physics, chemistry & mathematics - career …...write your name & roll no. in the space...

TEST NO. # 7 (Take Home) Rev.7- (TR-4)

PHYSICS, CHEMISTRY & MATHEMATICS TOPIC

Physics : Unit - 10 Chemistry : Unit - 11 Mathematics : Unit - 12

Duration : 3 Hrs. Max. Marks : 480 Date :19 / 03 / 2011 Name :________________________________________________________ Roll No. : ________________________

Target / Fresher / Compact / Path Finder

CPtriple E Target 2011

SEA

L

Instructions to Candidates GENERAL:

1. This paper contains 120 Qs. in all. All questions are compulsory. 2. There is Negative Marking. Guessing of answer is harmful. 3. Write your Name & Roll No. in the space provided on this cover page of question paper. 4. The question paper contains blank for your rough work. No additional sheet will be provided

for rough work. 5. The answer sheet, machine readable Optical Mark Recognition (OMR) is provided separately. 6. Do not break the seals of the question paper booklet before being instructed to do so by the

invigilator. 7. Blank papers, Clipboards, Log tables, Slide Rule, Calculators, Cellular Phones, Pagers and

Electronic Gadgets in any form are not allowed to be carried inside the examination hall.

MARKING SCHEME:

1. Each Question has four options, only one option is correct. For each incorrect response, one-fourth of the weightage marks allotted to the question would be deducted.

2. In Physics : Q. 1 - 40 carry 4 marks each, In Chemistry : Q. 41 - 80 carry 4 marks each, In Mathematics : Q. 81 - 120 carry 4 marks each

Now, Schedule practice questions are available on internet also, Visit www.examtayari.com

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph: 0744-3040000, Fax (0744) 3040050

email : [email protected] ; Website : www.careerpointgroup.com

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 || CPtripleE Target || Page # 2

Space for rough work

PHYSICS

Statement Based Questions : (1) Both Statement-I and Statement-II are true

but Statement-II is not the correct explanationof Statement-I

(2) Both Statement-I and Statement-II are true and Statement-II is the correct explanationof Statement-I

(3) Statement-I is true but Statement-II is false (4) Statement-I is false but Statement-II is true

Q.1 Statement – I : Lenz's law violates the principle ofconservation of energy.

Statement – II : Induced emf always opposes thechange in magnetic flux responsible for itsproduction.

Q.2 Statement – I : When number of turns in a coil isdoubled, coefficient of self-inductance of the coilbecomes 4 times.

Statement – II : This is because L ∝ N2. Q.3 Statement – I : The growth of current in RL circuit

is uniform. Statement – II : L opposes the growth of current. Q.4 Statement – I : The induced emf and current will

be same in two identical loops of copper andaluminium, when rotated with same speed in thesame magnetic field.

Statement – II : Induced emf is proportional torate of change of magnetic field while inducedcurrent depends on resistance of wire.

Q.5 Statement – I : Rapid removal of a coil from amagnetic field induces more emf.

Statement – II : Flux in an area due to magnetic

field is φ = ∫→→A.dB .

dFkuksa ij vk/kkfjr iz'u (1) dFku I rFkk dFku II nksuksa lR; gSa ijUrq dFku II,

dFku I dk lgh Li"Vhdj.k ugha gSA (2) dFku I rFkk dFku II nksuksa lR; gSa rFkk dFku

II, dFku I dk lgh Li"Vhdj.k gSA (3) dFku I lR; gS ijUrq dFku II vlR; gSA (4) dFku I vlR; gS ijUrq dFku II lR; gSA Q.1 dFku – I : ysat dk fu;e ÅtkZ ds laj{k.k ds fl)kar

ij vk/kkfjr ugha gksrk gSA dFku – II : izsfjr fo. ok. cy blds mRiUu gksus ds

fy, mÙkjnk;h pqEcdh; ¶yDl esa ifjorZu dk fojks/k djrk gSA

Q.2 dFku – I : tc ,d dq.Myh esa Qsjkas dh la[;k nqxquh dj nh tkrh gS] rks dq.Myh dk Loizsj.k xq.kkad 4 xquk gks tkrk gSA

dFku – II : ;g blfy, D;ksafd L ∝ N2 gksrk gSA

Q.3 dFku – I : RL ifjiFk esa /kkjk esa of̀) ,dleku gksrh gSA dFku – II : L /kkjk esa of̀) dk fojks/k djrk gSA Q.4 dFku – I : dkWij ,oa ,Y;qfefu;e ds nks le:i ywiksa

esa izsfjr fo. ok. cy ,oa /kkjk leku gksxhA tc leku pky ls bls leku pqEcdh; {ks=k esa ?kqf.kZr fd;k tkrk gSA

dFku – II : izsfjr fo. ok. cy pqEcdh; {ks=k ds ifjorZu dh nj ds lekuqikrh gksrh gS tcdh izsfjr /kkjk rkj ds izfrjks/k ij fuHkZj djrh gSA

Q.5 dFku – I : ,d pqEcdh; {ks=k ls ,d dq.Myh dks

'kh?kzrk ls gVkus ij vf/kd fo. ok. cy izsfjr gksrk gSA dFku – II : pqEcdh; {ks=k ds dkj.k ¶yDl

φ = ∫→→A.dB gSA



Q.6 Statement – I : The magnetic flux through a closed surface is zero.

Statement – II : Gauss's law applies in the case of electric flux only.

Q.7 Statement – I : Farady's law are consequence of

conservation of energy. Statement – II : In a purely resistive AC circuit,

the current lags behind the emf in phase. Q.8 In Young's experiment, the slit widths are in the

ratio 1 : 9. The ratio of the intensity at minima to that at maxima is -

(1) 1 : 4 (2) 1 : 3 (3) 1 : 9 (4) 1 : 1

Q.9 A monochromatic plane wave of speed c and wavelength λ is diffracted at a small aperture. The diagram illustrates successive wavefronts. Afterwhat time will some portion of the wavefront XY reach P ?

X

Y

P

(1) c2

3λ (2) cλ2 (3)

cλ3 (4)

cλ4

Q.10 In Young's double slit experiment, the two slits are at a distance d apart. Interference pattern is observed on the screen at a distance D from the slits. At a point on the screen directly opposite one of the slits, a dark fringe is observed. The wavelength of wave is nearly -

(1) Dd (2)

dD (3)

Dd 2

(4) 2dD

Q.11 A vernier calliper can be used to find the following dimension/ dimensions of a calorimeter -

(1) internal diameter (2) external diameter (3) depth (4) all the above

Q.6 dFku – I : ,d ifjc) lrg ls izokfgr pqEcdh; ¶yDl 'kwU; gksrk gSA

dFku – II : xkWml dk fu;e dsoy fo|qr ¶yDl dh fLFkfr esa ykxq gksrk gS A

Q.7 dFku – I : QSjkMs dk fu;e ÅtkZ ds laj{k.k dh vfHk/kkj.kk gSA

dFku – II : ,d 'kq) izfrjks/k AC ifjiFk esa, /kkjk dyk esa fo|qr okgd cy ds ihNs gksrh gSA

Q.8 ;ax ds iz;ksx esa] fLyV pkSM+kbZ esa vuqikr 1 : 9 gSA fufEu"B o mfPp"B ij rhozrk esa vuqikr gS -

(1) 1 : 4 (2) 1 : 3 (3) 1 : 9 (4) 1 : 1

Q.9 c pky o λ rjaxnS/;Z dh ,do.khZ lery rajx ,d NksVs }kjd ij foofrZr gSA fp=k Øekxr rjaxkxzks dks n'kkZrk gSA rjaxkxz XY dk dqN Hkkx P rd fdrus le; esa igq¡psxk ?

X

Y

P

(1) c2

3λ (2) cλ2 (3)

cλ3 (4)

cλ4

Q.10 ;ax ds f}&fLyV iz;ksx esa] nks fLyVsa ,d nwljs ls d

nwjh ij gSA O;frdj.k izk:i fLyVks ls D nwjh ij insZ ij izsf{kr gksrk gSA insZ ij fdlh ,d fLyV ds Bhd lkeus fdlh fcUnq ij ,d dkyh fÝat izsf{kr gksrh gSA rjx dh rjaxnS/;Z yxHkx gS -

(1) Dd (2)

dD (3)

Dd 2

(4) 2dD

Q.11 ,d ofuZ;j dsfyij dk mi;ksx dSyksjhekih dh fuEu eki Kkr djus ds fy, mi;ksx esa yk;k tk ldrk gS

(1) vkarfjd O;kl (2) ckâ; O;kl (3) xgjkbZ (4) mijksDr lHkh



Q.12 m vernier scale division coincides with (m – 1) main scale division. Main scale is graduated in cm and each cm has n division, then 1 VSD -

(1)

−

m1m

n1 cm (2)

−

n1n

m1 cm

(3) nm cm (4)

mn cm

Q.13 A pendulum of length 200 cm is horizontal when its

bob of mass 200 gm is released. It has 90% of

initial energy when the bob reaches the lower most

point. Speed of the bob at this point -

(1) 8 m/s (2) 4.5 m/s (3) 5 m/s (4) 6 m/s

Q.14 Even a rectangular base of a balance is provided with 3 leveling & screws only because -

(1) Stable equilibrium can be brought by 2 screws, so providing 1 more will supplement the purpose.

(2) Stable equilibrium can be provided by at least 3 screws so three will suffice the purpose.

(3) For representing a plane at least three points are required.

(4) None

Q.15 A 1m long pipe is closed at one end. If sound velocity in air is 320 m/s and end correction are to be neglected then frequency in Hz for air column to resonate in the pipe is -

(1) 500 (2) 600 (3) 400 (4) 700

Q.12 m ofuZ;j Ldsy Hkkx] (m – 1) eq[; Ldsy Hkkx ds

lEikrh gSA eq[; Ldsy cm esa funsZf'kr gSa rFkk izR;sd

cm, n [kkus j[krk gS] rks 1 VSD &

(1)

−

m1m

n1 cm (2)

−

n1n

m1 cm

(3) nm cm (4)

mn cm

Q.13 200 cm yEckbZ dk ,d yksyd {kSfrt gS tc bldk

200 gm nzO;eku okyk xksyd NksM+k tkrk gSA ;g

izkjfEHkd ÅtkZ dk 90% j[krk gS] tc xksyd fuEure

fcUnq rd igqaprk gSA bl fcUnq ij xksyd dh pky

gS & (1) 8 m/s (2) 4.5 m/s (3) 5 m/s (4) 6 m/s Q.14 ;fn ,d vk;rkdkj vk/kkj 3 ysofyax LØq }kjk

lUrqyu dh fLFkfr esa gSa] D;ksafd & (1) LFkk;h lUrqyu 2 LØq }kjk izkIr fd;k tk ldrk

gS vr% 1 LØw dk;Z dks iw.kZ djus esa iwjd dh rjg gksrk gS

(2) LFkk;h lUrqyu de ls de rhuksa LØqvksa }kjk fn;k tk ldrk gS blfy, rhuksa dk;Z dks iw.kZ djsaxs

(3) ,d ry dks iznf'kZr djus ds fy;s de ls de rhu fcUnqvksa dh vko';drk gksrh gSA

(4) dksbZ ugha

Q.15 ,d 1m yEck ikbZi ,d fljs ij can gSaA ;fn ok;q esa /ofu dk osx 320 m/s gS rFkk fljk la'kks/ku ux.; gS rks ikbi esa vuqukfnr ok;q LrEHk dh vkof̀Ùk ¼Hz esa½ gS &

(1) 500 (2) 600 (3) 400 (4) 700



Q.16 The temperature gradient in the earths crust is 32ºC/km and mean conductivity of the rocks is 0.008 cgs units. If the radius of the earth be taken as 6000 km, daily loss of energy by the earth is about -

(1) 1040 cal (2) 1030 cal (3) 1020 cal (4) 1010 cal

Q.17 An ideal gas is expanding such that PT2 = constant

the coefficient of volume expansion of the gas is -

(1) T1 (2)

T2 (3)

T3 (4)

T4

Statement Based Questions : (1) Both Statement-I and Statement-II are true

but Statement-II is not the correct explanationof Statement-I

(2) Both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I

(3) Statement-I is true but Statement-II is false (4) Statement-I is false but Statement-II is true

Q.18 Statement – I : At a given wavelength a good emitter behave like a good absorber

Statement – II : According to Kirchoff's is law emissive power is proportional to absorption power.

Q.19 Statement – I : Wavelength associate with maximum energy decreases with rise in temperature.

Statement – II : Wavelength associate with maximum energy is directly proportional to the fourth power of temperature.

Q.20 Statement – I : Every object emit heat at all temperature above absolute zero.

Statement – II : Rate of emission is directly proportional to forth power of absolute temperature.

Q.16 iF̀oh dh Åijh lrg esa rki izo.krk 32ºC/km rFkk pêkuksa dh ek/; pkydrk 0.008 cgs bdkbZ gSA ;fn iF̀oh dh f=kT;k 6000 km yh tk,] rks izfrfnu iF̀oh }kjk O;f;r ÅtkZ yxHkx gksxh &

(1) 1040 cal (2) 1030 cal (3) 1020 cal (4) 1010 cal Q.17 ,d vkn'kZ xSl dks izlkfjr fd;k tkrk gS rkfd

PT2 = fu;rkad gS] rks xSl dk vk;ru izlkj xq.kkad gSa&

(1) T1 (2)

T2 (3)

T3 (4)

T4

dFkuksa ij vk/kkfjr iz'u

(1) dFku I rFkk dFku II nksuksa lR; gSa ijUrq dFku II,

dFku I dk lgh Li"Vhdj.k ugha gSA

(2) dFku I rFkk dFku II nksuksa lR; gSa rFkk dFku

II, dFku I dk lgh Li"Vhdj.k gSA

(3) dFku I lR; gS ijUrq dFku II vlR; gSA

(4) dFku I vlR; gS ijUrq dFku II lR; gSA Q.18 dFku – I : nh xbZ rjaxnS/;Z ij] ,d vPNk mRltZd

,d vPNs vo'kks"kd dh Hkkafr dk;Z djrk gSA dFku – II : fdjpkWQ ds fu;e ds vuqlkj] mRltZu

{kerk] vo'kks"k.k {kerk ds lekuqikrh gSaA Q.19 dFku – I : vf/kdre ÅtkZ ls lEcfU/kr rjaxnS/;Z esa

rki esa of̀) ds lkFk deh gksrh gSA dFku – II : vf/kdre ÅtkZ ls lEcfU/kr rjaxnS/;Z rki

dh prqFkZ ?kkr ds lekuqikrh gksrh gSA

Q.20 dFku – I : izR;sd oLrq ije 'kwU; rki ls vf/kd lHkh rki ij Å"ek mRlftZr djrh gSA

dFku – II : mRltZu nj ije rki dh prqFkZ ?kkr ds lekuqikrh gksrh gSA



Q.21 Statement – I : All bodies of black colour can be considerd as ideal black body.

Statement – II : Black colour is good absorber of heat.

Q.22 Statement – I : With increase in temperature resistance of semiconductor decreases.

Statement – II : Thermal Resistive coefficient is positive for semiconductor.

Q.23 Statement – I : After knee voltage there is large

current in forward bias. Statement – II : Potential barrier is zero at knee

voltage Q.24 Statement – I : Resistance of semiconductor can

be changed with impurity. Statement – II : Number of e– & hole increases

with impurity. Q.25 Statement – I : Resistance in reverse bias is high. Statement – II : Current in reverse bias is low as

compare to forward bias. Q.26 Statement – I : Smaller the orbit of the planet

around the sun, shorter is the time it takes to complete one revolution.

Statement – II : According to Kepler's third law of

planetary motion, square of time period is

proportional to cube of mean distance from sun.

Q.27 Statement – I : Gravitational force between two

particles is negligibly small compared to the electrical force.

Statement – II : The electrical force is experienced by charged particles only.

Q.21 dFku – I : dkys jax dh lHkh oLrq,sa vkn'kZ df̀".kdk ekuh tk ldrh gSA

dFku – II : dkyk jax Å"ek dk vPNk vo'kks"kd gksrk gSA

Q.22 dFku – I : rki esa of̀) ds lkFk ,d v)Zpkyd dk izfrjks/k de gksrk gSA

dFku – II : ,d v)Zpkyd ds fy, rkih; izfrjks/k xq.kkad /kukRed gksrk gSA

Q.23 dFku – I : uh-oksYVrk ds i'pkr~ vxz vfHkurh esa /kkjk

izokg vf/kd gksrk gSA dFku – II : uh-oksYVrk ij foHkojksf/kdk 'kwU; gSA Q.24 dFku – I : v)Zpkyd dk izfrjks/k v'kqf) ds lkFk

ifjofrZr gks ldrk gSA dFku – II : bysDVªkWu o gksy dh la[;k v'kqf) ds

lkFk c<+rh gSA

Q.25 dFku – I : mRØe vfHkurh esa izfrjks/k mPp gksrk gSA dFku – II : mRØe vfHkurh es /kkjk vxz vfHkurh dh

rqyuk esa fuEu gksrh gSA

Q.26 dFku – I : lw;Z ds pkjksa vksj ifjØe.k djrs fdlh xzg dh d{kk ftruh NksVh gksxh] ,d ifjØe.k djus esa yxus okyk le; Hkh mruk gh de gksxkA

dFku – II : dsiyj ds xzgksa dh xfr ds rr̀h; fu;e ds vuqlkj] xzgksa ds vkorZdky dk oxZ lw;Z ls xzg dh ek/; nwjh ds ?ku ds lekuqikrh gksrk gSA

Q.27 dFku – I : nks d.kksa ds chp yxus okyk xq:Rokd"kZ.k cy oS|qr cy dh rqyuk esa ux.; gksrk gSA

dFku – II : oS|qr cy dsoy vkosf'kr d.kksa }kjk gh vuqHko fd;k tk ldrk gSA



Q.28 Statement – I : An astronaut in an orbiting spacestation above the Earth experiencesweightlessness.

Statement – II : An object moving around theEarth under the influence of Earth's gravitationalforce is in a state of 'free-fall'.

Q.29 An electric bulb of 100 W – 300 V is connected

with an AC supply of 500 V and (150/π) Hz. The required inductance to save the electric bulb is -

(1) 2H (2) 21 H (3) 4 H (4)

41 H

Q.30 The impedance of a circuit consists of 3 ohm

resistance and 4 ohm reactance. The power factorof the circuit is -

(1) 0.4 (2) 0.6 (3) 0.8 (4) 1.0 Q.31 A direct current of 2 A and an alternating current

having a maximum value of 2 A flow through twoidentical resistances. The ratio of heat produced inthe two resistances will be -

(1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) 4 : 1 Q.32 In non-resonant circuit, what will be the nature of

the circuit for frequencies higher than the resonantfrequency ?

(1) resistive (2) capacitive (3) inductive (4) none of these Q.33 A particle starts S.H.M from the mean position,. Its

amplitude is A and time period is T. At the timewhen its speed is half of the maximum speed, itsdisplacement y is -

(1) 2A (2)

2A (3)

23A (4)

32A

Q.34 Two equations of two S.H.M. are y = a sin(ωt – α)

and y = b cos (ωt – α). The phase differencebetween the two is -

(1) 0º (2) αº (3) 90º (4) 180º

Q.28 dFku – I : iF̀oh dk ifjØe.k dj jgs fdlh df̀=ke mixzg esa cSBk gqvk O;fDr Hkkjghu eglwl djrk gSA

dFku – II : iF̀oh ds xq:Rokd"kZ.k cy ds izHkko esa iF̀oh ds pkjksa vksj xfreku ,d oLrq eqDr :i ls fxjus dh voLFkk esa gksrh gSA

Q.29 100 W – 300 V dk ,d fo|qr cYc 50 V o (150/π) Hz ds AC lIykbZ ls tqM+k gqvk gSA fo|qr cYc dks lqjf{kr j[kus ds fy, vko';d izsjdRo gS -

(1) 2H (2) 21 H (3) 4 H (4)

41 H

Q.30 ,d ifjiFk dh izfrck/kk 3 ohm izfrjks/k o 4 vkse izfr?kkr ls ;qDr gSA ifjiFk dk 'kfDr xq.kkad gS -

(1) 0.4 (2) 0.6 (3) 0.8 (4) 1.0

Q.31 nks le:i izfrjks/kksa ls 2 A fn"V /kkjk ,oa 2 A vf/kdre eku dh izR;korhZ /kkjk izokfgr gSA nksuksa izfrjks/kksa esa mRiUUk Å"ek dk vuqikr gksxk -

(1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) 4 : 1 Q.32 vuquknjfgr ifjiFk esa] vuquknh vkof̀Ùk dh rqyuk esa

mPp vkof̀Ùk;ksa ds fy, ifjiFk dh izdf̀r D;k gksxh ? (1) izfrjks/kd (2) /kkfjrh; (3) izsj.kh; (4) buesa ls dksbZ ugha Q.33 ,d d.k ek/; fLFkfr ls ljy vkorZ xfr izkjEHk

djrk gS ftldk vk;ke A rFkk vkorZdky T gSA fdlh ,d le; bldh pky vf/kdre xfr dh vk/kh gksrh gSA d.k dk foLFkkiu y gksxk -

(1) 2A (2)

2A (3)

23A (4)

32A

Q.34 nks ljy vkorZ xfr;k¡ ftuds lehdj.k y = a sin(ωt – α) rFkk y = b cos (ωt – α) }kjk iznf'kZr dh tkrh gSA buds chp dykUrj gksxk -

(1) 0º (2) αº (3) 90º (4) 180º



Q.35 A particle is oscillating according to the equation X = 7 cos 0.5 πt, where t is in second. The point moves from the position of equilibrium to maximum displacement in time -

(1) 4.0 s (2) 2.0 s (3) 1.0 s (4) 0.5 s

Q.36 Which of the following equations does not

represent a simple harmonic motion -

(1) y = a sin ωt (2) y = a cos ωt (3) y = a sin ωt+b cosωt (4) y = a tan ωt Q.37 A brass rod of cross-sectional area 1 cm2 and

length 0.2 m is compressed lengthwise by a weight of 5 kg. If Young's modulus of elasticity of brass is 1 × 1011 N/m2 and g = 10 m/sec2, then increase in the energy of the rod will be -

(1) 10–5 J (2) 2.5 × 10–5 J

(3) 5 × 10–5 J (4) 2.5 × 10–4 J

Q.38 Work done in splitting a drop of water of 1 mm radius into 106 droplets is (Surface tension of water = 72 × 10–3 J/m2) –

(1) 9.58 × 10–5 J (2) 8.95 × 10–5 J

(3) 5.89 × 10–5 J (4) 5.98 × 10–6 J

Q.39 In the figure, the velocity V3 will be – A2 = 0.2 m2

V1= 4ms–1

A1 = 0.2m2

V2 = 2ms–1

V3 A3= 0.4m2

(1) zero (2) 4ms–1 (3) 1 ms–1 (4) 2ms–1

Q.40 If the terminal speed of a sphere of gold (density = 19.5 kg/m3) is 0.2m/s in a viscous liquid (density = 1.5 kg/m3), find the terminal speed of a sphere of silver (density = 10.5 kg/m3) of the same size in the same liquid -

(1) 0.133 m/s (2) 0.1 m/s (3) 0.2 m/s (4) 0.4 m/s

Q.35 ,d d.k fuEu lehdj.kkuqlkj nksyu dj jgk gSA X = 7 cos 0.5 πt tgk¡ t lSd.M esa gS] rks fcUnq lUrqyu dh fLFkfr ls vf/kdre foLFkkiu dh fLFkfr rd igq¡pus esa le; ysxk -

(1) 4.0 s (2) 2.0 s (3) 1.0 s (4) 0.5 s

Q.36 fuEu lehdj.kksa esa ls dkSulk lehdj.k ljy vkorZ

xfr dks iznf'kZr ugha djrk gS - (1) y = a sin ωt (2) y = a cos ωt (3) y = asin ωt + bcosωt (4) y = a tan ωt

Q.37 1.0 lseh2 vuqizLFk dkV ds {ks=kQy rFkk 0.2 m yEch ihry dh NM+ dks 5 fdxzk Hkkj ds cy ls yEckbZ dh fn'kk esa nck;k tkrk gSA ;fn ihry dk ;ax izR;kLFkrk ekikad 1 × 1011 N/m2 rFkk g = 10 m/sec2, gks] rks NM+ dh ÅtkZ esa of̀) gksxh -

(1) 10–5 J (2) 2.5 × 10–5 J

(3) 5 × 10–5 J (4) 2.5 × 10–4 J

Q.38 1 mm f=kT;k okyh ikuh dh ,d cw¡n dks 106 cwanks esa foHkDr gksus esa fd;k x;k dk;Z gksxk (ikuh dk i"̀Bruko = 72 × 10–3 J/m2) –

(1) 9.58 × 10–5 J (2) 8.95 × 10–5 J

(3) 5.89 × 10–5 J (4) 5.98 × 10–6 J

Q.39 fuEu fp=k esa] osx V3 gksxk – A2 = 0.2 m2

V1= 4ms–1

A1 = 0.2m2

V2 = 2ms–1

V3 A3= 0.4m2

(1) zero (2) 4ms–1 (3) 1 ms–1 (4) 2ms–1

Q.40 ;fn ,d ';ku nzo esa Lo.kZ ds ,d xksys (?kuRo = 19.5 kg/m3) dk lhekUr osx 0.2m/s gS] rks leku nzo esa mlh vkdkj okys jtr (?kuRo = 1.5 kg/m3), ds ,d xksys dh lhekar pky Kkr dhft, (?kuRo = 10.5 kg/m3) -

(1) 0.133 m/s (2) 0.1 m/s (3) 0.2 m/s (4) 0.4 m/s



CHEMISTRY

Q.41 H3C–C≡CH → MgBrCH3 CH4 + A OH

CO

3

2⊕ → B

B will be

(1) (2) CMgBr

(3) O

OH (4)

O

OH

Q.42 COOH 1.ND3.∆

2. KOH, Br21.NH3,∆ 2.KOD,Br2

'X' :

'Y'

(1) 'X' is NH2 and Y is ND2

(2) 'X' is ND2 and Y is NH2

(3) Both are NH2

(4) Both are ND2

Q.43 CN CN CN CN

1. H3O⊕

2. ∆ (P);Product P will be -

(1) COOH

COOH

(2) O

O

O

(3) COOH COOH COOH COOH

(4) O

O

Q.41 H3C–C≡CH → MgBrCH3 CH4 + A OH

CO

3

2⊕ → B

B gksxk &

(1) (2) CMgBr

(3) O

OH (4)

O

OH

Q.42 COOH1.ND3.∆

2. KOH, Br21.NH3,∆2.KOD,Br2

'X' :

'Y'

(1) 'X', NH2 rFkk Y, ND2 gS

(2) 'X', ND2 rFkk Y, NH2 gS

(3) nksuksa NH2 gS

(4) nksuksa ND2 gS

Q.43 CNCNCNCN

1. H3O⊕

2. ∆ (P);mRikn P gksxk &

(1) COOH

COOH

(2) O

O

O

(3) COOHCOOHCOOHCOOH

(4) O

O



Q.44

CH3CCH2CCHO

O CH3

CH3

KOH, H2O Product (C7H10O) is -

(1) O

CH3 CH3

(2)

O

CH3

CH3

(3) O CH3

CH3 (4)

O

CH3 CH3

Q.45

O

O

H ⊕

A acid orbase B ∆ C , What are

A, B and C in the following -

(1)

O

OH

,

O

OH ,

O

(2) OH

,

O

O ,

(3) OH

, OH

,

O

(4) None of the above Q.46 The products of the reaction of HCHO and PhCHO,

in the presence of concentrated base are - (1) CH3OH + PhCOO– (2) HCOO– + PhCH2OH (3) PhCOOCH3 (4) HOOCCH2Ph

Q.44

CH3CCH2CCHO

O CH3

CH3

KOH, H2O mRikn (C7H10O) gS &

(1) O

CH3 CH3

(2)

O

CH3

CH3

(3) OCH3

CH3 (4)

O

CH3 CH3

Q.45

O

O

H⊕

A vEy ;k{kkj

B ∆ C , fuEu esa

A, B rFkk C D;k gS &

(1)

O

OH

,

O

OH ,

O

(2) OH

,

O

O,

(3)

OH

, OH

,

O

(4) mijksä esa ls dksbZ ugha Q.46 lkfUnzr {kkj dh mifLFkfr esa HCHO rFkk PhCHO

dh vfHkfØ;k ds mRikn gS &

(1) CH3OH + PhCOO– (2) HCOO– + PhCH2OH (3) PhCOOCH3 (4) HOOCCH2Ph



Q.47 Which of the following carbonyl compounds when treated with dilute acid forms a stable cation ?

(1) H3C–C–CH3

O

(2)

O

(3)

O

(4)

C

O

Q.48 Dimethyl glyoxime is used for the estimation of - (1) Nickel (2) Cobalt (3) Zinc (4) Manganese Q.49 Which of the following biomolecules contain non-

transition metal ion ? (1) Vitamin B12 (2) Chlorophyll (3) Haemoglobin (4) Insulin Q.50 Which of the following will not show geometrical

isomerism ? (1) [Cr(NH3)4Cl2] Cl (2) [Co(en2)Cl2]Cl (3) [Co(NH3)5NO2]Cl2 (4) [Pt(NH3)2Cl2] Statement based Question (Q. 51 to Q.52) The following question consists of an

"Statement-1" and "Statement-2" Type question. Use the following Key to choose the appropriate answer.

(A) Both Statement 1 & Statement 2 are correct & Statement 2 is not correct explanation of Statement 1

(B) Both Statement 1 & Statement 2 are correct & Statement 2 is correct explanation of Statement 1

(C) Statement 1 is correct & Statement 2 is wrong

(D) Statement 2 is correct & Statement 1 is wrong

Q.47 fuEu esa ls dkSuls dkcksZfuy ;kSfxd tc ruq vEy ds

lkFk fØ;k djrs gS rks ,d LFkk;h /kuk;u cukrs gS \

(1) H3C–C–CH3

O (2)

O

(3)

O

(4)

C

O

Q.48 MkbesfFky XykbvkWfDle fuEu dh tkap gsrw iz;qDr fd;k tkrk gS &

(1) fudy (2) dksckYV (3) ftad (4) eSaxuht

Q.49 fuEu esa ls dkSuls tSov.kq esa vlaØe.k /kkrq vk;u mifLFkr gS \

(1) foVkfeu B12 (2) DyksjksfQy (3) gheksXyksfcu (4) bUlqfyu

Q.50 fuEu esa ls dkSulk T;kferh; leko;ork ugha n'kkZ,xk \ (1) [Cr(NH3)4Cl2] Cl (2) [Co(en2)Cl2]Cl (3) [Co(NH3)5NO2]Cl2 (4) [Pt(NH3)2Cl2] dFkuksa ij vk/kkfjr iz'u (Q.51 ls Q.52)

fuEufyf[kr iz'u "dFku-1" rFkk "dFku-2" izdkj

dk iz'u gSaA vr% mfpr mÙkj dk p;u djus ds

fy;s fuEu fodYi dk mi;ksx dhft;sA

(A) dFku-1 vkSj dFku-2 nksukas lgh gS rFkk dFku-2,

dFku-1 dk lgh Li"Vhdj.k ugha gS

(B) dFku-1 vkSj dFku-2 nksukas lgh gS rFkk dFku-2 ,

dFku-1 dk lgh Li"Vhdj.k gS

(C) dFku-1 lgh gS rFkk dFku-2 xyr gS

(D) dFku-2 lgh gS rFkk dFku-1 xyr gS



Q.51 Statement-1 : Mohr's salt is a double salt. Statement-2 : Mohr's salt does not give tests of

Fe2+, NH4+ and SO4

2– ions. Q.52 Statement I : Crystals exhibiting Frenkel type

defects do not show any change in density due to defect.

Statement II : In Frenkel defect the cation and anion vacancies are equal in number.

Q.53 Of the following outer electronic configuration of

the atoms, the highest oxidation state is exhibited by which one of them -

(1) (n – 1) d8ns2 (2) (n – 1) d5ns–1 (3) (n – 1) d3 ns2 (4) (n – 1) d5 ns2

Q.54 Transition metals, despite high Eº oxidation are

poor reducing agents. The incorrent reason is – (1) high heat of vaporization (2) high ionization energies (3) low heats of hydration (4) complex forming nature Q.55 Which of the following is not correct about

transition metals ?

(1) Their melting and boiling points are low

(2) Their compounds are generally coloured and

paramagnetic

(3) They form ionic compounds in lower oxidation

state

(4) They exhibit variable valency

Q.51 dFku-1 : eksgj yo.k ,d nqxquk yo.k gSA dFku-2 : eksgj yo.k Fe2+, NH4

+ rFkk SO42– vk;uksa

ds ijh{k.k ugha nsrk gSA

Q.52 dFku I : Ýsady izdkj dh =kqfV n'kkZus okys fØLVy =kqfV ds dkj.k ?kuRo ds dksbZ ifjorZu iznf'kZr ugha djrsA.

dFku II : Ýsady nks"k esa /kuk;u rFkk _.kk;u fjfDr;ksa dh la[;k leku gksrh gSA

Q.53 ijek.kqvksa ds fuEu ckg~;re bysDVªkWfud foU;kl esa ls

fdl ,d ds }kjk vf/kdre vkWDlhdj.k voLFkk

iznf'kZr gksrh gS & (1) (n – 1) d8ns2 (2) (n – 1) d5ns–1 (3) (n – 1) d3 ns2 (4) (n – 1) d5 ns2

Q.54 laØe.k /kkrq,a mPp Eº vkWDlhdj.k ds ckotwn nqcZy

vipk;d gksrh gSA vlR; dkj.k gS &

(1) ok"ihdj.k dh mPp Å"ek (2) mPp vk;uhdj.k ÅtkZ,sa (3) ty;kstu dh fuEu Å"ek,a (4) ladqy fuekZ.k izdf̀r

Q.55 fuEu esa ls dkSulk laØe.k /kkrqvksa ds ckjs esa lgh

ugha gS \

(1) buds xyukad o DoFkukad fuEu gksrs gS

(2) buds ;kSfxd izk;% jaxhu o vuqpqEcdh; gksrs gS

(3) ;s fuEu vkWDlhdj.k voLFkk esa vk;fud ;kSfxd

cukrs gS

(4) ;s ifjorZu'khy la;kstdrk n'kkZrs gS



Q.56 Which of the following oxides of chromium is amphoteric in nature ?

(1) CrO (2) Cr2O3

(3) CrO3 (4) CrO5

Q.57 Phenol → dustZn A 42

3

SOH

HNOCon

+ → B

NaOH

Zn→ C

In the above reaction A, B and C are - (1) Benzene, Nitrobenene and Aniline (2) Benzene, dinitrobenzene and m-nitroaniline (3) Benzene, nitrobenzene, and hydrazobenzene (4) Toluene, m-nitrobenzene and m-toludine Q.58 Which of the following reactions will not result in

the formation of anisole ?

(1) OH+ (CH3)2 SO4 →NaOH

(2) OH+ CH3I →

(3) OH+ CH2 N2 →

(4) OH+ CH3 – Mg – I →

Q.59 Match the column : Column I Column II

(A) C=O

H

(P) Positive Iodoform test

(B)

CH=CH2

OH

(Q) Semicarbazide (C)NH2–NH–CO–NH2 (R) Brady's reagent

(D) NH–NH2

NO2

NO2

(S) Cannizzaro's reaction

Q.56 fuEu esa ls dkSulk Øksfe;e vkWDlkbM izdf̀r esa

mHk;/kehZ gS \ (1) CrO (2) Cr2O3

(3) CrO3 (4) CrO5 Q.57 fQukWy → dustZn A

42

3

SOH

HNOCon

+ → B

NaOH

Zn→ C

mijksDr vfHkfØ;k esa A, B rFkk C gS & (1) csathu] ukbVªkscsathu rFkk ,fuyhu

(2) csathu] MkbukbVªkscsathu rFkk m-ukbVªks,uhyhu (3) csathu] ukbVªkscsathu o gkbMªstkscsathu (4) VkWywbZu] m-ukbVªkscsathu rFkk m-VkWywMhu

Q.58 fuEu esa ls dkSulh vfHkfØ;kvksa esa ,uhlksy dk fuekZ.k ugha gksxk \

(1) OH+ (CH3)2 SO4 →NaOH

(2) OH+ CH3I →

(3) OH+ CH2 N2 →

(4) OH+ CH3 – Mg – I →

Q.59 LrEHk dk feyku dhft,:

LrEHk I LrEHk II

(A) C=O

H

(P) /kukRed vk;ksMksQkeZ ijh{k.k

(B)

CH=CH2

OH

(Q) lsehdkcsZtkbM

(C)NH2–NH–CO–NH2 (R) czkMh vfHkdeZd

(D) NH–NH2

NO2

NO2

(S) dsfutkjks vfHkfØ;k



(1) A→P; B→S; C→R; D→Q (2) A→S; B→P; C→R; D→Q (3) A→S; B→P; C→Q; D→R (4) A→P; B→S; C→Q; D→R Q.60 izcyre {kkj gS &

(1) H2NH2N

C=NH (2) H2N H2N

C=NH2⊕

(3) H2NH2N

C=O (4) H2N H2N

C=OH⊕

Q.61 fuEu vfHkfØ;k esa X rFkk Y ds eku gS &

CH3–C–N

OCH3

CH3

H2O X + Y

(1) CH3CHO + (CH3)2NH (2) CH3CO2H + (CH3)2NH (3) CH3COCH3 + CH3NH2 (4) CH3CONH2 + CH3NH2

Q.62

CH3CH2–N–C3H7

CH3

CH3

⊕

OH

dk rkih; vi?kVu nsrk gS &

(1) CH3CH=CH2 (2) CH2=CH2

(3) CH3–CH3 (4) CH3CH2CH3

Q.63 'A' D;k gS] tks vfHkfØ;k esa yky jax nsrk gS &

'A' NaOH)ii(

HNO)i( 2 → yky jax

(1) CH3CH2NO2 (2) (CH3)2CHNO2

(3) (CH3)3CNO2 (4)

NO2

(1) A→P; B→S; C→R; D→Q (2) A→S; B→P; C→R; D→Q (3) A→S; B→P; C→Q; D→R (4) A→P; B→S; C→Q; D→R

Q.60 Strongest base is -

(1) H2N H2N

C=NH (2) H2N H2N

C=NH2⊕

(3) H2N H2N

C=O (4) H2N H2N

C=OH⊕

Q.61 In the reaction, the value of X and Y are :

CH3–C–N

O CH3

CH3

H2O X + Y

(1) CH3CHO + (CH3)2NH (2) CH3CO2H + (CH3)2NH (3) CH3COCH3 + CH3NH2 (4) CH3CONH2 + CH3NH2

Q.62 Thermal decomposition of

CH3CH2–N–C3H7

CH3

CH3

⊕

OH

gives : (1) CH3CH=CH2 (2) CH2=CH2 (3) CH3–CH3 (4) CH3CH2CH3 Q.63 What is 'A' that gives red colour in the reaction :

'A' NaOH)ii(

HNO)i( 2 → red colour -

(1) CH3CH2NO2 (2) (CH3)2CHNO2

(3) (CH3)3CNO2 (4)

NO2



Q.64 In the given reaction sequence

C6H5NO2 Sn/HCl A

NaNO2 + HCl B

CuCN/HCN

C

H2O/H⊕∆

D Which one is not correct ? (1) A is C6H5NH2

(2) B is C6H5–N≡N Cl ⊕

(3) C is C6H5CH2CN (4) D is C6H5COOH Q.65 Which of the following compounds will not give

Friedel-Crafts reaction ?

(1)

NO2

(2)

NH3

⊕

(3)

NH2

(4) All of these

Q.66 Consider the reaction :

C–H

O

HCN 'A' Compound (A) is -

(1) C–CH3

CN

OH

(2) CO–CH3

CN

(3) CH–CN

OH

(4)

C–CH2CNO

Q.64 fn, x, vfHkfØ;k Øe esa &

C6H5NO2Sn/HCl A

NaNO2 + HCl B

CuCN/HCN

C

H2O/H⊕∆

D dkSulk dFku lgh ugha gS ? (1) A, C6H5NH2 gS

(2) B, C6H5–N≡N Cl⊕

gS (3) C, C6H5CH2CN gS (4) D, C6H5COOH gS

Q.65 fuEu esa ls dkSulk ;kSfxd fÝMy Øk¶V vfHkfØ;k ugha nsxk \

(1)

NO2

(2)

NH3

⊕

(3)

NH2

(4) mijksä lHkh

Q.66 vfHkfØ;k ij fopkj dhft, &

C–HO

HCN 'A' ;kSfxd (A) gS &

(1) C–CH3

CN

OH

(2) CO–CH3

CN

(3) CH–CNOH

(4)

C–CH2CNO



Q.67 Which of the following will not produce benzoic acid by oxidation with alkaline KMnO4 ?

(1) Ph–CH3 (2) Ph–CH2–Cl

(3) Ph–CH2CH3 (4) Ph–C–CH3

CH3

CH3

Q.68 In which of the following reactions tertiary butyl

benzene is formed -

(1)

+ Me3C–OH+BF3

(2)

+ Me2C=CH2+H2SO4

(3)

+ Me2CH–CH2Cl + AlCl3

(4) All of these

Q.69 Cl H3C HNO3+H2SO4

? identify

the major product -

(1) ClCH3

NO2

(2) ClCH3

NO2

(3) ClCH3

NO2

(4) ClCH3

NO2

Q.67 fuEu esa ls dkSulk {kkjh; KMnO4 ds lkFk

vkWDlhdj.k }kjk csatksbd vEy ugha cuk,xk ? (1) Ph–CH3 (2) Ph–CH2–Cl

(3) Ph–CH2CH3 (4) Ph–C–CH3

CH3

CH3

Q.68 fuEu esa ls dkSulh vfHkfØ;kvksa esa rr̀h;d C;qfVy csUthu fufeZr gksrk gS &

(1)

+ Me3C–OH+BF3

(2)

+ Me2C=CH2+H2SO4

(3)

+ Me2CH–CH2Cl + AlCl3

(4) mijksä lHkh

Q.69 Cl H3CHNO3+H2SO4

? eq[;

mRikn Kkr dhft, &

(1) ClCH3

NO2

(2) ClCH3

NO2

(3) ClCH3

NO2

(4) ClCH3

NO2



Q.70 For the reaction ? →

C6H5

Cl

O

, The best

reactants are -

(1) C6H5Cl +

O

C6H5–C–Cl , AlCl3

(2)

O

C6H5–C–C6H5 + Cl2, FeCl3

(3) C6H5–CH2–C6H5+Cl2, FeCl3 followed by oxidation

(4) None of these yield the desired product

Q.71 Ice crystallises in a hexagonal lattice having

volume of unit cell as 132×10–24 cm3. If density is 0.92 g cm–3 at a given temperature, the number of H2O molecules per unit cell is

(1) 1 (2) 2 (3) 3 (4) 4 Q.72 A compound with formula AB2O4 adopts a crystal

structure in which O2– ions forms ccp arrangement. A2+ ions occupy octahedral void which B3+ ions are distributed equally between tetrahedral and octahedral voids. The fraction of octahedral and tetrahedral voids occupied are respectively -

(1) 12.5 %, 50% (2) 50%, 12.5% (3) 25%, 50% (4) 50%, 25% Q.73 Which of the following is non-conducting ? (1) CoCl3 . 3NH3 (2) CoCl3 . 6NH3

(3) CoCl3 . 4NH3 (4) CoCl3 . 5NH3

Q.70 vfHkfØ;k] ? →

C6H5

Cl

O

, ds fy, lcls

vPNs vfHkdkjd gS &

(1) C6H5Cl +

O

C6H5–C–Cl , AlCl3

(2)

O

C6H5–C–C6H5 + Cl2, FeCl3 (3) C6H5–CH2–C6H5+Cl2, FeCl3 }kjk vkWDlhdj.k

gksrk gS

(4) buesa ls fdlh ls Hkh bfPNr mRikn dh yfC/k ugha

gksrh gSa

Q.71 "kVdks.kh; tkyd ;qDr fØLVyhdr̀ cQZ esa bdkbZ

dksf'kdk dk vk;ru 132×10–24 cm3 gksrk gSA fn, x,

rki ij ;fn ?kuRo 0.92 g cm–3 gS rks izfr bdkbZ

dksf'kdk esa H2O v.kqvksa dh la[;k gS & (1) 1 (2) 2 (3) 3 (4) 4

Q.72 lw=k AB2O4 ;qDr ,d ;kSfxd dh fØLVy lajpuk esa O2– vk;u ccp O;oLFkk cukrs gSaA A2+ vk;u v"VQydh; fNnz ?ksjrs gS tcfd B3+ vk;u prq"Qydh; rFkk v"VQydh; fNnzksa ds e/; leku :i ls forfjr jgrs gSA ?ksjs x, v"VQydh; rFkk prq"Qydh; fNnzksa dh fHkUu Øe'k% gS &

(1) 12.5 %, 50% (2) 50%, 12.5% (3) 25%, 50% (4) 50%, 25% Q.73 fuEu esa ls dkSulk vpkyd gS \ (1) CoCl3 . 3NH3 (2) CoCl3 . 6NH3

(3) CoCl3 . 4NH3 (4) CoCl3 . 5NH3



Q.74 The structure of TiCl is similar to CsCl. The radius ratio –ClTi r/r + is most likely to lie between -

(1) 0.155 – 0.225 (2) 0.225 – 0.414 (3) 0.414 – 0.732 (4) 0.73 – 1.00

Q.75 The time required to coat a metal surface of 80cm2

with 5 × 10–3 cm thick layer of silver (density 1.05 g cm–3) with the passing of 3A current through silver nitrate solution is -

(1) 115 s (2) 125 s (3) 135 s (4) 145 s

Q.76 Zn rod is placed in 100 mL of 1 M CuSO4 solution so that molarity of Cu2+ changes to 0.7 M. The

molarity of –24SO ions at this stage will be -

(1) 0.8 M (2) 1 M (3) 0.7 M (4) 1.8 M

Q.77 When electric current is passed through aqueous solution of sodium chloride what happens ?

(1) O2 is evolved at cathode (2) O2 is evolved at anode (3) pH of solution gradually decreases (4) pH of the solution gradually increases

Q.78 Which reagents will not liberate Br2 ? (1) KBrO3 + I2 (2) KBr + I2 (3) KBr + Cl2 (4) Both (a) and (b)

Q.79 In lead storage battery, the anode reaction is - (1) Pb2+ + 2e– → Pb (2) Pb + H2SO4 → PbSO4 + 2H+ + 2e– (3) PbO + H2SO4 → PbSO4+H2O (4) None of these

Q.80 The resistance of 1 N solution of acetic acid is 250 ohm, when measured in a cell of cell constant 1.15 cm–1. The equivalent conductance (in ohm–1 cm2

equiv–1) of 1 N acetic acid is - (1) 4.6 (2) 9.2 (3) 18.4 (4) 0.023

Q.74 TiCl dh lajpuk CsCl ds leku gksrh gSA f=kT;k vuqikr –ClTi r/r + fuEu ds e/; jgrk gS &

(1) 0.155 – 0.225 (2) 0.225 – 0.414 (3) 0.414 – 0.732 (4) 0.73 – 1.00

Q.75 flYoj ukbVªsV foy;u esa ls 3A dh /kkjk dks izokfgr djus ij 80cm2 dh /kkrq lrg ij flYoj (?kuRo 1.05 g cm–3) dh 5 × 10–3 cm eksVh ijr p<+kus esa yxus okyk vko';d le; gS &

(1) 115 s (2) 125 s (3) 135 s (4) 145 s

Q.76 1 M CuSO4 foy;u ds 100 mL esa Zn NM+ dks j[kk tkrk gS rkfd Cu2+ dh eksyjrk 0.7 M gks tk;s] bl voLFkk esa –2

4SO vk;uksa dh eksyjrk gksxh & (1) 0.8 M (2) 1 M (3) 0.7 M (4) 1.8 M

Q.77 tc fo|qr /kkjk dks lksfM;e DyksjkbM ds tyh; foy;u esa izokfgr djrs gS rks D;k gksrk gS \

(1) dsFkksM ij O2 eqDr gksrh gS (2) ,uksM ij O2 eqDr gksrh gS (3) foy;u dh pH /khjs&/khjs ?kVrh gS (4) foy;u dh pH /khjs&/khjs c<+rh gS

Q.78 dkSulk vfHkdeZd Br2 fu"dkflr ugha djsxk ? (1) KBrO3 + I2 (2) KBr + I2 (3) KBr + Cl2 (4) (a) rFkk (b) nksuksa

Q.79 lhlk lapk;d lsy esa ,uksM vfHkfØ;k gS & (1) Pb2+ + 2e– → Pb (2) Pb + H2SO4 → PbSO4 + 2H+ + 2e– (3) PbO + H2SO4 → PbSO4+H2O (4) buesa ls dksbZ ugha Q.80 ,flfVd vEy ds 1 N foy;u dk izfrjks/k 250 ohm

gksrk gSA tc lsy dk lsy fLFkjkad 1.15 cm–1 ekik tkrk gSA 1N ,flfVd vEy dh rqY;kadh pkydrk (ohm–1 cm2 equiv–1 esa) gS &

(1) 4.6 (2) 9.2 (3) 18.4 (4) 0.023



MATHEMATICS

Q.81 Area common to the curve y = 2–9 x & x2 + y2 = 6x

in first quadrant is -

(1) 4

3+π (2) 4

3–π

(3)

+π

433 (4)

π

433–3

Q.82 If y = 2 sin x + sin 2x for 0 ≤ x ≤ 2π, then the area enclosed by the curve and the x-axis is -

(1) 9/2 (2) 8 (3) 9 (4) 4

Q.83 Spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop if the constant of proportional is K > 0, is -

(1) 0=+ Kdtdr (2) 0– =K

dtdr

(3) Krdtdr

= (4) none

Q.84 Solution of the differential equation dy = (4x + y + 1)2 dx is -

(1) cxyx+=

++

214tan

21 1–

(2) cxyx –2

14tan21 1– =

++

(3) cxyx+=

++

214sec

21 1–

(4) cxyx –2

14sec21 1– =

++

Q.81 oØksa y = 2–9 x ,oa x2 + y2 = 6x dk izFke prqFkkZa'k

esa mHk;fu"B {ks=kQy gksxk -

(1) 4

3+π (2) 4

3–π

(3)

+π

433 (4)

π

433–3

Q.82 ;fn y = 2 sin x + sin 2x, 0 ≤ x ≤ 2π gS, rc oØ ,oa

x-v{k ls ifjc) {ks=kQy gS - (1) 9/2 (2) 8 (3) 9 (4) 4

Q.83 xksykdkj cjlkr dh cwan blds i"̀Bh; {ks=kQy ds

vuqØekuqikrh nj ls ok"ihdr̀ gksrh gS] rc cwan dh

f=kT;k eas ifjorZu dh nj ds laxr vody lehdj.k]

;fn vuqØekuqikrh dk fu;rkad K > 0 gks] gksxh -

(1) 0=+ Kdtdr (2) 0– =K

dtdr

(3) Krdtdr

= (4) dksbZ ugha

Q.84 vody lehdj.k dy = (4x + y + 1)2 dx dk gy gS -

(1) cxyx+=

++

214tan

21 1–

(2) cxyx –2

14tan21 1– =

++

(3) cxyx+=

++

214sec

21 1–

(4) cxyx –2

14sec21 1– =

++



Q.85 The general solution of the differential equation y ′ + yf ′(x) – f (x). f ′(x) = 0 where f (x) is a known function is -

(1) y = ce–f (x) + f(x) – 1 (2) y = ce+f (x) + f (x) – 1 (3) y = ce–f (x) – f(x) + 1 (4) y = ce–f (x) + f (x) + 1

Q.86 A curve passes through the point

π

4,1 & its slope

at any point is given by

xy

xy 2cos– . Then the

curve has the equation -

(1) y = x tan–1 )(xenl (2) y = x tan–1 (ln + 2)

(3) y = )(tan1 1–

xen

xl (4) none

Q.87 If xyz = – 2007 and

∆ = 0=+

++

zcbacybacbxa

then the value of

ayz + bzx + cxy is - (1) 2007 (2) 0 (3) – 2007 (4) None of these

Q.88 If a, b, c are positive integers such that a > b > c

and 2–111

222=

cbacba Then 3a + 7b – 10c

equals - (1) 10 (2) 11 (3) 12 (4) 13

Q.85 vody lehdj.k y ′ + yf ′(x) – f (x). f ′(x) = 0 , tgk¡

f (x) Kkr Qyu gS] dk O;kid gy gS -

(1) y = ce–f (x) + f(x) – 1 (2) y = ce+f (x) + f (x) – 1 (3) y = ce–f (x) – f(x) + 1 (4) y = ce–f (x) + f (x) + 1

Q.86 ,d oØ fcUnq

π

4,1 ls xqtjrk gS ,oa bldk fdlh

fcUnq ij <+ky

xy

xy 2cos– }kjk fn;k tkrk gS] rc

oØ dk lehdj.k gS -

(1) y = x tan–1 )(xenl (2) y = x tan–1 (ln + 2)

(3) y = )(tan1 1–

xen

xl (4) dksbZ ugha

Q.87 ;fn xyz = – 2007 ,oa

∆ = 0=+

++

zcbacybacbxa

gS] rc ayz + bzx + cxy

dk eku gS -

(1) 2007 (2) 0

(3) – 2007 (4) bueas ls dksbZ ugha

Q.88 ;fn a, b, c /kukRed iw.kk±d bl izdkj gS fd a > b > c

,oa 2–111

222=

cbacba gS] rc 3a + 7b – 10c =

(1) 10 (2) 11 (3) 12 (4) 13



Q.89 If 1331212

51

23

2

+++=

NNnNNn

nU n , Then ∑

=

N

nnU

1

is

equal to -

(1) 2∑=

N

n

n1

(2) ∑=

N

n

n1

22 (3) ∑=

N

n

n1

2

21 (4) 0

Q.90 If f (x) = x

xlog .Then

201–

110log

2x

xx

xx

is -

(1) x3 f ′′(x) (2) x2 f ′′′(x) (3) x3 f ′′′(x) (4) None of these

Q.91 Let A + 2B =

135–33–6021

and

2A – B =

21061–251–2

Then tr(A) – tr(B) has the

value equal to - (1) 0 (2) 1 (3) 2 (4) None

Q.92 If A is non singular and (A – 2I)(A – 4I) = 0, Then

1–

34

6AA

+ is equal to -

(1) 0 (2) I (3) 2I (4) 6I

Q.93 If

251

0251

x=

2–

5–05

a

, Then the value of x is

(1) 1252a (2)

252a (3)

125a (4) None

Q.89 ;fn1331212

51

23

2

+++=

NNnNNn

nU n gS] rc ∑

=

N

nnU

1

=

(1) 2∑=

N

n

n1

(2) ∑=

N

n

n1

22 (3) ∑=

N

n

n1

2

21 (4) 0

Q.90 ;fn f (x) = x

xloggS] rc

201–

110log

2x

xx

xx

gS -

(1) x3 f ′′(x) (2) x2 f ′′′(x) (3) x3 f ′′′(x) (4) buesa ls dksbZ ugha

Q.91 ekuk A + 2B =

135–33–6021

,oa

2A – B =

21061–251–2

gS] rc tr(A) – tr(B) dk eku gS-

(1) 0 (2) 1 (3) 2 (4) dksbZ ugha

Q.92 ;fn A ,d O;qRkØe.kh; eSfVªDl gS ,oa (A – 2I)(A – 4I) = 0

gS] rc 1–

34

6AA

+ =

(1) 0 (2) I (3) 2I (4) 6I

Q.93 ;fn

251

0251

x=

2–

5–05

a

gS] rc x dk eku gS -

(1) 1252a (2)

252a (3)

125a (4) dksbZ ugha



Q.94 Match the column :

Let A =

α

α0tan

tan–0,

B(α) =

αααα

cossinsin–cos

(a) (I + A) (I – A)–1 (p) A – B (α) (b) (I – A) (I + A)–1 (q) B (2α) (c) B(α)2 (r) B (–2α) (d) B(α)–2 (s) AB (– α)

(1) a → q ; b → r ; c → q ; d → r (2) a → r ; b → p ; c → q ; d → p (3) a → q ; b → q ; c → p ; d → r (4) None of these

Passage : (Q.95 to Q.96) Let there are 'n' different things. Then the number

of ways in which 'r' different objects can be select

is nCr where rnr

nCr

n

–=

Q.95 A binary sequence is a sequence of 0's & 1′s. The number of 'n' digit binary sequences , which contain odd number of 0's is -

(1) 2n – 1 (2) 2n – 1 (3) 2n – 2 (4) None Q.96 The greatest possible number of points of

intersections of 8 straight lines & 4 circles is - (1) 32 (2) 64 (3) 76 (4) 104

Q.97 The exponent of 2 in x = 20.19.18….11 is - (1) 10 (2) 8 (3) 18 (4) None

Q.94 LrEHk feyku djks %

ekuk A =

α

α0tan

tan–0,

B(α) =

αααα

cossinsin–cos

(a) (I + A) (I – A)–1 (p) A – B (α) (b) (I – A) (I + A)–1 (q) B (2α) (c) B(α)2 (r) B (–2α) (d) B(α)–2 (s) AB (– α)

(1) a → q ; b → r ; c → q ; d → r (2) a → r ; b → p ; c → q ; d → p (3) a → q ; b → q ; c → p ; d → r (4) buesa ls dksbZ ugha

Xk|ka'k : (Q.95 to Q.96)

Ekkuk 'n' fHkUu oLrq,¡ gS] rc mu rjhdks dh la[;k ftuesa 'r' fHkUUk oLrqvksa dk p;u fd;k tk ldrk gks

nCr gSA tgk¡ rnr

nCr

n

–=

Q.95 ,d f}vk/kkjh vuqØe 0's o 1′s dk vuqØe gS] rc 'n'

vadks dh f}vk/kkjh vuqØeksa dh la[;k ftuesa 0's dh

fo"ke la[;k, gks] gS - (1) 2n – 1 (2) 2n – 1 (3) 2n – 2 (4) dksbZ ugha

Q.96 8 ljy js[kkvksa ,ao 4 oÙ̀kksa ds izfrPNsnu fcUnqvksa dh

vf/kdre laHko la[;k gS - (1) 32 (2) 64 (3) 76 (4) 104

Q.97 x = 20.19.18….11 esa 2 dh ?kkr gS - (1) 10 (2) 8 (3) 18 (4) dksbZ ugha



Q.98 The product of r consecutive positive integers, divided by r is -

(1) A proper fraction (2) Equal to r (3) A positive integer (4) None Q.99 Total number of 5 digit numbers having all

different digits & divisible by 4, that can be formed using digits {1, 2, 3, 6, 8, 9} is -

(1) 192 (2) 32 (3) 1152 (4) None

Q.100 The numbers of ways to select 3 numbers in AP, from first (2n + 1) natural numbers is given by

(1) 4

)1–( 2n (2) 5

2n

(3) n2 (4) none

Q.101 The numbers of times the digit 5 will be written when listing the integers from 1 to 1000 is -

(1) 271 (2) 272 (3) 300 (4) None

Q.102 Coefficient of x40 in expansion of

(1 +x2)40. 5–

22 2

1

++x

x is -

(1) 30C15 (2) 1 (3) 30C20 (4) None Q.103 If 'S' be the sum of coefficients in the expansion of

of (p x + q y + zr )n p, q , r > 0, then

=

+∞→

nnn

S

S

)1(

lim 1

(1) r

Pq (2) rPq

e (3) 0 (4) None

Q.98 r }kjk foHkkftr r Øekxr /kukRed iw.kk±dksa dk

xq.kuQy gS -

(1) mfpr fHkUu (2) r ds cjkcj

(3) ,d /kukRed iw.kk±d (4) dksbZ ugha

Q.99 vadks {1, 2, 3, 6, 8, 9} ds mi;ksx ls cukbZ tk ldus

okyh 4 ls HkkT; 5 vadks dh la[;kvksa dh la[;k]

ftueas lHkh vad fHkUu-fHkUu gks] gksxh - (1) 192 (2) 32 (3) 1152 (4) dksbZ ugha Q.100 izFke (2n + 1) izkdr̀ la[;kvksa esa ls l.Js. esa 3

la[;kvksa ds p;u ds rjhdksa dh la[;k gS -

(1) 4

)1–( 2n (2) 5

2n

(3) n2 (4) dksbZ ugha

Q.101 1 ls 1000 rd ds vadks dh lwph cukus ij 5 fdruh ckj fy[kk tkrk gS -

(1) 271 (2) 272 (3) 300 (4) dksbZ ugha

Q.102 (1 +x2)40. 5–

22 2

1

++x

x ds izlkj esa x40 dk xq.kkad gS

(1) 30C15 (2) 1

(3) 30C20 (4) dksbZ ugha

Q.103 ;fn (p x + q y + zr )n p, q , r > 0 ds izlkj esa

xq.kkadks dk ;ksxQy 'S' gS] rc =

+∞→

nnn

S

S

)1(

lim 1

(1) r

Pq (2) rPq

e (3) 0 (4) dksbZ ugha



Q.104 If an = ∑=

n

r rn C0

1 then ∑=

n

r rn C

r

0

=

(1) (n – 1)an (2) nan

(3) nan2

(4) None

Q.105 If (1 + x + 2x2)20 = a0 + a1x + a2x2 ……+ a40x40

Then a1 + a3 + a5 +……..+a37 = (1) 219(220 – 21) (2) 220(219 – 19) (3) 219(220 + 21) (4) None Q.106 In expansion of (41/3 + 6–1/4)20, consider following

statement : (i) The number of irrational terms = 18 (ii) The number of rational terms = 2 (iii) the middle term is irrational Then correct statements are - (1) (i), (iii) (2) (ii), (iii) (3) (i), (ii) (4) All three

Q.107 The value of ∫+ x

x

bea

e dx is

(1) Kbea x ++ (2) Kbeaab

x ++1

(3) Kbeaba x ++ (4) Kbea

bx ++

2

Q.108 The value of dxxecx∫ + cossec

1 is -

(1)

π

+82

tanlog.22

1–)cos–(sin21 xxx

(2) 2

tanlog22

1–)cos–(sin xxx

(3) 2

tan.log)cos(sin xxx ++

(4) None

Q.104 ;fn an = ∑=

n

r rn C0

1gS] rc∑

=

n

r rn C

r

0

=

(1) (n – 1)an (2) nan

(3) nan2

(4) dksbZ ugha

Q.105 ;fn (1 + x + 2x2)20 = a0 + a1x + a2x2 ……+ a40x40 gS] rc

a1 + a3 + a5 +……..+a37 = (1) 219(220 – 21) (2) 220(219 – 19) (3) 219(220 + 21) (4) dksbZ ugha

Q.106 (41/3 + 6–1/4)20 ds izlkj eas fUkEu dFkuksa ij fopkj dhft, :

(i) vifjes; inksa dh la[;k 18 gS (ii) ifjes; inksa dh la[;k 2 gS (iii) e/; in vifjes; gS dkSuls dFku lR; gS - (1) (i), (iii) (2) (ii), (iii) (3) (i), (ii) (4) lHkh rhuksa

Q.107 ∫+ x

x

bea

e dx dk eku gS

(1) Kbea x ++ (2) Kbeaab

x ++1

(3) Kbeaba x ++ (4) Kbea

bx ++

2

Q.108 dxxecx∫ + cossec

1 dk eku gS -

(1)

π

+82

tanlog.22

1–)cos–(sin21 xxx

(2) 2

tanlog22

1–)cos–(sin xxx

(3) 2

tan.log)cos(sin xxx ++

(4) dksbZ ugha



Q.109 The integral xdxx∫ + 22 tansin

1 is

(1)

+ )(tantan

21cot

21– 1– xxc

(2)

)tan2(tan

21–cot

21– 1– xxc

(3)

++ )(tantan.

21tan

21 1– xxc

(4) None

Q.110 The integral =+

∫ dxx

xx7

6 log6–)1(log

(1) cxx +++ )1(log)1(61 6–6–

(2) cx ++ )1(61 6–

(3) cxx +++ )}1(log–1{)1(61 6–6–

(4) None

Q.111 ∫

+ xx

3–3tan2sin 1– dx =

(1)

+

3sin.

43

3cos.

23 1– xx

(2)

+

3cossin.

43

3cos.

23 1– xx

(3)

+

3cos2sin

43

3cos 1–1– xx

(4)

+

3cos2sin

43

3cos

23– 1–1– xx

Q.109 lekdyu xdxx∫ + 22 tansin

1 gS -

(1)

+ )(tantan

21cot

21– 1– xxc

(2)

)tan2(tan

21–cot

21– 1– xxc

(3)

++ )(tantan.

21tan

21 1– xxc

(4) dksbZ ugha

Q.110 lekdyu =+

∫ dxx

xx7

6 log6–)1(log

(1) cxx +++ )1(log)1(61 6–6–

(2) cx ++ )1(61 6–

(3) cxx +++ )}1(log–1{)1(61 6–6–

(4) dksbZ ugh

Q.111 ∫

+ xx

3–3tan2sin 1– dx =

(1)

+

3sin.

43

3cos.

23 1– xx

(2)

+

3cossin.

43

3cos.

23 1– xx

(3)

+

3cos2sin

43

3cos 1–1– xx

(4)

+

3cos2sin

43

3cos

23– 1–1– xx



Q.112 Column Match : Column I Column II

(a) dxxx∫ + 2sinsin

12

(p) x tan x – log sec x

(b) dxxx

x∫ ++ 5sin4sin

cos2

(q) tan–1(sin x + 2)

(c) dxx∫ 1–sin (r) x sin–1x + 2–1 x

(d) dxxx∫ 2sec. (s)

+ 2tantanlog

21

xx

(1) a → s ; b → r ; c → r ; d → p

(2) a → q ; b → r ; c → q ; d → p (3) a → s ; b → q ; c → r ; d → p (4) None

Q.113 The value of { }∫][

0

][–x

dxxx is, [] is GIF.

(1) [x] (2) 2 [x] (3) ][21 x (4)

][21x

Q.114 ∫π

π

=20

20–

|cos| dxx

(1) 20 (2) 40 (3) 80 (4) 100

Q.115 The value of =∫ dxx2

2/110 |log|

(1) )/8(log21

10 e (2) )/8(log10 e

(3) )/8(log41

10 e (4) None

Q.112 LrEHk feyku: LrEHk I LrEHk II

(a) dxxx∫ + 2sinsin

12

(p) x tan x – log sec x

(b) dxxx

x∫ ++ 5sin4sin

cos2

(q) tan–1(sin x + 2)

(c) dxx∫ 1–sin (r) x sin–1x + 2–1 x

(d) dxxx∫ 2sec. (s)

+ 2tantanlog

21

xx

(1) a → s ; b → r ; c → r ; d → p

(2) a → q ; b → r ; c → q ; d → p (3) a → s ; b → q ; c → r ; d → p (4) dksbZ ugha

Q.113 { }∫][

0

][–x

dxxx dk eku] tgk¡ [] egÙke iw.kkZad Qyu gS]

gksxk.

(1) [x] (2) 2 [x] (3) ][21 x (4)

][21x

Q.114 ∫π

π

=20

20–

|cos| dxx

(1) 20 (2) 40 (3) 80 (4) 100

Q.115 dx|xlog|2

2/110∫ dk eku gS &

(1) )/8(log21

10 e (2) )/8(log10 e

(3) )/8(log41

10 e (4) dksbZ ugha



Q.116 The value of dxx

x∫

π

π

π+4/

4/–2cos–24/ =

(1) 3

2π (2) 36

2π (3) 6

2π (4) None

Q.117 The value of ∫π

π+

2/

2/–sin 1

1 dxe x is -

(1) 2π (2)

4π (3)

8π (4)

2πe

Q.118 The value of ∫

2

2/1

1–sin.1 dxx

xx

(1) 1 (2) – 1 (3) 0 (4) None

Q.119 The value of ∫π

+0

2cos1dx

xx is

(1) 2

2π (2) 22

2π (3) 2

2π (4) None

Q.120 If f (a – x) = – f (a + x) then dxxafa

∫2

0

)–2( is -

(1) a (2) 2a (3) – a (4) 0

Q.116 dxx

x∫

π

π

π+4/

4/–2cos–24/ dk eku gS -

(1) 3

2π (2) 36

2π (3) 6

2π (4) dksbZ ugha

Q.117 ∫π

π+

2/

2/–sin 1

1 dxe x dk eku gS -

(1) 2π (2)

4π (3)

8π (4)

2πe

Q.118 ∫

2

2/1

1–sin.1 dxx

xx

dk eku gS -

(1) 1 (2) – 1 (3) 0 (4) dksbZ ugha

Q.119 ∫π

+0

2cos1dx

xx

dk eku gS -

(1) 2

2π (2) 22

2π (3) 2

2π (4) dksbZ ugha

Q.120 ;fn f (a – x) = – f (a + x) gS] rc dxxafa

∫2

0

)–2( gS -

(1) a (2) 2a (3) – a (4) 0



HkkSfrd foKku : Unit - 10 jlk;u foKku : Unit - 11 xf.kr : Unit - 12

Duration : 3 Hrs. Max. Marks : 480 Date :19 / 03 / 2011 fuEu funsZ'kksa dks /;kuiwoZd if<+;s:

Now, Schedule practice questions are available on internet also, Visit www.examtayari.com

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph: 0744-3040000, Fax (0744) 3040050

email : [email protected] ; Website : www.careerpointgroup.com

ijh{kkfFkZ;ksa ds fy;s funsZ'k

1. bl iz'u i=k esa dqy 120 iz'u gSaA lHkh iz'u gy djus vfuok;Z gSaA

2. blesa _.kkRed vadu gS vr% mÙkj vuqekfur djuk gkfudkjd gks ldrk gSA

3. bl iz'u i=k ds doj ist ij fn;s x;s LFkku esa viuk uke rFkk jksy uEcj fyf[k;sA

4. bl iz'u i=k esa gh jQ odZ ds fy, [kkyh LFkku fn;k x;k gSA jQ odZ ds fy, dksbZ vfrfjDr 'khV ugha nh tk,sxhA

5. mÙkj O.M.R.(Optical Marks Recognisation) 'khV esa vafdr djus gSaA ;g vyx ls nh xbZ gSA

6. iz'u i=k dh lhy rc rd u [kksysa tc rd ,slk djus ds fy, ifjoh{kd }kjk dgk u tk,sA

7. [kkyh dkx+t] fDyi cksMZ] ykWx lkj.kh] LykbM :y] dsYdqysVj] lsY;qyj Qksu] istj ;k fdlh Hkh izdkj dk vU; bysDVªkWfud midj.k

fdlh Hkh :i esa ijh{kk gkWy ds vUnj ys tk;s tkus dh vuqefr ugha gSA

vadu i)fr :

1. izR;sd iz'u esa pkj fodYi fn;s x;s gSa] dsoy ,d fodYi lgh gSA izR;sd xyr mÙkj ds fy, ml iz'u ds fy, fu/kkZfjr vadksa esa ls

,d&pkSFkkbZ dkV fy, tk,saxsA

2. HkkSfrd foKku esa : Q. 1 - 40 izR;sd ds fy, 4 vad, jlk;u foKku esa : Q. 41 - 80 izR;sd ds fy, 4 vad, Xkf.kr esa : Q. 81 - 120 izR;sd ds fy, 4 vad,

SEA

L

physics, chemistry & mathematics - career …...write your name & roll no. in the space...

Documents