B JEE (Main) 2018 - Matrix JEE AcademyMatrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911,...

Date: 08/April/2018

Time: 3 Hours. Max. Marks: 360

INSTRUCTIONS

1. The test is of 3 hours duration.

2. The Test Booklet consists of 90 questions. The maximum marks are 360.

3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics

having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct

response.

4. Candidates will be awarded marks as stated above in instruction No. 3 for correct response of each

question. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted for indicating

incorrect response of each question. No deduction from the total score will be made if no response is

indicated for an item in the answer sheet.

5. There is only one correct response for each question. Filling up more than one response in any question will

be treated as wrong response and marks for wrong response will be deducted accordingly as per instruc-

tion 4 above.

6. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the

Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.

7. Do not fold or make any stray mark on the Answer Sheet

USEFUL DATA

Atomic weights: Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14,Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.5, Ti = 48,Ba = 137, U = 238, Co= 59, B =11, F = 19, He = 4, Ne = 20, Ar = 40 , Mo = 96, Ni = 58.5, Sr = 87.5,Hg = 200.5 , Tl = 204, Pb = 207 [Take : ln 2 = 0.69, ln 3 = 1.09, e = 1.6 × 10–19, m

e= 9.1 × 10–31 kg ]

Take g = 10 m/s2 unless otherwise stated

BCode

JEE (Main) 2018

2Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

JEE (MAIN) 2018MatrixJEE Academy

PHYSICS

PART-APHYSICS

SINGLE CORRECT CHOICE TYPE

Q.1 to Q.30 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

Modern Physics : Atomic physics

1. If is found that if a neutron suffers an eleastic collinear collision with deuterium at rest, fractional loss of its

energy is pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is p

c. The value

of pd and p

c are respectively :

;fn ,d U;wVªkWu dh ,d fLFkj voLFkk ds M;wVhfj;e ls izR;kLFk ,d js[kh; la?kV~V gksrh gS rks mldh ÅtkZ dk vkaf'kd

{k; pd ik;k tkrk gSA mlds fLFkj voLFkk ds dkcZu ukfHkd ls le:i la?kV~V esa ÅtkZ dk vkaf'kd {k; p

c ik;k tkrk gSA

pd rFkk p

c ds eku Øe'k% gksaxsa &

(1) (0, 0) (2) (0, 1) (3) (0.89, 0.28) (4) (0.28, 0.89)

Ans. 3

Sol. m v 2m

mU mv mv 1 22

ev v

v

2 1 1

v v2 1 = v

2 2 1v v = v

2v2 = 2v

v2 =

2

3

v

vv

1

2

3 – v

v

3

pmv

mv

a

mvd

1

2

1

21

2

22

2

8

90888.

pmv m

v

mvc

1

2

1

2

121

1691

2

22

2

48

1690 28.

m v 12m



PHYSICS

mU = mv1 + 12 mv

2

v v v1 212

v v v2 1

13 22v v

vv

2

2

13

vv

v1

2

3

110

13

Modern Physics : Atomic Physics

2. The mass of a hydrogen molecule is 3.32×10–27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall

of area 2 cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the

pressure on the wall is nearly :

,d gkbMªkstu v.kq dk nzO;eku 3.32×10–27 kg gSA 2 cm2 {ks=kQy dh ,d nhokj ij 1023 izfr lsd.M dh nj ls gkbMªkstu

v.kq ;fn vfHkyEc ls 45° ij izR;kLFk VDdj djds 103 m/s dh xfr ls ykSVrs gS] rks nhokj ij yxs nkc dk fudVre eku

gksxk &

(1) 2.35 × 102 N/m2 (2) 4.70 × 102 N/m2 (3) 2.35 × 103 N/m2 (4) 4.70 × 103 N/m2

Ans. 3

Sol. PF

A

F = 2mv

F = 2×n×mH×v

F = 2×1023×3.32×10–27×103

P = 2 10 332 10 10

2 10

23 27 3

4

. = 2.35 × 103 N/m2

Properties of matter

3. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical

container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of

cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional

decrement in the radius of the sphere, dr

r

FHGIKJ , is :

fdlh eqyk;e inkFkZ }kjk cus gq, r f=kT;k dk ,d Bksl xksyk ftldk vk;ru izR;kLFkrk xq.kkad K gS] ,d csyukdkj crZu



PHYSICS

esa fdlh nzo }kjk f?kjk gqvk gSA a {kS=kQy dk ,d nzO;ekufoghu fiLVu csykukdkj crZu ds laiw.kZ vuqizLFkdkV dks <+drs

gq, nzo ds lrg ij rSjrk gSA nzo ds laihMu gsrq tc fiLVu ds lrg ij nzO;eku m j[kk tkrk gS] rks xksys dh f=kT;k esa

gksus okyk vkaf'kd ifjorZu dr

r

FHGIKJ gksxk &

(1) mg

Ka3(2)

mg

Ka(3)

Ka

mg(4)

Ka

mg3

Ans. 1

Sol.

P KV

V

V

V

dr

rFHGIKJ

3

Pmg

a

dr

r

mg

Ka

FHGIKJ 3

Current Electricity

4. Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 . The internal

resistances of the two batteries are 1 and 2 respectively. The voltage across the load lies between :

(1) 11.4 V and 11.5 V (2) 11.6 V and 11.7 V

(3) 11.6 V and 11.7 V (4) 11.5 V and 11.6 V

12 V rFkk 13 V fo|qr okgd cy dh nks cSVjh dks lekarj Øe esa ,d 10 ds yksM izfrjks/k ds lkFk tksM+k x;k gSA nksuksa

cSVjh ds vkarfjd izfrjks/k Øe'k 1 rFkk 2 gSA yksM izfrjks/k ds fljksa dk foHko fuEu esa ls fdu ekuksa ds chp gksxk

&

(1) 11.4 V rFkk 11.5 V (2) 11.6 V rFkk 11.7 V

(3) 11.6 V rFkk 11.7 V (4) 11.5 V rFkk 11.6 V

Ans. 4

Sol.V

r r

r r r

1

1

2

2

1 2

1 1 1

V

12

1

13

21

1

1

2

1

10

= V = 11.56

So the voltage is between 11.5 V and 11.6 V



PHYSICS

Work, Power and Energy

5. A particle is moving in a circular path of radius a under the action of an attractive potential Uk

r

2 2 . Its total

energy is :

,d d.k fdlh ,d vkd"kZ.k foHko Uk

r

2 2 ds varxZr f=kT;k a ds ,d xksykdj iFk esa py jgk gS mldh dqy ÅtkZ

gksxh &

(1) Zero (2) 3

2 2

k

a(3)

k

a4 2 (4) k

a2 2

Ans. 1

Sol. FdU

dr

Fk

r

3 = mv

r

2

K E. mvk

r.

1

2 22

2

T E. P E. K E.. . .

T.E. = k

r2 2 + k

r2 2 = 0

Newton's law of motion

6. Two masses m1 = 5 kg and m

2 = 10 kg, connected by an inestensible string over a frictionless pulley, are

moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m

that should be put on top of m2 to stop the motion is :

m1 = 5 kg rFkk m

2 = 10 kg ds nks nzO;eku ,d vforkU; Mksjh }kjk ,d ?k"kZ.k jfgr f?kjuh ds Åij ls tqM+s gq, gS tSlk

fd fp=k esa n'kkZ;k x;k gSA {kSfrt lrg dk ?k"kZ.k xq.kkad 0.15 gSA og U;wure nzO;eku m ftldks nzO;eku m2 ds Åij j[kus

ls xfr :d tk;s] gksuk pkfg,A

m

m2

m1

m1g

T

T

(1) 43.3 kg (2) 10.3 kg (3) 18.3 kg (4) 27.3 kg



PHYSICS

Ans. 4

Sol. Balancing the forces on m1 for equilibrium

T = m1g = 5g

Balancing the forces on m2 for equilibrium

T = N m m g015 2. b g5g = 015 2. m m gb gm = 23.33 kg

so the minimum mass from the given options will be 27.3 kg


7. If the series limit frequency of the Lyman series is L , then the series limit frequency of the Pfund series is :

;fn ykbeu Js.kh dh lhek vko`fr L gS tks Qq.M Js.kh dh lhek vko`fr gksxh &

(1) L /16 (2) L /25 (3) 25 L (4) 16 L

Ans. 2

Sol. Series limit : hn n

FHG

IKJ136

1 1

12

22

.

For lyman : h L 13 6. (here n1 = 1, n

2 = )

For Pfund : h f

LNM

OQP13 6

1

5

12

. (here n1 = 5, n

2 = )

v

vL

f

25

vv

fL

25

Wave optics

8. Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed

behing A. The intensity of light beyond B is found to be I

2. Now another identical polarizer C is placed

between A and B. The intensity beyond B is now found to be I

8. The angle between polarizer A and C is :

rhozrk I dk v/kzqfor izdk'k dk ,d vkn'kZ iksyjkWbM A ls xqtjrk gSA blh rjg dk ,d vkSj iksyjkWbM B dks iksyjkWbM

A ds ihNs j[kk x;k gSA iksyjkWbM B ds i'pkr~ izdk'k dh rhozrk I

2 ik;h tkrh gSA vc ,d vkSj mlh rjg ds iksyjkWbM

C dks A rFkk B ds chp j[kk tkrk gS ftlls B ds i'pkr~ rhozrk I

8 ik;h tkrh gSA iksyjkWbM A rFkk C ds chp dk dksa.k

gksxk &

(1) 45° (2) 60° (3) 0° (4) 30°



PHYSICS

Ans. 1

Sol.

unpolarizedlight

(I )0

I /20

I /20

A C B

Here from both A and B, intensity becomes

2 so both have parallel pass axis.

B Angle between A and C

Intensity after crossing C I0 2

2cos , Intensity after crossing B

I0 2

2cos FHG

IKJ cos2 =

I0

8

cos1

2

45


9. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let

n g, be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let n be

the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B

are constatns)

,d bysDVªkWu fdlh gkbMªkstu ijek.kq ds fofHkUu mÙksftr voLFkkvksa ls fofdj.k mRlftZr djds fuEure voLFkk esa vk tkrk

gSA ekuk fd n g, , noha voLFkk rFkk fuEure voLFkk esa bysDVªkWu dh Mh czksXyh rajxnS/;Z gSA ekuk n oha voLFkk ls fuEure

voLFkk esa laØe.k }kjk mRlftZr QksVksu dh rjaxnS/;Z n gSA n ds cM+s eku ds fy, (;fn A rFkk B fLFkjkad gS)

(1) n nA B2 2 (2) n2

(3) n

n

AB

2 (4) n nA B

Ans. 3



PHYSICS

Sol.1

11

2 FHGIKJR

nE

n

13 62

.

FHGIKJ

11

12R n

P

m n

2

22

136

.

FHGIKJ

11

12R n

h

m nn

2

2 22

136

.

1 1

2R Rnn

m

hn

2

2

227 2FHGIKJ

.

Ah

R m n

2

227 2.

AB

n2

Semiconductors

10. The reading of the ammeter for a silicon diode in the given circuit is :

fn;s x;s ifjiFk esa silicon Mk;ksM ds fy, vehVj dk ikB~;kad gksxk &

3V

(1) 11.5 mA (2) 13.5 mA (3) 0 (4) 15 mA

Ans. 1

Sol. As the voltage drop across a silicon diode in forward bias is 0.7 V so the voltage drop across the 200resistor will be 2.3 V.

i = 2 3

200

. = 11.5 mA

EMF : Electromagnetic field and forces

11. An electron, a proton and an alpha particle having the same kinetic energy are moving in circuilar obrits of radii

r r re p, , respectively in a uniform magnetic field B. The relation between r r re p, , is :

,d xfrt ÅtkZ ds ,d bysDVªkWu ,d izksVkWu ,oa ,d vYQk d.k fdlh ,dleku pqEcdh; {ks=k B esa Øe'k% r r re p, , f=kT;k

dh xksykdkj d{kk esa ?kwe jgs gSaA r r re p, , ds chp laca/k gksxk &

(1) r r re p (2) r r re p

(3) r r re p (4) r r re p



Ans. 4

Sol. F = qVB = mV

R

2

As K.E. and magnetic field B are constant

So, R = C

qV (C = constant)

K.E. = 1

22mV

VC

m 1

R = C m

q

qe = 1, q

p = 1, q = 2

mp > m

e and m = 4×m

p

So, r r re p

Capacitance

12. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of

dielectric constant K 5

3 is inserted between the plates, the magnitudc of the induced charge will be :

90 pF /kkfjrk ds ,d lekUrj IysV la/kkfj=k dks 20 V fo|qr okgd cy dh ,d cSVjh ls tksM+rs gSA ;fn K 5

3 ijkoS|qr

inkFkZ IysVksa ds chp izfo"V fd;k tkrk gS rks izsfjr vkos'k dk ifjek.k gksxk &

(1) 2.4 nC (2) 0.9 nC (3) 1.2 nC (4) 0.3 nC

Ans. 3

Sol.

Q = CV

Q = 90 × 10–12 × 20

= 18 × 10– 10 C

Q = 1.8 nC

– Due to slab

Q1 = kQ

Qln = Q(K – 1)

PHYSICS



= 1.8 × 5

31

FHGIKJ

= 182

3. nC

= 1.2 nC

Alternating current

13. For an RLC circuit driven with voltage of amplitude m and frequency 0

1

LC the current exibits resonance.

The quality factor, Q is given by :

m vk;ke rFkk 0

1

LC vkofr ds foHko }kjk pfyr ,d RLC ifjiFk vuqukfnr gksrk gSA xq.krk dkjd Q dk eku

gksxk &

(1) R

C0b g (2) CR

0(3)

0L

R(4)

0R

L

Ans. 3

Sol. Quality factor Q is given by

Q = X

R

L

RL

0

Electro Magnetic Waves

14. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for

transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a

bandwidth of 5 kHz :

,d VsyhQksu lapj.k lsok] okgd vko`fr 10 GHz ij dke djrh gSA bldk dsoy 10% lapkj ds fy;s mi;ksx fd;k tkrk

gSA ;fn izR;sd pSuy dh cSaM pkSM+kbZ 5 kHz gks rks ,d lkFk fdrus VsyhQksfud pSuy lapkfjr fd;s tk ldrs gS \

(1) 2 × 105 (2) 2 × 106 (3) 2 × 103 (4) 2 × 104

Ans. 1

Sol. (10 × 109) × 10

100 = n × 5 × 103 {n = no. of channel}

n

10

5 10

9

3

n = 2 × 105

String waves

15. A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density

of granite is 2.7×103 kg/m3 and its Young's modulus is 9.27×1010 Pa. What will be the fundamental frequency

PHYSICS



of the lognitudinal vibrations :

60 cm yEckbZ dh xzsukbZV dh ,d NM+ dks mlds e/; ls ifjc) djds mlesa vuqnS/;Z dEiUu mRiUu fd;s tkrs gSA xzsukbZV

dk ?kuRo 2.7×103 kg/m3 rFkk ;ax izR;kLFkrk xq.kkad 9.27×1010 Pa gSA vuqnS/;Z dEiUu dh ewy vko`fr D;k gksxh \

(1) 10 kHz (2) 7.5 kHz (3) 5 kHz (4) 2.5 kHz

Ans. 3

Sol. F = 1

2l

Y

=1

2 0.6

9 27 10

2 7 10

10

3

.

.

= 5 kHz

Rotational Motion

16. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The

moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is :

fp=kkuqlkj lkr ,d tSlh o`rkdkj lrey fMLdksa] ftuesa izR;sd dk nzO;eku M rFkk f=kT;k R gS] dks lefer :i ls tksM+k

tkrk gSA lery ds yEcor~ rFkk P ls xqtjus okyh v{k ds lkis{k] bl la;kstu dk tM+Ro vk?kw.kZ gSA

P

O

(1) 273

MR2

(2) 2181

MR2

(3) 219

MR2

(4) 255

MR2

Ans. 2

Sol. Icom

= MR MR

MR2 2

2

66

24

FHG

IKJ

55

2

2MR

IP = I

com + 7m (3R)2

= 181

2

2MR

PHYSICS



Electrostatics

17. Three concentric metal shells A, B and C of respective radii a, b and c(a < b < c) have surface charge densities

+, – and +respectively. The potential of shell B is :

rhu ladsUnzh /kkrq dks"k A, B rFkk C ftudh f=kT;k;sa Øe'k% a, b rFkk c(a < b < c) gSa] dk i`"B&vkos'k&?kuRo Øe'k% +,

– rFkk +gSA dks"k B dk foHko gksxk &

(1)

2 2

0

b ca

b

(2)

2 2

0

b ca

c

(3)

2 2

0

a bc

a

(4)

2 2

0

a bc

b

Ans. 4

Sol.

A

PA

C

V Ka

b

b

b

c

cP LNM

OQP

4 4 42 2 2

=

LNM

OQP0

2a

bb c

Va b

bcP

LNM

OQP

0

2 2

Current Electricity

18. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals

of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 ,

a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.

,d foHkoekih iz;ksx ds nkSjku ik;k x;k fd tc lsy fljksa dks foHkoekih rkj ds 52 cm yEckbZ ds nksuksa rjQ tksM+k tkrk

gS rks xSYouksehVj esa dksbZ /kkjk dk izokg ugha gksrk gSA ;fn lsy dks 5 , izfrjks/k }kjk 'kaV dj fn;k tk;s rks lsy ds fljksa

dks rkj ds 40 cm yEckbZ ds nksuksa rjQ tksM+us ls larqyu izkIr gks tkrk gSA lsy dk vkarfjd izfrjks/k gksxk &

(1) 2 (2) 2.5 (3) 1 (4) 1.5

Ans. 4

Sol. volt meter/

PHYSICS



52 cm

E

E = 52.0 ......(1)

E

r

5

5b g = 42 .....(2)

52 R = 40 R + 40 r

740r + 12 R

r = 12

405 =

12

8

3

215 .

Electro Magnetic waves

19. An EM wave from air enters a medium. The electric fields are

1 01

zÊ E x cos 2 v t

c

in air and 2 02

Ê E x cos k 2z ct

in medium, where the wave number k

and frequency v refer to their values in air. The medium is non-magnetic. If 1r

and 2r

refer to relative permittivities

of air and medium respectively, which of the following options is correct ?

,d foHko pqacdh; rjax gok ls fdlh ek/;e esa izos'k djrh gSA muds oS|qr {ks=k 1 01

zÊ E x cos 2 v t

c

gok esa

,oa 2 02Ê E x cos k 2z ct

ek/;e esa gSa] tgk¡ lapj.k la[;k k rFkk vko`fr v ds eku gok esa gSA ek/;e vpqEcdh;

gSA ;fn 1r

rFkk 2r

Øe'k% gok ,oa ek/;e dh lkis{k fo|qr'khyrk gks rks fuEu esa ls dkSu lk fodYi lR; gksxk &

(1) 1

2

r

r

1

4

(2) 1

2

r

r

1

2

(3) 1

2

r

r

4

(4)

1

2

r

r

2

Ans. 1

PHYSICS



Sol. for E1 = E

01 cosx v

z

ct2

FHGIKJ

LNM

OQP wave speed (v

1) = C

for E2 = E

02 cosx k z ct2 b g wave speed (v

2) = C/2

vr

1

1 0

1

....(1)

vr

2

2 0

1

....(2)

From (1) and (2)

2 2

1

r

r =

r

r

1

2

1

4

Wave optics

20. The angular width of the central maximum in a single slit diffraction pattern is 60º. The width of the slit is 1 µm.

The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's

fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1

cm, what is slit separation distance ? (i.e. distance between the centres of each slit.)

fdlh ,dy f>jh foorZu iSVuZ ds dsanzh; mfPp"B dh dk.kh; pkSM+kbZ 60º gSA f>jh dh pkSM+kbZ 1 µm gSA f>jh dks ,do.khZ;

lery rjax ls izdkf'kr djrs gSA ;fn mlh pkSM+kbZ dh ,d u;h f>jh ds ikl cuk nh tk;s rks f>fj;ksa ls 50 cm nwj j[ks

insZ ij ;ax fÝats ns[kh tk ldrh gSA ;fn fÝatks dh pkSM+kbZ 1 cm gks rks f>fj;ksa ds dsUnzksa ds chp dh nwjh gksxhA

(1) 75 µm (2) 100 µm (3) 25 µm (4) 50

Ans. 3

Sol.a

230

2sin

a

2

fringe width

0

2

(10–2) = 10

2

0 56

.

d

d = 0.25 × 10–4

25 µm

SHM

21. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec.

What is the force constant of the bonds connecting one atom with the other ? (Mole wt. of silver = 108 and

Avagadro number = 6.02×1023 gm mole–1)

PHYSICS



fdlh Bksl eas pknh dk ,d ijek.kq 1012/sec dh vko`fr ls fdlh fn'kk eas ljy vkorZ xfr djrk gSA ,d ijek.kq dks nwljs ijek.kq

ls tksMusa okys ca/k dk cy fu;arkd fdruk gksxk

(pkanh dk vkf.od Hkkj =108 vkSj vokxzknh (Avagadro) la[;k = 6.02 × 1023 gm mole–1)

(1) 2.2 N/m (2) 5.5 N/m (3) 6.4 N/m (4) 7.1 N/m

Ans. 4

Sol.

1

2

K

m

K m 4 2 2

= 4 × 10 × 1024 × 108 10

6 10

3

23

K = 7.1 N/m

Rotational Dynamics

22. From a uniform circular disc of radius R and mass 9 M, a small disc of radius R

3 is removed as shown in the

figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and

passing through centre of disc is :

R f=kT;k rFkk 9 M nzO;eku ds ,dleku xksykdkj fMLd R

3 f=kT;k dk ,d NksVk xksykdj fMLd dkV dj fudky fy;k

tkrk gS] tSlk fd fp=k esa n'kkZ;k x;k gSA fMLd ds lrg ds yEcor~ ,oa mlds dsUnz ls xqtjus okys v{k ds lkis{k cph gqbZ

fMLd dk tM+Ro vk?kw.kZ gksxkA

2R3

R

(1) 10 MR2 (2) 37

9MR2 (3) 4 MR2 (4)

40

9MR2

Ans. 3

PHYSICS



Sol. Mass of removed disc is

r M2 9

So, r

M2

9

Mass of removed disc is M

I = 9

2 2 92 3

2 22MR MR

M R

FHG

IKJ/b g

I = 4 MR2

COM and Collision

23. In a collinear collsion, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final

total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between

the two particles, after collision, is :

,d ,djs[kh; la?kV~V (collinear collsion) esa] vkjfEHkd pky v0 dk ,d d.k leku nzO;eku ds ,d nqljs :ds gq, d.k

ls Vdjkrk gSA ;fn dqy vafre xfrt ÅtkZ] vkjfEHkd xfrt ÅtkZ 50% T;knk gks rks VDdj ds ckn nksuksa d.kksa ds lkis{k

xfr dk ifj.kke gksxkA

(1) 0

2

(2)

0

2

(3)

0

4

(4)

02

Ans. 4

Sol. As the K.E. after collision is increasing so the velocity of seperation after the collision will be greater then

velocity of approach, So, from the given options only 02 is greater then V

0.

Magnetic field

24. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop

is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of

the loop is B2. The ratio

1

2

B

B is :

/kkjk I okys ,d orkdkj ik'k dk f}/kzqo vk?kw.kZ m rFkk mlds dsUnz ij pqEcdh; {ks=k B1 gSA /kkjk fLFkj j[krs gq, f}/kzqo vk?kw.kZ

dks nksxquk djus ij ik'k ds dsUnz ij pqEcdh; {ks=k B2 gks tkrk gSA vuqikr

1

2

B

B gksxkA

(1) 2 (2) 1

2(3) 2 (4) 3

Ans. 1

Sol. M1 = I r 1

2 as M2 = 2M

1

M2 = I r 2

2 so, r2 = 2 1r

PHYSICS



Also Bcenter

= 0

2

I

r

So, Br

1

B

B

r

r1

2

2

1

2

Error and Measurements and instruments

25. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass.

If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in

determining the density is :

?ku dh vkd`fr okys fdlh inkFkZ dk ?kuRo] mldh rhu Hkqtkvksa ,oa nzO;eku dks eku dj] fudkyk tkrk gSA ;fn nzO;eku

,oa yEckbZ dks ekius esa lkis{k =kqfV;k¡ Øe'k% 1.5% rFkk 1% gks rks ?kuRo dks ekius esa vf/kdre =kqfV gksxhA

(1) 4.5% (2) 6% (3) 2.5% (4) 3.5%

Ans. 1

Sol. M

V

M

b hl

Error in density = Error in mass + (Error in sides of cube)= 1.5 + 3 × 1

= 4.5 %

Current Electricity

26. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance

of their series combination is 1 k. How much was the resistance on the left slot before interchanging the

resistances ?

izfrjks/k dks cnyus ls ehVj lsrq dk larqyu fcUnq 10 cm ck¡;j rjQ f[kld tkrk gSA muds Js.kh Øe la;kstu dk izfrjks/k

1 kgSA izfrjks/kksa dks cnyus ls igys ck¡;s rjQ ds [kk¡ps dk izfrjks/k fdruk Fkk \

(1) 550 (2) 910 (3) 990 (4) 505

Ans. 1

Sol.

R

R

L

L1

2 100

.....(1)

Now, after interchange R

R

L

L2

1

10

100 10

b g ....(2)

from (1) and (2)

100 L

L =

L

L

10

110

L = 55 cm

Also R R1 2 1000

PHYSICS



Now, R

R1

11000

55

45

11

9

9R1 = 11000 – 11R

1

R1 = 550

Alternating current (AC)

27. In an a.c. circuit, the instantaneous e.m.f. and current are given by

e = 100 sin 30 t

i 20sin 30t4

In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively :

,d a.c. ifjiFk ds fo|qr okgd cy rFkk /kkjk dk rkR{kf.kd eku fuEufyf[kr lehdj.kksa ls fn;k x;k gSA

e = 100 sin 30 t

i 20sin 30t4

a.c. ds ,d iw.kZ pØ esa ifjiFk }kjk vkSlr 'kfDr O;; rFkk okVghu /kkjk ds eku] Øe'k gS &

(1) 50

,02

(2) 50, 0 (3) 50, 10 (4) 1000

,102

Ans. 4

Sol. P V Iavg rms rmsFHGIKJcos

4

100

2

20

2

1

2

1000

2

Iwattless

= Irms sin

= 10 21

2:d i = 10

Kinamatics

28. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up :

fn;s x;s lkjs xzkQ ,d gh xfr n'kkZrs gSA dksbZ ,d xzkQ ml xfr dks xyr rjhds ls n'kkZrk gSA og xzkQ gSA

(1) Time

Position

(2) Time

Velocity

PHYSICS



(3) Position

Velocity

(4) Time

Distance

Ans. 4

Sol. (1) Time

Position

Its equation can be return as t2 = – C1X + C

2 (X is position)

It can be seen that acceleration is negativeIt can be seen in option (2)

(2) Time

Velocity

= slope is negative and slope represents acceleration

So acceleration is negative

(3) Position

Velocity

V2 = U2 – 2aX

Its general equation for velocity - position graph with negative acceleration

PHYSICS



(4) Time

Distance

Here distance - time graph does not represent any of the conditions shown in above options

Thermodynamics

29. Two moles of an ideal monoatomic gas occupies a volume V at 27ºC. The gas expands adiabatically to a

volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

fdlh ,dijek.kq vkn'kZ xSl ds 2 eksy 27ºC rkieku ij V vk;ru ?ksjrs gSA xSl dk vk;ru :)ks"e izØe }kjk QSy dj

2 V gks tkrk gSA xSl ds (a) vafre rkieku dk eku ,oa (b) mldh vkarfjd ÅtkZ esa ifjorZu dk eku gksxkA

(1) (a) 189 K (b) –2.7 kJ (2) (a) 195 K (b) 2.7 kJ

(3) (a) 189 K (b) 2.7 kJ (4) (a) 195 K (b) –2.7 kJ

Ans. 1

Sol. (a) T2 = T

1 ×

V

V1

2

1FHGIKJ –

T2 =

300

22 3 = 189

(b) U nC Tv

23

2189 300

R b g= –2.7 kJ

Circular motion

30. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional

to the nth power of R. If the period of rotation of the particle is T, then :

(1) T R(n+1)/2 (2) T Rn/2

(3) T R3/2 for any n (4) n

12T R

,d d.k R f=kT;k ds ,d o`rkdkj iFk ij fdlh ,d dsfUnz; cy tks fd R dh n oha ?kkr ds O;qRØekuqikrh gS] ds vraxZr

?kwerk gSA ;fn d.k dk vkorZ dky T gks] rks

(1) T R(n+1)/2 (2) T Rn/2

PHYSICS



(3) T R3/2 (n ds fdlh Hkh eku ds fy,) (4) n

12T R

Ans. 1

Sol. FmV

R

K

Rn

2

V R n 1 2b g and TR

V

So, T R(n+1)/2



CHEMISTRY

PART - CCHEMISTRY



61. Total number of lone pair of electrons in 3I ion is

3I vk;u eas bysDVªkWUkksa ds ,dkdh ;qXe dh dqy la[;k gksxh

(1) 9 (2) 12 (3) 3 (4) 6

Ans. 1

Sol. Structure of 3I is

I — I — I : :.. .. ..

.. .. ... . . .

..

Each iodine atom contains 3 lone pairs, hence total number of lone pairs is 9

62. Which of the following satls is the most basic in aqueous solution

fuEu yo.kksa esa dkSu lk tyh; foy;u esa lokZf/kd {kkfj; gSaA

(1) FeCl3

(2) Pb(CH3COO)

2(3) Al(CN)

3(4) CH

3COOK

Ans 4

Sol. FeCl3 is a salt of weak base and strong acid

Pb(CH3COO)

2 is a salt of weak acid and weak base

Al(CN)3 is a salt of weak base and weak acid

CH3COOK is a salt of weak acid and strong base hence its aqueous solution will be most basic.

63. Phenol reacts with methyl chloroformate in the presence of NaOH to form proudct

A. A react with Br2 to from product B. A and B are respectively

NaOH dh mifLFkfr eas fQukWy] esfFky DyksjksQkWesZV ls vfHkfØ;k djds A mRikn cukrk gSaA A ,Br2 ds lkFk vfHkfØ;k

djds mRikn B nsrk gSaA A rFkk B Øe'k% gSa

(1)

(2)

(3)



CHEMISTRY

(4)

Ans. 1

Sol.

OH

Cl�C OCH3�

NaOH

O

O C OCH� 3�

O

2Br

O C OCH� 3�

O

Br

First reaction follows SNth mechanism while second reaction

follows electrophilic substitution reaction

64. The increasing order of basicity of the following compounds is

fuEu ;kSfxdksa dh {kkfj;rk dk c<rk Øe gSa

(a)

(b)

(c)

(d)

(1) (b) < (a) <(d) < (c)

(2) (d) < (b) <(a) < (c)

(3) (a) < (b) <(c) < (d)

(4) (b) < (a) <(c) < (d)

Ans. 1

Sol. To find out basic strength always think for available electron density so order will be

> > >

3Na/liq NH



CHEMISTRY

65. An alkali is titrated against an acid with methyl orange as indicator which of the following is a correct

combination

Base Acid End point

(1) Weak Strong Yellow to pinkish red

(2) Strong Strong Pink to colourless

(3) Weak Strong Colourless to pink

(4) Strong Strong Pinkish red to yellow

esfFky vkjsUTk dks ,d lwpd ds :i eas iz;ksx djds ,d {kkj dks ,d vEy ds fo:) vuqekfir fd;k tkrk gSA fuEu

esa ls dkSu lk ,d lgh la;ksx gS\

{kkj vEy vUR; fcUnq

(1) nqcZy izcy ihys ls xqykch yky

(2) izcy izcy xqykch ls jaxghu

(3) nqcZy izcy jaxghu ls xqykch

(4) izcy izcy xqykch yky ls ihyk

Ans. 1

Sol. Color transition pH range of methyl orange is

Pinkish red 3.1 4.1

yellow

........................ alkali

acid

If initially weak base is present in beaker than pH is greater than 4 and color is yellow and just after

equivalance point due to presence of strong acid, pH will be less than 3.1 and color will change to pinkish

red.

66. The trans alkenes are formed by the reducation of alkynes with

fuEu eas ls fdlds lkFk ,Ydkbukas ds vip;u }kjk VªkUl ,YdhUl curs gSaA

(1) Na/liq.NH3

(2) Sn–HCl (3) H2-Pd/C,BaSO

4(4) NaBH

4

Ans. 1

Sol. Alkyne 3Na/liq NH R

R

(birch reducation)



CHEMISTRY

67. The ratio of mass percent of C and H of an organic compound (CxH

yO

z) is 6 : 1. If one molecule of the

above compound (CxH

yO

z) contains half as much oxygen as required to burn one molecules of compound

CxH

y completelyto CO

2 and H

2 O.The empirical formula of compound C

xH

yO

z is

,d dkcZfud ;kSfxd (CxH

yO

z) eas C rFkk H ds lagfr izfr'krrk dk vuqikr 6 :1 gSaA ;fn mijksDr ;kSfxd ds ,d

v.kq esa vkDlhtu dh ek=kk ;kSfxd CxH

yds ,d v.kq dks iw.kZ :i ls tykdj CO

2 rFkk H

2O eas cnyusa okyh

vkWDlhtu dh ek=kk dh vk/kh gSaA ;kSfxd CxH

yO

z dk ewykuqikrh lw=k gSa

(1) C3H

4O

2(2) C

2H

4O

3(3) C

3H

6O

3(4) C

2H

4O

Ans 2

Sol. CxH

y + 2 2 2

y yx O xCO H O

4 2

1 y zx

2 4 2

yx z

4

x 12 12x y 16z6

12x y 16z y

12x6

y

2x = y x = y/2

2xx z

4

3xz

4 ;

2zx

3

y 2zx

2 3 ;

x y z

1 2 3 / 2

x:y:z = 2:4:3

68. Hydrogen peroxide oxidises [Fe(CN)6]4– to [Fe(CN)

6]3–in acidic medium but reduces [Fe(CN)

6]3– to

[Fe(CN)6]4– in alkaline medium. The other products formed are respectively

gkbMªkstu ijkWDlkbM vEyh; ek/;e eas [Fe(CN)6]4– dks [Fe(CN)

6]3– easa mipf;r djrk gSa ijUrq {kkfj; ek/;e eas

[Fe(CN)6]3– dksa [Fe(CN)

6]4– eas vipf;r djrk gSaA vU; cuus okys mRikn Øe'k% gSa

(1) H2O and (H

2O+O

2) (2) H

2O and (H

2O+OH–)

(3) (H2O+O

2) and H

2O (4) (H

2O+O

2) and (H

2O+OH–)

Ans 1

Sol. 2[Fe(CN)6]4– + H

2O

2 + 2H+ 2[Fe(CN)

6]3– + 2H

2O



CHEMISTRY

2[Fe(CN)6]3– + H

2O

2 + 2OH– 2[Fe(CN)

6]4– +O

2 + 2H

2O

69. The major product formed in the following reaction is

fuEu vfHkfØ;k eas cuus okys eq[; mRikn gSa

(1) (2)

(3) (4)

Ans. 2

Sol.O

OHI

Heat (SN')

O

I2

HI

SN

OH

I

70. How long (approximate) should water be electrolyed by passing through 100 amperes current can

completely burn 27.66 g of diborane (Atomic weight of B = 10.8u)

100 ,fEi;j fo|qr /kkjk izokfgr djds ty dk yxHkx fdruh nsj rd fo|qrvi?kVu fd;k tk;s fd fudyus okyh

vkWDlhtu 27.66g Mkbcksjsu dks iw.kZ :i ls tyk lds

(B dk ijek.kq Hkkj = 10.8u)

(1) 3.2 hours (2) 1.6 hours (3) 6.4 hours (4) 0.8 hours

Ans. 1

Sol. B2H

6 + 3O

2 B

2O

3 + 3H

2O

Moles of B2H

6 =

27.661 mole

27.6

Hence moles of O2 required to completely born 27.66 gm of B

2H

6 is 3.

2H2O 2H

2 + O

2

for one mole of O2, 4 moles of electron are involved



CHEMISTRY

for 3 mole of O2 12 moles of electron are involved

12 × 96500 = 100 × t t = 12 × 965 sec.

So time in hr = 12 965

3.21 hr3600

71. Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an

exothermic reaction

,d Å"ek{ksih vfHkfØ;k ds fy, fuEu eas ls dkSu lh js[kk lkE;fLFkjkad K dh rki ij fuHkZjrk dks lgh :i ls

iznf'kZr djrk gSa

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

Ans. 3

Sol. lnkeq

= H 1 S

R T R

for exothermic reaction H = –ve

hence slope of lnkeq

vs 1

T, graph will be +ve

72. At 518°C the rate of decomposition of a sample of gseous acetaldehyde, initially at a pressure of 363 Torr,

was 1.00 Torr s–1 when 5 % had reacted and 0.5 Torr s–1 when 33% had reacted. The order of the

reaction is

518°C ij XkSlh; ,flVfYMgkbM ds ,d izfrn'kZ dh fo;kstu nj ftldk izkjfEHkd nkc 363 Vkj Fkk 5% vfHkfØ;k

dj ysus ij 1.00 Torr s–1 rFkk 33% vfHkfØ;k dj ysus ij 0.5 Torr s–1 ik;h xbZ vfHkfØ;k dh dksfV gSa

(1) 1 (2) 0 (3) 2 (4) 3

Ans. 3

Sol. r = K 3

n

CH CHOP

n

1

2

r 0.95 363

r 0.67 363

n1 95

0.5 67

n = 2



CHEMISTRY

73. Glucose on prolonged heating with HI gives

(1) Hexanoic acid (2) 6-iodohexanal (3) n-Hexane (4) 1-Hexene

Xywdkst dks HI ds lkFk yEcs le; rd xeZ djus ij izkIRk gksrk gSa

(1) gsDlkuksbd ,slhM (2) 6-vk;MksgsDlsuy (3) n-gsDlsu (4) 1-gsDlhu

Ans. 3

Sol.

HHO

HH

OHHOHOH

CH OH2

CHO

D-Glucose

HI/P

74. Consider the following reaction and statements

[Co(NH3)

4Br

2]+ + Br– [Co(NH

3)

3Br

3]+NH

3

(I) Two isomers are produced if the reactant complex ion is cis-isomer

(II) Two isomers are produced if the reactant complex ion is a trans isomer

(III) Only one isomer is produced if the reactant complex ion is a trans-isomer

(IV) Only one isomer is produced if the reactant complex ion is a cis-isomer

The correct statement are

fuEu vfHkfØ;k rFkk dFkukas ij fopkj dhft,

[Co(NH3)

4Br

2]+ + Br– [Co(NH

3)

3Br

3]+NH

3

(I) nks leko;oh curs gSa ;fn vfHkdkjd dkWeIysDl vk;u ,d fll leko;oh gS

(II) nks leko;oh curs gSa ;fn vfHkdkjd dkWeIysDl vk;u ,d Vªkal leko;oh gSaA

(III) ek=k ,d leko;oh curk gS ;fn vfHkdkjd dkWeIysDl vk;u ,d Vªkal leko;oh gSa

(IV) dsoy ,d leko;oh curk gSaA ;fn vfHkdkjd dkWeIysDl vk;u ,d fll leko;oh gSa

lgh dFku gSa

(1) (III) and (IV) (2) (II) and (IV) (3) (I) and (II) (4) (I) and (III)

Ans. 4

Sol.CO

NH3 NH3

NH3NH3

Br

Br

Br CO

NH3 Br

NH3NH3

Br

Br



Trans only one isomer is formed

CONH3 Br

H3N Br

NH3

NH3

Br CO

NH3 Br

NH3Br

Br

NH3

Cis

Br–

CONH3 Br

Br Br

NH3

NH3

Two isomers are formed

75. The major product of the following reaction is

fuEu vfHkfØ;k dk eq[; mRikn gSa

NaOme

MeOH

(1) (2)

(3) (4)

Ans. 4

Sol. NaOme

MeOH

This reaction proceed via E2 mechanism

CHEMISTRY



76. Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compund X as

the major product. X on treatment with (CH3CO)

2 O in the presence of catalytic amount of H

2SO

4

produces

NaOH dh mifLFkfr eas QsukWy CO2 ds lkFk vfHkfØ;r djus rnqijkUr vfEyr djus ij ,d ;kSfxd x eq[; mRikn

ds :i eas nsrk gSaA H2SO

4 dh mRizsfjdh; ek=kk eas mifLFkr jgus eas x dks (CH

3CO)

2 O ds lkFk vfHkfØf;r djus

ij izkIr gksxk

(1) (2)

CO H2

CO H2

CH3O

O

(3) CO H2

O

OCH3

(4)

O

O

CO H2

CH3

Ans. 3

Sol.

OH

2i CO /NaOH

ii H

OH

COOH 3 22

CH CO O/H OH

O C CH� 3�

O

Br

COOH

(Kolbe reaction) (X) Aspirin

77. An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1M solution of Na2SO

4

is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO

4 is 1

× 10–10. What is the original concentration of Ba2+ ?

,d tyh; foy;u esa Ba2+ gSa ftldh lkUnzrk vKkr gSaA mlesa 1 M Na2SO

4 ds 50 mL foy;u feykrs gh BaSO

4

dk vo{ksi cuuk 'kq: gks tkrk gSaA vafre vk;ru 500 mL gSaA BaSO4 dk foys;rk xq.kkad 1×10–10 gSaA Ba2+ dh ewy

lkUnzrk jgh gksxh

(1) 1.1 × 10–9 M (2) 1.0 × 10–10 M (3) 5 × 10–9 M (4) 2 × 10–9 M

Ans. 1

Sol. Ba+2 + SO4

2– BaSO4(s)

final conc. of 2 14

1 2

MV 1 50SO 0.1M

V V 500

Final conc. of [Ba+2] when BaSO4 start precipitating

KSP

= QSP

= [Ba+2] [SO42–]

CHEMISTRY



10–10 = [Ba+2] (0.1 M)[Ba+2] = 10–9 M

initial conc. [Ba+2]; Initial volume was = 500 – 50 = 450 ml

M1V

1 = M

2V

2

92 2

1

1

M V 10 500M

V 450

M1 = 1.1 × 10–9 M

78. Which of the following compounds will be suitabel for Kjeldahl's method for nitrogen estimation ?

ukbVªkstu vkdyu ds fy, dsYMky fof/k eas fuEu ;kSfxdksa esa ls dkSu mi;qDr gksxk

(1) (2)

(3) (4)

Ans. 4

Sol. Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and

nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to

ammonium sulphate under these conditions.

79. When metal 'M' is treated with NaOH, a white gelationous precipitate 'X' is obtained, which is soluble in

excess of NaOH. Compound 'X' when heated strongly gives an oxide which is used in chromatography as

an adsorbent. The metal 'M' is

tc ,d /kkrq 'M' dks NaOH ds lkFk vfHkfØf;r fd;k tkrk gSa tks ,d lQsn ftysfVul vo{ksi 'X' izkIr gksrk gS

tks NaOH ds vkf/kD; eas ?kqyu'khy gSaA ;kSfxd 'X' dks tc vf/kd xje fd;k tkrk gSa rks ,d vkWDlkbM izkIr gksrh gS

tks ØksesVksxzkQh eas ,d vf/k'kks"kd ds :i eas iz;qDr gksrh gSaA /kkrq 'M' gSa

Ans. 1

(1) Al (2) Fe (3) Zn (4) Ca

Sol. Al NaOH Excess3 4NaOH

Al(OH) Na[Al(OH) ]

white ppt. soluble

2Al(OH)3 Al

2O

3 + 3H

2O

Al2O

3 is used in column chromatography

80. An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the

formation of HS– from H2S is 1.0 × 10–7 and that of S2– from HS– ions is 1.2 × 10–13 then the concentration

of S2– ions in squeous solution is

CHEMISTRY



,d tyh; foy;u eas 0.10 M H2S rFkk 0.20 M HCl gSA ;fn H

2S ls cuus dk lkE; fLFkjkad 1.0 × 10–7 gks rFkk

HS– ls S2– cuus dk lkE; fLFkjkad 1.2 × 10–13 gks rks tyh; foy;u eas S2– vk;ukas dh lkUnzrk gksxh (1) 6 × 10–

21 (2) 5 × 10–19 (3) 5 × 10–8 (4) 3 × 10–20

Ans. 4

Sol. H2S (aq) H+ + HS–

0.1 0.11

2 + 0.1

1 +0.2

0.1(1–1) 0.1

10.1

1 (1–

2)

HS– H+ + S2–

0.11(1–

2) 0.1

1

2 + 0.1

1 +0.2 0.1

1

2

Approximations

1 – 1 1

1 – 2 1

0.11 + 0.1

1

2 + 0.2

0.2

10–7 = 10.1 0.2

0.1

67

1

105 10

2

13 1 2

1

0.2 0.11.2 10

0.1

1313

2

1.2 106 10

0.2

2 7 13S 0.1 5 10 6 10

= 3 × 10–20

81. The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluroide ion is required

to make teeth enamel harder by converting [3Ca3(PO

4)

2 . Ca(OH)

2] to

is;ty esa ¶yksjkbM vk;u dh vuq'kkaflr lkUnzrk 1ppm rd gSA pw¡fd nk¡r ,ukesy dks dBksj cukus eas ¶yksjkbM vk;u

dh vko';drk gksrh gSa tks [3Ca3(PO

4)

2 . Ca(OH)

2] dks fuEu esa cnydj djrh gSa

(1) [2Ca3(PO

4)

2 . CaF

2] (2) [3(Ca{OH)

2}. CaF

2]

(3) [CaF2] (4) [3(CaF

2) . Ca(OH)

2]

Ans. 1

Sol. [2Ca3(PO

4)

2 . CaF

2]

CHEMISTRY



82. The compound that does not produce nitrogen gas by the thermal decompositon is

og ;kSfxd tks rkih; fo?kVu }kjk ukbVªkstu xSl ugha mRiUu djrk gSa

(1) NH4 NO

2(2) (NH

4)

2SO

4(3) Ba(N

3)

2(4) (NH

4)

2Cr

2O

7

Ans. 2

Sol. NH4 NO

2

(NH4)

2SO

4

SO

4

Ba(N3)

2 Ba + N

2

(NH4)

2Cr

2O

7 N

2 + Cr

2O

3

83. The predominant form of histamine present in human blood is (pKa, Histidine = 6.0)

ekuo jDr esa mifLFkr fgLVkfeu dk izeq[k :i gSa (pKa, fgLVkfMu = 6.0)

(1) (2)

(3) (4)

Ans. 2

Sol. The predominant form of histamine in human blood H0P 7.4 is

N

H

NH3

+

N

as terminal have akP of ring is nearly 6. Mas ratio of C & H is 6 :1

There for ,X : Y is 2 : 1

84. The oxidation states of Cr in [Cr(H2O)

6]Cl

3, [Cr(C

6H

6)

2], and K

2[Cr(CN)

2(O)

2 (O)

2(NH

2)] respectively

are

[Cr(H2O)

6]Cl

3, [Cr(C

6H

6)

2], rFkk K

2[Cr(CN)

2(O)

2 (O)

2(NH

2)] esa Øksfe;e dh vkDlhdj.k voLFkk;sa Øe'k% gSS

respectively are

(1) +3, 0, and +6 (2) +3, 0, and +4 (3) +3, +4, and +6 (4) +3, +2 and +4

CHEMISTRY



Ans. 1

Sol. [Cr(H2O)

6]Cl

3x + 0 = + 3

x = +3

[Cr(C6H

6)

2] y = 0

K2[Cr(CN)

2(O)

2 (O)

2(NH

2)]

z – 2 + 2 (–2) + (–2) + 0 = – 2

z = +6

85. Which type of 'defect' has the presence of cations in the interstitial sites ?

fdlh rjg dh =kqfV eas vrajdk'kh LFkku eas /kuk;u (dSVk;u) dh mifLFkfr gksrh gSa

(1) Frenkel defect/Ýasdy =kqfV (2) Metal deficiency defect//kkrq ghurk =kqfV

(3) Schottky defect/lkV~dh =kqfV (4) Vacancy defect/fjfDrdk =kqfV

Ans. 1

Sol. In Frenkel defect smaller ion (generally cations) are dislocated to interstitial sites

86. The combustion of benzene (l) gives CO2(g) and H

2O(l). Given that heat of combustion of benzene at

constant volume is – 3263.9 kJ mol–1 at 25ºC; heat of combustion (in kJ mol–1) of benzene at constant

pressure will be

(R = 8.314 JK–1 mol–1)

casthu ds ngu djus ij CO2(g) rFkk H

2O(l) izkIRk gksrh gSaA fLFkj vk;ru ij casthu (l) dh ngu Å"ek 25°C ij –

3263.9 kJ mol–1 gSaA fLFkj nkc ij casthu dh ngu Å"ek (kJ mol–1) dk eku gksxk

(R = 8.314 JK–1 mol–1)

(1) 3260 (2) –3267.6 (3) 4152.6 (4) –452.46

Ans. 2

Sol. C6H

6(l) + 2

15O (g)

2 6CO

2(g) + 3H

2O(l)

U = –3263.9 kJ/mole

H = U + ngRT

H = –3263.9 + 1.5 8.314 298

1000

; H = –3267.6 kJ/mol

87. Which of the follwoing are lewis acids

(1) PH3 and SiCl

4(2) BCl

3 and AlCl

3(3) PH

3 and BCl

3(4) AlCl

3 and SiCl

4

CHEMISTRY



fuEu eas ls dkSu ywbZl vEy gSa

(1) PH3 rFkk SiCl

4(2) BCl

3 rFkk AlCl

3(3) PH

3 rFkk BCl

3(4) AlCl

3 rFkk SiCl

4

Ans. 2

Sol. BCl3 and AlCl

3 both are electron deficient species so they can act as lewis acid.

88. Which of the following compounds contain(s) no covalent bond(s) ?

fuEu ;kSfxdksa esa ls fdlesa lgla;kstd vkcU/k ugha gSa

(1) KCl (2) KCl, B2H

6(3) KCl, B

2H

6, PH

3(4) KCl, H

2SO

4

Ans. 1

Sol. K+Cl– has only ionic bond

89. For 1 molar aqueous solution of the following compounds, which one will show the highest freezing point ?

fuEu ;kSfxdks ds 1 eksyy tyh; foy;u ysus ij fdldk fgekad mPpre gksxk

(1) [Co(H2O)

4Cl

2]Cl . 2H

2O (2) [Co(H

2O)

3Cl

3] . 3H

2O

(3) [Co(H2O)

6]Cl

3(4) [Co(H

2O)

5Cl]Cl

2 . H

2O

Ans. 2

Sol. Tf = 2 × K

f × 1 for 1

Tf = 1 × K

f × 1 for 2

Tf = 4 × K

f × 1 for 3

Tf = 3 × K

f × 1 for 4

for [Co(H2O)

3Cl

3] . 3H

2O , value of T

f is lowest hence it freezing point will be highest.

90. According to molecular orbital theory, which of the following will not be a viable molecule ?

v.kqd{kd fl}kUr ds vqulkj fuEu esa ls dkSu lk v.kq O;ogk;Z ugha gksxk

(1) 2H (2) 22H (3) 2

2He (4) 2He

Ans. 2

Sol. 22H have bond order zero do not exist

CHEMISTRY

21Matrix JEE Academy : Opposite Reliance Petrol Pump, Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999


MATHS

PART - BMATHEMATICS



31. If the tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 then the value

of c is

;fn oØ x2 = y – 6 ds fcUnq (1, 7) ij cuh Li'kZjs[kk or x2 + y2 + 16x + 12y + c = 0 dks Li'kZ djrh gS] rks c dk

eku gS &

(1) 85 (2) 95 (3) 195 (4) 185

Ans. 2

Sol. x2 = y – 6

(1,7)

dy

dx

= 2

T (1, 7) = 2x – y + 5 = 0

distance of center from tangent = radius

16 6 5r 5

5

2 2g f c 5

c = 95

32. If L1 is the line of intersection of the planes 2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 and L

2 is the line of

intersection of the planes x + 2y – z – 3 = 0, 3x – y + 2z – 1 = 0, then the distance of the origin from the

plane containing the lines L1 and L

2 is

;fn leryksa 2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 dh ifjPNsnh js[kk L1 gS rFkk leryksa x + 2y – z – 3 = 0,

3x – y + 2z – 1 = 0 dh ifjPNsnh js[kk L2 gS] rks ewy fcUnq dh nwjh ml leery ls tks js[kkvksa L

1 vkSj L

2 dk

varfoZ"V djrk gS] gS &

(1) 1

2 2(2)

1

2(3)

1

4 2(4)

1

3 2

Ans. 4

Sol. required plane will pass though line of intersection of planes so equation of required plane will be

x + 2y – z – 3 + (3x – y + 2z – 1) = 0

take a fixed point on L1 (y = 0)

P(–5, 0, 4)

point P will lie on required plane



MATHS

= –3

2

So equation of required plane will be

7x – 7y + 8z + 3 = 0

distance from origin = 3 1

162 3 2

33. If C are the distinct roots, of the equation x2 – x + 1 = 0, then 101 + 107 is equal to

;fn C lehdj.k x2 – x + 1 = 0 ds fofHkUu ewy gS] rks 101 + 107 cjkcj gS &

(1) 1 (2) 2 (3) –1 (4) 0

Ans. 1

Sol. Roots of x2 – x + 1 = 0 are ––2

101 + 107 = –101 – 214

–2 – = 1

34. Tangent are drawn to the hyperbola 4x2 – y2 = 36 at the points P and Q. If these tangents intersect at the

point T(0, 3) then the area (in sq. units) of PTQ is

,d vfrijoy; 4x2 – y2 = 36 ds fcanqvksa P rFkk Q ij Li'kZ js[kk,¡ [khaph tkrh gSA ;fn ;g Li'kZjs[kk,¡ fcanq T(0, 3)

ij dkVrh gS] rks PTQ dk {ks=kQy (oxZ bdkb;ksa esa) gS &

(1) 60 3 (2) 36 5 (3) 45 5 (4) 54 3

Ans. 3

Sol.

T

PQ

Eqaution of tangent at P

x sec y tan1

3 6

it will pass though (0, 3)

3 tan0 1

6

tan = –2 sec = 5

area = 1

6 sec | 6 tan | 32

= 45 5



MATHS

35. If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is

;fn oØ y2 = 6x, 9x2 + by2 = 16 ledks.k ij izfrPNsn djrs gSa] rks b dk eku gS &

(1) 4 (2) 9

2(3) 6 (4)

7

2

Ans. 2

Sol. y2 = 6x

2yy' = 6

y' = 1

3

y

9x2 + by2 = 1618x + 2byy' = 0

y' = 1

1

9x

by

m1m

2 = –1

27x1 = by

12 ..... (i)

6x1 = y

12 ..... (ii)

(i) (ii)

b = 9

2

36. If the system of linear equations

x + ky + 3z = 0

3x + ky – 2z = 0

2x + 4y – 3z = 0

has a non-zero solution (x, y, z), then 2

xz

y is equal to

;fn jSf[kd lehdj.k fudk;

x + ky + 3z = 0

3x + ky – 2z = 0

2x + 4y – 3z = 0

dk ,d 'kwU;srj gy (x, y, z) gS] rks 2

xz

y cjkcj gS &

(1) –30 (2) 30 (3) –10 (4) 10

Ans. 4

Sol. x + ky + 3z = 0 ...... (i)

3x + ky – 2z = 0 .... (ii) system will have solutions

2x + 4y – 3z = 0 .... (iii)



MATHS

(i) – (ii)

x = 5z

2 then y =

z

2

2

xy10

z

37. Let S = {x R: x 0 and 2| x 3 | x ( x 6) 6 0 }. Then S :

(1) contains exactly two elements (2) contains exactly four elements

(3) is an empty set (4) contains exactly one element

ekuk S = {x R: x 0 rFkk 2| x 3 | x ( x 6) 6 0 } rks S

(1) esa ek=k nks vo;o gSa (2) esa ek=k pkj vo;o gSa

(3) ,d fjDr leqPp; gSa (4) esa ek=k ,d gh vo;o gSa

Ans. 1

Sol. 2 x 3 x 6 x 6 0

Put x 3 t

2t + t2 – 3 = 0t = 1 t = – 3

x 3 1 x 3 3

x 4, x 2 x =

x = 16, x = 4

38. If sum of all the solution of the equation 1

8cos x. cos x .cos x 16 6 2

in [0, ] is k, then k

is equal to

;fn lehdj.k 1

8cos x. cos x .cos x 16 6 2

ds varjky [0, ] esa lHkh gyksa dk ;ksx kgS] rks k

cjkcj gS&

(1) 8

9(2)

20

9(3)

2

3(4)

13

9

Ans. 4

Sol. 8cosx .cos x cos x 4cos x 16 6

238cosx sin x 4cos x 1

4

24cos x 12cosx 4cos x 1

4



38cos x 2cos x 4cos x 1 8cos3x – 6cosx = 1cos3x = 1/2

3 x 2n3

5 7x , ,

9 9 9

Sum = 13

9

39. A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and

this ball along with two additional balls of the same colour are returened to the bag. If now a ball is drawn at

random from the bag, then the probability that this drawn ball is red, is

,d FkSys esa 4 yky rFkk 6 dkyh xsansa gSA FkSys esa ls ;kn`PN;k ,d xsan fudkyh x;h] rFkk mldk jax ns[kdj] ml xsan

dks] nks vU; mlh jax dh xsanksa ds lkFk okfil FkSys esa Mky fn;k x;kA vc ;fn FkSys esa ls ;kn`PN;k ,d xsan fudkyh

tk,] rks izkf;drk fd ml xsan dk jax yky gS] gS &

(1) 1

5(2)

3

4(3)

3

10(4)

2

5

Ans. 4

Sol. = P (Red in Ist turn & Red in IInd turn) + P(Black in Ist turn & Red in IInd turn)

4 6

10 12

+

6 4 2

10 12 5

40. Let 2

2

1f (x) x

x and

1g(x) x

x , x R– {–1, 0, 1}. If

f (x)h(x)

g(x) , then the local minimum value

of h(x) is

ekuk 2

2

1f (x) x

x rFkk

1g(x) x

x , x R– {–1, 0, 1} gSaA ;fn

f (x)h(x)

g(x) gS] rks h(x) dk LFkkuh;

U;u`re eku gS &

(1) 2 2 (2) 2 2 (3) 3 (4) –3

Ans. 2

Sol.

2

2

1x

xh(x)1

1x

2t 2 2h(x) t

t t

where

1t x

x

h'(x) 2

21

t

dt

dx

MATHSMATHSMATHS



2

2

t 2 1h '(x) 1

t x

Minima at t = 2

Min

2 2h(x) | 2 2

2

41. Two sets A and B are as under :

A = { (a, b) R × R : |a – 5| < 1 and |b – 5| < 1}: B {(a, b) R × R: 4(a – 6)2 + 9(b – 5)2 36}. Then

(1) A B = (an empty set) (2) neither A B not B A

(3) B A (4) A B

nks leqPp; A rFkk B fuEu izdkj ds gS &

A = {(a, b) R × R : |a – 5| < 1 rFkk |b – 5| < 1}: B {(a, b) R × R: 4(a – 6)2 + 9(b – 5)2 36}. rks

(1) A B = (,d fjDr leqPp;) (2) u rks A B vkSj u gh B A

(3) B A (4) A B

Ans. 4

Sol. Let a – 6 = x b – 5 = y

|x + 1| < 1 |y| < 1 4x2 + 9y2 36

–2 < x < 0 –1 < y < 1

Boundries of square are lying in ellipse

A B

42. The Boolean expression ~(p q) (~p q) is equivalent to

cwys ds O;atd ~(p q) (~p q) ds lerqY; gS &

(1) q (2) ~q (3) ~p (4) p

Ans. 3

Sol.

p q p q ~ (p q) ~(p) (~p) q (~(p q)) ((~p)q))

T T T F F F F

T F T F F F F

F T T F T T T

F F F T T F T

MATHSMATHSMATHS



43. Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the

parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and CPB =

, then a value of tan is

ijoy; y2 = 16x ds ,d fcUnq P(16, 16) ij Li'kZ js[kk rFkk vfHkyEc [khaps tkrs gS rks ijoy; ds v{k dks fcUnqvksa

Øe'k% A rFkk B ij izfrPNsn djrs gSaA ;fn fcUnqvksa P, A rFkk B ls gksdj tkus okys or dk dsUnz C gS rFkk CPB

= rks tan dk ,d eku gS &

(1) 3 (2) 4

3(3)

1

2(4) 2

Ans. 4

Sol.

��

A C B(4,0)

P

dxt

dy

= – 2

Slope of normal at P = – tan = – 2

tan = 2

44. If 2

x 4 2x 2x

2x x 4 2x A Bx x A

2x 2x x 4

, then the ordered pair (A, B) is equal to

;fn 2

x 4 2x 2x

2x x 4 2x A Bx x A

2x 2x x 4

rks Øfer ;qXe (A, B) cjkcj gS &

(1) (–4, 5) (2) (4, 5) (3) (–4, –5) (4) (–4, 3)

Ans. 1

Sol.

a b c

4x c a b

b c a = (a + b +c) (a2 + b2 + c2 – ab – bc – ca)

2

x 4 2x 2x

2x x 4 2x (5x 4)(x 4)

2x 2x x 4

A = – 4, B = 5

MATHSMATHSMATHS



45. The sum of the co-efficients of all odd degree terms in the expansion of 5 5

3 3x x 1 x x 1 , (x

> 1) is

5 5

3 3x x 1 x x 1 , (x > 1) ds izlkj esa lHkh fo"ke ?kkrksa okys inksa ds xq.kkadks dk ;ksx gS &

(1) 1 (2) 2 (3) –1 (4) 0

Ans. 2

Sol. 5 5

3 3f (x) x x 1 x x 1

= 2(x5 + 10x3 (x3 – 1) + 5x(x3 – 1)2)

Sum of coefficients off add degree = f (1) f ( 1)

22

46. Let a1, a

2, a

3, .......... a

49 be in A.P. such that

12

4k 1k 0

a 416

and a9 + a

43 = 66. If

2 2 21 2 17a a ....... a 140m , then m is equal to

ekuk a1, a

2, a

3, .......... a

49 ,d lekarj Js<+h esa ,sls gS fd

12

4k 1k 0

a 416

rFkk a9 + a

43 = 66 gSA ;fn

2 2 21 2 17a a ....... a 140m gS] rks m cjkcj gS &

(1) 34 (2) 33 (3) 66 (4) 68

Ans. 1

Sol. Let a1 = a common difference = d

12

4k 1k 0

a 416

a + 24d = 32

a9 + a

43 = 66 a + 25d = 33

a = 8

d = 1

2 2 2 2 2 21 2 17a a ..... a 8 9 ...... 24

= (12 + 22 + ...... + 242) – (12 + 22 + ... + 72)

4760 = 140 m

m = 34

47. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is

the origin and the rectangle OPRQ is completed, then the locus of R is

,d ljy js[kk] tks ,d vpj fcUnq (2, 3) ls gksdj tkrh gS] funsZ'kkad v{kksa dks nks fofHkUu fcUnqvksa P rFkk Q ij

izfrPNsn djrh gSA ;fn O ewy fcUnq gS rFkk vk;r OPRQ dks iwjk fd;k tkrk gS rks R dk fcUnqiFk gS &

(1) 3x + 2y = xy (2) 3x + 2y = 6xy (3) 3x + 2y = 6 (4) 2x + 3y = xy

MATHSMATHSMATHS



Ans. 1

Sol.

o P(h,0)

R(h,k)

A(2,3)

Q(0,k)

P, A, Q will be collinear

h 0 1

2 3 1 0

0 k 1

3h + 2k = hklocus of R (h, k) will be

3x + 2y = xy

48. The value of

22

x

2

sin xdx

1 2

is

22

x

2

sin xdx

1 2

dk eku gS &

(1) 4 (2*) 4

(3)

8

(4)

2

Ans. 2

Sol.

22

x

2

sin xI dx

1 2

Use king & Add

22

2

2I sin x dx

= 22

2

0

s in x d x

I = 4

MATHSMATHSMATHS



49. Let g(x) = cos x2, f(x) = x , and () be the roots of the quadratic equation 18x2 – 9x + 2 =

0. Then the area (in sq. unit) bounded by the curve y = (gof) (x) and the lines x = , x = and y = 0 is

ekuk g(x) = cos x2, f(x) = x , rFkk () f}?kkrh lehdj.k 18x2 – 9x + 2 = 0 ds ewy gSaA rks oØ y

= (gof) (x) rFkk js[kkvksa x = , x = rFkk y = 0 }kjk f?kjs {ks=k dk {ks=kQy (oxZ bdkb;ksa esa) gS&

(1) 13 2

2 (2) 1

2 12

(3) 13 1

2 (4) 1

3 12

Ans. 3

Sol. gof (x) = cosx

18x2 – x + 2 = 0

(6x – ) (3x – ) = 0

x , x6 3

3

6

3 1A cos x dx

2

50. For each t R, let [t] be the greatest integer less than or equal to t. Then x 0

1 2 15lim x ......

x x x

(1) is equal to 120 (2) does not exist (in R)

(3) is equal to 0 (4) is equal to 15

izR;sd t R ds fy, ekuk [t], t ls NksVk egre iw.kkZad gS] rks x 0

1 2 15lim x ......

x x x

(1) 120 ds cjkcj gS (2) (R esa) bldk vfLrRo ugha gS

(3) 0 ds cjkcj gS (4) 15 ds cjkcj gS

Ans. 1

Sol. x – 1 < [x] x

1 1 11

x x x

1 1 xx 1 x

x x x

use sandwich

MATHSMATHSMATHS



x 0

1lim x 1

x

Simillarly x 0

2lim x 2

x

x 0

15lim x 15

x

Ans = 1 + 2 + ..... + 15 = 120

51. If 9

ii 1

x 5 9

and 9

2

ii 1

x 5 45

, then the standard deviation of the 9 items x1, x

2, ......, x

9 is

;fn 9

ii 1

x 5 9

rFkk 9

2

ii 1

x 5 45

gS] rks ukS izs{k.kksa x1, x

2, ......, x

9 dk ekud fopyu gS &

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

Ans. 1

Sol. There is no effect on S.D. on shiffing of origin

xi – 5 = x

i

ix 9

2

ix 45

22i i

x xS.D.

9 9

=

245 9

29 9

52. The integral

2 2

25 3 2 3 2 5

sin x cos xdx

sin x cos x sin x sin x cos x cos x is equal to

lekdyu

2 2

25 3 2 3 2 5

sin x cos xdx

sin x cos x sin x sin x cos x cos x cjkcj gS &

(1) 3

1C

1 cot x

(2) 3

1C

1 cot x

(3) 3

1C

3(1 tan x)

(4) 3

1C

3(1 tan x)

(Where C is a constant of integration)

MATHS



(tgk¡ C ,d lekdyu vpj gS)

Ans. 4

Sol. Divide numberator & denominator by cos10x

2 6

5 2 3 2

tan x sec x dx

(tan x tan x tan x 1)

2 6

2 23 2

tan x sec x dx

1 tan x 1 tan x

2 6

23

tan x sec x dx

1 tan x

Put 1 + tan3x = t

2

1 dtI

3 t

1C

3t

3

1C

3(1 tan x)

53. Let S = {t R: f(x) = |x – | . (e|x| – 1) sin |x| is not differentiable at t}. Then the set S is equal to

(1) {} (2) {0, } (3) (an empty set) (4) {0}

ekuk S = {t R: f(x) = |x – | . (e|x| – 1) sin |x| tks ij vodyuh; ugha gS}, rsk leqPp; S cjkcj gS &

(1) {} (2) {0, } (3) (,d fjDr leqPp;) (4) {0}

Ans. 3

Sol. We will check at x = 0,

at x = 0f'(0+) = 0f'(0–) = 0

at x =

f'(+) = 0

f'(–) = 0

It is differentiable at x = 0,

S = { } =

MATHS



54. Let y = y(x) be the solution o f the differential equation dy

sin x y cos x 4xdx

, x (0, ) . If y 02

,

then y6

is equal to

ekuk vody lehdj.k dy

sin x y cos x 4xdx

, x (0, ) dk y = y(x) ,d gy gSA ;fn y 02

gS] rks

y6

cjkcj gS &

(1) 28

9 (2)

24

9 (3)

24

9 3 (4)

28

9 3

Ans. 1

Sol.dy

sin x 4cos x 4xdx

d(ysin x) 4x dx 4 sinx = 2x2 + C

x = 0, y = 2

C = –

2

2

Put x = 6

2 21y 2

2 36 2

28y

9

55. Let u be a vector coplanar with the vectors ˆ ˆ â 2i 3j k

and ˆ ˆb j k

. If u is perpendicular to a

and

u.b 24 , then 2| u |

is equal to

ekuk u ,d ,slk lfn'k gS tks lfn'kksa ˆ ˆ â 2i 3j k

rFkk ˆ ˆb j k

ds lkFk leryh; gSA ;fn u

, a

ij yacor~

gS rFkk u.b 24 gS] rks 2| u |

cjkcj gS &

(1) 256 (2) 84 (3) 336 (4) 315

Ans. 3

Sol. u a a b

a.b a a.a b

2a 14b

ˆ ˆ û 4i 8j 16k

MATHS



u.b 24 1

ˆ ˆ ˆu 4i 8 j 16k

2| u | 16 64 256 336

56. The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane, x +

y + z = 7 is

fcUnqvksa (5, –1, 4) rFkk (4, –1, 3) dks feykus okys js[kk[k.M dk lery x + y + z = 7 ij Mkys x, iz{ksi dh yEckbZ

gS &

(1) 1

3(2)

2

3(3)

2

3(4)

2

3

Ans. 2

Sol. �A

B

A (5,–1,4)

B (4,–1,3)

= angle between plane & line

ˆ ˆ ˆAB. i j k 2cos 90

ˆ ˆ ˆ 6AB i j k

Req. length = | AB | cos

2 22

36

57. PQR is a triangular park with PQ = PR = 200m. A T.V. tower stands at the mid-point of QR. If the angles

of elevation of the top of the tower at P, Q and R are respectively 45º and 30º and 30º, then the height of

the tower (in m) is

PQR ,d f=kdks.kkdkj ikdZ gS ftlesa PQ = PR = 200m gSA QR ds e/; fcUnq ij ,d Vhoh Vkoj fLFkr gSA ;fn

fcUnqvksa P, Q, R ls Vkoj ds f'k[kj ds mUu;u dks.k Øe'k% 45º, 30º rFkk 30º gS rks Vkoj dh Å¡pkbZ (m esa) gS &

(1) 100 3 (2) 50 2 (3) 100 (4) 50

Ans. 3

MATHS



Sol.h

P

Q R

200 200

3hS

3h

height of tower be h

2

2 23h (h) (200)

h = 100

58. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and

arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements

is

(1) at least 500 but less than 750 (2) at least 750 but less than 1000

(3) at least 1000 (4) less than 500

6 fHkUu miU;klksa rFkk 3 fHkUu 'kCndksa'kksa esa ls 4 miU;klksa rFkk 1 'kCndks'k dks pqudj ,d iafDr esa ,d 'kSYQ ij bl

izdkj ltk;k tkuk gS fd 'kCn dks'k lnk e/; esa gksA bl izdkj ds foU;klksa dh la[;k gS &

(1) de ls de 500 ysfdu 750 ls de (2) de ls de 750 ysfdu 1000 ls de

(3) de ls de 1000 (4) 500 ls de

Ans. 3

Sol. 6C4 3C

1 × 4!

= 1080

59. Let A be the sum of the first 20 terms and B be sum of the first 40 terms of the series 12 + 2.22 + 32 + 2.42

+ 52 + 2.62 + ...... If B – 2A = 100, then is equal to

ekuk Js.kh 12 + 2.22 + 32 + 2.42 + 52 + 2.62 + ..... ds izFke 20 inksa dk ;ksx A gS rFkk izFke 40 inksa dk ;ksx B

gSA ;fn B – 2A = 100, rks cjkcj gS &

(1) 464 (2) 496 (3) 232 (4) 248

Ans. 4

Sol. f(x) = 12 + 2(2)2 + 32 + .......... + 2(x)2

(12 + 22 + ....... + n2) + 2(22 + 42 + ........ + n2)

1 n

1 n 12 2n(n 1)(2n 1)

46 6

2n(n 1)f (x)

2

; B – 2A = f(40) – 2f(20)

MATHS



2 240 (41) 20 (21)2

2 2

; 100 = 24800

= 248

60. Let the orthocentre and centroid of a triangle be A(–3, 5) and B(3, 3) respectively. If C is the circumcentre

of this triangle, then the radius of the circle having line segment AC as diameter, is

ekuk ,d f=kHkqt dk yEc dsUnz rFkk dsUnzd Øe'k% A(–3, 5) rFkk B(3, 3) gSA ;fn bl f=kHkqt dk ifjdsUnz C gS] rks

js[kk[k.M AC dks O;kl eku dj cuk, tkus okus o`r dh f=kT;k gS &

(1) 5

32

(2) 3 5

2(3) 10 (4) 2 10

Ans. 1

Sol.

B divides AC in 2 : 1

3ABAC

2 ;

AC 3ABr

2 4

3 40

4 ;

53

2

MATHS

B JEE (Main) 2018 - Matrix JEE AcademyMatrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911,...

Documents

Transcript of B JEE (Main) 2018 - Matrix JEE AcademyMatrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911,...