UPSEE 2017 Paper 1 PCM Code AD

 PAPER-1 PCM àíZnwpñVH$m H«$‘m§H$ àíZnwpñVH$m H$moS>

AD
Question Booklet Sr. No.
AZwH«$‘m§H$ / Roll No.

Q. Booklet Code

CÎma-erQ> H«$‘m§H$ / OMR Answer Sheet No.

KmofUm : / Declaration :
‘¢Zo n¥îR> g§»¶m 1 na {X¶o J¶o {ZX}em| H$mo n‹T>H$a g‘P {b¶m h¡& narjm Ho$ÝÐmܶj H$s ‘moha
I have read and understood the instructions given on page No. 1 Seal of Superintendent of Examination Centre

narjmWu H$m hñVmja /Signature of Candidate
(AmdoXZ nÌ Ho$ AwZgma /as signed in application) H$j {ZarjH$ Ho$ hñVmja /Signature of the Invigilator

narjmWu H$m Zm‘/
Name of Candidate :

narjmWu H$mo {X¶o n¡amJ«m’$ H$s ZH$b ñd¶§ H$s hñV{b{n ‘| ZrMo {X¶o J¶o [a³V ñWmZ na ZH$b (H$m°nr) H$aZr h¡&
""Amn ghr ì¶dgm¶ ‘| h¢, ¶h Amn V^r OmZ|Jo O~ : Amn H$m‘ na OmZo Ho$ {bE qM{VV h¢, Amn {Z˶ AnZm H$m‘ g~go AÀN>m H$aZm MmhVo h¢, Am¡a Amn AnZo H$m¶© Ho$
‘hËd H$mo g‘PVo h¢&'' AWdm / OR
To be copied by the candidate in your own handwriting in the space given below for this purpose is compulsory.
‘‘You will know you are in the right profession when : you wake anxious to go to work, you want to do your best daily, and you know your work is
important.”

* Bg n¥îR> H$m D$nar AmYm ^mJ H$mQ>Zo Ho$ ~mX drjH$ Bgo N>mÌ H$s OMR sheet Ho$ gmW gwa{jV aIo&
* After cutting half upper part of this page, invigilator preserve it along with student’s OMR sheet.

 
nwpñVH$m ‘| ‘wIn¥îR> g{hV n¥îR>m| H$s g§»¶m g‘¶ 3 K§Q>o A§H$ / Marks nwpñVH$m ‘| àíZm| H$s g§»¶m
No. of Pages in Booklet including title
36 Time 3 Hours 600 No. of Questions in Booklet
150

PAPER-1 PCM àíZnwpñVH$m H«$‘m§H$/ Question Booklet Sr. No.

AZwH«$‘m§H$ / Roll No.
H$j {ZarjH$ Ho$ hñVmja /Signature of the Invigilator
àíZnwpñVH$m H$moS>
narjmWu H$m Zm‘/
Name of Candidate : AD
Q. Booklet Code
narjm{W©¶m| Ho$ {bE {ZX}e /INSTRUCTIONS TO CANDIDATE
Aä¶{W©¶m| hoVw Amdí¶H$ {ZX}e : Instructions for the Candidate :
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Zrbo ¶m H$mbo ~mb ßdmB§Q> noZ H$m hr Cn¶moJ H$a|& the bubbles in the OMR Answer Sheet.
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your Name, Roll Number ( In figures), OMR Answer-sheet Number in
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‘yë¶m§H$Z Zhr hmo gHo$Jm VWm Eogo Aä¶Wu A¶mo½¶ Kmo{fV hmo Om¶|Jo& evaluated leading the disqualification of the candidate.
3. à˶oH$ àíZ Mma A§H$m| H$m h¡& {Og àíZ H$m CÎma Zht {X¶m J¶m h¡, Cg na H$moB© 3. Each question carries FOUR marks. No marks will be awarded for
A§H$ Zht {X¶m Om¶oJm& JbV CÎma na A§H$ Zht H$mQ>m OmEJm& unattempted questions. There is no negative marking on wrong answer.
4. Each multiple choice questions has only one correct answer and marks
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H$m à¶moJ d{O©V h¡& slide rule etc. is strictly prohibited.
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Om¶oJr& examination time period only.
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ghm¶Vm bo gH$Vo/gH$Vr h¢, nm¶r Om¶oJr, Vmo Cgo A¶mo½¶ Kmo{fV H$a {X¶m Om or written material from which he/she might derive assistance, he/she
gH$Vm h¡& Bgr àH$ma, ¶{X H$moB© Aä¶Wu {H$gr ^r àH$ma H$s ghm¶Vm {H$gr ^r is liable to be treated at disqualified. Similarly, if a candidate is found
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source, he/she is liable to be disqualified.
Vmo Cgo ^r A¶mo½¶ Kmo{fV {H$¶m Om gH$Vm h¡&
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‘mZm Om¶oJm& final to resolve any ambiguity.
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Paper H$s grb Ho$db nona ewé hmoZo Ho$ g‘¶ na hr Imobm Om¶oJm& paper but seal of paper must be opened only at the start of paper.

PAPER-1
Physics : Q. 1 to Q. 50
Chemistry : Q. 51 to Q. 100
Mathematics : Q. 101 to Q. 150

PHYSICS / ^m¡{VH$emó

001. In Young’s double slit experiment, the path 001. `§J Ho$ {Û{N>Ð à`moJ ‘o§ nX} na EH$ {~ÝXþ na ì`{VH$aU
difference between two interfering waves at a point H$aZo dmbr Xmo Va§Jm| Ho$ ‘Ü` nWm§Va Va§JX¡Ü`© H$m 13.5
on screen is 13.5 times the wavelength. The point is: JwUm h¡ Vmo {~ÝXþ hmoJm :
(A) central bright (A) Ho$ÝÐr` Xrá
(B) dark (B) AXrá
(C) bright but not central bright
(C) Xrá naÝVw Ho$ÝÐr` Xrá Zht
(D) neither bright nor dark
(D) Z Vmo Xrá Z hr AXrá
002. A ball having velocity v towards right and having 002. EH$ J|X {OgH$m X{jUmdV© H$moUr` doJ h¡, `h Xm`t Va’$
angular velocity clockwise approaches the wall. It doJ v go EH$ Xrdma H$s Va’$ J{V H$a ahr h¡& Xrdma go
collides elastically with wall and moves towards `h àË`mñW Q>¸$a H$aVr h¡ VWm `h ~m`t Va’$ bm¡Q>Vr h¡ &
left. Ground and wall are frictionless . Select the O‘rZ d Xrdma Kf©Ua{hV h¡ & Xrdma Ho$ gmW Q>¸$a Ho$ ~mX
correct statement about angular velocity of the ball J|X Ho$ H$moUr` doJ Ho$ ~mao ‘| ghr H$WZ M`Z H$amo -
after collision.

(A) H$moUr` Mmb KQ>Vr h¡
(A) Angular speed decreases
(B) X{jUmdV© hmoJr
(B) It will be clockwise
(C) dm‘mdV© hmoJr
(C) It will be anticlockwise
(D) It becomes zero (D) `h eyÝ` hmo OmVr h¡

1-AD ] [2] [ Contd...

003. Which of the following particle will describe the 003. {ZåZ H$Um| ‘| go H$m¡Zgm H$U g~go N>moQ>r {ÌÁ`m H$m d¥Îm
smallest circle when projected with same velocity ~ZmEJm O~ `h Mwå~H$s` joÌ Ho$ bå~dV g‘mZ doJ go
perpendicular to magnetic field ? àjo{nV {H$`m OmVm h¡ ?
(A) Li+ (B) electron (A) Li+ (B) BboŠQ´moZ
(C) proton (D) He+
(C) àmoQ>moZ (D) He+
004. A loop PQR carries a current of 2A as shown. A 004. {MÌmZwgma EH$ byn PQR ‘| Ymam 2A h¡ & EH$ g‘mZ
uniform magnetic field (B=2T) is parallel to plane Mwå~H$s` joÌ (B=2T) byn Ho$ Vb Ho$ g‘mÝVa h¡ & byn
of the loop. The magnetic torque on the loop is : na Mwå~H$s` AmKyU© h¡ :

(A) zero (B) 4 Nm (A) eyݶ (B) 4 Nm
(C) 16 Nm (D) 8 Nm (C) 16 Nm (D) 8 Nm

005. The sides of a rectangle are 7.01 m and 12 m. Taking 005. EH$ Am`V H$s ^wOmE± 7.01 m VWm 12 m h¡ & gmW©H$ A§H$mo
the significant figures into account , the area of the H$mo boVo hþE Am`V H$m joÌ’$b hmoJm :
rectangle is :
(A) 84.12 m2 (B) 84 m2
(A) 84.12 m2 (B) 84 m2
(C) 84.1 m2 (D) 84.00 m2 (C) 84.1 m2 (D) 84.00 m2

006. In steady state, charge on 3nF capacitor is : 006. ñWm`r AdñWm ‘o 3nF g§Ym[aÌ na Amdoe hmoJm:

(A) 18 nC (B) 54 nC (A) 18 nC (B) 54 nC
(C) 36 nC (D) 27 nC (C) 36 nC (D) 27 nC

007. Consider one dimensional motion of a particle. 007. EH$ H$U Ho$ {bE EH$ {d‘r` J{V br{OE & `hm± doJ v VWm
Velocity v versus time t graph is shown. Which graph g‘` t Ho$ ‘Ü` J«m’$ Xem©`m J`m h¡& H$m¡Zgm J«m’$ g‘` t
is most appropriate for displacement x versus time t ? Ho$ gmnoj {dñWmnZ x H$mo g~go Cn`wº$ ê$n go Xem©Vm h¡ ?

(A) (B) (A) (B)

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

. 1-AD ] [3] [ P.T.O.

008. An object of mass 26kg floats in air and it is in 008. Ðì`‘mZ 26 kg H$s dñVw hdm ‘| V¡aVr hþB© gmå`dñWm
equilibrium state. Air density is 1.3 kg/m3 . The pñW{V ‘| h¡ & hdm H$m KZËd 1.3 kg/m3 h¡ & dñVw H$m
volume of the object is : Am`VZ hmoJm :
(A) 13 m3 (B) 26 m3 (A) 13 m3 (B) 26 m3
(C) 10 m3 (D) 20 m3 (C) 10 m3 (D) 20 m3

009. In the given circuit cell E has internal resistance of 009. {XE JE n[anW ‘| gob E H$m Am§V[aH$ à{VamoY r = 2X
r = 2X .What is the value of resistance R so that h¡& à{VamoY R H$m ‘mZ Š`m hmoZm Mm{hE Vm{H$ à{VamoY R
power delivered to resistor R is maximum ? H$mo àXmZ H$s JB© e{º$ A{YH$V‘ hmoJr ?

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

010. Two cylindrical rods A and B have same resistivities 010. Xmo ~obZmH$ma N>‹S>o§ A VWm B H$s à{VamoYH$Vm g‘mZ h¡ VWm
and same lengths . Diameter of rod A is twice the bå~mB© ^r g‘mZ h¡ & N>‹S> A H$m ì`mg N>‹S> B Ho$ ì`mg
diameter of the rod B. Ratio of voltage drop across H$m XþJwZm h¡ & N>‹S> A na dmoëQ>Vm H$m N>‹S> B na dmoëQ>Vm
rod A to rod B is : Ho$ gmW AZwnmV Š`m h¡ ?

1 1
(A) 4 (B) (A) 4 (B)
4 4
1 1
(C) (D) 2 (C) (D) 2
2 2
011. Which of the following material is not ferromagnetic 011. {ZåZ ‘| go H$m¡Zgm nXmW© bm¡ôMwå~H$Ëd àH¥${V H$m Zht h¡?
in nature ?
(A) Ni (B) Al (A) Ni (B) Al
(C) Fe (D) Co (C) Fe (D) Co

012. Three small balls of masses 1kg , 2kg and 3kg are 012. Ðì`‘mZ 1kg, 2kg VWm 3kg H$s VrZ N>moQ>r J|Xo EH$ hr
moving in a plane and their velocities are 1  m/s, Vb ‘| doJ H«$‘e… 1  m/s, 2  m/s VWm 3  m/s go {MÌmZwgma
2  m/s and 3 m/s respectively as shown. The total
J{V H$a ahr h¢ & {XE JE jU na {~ÝXþ P Ho$ gmnoj VrZmo
angular momentum of the system of the three balls
about point P at given instant of time is :
J|Xm| Ho$ {ZH$m` H$m Hw$b H$moUr` g§doJ h¡ :

(A) 36 kgm2s–1 (A) 36 kgm2s–1
(B) 7 kgm2s–1 (B) 7 kgm2s–1
(C) 8 kgm2s–1 (C) 8 kgm2s–1
(D) 9 kgm2s–1 (D) 9 kgm2s–1

1-AD ] [4] [ Contd...

013. Three identical resistors each of resistance R are 013. VrZ EH$ g‘mZ à{VamoY {OZ‘| àË`oH$ H$m à{VamoY R h¡ H$mo
m connected to an ideal cell of voltage V as shown . V dmoëQ> Ho$ AmXe© gob go {MÌmZwgma Omo‹S>m OmVm h¡ Vmo BZ
Total power dissipated in all three resistors is :
VrZ à{VamoYm| ‘o§ Hw$b ì`{`V e{º$ hmoJr:

R
V2 2V 2 V2 2V 2
(A) (B) (A) (B)
3R 3R 3R 3R
3V 2 3V 2 3V 2 3V 2
(C) (D) (C) (D)
2R R 2R R
014. For given logic diagram , output F=1, then inputs are: 014. {XE JE VH©$ n[anW ‘| {ZJ©V F=1, V~ {Zdoer h¡:

(A) A = 1, B = 1, C = 1 (A) A = 1, B = 1, C = 1
(B) A = 0, B = 0, C = 1 (B) A = 0, B = 0, C = 1
(C) A = 0, B = 0, C = 0 (C) A = 0, B = 0, C = 0
m
(D) A = 0, B = 1, C = 0 (D) A = 0, B = 1, C = 0
015. Consider two polaroids A and B as shown. 015. Xem©E AZwgma Xmo nmoboamoBS> A VWm B na {dMma H$s{OE&
Unpolarized light is incident on polaroid A. Now
both the polaroids are rotated simultaneously by
AY«w{dV àH$me nmoboamoBS> A na Amn{VV hmoVm h¡ & A~
180° in same sense of rotation such that at every XmoZm| nmoboamoBS> H$mo EH$ gmW 180° KyU©Z EH$ hr {Xem ‘|
instant, their pass(transmission) axes always Bg àH$ma go Ky{U©V {H$`m OmVm h¡ {H$ àË`oH$ jU XmoZm| H$s
remain parallel to each other. During the rotation, nmaJ‘Z Aj h‘oem EH$ Xygao Ho$ g‘mÝVa ahVo h¢& KyU©Z Ho$
intensity of transmitted light through polaroid B : Xm¡amZ nmoboamoBS> B go nmaJ{‘V àH$me H$s Vrd«Vm :

?
(A) remains same (A) g‘mZ ahVr h¡
(B) decreases continuously (B) bJmVma KQ>Vr h¡
(C) increases continuously (C) bJmVma ~‹T>Vr h¡
(D) first increases then decreases (D) nhbo KQ>Vr h¡ {’$a ~‹T>Vr h¡
r
016. Activity of a radioactive substance becomes from 016. EH$ ao{S>`mog{H«$` nXmW© H$s g{H«$`Vm 8000Bq go 1000Bq
8000Bq to 1000Bq in 12 Days. What is the half life VH$ 12 {XZm| ‘| hmo OmVr h¡ & ao{S>`mog{H«$` nXmW© H$s
of the radioactive substance ? AÕ©Am`w Š`m h¡?
(A) 2 days (B) 3 days (A) 2 {XZ (B) 3 {XZ
(C) 4 days (D) 6 days (C) 4 {XZ (D) 6 {XZ

017. The energy levels of a hypothetical one electron 017. EH$ H$mën{ZH$ EH$ BboŠQ´mZ na‘mUw {ZH$m` Ho$ D$Om© ñVa
16 16
atom system are given by E n = - 2 eV , where E n = - 2 eV h¡ Ohm± (n = 1, 2, 3,….) h¡ & O~ `h
n n
n = 1, 2, 3,….The wavelength of emitted photon àW‘ CÎmo{OV AdñWm go ‘yb ñVa ‘o§ g§H«$‘U H$aVm h¡ V~
corresponding to transition from first excited level CËg{O©V ’$moQ>moZ H$s Va§JX¡Ü`© bJ^J hmoJr :
to ground level is about :
(A) 3650 A° (B) 690 A° (A) 3650 A° (B) 690 A°
(C) 1035 A° (D) 1220 A° (C) 1035 A° (D) 1220 A°

1-AD ] [5] [ P.T.O.

018. What is the voltage across an ideal PN junction 018. {MÌmZwgma n[anW ‘| {XE JE AmXe© PN g§{Y S>m`moS> na
diode for shown circuit ?
dmoëQ>Vm Š`m hmoJr ?

(A) 2V (B) 0V (A) 2V (B) 0V
(C) 0.7V (D) 1V (C) 0.7V (D) 1V

019. Power emitted by a black body at temperature 50°C 019. Vmn 50°C na EH$ H¥$îUrH$m Ûmam CËg{O©V e{º$ P h¡ &
is P. Now temperature is doubled i.e. temperature of A~ H¥$îUrH$m Vmn XþJwZm AWm©V 100°C H$a {X`m OmVm h¡
black body becomes 100°C. Now power emitted is : Vmo A~ CËg{O©V e{º$ hmoJr:
(A) P (A) P
(B) 16 P (B) 16 P
(C) greater than P but less than 16P (C) P go A{YH$ naÝVw 16 P go H$‘
(D) greater than 16P (D) 16 P go A{YH$

020. An experimenter needs to heat a small sample to 020. EH$ à`moJ{dX EH$ N>moQ>o à{VXe© (sample) H$mo 900K Vmn
temperature 900K, but the only available large VH$ J‘© H$aZm MmhVm h¡ naÝVw ~‹S>r dñVw (object) H$m
object has maximum temperature of 600K. Could CnbãY A{YH$V‘ Vmn Ho$db 600K h¡ & Š`m à`moJ{dX
the experimenter heat the sample to 900K by using
a large lens to concentrate the radiation from the
Ûmam {MÌmZwgma ~‹S>r dñVw go {d{H$aU H$mo EH$ ~‹S>o b|g
large object onto the sample as shown below ? Ûmam à{VXe© na H|${ÐV H$a à{VXe© H$m 900K Vmn {H$`m
Om gH$Vm h¡ ?

(A) It is not possible
(B) Yes, if the volume of the large object is at least (A) `h g§^d Zht h¡ &
1.5 times the volume of the sample. (B) hm±, `{X ~‹S>r dñVw H$m Am`VZ à{VXe© Ho$ Am`VZ
(C) Yes, if the front area of the large object is H$m 1.5 JwUm H$a {X`m OmE &
at least 1.5 times the area of the front of the (C) 
hm±, `{X ~‹S>r dñVw H$m gå‘wI joÌ’$b à{VXe© Ho$ gå‘wI
sample. joÌ’$b H$m H$‘ go H$‘ 1.5 JwUm H$a {X`m OmE&
(D) Yes, if the sample is placed at the focal point
(D) hm±, `{X à{VXe© H$mo b|g Ho$ ’$moH$g {~ÝXþ na aIm OmE&
of the lens.

021. Consider a small electric dipole with magnitude of 021. EH$ N>moQ>o {dÚwV {ÛY«wd {OgH$m {ÛY«wd AmKyU© H$m n[a‘mU
dipole moment p which is placed far away from point p h¡ BgH$mo {~ÝXþ A go H$m’$s Xya {MÌmZwgma aIm OmVm h¡ &
A as shown. The electric potential at the point A is : {~ÝXþ A na {dÚwV {d^d h¡ :

kp kp
(A) (B) exactly zero (A) (B) nyU©V`m eyÝ`
r r
kp -k p kp -k p
(C) (D) (C) (D)
r2 r2 r2 r2

1-AD ] [6] [ Contd...

022. A conducting loop (as shown) has total resistance 022. {MÌmZwgma EH$ MmbH$ byn H$m Hw$b à{VamoY R h¡ & byn
R. A uniform magnetic field B = γt is applied
Ho$ Vb Ho$ bå~dV EH$g‘mZ Mwå~H$s` joÌ B = γ t H$mo
perpendicular to plane of the loop where γ is a
constant and t is time. The induced current flowing
Amamo{nV {H$`m OmVm h¡ Ohm± γ AMa h¡ VWm t g‘` h¡&
through loop is : byn go àdm{hV ào[aV Ymam hmoJr:

(b 2 - a 2) ct (b 2 + a 2) c (b 2 - a 2) ct (b 2 + a 2) c
(A) (B) (A) (B)
R R R R
(b 2 + a 2) ct (b 2 - a 2) c (b 2 + a 2) ct (b 2 - a 2) c
(C) (D) (C) (D)
R R R R

023. A uniform disc of mass M and radius R is hinged at 023. Ðì`‘mZ M d {ÌÁ`m R H$s EH$g‘mZ MH$Vr BgHo$ Ho$ÝÐ C na
its centre C. A force F is applied on the disc as shown. H$sb{H$V h¡ & EH$ ~b F H$mo MH$Vr na {MÌmZwgma Amamo{nV
At this instant, angular acceleration of the disc is : {H$`m OmVm h¡ & Bg g‘` MH$Vr H$m H$moUr` ËdaU h¡:

2 F F 2 F F
(A) (B) (A) (B)
3 MR 2MR 3 MR 2MR
F F F F
(C) 3 (D) (C) 3 (D)
MR MR MR MR

024. The velocity of a particle is zero at time t = 2 , then 024. g‘` t = 2 na H$U H$m doJ eyÝ` h¡ Vmo
(A) velocity must be zero for t > 2 (A) t > 2 Ho$ {bE doJ eyÝ` hr hmoJm &
(B) acceleration must be zero at t = 2
(B) t = 2 na ËdaU eyÝ` hr hmoJm &
(C) displacement must be zero in the interval
t = 0 to t = 2. (C) t = 0 go t = 2 A§Vamb ‘| {dñWmnZ eyÝ` hr hmoJm &
(D) acceleration may be zero at t = 2 (D) t = 2 na ËdaU eyÝ` hmo gH$Vm h¡ &

025. A ball moving in xy plane, has velocity (4 it - 4tj ) m/s 025. EH$ J|X xy Vb ‘| J{V H$aVr h¡ d O‘rZ go Q>¸$a go
just before the collision with ground. Coefficient of
1 R>rH$ nyd© doJ (4 it - 4tj ) m/s h¡& Q>¸$a Ho$ {bE àË`mdñWZ
restitution for collision is e = . What will be velocity
2
JwUm§H$ e = 12 h¡& O‘rZ go Q>¸$a Ho$ R>rH$ nümV J|X H$m
of the ball just after the collision with ground?
doJ Š`m hmoJm ?

(A) (2 it + 4tj) m/s
(A) (2 it + 4tj) m/s
(B) (4 it + 4tj) m/s
(B) (4 it + 4tj) m/s
(C) (2 it + 2tj) m/s
(C) (2 it + 2tj) m/s
(D) (4 it + 2tj) m/s
(D) (4 it + 2tj) m/s

1-AD ] [7] [ P.T.O.

026. A light ray moving in medium- I (of refractive 026. EH$ àH$me {H$aU AndV©Zm§H$ n1 Ho$ ‘mÜ`‘-I
index n1) is incident on interface of two media ‘| J{V H$aVr hþB© XmoZm| ‘mÜ`‘m| H$s A§Vg©Vh
and it is totally internally reflected at the na Amn{VV hmoVr h¡ VWm A§Vg©Vh na nyU©V`m
interface. Now refractive index n2 of medium-II Am§V[aH$ namd{V©V hmoVr h¡ & A~ ‘mÜ`‘-II H$m
is decreased, then AndV©Zm§H$ n2 H$m ‘mZ KQ>m`m OmVm h¡ Vmo -

(A)  ray will be totally transmitted into medium-II (A) {H$aU ‘mÜ`‘-II ‘| nyUV© `m nmaJ{‘V Ho$db V^r
only if angle of incidence is increased. hmoJr O~ AmnVZ H$moU ~‹T>m`m OmVm h¡&
(B)  ray will be totally transmitted in medium-II. (B) {H$aU nyU©V`m ‘mÜ`‘-II ‘| nmaJ{‘V hmoVr h¡&
(C)  ray will move completely parallel to the (C) {H$aU A§Vg©Vh Ho$ nyU©V`m g‘mÝVa Om`oJr &
interface .
(D) {H$aU A~ ^r A§Vg©Vh na nyU©V`m Am§V[aH$
(D) ray will be still totally internally reflected at namd{V©V hmoJr &
interface.

027. A light beam consists of two types of photons. 027. EH$ àH$me {H$aU ‘| Xmo àH$ma Ho$ ’$moQ>moZ h¡& EH$
In one type each photon has energy 2eV and in Vah ‘| àË`oH$ ’$moQ>moZ H$s D$Om© 2eV h¡ VWm Xygao
other type each photon has energy 3eV. The light Vah ‘| àË`oH$ ’$moQ>moZ H$s D$Om© 3eV h¡& àH$me
beam is incident on a photoelectric material of {H$aU EH$ àH$me {dÚwVnXmW© {OgH$m H$m`©’$bZ
work function 1eV. The maximum kinetic energy 1eV h¡ Cg na {JaVr h¡& CËg{O©V ’$moQ>moBboŠQ´moZ
of emitted photoelectron is : H$s A{YH$V‘ J{VO D$Om© h¡ :
(A) 4eV (B) leV (A) 4eV (B) leV
(C) 2eV (D) 3eV (C) 2eV (D) 3eV

028. A light beam parallel to axis is incident on 028. Mma CÎmb b|gm| A, B, C VWm D Ho$ {ZH$m` na
the system of four convex lenses A, B, C and Aj Ho$ g‘mÝVa àH$me {H$aU nw§O Amn{VV hmoVm
D. Focal lengths of A, B, C and D are 30cm,
h¡& boÝg A, B, C VWm D H$s ’$moH$g bå~mB©`m§
10cm, 30cm and 10cm respectively as shown.
H«$‘e… 30cm, 10cm, 30cm VWm 10cm h¡§& `hm±
Here fixed distance BC=20cm. What should be
pñWa Xyar BC = 20cm h¡& b|g A VWm b|g D
the distance between the lens A and lens D so
Ho$ ‘Ü` Xyar {H$VZr hmoZr Mm{hE Vm{H$ AndV©Z Ho$
that after refractions, rays will be parallel to
axis in regions I, III and V?
nümV {H$aUo§ (region) joÌ I, III VWm V ‘| Aj
Ho$ g‘mÝVa hmo OmE±:

(A) 80 cm (B) 20 cm (A) 80 cm (B) 20 cm
(C) 40 cm (D) 100 cm (C) 40 cm (D) 100 cm

1-AD ] [8] [ Contd...

029. A long silver tea spoon is placed in a cup filled 029. EH$ bå~o Mm§Xr Ho$ Mm` Må‘M H$mo J‘© Mm` go ^ao
with hot tea. After some time, the exposed end H$n ‘o§ aIm OmVm h¡& Hw$N> g‘` ~mX Må‘M H$m
(the end which is not dipped in tea) of the
spoon becomes hot even without a direct contact
Iwbm {gam (Omo Mm` ‘o Zht Sy>~m h¡) J‘© hmo OmVm
with the tea. This phenomenon can be explained h¡ `Ú{n `h Mm` Ho$ grYo g§nH©$ ‘o Zht Wm& `h
mainly by: à^md ‘w»` ê$n go {ZåZ go g‘Pm Om gH$Vm h¡:
(A) radiation (A) {d{H$aU
(B) thermal expansion (B) D$î‘r` àgma
(C) conduction (C) MmbZ
(D) reflection (D) namdV©Z
030. Figure shows a nonconducting semicircular rod 030. {MÌ ‘| xy Vb ‘| EH$ AMmbH$ AY© d¥ÎmmH$ma N>S‹ >
in xy plane. Top half (quarter circle) has uniform Xem©`r JB© h¡& D$nar AmYo ^mJ (MVwWmªe d¥Îm) ‘o§
linear charge density - m whereas remaining half EH$ g‘mZ aoIr` Amdoe KZËd - m h¡ O~{H$ eof
has uniform linear charge density + m . What is the AmYo ^mJ ‘| EH$ g‘mZ aoIr` Amdoe KZËd + m   h&¡
direction of the net electric field at point P? {~ÝXþ P na n[aUm‘r {dÚwV joÌ H$s {Xem Š`m hmoJr?
-m -m

+m +m

(A) along the bisector of x axis and y axis. (A) x Aj d y Aj Ho$ AÕ©^mOH$ Ho$ AZw{Xe
(B) along +x axis (B) +x Aj Ho$ AZw{Xe
(C) along +y axis (C) +y Aj Ho$ AZw{Xe
(D) electric field is zero at point P, so direction (D)  {~ÝXþ P na {dÚwV joÌ eyÝ` h¡ AV… {Xem kmV Zht
cannot be determined.
H$s Om gH$Vr h¡
031. A bead of mass m can slide without friction on a 031. EH$ O‹S>dV d¥ÎmmH$ma j¡{VO db` {OgH$s {ÌÁ`m
fixed circular horizontal ring of radius 3R having 3R h¡ d Ho$ÝÐ C na h¡, Cg na m Ðì`‘mZ H$m
centre at the point C. The bead is attached to one EH$ ‘ZH$m {~Zm Kf©U Ho$ {’$gb gH$Vm h¡& ‘ZHo$
of the ends of spring of spring constant k. Natural H$mo EH$ pñà§J Ho$ EH$ {gao go ~m±Ym OmVm h¡ &
length of spring is R and the other end of the Cg pñà§J H$m pñà§J {Z`Vm§H$ k h¡ VWm pñà§J H$s
spring is fixed at point O as shown in figure. Bead àmH¥${VH$ bå~mB© R h¡ VWm pñà§J H$m Xygam {gam
is released from position A, what will be kinetic {MÌmZwgma {~ÝXþ O na O‹S>dV h¡ & ‘ZHo$ H$mo pñW{V
energy of the bead when it reaches at point B ? A go ‘wº$ {H$`m OmVm h¡ Vmo O~ `h pñW{V B
na nhþ§MVm h¡ V~ ‘ZHo$ H$s J{VO D$Om© hmoJr:

(A) 8kR2 (B) 12 kR 2 (A) 8kR2 (B) 12 kR 2
25 9 25 9
(C) kR 2 (D) kR2 (C) kR 2 (D) kR2
2 2 2 2

1-AD ] [9] [ P.T.O.

032. The total electrostatic energy stored in both the 032. XmoZm| g§Ym[aÌ ‘o g§J«{hV Hw$b pñWa {dÚwV D$Om© h¡ :
capacitors is :

(A) 13.5 nJ (B) 18 nJ (A) 13.5 nJ (B) 18 nJ
(C) 9 nJ (D) 40.5 nJ (C) 9 nJ (D) 40.5 nJ

033. Gravitational force acts on a particle due to fixed 033. EH$ g‘mZ R>mog O‹S>dV Jmobo Ho$ H$maU EH$ H$U na
uniform solid sphere. Neglect other forces. Then
particle :
Jwê$Ëdr¶ ~b bJVm h¡, AÝ` ~b ZJÊ` h¡& V~ `h H$U:
(A) always moves in circular orbit. (A) h‘oem d¥Îmr` J{V H$aoJm&
(B) experiences a force directed along the radial (B) Ho$db {ÌÁ`r` {Xem Ho$ AZw{Xe hr ~b AZw^d H$aoJm&
direction only. (C) h‘oem {ÌÁ`r` {Xem Ho$ bå~dV J{V H$aoJm&
(C) always moves normal to the radial direction
(D) always moves in the radial direction only. (D) h‘oem {ÌÁ`r` {Xem Ho$ AZw{Xe J{V H$aoJm&

034. A block performs simple harmonic motion with 034. EH$ ãbm°H$ gmå`mdñWm {~ÝXþ x =0 Ho$ gmnoj gab Amd¥{V
equilibrium point x = 0. Graph of acceleration of the J{V H$aVm h¡ & ãbm°H$ Ho$ ËdaU H$mo g‘` Ho$ ’$bZ Ho$ ê$n
block as a function of time is shown. Which of the ‘o J«m’$ Xem©`m J`m h¡ & ãbm°H$ Ho$ ~mao ‘o§ H$m¡Zgm H$WZ
following statement is correct about the block?
gË` h¡ ?

(A) speed is minimum at t = 2s. (A) t = 2s na Mmb Ý`yZV‘ h¡&
(B) speed is maximum at t = 3s.
(B) t = 3s na H$U H$s Mmb A{YH$V‘ h¡&
(C)  displacement from equilibrium is maximum
at t = 4s. (C) t = 4s na H$U H$m gmå`mdñWm go {dñWmnZ A{YH$V‘ h¡&
(D) speed is maximum at t = 4s. (D) t = 4s na Mmb A{YH$V‘ h¡ &

035. There are two identical springs each of spring 035. `hm± Xmo EH$g‘mZ pñà§J h¡§ d àË`oH$ H$m pñà§J {Z`Vm§H$ k
constant k. Here springs, pulley and rods are h¡ & `hm± ãbm°H$ H$m Ðì`‘mZ m h¡ VWm pñà§J, nybr VWm
massless and block has mass m. What is the
N>‹S>o§ (rods) Ðì`‘mZhrZ h¡& gmå`mdñWm ‘§o àË`oH$ pñà§J H$m
extension of each spring at equilibrium ?
{dñVma Š`m hmoJm ?

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

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036. Two tuning forks A and B produce 4 beats/sec. 036. A VWm B Xmo ñd[aÌ 4 {dñn§X /goH$ÊS> CËnÞ H$aVo h¢ & B
Forks B and C produce 5 beats/sec. Forks A and C VWm C ñd[aÌ 5 {dñn§X /goH$ÊS> CËnÞ H$aVo h¢ Vmo A VWm
may produce ……. beats/sec. C ñd[aÌ ......... {dñn§X /goH$ÊS> CËnÞ H$a gH$Vo h¢&
(A) 20 (B) 2 (A) 20 (B) 2
(C) 5 (D) 9 (C) 5 (D) 9

037. A 10gm bullet moving directly upward at 1000 m/s 037. EH$ 10gm H$s Jmobr 1000 m/s go grYr D$na J{V H$aVr
strikes and passes through the center of mass of hþE {dam‘ ‘| n‹S>o 10 kg Ðì`‘mZ Ho$ ãbm°H$ go Q>H$amVr h¡
a 10 kg block initially at rest .The bullet emerges VWm CgHo$ Ðì`‘mZ Ho$ÝÐ go JwOaVr h¡& Jmobr grYo D$na
from the block moving directly upward at 400 m/s. H$s Va’$ 400 m/s go ãbm°H$ ‘| go ~mha {ZH$bVr h¡ &O~
What will be velocity of the block just after the Jmobr ãbm°H$ go R>rH$ ~mha {ZH$bVr h¡ Cg jU ãbm°H$ H$m
doJ Š`m hmoJm ?
bullet comes out of it ?

(A) 1.4 m/s (B) 0.6 m/s (A) 1.4 m/s (B) 0.6 m/s
(C) 1 m/s (D) 0.4 m/s (C) 1 m/s (D) 0.4 m/s

038. Two identical balls P and Q are projected with 038. Xmo EH$g‘mZ J|Xo P VWm Q EH$ hr g‘mZ {~ÝXþ O go
same speeds in vertical plane from same point O
CÜdm©Ya Vb ‘| g‘mZ Mmb go jo{VO Ho$ gmW àjonU H$moU
with making projection angles with horizontal 30°
and 60° respectively and they fall directly on plane
H«$‘e… 30° d 60° na àjo{nV H$s OmVr h¡ VWm do grYo
AB at points P' and Q' respectively. Which of the hr Vb A­ B na H«$‘e… {~ÝXþ P' d Q' na {JaVr h¡& Xyar Ho$
following statement is true about distances as given gå~ÝY ‘§o H$m¡Zgm {dH$ën gË` h¡ ?
in options?

(A) AP' ≤ AQ' (A) AP' ≤ AQ'
(B) AP' = AQ' as there are complimentary (B) AP' = AQ' Š`m|{H$ CZHo$ àjonU H$moU nyaH$ H$moU h¡§
projection angles. (C) AP' > AQ'
(C) AP' > AQ' (D) AP' < AQ'
(D) AP' < AQ'

039. A string has a length of 5m between fixed points 039. Xmo pñWa {~ÝXþAmo§ Ho$ ‘Ü` EH$ añgr H$s bå~mB© 5m h¡ VWm
and has fundamental frequency of 20 Hz. What is BgH$s ‘yb^yV Amd¥{V 20 Hz h¡ Vmo {ÛVr` A{Yñda H$s
the frequency of the second overtone ? Amd¥{V Š`m hmoJr ?
(A) 60 Hz (A) 60 Hz
(B) 30 Hz (B) 30 Hz
(C) 40 Hz (C) 40 Hz
(D) 50 Hz (D) 50 Hz

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040. Displacement x versus t2 graph is shown for a 040. EH$ H$U Ho$ {dñWmnZ x H$m t2 Ho$ gmW J«m’$ ~Vm`m J`m
particle. The acceleration of the particle is : h¡& H$U H$m ËdaU h¡ :

(A) zero (B) 2m/s2 (A) eyݶ (B) 2m/s2
(C) 4m/s2 (D) 8m/s2 (C) 4m/s2 (D) 8m/s2

041. For given LR circuit, growth of current as function 041. {XE JE LR n[anW ‘| Ymam H$s d¥{Õ H$mo g‘` t Ho$ ’$bZ
of time t is shown in graph. Which of the following
Ho$ ê$n ‘o Xem©`m J`m h¡& {ZåZ ‘| go H$m¡Zgm {dH$ën n[anW
option represents value of time constant most
Ho$ {bE H$mb {Z`Vm§H$ Ho$ ‘mZ Ho$ g~go ZOXrH$ h¡ ?
closely for the circuit?

(A) 2.4 s (B) 0.4 s (A) 2.4 s (B) 0.4 s
(C) 0.7 s (D) 1 s (C) 0.7 s (D) 1 s

042. Radii of two conducting circular loops are b and a 042. Xmo d¥ÎmmH$ma MmbH$ bynmo§ H$s {ÌÁ`mE± b VWm a Ohm±
respectively where b > > a. Centers of both loops
b > > a, XmoZm| Ho$ Ho$ÝÐ gånmVr h¡§ bo{H$Z XmoZm| bynmo§ Ho$
coincide but planes of both loops are perpendicular
to each other. The value of mutual inductance for
Vb nañna bå~dV h¡§ & BZ bynm§o Ho$ {bE AÝ`moÝ` àoaH$Ëd
these loops : H$m ‘mZ h¡ :
n0 rab n0 ra 2 n0 rab n0 ra 2
(A) (B) (A) (B)
2 (a + b) 2b 2 (a + b) 2b
n0 rb 2 n0 rb 2
(C) (D) zero (C) (D) eyݶ
2a 2a

043. A block of mass of 1kg is moving on the x axis. A 043. Ðì`‘mZ 1kg H$m EH$ ãbm°H$ x Aj na J{V‘mZ h¡ Bg na
force F acting on the block is shown. Velocity of H$m`©aV ~b F {MÌmZwgma h¡ & g‘` t = 2s na ãbm°H$ H$m doJ
the block at time t = 2s is - 3m/s . What is the speed
- 3m/s h¡ Vmo g‘` t = 4s na ãbm°H$ H$s Mmb Š`m hmoJr?
of the block at time t = 4s ?

(A) 3 m/s (B) 5 m/s (A) 3 m/s (B) 5 m/s
(C) 8 m/s (D) 2 m/s (C) 8 m/s (D) 2 m/s

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044. Two particles P and Q are moving on a circle. 044. Xmo H$U P VWm Q EH$ d¥Îm na J{V H$a aho h¡§& {H$gr jU
At a certain instant of time both the particles
XmoZm| H$U ì`mgV…{dnarV h¡§ VWm P H$m ñne©aoIr` ËdaU
are diametrically opposite and P has tangential
8 m/s2 VWm A{^H|${Ð` ËdaU 5 m/s2 h¡ O~{H$ Q Ho$db
acceleration 8 m/s2 and centripetal acceleration
5 m/s2 whereas Q has only centripetal acceleration A{^H|${Ð` ËdaU 1 m/s2 aIVm h¡ & {XE JE jU na Q
of 1 m/s2. At that instant acceleration (in m/s2) of P Ho$ gmnoj P H$m ËdaU (m/s2 ‘|) h¡ :
with respect to Q is :
(A) 10 (B) 12 (A) 10 (B) 12
(C) 14 (D) 80 (C) 14 (D) 80

045. In the given figure, atmospheric pressure P0 = 1 atm 045. {MÌ ‘| dm`w‘§S>br` Xm~ P0 = 1 atm VWm nmao ñV§^ H$s
and mercury column length is 9cm. Pressure P of bå~mB© 9cm h¡ & Zbr ‘| n[a~Õ J¡g H$m Xm~ P Š`m
the gas enclosed in the tube is : hmoJm?

(A) pressure of 78cm of Hg (A) 78cm Hg ñV§^ Xm~
(B) pressure of 85cm of Hg (B) 85cm Hg ñV§^ Xm~
(C) pressure of 67cm of Hg (C) 67cm Hg ñV§^ Xm~
(D) pressure of 90cm of Hg (D) 90cm Hg ñV§^ Xm~

046. PV diagram of an ideal gas is shown. The gas 046. EH$ AmXe© J¡g H$m PV AmaoI Xem©`m J`m h¡ & J¡g H$s
undergoes from initial state A to final state B such àmapå^H$ AdñWm A go A§{V‘ AdñWm B VH$ àH«$‘ Bg
that initial and final volumes are same . Select the àH$ma h¡ {H$ àma§{^H$ Am`VZ d A§{V‘ Am`VZ g‘mZ h¡&
correct alternative for given process AB. {XE JE AB àH«$‘ Ho$ {bE ghr {dH$ën M`Z H$amo :

(A) temperature of gas increases continuously (A) J¡g H$m Vmn bJmVma ~‹T>Vm h¡
(B) process is isochoric (B) àH«$‘ g‘Am`VZr h¡
(C) work done by gas is positive (C) J¡g Ûmam H$m`© YZmË‘H$ h¡
(D) work done by gas is negative (D) J¡g Ûmam H$m`© F$UmË‘H$ h¡

1-AD ] [ 13 ] [ P.T.O.

047. A small object of mass of 100gm moves in a circular 047. EH$ N>moQ>r dñVw {OgH$m Ðì`‘mZ 100gm h¡,`h EH$
path. At a given instant velocity of the object is d¥ÎmmH$ma nW ‘o§ J{V H$aVr h¡& {H$gr jU na Bg dñVw H$m
10it m/s and acceleration is (20it + 10tj ) m/s 2 . At doJ 10it m/s VWm ËdaU (20it + 10tj ) m/s 2 h¡ & Bg jU
this instant of time, rate of change of kinetic energy
na dñVw H$s J{VO D$Om© ‘| n[adV©Z H$s Xa hmoJr :
of the object is :
(A) 10000 kgm2 s–3 (B) 20  kgm2 s–3 (A) 10000 kgm2 s–3 (B) 20  kgm2 s–3
(C) 200  kgm2 s–3 (D) 300  kgm2 s–3 (C) 200  kgm2 s–3 (D) 300  kgm2 s–3

048. A time varying horizontal force (in Newton) 048. g‘` n[adVu j¡{VO ~b (Ý`yQ>Z ‘|) F = 8 sin (4rt)
F = 8 sin (4rt) is acting on a stationary block of EH$ {dam‘ ‘| aIo 2kg Ho$ ãbm°H$ na {MÌmZwgma bJVm h¡&
mass 2kg as shown. Friction coefficient between the `hm± ãbm°H$ VWm O‘rZ Ho$ ‘Ü` Kf©U JwUm§H$ n = 0.5 VWm
block and ground is n = 0.5 and g = 10m/s 2 . Then
g = 10m/s 2 h¡& ãbm°H$ H$s n[aUm‘r J{V hmoJr…
resulting motion of the block will be :

(A) It moves towards left (A) ~m§`r Va’$ J{V H$aoJm
(B) It moves towards right (B) Xm`t Va’$ J{V H$aoJm
(C) It will oscillate (C) XmobZ H$aoJm
(D) It remains stationary (D) {dam‘ ‘| hr ahoJm

049. Take Bulk modulus of water B = 2100MPa . What 049. `hm± Ob H$m Am`VZ àË`mñWVm JwUm§H$ B = 2100MPa
increase in pressure is required to decrease the br{OE & Ob Ho$ 200 brQ>a Am`VZ H$mo 0.004 à{VeV
volume of 200 liters of water by 0.004 percent ? KQ>mZo Ho$ {bE {H$VZm Xm~ n[adV©Z Amdí`H$ h¡?
(A) 8400 kPa (B) 84 kPa (A) 8400 kPa (B) 84 kPa
(C) 210 kPa (D) 840 kPa (C) 210 kPa (D) 840 kPa

050. Thin semicircular part ABC has mass m1 and 050. nVbo AÕ© d¥ÎmmH$ma ^mJ ABC H$m Ðì`‘mZ m1 h¡ VWm
diameter AOC has mass m2. Here axis passes ì`mg ­AOC H$m Ðì`‘mZ m2 h¡ &`hm± ì`mg Ho$ ‘Ü` {~ÝXþ
through mid point of diameter and the axis is
go Aj JwOaVm h¡ VWm Vb ­ABC Ho$ bå~dV Aj h¡ VWm
perpendicular to plane ABC. Here AO = OC = R. The
AO = OC = R h¡ & Bg g§`wº$ {ZH$m` H$s Cg Aj (axis)
moment of inertia of this composite system about
the axis is: Ho$ gmnoj O‹S>Ëd AmKyU© hmoJm :

m2 R 2 m2 R 2 m2 R 2 m2 R 2
(A) m1 R 2 + (B) m1 R 2 + (A) m1 R 2 + (B) m1 R 2 +
3 12 3 12
2
m1 R 2 m2 R 2 m1 R 2 m2 R 2 m1 R m R2 m1 R 2
m R2
(C) + (D) + (C) + 2 (D) + 2
2 3 2 6 2 3 2 6

1-AD ] [ 14 ] [ Contd...

CHEMISTRY / agm¶Zemó

051. The number of moles of Grignard reagent consumed 051. ZrMo {XE JE `m¡{JH$ ‘| à{V ‘mob Cn^moJ hmoZo dmbo {J«¾mS©>
per mole of the compound : A{^H$‘©H$ Ho$ {H$VZo ‘mob hm|Jo :

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

052. The paramagnetic species is : 052. {ZåZ ‘| go AZwMwåãH$s` h¡ :
(A) BaO2 (B) KO2 (A) BaO2 (B) KO2
(C) SiO2 (D) TiO2 (C) SiO2 (D) TiO2

053. Which one of the following has the highest 053. {ZåZ ‘| go {H$gH$s Zm{^H$ ñZo{hVm A{YH$V‘ h¡?
Nucleophilicity ?
(A) NH2 F–
-
(A) NH2
-
(B) F– (B)
(C) OH – (D) CH-3 (C) OH – (D) CH-3

054. In view of ∆rG0 for the following reactions : 054. {ZåZ A{^{H«$`mAm| Ho$ {bE ∆rG0 H$mo Ü`mZ ‘| aIVo
PbO 2 + Pb " 2PbO, Dr G0 < 0 hþE b¡S> (grgo) Am¡a {Q>Z Ho$ {bE H$m¡Zgr Am°ŠgrH$aU
SnO 2 + Sn " 2SnO, Dr G0 > 0 AdñWmE§ A{YH$ A{^bmj{UH$ h¢?
Which oxidation state is more characteristic for PbO 2 + Pb " 2PbO, Dr G0 < 0
lead and tin ? SnO 2 + Sn " 2SnO, Dr G0 > 0
(A) For lead +2, for tin +4 (A) b¡S> Ho$ {bE +2, {Q>Z Ho$ {bE +4
(B) For lead +4, for tin +2 (B) b¡S> Ho$ {bE +4, {Q>Z Ho$ {bE +2
(C) For lead +2, for tin +2 (C) b¡S> Ho$ {bE +2, {Q>Z Ho$ {bE +2
(D) For lead +4, for tin +4 (D) b¡S> Ho$ {bE +4, {Q>Z Ho$ {bE +4

055. Which of the following compounds will exhibit 055. {ZåZ ‘| go H$m¡Zgm `m¡{JH$ Á`m{‘Vr` g‘d`mdVm àX{e©V
geometrical isomerism? H$aoJm?
(A) 1,1–Diphenyl–1–propene (A) 1,1-S>mB© {’$ZmBb-1-àmonrZ
(B) 1–Phenyl–2–butane (B) 1-{’$ZmBb-2-ã`yQ>oZ
(C) 3–Phenyl–1–butene (C) 3 -{’$ZmBb-1-ã`yQ>rZ
(D) 2–Phenyl–1–butene (D) 2-{’$ZmBb-1-ã`yQ>rZ

056. At Critical Micell Concentration (CMC), the 056. H«$m§{VH$ {‘gob gm§ÐVm na gµ’$}ŠQ>oÝQ> AUw :
surfactant molecules:
(A) nyU©V`m KwbZerb hmoVo h¡§
(A) become completely soluble
(B) AnK{Q>V hmoVo h¢
(B) decompose
(C) dissociate (C) {d`mo{OV hmoVo h¡§
(D) associate (D) g§`mo{OV hmoVo h¡§

1-AD ] [ 15 ] [ P.T.O.

057. Which one of the following will be reactive for 057. n{H©$Z g§KZZ A{^{H«$`m Ho$ {bE {ZåZ ‘| go H$m¡Z {H«$`merb
Perkin condensation ?
hmoJm?
(A) O2N CHO (A) O2N CHO
(B) C6H5 CHO (B) C6H5 CHO
(C) CH3     O CHO (C) CH3     O CHO
(D) CH3 CHO (D) CH3 CHO

058. The pair of metal carbonyl complexes that are 058. YmVw H$m~m}{Zb g§Hw$b `m¡{JH$ H$m H$m¡Zgm `w½‘ g‘BboŠQ´mZ
isoelectronic is : h¡:
(A) [Fe(CO)4]– and Cr(CO)6 (A) [Fe(CO)4]– Am¡a Cr(CO)6
(B) [Co(CO)4]– and Ni(CO)4 (B) [Co(CO)4]– Am¡a Ni(CO)4
(C) Ni(CO)4 and V(CO)6 (C) Ni(CO)4 Am¡a V(CO)6
(D) [Cr(CO)6] and V(CO)6
(D) [Cr(CO)6] Am¡a V(CO)6
059. Which one of the following has (have) octahedral 059. {ZåZ ‘| go {H$gH$s /{H$ZH$s Aï>’$bH$s` Á`m{‘{V h¡ ?
geometry ?
(i) SbCl-6 (ii) SnCl62- (i) SbCl-6 (ii) SnCl62-
(iii) XeF6 (iv) IO65- (iii) XeF6 (iv) IO65-
(A) All of these (B) (i), (ii) & (iii) (A) ¶o g^r (B) (i), (ii) & (iii)
(C) (i), (ii) & (iv) (D) (ii), (iii) & (iv) (C) (i), (ii) & (iv) (D) (ii), (iii) & (iv)
060. In terms of polar character which one of the 060. Y«wdr` àH¥${V Ho$ g§X^© ‘| {ZåZ ‘| H$m¡Zgm H«$‘ ghr h¡?
following orders is correct?
(A) HF < H2O < NH3 <H2S (A) HF < H2O < NH3 <H2S
(B) NH3 < H2O < HF <H2S (B) NH3 < H2O < HF <H2S
(C) H2S < NH3 < H2O < HF (C) H2S < NH3 < H2O < HF
(D) H2O < NH3 < H2S < HF (D) H2O < NH3 < H2S < HF

061. Among the following compounds of Boron, the 061. ~moamZ Ho$ {ZåZ{bpIV `m¡{JH$m| ‘| go H$m¡Z σ – ~§Ymo§ Ho$ gmW
species which also forms π – bond in addition to
σ – bonds is:
gmW π – ~§Y ^r ~ZmVm h¡ :
(A) BF3 (B) BF 4-
(A) BF3 (B) BF 4-
(C) BH3 (D) B2H6
(C) BH3 (D) B2H6

062. Identify the Brönsted acid in the following equation: 062. {ZåZ g‘rH$aU ‘| ~«m|gQ>oS> Aåb H$mo nhMmZ|:
PO34- + H 2 O (l) " HPO 24- (aq) + OH- (aq) PO34- + H 2 O (l) " HPO 24- (aq) + OH- (aq)
(A) H2O (B) OH- (A) H2O (B) OH-
(C) PO3-
4 (D) HPO4 (C) PO3-
4 (D) HPO4
063. The number of grams/weight of NH4Cl required 063. Vmn 298K na 9.45 pH Ho$ ~’$a {d{b`Z H$mo V¡`ma H$aZo Ho$
to be added to 3 liters of 0.01M NH3 to prepare the {bE NH4Cl Ho$ {H$VZo J«m‘/^ma H$mo 3 brQ>a 0.01M   NH3
buffer of pH=9.45 at temperature 298K
Ho$ {d{b`Z ‘| {‘bm`m Om`oJm ?
(Kb for NH3 is 1.85×10–5)
(`hm± NH3 Ho$ {bE Kb =1.85×10–5)
(A) 0.455gm (B) 3.53 gm
(A) 0.455gm (B) 3.53 gm
(C) 0.354 gm (D) 4.55 gm (C) 0.354 gm (D) 4.55 gm

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064. For the reaction 2HI (g) H 2 (g) + I 2 (g) the 064. A{^{H«$`m 2HI (g) H 2 (g) + I 2 (g) H$s {d`moOZ
degree of dissociation (α) of HI(g) is related to H$s H$mo{Q> (α) gmå`mdñWm pñWam§H$ Kp ‘| gå~ÝY h¡ :
equilibrium constant Kp by the expression:
2 Kp 1+ 2 Kp
2 Kp 1+ 2 Kp (A) (B)
(A) (B) 1 + 2 Kp 2
1 + 2 Kp 2
1 + 2K p 2K p 1 + 2K p 2K p
(C) (D) (C) (D)
2 1 + 2K p 2 1 + 2K p

065. A 6% solution of sucrose C22H22O11 is isotonic with 065. gwH«$moO C22H22O11 H$m 6% {db`Z EH$ AkmV H$m~©{ZH$
3% solution of an unknown organic substance. The nXmW© Ho$ 3% {db`Z Ho$ gmW g‘namgmar h¡& AkmV
molecular weight of unknown organic substance H$m~©{ZH$ nXmW© H$m AmU{dH$ ^ma hmoJm:
will be:
(A) 100 (B) 342 (A) 100 (B) 342
(C) 684 (D) 171 (C) 684 (D) 171

066. The enthalpy of the formation of CO2 and H2O are 066. CO2 Am¡a H2O Ho$ g§^dZ H$s D$î‘m H$m ‘mZ H«$‘e…
– 395 kJ and – 285 kJ respectively and the enthalpy -395 kJ Am¡a -285 kJ h¡ Am¡a E{g{Q>H$ E{gS> Ho$ XhZ H$s
of combustion of acetic acid is 869 kJ. The enthalpy D$î‘m 869 kJ h¡& E{g{Q>H$ E{gS> Ho$ g§^dZ H$s D$î‘m h¡:
of formation of acetic acid is:
(A) 491 kJ (B) 235 kJ (A) 491 kJ (B) 235 kJ
(C) 340 kJ (D) 420 kJ (C) 340 kJ (D) 420 kJ

067. Which of the following is a lyophobic colloid : 067. {ZåZ ‘| go H$m¡Zgm EH$ Ðd{damJr H$mobmBS> h¡ :
(A) Gum Arabica (B) Gelatin (A) J‘ Aao{~H$ (B) {OboQ>rZ
(C) Sulphur (D) Starch (C) gë’$a (D) ñQ>mM©

068. For car battery which one is correct statement ? 068. H$ma H$s ~¡Q>ar Ho$ {bE H$m¡Zgm H$WZ gË` h¡ ?
(A) Cathode is Copper (Cu) and anode is Lead (Pb) (A) H¡$WmoS> H$m°na (Cu) Ed§ EZmoS> boS> (Pb) hmoVm h¡
(B) Cathode is Lead dioxide (PbO2) and anode is (B) H¡$WmoS> boS> S>mBAm°ŠgmBS> (PbO2) Ed§ EZmoS> boS>
Lead (Pb) (Pb) hmoVm h¡
(C) Cathode is Lead dioxide (PbO2) and anode is (C) H¡$WmoS> boS> S>mBAm°ŠgmBS> (PbO2) Ed§ EZmoS> H$m°na
Copper (Cu) (Cu) hmoVm h¡
(D) Cathode is Copper (Cu) and anode is Lead (D) H¡$WmoS> H$m°na (Cu) Ed§ EZmoS> boS> S>mBAm°ŠgmBS>
dioxide (PbO2) (PbO2) ) hmoVm h¡

069. Considering entropy(s) as a thermodynamic 069. E§Q´monr H$mo D$î‘mJ{VH$s àmMb ‘mZVo hþE {H$gr ñdV…
parameter, the criterion for the spontaneity of any àd{V©V àH«$‘ Ho$ {bE E§Q´monr n[adV©Z hmoJm:
process the change in entropy is :
(A) (∆ Ssystem + ∆Ssurrounding ) > 0 (A) (∆SV§Ì + ∆Sn[adoe ) > 0

(B) (∆ Ssystem – ∆Ssurrounding ) > 0 (B) (∆ SV§Ì – ∆Sn[adoe ) > 0
(C) ∆Ssystem > 0 only (C) Ho$db ∆SV§Ì > 0
(D) ∆S surrounding > 0 only (D) Ho$db ∆S n[adoe > 0

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070. At low pressure and high temperature, the Vander 070. H$‘ Xm~ Am¡a Cƒ Vmn‘mZ na, dm§S>a dmb g‘rH$aU H$m
Waal’s equation is finally reduced (simplified) to : A§{V‘ gabrH¥$V n[ad{V©V ê$n hmoJm:
a a
(A) c P + 2 m Vm = RT (A) c P + m V = RT
Vm V m2 m
(B) PVm = RT (B) PVm = RT
a a
(C) c P + m (Vm - b) = RT (C) c P + m (Vm - b) = RT
V m2 V m2
(D) P(Vm – b) = RT (D) P(Vm – b) = RT

071. Which graph represents the zero order reaction 071. {ZåZ ‘| go H$m¡Zgm J«m’$ eyÝ` H$mo{Q> A{^{H«$`m
[A (g) " B (g)]
[A (g) " B (g)] H$mo àX{e©V H$aVm h¡ :

(A) (B)
(A) (B)

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

072. Which of the following compounds is insoluble 072. {ZåZ ‘| go H$m¡Zgm `m¡{JH$ J‘© gmÝÐ H2SO4 ‘| ^r A{dbo`
even in hot concentrated H2SO4? h¡ ?
(A) Aniline (B) Ethylene (A) E{ZbrZ (B) E{WbrZ
(C) Benzene (D) Hexane (C) ~|OrZ (D) hoŠgoZ

073. The half life of Th232 is 1.4 × 1010 years and that 073. Th232 H$s AY© Am`w H$m ‘mZ 1.4 × 1010 df© h¡ Am¡a
of its daughter element Ra238 is 7 years. What Bggo CËnÞ nwÌr VËd Ra238 H$s AY© Am`w 7 df© h¡ &
amount (most nearly) weight of Ra238 will be in Ra238 H$s {H$VZr (g~go g‘rnV‘) ‘mÌm Th232 H$s
equilibrium with 1gm of Th232 ? 1gm ‘mÌm Ho$ gmW gmå` ‘| hmoJr ?
(A) 2 × 10–10 gm (B) 5 × 10–10gm (A) 2 × 10–10 gm (B) 5 × 10–10gm
(C) 5.0 gm (D) 1.95 × 10–9 gm (C) 5.0 gm (D) 1.95 × 10–9 gm

074. Which of the following electron has minimum energy? 074. {ZåZ{bpIV ‘| go H$m¡Zgm BboŠQ´m°Z Ý`yZV‘ D$Om© aIVm h¡?
1 1
(A) n = 5, l = 0, m = 0, s = + (A) n = 5, l = 0, m = 0, s = +
2 2
1 1
(B) n = 3, l  =  2, m = –2, s = + (B) n = 3, l  =  2, m = –2, s = +
2 2
1 1
(C) n = 4, l =  0, m = 0, s = + (C) n = 4, l =  0, m = 0, s = +
2 2
1 1
(D) n = 4, l = 1, m = +1, s = + (D) n = 4, l = 1, m = +1, s = +
2 2
075. Total number of stereoisomers of the following 075. {ZåZ `m¡{JH$m| Ho$ {Ì{d‘ g‘md`dr`m| H$s g§»`m H«$‘e… h¢:
compounds are respectively :

(i) (ii) (i) (ii)

(A) 8, 8 (B) 4, 6 (A) 8, 8 (B) 4, 6
(C) 8, 0 (D) 6, 6 (C) 8, 0 (D) 6, 6

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076. Which of the following is a monomer of Dacron: 076. {ZåZ ‘| go H$m¡Zgm So>H«$moZ H$m EH$bH$ h¡ ?
(A) HOH 2 C - CH 2 OH (A) HOH 2 C - CH 2 OH
(B) CH 2 CH - CH CH 2 (B) CH 2 CH - CH CH 2
Cl Cl

(C) H 2 C C - CH CH 2 (C) H 2 C C - CH CH 2

(D) COOH COOH (D) COOH COOH

077. Which of the following is a meso compound ? 077. {ZåZ ‘| go H$m¡Zgm {‘gmo `m¡{JH$ h¡ ?
(A) cis–1, 4–dimethylcyclohexane (A) {gg -1, 4- S>mB©‘o{WbgmBŠbmohoŠgoZ
(B) trans–1, 4–dimethylcyclohexane (B) Q´m§g -1, 4- S>mB©‘o{WbgmBŠbmohoŠgoZ
(C) cis–1, 3–dimethylcyclohexane (C) {gg -1, 3- S>mB©‘o{WbgmBŠbmohoŠgoZ
(D) trans–1, 3–dimethylcyclohexane (D) Q´m§g -1, 3- S>mB©‘o{WbgmBŠbmohoŠgoZ

078. IUPAC name of the following is : 078. {ZåZ H$m IUPAC Zm‘ h¡ :
CH3 CH 2 CH CHCH 2 CH3 CH3 CH 2 CH CHCH 2 CH3

CH3 CHO CH3 CHO
(A) 8 methyl– 2 ethyl pentanal (A) 8 ‘o{Wb, 2 E{Wb n|Q>oZb
(B) 2,5 Butyl butenal (B) 2, 5 ã`w{Q>b ã`yQ>oZb
(C) 2,3 di ethyl butenal (C) 2, 3 S>mB E{Wb ã`yQ>oZb
(D) 2 ethyl–3 methyl pentanal (D) 2 E{Wb, 3 ‘o{Wb n|Q>oZb

079. Which of the following is Reimer - Tieman reaction? 079. {ZåZ ‘| go H$m¡Zgr ar‘a Q>r‘mZ A{^{H«$`m h¡ ?
OC2H5 OC2H5
Conc.H2 SO4 (A) Conc.H2 SO4
(A) Conc.HNO3
Conc.HNO3

OH OH

(B) + CHCl3 + aq. NaOH (B) + CHCl3 + aq. NaOH

OH OH

(C) + CHCl3 + alcoh. NaOH (C) + CHCl3 + alcoh. NaOH

OCH3 OCH3
anhy. AlCl3 anhy. AlCl3
(D) + CH3 COCl (D) + CH3 COCl

080. The increasing order of the first ionization enthalpies 080. B, P, S Am¡a F VËdm| H$s àW‘ Am`ZZ EÝWopën`m| H$m
of the elements B, P, S and F is: ~‹T>Vm hþAm H«$‘ h¡ :
(A) P < S < B < F (A) P < S < B < F
(B) B < P < S < F (B) B < P < S < F
(C) B < S < P < F (C) B < S < P < F
(D) F < S < P < B (D) F < S < P < B

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081. Some pairs of ions are given below. In which pair, 081. ZrMo Hw$N> Am`Zm| Ho$ `w½‘ {XE JE h¢, BZ‘o§ go {H$g‘o àW‘
first ion is more stable than second ion ? Am`Z Xÿgao Am`Z go A{YH$ ñWm`r h¡?
H3 C - CH - CH3 H3 C - N - CH3 H3 C - CH - CH3 H3 C - N - CH3

(A) and (A) VWm
H 2 C - C - CH3 H3 C - C - CH3 H 2 C - C - CH3 H3 C - C - CH3
5 5 5
5 5 5
5 5 p
(B) H3 C - CH - CH3 VWm - CH - OCH 3
p
(B) H3 C - CH - CH3 and - CH - OCH 3
5
5 (C) H3 C - CH 2 - CH - CH3 VWm
(C) H3 C - CH 2 - CH - CH3 and 5
5 H 2 C - CH 2 - CH - CH 2
H 2 C - CH 2 - CH - CH 2 5 5
5 5 CH 2 CH 2
CH 2 CH 2 (D) VWm
(D) and

082. Which alkaline earth metal compound is volatile ? 082. {ZåZ ‘| go H$m¡Zgm jmar` ‘¥Xm YmVw `m¡{JH$ dmînerb h¡ ?
(A) None of the options (B) Be3N2 (A) BZ‘o go H$moB© {dH$ën Zht (B) Be3N2
(C) Mg3N2 (D) Ca3N2 (C) Mg3N2 (D) Ca3N2

083. What is the name of the following reaction? 083. {ZåZ A{^{H«$`m H$m Zm‘ Š`m h¡ ?
NaOH NaOH
HCHO + HCHO CH3OH + HCOONa HCHO + HCHO CH3OH + HCOONa
T T
(A) None of the options (A) H$moB© ^r {dH$ën Zht
(B) Hell–Volhard reaction (B) hob dmoëhmS©> A{^{H«$`m
(C) Clemmensen reaction (C) Šbo‘|gZ A{^{H«$`m
(D) Cannizzaro reaction (D) H$m{ZµOamo A{^{H«$`m
084. Inorganic graphite is:
084. AH$m~©{ZH$ J«o’$mBQ> h¡ :
(A) BF3 (B) B2N3H6 .5
(A) BF3 (B) B2N3H6
(C) B2H6 (D) BN
(C) B2H6 (D) BN
085. Rank the following in decreasing order of basic strength:
085. {ZåZ H$s jmar` gm‘Ï`© H$m KQ>Vm hþAm H«$‘ h¡ :
(i) CH3 - CH 2 - C / C-
(i) CH3 - CH 2 - C / C-
(ii) CH3 - CH 2 - S-
(ii) CH3 - CH 2 - S-
(iii) CH3 - CH 2 - CO-2
(iii) CH3 - CH 2 - CO-2
(iv) CH3 - CH 2 - O-
(iv) CH3 - CH 2 - O-
(A) i > iv > iii > ii (B) ii > i > iv > iii
(A) i > iv > iii > ii (B) ii > i > iv > iii
(C) iv > i > ii > iii (D) i > iv > ii > iii
(C) iv > i > ii > iii (D) i > iv > ii > iii
086. Among the given compound choose the two that 086. {ZåZ ‘| go Xmo `m¡{JH$m| H$m M`Z H$a| Omo {H$ Am`ZZ Ho$ ~mX
yield same carbocation on ionization. g‘mZ H$m~© YZAm`Z ~Zm`|Jo-
Br Br Br Br
Br Br
Br Br
(i) (ii) (iii) (iv) (i) (ii) (iii) (iv)
(A) (ii),(iii) (B) (i),(iii) (A) (ii),(iii) (B) (i),(iii)
(C) (ii),(iv) (D) (i),(ii) (C) (ii),(iv) (D) (i),(ii)

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087. Increasing order of acidic strength of given 087. {ZåZ `m¡{JH$m| H$s Aåbr` gm‘Ï`© H$m ~‹T>Vm hþAm H«$‘ h¡:
compounds is :
OH OH OH OH OH OH OH
OH

CN OCH3 Cl CN OCH3 Cl
(i) (ii) (iii) (iv) (i) (ii) (iii) (iv)
(A) i < iii < ii < iv (B) iii < i < iv < ii (A) i < iii < ii < iv (B) iii < i < iv < ii
(C) ii < i < iv < iii (D) i < iii < iv < ii (C) ii < i < iv < iii (D) i < iii < iv < ii

088. Which of the following effects of –NO2 group 088. ZrMo {XE JE AUw ‘| {ZåZ ‘| go H$m¡Zgm à^md
operates on –NH2 group in this molecule ?
–NO2 g‘yh –NH2 g‘yh na à^mdr hmoJm ?
NH2
NH2

Me Me
NO2 Me Me
NO2
(A) Both –I and –M effect (A) XmoZm| –I Am¡a –M à^md
(B) Only –I effect (B) Ho$db –I à^md
(C) Only +M effect (C) Ho$db +M à^md
(D) Only –M effect (D) Ho$db –M à^md

089. Which of the following material is known as lunar 089. {ZåZ ‘| go H$m¡Zgm nXmW© byZa H$mpñQ>H$ Ho$ Zm‘ go OmZm
caustic ? OmVm h¡ ?
(A) NaOH (B) NaNO3 (A) NaOH (B) NaNO3
(C) AgCl (D) AgNO3 (C) AgCl (D) AgNO3

090. Provide an acceptable name for the alkane shown 090. ZrMo Xr J`r EëHo$Z H$m ñdrH$m`© Zm‘ ~VmB`o:
below :
H CH 2 CH 2 CH (CH3) 2 H CH 2 CH 2 CH (CH3) 2

CH3 CH 2 CH 2 CH 2 - C C - CH 2 CH 2 CH3 CH3 CH 2 CH 2 CH 2 - C C - CH 2 CH 2 CH3

CH 2 CH3 H CH 2 CH3 H
(A) 2–ethyl–6–methyl–5–propyldecane (A) 2–E{Wb–6–‘o{Wb–5–àmo{nbSo>Ho$Z
(B) 6–ethyl–2–methyl–5–propyldecane (B) 6–E{Wb–2–‘o{Wb–5–àmo{nbSo>Ho$Z
(C) 5–ethyl–6–methyl–2–propyldecane (C) 5–E{Wb–6–‘o{Wb–2–àmo{nbSo>Ho$Z
(D) 2–ethyl–6–methyl–2–propyldecane (D) 2–E{Wb–6–‘o{Wb–2–àmo{nbSo>Ho$Z

HO - HO - HO -
½byH$moO HO
-
091. D – Mannose D – glucose (A) 091. D – ‘oZmoO D– (A)
Product (A) of above reaction is: Cnamoº$ A{^{H«$`m H$m CËnmX (A) h¡:
(A) D–Idose (B) D–glucose (A) D – Am`moS>moO (B) D – ½byH$moO
(C) D–fructose (D) D–Talose (C) D – ’«$ŠQ>moO (D) D – Q>obmoO

1-AD ] [ 21 ] [ P.T.O.

092. What is the product in the following reaction ? 092. {ZåZ A{^{H«$`m H$m CËnmX hmoJm ?
OH OH
(NH4) 2 Cr2 O7 (NH4) 2 Cr2 O7
H2SO4 H2SO4
(A) Benzoic sulphate (A) ~oݵOmoBH$ gë’o$Q>
(B) Benzoic Acid (B) ~oݵOmoBH$ Aåb
(C) Benzoquionone (C) ~oÝOmo{¹$Zm|Z
(D) Cyclohexane-1-one (D) gm`ŠbmohoŠgoZ-1-AmoZ
093. How many bonds are there in : 093. {XE JE AUw ‘| Hw$b {H$VZo ~§Y h¢ ?

(A) 14σ, 2p (B) 14σ, 8p (A) 14σ, 2p (B) 14σ, 8p
(C) 18σ, 8p (D) 19σ, 4p (C) 18σ, 8p (D) 19σ, 4p
094. Which of the following molecules is optically 094. {ZåZ ‘| go H$m¡Zgo AUw àH$m{eH$ g{H«$` h¡ ?
active ?

(A) (i), (ii) and (iii) (B) (i) and (ii) (A) (i), (ii) Am¡a (iii) (B) (i) Am¡a (ii)
(C) (i) and (iii) (D) (ii) and (iii) (C) (i) Am¡a (iii) (D) (ii) Am¡a (iii)
095. Which of the following statement is correct ? 095. {ZåZ ‘| go H$m¡Zgm H$WZ gË` h§¡ ?
(A) Both BCl3 and AlCl3 are not Lewis acids (A) BCl3 Am¡a A ­ lCl3 XmoZm| hr bwB©g Aåb Zht h¢
(B) BCl3 and AlCl3 are both Lewis acids and BCl3 (B) BCl3 Am¡a A ­ lCl3 XmoZm| bwB©g Aåb h¢ Ed§ BCl3,
is stronger than AlCl3 AlCl3 go e{º$embr h§¡
(C) BCl3 and AlCl3 are both Lewis acids and (C) BCl3 Am¡a AlCl3 XmoZm| bwB©g Aåb h¢ Ed§ ­AlCl3,
AlCl3 is stronger than BCl3 BCl3 go e{º$embr h¢
(D) BCl3 and AlCl3 are both equally strong Lewis (D) BCl3 Am¡a ­AlCl3 XmoZm| g‘mZ e{º$embr bwBg
© Aåb h¢
acid
096. Consider the following compounds. 096. ZrMo {XE JE `m¡{JH$m| ‘| go {H$Z `m¡{JH$m| H$mo àmá H$aZo Ho$ {bE
’«$sS>b H«$mâQ> E{g{bH$aU H$m Cn`moJ {H$`m Om gH$Vm h¡:

(I) (II)
(I) (II)

(III) (IV)
Friedel–Crafts acylation can be used to obtain: (III) (IV)
(A) I, II, III (B) I, III, IV (A) I, II, III (B) I, III, IV
(C) II, III, IV (D) I, II, IV (C) II, III, IV (D) I, II, IV

1-AD ] [ 22 ] [ Contd...

097. Provide the systematic name of the compound 097. ZrMo àX{e©V `m¡{JH$ H$m ì`dpñWV Zm‘ Xr{O`o:
shown:

(A) 2 – butyl – 4 – ethyl – 1 – methylcycloheptane (A) 2- ã`w{Q>b -4-E{Wb -1-‘o{WbgmBŠbmohoßQ>oZ
(B) 4 – butyl – 1 – ethyl – 2 – methylcycloheptane (B) 4-ã`w{Q>b - 1- E{Wb - 2 - ‘o{WbgmBŠbmohoßQ>oZ
(C) 4 – butyl – 2 – ethyl – 1 – methylcycloheptane (C) 4- ã`w{Q>b -2- E{Wb -1- ‘o{WbgmBŠbmohoßQ>oZ
(D) 1 – butyl – 4 – ethyl – 3 – methylcycloheptane (D) 1- ã`w{Q>b -4-E{Wb - 3 -‘o{WbgmBŠbmohoßQ>oZ

098. Give the IUPAC name for the following structure: 098. {ZåZ ga§MZm H$m IUPAC Zm‘ Xr{O`o::

(A) 5 – chloro – 2 – methylcyclohexanol (A) 5 - Šbmoamo - 2 - ‘o{WbgmBŠbmohoŠgmZmob
(B) 3 – chloro – 2 – methylcyclohexanol (B) 3 - Šbmoamo - 2 - ‘o{WbgmBŠbmohoŠgmZmob
(C) 2 – methyl – 5 – chlorocyclohexanol (C) 2 - ‘o{Wb - 5 - ŠbmoamogmBŠbmohoŠgmZmob
(D) 1 – chloro – 4 – methylcyclohexanol (D) 1 - Šbmoamo - 4 - ‘o{WbgmBŠbmohoŠgmZmob

099. In aldol addition reaction product is always: 099. EëS>mob `moJmË‘H$ A{^{H«$`m ‘| CËnmX h‘oem hmoJm :
(A) a, b – unsaturated ketone (A) a, b – Ag§V¥á H$sQ>moZ
(B) b – hydroxyaldehyde (B) b – hmB©S´moŠgrEëS>rhmB©S>
(C) b – hmB©S´moŠgrH$sQ>moZ
(C) b – hydroxyketone
(D) a, b – Ag§V¥á EëS>rhmB©S>
(D) a, b – unsaturated aldehyde

100. Which one of the following compounds will have 100. {ZåZ ‘| go H$m¡Zgo `m¡{JH$ Ho$ {bE {XY«wd AmKyU© H$m ‘mZ
the highest dipole moment ? A{YH$V‘ hmoJm ?

(A) (A)

(B) (B)

(C) (C)

(D) (D)

1-AD ] [ 23 ] [ P.T.O.

MATHEMATICS / J{UV

7 2x 3 7
101. If and 1 are the roots of the equation 7
2 101. `{X g‘rH$aU 2 2x 2 = 0 Ho$ ‘yb VWm 1
2
2x 3 7 7 6 2x
2 2x 2 = 0 then the third root is : h¡ Vmo Vrgam ‘yb hmoJm :
7 6 2x
(A) – 3/2 (B) – 5/2 (A) – 3/2 (B) – 5/2
(C) – 7/2 (D) – 9/2 (C) – 7/2 (D) – 9/2

102. If cos (log i 4i) = a + i b , then 102. ¶{X cos (log i 4i) = a + i b hmo V~
(A) a = 1 , b = 0 (A) a = 1 , b = 0
(B) a = 1 , b = 2 (B) a = 1 , b = 2
(C) a = 1 , b = − 1 (C) a = 1 , b = − 1
(D) a =− 1 , b = 1 (D) a =− 1 , b = 1

103. The function y = 2x - x 2 103. ’$bZ y = 2x - x 2
(A) Increases in ( 1 , 2 ) but decreases in ( 0 , 1 ) (A) (1, 2) ‘| ~‹T>Vm h¡ naÝVw (0 , 1) ‘| KQ>Vm h¡
(B) increases in ( 0 , 2 ) (B) (0, 2) ‘| ~‹T>Vm h¡
(C) increases in ( 0 , 1 ) but decreases in ( 1 , 2 ) (C) (0, 1) ‘| ~‹T>Vm h¡ naÝVw (1 , 2) ‘| KQ>Vm h¡
(D) Decreases in ( 0 , 2 ) (D) (0, 2) ‘| KQ>Vm h¡

104. `{X {~ÝXþ (a , a ) aoImAmo 2x + y = 5 Ho$ ‘Ü` pñWV h¡
104. If the point (a , a ) lies between the lines 2x + y = 5
then select one of the most appropriate option: V~ g~go Cn`wº$ EH$ {dH$ën M`Z H$amo :
11 5 11 5
(A) a < (B) a < (A) a < (B) a <
3 2 3 2
5 7 5 7
(C) (D) (C) a < (D) a <
a <
3
a <
2 3 2

`{X log sin r ' 3 zz -22 -31 1 >1 hmo Vmo
- +
z- 2 +3 105.
105. If log sin r ' 1 >1 , then
6 3 z- 2 -1 6

(A) z - 2 <6 (B) z - 2 >7
(A) z - 2 <6 (B) z - 2 >7
(C) z - 2 <7 (D) z - 2 <3
(C) z - 2 <7 (D) z - 2 <3

106. The nth term of the series 106. Xr JB© loUr H$m nth nX hmoJm :
1 + 4 + 13 + 40 + 121 + 364 + …… , is : 1 + 4 + 13 + 40 + 121 + 364 + ……
n
2 + 1j
(A) `
n
2 + 1j
2 (A) `
2
n
(B) 3 - 1 n
(B) 3 - 1
1 n+ 1 n+
(C) (3 1) (C) (3 1)
2 2
1 n- 1 n-
(D) (3 1) (D) (3 1)
2 2

1-AD ] [ 24 ] [ Contd...

107. The interval in which the function y = x - 2 sin x; 107. dh A§Vamb Š`m hmoJm {Og‘o ’$bZ y = x - 2 sin x;
0 # x # 2r increases throughout is : 0 # x # 2r ewê$ go AÝV VH$ ~‹T>Vm h¡ :
r 5r j
r 5r j
(A) ` , (B) `0, r j (A) ` , (B) `0, r j
3 3 4 3 3 4
5
5 (C) ` , 2r j (D) `0, j
r r
(C) ` , 2r j (D) `0, j
r r
3 3 3 3

{ÛnX {dñVma c2 1 3 + 11 m Ho$ ewéAmµV go gmVd| nX
x
108. If the ratio of the seventh term from the beginning 108.
1 1 x 3 3
of the binomial expansion of c 2 3 + 1 m to the
3 3 d AÝV go gmVd| nX H$m AZwnmV 1/6 h¡ Vmo x H$m
seventh term from its end is 1/6 , then the value of x is: ‘mZ h¡ :
(A) 9 (B) 7 (A) 9 (B) 7
(C) 5 (D) 11 (C) 5 (D) 11

109. Let A={ u, v, w, z } and B= { 3 , 5 } , then the 109. ‘mZm A={ u, v, w, z } VWm B= { 3 , 5 } , V~ A go B H$mo
number of relations from A to B is : gå~ÝYm| H$s g§»¶m hmoJr:
(A) 512 (B) 64 (A) 512 (B) 64
(C) 256 (D) 1024 (C) 256 (D) 1024

110. Given y = x 2 . As x " 2, y " 4 what must the value 110. y = x 2 {X`m h¡ O~ x " 2, y " 4 hmo Vmo δ H$m ‘mZ Š`m
of δ be for which from | x – 2 |< δ it follows that hmoZm Mm{hE {Oggo {H$ | x – 2 |< δ go | y – 4 | < ∈ = 0.001
| y – 4 | < ∈ = 0.001 ? AZwgaU hmoVm h¡ :
(A) 0.4< δ < 0.5 (B) 0 < δ < 0.00025 (A) 0.4< δ < 0.5 (B) 0 < δ < 0.00025
(C) 0.03 < δ < 0.05 (D) 0.2 < δ < 0.25 (C) 0.03 < δ < 0.05 (D) 0.2 < δ < 0.25

111. Given that f(0)= 0 and lim f (x) exists, say L. 111. {X`m h¡ {H$ f(0)= 0 h¡ VWm lim f (x) {dÚ‘mZ h¡ ‘mZm
x"0 x x"0 x
Here f l (0) denotes the derivative of f w. r. t. x at {H$ `h L h¡& `hm± f H$m x Ho$ gmnoj AdH$bZ Ho$ ‘mZ H$mo
x = 0 na f l (0) Ûmam àX{e©V {H$`m OmVm h¡ V~ L hmoJm:
x = 0. Then L is :
(A) f l (0) (B) 0 (A) f l (0) (B) 0
(C) 2f l (0) - 6 (D) 2f l (0) - 5 (C) 2f l (0) - 6 (D) 2f l (0) - 5
2x x
112. The inverse of the function y =
1+ 2x
is: 112. ’$bZ y = +2 x H$m ì`wËH«$‘ h¡ :
1 2
1 m 1 m
(A) x = log 2 c (A) x = log 2 c
1- y 1- y
y y
(B) x = log 2 (B) x = log 2
1- y 1- y
1 1
(C) x = log 2 (C) x = log 2
1- 2 y 1 2y
-
1 1
(D) x = log 2 c1 - m (D) x = log 2 c1 - m
y y

1-AD ] [ 25 ] [ P.T.O.

113. The domain of the definition of the function 113. {ZåZ {XE JE ’$bZ H$s n[a^mfm H$m àmÝV h¡ :
1 1
y= + (x + 2) is : y= + (x + 2)
log10 (1 - x) log10 (1 - x)
(A) - 2 # x <0 (B) - 2 # x <1 (A) - 2 # x <0 (B) - 2 # x <1
(C) x $ - 2 (D) - 3 < x # - 2 (C) x $ - 2 (D) - 3 < x # - 2

Z Z
]- 2 sin x r r
if x # - ]- 2 sin x if x # -
]] 2 ]] 2
r r r r
114. Let f (x) = [ A sin x + B if - <x < ; 114. ‘mZm f (x) = [ A sin x + B if - <x < ;
2 2 2 2
] r ]
] cos x if x $ r
2 ] cos x if x $
\ 2
\
For what values of A and B, the function f  (x) is ’$bZ f  (x) Ho$ nyar dmñV{dH$ aoIm na gVV hmoZo Ho$ {bE
continuous throughout the real line ? A VWm B Ho$ Š`m ‘mZ hmoZo Mm{hE ?
(A) A = 1, B = − 1 (B) A = 1, B = 1 (A) A = 1, B = − 1 (B) A = 1, B = 1
(C) A = − 1, B = 1 (D) A  =  − 1, B = − 1 (C) A = − 1, B = 1 (D) A  =  − 1, B = − 1

rx
115. Let f (x) = *
a (x) sin for x ! 0; rx
2 115. ‘mZm f (x) = *a (x) sin 2 for x ! 0;
1 for x = 0 1 for x = 0
where a (x) is such that lim a (x) = 3 Ohm± a (x) Bg Vah h¡ {H$lim a (x) = 3
x"0 x"0
Then the function f(x) is continuous at x = 0 if V~ x = 0 na ’$bZ f(x) gVV hmoJm `{X a (x) {ZåZ Vah
a (x) is chosen as :
go {b`m OmE :
2
(A) 2
rx 2 (A)
rx 2
1
(B) 1
x (B)
x
2
(C) 2
rx (C)
rx
1
(D) 1
x2 (D)
x2

lim y- a ry is : lim y- a ry H$m ‘mZ h¡ :
116. The $` j $ ` tan j. 116. $`sin j $ ` tan j.
y " a sin 2 2a y"a 2 2a
a a
(A) - (A) -
r r

a a
(B) (B)
2r 2r

2a 2a
(C) (C)
r r

a a
(D) (D)
r r

1-AD ] [ 26 ] [ Contd...

2 n + (- 2) n 2 n + (- 2) n 2 n + (- 2) n 2 n + (- 2) n
117. Let , n = n and Ln = then 117. ‘mZm , n = VWm Ln = V~
2 3n 2 n
3n
as n " 3 n " 3 OmZo na:

(A) Both the sequences do not have limits. (A) XmoZm| AZwH«$‘mo§ H$s gr‘mE± Zht h¡§.

(B) Both the sequences have limits (B) XmoZm| AZwH«$‘m§o H$s gr‘mE± hm§oJr
(C) lim , n exists but  lim Ln does not exist (C) lim , n {dÚ‘mZ h¡§ naÝVw lim Ln {dÚ‘mZ Zht h¡
n"3 n"3 n"3 n"3

(D)  lim , n does not exist but lim Ln exists (D) lim , {dÚ‘mZ Zht h¡ naÝVw lim L {dÚ‘mZ h¡
n n
n"3 n"3 n"3 n"3

118. For what interval of variation of x, the identity 118. Ma x Ho$ n[adV©Z Ho$ {H$VZo A§Vamb Ho$ {bE {ZåZ gd©g{‘H$m
1 - x2 = - x =- 2
arc cos
1 + x2
2arc tan x is true? gË` h¡ ? arc cos 1 -
+ 2
2arc tan x
1 x
(A) 0 # x # 1 (B) 0 # x <3 (A) 0 # x # 1 (B) 0 # x <3
(C) - 3 < x # 0 (D) 1 < x <3 (C) - 3 < x # 0 (D) 1 < x <3

119. The points of the curve y = x3 + x - 2 at which its 119. dH«$ y = x3 + x - 2 Ho$ do {~ÝXþ Š`m h¢ {Og na IrMt
tangents are parallel to the straight line y = 4x – 1 JB© ñne© aoImE± Xr JB© gab aoIm y = 4x – 1 Ho$ g‘mÝVa
are : h¡ :
(A) `- 2 3 , - 2 3 j, (0, - 4)
1 1
(A) `- 2 3 , - 2 3 j, (0, - 4)
1 1

(B) ( 1, 0 ), ( – 1, – 4 ) (B) ( 1, 0 ), ( – 1, – 4 )
(C) ( 2 , 7 ) , ( – 2 , – 11 ) (C) ( 2 , 7 ) , ( – 2 , – 11 )
(D) ( 0 , – 2 ) , ` j
1 1
(D) ( 0 , – 2 ) , `2 3 , 2 3 j
1 1
2 3, 2 3

120. If a , b , c are three vectors such that 6 a b c @ = 5 120. `{X a , b , c VrZ g{Xe Bg àH$ma h¢ {H$ 6 a b c @ = 5
then the value of 6 a # b , b # c , c # a @ is : V~ 6 a # b , b # c , c # a @ H$m ‘mZ h¡ :
(A) 20 (B) 10 (A) 20 (B) 10
(C) 15 (D) 25 (C) 15 (D) 25

121. A chord of the parabola y = x 2 - 2x + 5 joins the 121. nadb` y = x 2 - 2x + 5 H$s Ordm nadb` Ho$ {~ÝXþAm|
point with the abscissas x1 = 1, x2 = 3 Then the
x1 = 1, x2 = 3 H$mo Omo‹S>Vr h¡ Vmo Bg Ordm Ho$ g‘mÝVa
equation of the tangent to the parabola parallel to
the chord is :
nadb` H$s ñne© aoIm H$m g‘rH$aU hmoJm :
5= 5=
(A) 2x + y + 1 = 0 (B) 2x - y + 0 (A) 2x + y + 1 = 0 (B) 2x - y + 0
4 4
(C) 2x – y + 2 = 0 (D) 2x – y + 1 = 0 (C) 2x – y + 2 = 0 (D) 2x – y + 1 = 0

1-AD ] [ 27 ] [ P.T.O.

x

122. The point of inflection of the function 122. ’$bZ y = w ^t 2 - 3t + 2h dt H$m Z{V n[adV©Z
x 0

w
y = ^t - 3t + 2h dt is :
2
{~ÝXþ hmoJm :
0

1 3 `1 , 3 j
(A) `- , - j (B) 1 3
(A) `- , - j (B) `1 , 3 j
2 2 2 2 2 2 2 2
3 3 `- 3 , - 3 j
(C) ` , j (D) 3 3
(C) ` , j (D) `- 3 , - 3 j
2 4 2 4 2 4 2 4

lim $2x tan x - r . is : 123. lim $2x tan x - r . H$m ‘mZ h¡ :
123. The r cos x
x"
r cos x x"
2 2
(A) 0 (B) –1 (A) 0 (B) –1
(C) – 3 (D) – 2 (C) – 3 (D) – 2

124. The equation of the normal to the curve 124. àW‘ MVwWmªe Ho$ AY©^mOH$ Ho$ gmW dH«$ y = - x + 2
y = - x + 2 at the point of its intersection with Ho$ H$Q>mZ {~ÝXþ na dH«$ Ho$ A{^bå~ H$m g‘rH$aU Š`m
the bisector of the first quadrant is : hmoJm ?
(A) 2x – y – 1 = 0 (B) 2x – y + 1 = 0 (A) 2x – y – 1 = 0 (B) 2x – y + 1 = 0
(C) 4x – y + 16 = 0 (D) 4x – y = 16 (C) 4x – y + 16 = 0 (D) 4x – y = 16

125. Let the equation of a curve is given in implicit form 125. ‘mZm dH«$ H$m g‘rH$aU H$m àmê$n y = tan(x + y) h¡ V~
d2 y d2 y
as y = tan(x + y). Then in terms of y is : y Ho$ nXm| ‘| hmoJm :
dx 2 dx 2
- 2 (1 + y 2) 2 (1 + y 2) 2 - 2 (1 + y 2) 2 (1 + y 2) 2
(A) (B) (A) (B)
y5 y5 y5 y5
2 (1 + y 2) - 2 (1 + y 2) 2 (1 + y 2) - 2 (1 + y 2)
(C) (D) (C) (D)
y6 y6 y6 y6

126. Suppose the area of the Δ ABC is 10 3 . Length of 126. ‘mZm {H$ {Ì^wO ­Δ ABC H$m joÌ’$b 10 3 h¡ IÊS> A ­C
segments AC and AB be 5 and 8 respectively. Then
VWm ­AB H$s bå~mB©`m§ H«$‘e… 5 VWm 8 h¡ Vmo H$moU A h¡:
the angle A is (are) :
(A) 60° or 120° (B) 45° or 135° (A) 60° ¶m 120° (B) 45° ¶m 135°
(C) 30° or 150° (D) 90° (C) 30° ¶m 150° (D) 90°

127. The angle at which the curve y = x 2 and the curve 127. dh H$moU {Og na dH«$ y = x 2 VWm dH«$
5 5 5 5
x = cos t, y = sin t intersect is : x= cos t, y = sin t H$mQ>Vo h¢ :
3 4 3 4
2 41 2 41
(A) - tan- 1 (B) 2 tan- 1 (A) - tan- 1 (B) 2 tan- 1
41 2 41 2
2 41 2 41
(C) tan- 1 (D) tan- 1 (C) tan- 1 (D) tan- 1
41 2 41 2

1-AD ] [ 28 ] [ Contd...

128. The maximum value of the function
128. ’$bZ y = 2 tan x - tan 2 x H$m A§Vamb 80, r2 B na
y = 2 tan x - tan 2 x over 80, B is :
r
2 A{YH$V‘ ‘mZ hmoJm :
(A) 3 (B) 2
(A) 3 (B) 2
(C) ∞ (D) 1
(C) ∞ (D) 1

129. Let O  =  (0, 0), A  =  (a, 11) and B = (b , 37) are the 129. EH$ g‘~mhþ {Ì^wO OAB Ho$ O = ( 0 , 0 ) ,
vertices of an equilateral triangle OAB, then a and b A = ( a , 11 ) VWm B = (b , 37) erf© h¢ Vmo a VWm b {ZåZ
satisfy the relation : gå~ÝY g§Vwï> H$a|Jo :
(A) (a 2 + b 2) + 3ab = 130 (A) (a 2 + b 2) + 3ab = 130
(B) (a 2 + b 2) - 3ab = 138 (B) (a 2 + b 2) - 3ab = 138
(C) (a 2 + b 2) - 4ab = 138 (C) (a 2 + b 2) - 4ab = 138
(D) (a 2 + b 2) - ab = 124 (D) (a 2 + b 2) - ab = 124

130. Let f be an odd function defined on the real 130. ‘mZm {H$ f EH$ {df‘ ’$bZ dmñV{dH$ g§»`mAmo§ na Bg àH$ma
numbers such that f (x) = 3 sin x + 4 cos x, for n[a^m{fV h¡ {H$ x $ 0, Ho$ {bE f (x) = 3 sin x + 4 cos x,
x $ 0, then f (x) for x < 0 is : V~ x < 0 Ho$ {bE f (x) hmoJm:
(A) 3 sin x + 4 cos x (A) 3 sin x + 4 cos x
(B) 3 sin x − 4 cos x (B) 3 sin x − 4 cos x
(C) −3 sin x +4 cos x (C) −3 sin x +4 cos x
(D) − 3 sin x − 4 cos x (D) − 3 sin x − 4 cos x

1
131. The function f (x) = x tan- 1
x
for x ! 0, 131. ’$bZ f (x) = x tan- 1 1x for x ! 0, f (0)  = 0 h¡ `h ’$bZ
f (0)  = 0 is: hmoJm :
(A) Not continuous at x = 0
(A) x = 0 na gVV Zht h¡
(B) continuous at x = 0 but not differentiable at
(B) x = 0 na gVV h¡ naÝVw x = 0 na AdH$bZr` Zht hmoJm
x=0
(C) x = 0 na AdH$bZr` hmoJm
(C) Differentiable at x = 0
(D) Neither continuous at x = 0 nor differentiable (D) Z Vmo x = 0 na gVV h¡ Z hr x = 0 na AdH$bZr`
at x = 0 h¡

132. Let  a and  b be two numbers where a  < b The 132. ‘mZm a VWm b Xmo g§»`mE± h¡§ Ohm± a  < b h¡ & BZ Xmo
geometric mean of these numbers exceeds the g§»`mAmo§ H$m JwUmoÎma ‘mÜ` N>moQ>r g§»`m a go 12 A{YH$
smaller number a by 12 and the arithmetic mean
h¡ VWm BÝht Xmo g§»`mAmo§ H$m g‘mÝVa ‘mÜ` ~‹S>r g§»`m b
of the same number is smaller by 24 than the larger
number b , then the value of b - a is :
go 24 N>moQ>m h¡ Vmo b - a H$m ‘mZ hmoJm :
(A) 44 (A) 44
(B) 27 (B) 27
(C) 48 (C) 48
(D) 45 (D) 45

1-AD ] [ 29 ] [ P.T.O.

133. The values of a and b for which the function 133. a VWm b Ho$ Š`m ‘mZ hm|Jo {OgHo$ {bE ’$bZ
y = a log e x + bx 2 + x, has extremum at the points y = a log e x + bx 2 + x, {~ÝXþAmo x1 = 1 VWm x2 = 2
x1 = 1 and x2 = 2 are : na Ma‘ ‘mZ aIVm h¡ :
2 =1 2 =1
(A) a = - ,b (A) a = - ,b
3 6 3 6
1 1 1 1
(B) a = - , b = - (B) a = - , b = -
3 6 3 6
2 =- 1 2 =- 1
(C) a = ,b (C) a = ,b
3 6 3 6
2 =- 1 2 =- 1
(D) a = - ,b (D) a = - ,b
3 6 3 6
134. A point p is selected randomly from the interior 134. EH$ {~ÝXþ p H$mo EH$ d¥Îm Ho$ ^rVar ^mJ go `mÑpÀN>H$ ê$n
of the circle, then the probability that it is closer go M`Z {H$`m OmVm h¡ Vmo dh àm{`H$Vm Š`m hmoJr {Og‘o
to the center of the circle rather than its boundary `h dñVwV… d¥Îm H$s n[agr‘m Ho$ ZOXrH$ hmoZo H$s ~OmE
is : d¥Îm Ho$ Ho$ÝÐ Ho$ A{YH$ ZOXrH$ hmoJm:
3 1 3 1
(A) (B) (A) (B)
4 3 4 3
2 1 2 1
(C) (D) (C) (D)
3 4 3 4

135. If the letters of the word ASHOKA are written 135. `{X eãX A ­ SHOKA Ho$ Ajam| H$mo `mÑpÀN>H$ ê$n go
down at randomly, then the chance that all A’s are {bIm OmE Vmo g^r A Ho$ H«$‘mJV (H«$‘ go bJmVma) hmoZo
consecutive is : H$s àm{`H$Vm Š`m hmoJr?
2 1 2 1
(A) (B) (A) (B)
3 2 3 2
1 1 1 1
(C) (D) (C) (D)
3 4 3 4

136. In a triangle Δ ABC 136. ¶{X {Ì^wO Δ ABC ‘|
3 sin A + 4 cos B = 6 and 3 sin A + 4 cos B = 6 VWm
4 sin B + 3cos A = 1, then the angle C is : 4 sin B + 3cos A = 1, Vmo H$moU C hmoJm:
(A) 60° (B) 30° (A) 60° (B) 30°
(C) 150° (D) 45° (C) 150° (D) 45°

dx
137. The value of the integral w x xdx- a is equal to: 137. g‘mH$b w H$m ‘mZ h¡ :
2 2 x x2 - a2
a + a +
(A) sin- 1 c (A) sin- 1 c
|x | |x |
1 a 1 a
(B) c + sin- 1 (B) c + sin- 1
a |x | a |x |

1 a 1 a
(C) c - sin- 1 (C) c - sin- 1
a |x | a |x |

1 a 1 a
(D) c - cos- 1 (D) c - cos- 1
a |x | a |x |

1-AD ] [ 30 ] [ Contd...

138. The function y specified implicitly by the relation 138. EH$ ’$bZ y, gå~ÝY
y x y x

w e dt + w cos t dt = 0 satisfies
t
the differential w e dt + w cos t dt = 0 Ûmam {Z{X©ï> {H$`m J`m h¡ `h
t

0 0 0 0

equation : ’$bZ {ZåZ AdH$b g‘rH$aU H$mo g§Vwï> H$aoJm :
d2 y dy 2 d2 y dy 2
(A) e 2 2 + c m m = sin x
yc (A) e y c 2 + c m m = sin x
dx dx dx 2 dx

ycd y 2
dy 2 d2 y dy 2
(B) e c m m = sin x
2 + dx
(B) e y c + c m m = sin x
dx dx 2 dx

2y c d y 2
dy 2 d2 y dy 2
(C) e c m m = sin x
2 + dx
(C) e 2y c + c m m = sin x
dx dx 2 dx

n ycd y 2
dy 2 d2 y dy 2
(D) e c m m = sin 2x
2 + dx
(D) e y c + c m m = sin 2x
o dx dx 2 dx

139. Let a and b be real numbers such that 139. ‘mZm a VWm b dmñV{dH$ g§»`mE± Bg Vah h¢ {H$
sin a + sin b =
1
and cos a + cos b =
6
then sin a + sin b =
1
VWm cos a + cos b = 26 Vmo
2 2 2
the value of sin (a + b) is : sin (a + b) H$m ‘mZ Š`m hmoJm :
2 1 2 1
(A) (B) (A) (B)
3 2 2 3 2 2
1 3 1 3
o (C) (D) (C) (D)
3 2 3 2

140. The tangent to the graph of a continuous function 140. EH$ gVV ’$bZ y = f (x) Ho$ J«m’$ na EH$ {~ÝXþ ({OgH$m x
y = f (x) at the point with abscissa x = a forms {ZX}em§H$ x = a h¡ ) na ñne© aoIm x Aj Ho$ gmW r3 H$moU
r ~ZmVr h¡ VWm Xygao {~ÝXþ ({OgH$m x {ZX}em§H$ x = b h¡ )
with the x axis an angle of and at the point with
3
r
abscissa x = b an angle of , then what is the value na ItMr JB© ñne© aoIm x Aj Ho$ gmW r4 H$moU ~ZmVr h¡
4 b

Vmo g‘mH$b w e x {f l (x) + f m (x)} dx H$m ‘mZ Š`m hmoJm?
b

of the integral w e {f l (x) + f m (x)} dx ?
x
a
a
(`hm± f l (x) ’$bZ f H$m x Ho$ gmW àW‘ AdH$b h¡ Omo
(where f l (x) the derivative of f w.r.to x which is
assumed to be continuous and similarly f m (x) the {H$ gVV h¡ d Cgr àH$ma f m (x) ’$bZ f H$m x Ho$ gmW
double derivative of f w.r.to x) {ÛVr` AdH$b h¡ )
(A) eb + 3e a (B) - eb + 3e a (A) eb + 3e a (B) - eb + 3e a
(C) eb + 3 e a (D) eb - 3 e a (C) eb + 3 e a (D) eb - 3 e a

1 -1 2 x 3 1 -1 2 x 3
141. The system f3 5 - 3 p f y p = fb p has no 141. {ZH$m` f3 5 - 3 p f y p = fb p H$m hb Zht
2 6 a z 2 2 6 a z 2
solution if hmoJm `{X
(A) a ! - 5, b = 5 (B) a ! - 5, b ! 5 (A) a ! - 5, b = 5 (B) a ! - 5, b ! 5
(C) a = - 5, b ! 5 (D) a = - 5, b = 5 (C) a = - 5, b ! 5 (D) a = - 5, b = 5

1-AD ] [ 31 ] [ P.T.O.

142. Let a , b be the roots of x 2 + 3x + 5 = 0 then the 142. ‘mZm g‘rH$aU x 2 + 3x + 5 = 0 Ho$ ‘yb a , b h¡ Vmo dh
equation whose roots are -
1
a
1
and - is :
b
g‘rH$aU Š`m hmoJm {OgHo$ ‘yb - a1 VWm - 1b h¡ :
(A) 5x 2 + 3x - 1 = 0 (A) 5x 2 + 3x - 1 = 0
(B) 5x 2 - 3x + 1 = 0 (B) 5x 2 - 3x + 1 = 0
(C) 5x 2 + 3x - 4 = 0 (C) 5x 2 + 3x - 4 = 0
(D) 5x 2 - 3x + 4 = 0 (D) 5x 2 - 3x + 4 = 0

143. A closed figure S is bounded by the 143. EH$ ~§X AmH¥${V S, A{Vnadb` x 2 - y 2 = a 2 VWm
hyperbola x - y 2 = a 2 and the straight line
2
gab aoIm x = a + h; (h > 0, a > 0) Ûmam n[a~Õ h¡ &
x = a + h; (h > 0, a > 0) . This closed figure is Bg ~§X AmH¥${V S H$mo x-Aj Ho$ n[aV…Ky{U©V {H$`m OmVm h¡
rotated about the x-axis. Then the volume of the Vmo Bg ~§X AmH¥${V Ho$ n[a^«‘U Ho$ R>mog H$m Am`VZ hmoJm:
solid of revolution is :
rh 2 rh 2 rh 2 rh 2
(A) (3a + h) (B) (3a + h) (A) (3a + h) (B) (3a + h)
3 2 3 2
rh 2 rh 2
(C) rh 2 (3a + h) (D) (3a + h) (C) rh 2 (3a + h) (D) (3a + h)
6 6

144. The general solution of the equation 144. {ZåZ g‘rH$aU H$m ì`mnH$ hb hmoJm :
dy y2 - x dy y2 - x
= is : =
dx 2y (x + 1) dx 2y (x + 1)
c c
(A) y 2 = (1 - x) log -1 (A) y 2 = (1 - x) log -1
(1 + x) (1 + x)
c - c -
(B) y 2 = (1 + x) log 1 (B) y 2 = (1 + x) log 1
1+ x 1+ x
(C) y 2 = (1 + x) log (1 + x) - c (C) y 2 = (1 + x) log (1 + x) - c

c c
(D) y 2 = (1 + x) log -1 (D) y 2 = (1 + x) log -1
^1 - xh ^1 - xh

145. The equation of displacement of a particle is 145. EH$ H$U H$m {dñWmnZ x (t) = 5t 2 - 7t + 3 h¡& O~
x (t) = 5t 2 - 7t + 3 . The acceleration at the moment BgH$m doJ 5m / sec hmo OmVm h¡ Cg jU ËdaU {H$VZm
when its velocity becomes 5m / sec is : hmoJm ?:
(A) 10m / sec2 (B) 8m / sec2 (A) 10m / sec2 (B) 8m / sec2
(C) 3m / sec2 (D) 7m / sec2 (C) 3m / sec2 (D) 7m / sec2

146. If 5p 2 - 7p - 3 = 0 and 5q 2 - 7q - 3 = 0, 146. `{X 5p 2 - 7p - 3 = 0 VWm
p ! q , then the equation whose roots are 5p – 4q and 5q 2 - 7q - 3 = 0, p ! q , h¡ Vmo dh g‘rH$aU Š`m
5q – 4p is : hmoJm {OgHo$ ‘yb 5p – 4q VWm 5q – 4p h¢ :
(A) 5x 2 + 7x + 439 = 0 (A) 5x 2 + 7x + 439 = 0
(B) 5x 2 + x - 439 = 0 (B) 5x 2 + x - 439 = 0
(C) 5x 2 + 7x - 439 = 0 (C) 5x 2 + 7x - 439 = 0
(D) 5x 2 - 7x - 439 = 0 (D) 5x 2 - 7x - 439 = 0

1-AD ] [ 32 ] [ Contd...

147. The range of x for which the formula 147. dh x H$s namg Š`m hmoJr {OgHo$ {bE gyÌ
3 sin x = sin- 1 6 x^3 - 4x 2h@ hold is :
-1
3 sin-1
x = sin- 1 6 x^3 - 4x 2h@ ‘mÝ` ahVm h¡:
1 - 2 #x# 2 1
(A) -
3
# x #1 (B)
3 3 (A) - # x #1 (B) - 2 #x# 2
3 3 3
1 1 1 2 1 1 1 2
(C) - # x # (D) - # x # (C) - # x # (D) - # x #
2 2 4 3 2 2 4 3

148. The equation of the ellipse, whose focus is the 148. Cg XrK©d¥Îm H$m g‘rH$aU Š`m hmoJm {OgH$s Zm{^ {~ÝXþ
point ( – 1 , 1 ), whose directrix is the straight line
(–1, 1 ), h¡ VWm {OgH$s {Z`Vm gab aoIm x – y + 3 = 0
x – y + 3 = 0 and whose eccentricity is 1/2 is :
h¡ VWm {OgH$s CËHo$ÝÐVm 1/2 h¡ :
1 - + 2
(A) (x + 1) 2 + (y - 1) 2 = (x y 3) 1 - + 2
6 (A) (x + 1) 2 + (y - 1) 2 = (x y 3)
1 6
(B) (x + 1) 2 + (y - 1) 2 = (x - y + 3) 2 1
2 (B) (x + 1) 2 + (y - 1) 2 = (x - y + 3) 2
2
1 - + 2
(C) (x + 1) 2 + (y - 1) 2 = (x y 3) 1 - + 2
8 (C) (x + 1) 2 + (y - 1) 2 = (x y 3)
8
1 - + 2
(D) (x + 1) 2 + (y - 1) 2 = (x y 1) 1 - + 2
8 (D) (x + 1) 2 + (y - 1) 2 = (x y 1)
8
2 2
149. The mean value of the function f (x) = on 149. ’$bZ f (x) = H$m ‘mÜ` ‘mZ A§Vamb
ex + 1 ex + 1
the interval [ 0 , 2 ] is : [ 0 , 2 ] na hmoJm :
2 2
(A) 2 + log e c 2 m (A) 2 + log e c 2 m
e -1 e -1
2 2
(B) - 2 + log e c 2 m (B) - 2 + log e c 2 m
e -1 e -1
2 2
(C) 2 - log e c 2 m (C) 2 - log e c 2 m
e +1 e +1
2 2
(D) 2 + log e c 2 m (D) 2 + log e c 2 m
e +1 e +1

150. The general solution of the differential equation 150. AdH$b g‘rH$aU
dy x+ y x- y dy x+ y x - y H$m ì`mnH$ hb hmoJm :
+ sin = sin is : + sin = sin
dx 2 2 dx 2 2
y
y
= - sin x + c (A) log e tan = - sin x + c
(A) log e tan 2 2
2 2
y
y
= - 2 sin x + c (B) log e tan = - 2 sin x + c
(B) log e tan 4 2
4 2
y
y
= - 2 sin x + c (C) log e tan = - 2 sin x + c
(C) log e tan 2 2
2 2
y
y
= 2 sin x + c (D) log e tan = 2 sin x + c
(D) log e tan 4 2
4 2

1-AD ] [ 33 ] [ P.T.O.

SPACE FOR ROUGH WORK / H$ÀMo H$m‘ Ho$ {b¶o OJh

1-AD ] [ 34 ]

SPACE FOR ROUGH WORK / H$ÀMo H$m‘ Ho$ {b¶o OJh

1-AD ] [ 35 ]

SPACE FOR ROUGH WORK / H$ÀMo H$m‘ Ho$ {b¶o OJh

1-AD ] [ 36 ]