, what-soever be the value of angle
~mha Z Am`o & AmnVZ H$moU Ho$ g^r _mZ Ho$ {b`o
of incidence ?
(A) n > 2 (B) n = 1 (A) n > 2 (B) n = 1
(C) n = 1.1 (D) n = 1.3 (C) n = 1.1 (D) n = 1.3
46. A concave mirror is held in liquid. What 46. EH$ AdVb Xn©U H$mo EH$ Ðd _| Sw>~m`m OmVm h¡,
should be the change in focal length Vmo CgHo$ ‹>\$moH$g Xÿar _| Š`m n[adV©Z hmoJm ?
of the mirror ?
(A) increases (A) ~‹T>Vm h¡
(B) decreases (B) KQ>Vm h¡
(C) depends upon the refractive index (C) Ðd Ho$ AndV©Zm§H$ na {Z^©a H$aVm h¡
of liquid (D) H$moB© n[adV©Z Zht hmoVm
(D) no change
47. How much intensity of the image is 47. EH$ à{V{~å~ H$s Vrd«Vm {H$VZr ~‹T>oJr `{X XÿaXeu
increased if the diameter of the objective Ho$ A{^Ñí` b|g Ho$ ì`mg H$mo XþJwZm H$a X| ?
lens of a telescope is doubled ?
(A) No change (A) H$moB© n[adV©Z Zht
(B) Two times (B) Xmo JwZm
(C) Four times (C) Mma JwZm
(D) Sixteen times (D) gmobh JwZm

-18- Set-A

constan t is called 48. λm =
{Z`Vm§H$ Š`m H$hbmVm h¡ ?
48. λm =
T T
(A) Kirchhoff’s law (A) {H$aMm¡\$ H$m {Z`_
(B) Newton’s law of cooling (B) Ý`yQ>Z H$m erVbZ {Z`_

(C) Stefan’s law (C) ñQ>r\$Z H$m {Z`_
(D) Wien’s displacement law (D) drZ H$m {dñWmnZ {Z`_

49. A spring of force constant k is cut 49. EH$ pñà¨J {OgH$m ~b {Z`Vm§H$ k h¡, H$mo VrZ
into lengths of ratio 1 : 2 : 3. They are ^mJm| _| Bg Vah H$mQ>m OmVm h¡ {H$ CgH$s bå~mB©
connected in series and the new force H$m AZwnmV 1 : 2 : 3 h¡ & BZ H$Q>o ^mJm| H$mo loUr
constant is k′. Then they are connected H«$_ _| Omo‹S>Zo na ~b {Z`Vm§H$ k′ VWm g_mZm§Va
in parallel and force constants is k″. H«$_ _| Omo‹S>Zo na ~b {Z`Vm§H$ k″ àmßV hmoVm h¡, Vmo
Then k′ : k″ is k′ : k″ H$m _mZ hmoJm
(A) 1 : 9 (A) 1 : 9
(B) 1 : 11 (B) 1 : 11
(C) 1 : 14 (C) 1 : 14
(D) 1 : 6 (D) 1 : 6

50. The velocity
 vector v and displacement EH$ gab AmdVu J{V H$aVo H$U Ho$ doJ g{Xe
50.
vector x of a particle executing shm  
dv v VWm {dñWmnZ g{Xe x _| {ZåZ g§~§Y h¡
are related as v = – ω2x with the dv
dx v = – ω2x VWm àma§{^H$ _mZ x = 0 na
initial condition v = v0 at x = 0. The dx
velocity v, when displacement x is v = v0 h¡, Vmo {dñWmnZ x na v H$m _mZ hmoJm

2 2 2
(A) v = v 0 + ω x 2 2 2
(A) v = v 0 + ω x

(B) v = v 20 − ω 2 x 2 (B) v = v 20 − ω 2 x 2

(C) v = 3 v 30 + ω 2 x 2 (C) v = 3 v 30 + ω 2 x 2

( ) ( )
1 1
3 3 3 3
(D) v = v 0 − ω 2 x 3ex (D) v = v 0 − ω 2 x 3ex

-19- Set-A

PART – II
chemistry agm`Z emñÌ
51.
Cu2+ ions react with Fe2+ ions according 51. Cu 2+ Am`Z Fe 2+ Am`Z Ho$ gmW {ZåZmZwgma
to the following reaction : A{^{H«$`m H$aVm h¡ &
Cu2+ + 2Fe2+  Cu + 2Fe3+ Cu2+ + 2Fe2+  Cu + 2Fe3+
At equilibrium, the concentration of
gmå`mdñWm na, Cu2+ Am`Z H$m gmÝÐU {ZåZ _| {H$gr
Cu2+ ions is not changed by the addition
of EH$ Ho$ {_bmZo na n[ad{V©V Zht hmoVm
(A) Cu (B) Cu2+ (A) Cu (B) Cu2+

(C) Fe2+ (D) Fe3+ (C) Fe2+ (D) Fe3+

52. A polymer of prop-z-ene nitrile is 52. àmon-z-B©Z ZmB©Q´>mB©b H$m ~hþbH$ H$hbmVm h¡
called
(A) Saran (A) gamZ
(B) Orlon (B) Amabm°Z
(C) Dacron (C) S>oH«$m°Z
(D) Teflon (D) Q>oâbm°Z

53. The solubility of calcium phosphate 53. 25°C na Ob _| Ho$pëg`_ \$mñ\o$Q> (AUw^ma, M)
(molar mass, M) in water is Wg per H$s {dbo`Vm Wg à{V 100 {_{b h¡ & 25°C na
100 ml at 25°C. Its solubility product
at 25°C will be approximately
BgH$m {dbo`Vm JwUZ\$b bJ^J hmoJm
5 5
3W 3W
(A) 10   (A) 10  
M M
5 5
5W 5W
(B) 10   (B) 10  
M M
5 5
(C) 107  
W
(C) 107  
W
M M
5 5
(D) 109  W  9W
(D) 10  
M M

-20- Set-A

SOCl Benzene
CH3COOH 
2
→ a  →b 54. SOCl2 ~oÝOrZ
54. Anhydrous AlCl3 CH3COOH  → a 
AZmÐ AlCl3 → b
HCN HOH
→ c → d HCN HOH
compound d is → c → d
`m¡{JH$ d h¡
COOH COOH
CH2 – C – CH3 CH2 – C – CH3
OH OH
(A) (A)

CN CN

(B) C – CH3 (B) C – CH3
OH OH

OH OH
CH2 – C – CH3 CH2 – C – CH3
(C) (C)
CN CN

COOH COOH

HO – C – CH3 HO – C – CH3

(D) (D)

-21- Set-A

 CH3 CH3   CH3 CH3 
55. | |
 55.  | |

 − CH2 − C − CH2 − C −  − CH2 − C − CH2 − C −
| | | |
 CH3 CH3 
n

 CH 3 CH 
3 
n

is a polymer having monomer units EH$ ~hþbH $h¡ {Og_| _moZmo_a (EH$bH$) BH$mB©`m±
hmoVr h¡
(A) (B) (A) (B)
H H

(C) (D) (C) (D)
H H

56. Match List – I with List – II and select 56. gyMr – I H$mo gyMr – II go gw_o{bV H$a {ZåZ H$moS>
the correct answer using following H$s ghm`Vm go ghr CÎma M`{ZV H$s{O`o&
codes.
List – I List – II gyMr – I gyMr – II
(Complexes) (Isomerism) (g§H$a `m¡{JH$) (g_md`dVm)
a.[Co(NH3)4Cl2] 1. Optical a. [Co(NH3)4Cl2] 1. àH$mer`
isomerism g_md`dVm
b. Cis – [Co(en)3Cl2] 2. Ionization b. Cis – [Co(en)3Cl2] 2. Am`ZrH$aU
isomerism g_md`dVm
c. [Co(en)2(No2)Cl]SCN 3. Coordination c. [Co(en)2(No2)Cl]SCN 3. g_Ýd`
isomerism g_md`dVm
d. [Co(NH3)6] [Co(CN)6] 4. Geometrical d. [Co(NH3)6] [Co(CN)6] 4. Á`m{_{V`
isomerism g_md`dVm
e. [Co(NH3)5 (ONO)]Cl2 5. Linkage e. [Co(NH3)5 (ONO)]Cl2 5. qbHo$O
isomerism g_md`dVm
a b c d e a b c d e
(A) 5 3 2 1 4 (A) 5 3 2 1 4
(B) 1 4 2 3 5 (B) 1 4 2 3 5
(C) 4 1 2 3 5 (C) 4 1 2 3 5
(D) 5 4 3 2 1 (D) 5 4 3 2 1

-22- Set-A

57. Among the following which mismatched ? 57. {ZåZ Ho$ _Ü` H$m¡Z gw_o{bV Zht h¡ ?
I. [Mn(CN)6]4– – d2sp3 I. [Mn(CN)6]4– – d2sp3
II. [CuCl4]2– – dsp2 II. [CuCl4]2– – dsp2
III. [Fe(CO)5] – dsp3 III. [Fe(CO)5] – dsp3
IV. [Fe(CN)6]3– – sp3d2 IV. [Fe(CN)6]3– – sp3d2
V. [NiCl4]2– – sp3
V. [NiCl4]2– – sp3
(A) I Ed§ III
(A) I and III
(B) II Ed§ IV
(B) II and IV
(C) III and V (C) III Ed§ V
(D) I and V (D) I Ed§ V

58. Consider following reactions. {ZåZ{b{IV A{^{H«$`m na {dMma H$s{O`o&
58.
[A] + H2SO4 → [B] (a colourless and [A] + H2SO4 → [B] (a§JhrZ Ed§ VrúUJ§Y J¡g)
irritating gas) [B] + K 2Cr 2O 7 + H 2SO 4 → hao a§ J H$m
[B] + K 2 Cr 2 O 7 + H 2 SO 4 → Green {db`Z
coloured solution
[A] and [B] are [A] Ed§ [B] h¢

(A) CO32–, CO2 (A) CO32–, CO2
(B) S2–, H2S (B) S2–, H2S
– –
(C) Cll , HCl (C) Cll , HCl
2– 2–
(D) SO3 , SO2 (D) SO3 , SO2

59. In a hydrogen-oxygen fuel cell, 59. EH$ hmBS´>mOo Z-Am°ŠgrOZ BªYZ gob _|, hmBS´>mOo Z H$m
combustion of hydrogen occurs to XhZ ________ Ho$ {bE hmoVm h¡ &
(A) Generate heat (A) D$î_m CËnÞ H$aZo
(B) Remove absorbed oxygen from (B) BboŠQ´>mS o > n¥îR>m| go Ademo{fV Am°ŠgrOZ H$mo
electrode surfaces
hQ>mZm
(C) Produce high purity water (C) Cƒ ewÕVm Ob CËnmXZ
(D) Create potential difference (D) Xmo BboŠQ´>mSo > Ho$ ~rM {d^dm§Va {Z{_©V$ H$aZm
between two electrodes

-23- Set-A

60. In which one of the following properties 60. {ZåZ{b{IV _| go H$m¡Z-gr EH$ {deofVm go nm`g
emulsions differ from colloidal sols. ? H${bbr` {db`Zm| go {^Þ h¡ ?
(A) Tyndall effect (A) qQ>>S>b à^md
(B) Brownian movement (B) ~«mC{Z`Z J{V
(C) Electrophoresis (C) d¡ÚV w H$U g§MbZ
(D) Size of the particles of the (D) {dg[aV àmdñWm Ho$ H$Um| H$m AmH$ma
dispersed phase
61. Pb (lead) is extracted from its chief ore 61. Pb (grgm) AnZo _w»` A`ñH$ _______go Ûmam
by {ZîH${f©V {H$`m OmVm h¡ &
(A) carbon reduction (A) H$m~©Z AnM`Z
(B) self reduction (B) ñd AnM`Z
(C) electrolysis (C) {dÚwV AnKQ>Z
(D) carbon reduction and electrolysis (D) H$m~©Z AnM`Z Am¡a {dÚwV AnKQ>Z XmoZm|
both

62. The basic nature of transition metal 62. g§H«$_U VËdm| Ho$ _moZmoAmŠgmBS> Ho$ Xmar` àH¥${V
monoxides follows the order H$m ghr H«$_ h¡
(A) CrO > VO > FeO > TiO (A) CrO > VO > FeO > TiO
(B) TiO > FeO > VO > CrO (B) TiO > FeO > VO > CrO
(C) TiO > VO > CrO > FeO (C) TiO > VO > CrO > FeO
(D) VO > CrO > TiO > FeO (D) VO > CrO > TiO > FeO

63. Consider following ionic reaction 63. {ZåZ Am`{ZH$ A{^{H«$`m na {dMma H$s{O`o
Cr2O72–+[X] H++ [Y]I–→2Cr3++ [Z]I2 + 7H2O Cr2O72–+[X] H++ [Y]I–→2Cr3++ [Z]I2 + 7H2O
The values of coefficients [X], [Y] and JwUm§H$ [X], [Y] Ed§ [Z] Ho$ _mZ h¢
[Z] are
[X] [Y] [Z] [X] [Y] [Z]
(A) 16 3 2 (A) 16 3 2
(B) 12 7 6 (B) 12 7 6
(C) 14 6 3 (C) 14 6 3
(D) 2 6 3 (D) 2 6 3

-24- Set-A

64. There are four complexes species of 64. {ZH$b Ho$ Mma g§H$a ñno{gO ZrMo {X`o J`o h¢
Nickel are given below

i. [Ni(CN)4]2– i. [Ni(CN)4]2–

ii. [NiCl4]2– ii. [NiCl4]2–

iii. [Ni(Co)4] iii. [Ni(Co)4]

iv. [Ni(H2O)6]2+ iv. [Ni(H2O)6]2+
Complexes which are attracted by the g§H$a Omo Mwå~H$s` joÌ Ûmam AmH${f©V hmoVo h¢
magnetic field
(A) i only (A) Ho$db i
(B) ii and iii (B) ii Ed§ iii
(C) ii, iii and iv (C) ii, iii Ed§ iv
(D) ii and iv (D) ii Ed§ iv

65. Perovskite is a mineral with formula 65. noamodñH$mBQ> EH$ I{ZO hmoVm h¡, {OgH$m gyÌ
CaTiO3. Which of the positive ions in CaTiO3 h¡ & AîQ>\$bH$s` {N>Ð _| H$m¡Z-gm
the crystal is more likely to be packed
YZm`Z {H«$ñQ>b _| Á`mXmVa ^am hmoVm h¡ ?
in the octahedral holes ?
(A) Ca2+ (B) O+2 (A) Ca2+ (B) O+2
(C) Ti2+ (D) Ti4+ (C) Ti2+ (D) Ti4+
235
66. Which of the following elements is an 66. {ZåZ _| 92 U H$m H$m¡Z-gm VËd AmBgmoS>m`\$a h¡ ?
235
isodiapher of 92 U ? 231
231 (A) Th
(A) Th 90
90
231
231 (B) Pa
(B) Pa 91
91
212
(C)
212 (C) Pb
82
Pb 82

209
209
(D) bi (D) 83
bi
83

-25- Set-A

67. Half life of a radioactive sample is 67. EH$ ao{S>`mog{H«$` Z_yZo H$s AÕ©Am`w 2x df© h¡ &
2x years, what fraction of this sample x df© níMmV Bg Z_yZo H$m {H$VZm A§e A{dK{Q>V
will remain undecayed after x years ?
ah Om`oJm ?
(A) 2 (A) 2

1 1
(B) (B)
3 3

1 1
(C) (C)
2 2

1 1
(D) (D)
2 2

68. The correct order of increasing ionic 68. ~‹T>Vr Am`Zr` {deofVm H$m ghr H«$_ h¡
character
(A) BeCl2 < MgCl2 < CaCl2 < BaCl2 (A) BeCl2 < MgCl2 < CaCl2 < BaCl2
(B) BeCl2 < MgCl2 < BaCl2< CaCl2 (B) BeCl2 < MgCl2 < BaCl2< CaCl2

(C) BeCl2 < BaCl2< MgCl2 < CaCl2 (C) BeCl2 < BaCl2< MgCl2 < CaCl2

(D) BaCl2< CaCl2< MgCl2< BeCl2 (D) BaCl2< CaCl2< MgCl2< BeCl2

69. The ions O2–, F–, Na+, Mg2+ and Al3+ are 69. O2–, F–, Na+, Mg2+ Am¡a Al3+ Am`Z
isoelectronic. Their ionic radii show g_BboŠQ´>m{° ZH$ h¡ & BZH$s Am`Zr` {ÌÁ`m Xem©Vr h¡
(A) A decrease from O2– to F– and then (A) O2– go F– _| H$_r Am¡a {\$a Na+ go Al3+ H$s
increase from Na+ to Al3+ Amoa ~‹T>V
(B) A significant increase from O2– to (B) O2– go Al3+ H$s Amoa EH$ _hËdnyU© ~‹T>V
Al3+
(C) A significant decrease from O2– to (C) O2– go Al3+ H$s Amoa EH$ _hËdnyU© KQ>V
Al3+
(D) An increase from O2– to F– and (D) O2– go F– H$s Amoa EH$ ~‹T>V Am¡a Na+ go
then decrease from Na+ to Al3+ Al3+ H$s Amoa KQ>V

-26- Set-A

70. What are the products of following 70. {ZåZ{b{IV A{^{H«$`m Ho$ CËnmX Š`m h¢ ?
reaction ?

OCH2CH2OH OCH2CH2OH

HBr inexcess
→
A{YH$Vm _§o HBr
→
Heat J_©

(A) Br OCH2CH2Br (A) Br OCH2CH2Br

(B) OH + BrCH2CH2Br (B) OH + BrCH2CH2Br

(C) Br OH + BrCH2CH2Br (C) Br OH + BrCH2CH2Br

(D) Br+BrCH2CH2OH (D) Br+BrCH2CH2OH

-27- Set-A

71. OH 71. OH

Zn dust
→ X 
3
→
CH Cl

Zn
→ X 
3 Yyb
→
CH Cl
Anhydrous AlCl3 AZmÐ AlCl3
alkaline
Y → Z Y 
jmar`
→Z
kMnO4 kMnO 4

product Z is CËnmX Z h¡
(A) Toluene (A) Q>mbwB©Z
(B) Benzaldehyde (B) ~oÝOmpëS>hmBS>
(C) Benzoic acid (C) ~oÝOmoB©H$ Aåb
(D) Benzene (D) ~oÝOrZ

72. NH NaNO 2 / HCl
2 
CuCN / HCN
→ X → 72. NH NaNO 2 / HCl
2 
CuCN / HCN
→ X →
273k 273k

Sn / HCl
Y → Z Sn / HCl
Y → Z
Z is
Z h¡
(A) CHO (A) CHO

(B) COOH (B) COOH

(C) CH2NH2 (C) CH2NH2

(D) (D)

-28- Set-A

73. In the reaction 73. A{^{H«$`m
P4 + 3KOH + 3H2O → PH3 + 3KH2PO2, P4 + 3KOH + 3H2O → PH3 + 3KH2PO2,
phosphorus is _| \$m°ñ\$moag h¡
(A) Reduced (A) AnM{`V
(B) Oxidised (B) CnM{`V
(C) Both oxidised and reduced (C) CnM{`V Am¡a AnM{`V XmoZm|
(D) Neither oxidised nor reduced (D) Z Vmo CnM{`V Z hr AnM{`V
–
74. The tri-iodide ion I3 is formed by dissolving 74. Obr` KI {db`Z _| Am`moS>rZ {dbo` H$a
I2 in aqueous KI solution. The hybridization Q´>mB -Am`moS>mB©S> Am`Z I3– {Z{_©V hmoVm h¡& I3–
–
and geometry of I3 ion is Am`Z _| g§H$aU Ed§ Á`m{_{V h¡
(A) sp2, triangular (A) sp2, {ÌH$moUr`
(B) sp3, tetrahedral (B) sp3, MVwî\$bH$s`
(C) sp3d, trigonal bipyramidal (C) sp3d, {Ì^wOr` {Û{nam{_{S>`
(D) sp3d, linear (D) sp3d, a¡{IH$
75. Arrange the following ions as per 75. {ZåZ Am`Zm| H$mo A`wp½_V BboŠQ´>mZm| H$s g§»`m Ho$
decreasing order of number of AmYma na KQ>Vo H«$_ _| ì`dpñWV H$s{O`o&
unpaired electrons.
Co2+, Fe2+, Cu2+, Mn2+, Ti4+
Co2+, Fe2+, Cu2+, Mn2+, Ti4+
(A) Fe2+ > Mn2+ >Ti4+ > Co2+ > Cu2+
(A) Fe2+ > Mn2+ >Ti4+ > Co2+ > Cu2+
(B) Mn2+ > Fe2+ > Co2+ > Cu2+ > Ti4+
(B) Mn2+ > Fe2+ > Co2+ > Cu2+ > Ti4+
(C) Ti4+ > Cu2+ > Co2+ > Fe2+ > Mn2+ (C) Ti4+ > Cu2+ > Co2+ > Fe2+ > Mn2+
(D) Cu2+ > Co2+ > Fe2+ > Mn2+ > Ti4+ (D) Cu2+ > Co2+ > Fe2+ > Mn2+ > Ti4+

76. What term is used to describe the 76. S>rEZE I§S> H$s ZH$b go CËnÞ g§XoedmhH$
process by which a segment of DNA AmaEZE AUw Ho$ ~ZZo H$s {H«$`m Ho$ dU©Z hoVw Š`m
is copied to produce a molecule of eãX Cn`moJ {H$`m OmVm h¡ ?
messenger RNA ?
(A) Reproduction (A) àOZZ
(B) Replication (B) à{VH¥${V
(C) Translation (C) AZwdmX
(D) Transcription (D) AZwboIZ

-29- Set-A

77. Which amino acid can form disulphide 77. H$m¡Z-gm A_rZmo Aåb S>mB©gë\$mBS> ~ÝY ~ZmVm h¡ ?
bonds ?
(A) Proline (B) Leucine (A) àmobmBZ (B) ë`ygrZ
(C) Cysteine (D) Valine (C) {gñQ>rZ (D) d¡brZ
78. The atomic transition gives rise to the 78. na_mUw{dH$ g§H$« _U Ho$ Ûmam 104 MHz Amd¥{Îm H$s
radiation of frequency 104 MHz. The {d{H$aU àmßV hmoVr h¡ & na_mUw Ho$ D$Om© à{V_mob
change in energy per mole of atoms
taking place would be
_| n[adV©Z hmoJm
(A) 6.62 × 10–24 J (A) 6.62 × 10–24 J
(B) 3.99 × 10–6 J (B) 3.99 × 10–6 J
(C) 3.99 J (C) 3.99 J
(D) 6.62 × 10–30 J (D) 6.62 × 10–30 J
79. The energy of second orbit of hydrogen 79. {H$gH$s D$Om© hmBS´>moOZ Ho$ {ÛVr` H$jm H$s D$Om©
is equal to the energy of Ho$ ~am~a hmoJm ?
(A) Second orbit of Li2+ (A) Li2+ H$s {ÛVr` H$jm
(B) Fourth orbit of Li2+ (B) Li2+ H$s MVwW© H$jm
(C) Fourth orbit of He+ (C) He+ H$s MVwW© H$jm
(D) Second orbit of He+ (D) He+ H$s {ÛVr` H$jm

80. Which is the end product [B] of 80. {ZåZ A{^{H«$`m Ho$ A§{V_ CËnmX [B] H$m¡Z h¡ ?
following reactions ?

br / CCl NaNH
 →[ A ] 
∆ →[B]
br / CCl NaNH 2 4 2
 →[ A ] 
∆ →[B]
2 4 2
Ph Ph

(A) C6H5 − CH − CH2 (A) C6H5 − CH − CH2
| | | |
br br br br

(B) C6H5CH2CH3 (B) C6H5CH2CH3

(C) C6H5C ≡ CH (C) C6H5C ≡ CH

(D) C6H5CH3 (D) C6H5CH3

-30- Set-A

81. Carnallite’s solution in H2O, shows the 81. H 2O _| H$mZm©bmBQ> {db`Z ________ H$s
properties of {deofVmE± Xem©Vm h¡ &
(A) K , Mg , Cl , Br
+ 2+ – –
(A) K+, Mg2+, Cl–, Br –
(B) K+, Mg2+, CO2−
3 (B) K+, Mg2+, CO2−
3
(C) K+, Cl–, SO2−
4 , Br
–
(C) K+, Cl–, SO2−
4 , Br
–

(D) K+, Mg2+, Cl – (D) K+, Mg2+, Cl –

82. Which one of the following reactions is 82. {ZåZ{b{IV _| go H$m¡Z-gr A{^{H«$`m H¡$ëgrH$aU
an example for calcination process ? à{H«$`m H$m EH$ CXmhaU h¡ ?
(A) 2Ag + 2HCl + [O] → 2 AgCl + H2O (A) 2Ag + 2HCl + [O] → 2 AgCl + H2O
(B) 2Zn + O2 → 2ZnO (B) 2Zn + O2 → 2ZnO
(C) 2ZnS + 3O2 → 2ZnO + 2SO2 (C) 2ZnS + 3O2 → 2ZnO + 2SO2
(D) MgCO3 → MgO + CO2 (D) MgCO3 → MgO + CO2

CH3 CH3
| | NaOH
83. H2C = CH − C − COOH 
NaOH
→X 83. H2C = CH − C − COOH 
CaO / ∆
→X
| CaO / ∆ |
CH3 CH3
X will be X hmoJm
CH3 CH3
| |
(A) H3C − CH = C − CH3 (A) H3C − CH = C − CH3

CH3 CH3
| |
(B) H2C = CH − C − CH3 (B) H2C = CH − C − CH3
| |
H H
CH3 CH3
| |
(C) H3C − CH = C − CH2OH (C) H3C − CH = C − CH2OH

CH3 CH3
| |
(D) H2C = CH − C − COONa (D) H2C = CH − C − COONa
| |
CH3 CH3

-31- Set-A

CN CN
84. 84.
CH Mgbr H3+O

3
→ X → Y CH Mgbr H+O
OCH3 OCH3
3
→ X 
3
→Y


compound Y is `m¡{JH$ Y h¡
O O

(A) (A)

OCH3 OCH3

OH OH

(B) (B)

OCH3 OCH3

(C) (C)

OCH3 OCH3

O O

(D) (D)

-32- Set-A

O kMnO4 O
85.
kMnO
Alkene X 4
→ + 85. EëH$sZ X → +
∆ ∆
OH OH

Alkene X is EëH$sZ X h¡

(A) (A)

(B) (B)

(C) (C)

(D) (D)

86. The IUPAC name of the compound is 86. `m¡{JH$ H$m IUPAC Zm_H$aU h¡

CH2CH3 Cl CH2CH3
Cl

C=C C=C

I H3C I
H3C

(A) trans-3-iodo-4-chloro-3-pentene (A) Q´>mÝg-3-Am`moS>mo-4-Šbmoamo-3-noÝQ>rZ

(B) cis-2-chloro-3-iodo pentene (B) {gg-2-Šbmoamo-3-Am`moS>mo noÝQ>rZ

(C) trans-2-chloro-3-iodo-2-pentene (C) Q´>mÝg-2-Šbmoamo-3-Am`moS>mo-2-noÝQ>rZ

(D) cis-3-iodo-4-chloro-3-pentene (D) {gg-3-Am`moS>mo-4-Šbmoamo-3-noÝQ>rZ

-33- Set-A

87. Rate of SN1 reaction in the following 87. {ZåZ{b{IV `m¡{JH$m| _| SN1 {H«$`m H$s Xa hmoJr
compounds will be

I. Br I. Br

II. CH2Br II. CH2Br

III. CH2CH2Br III. CH2CH2Br

CH3 CH3
IV. CHBr IV. CHBr

(A) IV > I > III > II (A) IV > I > III > II

(B) II > III > I > IV (B) II > III > I > IV

(C) I > III > II > IV (C) I > III > II > IV

(D) IV > II > III > I (D) IV > II > III > I

-34- Set-A

88. A reactant (A) forms two products A{^H$maH$ (A) Xmo CËnmX ~ZmVm h¡
88.
A  k1
→ B , activation energy (Ea1) A  k1
→ B , g{H«$`U D$Om© (Ea1)
k2
A → C , activation energy (Ea2) A  k2
→ C , g{H«$`U D$Om© (Ea2)
If Ea2 = 2Ea1, then k1 and k2 will be
`{X Ea2 = 2Ea1, V~ k1 VWm k2 g§~{§ YV hm|Jo
related as
(A) k1 = Ak2eEa1/RT (A) k1 = Ak2eEa1/RT
(B) k2 = k1 eEa1/RT (B) k2 = k1 eEa1/RT
(C) k2 = k1eEa2/RT (C) k2 = k1eEa2/RT
(D) k1 = 2k2eEa2/RT (D) k1 = 2k2eEa2/RT

89. A chemical reaction was carried out at 89. EH$ amgm`{ZH$ A{^{H«$`m 300 K Am¡a 280 K na
300 K and 280 K. The rate constants H$s JB© Wr & Xa {Z`Vm§H$ k1 Am¡a k2 H«$_e: nmE JE&
were found to be k1 and k2 respectively.
Vmo
Then
(A) k2 = 4 k1 (B) k2 = 2 k1 (A) k2 = 4 k1 (B) k2 = 2 k1
(C) k2 = 0.25 k1 (D) k2 = 0.5 k1 (C) k2 = 0.25 k1 (D) k2 = 0.5 k1

90.
The half cell reactions for the corrosion 90. g§jmaU Ho$ {bE AY© gob A{^{H«$`m h¡
are
2H+ + ½O2 + 2e–→ H2O; E0 = –1.23 V
2H+ + ½O2 + 2e–→ H2O; E0 = –1.23 V
Fe2+ + 2e– → Fe(s); E0 = – 0.44 V
Fe2+ + 2e– → Fe(s); E0 = – 0.44 V
Find the ∆G° (in kJ) for the overall g_J« A{^{H«$`m Ho$ {bE ∆G° (kJ _|) kmV H$a| &
reaction.
(A) –76 (B) –322 (A) –76 (B) –322
(C) –161 (D) –152 (C) –161 (D) –152

91. Mesotartaric acid is optically inactive 91. _ogmoQ>mQ>©[aH$ Aåb àH$mer` A{H«$` h¡ BgH$m H$maU
due to presence of CnpñWV h¡
(A) Two asymmetric carbon atom (A) Xmo Ag_{_V H$m~©Z na_mUw
(B) Molecular asymmetry (B) AmU{dH$ Ag_{_{V
(C) External compensation (C) ~mø H$ånZeogZ
(D) Internal compensation (D) AmÝV[aH$ H$ånZeogZ

-35- Set-A

92. The incorrect statement among the 92. {ZåZ{b{IV _| go JbV H$WZ h¡
following is
(A) The first ionisation potential of (A) Al H$m àW_ Am`ZrH$aU {d^d Mg Ho$ àW_
Al is less than the first ionisation Am`ZrH$aU {d^d go H$_ h¡ &
potential of Mg.
(B) The second ionisation potential (B) Mg H$m {ÛVr` Am`ZrH$aU {d^d Na Ho$
of Mg is greater than the second
ionization potential of Na.
{ÛVr` Am`ZrH$aU {d^d go A{YH$ hmoVm h¡ &
(C) The first ionisation potential of (C) Na H$m àW_ Am`ZrH$aU {d^d Mg Ho$ àW_
Na is less than the first ionization Am`ZrH$aU {d^d go H$_ h¡ &
potential of Mg.
(D) The third ionisation potential of Mg (D) Mg H$m V¥Vr` Am`ZrH$aU {d^d Al Ho$ V¥Vr`
is greater than the third ionization Am`ZrH$aU {d^d go A{YH$ hmoVm h¡ &
potential of Al.

93. The solubilities of carbonates decrease 93. H$m~m}ZQo > H$s KwbZerbVm _¡½Zrer`_ g_yh _| ZrMo H$s
down the magnesium group due to a Amoa ________ _| EH$ KQ>V Ho$ H$maU KQ>Vr h¡ &
decrease in
(A) F$U Am`Z H$s Ob`moOZ D$Om©
(A) Hydration energies of cations
(B) Inter-ionic interaction (B) A§Va-Am`Zr` AZÝ`mo{H«$`m
(C) Entropy of solution formation (C) {db`Z {Z_m©U H$m CËH«$_ _mn
(D) Lattice energies of solids (D) R>mg o m| H$s OmbH$ D$Om©

94. Which one of the following exhibits 94. {ZåZ _| go H$m¡Z EH$ àoa{UH$,_ogmo_o[aH$ Ed§
inductive, mesomeric and A{Vg§`w½_Z à^md àX{e©V H$aVm h¡ ?
hyperconjugation effects ?

(A) CH3Cl (A) CH3Cl

(B) CH3 – CH = CH2 (B) CH3 – CH = CH2
O O
|| ||
(C) CH − CH = CH − C − CH (C) CH3 − CH = CH − C − CH3
3 3

(D) CH2 = CH – CH = CH2 (D) CH2 = CH – CH = CH2

-36- Set-A

95. A hydrocarbon reacts with HI to give 95. EH$ hmBS´>moH$m~©Z HI go {H«$`m H$a (X) XoVm h¡ Omo
(X) which on reacting with KOH(aq.) {H$ Obr` KOH go {H«$`m H$a (Y) {Z{_©V H$aVm
forms (Y). Oxidation of (Y) gives h¡& (Y) Ho$ AmåbrH$aU go 3-{_W¡b-2-ã`yQ>oZm°Z
3-methyl-2-butanone, the hydrocarbon is àmßV hmoVm h¡, Vmo hmBS´>moH$m~©Z h¡
CH3 CH3

(A) CH3 – CH = C (A) CH3 – CH = C
CH3 CH3

CH3 CH3

(B) CH2 = CH – CH (B) CH2 = CH – CH
CH3 CH3

(C) CH3 − CH2 − C = CH2 (C) CH3 − CH2 − C = CH2
| |
CH3 CH3

CH3 CH3

(D) HC ≡ C – CH (D) HC ≡ C – CH
CH3 CH3

96. Van’t Hoff factor of mercurous chloride in 96. _Š`©yag ŠbmoamBS> Ho$ Obr` {db`Z _| CgH$m
its aqueous solution will be (mercurous dmÝQ> hm\$ JwUH$ hmoJm (Obr` {db`Z _| _Š`y©ag
chloride is 90% ionised in solution)
ŠbmoamBS> 90% Am`{ZH¥$V h¡ )
(A) 1.8 (B) 3.7 (A) 1.8 (B) 3.7
(C) 2.8 (D) 3.8 (C) 2.8 (D) 3.8

97. Among KO2, AIO–2 , BaO2 and NO+2 , 97. KO2, AIO–2 , BaO2 VWm NO+2 _| {H$g_o
unpaired electron is present in A`wp½_V BboŠQ´>mZ CnpñWV h¡ ?
(A) KO2 (B) BaO2 (A) KO2 (B) BaO2
(C) KO2 and AIO–2 (D) NO+2 and KO2 (C) KO2 VWm AIO–2 (D) NO+2 VWm KO2

-37- Set-A

98. Sodium metal exists in BCC unit cell. 98. gmo{S>`_ YmVw BCC EH$H$ gob _| hmoVm h¡ &
The distance between nearest sodium g~go ZOXrH gmo{S>`_ na_mUwAm| Ho$ ~rM H$s
atoms is 0.368 nm. The edge length of Xÿar 0.368 nm h¡ & EH$H$ gob Ho$ {H$Zmao H$s
the unit cell is bå~mB© Š`m hmoJr ?
(A) 0.184 nm (A) 0.184 nm
(B) 0.425 nm (B) 0.425 nm
(C) 0.368 nm (C) 0.368 nm
(D) 0.575 nm (D) 0.575 nm

99. A gas expands adiabatically at constant 99. pñWa X>m~ na EH$ J¡g H$m éÕmoî_ àgma {ZåZ àH$ma go
1
pressure such that T α . hmoVm h¡ T α 1 ,
V V
The value of γ i.e. (Cp/Cv) of the gas J¡g Ho$ γ H$m _mZ (Cp/Cv) hmoJm
will be
(A) 1.5 (A) 1.5
(B) 1.7 (B) 1.7
(C) 1.3 (C) 1.3
(D) 2.0 (D) 2.0

100. At 27°C, the heat of combustion of 100. 27°C na, R>mg
o ~oÝOmoBH$ Aåb H$s XhZ D$î_m pñWa
solid benzoic acid at constant volume Am`VZ na – 321.30 kJ h¡ & 27°C Vmn_mZ VWm
is – 321.30 kJ. The heat of combustion at pñWa Xm~ na BgH$s XhZ D$î_m hmoJr
constant pressure and 27°C temperature
will be
(A) –321.30 + 900 R (A) –321.30 + 900 R

(B) –321.30 + 300 R (B) –321.30 + 300 R

(C) –321.30 – 300 R (C) –321.30 – 300 R

(D) –321.30 – 150 R (D) –321.30 – 150 R

-38- Set-A

PART – III (A)

mathematics J{UV
101. If area of triangle is 35 sq. units with 101. `{X erfm] (2, –6), (5, 4) Am¡a (k, 4) dmbo {Ì^wO
vertices (2, – 6), (5, 4) and (k, 4), then H$m joÌ\$b 35 dJ© BH$mB© hmo, Vmo k hmoJm
k is
(A) 12 (B) – 2 (A) 12 (B) – 2
(C) –12, – 2 (D) 12, – 2 (C) –12, – 2 (D) 12, – 2
102. Consider the system of linear equations 102. a¡{IH$ g_rH$aUm| Ho$ V§Ì
y1 + 2y2 + y3 = 3 y1 + 2y2 + y3 = 3
2y1 + 3y2 + y3 = 3 2y1 + 3y2 + y3 = 3
3y1 + 5y2 + 2y3 = 1 has 3y1 + 5y2 + 2y3 = 1 H$m hb hmoJm
(A) exactly 3 solutions (A) R>rH$ 3 hb
(B) a unique solution (B) EH$ A{ÛVr` hb
(C) no solution (C) H$moB© hb Zht
(D) infinite number of solutions (D) AZ§V hb

103. The region represented by joÌ Omo 2x + 3y – 5 ≤ 0 VWm
103.
2x + 3y – 5 ≤ 0 and 4x – 3y + 2 ≤ 0, is 4x – 3y + 2 ≤ 0 go àX{e©V h¢ , hmoJm
(A) Not in first quadrant (A) àW_ MVwWmªe _|o Zht h¡
(B) Bounded in first quadrant (B) àW_ MVwWmªe H$s gr_m _| h¡
(C) Unbounded in first quadrant (C) àW_ MVwWmªe H$s gr_m _| Zht h¡
(D) None of these (D) BZ_| go H$moB© Zht

104. If 3x1 + 5x2 ≤ 15 104. `{X 3x1 + 5x2 ≤ 15
6x1 + 2x2 ≤ 10 6x1 + 2x2 ≤ 10
x1, x2 ≥ 0 x1, x2 ≥ 0
then the maximum value of 5x1 + 3x2 V~ 5x1 + 3x2 H$m J«m\$s` {d{Y go
by graphical method is _hÎm_ _mZ hmoJm
7 1 7 1
(A) 12 (B) 12 (A) 12 (B) 12
19 7 19 7
3 3
(C) 12 (D) 12 (C) 12 (D) 12
5 5

-39- Set-A

105. ∫x tan–1x dx is equal to 105. ∫x tan–1x dx ~am~a h¡
1 2 1 2
(A) (x + 1) tan– 1x – x + c (A) (x + 1) tan– 1x – x + c
2 2
1 2 1 2
(B) (x + 1) tan– 1x + x + c (B) (x + 1) tan– 1x + x + c
2 2
1 2 1 1 2 1
(C) (x + 1) tan– 1x – x+c (C) (x + 1) tan– 1x – x+c
2 2 2 2
1 2 1 1 2 1
(D) (x – 1) tan– 1x – x+c (D) (x – 1) tan– 1x – x+c
2 2 2 2

x5 x5
106. ∫ dx equals 106. ∫ dx ~am~a h¡
1+ x 3 1+ x 3

(A) 2 (x 3 − 2) 1 + x 3 + c (A) 2 (x 3 − 2) 1 + x 3 + c
9 9

(B) 2 (x 3 + 2) 1 + x 3 + c (B) 2 (x 3 + 2) 1 + x 3 + c
9 9
3 3 3 3
(C) (x + 2) 1 + x + c (C) (x + 2) 1 + x + c
(D) none of these (D) BZ_| go H$moB© Zht
107. The locus of the middle-points of the 107. r {ÌÁ`m dmbo d¥Îm H$s OrdmAmo§ Omo Ho$ÝÐ na g_H$moU
chords of a circle with radius r which AÝV[aV H$aVm h¢, Ho$ _Ü` {~ÝXþ H$m {~ÝXþnW EH$$
subtend a right angle at the centre of g_Ho$ÝÐr`$ d¥Îm hmoJm {OgH$s R Eogr hmoJr {H$
the circle is a concentric circle where
radius R is such that
(A) R = r (A) R = r

1 1
(B) R = r (B) R = r
2 2
(C) R = 2r (C) R = 2r
1 1
(D) R = r (D) R = r
2 2

-40- Set-A

108. The equations of the normals at the 108. nadb` Ho$ Zm^r` Ordm Ho$ {gam§o go IrMr JB©
ends of the latus rectum of the parabola A{^bå~m| H$m g_rH$aU hmoJm
are given by
(A) x2 – y2 – 6ax + 9a2 = 0 (A) x2 – y2 – 6ax + 9a2 = 0
(B) x2 – y2 – 6ax – 6ay + 9a2 = 0 (B) x2 – y2 – 6ax – 6ay + 9a2 = 0
(C) x2 – y2 – 6ay + 9a2 = 0 (C) x2 – y2 – 6ay + 9a2 = 0
(D) x2 – y2 – 6ax + 6ay – 9a2 = 0 (D) x2 – y2 – 6ax + 6ay – 9a2 = 0
  
109.
The shortest distance   between the 109. pñH$`y$ aoI mAm| l1: r = a1 + λb1 VWm
skew lines l1 : r = a1 + λb1 and l2 : r = a 2 + mb2 Ho$ _Ü` Ý`yZV_ Xÿar h¡
  
l2 : r = a 2 + mb2 is
       
(a 2 − a1). b1 × b2 (a 2 − a1). b1 × b2
(A)   (A)  
b1 × b2 b1 × b2
       
(a1 − b1). a 2 × b2 (a1 − b1). a 2 × b2
(B)    
b1 × b2 (B)
b1 × b2

       
(a 2 − b2 ). a1 × b1 (a 2 − b2 ). a1 × b1
(C)   (C)  
b1 × b2 b1 × b2
       
(D) (a 2 − b1). b1 × a 2 (D) (a 2 − b1). b1 × a 2
   
b1 × b2 b1 × b2

110. The angle between the straight lines 110. gab aoImAm§o x = 1, y = 2 VWm y = – 1, z = 0
x = 1, y = 2 and y = – 1, z = 0 is Ho$ ~rM H$m H$moU
(A) 90° (A) 90°

(B) 30° (B) 30°

(C) 60° (C) 60°

(D) 0° (D) 0°

-41- Set-A

111. The sum of n terms of the series 111. loUr 12 + (12 + 32) + (12 + 32 + 52) + . . . .
12 + (12 + 32) + (12 + 32 + 52) + . . . . is Ho$ n nXm| H$m `moJ hmoJm
1 4 1 4
(A) (n + 2n2) (A) (n + 2n2)
3 3
1 1
(B) n (n + 1) (2n2 + 2n – 1) (B) n (n + 1) (2n2 + 2n – 1)
6 6
1 3 1 3
(C) (n + 3n2 – n) (C) (n + 3n2 – n)
3 3
(D) none of these (D) BZ_| go H$moB© Zht

112. Let α, β are roots of the equation 112. `{X α Am¡a β g_rH$aU (x – a) (x – b) = c,
(x – a) (x – b) = c, c ≠ 0, then the roots of c ≠ 0 Ho$ _yb h¢, Vmo g_rH$aU (x – α) (x – β) +
the equation (x – α) (x – β) + c = 0 are c = 0 Ho$ _yb hm|Jo
(A) a, c (B) b, c (A) a, c (B) b, c
(C) a + c, b + c (D) a, b (C) a + c, b + c (D) a, b

113. A man has 10 friends. In how many 113. EH$ ì`pŠV Ho$ 10 XmoñV h¢& {H$VZo VarHo$ go dh
ways he can invite one or more of them EH$ `m Á`mXm XmoñVm| H$mo EH$ nmQ>u _| Am_§{ÌV H$a
to a party ? gH$Vm h¡ ?
(A) 10 (B) 210 (A) 10 (B) 210
(C) 210 – 1 (D) 10 – 1 (C) 210 – 1 (D) 10 – 1

sin−1 x
114. The domain of the function 114. \$bZ f(x) = H$m àmÝV h¡
−1
[x]
sin x
f( x ) = is
[x]
(A) [–1, 1] – {0} (A) [–1, 1] – {0}

(B) [–1, 0) (B) [–1, 0)

(C) [–1, 0) ∪ {1} (C) [–1, 0) ∪ {1}

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

-42- Set-A

115. Let f(x) = 210. x + 1 and g(x) = 310. x – 1. 115. _mZ bmo f(x) = 210. x + 1 Am¡a g(x) = 310. x – 1,
If (fog) (x) = x, then x is equal to `{X (fog) (x) = x, Vmo x ~am~a h¡
310 − 1 210 − 1 310 − 1 210 − 1
(A) (B) 10 (A) (B) 10
310 − 2−10 2 − 3−10 310 − 2−10 2 − 3−10
1 − 2−10 1 − 3−10 1 − 2−10 1 − 3−10
(C) 10 (D) 10 (C) 10 (D)
3 − 2−10 2 − 3−10 3 − 2−10 210 − 3−10

 x 2 − 4x + 3  x 2 − 4x + 3
, x ≠1
116. Let f(x ) =  x 2 + 2x − 3 116. _mZ bmo f(x) =  x 2 + 2x − 3 , x ≠ 1
 k , x =1  k , x =1
 
if f(x) is continuous at x = 1, then the `{X f(x), x = 1 na gVV h¡, Vmo k H$m _mZ hmoJm
value of k will be
1 1
(A) 1 (B) (A) 1 (B)
2 2
1 1
(C) –1 (D) − (C) –1 (D) −
2 2
−1  1 − x   1− x 
117. If y = sin   , then differential 117. `{X y = sin−1   , Vmo x Ho$ gmnoj
 1+ x   1+ x 
coefficient w.r.t. x is AdH$b JwUm§H$ h¡
−2 −2
(A) (B) x (A) (B) x
1+ x 1+ x
2 2
(C) (D) 1 (C) (D) 1
x x
118. If a differentiable function f(x) has a 118. `{X EH$ AdH$bZr` \$bZ f(x), x = 0 na Ý`yZV_
minimum at x = 0, then function h¡, Vmo \$bZ g(x) = f(x) + ax + b ^r x = 0
g(x) = f(x) + ax + b will also have a na Ý`yZV_ hmoJm
minimum at x = 0
(A) for all values of a and b (A) a Am¡a b Ho$ g^r _mZm| Ho$ {b`o
(B) for all values of b if a = 0 (B) b Ho$ g^r _mZm| Ho$ {b`o `{X a = 0
(C) for all positive values of b (C) b Ho$ g^r YZmË_H$ _mZm| Ho$ {b`o
(D) for all positive values of a (D) a Ho$ g^r YZmË_H$ _mZm| Ho$ {b`o

-43- Set-A

119. If the Rolle’s theorem holds for the `{X A§Vamb [–1, 1] _| \$bZ
119.
function f(x) = 2x3 + ax2 + bx in the f(x) = 2x3 + ax2 + bx Am¡a c = 1 Ho$ {b`o amoobr
interval [–1, 1] for the point c = 1 , then 2
the value of 2a + b is 2 H$m à_o` g§VwîQ> hmoVm h¡, Vmo 2a + b H$m _mZ h¡
(A) 1 (B) –1 (A) 1 (B) –1
(C) 2 (D) –2 (C) 2 (D) –2

120. The binomial distribution for which 120. {ÛnX ~§Q>Z {OgHo$ {bE _mÜ` = 6 VWm
mean = 6 and variance = 2 is àgaU = 2 h¡, hmoJm
6
 2 1  2 1
6
(A)  +  (A)  + 
3 3 3 3
9
 2 1 9
(B)  +   2 1
3 3 (B)  + 
3 3
6
 1 2
(C)  +  6
 1 2
(C)  + 
3 3
3 3
(D) None of the above
(D) CnamoŠV _| go H$moB© Zht
121. Two lines of regression are 3x + 4y – 7 = 0 Xmo g_ml`U aoIm`o§ 3x + 4y – 7= 0 VWm
121.
and 4x + y – 5 = 0. Then correlation 4x + y – 5 = 0 h¢, Vmo x VWm y Ho$ ~rM
coefficient between x and y is gh-g§~§Y JwUm§H$ h¡
3 − 3 3 − 3
(A) (B) (A) (B)
4 4 4 4
3 3 3 3
(C) (D) − (C) (D) −
16 16 16 16
7 dx 7 dx
122. By simpson’s rule, the value of ∫ x is 122. {gångZ {Z`_ go ∫ H$m _mZ hmoJm
1 1 x
(A) 1.358 (B) 1.958 (A) 1.358 (B) 1.958
(C) 1.625 (D) 1.458 (C) 1.625 (D) 1.458

-44- Set-A

123. The value of cos–1 (cos 12) – sin–1 (sin 12) is 123. cos–1 (cos 12) – sin–1 (sin 12) H$m _mZ hmoJm

(A) 0 (B) π (A) 0 (B) π
(C) 8π – 24 (D) 2π – 24 (C) 8π – 24 (D) 2π – 24

124. In ∆ ABC if angles A, B, C are 124. `{X {H$gr ∆ ABC _| H$moU A, B, C Bg àH$ma
tan A : tan B : tan C = 1 : 2 : 3, then sides h¡§ tan A : tan B : tan C = 1 : 2 : 3, Vmo
a : b : c is ^wOmE| a : b : c = ?
(A) 2 : 3 : 4 (B) 5: 8: 9 (A) 2 : 3 : 4 (B) 5: 8: 9

(C) 1 : 2 : 3 (D) 7 : 9 : 11 (C) 1 : 2 : 3 (D) 7 : 9 : 11

π/4 π/4
125. ∫ log(1 + tan x ) dx equals 125. ∫ log(1 + tan x ) dx ~am~a h¡
0 0
 π  π
(A)   log 2 (A)   log 2
 2  2

 π  π
(B)   log 2 (B)   log 2
 4  4

 π  1  π  1
(C)  4  log  2  (C)  4  log  2 

 π  π
(D)   log 2 (D)   log 2
8 8

126. The area between the parabolas 126. nadb` y2 = 4ax Am¡a x2 = 4ay Ho$ ~rM H$m
y2 = 4ax and x2 = 4ay is joÌ\$b h¡
(B)   a2
16
(B)   a2
16 (A)  8  a2
(A)  8  a2  
   3 3  3
3

 10  2  32  2  10  2  32  2
(C)   a (D)   a (C)   a (D)   a
 3 3  3 3

-45- Set-A

127. If p and q be respectively order and 127. `{X AdH$b g_rH$aU
degree of differential equation
2
 d2 y 
2
 dy   d2 y   dy 
y 2  + 3x   + x 2 y 2 = sinx , y  2  + 3x   + x 2 y 2 = sinx Ho$
2
2   dx 
 dx   dx   dx 
then H$mo{Q> Am¡a KmV H«$_e: p Am¡a q h¢, Vmo
(A) p < q (A) p < q
(B) p > q (B) p > q
(C) p = q (C) p = q
(D) 2p = q (D) 2p = q

128. The root of the equation x3 – 6x + 1 = 0 128. g_rH$aU x3 – 6x + 1 = 0 Ho$ _yb AÝVamb _|
lies in the interval hm|Jo
(A) (2, 3) (B) (3, 4) (A) (2, 3) (B) (3, 4)

(C) (3, 5) (D) (4, 6) (C) (3, 5) (D) (4, 6)

1
129. By trapezoidal rule, the value of 129. Q´>onoÁdmBS>b Ho$ {Z`_ go ∫0 x 3 dx H$m _mZ hmoJm
1

∫ x dx considering five subintervals is O~ {H$ AÝVamb H$mo 5 Cn^mJm| _| ~m§Q>m OmVm h¡
3
0

(A) 0.21 (B) 0.23 (A) 0.21 (B) 0.23
(C) 0.24 (D) 0.26 (C) 0.24 (D) 0.26

130. The L.P. Problem Max. z = x1 + x2 such 130. a¡{IH$ àmoJ«m{_H$ g_ñ`m Max. z = x1 + x2
that –2x1 + x2 ≤ 1, x1 ≤ 2, x1 + x2 ≤ 3 and Bg àH$ma h¢ {H$ –2x1 + x2 ≤ 1, x1 ≤ 2,
x1, x2 ≥ 0 has x1 + x2 ≤ 3 VWm x1, x2 ≥ 0 H$m h¡
(A) One solution (A) EH$ hb
(B) Three solutions (B) VrZ hb
(C) An infinite number of solutions (C) AZ§V hb
(D) None of these (D) BZ_| go H$moB© Zht

-46- Set-A

131. The solution of the differential equation dy
131. AdH$b g_rH$aU + y = cosx H$m hb h¡
dx
dy
+ y = cosx is
dx
1 1
(A) y = (cosx + sinx ) + ce− x (A) y = (cosx + sinx ) + ce− x
2 2
1
(B) y = (cosx − sinx ) + ce− x (B) y =
1
(cosx − sinx ) + ce− x
2 2
(C) y = cosx + sinx + ce–x (C) y = cosx + sinx + ce–x

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

132. If two events A and B are such that 132. `{X Xmo KQ>Zm`| A Am¡a B Bg àH$ma h¡ {H $
5 1 1 5 1 1
P(A + B) = , P(AB) = and P(A) = , P(A + B) = , P(AB) = and P(A) = ,
6 3 2 6 3 2
then the events A and B are Vmo A Am¡a B hm|Jo
(A) Independent (A) ñdV§Ì
(B) Mutually exclusive (B) nañna AndOu
(C) Independent and Mutually (C) ñdV§Ì VWm nañna AndOu
exclusive
(D) None of these (D) BZ_| go H$moB© Zht

133. For any event A 133. {H$gr KQ>Zm A Ho$ {bE
(A) P(A) + P(Ā) = 0 (A) P(A) + P(Ā) = 0

(B) P(A) + P(Ā) = 1 (B) P(A) + P(Ā) = 1

(C) P(A) > 1 (C) P(A) > 1

(D) P(Ā) < 1 (D) P(Ā) < 1

-47- Set-A

134. Let f : (–1, 1) → R be a differentiable _mZ bmo f : (–1, 1) → R EH$ AdH$bZr` \$bZ
134.
function with f(0) = – 1 and f′(0) = 1. h¡ VWm f(0) = – 1 Am¡a f′(0) = 1, _mZ bmo
Let g(x) = [ f (2f(x) + 2)]2. Then g′(0) = g(x) = [ f (2f(x) + 2)]2 , Vmo g′(0) =
(A) –4 (B) 0 (A) –4 (B) 0
(C) –2 (D) 4 (C) –2 (D) 4

log(3 + x ) − log(3 − x ) log(3 + x ) − log(3 − x )
135. If lim = k, 135. `{X lim = k , Vmo k
x →0 x x →0 x
then the value of k will be H$m _mZ hmoJm
1 1
(A) 0 (B) − (A) 0 (B) −
3 3
2 2 2 2
(C) − (D) (C) − (D)
3 3 3 3

136.
if the lines x + ay + a = 0, bx + y + b = 0, 136. `{X aoIm`| x + ay + a = 0, bx + y + b = 0,
and cx + cy + 1 = 0 (a, b and c being VWm cx + cy + 1 = 0 (Ohm± a, b VWm c AbJ
distinct ≠1) are concurrent, then the ({S>pñQ>ÝŠQ>) h¢, ≠1) g§Jm_r hmo,
value of a + b + c is Vmo a b c H$m _mZ hmoJm
a −1 b −1 c −1 + +
a −1 b −1 c −1
(A) –1 (B) 0 (A) –1 (B) 0
(C) 1 (D) abc (C) 1 (D) abc

137. If one of the lines given by 6x2 – xy + 137. `{X 6x2 – xy + 4cy2 = 0 Ûmam àmßV EH$ aoIm
4cy2 = 0 is 3x + 4y = 0, then c equals 3x + 4y = 0 h¡, Vmo c H$m _mZ hmoJm
(A) 3 (B) –1 (A) 3 (B) –1
(C) 1 (D) –3 (C) 1 (D) –3

138. The ratio in which the plane 2x – 1 = 0 138. g_Vb 2x – 1 = 0, {~ÝXþAm| (–2, 4, 7) VWm
divides the line joining (–2, 4, 7) and (3, –5, 8) H$mo {_bmZo dmbo aoImIÊS> H$mo {H$g
(3, –5, 8) is
AZwnmV _| {d^m{OV H$aVm h¡ ?
(A) 2 : 3 (B) 4 : 5 (A) 2 : 3 (B) 4 : 5
(C) 7 : 8 (D) 1 : 1 (C) 7 : 8 (D) 1 : 1

-48- Set-A

x−3 y−4 z−5 x−3 y−4 z−5
139. Straight line = = lies 139. gab aoIm = = , g_Vb
2 3 4 2 3 4
on the plane 4x + 4y – kz – d = 0. Then 4x + 4y – kz – d = 0 _| pñWV h¡& V~, k VWm
the value of k and d respectively is d Ho$ _mZ H«$_e: h¡
(A) 4, 8 (B) – 5, 3 (A) 4, 8 (B) –5, 3
(C) 5, 3 (D) –4, –8 (C) 5, 3 (D) –4, –8
 
140. The equation | r | = 5 represents 140. g_rH$aU | r | = 5 àX{e©V H$aVm h¡
(A) a circle (B) a straight line (A) EH$ d¥Îm (B) EH$ gab aoIm
(C) a sphere (D) none of these (C) EH$ Jmobm (D) BZ_| go H$moB© Zht

dy
141. Solution of the differential equation 141. AdH$b g_rH$aU 2x −y=3
dy dx
2x − y = 3 represents H$m hb Xem©Vm h¡
dx
(A) circles (B) straight lines (A) d¥Îmm| (B) gab aoImAm|
(C) ellipses (D) parabolas (C) XrK©d¥Îmm| (D) nadb`m|

dy y−x
142. The solution of differential equation 142. AdH$b g_rH$aU dx = y + x H$m hb h¡
dy y − x is
=
dx y+x

−1 y 2 2 −1 y
2 2
(A) log (x + y ) + 2tan +c (A) log (x + y ) + 2tan +c
2 2 x x
y x y2 x2
(B) + xy = xy − +c + xy = xy − +c
2 2 (B)
2 2
(C) y = x – 2 logy + c (C) y = x – 2 logy + c
(D) None of the above (D) CnamoŠV _| go H$moB© Zht
13
∑ (i n + i (n+1) ) , i = −1 H$m _mZ hmoJm
13
143. The value of ∑ (i + i
n (n+1)
) , i = −1 , 143.
n =1
n =1
equals
(A) 1 + i (B) –1 + i (A) 1 + i (B) –1 + i
(C) 1 (D) – i (C) 1 (D) – i

-49- Set-A

144. An infinite geometric progression has 144. `{X {H$gr AZ§V JwUmoÎma loUr H$m àW_ nX x Am¡a
first term x and sum 5, then x belongs to `moJ 5 hmo, Vmo x H$m _mZ hmoJm
(A) x < –10 (A) x < –10
(B) –10 < x < 0 (B) –10 < x < 0
(C) 0 < x < 10 (C) 0 < x < 10
(D) None of the above (D) CnamoŠV _| go H$moB© Zht
145. Suppose a, b, c are in arithmetic 145. `{X a, b, c g_mZm§Va loUr, a2, b2, c2 JwUmoÎma
progression and a2, b2, c2 are in geometric loUr _| hmo, a < b < c VWm a + b + c = 3 ,
progression, a < b < c, a + b + c = 3 , V~ a H$m _mZ hmoJm 2
then value of a is 2
1 1 1 1
(A) 2 2 (B) 2 3 (A) 2 2 (B) 2 3

1 1 1 1 1 1 1 1
(C) − (D) − (C) − (D) −
2 3 2 2 2 3 2 2

146. The value of x for which  the angle 146. x H$m dh _mZ, {OgHo$ {bE g{Xem|
 
ˆ ˆ ˆ a = − 3iˆ + xjˆ + kˆ VWm b = x ˆi + 2xjˆ + kˆ Ho$
 the vectors a = − 3i + xj + k
between

and b = x ˆi + 2xjˆ+ kˆ is acute and the ~rM Ý`yZ H$moU h¡ VWm b d x-Aj Ho$ ~rM H$moU
angle between b and the x-axis is lies π
π
d π Ho$ ~rM h¡
between and π 2
2
(A) x > 0 (B) x < 0 (A) x > 0 (B) x < 0
(C) only x > 1 (D) only x < –1 (C) Ho$db x > 1 (D) Ho$db x < –1
     
147. If a, b, c are three
  vectors such that 147. `{X VrZ g{Xe a,b, c Bg àH$ma h¡ {H$
     
a × b = c and b × c = a , then a × b = c VWm b × c = a , V~
     
(A) a, b, c are mutually orthogonal (A) a, b, c nmañn[aH$ bmpå~H$ h¢
     
(B) | a | = | b | = | c | (B) | a | = | b | = | c |
     
(C) | a | = | b | = | c | ≠ 1 (C) | a | = | b | = | c | ≠ 1
(D) none of these (D) BZ_| go H$moB© Zht

-50- Set-A

148.
If the pth, qth and rth terms of a G.P. are `{X JwUmoÎma loUr H$m p dm±, q dm± VWm r dm± nX a, b
148.
the positive numbers a, b and c. Then VWm c YZmË_H$ g§»`m h§¡, Vmo g{Xe
angle between the vectors log a 3 ˆi + logb3 ˆj + logc 3kˆ VWm
log a 3 ˆi + logb3 ˆj + logc 3kˆ and (q − r)iˆ + (r − p)jˆ + (p − q)kˆ Ho$ ~rM H$m
(q − r)iˆ + (r − p)jˆ + (p − q)kˆ is H$moU h¡
π π
(A) (A)
6 6
π π
(B) (B)
2 2
π π
(C) (C)
3 3
−1
 1  −1
 1 
(D) sin   (D) sin  
2 2 2 
 a +b +c   a 2 + b2 + c 2 

149. The angular elevation of a tower CD 149. {H$gr Q>mda CD H$m CÞ`Z H$moU CgHo$ {H$gr X{jU
at a point A due south of it is 60° and q~Xþ A go H$moUr` PwH$md 60° Am¡a {H$gr npíM_r
a point B due west of A the elevation q~Xþ B go PwH$md 30° h¡, `{X AB = 3 BH$mB© hmo,
is 30°. If AB = 3 units, then height of Vmo Q>mda H$s D§$MmB© hmoJr
the tower is
(A) 2 3 units (B) 2 6 units (A) 2 3 BH$mB© (B) 2 6 BH$mB©
3 3 3 3
(C) 3 3 units (D) units (C) 3 3 BH$mB© (D) BH$mB©
2 10 2 10

The most general value of θ satisfying the 1
150. 150. g_rH$aUm| tan θ = –1 Am¡a cos θ = H$mo
2
1
equations tan θ = –1 and cos θ = is g§VwîQ> H$aZo dmbm θ H$m g~go gm_mÝ` _mZ hmoJm
2
7π 7π 7π 7π
(A) nπ + (B) nπ + (−1)n (A) nπ + (B) nπ + (−1)n
4 4 4 4

7π 7π
(C) 2nπ + (D) no solution (C) 2nπ + (D) H$moB© hb Zht
4 4

-51- Set-A

PART – III (B)
Biology Ord {dkmZ
101. Which of the following is a role of 101. b¡H$ àMmbH$ _| boŠQ>moO H$s ^y{_H$m {ZåZ _| go Š`m
Lactose in lac operon ? h¡ ?
(A) Inhibitor (B) Inducer (A) _§XH$ (B) àoaH$
(C) Operator (D) Promoter (C) àMmbH$ (D) CÝZm`H$

102. Which of the following transcribes 102. {ZåZ _| go H$m¡Z amB~mogmo_b Ama.EZ.E. H$m
Ribosomal R.N.A. ? AZwboIZ H$aVm h¡ ?
(A) R.N.A. Polymerase I (A) Ama.EZ.E. nm°br_aoO I
(B) R.N.A. Polymerase II (B) Ama.EZ.E. nm°br_aoO II
(C) R.N.A. Polymerase I and III (C) Ama.EZ.E.nm°br_aoO I Am¡a III
(D) D.N.A. Polymerase (D) S>r.EZ.E. nm°br_aoO

103. Which species of Pinus seed is sold 103. nmBZg H$s H$m¡Z-gr àOm{V Ho$ ~rOmo§ ewîH$ \$b
as a dry fruit ? Ho$ ê$n _| ~oMm OmVm h¡ ?
(A) armandi (B) gerardiana (A) Aa_ÊS>mB© (B) {Oama{S>`mZm
(C) wallichiana (D) roxburghii (C) d¡{b{M`mZm (D) am°Šg~Km©B

104. In which part of selaginella, 104. {gbo{OZobm Ho$ {H$g ^mJ _| ½bmgmonmo{S>`_ H$m
Glossopodium is formed ? {Z_m©U hmoVm h¡ ?
(A) Root (B) Stem (A) O‹S> (B) VZm
(C) Leaf (D) Ligule (C) nÎmr (D) {b½`yb

105. Upper part of sea or aquatic ecosystem 105. g_wÐ `m Obr` nm[apñW{VH$ V§Ì H$s D$nar gVh
contains na nm`o OmVo h¢
(A) Planktons (A) ßb¡§ŠQ>m°Ýg
(B) Nektons (B) ZoŠQ>m°Ýg
(C) Benthos (C) ~oÝWmoO
(D) All the above (D) CnamoŠV g^r

-52- Set-A

106. The dominant second trophic level in 106. Prb Ho$ nm[apñW{VH$ V§Ì _| à^mdr {ÛVr` nmofU
a lake ecosystem is g§ñVa hmoVm h¡
(A) Phytoplankton (A) \$m`Q>moßb¡§ŠQ>m°Z
(B) Zooplankton (B) Oyßb¡§ŠQ>m°Z
(C) Plankton (C) ßb¢ŠQ>m°Z
(D) Benthos (D) ~oÝWmoO

107. Packaging of materials in the cell is a 107. H$mo{eH$m _| gm_J«r`m| H$m nwbÝXm ~ZmZm {ZåZ _| go
main function of which of the following {H$g H$mo{eH$m§J H$m à_wI H$m`© h¡ ?
cell organell ?
(A) Mitochondria (A) _mBQ>moH$m°§{S>´>`m
(B) Chloroplast (B) h[aVbdH$
(C) Golgibody (C) Jm°ëOrH$m`
(D) Nucleus (D) Ho$ÝÐH$

108. Dog flower is an example of which of 108. S>m°J âbmda {ZåZ _| go {H$gH$m CXmhaU h¡ ?
the following ?
(A) Incomplete dominance (A) AnyU© à^m{dVm
(B) Complete dominance (B) nyU© à^m{dVm
(C) Co-dominance (C) gh-à^m{dVm
(D) Multiple allelism (D) ~hþ AbrbVm

109. Example of most stable ecosystem is 109. gdm©{YH$ ñWm`r nm[apñW{VH$ V§Ì H$m CXmhaU h¡
(A) Ocean (B) Mountain (A) _hmgmJar` (B) nd©Vr`
(C) Forest (D) Desert (C) dZ (D) _éñWbr`

110. Coir is obtained from which part of the 110. Zm[a`b \$b Ho$ {H$g ^mJ go H$mo`a àmßV {H$`m
coconut fruit ? OmVm h¡ ?
(A) Epicarp (B) Mesocarp (A) BnrH$mn© (B) _rOmoH$mn©

(C) Seed coat (D) endocarp (C) ~rOmdaU (D) EÊS>moH$mn©

-53- Set-A

111. Match the following and choose correct {ZåZ{b{IV H$s OmoS‹ >r ~ZmB`o VWm {X`o J`o {dH$ënm|
111.
combination from the option given. _| go ghr {dH$ën Mw{Z`o&
List – I List – II gyMr – I gyMr – II
a. S 1. Chlorophyll a. S 1. Šbmoamo{\$b
b. Zn 2. Nitrogenase b. Zn 2. ZmBQ´>mo{OZoO
c. Mg 3. Mithionin c. Mg 3. {_{W`mo{ZZ
d. Md 4. Auxin d. Md 4. Am°pŠOZ
(A) a – 1, b – 2, c – 3, d – 4 (A) a – 1, b – 2, c – 3, d – 4
(B) a – 3, b – 4, c – 1, d – 2 (B) a – 3, b – 4, c – 1, d – 2
(C) a – 3, b – 1, c – 2, d – 4 (C) a – 3, b – 1, c – 2, d – 4
(D) a – 2, b – 4, c – 1, d – 3 (D) a – 2, b – 4, c – 1, d – 3

112. Guttation is the result of 112. {~ÝXþ òmd {H$g H$m n[aUm_ h¡ ?
(A) Transpiration (A) dmînmoËgO©Z
(B) Osmosis (B) namgaU
(C) Diffusion (C) {dgaU
(D) Root pressure (D) _yb Xm~

113. The drug used by ophthalmologists to 113. dh Am¡f{Y {OgH$m Cn`moJ ZoÌ-{deofkm| Ho$ Ûmam
enlarge the pupil of eyes is obtained Am±I H$s nwVbr H$mo ~‹S>m H$aZo Ho$ {bE {H$`m OmVm
from h¡, Cgo {H$ggo V¡`ma {H$`m OmVm h¡ ?
(A) Caffeine (B) Digitalis (A) H¡$\$sZ (B) {S>OrQ>°{bg

(C) Belladona (D) Ginseng (C) ~obmS>moZm (D) {OZg|J

114. Genetically engineered golden rice 114. AmZwd§{eH$ ê$n go A{^`m§{ÌV JmoëS>oZ Mmdb _|
synthesize large amount of {H$gH$m g§íbofU A{YH$ _mÌm _| hmoVm h¡ ?
(A) Vitamin K (A) {dQ>m{_Z K
(B) Beta carotene (B) ~rQ>m H¡$amoQ>rZ
(C) Vitamin C (C) {dQ>m{_Z C
(D) Beta galactosidase (D) ~rQ>m J¡boŠQ>mogmBS>oO

-54- Set-A

115. Micro-organism used in the production 115. {H$g gyú_Ord H$m Cn`moJ Zrbr nZra Ho$ CËnmXZ
of blue cheese is _| {H$`m OmVm h¡ ?
(A) Rhizopus stolonifer (A) amBOmong ñQ>mobmoZr\$a
(B) Lactobacillus bulgaricus (B) b¡ŠQ>mo~ogrbg ~ëJo[aH$g
(C) Penicillium roqueforti (C) noZrgr{b`_ am°H$\$moQ>u
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

116. Which of the following is found in 116. h[aVbdH$ Ho$ ñQ´>mo_m _| {ZåZ _| go Š`m nm`m
stroma of chloroplast ? OmVm h¡ ?
(A) D.N.A. (B) Enzymes (A) S>r.EZ.E. (B) {dH$a
(C) Ribosomes (D) All the above (C) amB~mogmoåg (D) CnamoŠV g^r

117. Which of the following is not found in an 117. _mBQ>moH$m±{S>´>`m Ho$ Am§V[aH$ H$moîR> _| {ZåZ _| go Š`m
inner compartment of Mitochondria ? Zht nm`m OmVm h¡ ?
(A) Enzymes of kreb cycle (A) Ho«$~ MH«$ Ho$ {dH$a
(B) Enzymes of respiratory chain (B) ídgZ ûm¥§Ibm Ho$ {dH$a
(C) D.N.A. (C) S>r.EZ.E.
(D) R.N.A. (D) Ama.EZ.E.

118. Which of the following is not an 118. {ZåZ _| go H$m¡Z Eë~w{_{Z`g ~rO H$m CXmhaU
example of Albuminous seed ? Zht h¡ ?
(A) Wheat (B) Maize (A) J|hÿ (B) _ŠH$m
(C) Pea (D) Sunflower (C) _Q>a (D) gyaO_wIr

119. Cork cambium and vascular cambium 119. H$mH©$ H¡$på~`_ Am¡a dmñŠ`wba H¡$på~`_ h¡
are
(A) Parts of secondary xylem and (A) {ÛVr`H$ OmBb_ Ed§ âbmo`_ H$m ^mJ
phloem
(B) Parts of pericycle (B) n[aaå^ H$m ^mJ
(C) Lateral meristem (C) nmíd© à{d^mOr
(D) Apical meristem (D) AJ«ñW à{d^mOr

-55- Set-A

120. Which of the following virus has 120. {ZåZ _| go {H$g {dfmUw H$m ~mh²` AmH$ma ~hþ\$bH$s`
polyhedral morphology ? hmoVm h¡ ?
(A) T.M.V. (B) Rabies (A) Q>r.E_.ìhr. (B) ao~rO
(C) Ebola (D) Polio (C) B~mobm (D) nmo{b`mo

121. Holozoic mode of nutrition is found in 121. àm{U g_^moOr nmofU H$s {d{Y {ZåZ _| go {H$g
which of the following kingdom ? OJV _| nmB© OmVr h¡ ?
(A) Protista (B) Plantae (A) àmo{Q>ñQ>m (B) ßbm§Q>r
(C) Animalia (D) Monera (C) Eo{Z_o{b`m (D) _m°Zoam
122. Organic cofactors that are tightly bound 122. H$m~©{ZH$ H$mo\¡$ŠQ>a Omo {H$ EÝO>mB_ Ho$ gmW _O~yVr
to the enzyme are called as go ~ÝYm ahVm h¡ Cgo H$hVo h¢
(A) Active enzymes (A) EpŠQ>>d EÝOmB_
(B) Coenzymes (B) H$moEÝOmB_
(C) Inactive enzymes (C) A{H«$` EÝOmB_
(D) Prosthetic group (D) àmoñWo{Q>H$ g_yh

123. Which plant hormone is helpful in H$m¡Z-gm nmXn hm_m}Z RNA Ed§ àmoQ>rZ ~ZmZo _|
123.
making RNA and Protein ? _XXH$mar hmoVm h¡ ?
(A) Gibberellins (B) Auxins (A) {O~ao{bÝg (B) Am°pŠOÝg
(C) Cytokinins (D) Ethylene (C) gm`Q>moH$mB{ZÝg (D) BWmBbrZ

124. Which of the following is a cause of nnrVo Ho$ ~§Mr Q>m°n amoJ H$m H$maH$ {ZåZ _| go
124.
Bunchy top disease of Papaya ? H$m¡Z h¡ ?
(A) Bacteria (B) Virus (A) OrdmUw (B) {dfmUw
(C) Fungi (D) Mycoplasma (C) H$dH$ (D) _mBH$moßbmÁ_m

125. Who amongst the following forms 125. {ZåZ _| go H$m¡Z hoQ>oamo{gñQ> H$m {Z_m©U H$aVm h¡ ?
heterocyst ?
(A) Bacteria (A) OrdmUw
(B) Virus (B) {dfmUw
(C) Cyanobacteria (C) gm`ZmoOrdmUw
(D) Mycoplasma (D) _mBH$moßbmÁ_m

-56- Set-A

126. Which of the following statement is false 126. ñVZr` ewH«$mUw H$s OrdZ j_Vm Ho$ {df` _|
in respect of viability of mammalian {ZåZ{b{IV _§o go H$m¡Z-gm EH$ H$WZ AgË` h¡ ?
sperm ?
(A) Sperm is viable only up to (A)ewH«$mUw Ho$db 24 K§Q>m VH$ OrdZ j_ ~Zm
24 hours. ahVm h¡&
(B) Survival of sperm depends on the (B) ewH«$mUw H$s CÎmaOr{dVm _mÜ`_ Ho$ pH na
pH of the medium and is more {Z^©a hmoVr h¡ Am¡a jmar` _mÜ`_ _| dh
active in alkaline medium. A{YH$ gH«$s` ~Zm ahVm h¡&
(C) Viability of sperm is determined (C) ewH«$mUw H$s OrdZ j_Vm CgH$s J{VerbVm
by its motility. Ûmam {ZYm©[aV hmoVr h¡&
(D) Sperm must be concentrated in (D) ewH«$mUwAm| H$m gm§ÐU EH$ Jm‹T> o {Zbå~ Ho$
thick suspension. ^rVa hmoZm Mm{hE&
127. Which of the following is viviparous ? 127. {ZåZ _| go H$m¡Z {d{dnoag h¡ ?
(A) Penguin (B) Ostrich (A) n|p½dZ (B) Am°pñQ´>M
(C) Albatross (D) None (C) Eë~mQ´>mg (D) H$moB© Zht
128. The shock absorber fluid of the developing 128. ^«yU H$mo ~mø YŠH$mo§ go gwajm àXmZ H$aZo dmbm Ðd
embryo is known as H$m¡Z-gm hmoVm h¡ ?
(A) Chorionic fluid (A) H$moarAmo{ZH$ Ðd
(B) Amniotic fluid (B) EåZrAmo{Q>H$ Ðd
(C) Allantoic fluid (C) EboZQ>moBH$ Ðd
(D) Coelomic fluid (D) XohJwhr` Ðd
129. Match the following. 129. {ZåZ H$mo gw_o{bV H$a| &
List – I List – II gyMr – I gyMr – II
a. XX-XO method of I. Heterogametic
a. XX-XO, qbJ I  {df_`w½_H$
sex determination
{ZYm©aU H$s {d{Y
b. 1.5X/A ratio II. Turner’s
syndrome b. 1.5X/A AZwnmV II. Q>Z©a {gÊS´>mo_
c. Karyotype 45 III. Hemiptera c. H¡$[a`moQ>mBn 45 III. ho{_ßQ>oam
d. ZW-ZZ method of IV. Metafemale d. qbJ {ZYm©aU H$s IV. _oQ>m{\$_ob
sex determination ZW-ZZ {d{Y
(A) a-I, b-IV, c-III, d-II (A) a-I, b-IV, c-III, d-II
(B) a-III, b-IV, c-II, d-I (B) a-III, b-IV, c-II, d-I
(C) a-IV, b-I, c-II, d-III (C) a-IV, b-I, c-II, d-III
(D) a-I, b-IV, c-II, d-III (D) a-I, b-IV, c-II, d-III

-57- Set-A

130. Match the item in List – I and List – II 130. gyMr-I VWm gyMr-II H$mo gw_o{bV H$a ghr {dH$ën
and choose the correct alternative. H$m M`Z H$s{OE&
List – I List – II gyMr – I gyMr – II
a. Sickle cell 1. 7th Chromosome a. {gH$b gob EZr{_`m 1. 7th JwUgyÌ
Anaemia
b. Phenylketonuria 2. 4th Chromosome b. {\$ZmBb{H$Q>moÝ`y[a`m 2. 4th JwUgyÌ
c. Cystic fibrosis 3. 11th Chromosome c. {gñQ>rH$ \$mB~«mo{gg 3. 11th JwUgyÌ
d. Huntington’s 4. X-Chromosome d. hpÊQ>JQ>Z amoJ 4. X-JwUgyÌ
disease e. dUm©ÝYVm 5. 12th JwUgyÌ
e. Colour blindness 5. 12th Chromosome
(A) a-1, b-3, c-4, d-2, e-5
(A) a-1, b-3, c-4, d-2, e-5
(B) a-2, b-3, c-4, d-5, e-1
(B) a-2, b-3, c-4, d-5, e-1
(C) a-2, b-1, c-3, d-5, e-4
(C) a-2, b-1, c-3, d-5, e-4
(D) a-3, b-5, c-1, d-2, e-4
(D) a-3, b-5, c-1, d-2, e-4
131. This one is a viral disease in silkworm 131. {gëH$d_© H$s dm`ab {S>grO h¡
(A) Flacherie (A) âboMoar
(B) Pebrine disease (B) no~arZ {S>grO
(C) Muscardine (C) _ñH$m{S>©Z
(D) Maggot disease (D) _oJQ> {S>grO
132. Which one is vector for hookworm 132. hþH$d_© {S>grO H$m dmhH$ h¡
disease ?
(A) Loa loa (B) Bugs (A)bmoAm bmoAm (B) ~½g
(C) Rickettsia (D) None (C)[aHo$Q²> {g`m (D) H$moB© Zht
133. What happened when two different 133. O~ Xmo {^ÝZ aŠV g_yh H$mo {_bm`m OmVm h¡, Vmo
blood groups mixed together ? hmoVm h¡
(A) Coagulation (A) H$moJyboeZ
(B) Agglutination (B) E½by{Q>ZoeZ
(C) Thrombus formation (C) W«moå~g \$°ma_oeZ
(D) Ebolism (D) B~m°{bÁ_
134. A mutation that changes a codon 134. Eogm å`yQ>oeZ Omo H$moS>mZ EH$ A{_Zmo Aåb H$mo
specifying one amino acid to a ñnogr\$m` H$aVm h¡ H$mo Q>a{_ZoeZ H$moS>mZ _| ~XbVm h¡
termination codon is called a
(A) Missense mutation (A) {_goÝg å`yQ>oeZ
(B) Transition mutation (B) Q´>m§{ggZ å`yQ>oeZ
(C) Nonsense mutation (C) Zm°ZgÝg å`yQ>oeZ
(D) Frameshift mutation (D) \«o$_{eâQ>> å`yQ>oeZ

-58- Set-A

135.
Match the following. gw_o{bV H$a§o &
135.
a. Outermost layer 1. Anal gland a. ~mhar AmdaU 1. EZb JǤWr
b. Gives scent 2. Melanocyte b. Iwe~y XoVm h¡ 2. _rboZmoogmBQ>
c. Protect from 3. Sebaceous c. UV {d{H$aU 3. gr~o{g`g JǤWr
UV rays gland go ~MmVm h¡
d. Produces sebum 4. Stratum d. gr~_ CËnÝZ 4. ñQ´>o Q>_ H$maZr`_
corneum H$aVm h¡
(A) a-1, b-2, c-3, d-4 (A) a-1, b-2, c-3, d-4
(B) a-4, b-1, c-2, d-3 (B) a-4, b-1, c-2, d-3
(C) a-4, b-1, c-3, d-4 (C) a-4, b-1, c-3, d-4
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

136.
Kupffer cells of liver are concerned with 136. `H¥$V H$m Hw$\$a H$mo{eH$m gå~§{YV h¡
(A) Secretion of heparin and (A) {hno[aZ VWm {hñQ>m_mBZ Ho$ òmdU go
histamine
(B) Deposition of fat (B) dgm Ho$ g§M`Z go
(C) Conversion of glucose into
(C) ½byH$moO Ho$ ½bmBH$moOZ _| n[adV©Z go
glycogen
(D) Ingesting RBC which have (D) {H«$`m{dhrZ bmbaŠV H$U Ho$ ^jU go
stopped function
137. Which part of brain is most affected by 137. _pñVîH$ H$m H$m¡Z-gm ^mJ EëH$mohmb go g~go
alcohol ? Á`mXm à^m{dV hmoVm h¡ ?
(A) Cerebrum (A) goar~«_
(B) Cerebellum (B) goar~ob_
(C) Medulla oblongata (C) _oS²>`ybm Amãbm§JoQ>m
(D) Pons Varolii (D) nmÝg doamobr
138. Periyar wildlife sanctuary is located in 138. no[a`ma dmBëSbmB\$ g|ŠMwdar pñWV h¡
(A) Kerala (A) Ho$ab
(B) Karnataka (B) H$Zm©Q>H$
(C) Tamil Nadu (C) V{_bZmSy>
(D) Andra Pradesh (D) Am§Y« àXoe
139. First vertebrate appeared in 139. àW_ H$eoéH$s Anr`a hþAm
(A) Permian (B) Silurian (A) na{_`Z (B) {gë`y[a`Z
(C) Ordovician (D) Cambrian (C) AmS>m}{d{e`Z (D) H¡$på~«`Z

-59- Set-A

140. Which of the following is an oncogenic 140. AmÝH$moOo{ZH$ dm`ag h¡
virus ?
(A) Herpes simplex II
(A) hanog {gåßboŠg II
(B) Papilloma
(B) nonrbmo_m
(C) Epstein-Barr (C) EnñQ>rZ-~ma
(D) All of these (D) CnamoŠV g^r
141.
AIDS is due to 141. AIDS {H$g H$maU hmoVm h¡ ?
(A) Reduction in number of helper (A) hoëna T H$mo{eH$m Ho$ g§»`m _| H$_r
T-cells
(B) Lack of interferon (B)B§Q>a\o$amZ H$s H$_r
(C) Reduction in number of killer (C) {H$ëba T H$mo{eH$m H$s g§»`m _| H$_r
T-cells
(D) Autoimmunity (D) AmQ>moBå`y{ZQ>r

142.
Match list I and II and choose correct 142. gyMr> I VWm II H$mo gw_o{bV H$ao§ VWm ghr CÎma
answer. MwZo§&
List – I List – II gyMr – I gyMr – II
a. Hypothalamus 1. Sperm lysine a. hmBnmoWob_g 1. ñn_© bmBgrZ
b. Acrosome 2. Estrogen b. EH«$mogmo_ 2. BñQ´>moOZ
c. Graafian follicle 3. Relaxin c. J«o{\$`Z \$m°{bH$b 3. [aboŠgrZ
d. Leydig cell 4. Gn RH d. bmB{S>J H$mo{eH$m 4. Gn RH
e. Parturition 5. Testosterone e. {eewOÝ_ 5. Q>oñQ>moñQ>oamoZ
(A) a-4, b-1, c-2, d-3, e-5 (A) a-4, b-1, c-2, d-3, e-5
(B) a-5, b-3, c-2, d-1, e-4 (B) a-5, b-3, c-2, d-1, e-4
(C) a-4, b-3, c-1, d-2, e-5
(C) a-4, b-3, c-1, d-2, e-5
(D) a-4, b-1, c-2, d-5, e-3
(D) a-4, b-1, c-2, d-5, e-3
143. Hyperglycemia is induced by all the 143. EH$ AndmX H$mo N>mo‹S>H$a {ZåZ{b{IV hma_moZ Ûmam
following hormones except hmBna½bmBgo{_`m (aŠV _| CÀM eH©$am) CËào[aV
hmoVm h¡
(A) epinephrine (B) thyroxin
(A) E{nZo{\«$Z (B) Wm`am°ŠgrZ
(C) glucagon (D) aldosterone (C) ½byH$mJm°Z (D) EëS>moñQ>oamoZ

-60- Set-A

Statements :
144. 144. H$WZ :
1. Iodine is very important for 1. Wm`am°ŠgrZ CËnmXZ Ho$ {bE Am`mo{S>Z A{V
production of thyroxin. _hËdnyU© h¡&
2. Vitamin B6 is also known as niacin 2. {dQm>{_Z B6, {Z`mgrZ AWdm {ZH$moQ>rZrH$
or nicotinic acid.
Aåb Ho$ ê$n _o§ OmZm OmVm h¡&
3. Fructose is hexose and
monosaccharide. 3. \«$ŠQ>moO hoŠgmoO VWm _moZmog¡H$amBS> h¡&
4. Globulin is a conjugated protein. 4. ½bmoã`w{bZ EH$ H§$OwJoQ>oS> àmoQ>rZ h¡&
Select correct statements. ghr H$WZ MwZo§&
(A) 1, 2 and 3 are correct, but 4 is (A) 1, 2 VWm 3 ghr h§¡ naÝVw 4 JbV h¡
wrong
(B) 1 and 3 are correct, but 4 and
2 are wrong (B) 1 VWm 3 ghr h¡§, naÝVw 4 Am¡a 2 JbV h¡§
(C) 1 and 2 are correct, but 3 and
4 are wrong (C) 1 Am¡a 2 ghr h¡§, naÝVw 3 VWm 4 JbV h§¡
(D) 1, 3 and 4 are correct, but 2 is
wrong (D) 1, 3 VWm 4 ghr h¡§, naÝVw 2 JbV h¡

145.
The theory of ageing holds that ageing 145. d`Vm Ho$ {gÕmÝV Ho$ AZwgma d`Vm hmoZo H$m H$maU h¡
is due to
(A) Random mutation in DNA of (A) X¡{hH$ H$mo{eH$m Ho$ DNA _| AmH$pñ_H$
somatic cell CËn[adV©Z
(B) Increased cross linkage of collagen (B) H$moboOZ VWm AÝ` àmo{Q>Z Ho$ A{YH$ H«$mg
and other proteins qbHo$O go
(C) Cumulative result of damage to (C) ñdV§Ì _ybH$ H$s {H«$`mAm| Ûmam D$ÎmH$ H$m
tissues by free radicals g§nyU© {dZme
(D) All of these (D) CnamoŠV g^r

146. Which of the following banding group is 146. {ZåZ _§o go {H$g ~|{S>§J g_yh (banding group)
used in staining both plant and animal H$m Cn`moJ XmoZm| nmXn VWm OÝVw JwUgyÌ Ho$
chromosome ? A{^a§OZ (staining) _o§ hmoVm h¡ ?
(A) C-group (B) G-group (A) C-g_yh (B) G-g_yh

(C) M-group (D) Q-group (C) M-g_yh (D) Q-g_yh

-61- Set-A

147. Which of the following is limited by 147. {ZåZ _| go Š`m nmnyboeZ H$s bm°{OpñQ>H$b J«moW H$s
the carrying capacity for a population H¡$[aq`J H¡$no{gQ>r go {Z`§{ÌV hmoVm h¡ ?
growing logistically ?
(A) Environmental resistance (A) EÝdm`aÝ_o§Q>b a{gñQ>oÝg
(B) Biotic potential (B) ~m`mo{Q>H$ nmoQ>opÝe`b
(C) Natality (C) ZoQ>o{bQ>r
(D) All of these (D) CnamoŠV g^r
148. Aldrin causes 148. EpëS´>Z H$aVm h¡
(A) Air pollution (A) dm`w àXÿfU
(B) Soil pollution (B) _¥Xm àXÿfU
(C) Sound pollution (C) Üd{Z àXÿfU
(D) None of these (D) CnamoŠV H$moB© Zht

149. Phylogenetic system of classification 149. dJuH$aU H$m \$m`bmoOoZo{Q>H$ {gñQ>_ {H$`m J`m
was given by
(A) Hutchinson (A) hQ>{MÝgZ
(B) Linnaeus (B) br{Z`g
(C) Bentham and Hooker (C) ~|W_ d hÿH$a
(D) Engler and Prantl (D) E§Jba d àmÝQ>b

150. The phosphogen that helps the 150. H$eoéH$s _| _m±gnoer g§Hw$MZ Ho$ Xm¡amZ ADP go
regeneration of ATP from ADP during ATP H$m nwZ{Z©_m©U H$aZo dmbm \$moñ\$moOZ hmoVm h¡
muscle contraction in vertebrate is
(A) Creatine phosphate (A) {H«$`oQ>rZ \$°mñ\o$Q>
(B) Arginine phosphate (B) AmOuZrZ \$°mñ\o$Q>
(C) ADP (C) ES>rnr
(D) Inositol phosphate (D) BZmogrQ>mob \$°mñ\o$Q>

-62- Set-A

-63- Set-A

SET – A

CÎma A§{H$V H$aZo H$m g_` : 3 K§Q>o A{YH$V_ A§H$ : 150
Time for making answers : 3 Hours Maximum Marks : 150

ZmoQ> :
1. Bg àíZ nwpñVH$m _| VrZ ^mJ - àW_ ^mJ ^m¡{VH$ emñÌ 50 àíZ, {ÛVr` ^mJ agm`Z emñÌ 50 àíZ,
V¥Vr` ^mJ - (A) J{UV 50 àíZ, (~) Ord {dkmZ 50 àíZ h¡ & ^mJ àW_ Ed§ {ÛVr` A{Zdm`© h¡ Am¡a ^mJ V¥Vr` (A)
VWm (~) _| go {H$gr EH$ ^mJ H$m hr M`Z H$a| & àË`oH$ àíZ 1 A§H$ H$m h¡ & Hw$b 150 àíZ H$aZo A{Zdm`© h¢ &
2. àíZm| Ho$ CÎma Xr JB© OMR CÎma-erQ> (Am§ga erQ>) 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$m à`moJ d{O©V h¡ Ÿ&
5. OMR CÎma-erQ> (Am§ga erQ>) H$m à`moJ H$aVo g_` Eogr H$moB© AgmdYmZr Z ~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 has 50 questions,
Second Part Chemistry has 50 questions, Third Part – (A) Mathematics has
50 questions, (B) Biology has 50 questions. Part First and Second are compulsory.
Candidates should attempt Any one Part from Part Third (A) and (B). Each question
carries 1 mark. All 150 questions are compulsory.
2. Indicate your answers on the OMR Answer-Sheet provided.
3. No negative marking will be done.
4. Use of any type of calculator or log table and mobile phone is prohibited.
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.

-64- Set-A