PAPERŒ2 - Art of Problem Solving

PAPER�2

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

1 to 5 SCQ (5 options) 5 3 -1 15

6 to 9 Comprehension (2 Com. × 2 Q.) 4 3 0 12

10 to 19 Double digit Integer Type 10 4 �1 40

20 to 24 SCQ (5 options) 5 3 -1 15



39 to 43 SCQ (5 options) 5 3 -1 15



57 Total 201

Paper-2

Total

Physics

Maths

Chemistry

SECTION - IStraight Objective Type

This section contains 5 multiple choice questions. Each question has choices (A), (B), (C), (D) and (E),out of which ONLY ONE is correct.

[k.M- Ilh/ks oLrqfu"B izdkj

bl [k.M esa 5 cgq&fodYih iz'u gSA izR;sd iz'u ds 5 fodYi (A), (B),(C), (D) rFkk (E) gSa, ftuesa ls flQZ ,d lgh gSA

1. Slider block A moves to the right with constant velocity 6 m/s in the arrangement as shown in the figure. C and

D are points on the string . If BV , CV and DV are the velocities of B, C and D respectively then, choose

incorrect option :

CykWd A nka;h vksj fu;r osx 6 m/s ls n'kkZ;s vuqlkj xfr dj jgk gSA C o D Mksjh ij nks fcUnq gSA ;fn B,C o D ds osx Øe'k%

BV , CV rFkk DV gS] rc xyr fodYi dk p;u dhft,A

A

6 m/s

C D

B

(A) BDC V4VV (B*) BDBCB VVVVV

(C) BDBCB V2VVVV (D) BDC V2VV

(E) 4VV DC

Sol.

A

6 m/s

C D

B

X3 X4X2

VB

X1

Clearly Li"Vr% , VC = 6 m/s

By string constraint, Mksjh ca/ku ls

0xxxx 4321

�6 + VB + V

B + V

B = 0 V

B = 2 m/s

Also, ;g Hkh BDc V

2

VV

VD = � 2 m/s

VD = 2 m/s

C D

VB

VDVC

Thus vr% s/m826VV DC

s/m4VV CB

s/m4VV DB

s/m426VV DC

2. A particle is moving in a straight line according to graph given below : (a is acceleration, v is velocity, x is position,t is time) :uhps fn;s x;s xzkQ ds vuqlkj ,d d.k ljy js[kk esa xfr'khy gS rks (a Roj.k gS, v osx gS, x fLFkfr gS, t le; gSA) :

(A*)

x

a

(B*) (C)

x

t

(D)

(E) None of these buesa ls dksbZ ugh

Sol. Given fn;k x;k gS v = x & t = 0v = 1, x = 1

dv dxa

dt dt &

dvv. x

dx

a = v = x

v v

1 1

v.dv x.dx

v v2 2v x1 1 v = x

dxx

dt

v v

1 0

dxdt

x nx t

tx e

3. The system is released from rest with both the springs in unstretched position. Mass of each block is 1 kg andforce constant of each spring is 10 N/m. Assume pulleys and strings are massless and all contacts are smooth.(g = 10 m/s2)viuh LokHkkfod yEckbZ esa fLFkr nks fLizaxksa dk ,d fudk; fojke ls NksM+k tkrk gSA izR;sd CykWd dk nzO;eku 1 kg rFkk izR;sdfLizax dk cy fu;rkad 10 N/m gSA ;g ekfu, fd f?kjfu;k¡ rFkk Mksfj;k¡ nzO;ekughu gS rFkk lHkh lEidZ fpdus gS(g = 10 m/s2)

(A) Extension of horizontal spring in equilibrium is 1/5 m(B*) Extension of vertical spring in equilibrium is 1/5 m(C) Extension of horizontal spring in equilibrium is 3/5 m(D) Extension of vertical spring in equilibrium is 3/5 m(E) Extension of vertical spring and horizontal spring in equilibrium are equal

(A) lkE;oLFkk esa {kSfrt fLizax dk foLrkj 1/5 m gSA (B*) lkE;oLFkk es Å /okZ/kj fLizax dk foLrkj 1/5 m gSA(C) lkE;oLFkk es {kSfrt fLizax dk foLrkj 3/5 m gSA (D) lkE;oLFkk es Å /okZ/kj fLizax dk foLrkj 3/5 m gSA(E) lkE;oLFkk es Å /okZ/kj fLçax rFkk {kSfrt fLçax dk foLrkj leku gksrk gSA

Sol.

T = 2kx5kx = mg

x = k5

mg =

105101

=

51

4. There is a trolley in which there is a fixed inclined surface on which a smooth block of mass 5 kg is placed.Two horizontal forces of magnitude F

1 and F

2 are applied on the trolley as shown to keep the trolley at

rest.The value of F1 � F

2 is: (Assume there is no friction between trolley and horizontal ground and g = 10

m/s2)fp=kkuqlkj ,d VªkWyh ds vUnj fLFkj ur lrg cuh gqbZ gS] ftl ij 5 kg nzO;eku dk ,d fpduk CykWd j[kk gSA F

1 o F

2 nks

{kSfrt cy fp=kkuqlkj VªkWyh ij VªkWyh dks fojkekoLFkk esa j[kus ds fy, yxk;s tkrs gSA F1 � F

2 dk eku gksxk : (;g ekfu, fd VªkWyh

rFkk {kSfrt lrg ds e/; dksbZ ?k"kZ.k ugha gS rFkk g = 10 m/s2)

(A) 0 N (B) 24N (C*) �24N (D) � 30N (E) 30 N

Sol.

F1 + N sin 37 = F

2

F1 � F

2 = � N sin 37

N = mg cos 37

F1 � F

2 = �mg cos 37 sin 37 = �5 × 10 ×

54

× 53

= �24N

5. A block weighing 10 N travels down a smooth curved track AB joined to a rough horizontal surface (figure). Therough surface has a friction coefficient of with the block. If the block starts slipping on the track from a point 1.0m above the horizontal surface, the distance it will move on the rough surface is 10 m, then is :

,d 10 N Hkkj dk xqVdk fpdus oØh; iFk AB tks [kqjnjh {kSfrt lrg ¼fp=k esa½ ls tqM+k gS ij uhps xfr djrk gSA [kqjnjh lrg

dk ?k"kZ.k xq.kkad xqVds ds lkFk gSA ;fn xqVdk {kSfrt lrg ls 1.0 eh- Å apkbZ ls iFk ij fQlyuk izkjEHk djs rks ;g [kqjnjh

lrg ij 10 m rd xfr djsxkA rc gksxkA

1.0m

A

B Rough ( )[kqjnjk

(A*) 0.1 (B) 0.25 (C) 0.4 (D) 0.35 (E) 0.45Sol. W

G + W

f = 0 � 0

10 × 1 + Wf = 0

10 � mg x = 0Put values = 0.1

SECTION - IIComprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 2 multiple choice questions have to beanswered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

[k.M - IIc) cks/ku çdkj

bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'u ds4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA

Paragraph for Question Nos. 6 to 7iz'u 6 ls 7 ds fy, vuqPNsn

Comprehension # 1 [M.Bank(07-08)_WPE_11.2(Q.1,2,3)]In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.

fp=k m = 2kg nzO;eku d s d .k d h fLFkfrt Å t kZ d k bld s x- funsZ'kkad d s lkFk ifjorZu iznf'kZr d jrk gSA laj{kh cyd s izHkko esa d .k x-fn'kk esa xfr d j jgk gS]

6. If the particle is released at the origin then :(A) it will move towards positive x-axis.(B*) it will move towards negative x-axis.(C) it will remain stationary at the origin.(D) its subsequent motion cannot be decided due to lack of information.

;fn d .k d ks ewy fcUnq ls NksM+k t k, rks(A) ;g /kukRed x-fn'kk esa xfr d jsxk(B*) ;g _ .kkRed x-fn'kk esa xfr d jsxk(C) ;g ewy fcUnq ij fLFkj jgsxk(D) bld h vkxs d h xfr lwpuk d h d eh d s d kj.k ugha crkbZ t k ld rh

Sol. If the particle is released at the origin, it will try to go in the direction of force. Here dxdu

is positive and

hence force is negative, as a result it will move towards � ve x-axis.

;fn d .k d ks ewy fcUnq ls NksMk t krk gS] rks ;s cy d h fn'kk esa t kus d k iz;kl d jsxkA ;gkW dxdu

/kukRed gS vr% cy

_ .kkRed gS] ifjek.kLo: i ;s � ve x-v{k d h rjQ xfr d jsxkA

7. If the particle is released at x = 2 + where 0 (it is positive) then its maximum speed in subsequentmotion will be :;fn bls x = 2 + ls NksM+k t k;s t gk¡ 0 ¼;g /kukRed gS½ gks rks bld h vkxs d h xfr esa vf/kd re pky gksxh &

(A) 10 m/s (B*) 5 m/s (C) 5 2 (D) 7.5 m/s

Sol. When the particle is released at x = 2 + it will reach the point of least possible potential energy (�15 J) where

it will have maximum kinetic energy.

tc d.k x = 2 + ls NksM+k tkrk gS] ;g U;wure laHko fLFkfrt Å tkZ ds fcUnq (�15 J) ij igqap tk;sxkA tgk¡ bldh xfrtÅ tkZ vf/kdre gksxhA

2maxvm

21

= 25 vmax

= 5 m/s


A small bead of mass 'm' is threaded onto a rough circular wire of radius 'a' with coefficient of friction 0.5. Thecircular wire frame is rotated about its vertical diameter as shown. Answer the following questions. (Assumeaccelertion due to gravity is g)

'm' nzO;eku dk ,d NksVk eksrh a f=kT;k rFkk ?k"kZ.k xq.kkad 0.5 ds oÙkh; [kqjnjs rkj ij fLFkr gSA ;g oÙkh; rkj Ýse blds

Å /okZ/kj O;kl ds ifjr% fp=kkuqlkj ?kw.kZu dj jgk gSA fuEu iz'uksa ds mÙkj nhft,A

(ekuk xq:Ro ds dkj.k Roj.k g gSA)

8. The angular speed required if the bead is to be made to move in a horizontal circle of radius 23a

if friction is

absent, is

?k"kZ.k dh vuqifLFkfr esa eksrh dks 23a

f=kT;k ds {kSfrt oÙkh; iFk esa pykus ds fy, vko';d dks.kh; pky D;k gksxhA

(A) 2/1

a

g

(B*)

2/1

a

g2

(C)

2/1

a2

g3

(D)

2/1

a

g3

Sol. N sin = mg

N cos = m2 23a

When friction in absent ;

tc ?k"kZ.k vuqifLFkr gSA

tan= 3a

g22

=

3

1

= 21

ag2

9. The least angular speed required if the bead is to be made to move in horizontal circle of radius 23a

if friction

is present, is :

?k"kZ.k cy dh mifLFkfr esa eksrh dks 23a

f=kT;k ds {kSfrt oÙkh; iFk esa pykus ds fy, vko';d U;wure~ dks.kh; pky D;k gksxh:

(A) 21

35(ag2

(B*)

21

3815(a3g2

(C)

21

3815(a3g2

(D)

21

310(a3g2

Sol. When friction is present

tc ?k"kZ.k mifLFkr gSA

As ¼pwfd½ = 60º

N sin� N cos = m2 23a

.............(1)

and ¼rFkk½, N sin + Ncos= mg ............(2)

1 ÷ 2

cossincossin

= g

2

23a

put = 60º and ¼rFkk½ = 0.5 j[kus ij

g23a 2

=

21

43

41

23

= 23

132

][g23a 2 =

1)32()132(

; = 21

)3815(a3g2

SECTION - IIIInteger Answer Type

This section contains 10 questions. The answer to each of the questions is adouble digit integer, ranging from 0 to 99. The appropriate bubble below therespective question number in the ORS have to be darkened.

[k.M - III

iw.kk±d mÙkj izdkjbl [k.M esa 10 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 99 rd ,d nks vad iw.kk±d gSA ORS

esa iz'u la[;k ds laxr uhps fn;s x;s cqYyksa esa ls lgh mÙkj okys cqYyksa dks dkyk fd;k tkuk

gSA

10. A particle is projected along a horizontal field whose coefficient of friction varies as = 2r

A where r is the

distance from the origin in meters and A is a positive constant whose value is 20 m2. The initial distance ofthe particle is 1 m from the origin and its velocity is radially outwards. The minimum initial velocity at thispoint so that particle never stops is 10x m/sec then x is : use g = 10 m/sec2

,d d .k d ks {kSfrt {ks=k d s vuqfn'k iz{ksfir fd ;k x;k gS] ft ldk ?k"kZ.k xq.kkad = 2r

A d s vuqlkj ifjofrZr gksrk gS t gk¡

r ewy fcUnq ls nwjh eh- es gS rFkk A ,d /kukRed fLFkjkad gS ft ld k eku 20 m2 gSA ewyfcUnq ls d .k d h izkjfEHkd nwjh 1eh- gS rFkk bld k osx f=kT;h; fn'kk esa ckgj d h vksj gSA bl fcUnq ij U;wure çkjfEHkd osx 10x m/sec gS ft lls d .kd Hkh u : d sA rc x gksxkA (g = 10 m/sec2 )Ans. 02

Sol. Work done against friction must equal the initial kinetic energy.?k"kZ.k ds fo:) fd;k x;k dk;Z çkjfEHkd xfrt Å tkZ ds cjkcj gksrk gSA

2mv

21

=

1

dxmg ; 2v2

=

12

dxx

1gA ;

2v2

= Ag Ag

1x1

v2 = 2gA v = Ag2 v = 400 = 20 hece x = 2

11. A balloon moves up with a velocity 5 m/s .A stone is thrown from it with a horizontal velocity 2 m/s relative to it.The stone hits the ground at a point 10 m horizontally away from it. The height of the balloon at the time ofthrowing the stone in meter is 20 x . Then find x . (Take g = 10 m/s2),d xqCckjk Å ij dh vksj 5 m/s ds osx ls xfr djrk gSA bl xqCckjs ls ,d iRFkj blds lkis{k 2 m/s ds {kSfrt osx ls Qsadktkrk gSA ;g iRFkj xqCckjs ls nwj 10 m {kSfrt nwjh ij tehu ij Vdjkrk gSA iRFkj Qsadus ds le; ij xqCckjs dh Å ¡pkbZ ehVjesa 20x gSA rc x Kkr d jksA (g = 10 m/s2)Ans. 05

Sol.

10 = 2tt = 5

H = ut + 21

at2 = 5 × 5 + 21

(�10) 52 = 25 � 125 = 100 m

12. A force of 1000 N acts on a particle parallel to its direction of motion which is horizontal. Its velocity increasesfrom 1 m s�1 to 10 m s�1, when the force acts through a distance of 4 metre. Calculate the mass of the particle.Given : a force of 10 Newton is necessary for overcoming friction.

,d d .k ij bld h xfr d h fn'kk t ks fd {kSfrt gS] d s vuqfn'k 1000 N d k ,d cy yxk;k t krk gSA t c cy 4 m d h

nwjh rd d k;Zjr jgrk gS rks bld k osx 1 m/s�1 ls 10 ms�1 gks t krk gSA d .k d k nzO;eku Kkr d hft ;sA fn;k x;k gS %

?k"kZ.k cy d s fo: ) pyus esa U;wure 10 N cy d h vko';d rk gksrh gSAAns. 80

Sol. By work energy theorem, d k;Z Å t kZ izes; lsW

F + W

f = K

1000 × 4 � 10 × 4 = 21

m(102 � 12)

m = 80 kg.

13. A frictionless track ABCDE ends in a circular loop of radius R. A body slides down the track from point A whichis at height h = 5 cm. Find the maximum value of R in cm for a body to complete the loop successfully.,d ?k"kZ.k jfgr iFk ABCDE d s var esa R f=kT;k d k oÙkkd kj ywi gSA iFk d s fcUnq A ls] ft ld h Å apkbZ h = 5 lseh- gS],d oLrq fQlyuk izkjEHk djrh gSA R dk og vf/kd re eku lseh0 esa Kkr d hft ;s ft ld s fy;s oLrq ywi dks lqxerk iwoZdiwjk d jysA

Ans. 02Sol. Using energy conservation :

Å t kZ laj{k.k d k mi;ksx d jrs gq,

2Bmv

21

= mgh

vB =

mmgh2

vB = hg2

.....(1)Also to complite vertical circleÅ /okZ/kj oÙk iwjk d jus d s fy, Hkh

vB = gR5 .....(2)

R = h52

= 2 cm

14. A particle is moving along a circular path of radius 5 m with a uniform speed 5 ms�1. If the magnitude of average

acceleration during the interval in which particle completes half revolution is 2s/m

x5

then find x.

,d d .k 5 eh f=kT;k okys oÙkkd kj iFk ij 5 ms�1. d h ,d leku pky ls xfr d j jgk gSA d .k }kjk

vk/kk pDd j iwjk d jus d s nkSjku vkSlr Roj.k d k ifjek.k 2s/mx5

gS rc x Kkr d jksA

Ans. 02Sol. Change in velocity when particle complete the half revolution : v = v

f � v

i = 2v

Time taken to complete the half revolution t = vR

average acceleration = tv

= v/R

v2

= Rv2 2

tc d.k v/kZoÙk iw.kZ dj ysrk gS rc osx esa ifjorZu : v = vf � v

i = 2v

v/kZoÙk iw.kZ djus esa fy;k x;k le; t = vR

vkSlr Roj.k = tv

= v/R

v2

= Rv2 2

= 552 2

=

10 m/s2 Ans.

15. An object O is kept in air and a lens of focal length 10 cm (in air) is kept at the bottom of a container which

is filled upto a height 44 cm by water. The refractive index of water is 4/3 and that of glass is3/2. The bottom of the container is closed by a thin glass slab of refractive index 3/2. Find the position ofthe final image in cm formed by the system (refer to figure shown below).,d oLrq O gok esa j[kh gqbZ gSaA 10 lseh- Qksdl nwjh ¼ok;q esa½ dk ,d ,d ySal 44 lseh- Å pk¡bZ rd ikuh ls Hkjs gq, ,d ik=kds isans esa fp=kkuqlkj j[kk gqvk gSaA ikuh dk viorZukad 4/3 o dk¡p dk viorZukad 3/2 gSA. 3/2 viorZUkkad dh dk¡p dh ,d iryhifêdk }kjk ik=k dk fupyk ry ¼isank½ cUn gSA fudk; }kjk cus vfUre çfrfcEc dh fLFkfr lseh0 esa Kkr dhft,A ¼uhps iznf'kZrfp=k dks nsf[k,½

Ans. 90Sol. For refraction through water surface, ty i"B ls viorZu ds fy,

u = � 12 cm, usingv2 �

u1 =

Rµ� 12 mi;ksx ls

121

v34 = 0 v = � 16 cm

Now, for lens, vc ySal ds fy,

af1

= ( g � 1)

21 R1

R1

wf1

=

1

w

g

21 R1

R1

fw = )1(

)1(

w

g

g

fa

= 11

3/42/3

23

× 10 = 40 cm

For refraction through the lens, ySal ls viorZu ds fy,u = � (16 + 44)cm = � 60 cmf = + 40 cm

v1

= u1

f1 =

601

401 =

1201

v = 120 cmFor refraction from water to glass slab, ty ls dkap ifêdk esa viorZu ds fy,

u = 120 cm 2 =

23

, 1 =

34

120v34

23

= 0. ( R = )

v = 135 cmAgain for refraction from glass to air, iqu% dkap ls ok;q esa viorZu ds fy,

u = 135 cm 2 = 1,

1 =

23

.

v1

� 135

23

= 0 v = 90 cm

16. A motor boat has 2 throttle position on its engine. The high speed position propels the boat at 10 kmh1 in still water and the low position gives half the higher speed. The boat travels from its dock downstreamon a river with the throttle at low position and returns to its dock with throttle at high position. The return triptook 25 % longer time than it did for the downstream trip. The velocity of the water current in the river is

3x

km/hr. Then x is :

,d eksVj cksV ds batu dh nks pky fLFkfr;ka gSA mPp pky fLFkfr esa cksV 10 km h�1 ls rFkk fuEu pky fLFkfr esa mPp pky

dh vk/kh pky ls 'kkar ty esa xfr dj ldrh gSA cksV unh ds cgko esa fuEu pky fLFkfr ls pyrk gqvk okil mPp pky fLFkfr

esa pydj çkjfEHkd LFkku ij igq¡prk gSA okil vkus esa yxk le;] tkus es yxs le; ls 25 % T;knk gSA unh ds ty çokg dk

osx 3x

km/hr gSA rc x gksxkA

Ans. 03Sol. Downstream v = 5 km/hr

Let u = velocity of water currentLet 'x' distance is covered in 't' time downstreamx = (v + u) t x = (5 + u) t ........... (1)Upstream V = 10 km

t' = 4t5

x = (V � u) t' = (10 � u) 4t5

.............. (2)

4(5 + u) = (10 � u) 5 (from ..(1) & ..(2))20 + 4 u = 50 � 5 u

9 u = 30

u = 3

10 m/s Ans.

Sol. cgko dh fn'kk esa v = 5 km/hr

ekuk u = unh cgko dk osx

ekuk cgko dh fn'kk esa 't' le; esa 'x' nwjh r; djrk gSA

x = (v + u) t x = (5 + 4) t ........... (1)

cgko ds foifjr fn'kk esa V = 10 km

t' = 4t5

x = (V � u) t' = (10 � u) 4t5

.............. (2)

4(5 + u) = (10 � u) 5 (from ..(1) & ..(2))20 + 4 u = 50 � 5 u

9 u = 30

u = 3

10 m/s Ans.

17. On a 2lane road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in oppositedirections with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, bothbeing 1 km, B decides to overtake A before C does. The minimum acceleration of car B is required to avoidan accident is x m/s2 . Therefore, find 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 dkjA 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 djukpkgrk gSA B dk U;wure Roj.k x m/s2 gS rkfd nq?kZVuk ls cpk tk ldsA rc x Kkr djksAAns. 01

Sol.

B

A

C

54 km/hr54 km/hr

36 km/hr

Time taken by C to overtake A is = h/km)5436(km1

=

901

hr = 32

× 60 sec. = 40 sec.

Let the minimum acceleration of B is a to overtake A before C then

C }kjk A ls vkxs fud yus d s fy, fy;k x;k le; = h/km)5436(km1

=

901

hr = 32

× 60 sec. = 40 sec.

ekuk B d k U;wure Roj.k a ft lls og A d kss ]C ls igys ikj d j t k;s] rc

S = ut + 21

at2

1 km = (54 � 36) × 901

+ 29090

1

× a

290901

× a = 1 �

9018

= 9072

a = 9072

× 90 × 90 × 2 km/h2

60606060100029070

m/sec2 = 1 m/sec2 .

18. A cannon fires successively two shells with velocity v0= 250 m/s; the first at the angle 1 = 53° and the second

at the angle 2 = 37° to the horizontal, in the same vertical plane, neglecting the air drag, find the time interval

in seconds between firings leading to the collision of the shells. (g = 10 m/s2).,d rksi }kjk v

0= 250 m/s; osx ls nks xksys Ø e'k% {kSfrt ls

1 = 53° rFkk

2 = 37° ij bl çd kj ç{ksfir fd ;s t krs gSa

fd nksuksa d k Å /ok/kZj ry leku gSA ok;q ?k"kZ.k ux.; gSA og le; vUrjky lSd .M esa Kkr djks tks nksauks xksyksa dh VDdjd s fy, vko';d gSA (g = 10 m/s2).Ans. 10

Sol. Along Horizontal direction

x = v

0 cos 53º t = v

0 cos 37º (t � t

0 )

{kSfrt fn'kk esa x = v0 cos 53º t = v

0 cos 37º (t � t

0 )

53

t =

54

(t � t0) 3t = 4 (t � t

0)...........(1)

Vertical direction ;Å /okZ/kj fn'kk esa

y = v0 sin53º t �

21

gt2 = v0 sin37º (t � t

0) �

21

g (t � t0)2 ..........(2)

v0 ×

54

t � 5t2 = v0 ×

53

× 16t9

210

�4t3 2

49

�45

tv0 = 5 t2

169

�1

47

5250

= 5t × 167

t = 40

so vr% t0 = 10 sec

19. Four blocks of masses 1 kg ,2kg,4kg and 8 kg are connected with three ideal springs. The whole system isreleased from rest on a frictionless horizontal surface .After some time the accelerations of 1 kg and 4 kg arefound to be zero while the middle spring is found to be compressed by 0.1 m. At the same instant of time the ratioof acceleration of 2 kg and 8 kg block is :1 kg , 2kg, 4kg rFkk 8 kg nzO;eku ds pkj CykWd rhu vkn'kZ fLçaxksa ds lkFk tqM+s gSA lEiw.kZ fudk; ?k"kZ.kjfgr {kSfrt lrg ijfojke ls NksM+k tkrk gSA dqN le; i'pkr~ 1 kg rFkk 4 kg ds Roj.k 'kwU; çkIr gksrs gS] tcfd e/; fLçax 0.1 m ls lEihfM+r izkIrgksrh gSA mlh {k.k ij 2 kg rFkk 8 kg CykWd ds Roj.k dk vuqikr gksxkA

Ans. 04Sol. As the acceleration of 1 kg block is zero this means the spring between 1 kg and 2 kg will relaxed.

The force acting on 2kg block is 200 × 0.1 = 20 N, so, acceleration of 2 kg block 10 m/s2 .

Similarly acceleration of 8 kg block = 820

m/s2.

Required ratio = 4.Sol. 1 kg CykWd dk Roj.k 'kwU; gSA bldk vFkZ gS fd 1 kg rFkk 2 kg nzO;eku ds e/; fLçax lkekU; voLFkk esa gSA

2 kg CykWd ij dk;Zjr cy 200 × 0.1 = 20 N gksxkA vr% 2 kg CykWd dk Roj.k 10 m/s2 gksxkA

blh çdkj 8 kg CykWd dk Roj.k = 820

m/s2 gksxkA

Required ratio = 4.

TEST PATTERN

ELPDCT-04 Date : 21.10.2012

Test Syllabus : Physical ChemistrySolid State, Ionic Equilibrium, Chemical Bonding, Basic Inorganic Nomenclature.

Test Syllabus : Organic ChemistryGeometrical and Optical isomerism, Conformational isomerism,


1 to 5 SCQ (5 options) 5 3 -1 15



20 to 24 SCQ (5 options) 5 3 -1 15



39 to 43 SCQ (5 options) 5 3 -1 15



57 Total 201

Paper-2

Total

Physics

Maths

Chemistry

Note :1. Paper-1 (JEE-Mains) should be lighter side. Should contain questions of AIEEE type.

2. Paper-2 (JEE-Advance) should be tough-fer side. Should contain questions of IIT-JEE type.

Paper-2SECTION - I

Straight Objective TypeThis section contains 5 questions. Each question has 5 choices (A), (B), (C), (D) and (E) for its answer, outof which ONLY ONE is correct.

[k.M- I

lh/ks oLrqfu"B izdkj

bl [k.M esa 5 iz'u gSaA izR;sd iz'u ds 5 fodYi (A), (B), (C), (D) rFkk (E) gSa] ftuesa ls flQ Z ,d lgh gSA

Physical SCQ (3) (5 options)20. An acid HA (Ka = 10�5) reacts with NaOH at 298 K. What would be the value of the rate constant of the

reverse reaction at the same temperature if the rate constant of the forward reaction is 10�11 mol�1 L sec�1 ?,d vEy HA (Ka = 10�5) 298 K ij NaOH ls fØ;k djrk gSA izrhi vfHkfØ;k ds osx fLFkjkad dk eku leku rki ijD;k gksxk ;fn vxz vfHkfØ;k dk osx fLFkjkad 10�11 mol�1 L sec�1 gksA(A) 10�9 (B) 109 (C) 10�5 (D*) 10�20

(E) 10�16

Sol. CH3COOH + NaOH CH3COONa + H2O

Equilibrium constant of the backward reaction izrhi vfHkfØ;k dk lkE; fLFkjkad

Kh = a

w

KK

Equilibrium constant of the forward reaction vxz vfHkfØ;k dk lkE; fLFkjkad

Kequilibrium lkE; = Kh

�1 = w

a

K

K = 109

109 = b

f

KK

= b

11

K10

Kb = 10�20

21. Which one is incorrect statement among the following ?

(A) 5PH , SCl6 and FCl

3 do not exist. (B) p d bond is present in SO

2 molecule.

(C) 12 P�O bonds are present in P4O

6 molecule. (D) Cation of PCl

5(s) has tetrahedral shape.

(E*) Bond angle in SiH4 less than that in CH

4.

fuEu esa ls dkSulk dFku lR; ugha gS \

(A) 5PH , SCl6 rFkk FCl

3 dk vfLrRo ugha gksrk gSA (B) SO

2 v.kq esa p d ca/k mifLFkr gksrk gSA

(C) P4O

6 v.kq esa 12 P�O ca/k mifLFkr gksrs gSaA (D) PCl

5(s) ds /kuk;u dh vkÑfr prq"Qydh; gksrh gSA

(E*) SiH4 dk cU/k dks.k CH

4 ds cU/k dks.k ls de gSA

Sol. (A) PH5 does not exist as there is large difference in energies of s, p and d orbitals and hence it does notundergo sp3d hybridisation.In SCl6, smaller S cannnot accomodate six larger Cl� ions. Fluorine can not expand its octet because itdoes not have vacant d-orbitals.

(B)

(C) P4O

6

(D)

gy (A) PH5 dk vfLrRo ugha gksrk gS D;ksafd ;gka s, p rFkk d d{kdksa dh Å tkZ esa vUrj vf/kd gksrk gSA rFkk bl izdkj ;g sp3d

ladj.k esa Hkkx ugha ysrs gSAASCl6 esa NksVs vkdkj dk S ijek.kq cM+s vkdkj ds Cl� vk;uksa dks vfHkxzfgr ugha djrk gSA ¶yksjhu dk v"Vd izlkj ugha gksldrk gS D;ksafd ¶yksjhu esa fjDr d-d{kd ugha gksrs gSaA

(B)

(C) P4O

6

(D)

22. We know that NH3 is a stronger base than CH

4. Which of the following is correct ?

(A*) NH3 is a stronger acid than CH

4. (B) NH

3 is a weaker acid than CH

4.

(C) NH4

+ is a weaker acid than CH4. (D) NH

2� is a stronger base than CH

3�.

(E) NH3 is a stronger base than CH

3�

NH3, CH

4 dh vis{kk ,d izcy {kkj gSA rc fuEu esa ls dkSulk dFku lgh gS \

(A*) NH3, CH

4 dh vis{kk ,d izcy vEy gSA (B) NH

3, CH

4 dh vis{kk ,d nqcZy vEy gSA

(C) NH4

+, CH4 dh vis{kk ,d nqcZy vEy gSA (D) NH

2�, CH

3� dh vis{kk ,d izcy {kkj gSA

(E) NH3, CH

3� dh vis{kk ,d izcy {kkj gSA

Organic SCQ (2)(5 options)23. Which of the following molecule is chiral :

fuEu esa ls dkSuls ;kSfxd fdjSy gS %

(A) (B*)

(C) (D)

(E)

Sol. (A) Plane of symmetry; (B) C2 axis symmetry; (C) Centre of symmetry ; (D) Plane of symmetry ;

(E) Plane of symmetry;

(A) lefefr ry ; (B) C2 lefefr v{k; (C) lefefr dsUæ ; (D) lefefr ry ;

(E) lefefr ry ;

24. The racemic mixture of Alanine can be resolved by using,

(1) (+)-2-Butanol (2) ()-2-Chlorobutanoic acid(3) (±) -2-Butanol (4) (d mix)-2-Chlorobutanoic acid(A*) 1 & 2 only (B) 1 & 3 only (C) 2 & 4 only (D) 3 & 4 only(E) 2, 3 & 4

,ykfuu ds jslsfed feJ.k dks fuEu esa ls fdlds }kjk iFkd fd;k tk ldrk gS

(1) (+)-2-C;wVsukWy (2) ()-2-DyksjksC;wVsuksbd vEy(3) (±)-2-C;wVsukWy (4) (d mix)-2-DyksjksC;wVsuksbd vEy(A*) dsoy 1 ,oa 2 (B) dsoy 1 ,oa 3 (C) dsoy 2 ,oa 4 (D) dsoy 3 ,oa 4(E) 2, 3 ,oa 4

Sol. + two diastereomeric esters

+ two diastereomeric amides

(�) �2�chlorobutanoicacid

Racemic mixture can be resolved by using optically active compounds.

Sol. + nks foofje leko;oh ,LVj

+ nks foofje leko;oh ,ekbM

(�) �2�DyksjksC;wVsukWbdvEy

izdkf'kd lfØ; ;kSfxdksa }kjk jslsfed feJ.k dks iFkd fd;k tkrk gSA

SECTION - IIComprehension Type

This section contains 2 paragraphs. Based upon each paragraph, there are 2 questions. Each question has4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

[k.M - IIcks/ku çdkj

bl [k.M esa 2 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 ç'u gSA çR;sd ç'u d s 4 fodYi(A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA


For any polyprotic acid, we always consider successive dissociation. The value of equilibrium consant ofsuccessive dissociation decreases due to common ion effect.For example :

H2A is a dibasic acid.

H2A H+ + HA� K

1 = ]AH[

]HA][H[

2

HA� H+ + AA� � K2 =

]HA[

]A][H[

K1 is greater than K

2.

fdlh iksyhizksfVd vEy esa ge ges'kk Øekxr fo;kstu dk voyksdu djrs gSaA levk;u izHkko ds dkj.k Øekxr fo;kstu dslkE; fu;rkad dk eku ?kVrk tkrk gSAmnkgj.k ds fy, :

H2A ,d f}{kkjh; vEy gSA

H2A H+ + HA� K

1 = ]AH[

]HA][H[

2

HA� H+ + AA� � K2 =

]HA[

]A][H[

K1dk eku K

2 ls vf/kd gksrk gSA

25. Find the pH of 0.1 M NaHCO3.

Use data (K1 = 4 × 10�7, K

2 = 4 × 10�11

for H

2CO

3, log 4 = 0.6) :

0.1 M NaHCO3 dk pH Kkr dhft;sA

vk¡dM+s iz;qDr dhft;s % (H2CO

3 ds fy, K

1 = 4 × 10�7, K

2 = 4 × 10�11, log 4 = 0.6) :

(A) 3.7 (B*) 8.4 (C) 9.6 (D) None of these buesa ls dksbZ ugha

Sol. [H+] = 21KK = 117 104104 = 4 × 10�9 M

26. Find the concentration of H+ ions in an aqueous solution which is saturated with H2S (0.1 M) as well as H

2CO

3

(0.2 M).Use data [K

1 = 10�7, K

2 = 10�14 for H

2S, K

1 = 4 × 10�7, K

2 = 4 × 10�11

for H

2CO

3] :

H2S (0.1 M) ds lkFk&lkFk (0.2 M) H

2CO

3 ds lkFk Hkh larIr tyh; foy;u esa H+ dh lkUnzrk Kkr dhft;sA

vk¡dM+s iz;qDr dhft;s % [K1 = 10�7, K

2 = 10�14 for H

2S, K

1 = 4 × 10�7, K

2 = 4 × 10�11

for H

2CO

3] :

(A*) 3 × 10�4 M (B) 3.83 × 10�4 M (C) 2.83 × 10�4 M (D) None of these buesa ls dkbZ ugha

Sol. [H+] = 2211 CKCK = 2.01041.010 77

= 3 × 10�4 M

Paragraph for Question Nos. 27 to 28

ç'u 27 ls 28 ds fy, vuqPNsn

An unsaturated compound X (C5H

10O) exists as four stereoisomers. Y, (structural isomer of X) is neither

unsaturated nor shows any stereoisomerism. X and Y have same functional group. The third isomer (Z) is apositional isomer of X, and has only two stereoisomers.

,d vlarIr ;kSfxd X (C5H

10O) ds pkj f=kfoe leko;oh lEHko gSA Y, (X dk lajpukRed leko;oh) uk rks vlarIr gS vkSj

uk gh f=kfoe leko;ork n'kkZ ldrk gSA X o Y esa leku fØ;kRed lewg ik;k tkrk gSA rhljk leko;oh (Z), X dk fLFkfrleko;oh gS rFkk dsoy nks f=kfoe leko;oh j[krk gSA

27. The compound Z cannot be ?;kSfxd Z ugha gks ldrk gS\

(A) CH2 = CH�CH

OH|�CH

2 CH

3(B) H

2C

= CH�CH

2�CH

OH|�CH

3

(C) CH3 � CH = CH�CH

2�CH

2�OH (D*)

|H C=C�CH�CH2 3

CH3

|OH

Sol. (Z) is not (Z) ugha gS |

H C=C�CH�CH2 3

CH3

|OH

28. Y can be :Y gks ldrk gS %

(A) O

(B)

OH|

(C)

CH OH2|

(D*) (B) and (C) both (B) rFkk (C) nksuksa

Sol. (Y) can be

OH|

and

CH OH2|

.

gy (Y) gks ldrk gS

OH|

rFkk CH OH2|

.

SECTION - IIIInteger Answer Type

This section contains 10 questions. The answer to each of the questionsis a double digit integer, ranging from 00 to 99. The appropriate bubblebelow the respective question number in the ORS have to be darkened.

[k.M - III

iw.kk±d mÙkj izdkjbl [k.M esa 10 iz'u gSaA izR;sd iz'u dk mÙkj 00 ls 99 rd iw.kk±d gSA ORS esa iz'u

la[;k ds laxr uhps fn;s x;s cqYyksa esa ls lgh mÙkj okys cqYyksa dks dkyk fd;k tkuk

gSA

Physical Double digit Integer Type (7)29. BrF3 is a liquid which considerably undergoes self ionization to form cationic and anionic species. Based on

VSEPR theory, number of 90 degree F�Br�F bond angles is ...................... in anionic species.

2BrF3 [BrF2]+ + [BrF4]�

BrF3 ,d nzo gS] ftlds Lor% vk;uu ij /kuk;fud ,oa _ .kk;fud Lih'kht curh gSaA VSEPR fl)kUr ds vk/kkj ij]_ .kk;fud Lih'kht esa 90 fMxzh ds F�Br�F ca/k dks.kksa dh la[;k ...................... gSA

2BrF3 [BrF2]+ + [BrF4]�

Ans. 04

2BrF3 [BrF2]+ + [BrF4]�

30. Number of shortest P � O bonds present in triphosphate ion P3O

105� is.

VªkbZQkLQsV vk;u P3O

105� esa lcls NksVs P � O cU/kksa dh la[;k gS %

Ans. 02

Sol.

31. Among the following molecular orbitals, how many have at least one nodal plane?1s,2s, *1s, *2p

z, 2p

x , *2p

y, 2p

z

fuEu esa ls fdrus vkf.od d{kdksa esa de ls de ,d uksMy ry gS \1s,2s, *1s, *2p

z, 2p

x , *2p

y, 2p

z

Ans. 05

Sol. *1s *2pz

2px or 2p

y *2p

x or *2p

y

2pz

32. In F.C.C. arrangement of identical spheres, distance between two nearest octahedral void is 8.51Å. The

distance between two nearest tetrahedral voids would be (in Å) ?

leku xksydksa dh F.C.C. O;oLFkk esa nks lehiLFk v"VQydh; fNnzksa ds e/; dh nwjh 8.51Å gS] rks nks lehiLFk prq"Qydh;fNnzksa ds e/; dh nwjh ¼Å esa½ Kkr dhft;s \

Ans. 06

Sol. distance between two octahedral void = 2

a = 8.51

a = 8.51 × 1.41 = 12Å

distance between two tetrahedral void = a/2 = 6Å

nks v"VQydh; fNnzksa ds e/; dh nwjh = 2

a = 8.51

a = 8.51 × 1.41 = 12Å

nks prq"Qydh; fNnzksa ds e/; dh nwjh = a/2 = 6Å

33. 18 ml of mixture of acetic acid and sodium acetate required 6ml of 0.1 M NaOH for neutalization of the acidand 12 ml of 0.1 MHCl for reaction with salt, separately. If pKa of the acid is 4.7, what is the pH of the mixture? [log2 = 0.3],flfVd vEy o lksfM;e ,flVsV ds 18 ml feJ.k dks] vEy ds mnklhuhdj.k ds fy, 6ml, 0.1 M NaOH rFkk yo.k lsfØ;k ds fy, vyx ls 12 ml, 0.1 M HCl dh vko';drk gSaA ;fn vEy dk pKa 4.7 gS] rks feJ.k dk pH D;k gSA [log2= 0.3]

Ans. 05

Sol. From the information, lwpukvksa ls ]acid[]salt[

= 2

so blfy, , pH = pKa + log ]acid[]salt[

= 05

34. Aniline behaves as a weak base. When 0.1 M , 50 ml solution of aniline was mixed with 0.1 M,25 ml solution of HCl the pH of resulting solution was 8. Then the pH of 0.01 M solution of aniliniumchloridewill be (K

w = 10�14)

,fuyhu nqcZy {kkj fd Hkkafr O;ogkj djrh gSA t c ,fuyhu ds 0.1 M , 50 ml foy;u dks HCl ds 0.1 M,

25 ml foy;u ds lkFk feyk;k x;k rks ifj.kkeh foy;u dh pH 8 FkhA rc ,fuyhfu;e DyksjkbM ds 0.01M foy;u dkpH D;k gksxk A (K

w = 10�14)

Ans. 05

Sol. POH

= pKb + log

basesalt

= pKb = 6 K

b = 10�6

so pH = 8Now for anilinium chloride solution we havevc ,fuyhfu;e DyksjkbM foy;u ds fy, ge tkurs gSa fd

+ H3O+ ; K

h =

b

w

KK

= )h1(

ch2

~ ch2

h = b

w

Kc

K so concentration of [H+] blfy, [H+] dh lkUnzrk = ch =

b

w

K

cK

[H+] = 6

142

10

1010

= 1010 = 10�5 pH = 5

35. If the solubility of Ag2SO4 in 10�2 M Na2SO4 solution be 2 × 10�8 M then Ksp of Ag2SO4 will be a × 10�18 thenfind out 'a' :;fn 10�2 M Na2SO4 foy;u esa Ag2SO4 dh foys;rk 2 × 10�8 M gS] rks Ag2SO4 dk Ksp, a × 10�18 gksxk rc 'a' Kkrdhft;sA

Ans. 16Sol. Ag2SO4 2Ag+ + SO4

�2

2s' (s' + 10�2) 0�2

Ksp = (2s')2 (10�2) = (2 × 2 × 10�8)2 (10�2) = 16 × 10�18

Organic Double digit Integer Type (3)36. Total number of stereoisomes possible for molecule A will be

v.kq A ds fy, dqy leko;oh;ksa dh la[;k crkb;s

Ans. 16

Sol.

* indicates stereocentre * f=kfoe dsUnz dks iznf'kZr djrk gSATotal isomers = (2)n dqy leko;oh = (2)n

24 = 16 24 = 16

37. How many number of chiral centres present in the following compounds ?fuEu ;kSfxdksa esa mifLFkr fdjSy dsUnzksa dh la[;k crkb;s ?

Ans. 12

Sol. = 12

38. How many monochloro isomeric products are formed when 2-Phenyl butane is treated with Cl2/h:

2-Qsfuy C;wVsu dk eksuksDyksjksuhdj.k Cl2/hdh mifLFkfr esa djkus ij fdrus eksuksDyksjks leko;oh mRikn izkIr gksxsa \

Ans. 10

Sol.

3

32

CH|

CHCHCHPh

hCl2

ClCH|

CHCHCHPh*

2

32 +

3

32

CH|

CHCHCPh *| Cl

+

3

3

CH|

CHCHCHPh|

*

*

Cl

d/ d/ (2)2 = 4

+

3

22

CH|

ClCHCHCHPh *

d/

Page # 1

TEST PATTERNCT-04 Date : 21-10-2012

Syllabus : Limits, Continuity & Derivability, Application of Derivatives, Sequence &Series, Solution of Triangle, Indefinite Integration (Up to Lecture # 4)


1 to 5 SCQ (5 options) 5 3 -1 15



20 to 24 SCQ (5 options) 5 3 -1 15



39 to 43 SCQ (5 options) 5 3 -1 15



57 Total 201

Maths

Chemistry

Paper-2

Total

Physics

Note :1. Paper-1 (JEE-Mains) should be lighter side. Should contain questions of AIEEE type.

2. Paper-2 (JEE-Advance) should be tough-fer side. Should contain questions of IIT-JEE type.

SCQ (5)39. Two curves C

1 : y = x2 � 3 and C

2 : y = kx2 (k 0) intersect each other at two different points. The tangent

drawn to C2 at one point of intersection A (a,y

1) , a > 0 meets C

1 again at B (1, y

2) (y

1y

2). The value of

'a' isnks oØ C

1 : y = x2 � 3 rFkk C

2 : y = kx2 (k 0) ,d nwljs dks nks vyx vyx fcUnqvksa ij izfrPNsn djrs gSA ,d izfrPNsn

fcUnq A (a,y1) , a > 0 ls C

2 ij Li'kZ js[kk [khaph tkrh gS rks iqu% C

1 dk fcUnq B(1, y

2) (y

1y

2) ij feyrh gS] a dk eku

gS&(A*) 3 (B) 2 (C) 1 (D) 4 (E) 5

Ans. 3Sol. Point A lies on C

1 and C

2fcUnq A, C

1 rFkk C

2 ij fLFkr gSA

y1 = a2 � 3

y1 = ka2

a2 � 3 = ka2 ..................(1)

Now vc y = kx2 dxdy

= 2 kx

)y,a( 1

dxdy

= 2ka =

a1yy 12

But y

2 = 1 �3 = �2

2 ka = a1a1 2

= 1+ a .............(2)

Substituting ?kVkus ij k = 2

2

a

3a

Page # 2

2

2

a

)3a(a2 = 1 + a a = +3 andvkSj a = � 2 Rejected vLohdk;Z

40. Let f (x) =

1x),2b(logx2

1x,5x10xx2

2

23

. The set of values of b for which f(x) has greatest value at x = 1, is

given by

ekuk f (x) =

1x),2b(logx2

1x,5x10xx2

2

23

. b ds ekuksa dk leqPp; ftlds fy, f(x), x = 1 ij vf/kdre eku j[krk gS] nh

tkrh gS&(A) 1 b 2 (B) b {1,2}

(C) b (�,�1) (D) b (�,� 2 ) ( 2 , )

(E*) None of these buesa ls dksbZ ugha

Sol. f(x) =

1x;)2b(logx2

1x;5x10xx2

2

23

f �(x) =

1x;2

1x;10x2x3 2

f(1) f(1+)1 � 1+10 � 5 > �2 + log

2(b2 � 2)

log2(b2 � 2) < 7

b2 �2 < 27 and vkSj b2 � 2 > 0 2 <b2 < 130

41. The graph of y = n

lim1�x

1xn2

n2

, n N is

y = n

lim1�x

1xn2

n2 , n N dk vkys[k gS&

(A) (B)

(C) (D*)

(E)

Sol. y = n

lim1�x

1xn2

n2

Page # 3

If ;fn x > 1 or ;k x < �1, f(x) = n

lim

n2

n2

x

1�1

x

11

= 1

If ;fn x = 1, � 1 f(x) = n

lim1�1

11n2

n2 limit does not existlhek fo|eku ugha gSA

f(1) & f(�1) are not defined ifjHkkf"kr ugha gSA

If ;fn 0 < x < 1 or;k �1 < x < 0, f(x) = n

lim1�x

1xn2

n2 =

1�010

= � 1

If ;fn x = 0, f(x) = 1�010

= � 1

42. If g(x) = 2f (2x3 � 3x2) + f (6x2 � 4x3 � 3), x R and f (x) > 0, x R, then g(x) > 0 for x belonging to

;fn g(x) = 2f (2x3 � 3x2) + f (6x2 � 4x3 � 3), x R ,oa f (x) > 0, x R gS] rks ftl vUrjky esa g(x) > 0 gks]og gS&

(A) 1

, 0,12

(B*) 1

,0 1,2

(C) 0, (D) ,1 (E)

0,

2

1�

Sol. Since f(x) > 0 pw¡fd f(x) > 0

f(x) is always increasing. f(x) lnSo o)Zeku gSg(x) = 2f (2x3 � 3x2) × (6x2 � 6x) + f (6x2 � 4x3 � 3) (12x � 12x2)= 12(x2 � x) . (f (2x3 � 3x2) � f (6x2 � 4x3 � 3))

= 12x (x � 1) [f (2x3 � 3x2) � f (6x2 � 4x3 � 3)]

For increasing g(x) > 0 o/kZeku gksus ds fy, g(x) > 0

CasefLFkfr-I x < 0 or x > 1 f (2x3 � 3x2) > f (6x2 � 4x3 � 3)

2x3 � 3x2 > 6x2 � 4x3 � 3 { f(x) is increasing} { f(x) o/kZeku gSA}

(x � 1)2

2

1x > 0 x > �

21

x

0,

2

1� (1, )

CasefLFkfr-II If 0 < x < 1f (2x3 � 3x2) < f (6x2 � 4x3 � 3)

(x � 1)2

21

x < 0 x < �21

, so there is no solution x < �21

, vr% dksbZ gy ugha gSA

Hence the values are x 1

,0 1,2

vr% x ds eku 1

,0 1,2

gSA

43. For which of the following functions Rolle's theorem is applicable ?(A) f(x) = 3x2 � 2 on [2, 3] (B) f(x) = |x � 1| on [0, 2]

(C) f(x) = |sgn x| on [�1, 1] (D*) f(x) = x + x1

on

3,

31

(E) none of these

fuEufyf[kr esa ls fdl Qyu ds fy, jksy izes; ykxw dh tk ldrh gS \(A) [2, 3] esa f(x) = 3x2 � 2 ij (B) [0, 2] esa f(x) = |x � 1| ij

(C) [�1, 1] esa f(x) = |sgn x| ij (D*)

3,

31

esa f(x) = x + x1 ij (E) buesa ls dksbZ ugha

Sol. For f(x) = 3x2 � 2 : f(2) f(3)For f(x) = |x � 1| : f (1) does not existFor f(x) = |sgn x| : f(x) is not continuous on [�1, 1]

Page # 4

For f(x) = x + x1

: f

3

1 = f(3), continuous in

3,

31

and differentiable in

3,

31

Hindi f(x) = 3x2 � 2 ds fy, f(2) f(3)

f(x) = |x � 1| ds fy, f (1) fo|eku ugha gSf(x) = |sgn x| ds fy, f(x) vUrjky [�1, 1] esa lrr~ ugha gS

f(x) = x + x1

ds fy, f

31

= f(3),

3,

31

esa lrr~ gS vkSj

3,

31

esa vodyuh; gS

Comprehension (2×2)


In a triangle if the sum of two sides is x and their product is y such that (x + z).(x - z) = y, where z is the thirdside of the triangle.,d f=kHkqt esa ;fn nks Hkqtkvksa dk ;ksx x rFkk mudk xq.kuQy y bl izdkj gS fd (x + z).(x - z) = y, tgk¡ z f=kHkqt dhrhljh Hkqtk gSA

44. Greatest angle of the triangle isf=kHkqt dk vf/kdre dks.k gS&(A) 105º (B*) 120º (C) 135º (D) 150º

Sol. (b + c + a)(b + c � a) = bc

2s(2s � 2a) = bc

orbc

)a�s(s=

21

2A

cos = cos 3

A = 32

= 120º

45. Area of the triangle isf=kHkqt dk {ks=kQy gS&

(A*) 43y

(B) 4

3zx(C)

43z

(D) 2xz

Sol. = 21

bc sinA

= 21

y. 23

= 43y

Paragraph for Question Nos. 46 to 47

If f(x) =

1x,b2xax

1x,bax3

is a differentiable function for all values of x, then

46. The ordered pair (a, b) is

(A)

1,

21

(B) (0, 1) (C*)

1,

21

(D)

0,

21

Page # 5

Sol. �a + b = �a � 1 + 2b using continuity

b = 1

f (x) =

1x,1ax3

1x,a2

a = 3a + 1 a = � 21

47. If x [�2, �1] (1, 2), then range of f(x) is

(A*)

25

,0 (B)

2,

23

(C)

25

,23

(D) none of these

Sol. y = f(x) =

1x,2x2x

1x,12x

3

For x [�2, �1], f(x) is decreasing

f(�2) = 2 and f(�1) = 23

f(x)

2,

2

3..... (i)

for x (1, 2), f(x) is decreasing

f(1) = 25

and f(2) = 0

f(x)

2

5,0 ..... (ii)

from (i) and (ii), we get f(x)

2

5,0

Hence the range is

2

5,0

iz'u 46 ls 47 ds fy, vuqPNsn

;fn x ds lHkh ekuksa ds fy, f(x) =

1x,b2xax

1x,bax3 ,d vodyuh; Qyu gS] rks

46. Øfer ;qXe (a, b) gS &

(A)

1,

21

(B) (0, 1) (C*)

1,

21

(D)

0,

21

Sol. �a + b = �a � 1 + 2b lrr~rk dk mi;ksx djus ij b = 1

f (x) =

1x,1ax3

1x,a2

a = 3a + 1 a = � 21

47. ;fn x [�2, �1] (1, 2) gks] rks f(x) dk ifjlj gS &

Page # 6

(A*)

2

5,0 (B)

2,

2

3(C)

2

5,

2

3(D) buesa ls dksbZ ugha

Sol. y = f(x) =

1x,2x2x

1x,12x

3

x [�2, �1] ds fy, f(x) âkleku gSA

f(�2) = 2 vkSj f(�1) = 23

f(x)

2,

2

3..... (i)

x (1, 2) ds fy, f(x) âkleku gSA

f(1) = 25

vkSj f(2) = 0

f(x)

2

5,0 ..... (ii)

(i) vkSj (ii) ls f(x)

25

,0

vr% ifjlj

2

5,0 gSA

Subjective -10 (Double Digit Answer)

48. Let k = n

lim

n2�3n2�n1n2n3nn 23 23, find the value of 18k

ekuk k = n

lim

n2�3n2�n1n2n3nn 23 23] rks 18k dk eku Kkr dhft,A

Ans. 12

Sol.n

lim

)1�n(�3n2�nn)1n(�1n2n3nn 23 23

= n

lim

23/1233/223

323

)1n()1n2n3n)(1n()1n2n3n(

)1n(�1n2n3nn

+

)1�n(3n2�n

)1�n(�3n2�nn2

22

= n

lim23/1233/223 )1n()1n2n3n)(1n()1n2n3n(

)n(�n

+

nlim

)1�n(3n2�n

)2(n2

= 111

1�

+ 11

2

= � 31

+ 1 = 32

= k

18k = 12

49. If 15th term of the sequence 1,0,1,8,29,80,193,..... is a16 � b, a, b N, then b � 100a is equal to

;fn vuqØe 1,0,1,8,29,80,193,..... dk 15ok¡ in a16 � b gS tgk¡ a, b N , rc b � 100a dk eku gS&Ans. 55Sol. Differences are vUrj fof/k ls

�1 1 7 21 51.......... 2 6 14 30 .............. 4 8 16 ....................Let ekuk T

n = a.2n � 1 + bn2 + cn +d

T1 = a + b + c + d = 1

Page # 7

T2 = 2a + 4b + 2c + d = 0

T3 = 4a + 9b + 3c + d = 1

T4 = 8a + 16b + 4c + d = 8

a = 4, b = �1, c = �2, d = 0

Tn = 4.2n � 1 � n2 � 2n = 2n + 1 � n2 � 2n.

T15

= 216 � 255

a = 2 , b = 255 b � 100a = 255 � 200 = 55

50. A function f(x) is defined as f(x) =

irrationalisx

rationalisx

,

,

bx2ax

1axx2

2

is continuous at x = 1 and 2, then find the value of 20(a + b).

,d Qyu f(x) ifjHkkf"kr gS f(x) =

vifjes; gSifjes; gS

x

x,

,

bx2ax

1axx2

2

x = 1 vkSj 2 ij lrr~ gks] rks 20(a + b) dk eku Kkr dhft,AAns. 10

Sol. f(x) =

irrationalisx

rationalisx

,

,

bx2ax

1axx2

2

is continuous at x = 1 and 2

f(x) =

vifjes; gSifjes; gS

x

x,

,

bx2ax

1axx2

2

x = 1 vkSj 2 ij lrr~ gSA

x = 1 and 2 are the roots of the equation x2 + ax + 1 = ax2 + 2x + bx = 1 vkSj 2 lehdj.k x2 + ax + 1 = ax2 + 2x + b ds ewy gSAor ;k (a � 1)x2 + (2 � a) x + b � 1 = 0

1�a2�a

= 3 andvkSj 1�a1�b

= 2

a = 21

andvkSj b = 0

20(a + b) = 10

51. Set of all possible values of a such that f(x) = e2x � (a + 1) ex + 2x is monotonically increasing for allx R, is (�, ]. Value of 4 is;fn a ds lHkh laHkkfor ekuksa dk leqPp; ftuds fy, f(x) = e2x � (a + 1) ex + 2x, x R ,dfn"V o/kZeku gS] (�,

] gS] rc 4 dk eku gS&Ans. 12Sol. f(x) = e2x � (a + 1) ex + 2x

f(x) = 2e2x � (a + 1) ex + 2Nowvc, 2e2x � (a + 1) ex + 2 0 x R

i.e. 2

xx

e

1e � (a + 1) 0 x R

i.e. 4 � (a + 1) 0i.e. a 3

Page # 8

52. If f(x) = sin�1

2x1

x2 then the value of

2

34

f43

f75

is equal to

;fn f(x) = sin�1

2x1

x2 rks

2

34

f43

f75

dk eku cjkcj gS&

Ans. 21

Sol. Letekuk x = tan, 2�

< < 2

f(x) = sin�1 (sin2)

=

2/4/

4/�4/�

4/�2/�

,

,

,

2�

2

2��

f(x) =

1x

1x1�

1�x

,

,

,

xtan2�

xtan2

xtan2��

1�

1�

1�

f

4

3+ f

3

4=

169

1

2

�

916

1

2

= 25

18�32 =

2514

2

34

f43

f75

= 21

53. If the sum of the series 3.1

12

+ 5.3

22

+ 7.5

32

+......+ 41.39

202

is 41.a21.20

, then find the value of 5a.

;fn Js.kh 3.1

12

+ 5.3

22

+ 7.5

32

+......+ 41.39

202

dk ;ksxQy 41.a21.20

gks] rks 5a dk eku Kkr dhft,A

Ans. 10

Sol. tr =

)1r2)(1�r2(r2

n

1rrt4 = n +

21

n

1r1r2

1�

1�r21

n

1rrt = )1n2(2

)1n(n

, now put n = 20 j[kus ij

54. The maximum volume of the cylinder which can be inscribed in a sphere of radius 5 3 cm is 10k cm3 .

Find the value of k.

,d 5 3 cm f=kT;k ds xksys ds vUnj cuk, tkus okys csyu dk vf/kdre vk;ru 10kcm3 gS] rks k dk eku Kkr

dhft,AAns. 50

Page # 9

Sol. Let ABCD be the cylinder inscribed in a sphere of radius = 5 3 cm

ekuk f=kT;k 5 3 cm ds xksys esa ,d csyu ABCD cuk;k tkrk gSA

Let its base radius be r and height h (in cm ) . Thenekuk blds vk/kkj dh f=kT;k r gS ,oa Å apkbZ h (cm esa ) gS rcOB2 = r2 + OP2

r2 = (5 3 )2 �

2

2h

= 75 �

41

h2 [2]

Now V = volume of the cylinder = r2hvc V = csyu dk vk;ru = r2h

V =

2h41

�75 h =

3h41

�h75

For maximum or minimum , hdVd

= 0 vf/kdre ,oa U;wure ds fy, hdVd

= 0

75 � 43

h2 = 0 , h2 = 100 , h = 10 cm

and vkSj r2 = 75 � 41

(10)2 = 50

For this value of h. h ds bl eku ds fy,

2

2

hd

Vd = �

23 × 10 = �15 < 0

V is maximum vf/kdre gSA

Maximum value of V is ( 75 × 10 � 41

× 103) V dk vf/kdre eku ( 75 × 10 � 41

× 103)

= 500 cm3

55. In a geometric progression with common ratio 'q' the sum of the first 109 terms exceeds the sum of

the first 100 terms by 12. If the sum of the first nine terms of the progression is 100q

, then the value

of is equal tolkoZvuqikr 'q' okyh ,d xq.kksÙkj Js<+h esa izFke 109 inksa d k ;ksx] izFke 100 inksa d s ;ksx ls 12 vf/kd gSA ;fn

Js<+h d s izFke ukS inksa d k ;ksx 100q

gS] rks d k eku cjkcj gS &

Ans. 12Sol. Let a be the first term, then

a

1q1q109

� a

1q1q100

= 12

1qa

(q109 � q100) = 12

a.q100 1q

)1q( 9

= 12

q100 100q

= 12 = 12

Hindi. ekukfd a izFke in gS] rks

a

1q1q109

� a

1q1q100

= 12

Page # 10

1qa

(q109 � q100) = 12

a.q100 1q

)1q( 9

= 12

q100 100q

= 12 = 12

56. If L =x

lim

)x(tan2

1e21

x

12

, then find the value of | 20L |.

;fn L =x

lim

)x(tan2

1e21

x

12

gS] rks | 20L | dk eku Kkr dhft,A

Ans. 10

Sol. x

lim

)x(tan2

1e21

x

12

= x

lim

21

x

1

xcot2

2

1e2

= x

lim

21

x

1

x

1tan2

1e2

= �

21

57. Given log(a + c), log (a + b), log (b + c) are in A.P. and a, c, b are in H.P., where a, b, c > 0. If a + b = 4kc

,

then find the value of |2k|.fn;k x;k gS log(a + c), log (a + b), log (b + c) lekUrj Js<+h esa gS vkSj a, c, b gjkRed Js<+h eas gS] tgk¡ a, b, c > 0.

;fn a + b = 4kc

gks] rks |2k| dk eku Kkr dhft,A

Ans. 16Sol. log (a + c) + log (b + c) = 2 log (a + b)

(a + c) (b + c) = (a + b)2

ab + c(a + b) + c2 = (a + b)2

2ab + 2c(a + b) + 2c2 = 2(a + b)2 ...... (i)

also iqu%c = ba

ab2

2ab = c(a + b) ..... (ii)

From (i) and (ii) (i) vkSj (ii) lsc(a + b) + 2c(a + b) + 2c2 = 2(a + b)2

2(a + b)2 � 3c(a + b) � 2c2 = 0

a + b = 4

c5c34

c16c9c3 22

= 2c or � 2c

a + b = 2c ( a, b, c > 0)

4k

= 2 k = 8

|2k| = 16

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