TEST SERIES FOR AIEEE (UNIT TEST) - Career Point...perpendicular to the plane of the loop as shown...

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TEST SERIES FOR AIEEE (UNIT TEST)

Time : 3 : 00 Hrs. MAX MARKS: 480

Name : ______________________________________Roll No. : ____________________ Date : _____________________

SEA

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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

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RTS-A-XIII-11-R-4



PHYSICS

Q.1 An aeroplane in which the distance between thetips of the wings is 50 m is flying horizontally fromeast to west with the speed of 360km/hr over aplace where the vertical component of the earth'smagnetic field is 2.0 × 10–4 Wb-m–2. The potentialdifference between the tips of the wings would be -

(1) 0.1 V (2) 0.01V (3) 0.2 V (4) 1.0 V

Q.2 A metallic square loop ABCD is moving in its own

plane with velocity v in a uniform magnetic field

perpendicular to its plane as shown in the figure.

An electric field is induced –

v

B

CD

A

(1) in AD, but not in BC

(2) in BC, but not in AD

(3) neither in AD nor in BC

(4) in both AD and BC

Q.3 A conducting square loop of side l and resistance R

moves in its plane with a uniform velocity v

perpendicular to one of its sides. A uniform and

constant magnetic field B exists along the

perpendicular to the plane of the loop as shown in

figure.The current induced in the loop is –

Q.1 ,d ok;q;ku ftlesa ia[kksa ds 'kh"kksZ ds e/; nwjh 50 m

gSa] ml LFkku tgk¡ iFoh ds pqEcdh; ks=k dk m/okZ/kj

?kVd 2.0 × 10–4 Wb-m–2 gS] ds Åij 360km/hr pky

ls iwoZ ls if'pe dh vksj kSfrtr% mM+ jgk gSA ia[kksa

ds 'kh"kksZ ds e/; foHkokUrj gksxk -

(1) 0.1 V (2) 0.01V (3) 0.2 V (4) 1.0 V

Q.2 ,d /kkfRod oxkZdkj ywi ABCD fp=k esa n'kkZ,

vuqlkj blds ry ds yEcor~ ,d ,dleku pqEcdh;

ks=k esa v osx ls ry esa xfr dj jgk gSA fo|qr ks=k

izsfjr gksrk gS –

v

B

C D

A

(1) AD esa, fdUrq BC esa ugha

(2) BC esa, fdUrq AD esa ugha

(3) uk rks AD uk gh BC esa

(4) AD o BC nksuksa esa

Q.3 l Hkqtk ,oa R izfrjks/k dk ,d pkyd oxkZdkj ywi

bldh fdlh ,d Hkqtk ds yEcor~ ,d ,dleku osx v

ls blds ry esa xfr djrk gSA ,d ,dleku ,oa

fu;r pqEcdh; ks=k B fp=k esa n'kkZ, vuqlkj ywi ds

ry ds vfHkyEcor~ gksrk gSA ywi esa izsfjr /kkjk

gS –



× × × × ×

× × × × ×

× × × × ×

× × × × ×

v

(1) R

Blv clockwise (2) R

Blv anticlockwise

(3) RBlv2 anticlockwise(4) zero

Q.4 In the circuit shown in the figure, the switch S is

closed at time t = 0. The current through the capacitor and inductor will be equal at time t equals

(Given : R = CL ) -

C R

R R

S

(1) RC (2) RC ln 2 (3) 2ln

1R

(4) LR

Q.5 In the circuit shown in the figure, what is the value of I1 just after pressing the key K ?

I2

I1 2 mH 6 Ω

I3

8 Ω8 Ω10 V

K

(1) 5/7 A (2) 5/11 A (3) 1 A (4) None of these

× × × × ×

× × × × ×

× × × × ×

× × × × ×

v

(1) R

Blv nfk.kkorZ (2) R

Blv okekorZ

(3) RBlv2 okekorZ (4) 'kwU;

Q.4 fp=k esa n'kkZ, ifjiFk esa] le; t = 0 ij fLop S can

gSA la/kkfj=k ,oa izsjd ls izokfgr /kkjk fdl le; t ij

cjkcj gksxh (fn;k gS : R = CL ) -

C R

R R

S

(1) RC (2) RC ln 2 (3) 2ln

1R

(4) LR

Q.5 fp=k esa n'kkZ, x, ifjiFk esa] Bhd dq¡th K dks nckus ds i'pkr~ I1 dk eku D;k gksrk gS]?

I2

I1 2 mH 6 Ω

I3

8 Ω8 Ω10 V

K

(1) 5/7 A (2) 5/11 A (3) 1 A (4) buesa ls dksbZ ugha



Q.6 An electron moves along the line AB which lies in

the same plane as a circular loop of conducting

wire as shown in figure. What will be the direction

of the current induced (if any) in the loop ?

–e v A B

(1) No current will induced

(2) The current will be clockwise

(3) The current will be anticlockwise

(4) The current will change direction as the electron

passes the through

Q.7 A vertical conducting ring of radius R falls

vertically in horizontal magnetic field of magnitude

B. The direction

⊗ BA D

C

E of B is ⊗. When the speed of the ring is v -

(1) current flows in the ring

(2) A and D at different potential

(3) C and E are at different potential

(4) None of these

Q.6 ,d bysDVªkWu AB js[kk tks fp=k esa n'kkZ, vuqlkj ,d

pkyd rkj ds oÙkkdkj ywi ds leku ry esa jgrh

gS] ds vuqfn'k xfr djrk gSA ywi esa izsfjr /kkjk (;fn dksbZ gS) dh fn'kk D;k gksxh ?

–e vA B

(1) dksbZ /kkjk izsfjr ugha gksxh (2) /kkjk nfk.kkorZ gksxh (3) /kkjk okekorZ gksxh (4) /kkjk dh fn'kk ifjofrZr gksxh tSls gh bysDVªkWu

blls xqtjrk gS

Q.7 R f=kT;k dh ,d m/okZ/kj pkyd oy; B ifjek.k

okyh kSfrt pqEcdh; ks=k esa m/okZ/kj :i ls vkifrr

gksrh gSA

⊗ B A D

C

E B dh fn'kk ⊗ gSA tc oy; dh pky v gS - (1) oy; esa dksbZ /kkjk gksxh (2) A o D ds e/; foHkokUrj gS (3) C o E ds e/; foHkokUrj gS (4) buesa ls dksbZ ugha



Q.8 In a Young's experiment on interference, the distance between the slits is d. The screen at a distance D from the slits. If a bright fringe is formed opposite to a slit, on the screen, then theorder of fringe is-

(1) D

dλ2

2 (2) 2

2d

Dλ

(3) Dd

λ (4)

Dd

λ2

Q.9 A light beam is incident on a rectangular glass plate

(µ = 1.54). The reflected light OB passes through a nicol prism. On observing the transmitted light while rotating the prism, it is seen that –

29º

A B

Glass O

Air

(1) intensity of light reduces to zero (2) intensity of light decreases and then increases (3) there is no change of intensity of light (4) None of these

Q.10 A piece of glass is heated to a high temperature and then allowed to cool. If it crack, a probable reason for this is the following property of glass -

(1) Low thermal conductivity (2) High thermal conductivity (3) high specific heat (4) high melting point

Q.8 ;ax ds O;frddj.k iz;ksx esa] fLyVksa ds e/; nwjh d gSA inkZ fLyVksa ls D nwjh ij gSA ;fn ,d pedhyh fÝat

insZ ij fLyV ds lEeq[k curh gS] rks fÝat dk Øe

gS -

(1) D

dλ2

2 (2) 2

2d

Dλ

(3) Dd

λ (4)

Dd

λ2

Q.9 ,d izdk'k iq¡t ,d vk;rkdkj dk¡p IysV (µ = 1.54)

ij vkifrr gksrk gSA ijkofrZr izdk'k OB fudkWy

fizTe ls xqtjrk gSA lapfjr izdk'k izsfkr djus ij

tcfd fizTe ?kw.kZu djrk gks] ;g ns[kk tkrk gS fd–

29º

AB

Glass O

Air

(1) izdk'k dh rhozrk 'kwU; rd de gksrh gSA (2) izdk'k dh rhozrk de gksrh gS o fQj c<+rh gSA (3) izdk'k dh rhozrk esa ifjorZu ugha gksxkA (4) buesa ls dksbZ ugha

Q.10 dk¡p ds ,d VqdM+s dks mPp rki ij xeZ fd;k tkrk gS ,oa fQj B.Mk gksus fn;k tkrk gSA ;fn ;g VwVrk gS] rks bldk dkj.k dk¡p dk fuEu esa ls dkSulk xq.k gksrk gS &

(1) fuEu Å"eh; pkydrk (2) mPp Å"eh; pkydrk (3) mPp fof'k"V Å"ek (4) mPp xyukad



Q.11 A cylinder of radius R made of a material ofthermal conductivity k1 is surrounded by acylindrical shell of inner radius R and outer radius 2R.Two ends of the combined system aremaintained at two different temperature. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is -

(1) k1 + k2 (2) 43 21 kk +

(3) 21

21

kkkk

+ (4)

43 21 kk +

Q.12 Distance between two consecutive threads taken

parallel to its axis and measured by distancethrough which the screw moves forward or backward when 1 full rotation is given to the cap is called -

(1) least count (2) least distance (3) sag (4) pitch Q.13 A screw gauge have 250 equal divisions marked

along the periphery of circular scale. If one full

rotation gives advancement of 0.625 cm. the least

count of the arrangement is -

(1) 2.5 × 10–4 cm (2) 2.5 × 10–2 cm

(3) 2.5 × 10–3 cm (4) none

Q.14 The resonance position with the change of liquid -

(1) will change

(2) will not change

(3) some time 1 some time 2

(4) Can't say

Q.11 k1 rkih; pkydrk ds inkFkZ ls cuk R f=kT;k dk csyu ,d vkUrfjd f=kT;k R o cká f=kT;k 2R ds ,d csyukdkj dks'k kjk ifjc) gSA la;qDr fudk; ds nksuksa fljsa nks fHkUu rkiksa ij larqfyr gSA csyukdkj lrg ds fljs ij Å"ek dk ákl ugha gksrk gS rFkk fudk; LFkk;h voLFkk esa gSaA fudk; dh izHkkoh rkih; pkydrk gS &

(1) k1 + k2 (2) 43 21 kk +

(3) 21

21

kkkk

+ (4)

43 21 kk +

Q.12 tc ,d LØw dh dsi dks vkxs ;k fiNs ,d iwoZ ?kw.kZu

fn;k tkrk gS] rc mlds vk ds lekUrj nks /kkxksa ds e/; ekih xbZ nwjh dgykrh gS -

(1) vYirekad (2) vYi nwjh (3) e/; esa ls >qduk (sag) (4) pwM+h varjky

Q.13 ,d LØqxst ds oÙkkdkj iSekus dh ifj/kh ds vuqfn'k

250 fpfUgr leku [kkus j[krk gSA ;fn ,d iw.kZ ?kw.kZu

djus ij 0.625 cm vkxs c<+ tkrk gS] rks O;oLFkk dk

vYirekad gksxk -

(1) 2.5 × 10–4 cm (2) 2.5 × 10–2 cm

(3) 2.5 × 10–3 cm (4) dksbZ ugha

Q.14 nzo ds ifjorZu ds lkFk vuqukn fLFkfr esa &

(1) ifjorZu gksxk

(2) ifjorZu ugha gksxk

(3) dHkh 1 dHkh 2

(4) dgk ugha tk ldrk



Q.15 Experimental verification of Newton's law ofcooling is valid for -

(1) large temperature difference i.e. 30° C to 85°Cbetween hot liquid and surrounding

(2) Very large difference i.e. 5°C to 95°C betweenhot liquid and surrounding

(3) Small temperature difference i.e. 30°C to 35°Cbetween hot liquid and surrounding

(4) Any temperature difference

Q.16 Value of Y of a wire depends upon -

(1) length of wire and change in length of wire

(2) force and area of wire

(3) 1 and 2 both

(4) none

Q.17 In the case of forward biasing of P-N junction,

which one of the following figures correctly depicts

the direction of flow of positive charge carriers ?

(1)

p

←

++++

– – – –

n

→

Vb

VF + –

(2)

p

→

++++

– – – –

n

→

Vb

VF + –

Q.15 U;wVu ds 'khryu fu;e dk izk;ksfxd fufjk.k ekU;

gS & (1) nh?kZ rkikUrj vFkkZr xeZ nzo o ifjos'k ds e/;

30° C ls 85°C ij (2) cgqr nh?kZ rkikUrj vFkkZr ifjos'k o xeZ nzo ds

e/; 5°C ls 95°C ij (3) y?kq rkikUrj vFkkZr xeZ nzo o ifjos'k ds e/;

30°C ls 35°C ij (4) fdlh Hkh rkikUrj ij

Q.16 ,d rkj dk Y dk eku fuHkZj djrk gS &

(1) rkj dh yEckbZ ,oa rkj dh yEckbZ esa ifjorZu ij

(2) cy o rkj ds ks=kQy ij

(3) 1 o 2 nksuksa ij

(4) dksbZ ugha

Q.17 P-N laf/k dh vxz vfHkurh dh fLFkfr esa] /kukRed

vkos'k okgdks ds izokg dh fn'kk dks fuEu esa ls

dkSulk fp=k lgh izdkj ls n'kkZrk gS ?

(1)

p

←

++++

––––

n

→

Vb

VF

+ –

(2)

p

→

++++

––––

n

→

Vb

VF

+ –



(3)

p

→

++++

– – – –

n

←

Vb

VF + –

(4)

p

←

++++

– – – –

n

←

Vb

VF + –

Q.18 The circuit has two oppositely connected ideal

diodes in parallel as shown in figure. What is the current flowing in the circuit ?

4Ω

2Ω3Ω 12V

D1 D2

(1) 1.33 A (2) 1.71 A (3) 2.00 A (4) 2.31 A

Q.19 In the transistor circuit shown here, the emitter, collector and base currents are Ie, Ic and Ibrespectively. The correct relation between them is -

Ib Ie Ic

(1) Ib > Ic > Ie (2) Ib < Ic < Ie (3) Ib < Ie < Ic (4) Ic < Ie < Ib

(3)

p

→

++++

––––

n

←

Vb

VF

+ –

(4)

p

←

++++

––––

n

←

Vb

VF

+ –

Q.18 ifjiFk fp=k esa n'kkZ, vuqlkj lekUrj Øe esa nks foijhr :i ls tqM+s vkn'kZ Mk;ksM dks n'kkZrk gS ? ifjFk esa izokfgr /kkjk D;k gksrh gS ?

4Ω

2Ω3Ω 12V

D1 D2

(1) 1.33 A (2) 1.71 A (3) 2.00 A (4) 2.31 A

Q.19 ;gk¡ n'kkZ, x, VªkwftLVj ifjiFk esa] mRltZd] laxzkgd o vk/kkj /kkjk,sa Øe'k% Ie, Ic o Ib gSA buds e/; lgh lEca/k gS -

Ib Ie Ic

(1) Ib > Ic > Ie (2) Ib < Ic < Ie (3) Ib < Ie < Ic (4) Ic < Ie < Ib



Q.20 A common emitter amplifier is designed with n-p-n transistor (α = 0.99). The input impedance is 1 kΩ and load is 10 KΩ. The voltage gain will be -

(1) 9.9 (2) 99 (3) 990 (4) 9900

Q.21 A mass M is split into two parts, m and (M–m), which are then separated by a certain distance. What ratio of m/M maximizes the gravitational force between the two parts -

(1) 1/3 (2) 1/2 (3) 1/4 (4) 1/5

Q.22 Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to -

(1)

+

21n

R (2)

21–n

R

(3) Rn (4)

22–n

R

Q.23 Two bodies of mass m1 and m2 are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is -

(1) 2/1

21 )–(2

rmmG (2)

2/1

21 )(2

+ mm

rG

(3) 2/1

21 )(2

mmG

r (4) 2/1

212

mm

rG

Q.20 ,d mHk;fu"B mRltZd izo/kZd n-p-n VªkaftLVj (α = 0.99) ds lkFk la:fir gSA vkarfjd izfrck/kk1 kΩ o yksM 10 KΩ gSA oksYVrk ykHk gS -

(1) 9.9 (2) 99 (3) 990 (4) 9900

Q.21 nzO;eku M nks Hkkxks m o (M–m) esa foHkDr gksrs gS

tks ,d fuf'pr nwjh ij iFkDdr gksrs gSA nksuksa Hkkxksa ds e/; xq:Roh; cy ds vf/kdre eku ds fy, m/M dk vuqikr gksxk -

(1) 1/3 (2) 1/2 (3) 1/4 (4) 1/5 Q.22 ekuk dh xq:Roh; cy O;qRØe :i ls ifjofrZr gksrk

gS tks nwjh dh nth ?kkr gSA rc lw;Z ds pkjksa vksj R f=kT;k dh oÙkkdkj dkk es ,d xzg dk vkorZdky lekuqikrh gksxk -

(1)

+

21n

R (2)

21–n

R

(3) Rn (4)

22–n

R

Q.23 izkjEHk esa m1 o m2 nzO;eku ds nks oLrq,sa ,dnwljs ls vuUr nwjh ij fojkekoLFkk esa gSA mUgsa fQj ijLij xq:Rokd"kZ.k ds v/khu ,d nwljs dh vksj xfr djus fn;k tkrk gSA buds e/; r iFkdj.k nwjh ij igq¡pus ij vkisfkd osx gS -

(1) 2/1

21 )–(2

rmmG (2)

2/1

21 )(2

+ mm

rG

(3) 2/1

21 )(2

mmG

r (4) 2/1

212

mm

rG



Q.24 The relation between γ, η and K for a elasticmaterial is -

(1) η1 =

γ31 +

K91

(2) K1 =

γ31 +

η91

(3) γ1 =

K31 +

η91

(4) γ1 =

η31 +

K91

Q.25 A vessel, whose bottom has round holes with

diameter of 0.1mm, is filled with water. themaximum height to which the water can be filledwithout leakage is :

(S.T. of water = 75 dyne/cm, g = 1000 cm/s2) (1) 100 cm (2) 75 cm (3) 50 cm (4) 30 cm Q.26 A cubical block is floating in a liquid with half of

its volume immersed in the liquid. When the wholesystem accelerates upwards with acceleration ofg/3, the fraction of volume immersed in the liquidwill be –

3g

(1) 21 (2)

83

(3) 32 (4)

43

Q.24 ,d izR;kLFk inkFkZ ds fy, γ, η o K ds e/; lEca/k gS -

(1) η1 =

γ31 +

K91

(2) K1 =

γ31 +

η91

(3) γ1 =

K31 +

η91

(4) γ1 =

η31 +

K91

Q.25 ,d crZu ftlds ry esa 0.1mm O;kl ds xksy Nsn

gSa] ikuh ls Hkjk gSA og vf/kdre Å¡pkbZ tgk¡ rd bl crZu esa ikuh Hkjk tk ldrk gS rkfd ;g yhd u gks] gksxh (;fn ikuh dk i"B ruko = 75 dyne/cm, g = 1000 cm/s2)

(1) 100 cm (2) 75 cm (3) 50 cm (4) 30 cm

Q.26 ,d ?kukdkj fi.M fdlh nzo esa bl izdkj rSj jgk gS fd mldk vk/kk vk;ru nzo esa Mwck gSaA ;fn lEiw.kZ fudk; Åij dh vksj g/3 Roj.k ls Rofjr gks] rks fi.M dk og Hkkx tks nzo esa Mwcsxk] gksxk –

3g

(1) 21 (2)

83

(3) 32 (4)

43



Q.27 A projectile is projected with velocity kve in vertically upward direction from the ground intothe space. (vc is escape velocity and k < 1). If airresistance is considered to be negligible then themaximum height from the centre of earth to whichif can go, will be : (R = radius of earth) -

(1) 12 +k

R (2) 1–2k

R

(3) 2–1 kR (4)

1+kR

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 trueand 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.28 Statement - I : Average energy in the interference

pattern is the same as it would be if there were nointerference.

Statement – II : Interference is the only rarephenomenon in which law of conservation ofenergy does not hold good.

Q.29 Statement I : Temperature of blue star is more than

red star.

Statement II : According to Wein's law T ∝ mλ

1

Q.30 Statement I : Absorption power of black body is

infinite. Statement II : Black body can emit all possible

wavelength.

Q.27 iFoh ry ls ,d izksI; dk vkdk'k esa m/okZ/kj Åij dh vksj kve osx ls izksfir fd;k tkrk gS (;gk¡ vc

iyk;u osx gS ,oa k < 1)A ;fn ok;q ?k"kZ.k dks ux.; ekusa rks izksI; iFoh ds dsUnz ls fdruh vf/kdre Å¡pkbZ rd igq¡psxk : (R = iFoh dh f=kT;k) -

(1) 12 +k

R (2) 1–2k

R

(3) 2–1 kR (4)

1+kR

dFkuksa ij vk/kkfjr iz'u

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

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

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

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

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

(4) ;fn dFku I vlR; gS ijUrq dFku II lR; gSA Q.28 dFku - I : O;frdj.k izk:Ik esa vkSlr ÅtkZ leku

gksrh gS tSls dksbZ O;frdj.k ugha gqvkA

dFku - II : O;frdj.k dsoy ,d ?kVuk gS ftlesa

ÅtkZ lajk.k dk fu;e ykxq ugha gksrk gSA

Q.29 dFku I : uhys rkjs dk rki yky rkjs ls vf/kd gksrk gSA

dFku II : ohu ds fu;e ds vuqlkj T ∝ mλ

1 gksrk

gSA

Q.30 dFku I : ,d d".k oLrq dh vo'kks"k.k kerk vuUr gksrh gSA

dFku II : d".k oLrq lHkh lEHko rjaxnS/;Z mRlftZr dj ldrh gSA



Q.31 Statement I : Rate at cooling of plate is more than a sphere of same material & same mass.

Statement II : Rate of cooling is directly proportional to the surface area.

Q.32 Statement I : Radiation is the fastest process of

heat transfer. Statement II : Radiation are electromagnetic wave. Q.33 Statement – I : Choke coil is used with a capacitor

to reduce AC in a circuit. Statement – II : Inductor acts as a leak-proof

wattless resistor. Q.34 Statement – I : The DC and AC both can be

measured by a hot wire instrument. Statement - II : The hot wire instrument is based

on the principle of magnetic effect of current. Q.35 Statement – I : A bulb connected in series with a

solenoid is connected to AC source. If a soft iron core is introduced in the solenoid, the bulb will glow brighter.

Statement – II : On introducing soft iron core in the solenoid, the inductance increases.

Q.36 Statement – I : In series L-C-R circuit resonance

can take place. Statement – II : Resonance takes place if inductive

and capacitive reactance are equal and opposite. Q.37 Statement – I : All oscillatory motions are

necessarily periodic motion but all periodic motion are not oscillatory.

Statement- II : Simple pendulum is an example of oscillatory motion.

Q.31 dFku I : IysV ds 'khryu dh nj leku inkFkZ ,oa leku nzO;eku okys ,d xksys ls vf/kd gksrh gSA

dFku II : 'khryu dh nj i"B ks=kQy ds lh/ks lekuqikrh gksrh gSA

Q.32 dFku I : fofdj.k Å"ek lapj.k dh rhozre izfØ;k gSA dFku II : fofdj.k fo|qr pqEcdh; rjax gSA Q.33 dFku – I : ifjiFk esa AC dks de djus ds fy,

la/kkfj=k ds lkFk pksd dq.Myh dk mi;ksx fd;k tkrk gSA

dFku – II : izsjdRo ,d kj.kjks/kh okWVghu izfrjks/k dh rjg dk;Z djrk gSA

Q34 dFku – I : DC o AC nksuksa ,d xeZ rkj midj.k kjk ekis tk ldrs gSA

dFku - II : xeZ rkj midj.k /kkjk ds pqEcdh; izHkko ds fl)kar ij vk/kkfjr gksrk gSA

Q.35 dFku – I : ifjufydk ls Js.khØe esa tqM+k gqvk ,d

cYc AC L=kksr ls la;ksftr gSA ;fn ,d ueZ yksg dksj ifjufydk esa izosf'kr dh tkrh gS] rks cYc pedsxkA

dFku – II : ueZ yksg dksj dks ifjukfydk esa izosf'kr djus ij izsjdRo esa o)h gksrh gSA

Q.36 dFku – I : L-C-R Js.kh ifjiFk esa] vuqukn gks ldrk gSA

dFku – II : vuqukn gksrk gS ;fn izsj.kh o /kkfjrh; izfr?kkr leku o foijhr gksrs gSA

Q.37 dFku – I : lHkh dkEifud xfr;k¡ vko';d :i ls

vkorhZ gksrh gS fdUrq lHkh vkorhZ xfr;k¡ dkEifud ugha gksrh gSA

dFku - II : ljy yksyd dkEifud xfr dk mnkgj.k gSA



Q.38 Statement – I : In simple harmonic motion, the velocity is maximum when acceleration is minimum.

Statement – II : Displacement and velocity of S.H.M. differ in phase by π/2.

Q.39 Statement – I Damped oscillation indicates loss of

energy. Statement – II : The energy loss in damped

oscillation may be due to friction, air resistance etc. Q.40 Statement – I : In S.H.M, the motion is 'to and fro'

and periodic. Statement – II : Velocity of the particle

(v) = ω 22 – xk (where x is the displacement and k is amplitude).

Q.38 dFku – I : ljy vkorZ xfr esa Roj.k U;wure gksus dh n'kk esa osx vf/kdre gksxkA

dFku – II : ljy vkorZ xfr eas foLFkkiu vkSj osx dh dyk ,dnwljs ls π/2 ls vyx gksrh gSA

Q.39 dFku – I : voefUnr nksyu ÅtkZ gkfu dks O;Dr

djrs gSA dFku – II : voefUnr nksyuksa esa ?k"kZ.k] ok;q izfrjks/k

bR;fn ds dkj.k ÅtkZ gkfu gksrh gSA Q.40 dFku – I : ljy vkorZ xfr esa oLrq esa] xfr nksyuh

rFkk vkorhZ gksrh gSA

dFku – II : d.k dk osx (v) = ω 22 – xk (tgk¡ x foLFkkiu ,oa k vk;ke gSA



CHEMISTRY

Q.41 Cu2+ is more stable than Cu+. This is due to - (1) extensive hydration of Cu2+ ion

(2) greater reduction potential of Cu2+ (3) both (1) and (2) (4) none of these Q.42 Arrange the following in decreasing order of their

oxidizing nature - (1) MnO4

– > Cr2O72– > VO2

+ (2) VO2

+ > Cr2O72– > MnO4

– (3) Cr2O7

2– > VO2+ > MnO4

– (4) MnO4

– > VO2+ > Cr2O7

– Q.43 Which of the following show d-d transition - (1) (NH4)2 [Ti F6] (2) [2n(NH3)4]2+ (3) [Ni(H2O)6] (4) [Cu(H2O)6]+

Q.44 Which has maximum paramagnetic nature ? (1) [MnI(H2O)6]2+ (2) [Cu(NH3)4]2+

(3) [Fe(CN)6]4– (4) [Cu(H2O)4]2+

Q.45 Which is an incorrect statement ? (1) Co(NH3)3Cl3 has fac and mer isomers (2) [PtrCl(NH2OH)(py)] has three geometrical

isomers (3) Square planar MA4 complex and octahedral

MA5B complex do not show geometricalisomerism

(4) Tetrahedral bis(glycinato) nickel (II) isoptically inactive

Q.46 The two complexes PtCl4 . 2NH3 and PtCl4 . 2KCl

do not give ppt of AgCl when treated with AgNO3. The conductance studies indicates zero and threeions per mole of the complex respectively in theirsolution. The structure of these complexes are -

(1) [Pt(NH3)2Cl4] and K2[PtCl6] (2) [Pt(NH3)2] Cl4 and K2 [PtCly] (3) [Pt(NH3)Cl2] Cl2 and K2 [PtCl4] Cl2

(4) [Pt(NH3)2 Cl4] and [K2PtCl6]

Q.41 Cu2+, Cu+ ls vf/kd LFkk;h gS] bldk dkj.k gS & (1) Cu2+ vk;u dk foLrh.kZ ty;kstu

(2) Cu2+ dk vf/kdre vip;u foHko (3) (1) rFkk (2) nksuksa (4) buesa ls dksbZ ugha

Q.42 fuEufyf[kr dks budh vkWDlhdkjd izdfr ds ?kVrs Øe eas O;ofLFkr dhft, &

(1) MnO4– > Cr2O7

2– > VO2+

(2) VO2+ > Cr2O7

2– > MnO4–

(3) Cr2O72– > VO2

+ > MnO4–

(4) MnO4– > VO2

+ > Cr2O7–

Q.43 fuEu esa ls dkSulk d-d laØe.k n'kkZrk gS & (1) (NH4)2 [Ti F6] (2) [2n(NH3)4]2+ (3) [Ni(H2O)6] (4) [Cu(H2O)6]+

Q.44 fdldh vf/kdre vuqpqEcdh; izdfr gksrh gS ? (1) [MnI(H2O)6]2+ (2) [Cu(NH3)4]2+

(3) [Fe(CN)6]4– (4) [Cu(H2O)4]2+

Q.45 dkSulk ,d vlR; dFku gS ? (1) Co(NH3)3Cl3 Qsl (fac) rFkk ej (mer) leko;oh gS (2) [PtrCl(NH2OH)(py)] ds rhu T;kferh; leko;oh

gksrs gS (3) oxkZdkj leryh; MA4 ladqy rFkk v"VQydh;

MA5B ladqy T;kferh; leko;ork ugha n'kkZrs gS (4) prq"Qydh; fcl (Xyk;flusVks) fudy (II) izdkf'kd

vlfØ; gS &

Q.46 nks ladqy PtCl4 . 2NH3 rFkk PtCl4 . 2KCl tc AgNO3 ds lkFk fØ;k djrs gS rks AgCl dk voksi ugha nsrs gS pkydrk dk v/;;u buds foy;u esa Øe'k% ladqy ds izfr eksy rhu vk;u rFkk 'kwU; vk;u dks iznf'kZr djrk gSA bu ladqyks dh lajpuk gS &

(1) [Pt(NH3)2Cl4] rFkk K2[PtCl6] (2) [Pt(NH3)2] Cl4 rFkk K2 [PtCly] (3) [Pt(NH3)Cl2] Cl2 rFkk K2 [PtCl4] Cl2

(4) [Pt(NH3)2 Cl4] rFkk [K2PtCl6]



Q.47 According to IUPAC nomenclature, sodium nitroprosside is named as -

(1) sodium nitro ferricyanide (2) sodium nitro ferrocyanide (3) sodium pentacyanonitrosoniumferrate (II) (4) sodium pentacyanonitrosoniumferrate(III) Q.48 The CFSE for an octahedral complex [ML6]n– of a

metal ion is 200 kJ mol–1. The CFSE for the tetrahedral complex of the same metal ion will beapproximately -

(1) 60 kJ mol–1 (2) 360 kJ mol–1

(3) 517 kJ mol–1 (4) 102 kJ mol–1

Q.49 In the process of extraction of gold,

Roasted gold ore + CN– + H2O → 2O [X] + OH– [X] + Zn → [Y] + Au [X] and [Y] are - (1) [X] =[Au(CN)2]–, [Y] =[Zn(CN)4]–2 (2) [X] = [Au(CN)4]–3, [Y] = [Zn(CN)4]–2 (3) [X] = [Au(CN)2]–, [Y] = [Zn(CN)6]–4 (4) [X] = [Au(CN)4]–, [Y] = [Zn(CN)4]–2

Q.50 OH

OH

O

H2SO4

∆ Product :

(1) H2C=CH–CH2–CH2–

OC –OH

(2) OH

O

(3) O O

(4) O O

Q.47 IUPAC ukedj.k ds vuqlkj lksfM;e ukbVªksizqlkbM dk uke fuEu izdkj gS &

(1) lksfM;e ukbVªksQsjhlk;ukbM (2) lksfM;e ukbVªksQsjkslk;ukbM (3) lksfM;e isUVklk;uks ukbVªkslksfu;eQsjsV (II) (4) lksfM;e isUVk lk;uksukbVªkslksfu;e QsjsV (III)

Q.48 ,d v"VQydh; ladqy [ML6]n– ds ,d /kkrq vk;u

ds fy,] CFSE 200 kJ mol–1 gSA prq"Qydh; ladqy

ds leku /kkrq vk;u ds fy, CFSE yxHkx gksxk & (1) 60 kJ mol–1 (2) 360 kJ mol–1

(3) 517 kJ mol–1 (4) 102 kJ mol–1

Q.49 xksYM ds fu"d"kZ.k ds izØe esa &

HkftZr xksYM v;Ld + CN– + H2O → 2O [X] + OH–

[X] + Zn → [Y] + Au [X] rFkk [Y] gS & (1) [X] =[Au(CN)2]–, [Y] =[Zn(CN)4]–2 (2) [X] = [Au(CN)4]–3, [Y] = [Zn(CN)4]–2 (3) [X] = [Au(CN)2]–, [Y] = [Zn(CN)6]–4 (4) [X] = [Au(CN)4]–, [Y] = [Zn(CN)4]–2

Q.50 OH

OH

O

H2SO4

∆ mRikn :

(1) H2C=CH–CH2–CH2–

OC –OH

(2) OH

O

(3) OO

(4) O O



Q.51 Hg+2, H2SO4 A HCN

OH3⊕ , ∆

B

The end product B will be -

(1) OH

C≡N (2)

C≡N

OH

(3)

OH H2C

O (4)

OH

O

OH

Q.52

C≡N 1. OH3

⊕ 2. SOCl2

X AlCl3 Y

The structure of compound 'Y' -

(1)

Ph–C–O–Ph

O (2)

Ph–C–NH–Ph

O

(3)

Ph–C–Ph

O

(4)

Ph

O

O

O

Ph

Q.53

O O

H

HO

HO (1 mole)

H⊕ CH3MgBr

OH3⊕

Product :

(1)

OH

OH

H (2)

O O

H

O

(3)

O H3C

H

OH

(4)

O O

H

O

Q.51 Hg+2, H2SO4 A HCN

OH3⊕ , ∆

B

vfUre mRikn B gksxk -

(1) OH

C≡N (2)

C≡N

OH

(3)

OHH2C

O (4)

OH

O

OH

Q.52

C≡N1. OH3

⊕ 2. SOCl2

X AlCl3 Y

;kSfxd 'Y' dh lajpuk gS-

(1)

Ph–C–O–Ph

O (2)

Ph–C–NH–Ph

O

(3)

Ph–C–Ph

O

(4)

Ph

O

O

O

Ph

Q.53

OO

H

HO

HO(1 mole)

H⊕ CH3MgBr

OH3⊕

mRikn :

(1)

OH

OH

H (2)

O O

H

O

(3)

OH3C

H

OH

(4)

O O

H

O



Q.54

O

OH OH Above conversion can be achieved by -

(1) Wolff-Kishner reduction

(2) Clemmensen reduction

(3) LiAlH4

(4) NaBH4

Q.55 O3

Zn–H2O A

KOH, ∆ B

Compound B is -

(1)

OH

CHO(2) O

C–H

(3)

CHO(4)

OH

CHO

Q.56

O

m-CPBA P ; 'P' will be -

(1)

O

O (2)

O

O

(3)

O

O

(4)

O

O

Q.54

O

OH OH mijksDr vUr% ifjorZu dks fuEu ds kjk izkIr fd;k

tk ldrk gS- (1) oqYQ – fd'puj vip;u (2) DyhesUlu vip;u (3) LiAlH4 (4) NaBH4

Q.55 O3

Zn–H2O A

KOH, ∆ B

;kSfxd B gS -

(1)

OH

CHO (2) O

C–H

(3)

CHO (4)

OH

CHO

Q.56

O

m-CPBA P ; 'P' gksxk -

(1)

O

O (2)

O

O

(3)

O

O

(4)

O

O



Q.57 O

C–CH3 I2 + NaOH

X + Y Ag, ∆

(yellow ppt.)Z

Identify final product 'Z' :

(1) CHI3 (2) HC ≡ CH

(3) O Ph–C–ONa

⊕ (4) H2C=CH2

Q.58 The product/s formed in the reaction

NH2

NaNO2

dil.HCl, 0°C

(1)

OH (2) CH3

OH

(3) OH

(4) All of the above

Q.59 Consider the following compounds

CH3–C NH

NH2 I

CH3CH2NH2

II

(CH3)2NH

III

CH3–CO

NH2

IV The correct order of basicity of the above

compounds is (1) II > I > III > IV (2) I > III > II > IV (3) III > I > II > IV (4) I > II > III > IV

Q.60 The presence of a primary amine can be confirmed by its reaction with -

(1) HNO2 (2) CHCl3 and KOH (3) CS2 and HgCl2 (4) both (b) and (c)

Q.57 O

C–CH3I2 + NaOH

X + Y Ag, ∆

(yellow ppt.)Z

vfUre mRikn 'Z' Kkr dhft, : (1) CHI3 (2) HC ≡ CH

(3) OPh–C–ONa⊕ (4) H2C=CH2

Q.58 vfHkfØ;k esa fufeZr mRikn gSA

NH2

NaNO2

dil.HCl, 0°C

(1)

OH (2) CH3

OH

(3) OH

(4) mijksDr lHkh

Q.59 fuEu ;kSfxdksa ij fopkj dhft, &

CH3–CNH

NH2

I

CH3CH2NH2

II

(CH3)2NH

III

CH3–CO

NH2

IV

mijksDr ;kSfxdksa dh kkjh;rk dk lgh Øe gS

(1) II > I > III > IV (2) I > III > II > IV

(3) III > I > II > IV (4) I > II > III > IV

Q.60 izkFkfed ,ehu dh mifLFkfr dks bldh fuEu ds lkFk

vfHkfØ;k kjk fuf'pr fd;k tk ldrk gSA (1) HNO2 (2) CHCl3 rFkk KOH (3) CS2 rFkk HgCl2 (4) (b) rFkk (c) nksuksa



Q.61 Match the column :

Column I Column II

(A) O

R–C–NH2 →R–CH2NH2 (P) Schmidt reaction

(B) O

R–C–NH2 →R–NH2 (Q) P2O5

(C) O

R–C–NH2 →R–CN (R) Hoffmann reaction

(D) O

R–C–N3 →RNH2 (S) LiAlH4

(1) A→S; B→R; C→P; D→Q

(2) A→S; B→R; C→Q; D→P

(3) A→R; B→S; C→P; D→Q

(4) A→R; B→S; C→Q; D→P

Q.62 Consider the following sequence of reactions –

conc. HNO3

conc. H2SO4 X

Sn, conc. HCl

heat Y

60°C The product (Y) is

(1)

NH2

NO2

(2)

NH2

(3)

NH2

NH2

(4)

NH2

NH2

Q.61 LrEHkksa dk feyku dhft,

LrEHk I LrEHk II

(A) O

R–C–NH2 →R–CH2NH2 (P) f'keV vfHkfØ;k

(B) O

R–C–NH2 →R–NH2 (Q) P2O5

(C) O

R–C–NH2 →R–CN (R) gkWQeku vfHkfØ;k

(D) O

R–C–N3 →RNH2 (S) LiAlH4

(1) A→S; B→R; C→P; D→Q

(2) A→S; B→R; C→Q; D→P

(3) A→R; B→S; C→P; D→Q

(4) A→R; B→S; C→Q; D→P

Q.62 fuEu vfHkfØ;k Øe ij fopkj dhft,–

conc. HNO3

conc. H2SO4X

Sn, conc. HClheat

Y

60°C

mRikn (Y) gS

(1)

NH2

NO2

(2)

NH2

(3)

NH2

NH2

(4)

NH2

NH2



Statement based Question : (Q.63 to Q.64) 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 wrongQ.63 Statement I : Aneline gives m-nitro Aneline with

conc. HNO3 & conc. H2SO4 (direct nitration) Statement II : Aneline converts Anelinium with

nitrifying mix. Q.64 Statement I : Conductivity of 0.1 M NH4OH

solution is less than that of 0.001 M NH4OH solution.

Statement II : Dilution increases the degree of ionization of NH4OH.

Q.65 Which of the following reaction sequences would

produce benzoic acid ?

(1) C6H6 1. CH3COCl, AlCl3

2. I2, NaOH, heat3. H3O+

(2) C6H5CH3 1. Cl2(excess), heat

2. KOH, heat3. H3O+

(3) C6H5Br 1. Mg, Et2O, heat

2. CO2

3. H3O+

(4) All the above

dFku ij vk/kkfjr iz'u (iz. 63 ls Q.64)

uhps fn;s x;s fuEufyf[kr iz'u dFku-I rFkk dFku-II

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

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

(A) ;fn dFku I rFkk dFku II nksuksa lR; gSa ijUrq dFku

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

(B) ;fn dFku I rFkk dFku II nksuksa lR; gSa rFkk dFku II,

dFku I dk lgh Li"Vhdj.k gSA

(C) ;fn dFku I lR; gS ijUrq dFku II vlR; gSA

(D) ;fn dFku I vlR; gS ijUrq dFku II lR; gSA

Q.63 dFku I : ,fuyhu lkUnz HNO3 o lkUnz H2SO4 ds lkFk

m-ukbVªks,fuyhu nsrk gS (izR;k ukbVªhdj.k) dFku II : ,uhyhu] ukbVªhdr feJ.k ds lkFk ,uhfyfu;e

esa ifjofrZr gksrk gSA

Q.64 dFku I : 0.1 M NH4OH foy;u dh pkydrk

0.001 M NH4OH foy;u ls de gksrh gSA

dFku II : ruqdj.k] NH4OH ds vk;uhdj.k dh ek=kk

c<+krk gSA

Q.65 fuEu vfHkfØ;k Øe esa dkSulk csatksbd vEy cuk,xk?

(1) C6H6 1. CH3COCl, AlCl3

2. I2, NaOH, heat3. H3O+

(2) C6H5CH3 1. Cl2(excess), heat

2. KOH, heat3. H3O+

(3) C6H5Br 1. Mg, Et2O, heat

2. CO2

3. H3O+

(4) mijksä lHkh



Q.66 The major product of the reaction (monobromination)

H3C HN O

CH3 Fe/Br2 is -

(1)

H3C HN O

CH3

Br

(2)

H3C HN O

CH3

Br

(3)

H3C HN O

CH3

Br

(4)

H3C HN O

CH3

Br

Q.67 An organic compound A on treatment with CHCl3

and KOH gives B and C, both of which gives thesame compound D when distilled with Zn dust :oxidation of D yields E of the formula C7H6O2. The Na salt of E on heating with sodalime gives Fwhich can also be obtained by distilling A with Zndust. Identify F.

(1)

OH

CHO

(2) OH

COONa

(3)

COOH (4)

Q.66 vfHkfØ;k dk eq[; mRikn gksxk (eksuksczksuhdj.k)

H3CHN O

CH3 Fe/Br2

(1)

H3CHN O

CH3

Br

(2)

H3CHN O

CH3

Br

(3)

H3CHN O

CH3

Br

(4)

H3CHN O

CH3

Br Q.67 ,d dkcZfud ;kSfxd A, CHCl3 rFkk KOH ds lkFk fØ;k

djus ij B rFkk C nsrs gSa] nksuksa dks tc Zn jt ds lkFk vklfor fd;k tkrk gS] rks leku ;kSfxd D izkIr gksrk gS D ds vkWDlhdj.k ls E dh yfC/k gksrh gSA ftldk lq=k C7H6O2 gS E ds Na yo.k dks lksMkykbe ds lkFk xeZ djus ij F nsrk gS ftls A dks Zn jt ds lkFk vklfor djds Hkh izkIr fd;k tk ldrk gSA F Kkr dhft,A

(1)

OH

CHO

(2) OH

COONa

(3)COOH

(4)



Q.68 Which of the following statement is correct- (1) Phenol is less acidic than ethanol (2) Phenol is more acidic than ethanol (3) Phenol is more acidic than p-nitrophenol (4) Phenol is more acidic than acetic acid

Q.69

OCH3

CH3

(CH3)2C=CH2

H2SO4 gives -

(1)

OCH3

CH3

C(CH3)3 (2)

OCH3

CH3

C(CH3)3

(3)

OCH2CH

CH3

CH3

CH3 (4)

O––C––CH3

CH3

CH3

CH3

Q.70 Br2

FeBr3 Product is

(1)

Br

(2)

Br

(3)

Br

(4)

Br

Q.68 fuEu esa ls dkSulk dFku lgh gS - (1) fQukWy, ,FksukWy ls de vEyh; gS (2) fQukWy] ,FksukWy ls vf/kd vEyh; gS (3) fQukWy P–ukbVªksfQukWy ls vf/kd vEyh; gS (4) fQukWy] ,flfVd vEy ls vf/kd vEyh; gS

Q.69

OCH3

CH3

(CH3)2C=CH2

H2SO4nsrk gS &

(1)

OCH3

CH3

C(CH3)3 (2)

OCH3

CH3

C(CH3)3

(3)

OCH2CH

CH3

CH3

CH3 (4)

O––C––CH3

CH3

CH3

CH3

Q.70 Br2

FeBr3 mRikn gS &

(1)

Br

(2)

Br

(3)

Br

(4)

Br



Q.71 +

O

O

O

AlCl3

H3O ⊕ (A) ?

(1) (CH2)3–COOH

O

(2) O–C–(CH2)2COOH

O

(3) CH2–CH2–C–CH3

O O

(4) CH2–CH2–C

O O

Q.72 How many unit cells are present in 39.0 g of

potassium that crystallizes in body centred cubic structure -

(1) NA (2) NA/4 (3) 0.5 NA (4) 0.75 NA

Q.73 How many such diagonal as ac can be present in a

cube ?

b

Z

O d

X Y

c

a

(1) 2 (2) 4 (3) 8 (4) 6

Q.71 +

O

O

O

AlCl3

H3O ⊕ (A) ?

(1) (CH2)3–COOH

O

(2) O–C–(CH2)2COOH

O

(3) CH2–CH2–C–CH3

O O

(4) CH2–CH2–C

O O

Q.72 39.0 g ikSVsf'k;e esa tks dk; dsfUnzr ?ku lajpuk esa

fØLVyhdr gksrk gS] fdruh bdkbZ dksf'kdk,a

mifLFkr gksrh gS & (1) NA (2) NA/4 (3) 0.5 NA (4) 0.75 NA

Q.73 ,d ?ku esa ac ds :i esa fdrus fod.kZ mifLFkr gks

ldrs gS \

b

Z

Od

X Y

c

a

(1) 2 (2) 4 (3) 8 (4) 6



Q.74 Perovaskite is a mineral containing calcium, oxygen and titanium. The unit cell of its crystalline structure is shown below :

CaO Ti

The formula of perovaskite and oxidation number

of titanium are respectively - (1) CaTiO2, + 2 (2) CaTiO2, + 2 (3) CaTiO3, + 4 (4) CaTiO6, + 4 Q.75 Which of the following unit cells has least

symmetry ? (1) Tetragonal (2) Triclinic (3) Monoclinic (4) Cubic Q.76 One litre of 1 M CuSO4 solution is electrolysed.

After passing 2F of electricity, molarity of CuSO4

will be - (1) M/2 (2) M/4 (3) M (4) Zero Q.77 In electrolysis of very dilute solution of caustic

soda using platinum electrodes, (1) H2 is evolved at anode (2) Cl2 is obtained at cathode (3) O2 is produced at anode (4) NH3 is evolved at both the electrodes Q.78 Unit of electrochemical equivalent of the element

is - (1) Gram (2) Gram ampere–1 (3) Gram coulomb–1 (4) Coulomb gram–1

Q.74 isjksofLdV] dSfY'k;e] vkWDlhtu rFkk VkbVsfu;e ;qDr ,d [kfut gSA bldh fØLVyh; lajpuk dh bdkbZ dksf'kdk uhps n'kkZbZ xbZ gS &

CaO Ti

isjksofLdV dk lw=k rFkk VkbVsfu;e dh vkWDlhdj.k

la[;k Øe'k% gS & (1) CaTiO2, + 2 (2) CaTiO2, + 2 (3) CaTiO3, + 4 (4) CaTiO6, + 4

Q.75 fuEu esa ls dkSulh bdkbZ dksf'kdkvksa esa lcls de

lefefr gksrh gS ? (1) prq"dks.kh; (2) f=kurkk (3) ,durkk (4) ?kuh;

Q.76 1 M CuSO4 foy;u ds ,d yhVj fo|qr vi?kfVr gksrs gSA 2F dh fo|qr /kkjk izokfgr djus ds ckn CuSO4 dh eksyjrk gksxh &

(1) M/2 (2) M/4 (3) M (4) 'kwU;

Q.77 dkfLVd lksMk ds vfr ruq foy;u ds fo|qr vi?kVu esa IysfVue bysDVªksMks dk mi;ksx djrs gS] rks &

(1) ,uksM ij H2 eqDr gksrh gS (2) dsFkksM ij Cl2 izkIr gksrh gS (3) ,uksM ij O2 mRiUu gksrh gS (4) nksuksa bysDVªksMks ij NH3 eqDr gksrh gS

Q.78 rRo ds fo|qr jklk;fud rqY;kad dh bdkbZ gS &

(1) xzke (2) xzke ,fEi;j–1

(3) xzke dwykWEc–1 (4) dwykWEc xzke–1



Q.79 The resistance of 0.0025 M solution of K2SO4 is

326 ohm. The specific conductance of the solutionis -

(1) 4.997 × 10–4 (2) 5.997 × 10–7

(3) 6.997 × 10–4 (4) unpredictable Q.80 The reaction M + 2H+ → M2+ + H2, most

suitably applies to the combination - (1) Cu and HCl (2) Ag and HCl (3) Cu and HNO3 (4) Mg and HCl

Q.79 K2SO4 ds 0.0025 M foy;u dk izfrjks/k 326 ohm

gSA foy;u dh fof'k"V pkydrk gS &

(1) 4.997 × 10–4 (2) 5.997 × 10–7

(3) 6.997 × 10–4 (4) Kkr ugha fd;k tk ldrk

Q.80 vfHkfØ;k M + 2H+ → M2+ + H2 fuEu ds la;kstu

ds fy, mi;qDr gksrh gS &

(1) Cu rFkk HCl (2) Ag rFkk HCl (3) Cu rFkk HNO3 (4) Mg rFkk HCl

CAREER POINT , CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 || CPtriple E Target || Page # 26


MATHEMATICS

Q.81 The area bounded by the curve y = x2 + 4x + 5, the axes of co-ordinates & the minimum ordinate is -

(1) 323 (2)

324

(3) 325 (4) none

Q.82 If y = e–x (A cos x + B sin x), then y satisfies -

(1) 022

2=+

dxdy

dxyd (2) 022–2

2=+ y

dxdy

dxyd

(3) 0222

2=++ y

dxdy

dxyd (4) 022

2=+ y

dxyd

Q.83 Consider the differential equation

xydxdy tan+ = x tan x + 1. Then -

(1) The curves satisfying the differential equation and given by y = x + c sin x. (2) The angle at which the curves cut the

y-axis is 2π .

(3) Tangents to all the curves at their point of intersecting with y-axis are parallel. (4) none of these Q.84 The degree of the differential equation satisfying

22 –1–1 yx + = a (x – y) is -

(1) 1 (2) 2 (3) 3 (4) none of these

Q.81 oØ y = x2 + 4x + 5, funsZ'kh vkksa ,oa U;wure dksfV ls

ifjc) ks=kQy gS -

(1) 323 (2)

324

(3) 325 (4) dksbZ ugha

Q.82 ;fn y = e–x (A cos x + B sin x) gS] rc y larq"V djrk gS-

(1) 022

2=+

dxdy

dxyd (2) 022–2

2=+ y

dxdy

dxyd

(3) 0222

2=++ y

dxdy

dxyd (4) 022

2=+ y

dxyd

Q.83 vody lehdj.k xtanydxdy

+ = x tan x + 1 Ikj fopkj]

dhft,A rc -

(1) oØ tks vody lehdj.k dks larq"V djrk gS ,oa

y = x + c sin x kjk fn;k tkrk gS

(2) og dks.k ftl ij oØ y-vk dks dkVrk gS 2π gS

(3) lHkh oØksa ds y-vk ds lkFk izfrPNsnu fcUnqvksa

ij Li'kZ js[kk,a lekUrj gSA

(4) buesa ls dksbZ ugha

Q.84 22 –1–1 yx + = a (x – y) dks larq"V djus okyh

vody lehdj.k dh ?kkr gS -

(1) 1 (2) 2

(3) 3 (4) buesa ls dksbZ ugha



Q.85 A curve is such that the area of the region bounded by the co-ordinate axes, the curve & the ordinate of any point on it is equal to the cube of that ordinate. The curve represents -

(1) a pair of straight lines (2) a circle (3) a parabola (4) an ellipse

Q.86 The area of the closed figure bounded by y = x,

y = – x & the tangent to the curve y = 5–2x at the point (3, 2) is -

(1) 5 (2) 52

(3) 10 (4) 25

Q.87 The value of the determinanta

aa

++

+

111111111

is -

(1)

aa 2–13 (2)

+

aa 313

(3)

aa 3–13 (4)

+

aa 213

Q.88 If a is a complex cube root of 2, then

123

41121

22

aaa

a=∆ is -

(1) 10 (2) 20 (3) 30 (4) 0

Q.89 If the value of the determinantc

ba

111111

is

positive, then - (1) abc > 1 (2) abc > – 8 (3) abc <– 8 (4) abc > – 2

Q.85 ,d oØ bl izdkj gS fd funs±'kh vkksa] oØ ,oa bl

ij fdlh fcUnq dh dksfV ls ifjc) ks=kQy ml dksfV

ds ?ku ds cjkcj gS] rc oØ fu:fir djrk gS -

(1) ,d ljy js[kkvksa dk ;qXe (2) ,d oÙk

(3) ,d ijoy; (4) ,d nh?kZoÙk

Q.86 y = x, y = – x ,oa oØ y = 5–2x ds fcUnq (3, 2) ij

LIk'kZ js[kk kjk ifjc) ks=kQy gS -

(1) 5 (2) 52

(3) 10 (4) 25

Q.87 lkjf.kd a

aa

++

+

111111111

dk eku gS -

(1)

aa 2–13 (2)

+

aa 313

(3)

aa 3–13 (4)

+

aa 213

Q.88 ;fn a,2 dk lfEeJ ?kuewy gS] rc -

123

41121

22

aaa

a=∆ gS -

(1) 10 (2) 20 (3) 30 (4) 0

Q.89 ;fn lkjf.kd c

ba

111111

dk eku /kukRed gS] rc -

(1) abc > 1 (2) abc > – 8 (3) abc <– 8 (4) abc > – 2



Q.90 If 0 ≤ [x] < 2, – 1 ≤ [y] < 1 and 1 ≤ [z] < 3 where [ ], denotes the greatest integer functions. Then the maximum value of ∆ where

∆ = 1][][][

][1][][][][1][

++

+

zyxzyxzyx

is -

(1) 2 (2) 4 (3) 6 (4) None of these

Q.91 If adj B = A , | P | = | Q | = 1. Then adj (Q–1 BP–1) is - (1) PQ (2) QAP

(3) PAQ (4) PA–1Q

Q.92 Adj

1202–11–

201=

b

a

2–2–0112–5

, Then

[a b] is equal to - (1) [– 4 1] (2) [– 4 –1] (3) [4 1] (4) [4 – 1] Q.93 The total number of metrices formed with the help

of 6 different numbers are - (1) 6 (2) 6

(3) 62 (4) 64

Q.94 If AB = BA and C2 = B Then [A–1 CA]2 is equal to - (1) C (2) B

(3) A (4) None of these

Q.95 The numbers of ways in which we can select 3 numbers from 1 to 20. so as to exclude every selection of 3 consecutive numbers is -

(1) 1140 (2) 1123 (3) 1122 (4) None of these

Q.90 ;fn 0 ≤ [x] < 2, – 1 ≤ [y] < 1 ,oa 1 ≤ [z] < 3 tgk¡ [ ] egÙke iw.kk±d Qyuksa dks fu:fir djrk gS] rc ∆ dk egÙke eku]

tgk¡ ∆ = 1][][][

][1][][][][1][

++

+

zyxzyxzyx

gS] gksxk -

(1) 2 (2) 4 (3) 6 (4) buesa ls dksbZ ugha

Q.91 ;fn adj B = A , | P | = | Q | = 1 gS] rc adj (Q–1 BP–1) gS - (1) PQ (2) QAP (3) PAQ (4) PA–1Q

Q.92 Adj

1202–11–

201=

b

a

2–2–0112–5

gS] rc [a b]

gS - (1) [– 4 1] (2) [– 4 –1] (3) [4 1] (4) [4 – 1]

Q.93 6 fofHkUu la[;kvksa dh lgk;rk ls cukbZ xbZ dqy

eSfVªDlks dh la[;k gS - (1) 6 (2) 6

(3) 62 (4) 64

Q.94 ;fn AB = BA ,oa C2 = B gS, rc [A–1 CA]2 = (1) C (2) B (3) A (4) buesa ls dksbZ ugha

Q.95 mu rjhdksa dh la[;k ftuls 1 ls 20 esa ls izR;sd Øekxr

3 la[;kvksa ds p;u NksM+rs gq, 3 la[;k, pqu ldrs gks]

gksxh - (1) 1140 (2) 1123 (3) 1122 (4) buesa ls dksbZ ugha



Q.96 If each of 4 points in one straight line be joined to each of 3 points in another by straights lines terminated by points then excluding given points, the numbers of points of intersection is equal to - (1) 18 (2) 12 (3) 6 (4) None

Q.97 A person writes letters to six friends & addresses corresponding envelops . Let 'x' be numbers of ways so that at least 2 letters are in wrong envelopes & 'y' be number of ways so that all letters are in wrong envelopes then x – y =

(1) 719 (2) 265 (3) 454 (4) None

Q.98 The number of integral solutions of the equation X + Y + Z + t = 20, where X, Y, Z, t are all ≥ – 1 is - (1) 20C4 (2) 23C3 (3) 27C4 (4)27C3

Q.99 Total number of six digits numbers that can be formed having the property that every successding digit is greater than preceding digit, is equal -

(1) 9C3 (2) 10C3 (3) 9P3 (4) 10P3

Q.100 There are ' p ' points in space, no four of which are in same plane with exception of ' q ' points which are all in same plane. The number of different planes determined by the points is -

(1) PC2 – qC2 + 1 (2) PC3– qC3

+ 1 (3) PC3 – qC3

(4) (p – q)C3

Q.101 Number of zeroes at the end of 300 is equal to -

(1) 75 (2) 89 (3) 74 (4) 98

Q.102 rr

r CC +=∑ 150

20050

0

100 . =

(1) 300C50 (2) 100C50 (3) 100C50 . 200C50 (4) None of these

Q.96 ;fn ,d ljy js[kk ij fLFkr 4 fcUnqvksa dks nwljh

ljy js[kk ij fLFkr 3 fcUnqvksa ls bu fcUnq ij lekIr

gksus okyh ljy js[kkvksa ls feyk;k tkrk gS rc bu

fcUnqvksa ds vfrfjDr izfrPNsnu fcUnqvksa dh la[;k gS - (1) 18 (2) 12 (3) 6 (4) dksbZ ugha

Q.97 ,d O;fDr vius N% fe=kksa dks i=k fy[krk gS ,ao muds

laxr fyQkQks ij irs fy[krk gSA ekuk 'x' mu rjhdksa

dh la[;k gS ftuls de ls de nks i=k xyr fyQkQs eas

gks ,oa 'y' mu rjhdksa dh la[;k gS ftuls lHkh i=k xyr

fyQkQksa eas gks] rc x – y = (1) 719 (2) 265 (3) 454 (4) dksbZ ugha

Q.98 lehdj.k X + Y + Z + t = 20 ds iw.kk±d gyksa dh la[;k]

tgk¡ X, Y, Z, t ≥ – 1gS] gksxh - (1) 20C4 (2) 23C3 (3) 27C4 (4)27C3

Q.99 N% vadks dh dqy la[;kvksa dh la[;k] ftuesa izR;sd

vad iwoZ vad ls cM+k gks] gksxh - (1) 9C3 (2) 10C3 (3) 9P3 (4) 10P3

Q.100 lef"V esa 'p' fcUnq gS ftueas ls 'q' fcUnq ,d gh lery

eas fLFkr gS] rFkk 'ks"k esa ls dksbZ Hkh pkj fcUnq ,d

lery esa fLFkr ugha gS] rc bu fcUnqvksa ds kjk

cuk;s x, fofHkUu leryksa dh la[;k gS - (1) PC2 – qC2

+ 1 (2) PC3– qC3 + 1

(3) PC3 – qC3 (4) (p – q)C3

Q.101 300 ds vUr eas 'kwU;ksa dh la[;k gS - (1) 75 (2) 89 (3) 74 (4) 98

Q.102 rr

r CC +=∑ 150

20050

0

100 . =

(1) 300C50 (2) 100C50 (3) 100C50 . 200C50 (4) buesa ls dksbZ ugha



Q.103 If there is a term containing x2r in 3–

21 n

xx

+ , then -

(1) (n –2r) is a positive integral multiple of 3. (2) (n – 2r) is even (3) (n –2r) is odd (4) None of these

Q.104 If Cr = nCr and

(C0 + C1) . (C1 + C2)…….(Cn–1 + Cn) = k n

n n)1( +

then k = (1) C0C1C2 .... Cn (2) (C1C2…..Cn)2 (3) C1 + C2…..+Cn (4) None of these Q.105 (1–2x)4

+ 4x2 (1–2x)3 + 6x4(1–2x)2 + 4x6 (1–2x)1 + x8 = (1) (1 – x)6 (2) (1 – x)8 (3) (1– x2) (4) None of these

Q.106 Match the columns (a) n+1C3 = 2.nC2 then n is - (p) 9 (b) In expansion of (1 +ax)n first three terms are 1 + 12x + 64x2 then n is

(q) 5

(c) Remainder when 599 is divided by 13 is (r) 8 (d) Sum of all coefficients in expansion of (x2 – x – 1)99

(s) 0

(t) – 1 (1) a → q ; b → p ; c → r ; d → t

(2) a → r ; b → t ; c → r ; d → t

(3) a → t ; b → p ; c → r ; d → t (4) None of these

Q.103 3–

21 n

xx

+ eas ,d in gS ftlesa x2r gS] rc -

(1) (n –2r) , ,d /kukRed iw.kk±d gS tks 3 dk xq.kt gS (2) (n – 2r) le gS (3) (n –2r) fo"ke gS (4) buesa ls dksbZ ugha

Q.104 ;fn Cr = nCr ,oa

(C0 + C1) . (C1 + C2)…….(Cn–1 + Cn) = k n

n n)1( +

gS] rc k =

(1) C0C1C2 .... Cn (2) (C1C2…..Cn)2

(3) C1 + C2…..+Cn (4) buesa ls dksbZ ugha

Q.105 (1–2x)4 + 4x2 (1–2x)3 + 6x4(1–2x)2 + 4x6 (1–2x)1 + x8 =

(1) (1 – x)6 (2) (1 – x)8

(3) (1– x2) (4) buesa ls dksbZ ugha

Q.106 LrEHk feyku djks %

(a) n+1C3 = 2.nC2 gS] rc n gS - (p) 9 (b) (1 +ax)n ds izlkj eas izFke rhu in 1 + 12x + 64x2 gS] rc n gS -

(q) 5

(c) 599 dks 13 ls foHkkftr djus ij 'ks"kQy gS -

(r) 8

(d) (x2 – x – 1)99 ds izlkj lHkh xq.kkdksa dk ;ksxQy gS

(s) 0

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



Q.107 The value of ∫ +θ−dx

xx 1cos21

2 is -

(1)

θθ

sincos–tan 1– x (2)

θθ

θ sincos–tan

sin1 1– x

(3)

θθ

sincos–cot 1– x (4)

θθ

θ sincos–cot.

sin1 1– x

Q.108 ∫ ++ dx

xxxxxx

)cos(sin.cos =

(1)

+ xxxcos

log (2)

+

xxx coslog

(3)

+ xxxsin

log (4)

+

xxx sinlog

Q.109 ∫ dxxx 53 cos.sin

1 =

(1) ( )3tan.32 2 −x (2)

xx

tan)3(tan.

32 2 −

(3) )3.(tan34 2 −x (4) None of these

Q.110 ∫ + xx eedx

–4 =

(1) log (ex + 4e–x) (2) log (e2x + 4)

(3) )2/(tan21 1– xe (4) None of these

Q.111 ∫ + )1( xxdx =

(1) 1)1(1–)1(

log++

+

xx

(2) 1–)1(1)1(

logxx

+

++

(3) 1)1(1–)1(

log21

++

+

xx

(4) None of these

Q.107 ∫ +θ−dx

xx 1cos21

2 dk eku gS -

(1)

θθ

sincos–tan 1– x (2)

θθ

θ sincos–tan

sin1 1– x

(3)

θθ

sincos–cot 1– x (4)

θθ

θ sincos–cot.

sin1 1– x

Q.108 ∫ ++ dx

xxxxxx

)cos(sin.cos =

(1)

+ xxxcos

log (2)

+

xxx coslog

(3)

+ xxxsin

log (4)

+

xxx sinlog

Q.109 ∫ dxxx 53 cos.sin

1 =

(1) ( )3tan.32 2 −x (2)

xx

tan)3(tan.

32 2 −

(3) )3.(tan34 2 −x (4) buesa ls dksbZ ugha

Q.110 ∫ + xx eedx

–4 =

(1) log (ex + 4e–x) (2) log (e2x + 4)

(3) )2/(tan21 1– xe (4) buesa ls dksbZ ugha

Q.111 ∫ + )1( xxdx =

(1) 1)1(

1–)1(log

++

+

x

x (2)

1–)1(

1)1(log

x

x

+

++

(3) 1)1(1–)1(

log21

++

+

xx

(4) buesa ls dksbZ ugha



Q.112 Column Match .

Column-I Column-II

(a) ∫+

+

2

)1(2

2

x

xd (p) x tan x + log |cos x|

(b) ∫ 1–

14xx

dx (q) cx ++ 22 2

(c) ∫ )()( xydxy (r) 21–sec21 x

(d) ∫ )cos/( 2 xx dx (s) 2

22 yx

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

Q.113 ∫2

1–

|| dxxx =

(1) 1 (2) 3 (3) Not exist (4) None of these

Q.114 ∫∞

0

–– )–( dxba xx =

(1) (1/log a) – (1/log b) (2) log a + log b (3) log a – log b (4) (1/log a) + (1/log b)

Q.115 If In = ∫π

θθ4/

0

tan dn then I8 + I6 =

(1) 1/4 (2) 1/5 (3) 1/6 (4) 1/7

Q.112 LrEHk feyku &

LrEHk-I LrEHk-II

(a) ∫+

+

2

)1(2

2

x

xd (p) x tan x + log |cos x|

(b) ∫ 1–

14xx

dx (q) cx ++ 22 2

(c) ∫ )()( xydxy (r) 21–sec21 x

(d) ∫ )cos/( 2 xx dx (s) 2

22 yx

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

Q.113 ∫2

1–

|| dxxx =

(1) 1 (2) 3

(3) fo|eku ugha (4) buesa ls dksbZ ugha

Q.114 ∫∞

0

–– )–( dxba xx =

(1) (1/log a) – (1/log b) (2) log a + log b (3) log a – log b (4) (1/log a) + (1/log b)

Q.115 ;fn In = ∫π

θθ4/

0

tan dn gS] rc I8 + I6 =

(1) 1/4 (2) 1/5 (3) 1/6 (4) 1/7



Q.116 If Un =∫ −1

0

1tan. dxxxn

then (n + 1) Un +(n –1)Un – 2 =

(1) n2π –

2 (2)

n3–

3π (3)

n1–

2π (4)

n1–

3π

Q.117 Let f : R → R be a continuous function defined on

R and f(x) = ∫x

dttf0

)( then f(ln 5) =

(1) 0 (2) 5 (3) e5 (4) 1

Q.118 I = ∫π

π

=π++π+2/–

2/3–

23 )3(cos dxxx

(1) 32

4π (2) 232

4 π+

π

(3) 2π (4) 1–

21

π

Passage : Let f(x) be a continuous periodic function defined

on R. Let f (x) has period T. Then

∫ ∫∫αα+

+=nTnT

dxxfdxxfdxxf0 00

)()()(

Q.119 =∫+

dxe xx2/150

0

][–

(1) 50e + e1/2 – 49 (2) 50e + e1/2 – 51 (3) 50e + e1/2 – 50 (4) None of these

Q.120 If g(x) = ∫x

dtt0

4cos Then g(x +π) =

(1) g(x) +g(π) (2) g(x) – g(π) (3) g(x) g(π) (4) g(x)/g(π)

Q.116 ;fn Un = ∫ −1

0

1tan. dxxxn gS]

rc (n + 1) Un +(n–1) Un – 2 =

(1) n2π –

2 (2)

n3–

3π (3)

n1–

2π (4)

n1–

3π

Q.117 ekuk f : R → R ,d lar~r Qyu gS] tks R ij

ifjHkkf"kr gS ,oa f(x) = ∫x

dttf0

)( gS] rc f(ln 5) =

(1) 0 (2) 5 (3) e5 (4) 1

Q.118 I = ∫π

π

=π++π+2/–

2/3–

23 )3(cos dxxx

(1) 32

4π (2) 232

4 π+

π

(3) 2π (4) 1–

21

π

x|ka'k: ekuk f(x) ,d lar~r Qyu gS tks R ij ifjHkkf"kr gSA

ekuk f (x) dk vkorZukad T gS] rc -

∫ ∫∫αα+

+=nTnT

dxxfdxxfdxxf0 00

)()()(

Q.119 =∫+

dxe xx2/150

0

][–

(1) 50e + e1/2 – 49 (2) 50e + e1/2 – 51 (3) 50e + e1/2 – 50 (4) buesa ls dksbZ ugha

Q.120 ;fn g(x) = ∫x

dtt0

4cos gS] rc g(x +π) =

(1) g(x) +g(π) (2) g(x) – g(π)

(3) g(x) g(π) (4) g(x)/g(π)



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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

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TEST SERIES FOR AIEEE (UNIT TEST) - Career Point...perpendicular to the plane of the loop as shown...

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