UPSEE 2017 Paper 1 PCM Code AB

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

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

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

AZwH«$‘m§H$ / Roll No.
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àíZnwpñVH$m H$moS>
narjmWu H$m Zm‘/
Name of Candidate : AB
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$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) 8 m/s (B) 2 m/s (A) 8 m/s (B) 2 m/s
(C) 3 m/s (D) 5 m/s (C) 3 m/s (D) 5 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 Ho$ gmnoj P H$m ËdaU (m/s2 ‘|) h¡ :
with respect to Q is :
(A) 14 (B) 80 (A) 14 (B) 80
(C) 10 (D) 12 (C) 10 (D) 12

1-AB ] [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 67cm of Hg (A) 67cm Hg ñV§^ Xm~
(B) pressure of 90cm of Hg (B) 90cm Hg ñV§^ Xm~
(C) pressure of 78cm of Hg (C) 78cm Hg ñV§^ Xm~
(D) pressure of 85cm of Hg (D) 85cm 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) work done by gas is positive (A) J¡g Ûmam H$m`© YZmË‘H$ h¡

(B) work done by gas is negative (B) J¡g Ûmam H$m`© F$UmË‘H$ h¡
(C) temperature of gas increases continuously (C) J¡g H$m Vmn bJmVma ~‹T>Vm h¡

(D) process is isochoric (D) àH«$‘ g‘Am`VZr 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) 200  kgm2 s–3
(A) 200  kgm2 s–3
(B) 300  kgm2 s–3
(B) 300  kgm2 s–3
(C) 10000 kgm2 s–3 (C) 10000 kgm2 s–3

(D) 20  kgm2 s–3 (D) 20  kgm2 s–3

. 1-AB ] [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 will oscillate (A) XmobZ H$aoJm
(B) It remains stationary (B) {dam‘ ‘| hr ahoJm
(C) It moves towards left (C) ~m§`r Va’$ J{V H$aoJm
(D) It moves towards right (D) Xm`t Va’$ 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) 210 kPa (B) 840 kPa (A) 210 kPa (B) 840 kPa
(C) 8400 kPa (D) 84 kPa (C) 8400 kPa (D) 84 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
Ho$ gmnoj O‹S>Ëd AmKyU© hmoJm :
about the axis is:

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

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$m 13.5
on screen is 13.5 times the wavelength. The point is: JwUm h¡ Vmo {~ÝXþ hmoJm :
(A) bright but not central bright (A) Xrá naÝVw Ho$ÝÐr` Xrá Zht
(B) neither bright nor dark (B) Z Vmo Xrá Z hr AXrá
(C) central bright (C) Ho$ÝÐr` Xrá
(D) dark (D) AXrá

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

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) proton (B) He+ (A) àmoQ>moZ (B) He+
(C) Li + (D) BboŠQ´moZ
(C) Li+ (D) electron

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

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) 84.1 m2 (B) 84.00 m2
(A) 84.1 m2 (B) 84.00 m2
(C) 84.12 m2 (D) 84 m2 (C) 84.12 m2 (D) 84 m2

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

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

1-AB ] [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) 10 m3 (B) 20 m3 (A) 10 m3 (B) 20 m3
(C) 13 m3 (D) 26 m3 (C) 13 m3 (D) 26 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) 2 W (B) 3 W (A) 2 W (B) 3 W
(C) 5 W (D) 1 W (C) 5 W (D) 1 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
(A) (B) 2 (A) (B) 2
2 2
1 1
(C) 4 (D) (C) 4 (D)
4 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) Fe (B) Co (A) Fe (B) Co
(C) Ni (D) Al (C) Ni (D) Al

1-AB ] [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«$‘e… 1  m/s, 2  m/s VWm 3  m/s go {MÌmZwgma
t 2  m/s and 3 m/s respectively as shown. The total
angular momentum of the system of the three balls J{V H$a ahr h¢ & {XE JE jU na {~ÝXþ P Ho$ gmnoj VrZmo
about point P at given instant of time is : J|Xm| Ho$ {ZH$m` H$m Hw$b H$moUr` g§doJ h¡ :

(A) 8 kgm2s–1 (B) 9 kgm2s–1 (A) 8 kgm2s–1 (B) 9 kgm2s–1
(C) 36 kgm2s–1 (D) 7 kgm2s–1 (C) 36 kgm2s–1 (D) 7 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
3V 2 3V 2 3V 2 3V 2
(A) (B) (A) (B)
2R R 2R R
V2 2V 2 V2 2V 2
(C) (D) (C) (D)
3R 3R 3R 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 = 0 (B) A = 0, B = 1, C = 0 (A) A = 0, B = 0, C = 0 (B) A = 0, B = 1, C = 0
(C) A = 1, B = 1, C = 1 (D) A = 0, B = 0, C = 1 (C) A = 1, B = 1, C = 1 (D) A = 0, B = 0, 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 XmoZm| nmoboamoBS> H$mo EH$ gmW 180° KyU©Z EH$ hr {Xem ‘|
180° in same sense of rotation such that at every
instant, their pass(transmission) axes always
Bg àH$ma go Ky{U©V {H$`m OmVm h¡ {H$ àË`oH$ jU XmoZm| H$s
remain parallel to each other. During the rotation, nmaJ‘Z Aj h‘oem EH$ Xygao Ho$ g‘mÝVa ahVo h¢& KyU©Z Ho$
m Xm¡amZ nmoboamoBS> B go nmaJ{‘V àH$me H$s Vrd«Vm :
intensity of transmitted light through polaroid B :

(A) increases continuously (A) bJmVma ~‹T>Vr h¡
(B) first increases then decreases (B) nhbo KQ>Vr h¡ {’$a ~‹T>Vr h¡
(C) remains same (C) g‘mZ ahVr h¡
(D) decreases continuously (D) bJmVma KQ>Vr h¡
024. Activity of a radioactive substance becomes from 024. EH$ 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) 4 days (B) 6 days (A) 4 {XZ (B) 6 {XZ
(C) 2 days (D) 3 days (C) 2 {XZ (D) 3 {XZ

1-AB ] [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) 1035 A° (B) 1220 A° (A) 1035 A° (B) 1220 A°
(C) 3650 A° (D) 690 A° (C) 3650 A° (D) 690 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) 0.7V (B) 1V (A) 0.7V (B) 1V
(C) 2V (D) 0V (C) 2V (D) 0V

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) greater than P but less than 16P (A) P go A{YH$ naÝVw 16 P go H$‘
(B) greater than 16P (B) 16 P go A{YH$
(C) P (C) P
(D) 16 P (D) 16 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 front area of the large object is
(A) 
hm±, `{X ~‹S>r dñVw H$m gå‘wI joÌ’$b à{VXe© Ho$ gå‘wI
at least 1.5 times the area of the front of the
sample. joÌ’$b H$m H$‘ go H$‘ 1.5 JwUm H$a {X`m OmE&
(B) Yes, if the sample is placed at the focal point (B) hm±, `{X à{VXe© H$mo b|g Ho$ ’$moH$g {~ÝXþ na aIm OmE&
of the lens. (C) `h g§^d Zht h¡ &
(C) It is not possible
(D) hm±, `{X ~‹S>r dñVw H$m Am`VZ à{VXe© Ho$ Am`VZ
(D) Yes, if the volume of the large object is at least
1.5 times the volume of the sample. H$m 1.5 JwUm H$a {X`m OmE &

1-AB ] [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 -k p kp -k p
(A) (B) (A) (B)
r2 r2 r2 r2
kp kp
(C) (D) exactly zero (C) (D) nyU©V`m eyÝ`
r 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$mo
perpendicular to plane of the loop where γ is a
constant and t is time. The induced current flowing
Amamo{nV {H$`m OmVm h¡ Ohm± γ AMa h¡ VWm t g‘` h¡&
through loop is : byn go àdm{hV ào[aV Ymam hmoJr:

(b 2 + a 2) ct (b 2 - a 2) c (b 2 + a 2) ct (b 2 - a 2) c
(A) (B) (A) (B)
R R R R
(b 2 - a 2) ct (b 2 + a 2) c
(b 2 - a 2) ct (b + a 2) c
2
(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) 3 (B) (A) 3 (B)
MR MR MR MR
2 F F 2 F F
(C) (D) (C) (D)
3 MR 2MR 3 MR 2MR

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) displacement must be zero in the interval (A) t = 0 go t = 2 A§Vamb ‘| {dñWmnZ eyÝ` hr hmoJm &
t = 0 to t = 2. (B) t = 2 na ËdaU eyÝ` hmo gH$Vm h¡ &
(B) acceleration may be zero at t = 2
(C) t > 2 Ho$ {bE doJ eyÝ` hr hmoJm &
(C) velocity must be zero for t > 2
(D) acceleration must be zero at t = 2 (D) t = 2 na ËdaU 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) (2 it + 2tj ) m/s (B) (4 it + 2tj ) m/s (A) (2 it + 2tj ) m/s (B) (4 it + 2tj ) m/s
(C) (2 it + 4tj ) m/s (D) (4 it + 4tj ) m/s (C) (2 it + 4tj ) m/s (D) (4 it + 4tj ) m/s

1-AB ] [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 move completely parallel to the (A) {H$aU A§Vg©Vh Ho$ nyU©V`m g‘mÝVa Om`oJr &
interface . (B) {H$aU A~ ^r A§Vg©Vh na nyU©V`m Am§V[aH$
(B) ray will be still totally internally reflected at namd{V©V hmoJr &
interface. (C) {H$aU ‘mÜ`‘-II ‘| nyU©V`m nmaJ{‘V Ho$db
(C)  ray will be totally transmitted into medium-II V^r hmoJr O~ AmnVZ H$moU ~‹T>m`m OmVm
only if angle of incidence is increased. h¡&
(D)  ray will be totally transmitted in medium-II. (D) {H$aU nyU©V`m ‘mÜ`‘-II ‘| nmaJ{‘V hmoVr 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 ’$moQ>moBboŠQ´moZ
of emitted photoelectron is :
H$s A{YH$V‘ J{VO D$Om© h¡ :
(A) 2eV (B) 3eV (A) 2eV (B) 3eV
(C) 4eV (D) leV (C) 4eV (D) leV

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

1-AB ] [ 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) conduction (B) reflection (A) MmbZ (B) namdV©Z
(C) radiation (D) thermal expansion (C) {d{H$aU (D) D$î‘r` àgma

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 +y axis (A) +y Aj Ho$ AZw{Xe
(B) electric field is zero at point P, so direction (B)  {~ÝXþ P na {dÚwV joÌ eyÝ` h¡ AV… {Xem kmV Zht
cannot be determined. H$s Om gH$Vr h¡
(C) along the bisector of x axis and y axis. (C) x Aj d y Aj Ho$ AÕ©^mOH$ Ho$ AZw{Xe
(D) along +x axis (D) +x Aj 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¡ &
Cg pñà§J H$m pñà§J {Z`Vm§H$ k h¡ VWm pñà§J H$s
length of spring is R and the other end of the
àmH¥${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 9 25 9
(A) kR 2 (B) kR2 (A) kR 2 (B) kR2
2 2 2 2
(C) 8kR2 (D) 12 kR 2 (C) 8kR2 (D) 12 kR 2

1-AB ] [ 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) 9 nJ (B) 40.5 nJ (A) 9 nJ (B) 40.5 nJ
(C) 13.5 nJ (D) 18 nJ (C) 13.5 nJ (D) 18 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$U:
(A) always moves normal to the radial direction (A) h‘oem {ÌÁ`r` {Xem Ho$ bå~dV J{V H$aoJm&
(B) always moves in the radial direction only. (B) h‘oem {ÌÁ`r` {Xem Ho$ AZw{Xe J{V H$aoJm&
(C) always moves in circular orbit.
(C) h‘oem d¥Îmr` J{V H$aoJm&
(D) experiences a force directed along the radial
(D) Ho$db {ÌÁ`r` {Xem Ho$ AZw{Xe hr ~b AZw^d H$aoJm&
direction only.

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)  displacement from equilibrium is maximum
(A) t = 4s na H$U H$m gmå`mdñWm go {dñWmnZ A{YH$V‘ h¡&
at t = 4s.
(B) speed is maximum at t = 4s. (B) t = 4s na Mmb A{YH$V‘ h¡ &
(C) speed is minimum at t = 2s. (C) t = 2s na Mmb Ý`yZV‘ h¡&
(D) speed is maximum at t = 3s. (D) t = 3s na H$U H$s Mmb A{YH$V‘ 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$m Ðì`‘mZ m h¡ VWm pñà§J, nybr VWm
massless and block has mass m. What is the
extension of each spring at equilibrium ?
N>‹S>o§ (rods) Ðì`‘mZhrZ h¡& gmå`mdñWm ‘§o àË`oH$ pñà§J H$m
{dñVma Š`m hmoJm ?

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

1-AB ] [ 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) 5 (B) 9 (A) 5 (B) 9
(C) 20 (D) 2 (C) 20 (D) 2

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
doJ Š`m hmoJm ?
bullet comes out of it ?

(A) 1 m/s (B) 0.4 m/s
(A) 1 m/s (B) 0.4 m/s
(C) 1.4 m/s (D) 0.6 m/s
(C) 1.4 m/s (D) 0.6 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
following statement is true about distances as given gå~ÝY ‘§o H$m¡Zgm {dH$ën gË` h¡ ?
in options?

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

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) 40 Hz (A) 40 Hz
(B) 50 Hz (B) 50 Hz
(C) 60 Hz (C) 60 Hz
(D) 30 Hz (D) 30 Hz

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

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

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) 0.7 s (A) 0.7 s
(B) 1 s (B) 1 s
(C) 2.4 s (C) 2.4 s
(D) 0.4 s (D) 0.4 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
Vb nañna bå~dV h¡§ & BZ bynm§o Ho$ {bE AÝ`moÝ` àoaH$Ëd
to each other. The value of mutual inductance for
H$m ‘mZ h¡ :
these loops :
n0 rb 2
n0 rb 2 (A)
(A) 2a
2a

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

1-AB ] [ 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 (iii) (A) (i) Am¡a (iii)

(B) (ii) and (iii) (B) (ii) Am¡a (iii)

(C) (i), (ii) and (iii) (C) (i), (ii) Am¡a (iii)

(D) (i) and (ii) (D) (i) Am¡a (ii)

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 (A) BCl3 Am¡a AlCl3 XmoZm| bwB©g Aåb h¢ Ed§ ­AlCl3,
AlCl3 is stronger than BCl3 BCl3 go e{º$embr h¢

(B) BCl3 and AlCl3 are both equally strong Lewis (B) BCl3 Am¡a ­AlCl3 XmoZm| g‘mZ e{º$embr bwB©g
acid Aåb h¢
(C) Both BCl3 and AlCl3 are not Lewis acids (C) BCl3 Am¡a A ­ lCl3 XmoZm| hr bwB©g Aåb Zht h¢
(D) BCl3 and AlCl3 are both Lewis acids and BCl3 (D) BCl3 Am¡a A ­ lCl3 XmoZm| bwB©g Aåb h¢ Ed§ BCl3,
is stronger than AlCl3 AlCl3 go e{º$embr 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) II, III, IV (A) II, III, IV
(B) I, II, IV (B) I, II, IV
(C) I, II, III (C) I, II, III
(D) I, III, IV (D) I, III, IV

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

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

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

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

056. In aldol addition reaction product is always: 056. EëS>mob `moJmË‘H$ A{^{H«$`m ‘| CËnmX h‘oem hmoJm :
(A) b – hydroxyketone (A) b – hmB©S´moŠgrH$sQ>moZ

(B) a, b – unsaturated aldehyde (B) a, b – Ag§V¥á EëS>rhmB©S>
(C) a, b – Ag§V¥á H$sQ>moZ
(C) a, b – unsaturated ketone
(D) b – hmB©S´moŠgrEëS>rhmB©S>
(D) b – hydroxyaldehyde

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) (A)

(B) (B)

(C) (C)

(D) (D)

1-AB ] [ 16 ] [ Contd...

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

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

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 +2, for tin +2 (A) b¡S> Ho$ {bE +2, {Q>Z Ho$ {bE +2
(B) For lead +4, for tin +4 (B) b¡S> Ho$ {bE +4, {Q>Z Ho$ {bE +4
(C) For lead +2, for tin +4 (C) b¡S> Ho$ {bE +2, {Q>Z Ho$ {bE +4
(D) For lead +4, for tin +2 (D) b¡S> Ho$ {bE +4, {Q>Z Ho$ {bE +2

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$aoJm?
(A) 3–Phenyl–1–butene (A) 3 -{’$ZmBb-1-ã`yQ>rZ
(B) 2–Phenyl–1–butene (B) 2-{’$ZmBb-1-ã`yQ>rZ
(C) 1,1–Diphenyl–1–propene (C) 1,1-S>mB© {’$ZmBb-1-àmonrZ
(D) 1–Phenyl–2–butane (D) 1-{’$ZmBb-2-ã`yQ>oZ

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

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) CH3     O CHO (B) CH3 CHO (A) CH3     O CHO (B) CH3 CHO
(C) O2N CHO (D) C6H5 CHO (C) O2N CHO (D) C6H5 CHO

1-AB ] [ 17 ] [ P.T.O.

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) Ni(CO)4 and V(CO)6 (A) Ni(CO)4 Am¡a V(CO)6
(B) [Cr(CO)6] and V(CO)6 (B) [Cr(CO)6] Am¡a V(CO)6
(C) [Fe(CO)4]– and Cr(CO)6 (C) [Fe(CO)4]– Am¡a Cr(CO)6
(D) [Co(CO)4]– and Ni(CO)4 (D) [Co(CO)4]– Am¡a Ni(CO)4

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) & (iv) (B) (ii), (iii) & (iv) (A) (i), (ii) & (iv) (B) (ii), (iii) & (iv)
(C) All of these (D) (i), (ii) & (iii) (C) ¶o g^r (D) (i), (ii) & (iii)

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) H2S < NH3 < H2O < HF
(A) H2S < NH3 < H2O < HF
(B) H2O < NH3 < H2S < HF
(B) H2O < NH3 < H2S < HF
(C) HF < H2O < NH3 <H2S
(C) HF < H2O < NH3 <H2S
(D) NH3 < H2O < HF <H2S
(D) NH3 < H2O < HF <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) BH3 (B) B2H6
(A) BH3 (B) B2H6
(C) BF3 (D) BF 4-
(C) BF3 (D) BF 4-
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) PO3-
4 (B) HPO4 (A) PO3-
4 (B) HPO4
(C) H2O (D) OH -
(C) H2O (D) OH-

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 H$aZo Ho$
be added to 3 liters of 0.01M NH3 to prepare the {bE NH4Cl Ho$ {H$VZo J«m‘/^ma H$mo 3 brQ>a 0.01M   NH3
buffer of pH=9.45 at temperature 298K Ho$ {d{b`Z ‘| {‘bm`m Om`oJm ?
(Kb for NH3 is 1.85×10–5) (`hm± NH3 Ho$ {bE Kb =1.85×10–5)
(A) 0.354 gm (B) 4.55 gm (A) 0.354 gm (B) 4.55 gm
(C) 0.455gm (D) 3.53 gm (C) 0.455gm (D) 3.53 gm

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 + 2K p 2K p 1 + 2K p 2K p
(A) (B) (A) (B)
2 1 + 2K p 2 1 2K p
+

2 Kp 1+ 2 Kp 2 Kp 1+ 2 Kp
(C) (D) (C) (D)
1 + 2 Kp 2 1 + 2 Kp 2

1-AB ] [ 18 ] [ Contd...

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) 684 (B) 171 (A) 684 (B) 171
(C) 100 (D) 342 (C) 100 (D) 342

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) 340 kJ (B) 420 kJ
(A) 340 kJ (B) 420 kJ
(C) 491 kJ (D) 235 kJ
(C) 491 kJ (D) 235 kJ

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

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¡$WmoS> boS> S>mBAm°ŠgmBS> (PbO2) Ed§ EZmoS> H$m°na
Copper (Cu) (Cu) hmoVm h¡
(B) Cathode is Copper (Cu) and anode is Lead (B) H¡$WmoS> H$m°na (Cu) Ed§ EZmoS> boS> S>mBAm°ŠgmBS>
dioxide (PbO2) (PbO2) ) hmoVm h¡
(C) Cathode is Copper (Cu) and anode is Lead (Pb) (C) H¡$WmoS> H$m°na (Cu) Ed§ EZmoS> boS> (Pb) hmoVm h¡
(D) Cathode is Lead dioxide (PbO2) and anode is (D) H¡$WmoS> boS> S>mBAm°ŠgmBS> (PbO2) Ed§ EZmoS> boS>
Lead (Pb) (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) Ho$db ∆SV§Ì > 0
(A) ∆Ssystem > 0 only (B) Ho$db ∆S n[adoe > 0
(B) ∆S surrounding > 0 only
(C) (∆SV§Ì + ∆Sn[adoe ) > 0
(C) (∆ Ssystem + ∆Ssurrounding ) > 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 a
(A) c P + m (Vm - b) = RT (A) c P + m (Vm - b) = RT
V m2 V m2
(B) P(Vm – b) = RT (B) P(Vm – b) = RT
a a
(C) c P + m V = RT (C) c P + m V = RT
V m2 m V m2 m
(D) PVm = RT (D) PVm = RT

1-AB ] [ 19 ] [ P.T.O.

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) Benzene (B) Hexane (A) ~|OrZ (B) hoŠgoZ
(C) Aniline (D) Ethylene (C) E{ZbrZ (D) E{WbrZ
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) 5.0 gm (B) 1.95 × 10–9 gm (A) 5.0 gm (B) 1.95 × 10–9 gm
(C) 2 × 10–10 gm (D) 5 × 10–10gm –10
(C) 2 × 10 gm (D) 5 × 10–10gm

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 = 4, l =  0, m = 0, s = + (A) n = 4, l =  0, m = 0, s = +
2 2
1 1
(B) n = 4, l = 1, m = +1, s = + (B) n = 4, l = 1, m = +1, s = +
2 2
1 1
(C) n = 5, l = 0, m = 0, s = + (C) n = 5, l = 0, m = 0, s = +
2 2
1 1
(D) n = 3, l  =  2, m = –2, s = + (D) n = 3, l  =  2, m = –2, 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) 8, 0 (B) 6, 6
(A) 8, 0 (B) 6, 6 (C) 8, 8 (D) 4, 6
(C) 8, 8 (D) 4, 6

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

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

(B) COOH COOH (B) COOH COOH

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

1-AB ] [ 20 ] [ Contd...

084. Which of the following is a meso compound ? 084. {ZåZ ‘| go H$m¡Zgm {‘gmo `m¡{JH$ h¡ ?
(A) cis–1, 3–dimethylcyclohexane (A) {gg -1, 3- S>mB©‘o{WbgmBŠbmohoŠgoZ
(B) trans–1, 3–dimethylcyclohexane (B) Q´m§g -1, 3- S>mB©‘o{WbgmBŠbmohoŠgoZ
(C) cis–1, 4–dimethylcyclohexane
(C) {gg -1, 4- S>mB©‘o{WbgmBŠbmohoŠgoZ
(D) trans–1, 4–dimethylcyclohexane
(D) Q´m§g -1, 4- S>mB©‘o{WbgmBŠbmohoŠgoZ
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) 2,3 di ethyl butenal
(A) 2, 3 S>mB E{Wb ã`yQ>oZb
(B) 2 ethyl–3 methyl pentanal
(B) 2 E{Wb, 3 ‘o{Wb n|Q>oZb
(C) 8 methyl– 2 ethyl pentanal
(C) 8 ‘o{Wb, 2 E{Wb n|Q>oZb
(D) 2,5 Butyl butenal
(D) 2, 5 ã`w{Q>b ã`yQ>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 + alcoh. NaOH (A) + CHCl3 + alcoh. NaOH

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

OC2H5 OC2H5
(C) Conc.H2 SO4 Conc.H2 SO4
Conc.HNO3 (C)
Conc.HNO3
OH OH

(D) + CHCl3 + aq. NaOH (D) + CHCl3 + aq. NaOH

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«$‘ h¡ :
(A) B < S < P < F (B) F < S < P < B (A) B < S < P < F (B) F < S < P < B
(C) P < S < B < F (D) B < P < S < F (C) P < S < B < F (D) B < P < S < F
088. Some pairs of ions are given below. In which pair, 088. ZrMo Hw$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
(A) H3 C - CH 2 - CH - CH3 and (A) 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 5
CH 2 CH 2 CH 2 CH 2
(B) and (B) VWm

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

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

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

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) Clemmensen reaction (A) Šbo‘|gZ A{^{H«$`m
(B) Cannizzaro reaction (B) H$m{ZµOamo A{^{H«$`m
(C) None of the options (C) H$moB© ^r {dH$ën Zht
(D) Hell–Volhard reaction (D) hob dmoëhmS©> A{^{H«$`m

091. Inorganic graphite is: 091. AH$m~©{ZH$ J«o’$mBQ> h¡ :
(A) B2H6 (B) BN (A) B2H6 (B) BN
(C) BF3 (D) B2N3H6 (C) BF3 (D) B2N3H6

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) iv > i > ii > iii (B) i > iv > ii > iii (A) iv > i > ii > iii (B) i > iv > ii > iii
(C) i > iv > iii > ii (D) ii > i > iv > iii (C) i > iv > iii > ii (D) ii > i > iv > iii

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) (ii),(iv) (B) (i),(ii) (A) (ii),(iv) (B) (i),(ii)
(C) (ii),(iii) (D) (i),(iii) (C) (ii),(iii) (D) (i),(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) ii < i < iv < iii (B) i < iii < iv < ii (A) ii < i < iv < iii (B) i < iii < iv < ii
(C) i < iii < ii < iv (D) iii < i < iv < ii (C) i < iii < ii < iv (D) iii < i < iv < ii

1-AB ] [ 22 ] [ Contd...

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
NO2 Me Me
NO2
(A) Only +M effect (A) Ho$db +M à^md
(B) Only –M effect (B) Ho$db –M à^md
(C) Both –I and –M effect (C) XmoZm| –I Am¡a –M à^md
(D) Only –I effect
(D) Ho$db –I à^md
096. Which of the following material is known as lunar 096. {ZåZ ‘| go H$m¡Zgm nXmW© byZa H$mpñQ>H$ Ho$ Zm‘ go OmZm
caustic ? OmVm h¡ ?
(A) AgCl (B) AgNO3 (A) AgCl (B) AgNO3
(C) NaOH (D) NaNO3 (C) NaOH (D) NaNO3
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) 5–ethyl–6–methyl–2–propyldecane (A) 5–E{Wb–6–‘o{Wb–2–àmo{nbSo>Ho$Z
(B) 2–ethyl–6–methyl–2–propyldecane (B) 2–E{Wb–6–‘o{Wb–2–àmo{nbSo>Ho$Z
(C) 2–ethyl–6–methyl–5–propyldecane (C) 2–E{Wb–6–‘o{Wb–5–àmo{nbSo>Ho$Z
(D) 6–ethyl–2–methyl–5–propyldecane (D) 6–E{Wb–2–‘o{Wb–5–àmo{nbSo>Ho$Z

HO -
½byH$moO HO
-
HO - HO - 098. D – ‘oZmoO D– (A)
098. D – Mannose D – glucose (A)
Product (A) of above reaction is: Cnamoº$ A{^{H«$`m H$m CËnmX (A) h¡:
(A) D–fructose (B) D–Talose (A) D – ’«$ŠQ>moO (B) D – Q>obmoO
(C) D–Idose (D) D–glucose (C) D – Am`moS>moO (D) D – ½byH$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) Benzoquionone (B) Cyclohexane-1-one (A) ~oÝOmo{¹$Zm|Z (B) gm`ŠbmohoŠgoZ-1-AmoZ
(C) Benzoic sulphate (D) Benzoic Acid (C) ~oݵOmoBH$ gë’o$Q> (D) ~oݵOmoBH$ Aåb
100. How many bonds are there in : 100. {XE JE AUw ‘| Hw$b {H$VZo ~§Y h¢ ?

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

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

MATHEMATICS / J{UV

101. Let a and b be real numbers such that 101. ‘mZm a VWm b dmñV{dH$ g§»`mE± Bg Vah h¢ {H$
6
sin a + sin b =
1
and cos a + cos b =
2
then sin a + sin b =
1
VWm cos a + cos b = 26 Vmo
2 2
the value of sin (a + b) is : sin (a + b) H$m ‘mZ Š`m hmoJm :
1 3 1 3
(A) (B) (A) (B)
3 2 3 2
2 1 2 1
(C) (D) (C) (D)
3 2 2 3 2 2

102. The tangent to the graph of a continuous function 102. EH$ gVV ’$bZ y = f (x) Ho$ J«m’$ na EH$ {~ÝXþ ({OgH$m x
y = f (x) at the point with abscissa x = a forms {ZX}em§H$ x = a h¡ ) na ñne© aoIm x Aj Ho$ gmW r3 H$moU
r
with the x axis an angle of and at the point with ~ZmVr h¡ VWm Xygao {~ÝXþ ({OgH$m x {ZX}em§H$ x = b h¡ )
3
r
abscissa x = b an angle of , then what is the value na ItMr JB© ñne© aoIm x Aj Ho$ gmW r4 H$moU ~ZmVr h¡
b
4 b

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

1 -1 2 x 3 1 -1 2 x 3
103. The system f3 5 - 3 p f y p = fb p has no 103. {ZH$m` f3 5 - 3 p f y p = fb p H$m hb Zht
2 6 a z 2 2 6 a z 2
solution if hmoJm `{X
(A) a = - 5, b ! 5 (B) a = - 5, b = 5 (A) a = - 5, b ! 5 (B) a = - 5, b = 5
(C) a ! - 5, b = 5 (D) a ! - 5, b ! 5 (C) a ! - 5, b = 5 (D) a ! - 5, b ! 5

104. Let a , b be the roots of x 2 + 3x + 5 = 0 then the 104. ‘mZm g‘rH$aU x 2 + 3x + 5 = 0 Ho$ ‘yb a , b h¡ Vmo dh
equation whose roots are -
1 1
and - is : g‘rH$aU Š`m hmoJm {OgHo$ ‘yb - a1 VWm - 1b h¡ :
a b
(A) 5x 2 + 3x - 4 = 0 (A) 5x 2 + 3x - 4 = 0
(B) 5x 2 - 3x + 4 = 0 (B) 5x 2 - 3x + 4 = 0
(C) 5x 2 + 3x - 1 = 0 (C) 5x 2 + 3x - 1 = 0
(D) 5x 2 - 3x + 1 = 0 (D) 5x 2 - 3x + 1 = 0

1-AB ] [ 24 ] [ Contd...

105. A closed figure S is bounded by the 105. EH$ ~§X AmH¥${V S, A{Vnadb` x 2 - y 2 = a 2 VWm
hyperbola x - y 2 = a 2 and the straight line
2
gab aoIm x = a + h; (h > 0, a > 0) Ûmam n[a~Õ h¡ &
x = a + h; (h > 0, a > 0) . This closed figure is
Bg ~§X AmH¥${V S H$mo x-Aj Ho$ n[aV…Ky{U©V {H$`m OmVm
rotated about the x-axis. Then the volume of the
h¡ Vmo Bg ~§X AmH¥${V Ho$ n[a^«‘U Ho$ R>mog H$m Am`VZ
solid of revolution is :
hmoJm:
(A) rh 2 (3a + h)
(A) rh 2 (3a + h)
rh 2
(B) (3a + h) rh 2
6 (B) (3a + h)
6
rh 2
(C) (3a + h) rh 2
3 (C) (3a + h)
3
rh 2 rh 2
(D) (3a + h) (D) (3a + h)
2 2

106. The general solution of the equation 106. {ZåZ g‘rH$aU H$m ì`mnH$ hb hmoJm :
2
dy y -x dy y2 - x
= is : =
dx 2y (x + 1) dx 2y (x + 1)
(A) y 2 = (1 + x) log (1 + x) - c (A) y 2 = (1 + x) log (1 + x) - c
c c
(B) y 2 = (1 + x) log -1 (B) y 2 = (1 + x) log -1
^1 - xh ^1 - xh
c c
(C) y 2 = (1 - x) log -1 (C) y 2 = (1 - x) log -1
(1 + x) (1 + x)
c - c -
(D) y 2 = (1 + x) log 1 (D) y 2 = (1 + x) log 1
1+ x 1+ x

107. The equation of displacement of a particle is 107. EH$ H$U H$m {dñWmnZ x (t) = 5t 2 - 7t + 3 h¡& O~
x (t) = 5t 2 - 7t + 3 . The acceleration at the moment BgH$m doJ 5m / sec hmo OmVm h¡ Cg jU ËdaU {H$VZm
when its velocity becomes 5m / sec is : hmoJm ?:
(A) 3m / sec2 (A) 3m / sec2
(B) 7m / sec2 (B) 7m / sec2
(C) 10m / sec2 (C) 10m / sec2
(D) 8m / sec2 (D) 8m / sec2

108. If 5p 2 - 7p - 3 = 0 and 5q 2 - 7q - 3 = 0, 108. `{X 5p 2 - 7p - 3 = 0 VWm
p ! q , then the equation whose roots are 5p – 4q and 5q 2 - 7q - 3 = 0, p ! q , h¡ Vmo dh g‘rH$aU Š`m
5q – 4p is : hmoJm {OgHo$ ‘yb 5p – 4q VWm 5q – 4p h¢ :
(A) 5x 2 + 7x - 439 = 0 (A) 5x 2 + 7x - 439 = 0
(B) 5x 2 - 7x - 439 = 0 (B) 5x 2 - 7x - 439 = 0
(C) 5x 2 + 7x + 439 = 0 (C) 5x 2 + 7x + 439 = 0
(D) 5x 2 + x - 439 = 0 (D) 5x 2 + x - 439 = 0

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

109. The range of x for which the formula 109. dh x H$s namg Š`m hmoJr {OgHo$ {bE gyÌ
3 sin x = sin- 1 6 x^3 - 4x 2h@ hold is :
-1
3 sin-1
x = sin- 1 6 x^3 - 4x 2h@ ‘mÝ` ahVm h¡:
1 1 -1 #x# 2 1 1
(A) -
2
#x#
2
(B)
4 3 (A) - #x# (B) -1 #x# 2
2 2 4 3
1 2 2 1 2 2
(C) - # x # 1 (D) - # x # (C) - # x # 1 (D) - # x #
3 3 3 3 3 3

110. The equation of the ellipse, whose focus is the 110. Cg XrK©d¥Îm H$m g‘rH$aU Š`m hmoJm {OgH$s Zm{^ {~ÝXþ
point ( – 1 , 1 ), whose directrix is the straight line (–1, 1 ), h¡ VWm {OgH$s {Z`Vm gab aoIm x – y + 3 = 0
x – y + 3 = 0 and whose eccentricity is 1/2 is :
h¡ VWm {OgH$s CËHo$ÝÐVm 1/2 h¡ :
1 - + 2
(A) (x + 1) 2 + (y - 1) 2 = (x y 3) 1 - + 2
8 (A) (x + 1) 2 + (y - 1) 2 = (x y 3)
8
1 - + 2
(B) (x + 1) 2 + (y - 1) 2 = (x y 1) 1 - + 2
8 (B) (x + 1) 2 + (y - 1) 2 = (x y 1)
8
1 - + 2
(C) (x + 1) 2 + (y - 1) 2 = (x y 3) 1 - + 2
6 (C) (x + 1) 2 + (y - 1) 2 = (x y 3)
6
1
(D) (x + 1) 2 + (y - 1) 2 = (x - y + 3) 2 1
2 (D) (x + 1) 2 + (y - 1) 2 = (x - y + 3) 2
2

2 2
111. The mean value of the function f (x) = x on 111. ’$bZ f (x) = H$m ‘mÜ` ‘mZ A§Vamb
e +1 ex + 1
the interval [ 0 , 2 ] is : [ 0 , 2 ] na hmoJm :
2
(A) 2 - log e c 2 m (A) 2 - log e c 2
2
m
e +1 e +1
2
(B) 2 + log e c 2 m (B) 2 + log e c 2
2
m
e +1 e +1
2
(C) 2 + log e c 2 m (C) 2 + log e c 2
2
m
e -1 e -1
2
(D) - 2 + log e c 2 m (D) - 2 + log e c 2
2
m
e -1 e -1

112. The general solution of the differential equation 112. AdH$b g‘rH$aU
dy x+ y x- y dy x+ y x - y H$m ì`mnH$ hb hmoJm :
+ sin = sin is : + sin = sin
dx 2 2 dx 2 2
y
(A) log e tan
y
= - 2 sin x + c (A) log e tan = - 2 sin x + c
2 2 2 2
y
(B) log e tan
y
= 2 sin x + c (B) log e tan = 2 sin x + c
4 2 4 2
y
(C) log e tan
y
= - sin x + c (C) log e tan = - sin x + c
2 2 2 2
y
(D) log e tan
y
= - 2 sin x + c (D) log e tan = - 2 sin x + c
4 2 4 2

1-AB ] [ 26 ] [ Contd...

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

114. If cos (log i 4i) = a + i b , then 114. ¶{X cos (log i 4i) = a + i b hmo V~
(A) a = 1 , b = − 1 (B) a =− 1 , b = 1 (A) a = 1 , b = − 1 (B) a =− 1 , b = 1
(C) a = 1 , b = 0 (D) a = 1 , b = 2 (C) a = 1 , b = 0 (D) a = 1 , b = 2

115. The function y = 2x - x 2 115. ’$bZ y = 2x - x 2
(A) increases in ( 0 , 1 ) but decreases in ( 1 , 2 ) (A) (0, 1) ‘| ~‹T>Vm h¡ naÝVw (1 , 2) ‘| KQ>Vm h¡
(B) Decreases in ( 0 , 2 ) (B) (0, 2) ‘| KQ>Vm h¡
(C) Increases in ( 1 , 2 ) but decreases in ( 0 , 1 ) (C) (1, 2) ‘| ~‹T>Vm h¡ naÝVw (0 , 1) ‘| KQ>Vm h¡
(D) increases in ( 0 , 2 ) (D) (0, 2) ‘| ~‹T>Vm h¡

116. If the point (a , a ) lies between the lines 2x + y = 5 116. `{X {~ÝXþ (a , a ) aoImAmo 2x + y = 5 Ho$ ‘Ü` pñWV h¡
then select one of the most appropriate option: V~ g~go Cn`wº$ EH$ {dH$ën M`Z H$amo :
5 7 5 7
(A) a < (B) a < (A) a < (B) a <
3 2 3 2
11 5 11 5
(C) a < (D) a < (C) a < (D) a <
3 2 3 2

z- 2 +3
117. If log sin r ' 1 >1 , then `{X log sin r ' 3 zz -22 -31 1 >1 hmo Vmo
- +
3 z- 2 -1 117.
6 6

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

118. The nth term of the series 118. Xr JB© loUr H$m nth nX hmoJm :
1 + 4 + 13 + 40 + 121 + 364 + …… , is : 1 + 4 + 13 + 40 + 121 + 364 + ……
1 n+ 1 n- 1 n+ 1 n-
(A) (3 1) (B) (3 1) (A) (3 1) (B) (3 1)
2 2 2 2
2n + 1 j
(C) ` (D) 3n - 1 (C) `
2n + 1 j
(D) 3n - 1
2 2

119. The interval in which the function y = x - 2 sin x; 119. dh A§Vamb Š`m hmoJm {Og‘o ’$bZ y = x - 2 sin x;
0 # x # 2r increases throughout is : 0 # x # 2r ewê$ go AÝV VH$ ~‹T>Vm h¡ :
5r 5r
(A) ` , 2r j (B) `0, r j (A) ` , 2r j (B) `0, r j
3 3 3 3
5r j r 5r j
(C) ` , (D) `0, j
r
(C) ` , (D) `0, j
r r
3 3 4 3 3 4

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

120. If the ratio of the seventh term from the beginning 120. {ÛnX {dñVma c2 1 3 + 1 mx Ho$ ewéAmµV go gmVd| nX
1 1 x 1
3 3
of the binomial expansion of c 2 3 + 1 m to the
3 3 d AÝV go gmVd| nX H$m AZwnmV 1/6 h¡ Vmo x H$m
seventh term from its end is 1/6 , then the value of x is: ‘mZ h¡ :
(A) 5 (B) 11 (A) 5 (B) 11
(C) 9 (D) 7 (C) 9 (D) 7

121. Let A={ u, v, w, z } and B= { 3 , 5 } , then the 121. ‘mZm A={ u, v, w, z } VWm B= { 3 , 5 } , V~ A go B H$mo
number of relations from A to B is :
gå~ÝYm| H$s g§»¶m hmoJr:
(A) 256 (B) 1024
(A) 256 (B) 1024
(C) 512 (D) 64
(C) 512 (D) 64
122. Given y = x 2 . As x " 2, y " 4 what must the value 122. y = x 2 {X`m h¡ O~ x " 2, y " 4 hmo Vmo δ H$m ‘mZ Š`m
of δ be for which from | x – 2 |< δ it follows that hmoZm Mm{hE {Oggo {H$ | x – 2 |< δ go | y – 4 | < ∈ = 0.001
| y – 4 | < ∈ = 0.001 ? AZwgaU hmoVm h¡ :
(A) 0.03 < δ < 0.05 (B) 0.2 < δ < 0.25 (A) 0.03 < δ < 0.05 (B) 0.2 < δ < 0.25
(C) 0.4< δ < 0.5 (D) 0 < δ < 0.00025 (C) 0.4< δ < 0.5 (D) 0 < δ < 0.00025

123. Given that f(0)= 0 and lim f (x) exists, say L. 123. {X`m h¡ {H$ f(0)= 0 h¡ VWm lim f (x) {dÚ‘mZ h¡ ‘mZm
x"0 x x"0 x
Here f l (0) denotes the derivative of f w. r. t. x at {H$ `h L h¡& `hm± f H$m x Ho$ gmnoj AdH$bZ Ho$ ‘mZ H$mo
x = 0 na f l (0) Ûmam àX{e©V {H$`m OmVm h¡ V~ L hmoJm:
x = 0. Then L is :
(A) 2f l (0) - 6 (B) 2f l (0) - 5 (A) 2f l (0) - 6 (B) 2f l (0) - 5
(C) f l (0) (D) 0 (C) f l (0) (D) 0

x
124. The inverse of the function y =
2x
is: 124. ’$bZ y = +2 x H$m ì`wËH«$‘ h¡ :
1+ 2x 1 2
1 1
(A) x = log 2 (A) x = log 2
1- 2 y 1- 2 y
1
1
(B) x = log 2 c1 - m (B) x = log 2 c1 - m
y y
1 m
(C) x = log 2 c
1 m (C) x = log 2 c
1- y 1- y
y y
(D) x = log 2 (D) x = log 2
1- y 1- y

125. The domain of the definition of the function 125. {ZåZ {XE JE ’$bZ H$s n[a^mfm H$m àmÝV h¡ :
1
1 y= + (x + 2)
y= + (x + 2) is : log10 (1 - x)
log10 (1 - x)
(A) x $ - 2 (A) x $ - 2
(B) - 3 < x # - 2 (B) - 3 < x # - 2
(C) - 2 # x <0 (C) - 2 # x <0
(D) - 2 # x <1 (D) - 2 # x <1

1-AB ] [ 28 ] [ Contd...

Z Z
]- 2 sin x r ]- 2 sin x r
if x # - if x # -
]] 2 ]] 2
r r r r
126. Let f (x) = [ A sin x + B if - <x < ; 126. ‘mZm f (x) = [ A sin x + B if - <x < ;
2 2 2 2
] r ] r
] cos x if x $ ] cos x if x $
2 2
\ \
For what values of A and B, the function f  (x) is ’$bZ f  (x) Ho$ nyar dmñV{dH$ aoIm na gVV hmoZo Ho$ {bE
continuous throughout the real line ? A VWm B Ho$ Š`m ‘mZ hmoZo Mm{hE ?
(A) A = − 1, B = 1 (B) A = − 1, B = − 1 (A) A = − 1, B = 1 (B) A = − 1, B = − 1
(C) A = 1, B = − 1 (D) A = 1, B = 1 (C) A = 1, B = − 1 (D) A = 1, B = 1

rx
‘mZm f (x) = *a (x) sin 2
rx
= *a (x) sin for x ! 0; for x ! 0;
127. Let f (x) 2 127.
1 for x = 0 1 for x = 0

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

lim lim y- a ry H$m ‘mZ h¡ :
128. The $`sin
y- a
j $ ` tan
ry is :
j. 128. $`sin j $ ` tan j.
y"a 2 2a y"a 2 2a
2a a 2a a
(A) (B) (A) (B)
r r r r
a a a a
(C) - (D) (C) - (D)
r 2r r 2r

2 n + (- 2) n 2 n + (- 2) n 2 n + (- 2) n 2 n + (- 2) n
129. Let , n = and L = then 129. ‘mZm , n = VWm Ln = V~
2n n
3n 2n 3n

as n " 3 n " 3 OmZo na:
(A) lim , {dÚ‘mZ h¡§ naÝVw lim L {dÚ‘mZ Zht h¡
(A) lim , n exists but  lim Ln does not exist n n
n"3 n"3 n"3 n"3
lim , {dÚ‘mZ Zht h¡ naÝVw lim L {dÚ‘mZ h¡
(B)  lim , n does not exist but lim Ln exists (B) n n
n"3 n"3 n"3 n"3

(C) Both the sequences do not have limits. (C) XmoZm| AZwH«$‘mo§ H$s gr‘mE± Zht h¡§.
(D) Both the sequences have limits (D) XmoZm| AZwH«$‘m§o H$s gr‘mE± hm§oJr

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

130. For what interval of variation of x, the identity 130. Ma x Ho$ n[adV©Z Ho$ {H$VZo A§Vamb Ho$ {bE {ZåZ gd©g{‘H$m
1 - x2 = - 2
arc cos 2arc tan x is true? x =-
1 + x2 gË` h¡ ? arc cos 1 -
+ 2
2arc tan x
1 x
(A) - 3 < x # 0 (B) 1 < x <3
(A) - 3 < x # 0 (B) 1 < x <3
(C) 0 # x # 1 (D) 0 # x <3
(C) 0 # x # 1 (D) 0 # x <3

131. The points of the curve y = x3 + x - 2 at which its 131. dH«$ y = x3 + x - 2 Ho$ do {~ÝXþ Š`m h¢ {Og na IrMt
tangents are parallel to the straight line y = 4x – 1 JB© ñne© aoImE± Xr JB© gab aoIm y = 4x – 1 Ho$ g‘mÝVa
are : h¡ :
(A) ( 2 , 7 ) , ( – 2 , – 11 ) (A) ( 2 , 7 ) , ( – 2 , – 11 )
(B) ( 0 , – 2 ) , `2 3 , 2 3 j
1 1

(B) ( 0 , – 2 ) , `2 3 , 2 3 j
1 1

(C) `- 2 3 , - 2 3 j, (0, - 4)
1 1

(C) `- 2 3 , - 2 3 j, (0, - 4)
1 1

(D) ( 1, 0 ), ( – 1, – 4 ) (D) ( 1, 0 ), ( – 1, – 4 )

132. If a , b , c are three vectors such that 6 a b c @ = 5 132. `{X a , b , c VrZ g{Xe Bg àH$ma h¢ {H$ 6 a b c @ = 5
then the value of 6 a # b , b # c , c # a @ is : V~ 6 a # b , b # c , c # a @ H$m ‘mZ h¡ :
(A) 15 (B) 25 (A) 15 (B) 25
(C) 20 (D) 10 (C) 20 (D) 10

133. A chord of the parabola y = x 2 - 2x + 5 joins the 133. nadb` y = x 2 - 2x + 5 H$s Ordm nadb` Ho$ {~ÝXþAm|
point with the abscissas x1 = 1, x2 = 3 Then the x1 = 1, x2 = 3 H$mo Omo‹S>Vr h¡ Vmo Bg Ordm Ho$ g‘mÝVa
equation of the tangent to the parabola parallel to
nadb` H$s ñne© aoIm H$m g‘rH$aU hmoJm :
the chord is :

(A) 2x – y + 2 = 0 (A) 2x – y + 2 = 0

(B) 2x – y + 1 = 0 (B) 2x – y + 1 = 0

(C) 2x + y + 1 = 0 (C) 2x + y + 1 = 0
5= 5=
(D) 2x - y + 0 (D) 2x - y + 0
4 4
x

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

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

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

1-AB ] [ 30 ] [ Contd...

135. The lim $2x tan x - r . is : 135. lim $2x tan x - r . H$m ‘mZ h¡ :
x"
r cos x x"
r cos x
2 2
(A) – 3 (B) – 2 (A) – 3 (B) – 2
(C) 0 (D) – 1 (C) 0 (D) – 1
136. The equation of the normal to the curve 136. àW‘ MVwWmªe Ho$ AY©^mOH$ Ho$ gmW dH«$ y = - x + 2
y = - x + 2 at the point of its intersection with Ho$ H$Q>mZ {~ÝXþ na dH«$ Ho$ A{^bå~ H$m g‘rH$aU Š`m
the bisector of the first quadrant is : hmoJm ?
(A) 4x – y + 16 = 0 (B) 4x – y = 16 (A) 4x – y + 16 = 0 (B) 4x – y = 16
(C) 2x – y – 1 = 0 (D) 2x – y + 1 = 0 (C) 2x – y – 1 = 0 (D) 2x – y + 1 = 0
137. Let the equation of a curve is given in implicit form 137. ‘mZm dH«$ H$m g‘rH$aU H$m àmê$n y = tan(x + y) h¡ V~
d2 y d2 y
as y = tan(x + y). Then in terms of y is : y Ho$ nXm| ‘| hmoJm :
dx 2 dx 2
2 (1 + y 2) - 2 (1 + y 2) 2 (1 + y 2) - 2 (1 + y 2)
(A) (B) (A) (B)
y6 y6 y6 y6
- 2 (1 + y 2) 2 (1 + y 2) 2 - 2 (1 + y 2) 2 (1 + y 2) 2
(C) (D) (C) (D)
y5 y5 y5 y5
138. Suppose the area of the Δ ABC is 10 3 . Length of
138. ‘mZm {H$ {Ì^wO ­Δ ABC H$m joÌ’$b 10 3 h¡ IÊS> A ­C
segments AC and AB be 5 and 8 respectively. Then
the angle A is (are) : VWm ­AB H$s bå~mB©`m§ H«$‘e… 5 VWm 8 h¡ Vmo H$moU A h¡:
(A) 30° or 150° (B) 90° (A) 30° ¶m 150° (B) 90°
(C) 60° or 120° (D) 45° or 135° (C) 60° ¶m 120° (D) 45° ¶m 135°

139. The angle at which the curve y = x 2 and the curve 139. dh H$moU {Og na dH«$ y = x 2 VWm dH«$
5 5 5 5
x = cos t, y = sin t intersect is : x= cos t, y = sin t H$mQ>Vo h¢ :
3 4 3 4
2 41 2 41
(A) tan- 1 (B) tan- 1 (A) tan- 1 (B) tan- 1
41 2 41 2
2 41 2 41
(C) - tan- 1 (D) 2 tan- 1 (C) - tan- 1 (D) 2 tan- 1
41 2 41 2
140. The maximum value of the function 140. ’$bZ y = 2 tan x - tan 2 x H$m A§Vamb 80, r2 B na
y = 2 tan x - tan 2 x over 80, B is :
r
2 A{YH$V‘ ‘mZ hmoJm :
(A) ∞ (B) 1 (A) ∞ (B) 1
(C) 3 (D) 2 (C) 3 (D) 2

141. Let O  =  (0, 0), A  =  (a, 11) and B = (b , 37) are the 141. EH$ g‘~mhþ {Ì^wO OAB Ho$ O = ( 0 , 0 ) ,
vertices of an equilateral triangle OAB, then a and b A = ( a , 11 ) VWm B = (b , 37) erf© h¢ Vmo a VWm b {ZåZ
satisfy the relation :
gå~ÝY g§Vwï> H$a|Jo :
(A) (a 2 + b 2) - 4ab = 138
(A) (a 2 + b 2) - 4ab = 138
2 2
(B) (a + b ) - ab = 124
(B) (a 2 + b 2) - ab = 124
(C) (a 2 + b 2) + 3ab = 130
(C) (a 2 + b 2) + 3ab = 130
(D) (a + b 2) - 3ab = 138
2
(D) (a 2 + b 2) - 3ab = 138

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

142. Let f be an odd function defined on the real 142. ‘mZm {H$ f EH$ {df‘ ’$bZ dmñV{dH$ g§»`mAmo§ na Bg àH$ma
numbers such that f (x) = 3 sin x + 4 cos x, for n[a^m{fV h¡ {H$ x $ 0, Ho$ {bE f (x) = 3 sin x + 4 cos x,
x $ 0, then f (x) for x < 0 is : V~ x < 0 Ho$ {bE f (x) hmoJm:
(A) −3 sin x +4 cos x (B) − 3 sin x − 4 cos x (A) −3 sin x +4 cos x (B) − 3 sin x − 4 cos x
(C) 3 sin x + 4 cos x (D) 3 sin x − 4 cos x (C) 3 sin x + 4 cos x (D) 3 sin x − 4 cos x

143. The function f (x) = x tan- 1
1
x
for x ! 0, 143. ’$bZ f (x) = x tan- 1 1x for x ! 0, f (0)  = 0 h¡ `h ’$bZ
f (0)  = 0 is: hmoJm :
(A) Differentiable at x = 0 (A) x = 0 na AdH$bZr` hmoJm
(B) Neither continuous at x = 0 nor differentiable (B) Z Vmo x = 0 na gVV h¡ Z hr x = 0 na AdH$bZr`
at x = 0
h¡
(C) Not continuous at x = 0
(D) continuous at x = 0 but not differentiable at (C) x = 0 na gVV Zht h¡
x=0 (D) x = 0 na gVV h¡ naÝVw x = 0 na AdH$bZr` Zht hmoJm

144. Let  a and  b be two numbers where a  < b The 144. ‘mZm a VWm b Xmo g§»`mE± h¡§ Ohm± a  < b h¡ & BZ Xmo
geometric mean of these numbers exceeds the g§»`mAmo§ H$m JwUmoÎma ‘mÜ` N>moQ>r g§»`m a go 12 A{YH$
smaller number a by 12 and the arithmetic mean h¡ VWm BÝht Xmo g§»`mAmo§ H$m g‘mÝVa ‘mÜ` ~‹S>r g§»`m b
of the same number is smaller by 24 than the larger go 24 N>moQ>m h¡ Vmo b - a H$m ‘mZ hmoJm :
number b , then the value of b - a is :
(A) 48 (B) 45
(A) 48 (B) 45
(C) 44 (D) 27
(C) 44 (D) 27

145. The values of a and b for which the function 145. a VWm b Ho$ Š`m ‘mZ hm|Jo {OgHo$ {bE ’$bZ
y = a log e x + bx 2 + x, has extremum at the points y = a log e x + bx 2 + x, {~ÝXþAmo x1 = 1 VWm x2 = 2
x1 = 1 and x2 = 2 are : na Ma‘ ‘mZ aIVm h¡ :
2 =- 1 2 =- 1
(A) a = ,b (A) a = ,b
3 6 3 6
2 1 2 1
(B) a = - , b = - (B) a = - , b = -
3 6 3 6
2 1 2 1
(C) a = - , b = (C) a = - , b =
3 6 3 6
1 1 1 1
(D) a = - , b = - (D) a = - , b = -
3 6 3 6

146. A point p is selected randomly from the interior 146. EH$ {~ÝXþ p H$mo EH$ d¥Îm Ho$ ^rVar ^mJ go `mÑpÀN>H$ ê$n
of the circle, then the probability that it is closer go M`Z {H$`m OmVm h¡ Vmo dh àm{`H$Vm Š`m hmoJr {Og‘o
to the center of the circle rather than its boundary `h dñVwV… d¥Îm H$s n[agr‘m Ho$ ZOXrH$ hmoZo H$s ~OmE
is : d¥Îm Ho$ Ho$ÝÐ Ho$ A{YH$ ZOXrH$ hmoJm:
2 1 2 1
(A) (B) (A) (B)
3 4 3 4
3 1 3 1
(C) (D) (C) (D)
4 3 4 3

1-AB ] [ 32 ] [ Contd...

a 147. If the letters of the word ASHOKA are written 147. `{X eãX A ­ SHOKA Ho$ Ajam| H$mo `mÑpÀN>H$ ê$n go
down at randomly, then the chance that all A’s are {bIm OmE Vmo g^r A Ho$ H«$‘mJV (H«$‘ go bJmVma) hmoZo
consecutive is :
H$s àm{`H$Vm Š`m hmoJr?
1 1
(A) (B) 1 1
3 4 (A) (B)
3 4
2 1
(C) (D) 2 1
3 2 (C) (D)
3 2

Z 148. In a triangle Δ ABC 148. ¶{X {Ì^wO Δ ABC ‘|
3 sin A + 4 cos B = 6 and 3 sin A + 4 cos B = 6 VWm
4 sin B + 3cos A = 1, then the angle C is : 4 sin B + 3cos A = 1, Vmo H$moU C hmoJm:
(A) 150° (B) 45° (A) 150° (B) 45°
(C) 60° (D) 30° (C) 60° (D) 30°

m
149. The value of the integral w x xdx- a is equal to:
2 2 149. g‘mH$b w dx
H$m ‘mZ h¡ :
x x2 - a2
1 a 1 a
$ (A) c - sin- 1 (A) c - sin- 1
a |x | a |x |

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

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

1 a 1 a
(D) c + sin- 1 (D) c + sin- 1
a |x | a |x |

150. The function y specified implicitly by the relation 150. EH$ ’$bZ y, gå~ÝY
y x y x
w et dt + w cos t dt = 0 satisfies the differential w e dt + w cos t dt = 0 Ûmam {Z{X©ï> {H$`m J`m h¡ `h
t
0 0 0 0
equation : ’$bZ {ZåZ AdH$b g‘rH$aU H$mo g§Vwï> H$aoJm :

d2 y dy 2 d2 y dy 2
(A) e 2y c + c m m = sin x (A) e 2y c c m m = sin x
2 + dx
dx 2 dx dx
d2 y dy 2 d2 y dy 2
n (B) e y c + c m m = sin 2x (B) e y c c m m = sin 2x
2 + dx
dx 2 dx dx
o
d2 y dy 2 d2 y dy 2
(C) e y c 2 c m m = sin x
2 + dx (C) e y c 2 c m m = sin x
2 + dx
dx dx

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

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

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

1-AB ] [ 34 ]

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

1-AB ] [ 35 ]

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

1-AB ] [ 36 ]