Imp problems

Corporate Office (New Campus) : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005

Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555

JRPT1130714C0-1

PT-01 Date: 10-08-2014 Syllabus:- Geometrical optics (upto Thin lens and lens formula), Rectilinear motion complete

S.No. Subject Nature of Questions No. of Questions Marks Negative Total

1 to 25 SCQ 25 3 ñ1 75

26 to 30 MCQ 5 4 0 20

31 to 36 Comprehension tough (3 Com. ◊ 2 Q.) 6 3 ñ1 18

37 to 37 Match the Column (4 Vs. 5) 1 8 0 8

38 to 45 Integer Type Questions (Single Digit Answer) 8 4 0 32

45 153

PT-1

Total Total

Maths/ Physics/

Chemistry

SECTION-1 : (Only One option correct type) [k.Mñ1 : (dsoy ,d lgh fodYi çdkj)

This section contains 25 multiple choice qustions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

bl [k.M esa 25 cgqfodYi ç'u gSA çR;sd ç'u esa pkj fodYi (A), (B), (C) vkSj (D) gS] ftuesa ls dsoy ,d lgh gSA

SCQ._(25)

1. A particle leaves the origin at t = 0 with an initial velocity 0àv 3V i

. It experiences a constant

acceleration 1 2à àa 2a i 5a j

. The time at which the particle reaches its maximum x coordinate in positive direction is :

(V0 , a1 and a2 are positive constants) [RM_AA]

,d d.k t = 0 ij ewy fcUnq ij gS rFkk mldk izkjfEHkd osx 0àv 3V i

gSA ;g ,d fu;r Roj.k

1 2à àa 2a i 5a j

ls xfreku gSA og le; D;k gksxk ftlesa d.k dk x funsZa'kkad /kukRed fn'kk esa vf/kdre~ gks%

(V0 , a1 rFkk a2 /kukRed fu;rakd gSa)

(A) 0

2

3V10a

(B) 0

1 2

3V2a 5a

(C*) 0

1

3V2a

(D) 0

1 2

3V2a 5a

Sol. For maximum x coordinate vf/kdre~ x funsZa'kkad ds fy, 0 = 3V0 ñ 2a1t

t = 0

1

3V2a

2. A particle moves in such a way that its position vector as function of time is given by

2à à àr i 4t j tk

, where r is in (m) and t is in(s). Then the trajectory of the particle is : [RM_DD]

,d d.k bl izdkj ls xfr djrk gS] fd mldk fLFkfr lfn'k le; ij fuEukuqlkj fuHkZj djrk gS 2à à àr i 4t j tk

, tgkW r ehVj esa gS ,oa t lSd.M esa gSA rc d.k dk iFk gksxk : (A*) A parabola (B) An ellipse (C) A circle (D) None



JRPT1130714C0-2

(A*) ,d ijoy; (B) ,d nh?kZ oÙk (C) ,d oÙk (D) buesa ls dksbZ ughaA Sol. x = 1 y = 4t2 z = t y = 4z2 Parabola passing from (1,0,0). ijoy; fcUnq (1,0,0) ls xqtjrk gSA

3. Acceleration of a particle moving in straight line is given by the relation a = ñ4x (ñ1 + 14

x2), where

a is acceleration in m/s2 and x is position in meter. If velocity v = 17 m/s when x = 0, then the velocity of the particle when x = 4 meter is : [RM_AA]

ljy js[kk esa xfreku ,d d.k dk Roj.k laca/k a = ñ4x (ñ1 + 14

x2), ds }kjk ifjHkkf"kr fd;k tkrk gSA tgk¡ a

Roj.k m/s2 esa gS rFkk x ehVj esa fLFkfr gSA ;fn x = 0 ij v = 17 m/s gks rks x = 4 m ij d.k dk osx gksxk : (A) 12 m/s (B*) 15 m/s (C) 20 m/s (D) 25 m/s Sol. a = ñ4x (ñ1 + 0.25 x2)

v 4

3

17 0

vdv (4x x )dx

12

(v2 ñ 172) = 2 40

4 (x )2

ñ 4 40

1 (x )4

v2 ñ172 = 4(42) ñ 41 (4 )2

= 64 ñ 128

V2 = 289 ñ 64 = 225 V = 15 m/sec. 4. The radius of curvature of a convex spher ical mirror is 1.2 m. How far away from the

m irror is an object of height 1.2 cm if the distance between its virtual image and the m irror is 0.35 m? What is the height of the image? [Apply formula for paraxial rays] [GO_SM]

,d mÙky xksyh; niZ.k dh oØrk f=kT;k 1.2 ehVj gSA 1.2 lseh- Å¡pkbZ dh ,d oLrq niZ.k ls fdruh nwjh ij

fLFkr gS ;fn blds vkHkklh çfrfcEc o niZ.k ds chp dh nwjh 0.35 ehVj gS ? çfrfcEc dh Å¡pkbZ D;k gS ? [v{k ds utnhd fdj.kksa ds fy, lw=k yxkus ij ]

(A*) 84 cm, 0.5 cm (B) 84 cm, 0.25 cm (C) 84 cm, 1.2 cm (D) 80 cm, 0.25 cm Ans. 84 cm, 0.5 cm

Sol.

m = 2

1

h ñv

h u h2 = ñv

u h1 = ñ 0.35 1.2 cm

ñ0.84

= 0.5 cm.

5. A point object is moving along principal axis of a concave mirror with uniform velocity towards pole.

Initially the object is at infinite distance from pole on right side of the mirror as shown. Before the object collides with mirror, the number of times at which the distance between object and its image is 80 cm are. [GO_SM]

,d fcUnq fcEc ,d vory ni Z.k dh e q[; v{k d s vuqfn'k /k z qo dh vk sj ,dleku o sx ls xfr dj jgk gSA i zkjEHk es a fcEc ni Z.k d s nk a;h vk sj /k z qo l s vuUr nwjh ij gS tSlk fd fn[kk;k x;k g SA fcEc ds niZ.k l s Vdjkus ls igys ] fdrus ckj fcEc rFkk blds iz frfcEc ds chp dh n wjh 80 cm g SA



JRPT1130714C0-3

(A*) more than two t imes (B) less than two t imes (C) two t imes (D) Data insuff icient (A) nk s ckj ls T;knk (B) nk s ckj ls de (C) nk s ckj (D) vk ¡dM+ s vi;k ZIr Sol. As the object moves from infinity to centre of curvature, the distance between object

and image reduces from infinity to zero. As the object moves from centre of curvature to focus, the distance between object

and image increases from zero to inf in ity. As the object moves from focus to po le, the distance between object and its image

reduces from infinity to zero. Hence the distance between object and its im age shall be 40 cm three t imes.

Sol. tc fcEc vuUr ls oØrk dsUnz rd xfr djrk g S] fcEc rFkk i zfrfcEc ds chp dh nwjh vuUr ls ' k wU; rd ?kV tkrh gSA

tc fcEc oØrk dsUnz ls Qksdl dh vksj xfr djrk gS] fcEc rFkk blds izfrfcEc ds chp dh nwjh 'kwU; ls vuUr rd c<+rh gSA

tgk a fcEc Qk sdl ls /k z qo dh vk sj xfr djrk gS ] fcEc vk Sj blds i zfrfcEc ds chp dh n wjh vuUr ls 'k w U; rd ?kVrh gSA blfy, fcEc rFkk blds i zfrfcEc ds chp dh nwjh 40 cm rhu ckj gk sxhA

6. A particle is moving with velocity 5 m/s towards east and its velocity changes to 5 m/s north in 10

sec. Find the average acceleration in 10 seconds. [RM_AV] ,d d.k 5 m/s ds osx ls iwoZ dh vksj xfr dj jgk gS rFkk 10 sec esa bldk osx ifjofrZr gksdj 5 m/s mÙkj

dh vksj gks tkrk gSA 10 lSd.M esa d.k dk vkSlr Roj.k Kkr djksA

(A) 2 N W (B*) 1 N W2

(C) 1 N E2

(D) 2 N ñ E

Sol.

a

= f i

t

V ñ V

a =

à à5 j ñ 5i

10



JRPT1130714C0-4

| a

| = 5 210

= 12

m/s2

Direction north west fn'kk mÙkj if'pe 7. In the figure shown find the total magnification of the image formed after two successive reflections

first on M1 and then on M2 . (Assume paraxial rays only) [RM_SM]

n'kkZ;s x;s fp=k esa igys M1 o fQj M2 ij gksus okys nks mÙkjksÙkj ijkorZuksa ds ckn cus izfrfcEcksa dk dqy vko/kZu Kkr dhft,A (dsoy lek+{kh; fdj.ksa ekusa)

(A) + 6 (B*) ñ 6 (C) + 3 (D) ñ3

Sol. For M1 ds fy, : v1 = uf (ñ30) (ñ20)

u ñ f (ñ30) ñ (ñ20)

= ñ 60

M = ñ 1vu

= ñ 2.

For M2 ds fy, : u = + 20. f = 30

1v

+ 120

= 130

v = ñ60

m2 = ñ 6020

= 3 m = m1 ◊ m2 = ñ 6

8. A body moves with constant velocity of 50 m/sec from the point (3, 4) m in a direction of à à3 i 4 j .

The position vector of the body at t = 3 sec. is : [RM_DD] ,d oLrq fu;r osx 50 m/sec ls xfreku gS] ;g fcUnq (3, 4) m ls izkjEHk gksdj fn'kk à à3 i 4 j esa xfreku gSA

t = 3 sec. ds i'pkr~ oLrq dk fLFkfr lfn'k D;k gksxk : (A) ( à à51i 104 j )m (B) ( à à5i 117 j )m (C*) ( à à93i 124 j )m (D) ( à à2i 10 j )m Sol. f is r r vt

fà àr (3i 4 j)

= 50à à(3i 4 j)

5

◊ 3 fà àr 93i 124 j

9. From a building two balls A and B are thrown such that A is thrown upwards and B downwards

(both vertically with same speed). If VA and VB are their respective velocities on reaching the ground, then [RM_AA]



JRPT1130714C0-5

(1) vB> vA (2*) vA =vB (3) vA> vB (4) their velocities depends on their masses ,d bekjr ls nks xsansa A o B ,d leku pky ls Qsadh tkrh gSa, A dks Å/okZ/kj Åij dh vksj o B dks Å/okZ/kj

uhps dh vksj Qsadrs gSaA ;fn Hkwfe ij igqWpus ij nksuksa ds osx Øe'k% VA o VB gSa, rks % (1) vB> vA (2*) vA =vB (3) vA> vB (4) muds osx nzO;eku ij fuHkZj djrs gSa 10. The speed of a particle moving along straight line becomes half after every next second. The initial

speed is v0. The total distance travelled by the particle will be - [RM_DD] ,d lh/kh js[kk esa xfr'khy ,d d.k dh pky çR;sd vxys lSd.M ds ckn vk/kh gks tkrh gSA çkjfEHkd pky v0

gSA d.k }kjk r; dh xbZ dqy nwjh gksxh & [P&R] (A) v0 (B*) 2v0 (C) (D) None buesa ls dksbZ

ughaA

Sol. Distance nwjh = v0 ◊ 1 + 0v

2 ◊ 1 +

0v4

◊ 1 + ............

= v0

1 11 ..............2 4

= 2v0

11. A particle moves from xA = 0.5 m to positon xB = ñ1.5 m in 2s, then in another 4s it moves from xB =

ñ1.5 m to xC = 2.5 m, then ratio of magnitude of it's average velocity to average speed for whole time. [RM_AV]

,d d.k xA = 0.5 m ls xB = ñ1.5 m rd 2s esa pyrk gS fQj vxys 4s esa xB = ñ1.5 m ls xC = 2.5 m rd pyrk gS rks d.k ds vkSlr osx ds eku ,oa vkSlr pky ds eku dk vuqikr (iwjs le;kUrjky esa) D;k gksxk &

(A) 16

(B*) 13

(C) 12

(D) 23

Sol.

speed

avg

avg

2V66V6

ratio = 13

12. I t is found that all electromagnetic signals sent from A towards B reach point C. The

speed of electromagnetic s ignals in g lass can not be : [GO_RP] ;g ik;k tkrk g S fd fo|qr pqEcdh; lan s' k tk s fd fcUnq A ls B dh rjQ Hk st s x; s gS a ] fcUn q

Cij igq ¡pr s g S a rk s dk ¡p e s a fo|qr pqEcdh; lan s'kk s a dh pky ugh a gk s ldrh & C

B

A

dkap

fuokZr

(A) 1.0 ◊ 10

8 m/s (B*) 2.4 ◊ 108 m/s

(C) 2 ◊ 107 m/s (D) 1.4 ◊ 10

7 m/s Sol. This is a case of tota l internal ref lection. ;g iw.kZ vkUrfjd ijkorZu dh fLFkfr gSa



JRPT1130714C0-6

C

B

A

dkap

fuokZr

C

B

A

glass

Vaccum

> C (= s inñ1 1

)

1

< sin

1

< sin 45∫

µ > 1/s in 45∫ µ > 2

v = c

v < c2

= 83 10

2

v < 2.1 ◊ 108

only (B) is not possible. dsoy (B) laHko ugha gSA 13. The reflecting surface of a plane mirror is vertical. A particle is projected in a vertical plane which

is also perpendicular to the mirror. The initial speed of the particle is 10 m/s and the angle of projection is 60∞ from the normal of the mirror. The point of projection is at a distance 5m from the

mirror. The particle moves towards the mirror. Just before the particle touches the mirror, the velocity of approach (the rate at which separation is decreasing) of the particle and the image is ;

lery niZ.k dh ijkorZd lrg Å/okZ/kj gSA ,d d.k Å/okZ/kj ry esa iz{ksfir fd;k tkrk gS rFkk ;g Å/okZ/kj ry niZ.k ds yEcor~ gSA d.k dh izkjfEHkd pky 10 m/s rFkk niZ.k ds vfHkyEc ds lkFk iz{ksi.k dks.k 60∫ gSA iz{ksi.k fcUnq niZ.k ls 5m dh nwjh ij gSA d.k niZ.k dh rjQ xfr djrk gSA d.k ds niZ.k dks Li'kZ djus ds Bhd igys d.k rFkk izfrfcEc dk lkehI; osx ¼vFkkZr~ og nj ftlls d.k rFkk izfrfcEc ds e/; nwjh de gks jgh gS½ gksxk ; [GO_PM]

(A*) 10 m/s (B) 5 m/s (C) 10 3m / s (D) 5 3m / s Sol.

(Vi,m)x = ñ (Vo,m)x (Vi ñ Vm)x = ñ (Vo ñ Vm)x Vix ñ O = Vox Vix = ñ 5m/s Vapp = 5 ñ (ñ5) = 10 m/s



JRPT1130714C0-7

14. Given that, velocity of light in quartz = 1.5 108 m/s and velocity of light in glycerine = (9/4) 108 m/s. Now a slab made of quartz is placed in glycerine as shown. The shift in the position of the object produced by slab is [GO_RP]

DokVZt esa izdk'k dk osx = 1.5 108 eh@ls- ,ao fXyljhu esa izdk'k dk osx = (9/4) 108 eh@ls- fn;k x;k gSA vc fp=kkuqlkj] DokV~Zt dh ,d iV~Vh fXyljhu esa j[kh tkrh gSA iV~Vh }kjk mRiUu oLrq dh fLFkfr esa foLFkkiu gSA

(A*) 6 cm (B) 3.55 cm (C) 9 cm (D) 2 cm

Sol. n qua r tzDokVZt = 2 ; ng ly ce r ine fXyljhu = 43

quartz

glycerine

nn

DokVZt

fXyljhu

= 24 / 3

= 32

= µ r e l

shift foLFkkiu = t rel

11

= 18 113 / 2

= 6 cm

15. The figure shows a parallel slab of refractive index n2 which is surrounded by media of refractive

indices n1 and n3. Light is incident on the slab at angle of incidence ( 0). The time taken by the ray to cross the slab is ët1í if incidence is from ën1í and it is ët2í if the incidence is from ën3í. Then

assuming that n2 > n1, n2 > n3 and n3 > n1, value of t1/t2 is [GO_RP]

fp=k esa ,d lekUrj ifêdk ftldk viorZuk ad n2 gS] nks ek/;eksa ls f?kjh gqbZ gS ftlesa ,d dk viorZuk ad n1 vkSj nwljs dk n3 gSA izdk'k ( 0) vkiru dks.k ij ifêdk ij vkifrr gksrk gSA ifêdk ls xqtjus esa izdk'k fdj.k }kjk fy;k x;k le; ët1í gS vxj vkiru ën1í ls gk srk gS vkSj ;g ët2í gS vxj vkiru ën3í ls g k srk gS rk s ;g ekfu, fd n2 > n1, n2 > n3 vkSj n3 > n1, t1/t2 dk eku gksxk

(A) = 1 (B) > 1 (C*) < 1 (D) cannot be decided

fu.k Z; ugh dj ldrs



JRPT1130714C0-8

Sol. Time taken by ray to cross the slab. ifêdk dks ikj djus esa izdk'k fdj.k }kjk fy;k x;k le;

t = (d / cosr)(V / )

= dV cosr = 2

dsin iV cos r

t 21

cos r

Given fn;k gS n3 > n1

2 2

1 3

n nn n

r1 < r2 (r1 and r2 are angles of refraction in first and second case respectively) r1 < r2 (izFke rFkk f}rh; fLFkfr esa r1 rFkk r2 Øe'k% viorZu dks.k gS ) cos r1 > cos r2. t1 < t2

1

2

t 1t

16. A prism having refractive index 2 and refracting angle 30∫, has one of the

refracting surfaces pol ished. A beam of light incident on the other refracting surface wi ll retrace its path if the angle of inc idence is: [GO_PR]

30∫ viorZd dk s.k o 2 viorZuk ad okys ,d fi zTe dk dk sb Z ,d viorZd i "B ik Wfy'k fd;k x;k g SA n wlj s viorZd i "B ij vkifrr ,d izdk'k iq at blds iFk dk s iqu% vuqj s f[kr ( retrace) djsxk ;fn vkiru dk s.k g S %

(A) 0∫ (B) 30∫ (C*) 45∫ (D) 60∫ Sol.

sin i 2sin30∫

sin i = 1 122 2

i = 45∫ .



JRPT1130714C0-9

17. In the f igure ABC is the cross section of a right angled prism and BCDE is the cross section of a glass s lab. The possible value of so that l ight incident normally on the face AB does not cross the face BC is (given s inñ1 (3/5) = 37∫) [GO_PR]

fp=k esa ABC ledks.k fçTe dk vuqizLFk dkV rFkk BCDE Xykl ifêdk dk vuqizLFk dkV gSA dk eku ftlds fy, fdj.k lrg AB ij yEcor~ vkifrr gks ijUrq lrg BC ls ikj u dj ldsA ( fn;k gS sinñ1 (3/5) = 37∫)

(A) = 45∫ (B*) = 30∫ (C) = 60∫ (D) = 53∫

Sol. A = 90 ñ r2 = A = 90 ñ > C

cos > s inC = 6 / 53 / 2

= 45

< cos ñ1 45

= 37∞ .

18. An insect at point ëPí sees its two images in the water mirror system as shown in the figure. One

image is formed due to direct reflection from water surface and the other image is formed due to refraction, reflection & again refraction by water mirror system in order. The separation between the two images is. (Mirror has focal length 60 cm.) (nw = 4/3) [GO_RP]

fcUnq ëPí ij ,d dhM+k fp=k esa iznf'kZr ikuh niZ.k fudk; esa vius nks izfrfcEc ns[krk gSA ,d izfrfcEc ikuh dh lrg ls

lh/ks ijkorZu ds }kjk curk gS rFkk nwljk izfrfcEc ikuh niZ.k fudk; }kjk Øe'k% viorZu] ijkorZu rFkk iqu% viorZu ds Øe esa curk gSA nksuksa izfrfcEcksa ds e/; nwjh Kkr djksA niZ.k dh Qksdl nwjh 60 cm gSA (nw = 4/3)

(A) 12 cm (B*) 24 cm (C) 36 cm (D) 48 cm Sol. Due to reflection at plane water surface and other image formation is shown in the figure.



JRPT1130714C0-10

ikuh dh lrg ij ijkorZu ds dkj.k vU; i zfrfcEc fp=kku qlkj curk gSA

due to refraction at water d1 = 121

4 /3

= 16 cm. for m, P1 is an object.

for this 1v

+ 1ñ40

= 160

v = 24 cm this is at P2.

It will act as object for water surface which makes image at P.

d2 = 24 244 /3 = 36 cm

final images are P and P

distance P P = 36 ñ 12 = 24 cm. Ans.

i kuh ij viroZu ds dkj.k d1 = 121

4 /3

= 16 cm. m ds fy, , P1 oLrq g S

blls 1v

+ 1ñ40

= 160

v = 24 cm ;g P2 ij gSA

;g ikuh lrg ds fy, oLrq dh rjg dk;Z dj sxh rFkk P ij iz frfcEc curk g S

d2 = 24 244 / 3 = 36 cm

vfUre iz frfcEc P rFkk Pij curk g SA

n wjh P P = 36 ñ 12 = 24 cm. Ans. 19. A uniform, horizontal paralle l beam of light is incident upon a pr ism. The prism is in

the shape of a quarter cyl inder, of radius R = 5 cm, and has the index of refraction n = 5/3. The shortest d istance of the point of convergence at which the incident rays after normal incidence on plane surface and subsequent refraction at curved surface in tersect on x ax is from O as shown in f igure is : (Neglect the ray which travels a long x-ax is as shown in f igure) [GO_RP]

,d ,dleku izdk'k dk {kSfrt lekUrj fdj.k iqat fizTe ij vkifrr gksrk gS tSlk fd fn[kk;k x;k gSA fizTe fp=kkuqlkj pkSFkkbZ csyu ds vkdkj dk gS] ftldh f=kT;k R = 5 cm rFkk viorZukad n = 5/3 gSA lery lrg ls vfHkyEcor~ vkiru rFkk fQj oØh; lrg ij viorZu ds i'pkr~ x v{k ij fdj.ksa ftl fcUnq ij vfHklkfjr gksrh gS] ml fcUnq dh O fcUnq ls y?kqÙke nwjh gksxhA ¼og fdj.k tks x-v{k ds vuqfn'k gS mls NksM+ nhft, tSlk fd fp=k esa fn[kk;k x;k gS½

(A) 4 cm (B) 5/4 cm (C) 9/4 cm (D*) 25/4cm



JRPT1130714C0-11

Sol.

The crit ical angle for a ir-glass interface is = sinñ1 35

= 37∞ .

The rays above the ray incident on curved surface at i = 37∞ shall suffer TIR and need not be considered. The ray incident as curved surface at = 37∞ after refraction intersect curved surface at (B) a distance x2 from O. In PCB

2

31 x

= tan53∞ =

43

or x2 = 54

the require distance OB = 554

= 254

gok&dk ap ds vUrjki "B ds fy, Øk a frd dk s.k = 35

sinñ1 = 37∞ ,

i = 37∞ ij oØ lrg ij vkifrr fdj.k d s Åij fdj.kk s a dk i w.k Z vkUrfjd ijkor Zu (TIR) gk sxk rFkk mUgs a y su s dh vko';drk ugh a g SA = 37∞ ij oØ lrg ij vkifrr fdj.k viorZu ds ckn oØ lrg ij (B) ij O ls x2 nwjh ij iz frPNsn djrh g SA PCB e s a ]

2

31 x

= tan53∞ =

43

or x2 = 54

vko';d pk SM +kb Z x = x 1 ñ x2 = 15 52 4 = 25

4 cm

20. In the shown figure find the value of so that final image is 6 cm behind the silivered face : (Consider only near normal incidence) [GO_RP]

iznf'kZr fp=k esa dk eku Kkr dhft,] rkfd vfUre izfrfcEc ikWfy'k dh xbZ lrg ls 6 cm ihNs cus: (fdj.ksa dsoy vfHkyEc vkiru ds lehi ekfu;saA)

(A) 4 (B*) 2 (C) 6 (D) None dksbZ ugha

Sol.

6 2 2 8

= 2



JRPT1130714C0-12

21. A small rod ABC is put in water mak ing an angle 6∞ with vert ical. If it is viewed

(paraxially) from above, it wi ll look l ike bent shaped ABC'. The angle of bending

( CBC' ) will be in degree ... .. . . .. . w4n3

[GO_RP]

,d Nk sVh NM+ ABC dks m/ok Z/kj ls 6∞ ds dk s.k ij ikuh es a j[kk x;k g SA ;fn bls bldh v{k d s utnhd fdj.kk s a (paraxially) ls Åij ls n s[kk tkrk g S] rk s ;g eqM + s vkdkj ABC' tSlh

i zrhr gk srh gSA eqM +u s dk dk s.k ( CBC' ) fMxzh es a gk sxk .. . .. .. . w4n3

C

C'B

A6∞

(A*) 2∞ (B) 3∞ (C) 4∞ (D) 4.5∞

Sol. = xOC

= xOC'

= OC 'OC

= 1

C

C'B

A

6∞

x

= µ = 43

(6∞) = 8∞

bending angle = ñ = 2∞

Sol. = xOC

= xOC'

= OC 'OC

= 1

C

C'B

A

6∞

x

= µ = 4

3

(6∞) = 8∞

?k qeko dk s.k = ñ = 2∞ 22. A spher ical surface of radius of curvature R separates air (refractive index 1.0) from

glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in a ir is found to have a rea l image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to: [GO_RS]



JRPT1130714C0-13

R oØrk f=kT;k dk xksyh; i"B gok (viorZukad = 1.0) dks dk¡p (viorZukad = 1.5) ls iFkd djrk gSA oØrk dsUnz dk¡p esa fLFkr gSA gok esa fLFkr ,d fcUnqor~ oLrq P dk okLrfod çfrfcEc Q, dk¡p esa curk gSA js[kk PQ i"B dks O ij dkVrh gS rFkk PO = OQA nwjh PO cjkcj gS :

(A*) 5 R (B) 3 R (C) 2 R (D) 1.5 R Sol. Let us say ekuk PO = OQ = X

Applying 2 1

v u

= 2 1

R i z; qDr djus ij

Substituting the values with sign

mi;qDr fpUg ds lkFk eku j[kus ij 1.5 1.0X X

= 1.5 1.0R

(Distances are measured from O and are taken as positive in the direction of ray of light) (nwfj;k¡ O ls ekih xbZ gS o izdk'k lapj.k dh fn'kk esa /kukRed gSA½

2.5X

= 0.5R

X = 5 R 23. A prism of refractive index 2 has refracting angle 60∫. Angle of maximum deviation is : ,d 60∫ viorZu dks.k okys fizT+e dk viorZukad 2 gSA egÙke fopyu dks.k gS % [GO_PR] (A) 450 (B) sin-1 ( 2 sin15∫ ) (C*) 30∫ + sin-1 ( 2 sin15∫ ) (D) none dksbZ ugha Sol. for = max ds fy,

e = 90∫ r2 = sinñ1 1

r2 = sinñ1 12

= 45∫ r1 = A ñ r2 = 15∫

sini 2sin15∫

sin i = 2 sin 15∫ i = sinñ1 ( 2 sin 15∫) max = i + e ñ A = 30∫ + sin

ñ1 ( 2 sin 15∫) 24. A balloon is moving upwards with velocity 10 msñ1. It releases a stone which comes down to the

ground in 11 s. The height of the balloon from the ground at the moment when the stone was dropped is : [RM_DD]

,d xqCckjk 10 eh0@lS0 osx ls tehu ls Åij dh vksj tk jgk gSA blls ,d iRFkj fxjk;k tkrk gS ] tks tehu ij 11 lSd.M esa igq¡prk gSA tc iRFkj fxjk;k x;k Fkk] ml le; xqCckjs dh tehu ls Åpk¡bZ D;k Fkh\

(A*) 495 m (B) 592 m (C) 460 m (D) 500 m



JRPT1130714C0-14

Sol.

A

B

10 m/s

t = 11 secH

As pwafd s = ut + 12

at2

ñ H = 10 ◊ 11 ñ 5 ◊ (11)2

ñ H = 110 ñ 605 H = 495 m Aliter : oSdfYid fof/k

At the time of release, velocity of stone will be same as that of balloon, hence eqDr djus ds le; iRFkj dk osx xqCckjs ds osx ds rqY; gksxk vr% u = ñ 10 msñ1 , t = 11 s xqCckjs dh Å¡pkbZ gksxh

h = ut + 12

gt2

= (ñ 10) ◊ 11 + (10) (11)2 = ñ 110 + 605 = 495 m ìAî Ans

25. A ray of light travelling in the direction 1à ài 3 j

2 is incident on a plane mirror. After reflection, it

travels along the direction 1à ài 3 j

2 . The angle of incidence is : [GO_PM]

,d lery niZ.k ij vkifrr çdk'k fdj.k dh çxkeh fn'kk 1à ài 3 j

2 gSA ijkorZu ds ckn çxkeh fn'kk

1à ài 3 j

2 gks tkrh gSA fdj.k dk vkiru dks.k gS :

(A*) 30∫ (B) 45∫ (C) 60∫ (D) 75∫



JRPT1130714C0-15

SECTION-2 : (One or more option correct type) [k.Mñ2 : (,d ;k vf/kd lgh fodYi çdkj)

This section contains 5 multiple choice qustions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

bl [k.M esa 5 cgqfodYi ç'u gSA çR;sd ç'u esa pkj fodYi (A), (B), (C) vkSj (D) gS] ftuesa ls dsoy ,d ;k vf/kd lgh gSA

MCQ._(5) 26. W hich of the fol lowing statements are incorrect for spher ical mirrors. fuEu esa ls dkSulk dFku xksyh; niZ.k ds fy, xyr gS & [GO_SM] (A*) a concave mirror forms only virtua l images for any position of real object ,d vory niZ.k okLrfod oLrq dh fdlh Hkh fLFkfr ds fy, dsoy vkHkklh çfrfcEc cukrk gSA (B) a convex mirror forms only virtual images for any posit ion of a real object ,d mÙky niZ.k okLrfod oLrq dh fdlh Hkh fLFkfr ds fy, dsoy vkHkklh çfrfcEc cukrk gSA (C*) a concave mirror forms only a vir tual diminished image of an object placed

between its pole and the focus ,d vory niZ.k] /kzqo o Qksdl ds chp j[kh oLrq dk dsoy vkHkklh o NksVk çfrfcEc cukrk gSA (D*) a convex mirror forms a virtual enlarged image of an object if it l ies between its

pole and the focus. ;fn oLrq /kzqo rFkk Qksdl ds e/; gks rks mÙky niZ.k vkHkklh rFkk vkof/kZr çfrfcEc cukrk gSA Sol. (A) No, when object is between infinite and focus ,image is real. ugh, tc oLrq vuUr rFkk Qk sdl ds e/; g S] i z frfcEc okLrfod g S (C) when object is between pole and focus, image is magnified. tc oLrq /k z qo rFkk Qk sdl ds e/; g S ] iz frfcEc vkof/k Zr gk sxk (D) when object is between pole and focus image formed by convex mirror is real. tc oLrq /k z qo rFkk Qk sdl ds e/; g S ] mÙky niZ.k }kjk cuk i zfrfcEc okLrfod gk sxk 27. In the shown figure two long plane mirror are kept parallel to each other with their reflecting side

facing one another. A point object 'O' is situated at the distance of 20 cm from A then : [GO_PM] iznf'kZr fp=k esa nks yEcs lery niZ.k ,d nwljs ds lekUrj j[ks gq, gSA rFkk muds ijkorZd i"B ,d nwljs ds

vkeus lkeus gSA ,d fcUnq oLrq 'O'] A ls 20 cm dh nwjh ij fLFkr gS] rc %

(A*) distance of first image formed in A is 50 cm from B.

(B*) distance between fifth image formed in the two mirrors is 3 m (C) distance between fifth image formed in the two mirrors is 240 cm



JRPT1130714C0-16

(D*) total of infinite images are formed in the two mirrors (A*) A esa cus izFke izfrfcEc dh B ls nwjh 50 cm gksxhA

(B*) nksuksa niZ.kksa esa cus ik¡posa izfrfcEc ds e/; nwjh 3 m gksxhA (C) nksuksa niZ.kksa esa cus ik¡posa izfrfcEc ds e/; nwjh 240 cm gksxhA (D*) nksuksa niZ.kksa ds e/; vuUr izfrfcEc cusxsaA

Sol.

28. Two plane mirrors are inclined to each other with their reflecting faces making acute angle. A light ray is incident on one plane mirror. The total deviation after two successive reflections is:

[GO_PM] nks lery niZ.k vius ijkorhZ i"Bksa ds lkFk U;wu dks.k (acute angle) cukrs gq, ,d nwljs ds lkFk >qds gq,

gSA ,d çdk'k fdj.k fdlh ,d lery niZ.k ij vkifrr gksrh gSA nks mÙkjksÙkj ijkorZuksa ds ckn dqy fopyu& (A*) independent of the initial angle of incidence çkjfEHkd vkiru dks.k ij fuHkZj ugha djrk gSA (B) independent of the angle between the mirrors niZ.kksa ds chp dks.k ij fuHkZj ugha djrk gSA (C) dependent on the initial angle of incidence çkjfEHkd vkiru dks.k ij fuHkZj djrk gSA (D*) dependent on the angle between the mirrors. niZ.kksa ds chp dks.k ij fuHkZj djrk gSA Sol.

i r2 2

i + r = .. . .. . .. .( i) = 2i + 2r = 2 (Anticlockwise okekorZ) 29. Two plane mirrors are inclined to each other at 300. A ray is incident on M1 at angle of incidence

40∫. Find deviation produced in it by three successive reflections due to mirrors. [GO_PM] nks lery niZ.k ijLij 300 >qdko ij gSaA ,d fdj.k M1 ij 40∫ ij vkifrr gksrh gSA niZ.kkssa ls rhu Øekxr

ijkorZuksa ds ckn fdj.k dk fopyu dks.k D;k gksxkA

(A*) 160∞ Clockwise (B) 200∞ Clockwise

(C*) 200∞Anticlockwise (D) 160∞ Anticlockwise



JRPT1130714C0-17

Sol.

So = 160∫ clockwise = (360∫ ñ 160∫) Anticlockwise = 200∫ Anticlockwise 30. Position time graph for a particle moving on straight line is shown in figure. Assume same slope of

xñt graph at t = 0 and t = 20 s. Select correct alternative/s : [RM_AA] ljy js[kk esa xfr dj jgs ,d d.k ds fy, fLFkfr le;&xzkQ fp=kkuqlkj gSA ;g ekfu, fd t = 0 rFkk t = 20 s

ij xñt xzkQ dk <ky leku gSA lgh fodYi@fodYiksa dk p;u dhft, :

(A*) Average velocity of particle from t = 0 to t = 20 sec. is zero. (B*) Acceleration of particle from t = 10 sec. to t = 15 sec. is positive. (C*) Average acceleration of particle from t = 0 to t = 20 is zero. (D) Average velocity & instantaneous velocity becomes same in magnitude and direction more

than once from t = 0 to t = 20 sec. (A*) t = 0 ls t = 20 sec esa d.k dk vkSlr osx 'kwU; gksxkA (B*) t = 10 sec ls t = 15 sec esa d.k dk Roj.k /kukRed gksxkA (C*) t = 0 ls t = 20 sec esa d.k dk vkSlr Roj.k 'kwU; gksxkA (D) t = 0 ls t = 20 sec ds nkSjku d.k dk vkSlr osx rFkk rkR{kf.kd osx ifjek.k rFkk fn'kk esa ,d ls vf/kd

ckj leku gks tk,xkA



JRPT1130714C0-18

SECTION ñ 3 : (Paragraph Type) [k.M ñ 3 : (vuqPNsn çdkj)

This section contains 3 paragraphs each describing theory, experiment, data etc. Six questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D).

bl [k.M esa fl)karksa] ç;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 3 vuqPNsn gSA pkjksa vuqPNsnksa ls lacaf/kr N% ç'u gSa] ftuesa ls gj vuqPNsn ij nks ç'u gSaA fdlh Hkh vuqPNsn esa gj ç'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls dsoy ,d gh lgh gSA

Comp._(3)_(2 Ques.)

Paragraph for Questions 31 and 32 iz'u 31 vkSj 32 ds fy, vuqPNsn

COMPREHENSION # 1 A particle moves in x-y plane with constant acceleration à àa 6i 8 j

( in m/s2). At time t = 0, the

particle is at point having coordinates (0, 20 metre) and its initial velocity is à àu 12i 16 j

(in m/s).

,d d.k x-y ry es a fu;r Roj.k à àa 6i 8 j

(m/s2 es a) ls xfreku g SA t = 0 le; ij d.k ds

funs Z'kk ad (0, 20 metre) vk Sj iz kjfEHkd o sx à àu 12i 16 j

(m/s es a) g SA [RM_AA] 31. The instant of time when speed of the particle is zero will be : fdl le; ij d.k dh pky 'k w U; gk sxh & (A) 1 sec (B*) 2 sec (C) 3 sec (D) 4 sec Sol. The velocity vector at time t is v u at

= à à à à12i 16 j (6 i 8 j)t .... (1) solving for v 0

we get t = 2 sec. Ans. 32. The speed of the particle at the instant, position vector and velocity vector of the particle are

mutually perpendicular will be : ftl le; ij d.k dk fLFkfr lfn'k vk Sj o sx lfn'k ijLij yEcor~ gk srs gS] ml le; d.k dh

pky gk sxh & (A) 4 5 m/s (B) 20 m/s (C*) 12 5 m/s (D) None of these bues a

l s dk sb Z ugh a Sol. The velocity vector at time t is v u at

= à à à à12i 16 j (6 i 8 j)t .... (1) solving for v 0

we get t = 2 sec. Ans. The position vector at any time t is 2

012

r r ut at

= 212

à à à à( 12i 16 j)t (6i 8 j)t .... (2)

solving for r.v 0

we get t = 83 5 5

sec.

putting t in equation (1) we get | v | 12 5

m/s Ans.



JRPT1130714C0-19

gy% fdlh le; t ij osx lfn'k v u at

= à à à à12i 16 j (6 i 8 j)t .... (1) v 0

ds fy, gy djus ij ge iz kIr djrs gS t = 2 sec. Ans. fdlh le; t ij fLFkfr lfn'k

20

12

r r ut at

= 212

à à à à( 12i 16 j)t (6i 8 j)t .... (2)

r.v 0

ds fy, gy djus ij t = 83 5 5

sec.

lehdj.k (1) es a t dk eku j[ku s ij ge iz kIr djrs gSA | v | 12 5

m/s Ans. vr% le; ds Qyu ds :i es a x-funs Z'kk ad

x = x0 + ux t + 12

axt2 = ñ 12 t + 3t2

x = 0 ds fy, gy djus ij t = 5 sec. Ans.

Paragraph for Questions 33 and 34 iz'u 33 vkSj 34 ds fy, vuqPNsn

COMPREHENSION # 2 An object O is placed 10 cm above the face AB of a homogenuous glass slab of refractive index

= 32

with lower face CD silivered as shown then answer following questions

(assume only paraxial rays) : [GO_RP]

,d oLrq O] = 32viorZukad dh ,d lekax dk¡p dh iêhdk dh lrg AB ls 10 cm Åij fLFkr gSA

fupyh lrg CD dks fp=kkuqlkj ikWfy'k fd;k x;k gSA rc fuEu iz'uksa ds mÙkj nhft,A (lekf{k; fdj.ksa ekfu;saA)

D

33. Find the distance from CD of the image formed by the reflection of light from face CD : (A) 11 cm (B*) 18 cm (C) 14 cm (D) None lrg CD izdk'k ds ijkorZu }kjk cus izfrfcEc dh CD ls nwjh Kkr dhft,A (A) 11 cm (B*) 18 cm (C) 14 cm (D) dksbZ ugha 34. Find the distance of the final image formed by the slab from CD : (A*) 11 cm (B) 18 cm (C) 14 cm (D) None lrg CD ls iêhdk }kjk cus vfUre izfrfcEc dh nwjh Kkr dhft,A (A*) 11 cm (B) 18 cm (C) 14 cm (D) dksbZ ugha



JRPT1130714C0-20

Paragraph for Questions 35 and 36

iz'u 35 vkSj 36 ds fy, vuqPNsn COMPREHENSION # 3

A concave mirror of radius of curvature 20 cm is shown in the figure. A circular disc of diameter 1 cm is placed on the principal axis of mirror with its plane perpendicular to the principal axis at a distance 15 cm from the pole of the mirror. The radius of disc starts increasing according to the law r = (0.5 + 0.1 t) cm/sec where t is time is second. [GO_SM]

,d vory niZ.k dh oØrk f=kT;k fp=kkuqlkj 20 cm gSA ,d o Ùkkdkj pdrh ftldk O;kl 1 cm g S] dk s ni Z.k ds eq[; v{k ij eq[; v{k d s yEcor~ niZ.k ds /k z qo (pole) ls 15 cm dh nwjh ij fp=kku qlkj j[kk tkrk gSA vc pdrh dh f=kT;k fu;e r = (0.5 + 0.1 t) cm/sec ds vuqlkj c<+uk i zkjEHk djrh g S tgk ¡ t le; lsd.M es a g SA

35. The area of image of the disc at t = 1 second is : pdrh ds i zfrfcEc dk t = 1 lsd.M ij {k s=kQy gk sxk & (A) 1.2 cm2 (B*) 1.44 cm2 (C) 1.52 cm2 (D) none of these bue s a

l s dk sb Z ugh a Sol. (M oderate) At t = 1 sec. r = 0.5 t + 0.1 t = 0.6 cm

m = ff u

= 1010 15

= ñ 2

Radius of image = 2r = 1.2 cm Area of image = (1.2)2 = 1.44 cm2 . i zfrfcEc dh f=kT;k = 2r = 1.2 cm i zfrfcEc dk {k s=kQy = (1.2)2 = 1.44 cm2 . 36. What will be the rate at which the radius of image will be changing (A*) 0.2 cm/sec increasing (B) 0.2 cm/sec decreasing (C) 0.4 cm/sec increasing (D) 0.4 cm/sec decreasing i zfrfcEc dh f=kT;k ds ifjorZu dh nj D;k gk sxh & (A*) 0.2 cm/sec c<+rh gqb Z (B) 0.2 cm/sec ?kVrh g qb Z (C) 0.4 cm/sec c<+rh gqb Z (D) 0.4 cm/sec ?kVrh gqb Z

Sol. (E as y) drdt

= 0.1

rimage = |m|robject = 2robject

imagedrdt

= 2. drdt

= 0.2 = 0.2 cm/sec.



JRPT1130714C0-21

SECTION-4 : (Matrix - Match Type) [k.Mñ4 : (eSfVªDl&lqesy izdkj)

This section contains 1 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example. If the correct matches are A-p, s and t ; B-q and r; C-p and q; and D-s and t; then the correct darkening of bubbles will look like the following :

bl [k.M esa 1 iz'u gSaA izR;sd iz'u esa nks dkWye esa oäO; (statements) fn;s gq, gSa ftudk lqesy (match) djuk gSA dkWye-I (Column-I) ds oäO;ksa dks A, B, C rFkk D ukfer fd;k x;k gSa tcfd dkWye-II (Column-II) ds oäO;ksa dks p, q, r, s rFkk t ukfer fd;k x;k gSaA dkWye-I (Column-I) esa fn, x, dksbZ ,d oäO;

dkWye-I (Column-II) ds ,d ;k ,d ls vf/kd oäO; ¼oäO;ksa½ ls lgh lqesy

djrk gSA bu iz'uksa ds mÙkj fuEufyf[kr mnkgj.k ds vuqlkj mfpr cqYyksa dks dkyk djds n'kkZuk gSA

;fn lgh leqsy A-p, s vkSj t ; B-q vkSj r; C-p vkSj q; rFkk D-s vkSj t gS; rks lgh fof/k ls dkys fd, x,

cqYys ,sls fn[krs gSa tks fuEufyf[kr gSaA

MTC_(1)_(4 to 5) 37. The veloc ity t ime graph for a part icle moving along a straight l ine is given in each

s ituation of column-I . In the t ime interval 0 < t < , match the graph in column-I with corresponding statements in column-I I . [RM_AA]

lwph-I dh izR;sd fLFkfr esa ,d ljy js[kk ds vuqfn'k xfr djrs gq;s ,d d.k ds fy, osx≤ xzkQ fn;s x;s gSA le;kUrjky 0 < t < esa] lwph-I ds xzkQksa dks lwph-I I esa laxr dFkuksa ls lqesfyr dhft;sA

List List lwph lwph

(A) v

t

(p) speed of part ic le is continously

decreasing. d.k dh pky lrr :i ls ?kV jgh gSA



JRPT1130714C0-22

(B) v

t

(q) magnitude of acceleration of par t ic le is fi rst

decreasing then increasing. d.k ds Roj.k dk ifjek.k igys ?kV jgk gS rFkk fQj c<+ jgk gSA

(C) v

t

(r) speed of the particle continuously increasing

d.k dh pky yxkrkj c<+ jgh gSA

(D) v

t (s) magnitude of acceleration of part ic le does not

change. d.k ds Roj.k dk ifjek.k ifjofrZr ugh gksrk gSA

(t) magnitude of acceleration of particle is first increasing then decreasing.

d.k ds Roj.k dk ifjek.k igys c<+ jgk g S] rFkk fQj ?kV jgk g SA

Ans. (A) ñs ; (B) ñs ; (C) ñ p ; (D) ñ p Sol. In case P and Q acceleration is constant but speed f irst decreases and then

increases. In case R and S, the velocity does not change sign hence direction of acceleration is

constant. Speed and magnitude of acceleration decreases with t ime. fLFkfr P rFkk Q esa Roj.k fu;r gS ysfdu pky igys ?kVrh gS rFkk fQj c<+rh gSA fLFkfr R rFkk S esa osx dk

fpUg ugh cnyrk gS blfy;s Roj.k dh fn'kk fu;r jgrh gSA Roj.k dk ifjek.k rFkk pky le; ds lkFk ?kVrs gSaA



JRPT1130714C0-23

SECTION-5 : (Integer value correct Type) [k.M ñ 5 : (iw.kk±d eku lgh çdkj)

This section contains 7 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive)

bl [k.M esa 7 ç'u gSA çR;sd ç'u dk mÙkj 0 ls 9 rd ¼nksauks 'kkfey½ ds chp dk ,dy vadh; iw.kkZad gSA

Integer_(8)_(Single Digit) 38. A small piece of wood is floating on the surface of a m deep lake. When the sun is vertically

above the piece its shadow is formed at A. When the sun is just setting the shadow of the piece is formed at B. If the refractive index of water is 4/3 then find the distance between A and B in metre. [GO_RP]

ehVj xgjh ,d >hy ds i"B ij ydM+h dk ,d NksVk VqdM+k rSj jgk gSA tc lw;Z VqdM+s ds m/okZ/kj Åij gksrk gS rks bldh ijNkbZ A ij curh gSA tc lw;Z Bhd vLr gksus dh fLFkfr esa gksrk gS rc VqdM+s dh ijNkbZ B ij curh gSA ;fn ikuh dk viorZukad 4/3 gks rks A o B ds chp dh nwjh ehVj esa Kkr dhft,A

Ans. 3

sin ic = 1

= 1(4 / 3)

= 34

sin ic = 34

= ABOB

3

7

ic

4

tan ic = ABOA

= 37

AB = 7 ◊

37

= 3m

39. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is

observed to move from 253

m to 507

m in 30 seconds. What is the speed of the object in km per

hour. [GO_SM] 20 m oØrk f=kT;k ds ,d mÙky niZ.k dh vksj çdkf'kd v{k ij tkrs gq;s ,d fcac dk çfrfcac 30 sec. esa

253

m ls 507

m rd f[kldrk gSA fcac dh pky km/hour esa D;k gksxh \



JRPT1130714C0-24

Ans. 3 Sol. R = 20 m, f = 10 m For mirror, niZ.k ds fy,

1 1 1v u f

1

1 1 125 / 3 u 10

1

1 1 3u 10 25 = 1

50 u1 = ñ 50 cm

& rFkk 2

1 150 / 7 u

= 110

2

1u

= 125

u2 = ñ25 cm

So, speed = ut

= 2530

m/sec. 56

= m/sec.

vr% pky = ut

= 2530

m/sec. = 56

m/sec.

& rFkk in km/hr esa = 56

◊

185

= 3 km/hr.

40. On a 2 lane road, car A is travel ling with a speed of 36 km/h. Two cars B and C

approach car A in opposite directions with a speed of 54 km/h each. At a certa in instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. The minimum acceleration of car B is required to avoid an accident is x m/s2 . Therefore, f ind x.

,d nks ysu lM+d ij dkj A 36 km/h dh pky ls py jgh gSA nks dkj B vkSj C, izR;sd 54 km/h dh pky ls dkj A dh vksj foijhr fn'kkvksa ls vk jgh gSA fdlh fuf'pr {k.k ij nwjh AC = AB = 1km gS ,oa B, A dks C ls igys ikj djuk pkgrk gSA B dk U;wure Roj.k x m/s2 gS rkfd nq?kZVuk ls cpk tk ldsA rc x Kkr djksA

Ans. 1

Sol.

B

A

C

54 km/hr54 km/hr

36 km/hr

T ime taken by C to overtake A is

= 1km(36 54)km /h

= 190

hr = 23

◊ 60 sec. = 40 sec.

Let the minimum acceleration of B is a to overtake A before C then C }kjk A ls vkx s fudyus ds fy, fy;k x;k le;

= 1km(36 54)km /h

= 190

hr = 23

◊ 60 sec. = 40 sec.

ekuk B dk U;wure Roj.k a ftlls og A dks s ]C ls igys ikj dj tk;s] rc

S = ut + 12

at2

1 km = (54 ñ 36) ◊

190

+ 190 90 2

◊ a

190 90 2

◊ a = 1 ñ 1890

= 7290



JRPT1130714C0-25

a = 7290

◊ 90 ◊ 90 ◊ 2 km/h2

70 90 2 100060 60 60 60

m/sec2 = 1 m/sec2 .

41. A part icle is projected from ground in ver tical direction at t = 0. At t = 0.8 sec, it

reaches h = 14m. It wi ll again come to same height at t = x2

sec. after the motion

begins, then f ind x : [ g = 10 m/s≤ ] [RM_AA] ,d d.k dk s i Foh ry ls t = 0 ij m/ok Z/kj Åij dh vk sj iz{k sfir fd;k tkrk gSA t = 0.8 sec

ij ;g d.k /kjrh ls h = 14m Åapkb Z ij g ksrk g SA ;g nqckjk i Foh ls mruh Åapkb Z ij t = x2

lSd.M ij vkrk g S] rc x Kkr djk sA [ g = 10 m/s≤ ] Ans. 7 Sol. Tak ing upward direction as posit ive in it ia l velocity can be obtained by II equ. of motion

i.e. s = ut + 1/2 at2 consider ing motion from C to A

14 = u ◊ 0.8 ñ 12◊ 10 ◊ 0.8

2

so, u = 432

m/s

Let velocity magnitude at point A = v so, v = u ñ gt

v = 432

ñ 10 ◊ 0.8 =

272

m/s

Hence t ime taken from A to B i.e. t il l same level = 2vg

= 2.7 s

Hence the time instant at which the part icle comes to same level = 0.8 + 2.7 = 3.5 s Ans. Sol. Åij dh vk sj fn'kk dk s /kukRed ysr s g q, xfr ds n wlj s lehdj.k ls çkjfEHkd o sx u Kkr fd;k tk ldrk g S%

i.e. s = ut + 1/2 at2 C ls A rd dh xfr ds fy,

14 = u ◊ 0.8 ñ 12◊ 10 ◊ 0.8

2

vr%, u = 432

m/s

ekuk fd fcUnq A ij osx dk ifjek.k = v vr%, v = u ñ gt

v = 432

ñ 10 ◊ 0.8 =

272

m/s

vr% A ls B rd fy;k x;k le; vFkk Zr ~ leku ry rd = 2vg

= 2.7 s

vr% og {k.k tc d.k leku Å¡pkb Z ij gk sr k g S = 0.8 + 2.7 = 3.5 s Ans.



JRPT1130714C0-26

42. A man is sitting in a room at 2 m from a wall W1 wants to see the full height of the wall W2 behind him 4 m high and 6 m away from the facing wall W1. What is the minimum vertical length in meter of mirror on the facing wall required for the purpose ? [GO_PM]

,d dejs esa ,d O;fDr nhokj W1 ls 2 m nwjh ij cSBk gSA ;g vius ihNs 4m Å¡ph ,oa lkeus dh nhokj W1 ls 6m nwjh ij fLFkr W2 nhokj dh iwjh Å¡pkbZ ns[kuk pkgrk gSA bl m)s'; ds fy, lkeus dh nhokj W1 ij yxs niZ.k dh Å/okZ/kj yEckbZ de ls de fdruh gksuh pkfg, ?

Ans. 1

Sol.

Let the min. length of mirror be L

L4

= 28

L = 1 m

43. The graph shows the variation of 1V

(where V is the velocity of the particle) with respect to time.

Then find the value of acceleration at t = 3 sec in meter/sec2. [RM_AA]

xzkQ esa 1V

(tgk¡ V d.k dk osx gS) dk le; ds lkFk ifjorZu çnf'kZr gS rks d.k dk Roj.k t = 3 sec ij m/s2

D;k gksxk&

Ans. 3 Sol. At t = 3 ij

Slope <ky =

1dV

dt

= ñ1

ñ2

1V

dVdt

= ñ1

213

dVdt

= 1 a = dVdt

= 3 m/s2

44. There is a small air bubble inside a glass sphere (µ = 1.5) of radius 5 cm. The

bubble is at 'O' at 7.5 cm below the surface of the glass. The sphere is placed inside

water (µ =

43

) such that the top surface of glass is 10 cm below the surface of water .

The bubble is viewed normally from air. The apparent depth of the bubble is 27x

cm,

f ind x. [GO_RS]



JRPT1130714C0-27

5 lseh- f=kT;k ds dk¡p (µ = 1.5) ds ,d xksys ds vUnj gok dk ,d NksVk cqycqyk bl xksys ds i"B ls 7.5

lseh- uhps fcUnq 'O ij fLFkr gSA xksys dks ikuh (µ =

43

) ds vUnj bl çdkj j[kk tkrk gS fd dk¡p dk Åijh

i"B ikuh dh lrg ls 10 lseh- uhps gSA cqycqys dks gok ls vfHkyEcor~ ns[kus ij cqycqys dh vkHkklh xgjkbZ 27x

cm gS] rc x Kkr dhft,A

water ( )ikuh

observer ( )çs{kd

10cm

Cglass( )dk¡p

O

Ans. 2

Sol.

observer ( )çs{kd

10cm

COwater

( )ikuh

glass( )dk¡p

For f irs t refraction, (at the glass-water interface) izFke viorZu ds fy, (dkWp ikuh vUr% lrg ij)

4 / 3 3 / 2v 7.5

= 4 /3 3 / 25

4 33v 15 = 1

30

43v

= 530 v = ñ8 cm

For second refraction : (at a ir-water interface) f}rh; viorZu ds fy, (gok ikuh vUr% lrg ij )

Apparent depth vkHkklh xgjkbZ = (10 8)4 / 3

= 18 34 = 54

4 = 27

2 cm.

45. The position vector of a particle is given as 2 2à àr (t 4t 6) i (t ) j

. The time in second, after which the velocity vector and acceleration vector becomes perpendicular to each other is :

[RM_AA] d.k dk fLFkfr lfn'k 2 2à àr (t 4t 6) i (t ) j

}kjk fn;k tkrk gSA og le; lSd.M esa Kkr dhft,]

ftlds i'pkr~ osx lfn'k rFkk Roj.k lfn'k ,d nwljs ds yEcor~ gksA Ans. 1



JRPT1130714C0-28

Sol. 2 2à àr (t 4t 6) i t j

;

drvdt

= à à(2t 4) i 2t j = (2t ñ 4) + 2t ,

dvà àa 2i 2 j

dt

if a and v

are perpendicular ;fn a

rFkk v ijLij yEcor~ gS

a. v 0

à à à à2i 2 j).((2t 4) i 2t j) 0 8t ñ 8 = 0 t = 1 sec. Ans. t = 1 sec.

Page # 1

DEPARTMENT OF MATHEMATICS TEST PATTERN

BATCH : JR- PT-1 TEST DATE : 10-AUG-2014

SYLLABUS : Fundamentals of Mathematics (Graphs of polynom ials, Wavy Curve I nequalit ies and Modulus, Modulus-equat ions and inequat ions, Graphical Transform at ion,

I r rarat ional ineqaulit y, log equat ions, Logar ithm ic graphs and inequalit ies, GIF- equat ions and inequat ions, GIF-graphs and sgn (Fx) , Trigonom et ric-Equat ions, Trigonom et ric- I nequat ions)


1 to 25 SCQ 25 3 ñ1 75

26 to 30 MCQ 5 4 0 20




45 153

PT-1

Total Total

Maths/ Physics/

Chemistry

1. If sin < 0, cos < ñ 12

and 0 < < 2 , then 2sin3

lies in the interval [BAMS]

;fn sin < 0, cos < ñ 12rFkk 0 < < 2 , rc 2sin

3 vUrjky eas fLFkr gS&

(A) [ñ2, 2] (B*) [ñ2, ñ 3 ) (C) 1ñ1,2

(D) [ñ1, ñ 3 )

Sol. Let ekuk y = 2sin3

sin < 0, cos < ñ 12

in (0, 2 )

< < 43

43

< + 3

< 53

ñ1 sin3

< ñ 32

ñ 2 2sin3

< ñ 3

2. If ñ 4 x < 2, then ||x + 2| ñ 5| lies in the interval : [BAMD] ;fn ñ 4 x < 2, rc ||x + 2| ñ 5| vUrjky esa fLFkr gS&

(A*) (1, 5] (B) [1, 5] (C) (2, 6] (D) [5, 8)

Page # 2

Sol. ñ 4 x < 2 ñ 2 x + 2 < 4 0 |x + 2| < 4 ñ 5 |x + 2| ñ 5 < ñ 1 1 < ||x + 2| ñ 5| 5

3. Number of negative integral solutions of the in-equation 3 2

4 x

(x 1) (x 1)x (x 2)(e e)

0 is

vlfedk 3 2

4 x

(x 1) (x 1)x (x 2)(e e)

0 ds _.kkRed iw.kk±d gyksa dh la[;k gS & [BARI]

(A) 0 (B*) 1 (C) 2 (D) 3

Sol.

negative integral solution is ñ1 only.

Hindi :

_.kkRed iw.kk±d gy dsoy ñ 1 gSA

4. Number of integers greater than ñ8 satisfying the inequation 3 4 5

2

(x ñ 2) (x 5) (x 8)(x ñ 10) (x ñ 12)

0 are : [BARI]

vlfedk 3 4 5

2

(x ñ 2) (x 5) (x 8)(x ñ 10) (x ñ 12)

0 dks larq"B djus okys ñ8 ls cMs iw.kk±dksa dh la[;k gS&

(A) 8 (B) 9 (C*) 10 (D) 12

Sol. Solution can be represented as gy fu:fir fd;k tk ldrk gS&

ñ8 ñ5 2 10 12 integers greater than ñ 8 are ñ8 ls cMs iw.kk±d gS

x = ñ 5, 2, 3, 4, 5, 6, 7, 8, 9, 11

5. What is the most general value of which satisfies both of the equations cos = 12

and

tan = 1 ? [BAMS]

dk O;kid eku gksxk tcfd nksuksa lehdj.k cos = 12

vkSj tan = 1 dks larq"B djrk gSA

(A) n + 4

, n (B) n + 34

, n (C*) 2n + 54

, n (D) 2n + 34

, n

Page # 3

Sol. In interval 0 2

cos = 12

= 3 5,4 4

and vkSj tan = 1 = 5,4 4

general value of is = 2n + 54

, n

dk O;kid eku = 2n + 54

, n

6. Complete set of values of x satisfying inequality ||x ñ 1| ñ 5| < 2x ñ 5, is [BAMD]

vlfedk ||x ñ 1| ñ 5| < 2x ñ 5 dks larq"V djus okys x ds ekuksa dk lEiw.kZ leqPp; gS -

(A) 5 ,2

(B*) 11,3

(C) 1, (D) 11,3

Sol. 2x ñ 5 > 0 x > 52

Ö (1)

ñ 2x + 5 < |x ñ 1| ñ 5 < 2x ñ 5 ñ2x + 10 < |x ñ 1| < 2x

ñ2x + 10 < x ñ 1 < 2x (aspwfd x > 52

)

CasefLFkfr-1 : x ñ 1 < 2x ñ x < ñ 1 x > ñ 1 Ö (2)

CasefLFkfr-2 : x ñ 1 > ñ 2x + 10 3x > 11

x > 113

Ö (3)

By (1) (2) (3) x 11,3

7. The expression cos( n x).cos( n y) ñ 1 xcos n cos( n(xy)2 y

is simplified to ñ [BAMS]

O;atd cos( n x).cos( n y) ñ 1 xcos n cos( n(xy)2 y

dks ljy djus ij

(A) ñ1 (B*) 0 (C) 1 (D) ñ2

Sol. E = cos ( nx) cos ( ny) ñ 12

(cos ( nx ñ ny) + cos ( nx + ny))

= cos ( nx) cos ( ny) ñ 12

◊ 2 cos ( nx) cos ( ny)

= 0 8. The complete solution set of the equation |x2 ñ 5x + 6| + |x2 + 12x + 27| =|17x + 21| is - [BAMD] (A) x [ñ9, 3] (B) x [ñ3, 2) (2, 3] (C*) x [ñ9, ñ3] [2, 3] (D) x (ñ2, 3)

lehdj.k |x2 ñ 5x + 6| + |x2 + 12x + 27| =|17x + 21| dk lEiw.kZ gy leqPP; gS &

(A) x [ñ9, 3] (B) x [ñ3, 2) (2, 3] (C*) x [ñ9, ñ3] [2, 3] (D) x (ñ2, 3)

Page # 4

Sol. |a| + |b| = |a ñ b| a2 + b2 + 2 |ab| = a2 + b2 ñ 2ab or |ab| = ñab ab 0 (x2 ñ 5x + 6) (x2 + 12x + 27) 0 (x ñ 2) (x ñ 3) (x + 3) (x + 9) 0 x [ñ9, ñ3] [2, 3] 9. The correct graph of curve y = x3 ñ x2 ñ 8x + 5 is [BAMS]

oØ y = x3 ñ x2 ñ 8x + 5 dk lgh vkjs[k gS&

(A) (B)

(C) (D*)

10. The sum of all the values of x in the interval [ñ2 , ) satisfying the equation [BAMS] cos 5x . tan(6 |x|) + sin 5x = 0 is ñ

vUrjky [ñ2 , ) esa x ds lHkh ekuksa dk ;ksxQy tks lehdj.k cos 5x . tan(6 |x|) + sin 5x = 0 dks larq"B djrs gS ñ

(A*) 2 (B) 3 (C) 0 (D) 4Sol. cos 5x tan 6|x| + sin 5x = 0

Case fLFkfr x 0

cos 5x tan 6x + sin 5x = 0 cos 5x sin 6x + sin 5x cos 6x = 0 sin 11x = 0 11x = n

x = n11

, n w

x = 0, 2 3 4 5 6 7 8 9 10, , , , , , , , ,,11 11 11 11 11 11 11 11 11 11

CasefLFkfr x < 0

ñ cos 5x tan 6x + sin 5x = 0 ñ cos 5x sin 6x + cos 6x sin 5x = 0 sin (5x ñ 6x) = 0 sinx = 0 x = m , m ñ x = ñ , ñ2

sum ;ksxQy = 2

Page # 5

11. Set of all values of x satisfying the inequation 216 2x 3x < 1 + x is ñ [BAIR]

vlfedk 216 2x 3x < 1 + xdks lUrq"V djus okys x ds lHkh ekuksa dk leqPp; gSa (A) (ñ , ñ5/2) (B) (ñ1, 3/2)

(C) (ñ , ñ5/2) (3/2, ) (D*) 3 , 22

Sol. 16 ñ 2x ñ 3x2 0 3x2 + 2x ñ 16 0 (3x + 8) (x ñ 2) 0

x 8 , 23

..... (i)

Case fLFkfr I 1 + x < 0 solution not possible Case fLFkfr I 1 + x < 0 gy lEHko ugha gS Case fLFkfr I I 1 + x 0 x [ñ1, ) .... (ii) squaring both side nksuksa i{kksa dk oxZ djus ij 16 ñ 2x ñ 3x2 < x2 + 2x + 1 4x2 + 4x ñ 15 > 0

x 5,2

3 ,2

.... (iii)

from (i), (ii) and (iii) we get 3 , 22

(i), (ii) vkSj (iii) ls 3 , 22

12. Set of all the values of x satisfying |x2 + x ñ 12| + |sin x| = |(x2 + x ñ 12) + sin x|, where ñ x , is

|x2 + x ñ 12| + |sin x| = |(x2 + x ñ 12) + sin x|, tgk¡ ñ x dks lUrq"V djus okys x ds lHkh ekuksa

dk leqPp; gSa& [BAMD]

(A) [ñ3, 3] {ñ , } (B*) [ñ , 0] [3, ] (C) [ñ3, 0] [3, ] (D) [3, ] Sol. (x2 + x ñ 12) (sin x) 0 (x + 4) (x ñ 3) sin x 0 clearly x = 3, ñ , 0, are solutions .........(i) {ñ4 [ñ , ] } Li"Vr;k x = 3, ñ , 0, gy gS .........(i) {ñ4 [ñ , ] } Case I when tc 0 x then rks sin x 0 (x + 4) (x ñ 3) 0 i.e x ñ4 or x 3 [3, ] are solutions [3, ] gy gS .........(ii) Case II when tc ñ x 0, then rc sin x 0 (x + 4) (x ñ 3) 0 i.e ñ4 x 3

[ñ , 0] are solution [ñ , 0] gy gS .........(iii)

set of all solution is [ñ , 0] [3, ] lHkh gyksa dk leqPp; [ñ , 0] [3, ] 13. Number of integral values of x satisfying the inequation |x + 11| + |x + 7| < 6, is [BAMD]

vlfedk |x + 11| + |x + 7| < 6 dks lUrq"V djus okys x ds iw.kk±d ekuksa dh la[;k gS &

(A) 6 (B*) 5 (C) 7 (D) 3

Page # 6

Sol. Case I when tc x < ñ11 ñx ñ 11 ñ x ñ 7 < 6 i.e 2x > ñ24 i.e x > ñ12 no integral solutions dksbZ iw.kk±d gy ugha Case II when tc ñ11 x < ñ 7 x + 11 ñ x ñ 7 < 6 i.e 4 < 6 true for all x x + 11 ñ x ñ 7 < 6 i.e 4 < 6, x ds lHkh ekuksa ds fy, lR; gS integral values ñ11, ñ10, ñ9, ñ 8 are solutions. There are 4 integral solutions iw.kk±d eku ñ11, ñ10, ñ9, ñ 8 gy gSA dqy 4 iw.kk±d gy gSA Case II I when tc x ñ7 x + 11 + x + 7 < 6 i.e 2x < ñ12 x < ñ6 ñ7 x < ñ6 there is only one integral solution viz : ñ7 ;gk¡ dsoy ,d iw.kk±d gy gS viz : ñ7

total number of integral solutions is 5. dqy iw.kk±d gyksa dh la[;k = 5.

14. Set of all values of x for which 23xsin

23xsinlog 2

1.0 is defined, is [BAMS]

23xsin

23xsinlog 2

1.0 ds ifjHkkf"kr gksus ds fy, x ds lHkh ekuksa dk leqPp; gS &

(A) 6

5,6

(B*) 6

5n2,6

n2

(C) 6

)1n2(,6

n2 (D) no solution dksbZ gy ugha

Sol. log0.1 23xsin

23xsin2 0

0 < sin2 x ñ23 sin x +

23 1 ..... (i)

(i) sin2 x ñ 23 sin x +

23 = sin2 x ñ

23 sin x +

169 +

23 ñ

169

= 2

43xsin +

1615 > 0 x R

Inequality vlfedk (i) sin2 x ñ 23 sin x +

23 ñ 1 0

2 sin2 x ñ 3 sin x + 1 0 (sin x ñ 1) (2 sin x ñ 1) 0

21 sin x 1

2n +6

x 2n +6

5

15. Number of solutions of 1 + tan2 x ñ sec x tan x < 0 is [BAMS] 1 + tan2 x ñ sec x tan x < 0 ds gyksa dh la[;k gS & (A*) 0 (B) 2 (C) 3 (D) infinite vuUr

Page # 7

Sol. 1 + tan2 x ñ sec x tan x < 0 sec2 x ñ sec x tan x < 0 sec x(sec x ñ tan x) < 0

xtanxsec

xsec < 0

xsin1

1 < 0

no solution { 1 + sin x 0} dksbZ gy ugha { 1 + sin x 0} 16. Number of solutions of [x] + 2x = 5 + 4 {x}, which lie in the interval [1, 5] [BAGI] (where [.] and {.} represent greatest integral function and fractional part function) vUrjky [1, 5] esa lehdj.k [x] + 2x = 5 + 4 {x} ds gyksa dh la[;k gS & (;gk¡ [.] rFkk {.} Øe'k% egÙke iw.kk±d Qyu rFkk fHkUukRed Hkkx Qyu dks iznf'kZr djrs gSA) (A) 0 (B*) 1 (C) 2 (D) 3 Sol. [x] + 2x = 5 + 4 {x} Let x = [x] + {x} = I + f

I + 2I + 2f = 5 + 4f 3I ñ 5 = 2f

0 I3 52

< 1 3I ñ 5 < 2

5 3I < 7 53

I < 73

I = 2, f = 12

x = 52

17. Set of all real values of x satisfying inequation (x2 + 3x) (2x + 3) ñ16 x3x

3x22 0 is [BARI]

vlfedk (x2 + 3x) (2x + 3) ñ16 x3x

3x22 0 dks lUrq"V djus okys x ds lHkh okLrfod ekuksa dk leqPp; gS&

(A) x (ñ , ñ4] (ñ3, ñ3/2] (0, 1] (B*) x [ñ4, ñ3) [ñ3/2, 0) [1, ) (C) x [ñ4, ñ3] [ñ3/2, 0] [1, 2] (D) x (ñ4, ñ3/2] [1, )

Sol. (2x + 3) x3x

16x3x 22 0

x3x

16)x3x()3x2(2

22 0

)3x(x

)4x3x()4x3x()3x2( 22 0

ñ + ñ + ñ +

ñ4 ñ3 0 1 ñ 32

)3x(x

)4x()1x()3x2( 0

x [ñ4, ñ3) [ñ3/2, 0) [1, )

Page # 8

18. If x 0, y 0, then area of the region bounded by [x] + [y] = 99 is [BAGI]

(where [.] represent greatest integral function)

(A) 99 sq. unit (B*) 100 sq. unit (C) 101 sq. unit (D) 120 sq. unit

;fn x 0, y 0 gks] rks [x] + [y] = 99 }kjk ifjc) {ks=k dk {ks=kQy gS &

(;gk¡ [.] egÙke iw.kk±d Qyu dks iznf'kZr djrk gSA)

(A) 99 oxZ bdkbZ (B*) 100 oxZ bdkbZ (C) 101 oxZ bdkbZ (D) 120 oxZ bdkbZ

Sol. If [x] = 0, then [y] = 99 and the area enclosed = 1 unit ;fn [x] = 0 gks] rks [y] = 99 vkSj ifjc) {ks=kQy = 1 bdkbZ If [x] = 1, then [y] = 98 and the area enclosed = 1 unit ;fn [x] = 1 gks] rks [y] = 98 vkSj ifjc) {ks=kQy = 1 bdkbZ If [x] = 2, then [y] = 97 and the area enclosed = 1 unit ;fn [x] = 2 gks] rks [y] = 97 vkSj ifjc) {ks=kQy = 1 bdkbZ If [x] = 99, then [y] = 0 and the area enclosed = 1 unit ;fn [x] = 99 gks] rks [y] = 0 vkSj ifjc) {ks=kQy = 1 bdkbZ the total enclosed area = 100 units

dqy ifjc) {ks=kQy = 100 bdkbZ

19. If tan2x = cosec x ñ sin x, where x n then tan2

2x is equal to [BAMS]

;fn tan2x = cosec x ñ sin x, tgk¡ x n gks] rks tan2

2x =

(A) 2 ñ 5 (B) 5 + 2

(C*) (9 ñ 4 5 ) (2 + 5 ) (D) (9 + 4 5 ) (2 ñ 5 )

Sol. tan2x =

xsin1 ñ sin x

tan2x =

2xtan1

2xtan2

2xtan2

2xtan1

2

2

let y = tan2 2x

y2 + 4y ñ 1 = 0 y = 5 ñ 2 = (9 ñ 4 5 ) (2 + 5 )

20. If a > 5 12

, then 1a7

a2alog4

logalog

49

324/1 1a272 32 + a simplifies to : [BALG]

;fn a > 5 12

, rks 1a7

a2alog4

logalog

49

324/1 1a272 32 + a ds ljyhdj.k ls izkIr gksrk gS &

(A) a2 ñ a + 1 (B) (a ñ 1)2 (C) a2 + a + 1 (D*) (a + 1)2

Page # 9

Sol. 1ñañ7

a2ñ3ñ2alog4

logalog

49

3)12a(274/12

+ a

=1ñañ7

a2ñ3ñ22

7

23

42

alog

)1a(logalog + a

= 1ñaña

a2ñ)1a(ña2

24+ a

= 1ñaña)1a(ña

2

24 + a = a2 + 2a + 1 = (a + 1)2

21. If cot + tan = x, sec ñ cos = y, 0 < < 2

, then [BAMS]

;fn cot + tan = x, sec ñ cos = y, 0 < <2

gks] rks

(A) (xy2)2/3 ñ (x2y)2/3 = 1 (B*) (x2y)2/3 ñ (xy2)2/3 = 1 (C) x2/3 + y2/3 = 1 (D) x2/3 ñ y2/3 = 1

Sol. cot + tan = x i.e. cossin1 = x

sec ñ cos = y i.e. sin2 = y cos

x2 cos2 . y cos = 1 cos3 = yx

12

and sin2 = y sinx1 . sin3 = y

x

1 = 3/2

2yx1 +

3/2

xy = 3/23/4 yx

1 + 3/2

3/2

xy = 3/23/4

3/43/2

yxyx1

x4/3 y2/3 ñ x2/3 y4/3 = 1 (x2y)2/3 ñ (xy2)2/3 = 1

22. The complete solution set of inequality 2010532006

20082005

)9x()6x(.)3ñx(.)2x(x)1ñx()8x(.)5x( 0 is [BARI]

vlfedk 2010532006

20082005

)9x()6x(.)3ñx(.)2x(x)1ñx()8x(.)5x( 0 dk lEiw.kZ gy leqPp; gSA

(A) (ñ , ñ9) (ñ8, 0) (0, 1) (2, 3) [5, 6) (B) (ñ , ñ9) (ñ9, 0) (0, 1) (2, 3) (5, 6) (C*) (ñ , ñ9) (ñ9, 0) (0, 1] (2, 3) [5, 6) (D) (ñ , 0) (0, 1] (2, 3) [5, 6) Sol. x (ñ , ñ9) (ñ9, 0) (0, 1] (2, 3) [5, 6)

23. If sin A =53 , cos B =

1312 , A ,

2and B

2,0 , then [BAMS]

;fn sin A =53 , cos B =

1312 , A ,

2 rFkk B

2,0 , gks] rks

(A) < A + B < 2

3 and sin (A + B) = 6516

(B*) 2

< A + B < and sin (A + B) =6516

(C) 2

< A + B < and cos (A + B) = ñ 6533

(D) < A + B < 2

3 and cos (A + B) = ñ 6563

Page # 10

Sol. cos A = ñ 54 , sin B =

135 sin (A + B) =

6536 ñ

6520 =

6516

and cos (A + B) = ñ 6548 ñ

6515 = ñ

6563

24. If in triangle ABC : ∫45C , then all the values of sin2A + sin2B lie in the interval [BAMS] ;fn f=kHkqt ABC esa ∫45C gks] rks (sin2A + sin2B) ds lHkh eku fuEu varjky esa gksxsa!

(A) [0, 1] (B) (0, 1) (C*) 2

21,21 (D) 1 1ñ ,

2 2

Sol. = sin2A + sin2 (135∫ ñ A) = 1 + sin2A ñ sin2(45∫ ñ A)

= 1 + sin 45∫ ◊ sin(2A ñ 45∫) = 1 + 12

sin(2A ñ 45∫)

0 < A < 135∫ 0 < 2A < 270∫ ñ45∫ < 2A ñ 45∫ < 225∫

ñ 21 < sin (2A ñ 45∫) < 1

21 < sin2A + sin2B <

221

25. Number of real roots of equation )3x4ñx(log 233 = x ñ 3 is - [BALG]

lehdj.k )3x4ñx(log 233 = x ñ 3 ds okLrfod ewyksa dh la[;k gS &

(A) 2 (B) 1 (C*) 0 (D) infinite vuUr

Sol. x2 ñ 4x + 3 = x ñ 3 x2 ñ 5x + 6 = 0 x = 3, 2 and vkSj x2 ñ 4x + 3 > 0 (x ñ 3) (x ñ 1) > 0 x > 3 or ;k x < 1 .......(i) by (i) ls x = 3, 2 (3, )

MCQ 26. If the solution set of inequation 1/ 21 log ( x)

2 6x < 0 is (a, b) then ñ [BAIR]

(A*) a and b are rationals (B*) 2a ñ 3b = 0

(C*) 36 6

2log (b a)5

(D) sec can take value between a and b for some

;fn vlfedk 1/ 21 log ( x)2 6x

< 0 dk gy leqPp; (a, b) gS] rc ñ

(A*) a vkSj b ifjes; gSA

(B*) 2a ñ 3b = 0

(C*) 36 6

2log (b a)5

(D) sec fdlh ds fy, a vkSj b ds e/; eku ys ldrk gS

Page # 11

Sol. 1/ 21 log ( x) 06x 2

ñ6x ñ 2 > 0 x < ñ 13

Ö (1)

and vkSj 1 ñ log1/2 (ñx) < 0

log1/2 (ñx) > 1

ñx < 12

x > ñ 12

Ö (2)

Also ñ x > 0 x < 0 .............(3)

intersection of (1), (2), (3) gives x 1 1,2 3

(1), (2), (3) ds çfrPNsnu ls x 1 1,2 3

b ñ a = 1 1 13 2 6

36 6

1log6

= y y 1(36 6 ) 6

y = 25

27. If S and P are sum and product respectively of all real values of x satisfying equation

|4 ñ |x ñ 1|| = 3, then [BAMD] ;fn lehdj.k |4 ñ |x ñ 1|| = 3 dks larq"V djus okys x ds lHkh okLrfod ekuksa dk ;ksx rFkk xq.kuQy Øe'k%

S rFkk P gS] rks & (A*) S + P = 4 (B*) S ñ P = 4 (C*) S2 ñ P2 = 16 (D*) S2 + P2 = 16 Sol. ||x ñ 1| ñ 4| = 3 |x ñ 1| = 1, 7 x = 0, 2, ñ6, 8 S = 0 + 2 ñ 6 + 8 = 4, P = 0

28. If tan = tan and ,2

and (0, 4 ), then which of following can be possible ? [BAMS]

;fn tan = tan vkSj ,2

rFkk (0, 4 ), gks] rks fuEu esa ls dkSu&ls laHko gS&

(A*) = (B*) = + (C) = + (D*) = 2 + Sol. = n + :n = , + , 2 + 29. If [ . ] represents greatest integer function, then 2[x]2 ñ 3[x] + 1 = 0 is true for [BAGI]

(A) no real value of x (B*) 1 x < 2 (C) x R (D*) x 2,e

;fn [ . ] egÙkeiw.kk±d Qyu dks n'kkZrk gS] rks 2[x]2 ñ 3[x] + 1 = 0 lR; gksus ds fy, (A) x dk dksbZ okLrfod eku ugha (B*) 1 x < 2

(C) x R (D*) x 2,e

Page # 12

Sol. 2[x]2 ñ 3[x] + 1 = 0 Let [x] = y 2y2 ñ 3y + 1 = 0 (2y ñ 1) (y ñ 1) = 0 y = 1 or ;k y = 1/2 [x] = 1 1 x < 2

30. The least positive solution of cot x2sin33

= 3 lies in [BAMS]

cot x2sin33

= 3 dk U;wure /kukRed gy] fuEu varjky esa gS&

(A*) 6

,0 (B*) 6

,9

(C*) 3

,9

(D) 2

,3

Sol. cot x2sin33

= 3 33

sin 2x = n + 6

sin 2x = 3 3 n +23 n = 0

sin 2x = 23 x =

6

Comprehension # 1 Paragraph for Question Nos. 31 to 32

Consider the graph f(x) = 13x27x21x2

ekukfd f(x) = 13x27x21x2 gSA [BAMD]

31. If ñ15 < f(x) < 15, then complete set of values of x is ;fn ñ15 < f(x) < 15 gks] rks x ds lEiw.kZ ekuksa dk leqPp; gS&

(A*) (2, 5) (B) (3, 6) (C) (3, 5) (D) 2

13,21

Sol. For ñ15 < f(x) < 15, the line y = 15 intersects the lines y = ñ2x + 19 and y = 2x + 5 at x = 2 and x = 5 respectively ñ15 < f(x) < 15 if x (2, 5)

ñ15 < f(x) < 15 ds fy, js[kk y = 15 js[kkvksa y = ñ2x + 19 vkSj y = 2x + 5 dks Øe'k% x = 2 vkSj x = 5 ij izfrPNsn djrh gSA

ñ15 < f(x) < 15 ;fn x (2, 5)

12

72

13 2

12

18

f(x) = ñ6x + 21

f(x)=ñ2x+19 f(x)=2x+5

f(x) =

6x

ñ 21

x

y

Page # 13

Case I when x < 21

f(x) = ñ2x + 1 ñ 2x + 7 ñ 2x + 13 = ñ6x + 21

Case II when21 x <

27

f(x) = 2x ñ 1 ñ 2x + 7 ñ 2x + 13 = ñ2x + 19

Case II I when27 x <

213

f(x) = 2x ñ 1 + 2x ñ 7 ñ 2x + 13 = 2x + 5

Case IV when 2

13 x,

32. If 13 < f(x) < 17, then complete set of values of x is ñ

;fn 13 < f(x) < 17 gks] rks x ds lEiw.kZ ekuksa dk leqPp; gS&

(A) (3, 4) (B) 25,

23 (4, 6)

(C) (1, 6) (D*) (1, 3) (4, 6) Sol. For ñ15 < f(x) < 15, the line y = 15 intersects the lines y = ñ2x + 19 and y = 2x + 5 at x = 2 and x = 5 respectively ñ15 < f(x) < 15 if x (2, 5)

For 13 < f(x) < 17, the line y = 13 intersects the lines y = ñ2x + 19 and y = 2x + 5 at x = 3 and x = 4 respectively Again y = 17 intersects the lines y = ñ2x + 19 and y = 2x + 5 at x = 1 and x = 6 respectively

13 < f(x) < 17 if x (1, 3) (4, 6) 13 < f(x) < 17 ds fy, js[kk y = 13 js[kkvksa y = ñ2x + 19 vkSj y = 2x + 5 dks Øe'k% x = 3 vkSj x = 4 izfrPNsn djrh gSA

iqu% y = 17 js[kkvksa y = ñ2x + 19 vkSj y = 2x + 5 dks Øe'k% x = 1 vkSj x = 6 ij izfrPNsn djrh gSA 13 < f(x) < 17 ;fn x (1, 3) (4, 6)

12

72

13 2

12

18

f(x) = ñ6x + 21

f(x)=ñ2x+19 f(x)=2x+5

f(x) =

6x

ñ 21

x

y

Case I when x < 21

f(x) = ñ2x + 1 ñ 2x + 7 ñ 2x + 13 = ñ6x + 21

Case II when21 x <

27

f(x) = 2x ñ 1 ñ 2x + 7 ñ 2x + 13 = ñ2x + 19

Case II I when27 x <

213

Page # 14

f(x) = 2x ñ 1 + 2x ñ 7 ñ 2x + 13 = 2x + 5

Case IV when 2

13 x,

f(x) = 2x ñ 1 + 2x ñ 7 + 2x ñ 13 = 6x ñ 21

Comprehension # 2

Paragraph for Question Nos. 33 to 34 Given that N = 900log497 A = 3log2log4log4log 2222 4432 [BALG] D = (log5 49) (log7 125) Then answer the following questions : (Using the values of N, A, D)

fn;k x;k gS fd N = 900log497

A = 3log2log4log4log 2222 4432 D = (log5 49) (log7 125) rks fuEufyf[kr iz'uksa ds mÙkj nhft,A (N, A, D ds ekuksa dk mi;ksx djrs gq,) 33. If logA D = a, then the value of log6 12 is (in terms of a) ;fn logA D = a gks] rks log6 12 dk eku (a ds inksa esa) gS -

(A*) a3

a31 (B) a3

a21 (C) a2

a21 (D) a2

a31

Sol. N = 30

A = 8 D = 6 log8 6 = a

2log3

3log2log = a

(log 2) (1 ñ 3a) + log 3 = 0 log 3 = (3a ñ 1) log 2

Now log6 12 = 6log

12log

1a311a32

3log2log3log2log2 =

a3a31

34. The value of

10NA

log |N + A + D + 6| ñ log5 2 =

(A) 1 (B*) 2 (C) 3 (D) 4 Sol. N = 30 A = 8 D = 6

10NA

log |N + A + D + 6| ñ log5 2

log5 50 ñ log5 2 = 2

Page # 15

Comprehension # 3

Paragraph for Question Nos. 35 to 36 If f(x) = g(x), then number of solutions can be evaluated by studying number of intersection points of

y = f(x) and y = g(x). [BAMS] ;fn f(x) = g(x) rc y = f(x) vkSj y = g(x) ds izfrPNsn fcUnqvksa dh la[;k dk v/;;u djds gyksa dh la[;k Kkr dh tk

ldrh gSA 35. Complete set of values of for which there exist three real and distinct solutions of x3 ñ 3x + ñ 1 = 0 ds ekuksa dk lEiw.kZ leqPp; gksxk ftlds fy, lehdj.k x3 ñ 3x + ñ 1 = 0 ds rhu okLrfod vkSj fofHkUu gy fo|eku

gS - (A) (ñ3, 2) (B) (ñ1, 1) (C) (1, 5) (D*) (ñ1, 3) Sol. x3 ñ 3x ñ 1= ñ Let f(x) = x3 ñ 3x ñ 1 f ' (x) = 3x2 ñ 3 = 3 (x + 1) (x ñ 1)

+ ñ +

ñ1 1 (maxm) (minm)

f(1) = ñ3

ï

ï

f(ñ1) = 1

ñ1 1(0,ñ1)

(1,ñ3)

(ñ1,1)

For three real and distinct solution ñ (ñ3, 1) (ñ1, 3) rhu okLrfod vkSj fofHkUu gyksa ds fy, ñ (ñ3, 1) (ñ1, 3) 36. Number of solutions of 10 sin x = 5 ñ x lehdj.k 10 sin x = 5 ñ x ds gyksa dh la[;k gS - (A) 6 (B) 5 (C*) 7 (D) 8 Sol. 10 sin x = 5 ñ x

32 2

72

92

112

ñ2

ï ïï ï ï ïï ï ï ï

ñ3

2 4

5

6 7

(0,10)

(0,ñ10)

3 3 1052 2

=

Total number of solution gyksa dh dqy la[;k 7 Match the column

Page # 16

37. If graph of y = f(x) is given below : [BAMS] ;fn y = f(x) dk vkjs[k fn;k x;k gS&

x-axis

y-axis

Column ñ I Column ñ II LrEHk ñ I LrEHk ñ II

(A) y = f(|x|) (p)

x-axis

y-axis

(B) y = f(ñ|x|) (q)

x-axis

y-axis

(C) y = |)x(|f (r)

x-axis

y-axis

(D) y = |x|f (s)

x-axis

y-axis

Ans. (A) (r), (B) (s), (C) (r), (D) (q) Sol. Using shifting of graph vkjs[k ds :ikUrj.k ls

Integer

38. If solution set of the inequation log3

2

2

| x ñ 4x | 3x | x ñ 5 |

0, is (ñ , a] [b, c], then find the value of

(3a + 2b + c). [BAMD]

;fn vlfedk log3

2

2

| x ñ 4x | 3x | x ñ 5 |

0, dk gy leqPp; (ñ , a] [b, c], gS ] rc (3a + 2b + c) dk eku Kkr

dhft,A

Ans. 1

Page # 17

Sol. 2

2

| x ñ 4x | 3x | x ñ 5 |

1

|x2 ñ 4x| + 3 x2 + |x ñ 5| |x(x ñ 4)| + 3 x2 + |x ñ 5| x 5 x2 ñ 4x + 3 x2 + x ñ 5 ñ 5x ñ 8

x 85

x 4 x < 5 x2 ñ 4x + 3 x2 ñ x + 5 ñ 3x 2

x ñ 23

x 0 x < 4 ñ x2 + 4x + 3 x2 ñ x + 5 2x2 ñ 5x + 2 0

12

x 2

IV x < 0 x2 ñ 4x + 3 x2 ñ x + 5 ñ3x 2

x 23

x 2 1, ,23 2

3a + 2b + c = 1

39. If 2

a1a = 5, then find the value of 4

4

a1a . [BAMD]

;fn 2

a1a = 5 ] gks] rks 4

4

a1a dk eku Kkr dhft,A [New, M]

Ans. 7

Sol. 44

a1a = 2ñ

a1a

2

22 = 2ñ2ñ

a1a

22

= 2ñ)2ñ5( 2 = 7

40. If asin + bcos = a cosec ñ b sec = 1. Then find the value of a2+ b2 + b4/3 ñ b2/3 [BAMS]

;fn asin + bcos = a cosec ñ b sec = 1 gks rc a2+ b2 + b4/3 ñ b2/3 dk eku Kkr dhft,A

Ans. 1

Page # 18

Sol. a sin + b cos = 1 Ö (1) and a cosec ñ b sec =1 acos ñ b sin = cos sin Ö (2)

Squaring and adding (1) and (2), (1) o (2) dk oxZ djus ij

a2 + b2 = 1 + cos2 sin2 = 1 + cos2 (1 ñ cos2 ) = 1+ cos2 ñ cos4 Ö (3)

(1) ◊ cos ñ (2) ◊ sin gives fn;k x;k gS fd

acos sin + b cos2 ñ a cos sin + b sin2 = cos ñ cos sin2 b cos2 + b sin2 = cos (1 ñ sin2 ) cos3 = b cos = b1/3

(3) becomes (3) ls

a2 + b2 = 1 + b2/3 ñ b4/3

a2 + b2 + b4/3 ñ b2/3 = 1

41. 2

3 65 3 2 12 32 50

is simplified to ñ [BAIR]

2

3 65 3 2 12 32 50

ljy djus ij

Ans. 3

Sol. Let ekuk E = 3 65 3 4 3 4 2 5 2

= 3 63 2

= (3 6 )( 3 2)1

= 3 3 3 2 18 12

= 3 3 3 2 3 2 2 3

= 3

The value of the given expression = 2( 3 ) = 3 fn, x, O;atd dk eku = 2( 3 ) = 3

42. If x = 2 + 22/3 + 21/3, then the value of x3 ñ 6x2 + 6x = [BAMS] ;fn x = 2 + 22/3 + 21/3 gk s ] rk s x3 ñ 6x2 + 6x dk eku gS&

Ans. 2 Sol. x ñ 2 = 21/3 + 22/3

cubing both sides pw ¡ fd (x ñ 2)3 = (21/3 + 22/3)3 dk U;wure eku

x3 ñ 8 ñ 6x2 + 12x = 2 + 4 + 3 ◊ 21/3 ◊ 22/3 (21/3 + 22/3) x3 ñ 6x2 + 12x ñ 8 = 6 + 6(x ñ 2) x3 ñ 6x2 + 12x ñ 8 = 6 + 6x ñ 12 x3 ñ 6x2 + 6x = 2 43. Find the least value of the expression E = 2log10x ñ logx (0.01) for x > 1. [BALG] x > 1 ds fy, O;atd E = 2log10x ñ logx (0.01) dk U;wure eku Kkr dhft,A

Ans. 4

Page # 19

Sol. E = 2 log10x ñ logx (10ñ2) = 2log10x + 2 logx 10

= 2 1010

1log xlog x

Since x > 1, log10x > 0 least value of log10x + 10

1log x

= 2

Emin = 2(2) = 4

44. Find the number of solutions of the equation sin2 2x ñ cos2 8x = 12

cos 10x lying in the interval 0,2

.

[BAMS]

vUrjky 0,2

es a fLFkr lehdj.k sin2 2x ñ cos2 8x = 12

cos 10x ds gyk s a dh la[;k g S&

Ans. 8

Sol. sin22x ñ cos28x = 12

cos 10x

2 sin2 2x ñ 2 cos2 8x = cos 10x 1 ñ cos 4x ñ 1 ñ cos 16x = cos 10x cos 14x + cos 10x + cos 16x = 0 2 cos 10x cos 6x + cos 10x = 0 cos 10x (2 cos 6x + 1) = 0

10x = (2n +1) 2

, n I

x = (2n + 1) 20

, n I ; x 0,2

, x = 3 5 7 9, , , ,20 20 20 20 20

and vk Sj cos 6x = ñ 12

6x = 2k ± 23

x = k3 9

x = (3k ± 1) 9

, k

In 0,2

, x = 4 2, ,9 9 9

no of solution are 8 gyk a s dh la[;k 8 gSA

45. If + + = 2, 2 + 2 + 2 = 6 [BAGQ]

1 1 1 12

then find the value of 3+ 3 + 3

;fn + + = 2, 2 + 2 + 2 = 6

1 1 1 12

gk s ] rk s 3+ 3 + 3 dk eku Kkr dhft,A

Ans. 8

Page # 20

Sol. 3+ 3 + 3 = 3 + ( ) ( 2 + 2 + 2 ñ ñ ñ ) = 3 + ( + + ) (( + + )2 ñ 3( + + )) = 3 + 2 (4 ñ 3( + + ) Ö (1)

Also rFkk, 2 + 2 + 2 = 6

( + + )2 ñ 2( + + ) = 6 4 ñ 2( + + ) = 6 + + = ñ 1 Ö (2)

Also rFkk, 1 1 1 12

12

1 12

= ñ2 = ñ2 Ö (3)

Putting the values from (2) and (3) in (1), (2) vk Sj (3) ls (1) es a eku j[kus ij

3 + 3 + 3 = 3(ñ2) + 2(4 ñ 3 ◊ (ñ1)) = ñ 6 + 14 = 8

Course : JR(ELPD)(PT1) Test Date : 10- 08-2014 Paper Time Duration : 2 Hrs.

PAPER LEVEL : MODERATE

Test Syllabus : Syllabus : Complete Mole Concept, Quantum Numbers.

Test Syllabus : IUPAC Nomenclature & Structure isomers,

Hydrogenation & Monochlorination

SYLLABUS SCHEDULED SR. NO.

TOPIC NAME SYLLABUS SCHEDULED WEIGHTAGE

(BY FC)

WEIGHTAGE IN PAPER (BY FACULTY)

1. IUPAC Nomenclature

2. Structure isomers

3. Hydrogenation

4. Monochlorination

Test Pattern :


1 to 25 SCQ 25 3 ñ1 75

26 to 30 MCQ 5 4 0 20




45 153

PT-1

Total Total

Maths/ Physics/

Chemistry

JEE (ADVANCED) CHEMISTRY PAPER SKELETON Test Name : JR (PT-1) Faculty preparing the TEST PAPER should fill it according to paper pattern and submit it with finalisaion of paper at SMD.

PAPER - 1 S.

No. TYPE (P) (I) (O) TOPIC(S) SUBTOPIC(S) DIFFICULTY LEVEL : Easy (E), Moderate (M), Tough (T)

1 SCQ (P) Mole concept Basics E

2 SCQ (P) Mole concept Basics E

3 SCQ (P) Mole concept Empirical formula M

4 SCQ (P) Mole concept Basics M

5 SCQ (P) Mole concept Limiting reagent T

6 SCQ (P) Mole concept Average molar masss M

7 SCQ (P) Mole concept Pv=nRT M

8 SCQ (P) Mole concept Dilution of sol. M

9 SCQ (P) Mole concept Mixture analysis M

10 SCQ (P) Mole concept Con of atoms M

11 SCQ (P) Atomic structure Electronic configuration M

12 SCQ (P) Atomic structure Quantum number M

13 SCQ (P) Atomic structure Nodes E

14 SCQ (O)

15 SCQ (O)

16 SCQ (O)

17 SCQ (O)

18 SCQ (O)

19 SCQ (O)

20 SCQ (O) 21 SCQ (O)

22 SCQ (O)

23 SCQ (O)

24 SCQ (O)

25 SCQ (O)

26 MCQ (P) Mole concept Basics M

27 MCQ (P) Mole concept Basics M

28 MCQ (P) Atomic structure Quantum

mechanical hundred atom

M

29 MCQ (O)

30 MCQ (O)

31 Comp. 1 (Q.1) (P) Mole concept Application of

gas law M

32 Comp. 1 (Q.2) (P) Mole concept Application of

gas law M

33 Comp. 2 (Q.1) (P) Mole concept Stoichionetry of

chemical r x n M

34 Comp. 2 (Q.2) (P) Mole concept Stoichionetry of

chemical r x n M

35 Comp. 3 (Q.1) (O)

36 Comp. 3 (Q.2) (O)

37 MTC (P) Mole concept

Inter conversion of atoms,

molecules & moles

M

38 Single Integer Type

(P) Mole concept M


(P) Mole concept Stoichionuly of chemical r x n M


(P) Mole concept

Limits of concentration measurement/ Interconversion

M


(P) Mole concept Average molecular mass M


(O)


(O)


(O)


(O)

Physical Chemistry Paper

SCQ (13) 1.

Final Q. (SID sir) 1. A compact fluorescent lamp (CFL) was filled with low pressure mixture of argon gas & small amount of

Mercury vapours. The volume of argon gas (At. mass = 40) at 0∫C and 1 atm pressure is found to be 100 mL. The number of atoms of argon present in the bulb is : (Mole-1(P))(E) ,d Bksl izfnIr ysEi (CFL) esa vkWxZu xSl o FkksM+h ek=kk dh edZjh ok"i ds fuEu nkc feJ.k ls Hkjk x;k gSA 0∫C o 1 atm nkc ij vkWxZu xSl (ijek.kq Hkkj = 40) dk vk;ru 100 mL ls izkIr gqvkA cYc esa mifLFkr vkWxZu ds ijek.kqvksa dh la[;k fuEu gSA

(A) 40000 (B*) 2.68 ◊ 1021 (C) 3.3 ◊ 1024 (D) 5.4 ◊ 1021

Sol. Number of Ar atoms = 10022400

◊ NA

Ar ijek.kqvksa dh la[;k = 10022400

◊ NA

= 21602 10

224

= 2.68 ◊ 1021 2. A mixture of HCl(g), PCl3(g) and PCl5(s), each have equal number of Cl-atom then their molar ratio in the

mixture is : (Mole-1(P))[E]

,d HCl(g), PCl3(g) rFkk PCl5(s), ds ,d feJ.k esa izR;sd Cl-ijek.kq dh leku la[;k j[krs gS rks feJ.k esa budk eksyj vuqikr Kkr djks &

(A) 1 : 3 : 5 (B) 1 : 1 : 1 (C) 15 : 2 : 3 (D*) 15 : 5 : 3

Sol. Let their are N atom in each then moles of Cl-atoms = A

NN

;fn izR;sd esa N ijek.kq gSa rks Cl-ijek.kqvksa ds eksy = A

NN

HCl : PCl3 : PCl5

A

NN

: A

1 N3 N

: A

1 N5 N

15 : 5 : 3 3. In a hydrocarbon, the mass % of C and H are 90% and 10% respectively. If 20 g of given hydrocarbon in

vapour phase occupies a volume of 5.6 litre at STP , then the molecular formula of the hydrocarbon is : (Take molar volume at STP as 22.4 litre for an ideal gas) (Mole-1(P))[M]

,d gkbMªksdkcZu esa dkcZu o gkbMªkstu dh izfr'krrk Øe'k% 90% rFkk 10% gSA ;fn ok"i izkoLFkk esa fn;s x;s gkbMªksdkcZu ds 20 g dk vk;ru STP ij 5.6 yhVj gS rks gkbMªksdkcZu dk v.kqlw=k gksxk (STP ij ,d vkn'kZ xSl ds 1 eksy dk vk;ru 22.4 yhVj gS)

(A) C3H4 (B) C5H20 (C*) C6H8 (D) C9H12 Sol. C H Mass % 90 10

mole % 9012

101

Ratio 3 : 4 Empirical formula = C3H4 5.6 litre at NTP weighs 20 g 22.4 at NTP weighs 80 g molar mass = 80 n = 2 and molecular formula = (C3H4)2 = C6H8

Sol. C H nzO;eku% 90 10

eksy % 9012

101

vuqikr 3 : 4 ewykuqikrh lw=k = C3H4 5.6 yhVj dk NTP ij Hkkj 20 g gSA 22.4 yhVj dk NTP ij Hkkj 80 g gSA eksyj nzO;eku = 80 n = 2 rFkk v.kqlw=k = (C3H4)2 = C6H8

4. A graph is plotted for an element, by putting its weight (in gm) on X-axis and the corresponding number of

atoms on Y-axis. The atomic weight of element (in amu) for which the graph is plotted would be: (Mole-1(P))(M)

(A) 16 (B*) 20 (C) 35 (D) 40

,d rRo ds fy, X-v{k ij Hkkj ¼xzke esa½ rFkk Y-v{k ij lEcfU/kr ijek.kqvksa dh la[;k ds e/; xzkQ vkysf[kr fd;k x;k gSA rRo] ftlds fy, xzkQ fn;k x;k gS] dk ijek.kq Hkkj (amu esa) gksuk pkfg,A

(A) 16 (B*) 20 (C) 35 (D) 40

Sol. No. of atoms = Awt N

Mol.wt

Y = X A

0

NM

slope = A

0

NM = 3 ◊ 1022

M0 = 22

23

103106

= 20

gy- ijek.kqvksa dh la[;k = Awt N

Mol.wt

Y = X A

0

NM

<ky = A

0

NM = 3 ◊ 1022 ; M0 = 22

23

103106

= 20

5. The number of moles of chromite ore (FeCr2O4) that can be produced by allowing 0.2 moles of Fe, 0.3

moles of Cr and 0.4 moles of O2 to combine according to the following reaction : (Mole-1(P))(T) Fe + 2Cr + 2O2 FeCr2O4

fuEu vfHkfØ;k ds vuqlkj Fe ds 0.2 eksy] Cr ds 0.3 eksy o O2 ds 0.4 eksy la;qDr gksdj ØksekbV v;Ld (FeCr2O4) ds fdrus eksy mRiUu dj ldrs gS\

Fe + 2Cr + 2O2 FeCr2O4

(A) 0.2 (B*) 0.15 (C) 0.9 (D) 0.3 Sol. Fe + 2Cr + 2O2 FeCr2O4 Mole 0.2 0.3 0.4

L.R. 0.3 12

=0.15 mole

Sol. Fe + 2Cr + 2O2 FeCr2O4 eksy 0.2 0.3 0.4

L.R. 0.3 12

=0.15 eksy

6. Which of the following statements is correct : (Mole-1(P)) [M] (A) A compound upon analysis was found to contain 2% Oxygen by mass. Then, the minimum molecular

mass of given compound (in amu) is 1600. (B*) If the density of a gas at NTP is 0.0025 g/cm3, then the gas has same molecular mass as Butene. (C) Carbon and Oxygen always combine in the ratio 3 : 8 by mass, according to law of constant

proportions. (D) Average molecular mass of CO2 and SO3 mixture always more than 60. fuEu esa ls dkSulk dFku lgh gS %

(A) ,d ;kSfxd dk fo'ys"k.k djus ij Kkr gqvk fd og vius nzO;eku dk 2 %, vkWDlhtu j[krk gSA rc fn;s x;s ;kSfxd dk U;wure vkf.od nzO;eku (amu esa) 1600 gSA

(B*) ;fn NTP ij ,d xSl dk ?kuRo 0.0025 g/cm3 g S] rc xSl dk vkf.od nzO;eku] C;wVhu ds leku gSA (C) fLFkj vuqikr ds fu;e ds vuqlkj] dkc Zu rFkk vk WDlhtu ges'kk n zO;eku ds 3 : 8 vuqikr es a la;k s ftr gk sr s gS aA (D) CO2 o SO3 feJ.k dk vk Slr vkf.od nzO;eku ges'kk 60 ls vf/kd gk srk gSA Sol. (A) Let mol. formula of compound is ...............O.

Mass % of O in compound = 1 16 100Mol. mass

= 2

So, minimum molecular mass of given compound = 800 amu.

(B) (dgas)NTP = gasGMM22.4

2.5 = gasGMM22.4

GMMgas = 56

Mol. Mass of Butene (C4H8) = 4 ◊ 12 + 8 ◊ 1 = 56 So, the gas has same molecular mass as Butene. (C) Whenever Carbon and Oxygen combine to form CO2, they always

combine in the ratio 3 : 8 by mass, according to law of constant proportions. gy (A) ekuk fd ;kSfxd dk v.kqlw=k ...............O gSA

;kSfxd esa O dk nzO;eku % =vkf.od nzO;eku

1 16 100 = 2

blfy,, fn;s x;s ;kSfxd dk U;wure vkf.od nzO;eku = 800 amu.

(B) (dx Sl)NTP = GMM

22.4xSl

2.5 = GMM

22.4xSl GMMxSl = 56

C;wVhu (C4H8) dk vkf.od nzO;eku = 4 ◊ 12 + 8 ◊ 1 = 56 vr%] x Sl dk vkf.od nzO;eku] C;wVhu ds leku g SA (C) tc Hkh dkcZu o vk WDlhtu la;k sftr gksdj CO2 cukr s g S a ] os ges'kk fLFkj vuqikr ds fu;e ds

vu qlkj] n z O;eku ds 3 : 8 vuqikr e s a l a;k sftr gk sr s g SA 7. 5.6 L of a gaseous alkane at STP, one molecule of which contains 6 H-atoms, is allowed to burn in

presence of 80 g air (containing 20% oxygen gas by mass). The maximum mass of CO2 that can be

produced is : (Mole-1(P)) [M] (A) 22 g (B*) 12.57 g (C) 14 g (D) 20 g STP ij ,d 5.6 L xSlh; ,Ydsu] ftldk 1 v.kq 6 H-ijek.kq ;qDr gS] dks 80 g ok;q (tks Hkkj ls 20% vkWDlhtu xSl

j[krh gS) dh mifLFkfr esa tyk;k tkrk gSA CO2 dk vf/kdre nzO;eku] tks mRikfnr gks ldrk gS] fuEu gS % (A) 22 g (B*) 12.57 g (C) 14 g (D) 20 g Sol. General formula of alkane = CnH2n + 2 If 1 molecule contains 6 H-atoms, then 2n + 2 = 6. n = 2 Alkane = C2H6

Combustion reaction is :

C2H6 + 72

O2 2CO2 + 3H2O

mole 5.622.4

20100 80

32

= 0.25 = 0.5

molest. coeff.

0.251 0.5

7 / 2 = 17

(LR)

Moles of CO2 produced = 0.5 27 / 2

= 27

Mass of CO2 produced = 27

◊ 44 = 12.57 g.

Sol. ,Ydsu dk lkekU; lw=k = CnH2n + 2 ;fn 1 v.kq] 6 H-ijek.kq j[krk gS] rc 2n + 2 = 6. n = 2 ,Ydsu = C2H6

ngu vfHkfØ;k fuEu gS %

C2H6 + 72

O2 2CO2 + 3H2O

eksy 5.622.4

20100 80

32

= 0.25 = 0.5

eksyjllehdj.kferhxq.kkad

0.251 0.5

7 / 2 = 17

(LR)

mRikfnr CO2 ds eksy = 0.5 27 / 2

= 27

mRikfnr CO2 dk nzO;eku = 27

◊ 44 = 12.57 g.

8. How much volume of 63% w/w aq. HNO3 solution (d = 1.5 g/ml) is diluted with sufficient water to prepare

1 L of 3 M HNO3 solution (Mole-1(P)) [M]

63% w/w tyh; HNO3 foy;u (d = 1.5 g/ml) dk fdruk vk;ru i;k ZIr ty ls ruq fd;k tk;s fd 3 M

HNO3 dk 1 L foy;u i zkIr gk s tk;sA (A*) 200 ml (B) 300 ml (C) 120 ml (D) 150 ml Ans. 200 ml 63 gm HNO3 is present in 100 gm solution

63 gm HNO3 is present in 1001.5

ml solution

M = 63 1.5 1063

so M1V1 = M2V2

15 V1 = 3 1 V1 = 0.2 Litre 100 xzke foy;u es a 63 gm HNO3 mifLFkr g SA

1001.5

ml foy;u es a 63 gm HNO3 mifLFkr gSA

M = 63 1.5 1063

vr% M1V1 = M2V2

15 V1 = 3 1 V1 = 0.2 Litre 9. 284 g of mixture of CaCO3 and MgCO3 is heated to a constant weight. If total volume of CO2 produced is

67.2L at STP, then find the mass of residue left. (Mole-1(P)) [M] (A) 284 g (B) 132 g (C) 240 g (D*) 152 g. CaCO3 o MgCO3 ds 284 g feJ.k dks ,d fu;r Hkkj rd xeZ fd;k tkrk gSA ;fn STP ij mRikfnr CO2 dk dqy

vk;ru 67.2L gS] rks 'ks"k cps vof'k"V dk nzO;eku Kkr djksA (A) 284 g (B) 132 g (C) 240 g (D*) 152 g.

Sol. Moles of CO2 evolved = 67.222.4

= 3 moles

Mass of CO2 evolved = 3 ◊ 44 = 132 g mass of residue left = 284 ñ 132 = 152 g Total mass = 152 g.

gy fu"dkflr CO2 ds eksy = 67.222.4

= 3 moles

fu"dkflr CO2 dk nzO;eku = 3 ◊ 44 = 132 g 'ks"k cps vof'k"V dk nzO;eku = 284 ñ 132 = 152 g

dqy nzO;eku = 152 g. 10. Similar to the % labelling of oleum, a mixture of H3PO4 and P4O10 is labelled as (100 + x) % where x is the

maximum mass of water which can react with P4O10 present in 100 g mixture of H3PO4 and P4O10. If such a mixture is labelled as 127 %. Mass of P4O10 in 100 gm of mixture, is (Mole-1(P))(M)

vkWfy;e ds % vadu ds leku H3PO4 o P4O10 ds ,d feJ.k dk vadu (100 + x) % ds :i essa fd;k tkrk gS tgk¡ ty dk vf/kdre nzO;eku x gS tks fd H3PO4 o P4O10 ds 100 g esa mifLFkr P4O10 ds lkFk fØ;k dj ldrk gSA ;fn

bl izdkj ds ,d feJ.k dk vadu 127 % gSA 100 g feJ.k esa P4O10 dk nzO;eku fuEu gS& (A*) 71 gm (B) 47 gm (C) 83 gm (D) 35 gm Sol. P4O10 + 6H2O 4H3PO4

1 276 18

2718

moles of P4O10 = 14

mass of P4O10 in 100 gm mixture = 14◊ 284 = 71 gm

Sol. P4O10 + 6H2O 4H3PO4

1 276 18

2718

P4O10 ds eksy = 14

100 g feJ.k esa P4O10 dk nzO;eku = 14◊ 284 = 71 gm

11. If the value of total spin for Xa+ ion is found to be ±2, then the possible values of Z (atomic number) and 'a'

are respectively : (ATS(P))(M) ;fn Xa+ vk;u ds fy, dqy pØ.k dk eku ±2 ik;k tkrk gS] rc Z (ijek.kq Øekad) rFkk 'a' ds lEHkkfor eku Øe'k% gSa % (A) 26 & 3 (2*) 24 & 2 (3) 27 & 2 (4) 24 & 1

Sol. Total spin = ± n ◊ 12

= ± 2.

no. of unpaired electron = n = 4. 26X3+ 1s2 2s22p6 3s2 3p6 3d5 (5 unpaired electron). 24X2+ 1s2 2s22p6 3s2 3p6 3d4 (4 unpaired electron). 27X2 1s2 2s22p6 3s2 3p6 3d7 (3 unpaired electron). 24X+ 1s2 2s22p6 3s2 3p6 3d5 (5 unpaired electron).

gy. dqy pØ.k = ± n ◊ 12

= ± 2.

v;qfXer bysDVªkWu dh la[;k = n = 4. 26X3+ 1s2 2s22p6 3s2 3p6 3d5 (5 v;qfXer bysDVªkWu). 24X2+ 1s2 2s22p6 3s2 3p6 3d4 (4 v;qfXer bysDVªkWu). 27X2 1s2 2s22p6 3s2 3p6 3d7 (3 v;qfXer bysDVªkWu). 24X+ 1s2 2s22p6 3s2 3p6 3d5 (5 v;qfXer bysDVªkWu).

12. Let the total number of orbitals in ath shell be 16 and value of azimuthal quantum number for the unpaired

electron in vanadium atom (Z = 23) be 'b', then find the sum (a + b). (ATS(P))(M) ekuk fd ath dks'k esa d{kdksa dh dqy la[;k 16 gS rFkk osusfM;e ijek.kq (Z = 23) esa v;qfXer bysDVªkWu ds fy, f}xa'kh

DokaVe la[;k dk eku 'b' gS rc (a + b) ds ;ksx dk eku Kkr dhft,A (A) 4 (2) 5 (3*) 6 (4) 3 Sol. Total number of orbitals in ath shell = a2 = 16 a = 4. value of for unpaired electron in V atom = 2. ( 3d orbital) b = 2 (a + b) = 4 + 2 = 6. gy. ath dks'k esa d{kdksa dh dqy la[;k = a2 = 16 a = 4. V ijek.kq esa] v;qfXer bysDVªkWu ds fy, dk eku = 2. ( 3d d{kd) b = 2 (a + b) = 4 + 2 = 6 13. Which of the following contain maximum number of radial nodes ? (ATS(P))(E) fuEu esa ls f=kT;h; uksM (radial nodes) dh vf/kdre la[;k dkSu j[krk gS\

(A*) 5s (B) 5p (C) 6d (D) 4f SCQ (12) 14. Professor I.L. Finar instructed his student to convert the -Amino acid (X) into a new compound (Y) in

which the ó NH2 group is replaced by ó COOH group. The student was confused in writing the correct IUPAC names of these two compounds. The correct IUPAC names of X and Y will be respectively.

(X) =

(IUPAC Nomenclature(O))

(A) 2-Ethyl-2-amino ethanoic acid, 2-Ethyl-2-carboxy ethanoic acid (B*) 2-Aminobutanoic acid, Ethylpropanedioic acid (C) 2-Aminobutanoic acid, 2-Carboxybutanoic acid (D) 2-Carboxypropan-1-amine, Propane-1, 1-dicarboxylic acid. izksQslj I.L. Finar us vius fo|kFkhZ dks ,d mnkgj.k fn;k ftlesa -vehuks vEy (X) dks ,d u;s ;kSfxd (Y) ftlesa

ó NH2 lewg dks ó COOH lewg }kjk izfrLFkkfir fd;k x;k Fkk] esa ifjofrZr djus ds fy, fn;kA bu nksuksa ;kSfxd ds IUPAC uke fy[kus esa fo|kFkhZ dks my>u (confused) gqvkA Øe'k% X vkSj Y ds lgh IUPAC uke gSA

(X) =

(A) 2-,fFky-2-,ehuks ,FksukWbd vEy, 2-,fFky-2-dkcksZDlh ,FksukWbd vEy (B*) 2-,ehuksC;wVsuksbd vEy, ,fFkyizksisuMkbZvkWbd vEy (C) 2-,ehuksC;wVsuksbd vEy, 2-dkcksZDlhC;wVsuksbd vEy (D) 2-dkcksZDlhizksisu-1-,sehu, izksisu-1, 1-MkbZdkWcksZfDlfyd vEy

Sol. X = Y =

2-Aminobutanoic acid Ethylpropanedioic acid 2-,ehuksC;wVsuksbd vEy ,fFkyizksisuMkbZvkWbd vEy

15. In the following compound

O

CH O

CH

R2 R1

which of the following R1 and R2 functional groups will change the word root. (Made by SSS SIR)

(IUPAC Nomenclature(O))

(A) ñ CN and ñ OH (B) ñ S2OH and ñ OCH3 (C*) ñ CN and ñ CH2 ñ CH = CH2 (D) ñ COOH and ñ CH3

O

CH O

CH

R2 R1 fuEufyf[kr ;kSfxd esa]

fuEu esa ls fdlesa R1 rFkk R2 fØ;kRed lewg ds dkj.k ewy 'kCn ifjofrZr gks tk;sxk\ (A) ñ CN rFkk ñ OH (B) ñ S2OH rFkk ñ OCH3 (C*) ñ CN rFkk ñ CH2 ñ CH = CH2 (D) ñ COOH rFkk ñ CH3 Sol. CN and ñ CH2 ñ CH = CH2 will give non as a word root. CN rFkk ñ CH2 ñ CH = CH2 ewy 'kCn ds :i esa iz;qDr ugha gksxsaA 16. Select the structures with correct numbering for IUPAC name of the compound. (IUPAC Nomenclature(O)) fuEu ;kSfxdksa ds IUPAC ukedj.k ds fy;s lgh Øekadu okyh lajpukvksa dk p;u dhft;sA

(A) (B*) (C) (D)

Sol.

&

represent correct numbering

Sol.

rFkk

esa lgh Øekadu fd;k x;k gSA

17. In which of the following side chain (acyclic chain) is the main chain ? (IUPAC(O)) fuEu esa dkSulh ik'oZ Ja[kyk ¼vpØh; Ja[kyk½] eq[; Ja[kyk gS ?

(A) (B) (C*) (D)

18. Which of the following substituent is absent on the parent chain of the given compound ? (IUPAC(O))

(A) Ethyl (B) Methyl (C*) Tertiarybutyl (D) None of these uhps fn;s x;s ;kSfxd dh iSrd Ja[kyk eas fuEu esa ls dkSulk izfrLFkkih mifLFkr ugha gS \

(A) ,fFky (B) esfFky (C*) rrh;d C;wfVy (D) buesa ls dksbZ ugha Sol. Tertiarybutyl group is absent. rrh;d C;wfVy lewg mifLFkr ugha gSA

19.

OCOOHCOOH

The IUPAC name of this compound is :

(IUPAC(O))

(A) Decanoic anhydride (B) 2,8-Dicarboxyoxacyclononane (C*) 2,9-Epoxydecanedioic acid (D) 2,9-Epoxydecanedicarboxylic acid.

O

COOHCOOH

mijksDr ;kSfxd dk lgh IUPAC uke gS %

(A) Msdsuksbd ,sugkbMªkbM (B) 2,8-MkbdkcksZDlhvkWDlklkbDyksuksusu (C*) 2,9-bikWDlhMsdsuMkbvksbd vEy (D) 2,9-bikWDlhMsdsuMkbdkcksZfDlfyd vEy

Sol.

OCOOHCOOH

5

67

89 10

1 2 3

4

2,9-Epoxydecanedioic acid 2,9- bikWDlhMsdsuMkbvksbd vEy

20. 1-(1-Chloropropyl)-4-(2-chloropropyl)cyclohexane and 1-(2-Chloropropyl)-4-(3-chloropropyl)cyclohexane

are (Isomerism(O)) (A) Chain isomers (B*) Position isomers (C) Metamer isomers (D) Identical compounds 1-(1-Dyksjksizksfiy)-4-(2-Dyksjksizksfiy)lkbDyksgsDlsu ,oa 1-(2-Dyksjksizksfiy)-4-(3-Dyksjksizksfiy)lkbDyksgsDlsu gSA (A) Ja[kyk leko;oh (B*) fLFkfr leko;oh (C) e/;ko;oh leko;oh (D) le:i ;kSfxd

Sol. and (,oa) are position isomers . (fLFkfr leko;oh gS)

21. Which amongs the fol lowing can be the structure of molecular formula C5H8O 2. v.kqlw=k C5H8O2 okys ;kSfxd dh nh xbZ lajpukvksa esa ls dkSulh lajpuk ;kSfxd dh gS\ (Isomerism(O))

(A) (2) (3) (4*)

Sol. C5H8O 2 has two D.U. D is correct. C5H8O 2 ds fy, D.U. dk eku nks gSA D lgh fodYi gSA

22. NH ñ CHO ; H2N CHO

I II Which type of isomerism is observed between I and II. (Isomerism(O)) (A) Chain isomers (B) Position isomers (C*) Functional isomers (D) Metamer

NH ñ CHO ; H2N CHO

I II I rFkk II ds e/; dkSulh leko;ork ik;h tkrh gS\ (A) Ja[kyk leko;oh (B) fLFkfr leko;oh (C*) fØ;kRed leko;oh (D) e/;ko;oh

Sol. ;

Amide functional group Aldehyde and 1∫ amine

;

,sekbZM fØ;kRed lewg ,fYMgkbM rFkk 1∫ ,ehu 23. How many position of the skeleton of 3-methylhexane at which double bond is placed and that gives

positional isomers only. (Isomerism(O)) 3-esfFkygsDlsu dh lajpuk esa fdruh fLFkfr ij f}ca/k mifLFkr gS tks dsoy fLFkfr leko;oh nsrk gS % (A) 4 (B*) 5 (C) 6 (D) 7

Sol. 46

5 1

2

3

, 46

5 1

2

3

, 31

2 65

4

, 31

2 6

5

4

, 31

2 6

5

4

24. The incorrect combination of names for isomeric alcohols with molecular formula C4H10O is: (Isomerism(O)) (A) tert-butanol and 2-methylpropan-2-ol (B*) tert-butanol and 1, 1-dimethylethan-1-ol (C) n-butanol and butan-1-ol (D) isobutyl alcohol and 2-methylpropan-1-ol vk.kfod lw=k C4H10O okys leko;oh (isomeric) ,sYdksgkWyksa ds xyr ukeksa ds la;qDr gS ¼gSa½ % (A) rrh;d-C;wVsukWy (tert-butanol) ,oa 2-esfFkyizksisu-2-vkWy (B*) rrh;d-C;wVsukWy ,ao 1, 1-MkbesfFky,sFksu-1-vkWy (C) n-C;wVsukWy ,ao C;wVsu-1-vkWy (D) vkblksC;wfVy ,YdksgkWy ,oa 2-esfFkyizksisu-1-vkWy Sol. Alcohols with formula C4H10O are -

gy- lw=k C4H10O ds ,YdkWgy fuEu gS %

OH

1-C;wVsukWy OH

2-C;wVsukWy

OH

2- -2- (1,1- -1- )

rrh; C;wfVy ,Ydksgy ;kesfFkyC;wVsu vkWy

MkbesfFky,sFksu vkWy ugha

25. Arrange the following molecules in increasing order of to bond ratio: (IUPAC(O)) fuEufyf[kr esa ls dkSulk Øe mijksDr v.kqvksa ds o ca/k vuqikr dk c<+rk gqvk lgh Øe gS %

N NH

O

(I) (II) (III) (IV) (A) I < II < III < IV (B) I < II < IV < III (C*) II < I < IV < III (D) IV < II < I < III MCQ (3) 26. Ammonia (NH3) gas combines with oxygen gas over Pt catalyst to produce Nitric oxide (NO) and water. If

13.6 g of NH3 gas is taken initially, then : (Mole-1(P))(M) (A*) Volume of oxygen gas required at NTP is 22.4 L (B*) Volume of H2O( ) produced at 4∞ C (assuming density of water as 1000 Kg/m3) is 21.6 mL. (C*) Total mass of products obtained is 45.6 g. (D*) Number of moles of NO produced is 0.8. Pt mRiszjd dh mifLFkfr esa veksfu;k xSl (NH3), vkWDlhtu xSl (O2) ls fØ;k dj ukbfVªd vkWDlkbM (NO) ,oa ty

mRikfnr djrh gSA ;fn izkjEHk esa 13.6 g NH3 xSl yh tk;s] rkss % (A*) NTP ij vkWDlhtu xSl ds 22.4 L vk;ru dh vko';drk gksxhA (B*) 4∞C rki ij mRikn H2O( ) dk vk;ru 21.6 mL gSA ¼ty dk ?kuRo 1000 Kg/m3 ekudj½ (C*) izkIr mRiknksa dk dqy Hkkj 45.6 g gSA (D*) NO ds mRikfnr eksyksa dh la[;k 0.8 gSA

Sol. 4NH3 + 5O2 4NO + 6H2O

Volume of oxygen required at NTP = 13.617

54

22.4 = 22.4 L

Mass of water produced = 13.617

64

18 = 21.6 g

volume of H2O ( ) = 21.6 mL ( dH2O = 1000 Kg/m3 = 1 g/mL)

moles of NO produced = 13.617

= 45

= 0.8

Mass of NO produced = 0.8 30 = 24 g Total mass of products = mNO +

2H Om = 24 + 21.6 = 45.6 g Sol. 4NH3 + 5O2 4NO + 6H2O

NTP ij O2 dk vko';d vk;ru = 13.617

54

22.4 = 22.4 L

ty dk mRikfnr Hkkj = 13.617

64

18 = 21.6 g

H2O ( ) dk vk;ru = 21.6 mL ( dH2O = 1000 Kg/m3 = 1 g/mL)

NO ds mRikfnr eksy = 13.617

= 45

= 0.8

NO dk mRikfnr Hkkj = 0.8 30 = 24 g mRiknksa dk dqy Hkkj = mNO +

2H Om = 24 + 21.6 = 45.6 g 27. Which of the following statements is/are correct regarding 6.4g of 216

8 O ion : (Mole-1(P))(M) (A*) Total number of proton present is about 1.924 ◊ 1024. (B*) Total number of electrons present is 4NA (C) Magnitude of net charge on sample is 0.4 NAC (Coulomb.) (D*) Total number of neutrons present is 3.2 NA

2168 O vk;u ds 6.4 xzke ds lEcU/k esa fuEu esa ls dkSulk@dkSuls dFku lR; gS@gSa?

(A*) mifLFkr izksVkWuksa dh dqy la[;k 1.924 ◊ 1024 gSA (B*) mifLFkr bysDVªkWuksa dh dqy la[;k 4NA gSA (C) uewus ij dqy vkos'k dk ifjek.k 0.4 NAC (dqykEc) gSA (D*) mifLFkr U;wVªkWuksa dh dqy la[;k 3.2 NA gSA

Sol. Total number of proton present = 8N16

4.6A = 3.2 NA = 1.926 ◊ 1024

mifLFkr izksVªkWuksa dh dqy la[;k= 8N16

4.6A = 3.2 NA = 1.926 ◊ 1024

Total number of electrons present = 16

4.6 ◊ NA ◊ 10 = 4NA

mifLFkr bysDVªkWuksa dh dqy la[;k= 16

4.6 ◊ NA ◊ 10 = 4NA

Magnitude of net charge on sample = 16

4.6 ◊ NA ◊ 2e = 0.8 NAe

uewus ij dqy vkos'k dk ifjek.k = 16

4.6 ◊ NA ◊ 2e = 0.8 NAe

Total number of neutrons present = 16

4.6 ◊ NA ◊ 8 = 3.2 NA

mifLFkr U;wVªkWuksa dh dqy la[;k = 16

4.6 ◊ NA ◊ 8 = 3.2 NA

28. Which of the following statement is correct for 3dxy orbitals? (ATS(P))(M)

(A*) The orbitals drawn has two nodal planes, xz and yz. (B) The minimum probability point lie along = 45∫ . (C*) + ve and ñ ve signs represent sign of amplitude of electron wave. (D*) It is a non-axial orbital. 3dxy d{kdksa ds lUnHkZ esa dkSulk dFku lgh gS ?

(A*) n'kkZ;k x;k d{kd nks uksMy ry xz o yz j[krk gSA (B) = 45∫ ds vuqfn'k U;wure izkf;drk fcUnq gksrk gSA (C*) + ve o ñ ve fpUg bysDVªkWu rjax ds vk;ke (amplitude) ds fpUg dks iznf'kZr djrs gSA (D*) ;g ,d fuj{kh; d{kd gSA Sol. These are the fact. ;g rF; gSaA MCQ (2) 29. Which of the following relation is correct : (Isomerism(O))

fuEu esa ls dkSulk laca/k lgh gS %

(A*) and are chain iosmers Ja[kyk leko;oh gSaA

(B) and are positional isomers fLFkfr leko;oh gSaA

(C*) and are metamers e/;ko;oh gSaA

(D*) and are functional group isomers fØ;kRed lewg leko;oh gSaA

30. Which of the following compounds is/are a pair of isomers ? (Isomerism(O))

fuEu esa ls dkSulk@dkSuls leko;oh;ksa dk ;qXe gS \

(A*) and (B) and

(C*) and (D) and

Sol. (A) and are structural isomers.

(B) These compounds are identicle (C) These are chain isomers (D) These are homologs. Sol. (A) rFkk lajpukRed leko;oh gSaA

(B) ;s ;kSfxd le:ih gSA (C) ;s Jaa[kyk leko;oh gSA (D) ;s letkr gSA Comp.(3 x 2Q) 2

Paragraph for Question Nos. 31 to 32 iz'u 31 ls 32 ds fy, vuqPNsn

A gas X is introduced in a vessel of volume 60 Lt. at temperature 300K. When another gas Y along with

pervious gas X is added in the vessel of double capacity at 150 K temperature, the pressure becomes half. On this basis answer the following. (Mole-1(P))(M)

300 K rki ij 60 Lt. vk;ru okys ik=k esa ,d xSl X yh tkrh gSA tc 150K rki ij nksxquh {kerk ds ik=k esa mijksDr xSl X dks vU; xSl Y ds lkFk feyk;k tkrk gS] rks nkc vk/kk gks tkrk gSA bl vk/kkj ij fuEu iz'uksa ds mRrj nhft;sA

31. The relation between moles of gas X and gas Y present finally in the mixture is - (Mole-1(P))(M) vUrr% feJ.k esa xSl X rFkk xSl Y ds eksyks ds e/; lEcU/k fuEu gS %

(A) 1 : 3 (B*) 2 : 2 (C) 3 : 4 (D) 4 : 3 32. If mass of gas X added in 10 times the mass of Y added then the ratio molar mass of gas X and Y is ñ ;fn feyk;h x;h xSl X dk nzO;eku] feyk;h x;h xSl Y ds nzO;eku ls 10 xquk gS] rc xSl X rFkk xSl Y ds eksyj

nzO;ekuksa dk vuqikr fuEu gS : (Mole-1(P))(M) (A) 1 : 5 (B) 0.2 : 0.25 (C*) 10 : 1 (D) 1 : 3 Sol.

1 2

X X YV 60Lt. V 120Lt.T 300K T 150KP P P P / 2

12X

= 1X Y

X = Y Mass of X = 10 gm X dk nzO;eku = 10 gm Mass of Y = X Y dk nzO;eku = X

Moles of X = X

10aM

; Moles of Y = y

aM

X ds eksy = X

10aM

; Y ds eksy = y

aM

X

Y

10a /Ma /M

= 1 Y

X

M 1M 10

X

Y

M 10M 1

Paragraph for Question Nos. 33 to 34

iz'u 33 ls 34 ds fy, vuqPNsn

Following reaction sequence is given : 2A + B A2B (A) 2A2B + B A4B3 (2) (2) reaction starts after (A) reaction is finished. fuEu vfHkfØ;k vuqØe fn;s x;s gS % 2A + B A2B (A) 2A2B + B A4B3 (2) vfHkfØ;k (A) ds iw.kZ gksus dss i'pkr~ vfHkfØ;k (2) izkjEHk gksrh gSA (Mole-1(P))(M) 33. If 10 moles of A and 7.5 moles of B are taken in a flask, then moles of A4B3 produced in reaction are : (Mole-1(P))(M) ;fn ,d ¶ykLd esa 10 eksy A o 7.5 eksy B fy;s tkrs gSa] rks vfHkfØ;k esa mRikfnr A4B3 ds eksy fuEu gSa % (A) 1 (B) 1.5 (C) 2 (D*) 2.5 34. Mass of A2B left in above reaction sequence (molar mass of A = 24 g/mol and molar mass of B = 16

g/mol.) (Mole-1(P))(M) mijksDr vfHkfØ;k vuqØe esa 'ks"k cps A2B dk nzO;eku fuEu gS % (A dk eksyj nzO;eku = 24 g/mol rFkk B dk eksyj nzO;eku = 16 g/mol.) (A*) 0 (B) 1 (C) 1.5 (D) 2 Sol. L.R. 2A + B A2B 10 7.5 5 mole B left = 7.5 ñ 5 =2.5 2A2B + B A4B3 (No L.R.) 5 2.5 2.5 moles of A4B3 produced = 2.5

moles of A2B left = 0

mRikfnr A4B3 ds eksy = 2.5 'ks"k cps A2B ds eksy = 0

Comp.(3 x 2Q) 1

Paragraph for Question Nos. 35 to 36 iz'u 35 ls 36 ds fy, vuqPNsn

The analgesic drug ibuprofen (P) exist in two isomeric form. One isomer is Physiologically active, while the

other is inactive. The structure of ibuprofen is given below.

nnZfuokjd vkS"k/kh vkbczwQsu (P) ds nks leko;oh :i laHko gS ftuesa ls dsoy ,d leko;oh gh euksjksxfpfdRlk esa

lfØ; gS tcfd nwljk leko;oh vlfØ; gksrk gSA vkbczwQsu dh lajpuk uhps nh xbZ gS %

Me COOH

(P)

(IUPAC Nomenclature (O))

35. The IUPAC name of (P) is (A) 3-[para-Isobutylphenyl] propanoic acid (B*) 2-[para-Isobutylphenyl] propanoic acid (C) 3-[para-sec-Butylphenyl] propanoic acid (D) 2-[para-sec-Butylphenyl] propanoic acid

;kSfxd (P) dk lgh IUPAC uke gS %

(A) 3-[iSjk-vkblksC;wfVyQsfuy] izksisuksbd vEy

(B*) 2-[iSjk-vkblksC;wfVyQsfuy] izksisuksbd vEy

(C) 3-[iSjk-f}rh;d-C;wfVyQsfuy] izksisuksbd vEy

(D) 2-[iSjk-f}rh;d-C;wfVyQsfuy] izksisuksbd vEy

Sol.

Me COOH

(P)

1

2

3

2-[para-Isobutylphenyl] propanoic acid

2-[iSjk-vkblksC;wfVyQsfuy] izksisuksbd vEy

36. The number of sp3 hybridised carbon atom in (P) is : ;kSfxd (P) esa fdrus ladfjr dkcZu ijek.kq mifLFkr gS\ (IUPAC Nomenclature(O)) (A) 3 (B) 4 (C*) 6 (D) 5 Sol. Number of sp3 hybridised carbon atom is 6.

mijksDr ;kSfxd esa 6 sp3 ladfjr dkcZu ijek.kq mifLFkr gS\

MTC (4 Vs. 5) 1 37. Column-I Column-II (Mole-1(P))(M)

(A) 64 g of SO2 (p) 6.02 ◊ 1023 atoms of Sulphur

(B) 6.02 ◊ 1023 molecules of CO2 (q) 2 gm atoms of Oxygen

(C) 1 gm molecule of SO3 (r) 18.06 ◊ 1023 total atoms.

(D) 22.4 L of CO at 1 atm & 273 K (s) 1 mole of given compound.

(t) 14 NA number of electrons.

LrEHk-I LrEHk-II

(A) SO2 ds 64 xzke (p) lYQj ds 6.02 ◊ 1023 ijek.kq

(B) CO2 ds 6.02 ◊ 1023 v.kq (q) vkWDlhtu ds 2 xzke ijek.kq

(C) SO3 ds 1 xzke v.kq (r) dqy ijek.kq 18.06 ◊ 1023

(D) 1 atm o 273 K ij CO ds 22.4 yhVj (s) fn;s x;s ;kSfxd ds 1 eksy

(t) bysDVªkWu dh la[;k 14 NA

Ans. (A) p, q, r, s ; (B) q, r, s ; (C) p, s ; (D) s, t

Sol. (A) 64 g of SO2

Moles of SO2 = 6464 = 1

Number of molecules = 1 NA

Number of ëSí atoms = 1 NA

Number of ëOí atoms = 2 NA

Total number of atoms = 3 NA , No. of eñ = 32 NA.

(B) 6.02 ◊ 1023 molecules of CO2

moles of CO2 = 23

23

1002.61002.6 = 1

Number of molecules = 1 NA.

Number of ëCí atoms = 1 NA.

Number of ëOí atoms = 2 NA.

Total number of atoms = 3 NA.

Number of eñ = 22 NA.

(C) 1 gm molecule of SO3

moles of SO3 = 1


Number of ëSí atoms = 1 NA.

Number of ëOí atoms = 3 NA.



(D) moles of CO = RTPV =

2730821.04.221 = 1


Number of ë C í atoms = 1 NA.

Number of ë O í atoms = 1 NA.



(A) SO2 ds 64 xzke

SO2 ds eksy = 6464 = 1

v.kqvksa dh la[;k = 1 NA

ëSí ijek.kqvksa dh la[;k = 1 NA

ëOí ijek.kqvksa dh la[;k = 2 NA

ijek.kqvksa dh dqy la[;k = 3 NA , No. of eñ = 32 NA.

(B) CO2 ds 6.02 ◊ 1023 v.kq

CO2 ds eksy = 23

23

1002.61002.6 = 1

v.kqvksa dh la[;k = 1 NA.

ëCí ijek.kqvksa dh la[;k = 1 NA.

ëOí ijek.kqvksa dh la[;k = 2 NA.

ijek.kqvksa dh dqy la[;k = 3 NA.

bySDVªkWuksa dh la[;k = 22 NA.

(C) SO3 ds 1 xzke v.kq

SO3 ds eksy = 1

v.kqvksa dh la[;k = 1 NA.

ëSí ijek.kqvksa dh la[;k = 1 NA.

ëOí ijek.kqvksa dh la[;k = 3 NA.

ijek.kqvksa dh dqy la[;k = 4 NA.


(D) CO ds eksy = RTPV =

2730821.04.221 = 1

v.kqvksa dh la[;k = 1 NA. ëCí ijek.kq dh la[;k = 1 NA. ëOí ijek.kq dh la[;k = 1 NA. ijek.kq dh dqy la[;k = 2 NA.


Integer(Single digit) 4 38. Consider the reaction

2As2O3 + 3UO2 (NO3)2 U3O8 + 6NO2 + 2As2O5 (U = 238, As = 75, N = 14, O = 16) How many of the following conclusions are correct based on this equation ?

(Made by SM SIR On July2014) (Mole-1(P))(M)

(A) Mass ratio of As2O3 and UO2(NO3)2 present is 2 : 3 (B) Mole ratio of As2O3 and UO2(NO3)2 present is 2 : 3 (C) Mass ratio of As2O3 and UO2(NO3)2 reacted is 2 : 3 (D) Mole ratio of UO2(NO3)2 and U3O8 in reaction mixture is 3 : 1 (E) Mass ratio of UO2(NO3)2 and U3O8 in reaction mixture is 3 : 1 (F) Mole ratio of NO2 and U3O8 formed is 6 : 1 (G) Mole ratio of As2O5 and As2O3 present in reaction mixture is 1 : 1 (H) Sum of number of moles of As2O3 and As2O5 in the reaction mixture is always constant. (POAC) (I) Sum of number of moles UO2(NO3)2 and U3O8 present in the reaction mixture is always constant.

(POAC) fuEu vfHkfØ;k dk voyksdu dhft,

2As2O3 + 3UO2 (NO3)2 U3O8 + 6NO2 + 2As2O5

(U = 238, As = 75, N = 14, O = 16)

mijksDr lehdj.k ds vk/kkj ij fdrus fu"d"kZ fuEu esa ls lgh gS\

(A) As2O3 o UO2(NO3)2 dk nzO;eku vuqikr 2 : 3 gSA

(B) As2O3 o UO2(NO3)2 dk eksy vuqikr 2 : 3 eksy gSA

(C) fØ;k'khy As2O3 o UO2(NO3)2 dk nzO;eku vuqikr 2 : 3 gSA

(D) vfHkfØ;k feJ.k esas UO2(NO3)2 o U3O8 dk eksy vuqikr 3 : 1 gSA

(E) vfHkfØ;k feJ.k esas UO2(NO3)2 o U3O8 dk nzO;eku vuqikr 3 : 1 gSA

(F) cuk;s x;s NO2 o U3O8 dk eksy vuqikr 6 : 1 gSA

(G) vfHkfØ;k feJ.k esa mifLFkr As2O5 o As2O3 dk eksy vuqikr 1 : 1 gSaA

(H) vfHkfØ;k feJ.k esas mifLFkr As2O3 o As2O5 ds eksyksa dh la[;k dk ;ksx ges'kk fu;r jgrk gSA (POAC)

(I) vfHkfØ;k feJ.k esas mifLFkr UO2(NO3)2 o U3O8 ds eksyksa dh la[;k dk ;ksx ges'kk fu;r jgrk gSA (POAC)

Ans. 3 Sol. (C), (F) & (H) are correct statements. Rest are wrong. (C), (F) o (H) lgh dFku gSaA 'ks"k xyr gSaA 39. 27 kg of SO2Cl2 is reacted with excess of NaOH completely. If the difference between masses of Na2SO4

and NaCl produced is x kg, then x is : (Mole-1(P))(M) (SO2Cl2 + NaOH Na2SO4 + NaCl + H2O) 27 kg SO2Cl2, NaOH ds vkf/kD; ds lkFk iw.kZr% vfHkfØ;k djrk gSA ;fn mRIkkfnr Na2SO4 o NaCl ds nzO;ekuksa ds

e/; vUrj x kg gS, rc x dk eku Kkr dhft, : (SO2Cl2 + NaOH Na2SO4 + NaCl + H2O) Ans. 5 Kg Sol. SO2Cl2 + 4NaOH Na2SO4 + 2NaCl + 2H2O

moles = 27000135

=200 200 moles 400 moles

m = 200 ◊ 142 g m = 400 ◊ 58.5 g = 28.400 kg = 23.4 kg x = 28.4 ñ 23.4 = 5 kg. 40. Mole fraction of solute in solution is 0.25. What is the moles of solvent in solution having 1 mole of solute?

(Mole-1(P))(M) ,d foy;u esa foys; dh eksy fHkUu 0.25 gSA 1 eksy foys; j[kus okys foy;u esa foyk;d ds eksyksa dh la[;k fdruh

gksxh \ Ans. 3

Sol. nsolvent = 0.750.25

◊ 1 = 3

41. Find the weight of a sample(in g) which contains 0.5 mole of Ne gas and about 1.505 ◊1023 molecules of

an unknown gas X. If the sample has average molar mass 100 g/mole. Report your answer dividing the weight of sample by 15. (Mole-1(P))(M)

,d izkn'kZ dk Hkkj (xzke esa) Kkr dhft,] ftlesa Ne xSl ds 0.5 eksy rFkk ,d vKkr xSl X ds yxHkx 1.505 ◊1023

v.kq gS ;fn izkn'kZ dk vkSlr eksyj nzO;eku 100 g/mol gS rks viuk mÙkj] izkn'kZ ds Hkkj dks 15 ls Hkkx nsus ds i'pkr~

nsaA

Ans. 5

Sol. Average Molar mass = molestotalweighttotal

100 =

23

23

10023.610505.15.0

weighttotal = 25.05.0

weighttotal

100 = 75.0

weighttotal total weight = 100 ◊ 0.75 = 75

1575 = 5

vkSlr eksyj nzO;eku = eksydqy Hkkjdqy

100 =

23

23

10023.610505.15.0

Hkkjdqy =

25.05.0 Hkkjdqy

100 = 75.0 Hkkjdqy

dqy Hkkj = 100 ◊ 0.75 = 75

1575 = 5

Integer(Single digit) 4

42. How many positional isomers of dichloronaphthalene (C10H6Cl2) are possible in which each benzene ring contains only one chlorine atom. (Monochlorination(O))

MkbDyksjksuS¶FkSyhu (C10H6Cl2) ds fdrus fLFkfr leko;oh laHko gS ftlessa izR;sd csUthu oy; ij dsoy ,d Dyksjhu ijek.kq ekStwn gS\

Ans. 6

Sol.

Cl Cl

,

ClCl

, Cl Cl

,

Cl

Cl,

Cl

Cl

,

Cl

Cl

43. Consider all possible isomeric alcohols of MW = 102. How many of them are secondary alcohols : (Structure isomer(O)) v.kqHkkj = 102 okys ;kSfxd ds lHkh lEHkkfor ,YdksgkWy ij fopkj dhft,s] buesa ls fdrus ,YdksgkWy] f}rh;d ,YdksgkWy

gSA Ans. 6

Sol. CñCñCñCñCñC

|OH ,

CñCñCñCñCñC|OH ,

CñCñCñCñC|OHC

| , CñCñCñCñC

|OH

C| ,

CñCñCñCñC|OH

C| ,

CñCñCñC|C

OH|

C|

44. How many substi tuted phenol are poss ible by the molecular formula C10H10O. v.kqlw=k C10H10O ds fdrus izfrLFkkih fQukWy lEHko gS\ (Hydrogenation(O)) Ans. 9

Sol.

C H2 5

OH

,

C H2 5

OH,

C H2 5

OH,

CH3

CH3

OH,

CH3

CH3

OH,

CH3

CH3

OH

,

CH3

CH3

OH

,

CH3

H3C

OH

,

OH

H3C CH3

45.

O

COOH

HN

Ph

O NH

CñOPh

O

The number of functional group in the above compound is : mijksDr ;kSfxd esa mifLFkr fØ;kRed lewgksa dh la[;k dk dqy ;ksx gS % (IUPAC Nomenclature (O))

Ans. 5 Sol. Function group = 5 Sol. fØ;kRed lewg = 5

Documents

Imp problems