CG PPT 2021 Question Paper

Question Booklet No.
SET – A


A Subject Code : 32102/UE – PT/ENT – M

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Number of Pages in this Question Booklet : 48 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 :- J{UV - à.g§. 101 – 150 (iii) Third Part : – Mathematics – Q. No. 101 – 150
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& 12. In case of any ambiguity in English version the Hindi version shall
be considered authentic.

-1-

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DO

-2- Set-A

PART – I
Physics ^m¡{VH$ emñÌ

1. A beaker is completely filled with water 1. EH$ ~rH$a 4°C nmZr go nyUV© : ^am hþAm h¡& Bg ~rH$a
at 4°C. It will overflow if go nmZr ~mha ~hZo bJoJm (overflow) `{X,
(A) Heated above 4°C (A) Bgo 4°C go A{YH$ Vmn VH$ J_© {H$`m OmVm h¡
(B) Cooled below 4°C (B) Bgo 4°C go H$_ Vmn VH$ R>ÊS>m {H$`m OmVm h¡
(C) Both heated and cooled above (C) `{X Bgo H«$_e: 4°C go A{YH$ Am¡a H$_ Vmn
and below 4°C respectively na J_© Am¡a R>ÊS>m {H$`m OmVm h¡
(D) None of these (D) CnamoŠV _| go H$moB© Zht

2. The acceleration of a particle performing 2. gab AmdÎm© J{V H$aVo hþE {H$gr H$U H$m _mÜ` pñW{V
simple harmonic motion is 12 cm/sec2 go 3 go_r H$s Xÿar na ËdaU H$m _mZ 12 go_r/go2 h¡&
at a distance of 3 cm from the mean
Bg gab AmdÎm© J{V H$m AmdV© H$mb H$m _mZ h¡
position. Its time period is
(A) 0.5 sec (A) 0.5 goH$ÊS>

(B) 1.0 sec (B) 1.0 goH$ÊS>

(C) 2.0 sec (C) 2.0 goH$ÊS>

(D) 3.14 sec (D) 3.14 goH$ÊS>

3. A ray of light is incident on a plane 3. EH$ g_Vb Xn©U _| àH$me {H$aU A{^bå~dV
mirror normally. The angle of reflection Amn{VV hmo ahr h¡& namdV©Z H$moU hmoJm
will be
(A) 0° (A) 0°

(B) 90° (B) 90°

(C) Will not be reflected (C) namd{V©V Zht hmoJr
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

-3- Set-A

4. A concave mirror of focal length 4. EH$ AdVb Xn©U {OgH$s \$moH$g Xÿar f
f(in air) is immersed in water (hdm _|) h¡, H$mo nmZr _| Sw>~m`m OmVm h¡ &
( )
its refractive index is µ = 4 the
3 ( )
{OgH$m AndV©Zm§H$ µ = 4 3 h¡ & nmZr _|
focal length of the concave mirror in AdVb Xn©U H$s \$moH$g Xÿar hmoJr
water will be
(A) f (B) 1.33 f (A) f (B) 1.33 f
(C) 0.75 f (D) 2f (C) 0.75 f (D) 2f

5. A substance of mass m1 with specific 5. m1 Ðì`_mZ VWm s1 {d{eîR> Cî_m Ho$ EH$ nXmW©
heat capacity s1 and initial temperature {OgH$m àmapå^H$ Vmn θ1 h¡, H$mo EH$ AÝ` m2
θ1 is mixed with another substance of
mass m2, specific heat s2 and initial Ðì`_mZ, s2 {d{eîQ>> Cî_m Ho$ nXmW© {OgH$m àmapå^H$
temperature θ2 respectively. Then, their Vmn θ2 h¡, _o§ {_bm`m OmVm h¡& V~ gmå`mdñWm Vmn
equilibrium temperature is (take θ1 > θ2) H$m _mZ hmoJm (θ1 > θ2 {b`m Om`o)
m1s1θ1 − m2 s 2 θ2 m1s1θ1 − m2 s 2 θ2
(A) (A)
m1s1 − m2 s 2 m1s1 − m2 s 2
m1s1θ1 − m2 s 2 θ2 m1s1θ1 − m2 s 2 θ2
(B) (B)
m1s1 + m2 s 2 m1s1 + m2 s 2
m2 s 2 θ2 − m1s1θ1 m2 s 2 θ2 − m1s1θ1
(C) (C)
(m1 + m2 ) (m1 + m2 )
m1s1θ1 + m2 s2 θ2 m1s1θ1 + m2 s2 θ2
(D) (D)
m1s1 + m2 s2 m1s1 + m2 s2
6. When a bar of iron 50.0 cm long at 6. 15°C Vmn H$s EH$ bmoho H$s N>S‹ > H$mo O~ EH$ ^Q²>Q>r
15°C is heated in a furnace, it becomes _§o J_© {H$`m OmVm h¡ V~ CgH$s bå~mB© 50 go_r. go
50.1 cm. If the coefficient of linear ~‹T> H$a 50.1 go_r. hmo OmVr h¡& `{X bmoho Ho$ nXmW©
expansion of iron is 0.000011/°C, then H$m aoIr` àgma JwUm§H$ 0.000011/°C h¡, V~ ^Q²>Q>r
the temperature of the furnace is (furnace) H$m Vmn kmV H$s{OE&
(A) 192°C (A) 192°C
(B) 182°C (B) 182°C
(C) 197°C (C) 197°C
(D) Cannot be determined (D) kmV Zht {H$`m Om gH$Vm

-4- Set-A

7. Heat required to convert 1 gm ice at 7. 1 J«m_ ~\©$ Omo 0°C na h¡ Bgo 100°C H$s ^mn _§o
0°C into steam at 100°C is ~XbZo Ho$ {bE Amdí`H$ D$î_m H$m _mZ h¡
(A) 100 Calorie (A) 100 H¡$bmoar
(B) 0.01 Kilo Calorie (B) 0.01 {H$bmo H¡$bmoar
(C) 716 Calorie (C) 716 H¡$bmoar
(D) 580 Calorie (D) 580 H¡$bmoar

8. A convex mirror has a focal length f. 8. EH$ CÎmb Xn©U H$s \$moH$g Xÿar f h¡& EH$ dmñV{dH$
A real object is placed at a distance d dñVw Xn©U Ho$$ Y«wd go d Xÿar na aIr JB© h¡& BgH$m
infront of the pole produces an image
at
à{Vq~~ {ZåZ Xÿar na ~ZoJm
(A) infinity (B) f (A) AZ§V (B) f

(C) f (D) 2f (C) f (D) 2f
2 2

9. For an isotropic medium B, µ, H and M 9. EH$ g_X¡{eH$ _mÜ`_ Ho$ {b`o B, µ, H VWm M
are related as (where B, µ0, H and M {H$g g_rH$aU Ûmam gå~§{YV h¡ ? (Ohm± {X`o J`o
have their usual meaning in the context
àVrH$m| Ho$ gm_mÝ` AW© h¢)
of magnetic material)
(A) µ0H = B – M (A) µ0H = B – M
(B) M = µ0(H + B) (B) M = µ0(H + B)
(C) H = µ0(M + B) (C) H = µ0(M + B)

(D) B = µ0(H + M) (D) B = µ0(H + M)

10. The mathematical equation for 10. Mwå~H$s` ~b aoImAm| Ho$ {b`o J{UVr` g_rH$aU h¡
magnetic lines of forces is
   
(A) ∇. B = 0 (A) ∇. B = 0
   
(B) ∇. B ≠ 0 (B) ∇. B ≠ 0
   
(C) ∇. B > 0 (C) ∇. B > 0
   
(D) ∇. B < 0 (D) ∇. B < 0

-5- Set-A

11. Solar storm can cause on the earth 11. gm¡a Vy\$mZ n¥Ïdr na {ZåZ CËnÞ H$aVm h¡
(A) Aurora borealis (A) CÎma Y«wdr` Á`mo{V
(B) Interference in satellite (B) CnJ«h g§Mma _| ì`{VH$aU
communication
(C) Alteration in plane routes (C) g_Vbr` _mJm] _| n[adV©Z
(D) All of the above (D) BZ_| go g^r
12. The angle of dip at a certain place is 12. EH$ ñWmZ na {S>n H$moU 30° h¡ & `{X n¥Ïdr Ho$
30°. If the horizontal component of the Mw§~H$s` joÌ H$m j¡{VO KQ>H$ H h¡, Hw$b Mw§~H$s`
earth’s magnetic field is H, the intensity
of the total magnetic field is
joÌ H$m Vrd«Vm h¡
2H 2H
(A) H 2 (B) (A) H 2 (B)
3 3

(C) H 2 (D) H 3 (C) H 2 (D) H 3


13. The electrostatic potential energy 13. 1 A Xÿar na aIo àmoQ>mZ VWm BboŠQ´>mZ H$s pñWa
between proton and electron separated {dÚwV D$Om© hmoJr
by 1Å is
(A) – 13.6 ev (B) – 27.2 ev (A) – 13.6 ev (B) – 27.2 ev
(C) –14.4 ev (D) –1.44 ev (C) –14.4 ev (D) –1.44 ev

14. A wire 100 cm long and 2mm diameter 14. EH$ Vma {OgH$s bå~mB© 100 cm VWm ì`mg
has a resistance of 1.4 ohm, the 2 mm h¡, H$m à{VamoY 1.4 Amo_ h¡& BgHo$ nXmW©
electrical resistivity of the material is
H$s {dÚwVr` à{VamoYH$Vm h¡
(A) 2.2 × 10–6 ohm.m (A) 2.2 × 10–6 Amo_._r

(B) 4.4 × 10–6 ohm.m (B) 4.4 × 10–6 Amo_._r

(C) 1.1 × 10–6 ohm.m (C) 1.1 × 10–6 Amo_._r
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

-6- Set-A

15. In the network of resistors shown in 15. {MÌ _| {X`o J`o à{VamoYm| Ho$ OmbH$ _| {~ÝXþ A VWm
the adjoining figure, the equivalent
B Ho$ _Ü` Vwë` à{VamoY hmoJm
resistance between A and B is

2Ω 2Ω 2Ω 2Ω 2Ω 2Ω 2Ω 2Ω
2Ω 2Ω 2Ω 2Ω
A 2Ω 2Ω 2Ω B A 2Ω 2Ω 2Ω B
2Ω 2Ω 2Ω 2Ω 2Ω 2Ω

(A) 24Ω (B) 12Ω (A) 24Ω (B) 12Ω
(C) 6Ω (D) 3Ω (C) 6Ω (D) 3Ω

16. In a vernier calipers, one main scale 16. EH$ d{Z©`a H¡${bng© Ho$ _w»` n¡_mZo Ho$ EH$ ImZo
division is x cm and n division of the
vernier scale coincide with (n – 1) H$m _mZ x go_r h¡ Am¡a d{Z©`a ñHo$b Ho$ n ImZo
divisions of the main scale. The least _w»` ñHo$b Ho$ (n – 1) ImZm| Ho$ ~am~a h¡& Bg
count of the calipers (in cm) is d{Z©`a H¡${bng© H$m AënV_m±H$ (go_r _|) h¡
(n − 1) nx (n − 1) nx
(A) x (B) (A) x (B)
n (n − 1) n (n − 1)
x x x x
(C) (D) (C) (D)
n (x − 1) n (x − 1)

17. If the density of a substance is ‘P’ g cm–3. 17. `{X {H$gr nXmW© H$m KZËd ‘P’ J«m_ go_r–3 h¡& Bg
Then its density in SI unit is nXmW© Ho$ KZËd H$m _mZ Eg AmB© (SI) BH$mB© _o§ h¡
(A) P kg m–3 (A) P kg m–3
(B) 100 P kg m–3 (B) 100 P kg m–3
(C) 1000 P kg m–3 (C) 1000 P kg m–3
(D) 10 P kg m–3 (D) 10 P kg m–3

18. A car travels a distance d on a road 18. EH$ H$ma EH$ grYr g‹S>H$ na d Xÿar H$mo Xmo KÊQ>o _o§
in two hours and then returns to the V` H$aVr h¡ VWm AJbo VrZ KÊQ>m| _§o H$ma nwZ:
starting point in next three hours. Its àmapå^H$ {~ÝXþ na bm¡Q> AmVr h¡ & BgH$s Am¡gV
average speed is Mmb h¡
d d d d
(A) 2 (B) (A) 2 (B)
5 5
2d 3d 2d 3d
(C) (D) (C) (D)
5 5 5 5

-7- Set-A

19. From a building two balls A and B are 19. EH$ B_maV go Xmo JoÝXo A Am¡a B Bg àH$ma \|$H$s OmVr
thrown such that A is thrown upwards h¡ {H$ JoÝX A H$mo CÜdm©Ya> D$na VWm JoÝX B H$mo
and B downwards with the same speed
(both vertically). If VA and VB are their CÜdm©Ya ZrMo H$s Amoa g_mZ Mmb go \|$H$m J`m h¡&
respective velocities on reaching the `{X VA Am¡a VB BZ JoÝXm| H$s n¥Ïdr Vb na nhþM ± Zo
ground then Ho$ doJ h¡ V~
(A) VB > VA (B) VA = VB (A) VB > VA (B) VA = VB
(C) VA > VB (D) VA = vB (C) VA > VB (D) VA = vB
2 2

20. The wavelength of light in two liquids 20. EH$ àH$me H$m Va§JX¡¿`© {H$gr Ðd ‘A’ _| 3500 A
  
‘A’ and ‘B’ is 3500 A and 7000 A , then VWm Ðd ‘B’ _| 7000 A h¡& ‘A’ H$m ‘B’ Ho$ gmnoj
the critical angle of ‘A’ relative to ‘B’ will
H«$mpÝVH$ H$moU hmoJm
be
(A) 60° (B) 45° (A) 60° (B) 45°
(C) 30° (D) 15° (C) 30° (D) 15°

21. A concave lens of focal length 25cm is 21. \$moH$g b§~mB© 25 go_r H$m EH$ AdVb b|g
in contact with a convex lens of focal 40 go_r Ho$ EH$ CÎmb b|g Ho$ g§nH©$ _| h¡ & b|g
length 40cm. The power of the lens will H$s j_Vm hmoJr
be
(A) – 1.5D (B) + 6.5D (A) – 1.5D (B) + 6.5D
(C) – 6.5D (D) + 4.5D (C) – 6.5D (D) + 4.5D

22. The far point of a myopic eye is 50cm. 22. {ZH$Q> ÑpîQ> Xmof go nr{‹S>V ZoÌ H$m gwXay {~ÝXþ 50 go_r
For removing this defect the power of h¡& Xmof {ZdmaU hoVw Amdí`H$ boÝg H$s j_Vm
lens required in dioptre will be
S>m`moßQ>>a _| hmoJr
(A) 50D (B) – 4D (A) 50D (B) – 4D
(C) – 2D (D) 0.25D (C) – 2D (D) 0.25D

23. If magnetic flux is expressed in 23. `{X Mwå~H$s` âbŠg H$mo do~a _| ì`ŠV {H$`m Om`
weber, the magnetic induction can Vmo Mwå~H$s` àoaU H$mo ì`ŠV {H$`m Om gH$Vm h¡
be expressed in
(A) Weber/m2 (B) Weber/m (A) do~a/ _r2 (B) do~a/_r
(C) Weber-m (D) Weber-m2 (C) do~a-_r (D) do~a-_r2

-8- Set-A

24. A liquid of specific heat 0.5 at 60°C is 24. 0.5 {d{eîR> D$î_m Ho$ EH$ Ðd {OgH$m Vmn 60°C
mixed with another liquid of specific h¡& Bg Ðd H$mo EH$ AÝ` Ðd {OgH$m Vmn 20°C
heat 0.3 at 20°C. After mixing, the Am¡a {d{eîR> D$î_m 0.3 h¡, _o§ {_bm`m OmVm h¡&
temperature of the mixture becomes Bg à{H«$`m _o§ {_lU H$m Vmn 30°C hmo OmVm h¡&
30°C. The liquids are mixed in XmoZmo§ Ðdm| H$m {_lU ~ZmZo _§o à`wŠV Ðì`_mZm| H$m
proportion by masses AZwnmV h¡
(A) 5 : 1 (B) 1 : 5 (A) 5 : 1 (B) 1 : 5
(C) 2 : 5 (D) 5 : 3 (C) 2 : 5 (D) 5 : 3

25. The aperture of objective lens of a 25. XÿaXeu Ho$ A{^Ñî`H$ H$m ÛmaH$ ~‹S>m ~Zm`m OmVm h¡
telescope is made large so as to
(A) Increase the magnifying power of (A) XÿaXeu H$s AmdY©Z j_Vm ~‹T>mZo Ho$ {b`o
telescope
(B) Increase the resolving power of (B) XÿaXeu H$s {d^oXZ j_Vm ~‹T> mZo Ho$ {b`o
telescope
(C) Reduce the aberration of image (C) à{V{~å~ H$m dU© {dnWZ H$_ H$aZo Ho$
{b`o
(D) Focus on distant objects (D) Xÿa pñWV dñVwAm§o H$mo \$moH$g H$aZo Ho$ {b`o

26. The net force acting on a drop of rain of 26. ~mare (Rain) H$s EH$ 5 {_brJ«m_ Ðì`_mZ H$s
mass 5 mg falling down with constant EH$ ~y§X {Z`V Mmb go {Ja ahr h¡ & Bg ~y§X na bJZo
speed is dmbo ZoQ> ~b H$m _mZ h¡
(A) 5 N (B) 5 mN (A) 5 Ý`yQ>Z (B) 5 {_br Ý`yQ>Z
(C) Zero (D) 5 dyne (C) eyÝ` (D) 5 S>mBZ

27.
The instantaneous value of current in 27. EH$ àË`mdVu Ymam n[anW _| Ymam H$m VmËj{UH$ _mZ
 π  π
Ac circuit is I = 2sin  100 πt +  . I = 2sin  100 πt +  h¡, Vmo àW_ ~ma Ymam
 3  3
The current will be maximum for the
H$m A{YH$V_ _mZ {ZåZ g_` na hmoJm
first time at
1 1 1 1
(A) t = s (B) t = s (A) t = s (B) t = s
100 200 100 200
1 1 1 1
(C) t = s (D) t = s (C) t = s (D) t = s
400 600 400 600

-9- Set-A

28. A worker lives at a distance of 1.32 km 28. EH$ l{_H$ H$m Ka CgHo$ H$m`© ñWb (\o$ŠQ>ar) go
from the factory. If the speed of sound 1.32 {H$_r H$s Xÿar na h¡& `{X dm`w _| Üd{Z H$s Mmb
in air be 330 m/s, how many seconds
330 _rQ>a/goH$ÊS> h¡ V~ \o$ŠQ>ar _§o ~Oo gmB©aZ H$s
will the sound of factory siren take to
reach the worker ? Üd{Z Cg l{_H$ H$mo {H$VZo goH$ÊS>m| Ho$ CnamÝV gwZmB©
XoJr ?
(A) 5 (B) 6 (A) 5 (B) 6
(C) 4 (D) 3 (C) 4 (D) 3

29. Heat is supplied to a certain homogeneous 29. EH$ g_m§Jr goånb nXmW© _| D$î_m H$s Amny{V© {Z`V Xa
sample of matter at a uniform rate. Its go hmo ahr h¡ & `{X Bg nXmW© Ho$ Vmn H$m g_` Ho$ gmW
temperature is plotted against time as
shown in figure. Which of the following
{dMaU {MÌ _o§ Xem©`m J`m h¡& Bg {MÌ Ho$ AmYma> na
conclusion can be drawn ? {ZåZ{b{IV H$m¡Z gm n[aUm_ {ZH$mbm Om gH$Vm h¡ ?

Temp. Vmn

Time g_`

(A) Its specific heat capacity is greater (A)nXmW© H$s {d{eîQ>> D$î_mYm[a>Vm R>mgo AdñWm _o§
in solid state than in the liquid state Ðd AdñWm H$s VwbZm _§o A{YH$ h¡
(B) Its specific heat capacity is greater in (B) nXmW© H$s {d{eîQ>> D$î_mYm[aVm Ðd AdñWm _|
the liquid state than in the solid state R>mgo AdñWm H$s Anojm A{YH$ h¡
(C) Its latent heat of vapourization is (C) nXmW© H$s dmînZ H$s JwßV D$î_m BgH$s JbZ
smaller than its latent heat of fusion H$s JwßV D$î_m go H$_ h¡
(D) None of these (D) CnamoŠV _o§ go H$moB© Zht
30. A particle is moving in a straight line. 30. EH$ H$U EH$ F¥$Ow aoIr` J{V H$a ahm h¡ & Bg H$U
It has negative velocity and negative
acceleration. The particle is
doJ VWm ËdaU F¥$UmË_H$ h¡ & `h H$U
(A) at rest (A) {dam_mdñWm _| h¡
(B) speeding up (B) H$s Mmb ~‹T> ahr h¡
(C) speeding down (C) H$s Mmb KQ> ahr h¡
(D) none of these (D) CnamoŠV _o§ go H$moB© Zht

-10- Set-A

31. A current of 6A enters one corner P 31. VrZ Vmamo§ Ho$ g_~mhw {Ì^wO PQR {OZ_| àË`oH$
of an equilateral triangle PQR having ^wOm 2Ω H$m h¡, BgHo$ P {~ÝXþ go 6A H$s Ymam
three wires of resistances 2Ω each
and leaves by the corner R. Then the
àdoe H$aVr h¡, Omo {~ÝXþ R go {ZJ©V hmoVr h¡& V~
current I1 and I2 are respectively Ymam I1 VWm I2 Ho$ _mZ H«$_e: h¢
6A 6A
2Ω 2Ω
P P
I1 I2 I1 I2
2Ω 2Ω 2Ω 2Ω

Q 2Ω R Q 2Ω R

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

32. A cell of internal resistance r is 32. Am§V[aH$ à{VamoY r dmbr EH$ gob ~mø à{VamoY
connected to an external resistance R go OwS>r h¡ & R _| {dÚwV A{YH$V_ hmoJr, `{X
R. The current will be maximum in R,
if
(A) R = r (B) R < r (A) R = r (B) R < r

(C) R > r (D) R = r (C) R > r (D) R = r
2 2

33. The minimum wavelength of X-rays 33. V dmoëQ> {d^dmÝVa na Ëd[aV BboŠQ´>mZmo§ Ûmam
produced by electrons accelerated by a CËnm{XV X-{H$aUm§o H$m Ý`yZV_ V§aJX¡¿`© hmoJm
potential difference of V volts is equal to
ev ev
(A) (A)
hc hc
cV cV
(B) (B)
eh eh
hc hc
(C) (C)
ev ev
(D) None of these (D) Bg_| go H$moB© Zht

-11- Set-A

34. The acceleration due to gravity on the 34. MÝЧ_m na JwéËdr` ËdaU H$m _mZ n¥Ïdr na JwéËdr`
moon is one sixth that of the earth. If
the earth and moon are assumed to
ËdaU Ho$ _mZ H$m 1/6 h¡& `{X n¥Ïdr Am¡a MÝЧ_m H$m
have the same density, the ratio of the KZËd EH$g_mZ _mZm Om`o V~ MÝЧ_m VWm n¥Ïdr
radii of moon and earth will be H$s {ÌÁ`mAm| H$m AZwnmV hmoJm
1 1 1 1
(A) 6 (B) 1 (A) 6 (B) 1
( 6) 3
( 6) 3
1 1 1 1
(C) (D) 2 (C) (D) 2
36 ( 6) 3 36 ( 6) 3
35. The ratio of wavelength and the distance 35. {H$gr Va§J _|, Va§JX¡¿`© Am¡a Cg Va§J _| CËnÞ
between compression and an adjacent gånrS>Z VWm g§b½Z {ZH$Q>dVu {dabZ Ho$ ~rM H$s
rarefaction is equal to
Xÿar H$m AZwnmV hmoVm h¡
(A) 4 (B) 2 (A) 4 (B) 2
(C) 3 (D) 5 (C) 3 (D) 5

36. A 20 kg block is initially at rest. A 50 N 36. EH$ 20 {H$J«m Ðì`_mZ H$m EH$ ãbm°H$ àmamå^
force is required to set the block in _§o {dam_mdñWm _§o h¡& EH$ 50 Ý`yQ>Z ~b H$s
motion. The coefficient of static friction Amdí`H$Vm Bg ãbm°H$ H$mo J{V àXmZ H$aZo Ho$ {bE
is (g = 10 m/s2)
hmoVr h¡& V~ ñW¡{VH$ Kf©U Ho$ JwUm§H$ H$m _mZ h¡
(g = 10 _rQ>a/go2)
(A) 0.50 (A) 0.50
(B) 0.10 (B) 0.10
(C) 0.25 (C) 0.25
(D) None of these (D) CnamoŠV _| go H$moB© Zht
37. Two bulbs are working in parallel order. 37. Xmo ~ë~ g_mZmÝVa H«$_ _| H$m`© H$a ah| h¢& ~ë~ A
Bulb A is brighter than bulb B. If RA and RB H$s M_H$ ~ë~ B go Á`mXm h¡& `{X RA VWm RB
are their resistances respectively, then
H«$_e: ~ë~ A VWm ~ë~ B Ho$ à{VamoY hmo Vmo
(A) RA > RB (B) RA < RB (A) RA > RB (B) RA < RB
(C) RA = RB (D) None of these (C) RA = RB (D) BZ_| go H$moB© Zht

-12- Set-A

38. Lorentz force can be calculated by the 38. bma|O ~b H$s JUZm {ZåZ gyÌ Ûmam H$s OmVr h¡
formula
       
(A) F = q(E + V × B) (A) F = q(E + V × B)
       
(B) F = q(E − V × B) (B) F = q(E − V × B)
       
(C) F = q(E + V . B) (C) F = q(E + V . B)
       
(D) F = q(E − V . B) (D) F = q(E − V . B)

39. An emf of 15V applied to a circuit 39. 15 dmoëQ> H$m EH$ {dÚwV dmhH$ ~b 5 hoZ²ar àoaH$Ëd
containing 5 henry inductance and VWm 10Ω à{VamoY dmbo n[anW na bJm`m J`m
10Ω resistance. The ratio of the flow h¡& g_` t = ∞ VWm t = 1 goH$ÊS> na n[anW _|
of currents in circuit at time t = ∞ and àdm{hV hmoZo dmbr YmamAmo§ H$m AZwnmV hmoJm
t = 1 second is

e2 e2
1 1
e2 e2
(A) (B) (A) (B)
e2 − 1 e2 − 1
1
e 2 −1
1
e 2 −1

(C) 1– e–1 (D) e–1 (C) 1– e–1 (D) e–1

40. The time period of a particle in simple 40. gab AmdV© J{V H$aVo hþE {H$gr H$U H$m AmdV©H$mb
harmonic motion is equal to the time Cg H$U Ûmam EH$ {ZpíMV {~ÝXþ na H«$_mJV AmZo _§o
between consecutive appearances of bJo g_`mÝVa Ho$ ~am~a hmoVm h¡& `h {deof {~ÝXþ
the particle at a particular point in its
hmoVm h¡
motion. This point is
(A) the mean position (A) _mÜ` pñW{V H$m {~ÝXþ
(B) any extreme position (B) H$moB© ^r XÿaV_ (Ma_) pñW{V H$m {~ÝXþ
(C) anywhere between the mean (C) _mÜ` pñW{V VWm YZmË_H$ XÿaV_ pñW{V Ho$
position and the positive extreme ~rM H$moB© ^r pñW{V H$m {~ÝXþ
position
(D) anywhere between the mean position (D) _mÜ` pñW{V VWm F$UmË_H$ XÿaV_ pñW{V Ho$
and the negative extreme position ~rM {H$gr ^r pñW{V H$m {~ÝXw

-13- Set-A

41. When a particle oscillates simple 41. gab AmdÎmu J{V _§o XmobZ H$aVo hþE {H$gr H$U H$s
harmonically, its kinetic energy varies J{VO D$Om© Ho$ _mZ _| AmdÎmu` ({Z`VH$m{bH$)
periodically. If frequency of the particle is
n[adV©Z hmoVm h¡& `{X H$U H$s Amd¥{Îm n h¡ V~ Bg
n, the frequency of the kinetic energy is
J{V _| J{VO D$Om© H$s Amd¥{Îm H$m _mZ h¡
(A) 4n (B) n (A) 4n (B) n
(C) n/2 (D) 2n (C) n/2 (D) 2n

42. A 40 kg body is pushed with just 42. 40 {H$J«m Ho$ EH$ {nÊS> H$mo Ho$db CVZo ~b go YHo$bm
enough force to start it moving across OmVm h¡, Omo Ho$db J{Verb H$aZo Ho$ {bE à`m©ßV h¡
a floor and the same force continues to VWm `hr ~b {nÊS> na BgHo$ ~mX ^r bJm ahVm h¡ &
act afterwards. The coefficient of static
friction and kinetic friction are 0.5 and
`{X ñW¡{VH$ Kf©U VWm J{VH$ Kf©U {Z`Vm§H$ Ho$ _mZ
0.4 respectively. The acceleration of H«$_e: 0.5 VWm 0.4 hmo, Vmo {nÊS> _§o CËnÞ ËdaU
the body is hmoJm
(A) 2 m/s2 (B) 1 m/s2 (A) 2 _rQ>a/go2 (B) 1 _rQ>a/go2
(C) 4 m/s2 (D) 5 m/s2 (C) 4 _rQ>a/go2 (D) 5 _rQ>a/go2

43. How does the proper inflation of tyre 43. Q>m`a H$m dmñV{dH$ \w$bmd (Inflation) {H$g H$maU
save fuel ? go BªYZ _o§ ~Mmd H$aVr h¡ ?
(A) Normal reaction decreases (A) A{^bå~ à{V{H«$`m KQ>Vr h¡
(B) Normal reaction increases (B) A{^bå~ à{V{H«$`m ~‹T>Vr h¡
(C) Sliding contact with the road (C) g‹S>H$ Ho$ gmW ñbmBqS>J gånH©$ KQ>Vm h¡
decreases
(D) Sliding contact with the road (D) g‹S>H$ Ho$ gmW ñbmBqS>J gånH©$ ~‹T>Vm h¡
increases

44. Work equal to 25 J is done on a mass 44. EH$ 2 {H$J«m Ho$ Ðì`_mZ H$mo J{V àXmZ H$aZo Ho$
of 2 kg to set it in motion. If whole of {bE 25 Oyb H$m`© H$s Amdí`H$Vm hmoVr h¡ & `{X
work is used to increase the kinetic {H$`m J`m gånyU© H$m`© Ðì`_mZ H$s J{VO D$Om©
energy, then velocity gained by the ~‹T>mZo _o§ à`moJ hmo, Vmo Bg Ðì`_mZ Ho$ doJ _o§ d¥{Õ
mass in m/s is
hmoJr (_r./go. _§o)
(A) 5 (B) 10 (A) 5 (B) 10
(C) 15 (D) 25 (C) 15 (D) 25

-14- Set-A

45. A convex lens has a focal length f. It 45. EH$ CÎmb boÝg H$s \$moH$g Xÿar f h¡& O¡go {MÌ _|
is cut into two parts along the dotted {XIm`m J`m h¡, Bgo Xmo ^mJm| _| {~ÝXþ aoIm Ho$ AZw{Xe
line as shown in the figure. The focal
length of each part will be
H$mQ>m J`m & àË`oH$ ^mJ H$s \$moH$g Xÿar h¡

(A) 0.5 f (A) 0.5 f
(B) f (B) f
(C) 1.5 f (C) 1.5 f
(D) 2 f (D) 2 f

46. A mass of 72 kg man runs up a staircase 46. 72 {H$J«m Ðì`_mZ H$m EH$ ì`pŠV {H$gr gr‹T>r na
in 12 seconds while a 60 kg man runs 12 goH$ÊS> _o§ M‹T>Vm h¡& O~{H$ 60 {H$J«m H$m ì`pŠV
up the same staircase in 11 seconds.
The ratio of the rate of doing their
Cgr gr‹T>r na 11 goH$ÊS> _§o M‹T>Vm h¡& Vmo CZHo$ Ûmam
work is {H$`o J`o H$m`m] H$s Xam| H$m AZwnmV h¡
(A) 11 : 10 (A) 11 : 10
(B) 10 : 11 (B) 10 : 11
(C) 12 : 11 (C) 12 : 11
(D) 11 : 12 (D) 11 : 12

47. The specific heat of a substance 47. {H$gr nXmW© H$s {d{eîR> Cî_m H$m _mZ {Z^©a H$aVm h¡
depends on its
(A) Nature (A)CgH$s àH¥${V na
(B) Mass (B)CgHo$ Ðì`_mZ na
(C) Rise in temperature (C) CgHo$ Vmn _o§ d¥{Õ na
(D) Both (B) and (C) (D) XmoZm| (B) VWm (C) na

-15- Set-A

48. A student measures the focal length 48. EH$ {dÚmWu {H$gr CÎmb boÝg H$s \$moH$g Xÿar _mn
of a convex lens by putting an object
{~å~ {nZ H$mo boÝg H$s Xÿar u na aI H$a VWm
pin at a distance u from the lens and
measuring the distance v of the image à{V{~å~ {nZ H$s Xÿar v H$mo _mn H$a H$aVm h¡& Bg
pin. The graph between u and v plotted {dÚmWu Ûmam u VWm v Ho$ ~rM IrMm J`m J«m\$
by the student should look like H¡$gm {XImB© XoJm ?
v(cm) v(cm)

(A) (A)
u(cm) u(cm)

v(cm) v(cm)

(B) (B)
u(cm) u(cm)

v(cm) v(cm)

(C) (C)
u(cm) u(cm)

v(cm) v(cm)

(D) (D)
u(cm) u(cm)

-16- Set-A

49. A rectangular tank of depth 12 meter 49. EH$ Am`VmH$ma Q>¢H$ {OgH$s JhamB© 12 _rQ>a h¡,
( )
is full of water µ = 4 3 , the bottom is ( )
Ob µ = 4 3 go ^am h¡& CgH$s Vbr H$s JhamB©
seen at the depth {XImB© XoJr
(A) 6 m (A) 6 m
(B) 9 m (B) 9 m
(C) 12 m (C) 12 m
(D) 16 m (D) 16 m

50. A force F acting on an object varies 50. EH$ {nÊS> na bJZo dmbo ~b F H$m Xÿar x Ho$ gmW
with distance x as shown in the figure. n[adV©Z {MÌ _§o Xem©`m J`m h¡ & `{X ~b Ý`yQ>Z
The force is in N and distance x is in m. VWm Xÿar x _rQ>a _§o ì`ŠV H$s Om`o, V~ {nÊS> H$mo
The work done by the force in moving x = 0 go x = 6 _rQ>a VH$ J{V H$amZo _o§ {H$`o OmZo
the object from x = 0 to x = 6 m is
dmbo H$m`© H$m _mZ h¡

F(N) F(N)

3 3

2 2

1 1

1 2 3 4 5 6 1 2 3 4 5 6
x in (m) x ($_r) _o§

(A) 13.5 J (A) 13.5 Oyb
(B) 9 J (B) 9 Oyb
(C) 27 J (C) 27 Oyb

(D) 18 J (D) 18 Oyb

-17- Set-A

PART – II
Chemistry agm`Z emñÌ
51. The azimuthal quantum number of an 51. {H$gr na_mUw H$m {XJ§er ŠdmÊQ>_ g§»`m g§~{§ YV h¡
atom is related to the
(A) size of the principal shell (A) _w»` H$jm H$m AmH$ma
(B) spin angular momentum (B) MH«$U H$moUr` g§doJ
(C) orbital angular momentum (C) H$jH$ H$moUr` g§doJ
(D) orientation of the orbital in space (D) H$jH$ H$m A§V[aj _| A{^{dÝ`mg

52. Which of the following is correct ? 52. {ZåZ _| go H$m¡Z-gm ghr h¡ ?
1 3 1 3
(A) 1
H and 2He are isotopes (A) 1H Ed§ 2He g_ñWm{ZH$ h¡
14 14 14 14
(B) 6
C and 7N are isotones (B) 6
C Ed§ 7N g_Ý`yQ´>m°{ZH$ h¡
39 40 39 40
(C) 19
K and 20Ca are isotones (C) 19
K Ed§ 20Ca g_Ý`yQ´>m°{ZH$ h¡
19 24 19 24
(D) 9
F and 11Na are isodisperse (D) 9
F Ed§ 11Na g_{S>ñ\$g© h¡

53. The number of elements in each of the 53. AmdV© gm[aUr Ho$ àË`oH$ XrK©- AmdV© _| VÎdmo§ H$s
long period in the periodic table is g§»`m h¡
(A) 2 (A) 2
(B) 8 (B) 8
(C) 18 (C) 18
(D) 32 (D) 32

54. In froth floatation process many 54. \o$Z CËßbdZ àH«$_ _| ~hþV go amgm`{ZH$ nXmW© O¡go
chemicals, i.e, frother, collector, \«$moWa, H$boŠQ>a, EŠQ>rdoQ>a Ed§ {S>{àgoÝQ> H$m Cn`moJ
activator and depressant are used, the
{H$`m OmVm h¡, H$boŠQ>a Ho$ ê$n _| amgm`{ZH$ nXmW©
chemical used as collector is
H$m Cn`moJ hmoZo dmbm h¡
(A) Copper sulphate (A) H$mna gë\o$Q>
(B) Sodium ethyl xanthate (B) gmo{S>`_ BW¡b OoÝWoQ>
(C) Pine oil (C) nmBZ Vob
(D) Sodium cyanide and alkali (D) gmo{S>`_ gm`ZmBS> Ed§ jma

-18- Set-A

55. The matte obtained by smelting copper 55. H$mna nm`amB©Q> H$mo H$mna, H$moH$ Ed§ aoV Ho$ gmW
pyrites with copper, coke and sand àJbZ H$aZo na _¡Q>o àmßV hmoVm h¡ Omo {H$ _w»`V:
contains mainly
hmoVm h¡
(A) FeS + ZnS (A) FeS + ZnS
(B) CuS + FeS2 (B) CuS + FeS2
(C) Cu2S + FeS (C) Cu2S + FeS
(D) ZnS + CuS (D) ZnS + CuS

56. The maximum temperature 1500° C is 56. bm¡h Ho$ {ZîH$f©U _| à`wŠV dmV ^Å>r _| _hÎm_
obtained in region of the blast furnace VmnH«$_ 1500° C Ho$ joÌ H$mo H$hVo h¡§
in the extraction of iron. The region is
called
(A) Reduction zone (A) AdH$aU joÌ
(B) Zone of fusion (B) JbZ joÌ
(C) Combustion zone (C) XhZ joÌ
(D) Slag formation zone (D) YmVw_b {Z_m©U joÌ

57.
For the manufacture of NH3 by the 57. N2(g) + 3H2(g) → 2NH3(g) + ∆H
reaction A{^{H«$`m Ho$ Ûma>m A_mo{Z`m Ho$ {Z_m©U Ho$ {b`o
N2(g) + 3H2(g) → 2NH3(g) + ∆H,
AZwHy$b n[apñW{V`m± h¡
the favourable conditions are
(A) Low temperature, low pressure (A) {ZåZ Vmn, {ZåZ X>m~ Ed§ CËàoaH$
and catalyst
(B) Low temperature, high pressure (B) {ZåZ Vmn, CÀM X>m~ Ed§ CËàoaH$
and catalyst
(C) High temperature, low pressure (C) CÀM Vmn, {ZåZ Xm~ Ed§ CËàoaH$
and catalyst
(D) High temperature, high pressure (D) CÀM Vmn, CÀM Xm~ Ed§ CËàoaH$
and catalyst
58. pH – higher than 7 means, hydronium gmV go A{YH$ pH Ho$ {b`o hmBS´>mo{Z`_ Am`Z
58.
ion concentration gmÝÐU h¡
(A) 10–7 M (B) < 10–7 M (A) 10–7 M (B) < 10–7 M
(C) > 10–7 M (D) > 10–7 M (C) > 10–7 M (D) > 10–7 M

-19- Set-A

59. Endothermic reaction is one in which 59. EH$ Cî_memofr A{^{H«$`m {Og_o§
(A) Heat is converted into electricity (A) Cî_m, {dÚwV _| n[ad{V©V hmoVr h¡
(B) Heat is absorbed (B) Cî_m, Ademo{fV hmoVr h¡
(C) Heat is given out (C) Cî_m, CËg{O©V hmoVr h¡
(D) Heat is converted into mechanical (D) Cî_m, `m§{ÌH$ H$m`© _| n[ad{V©V hmoVr h¡
work
60. A compound “AnBm” is formed,when ‘n’ 60. O~ ‘A’ H$m ‘n’ na_mUw Ed§ ‘B’ H$m ‘m’ na_mUw
atom of ‘a’ and ‘m’ atom of ‘B’combine g§`wŠV hmoH$a `m¡{JH$ “AnBm” ~ZmVo h¡§, `{X
together, if ‘a’ is the mass of one atom ‘A’ Ho$ EH$ na_mUw Ðì`_mZ ‘a’ Ed§ ‘B’ Ho$ EH$
‘A’ and ‘b’ is the mass of one atom na_mUw H$m Ðì`_mZ ‘b’ hmo, Vmo `m¡{JH$ “AnBm”
of ‘B’, then mass of the compound H$m Ðì`_mZ hmoJm
“AnBm” would be
na mb na mb
(A) (na + mb) (B) (na + mb) (A) (na + mb) (B)
(na + mb)
(na + mb)
(C) (na + mb) (D) (C) (na + mb) (D) (na + mb)
na na
61. The modern basis of expressing atomic 61. na_mÊdr` Ed§ AmpÊdH$ Ðì`_mZ H$mo A{^ì`ŠV
and molecular masses is based on H$aZo H$m AmYw{ZH$ AmYma, AmYm[aV h¡
(A) Oxygen – 16 (A) AmŠgrOZ – 16
(B) Hydrogen – 1 (B) hmBS´>moOZ – 1
(C) Carbon – 12 (C) H$m~©Z – 12
(D) Chlorine – 35.5 (D) ŠbmoarZ – 35.5

62. The olefin which on ozonolysis gives 62. Amo{b{\$Z Omo AmoOmoZH¥$V H$aZo na CH3CH2CHO
CH3CH2CHO and CH3CHO is Ed§ CH3CHO XoVm h¡
(A) 1-butene (B) 2-butene (A) 1-ã`yQ>rZ (B) 2-ã`yQ>rZ
(C) 1-pentene (D) 2-pentene (C) 1-noÝQ>rZ (D) 2-noÝQ>rZ

63. Which one of the following alkanes is 63. {ZåZ _| go H$m¡Z-gm EH$ EëHo$Ýg gmÝÐ HNO3 Ûmam
not nitrated by conc.HNO3 ? ZmB{Q´>H¥$V Zht hmoVm h¡ ?
(A) CH3.CH3 (B) CH3.CH2.CH3 (A) CH3.CH3 (B) CH3.CH2.CH3
(C) (CH3)4C (D) CH4 (C) (CH3)4C (D) CH4

-20- Set-A

64. Among the following which is/are 64. {ZåZ Ho$ _Ü` H$m¡Z-gm gw_o{bV Zht h¡/h¢ :
mismatched :
Gaseous fuel Composition J¡gr`BªYZ g§JR>Z
I. Semi water gas – CO + H2 + N2 I. go_r dmQ>a J¡g
– CO + H2 + N2
II. Producer gas – CO + NO II. àmoS>çyga J¡g
– CO + NO
III. Water gas – CO + H2 III. dmQ>a J¡g – CO + H2
IV. Coal gas – H2 + CH4 + CO + IV. H$mob J¡g – H2 + CH4 + CO +
Hydrocarbons hmBS´>moH$m~©Ýg
V. LPG – Butane + Methane
V. Eb.nr.Or. – ã`yQ>oZ + {_WoZ
(A) (I) and (II) (B) (III) and (IV) (A) (I) Ed§ (II) (B) (III) Ed§ (IV)
(C) (II) and (V) (D) (I) and (IV) (C) (II) Ed§ (V) (D) (I) Ed§ (IV)

65. Which of the following is the residual 65. gm~wZ Ho$ {Z_m©U _| {ZåZ _| go H$m¡Z-gm Ad{eîQ>>
product in the formation of soap ? nXmW© h¡ ?
(A) Glyceraldehyde (A) p½bgampëS>hmBS>
(B) Glycerol (B) p½b{gamb
(C) Glycol (C) ½bmBH$mb
(D) Acrylonitrile (D) EH«$mBbmoZmB{Q´>b

66. On heating a mixture of Cu2O and 66. Cu2O Ed§ Cu2S Ho$ {_lU H$mo J_© H$aZo na àmßV
Cu2S, the product obtained are CËnmX h¡
(A) CuO + CuS (B) Cu + SO2 (A) CuO + CuS (B) Cu + SO2
(C) Cu2SO3 (D) Cu + SO3 (C) Cu2SO3 (D) Cu + SO3

67. Covalently bonded molecules with 67. H$R>>moaV_ àH¥${V Ho$ ghg§`moOr Am~pÝYV AUw h¡
hardest nature are
(A) CaC2, SiC, B4C (A) CaC2, Sic, B4C
(B) Diamond, B4C, SiC (B) S>m`_§S>, B4C, SiC
(C) BN, Diamond, Mg3N2 (C) BN, S>m`_§S>, Mg3N2
(D) B4C, Mg3N2, Li3N (D) B4C, Mg3N2, Li3N

-21- Set-A

68. A hydrocarbon has molecular formula 68. EH$ hmBS´>moH$m~©Z H$m AmU{dH$ gyÌ C6H8 h¡& {ZåZ
C6H8. Among the following which does Ho$ _Ü` {H$g_| sp, sp2 Ed§ sp3 g§§H$[aV H$m~©Z
not contain sp, sp2 and sp3 hybrid state
of carbon atoms ? na_mUw CnpñWV Zht h¡ ?
(A) CH2 = CH – CH = CH – CH = CH2 (A) CH2 = CH – CH = CH – CH = CH2
(B) CH2 = CH – C ≡ C – CH2 – CH3 (B) CH2 = CH – C ≡ C – CH2 – CH3
(C) CH3 – CH = C = CH – CH = CH2 (C) CH3 – CH = C = CH – CH = CH2
(D) CH ≡ C – CH2 – CH2 – CH = CH2 (D) CH ≡ C – CH2 – CH2 – CH = CH2

69. Food chain is 69. ImÚ ûm¥§Ibm h¡
(A) Number of human beings forming (A) ImÚ Ho$ {b`o {Z{_©V _mZd ûm¥§Ibm
a chain for food
(B) Animals near a source of food (B) ImÚ òmoVm| Ho$ nmg OÝVwAm| H$m hmoZm
(C) Transfer of food energy from (C) ImÚ COm© H$m CËnmXH$ go Cn^moŠVm H$s Amoa
producers to consumers ñWmZmÝVa
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

70. Role of bacteria in carbon cycle is 70. H$m~©Z MH«$> _| ~oŠQ>o[a`m {ZYm©[aV H$aVm h¡
(A) Photosynthesis (A) àH$meg§íbofU
(B) Chemosynthesis (B) amgm`{ZH$g§íbofU
(C) Breakdown of organic compounds (C) H$m~©{ZH$ `m¡{JH$m| H$mo IpÊS>V H$aZm
(D) Assimilation of nitrogen compounds (D) ZmBQ´>moOZ `m¡{JH$m| H$m g_m§JrH$aU

71. The ionisation energy of nitrogen is more 71. ZmBQ´>moOZ H$s Am`ZZ D$Om© Am°ŠgrOZ go A{YH$
than that of oxygen is, because of h¡, Š`m|{H$
(A) the greater attraction of electron (A) Zm{^H$ Ho$ Ûmam BboŠQ´>m°Z H$m A{YH$
by nucleus AmH$f©U
(B) the extra stability of half filled (B) AY©nyU© P-H$jH$m§o H$m A{V[aŠV ñWm{`Ëd
P-orbitals
(C) the smallest size of nitrogen (C) ZmBQ´>moOZ H$m N>moQ>m AmH$ma
(D) more penetration effect (D) A{YH$ ^oXZ à^md

-22- Set-A

72. Which one of the following statement 72. g§H«$_U VÎdm| Ho$ {b`o {ZåZ _| go H$m¡Z-gm H$WZ
is not correct for transition element ? ghr Zht h¡ ?
(A) They form coloured compounds. (A) `h a§JrZ `m¡{JH$ ~ZmVo h¡§ &
(B) All compounds of them are (B) BZHo$ g^r `m¡{JH$ AZwMwå~H$s` hmoVo h§¡ &
paramagnetic.
(C) They show variable oxidation (C) n[adVu AmŠgrH$aU AdñWm`o hmoVr h¡ &
states.
(D) They have incomplete ‘d’-orbital. (D) BZHo$ ‘d’-H$jH$ AnyU© hmoVo h¡§ &

73. Carbon atoms in the diamond are 73. hrao _| H$m~©Z na_mUw {ZåZ {dÝ`mg _| ì`dpñWV
arranged in the following configuration ahVo h¢
(A) Tetrahedral (B) Planar (A) MVwî\$bH$s` (B) g_Vbr`
(C) Linear (D) Octahedral (C) a¡Ir` (D) AîQ>\$bH$s`
74. BOD5 is 74. BOD5 h¡
(A) Water decomposed in 5 days (A) 5 {XZ _| Ob H$m {dKQ>Z
(B) Oxygen used in 5 days (B) 5 {XZ _| AmŠgrOZ H$m Cn`moJ hmoZm
(C) Microorganism killed in 5 days (C) 5 {XZ _| _mBH«$moAmJ}{Zñ_ H$m _mam OmZm
(D) Dissolved oxygen left after 5 days (D) 5 {XZ ~mX Kw{bV AmŠgrOZ eof ahZm

75. Which of the following is non-renewable 75. {ZåZ _| go H$m¡Z-gm J¡a-ZdrZrH$aU COm© òmoV h¡ ?
source of energy ?
(A) Hydro power (A) Ob epŠV
(B) Tidal power (B) Q>mBS>b nmda
(C) Geothermal energy (C) {O`moW_©b COm©
(D) Nuclear energy (D) Zm{^H$s` COm©
76. An object is located at the height of 60 km 76. EH$ dñVw n¥Ïdr Ho$ gVh go pñWV 60 {H$bmo_rQ>a
from the surface of earth. The object is D±$MmB© na h¡ & dñVw dm`w_ÊS>b Ho$ {H$g ^mJ _|
located in which part of the atmosphere ?
pñWV h¡ ?
(A) Troposphere (B) Stratosphere (A) Q´>monmoñ\$s`a (B) ñQ´>oQ>moñ\$s`a
(C) Mesosphere (D) Ionosphere (C) _ogmoñ\$s`a (D) Am`moZmoñ\$s`a

-23- Set-A

77. Petrol is a mixture hydrocarbons from C6 77. noQ>´mb
o C6 go C8 VH$ hmBS´>mHo $m~©Z H$m {_lU h¡& noQ>´mb
o
to C8. The quality of petrol is determined H$s JwUdÎmm AmŠQ>Zo g§»`m Ho$ AmYma na {ZYm©[aV H$s
in terms of octane number. The correct
OmVr h¡& AmŠQ>Zo g§»`m H$m ghr H«$_ h¡
order of octane number is
(A) cycloalkanes < alkenes < alkanes < (A) gmBŠbmoEëHo$Ýg < EëH$sÝg < EëHo$Ýg <
aromatic hydrocarbons Eamo_o{Q>H$ hmBS´>moH$m~©Ýg
(B) alkenes < alkanes < aromatic (B) EëH$sÝg < EëHo$Ýg < Eamo_{o Q>H$ hmBS´>mHo $m~©Ýg <
hydrocarbons < cycloalkanes
gmBŠbmoEëHo$Ýg
(C) alkanes < aromatic hydrocarbons < (C) EëHo$Ýg < Eamo_o{Q>H$ hmBS´>moH$m~©Ýg <
cycloalkanes < alkenes gmBŠbmoEëHo$Ýg < EëH$sÝg
(D) alkanes < alkenes < cycloalkanes < (D) EëHo$Ýg < EëH$sÝg < gmBŠbmoEëHo$Ýg <
aromatic hydrocarbons
Eamo_o{Q>H$ hmBS´>moH$m~©Ýg
78. Maximum percentage of gas present 78. H$mob J¡g _| J¡g {OgH$s gdm©{YH$ à{VeV h¡
in coal gas is
(A) Methane (A) {_WoZ
(B) Oxygen (B) AmŠgrOZ
(C) Hydrogen (C) hmBS´>moOZ
(D) Carbon monoxide (D) H$m~©Z _moZmoAmŠgmBS>

79. A group of interconnected food chain 79. nañna Ow‹S>o hþ`o ImÚ ûm¥§Ibm Ho$ g_yh H$mo H$hVo h§¡
is called
(A) Pyramid of energy (A) COm© H$m {nam{_S>
(B) Complex food chain (B) O{Q>b ImÚ ûm¥§Ibm
(C) Food web (C) ImÚ do~
(D) Food cycle (D) ImÚ MH«$

-24- Set-A

80. Consider following statements. 80. {ZåZ H$WZm| na {dMma H$s{O`o &
I. Food chain occurs unidirectional I. ImÚ ûm¥§Ibm EH X¡{eH$ COm© H$m àdmh
flow of energy. hmoVm h¡ &
II. A food web is a simplified way II. ImÚ do~ dZñn{V Ed§ OÝVwAm| Ho$ _Ü` ImÚ
showing food relationship between g§~§Y àX{e©V H$aZo H$m gabV_ ê$n h¡ &
plants and animals.
III. An ecological pyramid of biomass ~m`mo_mg H$m nm[apñW{VH$ {nam{_S> CnpñWV
III.
shows relationship between ~m`mo_mg Ho$ COm© H$s _mÌm à{V Q´>mo{nH$ ñVa
energy amount of biomass present
per tropic level.
Ho$ _Ü` g§~§Y àX{e©V H$aVm h¡ &
IV. Fe, Zn, Cu are macronutrients for IV. Fe, Zn, Cu dZñn{V Ho$ ñWyb nmofH$
plants. VËd h¡ &
V. Biogeochemical cycle is the V. ~m`mo{O`moHo${_H$b MH«$ n`m©daU _| CnpñWV
cycling of chemical elements Or{dV Ed§ AOr{dV ^mJm| Ûmam Amdí`H$
required by living and nonliving
part of environment.
amgm`{ZH$ VËdm| Ho$ MH«$U h¡ &
Wrong statements are : JbV H$WZ h¡ :
(A) I and III (B) II and IV (A) I Ed§ III (B) II Ed§ IV
(C) II and III (D) IV and V (C) II Ed§ III (D) IV Ed§ V

81. Consider following statements. 81. {ZåZ H$WZm| na {dMma H$s{O`o &
I. Mesosphere and thermosphere are I. _ogmoñ\$s`a Ed§ W_m}ñ\$s`a H$mo g_yh ê$n go
collectively called ionosphere. Am`Zmoñ\$s`a H$hVo h¢ &
II. London smog is formed in summer II. b§XZ ñ_m°J J«rî_ F$Vw _| ~ZVm h¡ &
season.
III. Soil containing 34% air and 66% III. _¥Xm {Og_| 34% dm`w Ed§ 66% Ob hmo
water is considered to be best soil \$gbm| Ho$ {b`o gdm}Îm_ _¥Xm _mZm OmVm h¡&
for most of the crops.
IV. Presence of hydrocarbon is \$moQ>moHo${_H$b ñ_m°J Ho$ {b`o hmBS´>moH$m~©Z H$s
IV.
essential for photochemical smog. CnpñW{V AË`mdí`H$ h¡ &
V. The lung disease caused by V. Eg~oñQ>mg o go b§J {S>grg hmoZo H$mo {g{bH$mo{gg
asbestos is called silicosis. H$hVo h¢ &
Wrong statements are : JbV H$WZ h¡ :
(A) II and V (B) I and III (A) II Ed§ V (B) I Ed§ III
(C) III and IV (D) I and IV (C) III Ed§ IV (D) I Ed§ IV

-25- Set-A

82. Which one is electron deficient 82. {ZåZ _| go BboŠQ´>mZ Ý`yZ `m¡{JH$ h¡
compound ?
(A) ICI (B) NH3 (A) ICI (B) NH3
(C) BCl3 (D) PCl3 (C) BCl3 (D) PCl3

83. Which of the following property is true 83. ghg§`moOH$ `m¡{JH$m| Ho$ {b`o {ZåZ _| go H$m¡Z-gm
for covalent compounds ? bjU ghr h¡ ?
(A) They are soluble in water (A) `h Ob _| KwbZerb h¡
(B) They are insoluble in water (B) `h Ob _| AKwbZerb h¡
(C) They ionise in water (C) `h Ob _| Am`{ZV hmoVo h¢
(D) They hydrolyse in water (D) `h Ob _| Ob AnK{Q>V hmoVo h¢

84. With reference to wind energy, which 84. ndZ COm© Ho$ g§X^© _| H$m¡Z-gm H$WZ JbV h¡ ?
statement is wrong ?
(A) Wind energy mainly depends (A) ndZ COm© _w»`V: dm`w Ho$ VmnH«$_, Xm~ Ed§
upon temperature, pressure and KZËd na {Z^©a H$aVm h¡&
density of air.
(B) Wind energy is an indirect form of (B) ndZ COm©, gmoba COm© H$m EH$ AàË`jê$n
solar energy. h¡&
(C) Wind energy is ecofriendly and (C) ndZ COm© B©H$mo\«o$ÝS>br VWm àXÿfU _wŠV h¡&
pollution free.
(D) For wind energy the speed of air (D) ndZ COm© Ho$ {b`o dm`w H$m doJ 50 Km/h
should not be less than 50 Km/h. go H$_ Zht hmoZm Mm{h`o&
85. Nuclear fusion reaction takes place at 85. Zm{^H$s` g§b`Z {H«$`m CÀM Vmn na hmoVr h¡ Š`m|{H$
high temperature because CÀM Vmn na
(A) Atoms ionized at high temperature (A) CÀM Vmn na na_mUw Am`{ZV hmoVo h¢
(B) Kinetic energy is high enough (B) Zm{^H$m| Ho$ _Ü` {dH$f©U H$mo H$m~y nmZo n`m©ßV
to overcome repulsion between CÀM J{VO COm© hmoZm Mm{h`o
nuclei
(C) Nuclei breaks at high temperature (C) CÀM VmnH«$_ na Zm{^H$ {dIpÊS>V hmoVo h¡§
(D) Molecules breaks at high (D) CÀM Vmn na AUw IpÊS>V hmoVo h¡§
temperature

-26- Set-A

86. Consider following statements : 86. {ZåZ H$WZm| na {dMma H$s{O`o :
(I) Coal, petroleum and natural gas (I) H$mo`bm, noQ´>mo{b`_ Ed§ àmH¥${VH$ J¡g H$mo
are called as fossil fuels.
Ordmí_ BªYZ H$hm OmVm h¡&
(II) Ideal fuel undergoes combustion (II) AmXe© BªYZ H$m XhZ CÀM Xa go hmoVm h¡&
with high rate.
(III) Percentage efficiency of nuclear (III) Zm{^H$s` g§b`Z H$m à{VeV XjVm 0.38 h¡&
fusion is 0.38.
(IV) Calorific value of biogas is (IV) ~m`moJ¡g H$m H¡$bmoar{\$H$ _mZ 10 – 20
10 – 20 kJ/gm. {H$bmoOyb à{V J«m_ h¡&
(V) Internal temperature of solar cooker (V) gmoba> Hy$H$a H$m Am§V[aH$ VmnH«$_ 2-3 K§Q>mo§ _|
is 100° – 140° C within 2-3 hours.
100° – 140° C VH$ hmo OmVm h¡ &
Wrong statements are JbV H$WZ h¢
(A) (I) and (III) (A) (I) Ed§ (III)
(B) (III) and (V) (B) (III) Ed§ (V)
(C) (II) and (V) (C) (II) Ed§ (V)
(D) (II) and (IV) (D) (II) Ed§ (IV)

87. Portland cement does not contain 87. nmoQ>©boÊS> gr_oÊQ> _| Zht hmoVm h¡
(A) Ca3Al2O6 (B) Ca3SiO5 (A) Ca3Al2O6 (B) Ca3SiO5
(C) Ca2SiO4 (D) Ca3(PO4)2 (C) Ca2SiO4 (D) Ca3(PO4)2

88. The formula of washing soda is 88. Ymo~Z gmoS>m H$m gyÌ h¡
(A) Na2CO3 . 10H2O (A) Na2CO3 . 10H2O
(B) Na CO . 7H O
2 3 2
(B) Na CO . 7H O
2 3 2

(C) Na2CO3 . H2O (C) Na2CO3 . H2O
(D) Na2CO3 (D) Na2CO3

89. Power alcohol is nmda AëH$mohb h¡
89.
(A) Absolute alcohol and petrol 20 : 80 (A) A~gbyQ> AëH$mohb Ed§ noQ´>mob 20 : 80
(B) Absolute alcohol and petrol 80 : 20 (B) A~gbyQ> AëH$mohb Ed§ noQ´>mob 80 : 20
(C) Rectified spirit and benzene 20 : 80 (C) aopŠQ>\$mBS> pñn[aQ>> Ed§ ~oÝOrZ 20 : 80
(D) Rectified spirit and benzene 80 : 20 (D) aopŠQ>\$mBS> pñn[aQ> Ed§ ~oÝOrZ 80 : 20

-27- Set-A

90. Which of the following mixtures are 90. {ZåZ _| go H$m¡Z-gm {_lU g_m§Jr h¡ ?
homogeneous ?
(A) Wood (A) bH$‹S>r
(B) Tap water (B) Zb H$m nmZr
(C) Soil (C) _¥Xm
(D) Cloud (D) _oK

91. Which of the following characteristic is 91. {ZåZ _| H$m¡Z-go bjU AYmVw Ho$ {b`o gË`
not true for non-metals ? Zht h¡ ?
(A) Electronegative elements (A) {dÚwVF$UmË_H$ VÎd hmoVo h¡§
(B) Poor conductor of electricity and (B) {dÚwV Ed§ Cî_m Ho$ Xþ~©b MmbH$ hmoVo h¡§
heat
(C) Solid non-metals are brittle (C) R>mog AmYmVw ^§Jwa hmoVo h¡§
(D) Non-metals are malleable (D) AmYmVw AmKmVdY©Zr` hmoVo h¡§

92. Tyndall effect in colloidal solution is due to 92. H$mobmBS>b {db`Z _| {Q>ÊS>b à^md H$m H$maU h¡
(A) scattering of light (A) àH$me H$m àH$sU©Z
(B) reflection of light (B) àH$me H$m namdV©Z
(C) dispersion of light (C) àH$me H$m dU©{djonU
(D) refraction of light (D) àH$me H$m AndV©Z

93. Hydrogen is evolved by action of cold 93. R>§S>o Ed§ VZw HNO3 go {H«$`mH$a> H2 _wŠV H$aZo dmbm
and dilute HNO3 on YmVw h¡
(A) Zn (A) Zn
(B) Mg (B) Mg
(C) Cu (C) Cu
(D) Al (D) Al

-28- Set-A

94. Oxygen is more electronegative than 94. AmŠgrOZ, gë\$a> H$s Anojm A{YH$ {dÚwV
sulphur yet H2S is acidic while H2O is
F$UmË_H$ h¡ O~{H$ H2S Aåbr` Ed§ H2O EH$
neutral molecule because
CXmgrZ AUw h¡ Š`m|{H$ $
(A) Water is a highly associated (A) Ob EH$ g§`mo{OV Ðd h¡
liquid

(B) H – S bond is weaker than H – O (B) H – S Am~ÝY H – O Am~ÝY H$s Anojm
bond Xþ~©b h¡
(C) H 2 S is a gas while H 2 O is a (C) H2S EH$ J¡g h¡ O~{H$ H2O EH$ Ðd h¡
liquid

(D) Molecular mass of H2S is more (D) H2S H$m AmU{dH$ _mÌm H2O go A{YH$
than that of H2O hmoVm h¡

95. Pure N2 is obtained by the following 95. {dewÕ N2 {ZåZ Ho$ A{^{H«$`m go àmßV {H$`m OmVm h¡
reaction
(A) NH2CONH2 + HNO2 (A) NH2CONH2 + HNO2
(B) NH4Cl + NaNO2 (B) NH4Cl + NaNO2
(C) NH3 + CuO (C) NH3 + CuO
(D) (NH4)2Cr2O7 (D) (NH4)2Cr2O7

96. The reaction : 96. A{^{H«$`m :
Mg(s) + CuSO4 (aq) Mg(s) + CuSO4 (aq)
MgSO4 (aq) + Cu(s) MgSO4 (aq) + Cu(s)
is example of CXmhaU h¡
(A) Displacement reaction (A) {dñWmnZ A{^{H«$`m

(B) Combination reaction (B) g§`moOZ A{^{H«$`m
(C) Decomposition reaction (C) AnKQ>Z A{^{H«$`m
(D) Oxidation-reduction reaction (D) Am°ŠgrH$aU-AnM`Z A{^{H«$`m

-29- Set-A

97. The formula of sodium carbonate is 97. gmo{S>`_ H$m~m}ZoQ> H$m gyÌ Na2Co3 h¡, Vmo
Na2Co3 and that of calcium hydrogen
H¡$pëg`_ hmBS´>moOZ H$m~m}ZoQ> H$m h¡
carbonate is
(A) Ca(HCO3)2 (A) Ca(HCO3)2
(B) Ca2HCO3 (B) Ca2HCO3
(C) CaHCO3 (C) CaHCO3
(D) Ca(HCO3)3 (D) Ca(HCO3)3

98. An oxidising agent is capable of 98. EH$ AmŠgrH$maH$ gj_ h¡
(A) Giving electrons (A) BboŠQ´>m°Z XoZo Ho$ {bE
(B) Accepting electrons (B) BboŠQ´>m°Z J«hU H$aZo Ho$ {bE
(C) Producing cation (C) YZm`Z ~ZmZo Ho$ {bE
(D) Unpredictable (D) Anydm©Zw_o`

99. The molecular formula of a compound {H$gr `m¡{JH$ H$m AUwgyÌ, CgHo$ _wbmZwnmVr gyÌ
99.
is related with its empirical formula; go g§~§{YV h¡,
select the correct relation.
ghr g§~§Y H$m M`Z H$ao &
(A) empirical = n × molecular formula (A) _wbmZwnmVr gyÌ = n × AUwgyÌ
formula
(B) molecular = n × empirical formula (B) AUwgyÌ = n × _wbmZwnmVr gyÌ
formula
n
(C) empirical = n
(C) _wbmZwnmVr gyÌ =
formula molecular formula AUwgyÌ
(D) molecular = n n
formula empirical formula (D) AUwgyÌ =
_wbmZwnmVr gyÌ

100. The sum of the number of neutrons and 100. hmBS´>moOZ Ho$ EH$ g_ñWm{ZH$ _| Ý`yQ´>mZmo§ Ed§ àmoQ>mZm§o
protons in an isotope of hydrogen is H$s g§»`m H$m `moJ h¡
(A) 6 (B) 5 (A) 6 (B) 5
(C) 4 (D) 3 (C) 4 (D) 3

-30- Set-A

PART – III
Mathematics J{UV
101. 85 kg of a mixture contains milk and 101. 85 {H$bm| Ho$ {_lU _| 27 : 7 Ho$ AZwnmV _| XyY
water in the ratio 27 : 7. How much
Am¡a nmZr h¡ & Cg_| {H$VZm nmZr Am¡a {_bm`m Om`
more water is to be added to get a new
mixture containing milk and water in {H$ Z`o {_lU _| XyY Am¡a nmZr H$m AZwnmV 3 : 1
the ratio 3 : 1 ? hmo Om`o ?
(A) 5 kg (B) 6 kg (A) 5 {H$bmo (B) 6 {H$bmo
(C) 6.5 kg (D) None of these (C) 6.5 {H$bmo (D) BZ_| go H$moB© Zht

102. If x is a positive real number and a, b, c 102. `{X x EH$ YZmË_H$ dmñV{dH$ g§»`m
are rational numbers, then the value h¡ Am¡a a, b, c n[a_o` g§»`mE± h¡, Vmo
1 1 1 1
of b−a c −a
+ a −b +
1+ x +x 1+ x + xc −b 1+ x b−a
+x c −a
1+ x a −b
+ xc −b
1 1
+ b−c
is + H$m _mZ hmoJm
1+ x + xa − c 1 + xb − c + x a − c

(A) –1 (A) –1
(B) 0 (B) 0
(C) 1 (C) 1
(D) None of these (D) BZ_| go H$moB© Zht

103. In the following table the cumulative 103. {ZåZ{b{IV Vm{bH$m _| dJ© 31 – 35 H$s g§M`r
frequency for class 31 – 35 is ~maå~maVm h¡
Class 16 – 20 21 – 25 26 – 30 31 –35 36 – 40 41 – 45 dJ© 16 – 20 21 – 25 26 – 30 31 –35 36 – 40 41 – 45
Frequency 10 7 5 12 9 11 Amd¥{Îm 10 7 5 12 9 11

(A) 12 (A) 12
(B) 34 (B) 34
(C) 43 (C) 43
(D) None of these (D) BZ_| go H$moB© Zht

-31- Set-A

104. From the following frequency polygon, 104. {ZåZ{b{IV Amd¥{Îm ~hþ^wO go, dJ© 20 – 30 H$m
the frequency of class 20 – 30 is
Amd¥{Îm h¡

25 25

20 20
Frequency

Amd¥{Îm
15 15

10 10

5 5

0 10 20 30 40 50 Class 0 10 20 30 40 50 dJ©

(A) 25 (B) 20 (A) 25 (B) 20
(C) 10 (D) 5 (C) 10 (D) 5
2 2 2 2
105. 0 ≤ x ≤ π and 81sin x + 81cos x = 30, 105. 0 ≤ x ≤ π Am¡a 81sin x + 81cos x = 30, Vmo
then x is equal to x H$m _mZ hmoJm
π 5π π 5π
(A) , (A) ,
6 6 6 6
π 2π π 2π
(B) , (B) ,
3 3 3 3
(C) Neither option (A) nor option (B) (C) Z hr {dH$ën (A) ghr h¡ Am¡a Z hr {dH$ën
is true (B) ghr h¡
(D) Both options (A) and (B) are true (D) XmoZm| {dH$ën (A) Am¡a$ (B) ghr h¡

106. The general solution of the equation 106. g_rH$aU tan2α + 2 3 tanα = 1 H$m ì`mnH$
tan2α + 2 3 tanα = 1 is given by hb hmoJm
nπ  1 nπ  1
(A) (B)  n +  π (A) (B)  n +  π
12  2 12  2
π π
(C) (6n + 1) (D) none of these (C) (6n + 1) (D) BZ_| go H$moB© Zht
12 12

-32- Set-A

107. Solve the equation. 107. g_rH$aU hb H$amo &
tanθ + tan2θ + tan3θ = 0. Then θ =
tanθ + tan2θ + tan3θ = 0, Vmo θ =
π π
(A) nπ + (A) nπ +
4 4
(
(B) nπ + tan
−1
2 ) (B) nπ + tan (
−1
2 )
 −1 1   −1 1 
(C) nπ +  tan  (C) nπ +  tan 
 2  2
(D) None of these (D) BZ_| go H$moB© Zht

108. If 3, 3log yx, 3log zy and 7log xz are 108. `{X 3, 3logyx, 3logzy Am¡a 7logxz g_mÝVa loUr _|
in A.P. and x18 = y21 = z28, then the h¢ Am¡a x18 = y21 = z28, Vmo BZ nXm| H$m gdm©ÝVa hmoJm
common difference of the terms is
1 1
(A) (B) 0 (A) (B) 0
2 2
1 1
(C) – (D) None of these (C) – (D) BZ_| go H$moB© Zht
2 2
109. Ten years ago, A was half of B in age. 109. Xg gmb nhbo A H$s C_« B H$s C_« H$s AmYr Wr &
If the ratio of their present ages is 3 : 4, `{X CZH$s dV©_mZ Am`w H$m AZwnmV 3 : 4 hmo, Vmo
then the total of their present ages is dV©_mZ Am`w H$m `moJ hmoJm
(A) 20 years (A) 20 df©
(B) 45 years (B) 45 df©
(C) 30 years (C) 30 df©
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

110. Father is aged three times more than 110. {nVm H$s C_« CgHo$ ~oQo amhþb H$s C_« go VrZ JwZm
his son Rahul. After 8 years, he would A{YH$ h¡ & 8 df© níMmV² CgH$s C_« amhþb H$s C_«
be two and half times of Rahul’s age.
go T>mB© JwZm hmo Om`oJr Am¡a AmR> df© níMmV² {nVm
After further 8 years, how many times
would he be of Rahul’s age ? H$s C_« amhþb H$s C_« H$s {H$VZo JwZm hmoJr ?
(A) 2 times (B) 3 times (A) 2 JwZm (B) 3 JwZm
1 1
(C) 2 times (D) None of these (C) 2 JwZm (D) BZ_| go H$moB© Zht
2 2

-33- Set-A

111. In measuring the sides of a rectangle, 111. {H$gr Am`V H$s ^wOmAm| H$mo O~ _mnm J`m Vmo
one side is taken 5% in excess and EH$ ^wOm 5% A{YH$ _mnr J`r Am¡a EH$ _| 4%
the other 4% in deficit. Find the error
H$_ _mnm J`m & BZ _mnm| go _mno J`o joÌ\$b _|
percent in the area calculated by these
measurements. ÌwQ>r H$m à{VeV hmoJm
(A) 2% (B) 0.2% (A) 2% (B) 0.2%
(C) 0.8% (D) None of these (C) 0.8% (D) BZ_| go H$moB© Zht

112. If (p∧q ⇒q∨r)∧(p⇒r) is a statement, 112. `{X (p∧q ⇒q∨r)∧(p⇒r) EH$ H$WZ h¡, V~
then this statement is equivalent to `h H$WZ Vwë` h¡ __________Ho$&
(A) p∧q∧r (B) p⇒r (A) p∧q∧r (B) p⇒r
(C) p⇒q (D) q⇒r (C) p⇒q (D) q⇒r

113. For given two statements p and q 113. {X`o JE Xmo H$WZm| p VWm q Ho$ {bE {ZåZ _| go
which of these is correct ? H$m¡Z-gm ghr h¡ ?
(A) (p⇒q)≡(p⇔q) (A) (p⇒q)≡(p⇔q)
(B) ~p≡(~p⇔~q) (B) ~p≡(~p⇔~q)
(C) (p⇒q)≡ (q⇒p) (C) (p⇒q)≡ (q⇒p)
(D) none of these (D) BZ_| go H$moB© Zht
114. In the following figure OABC is a 114. {ZåZ{b{IV {MÌ _| OABC EH$ Am`V h¡, B Ho$
rectangle, coordinates of B is (2, 2) and {ZX}em§H$ (2, 2) h¢ VWm D′, D H$m Y-Aj Ho$
D′ is the image of D with respect to the gmnoj à{V{~å~ h¡, V~ D′ Ho$ {ZX}em§H$ h¢
Y-axis, then coordinates of D′ is
Y Y

C B(2, 2) C B(2, 2)

D D
D′ D′

O A X O A X

(A)  − , 1
1
(A)  − , 1
1 (B) (–1, –1)
(B) (–1, –1)
 2   2 
(C) (–1, 1) (D) None of these (C) (–1, 1) (D) BZ_| go H$moB© Zht

-34- Set-A

115. If (–1, –2) is incident to lines ax + by = 14 115. `{X (–1, –2) aoImAm| ax + by = 14 VWm
and ax – by = 6, then the values of ax – by = 6 na Amn{VV h¡, V~ a, b Ho$ _mZ
a, b are respectively
h¢ H«$_e:
(A) 10, 2 (B) –10, –2 (A) 10, 2 (B) –10, –2
(C) –10, 2 (D) –2, 10 (C) –10, 2 (D) –2, 10

116. The triangles formed by one set of 116. gab aoImAm| Ho$ g_wƒ` x – 2y = 0, 2x – y = 0,
straight lines x – 2y = 0, 2x – y = 0, x + y = 3 VWm Xÿgar gab aoImAm| Ho$ g_wƒ`
x + y = 3 and another set of straight
x + 2y = 0, 2x + y = 0, x – y = 3 Ûmam
lines x + 2y = 0, 2x + y = 0, x – y = 3
are {Z{_©V {Ì^wO h¡§
(A) isosceles and congruent (A) g_{Û~mhþ Ed§ gdmªJg_
(B) equilateral and congruent (B) g_~mhþ Ed§ gdmªJg_
(C) isosceles but not congruent (C) g_{Û~mhþ naÝVw gdmªJg_ Zht
(D) equilateral but not congruent (D) g_~mhþ naÝVw gdmªJg_ Zht
117. If the roots of the equation x2 + px – q = 0 117. `{X g_rH$aU x2 + px – q = 0 Ho$ _yb tan30°
are tan30° and tan15°, then the value Am¡a tan15° h¡, Vmo 2 + q – p H$m _mZ hmoJm
of 2 + q – p is
(A) 0 (B) 1 (A) 0 (B) 1
(C) 2 (D) 3 (C) 2 (D) 3

118. Which is not true in the following 118. {ZåZ {MÌ _| H$m¡Z-gm gË` Zht h¡ ?
figure ? A
A

F E
F E
120°
120° O
O 120°
120°
B C
B C

D
D
(A) ∆ABC ~ ∆DEF
(A) ∆ABC ~ ∆DEF
(B) ∆FOE ~ ∆BOC (B) ∆FOE ~ ∆BOC
(C) Fig. ACOF ~ fig. BCOF (C) {MÌ ACOF ~ {MÌ BCOF
(D) arc BDC ~ arc FAE (D) Mmn BDC ~ Mmn FAE

-35- Set-A

3   1  
1 1  3   1  
1 1 
119. 5 −  + 2 −  0.5 + −    is 119. 5 −  +  2 −  0 .5 + −    H$m
 2 
 4  6 7 
   2 
 4  6 7 
 
equal to _mZ hmoJm
61 47 61 47
(A) 2 (B) 2 (A) 2 (B) 2
84 84 84 84
23 23
(C) 2 (D) None of these (C) 2 (D) BZ_| go H$moB© Zht
84 84

120. Let F = 0.84181. Then if F is written 120. _mZm {H$ F = 0.84181, `{X F H$mo {ZåZV_ nXm|
as a fraction in lowest terms, the Ho$ {^Þ Ho$ ê$n {bIm OmE Vmo ha CgHo$ A§e go
denominator exceeds the numerator
{H$VZm ~S>m h¡ ?
by
(A) 87 (B) 81 (A) 87 (B) 81
(C) 29 (D) none of these (C) 29 (D) BZ_| go H$moB© Zht

 1  3 1 1

4 4 2 4
121. ì`§OH$  2 − 1  2 + 2 + 2 + 1 H$m
121. The value of the expression
   
 1  3 1 1

4 4 2 4
 2 − 1  2 + 2 + 2 + 1 is _mZ hmoJm
   
7 7
(A) 24 (B) 1 (A) 24 (B) 1
(C) 0 (D) None of these (C) 0 (D) BZ_| go H$moB© Zht

122. If coordinates of points P, Q, R and S 122. `{X {~ÝXþAm| P, Q, R VWm S Ho$ {ZX}em§H$ H«$_e:
are respectively (–3, 7), (1, 7), (–3, 7), (1, 7), (1, 2) VWm (3, 4) h¢ VWm `{X
(1, 2) and (3, 4) and if T is mid point of
T, RS H$m _Ü` {~ÝXþ h¡ Ed§ U, PQ na EH$ {~ÝXþ
RS and U is the point at PQ dividing
it internally in the ratio 1 : 3, then h¡ Omo Bgo 1 : 3 Ho$ AZwnmV _| AÝV{d©^ŠV H$aVm
distance between T and U is h¡, V~ T VWm U Ho$ ~rM H$s Xÿar h¡
(A) 32 (B) 2 8 (A) 32 (B) 2 8

(C) 16 2 (D) None of these (C) 16 2 (D) BZ_| go H$moB© Zht

-36- Set-A

123. If O(0, 0), A(3, 0) and B(0, 4) are 123. `{X O(0, 0), A(3, 0) VWm B(0, 4) ∆OAB
vertices of ∆OAB and if G is the
Ho$ erf© h¢ VWm `{X G Ho$ÝÐH$ h¡ Ed§ I AÝV: Ho$ÝÐ
centroid and I is the incentre of ∆OAB,
then area of ∆OIG = ? h¡ ∆OAB H$m, V~ ∆OIG H$m joÌ\$b = ?

(A) 1 (A) 1

1 1
(B) 4 (B) 4

1 1
(C) 3 (C) 3

(D) O, I, G are collinear (D) O, I, G EH$ a¡{IH$ h¢

124. In the following figure, ABCD and EBC 124. {ZåZ{b{IV {MÌ _|, ABCD VWm EBC H«$_e:
are respectively rectangle and triangle. Am`V VWm {Ì^wO h¢ & `{X AD = 10 go_r,
If AD = 10 cm, AB = 5 cm and E is any
AB = 5 go_r VWm E H$moB© {~ÝXþ h¡ AD na, V~
area ABCD
point on AD, then =? ABCD H$m joÌ\ b
area EBC =?
EBC H$m joÌ\ b
A E D A E D

B C B C

1 1
(A) (A)
2 2
1 1
(B) (B)
4 4
(C) 2 (C) 2
(D) Can’t be determined (D) {ZYm©[aV Zht {H$`m Om gH$Vm h¡

-37- Set-A

125. Cumulative frequency of class 30 – 40 125. {ZåZ{b{IV Am`V {MÌ _| dJ© 30 – 40 H$s g§M`r
in the following histogram is ~maå~maVm h¡

50 50

40 40
Frequency

Amd¥{Îm
30 30

20 20

10 10

10 20 30 40 50 60 Class 10 20 30 40 50 60
dJ©
(A) 50 (B) 80 (A) 50 (B) 80
(C) 140 (D) None of these (C) 140 (D) BZ_| go H$moB© Zht

126. Which of the following is not correct ? 126. {ZåZ{b{IV _| go H$m¡Z-gm ghr Zht h¡ ?
(A) 2 mean = 3 median – mode (A) 2 _mÜ` = 3 _mpÜ`H$m – ~hþbH$
(B) Median or mode is not affected by (B) àojUm| Ûmam _mpÜ`H$m `m _mÜ` à^m{dV Zht
observations hmoVm h¡
(C) 2 median = mode + 3(mean – (C) 2 _mpÜ`H$m = ~hþbH$ + 3(_mÜ` –
mode) ~hwbH$)
(D) Median = second quartile (D) _mpÜ`H$m = {ÛVr` MVwW©H$

127. Two points A and B are taken at 127. 60 go_r. bå~r EH$ gab aoIm PQ na Xmo {~ÝXþ
random on a straight line PQ of length A VWm B `mÑpÀN>H$ ê$n go {bE OmVo h¢ & V~
60 cm. Then the probability of the
distance AB exceeding given length
{X`o J`o bå~mB© 15 go_r. go Xÿar AB Ho$ ~‹T>Zo H$s
15 cm is àm{`H$Vm h¡
1 9 1 9
(A) (B) (A) (B)
16 16 16 16
3 3
(C) (D) None of these (C) (D) BZ_| go H$moB© Zht
4 4

-38- Set-A

128. If 6 cubes each of volume 64 cm3 are 128. `{X 6 KZm| H$mo àË`oH$ 64 go_r3 Am`VZ dmbo,
arranged as in the following figure, {ZåZ{b{IV {MÌ H$s Vah ì`dpñWV {H$`m J`m h¡, V~
then the outer surface area of the
cuboid so formed is Bg àH$ma ~Zo KZm^ Ho$ ~mhar gVhm| H$m joÌ\$b h¡


(A) 320 cm2 (B) 384 cm2 (A) 320 go_r2 (B) 384 go_r2
(C) 256 cm2 (D) none of these (C) 256 go_r2 (D) BZ_| go H$moB© Zht

129. In the following right circular cylinder of 129. 4 go_r {ÌÁ`m Ed§ 8 go_r D±$MmB© dmbo {ZåZ{b{IV
radius 4 cm and height 8 cm, a prism bå~d¥Îmr` ~obZ _|, EH$ {àÁ_ ABEFDC aIm
ABEFDC is placed, then volume of this
prism is
J`m h¡, V~ {àÁ_ H$m Am`VZ h¡
A D A D

E E
F F

B C B C

(A) 64π cm3 (B) 128 cm3 (A) 64π go_r3 (B) 128 go_r3
(C) 64 cm3 (D) none of these (C) 64 go_r3 (D) BZ_| go H$moB© Zht

130. If following data is arranged in 130. `{X {ZåZ{b{IV Am±H$‹S>m| H$mo dJ© AmH$ma Ho$ KQ>Vo
descending order of class size, then
what is the class mark of third class
H«$_ _| gOm`m Om`, V~ àmaå^ go Vrgao dJ© H$m
from beginning ? dJ© {M• Š`m hmoJm ?

Class 0 – 10 10 – 20 20 – 30 30 –45 45 – 85 85 – 135 135 – 180 dJ© 0 – 10 10 – 20 20 – 30 30 –45 45 – 85 85 – 135 135 – 180
Frequency 4 6 5 10 20 15 12 Amd¥{Îm 4 6 5 10 20 15 12

(A) 45 (B) 85 (A) 45 (B) 85
(C) 65 (D) none of these (C) 65 (D) BZ_| go H$moB© Zht

-39- Set-A

131. Which of the following is not included in the 131. {ZåZ{b{IV _| go H$m¡Z-gm ~hþdMZ Ho$ {dMma _|
definition of statistics in plural sense ? gm¨p»`H$s H$s n[a^mfm _| gpå_{bV Zht h¡ ?
(A) Analysis of data (A) Am±H$‹S> m| H$m {díbofU
(B) Collection of data in a (B) H«$_~Õ VarHo$ go Am±H$‹S>m| H$m g§H$bZ
systematic manner
(C) Data collection for (C) nyd© {ZYm©[aV CÚoí` Ho$ {bE Am±H$‹S>m| H$m
predetermined purpose g§H$bZ
(D) Collection of data for (D) VwbZm Ho$ {bE Am±H$‹S>m| H$m g§H$bZ
comparison

132. The fourth generation computers are 132. Mm¡Wr nr‹T>r Ho$ H§$ß`yQ>a {H$g na AmYm[aV h¢ ?
based on
(A) Transistor (A) Q´>m§{OñQ>a
(B) Integrated circuit (B) EH$sH¥$V g{H©$Q>
(C) VLSI microprocessor (C) drEbEgAmB© _mBH«$moàmogoga
(D) Diode (D) S>m`moS>

133. If α and β are the roots of the quadratic 133. `{X α Am¡a β {ÛKmV g_rH$aU
equation 2x2 + 2(m + n)x + (m2 + n2) = 0. 2x2 + 2(m + n)x + (m2 + n2) = 0 Ho$ _yb
Then the equation, whose roots are hmo, Vmo g_rH$aU, {OZHo$ _yb (α + β)2 Am¡a
(α + β)2 and (α –β)2 is (α –β)2 hmo, hmoJm
(A) x2 + 4mny + (m2 – n2)2 = 0 (A) x2 + 4mny + (m2 – n2)2 = 0
(B) x2 – 4mnx – (m2 – n2)2 = 0 (B) x2 – 4mnx – (m2 – n2)2 = 0
(C) x2 – (m2 – n2)x + 4mn = 0 (C) x2 – (m2 – n2)x + 4mn = 0
(D) None of these (D) BZ_| go H$moB© Zht

134. Which of the following retains the 134. {gñQ>_ H$mo nmda Am°\$ H$aZo na {ZåZ{b{IV _| go H$m¡Z
information it’s storing when the power AnZo ñQ>mao _| OmZH$mar H$mo ~aH$ama aIVm h¡ ?
to the system is turned off ?
(A) CPU (A) grnr`y
(B) ROM (B) amo_
(C) RAM (C) a¡_
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

-40- Set-A

135. If E denotes the set of books on 135. `{X E gm{hË` H$s nwñVH$m| Ho$ g_wƒ` H$mo gy{MV
literature and F denotes the books on
H$aVm h¡ VWm F _m¡{bH$ {dkmZ Ho$ nwñVH$m| H$mo gy{MV
basic sciences and numbers indicated
in the following venn diagram represent H$aVm h¡ VWm {ZåZ{b{IV doZ AmaoI _| Bª{JV
corresponding number of books, then g§»`mE± g§~§{YV nwñVH$m| H$s g§»`m Xem©Vm hmo, Vmo
the number of books which is not only CZ nwñVH$m| H$s g§»`m Omo Ho$db gm{hË` H$s Zht
of literature is
hmo, h¡
E F E F

40 20 50 40 20 50



(A) 20 (A) 20
(B) 40 (B) 40
(C) 70 (C) 70
(D) None of these (D) BZ_| go H$moB© Zht

136. If the equations k(6x2 + 3) + rx + 2x2 – 1 = 0 136. `{X g_rH$aU k(6x2 + 3) + rx + 2x2 – 1 = 0
and 6k (2x2 + 1) + px + 4x2 – 2 = 0 have Am¡a 6k (2x2 + 1) + px + 4x2 – 2 = 0 Ho$ _yb
both the roots common, find the value
of 2r – p.
g_mZ h¡, Vmo 2r – p H$m _mZ kmV H$a|&
(A) 1 (B) 0 (A) 1 (B) 0
(C) 2k (D) none of these (C) 2k (D) BZ_| go H$moB© Zht
137. If one root of the equation x2 – x – k = 0 137. `{X g_rH$aU x2 – x – k = 0 H$m EH$ _yb Xÿgao
be square of the other, then k = _yb H$m dJ© hmo, Vmo k =
(A) 2 ± 3 (B) 3 ± 2 (A) 2 ± 3 (B) 3 ± 2
(C) 2 ± 5 (D) None of these (C) 2 ± 5 (D) BZ_| go H$moB© Zht

138. (1100.11)2 = (?)10 138. (1100.11)2 = (?)10
(A) 12.55 (B) 12.75 (A) 12.55 (B) 12.75
(C) 12.525 (D) None of these (C) 12.525 (D) BZ_| go H$moB© Zht

-41- Set-A

139. (25.6)10 = (?)2 139. (25.6)10 = (?)2
(A) 11001.1001 (A) 11001.1001
(B) 10111. 1001 (B) 10111. 1001
(C) 11001. 0011 (C) 11001. 0011

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

140. If 0° < θ < 180°, then 140. `{X 0° < θ < 180°, Vmo
2 + 2 + 2 + ... + 2(1 + cos θ) 2 + 2 + 2 + ... + 2(1 + cos θ)

is equal to H$m _mZ hmoJm
θ θ θ θ
(A) 2 cos r −1 (B) 2 cos (A) 2 cos r −1 (B) 2 cos
2 r
2 2 2r
θ θ
(C) 2 cos (D) None of these (C) 2 cos r +1 (D) BZ_| go H$moB© Zht
2r +1 2

141. Read the following statements and find 141. {ZåZ{b{IV H$WZ n‹T> m§o Am¡a ghr {dH$ën MwZmo &
which of the option is true.
H$WZ I : 27cos2x 81sin2x H$m Ý`yZV_ _mZ
Statement I : The minimum value of 1
1 h¡ &
27cos2x 81sin2x is . 243
243
Statement II : The minimum value of
H$WZ II : a cosθ + b sinθ H$m Ý`yZV_ _mZ
– a2 + b2 h¡ &
a cosθ + b sinθ is – a2 + b2 .
(A) Statement I is true, statement II (A) H$WZ I gË` h¡, H$WZ II gË` h¡; H$WZ II,
is true; statement II is a correct H$WZ I H$m ghr ñnîQ>rH$aU h¡
explanation for statement I
(B) Statement I is true, statement II is (B) H$WZ I gË` h¡, H$WZ II gË` h¡; H$WZ II,
true; statement II is not a correct H$WZ I H$m ghr ñnîQ>rH$aU Zht h¡
explanation for statement I
(C) Statement I is true, statement II is (C) H$WZ I gË` h¡, H$WZ II AgË` h¡
false
(D) Statement I is false, statement II (D) H$WZ I AgË` h¡, H$WZ II gË` h¡
is true

-42- Set-A

142. The value of the expression
(243)0.13 × (243)0.07
(243)0.13 × (243)0.07 142. ì`§OH$ H$m
is (7)0.25 × (49)0.075 × (343)0.2
(7)0.25 × (49)0.075 × (343)0.2 _mZ hmoJm

7 3 7 3
(A) (B) (A) (B)
3 7 3 7
3 3
(C) 1 (D) None of these (C) 1 (D) BZ_| go H$moB© Zht
7 7
n
(243) × 32n+1
5
143. The value of the expression 143. ì`§OH$ H$m _mZ hmoJm
n 9n × 3n −1
(243) × 32n+1
5
is
9n × 3n −1
(A) 3 (A) 3
(B) 9 (B) 9
(C) 243 (C) 243
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

1 1 1 1
144. (n − m )
+ (m − n)
=? 144. + =?
(n − m )
1+ a 1+ a 1+ a 1 + a(m − n)
1 1
(A) 0 (B) (A) 0 (B)
2a 2a
(C) 1 (D) am+n (C) 1 (D) am+n

145. In a bag, there are coins of 25p, 10p 145. {H$gr W¡bo _| 25 n¡go, 10 n¡go Am¡a 5 n¡go Ho$ {g¸o$
and 5p in the ratio of 1 : 2 : 3. If there 1 : 2 : 3 Ho$ AZwnmV _| h¡& `{X Hw$b 30 é. hmo,
are Rs. 30 in all, how many 5p coins
are in the bag ?
Vmo W¡bo _| 5 n¡go Ho$ {H$VZo {g¸o$ h¡ ?

(A) 100 (B) 50 (A) 100 (B) 50
(C) 150 (D) None of these (C) 150 (D) BZ_| go H$moB© Zht

-43- Set-A

146. A circular ring of radius 5cm is 146. EH$ d¥ÎmmH$ma [a¨J {OgH$s {ÌÁ`m 5 go_r h¡ H$mo j¡{VO
suspended horizontally from a point
AdñWm _| 6 YmJm| Ûmam Omo CgHo$ n[a{_{V na ~am~a
12 cm vertically above its centre by
6 strings attached to its circumference at AÝVa go ~ÝYo h¡ & EH$ {~ÝXþ go Q>mJ§ m J`m h¡, Omo Cg
equal intervals. The cosine of the angle [a¨J Ho$ Ho$ÝÐ go 12 go_r H$s CÜdm©Ya Xyar na h¡ & Xmo
between two consecutive strings is bJmVma YmJm| Ho$ ~rM Ho$ H$moU H$s H$moÁ`m hmoJr
12 313 12 313
(A) 13 (B) (A) 13 (B)
338 338
1 1
(C) (D) None of these (C)
5
(D) BZ_| go H$moB© Zht
5

147. AB is a vertical pole. The end A is on 147. AB EH$ IS>m (CÜdm©Ya) I§^m h¡ {OgH$m {gam
the level ground. C is the middle point A O_rZ na h¡ & C,$ AB H$m _Ü` {~ÝXþ h¡, P
of AB, P is a another point on the level
O_rZ na Xygam {~ÝXþ h¡ & BC , {~ÝXþ P na β
ground. The portion BC subtends an
angle β at P. If AP = nAB, then tanβ H$moU ~ZmVm h¡ & `{X AP = nAB, Vmo tanβ
is equal to H$m _mZ hmoJm
n n
(A) n + 1 (A) n + 1
n n
(B) 2 (B) 2
n +1 n +1
n n
(C) 2 (C) 2
2n + 1 2n + 1
(D) none of the above (D) CnamoŠV _| go H$moB© Zht

148. The value of the expression  x y  y z  z x
 x y  y z  z x  y − x   z − y   x − z 
 y − x   z − y   x − z  148. ì`§OH$
is  1 1 1 1 1 1
 1  2 − 2  2
− 2 
− 2
1 1 1 1 1 x y y z z 2
x 
 2 − 2   2 − 2   2 − 2 
x y y z z x H$m _mZ hmoJm

(A) x2y2z2 (B) – x2y2z2 (A) x2y2z2 (B) – x2y2z2
(C) xyz (D) 1 (C) xyz (D) 1

-44- Set-A

149. In the following figure O is centre and 149. {ZåZ{b{IV {MÌ _| O Ho$ÝÐ h¡ VWm OB = 2 go_r
OB = 2 cm is radius of circle APBQA. 
APB 1
 1
APB d¥Îm APBQA H$s {ÌÁ`m h¡ & `{X  = ,
If = , then the sectorial area AQB 5
 5
AQB V~ {ÌÁ`-IÊS>r` joÌ\$b OAPBO = ?
OAPBO = ?
Q Q

O O

A B A B
P P

2π π 2π π
(A) cm2 (B) cm2 (A)
3
go_r2 (B)
4
go_r2
3 4
4 4
(C) cm2 (D) none of these (C) go_r2 (D) BZ_| go H$moB© Zht
5 5

150. If four cubes each of edges 2 cm 150. `{X Mma KZ àË`oH$ 2 go_r ^wOmdmbr H$mo
arranged as in the following figure, {ZåZ{b{IV {MÌ H$s Vah ì`dpñWV {H$`m J`m h¡,
then area of the outer six walls of this V~ Bg {MÌ Ho$ N>: ~mhar Xrdmam| H$m joÌ\$b h¡
figure is



(A) 64 cm2 (B) 48 cm2 (A) 64 go_r2 (B) 48 go_r2
(C) 132 cm2 (D) none of these (C) 132 go_r2 (D) BZ_| go H$moB© Zht

-45- Set-A

-46- Set-A

-47- 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$m _| 150 àíZ h¡ & àË`oH$ àíZ 1 A§H$ H$m h¡ & g^r àíZ hb H$aZm A{Zdm`© h¡ &
2. àíZm| Ho$ CÎma Xr JB© OMR CÎmaerQ> (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$m à`moJ d{O©V h¡ Ÿ&
5. OMR CÎmaerQ> (Am§ga erQ>) H$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 150 questions. Each question carries 1 mark. All
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.

-48- Set-A