UPSEE 2017 Paper 2 PCB Code BB

 PAPER-2 PCB àíZnwpñVH$m H«$‘m§H$ àíZnwpñVH$m H$moS>

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

Q. Booklet Code

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

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I have read and understood the instructions given on page No. 1 Seal of Superintendent of Examination Centre

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narjmWu H$m Zm‘/
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narjmWu H$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¡&
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To be copied by the candidate in your own handwriting in the space given below for this purpose is compulsory.
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 
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-2 PCB 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$m Zm‘/
Name of Candidate : BB
Q. Booklet Code
narjm{W©¶m| Ho$ {bE {ZX}e /INSTRUCTIONS TO CANDIDATE
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PAPER-2
Physics : Q. 1 to Q. 50
Chemistry : Q. 51 to Q. 100
Biology : Q. 101 to Q. 150

PHYSICS / ^m¡{VH$emó

001. A block of mass of 1kg is moving on the x axis. A 001. 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$m doJ
the block at time t = 2s is - 3m/s . What is the speed
of the block at time t = 4s ? - 3m/s h¡ Vmo g‘` t = 4s na ãbm°H$ H$s Mmb Š`m hmoJr?

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

002. Two particles P and Q are moving on a circle. 002. 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
acceleration 8 m/s2 and centripetal acceleration 8 m/s2 VWm A{^H|${Ð` ËdaU 5 m/s2 h¡ O~{H$ Q Ho$db
5 m/s2 whereas Q has only centripetal acceleration A{^H|${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
with respect to Q is :
Ho$ gmnoj P H$m ËdaU (m/s2 ‘|) h¡ :
(A) 12 (B) 14 (A) 12 (B) 14
(C) 80 (D) 10 (C) 80 (D) 10

2-BB ] [2] [ Contd...

003. In the given figure, atmospheric pressure P0 = 1 atm 003. {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 85cm of Hg (A) 85cm Hg ñV§^ Xm~
(B) pressure of 67cm of Hg (B) 67cm Hg ñV§^ Xm~
(C) pressure of 90cm of Hg (C) 90cm Hg ñV§^ Xm~
(D) pressure of 78cm of Hg (D) 78cm Hg ñV§^ Xm~

004. PV diagram of an ideal gas is shown. The gas 004. 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) process is isochoric (A) àH«$‘ g‘Am`VZr h¡
(B) work done by gas is positive (B) J¡g Ûmam H$m`© YZmË‘H$ h¡
(C) work done by gas is negative (C) J¡g Ûmam H$m`© F$UmË‘H$ h¡
(D) temperature of gas increases continuously (D) J¡g H$m Vmn bJmVma ~‹T>Vm h¡

005. A small object of mass of 100gm moves in a circular 005. 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) 20  kgm2 s–3
(A) 20  kgm2 s–3
(B) 200  kgm2 s–3
(B) 200  kgm2 s–3
(C) 300  kgm2 s–3
(C) 300  kgm2 s–3
(D) 10000 kgm2 s–3
(D) 10000 kgm2 s–3

. 2-BB ] [3] [ P.T.O.

006. A time varying horizontal force (in Newton) 006. 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 right (A) Xm`t Va’$ J{V H$aoJm
(B) It will oscillate (B) XmobZ H$aoJm
(C) It remains stationary (C) {dam‘ ‘| hr ahoJm
(D) It moves towards left (D) ~m§`r Va’$ J{V H$aoJm

007. Take Bulk modulus of water B = 2100MPa . What 007. `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) 84 kPa (B) 210 kPa (A) 84 kPa (B) 210 kPa
(C) 840 kPa (D) 8400 kPa (C) 840 kPa (D) 8400 kPa

008. Thin semicircular part ABC has mass m1 and 008. 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.
AO = OC = R h¡ & Bg g§`wº$ {ZH$m` H$s Cg Aj (axis)
The moment of inertia of this composite system
about the axis is:
Ho$ gmnoj O‹S>Ëd AmKyU© hmoJm :

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

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

009. In Young’s double slit experiment, the path 009. `§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$m 13.5
on screen is 13.5 times the wavelength. The point is: JwUm h¡ Vmo {~ÝXþ hmoJm :
(A) dark (A) AXrá
(B) bright but not central bright (B) Xrá naÝVw Ho$ÝÐr` Xrá Zht
(C) neither bright nor dark (C) Z Vmo Xrá Z hr AXrá
(D) central bright (D) Ho$ÝÐr` Xrá

2-BB ] [4] [ Contd...

010. A ball having velocity v towards right and having 010. 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$a ahr h¡& Xrdma go
collides elastically with wall and moves towards
left. Ground and wall are frictionless . Select the
`h àË`mñW Q>¸$a H$aVr h¡ VWm `h ~m`t Va’$ bm¡Q>Vr h¡ &
correct statement about angular velocity of the ball O‘rZ d Xrdma Kf©Ua{hV h¡ & Xrdma Ho$ gmW Q>¸$a Ho$ ~mX
after collision. J|X Ho$ H$moUr` doJ Ho$ ~mao ‘| ghr H$WZ M`Z H$amo -

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

011. Which of the following particle will describe the 011. {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) electron (B) proton (A) BboŠQ´moZ (B) àmoQ>moZ
+ (C) He+ (D) Li+
(C) He (D) Li+

012. A loop PQR carries a current of 2A as shown. A 012. {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) 4 Nm (B) 16 Nm (A) 4 Nm (B) 16 Nm
(C) 8 Nm (D) zero (C) 8 Nm (D) eyݶ

013. The sides of a rectangle are 7.01 m and 12 m. Taking 013. 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 m2 (B) 84.1 m2
(A) 84 m2 (B) 84.1 m2
(C) 84.00 m2 (D) 84.12 m2 (C) 84.00 m2 (D) 84.12 m2

014. In steady state, charge on 3nF capacitor is : 014. ñWm`r AdñWm ‘o 3nF g§Ym[aÌ na Amdoe hmoJm:

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

2-BB ] [5] [ P.T.O.

015. Consider one dimensional motion of a particle. 015. 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 g‘` t Ho$ ‘Ü` J«m’$ Xem©`m J`m h¡& H$m¡Zgm J«m’$ g‘` t
graph 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)

016. An object of mass 26kg floats in air and it is in 016. 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) 26 m3 (B) 10 m3 (A) 26 m3 (B) 10 m3
(C) 20 m 3 (D) 13 m3 (C) 20 m3 (D) 13 m3

017 In the given circuit cell E has internal resistance of 017. {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) 1 W (B) 2 W (A) 1 W (B) 2 W
(C) 3 W (D) 5 W (C) 3 W (D) 5 W
018. Two cylindrical rods A and B have same resistivities 018. 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 1 1
(A) (B) (A) (B)
4 2 4 2
(C) 2 (D) 4 (C) 2 (D) 4

019. Which of the following material is not ferromagnetic 019. {ZåZ ‘| go H$m¡Zgm nXmW© bm¡ôMwå~H$Ëd àH¥${V H$m Zht h¡?
in nature ?
(A) Al (B) Fe (A) Al (B) Fe
(C) Co (D) Ni (C) Co (D) Ni

2-BB ] [6] [ Contd...

020. Three small balls of masses 1kg, 2kg and 3kg are 020. 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«$‘e… 1  m/s, 2  m/s VWm 3  m/s go {MÌmZwgma
t 2  m/s and 3 m/s respectively as shown. The total J{V H$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) 7 kgm2s–1 (B) 8 kgm2s–1 (A) 7 kgm2s–1 (B) 8 kgm2s–1
(C) 9 kgm2s–1 (D) 36 kgm2s–1 (C) 9 kgm2s–1 (D) 36 kgm2s–1

021. Three identical resistors each of resistance R are 021. VrZ EH$ g‘mZ à{VamoY {OZ‘| àË`oH$ H$m à{VamoY R h¡ H$mo
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:

m

2V 2 3V 2 2V 2 3V 2
(A) (B) (A) (B)
3R 2R 3R 2R
3V 2 V2 3V 2 V2
(C) (D) (C) (D)
R 3R R 3R
R
022. For given logic diagram , output F=1, then inputs are: 022. {XE JE VH©$ n[anW ‘| {ZJ©V F=1, V~ {Zdoer h¡:

(A) A = 0, B = 0, C = 1 (B) A = 0, B = 0, C = 0 (A) A = 0, B = 0, C = 1 (B) A = 0, B = 0, C = 0
(C) A = 0, B = 1, C = 0 (D) A = 1, B = 1, C = 1 (C) A = 0, B = 1, C = 0 (D) A = 1, B = 1, C = 1
023. Consider two polaroids A and B as shown. 023. Xem©E AZwgma Xmo nmoboamoBS> A VWm B na {dMma H$s{OE&
Unpolarized light is incident on polaroid A. Now AY«w{dV àH$me nmoboamoBS> A na Amn{VV hmoVm h¡ & A~
both the polaroids are rotated simultaneously by
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$s
m remain parallel to each other. During the rotation, nmaJ‘Z Aj h‘oem EH$ 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) decreases continuously (A) bJmVma KQ>Vr h¡
(B) increases continuously (B) bJmVma ~‹T>Vr h¡
(C) first increases then decreases (C) nhbo KQ>Vr h¡ {’$a ~‹T>Vr h¡
(D) remains same (D) g‘mZ ahVr h¡
024. Activity of a radioactive substance becomes from 024. EH$ 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) 3 days (B) 4 days (A) 3 {XZ (B) 4 {XZ
(C) 6 days (D) 2 days (C) 6 {XZ (D) 2 {XZ

2-BB ] [7] [ P.T.O.

025. The energy levels of a hypothetical one electron 025. 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) 690 A° (B) 1035 A° (A) 690 A° (B) 1035 A°
(C) 1220 A° (D) 3650 A° (C) 1220 A° (D) 3650 A°

026. What is the voltage across an ideal PN junction 026. {MÌmZwgma n[anW ‘| {XE JE AmXe© PN g§{Y S>m`moS> na
diode for shown circuit ? dmoëQ>Vm Š`m hmoJr ?

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

027. Power emitted by a black body at temperature 50°C 027. 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) 16 P (A) 16 P
(B) greater than P but less than 16P (B) P go A{YH$ naÝVw 16 P go H$‘
(C) greater than 16P (C) 16 P go A{YH$
(D) P (D) P

028. An experimenter needs to heat a small sample to 028. 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 Ûmam {MÌmZwgma ~‹S>r dñVw go {d{H$aU H$mo EH$ ~‹S>o b|g
a large lens to concentrate the radiation from the Ûmam à{VXe© na H|${ÐV H$a à{VXe© H$m 900K Vmn {H$`m
large object onto the sample as shown below ? Om gH$Vm h¡ ?

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

2-BB ] [8] [ Contd...

029. Consider a small electric dipole with magnitude of 029. 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) exactly zero (B) (A) nyU©V`m eyÝ` (B)
r2 r2
-k p kp -k p kp
(C) (D) (C) (D)
r2 r r2 r

030. A conducting loop (as shown) has total resistance 030. {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$mo
perpendicular to plane of the loop where γ is a Amamo{nV {H$`m OmVm h¡ Ohm± γ AMa h¡ VWm t g‘` h¡&
constant and t is time. The induced current flowing
through loop is : byn go àdm{hV ào[aV Ymam hmoJr:

(b 2 + a 2) c (b 2 + a 2) ct (b 2 + a 2) c (b 2 + a 2) ct
(A) (B) (A) (B)
R R R R
(b 2 - a 2) c (b 2 - a 2) ct (b 2 - a 2) c (b 2 - a 2) ct
(C) (D) (C) (D)
R R R R
031. A uniform disc of mass M and radius R is hinged at 031. 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¡:

F F F F
(A) (B) 3 (A) (B) 3
2MR MR 2MR MR
F 2 F F 2 F
(C) (D) (C) (D)
MR 3 MR MR 3 MR
032. The velocity of a particle is zero at time t = 2 , then 032. g‘` t = 2 na H$U H$m doJ eyÝ` h¡ Vmo
(A) acceleration must be zero at t = 2 (A) t = 2 na ËdaU eyÝ` hr hmoJm &
(B) displacement must be zero in the interval (B) t = 0 go t = 2 A§Vamb ‘| {dñWmnZ eyÝ` hr hmoJm &
t = 0 to t = 2. (C) t = 2 na ËdaU eyÝ` hmo gH$Vm h¡ &
(C) acceleration may be zero at t = 2
(D) velocity must be zero for t > 2 (D) t > 2 Ho$ {bE doJ eyÝ` hr hmoJm &

033. A ball moving in xy plane, has velocity (4 it - 4tj ) m/s 033. 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 R>rH$ nyd© doJ (4 it - 4tj ) m/s h¡& Q>¸$a Ho$ {bE àË`mdñWZ
1
restitution for collision is e = . What will be velocity JwUm§H$ e = 12 h¡& O‘rZ go Q>¸$a Ho$ R>rH$ nümV J|X H$m
2
of the ball just after the collision with ground? doJ Š`m hmoJm ?

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

2-BB ] [9] [ P.T.O.

034. A light ray moving in medium- I (of refractive 034. 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 in medium-II.
(A) {H$aU nyU©V`m ‘mÜ`‘-II ‘| nmaJ{‘V hmoVr h¡&
(B)  ray will move completely parallel to the (B) {H$aU A§Vg©Vh Ho$ nyU©V`m g‘mÝVa Om`oJr &
interface .
(C) {H$aU A~ ^r A§Vg©Vh na nyU©V`m Am§V[aH$
(C) ray will be still totally internally reflected at namd{V©V hmoJr &
interface. (D) {H$aU ‘mÜ`‘-II ‘| nyU©V`m nmaJ{‘V Ho$db
(D)  ray will be totally transmitted into medium-II V^r hmoJr O~ AmnVZ H$moU ~‹T>m`m OmVm
only if angle of incidence is increased. h¡&
035. A light beam consists of two types of photons. 035. EH$ 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 ’$moQ>moBboŠQ´moZ
of emitted photoelectron is :
(A) leV (B) 2eV
H$s A{YH$V‘ J{VO D$Om© h¡ :
(C) 3eV (D) 4eV (A) leV (B) 2eV
(C) 3eV (D) 4eV
036. A light beam parallel to axis is incident on 036. 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 nümV {H$aUo§ (region) joÌ I, III VWm V ‘| Aj
axis in regions I, III and V? Ho$ g‘mÝVa hmo OmE±:

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

2-BB ] [ 10 ] [ Contd...

037. A long silver tea spoon is placed in a cup filled 037. EH$ bå~o Mm§Xr Ho$ Mm` Må‘M H$mo J‘© Mm` go
with hot tea. After some time, the exposed end ^ao H$n ‘o§ aIm OmVm h¡ & Hw$N> g‘` ~mX Må‘M
(the end which is not dipped in tea) of the H$m Iwbm {gam (Omo Mm` ‘o Zht Sy>~m h¡) J‘© hmo
spoon becomes hot even without a direct contact
OmVm h¡ `Ú{n `h Mm` Ho$ grYo g§nH©$ ‘o Zht
with the tea. This phenomenon can be explained
Wm & `h à^md ‘w»` ê$n go {ZåZ go g‘Pm Om
mainly by:
gH$Vm h¡ :
(A) thermal expansion (B) conduction
(A) D$î‘r` àgma (B) MmbZ
(C) reflection (D) radiation
(C) namdV©Z (D) {d{H$aU

038. Figure shows a nonconducting semicircular rod 038. {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 +x axis
(A) +x Aj Ho$ AZw{Xe
(B) along +y axis
(B) +y Aj Ho$ AZw{Xe
(C) electric field is zero at point P, so direction
(C)  {~ÝXþ P na {dÚwV joÌ eyÝ` h¡ AV… {Xem kmV Zht
cannot be determined.
H$s Om gH$Vr h¡
(D) along the bisector of x axis and y axis. (D) x Aj d y Aj Ho$ AÕ©^mOH$ Ho$ AZw{Xe

039. A bead of mass m can slide without friction on a 039. 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$s
àmH¥${VH$ bå~mB© R h¡ VWm pñà§J H$m Xygam {gam
spring is fixed at point O as shown in figure. Bead
{MÌmZwgma {~ÝXþ O na O‹S>dV h¡ & ‘ZHo$ H$mo pñW{V
is released from position A, what will be kinetic A go ‘wº$ {H$`m OmVm h¡ Vmo O~ `h pñW{V B
energy of the bead when it reaches at point B ? na nhþ§MVm h¡ V~ ‘ZHo$ H$s J{VO D$Om© hmoJr:

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

2-BB ] [ 11 ] [ P.T.O.

040. The total electrostatic energy stored in both the 040. XmoZm| g§Ym[aÌ ‘o g§J«{hV Hw$b pñWa {dÚwV D$Om© h¡ :
capacitors is :

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

041. Gravitational force acts on a particle due to fixed 041. EH$ g‘mZ R>mog O‹S>dV Jmobo Ho$ H$maU EH$ H$U na
uniform solid sphere. Neglect other forces. Then Jwê$Ëdr¶ ~b bJVm h¡, AÝ` ~b ZJÊ` h¡ & V~ `h
particle : H$U:
(A) experiences a force directed along the radial (A) Ho$db {ÌÁ`r` {Xem Ho$ AZw{Xe hr ~b AZw^d H$aoJm&
direction only. (B) h‘oem {ÌÁ`r` {Xem Ho$ bå~dV J{V H$aoJm&
(B) always moves normal to the radial direction
(C) h‘oem {ÌÁ`r` {Xem Ho$ AZw{Xe J{V H$aoJm&
(C) always moves in the radial direction only.
(D) always moves in circular orbit. (D) h‘oem d¥Îmr` J{V H$aoJm&

042. A block performs simple harmonic motion with 042. 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 maximum at t = 3s.
(A) t = 3s na H$U H$s Mmb A{YH$V‘ h¡&
(B)  displacement from equilibrium is maximum
(B) t = 4s na H$U H$m gmå`mdñWm go {dñWmnZ A{YH$V‘ h¡&
at t = 4s.
(C) speed is maximum at t = 4s. (C) t = 4s na Mmb A{YH$V‘ h¡ &
(D) speed is minimum at t = 2s. (D) t = 2s na Mmb Ý`yZV‘ h¡&

043. There are two identical springs each of spring 043. `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 ?

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

2-BB ] [ 12 ] [ Contd...

044. Two tuning forks A and B produce 4 beats/sec. 044. 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) 2 (B) 5
(A) 2 (B) 5
(C) 9 (D) 20
(C) 9 (D) 20

045. A 10gm bullet moving directly upward at 1000 m/s 045. 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
bullet comes out of it ? doJ Š`m hmoJm ?

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

046. Two identical balls P and Q are projected with 046. 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$moU
with making projection angles with horizontal 30° H«$‘e… 30° d 60° na àjo{nV H$s OmVr h¡ VWm do grYo
and 60° respectively and they fall directly on plane hr Vb A­ B na H«$‘e… {~ÝXþ P' d Q' na {JaVr h¡& Xyar Ho$
AB at points P' and Q' respectively. Which of the gå~ÝY ‘§o H$m¡Zgm {dH$ën gË` h¡ ?
following statement is true about distances as given
in options?

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

2-BB ] [ 13 ] [ P.T.O.

047. A string has a length of 5m between fixed points 047. 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$s
the frequency of the second overtone ? Amd¥{V Š`m hmoJr ?
(A) 30 Hz (B) 40 Hz (A) 30 Hz (B) 40 Hz
(C) 50 Hz (D) 60 Hz (C) 50 Hz (D) 60 Hz
048. Displacement x versus t2 graph is shown for a 048. EH$ 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) 2m/s2 (B) 4m/s2 (A) 2m/s2 (B) 4m/s2
(C) 8m/s2 (D) zero (C) 8m/s2 (D) eyݶ

049. For given LR circuit, growth of current as function 049. {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) 0.4 s (B) 0.7 s (A) 0.4 s (B) 0.7 s
(C) 1 s (D) 2.4 s (C) 1 s (D) 2.4 s

050. Radii of two conducting circular loops are b and a 050. 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 ra 2 n0 ra 2
(A) (A)
2b 2b

n0 rb 2 n0 rb 2
(B) (B)
2a 2a
(C) zero (C) eyݶ
n0 rab
(D) n0 rab
2 (a + b) (D)
2 (a + b)

2-BB ] [ 14 ] [ Contd...

CHEMISTRY / agm¶Zemó

051. Which of the following molecules is optically 051. {ZåZ ‘| go H$m¡Zgo AUw àH$m{eH$ g{H«$` h¡ ?
active ?

(A) (i) and (ii) (B) (i) and (iii) (A) (i) Am¡a (ii) (B) (i) Am¡a (iii)
(C) (ii) and (iii) (D) (i), (ii) and (iii) (C) (ii) Am¡a (iii) (D) (i), (ii) Am¡a (iii)
052. Which of the following statement is correct ? 052. {ZåZ ‘| go H$m¡Zgm H$WZ gË` h§¡ ?
(A) BCl3 and AlCl3 are both Lewis acids and BCl3 (A) BCl3 Am¡a A ­ lCl3 XmoZm| bwB©g Aåb h¢ Ed§ BCl3,
is stronger than AlCl3 AlCl3 go e{º$embr h§¡
(B) BCl3 and AlCl3 are both Lewis acids and (B) BCl3 Am¡a AlCl3 XmoZm| bwB©g Aåb h¢ Ed§ ­AlCl3,
AlCl3 is stronger than BCl3 BCl3 go e{º$embr h¢
(C) BCl3 and AlCl3 are both equally strong Lewis (C) BCl3 Am¡a ­AlCl3 XmoZm| g‘mZ e{º$embr bwB©g
acid Aåb h¢
(D) Both BCl3 and AlCl3 are not Lewis acids (D) BCl3 Am¡a A ­ lCl3 XmoZm| hr bwB©g Aåb Zht h¢
053. Consider the following compounds. 053. ZrMo {XE JE `m¡{JH$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)
(III) (IV)
Friedel–Crafts acylation can be used to obtain:
(A) I, III, IV (B) II, III, IV
(A) I, III, IV (B) II, III, IV
(C) I, II, IV (D) I, II, III
(C) I, II, IV (D) I, II, III

054. Provide the systematic name of the compound 054. ZrMo àX{e©V `m¡{JH$ H$m ì`dpñWV Zm‘ Xr{O`o:
shown:

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

2-BB ] [ 15 ] [ P.T.O.

055. Give the IUPAC name for the following structure: 055. {ZåZ ga§MZm H$m IUPAC Zm‘ Xr{O`o::

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

056. In aldol addition reaction product is always: 056. EëS>mob `moJmË‘H$ A{^{H«$`m ‘| CËnmX h‘oem hmoJm :
(A) b – hydroxyaldehyde (A) b – hmB©S´moŠgrEëS>rhmB©S>
(B) b – hydroxyketone (B) b – hmB©S´moŠgrH$sQ>moZ
(C) a, b – unsaturated aldehyde (C) a, b – Ag§V¥á EëS>rhmB©S>
(D) a, b – unsaturated ketone (D) a, b – Ag§V¥á H$sQ>moZ

057. Which one of the following compounds will have 057. {ZåZ ‘| go H$m¡Zgo `m¡{JH$ Ho$ {bE {XY«wd AmKyU© H$m ‘mZ
the highest dipole moment ? A{YH$V‘ hmoJm ?

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

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

058. The number of moles of Grignard reagent consumed 058. 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) 4 (B) 2 (A) 4 (B) 2
(C) 3 (D) 1 (C) 3 (D) 1

059. The paramagnetic species is : 059. {ZåZ ‘| go AZwMwåãH$s` h¡ :
(A) KO2 (B) SiO2 (A) KO2 (B) SiO2
(C) TiO2 (D) BaO2 (C) TiO2 (D) BaO2
060. Which one of the following has the highest 060. {ZåZ ‘| go {H$gH$s Zm{^H$ ñZo{hVm A{YH$V‘ h¡?
Nucleophilicity ?
(A) F – (B) OH – (A) F – (B) OH –
(C) CH3 (D) NH2
- -
(C) CH3 (D) NH2
- -

2-BB ] [ 16 ] [ Contd...

061. In view of ∆rG0 for the following reactions : 061. {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 +4, for tin +2 (A) b¡S> Ho$ {bE +4, {Q>Z Ho$ {bE +2
(B) For lead +2, for tin +2 (B) b¡S> Ho$ {bE +2, {Q>Z Ho$ {bE +2
(C) For lead +4, for tin +4 (C) b¡S> Ho$ {bE +4, {Q>Z Ho$ {bE +4
(D) For lead +2, for tin +4 (D) b¡S> Ho$ {bE +2, {Q>Z Ho$ {bE +4

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

063. At Critical Micell Concentration (CMC), the 063. H«$m§{VH$ {‘gob gm§ÐVm na gµ’$}ŠQ>oÝQ> AUw :
surfactant molecules: (A) AnK{Q>V hmoVo h¢
(A) decompose (B) {d`mo{OV hmoVo h¡§
(B) dissociate (C) g§`mo{OV hmoVo h¡§
(C) associate
(D) nyU©V`m KwbZerb hmoVo h¡§
(D) become completely soluble

064. Which one of the following will be reactive for 064. n{H©$Z g§KZZ A{^{H«$`m Ho$ {bE {ZåZ ‘| go H$m¡Z {H«$`merb
Perkin condensation ? hmoJm?
(A) C6H5 CHO (B) CH3     O CHO (A) C6H5 CHO (B) CH3     O CHO

(C) CH3 CHO (D) O2N CHO (C) CH3 CHO (D) O2N CHO

065. The pair of metal carbonyl complexes that are 065. YmVw H$m~m}{Zb g§Hw$b `m¡{JH$ H$m H$m¡Zgm `w½‘ g‘BboŠQ´mZ
isoelectronic is : h¡:
(A) [Co(CO)4]– and Ni(CO)4 (A) [Co(CO)4]– Am¡a Ni(CO)4
(B) Ni(CO)4 and V(CO)6 (B) Ni(CO)4 Am¡a V(CO)6
(C) [Cr(CO)6] and V(CO)6 (C) [Cr(CO)6] Am¡a V(CO)6
(D) [Fe(CO)4]– and Cr(CO)6 (D) [Fe(CO)4]– Am¡a Cr(CO)6

066. Which one of the following has (have) octahedral 066. {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) (i), (ii) & (iii) (B) (i), (ii) & (iv) (A) (i), (ii) & (iii) (B) (i), (ii) & (iv)
(C) (ii), (iii) & (iv) (D) All of these (C) (ii), (iii) & (iv) (D) ¶o g^r

2-BB ] [ 17 ] [ P.T.O.

067. In terms of polar character which one of the 067. Y«wdr` àH¥${V Ho$ g§X^© ‘| {ZåZ ‘| H$m¡Zgm H«$‘ ghr h¡?
following orders is correct?
(A) NH3 < H2O < HF <H2S (A) NH3 < H2O < HF <H2S
(B) H2S < NH3 < H2O < HF (B) H2S < NH3 < H2O < HF
(C) H2O < NH3 < H2S < HF (C) H2O < NH3 < H2S < HF
(D) HF < H2O < NH3 <H2S (D) HF < H2O < NH3 <H2S

068. Among the following compounds of Boron, the 068. ~moamZ Ho$ {ZåZ{bpIV `m¡{JH$m| ‘| go H$m¡Z σ – ~§Ymo§ Ho$ gmW
species which also forms π – bond in addition to gmW π – ~§Y ^r ~ZmVm h¡ :
σ – bonds is:
(A) BF 4- (B) BH3 (A) BF 4- (B) BH3
(C) B2H6 (D) BF3 (C) B2H6 (D) BF3

069. Identify the Brönsted acid in the following equation: 069. {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) OH- (B) PO3-
4 (A) OH- (B) PO3-
4
(C) HPO4 (D) H2O (C) HPO4 (D) H2O

070. The number of grams/weight of NH4Cl required to 070. Vmn 298K na 9.45 pH Ho$ ~’$a {d{b`Z H$mo V¡`ma
be added to 3 liters of 0.01M NH3 to prepare the H$aZo Ho$ {bE NH4Cl Ho$ {H$VZo J«m‘/^ma H$mo 3 brQ>a
buffer of pH=9.45 at temperature 298K 0.01M   NH3 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) 3.53 gm (B) 0.354 gm
(A) 3.53 gm (B) 0.354 gm
(C) 4.55 gm (D) 0.455gm
(C) 4.55 gm (D) 0.455gm

071. For the reaction 2HI (g) H 2 (g) + I 2 (g) the 071. 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:
1+ 2 Kp 1 + 2K p 1+ 2 Kp 1 + 2K p
(A) (B) (A) (B)
2 2 2 2
2K p 2 Kp 2K p 2 Kp
(C) (D) (C) (D)
1 + 2K p 1 + 2 Kp 1 + 2K p 1 + 2 Kp

072. A 6% solution of sucrose C22H22O11 is isotonic with 072. 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) 342 (B) 684 (A) 342 (B) 684
(C) 171 (D) 100 (C) 171 (D) 100

073. The enthalpy of the formation of CO2 and H2O are 073. 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) 235 kJ (B) 340 kJ (A) 235 kJ (B) 340 kJ
(C) 420 kJ (D) 491 kJ (C) 420 kJ (D) 491 kJ

2-BB ] [ 18 ] [ Contd...

074. Which of the following is a lyophobic colloid : 074. {ZåZ ‘| go H$m¡Zgm EH$ Ðd{damJr H$mobmBS> h¡ :
(A) Gelatin (B) Sulphur (A) {OboQ>rZ (B) gë’$a
(C) Starch (D) Gum Arabica
(C) ñQ>mM© (D) J‘ Aao{~H$

075. For car battery which one is correct statement ? 075. H$ma H$s ~¡Q>ar Ho$ {bE H$m¡Zgm H$WZ gË` h¡ ?
(A) Cathode is Lead dioxide (PbO2) and anode is (A) H¡$WmoS> boS> S>mBAm°ŠgmBS> (PbO2) Ed§ EZmoS> boS>
Lead (Pb) (Pb) hmoVm h¡
(B) Cathode is Lead dioxide (PbO2) and anode is (B) H¡$WmoS> boS> S>mBAm°ŠgmBS> (PbO2) Ed§ EZmoS> H$m°na
Copper (Cu) (Cu) hmoVm h¡
(C) Cathode is Copper (Cu) and anode is Lead (C) H¡$WmoS> H$m°na (Cu) Ed§ EZmoS> boS> S>mBAm°ŠgmBS>
dioxide (PbO2) (PbO2) ) hmoVm h¡
(D) Cathode is Copper (Cu) and anode is Lead (Pb) (D) H¡$WmoS> H$m°na (Cu) Ed§ EZmoS> boS> (Pb) hmoVm h¡

076. Considering entropy(s) as a thermodynamic 076. 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) (∆ SV§Ì – ∆Sn[adoe ) > 0
(A) (∆ Ssystem – ∆Ssurrounding ) > 0
(B) ∆Ssystem > 0 only (B) Ho$db ∆SV§Ì > 0
(C) ∆S surrounding > 0 only (C) Ho$db ∆S n[adoe > 0
(D) (∆ Ssystem + ∆Ssurrounding ) > 0 (D) (∆SV§Ì + ∆Sn[adoe ) > 0

077. At low pressure and high temperature, the Vander 077. 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$n hmoJm:
(A) PVm = RT (A) PVm = RT
a a
(B) c P + m (Vm - b) = RT (B) c P + m (Vm - b) = RT
V m2 V m2
(C) P(Vm – b) = RT (C) P(Vm – b) = RT
a a
(D) c P + m V = RT (D) c P + m V = RT
V m2 m V m2 m
078. Which graph represents the zero order reaction 078. {ZåZ ‘| go H$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)

079. Which of the following compounds is insoluble 079. {ZåZ ‘| go H$m¡Zgm `m¡{JH$ J‘© gmÝÐ H2SO4 ‘| ^r A{dbo`
even in hot concentrated H2SO4? h¡ ?
(A) Ethylene (B) Benzene (A) E{WbrZ (B) ~|OrZ
(C) Hexane (D) Aniline (C) hoŠgoZ (D) E{ZbrZ

2-BB ] [ 19 ] [ P.T.O.

080. The half life of Th232 is 1.4 × 1010 years and that 080. 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) 5 × 10–10gm (B) 5.0 gm (A) 5 × 10–10gm (B) 5.0 gm
(C) 1.95 × 10–9 gm (D) 2 × 10–10 gm (C) 1.95 × 10–9 gm (D) 2 × 10–10 gm

081. Which of the following electron has minimum energy? 081. {ZåZ{bpIV ‘| go H$m¡Zgm BboŠQ´m°Z Ý`yZV‘ D$Om© aIVm h¡?
1 1
(A) n = 3, l  =  2, m = –2, s = + (A) n = 3, l  =  2, m = –2, s = +
2 2
1 1
(B) n = 4, l =  0, m = 0, s = + (B) n = 4, l =  0, m = 0, s = +
2 2
1 1
(C) n = 4, l = 1, m = +1, s = + (C) n = 4, l = 1, m = +1, s = +
2 2
1 1
(D) n = 5, l = 0, m = 0, s = + (D) n = 5, l = 0, m = 0, s = +
2 2
082. Total number of stereoisomers of the following 082. {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) 4, 6 (B) 8, 0 (A) 4, 6 (B) 8, 0
(C) 6, 6 (D) 8, 8 (C) 6, 6 (D) 8, 8

083. Which of the following is a monomer of Dacron: 083. {ZåZ ‘| go H$m¡Zgm So>H«$moZ H$m EH$bH$ h¡ ?
(A) CH 2 CH - CH CH 2 (A) CH 2 CH - CH CH 2
Cl Cl

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

(C) COOH COOH (C) COOH COOH

(D) HOH 2 C - CH 2 OH (D) HOH 2 C - CH 2 OH

084. Which of the following is a meso compound ? 084. {ZåZ ‘| go H$m¡Zgm {‘gmo `m¡{JH$ h¡ ?
(A) trans–1, 4–dimethylcyclohexane (A) Q´m§g -1, 4- S>mB©‘o{WbgmBŠbmohoŠgoZ
(B) cis–1, 3–dimethylcyclohexane (B) {gg -1, 3- S>mB©‘o{WbgmBŠbmohoŠgoZ
(C) trans–1, 3–dimethylcyclohexane (C) Q´m§g -1, 3- S>mB©‘o{WbgmBŠbmohoŠgoZ
(D) cis–1, 4–dimethylcyclohexane (D) {gg -1, 4- S>mB©‘o{WbgmBŠbmohoŠgoZ

2-BB ] [ 20 ] [ Contd...

085. IUPAC name of the following is : 085. {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) 2,5 Butyl butenal
(A) 2, 5 ã`w{Q>b ã`yQ>oZb
(B) 2,3 di ethyl butenal (B) 2, 3 S>mB E{Wb ã`yQ>oZb
(C) 2 ethyl–3 methyl pentanal (C) 2 E{Wb, 3 ‘o{Wb n|Q>oZb
(D) 8 methyl– 2 ethyl pentanal (D) 8 ‘o{Wb, 2 E{Wb n|Q>oZb

086. Which of the following is Reimer - Tieman reaction? 086. {ZåZ ‘| go H$m¡Zgr ar‘a Q>r‘mZ A{^{H«$`m h¡ ?
OH OH

(A) + CHCl3 + aq. NaOH (A) + CHCl3 + aq. NaOH

OH OH

(B) + CHCl3 + alcoh. NaOH (B) + CHCl3 + alcoh. NaOH

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

OC2H5 OC2H5
Conc.H2 SO4 (D) Conc.H2 SO4
(D) Conc.HNO3
Conc.HNO3
087. The increasing order of the first ionization enthalpies 087. 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) B < P < S < F (B) B<S<P<F (A) B < P < S < F (B) B < S < P < F
(C) F < S < P < B (D) P < S < B < F (C) F < S < P < B (D) P < S < B < F

088. Some pairs of ions are given below. In which pair, 088. 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¡?
5 5
5 5 p
p
(A) H3 C - CH - CH3 and - CH - OCH (A) H3 C - CH - CH3 VWm - CH - OCH 3
3

5 5
(B) H3 C - CH 2 - CH - CH3 and (B) H3 C - CH 2 - CH - CH3 VWm
5 5
H 2 C - CH 2 - CH - CH 2 H 2 C - CH 2 - CH - CH 2
5 5
5
CH 2
5
CH 2 CH 2 CH 2
(C) and (C) VWm

H3 C - CH - CH3 H3 C - N - CH3 H3 C - CH - CH3 H3 C - N - CH3
(D) and (D) VWm
H 2 C - C - CH3 H3 C - C - CH3 H 2 C - C - CH3 H3 C - C - CH3
5 5 5 5

089. Which alkaline earth metal compound is volatile ? 089. {ZåZ ‘| go H$m¡Zgm jmar` ‘¥Xm YmVw `m¡{JH$ dmînerb h¡ ?
(A) Be3N2 (B) Mg3N2 (A) Be3N2 (B) Mg3N2
(C) Ca3N2 (D) None of the options (C) Ca3N2 (D) BZ‘o go H$moB© {dH$ën Zht

2-BB ] [ 21 ] [ P.T.O.

090. What is the name of the following reaction? 090. {ZåZ A{^{H«$`m H$m Zm‘ Š`m h¡ ?
NaOH NaOH
HCHO + HCHO CH3OH + HCOONa HCHO + HCHO CH3OH + HCOONa
T T
(A) Hell–Volhard reaction (A) hob dmoëhmS©> A{^{H«$`m
(B) Clemmensen reaction (B) Šbo‘|gZ A{^{H«$`m
(C) Cannizzaro reaction (C) H$m{ZµOamo A{^{H«$`m
(D) None of the options (D) H$moB© ^r {dH$ën Zht
091. Inorganic graphite is: 091. AH$m~©{ZH$ J«o’$mBQ> h¡ :
(A) B2N3H6 (B) B2H6 (A) B2N3H6 (B) B2H6
(C) BN (D) BF3 (C) BN (D) BF3
092. Rank the following in decreasing order of basic strength: 092. {ZåZ H$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) ii > i > iv > iii (B) iv > i > ii > iii (A) ii > i > iv > iii (B) iv > i > ii > iii
(C) i > iv > ii > iii (D) i > iv > iii > ii (C) i > iv > ii > iii (D) i > iv > iii > ii
093. Among the given compound choose the two that 093. {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) (i),(iii) (B) (ii),(iv) (A) (i),(iii) (B) (ii),(iv)
(C) (i),(ii) (D) (ii),(iii) (C) (i),(ii) (D) (ii),(iii)
094. Increasing order of acidic strength of given 094. {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) iii < i < iv < ii (B) ii < i < iv < iii (A) iii < i < iv < ii (B) ii < i < iv < iii
(C) i < iii < iv < ii (D) i < iii < ii < iv (C) i < iii < iv < ii (D) i < iii < ii < iv

095. Which of the following effects of –NO2 group 095. 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 Me Me
NO2 NO2
(A) Only –I effect (A) Ho$db –I à^md
(B) Only +M effect (B) Ho$db +M à^md
(C) Only –M effect (C) Ho$db –M à^md
(D) Both–I   and  –M effect (D) XmoZm|–I  Am¡a  –M à^md

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096. Which of the following material is known as lunar 096. {ZåZ ‘| go H$m¡Zgm nXmW© byZa H$mpñQ>H$ Ho$ Zm‘ go OmZm
caustic ? OmVm h¡ ?
(A) NaNO3 (B) AgCl (A) NaNO3 (B) AgCl
(C) AgNO3 (D) NaOH (C) AgNO3 (D) NaOH

097. Provide an acceptable name for the alkane shown 097. 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) 6–ethyl–2–methyl–5–propyldecane (A) 6–E{Wb–2–‘o{Wb–5–àmo{nbSo>Ho$Z
(B) 5–ethyl–6–methyl–2–propyldecane (B) 5–E{Wb–6–‘o{Wb–2–àmo{nbSo>Ho$Z
(C) 2–ethyl–6–methyl–2–propyldecane (C) 2–E{Wb–6–‘o{Wb–2–àmo{nbSo>Ho$Z
(D) 2–ethyl–6–methyl–5–propyldecane (D) 2–E{Wb–6–‘o{Wb–5–àmo{nbSo>Ho$Z

HO - HO - HO - HO -
098. D – Mannose D – glucose (A) 098. D – ‘oZmoO D– ½byH$moO (A)
Product (A) of above reaction is: Cnamoº$ A{^{H«$`m H$m CËnmX (A) h¡:
(A) D–glucose (A) D – ½byH$moO
(B) D–fructose (B) D – ’«$ŠQ>moO
(C) D–Talose (C) D – Q>obmoO
(D) D–Idose (D) D – Am`moS>moO

099. What is the product in the following reaction ? 099. {ZåZ A{^{H«$`m H$m CËnmX hmoJm ?
OH OH
(NH4) 2 Cr2 O7 (NH4)2 Cr2 O7
H2SO4 H2SO4

(A) Benzoic Acid (A) ~oݵOmoBH$ Aåb
(B) Benzoquionone (B) ~oÝOmo{¹$Zm|Z
(C) Cyclohexane-1-one (C) gm`ŠbmohoŠgoZ-1-AmoZ
(D) Benzoic sulphate (D) ~oݵOmoBH$ gë’o$Q>

100. How many bonds are there in : 100. {XE JE AUw ‘| Hw$b {H$VZo ~§Y h¢ ?

(A) 14σ, 8p (A) 14σ, 8p
(B) 18σ, 8p (B) 18σ, 8p
(C) 19σ, 4p (C) 19σ, 4p
(D) 14σ, 2p (D) 14σ, 2p

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Biology / Ordemó

101. Juvenile harmone is secreted by : 101. {H$emoa hm‘m}Z H$m ódU BgHo$ Ûmam hmoVm h¡ :
(A­) Thyroid gland (A) Wm`amoBS> J«§{W
(B) Thymus gland (B) Wm`‘g J«§{W
(C) ­Adrenal gland (C) E{S´Zb J«§{W
(D) Carpora allata (D) H$manmoam EbmQ>m

102. Which organ is concerned with the formation 102. Iaho ‘| H$m¡Zgm A§J `y[a`m {Z‘m©U go gå~§{YV
of urea in rabbit ? h¡?
(A­) Blood (A) aº$
(B) Kidney (B) d¥³H$
(C) Spleen (C) ßbrhm
(D) Liver (D) `H¥$V

103. In mammals which organ acts as blood 103. ñVZYm[a`m| ‘| H$m¡Zgm A§J ãbS> ~¢H$ H$s Vah
bank ? H$m`© H$aVm h¡?
(­A) Heart (A) öX`
(B) Lung (B) ’o$’$‹S>o
(C) Liver (C) brda
(D) Spleen (D) ßbrhm

104. What is chemical composition of fertilizin? 104. ’$Q>ubmBOrZ H$m amgm`{ZH$ g§KQ>Z Š`m h¡?
(­A) Lipoprotein (A) {bnmoàmoQ>rZ
(B) Glycoprotein (B) ½bmBH$moàmoQ>rZ
(C) Mucoprotein (C) å`wH$moàmoQ>rZ
(D) Chromoprotein (D) H«$mo‘moàmoQ>rZ

105. Rh factor is present in : 105. Rh factor CnpñWV hmoVm h¡ :
(­A)­ All vertebrates (A) g^r H$eoé{H$`m| ‘|
(B) ­All mammals (B) g^r ñVZYm[a`m| ‘|
(C) A­ll reptiles (C) g^r garg¥nm| ‘|
(D) Man and Rhesus monkey only. (D) {g’©$ ‘Zwî` Ed§ argg ~ÝXa ‘|

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106. Which of the following characteristic was not 106. ‘Q>a Ho$ nm¡Ym| na {H$E JE à`moJm| ‘| ‘|S>b
used by Mendel in his experiment on Pea Zo {ZåZ ‘| go {H$g bjU H$m Cn`moJ Zht
plant? {H$`m?
(A­) Flower colour (A) ’y$b H$m a§J
(B) Seed coat colour (B) ~rOmdaU H$m a§J
(C) Pod colour (C) ’$br H$m a§J
(D) Fruit shape (D) ’$b H$s AmH¥${V

107. A­scaris protects itself against digestive enzymes 107. EñHo$[ag ñd`§ H$mo nanmofr Ho$ nmMH$ E§OmB‘mo go
of the host by: BgHo$ Ûmam ~MmVm h¡ :
(­A) Mucus (A) ûcoî‘m
(B) A­ntienzyme (B) EÝQ>rE§OmB‘
(C) ­Antienzyme and cuticle (C) EÝQ>rE§OmB‘ Ed§ Š`y{Q>H$b
(D) Cuticle (D) Š`y{Q>H$b

108. HIV that causes A­IDS first starts destroying : 108. AIDS CËnÞ H$aZo dmbm HIV g~go nhbo Zï>
(A­) Leucocytes H$aVm h¡ :
(A) ë`yH$mogmBQ>mo H$mo
(B) Helper T lymphocytes
(B) hoëna T-{bå’$mogmBQ>mo
(C) Thrombocytes
(C) W«moå~mogmBQ>mo
(D) b  -  Lymphocytes (D) b  -  {bå’$mogmBQ>mo

109. Which one of the following is absent in 109. ñnÝOmo ‘| {ZåZ{bpIV ‘| go AZwnpñWV hmoVo h¢?
sponges?
(A) V§{ÌH$m H$mo{eH$m
(­A) Nerve cell
(B) g§doXr H$mo{eH$m
(B) Sensory cell
(C) JǤ{W H$mo{eH$m
(C) Gland cell
(D) All of the options (D) {XE JE g^r {dH$ën

110. The signals for parturition originate from : 110. àgd {H«$`m hoVw g§Ho$Vm| H$m AmZm Amaå^ hmoVm h¡  :
(­A) Fully developed foetus only (A) Ho$db nyU© {dH${gV J^© go

(B) Placenta only (B) Ho$db Anam go

(C) Placenta and fully developed foetus (C) Anam VWm nyU© {dH${gV J^© go

(D) Oxytocin harmone (D) Am°ŠgrQ>mo{gZ hm‘m}Z

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111. Kala-azar is caused by : 111. H$mbm AOa H$m H$maH$ h¡ :
(A­) Trypanosma cruzi (A) {Q´nZmogmo‘m H«y$Or
(B) Leishmania donovani (B) boe‘m{Z`m S>moZmodoZr
(C) Trypanosoma brucei (C) {Q´nZmogmo‘m ~«ygr
(D) Trypanosoma garnbeinse (D) {Q´nZmogmo‘m Joa~|gr

112. Hydra can not digest : 112. hmBS´m Zht nMm gH$Vm h¡ :
(A­) Proteins (A) àmoQ>rÝg
(B) Fats (B) dgmE±
(C) Starch (C) ñQ>mM©
(D) Sugars (D) eH©$am

113. A­n earthworm has : 113. EH$ H|$MwE ‘| hmoVo h¢ :
(A­) one pair of eyes (A) ZoÌm| H$s EH$ Omo‹S>r
(B) two pair of eyes (B) ZoÌm| H$s Xmo Omo‹S>r
(C) four pair of eyes (C) ZoÌm| H$s Mma Omo‹S>r
(D) no eyes (D) ZoÌ Zht hmoVo h¡

114. Which insect is useful for us? 114. H$m¡Zgm H$sQ> h‘mao {bE Cn`moJr h¡?
(A­) Periplaneta (A) no[aßboZoQ>m
(B) Musca (B) ‘ñH$m
(C) Bombyx (C) ~mpå~Šg
(D) Mosquitoes (D) ‘ÀN>a

115. The sequence of cell cycle is : 115. H$mo{eH$m MH«$ H$m H«$‘ h¡ :
(­A) S, M, G1 and G2 (­A) S, M, G1 Am¡a G2
(B) G1, G2, S and M (B) G1, G2, S Am¡a M
(C) G1, S, G2 and M (C) G1, S, G2 Am¡a M
(D) M, G1, G2 and S (D) M, G1, G2 Am¡a S

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116. “Green glands” found in some of A­rthropodes 116. Hw$N> AmW«monmoS>m OÝVwAm| ‘| nmB© OmZo dmbr ""J«rZ
are helpful in : ½b|S²g'' ‘XX H$aVr h¡ :
(A­) Respiration (A) œgZ ‘|
(B) Digestion (B) nmMZ ‘|
(C) Excretion (C) CËgO©Z ‘|
(D) Reproduction (D) àOZZ ‘|

117. One of the ex-situ conservation method for 117. g§H$Q>J«ñVOm{V`m| Ho$ na-ñWmZo g§ajU H$s EH$
endangered species is : {d{Y h¡ :
(A­) Wildlife sanctuarie (A) dÝ`Ord Aä`maÊ`
(B) Biosphere Reserve (B) O¡d‘§S>b [aOd©
(C) National park (C) amï´>r` nmH©$
(D) Cryopresevation (D) {ZåZ Vmn n[aajU

118. What is Epiblema? 118. E{nãbo‘m Š`m h¡?
(­A) Epidermis of root (A) ‘yb H$s A{YM‘©
(B) Epidermis of leaf (B) nÎmr H$s A{YM‘©
(C) Epidermis of stem (C) ñVå^ H$s A{YM‘©
(D) Epidermis of flower (D) nwîn H$s A{YM‘©

119. Widal test is carried out to test : 119. {dS>mb Q>oñQ> Ûmam nhMmZ H$s OmVr h¡ :
(A­) A­IDS (A) A­IDS H$s
(B) Typhoid fever (B) Q>m`’$mBS> Áda H$s
(C) Malaria (C) ‘bo[a`m H$s
(D) Diabetes (D) ‘Yw‘oh H$s

120. Which one feature is common to leech, 120. H$m¡Zgm EH$ bjU Om|H$, H$m°H$amoM VWm {~ÀNy> ‘|
cockroach and scorpion ? g‘mZ h¡?
(­A) Nephridia (A) d¥³H$H$
(B) Ventral Nerve cord (B) AYa V§{ÌH$m a‚mw
(C) Cephaliztion (C) {eamo^dZ
(D) A­ntennal (D) l§{JH$mE

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121. Which of the following group of plants exhibit 121. {ZåZ ‘| go H$m¡Zgo nmXn g‘yh ‘| Om{V {d{dYVm
more species diversity? A{YH$ nmB© OmVr h¡?
(­A) A­ngiosperms (A) AmdV©~rOr
(B) ­Algae (B) e¡dmb
(C) Bryophyta (C) ~«m`mo’$mBQ>m
(D) Fungi (D) H$dH$

122. Spore bearing leaf is called 122. ~rOmUw YmaU H$aZo dmbr nÎmr H$hbmVr h¡ :
(A­) Sorus (A) gmoag
(B) Indusium (B) BÝSy>{e`‘
(C) Ramentum (C) ao‘oÝQ>‘
(D) Sporophyll (D) ñnmoamo{’$b

123. Who proposed the natural system of plant 123. nmXn dJuH$aU H$s àmH¥${VH$ nÕ{V {H$gZo àñVm{dV
classification? H$s Wr?
(­A) Carolois Linnaeus (A) H¡$amobg {b{Z`g
(B) John Hutchinson (B) Om°hZ h{MgZ
(C) Bentham and Hooker (C) ~oÝW‘ Ed§ hÿH$a
(D) Oswald Tippo (D) Am°ñdmbS> {Q>ßnmo

124. Coacervates are : 124. H$moEg©doQ²g hmoVo h¢ :
(A) Lipoproteins
(A) {bnmoàmoQ>rÝg
(B) Mixture of ­Ammonia, carbohydrates and
(B) A‘mo{Z`m, H$m~m}hmBS´oQ> Ed§ Ob H$m {‘lU
water
(C) Colloidal suspensions (C) H$mobmoBS>b {Zbå~Z
(D) Fatty acids and nitrogenous compound (D) dgm Aåb Ed§ ZmB©Q´moOrZr `m¡{JH$

125. Medulla oblongata is responsible for : 125. ‘oS>`wbm Am°ãbmJmÝQ>m CÎmaXm`r h¡ :
(A­) Thermaregulation (A) Vmn {Z`§ÌU Ho$ {bE
(B) vision (B) Ñ{ï> Ho$ {bE
(C) memory (C) ñ‘¥{V Ho$ {bE
(D) balance (D) g§VwbZ Ho$ {bE

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126. The cockroach of genus Blatta is also called : 126. ãbmQ>m d§e Ho$ H$m°H$amoO ({VbMÅ>o) H$mo `h ^r
(A­) German cockroach H$hm OmVm h¡ :
(B) A­ustralian cockroach (A) O‘©Z H$m°H$amoO
(B) Am°ñQ´o{b`Z H$m°H$amoO
(C) Oriental cockroach
(C) Amo[a`§Q>b H$m°H$amoO
(D) ­American cockroach
(D) A‘o[aH$Z H$m°H$amoO

127. Groups of five kingdom scheme are : 127. n§M OJV n[aH$ënZm ‘| gpå‘{bV g‘yh h¡ :
(­A) Virus, Bacteria, Fungi, Plantae and Animalia
(A) {dfmUw, OrdmUw, H$dH$, nbmÝQ>r Ed§ E{Z‘o{b`m
(B) Monera, Protista, Fungi, Plantae and
(B) ‘moZao m, àmo{Q>ñQ>m, H$dH$, nbmÝQ>r Ed§ E{Z‘o{b`m
­Animalia
(C) Bacteria, Fungi, Plant and A­nimal (C) OrdmUw, H$dH$, nmXn Ed§ OÝVw
(D) Fungi, Protista, Plant and ­Animal (D) H$dH$, àmo{Q>ñQ>m, nmXn Ed§ OÝVw

128. Hygroscopic roots are found in :
128. AmЩVmJ«mhr O‹S>| nmB© OmVr h¢ :
(A) Trapa (B) Hydrilla
(A) Q´mnm ‘| (B) hmB{S´bm ‘|
(C) Orchid (D) Typha (C) Am°{H©$S> ‘| (D) Q>mB’$m ‘|

129. Fruit of Mango is a : 129. Am‘ H$m ’$b h¡ :
(­A) Pepo (B) Pome (A) nrnmo (B) nmo‘
(C) Berry (D) Drupe (C) ~oar (D) S¯>n

130. Passage cells are found in tissue : 130. ‘mJ© H$mo{eH$mE± Bg D$ÎmH$ ‘| nmB© OmVr h¢ :
(A­) Epidermis (A) A{YM‘©
(B) Xylem (B) OmBb‘
(C) Endodermis (C) AÝVü‘©
(D) Pericycle (D) n[aaå^

131. ­Adenosine is a : 131. E{S>Zmo{gZ h¡ EH$ :
(A­) Nitrogenous base (A) ZmBQ´moOrZr jma

(B) Nucleoside (B) Ý`ypŠb`mogmBS>

(C) Nucleotide (C) Ý`ypŠb`moQ>mBS>

(D) Nucleic acid (D) Ý`ypŠbH$ Aåb

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132. In blue green algae, photo synthesis takes 132. Zrb h[aV e¡dmb ‘| àH$me g§ûcofU H$s {H«$`m
place in : Bg‘| hmoVr h¡ :
(A­) Chloroplast (B) Lamellae (A) ŠbmoamoßbmñQ (B) b¡{‘br
(C) Heterocyst (D) Carotene (C) hoQ>oamo{gñQ (D) H¡$amoQ>rZ

133. What is plasmid? 133. ßbmpÁ‘S> Š`m hmoVm h¡?
(­A) Bacteria (A) OrdmUw
(B) Virus (B) {dfmUw
(C) Chromosomal DNA­ segment (C) JwUgyÌr` DNA I§S>
(D) Extra chromosomal DNA­ segment (D) A{V[aº$ JwUgyÌr` DNA­ I§S>

134. In Bryophyta, a specialized organ of the 134. ~«m`mo’$mBQ>m ‘| ~rOmUwX²{^X H$m EH$ {deof
sporophyte for attachment to the gametophyte A§J Omo Bgo `w½‘H$moX²{^X go Omo‹S>Vm h¡ H$hbmVm
is called : h¢ :
(A­) Stalk (B) Foot (A) ñQ>m°H$ (B) ’w$Q>
(C) Seeta (D) Rhizoids (C) grQ>m (D) amB©OmoBS²g

135. Inverted omega shaped vascular bundles are 135. CbQ>o Amo‘oJm AmH$ma Ho$ g§dhZr ~ÊS>b nmE OmVo
found in : h¡ :
(A­) Cycas corolloid root (A) gmBH$g àmdbm^ ‘yb
(B) Cycas stem (B) gmBH$g VZm
(C) Cycas rachis (C) gmBH$g ao{H$g
(D) Cycas leaflet (D) gmBH$g nU©H$

136. A­pogamy results in formation of : 136. An`w½‘Z Ho$ n[aUm‘ñdê$n ~ZVm h¡ :
(­A) Diploid Sparophyte (A) {ÛJw{UV ~rOmUwX²{^X
(B) Diploid Gametophyte (B) {ÛJw{UV `w½‘H$moX²{^X
(C) Haploid Sparophyte (C) AJw{UV ~rOmUwX²{^X
(D) Haploid Gametophyte (D) AJw{UV `w½‘H$moX²{^X

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137. Cuscuta is a : 137. H$ñHw$Q>m h¡ EH$ :
(­A) Total stem parasite (A) nyU© ñVå^ naOrdr
(B) Partial stem parasite (B) Am§{eH$ ñVå^ naOrdr
(C) Total root Parasite (C) nyU© ‘yb naOrdr
(D) Partial root Parasite (D) Am§{eH$ ‘yb naOrdr

138. Hydathodes are meant for : 138. Oba§Y«m| H$m H$m`© h¡ :
(­A) Respiration (A) œgZ
(B) Guttation (B) {~ÝXwómd
(C) Transpiration (C) dmînmoËgO©Z
(D) Photo respiration (D) àH$mer` œgZ

139. A molecule of A­TP is structurally similar to a 139. EH$ ­A­TP AUw g§aMZmË‘H$ ê$n go BgHo$ g‘mZ
molecule of : hmoVm h¡ :
(­A) RNA­ molecule (A) RNA­ AUw
(B) DN­A molecule (B) DN­A AUw
(C) A­mino acid (C) E{‘Zmo Aåb
(D) Fatty acid (D) dgm Aåb

140. Which colour of light is most effective for 140. àH$me H$m H$m¡Zgm a§J àH$me g§ûcofU Ho$ {bE
photo synthesis? gdm©{YH$ à^mdr h¡?
(­A) Blue (A) Zrbm
(B) Green (B) ham
(C) Red (C) bmb
(D) Violet (D) ~¢JZr

141. Imbibtion is a : 141. AÝV…MyfU h¡ EH$ :
(­A) Chemical Process (A) amgm`{ZH$ à{H«$`m
(B) Biological Process (B) O¡{dH$ à{H«$`m
(C) Physical Process (C) ^m¡{VH$ à{H«$`m
(D) Biochemical Process (D) O¡damgm`{ZH$ à{H«$`m

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142. Which of the plant harmone is extracted 142. H$m¡Zgm nmXn hm‘m}Z H$dH$ go {ZîH${f©V {H$`m J`m
from fungus? h¡?
(­A) ­Auxin (A) Am°ŠgrZ
(B) Gibberellin (B) {O~«o{bZ
(C) Cytokinin (C) gmBQ>moH$mB{ZZ
(D) 2, 4-D (D) 2, 4-D

143. What is emasculation? 143. nwÝgËdhaU Š`m h¡?
(­A) Removal of petals (A) ~møXbm| H$mo hQ>mZm
(B) Removal of sepals (B) Xbm| H$mo hQ>mZm
(C) Removal of anthers (C) namJH$moem| H$mo hQ>mZm
(D) Removal of ovary (D) A§S>me` H$mo hQ>mZm

144. Epinasty movement is found in : 144. AYmoHw§$MZ J{V nmB© OmVr h¡ :
(A­) Leaf sensitivity in Mimosa pudica (A) Nw>B© ‘wB© ‘| nU© Hw§$MZ
(B) Drooping of bud (B) H${bH$m H$m ZrMo bQ>H$Zm
(C) Opening of flower (C) nwîn H$m pIbZm
(D) Young leaves of Fern (D) ’$Z© H$s VéU n{Îm`m±

145. Genes do not pairs in : 145. OrZ `w½‘ ‘| Zht hmoVo h¢ :
(­A) Somatic cells (A) H$m{`H$ H$mo{eH$m ‘|
(B) Gamete (B) `w½‘H$ ‘|
(C) Fertilized egg (C) {Zfo{MV A§S> ‘|
(D) Zygote (D) `w½‘ZO ‘|

146. Ultimate source of genetic variability are : 146. AmZwdm§{eH$ {d{^ÞVm H$m na‘ ómoV h¡ :
(A­) Mutation (A) CËn[adV©Z
(B) Genetic drift (B) AmZwdm§{eH$ Andmh
(C) Gene flow (C) OrZ àdmh
(D) Transformation and translation (D) ê$nmÝVaU Ed§ AZwdmXrH$aU

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147. Which technique is used for the test tube 147. naIZbr {eew H$m`©H«$‘ Ho$ {bE H$m¡Zgr VH$ZrH$
baby program? BñVo‘mb H$s OmVr h¡?
(­A) Gameti intra fallopian transfer (A) `w½‘H$ AÝV… ’¡$bmonr ñWmZmÝVaU
(B) Zygoti intra fallopian transfer (B) `w½‘ZO AÝV… ’¡$bmonr ñWmZmÝVaU
(C) Intra cytoplasmic sperm injection (C) AÝV… H$mo{eH$m Ðì``r ewH«$mUw B§OoŠeZ
(D) Intra uterine insemination (D) AÝV… J^m©e`r dr`©goMZ

148. Main function of tapetum is : 148. Q>onrQ>‘ H$m ‘w»` H$m`© h¡ :
(­A) Protection (A) gwajm
(B) Nutrition (B) nmofU
(C) Pollination (C) namJU
(D) Fertilization (D) {ZfoMZ

149. Dinosaurs disappeared during : 149. S>m`Zmgmoa Bg Xm¡amZ bwá hþE :
(­A) Cretaceous (A) {H«$Q>o{e`g
(B) Permian (B) na{‘`Z
(C) Jurassic (C) Owam{gH$
(D) Triassic (D) Q´mBE{gH$

150. Maximum absorption of water occurs in : 150. Ob H$m A{YH$V‘ AdemofU hmoVm h¡ :
(­A) Colon (A) H$mobZ ‘|
(B) Rectum (B) ‘bme` ‘|
(C) Stomach (C) Am‘me` ‘|
(D) Small Intestine (D) N>moQ>r Am§V ‘|

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

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

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

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