UPSEE 2016 Question Paper 1

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

AA
Question Booklet Sr. No.
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Q. Booklet Code

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

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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
32 Time 3 Hours 600 No. of Questions in Booklet
150

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

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àíZnwpñVH$m H$moS>
narjmWu H$m Zm‘/
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Q. Booklet Code
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PAPER-1
Physics : Q. 1 to Q. 50
Chemistry : Q. 51 to Q. 100
Mathematics : Q. 101 to Q. 150

PHYSICS / ^m¡{VH$emñÌ
001. A small bead of mass M slides on a 001. EH$ M Ðì`‘mZ H$m N>moQ>m ‘ZH$m EH$
smooth wire that is bent in a circle of {MH$Zo Vma na {’$gbVm h¡& `hm± Vma EH$
radius R. It is released at the top of R {ÌÁ`m Ho$ d¥Îm Ho$ ^mJ Ho$ ê$n ‘| ‘w‹S>m
the circular part of the wire (point A hþAm h¡& ‘ZHo$ H$mo d¥{Îm` ^mJ Ho$ {eIa
in the figure) with a negligibly small
velocity. Find the height H where the
({MÌ ‘| q~Xþ A­) go ZJÊ` doJ go ‘wº$
bead will reverse direction. {H$`m OmVm h¡& dh D±$MmB© H kmV H$amo
Ohm± ‘ZH$m AnZr {Xem nbQ>Vm h¡&

3R 5R 3R 5R
(A) (B) (A) (B)
2 2 2 2
(C) R (D) 2R (C) R (D) 2R
002. Two persons A and B start from the same 002. Xmo ì`{º$ A
­ VWm B EH$ hr OJh go EH$
location and walked around a square in
opposite directions with constant speeds. dJ© na {dnarV {XemAm| ‘| AMa Mmbmo§ go
The square has side 60m. Speeds of A MbZm àmaå^ H$aVo h¢& dJ© H$s ^wOm 60m
and B are 4m/s and 2m/s respectively. h¡, A VWm B H$s Mmb| H«$‘e… 4m/s VWm
When will they meet first time? 2m/s h¡& do nhbr ~ma H$~ {‘b|Jo ?
(A) 10 sec (B) 20 sec (A) 10 sec (B) 20 sec
(C) 30 sec (D) 40 sec (C) 30 sec (D) 40 sec

1-AA ] [2] [ Contd...

003. A tire of radius R rolls on a flat surface 003. EH$ R {ÌÁ`m H$m n{h`m g‘Vb gVh na
with angular velocity ω and velocity ν H$moUr` doJ ω VWm doJ ν go {MÌmZwgma
as shown in the diagram. If ν  >  ωR, in
which direction does friction from the bw‹S>H$ ahm h¡& `{X ν  >  ωR Vmo Q>m`a Ûmam
tire act on the road ? g‹S>H$ na Kf©U {H$g {Xem ‘| bJoJm?

(A) Towards the left (A) ~m`t Va’$ (B) Xm`r Va’$
(B) Towards the right (C) ZrMo H$s Va’$ (D) D$na H$s Va’$
(C) Towards downwards
(D) Towards upwards
004. Consider one dimensional motion of 004. EH$ m Ðì`‘mZ Ho$ H$U H$s EH$ {d‘r`
a particle of mass m. It has potential J{V na {dMma H$s{OE & BgH$s pñW{VO
energy U = a + bx2 where a and D$Om© U = a + bx2 h¡ Ohm± a VWm b
b are positive constants. At origin YZmË‘H$ {Z`Vm§H$ h¢& ‘yb {~ÝXþ (x = 0)
(x = 0) it has initial velocity ν0. It na BgH$m àmapå^H$ doJ ν0 h¡ & `h gab
performs simple harmonic oscillations. Amd¥{V J{V H$aVm h¡ {OgH$s Amd¥{V {ZåZ
The frequency of the simple harmonic na {Z^©a H$aVr h¡
motion depends on (A) Ho$db b na
(A) b alone (B) Ho$db b VWm a na
(B) b and a alone (C) Ho$db b VWm m na
(C) b and m alone (D) Ho$db b, a VWm m na
(D) b, a and m alone
005.
The postulate on which the photoelectric 005. àH$me {dÚwV g‘rH$aU {ZåZ ‘| go {Og A{^J¥hrV
equation is derived is (H$ënZm) na ì`wËnÝZ H$s JB© h¡ dh h¡:
(A) electrons are restricted to orbits of (A) BboŠQ´moZ Ho$db CÝht H$jH$m| ‘| ah gH$Vo
h h
angular momentum n
2π
where n h¢ {OZ‘| H$moUr` g§doJ n 2π hmo VWm
is an integer. n EH$ nyUmªH$ h¡&
(B) electrons are associated with wave (B) BboŠQ´moZ go g§~Õ Va§J H$s Va§JX¡Ü`©
h h
of wavelength λ = where p is λ= h¡ Ohm± p g§doJ h¡ &
p p
momentum.
(C) àH$me V^r CËnÞ hmoVm h¡ O~ BboŠQ´moZ
(C) light is emitted only when electrons
jump between orbits.
EH$ H$jH$ go Xÿgao ‘| Hy$XVm h¡ &
(D) àH$me H$m AdemofU D$Om© Ho$ ³dm§Q>m
(D) light is absorbed in quanta of
energy E = hυ E = hυ Ho$ ê$n ‘| hmoVm h¡&

006. A layer of oil with density 724 kg/m3 006. EH$ Vob H$s naV {OgH$m KZËd 724 kg/m3
floats on water of density 1000 kg/m3. h¡& `h 1000 kg/m3 KZËd dmbo Ob Ho$ D$na
A block floats at the oil-water interface V¡a ahr h¡& EH$ ãbm°H$ Vob-Ob AÝVg©Vh na
with 1/6 of its volume in oil and 5/6 {MÌmZwgma Bg àH$ma V¡a ahm h¡ {H$ BgH$m 1/6
of its volume in water, as shown in the Am`VZ Vob ‘| VWm 5/6 Am`VZ Ob ‘| h¡
figure. What is the density of the block? Vmo ãbm°H$ H$m KZËd Š`m hmoJm?

(A) 776 kg/m3 (B) 954 kg/m3 (A) 776 kg/m3 (B) 954 kg/m3
(C) 1024 kg/m3 (D) 1276 kg/m3 (C) 1024 kg/m3 (D) 1276 kg/m3

1-AA ] [3] [ PTO

007. A string fixed at both ends has a standing 007. EH$ añgr XmoZm| {gam| go O‹S>dV h¡ VWm EH$
wave mode for which the distances AàJm‘r Va§J {dYm ‘| H«$‘mJV {ZñnÝXm| Ho$
between adjacent nodes is 18cm. For ‘Ü` Xÿar 18cm h¡& AJbr H«$‘mJV AàJm‘r
the next consecutive standing wave Va§J {dYm ‘| H«$‘mJV {ZñnÝXm| Ho$ ‘Ü` Xÿar
mode distances between adjacent nodes 16cm h¡& añgr H$s Ý`yZV‘ bå~mB© hmoJr
is 16cm. The minimum possible length (A) 288 cm (B) 72 cm
of the string is (C) 144 cm (D) 204 cm
(A) 288 cm (B) 72 cm
(C) 144 cm (D) 204 cm
008. EH$ Vma H$m byn Omo {H$ 20cm2 H$m
008. A wire loop that encloses an area of joÌ’$b n[a~Õ H$aVm h¡ VWm BgH$m à{VamoY
20cm2 has a resistance of 10Ω. The 10Ω h¡& Bg byn H$mo 2.4T Ho$ Mwå~H$s`
loop is placed in a magnetic field of joÌ ‘| Bg àH$ma aIm OmVm h¡ {H$ BgH$m
2.4T with its plane perpendicular to the Vb Mwå~H$s` joÌ Ho$ bå~dV hmo& A~
field .The loop is suddenly removed byn H$mo Mwå~H$s` joÌ ‘| go EH$mEH$ hQ>m
from the field. How much charge flows {X`m OmVm h¡ Vmo Vma (byn)Ho$ {H$gr q~Xþ
past a given point in the wire? go {H$VZm Amdoe àdm{hV hmoVm h¡?
(A) 4.8 × 10– 4C (B) 2.4 × 10– 3C (A) 4.8 × 10– 4C (B) 2.4 × 10– 3C
(C) 1.2 × 10– 4C (D) 10– 1C (C) 1.2 × 10– 4C (D) 10– 1C
009. A right isosceles triangle of side a has 009. EH$ g‘H$moU `wº$ g‘{Û~mhþ {Ì^wO {OgH$s
charges q, + 3q and – q arranged on {MÌmZwgma ^wOm a h¡ VWm Bg na Amdoe
its vertices as shown in the figure . q, + 3q VWm – q BgHo$ erfm] na {MÌmZwgma
What is the electric potential at point ì`dpñWV h¡& Amdoe +q VWm – q H$mo OmoS>Zo
P midway between the line connecting
the + q and – q charges ? dmbr aoIm H$m ‘Ü` q~Xþ P h¡ Vmo q~Xþ
P na {dÚwV {d^d {H$VZm hmoJm?

q 3q
(A) (B) q 3q
πε0 a 2 2 πεo a (A) (B)
3q 3q πε0 a 2 2 πεo a
(C) (D) 3q 3q
πεo a 2 πεo a (C) (D)
πεo a 2 πεo a
010. Shown below is a graph of current
010. ZrMo {X`m J`m J«m’$ S>m`moS> Ho$ {bE Ymam
versus applied voltage for a diode.
Approximately what is the resistance
(current) VWm Amamo{nV dmoëQ>Vm (voltage)
of the diode for an applied voltage of
Ho$ ‘Ü` ~Zm`m J`m h¡& Amamo{nV dmoëQ>Vm
−1.5V Ho$ {bE S>m`moS> H$m à{VamoY bJ^J
−1.5V?
{H$VZm hmoJm?

(A) Zero (B) 1Ω (A) eyÝ` (B) 1Ω
(C) 2Ω (D) ∞ (C) 2Ω (D) ∞
1-AA ] [4] [ Contd...

011. A sound wave is generated by the howl 011. EH$ ^o{‹S>`o H$s VoO AmdmO Ûmam am{Ì ‘|
of a wolf in the night. How would we EH$ Üd{Z Va§J CËnÝZ H$s OmVr h¡ (`hm±
describe the motion of a particular air J¡g AUwAm| Ho$ `mÑpÀN>H$ ^«‘U H$s Cnojm
molecule near the ground, a mile away H$aVo hþE) ^o{‹S>`o go EH$ ‘rb Xÿa O‘rZ
from the wolf, on average (i.e. ignoring na pñWV EH$ hdm Ho$ H$U H$s J{V Am¡gV
the random wandering of gas molecules)? ê$n go {H$g àH$ma àX{e©V hmoJr ?
(A) It moves up and down in an (A) `h D$na ZrMo EH$ XmobZr ê$n ‘|
oscillating fashion J{V H$aoJm &
(B) It moves away from the wolf at (B) `h ^o{‹S>`o go Xÿa H$s Va’$ Üd{Z H$s
the speed of sound Mmb go J{V H$aoJm&
(C) It moves back and forth (oscillating) (C) `h ^o{S‹ >`o H$s Va’$ AmJo nrN>o (XmobZr)
towards the wolf J{V H$aoJm &
(D) It moves in the horizontal circle. (D) `h EH$ jo{VO d¥Îm ‘| J{V H$aVm h¡&

012. Which of the following Material has 012. {ZåZ ‘| go g~go H$‘ à{VamoYH$Vm dmbm
lowest resistivity ? nXmW© h¡
(A) Constantan (B) Silver (A) H$m|ñQ>oZZ (B) Mm§Xr
(C) Manganin (D) Copper (C) ‘|¾tZ (D) Vmå~m

013. An incompressible non viscous fluid 013. EH$ Ag§nrS>ç Aí`mZ Ðd EH$ ~obZmH$ma
flows steadily through a cylindrical nmBn ‘| go gVV ê$n go ~h ahm h¡& BgHo$
pipe which has radius 2R at point A ~hmd H$s {Xem Ho$ AZw{Xe q~Xþ A­ na
and radius R at point B farther along Ðd H$m doJ V h¡& q~Xþ ­A na nmB©n H$s
the flow direction. If the velocity of {ÌÁ`m 2R h¡ VWm Ðd àdmh H$s {Xem ‘|
the fluid at point A is V, its velocity XÿañW q~Xþ B na nmB©n H$s {ÌÁ`m R h¡
at the point B will be Vmo q~Xþ B na Ðd H$m doJ Š`m hmoJm?
(A) 2V (B) V (A) 2V (B) V
(C) V/2 (D) 4V (C) V/2 (D) 4V

014. In a room where the temperature is 014. EH$ H$‘ao H$m Vmn 30°C h¡ Bg‘| EH$
30°C a body cools from 61°C to 59°C dñVw H$mo 61°C go 59°C VH$ R>ÊS>r hmoZo
in 4 minutes. The time taken by the ‘| bJm g‘` 4 {‘ZQ> h¡ & dñVw H$mo
body to cool from 51°C to 49°C will 51°C go 49°C VH$ R>ÊS>r hmoZo ‘| bJm
be about g‘` bJ^J hmoJm
(A) 4 minutes (B) 6 minutes (A) 4 {‘ZQ> (B) 6 {‘ZQ>
(C) 5 minutes (D) 8 minutes (C) 5 {‘ZQ> (D) 8 {‘ZQ>

015. A student’s 9.0 V, 7.5W portable radio 015. EH$ N>mÌ H$m 9.0 V Ed§ 7.5W H$m EH$
was left on from 9:00 P.M. until 3:00 ao{S>`mo 9:00 P.M go 3:00 A.M. VH$ Mmby
A.M. How much charge passed through ahVm h¡ Vmo Vma Ûmam {H$VZm Amdoe àdm{hV
the wires? hþAm?
(A) 6000C (B) 12000C (A) 6000C (B) 12000C
(C) 18000C (D) 24000C (C) 18000C (D) 24000C

1-AA ] [5] [ PTO

016. A conducting wheel rim in which there 016. EH$ n{hE H$s MmbH$ n[a{Y na {MÌmZwgma
are three conducting rods of each of VrZ MmbH$ N>‹S>o EH$ g‘mZ Mwå~H$s` joÌ
length l is rotating with constant angular
B ‘| AMa H$moUr` doJ ω go KyU©Z H$a
velocity ω in a uniform magnetic field
B as shown in figure. The induced ahr h¡ & àË`oH$ N>S> H$s bå~mB© l h¡ &
potential difference between its centre n{h`o H$s n[a{Y d H|$Ð Ho$ ‘Ü` CËnÝZ
and rim will be ào[aV {d^dmÝVa hmoJm

Bωl 2 Bωl 2
(A) 0 (B) (A) 0 (B)
2 2
2 3 2 3
(C) Bωl (D) Bωl 2 (C) Bωl (D) Bωl 2
2 2

017. An imaginary, closed spherical surface 017. EH$ H$mën{ZH$ JmobmH$ma ~§X gVh S H$s
S of radius R is centered on the origin. {ÌÁ`m R h¡ {OgH$m H|$Ð ‘yb q~Xþ na h¡&
A positive charge +q is originally at nhbo EH$ YZmË‘H$ Amdoe +q ‘yb q~Xþ
the origin and electric flux through the na aIm hþAm Wm VWm gVh go nm[aV
surface is ΦE. Three additional charges {dÚwV âbŠg ΦE Wm& A~ VrZ A{V[aº$
are now added along the x axis: −3q Amdoe x Aj Ho$ AZw{Xe {ZåZ Vah go
R R
at x = −   , + 5q at x = and 4q at
2 2
aIo OmVo h¢ −3q Amdoe x = −   R2   na,
3R R
x= . The flux through S is now +5q Amdoe x = na VWm 4q Amdoe
2 2
(A) 3ΦE (B) 4ΦE na h¡& A~ gVh S go nm[aV âbŠg hmoJm
(A) 3ΦE (B) 4ΦE
(C) 6ΦE (D) 7ΦE
(C) 6ΦE (D) 7ΦE

018. An 1800 W toaster, a 1.3KW electric 018. EH$ 1800 W H$m Q>moñQ>a, EH$ 1.3KW
fan and a 100W lamp are plugged in H$m {dÚwV n§Im d EH$ 100W H$m ~ë~
the same 120V circuit i.e. all the three H$mo 120V Ho$ EH$ hr n[anW ‘| bJm`m
devices are in parallel. What is the OmVm h¡ AWm©V `o g^r VrZm| `w{º$`m±
approximate value of the total current g‘mÝVa H«$‘ h¢& n[anW go Hw$b àdm{hV
(i.e. sum of the current drawn by the Ymam (AWm©V VrZm| `w{º$`m| Ûmam br JB©
three devices) through circuit ? YmamAm| H$m `moJ) H$m ‘mZ bJ^J hmoJm?
(A) 18A (B) 27A (A) 18A (B) 27A
(C) 40A (D) 120A (C) 40A (D) 120A

1-AA ] [6] [ Contd...

019. Four very long current carrying wires 019. Mma bå~o Ymamdmhr Vma EH$ hr Vb ‘|
in the same plane intersect to form a h¢ VWm EH$ dJ© H$s àË`oH$ ^wOm 40cm
square 40.0cm on each side as shown ~ZmVo hþE {MÌmZwgma à{VÀN>oX H$aVo h¢&
in the figure. What is the magnitude dJ© Ho$ H|$Ð na Mwå~H$s` joÌ eyÝ` hmoZo
of current I so that the magnetic field Ho$ {bE Ymam I H$m n[a‘mU {H$VZm hmoZm
at the centre of the square is zero?
Mm{hE?

(A) 2A (B) 18A (A) 2A (B) 18A
(C) 22A (D) 38A (C) 22A (D) 38A

020. If the current in the toroidal solenoid 020. EH$ Q>moamoBS>Zw‘m n[aZm{bH$m ‘| Ymam EH$
increases uniformly from zero to 6.0A g‘mZ ê$n go eyÝ` go 6.0A VH$ 3.0μs
in 3.0μs. Self inductance of the toroidal ‘| ~‹T>Vr h¡& Q>moamoBS>Zw‘m n[aZm{bH$m H$m
solenoid is 40μH. The magnitude of self ñdàoaH$Ëd 40μH h¡& ñd ào[aV {dÚwV dmhH$
induced emf is ~b H$m n[a‘mU h¡
(A) 24V (B) 48V (A) 24V (B) 48V
(C) 80V (D) 160V (C) 80V (D) 160V

021. An electron is at ground state of the 021. EH$ H na‘mUw Ho$ ‘yb ñVa ‘| EH$ BboŠQ´mZ
H atom. Minimum energy required to h¡& H na‘mUw H$mo {ÛVr` CÎmo{OV AdñWm
excite the H atom into second excited ‘| CÎmo{OV H$aZo Ho$ {bE Ý`yZV‘ {H$VZr
state is D$Om© H$s Amdí`H$Vm hmoJr ?
(A) 10.2eV (B) 3.4eV (A) 10.2eV (B) 3.4eV
(C) 13.6eV (D) 12.1eV (C) 13.6eV (D) 12.1eV

022. A particle enters uniform constant 022. EH$ H$U EH$ g‘mZ Mwå~H$s` joÌ ‘|
magnetic field region with its initial Mwå~H$s` joÌ H$s {Xem Ho$ AZw{Xe àmapå^H$
velocity parallel to the field direction. doJ go àdoe H$aVm h¡& BgHo$ doJ Ho$ ~mao
Which of the following statements ‘| H$m¡Zgm H$WZ gË` hmoJm? (AÝ` joÌm|
about its velocity is correct? (neglect Ho$ à^mdm| H$mo ZJÊ` ‘m{ZE)
the effects of other fields) (A) Ho$db n[a‘mU ‘| n[adV©Z hmoJm
(A) There is change only in magnitude (B) Ho$db {Xem ‘| n[adV©Z hmoJm
(B) There is change only in direction
(C) n[a‘mU d {Xem XmoZm| ‘| n[adV©Z
(C) There is change in both magnitude
and direction
hmoJm
(D) There is no change (D) H$moB© n[adV©Z Zht hmoJm

023. Magnetic susceptibility of diamagnetic 023. à{VMwå~H$s` nXmW© H$s Mwå~H$s` àd¥{V H$s
materials is of the order of (SI units) H$mo{Q> (SI BH$mB© ‘|) hmoJr
(A) +10 – 5 (B) –10 – 5 (A) +10 – 5 (B) –10 – 5
(C) +10 5 (D) +10 – 4 to +10 – 2 (C) +10 5 (D) +10 – 4 to +10 – 2

1-AA ] [7] [ PTO

024. Magnitude of binding energy of satellite 024. goQ>obmB©Q> H$s ~§YZ D$Om© H$m n[a‘mU E
is E and kinetic energy is K .The ratio h¡ VWm CgH$s J{VO D$Om© H$m ‘mZ K h¡
E/K is Vmo AZwnmV E/K hmoJm
(A) 1 (B) 1/2 (A) 1 (B) 1/2
(C) 2/1 (D) 1/4 (C) 2/1 (D) 1/4

025. Figure shows the total acceleration 025. {MÌ ‘| {ÌÁ`m R=1m Ho$ d¥Îm ‘| X{jUmdV©
a  =  32m/s2 of a moving particle moving Ky‘Vo hþE H$U H$m Hw$b ËdaU  a  =  32m/s2
clockwise in a circle of radius R=1m. h¡ Vmo H$U H$m A{^Ho$ÝÐr` ËdaU d H$U
What are the centripetal acceleration and H$s Mmb ν {XE JE jU na Š`m hmoJr?
speed v of the particle at given instant?

(A) 16m/s2, 16m/s
(A) 16m/s2, 16m/s (B) 16m/s2, 4m/s
(B) 16m/s2, 4m/s (C) 16 3 m/s2, 4 3 m/s
(C) 16 3 m/s2, 4 3 m/s (D) 16 3 m/s2, 4m/s
(D) 16 3 m/s2, 4m/s

026. A force F  =  75N is applied on a block 026. EH$ ~b F  =  75N H$mo 5kg Ðì`‘mZ Ho$
of mass 5kg along the fixed smooth ãbm°H$ na {MÌmZwgma pñWa {MH$Zo ZV Vb
incline as shown in figure. Here Ho$ AZw{Xe bJm`m OmVm h¡& `hm± JwéËdr`
gravitational acceleration g  =  10m/s2. ËdaU g  =  10m/s2 h¡& ãbm°H$ H$m ËdaU hmoJm
The acceleration of the block is

m m
(A) 5 2 downwards the incline (A) 5 2 ZV Vb Ho$ AZw{Xe ZrMo H$s Amoa
s s
m m
(B) 5 2 upwards the incline (B) 5 2 ZV Vb Ho$ AZw{Xe D$na H$s Amoa
s s
m m
(C) 10 2 downwards the incline (C) 10 2 ZV Vb Ho$ AZw{Xe ZrMo H$s Amoa
s s
m m
(D) 10 2 upwards the incline (D) 10 2 ZV Vb Ho$ AZw{Xe D$na H$s Amoa
s s
027. A 3kg object has initial velocity 027. EH$ 3kg H$s dñVw H$m àmapå^H$ doJ
^6it - 2tjh m/s. The total work done on ^6i - 2tjh m/s h¡ & `{X dñVw H$m do J
t
the object if its velocity changes to ^8it + 4tjh m/s hmo OmVm h¡ V~ VH$ dñVw
^8it + 4tjh m/s is na {H$`m J`m Hw$b H$m`© hmoJm
(A) 60J (B) 120J (A) 60J (B) 120J
(C) 216J (D) 44J (C) 216J (D) 44J

1-AA ] [8] [ Contd...

028. A heat engine absorbs 360J of energy 028. EH$ D$î‘m B§OZ àË`oH$ MH«$ ‘| 360J D$î‘m
by heat and performs 25J of work in H$m AdemofU H$aVm h¡ VWm 25J H$m`©
each cycle. The energy expelled to the àË`oH$ MH«$ ‘| H$aVm h¡& àË`oH$ MH«$ ‘|
cold reservoir in each cycle is R>ÝSo> hm¡O H$mo Xr JB© D$Om© hmoJr
(A) 360J (B) 385J (A) 360J (B) 385J
(C) 335J (D) 14.4J (C) 335J (D) 14.4J

029. Three nonconducting large parallel plates 029. {MÌmZwgma VrZ AMmbH$ ~‹S>r g‘mÝVa ßboQ>mo
have surface charge densities σ,−2σ and Ho$ n¥ð> Amdoe KZËd H«$‘e… σ,−2σ VWm
4σ respectively as shown in figure. The
electric field at the point P is 4σ h¡& q~Xþ P na {dÚwV joÌ h¡

3σ 3σ 3σ 3σ
(A) (B) (A) (B)
2ε0 ε0 2ε0 ε0
σ σ σ σ
(C) (D) (C) (D)
ε0 2ε0 ε0 2ε0
030. A battery of constant voltage is 030. EH$ AMa dmoëQ>Vm H$s ~¡Q>ar CnbãY h¡&
available. How to adjust a system of VrZ EH$g‘mZ g§Ym[aÌm| Ho$ {ZH$m` go Cƒ
three identical capacitors to get high pñWa {dÚwV D$Om©dmbr pñW{V àmá H$aZo Ho$
electrostatic energy with the given
{bE BÝh| H¡$go g§`mo{OV H$aZm Mm{hE
battery
(A) Xm| g‘mÝVa H«$‘ ‘| d EH$ loUr H«$‘
(A) Two parallel and one in series
(B) Three in series H$m g§`moOZ
(C) Three in parallel (B) VrZm| loUr H«$‘ ‘|
(D) Whatever may be combination, it (C) VrZm| g‘mÝVa H«$‘ ‘|
will always have same electrostatic (D) {H$gr ^r Vah H$m g§`moOZ hmo pñWa
energy
{dÚwV D$Om© h‘oem g‘mZ hmoJr
031. Five resistances are connected as shown 031. nm±M à{VamoY {MÌmZwgma Ow‹S>o h¢& q~Xþ A
in the figure. The equivalent resistance VWm q~Xþ C Ho$ ‘Ü` Vwë` à{VamoY hmoJm
between points A and C is

(A) 21.2 Ω (B) 30 Ω (A) 21.2 Ω (B) 30 Ω
20 20
(C) 44 Ω (D) Ω (C) 44 Ω (D) Ω
3 3

1-AA ] [9] [ PTO

032. The frequencies of X rays, Gamma rays 032. X {H$aUm|, Jm‘m {H$aUm| VWm Ñí` àH$me
and visible light waves rays are a, b Va§J {H$aUm| H$s Amd¥{V`m± H«$‘e… a, b
and c respectively, then VWm c h¢ V~
(A) a > b > c
(A) a > b > c (B) a > b, b < c (B) a > b, b < c
(C) a < b, b > c (D) a < b, b < c (C) a < b, b > c
(D) a < b, b < c
033. An equiconvex (biconvex) lens has
033. EH$ g‘ CÎmb b|g (C^`m|Îmb) H$s ’$moH$g
focus length f. It is cut into three parts Xÿar f h¡& BgH$mo {MÌmZwgma VrZ ^mJm| ‘|
as shown in the figure. What is the {d^m{OV {H$`m OmVm h¡ Vmo H$mQ>o JE ^mJ
focal length of Cut part I ? I H$s ’$moH$g bå~mB© Š`m hmoJr?

f
f (A) (B) 2f
2
(A) (B) 2f f
2 (C) 3f (D)
f 3
(C) 3f (D)
3

034. A cell has terminal voltage 2V in open 034. Iwbo n[anW ‘| EH$ gob H$s {gam| H$s
circuit and internal resistance of the dmoëQ>Vm 2V h¡ VWm {XE JE gob H$m
given cell is 2Ω. If 4A of current is Am§V[aH$ à{VamoY 2Ω h¡ & `{X 4A H$s
flowing between points P and Q in the Ymam q~XþAm| P VWm Q Ho$ ‘Ü` n[anW ‘|
circuit and then the potential difference ~h ahr h¡ {~ÝXþAm| P VWm Q Ho$ ‘Ü`
between P and Q is {d^dmÝVa h¡

(A) 30V (B) 26V (A) 30V (B) 26V

(C) 22V (D) 24V (C) 22V (D) 24V

035. A Proton and an alpha particle both are 035. EH$ àmoQ>moZ Ed§ EH$ Aë’$m H$U XmoZm| H$mo
accelerated through the same potential g‘mZ {d^dmÝVa Ûmam Ëd[aV {H$`m OmVm
difference. The ratio of corresponding h¡& CZH$s g§JV S>r ~«mo½br Va§JX¡Y`m} H$m
de-Broglie wavelengths is AZwnmV h¡
(A) 2 (B) 2 (A) 2 (B) 2
1 1
(C) 2 2 (D) (C) 2 2 (D)
2 2 2 2

1-AA ] [ 10 ] [ Contd...

036. Two balls of mass m and 4m are 036. Xmo J|Xo {OZH$m Ðì`‘mZ m VWm 4m h¢
connected by a rod of length L. The BZH$mo L bå~mB© H$s N>‹S> Ûmam Omo‹S>m OmVm
mass of the rod is small and can be h¡& N>‹S> H$m Ðì`‘mZ ZJÊ` h¡ VWm J|Xm|
treated as zero. The size of the balls can H$m AmH$ma ^r ZJÊ` h¡& h‘ `h ^r ‘mZVo
also can be neglected. We also assume h¢ {H$ N>‹S> H$m Ho$ÝÐ H$sb{H$V {H$`m OmVm
the centre of the rod is hinged, but the h¡ naÝVw N>‹S> D$Üdm©Ya Vb ‘| {~Zm Kf©U
rod can rotate about its centre in the Ho$ BgHo$ Ho$ÝÐ Ho$ gmnoj Ky{U©V hmo gH$Vr
vertical plane without friction. What is h¡& O~ N>‹S> H$m D$Üdm©Ya aoIm Ho$ gmW
the gravity induced angular acceleration {MÌmZwgma H$moU θ hmo Vmo Cg g‘` JwéËd
of the rod when the angle between the O{ZV N>‹S> H$m H$moUr` ËdaU Š`m hmoJm?
rod and the vertical line is θ as shown.

6g g 6g g
(A) sinθ (B) sinθ (A) sinθ (B) sinθ
5L 3L 5L 3L
5g g 5g g
(C) sinθ (D) cosθ (C) sinθ (D) cosθ
6L 6L 6L 6L

037. A projectile is projected with an initial 037. EH$ àjoß` H$mo àmapå^H$ doJ  ^4it + 5tjhm/s
velocity ^4it + 5tjh m/s. Here tj is the Ho$ gmW àjo{nV {H$`m OmVm h¡& `hm± tj
unit vector directed vertically upwards BH$mB© g{Xe D$Üdm©Ya D$na H$s Amoa h¡
and unit vector it is in the horizontal VWm it BH$mB© g{Xe jo{VO {Xem ‘| h¡&
direction .Velocity of the projectile (in àjoß` H$s O‘rZ go Q>³H$a go R>rH$ nyd©
m/s) just before it hits the ground is CgH$m doJ (‘r./go.) hmoJm
(A) 4it + 5tj (B) - 4it + 5tj (A) 4it + 5tj (B) - 4it + 5tj
(C) 4it - 5tj (D) - 4it - 5tj (C) 4it - 5tj (D) - 4it - 5tj

038. What is the approximate percentage 038. EH$ gab bmobH$ Ho$ AmdV©H$mb Ho$ ‘mnZ
error in the measurement of time period ‘| bJ^J à{VeV Ìw{Q> {H$VZr hmoJr `{X
of a simple pendulum if maximum bå~mB© l VWm JwéËdr` ËdaU g ‘mnZ ‘|
errors in the measurement of length l
A{YH$V‘ Ìw{Q> H«$‘e… 3% VWm 7% h¡
and gravitational acceleration g are 3%
(A) 2 %
and 7% respectively ?
(B) 3 %
(A) 2 % (B) 3 %
(C) 5 %
(C) 5 % (D) 10 %
(D) 10 %

1-AA ] [ 11 ] [ PTO

039. A gas undergoes the cyclic process 039. EH$ J¡g EH$ MH«$s` àH«$‘ ‘| {MÌmZwgma
shown in figure .The cycle is repeated AZwgaU H$aVr h¡ & Bg MH«$ H$s à{V
100 times per minute. The power {‘ZQ> 100 ~ma nwZamd¥{Îm H$s OmVr h¡ &
generated is
CËnÝZ e{º$ hmoJr

(A) 60W (B) 120W
(A) 60W (B) 120W
(C) 240W (D) 100W
(C) 240W (D) 100W

040. Three charges lie on the frictionless 040. VrZ Amdoe EH$ Kf©Ua{hV j¡{VO gVh na
horizontal surface at the vertices of EH$ g‘~mhþ {Ì^wO Ho$ erfm} na {MÌmZwgma
equilateral triangle as shown in figure. h¢& BZ‘| go Xmo Amdoe X VWm Y O‹S>dV
Two charges X and Y are fixed whereas (fixed) h¢ VWm Vrgam Amdoe Z ‘wº$ {H$`m
third charge Z is released. Which path OmVm h¡ Vmo ‘wº$ H$aZo Ho$ Cnam§V Amdoe Z
will charge Z take upon release ? Ûmam H$m¡Zgm nW (path) AnZm`m OmVm h¡?

(A) Path – I (B) Path – II (A) nW – I (B) nW – II

(C) Path – III (D) Path – IV (C) nW – III (D) nW – IV

041. There are two waves having wavelengths 041. Xmo Va§J| {OZH$s Va§JX¡Ü`© 100cm VWm
100cm and 101cm and same velocity 101cm h¡ VWm g‘mZ doJ 303m/s h¡&
303m/s. The beat frequency is {dñn§X Amd¥{V hmoJr
(A) 3Hz (B) 2Hz (A) 3Hz (B) 2Hz
(C) 4Hz (D) 1Hz (C) 4Hz (D) 1Hz

1-AA ] [ 12 ] [ Contd...

042. Two polaroids A and B are placed with 042. Xmo nmobamoBS> (Y«wdH$) A VWm B EH$ Xÿgao
their polaroid axes 30° to each other as go {MÌmZwgma Bg àH$ma aIr OmVr h¡ {H$
shown in the figure. A plane polarized CZH$s nmobamoBS> Ajm| Ho$ ‘Ü` H$moU 30°
light passes through the polaroid A and h¡ nmobamoBS> A go JwOaZo Ho$ nümV g‘Vb
after passing through it, intensity of Y«w{dV àH$me H$s Vrd«Vm I0 hmo OmVr h¡
light becomes I0.What is the intensity nmobamoBS> B go JwOaZo Ho$ nümV A§{V‘ ê$n
of finally transmitted light after passing go nmaJ{‘V àH$me H$s Vrd«Vm Š`m hmoJr?
through the polaroid B ?

(A) 0.25I0 (B) 0.5I0
(A) 0.25I0 (B) 0.5I0
(C) 0.75I0 (D) 0.866I0
(C) 0.75I0 (D) 0.866I0

043. Laser light has following property 043. boOa àH$me {ZåZ JwU aIVm h¡
(A) laser light is white light (A) boOa àH$me œoV hmoVm h¡
(B) laser light is highly coherent (B) boOa àH$me AË`{YH$ H$bmgå~Õ hmoVm h¡
(C) laser light always lies in X-rays (C) boOa àH$me h‘oem EŠg {H$aU joÌ
region ‘| hmoVm h¡
(D) laser light does not have directionality (D) boOa àH$me ‘| {XemË‘H$ JwU Zht
property
hmoVm h¡
044. A particle is moving in translatory 044. EH$ H$U ñWmZmÝVaU J{V H$a ahm h¡ &
motion. If momentum of the particle `{X H$U H$m g§doJ 10% KQ>Vm h¡ Vmo
decreases by 10%, kinetic energy will
BgH$s J{VO D$Om© KQ>oJr
decrease by
(A) 20% (B) 19%
(A) 20% (B) 19%
(C) 10% (D) 5% (C) 10% (D) 5%

045. Which of the statement is incorrect 045. gmYmaU(gab) gyú‘Xeu Ho$ ~mao ‘| H$m¡Zgm
about the simple microscope? H$WZ AgË` h¡ ?
(A) Magnification of microscope is (A) gyú‘Xeu H$m AmdY©Z {d^oÚ  (ñnï>)
inversely proportional to the least Ñ{ï> Ho$ Ý`yZV‘ ‘mZ Ho$ ì`wËH«$‘mZwnmVr
distance of distinct vision.
hmoVr h¡
(B) A convex lens of microscope with
(B) gyú‘Xeu Ho$ H$‘ ’$moH$g Xÿar Ho$ CÎmb
shorter focal length yields higher
magnification. b|g go A{YH$ AmdY©Z àmá hmoVm h¡
(C) Biology students use to see the (C) Ord {dkmZ Ho$ {dÚmWu ñbmBS> H$mo
slides. XoIZo ‘| H$m‘ ‘| boVo h¢&
(D) It is not used for magnification (D) àojH$ go Xÿa pñWV dñVw Ho$ AmdY©Z
of an object at far away from the Ho$ {bE `h Cn`moJ ‘| Zht AmVm h¡
observer.

1-AA ] [ 13 ] [ PTO

046. Surface tension of the liquid is S. 046. EH$ Ðd H$m n¥ð> VZmd S h¡& {H$gr {XE
Work done in increasing the radius of JE Vmn na EH$ gm~wZ Ho$ ~wb~wbo H$mo
soap bubble from R to 3R at given {ÌÁ`m R go 3R H$aZo ‘| {H$`m J`m H$m`©
temperature will be hmoJm
(A) 8πSR2 (B) 16πSR2 (A) 8πSR2 (B) 16πSR2
18πSR 2 18πSR 2
(C) 64πSR2 (D) (C) 64πSR2 (D)
3 3

047. Suppose you drive to Delhi (200 km 047. `h ‘m{ZE {H$ AmnH$mo 200 km Xÿa {X„r
away) at 400 km/hr and return at 200 H$mo 400 km/hr go OmZm h¡ VWm 200
km/hr. What is yours average speed for km/hr go bm¡Q>Zm h¡& AmnHo$ Bg Xm¡ao H$s
the entire trip? Am¡gV Mmb Š`m hmoJr?
(A) Zero (A) eyÝ`
(B) 300 Km/hr (B) 300 Km/hr
(C) Less than 300 km/hr (C) 300 Km/hr go H$‘
(D) More than 300 km/hr (D) 300 Km/hr go A{YH$

048. A system undergoes a reversible adiabatic 048. EH$ {ZH$m` EH$ CËH«$‘Ur` éÕmoî‘ àH«$‘ go
process. The entropy of the system JwOaVm h¡ & {ZH$m` H$s E§Q´monr (entropy)
(A) increases (A) ~‹T>oJr
(B) decreases (B) KQ>oJr
(C) remains constant (C) AMa ahVr h¡
(D) may increase or may decrease (D) ~‹T> `m KQ> gH$Vr h¡
049. For the combination of gates shown 049. ZrMo {XE JE VH©$ Ûmam| Ho$ g§`moOZ Ho$
here, which of the following truth table {bE {ZåZ gË` gmaUr H$m H$m¡Zgm ^mJ
part is not true gË` Zht h¡

(A) A = 1, B  = 1, C = 1 (A) A = 1, B  = 1, C = 1
(B) A = 1, B  = 0, C = 1
(B) A = 1, B  = 0, C = 1
(C) A = 0, B  = 1, C = 1
(C) A = 0, B  = 1, C = 1
(D) A = 0, B  = 0, C = 0
(D) A = 0, B  = 0, C = 0

050. A narrow white light beam fails to 050. EH$ œoV àH$me g§H$sU© {H$aU EH$ A{^gmar
converge at a point after going through a b|g go JwOaZo Ho$ nümV EH$ hr q~Xþ na
converging lens. This defect is known as A{^gm[aV hmoZo ‘| Ag’$b hmoVr h¡ `h
(A) polarization Xmof {ZåZ H$hbmVm h¡
(B) spherical aberration (A) Y«wdU
(C) chromatic aberration (B) Jmobr` {dnWZ
(C) dUu` {dnWZ
(D) diffraction
(D) {ddV©Z

1-AA ] [ 14 ] [ Contd...

CHEMISTRY / agm¶ZemñÌ
051. The one electron species having 051. EH$ BboŠQ´moZ ñnrerO {OgHo$ Am`ZZ D$Om©
ionization energy of 54.4 eVs 54.4 BboŠQ´moZ dmoëQ> h¡ -
(A) Be+2 (B) Be+3 (A) Be+2 (B) Be+3
(C) He + (D) H (C) He+ (D) H

052. Which of the following set of quantum 052. {ZåZ ‘| go H$m¡Zgo ³dm§Q>‘ g§»`mAmo H$m
numbers represents the highest energy g‘yh na‘mUw H$s CƒV‘ D$Om© H$mo {Zé{nV
of an atom ? H$aVm h¡
1 1
(A) n = 3, l = 0, m = 4, s = + (A) n = 3, l = 0, m = 4, s = +
2 2
1 1
(B) n = 3, l = 1, m = 1, s = + (B) n = 3, l = 1, m = 1, s = +
2 2
1
1 (C) n = 3, l = 2, m = 1, s = +
(C) n = 3, l = 2, m = 1, s = + 2
2
1
1 (D) n = 4, l = 0, m = 0, s = -
(D) n = 4, l = 0, m = 0, s = - 2
2
053. In OF2,  oxygen has hybridization of 053. OF2 ‘| Am°ŠgrOZ H$m g§H$aU h¡
(A) sp (B) sp2 (A) sp (B) sp2
(C) sp3 (D) None of the options
(C) sp3 (D) BZ‘| go H$moB© {dH$ën Zht

3- 2- 3- 2- 2-
A m o n g s t NO3 , AsO3 , CO3 ,
-
054. NO3 , AsO3 , CO3 , ClO3 , SO3
- -
054. Am¡a
2- 3-
ClO3 , SO3 and BO3 the non-planar
- 3-
BO3 ‘| go Ag‘Vb ñnrerO h¡
species are 2- 2-
(A) CO3 , SO3 VWm BO3
3-
2- 2- 3-
(A) CO3 , SO3 and BO3 3- 2-
(B) AsO3 , CO3 VWm SO3
2-
3- 2- 2-
(B) AsO3 , CO3 and SO3 - 2- 3-
- 2- 3- (C) NO3 , CO3 VWm BO3
(C) NO3 , CO3 and BO3 2- - 3-
2- - 3- (D) SO3 , ClO3 VWm BO3
(D) SO3 , ClO3 and BO3
055. The Lewis acidity of BF3 is less than 055. BF3 H$s bwB©g Aåbr`Vm BCl3 go H$‘ h¡
BCl 3 even though fluorine is more O~{H$ âbmo[aZ H$s {dÚwV F$UVm ŠbmoarZ
electronegative than chlorine. It is due to go A{YH$ h¡ & BgH$m H$maU h¡ -
(A) stronger 2p(B)–2p (F) σ - bonding (A) à~b 2p(B)–2p (F) σ - ~ÝYZ
(B) stronger 2p(B)–2p(F) π - bonding (B) à~b 2p(B)–2p(F) π - ~ÝYZ
(C) stronger 1p(B)–3p (Cl) σ - bonding (C) à~b 1p(B)–3p (Cl) σ - ~ÝYZ
(D) stronger 2p(B)-3p(Cl) π - bonding (D) à~b  2p(B)-3p(Cl) π - ~ÝYZ

056. The IUPAC name of the compound is: 056. `m¡{JH$ H$m AmB©.`y.nr.E.gr.Zm‘ h¡

(A) 2-methyl-6-oxohex-3-enamide (A) 2-‘o{Wb-6 Am°ŠgmohoŠg-3-BZm‘mBS>
(B) 6-keto-2-methyl hexamide (B) 6-H$sQ>mo-2-‘o{Wb hoŠgm‘mBS
(C) 2-carbamoylhexanal (C) 2-H$m~m}‘mo`bhoŠgoZob
(D) 2-carbamoylhex-3-enal (D) 2-H$m~m}‘mo`bhoŠg-3-BZob

1-AA ] [ 15 ] [ PTO

057. The IUPAC name of 057. {ZåZ H$m AmB©.`y.nr.E.gr. Zm‘ h¡

is is

(A) 1-Bromo-2-chloro-3-fluoro-6-iodo (A) 1-~«mo‘mo-2-Šbmoamo-3-âbmoamo-6-Am`S>mo
benzene ~oÝOrZ
(B) 2-Bromo-1-chloro-5-fluoro-3-iodo (B) 2-~«mo‘mo-1-Šbmoamo-5-âbmoamo-3-Am`S>mo
benzene ~oÝOrZ
(C) 4-Bromo-2-chloro-5-iodo-1-fluoro (C) 4-~«mo‘mo-2-Šbmoamo-5-Am`S>mo -1-âbmoamo
benzene ~oÝOrZ
(D) 2-carbamoylhex-3-enal (D) 2-H$m~m}‘mo`bhoŠg -3-BZob

058. Which of the following compounds 058. {ZåZ `m¡{JH$m| ‘| go {H$g‘o H$‘ go H$‘
contain at least one secondary alcohol? EH$ {ÛVr` EëH$mohb h¡?

(A) (i), (ii), (iv), (vi) (A) (i), (ii), (iv), (vi)
(B) (i), (ii), (iii) (B) (i), (ii), (iii)
(C) (i), (ii), (iii), (v) (C) (i), (ii), (iii), (v)
(D) (i), (iii), (v) (D) (i), (iii), (v)

059 Transition state 2 (T.S.2) is structurally 059 g§aMZmË‘H$ ê$n go g§H«$‘U AdñWm 2 (T.S.2)
most likely as: A{YH$ g‘mZ h¡

(A) intermediate 1 (A) ‘Ü`dVu 1 (intermediate 1)
(B) transition state 3(T.S.3) (B) g§H«$‘U AdñWm 3 (T.S.3)
(C) intermediate 2 (C) ‘Ü`dVu 2 (intermediate 2)
(D) product (D) CËnmX (product)

1-AA ] [ 16 ] [ Contd...

060. The decreasing order of electron affinity 060. BboŠQ´moZ AmË‘r`Vm (~§YwVm) H$m KQ>Vm hþAm
is: H«$‘ h¡-
(A) F > Cl > Br > I (A) F > Cl > Br > I
(B) Cl > F > Br > I (B) Cl > F > Br > I
(C) I > Br > Cl > F (C) I > Br > Cl > F
(D) Br > Cl > F > I (D) Br > Cl > F > I
061. The isomerism exhibited by following 061. AYmo{bpIV `m¡{JH$m| [Co(NH3)6][Cr(CN)6]
compounds [Co(NH3)6][Cr(CN)6] and VWm [Cr(NH3)6][Cr(CN)6] Ûmam g‘md`Vm
[Cr(NH3)6][Cr(CN)6] is àX{e©V hmo ahr h¡ -
(A) Linkage isomerism (A) ~ÝYZr g‘md`Vm
(B) Coordination isomerism (B) Cnghg§`moOZ g‘md`Vm
(C) Ionization isomerization (C) Am`ZZ g‘md`Vm
(D) Polymerisation isomerism (D) ~hþbH$sH$aU g‘md`Vm

062. For the reaction 062. A{^{H«$`m 2SO 2 + O 2 (excess) " 2SO3 Ho$
2SO 2 + O 2 (excess) " 2SO3 the order of
{bE O2 Ho$ gÝX^© (gmnoj ) ‘| A{^{H«$`m H$s
reaction with respect to O2 is
(A) zero (B) one
H$mo{Q> h¡
(A) eyÝ` (B) EH$
(C) two (D) three
(C) Xmo (D) VrZ
063. Friedel – Craft reaction is not related 063. ’«$sSo>b-H«$mâQ> A{^{H«$`m {ZåZ{bpIV ‘| go
with: gå~§{YV Zht h¡
(A) Sulphonation (B) Nitration (A) gë’$mo{ZH$aU (B) ZmBQ´rH$aU
(C) Acylation (D) Reduction (C) E{g{bH$aU (D) AnM`Z

064. Compound has the 064. `m¡{JH$ Ho$ {bE CngJ© h¡
following prefix
(A) E (B) Z (A) E (B) Z
(C) Q´m§g (D) EÝQ>r
(C) trans (D) Anti

065. The molecule C3O2 has a linear structure. 065. AUw C3O2 H$s g§aMZm a¡pIH$ h¡ & Bg `m¡{JH$ ‘|
This compound has (A) 4 σ VWm 4 π Am~ÝY
(A) 4 σ and 4 π bonds (B) 3 σ VWm 2 π Am~ÝY
(B) 3 σ and 2 π bonds
(C) 2 σ VWm 3 π Am~ÝY
(C) 2 σ and 3 π bonds
(D) 3 σ and 4 π bonds (D) 3 σ VWm 4 π Am~ÝY

066. The structure of XeF2 and NH3 066. XeF2 VWm NH3 H$s g§aMZmE± h¢ H«$‘e…
respectively are (A) ~§{H$V, MVwî’$bH$s`
(A) bent, tetrahedral
(B) a¡pIH$, {nar{‘{S>`
(B) linear, pyramidal
(C) linear, see-saw (C) a¡pIH$, T>ÝHw$br (gr gm°)
(D) bent, see-saw (D) ~§{H$V T>ÝHw$br (gr gm°)

1-AA ] [ 17 ] [ PTO

067. The number of lone pair(s) of electrons 067. 6 BrF4 @ , XeF6 VWm 6SbCl6 @3- Ho$ Ho$ÝÐr`
-

on the central atom in 6 BrF4 @ , XeF6 and
-
na‘mUw na EH$mH$s BboŠQ´moZ `w½‘m| H$s g§»`m h¡
6SbCl6 @3- are, respectively. H«$‘e…
(A) 2,0 and 1 (B) 1, 0 and 0 (A) 2,0 VWm 1 (B) 1, 0 VWm 0
(C) 2,1 and 1 (D) 2,1 and 0 (C) 2,1 VWm 1 (D) 2,1 VWm 0
068. Which one is not the property of crystalline 068. H$m¡Zgm EH$ {H«$ñQ>br` R>mogm| H$m JwU Zht h¡ ?
soild ? (A) g‘X¡{eH$
(A) isotropic (B) VrúU JbZm§H$ {~ÝXþ
(B) Sharp melting point
(C) A definite and regular geometry (C) {Z{üV Ed§ {Z`{‘V Á`m{‘Vr`
(D) High intermolecular forces (D) Cƒ AÝVampÊdH$ ~b

069. For a non-volatile solute: 069. EH$ Admînerb {dbo` Ho$ {bE
(A) vapour pressure of solute is zero (A) {dbo` H$m dmînXm~ eyÝ` hmoVm h¡
(B) vapour pressure of solvent is zero (B) {dbm`H$ H$m dmînXm~ eyÝ` hmoVm h¡
(C) vapour pressure of solution is more (C) {db`Z H$m dmînXm~ {dbm`H$ Ho$ dmînXm~
than vapour pressure of solvent go A{YH$ hmoVm h¡
(D) all of the options (D) {X¶o JE g^r {dH$ën ghr h¡
070. Micelles are: 070. {‘gob h¡
(A) gel (A) Oob
(B) associated colloids (B) ghMmar H$mobmBS>
(C) adsorbed catalyst (C) A{Yemo{fV CËàoaH$
(D) ideal solution (D) AmXe© {db`Z

071. Milk is an emulsion in which: 071. XÿY EH$ nm`g h¡ {Og‘|
(A) Milk fat is dispersed in water (A) XÿY dgm H$m Ob ‘| n[ajonU ahVm h¡
(B) a solid is dispersed in water (B) EH$ R>mog H$m Ob ‘| n[ajonU ahVm h¡
(C) a gas is dispersed in water (C) EH$ J¡g H$m Ob ‘| n[ajonU ahVm h¡
(D) lactose is dispersed in water (D) boŠQ>mog H$m Ob ‘| n[ajonU ahVm h¡

072. If enthalpies of formation for C2H4(g), 072. `{X C2H4(g), CO2(g) Am¡a H2O(l) Ho$ {bE
CO2(g) and H2O(l) at 25º C and 1 atm 25º C EH$ dm`w‘§S>br` Xm~ na {daMZ H$s
pressure be 52, –394 and –286 kJ mol–1 EÝWoënr H«$‘e… 52, –394 Am¡a –286 {H$bmo
respectively, enthalpy of combustion of Oyb ‘mob–1 h¡, C2H4 (g) Ho$ XhZ H$s EÝWoënr
C2H4 (g) will be hmoJr-
(A) +141.2 kJ mol–1 (B) +1412 kJ mol–1 (A) +141.2 kJ mol–1 (B) +1412 kJ mol–1
(C) –141.2 kJ mol–1 (D) –1412 kJ mol–1 (C) –141.2 kJ mol–1 (D) –1412 kJ mol–1
073. Which graph shows zero activation 073. A{^{H«$`m (reaction) Ho$ {bE H$m¡Zgm J«m’$
energy for reaction ? eyÝ` g{H«$`U D$Om© Xem©Vm h¡ ?

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

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

1-AA ] [ 18 ] [ Contd...

074. Which of the following is correct for a 074. àW‘ H$mo{Q> H$s A{^{H«$`m Ho$ {bE {ZåZ ‘| go
first order reaction ? H$m¡Zgm ghr h¡ ?
1 1
(A) t1/2 \ a (B) t1/2 \ (A) t1/2 \ a (B) t1/2 \
a a
0 2 0 2
(C) t1/2 \ a (D) t1/2 \ a (C) t1/2 \ a (D) t1/2 \ a

075. 8.50gm of NH3 is present in 250 ml 075. 250 ml ‘| 8.50 J«m‘ A‘mo{Z`m CnpñWV h¡ &
volume. Its active mass is: BgH$m g{H«$` Ðì`‘mZ h¡ -
(A) 1.0 ML–1 (B) 0.5 ML–1 (A) 1.0 ML–1 (B) 0.5 ML–1
(C) 1.5 ML–1 (D) 2.0 ML–1 (C) 1.5 ML–1 (D) 2.0 ML–1

076.
The equilibrium constants of the reaction 076. A{^{H«$`m
1 1
SO 2 (g) + O 2 (g) ? SO3 (g) SO 2 (g) + O 2 (g) ? SO3 (g) Am¡a
2 2
and 2SO 2 (g) + O 2 (g) ? 2SO3 (g) are K1 2SO 2 (g) + O 2 (g) ? 2SO3 (g) Ho$ amgm`{ZH$
and K2 respectively. The relationship gmå` pñWam§H$ H«$‘e… K1 Ed§ K2 h¡, K1 Am¡a K2
between K1 and K2 will be: ‘| gå~ÝY hmoJm?
3 3
(A) K1 = K2 (B) K 2 = K1 (A) K1 = K2 (B) K 2 = K1
2
(C) K1 = K 2 (D) K 2 = K1 (C) K12 = K 2 (D) K 2 = K1

077. 077.

pair is known as `w½‘ H$hbmVm h¡
(A) erythro stereoisomers (A) E[aW«mo {Ì{d‘ g‘md`r
(B) threo stereoisomers (B) {W«`mo {Ì{d‘ g‘md`r
(C) structure isomers (C) g§aMZm g‘md`r
(D) geometrical isomers (D) Á`m{‘{V g‘md`r

078. Which defect in any crystal lowers its 078. {H$gr {H«$ñQ>b ‘| H$m¡Zgr Ìw{Q> BgHo$ KZËd H$mo
density? H$‘ H$aVr h¡
(A) F centre (B) Frenkel (A) F Ho$ÝÐ (B) ’«|$Ho$b
(C) Schottky (D) Interstitial (C) emoQ>H$s (D) A§VamH$mer
079. The half life period of a radio active 079. EH$ ao{S>`mo g{H«$` VËd H$s AY© Am`w 30 {XZ h¡
element is 30 days, after 90 days the 90 {XZ ~mX CgH$s {ZåZ ‘mÌm eof ahoJr -
following quantity will be left
1 1 1 1
(A) (B) (A) (B)
8 4 8 4
1 1 1 1
(C) (D) (C) (D)
2 6 2 6

080. What is the number of atoms in the unit 080. H$m` H|${ÐV KZr` {H«$ñQ>b H$s EH$H$ H$mo{ð>H$m ‘|
cell of body centered cubic crystal ? na‘mUwAmo§ H$s g§»`m Š`m hmoVr h¡ ?
(A) 4 (B) 2 (A) 4 (B) 2
(C) 1 (D) 3 (C) 1 (D) 3

1-AA ] [ 19 ] [ PTO

081. When Grignard reagent reacts with 081. O~ {J«Ý`ma A{^H$‘©H$ H$sQ>m|Z go A{^{H«$`m
ketone it yields H$aVm h¡ Vmo àmá hmoVm h¡ -
(A) 1o alcohol (B) 2o alcohol (A) 1° EëH$mohb (B) 2° EëH$mohb
(C) 3o alcohol (D) Ethanol
(C) 3° EëH$mohb (D) EWoZmob

082. Formula of Bleaching powder is: 082. ãbrqMJ nmCS>a H$m gyÌ h¡
(A) CCl3CHO (B) CaOCl2 (A) CCl3CHO (B) CaOCl2
(C) Ca(OH)2 (D) CHCl3 (C) Ca(OH)2 (D) CHCl3

083. The geometry around the central atom in +
083. Cl F 4 ‘| Ho$ÝÐr` na‘mUw Ho$ Mmamo Amoa Á`m{‘{V
+
Cl F 4 is h¡ -
(A) square planar (A) dJ© g‘Vbr`
(B) square pyramidal (B) dJ© {nam{‘S>r`
(C) octahedral (C) Aï>’$bH$s`
(D) trigonal bipyramidal (D) {ÌH$moUr` {Û {nam{‘S>r`
084. Among the following, the equilibrium 084. Xm~ ~‹T>mZo na {ZåZ ‘| go H$m¡Zgm gmå` à^m{dV
which is NOT affected by an increase in Zht hmoVm h¡
pressure is
(A) 2SO3 (g) ? 2SO 2 (g) + O 2 (g)
(A) 2SO3 (g) ? 2SO 2 (g) + O 2 (g)
(B) H 2 (g) + I 2 (s) ? 2HI (g) (B) H 2 (g) + I 2 (s) ? 2HI (g)
(C) C (s) + H 2 O (g) ? CO (g) + H 2 (g) (C) C (s) + H 2 O (g) ? CO (g) + H 2 (g)
(D) 3Fe (s) + 4H 2 O (g) ? Fe3 O 4 (s) + 4H 2 (g) (D) 3Fe (s) + 4H 2 O (g) ? Fe3 O 4 (s) + 4H 2 (g)

085. In the manufacture of ammonia by 085. ho~a àH«$‘ Ho$ Ûmam A‘mo{Z`m Ho$ {Z‘m©U ‘|
Haber’s process N 2 (g) + 3H 2 (g) ? 2NH3 (g) + 92.3kJ
N 2 (g) + 3H 2 (g) ? 2NH3 (g) + 92.3kJ
{ZåZ ‘| go H$m¡Zgr eV© à{VHy$b h¡ ?
Which of the following conditions is
(A) Vmn ~‹T>Zm
unfavourable ?
(B) Xm~ H$m ~‹T>Zm
(A) Increasing the temperature
(B) Increasing the pressure (C) Vmn H$m KQ>Zm
(C) Reducing the temperature (D) A‘mo{Z`m Ho$ {Z‘m©U Ho$ gmW BgH$m
(D) Removing ammonia as it is formed {ZH$bZm

086. Which of the following compounds can 086. {ZåZ ‘| go H$m¡Zgm `m¡{JH$ Á`m{‘Vr` g‘md`Vm
exhibit both geometrical isomerism and VWm à{V{~å~ ê$nU (enantiomerism) XmoZm|
enantiomerism ? H$mo Xem©Vm h¡ ?
(A) CH3 - CH = CH - CH3 (A) CH3 - CH = CH - CH3
(B) (B)

(C) (C)

(D) CH3 - CHOH - COOH (D) CH3 - CHOH - COOH

1-AA ] [ 20 ] [ Contd...

087. Which of the following reacts fastest with 087. gmÝÐ HCl Ho$ gmW {ZåZ ‘| go H$m¡Zgm Vrd«V‘
conc. HCl ? ê$n go A{^{H«$`m H$aVm h¡
(A) (A)

(B) (B)
(C) (CH3)3COH (C) (CH3)3COH
(D) CH2 = CH–CH2OH (D) CH2 = CH–CH2OH

088. A polymer which is commonly used as a 088. ~hþbH$ Omo gm‘Ý`V`m nXmWm] H$s noqH$J ‘| H$m‘
packaging material is AmVm h¡
(A) Polythene (B) Polypropylene (A) nmobr{WZ (B) nmo{bàmonrbrZ
(C) PVC (D) Bakelite. (C) PVC (D) ~¡Ho$bmB©Q>

089. Which pair does not represent the cyclic 089. H$m¡Zgm `w½‘ C4H6 AUw gyÌ dmbo MH«$s` `m¡{JH$
compound of the molecular formula H$mo àX{e©V Zht H$aVm h¡
C4H6
(A)
(A)

(B)
(B)

(C)
(C)

(D)
(D)

090. 090.

Product P in the above reaction is: Cnamoº$ A{^{H«$`m ‘| CËnmX P h¡

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

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

1-AA ] [ 21 ] [ PTO

091. The structure of carboxylate ion is best 091. H$m~m}pŠgboQ> Am`Z H$s g§aMZm H$m g~go AÀN>m
represented as: {Zê$nU h¡-
(A) (B) (A) (B)

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

092. Which one of the following is not a unit 092. {ZåZ ‘| go H$m¡Zgr D$Om© H$s BH$mB© Zht h¡ ?
of energy ? (A) Nm (B) kg. ms–2
(A) Nm (B) kg. ms–2 (C) lit-atm (D) kg m2 s–2
(C) lit-atm (D) kg m2 s–2
093. When a liquid that is immiscible with 093. EH$ Ðd Omo Ob ‘| A{‘lUr` h¡ H$m ^mn AmgdZ
water was steam distilled at 95.2°C at a 95.2°C na VWm Hw$b Xm~ 99.652KPa na
total pressure of 99.652KPa. The distillate {H$`m J`m& AmgwV ‘| Ob Ho$ àË`oH$ J«m‘ Ho$
contained 1.27gm of the liquid per gram gmW Ðd H$m 1.27gm CnpñWV h¡& `{X Ob H$m
of water. What will be the molar mass of dmînXm~ 95.2°C na 85.140KPa h¡, Ðd H$m
the liquid if the vapour pressure of water ‘moba Ðì`‘mZ Š`m hmoJm ?
is 85.140KPa at 95.2°C ? (A) 134.1 gm mol–1
(A) 134.1 gm mol–1 (B) 105.74 gm mol–1
(B) 105.74 gm mol–1
(C) 99.65 gm mol–1 (C) 99.65 gm mol–1
(D) 18 gm mol–1 (D) 18 gm mol–1
094. What will happen if a cell is placed into 094. Š`m hmoVm h¡ `{X EH$ H$mo{eH$m H$mo 0.4% (Ðì`‘mZ
0.4% (mass/volume) NaCl solution /Am`VZ ) NaCl {db`Z ‘| aIm OmVm h¡?
(A) Cell will swell (A) H$mo{eH$m ’y${bV hmoJr
(B) Cell will shrink (B) H$mo{eH$m {gHw$‹S> Om`oJr
(C) there will be no change in cell volume (C) H$mo{eH$m Ho$ Am`VZ ‘o H$moB© n[adV©Z Zht hmoJm
(D) Cell will dissolve (D) H$mo{eH$m {db` hmo Om`oJr
-8 -8
095. What is pH of 2 # 10 molar HCl 095. 2 # 10 ‘moba HCl {db`Z H$s pH
solution? Here log2  =  0.301 and Š`m hmoJr? ¶hm± log2  =   0.301 Ed§
log3 = 0.477 log3 = 0.477
(A) 5.4 (B) 7.7 (A) 5.4 (B) 7.7
(C) 6.92 (D) 9.5 (C) 6.92 (D) 9.5
096. If at cubic cell, atom A present all corners 096. `{X EH$ KZr` H$mo{eH$m Ho$ g^r H$moZm| na ­A na‘mUw
and atom B at the centre of each face. CnpñWV h¡ Am¡a àË`oH$ ’$bH$ Ho$ Ho$ÝÐH$ na B
What will be the molecular formula of na‘mUw CnpñWV h¡ `{X EH$ H$m`{dH$U© na CnpñWV
the compounds, if all the atoms present g^r na‘mUwAm| H$mo na‘mUw C Ho$ Ûmam à{VñWm{nV
on one body diagonal are replaced by
atom C ? H$a {X`m OmE Vmo `m¡{JH$ H$m AUw gyÌ Š`m hmoJm?
(A) ABC3 (B) A3B12C4 (A) ABC3 (B) A3B12C4
(C) A3B12C (D) AB12C3 (C) A3B12C (D) AB12C3
097. If a compound is formed by X, Y and Z 097. `{X EH$ `m¡{JH$ na‘mUw X,Y Am¡a Z go {‘bH$a ~Zm
atoms and Z is present on the corners, hmo `{X Z na‘mUw H$moZm| na CnpñWV hmo, Y na‘mUw
Y is present 1 tetrahedral voids and X 1 1
2 2 MVwî’$bH$s` [ap³VH$mAm| ‘| Am¡a X na‘mUw 2
atom in 1 octahedral voids, which of the
2
following will be the molecular formula
AîQ>’$bH$s` [ap³VH$mAm| ‘| CnpñWV hmo Vmo `m¡{JH$
of the compound. H$m AUw gyÌ {ZåZ ‘| go H$m¡Zgm hmoJm?
(A) XYZ (B) X2ZY (A) XYZ (B) X2ZY
(C) X2Y4Z (D) XYZ4 (C) X2Y4Z (D) XYZ4

1-AA ] [ 22 ] [ Contd...

098. If an element A is placed in 098. `{X VËd A {dÚwV amgm`{ZH$ loUr ‘| VËd B go
electrochemicals series above element B D$na h¡ bo{H$Z VËd C go ZrMo CnpñWV h¡, VËdm|
but below element C, then the order of
oxidation power of elements
H$s Am°ŠgrH$aU j‘Vm H$m H«$‘ Š`m hmoJm?
(A) A > B > C (B) C > B > A (A) A > B > C (B) C > B > A
(C) C > A > B (D) B > A > C (C) C > A > B (D) B > A > C

099. What will be the decreasing order of 099. {ZåZ H$m~m}YZm`Zm§o Ho$ ñWm{`Ëd H$m KQ>Vm hþAm
stability of following carbocations ? H«$‘ hmoJm

(A) 3 > 5 > 4 > 1 > 2 (A) 3 > 5 > 4 > 1 > 2
(B) 1 > 2 > 3 > 5 > 4 (B) 1 > 2 > 3 > 5 > 4
(C) 5 > 4 > 3 > 2 > 1 (C) 5 > 4 > 3 > 2 > 1
(D) 1 > 2 >3 > 4 > 5 (D) 1 > 2 >3 > 4 > 5

100.
100.
In above reaction P and Q are Cnamo³V A{^{H«$`m ‘| P VWm Q h¡

(A) (A)

(B) (B)

(C) (C)

(D) (D)

1-AA ] [ 23 ] [ PTO

MATHEMATICS / J{UV
101. The resultant of two forces P and Q 101. `{X Xmo ~bm| P VWm Q Ho$ n[aUm‘r H$m n[a‘mU
is of magnitude P. If the force P is P h¡& `{X ~b P H$mo XþJwZm H$a {X`m OmE
doubled , Q remaining the same, then
angle between new resultant and the
d ~b Q H$mo An[ad{V©V aIm OmE Vmo ZE
force Q is n[aUm‘r VWm ~b Q Ho$ ‘Ü` H$moU hmoJm
(A) 30° (B) 45° (A) 30° (B) 45°
(C) 60° (D) 90° (C) 60° (D) 90°

102. The centre of gravity (centre of mass) 102. EH$ N>S‹ > bå~mB© L h¡ BgH$m aoIr` Ðì`‘mZ
of a rod (of length L) whose linear KZËd BgHo$ EH$ {gao go Xÿar Ho$ dJ© Ho$ AZwgma
mass density varies as the square of n[ad{V©V hmo ahm h¡& Bg N>S‹ > H$m JwéËd Ho$ÝÐ
the distance from one end is at
(Ðì`‘mZ Ho$ÝÐ) BgHo$ {gao go {ZåZ na hmoJm
L 3L L 3L
(A) (B) (A) (B)
3 4 3 4
3L 2L 3L 2L
(C) (D) (C) (D)
5 5 5 5
103. Three forces each of magnitude F are 103. VrZ ~b {OZH$m àË`oH$ H$m n[a‘mU F h¡
applied along the edges of a regular H$mo EH$ {Z`{‘V fQ²^wO Ho$ H$moamo§ ({H$Zmam|)
hexagon as shown in the figure. Each
side of hexagon is a. What is the
Ho$ AZw{Xe {MÌmZwgma Amamo{nV {H$`o OmVo
resultant moment (torque) of these h¡§& fQ²^wO H$s àË`oH$ ^wOm a h¡& Ho$ÝÐ
three forces about centre O? O Ho$ gmnoj BZ VrZ ~bm| H$m n[aUm‘r
AmKyU© Š`m hmoJm?

3 3
(A) 3aF (B) aF (A) 3aF (B) aF
2 2
3 3 1 3 3 1
(C) aF (D) aF (C) aF (D) aF
2 2 2 2
104. The coordinates of a moving point 104. EH$ Vb ‘| J{V‘mZ EH$ {~ÝXþ H$U H$m
particle in a plane at time t is given g‘` t na {ZX}em§H$,
b y x = a (t + sin t), y = a (1 - cos t) . T h e x = a (t + sin t), y = a (1 - cos t) h¡   Vmo H$U
magnitude of acceleration of the particle is Ho$ ËdaU H$m n[a‘mU h¡
(A) a (B) 3a (A) a (B) 3a
3 3
(C) 2 a (D) a (C) 2 a (D) a
2 2
105. A point particle moves along a straight 105. EH$ {~ÝXþ H$U EH$ gab aoIm ‘| x = t
line such that x = t where t is time. Ho$ AZwgma J{V H$a ahm h¡ Ohm± t g‘`
Then ratio of acceleration to cube of h¡& V~ H$U Ho$ ËdaU H$m doJ Ho$ KZ
the velocity is Ho$ gmW AZwnmV hmoJm
(A) − 3 (B) −  2 (A) − 3 (B) −  2
(C) − 1 (D) − 0.5 (C) − 1 (D) − 0.5

1-AA ] [ 24 ] [ Contd...

106. A body of mass m falls from rest through 106. EH$ dñVw {OgH$m Ðì`‘mZ m h¡ {dam‘ go h
a height h under gravitation acceleration g D±$MmB© go JwéËdr` ËdaU g Ho$ A§VJ©V {JaVr
and is then brought to rest by penetrating
through a depth d into some sand. The
h¡ VWm `h aoV ‘| JhamB© d VH$ Y±gVr h¡&
average deceleration of the body during aoV ‘| Y±gZo Ho$ Xm¡amZ Am¡gV ‘ÝXZ hmoJm
penetration into sand is gh gd
gh gd (A) (B)
(A) (B) d h
d2 h 2 2
gh gh
2 gh gh
(C) (D) (C) 2 (D) 2
d
2
2d
2 d 2d
107. A normal is drawn at a point (x1, y1) of
2
107. nadb` y 2 = 16x Ho$ {~ÝXþ (x1, y1) na EH$
the parabola y = 16x and this normal A{^bå~ Ir§Mm OmVm h¡ `h A{^bå~
makes equal angle with both x and y XmoZm| Ajmo§ x VWm y Ho$ gmW ~am~a H$moU
axes. Then point (x1, y1) is ~ZmVm h¡ Vmo {~ÝXþ (x1, y1) h¡
(A) (4, – 4) (B) (2, – 8) (A) (4, – 4) (B) (2, – 8)
(C) (4, – 8) (D) (1, – 4) (C) (4, – 8) (D) (1, – 4)
108. Two vectors A = 3 and B = 4 are 108. Xmo g{Xe A = 3 VWm B = 4 nañna bå~dV
perpendicular. Resultant of both these
vectors is R. The projection of the
h¢& BZ XmoZm| g{Xemo§ H$m n[aUm‘r R h¡& g{Xe
vector B on the vector R is B H$m g{Xe R na àjon hmoJm
(A) 3.2 (B) 2.4 (A) 3.2 (B) 2.4
(C) 5 (D) 1.25 (C) 5 (D) 1.25
109. A vector R is given by R = A # _B # C i 109. EH$ g{Xe R {ZåZ Ûmam {X`m OmVm h¡
Which of the following is true? R A
= # _ B # C i Vmo {ZåZ ‘| go H$m¡Zgm
(A) R is parallel to A H$WZ gË` h¡?
(B) R must be parallel to B (A) g{Xe R g{Xe A Ho$ g‘mÝVa h¡
(C) R must be perpendicular to B (B) g{Xe R g{Xe B Ho$ g‘mÝVa hr hmoJm
(D) None of the options (C) g{Xe R g{Xe B Ho$ bå~dV hr hmoJm
(D) BZ‘o go H$moB© {dH$ën Zht
110. Solution of the differential equation dy x- y 2 -y
dy 110. AdH$b g‘rH$aU = 2e + x e H$m
= 2e + x e is
x- y 2 -y dx
dx 3 hb h¡
(A) e = 2e + x + c
-y x 3
(A) e = 2e + x + c
-y x
3
3 3
(B) e = 2e + x + c
y -x 3
(B) e = 2e + x + c
y -x
3
3 3
(C) e = 2e + x + c
y x 3
3 y x x
(C) e = 2e + + c
-3 3
(D) e = 2e + x + c
-y x -3
(D) e = 2e + x + c
-y x
3
3
111. Solution of the differential equation dy
111. AdH$b g‘rH$aU _ x + 2y 3 i =y H$m
dy dx
_ x + 2y 3 i = y is hb h¡
dx 3
3
(A) y + cy = x (B) x + 2y3 = y + c (A) y + cy = x (B) 3
x + 2y = y + c
4 4
3 xy 3 xy
(C) y + cx = y (D) + xy = cy (C) y + cx = y (D) + xy = cy
2 2

1-AA ] [ 25 ] [ PTO

112. Value of the following expression is 112. {ZåZ ì`§OH$ H$m ‘mZ h¡
lim 1 2 2 2 2 lim 1 2
(1 + 2 + 3 + ...... + n ) 2 2 2
3 (1 + 2 + 3 + ...... + n )
n " 3 n3 n"3 n
1 1 1 1
(A) (B) (A) (B)
3 6 3 6
1 2 1 2
(C) (D) (C) (D)
2 3 2 3
113. If function f (x) = * x sin a 1x k ; x ! 0 1
113. `{X ’$bZ f (x) = * x sin a x k ; x ! 0
a ; x= 0
a ; x= 0
is continuous at x = 0 , then value of
a is x = 0 , na gVV h¡ Vmo a H$m ‘mZ h¡
(A) 1 (B) – 1 (A) 1 (B) – 1
(C) 0 (D) None of the options (C) 0 (D) BZ‘o§ go H$moB© ^r {dH$ën Zht
sinx
114. The derivative of y = x is 114. y= x
sinx
H$m AdH$bO h¡
sin x - 1 sin x - 1
(A) cos x x (A) cos x x
sin 2x sin x - 1 sin 2x sin x - 1
(B) x (B) x
2 2
sin x
acos x log x + sinx x k
sinx
(C) x acos x log x + x k (C) x
sinx

sin x
(D) cos x log x + x sin x
(D) cos x log x + x
115. The tangents to curve 115. dH«$ y = x3 - 2x 2 + x - 2 na ItMr JB©
3 2
y = x - 2x + x - 2 which are ñne© aoImAmo§ Omo {H$ gab aoIm y = x Ho$
parallel to straight line y = x are
g‘mÝVa h¡ Ho$ g‘rH$aU h¡§
86
(A) x - y = 2 and x + y = 86
27 (A) x - y = 2 VWm x + y =
27
86
(B) x + y = 2 and x + y = 86
27 (B) x + y = 2 VWm x + y =
27
86
(C) x + y = 2 and x - y = 86
27 (C) x + y = 2 VWm x - y =
27
86
(D) x - y = 2 and x - y = 86
27 (D) x - y = 2 VWm x - y =
27
116. The value of lim cos h x - cos x is
x"0 x sin x 116. lim cos h x - cos x H$m ‘mZ h¡
x"0 x sin x
1 1
(A) 1 (B) (A) 1 (B)
2 2
1 1
(C) (D) 2 (C) (D) 2
3 3
x x

117.
1
Value of Maxima of a x k is 117. ’$bZ a 1x k H$m C{ƒîQ> ‘mZ h¡
a1 e k a1 e k
(A) e (B) e (A) e (B) e
1 e 1 e
e
(C) a e k (D) e
e
(C) a e k (D) e

1-AA ] [ 26 ] [ Contd...

1 1
2 -1 2 -1
sin x dx sin x dx
118. The value of the integral w 2 32
118. w 2 3
$Ho$ g‘mH$b H$m ‘mZ
(1 - x )
0 0 (1 - x ) 2
π 1 1 π 1 1
(A) + log 2 (B) π - log 2 (A) + log 2 (B) π - log 2
2 2 2 2 2 2
π π 1 π π 1
(C) - log 2 (D) - log 2 (C) - log 2 (D) - log 2
2 4 2 2 4 2
1 1
119. Integral of 119. H$m g‘mH$b h¡
2 cos x
+ 2 cos x
+
(A) - sin x log (2 + cos x) + c (A) - sin x log (2 + cos x) + c

(B) sin x log (2 + cos x) + c (B) sin x log (2 + cos x) + c

1 -1 1 1 -1 1
(C) tan a tan x k + c (C) tan a tan x k + c
3 2 3 2
2 -1 1 x 2 -1 1 x
(D) tan d tan n + c (D) tan d tan n + c
3 3 2 3 3 2
120. The eccentricity of an ellipse 120. {XE JE XrK©d¥V
2 2
9x + 16y = 144 is 2 2
9x + 16y = 144 H$s CËHo$ÝÐVm h¡
7 2 7
(A) (B) 2
4 5 (A) (B)
4 5
3 5 3 5
(C) (D) (C) (D)
5 3 5 3
121. Taking axes of hyperbola as coordinate 121. A{Vnadb` Ho$ Ajmo§ H$mo {ZX}e Aj ‘mZH$a
axes, find its equation when the distance A{Vnadb` H$m g‘rH$aU Š`m hmoJm, O~ {H$
between the foci is 16 and eccentricity
is 2
Zm{^`mo§ H$s Xÿar 16 h¡ VWm CËHo$ÝÐVm 2 h¡
2 2 2 2
2
(A) x - y = 8
2 2
(B) x - y = 16
2 (A) x - y = 8 (B) x - y = 16
2 2 2 2
2
(C) x - y = 32
2 2
(D) x - y = 64
2
(C) x - y = 32 (D) x - y = 64
2 2
122. For a circle x + y = 81, what is the 122. d¥ Î m x 2 + y 2 = 81, H$s Cg Ordm H$m
equation of chord whose mid point is g‘rH$aU Š`m hmoJm, {OgH$m ‘Ü` {~ÝXþ
(– 2, 3)
(– 2, 3) h¡
(A) 2x - 3y - 13 = 0 (A) 2x - 3y - 13 = 0
(B) 2x + 3y + 13 = 0 (B) 2x + 3y + 13 = 0
(C) 2x - 3y + 13 = 0 (C) 2x - 3y + 13 = 0
(D) 3x - 2y + 13 = 0 (D) 3x - 2y + 13 = 0

123. The condition so that the line 123. dh eV© Š`m hmoJr O~ aoIm
2
lx + my + n = 0 may touch the parabola lx + my + n = 0 nadb` y = 8x H$mo ñne©
2
y = 8x H$a gHo$
2 2
(A) m = 8l n (B) m = 2l n 2
(A) m = 8l n
2
(B) m = 2l n
2 2
(C) 8m = l n (D) 2m = l n 2
(C) 8m = l n
2
(D) 2m = l n

1-AA ] [ 27 ] [ PTO

124. The equation of that diameter of the 124. d¥Îm x 2 + y 2 - 6x + 2y - 8 = 0 H$m ì`mg
2 2
circle x + y - 6x + 2y - 8 = 0 which (Omo {H$ ‘yb {~ÝXþgo JwOaVm h¡) H$m
passes through the origin is g‘rH$aU Š`m hmoJm?
(A) 6x - y = 0 (B) 3x + 2y = 0 (A) 6x - y = 0 (B) 3x + 2y = 0
(C) x + 3y = 0 (D) 3x - y = 0 (C) x + 3y = 0 (D) 3x - y = 0

125. If z is a complex number then 125. ¶{X z EH$ gpå‘l g§»¶m h¡ Vmo
(z + 5) ( z + 5 ) is (z + 5) ( z + 5 ) ~am~a h¡
2 2 2
(A) (z + 5)
2
(B) z + 5 (A) (z + 5) (B) z + 5
2 2
(C) z + 5i
2
(D) z - 5
2
(C) z + 5i (D) z - 5

126. If z is a complex number then which 126. ¶{X z EH$ gpå‘l g§»¶m h¡ Vmo {ZåZ ‘|
of the following statement is true? go H$m¡Zgm H$WZ g˶ h¡ ?
(A) _ z - z i is purely real (A) _ z - z i {dewÕ dmñV{dH$ h¡
(B) _ z + z i is purely imaginary (B) _ z + z i {dewÕ H$mën{ZH$ h¡
(C) _ z z i is purely imaginary (C) _ z z i {dewÕ H$mën{ZH$ h¡
(D) _ z z i is nonnegative real (D) _ z z i AG$UmË‘H$ dmñV{dH$ h¡

127. If ω is the cubic root of unity, then value 127. ¶{X ω BH$mB© H$m KZ‘yb h¡ Vmo
of the (1 + ω - ω2) 2 + (1 - ω + ω2) 2 + 1 is 2 2 2 2
(1 + ω - ω ) + (1 - ω + ω ) + 1 H$m ‘mZ h¡
(A) 1 (B) − 3
(A) 1 (B) −3
(C) −1 (D) 7
(C) −1 (D) 7
12 12
128. If, _1 + i 3 i = a + ib, Here a and b 128. ¶{X _1 + i 3 i = a + ib h¡ a VWm b
are real, then the value of b is dmñV{dH$ h¢ Vmo b H$m ‘mZ h¡
(A) 0 12
(B) 1 12 (A) 0 (B) 1 12
12
(C) _ 3 i (D) _ 2 i (C) _ 3 i (D) _2i
2 2
129. If f (θ) = 2 (sec θ + cos θ), then its 129. ¶{X f (θ) = 2 (sec 2 θ + cos 2 θ), h¡ Vmo BgH$m
value always ‘mZ gX¡d
(A) f _θ i <2 (B) f _θ i = 2 (A) f _θ i <2 (B) f _θ i = 2
(C) 4 > f (θ) >2 (D) f (θ) $ 4
(C) 4 > f (θ) >2 (D) f (θ) $ 4
130. If cot x - tan x = 2 , then generalized 130. ¶{X cot x - tan x = 2 , h¡ Vmo ì`mnH$ hb
solution is (here n is integer) h¡ (`hm± n EH$ nyUmªH$ h¡)
(A) x = 2nπ + π (B) x = nπ + π (A) x = 2nπ + π (B) x = nπ + π
2 4 2 4
nπ π nπ π nπ π nπ π
(C) x = + (D) x = + (C) x = + (D) x = +
2 8 4 16 2 8 4 16
131. A plane is flying horizontally at a height 131. EH$ {d‘mZ O‘rZ go 1Km D±$MmB© na
of 1Km from ground. Angle of elevation j¡{VO {Xem ‘| C‹S> ahm h¡ & {H$gr jU
of the plane at a certain instant is 60°. na {d‘mZH$m CÝZ`Z H$moU 60° h¡& 20
After 20 seconds angle of elevation is goH$ÊS> ~mX CÝZ`Z H$moU 30° nm`m J`m
found 30°. The speed of plane is Vmo {d‘mZ H$s Mmb h¡
100 200 100 200
(A) m /s (B) m /s (A) m /s (B) m /s
3 3 3 3
(C) 100 3 m/s (D) 200 3 m/s (C) 100 3 m/s (D) 200 3 m/s

1-AA ] [ 28 ] [ Contd...

2 3 4 2 3 4
132. sin θ cos θ - sin θ cos θ is equal 132. sin θ cos θ - sin θ cos θ ~am~a h¡
(A) 1 cos θ sin 4θ (B) 1
cos θ sin 4θ (A) 1 cos θ sin 4θ (B) 1
cos θ sin 4θ
2 4 2 4
(C) 1 sin 2 2θ (D) 1
sin θ sin 4θ (C) 1 sin 2 2θ (D) 1
sin θ sin 4θ
2 4 2 4

133. If 2 sin C cos A = sin B, then ∆ ABC is 133. ¶{X 2 sin C cos A = sin B,   h¡  Vmo ∆ ABC h¡
(A) Isosceles triangle (A) g‘{Û~mhþ {Ì^wO
(B) equilateral triangle (B) g‘~mhþ {Ì^wO
(C) right angle triangle (C) g‘H$moU `wº$ {Ì^wO
(D) none of the options (D) BZ‘o go H$moB© {dH$ën Zht

134. Value of the tan 9 1 cos- 1 a 2 kC is 134. {ZåZ tan 9 12 cos- 1 a 32 kC H$m ‘mZ h¡
2 3
(A) 5 (B) 5
5 5 1-
(A) (B) 1- 2 2
2 2
(C) 1 (D) 3
(C) 1 (D) 3 5 10
5 10

135.
2
If r = x + y + z and
2 2 2 135. ¶{X r 2 = x 2 + y 2 + z 2 VWm
-1 yz - 1 xz π -1
yz tan
tan
-1 - 1 xz π
xr + tan yr = 2 - tan φ then
-1
xr + tan yr = 2 - tan φ Vmo

x+ y x+ y yz xz
yz xz (A) φ = (B) φ = xr + yr
(A) φ = zr (B) φ = xr + yr zr
xy zr xy
zr (C) φ = xy (D) φ = zr
(C) φ = xy (D) φ = zr

136. Consider digits 1, 2, 3, 4, 5, 6 and 136. A§H$ 1, 2, 3, 4, 5, 6 VWm 7 br{OE& BZ
7. Using these digits, numbers of five A§H$mo§ H$m Cn`moJ H$aVo hþE nm±M A§H$mo§ H$s
digits are formed. Then probability of g§»`mE± ~ZmB© OmVr h¢ Vmo BZ nm±M A§H$mo§
these such five digit numbers that have H$s Eogr g§»`mAmo§ Ho$ XmoZm| {gam| na {df‘
odd digits at their both ends is A§H$ AmZo H$s àm{`H$Vm Š`m hmoJr?
1 2 1 2
(A) (B) (A) (B)
7 7 7 7
3 3
(C) (D) None of the options (C) (D) BZ‘o go H$moB© {dH$ën Zht
7 7
137. Out of 100 bicycles, ten bicycles have 137. gm¡ gmB{H$bm| ‘o go 10 gmB{H$bo§ n§Ma h¢¡
puncture. What is the probability of Vmo nm±M gmB{H$bm| Ho$ à{VXe© (goånb) ‘o
not having any punctured bicycle in a go {H$gr ^r gmB©{H$b ‘o§ n§Ma Zht hmoZo
sample of 5 bicycles ?
H$s àm{`H$Vm Š`m hmoJr?
1 1 1 1
(A) 5 (B) 5 (A) (B)
10 2 10
5
2
5
5 5
1 9 1 9
(C) 9 (D) d n (C) 9 (D) d n
2 10 10
2

1-AA ] [ 29 ] [ PTO

138. Probability of solving a particular 138. ì`{º$ A H$s {H$gr {d{eð> àý H$mo hb
question by person A is 1/3 and H$aZo H$s àm{`H$Vm 1/3 h¡ VWm Cgr àý
probability of solving that question by H$mo ì`{º$ B Ûmam hb H$aZo H$s àm{`H$Vm
person B is 2/5. What is the probability 2/5 h¡& CZ XmoZm| ‘o§ go H$‘ go H$‘ EH$
of solving that question by at least one
Ho$ Ûmam Cg àý H$mo hb H$aZo H$s àm{`H$Vm
of them ?
Š`m hmoJr?
(A) 2/5 (B) 2/3 (A) 2/5 (B) 2/3
(C) 3/5 (D) 7/9 (C) 3/5 (D) 7/9

139. Four men and three women are 139. Mma nwéf VWm VrZ ‘{hbmE± EH$ bmBZ
standing in a line for railway ticket. (n§{º$) ‘o aobdo {Q>H$Q> Ho$ {bE I‹S>o h¢
The probability of standing them in Vmo CZHo$ EH$m§Va H«$‘ ‘o I‹S>o hmoZo H$s
alternate manner is
àm{`H$Vm Š`m hmoJr?
(A) 1 (B) 1
(A) 1 (B) 1
35 33
35 33
(C) 1 (D) 1 1
84 7 (C) 1 (D)
84 7

140. log3 2, log6 2, log12 2 are in 140. log3 2, log6 2, log12 2 h¡
(A) A.P. (B) G.P. (A) A.P. Ԥo (B) G.P. Ԥo
(C) H.P. (D) None of the options (C) H.P. ‘§o (D) BZ‘o go H$moB© {dH$ën Zht

141. If p, q, r, s, t and u are in A.P. then 141. `{X p, q, r, s, t VWm u g‘mÝVa loUr
difference (t - r) is equal (A. P.) ‘| h¡§ Vmo AÝVa (t - r) ~am~a h¡
(A) 2 (s - p) (B) 2 (u - q) (A) 2 (s - p) (B) 2 (u - q)
(C) 2 (s - r) (D) (u - q) (C) 2 (s - r) (D) (u - q)

142. Value of 7_logb ai _log c bi _log a ciA 142. 7_logb ai _log c bi _log a ciA H$m ‘mZ h¡
(A) 0 (B) 1
(A) 0 (B) 1
(C) abc (D) log abc
(C) abc (D) log abc

1 + 1 +
143. If p =
log3 π log 4 π
1 then 143. ¶{X p = log1 π + log1 π + 1 Vmo
3 4
(A) 1.5  <   p   <   2 (A) 1.5  <   p   <   2
(B) 2   <   p   <   2.5 (B) 2   <   p   <   2.5
(C) 2.5  <   p   <   3 (C) 2.5  <   p   <   3
(D)  p   >   3
(D)  p   >   3
10
2 10
3x + 5 3x
2
5
144. In the expansion of f 2p 144. f 5 + 2 p Ho$ {dñVma ‘§o ‘ܶ nX h¡
5 3 x
midterm is 3x
(A) 252 (B) 284 (A) 252 (B) 284
(C) 291 (D) 242 (C) 291 (D) 242

1-AA ] [ 30 ] [ Contd...

2
145. If roots of equation of x + x + 1 = 0 145. ¶{X g‘rH$aU x 2 + x + 1 = 0 Ho$ ‘yb a, b
2
are a, b and roots of x + px + q = 0 h¡ VWm x 2 + px + q = 0 Ho$ ‘yb ba , ba Vmo
a b
are , a then value of p + q is p + q H$m ‘mZ h¡&
b
(A) – 1 (B) 1 (A) – 1 (B) 1
2 +1 2 +1
(C) 2 (D) (C) 2 (D)
2 2
3
1/a bc a
3 1/a bc a
3
146. The value of Determinant 1/b ca b
3 146. gma{UH$ 1/b ca b H$m ‘mZ h¡$
3
1/c ab c
3 1/c ab c
(A) 0 (A) 0
(B) (a - b) (b - c) (c - a) (B) (a - b) (b - c) (c - a)
2 2 2 2 2 2
(C) a b c (a - b) (b - c) (c - a) (C) a b c (a - b) (b - c) (c - a)
(D) None of the options (D) BZ‘o go H$moB© {dH$ën Zht

3 - 1 3x - 2x 8
147. If >
3 - 1 3x
H > H +>
- 2x 8
H = > H the 147. ¶{X > H > H +> H = > H h¡ Vmo
0 6 1 3 9 0 6 1 3 9
value of x is x H$m ‘mZ h¡
2 2
(A) 7 (B) - (A) 7 (B) -
9 9
3 3
(C) - (D) None of the options (C) - (D) BZ‘o go H$moB© {dH$ën Zht
8 8

148. Consider A and B two square matrices 148. EH$ hr H$mo{Q> H$s Xmo dJ© ‘o{Q´Šgmo A VWm
of same order. Select the correct B na {dMma H$s{OE& {ZåZ ‘o§ go H$m¡Zgm
alternative H$WZ gË` h¡
(A) A + B must be greater than A (A) A + B H$m  ‘mZ A   go  ~‹S>m hr hmoJm
(B) If AB = 0 either A or B must be (B) ¶{X AB = 0 V~ ¶m Vmo A ¶m B  eyݶ
zero matrix
(C) AB must be greater than A ‘¡{Q´>³g hr hmoJr&
(C) AB H$m ‘mZ A go ~‹S>m hr hmoJm
1 1
(D) > H is not unit matrix. 1 1
1 1 (D) > H BH$mB© ‘o{Q´>³g Zht h¡
1 1
149. Function f : N " N, f (x) = 2x + 3 is 149. ’$bZ f : N " N, f (x) = 2x + 3 h¡
(A) One-one Onto function
(A) EH¡$H$s AmÀN>mXH$
(B) One-one Into function
(B) EH¡$H$s AÝVj}nr
(C) Many- one Onto function
(D) Many -one Into function (C) ~hþEoH$s AmÀN>mXH$
(D) ~hþEoH$s AÝVj}nr
150. If domain of the function 2
2
f (x) = x - 6x + 7 is (- 3, 3) then its 150. ¶{X ’$bZ f (x) = x - 6x + 7
H$m àmÝV
range is (- 3, 3) h¡ Vmo BgH$m n[aga hmoJm
(A) (- 3, 3) (B) [- 2, 3) (A) (- 3, 3) (B) [- 2, 3)
(C) [- 2, 3] (D) (- 3, - 2) (C) [- 2, 3] (D) (- 3, - 2)

1-AA ] [ 31 ] [ PTO

UPSEE-2016
Answer Key Paper 1, Code AA

Physics Chemistry Mathematics
1 D 26 D 51 C 76 C 101 D 126 D
2 D 27 A 52 C 77 B 102 B 127 B
3 A 28 C 53 C 78 C 103 B 128 A
4 C 29 A 54 D 79 A 104 A 129 D
5 D 30 C 55 B 80 B 105 B 130 C
6 B 31 D 56 A 81 C 106 A 131 A
7 C 32 C 57 B 82 B 107 C 132 D
8 A 33 B 58 D 83 B 108 A 133 A
9 B 34 B 59 C 84 D 109 D 134 C
10 D 35 C 60 B 85 A 110 C 135 D
11 C 36 A 61 B 86 B 111 A 136 B
12 B 37 C 62 A 87 B 112 A 137 D
13 D 38 C 63 D 88 A 113 C 138 C
14 B 39 D 64 A 89 D 114 C 139 A
15 C 40 A 65 A 90 B 115 D 140 C
16 B 41 A 66 B 91 D 116 A 141 C
17 A 42 C 67 C 92 B 117 B 142 B
18 B 43 B 68 A 93 A 118 D 143 D
19 A 44 B 69 A 94 A 119 D 144 A
20 C 45 A 70 B 95 C 120 A 145 C
21 D 46 C 71 A 96 C 121 C 146 A
22 D 47 C 72 D 97 C 122 C 147 D
23 B 48 C 73 C 98 D 123 D 148 D
24 A 49 C 74 C 99 D 124 C 149 B
25 B 50 C 75 D 100 A 125 B 150 B

UPSEE 2016
Paper 1 Code AA Solutions
Physics
Ans.1: (D) 2R
By energy conservation between points A and B
1 1
Mg (2 R )  m(0)2  mgH  m(0) 2  H  2 R
2 2
Ans.2: (D) 40 sec
4t  2t  4(60)  t  40
Ans.3: (A) Towards the left
Point of contact of wheel has velocity towards left.
Ans.4: (C) b and m alone
dU 2b
F   2bx   
dx m
Ans.5: (D) light is absorbed in quanta of energy E  h
Ans.6: (B) 954 kg/m3
V 5V
Vg   724  g  1000  g    954 Kg / m3
6 6
Ans.7: (C) 144 cm
n(18)  l where length of string is l
( n  1)(16)  l
Gives n=8 and l=144cm
Ans.8:(A) 4.8 104 C
  20 10   2.4
4

Q   4.8 104 C
R 10
3q
Ans.9: (B)
2 2 0 a
kq k ( q ) k (3q) 1 3q 3q
V    
0 0
a sin 45 a sin 45 a cos 45 0
4 0 a 2 2 0 a
2
Ans.10:(D) 
dV 1 1 1
Resistance =    
dI  dI  Slope 0
 
 dV 
Ans.11: (C) It moves back and forth (oscillating) towards the wolf
Sound wave is longitudinal wave .
Ans.12:(B)Silver
Ans.13: (D) 4V
2
1 1  A2V2    2 R  V   R VB  4V  V B
2
AV
Ans.14: (B)6 minutes

d
  k  av   0 
dt
(59  61)  61  59 
 k  30 
4  2 
1 1
 k 30  k 
2 60
(49  51)  51  49 
 k  30 
t  2 
2
 k (20)  t  6
t
Ans.15:(C)18000C
7.5 7.5
i  Q  it  (6)(60)(60)  18000C
9 9
Bl 2
Ans.16: (B)
2
Ans.17: (A) 3 E
q q  ( 3q)  5q 3q
E      3 E
0 0 0
Ans.18:(B)27A
IV=P1+P2+P3
I(120)=1800+1300+100 ∴ I=26.67A
Ans.19: (A)2A
0
B 10  8  I  20   0 ∴I=2
2 (0.1)
Ans.20:(C)80V
di  60 
 L  (40  106 )  6 
 80V
dt  3  10 
Ans.21: (D) 12.1eV
E  E3  E1  1.5  (13.6)  12.1eV
Ans.22: (D) There is no change
  
F  qv  B  0
So velocity is constant
Ans.23: (B) 105
Ans.24: (A) 1
2
1 1  GM  1 GM
K .E  mv 2  m    m ,
2 2  r  2 r

GM GM GM GM
U  m  E  K .E  U  m m  m
r 2r r 2r
Alternative:
we know that E   K  E  K

Ans.25:(B) 16 m / s 2 , 4m / s
ac  32 cos 60 0  16 m / s 2
v2 v2
ac   16   v  4m / s
R 1
m
Ans.26:(D)10 2 upwards the incline
s
75  5 g sin 300
a  (75  25) / 5  10 m / s 2
5
Ans.27:(A) 60J
1 1 1
W  KE f  KEi  (3)(64  16)  (3)(36  4)  (3)(80  40)  60J
2 2 2
Ans.28: (C) 335J
W=QA−QR
25=360−QR ∴ QR=335J
3
Ans.29: (A)
2 0
 2 4 3
E   
2 0 2 0 2 0 2 0
Ans .30: (C) Three in parallel
1
U  CV 2 For U maximum, C must be maximum
2
20
Ans.31:(D) 
3

By Wheatstone bridge R eq 
 4  6 8  12  20 
 4  6   8  12 3
Ans.32: (C) a  b, b  c
Ans.33: (B) 2 f
1 1 1 
    1   
f  R R 
1 1 1
    1     f1  2 f
f1 R 
Ans.34: (B) 26V
V  (2  4)4  2  26 volt
Ans.35: (C) 2 2
h h h
  
p 2mKE 2mqV
1 m2 q2 4 m p 2e
  2 2
2 m1q1 m pe

Ans.36:(A) 6 g sin 
5L

2 2
 L L 5
I  4m    m    mL2
2 2 4
L L 3L
  4 mg sin   mg sin   mg sin 
2 2 2
 6g
  I     sin 
I 5L
Ans.37: (C) 4iˆ  5 ˆj
Horizontal component remains constant, whereas vertical component changes its sign.
Ans.38: (C) 5 %
l T l g T 3 7
T  2     %     %  5%
g T 2l 2 g T  2 2
Ans.39: (D) 100W
1 60  100
Work per cycle    30  10  8  2   60 J  P   100W
2 60
Ans.40: (A)Path –I
Ans.41: (A) 3Hz
1  30300 / 100  303Hz , 2  30300 / 101  300Hz  1   2  3Hz

Ans.42: (C) 0.75I 0
I  I 0 cos2 300  0.75 I 0
Ans.43: (B) laser light is highly coherent
Ans.44:(B) 19%
p22 (0.9 p ) 2 0.81 p 2
KE2   
2m 2m 2m
Ans.45: (A) Magnification of microscope is inversely proportional to the least distance of distinct vision.
D
Magnification M  1 
f
Ans.46: (C) 64 SR2
W  S 8 S (3R)2  8 S (R)2   64 S ( R)2
Ans.47: (C)Less than 300 km/hr
d  d 200  200 800
 v     267km / hr
t1  t2 200  200 3
400 200
Ans.48: (C) remains constant
dQ
dS   0 ∴S=constant
T
Ans.49:(C) A  0, B  1, C  1
Output C  A  AB
Ans.50: (C) chromatic aberration

Chemistry
Ans.51: (C) He+
Ionization Potential = E  – E1
54.4 = 0 – E1 or E1 = – 54.4 eV
2
But E1 = –13.6 × Z 2 eV or –54.4eV = –13.6 × Z2 or Z=2 ,So He+ ion
(1)

Ans.52:(C) n  3, l  2, m  1, s   1
2
Energy  (n  l )
For Options: (A) (n  l )  3  0  3
(B) (n  l )  3  1  4
(C) (n  l )  3  2  5
(D) (n  l )  4  0  4
So n  3, l  2, m  1, s   1 Set of quantum number has highest energy.
2
Ans.53: (C) sp3
OF2 :-
2 2 4
6O  1s 2s 2p

or 6 O

or

sp3, Two lone pairs of electron V-shape
Ans.54:. (D) SO32  , ClO3 and BO33
2
NO3  sp  Trigonal planar
3
AsO33  sp  Pyramidal (onelone pair)
CO32  sp 2  Trigonal planar
ClO3  sp2  Pyramidal(one lone pair)
SO32  sp 3  Pyramidal(one lone pair)
BO33  sp3  Pyramidal(one lone pair)
So SO32 , ClO3 & BO33 all are non-planar
Ans.55: (B) stronger 2p(B)–2p(F)   bonding

Size of Cl is more than the size of F so in case of BF3 strong 2p(B)–2p(F)  -bonding occurs so lewis acidity of
BF3 is less than BCl3 .

Ans.56: (A) 2-methyl-6-oxohex-3-enamide

or priority  Amide > Aldehyde

Ans.57: (B) 2-Bromo-1-chloro-5-fluoro-3-iodo benzene

Ans.58: (D)(i), (iii), (v)

So at least one 20- alcohol present in I, III & V
Ans.59: (C) intermediate 2
According to Hammonds Postulates the transition state resemble to that species which is energetically near to
it.
Ans.60: (B) Cl > F > Br > I
On moving up to down in the group. Electron affinity decrease due to decrease in size but chlorine has
high electron affinity fluorine due to presence of vacant d-orbitals.
Ans.61: (B) Coordination isomerism
Answer is (B) because of coordination isomerism is a form of structural isomerism in which the composition of
the complex ion varies. In a coordination isomer the total ratio of ligand to metal remains the same, but the
ligands attached to specific metal ion change.
Ans.62: (A) zero
Species which is excess in reaction mixture follow zero order kinetics, so order of reaction with respect to O2 is
zero
Ans.63: (D) Reduction
Friedel-Craft reaction is a aromatic electrophilic substitution. So reduction is not a fried-craft reaction.
Ans.64: (A) E
Higher priority group (*) are different side ,So prefix is (E)

Ans.65: (A) 4  and 4  bonds
4 & 4

Ans.66: (B) linear, pyramidal
XeF2 = sp3d hybridization, 3l.p. & 2 l.p.

NH3 = sp3 hybrid 1l.p. + 3b.p.

So
Ans.67: (C) 2,1 and 1
BrF4  sp3d 2  2 l.p.+ 4b.p.
XeF6  sp3d 3  1 l.p. + 6 b.p.
SbCl63  sp3d 3  1 l.p. + 6 b.p
Ans.68: (A) isotropic
Crystalline solids are anisotropic not isotropic
Ans.69: (A) vapour pressure of solute is zero
Non volatile solute is always have zero vapour pressure
Ans.70: (B) associated colloids
Micelles are associated colloids which are formed above the CMC (critical micelles concentration)
Ans.71: (A) Milk fat is dispersed in water
Emulsions are colloids in which both dispersed phase & dispersion medium are liquids. So milk is emulsion in
which liquid is dispersed in water.
Ans.72: (D) –1412 kJ mol–1
2C  2 H 2  C2 H 5 , H f  52 (1)
C  O2  CO2 , H f  394 (2)
1
H 2  O2  H 2 O, H f  286 (3)
2
(4)
C2 H 4  3O2  2CO2  2 H 2 O, H C  ?
But equ. 2× (equ-2) – 2 × (equ-3) – (equ-1) = equ-4
2 (–394) + 2(–286) – (52) = – 1412 KJmol–1

Ans.73: (C)
If the difference between energy of reactant & transition state is zero then activation energy is zero.
0
Ans.74: (C) t1/ 2  a
1
t1/ 2 
n 1
For first order reaction n = 1
1
So t1/ 2 
a0
Or t1/ 2  a 0 constant
Ans.75: (D) 2.0 ML–1
Active mass is concentration in mole litre–1 or concentration in molarity
So Molarity = 8.5  1000  2.0ML1
17 250
Ans.76: (C) K12  K 2
1 [ SO3 ]
SO2 ( g )  O2 ( g ) SO3 ( g ), K1 
2 [SO2 ][O2 ]1/ 2
[ SO3 ]2
2 SO2 ( g )  O2 ( g ) 2 SO3 ( g ), K 2 
[SO2 ]2 [O2 ]

[ SO3 ]2
K12   K2
[ SO2 ]2 [O2 ]
So K12  K 2
Ans.77: (B) threo stereoisomers
When same groups are present in opposite side called threo stereoisomer .
Ans.78: (C) Schottky
During the Schottky defects same number of cations & anions are missing from their lattice site so density is
decreased.
Ans.79: (A) 1
8
N  N 0 / 2n  N  N 0 / 23  N 0 / 8
Ans.80: (B) 2

= 1  8  1 1  2
8

Ans.81: (C) 30 alcohol

Ans.82: (B) CaOCl2
Bleaching powder is CaOCl2
Ans.83: (B) square pyramidal
3
ClF4  sp d /hybridization
4 b.p. of e–& 1 lone pair of e–& shape is square pyramidal
+
Cl
F F

F F

Ans.84: (D) 3Fe(s )  4 H 2O( g ) Fe3O4 (s )  4H 2 ( g )
If gaseous moles of reactant is equal to the gaseous moles of product then reaction is not affected by the
changing in pressure
So (A) 2SO3 ( g ) 2 SO2 ( g )  O2 ( g ), n  3  2  9
(B) H 2 ( g )  I 2 (s ) 2 HI ( g ), n  2  1  1
(C) C( s)  H 2O( g ) CO( g )  H 2 ( g ), n  2  1  1
(D) 3Fe( s )  4 H 2O( g ) Fe3O4( s )  4 H 2( g ) n  4  4  0
Ans.85: (A) Increasing the temperature
N 2 ( g )  3H 2 ( g ) 2 NH 3( g )  92.3 KJ
Reaction is exothermic so on increasing the temperature equilibrium shifted in backward direction

Ans.86: (B)
Compound

CH3

H3C CH2 C CH CH CH3

H gives geometrical isomerism & it is also give enantiomerism.
Ans.87: (B)

(a) (b)

(c) (d)

So compound give fastest reaction with conc. HCl

Ans.88: (A) Polythene

Ans.89: (D)
C4H6 Degree of unsaturation (DOU) = 10  6  2
2

So is not the pair of C4H6

Ans.90:(B)

Ans.91: (D)

Resonance in carboxylate ion

Ans.92: (B) kg. ms–2
E  mc 2  kg (ms 1 )2  Kgm2 s 2
–2
So kg.ms is not the unit of energy .
Ans.93: (A) 134.1 gm mol-1
Ptotal  99.652KPa
Pwater  85.140 KPa
Pliquid  (99.652  85.140) KPa  14.512 kPa
m A 1.27 g
And 
mB 1g
m A PA M A
We have 
mB PB M B
 1 
 m  P M 
or M A   A  B B  ∴ M A  (1.27)  85.140 KPa 18g mol  ≅ 134.1 g mol–1
 mB   PA   14.512 kPa 
Ans.94: (A) Cell will swell
Osmotic pressure
Ans.95: (C) 6.92
Solution is very dilute so concentration of H+ ions in HCl solution
= H+ ions in water + H+ is ion in HCl
= 1×10–7+ 2×10–8 = 12×10–8
So pH = – log(12 108 ) =  log(22  3 108 )
 2 log 2  log 3  8 log10  2(0.301)  0.477  8  6.92
Ans.96: (C) A3B12C
A B C
At corner At Centre of Each face At corner
1 1 1
6 6 2
8 2 8
3 1
3
4 4
3 12 1
So molecular formula = A3B12C,
Ans.97: (C) X2Y4Z
Z Y X
Corner in 1 Td in 1 Oh
2 2
Voids voids
1 1 1
8 8  1 4  1
8 2 2
1 4 2
So formula is X2Y4Z
Ans.98: (D) B > A > C
According to question the position of elements in electrochemical series is
C
A
B
Oxidizing power of elements increases in electrochemical series on moving up to down so decreasing order of
oxidizing power is B > A > C

Ans.99: (D) 1 > 2 > 3 > 4 > 5

So decreasing order of stability 1 > 2 > 3 > 4 > 5
−+
Ans.100: (A) OMgBr OH

CH3 CH3
CH3 CH3

(P) (Q)

Maths
0
Ans.101: (D) 90
Let the angle between P and Q be  .Then as resultant of P and Q is P.  P 2  P 2  Q 2  2 PQ cos 
0  Q  2P cos
Let  be the angle which the new resultant makes with Q
Then tan   2 P sin  / (Q  2 P cos  )  1/ 0  
   900
3L
Ans.102. (B) 4
L L L
2
 xdm  xdx  x x dx 3L
xcm  0L  0L  0L 
2 4
 dm
0
 dx
0
  x dx
0

3
Ans.103. (B) 2 aF
 3   3   3   3 
   aF  a F   a F   aF 3
 2   2   2   2  2
a

a
Ans.104: (A) a
d 2x d
ax  2   a(1  cos t )    a sin t
dt dt
2
d y d
a y  2   a sin t   a cos t
dt dt
acceleration  a x2  a 2y  a
Ans.105: (B) −2

dx 1  12
v  t
dt 2
dv 1  1   32 1  32
a     t  2 3 t  2 v 3
dt 2  2  2
gh
Ans.106: (A)
d
u  2 gh
 0  u 2  2ad
gh
2 gh  2ad  a 
d
Ans.107. (C) (4,  8)
0
Here a  4, m  tan 45  1
So required normal point  am , 2 am    4, 8 
2

Ans.108: (A) 3.2
P
tan   3 / 4  cos   4 / 5 3

Projection  4cos   3.2 
O 4
Ans.109: (D) None of the options
   

R must be perpendicular to A as well as perpendicular to B  C 
  
Let A  iˆ, B  iˆ  ˆj , C  kˆ
   
      
R  A  B  C  iˆ   iˆ  ˆj  kˆ   iˆ    ˆj  iˆ    kˆ
 
 
Hence R is neither parallel nor perpendicular to B
y x x3
Ans. 110.(C) e  2e   c
3
dy
ey  2e x  x 2
dx
 e dy    2e  x  dx
y x 2

x3
 e y  2e x  c
3
3
Ans.111:(A) y  cy  x
2 y 3dy  ydx  xdy
ydx  xdy x
2 ydy  2
d 
y  y
x
 y2  c 
y

1
Ans.112: (A)
3
1 1 n(n  1)(2n  1)
lim 3 (12  22  32  ......  n2 )  lim 3
n n n n 6
 1  1
1   2   2 1
n  n
 lim   
n  6 6 3
Ans.113: (C) 0
f (0)  lim f ( x)
x0
1
 a  lim x sin
x 0 x
 a  0  (1 to  1)  0

sin x sin x 
Ans.114: (C) x  cos x log x  
 x 
log y  log  x sin x   sin x log x
1 dy  sin x 
  cos x log x  
y dx  x 
dy  sin x 
  x sin x  cos x log x  
dx  x 
86
Ans.115: (D) x  y  2 and x  y 
27
Only in (D)option slopes of both lines(tangents) are 1 that is equal to slope of y=x line
Alternative method:
dy
 3x 2  4 x  1 …(1)
dx
dy
yx  1 …(2)
dx
From (1) and (2) 1  3x 2  4 x  1  x  0, 4 / 3
x  0 gives y  2 and x  4 / 3 gives y  50 / 27
Thus the tangents to the curve at the points  0, 2  and  4 / 3, 50 / 27  are parallel to line y  x .The
equations of these tangents are y  (2)  1( x  0) and y  (50 / 27)  1( x  4 / 3)
86
i.e. x  y  2 and x  y 
27
Ans.116: (A) 1
cosh x  cos x  cosh x  cos x  x   0 
lim  lim  2     form 
x 0 x sin x x0
 x  sin x   0 
 cosh x  cos x  0 
 lim  2
.1  form 
x 0
 x  0 

 sinh x  sin x  0 
 lim    form 
x 0
 2x  0 
 cosh x  cos x  1  1
 lim   1
x0
 2  2
Alternative:
 e x  e x  x2 
   1   ...  
cosh x  cos x  2  2 
lim  lim  3 
x 0 x sin x x  0
  x  
x  x   .. 
  3  
 x2   x2  
 1   ...    1   ...  2 
2 2   1  x  ...    
  
 
   
2  2   1 
 lim  3   lim  2  1
x 0
  x  
x  0
  1  x  ..  
x  x   .. 
  3     3 

 
 
1 
 
e
Ans.117: (B) e
x
f ( x)  1 / x 
x
f ( x)  1 / x    log x  1
f ( x )  0    log x  1  0  x  1 / e
1/ e
 1  1/ e
Maxima of function is    e
1/ e 
x 2 x 1
Here f ( x )  1 / x   log x  1  1 / x    
 x
At x  1/ e , f (1 / e)  0
x  1 / e पर f (1 / e)  0
 1
Ans.118: (D)  log 2
4 2
1 
Let x  sin   dx  cos  d  x  0,  0 and x  , 
2 4
 
4  4
I    sec2 d   tan     1.tan d
4
0
0 0

  
  tan  0.tan 0   logsec 04
4 4 

       1
4  4  4

    logsec  logsec0    log 2  log1 =  log 2
4 2

2  1 x
Ans.119: (D) tan 1  tan   c
3  3 2
2 x
1 dx dx sec
2 dx
 2  cos x dx   x

x x

x
1  2cos 2 3cos 2  sin 2 3  tan 2
2 2 2 2
x 1 2x
Let tan  t  sec dx  dt
2 2 2
dt 2 t 2  1 x
I  2 2
 tan 1 c  tan 1  tan   c
3 t 3 3 3  3 2
7
Ans.120.(A)
4
x2 y 2
 1
16 9
9 7
e  1 
16 4
Ans.121. (C) x  y 2  32
2

Distance between foci =2ae
and 16  2a 2  a  4 2
b 2  a 2 (e 2  1)  32(2  1), b  4 2
x2 y 2
Required equation  1
32 32
Ans.122: (C) 2 x  3 y  13  0
Equation of chord T  S 
T  2 x  3 y  81
S   4  9  81  68
Equation of chord 2 x  3 y  81  68
2 x  3 y  13  0
Alternative
xx1  yy1  x12  y12
2 2
2 x  3 y   2    3  13
Ans.123: (D) 2m 2  l n
Eliminating x between the given equations
 n  my 
y 2  8  2
  ly  8my  8n  0
 l 
Given straight line touches the parabola if roots of the equation are same

2
8m   4.l.8n  2m2  l n
Ans.124: (C) x  3 y  0
 1 1 
The centre C of the circle is given by    6  ,   2   or  3, 1
 2 2 
Required diameter is the line joining the origin (0,0)and the centre C(3,-1) and hence the requires equation is
1  0
y 0   x  0
3 0
3y  x  x  3y  0
2
Ans.125: (B) z  5
2
 z  5  z  5    z  5 z  5   z  5
 
Ans.126: (D) z z is nonnegative real
Ans.127: (B) −3
1     2  0
2 2 2 2
 1     2   1     2   1    2   2         1
 4 4  4 2  1  4    2   1  4( 1)  1  4  1  3
Ans.128: (A) 0
12
12
  
 
1  i 3  2  cos  i sin   212  cos 4  i sin 4   212  i.0
12

 3 3
Ans.129: (D) f ( )  4
2 2
f ( )  2   cos   sec    2 cos  sec     2  cos   sec    4   4
   
n 
Ans.130:(C) x  
2 8
cos x sin x
  2  cos 2 x  sin 2 x  2sin x cos x
sin x cos x
 cos 2 x  sin 2 x

 tan 2 x  1  2 x  n 
4
100
Ans.131: (A) m/s P Q
3
1 2
PQ  MN  LN  LM  1cot 300  1cot 60 0  3   km
3 3 N
L
2000 1 100 M
Speed   m/s
3 20 3
1
Ans.132: (D) sin  sin 4
4
sin  cos  cos 2   sin 2    sin 2  cos cos 2
2

1 1
 sin  sin 2 cos 2  sin  sin 4
2 4
Ans.133: (A) Isosceles triangle
sin B
2sin C cos A  sin B  2cos A 
sin C
2 2 2
b c a b
2   b 2  c 2  a 2  b2  c 2  a 2
2bc c
1
Ans.134:(C)
5
1  2   1  cos
tan  cos 1     tan 
2  3  2 sin 
2
1
1  cos 1  cos 3  1
  
1  cos 2  1  cos 2 5
1
3
xy
Ans.135: (D)  
zr
 yz xz   z 2 

1 yz 1 xz

1 xr

yr  
1 xyr
 x 2
 y 
tan  tan  tan  2 
 tan  2 
xr yr  1 z   1 z 
 r 2   r2 
 zr 2 2 
 xyr  x  y 2
   zr  xy  xy
 tan 1  2 2 2 2
  tan 1    cot 1     tan 1  
x  y  z  z   xy   zr  2  zr 
 
2
Ans.136: (B)
7
7
Total numbers P5  2520
4
Total ways of odd digits at both ends= P2  12
5
Total ways of writing digits at remaining 3 places= P3  60
Total favourable conditions= 12  60  720
720 2
Required probability= 
2520 7
5
 9
Ans.137: (D)  
 10 
 90 
Probability of cycle having no puncture is  
 100 

9 1
By binomial distribution p    , q 
 10  10
5
5 0 5 9
Required probability  C5 q p   
 10 
Ans.138: (C) 3/5
P ( E1  E2 )  P ( E1 )  P ( E2 )  P ( E1  E2 )
Since P ( E1 ) and P ( E2 ) are independent
P ( E1  E2 )  P ( E1 )  P ( E2 )  P ( E1 ) P ( E2 )
1 2 1 2 9 3
   .  
3 5 3 5 15 5
1
Ans.139: (A)
35
Exhaustive events= 7
Alternative manner
MWMWMWM
Total ways= 4  3
4 3 1
Required probability= 
7 35
Ans.140. (C) H.P.
log 2 6  log 2 (3)  log 2 2  log 2 3  1
log 2 12  log 2 (3)  log 2 4  log 2 3  2
Hence log 2 3 , log 2 6 , log 2 12 are in A.P.
1 1 1
or , , are in H.P.
log 2 3 log 2 6 log 2 12
Hence log 3 2 , log 6 2 , log12 2 are in H.P.
Ans.141.(C) 2( s  r )
r  t  2s  t  2s  r
t  r  2(s  r )
Ans.142: (B) 1
Ans.143: (D) p  3
p   log 3   log 4   1   log 12   1  (2  1)
log  12  log   2  log 12  2  log  2  2
Ans.144 (A) 252
Total terms=11
5 5
 3x 2   5 
10 10 10.9.8.7.6
T
Midterm= 6  C5   2    252
 5   3 x  5 5 5.4.3.2.1
Ans.145: (C) 2

From first eq. a  b  1, ab  1
a b a b
From second equation    p, .  q  1
b a b a
2
a 2  b 2  a  b   2 ab
p    ( 1) 2  2(1)  1  p  1
ab 1
pq2
Ans.146: (A) 0
1 abc a 4 1 1 a4
1
 1 bca b 4  1 1 b 4  0  C1  C2
abc
1 cab c 4 1 1 c 4
Ans.147: (D)None of the options
9 x  1  2 x  8 
 6    3   9 
     
7 x  1 8 9
 9   9   7 x  1  8  x  7
   
1 1
Ans.148: (D)   is not unit matrix.
1 1
Ans.149: (B) One-one Into function
For any x1 , x2  N
x1  x2  2 x1  3  2 x2  3
f ( x1 )  f ( x2 )
So function is one one function
y 3
y  2x  3  x 
2
x 3
f 1 ( x)   N (Domain) when x=1,2,3,..
2
So function is one one into function.
Ans.150: (B)  2,  
Let f ( x)  y  x 2  6 x  7  y  0
 x is real so B 2  4 AC  0
 36  4(7  y)  0  2  y  0
 y  2 Range   2,  