MONT 2

Code

®

P25-16

0

MAIN PATTERN PATTERN ONLINE ONLINE TEST-2 TEST-2 (MONT-2) (MONT-2) XI XI TARGET : JEE JEE (MAIN+ADVANC (MAIN+ADVANCED) ED) 2017

COURSE : VIJETA (JP), (JP), VISHWAAS (JF), (JF), ANOOP (EP), (EP), AKHIL (EF) Date (

) : 05-01-2017

Time: 3 Hours(

:3

)

Max. Marks (

) : 360

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. 5

INSTRUCT INSTRUCTIONS IONS / A.

General :

1.

A.

This This bookl booklet et is is your your Questi Question on Paper. Paper. Do not break break the seals of this booklet before being instructed to do so by the invigilators. Blan Blank k spac spaces es and and blan blank k page pages s are are provi provide ded d in the the question paper for your rough work. No additional sheets will be provided for rough work.

2.

1.

;g iqfLrdk vkidk iz'u&i=k gSA bldh eqgjsa rc rd u rksMsa tc rd fujh{kdks fujh{kdksa ds }kjk }kjk bldk funs 'k Z u fn;k tk;sA

2.

dPps dke ds fy;s [kkyh i`"B vkSj [kkyh txg bl iqfLrdk esa gh gh gSA dPps dke ds fy, dksbZ vfrfjDr dkxt ugha fn;k tk;sxkA dksjs dkxt] fDyi cksMZ (CLIP BOARD)] ykW x rkfydk] LykbM:y] dSYdqysVj] dSejk] lsyQksu] istj vkSj fdlh izdkj ds bys DVªkfud midj.k ijh{kk d{k esa vuqefr ugha gSA bl iqfLrdk ds fiNys fiNys i` "B ij fn, x, LFkku esesa viuk uke vkSj jksy uEcj fyf[k,A Åijh ewy i`"B ds cqycq yksa (BUBBLES) dks dkys ckWy IokbaV dye ls dkyk djsaA vks-vkj-,l- (ORS) ;k bl iqfLrdk esa gsj&Qsj@foÑfr u djs aA bl iq fLrdk fLrdk dh eq gjs a rks M+us ds i'pkr~ Ñi;k tk¡ p ys a fd bles ble as lHkh lHkh 90 iz 'u vkSj muds muds mÙkj mÙkj fodYi Bhd ls i<+ i<+ s tk ldrs ldrs gS gS aA lHkh [ka Mks a ds iz iz kjaHk es a fn;s fn;s gq gq , funs fun Zs' akks dks dks /;ku /;ku ls i<+ a sA

3.

Blank Blank papers papers,, clipb clipboar oards ds,, log log table tables, s, slide slide rules, rules, 3. calculators, cameras, cellular phones, pagers and electronic gadgets are NOT allowed inside the examination hall.

4.

Writ Write e you you nam name e and and roll roll num numbe berr in the the spac space e provided on the back cover of this booklet.

4.

5.

Using a black ball point pen, darken the bubbles on the upper original sheet.

5.

6.

DO NOT NOT TAM TAMPE PER R WITH WITH/M /MUT UTIL ILAT ATE E THE THE ORS ORS OR THE BOOKLET.

6.

7.

On breaki breaking ng the the seal seals s of of the the bookl booklet et chec check k that that it contains all the 90 questions and corresponding answer choices are legible. Read carefully the Instructions printed at the beginning of each section.

7.

B.

Filling the ORS Use only Black ball point pen only for filling the ORS.

B.

8.

Write Write your your Roll Roll no. no. in the boxes boxes give given n at at the the top top left left 8. corner of your ORS with black ball point pen. Also, darken the corresponding bubbles with Black ball point pen only. Also fill your roll no on the back side of your ORS in the space provided (if the ORS is both side printed). Fill your your Pape Paperr Code Code as mentio mentioned ned on the the Test Test Paper Paper 9. and darken the corresponding bubble with Black ball point pen.

9.

10.

:

(ORS)

ORS ORS

ORS ds

lcls Åij cka;s dksus esa fn, x, ck¡Dl esa viuk jksy uEcj dkys ck¡y ikbUV ls fyf[k, rFkk laxr xksys Hkh dsoy dkys isu ls Hkfj;sA ORS ds ihNs dh rjQ Hkh viuk ( ORS nks jksy uEcj fyf[k, ;fn uksa rjQ Nih Nih gqbZ gS gSA) ORS ij

viuk isij dksM fyf[k, rFkk laxr xksy aks dks dkys ck¡y isu ls dkys dhft,A dhft,A

If student student does not fill fill his/her his/her roll no. and and paper paper code code 10. ;fn fo|kFkhZ viuk jksy uEcj correctly and properly, then his/her marks will not be mfpr rjhds ugha Hkjrk gS rc displayed and 5 marks will be deducted (paper wise) tkosxk rFkk iziz'u&i=k esa izkIrka Irkad from the total. (P lease lease read the the las las t pag pag e of this book let let for res t of the ins tructions )

rFkk isij dksM lgh vkSj mldk ifj.kke jksd fy;k ls 5 vad dkV fy, tkosax as A

Resona Resonance nce Eduven Eduventure tures s Ltd. Ltd. CORPORATE CORPORATE OFFICE : CG Tower, Tower, A-46 & 52, IPIA, IPIA, Near Near City Mall, Mall, Jhalawar Jhalawar Road, Road, Kota (Raj.) (Raj.) - 324005 Ph.No. : +91-744-3012222, 6635555 | Toll Toll Fre Free e : 258 5555 Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 +91-022-39167222 P25-16 Websit Website e : www.resonance.ac.in | E-mail E-mail : [email protected] | CIN: CIN: U80302RJ2007PLC024029

R O T A L I G I V N I E H T Y B O S O D O T D E T C U R T S N I G N I E B T U O H T I W L A E S E H T K A E R B T O N O D

PHYSICS

PARTPA RT-II : PH PHYSI YSICS CS SECTION SECTION – I (

S C I S Y H P

- I)

Straight Objective Type (

)

This section contains 15 multiple choice questions. choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.

bl [k.M [k.M ese as 1.

15

iz 'u gS aA iz R;s d iziz'u ds 4 fodYi fodYi

rFkk (D) gS gS a] (A), (B), (C) rFkk

ftues a ls ls

gSA

A hollow cubical box P is moving on a smooth horizontal surface in the x–y plane with constant acceleration of a  3î  4ˆj m/s2 . A block Q of mass 2kg is at rest inside inside the cubical cubical box as shown 

in figure. lf the coefficient of fricition between the surface of the cube P and the block Q is 0.6. Then the force of fricition between P and Q is : (A) 5 N at an angle 53 0 wi with xx-axis (B) 8 N at an angle 530 with y-axis (C) 12 N at an angle 370 wi with x-axis (D) 10 N at an angle 37 0 with with y -axis -axis Z

P

Y Q

X

,d [kks[kyk ?kukdkj ?kukdkj ckWDl P, fu;r Roj.k a  3î  4ˆj m/s2 ls x–y ry esa ?k"kZ.kjfgr {kS frt lrg lrg ij xfr dj dj Space for Roug h Work WockWrk ) ij j[kk gqvk gSA ;fn ?ku P Q ?kukdkj jgk gSA 2kg nzO;eku dk ,d CykW d Rough Dl( ds vUnj fp=kkuqlkj fojke rFkk CykWd Q dh lrg ds e/; e/; ?k"kZ.k xq.kkad 0.6 gS rks P rFkk Q ds e/; ?k"kZ.k cy gksxk : dks.k ij dks.k ij (A) 5 N x–v{k ds lkFk 530 ds dks (B) 8 N y–v{k ds lkFk 530 ds dks 0 0 dks.k ij (C) 12 N x–v{k ds lkFk 37 ds dks (D) 10 N y–v{k ds lkFk 37 ds dks.k ij 

Space for Rough Roug h Work Wo rk (

®

)

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PJP/JFMONT2050117C0-1

PHYSICS

2.

For no friction to act on a block kept on an accelerating wedge as shown in figure, acceleration a of the wedge should be fp=k esa xqVds o ost ds e/; dksbZ ?k"kZ.k ugha gSA xqVds dks ost ij fLFkj jgus ds fy, os t dk Roj.k a gksuk

pkfg,A

(A)

3.

g 3

(B) g 3

(C)

g 2

(D) g

3 2

What is the minimum stopping distance for a vehicle of mass m moving with speed v along a level road. If the coefficient of friction between the tyres and the road is . (A)

v2 2g

m nzO;eku

(B)

2v 2 g

(C)

v2 g

(D) none of these

dk ,d okgu tks v pky ls lM+d ds lkFk xfr dj jgk gS U;wure fdruh nwjh ij jksd ldrs gS gSA

;fn okgu ds ifg;s ifg;s rFkk lM+d ds chp chp ?k"kZ.k xq.kkad  gksA (A)

v2 2g

(B)

2v 2 g


®

(C)

v2 g

(D) bueas

ls dksbZ ughaA

)



S C I S Y H P

PHYSICS

4.

The force acting on a body moving along x axis varies with the position of the particle as shown in S the fig. The body is in stable equilibrium at C I (A) x = x1 (B) x = x2 (C) both x1 and x2 (D) neither x1 nor x2 S Y H P

v{k ds vuq fn'k xfr'khy ,d oLrq ij dk;Z dk;Zjr cy d.k dh fLFkfr ds lkFk fp=kkuqlkj ifjofrZr gksrk gSA oLrq fdl fLFkfr ij LFkk;h lkE;koLFkk lkE;koLFkk esa gSA (A) x = x1 (B) x = x2 (C) x1 rFkk x2 nksauks (D) x1 rFkk x2 nksauks ugha x

5.

A particle of mass ’M’ is moved under constant power P 0. At some instant after the start, its speed is v and at a later instant, the speed is 2v. Neglecting friction, distance travelled by the particle as its speed increases from v to 2v is : ’M’ nzO;eku dk ,d d.k P0 fu;r 'kfDr ds vUrxZr xfr dj jgk gSA çkjEHk çkjEHk gksus ds dqN le; i'pkr~ bldh bldh

pky v gS rFkk blds dq dqN le; i'pkr~ pky pky gksxh tc bldh pky v ls 2v rd c<+rh gS & (A)

6.

12Mv 3 5P0

(B)

2v gSA

?k"kZ .k dks ux.; ekfu,A d.k }kjk r; dh xbZ nwjh D;k

8P0 v 3

(C)

M

9 Mv 3 5 P0

(D)

7Mv 3 3P0

A weight W attached to the end of a flexible rope of diameter d = 0.75cm is raised vertically by winding the rope on a reel as shown. If the reel is turned uniformly at the rate of 2 r.p.s. What is the tension in rope. The inertia of rope may be neglected. fp=kkuq lkj d = 0.75 cm O;kl dh yphyh jLlh ,d ,d jhy ij fyiVh fyiVh gqbZ gSA blds ,d fljs ij ij W Hkkj tqM k+ gSA

bl Hkkj ls jLlh Å/okZ/kj /kj uhps dh dh vks j vk jgh gSA ;fn jhy dks yisVk tk;s rks jLlh jLlh es a ruko D;k gS \ jLlh dk tM+Ro ux.; ux.; gSA


2 pDdj

çfr lSd.M dh ,d leku nj ls

)

d W

(A) 1.019W

(B) 0.51W


®

(C) 2.04W

(D) W

)



PHYSICS

7.

If the apparent weight of the bodies at the equator is to be zero, then the earth should rotate with S angular velocity C I

vxj fo"kq or js[kk ij oLrqvksa dk vkHkklh Hkkj 'kwU; gks rks i`Foh dks fdl dks.kh; osx ls ?kweuk gks xk & (A)

8.

g rad/sec R

(B)

2g rad/sec R

(C)

g rad/sec 2R

(D)

S Y H P

3g rad/sec 2R

A particle is tied to one end of a light inextensible string and is moving in a vertical circle, the other end of string is fixed at the centre. Then for complete motion in circle, which is correct. (air resistance is negligible) (A) Acceleration of the particle is directed towards the centre. (B) Total mechanical energy of the particle and earth remains constant (C) Tension in the string remains constant (D) Acceleration of the particle remains constant

,d d.k ,d gYdh vforkU; Mksjh ds ,d fljs ls ca/kk gqvk gS rFkk m/okZ/kj o` Ùk esa xfr dj jgk gSA Mksjh dk nqljk fljk dsUnz ij fLFkj ¼ fixed½ gS rks o`Ùk esa lEiw.kZ xfr ds fy, fuEu es a ls dkSulk fodYi lR; gSA ( ok;q iz frjks/k ux.; gSA) (A) d.k dk Roj.k dsUnz dh vksj gksxkA (B) d.k rFkk i`Foh dh dqy ;kaf=kd ÅtkZ fu;r gksxhA (C) Mkjh esa ruko fu;r jgrk gSA (D) d.k dk Roj.k fu;r jgrk gSA 9.

A small mass m is placed and released from the vertical portion of a quarter circular arc as shown in figure. As it slides, the acceleration of the mass M on the smooth horizontal surface : (A) Continuously increases (B) Continuously decreases (C) First increases then decreases (D) First decreases then increases

Space for Rough Work (

)

nzO;eku dks ,d pkSFkkbZ o`Ùkkdkj oØ ds 'kh"kZ ls fp=kkuqlkj NksM+k tkrk gS A tc CykWd nzO;eku okys CykWd dk fpduh {kSfrt lrg ij Roj.k : (A) yxkrkj c<+ (B) yxkrkj ?kVssxkA sxkA (C) igys c<+sxk fQj ?kVsxkA (D) igys ?kVsxk fQj c<+sxkA m


®

m fQlyrk

gS rks

M

)

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PHYSICS

10.

One end of a sting is attached to a bob and the other and is fiexed at a peg A. The bob is taken to a position where string makes an angle of 30 0 with the horizontal. On the circular path of the bob in vertical plane there is a peg ‘B at a symmetrical position with respect to the position of release as shown in the figure. If V c and Va be the minimum speeds in clockwise and anticlockwise directions respectively, given to the bob in order to hit the peg ‘B then ratio V c : Va is equal to :

,d jLlh ds ,d fljs ls ckW c tqM+k gS rFkk nwljk fljk [kwVh A ls tqM+k gSA ckW c dks {kSfrt ls 300 fLFkfr rd fp=kkuq lkj ys tk;k tkrk gS rFkk ;gka ls NksM+k tkrk gSA ckW c ds o`Ùkkdkj Å/okZ/kj iFk ij [kwVh ‘B leku ÅWpkbZ ij fLFkr gSA vc ckW c dks ;gka ls NksM+k tkrk gSA nf{k.kkorZ rFkk okekorZ fn'kkvksa ls ckWc ds [kwVh B ij Vdjkus ds fy, fn;s x;s U;wure osx Øe'k% Vc rFkk Va gks rks Vc : Va gS :

(A) 1 : 1 11.

(B) 1 :

2

(C) 1 : 2

(D) 1 : 4

Two particles tied to different strings are whirled in a horizontal circle as shown in figure. The ratio of lengths of the strings so that they complete their circular path with equal time period is: (A)

3 2

(B)

2 3

(C) 1


(D) None of these

)

jLlh ls cU/ks nks d.kksa dks fp=kkuq lkj {kSfrt o`Ùkkdkj iFk esa ?kq ek;k tkrk gSA jfLl;ksa dh yEckbZ dk vuqikr D;k gksxkA ftlls os viuk o`Ùkh; iFk leku le; vUrjky esa iw jk djrs gS & (A)

3 2

(B)

2 3


®

(D) buesa

(C) 1

ls dksbZ ughaA

)



S C I S Y H P

PHYSICS

S

12.

V I The figure shows a hollow cube of side ’a’ of volume V. There is a small chamber of volume in C 4 S Y

the cube as shown. This chamber is completely filled by m kg of water. Water leaks through a H hole H and spreads in the whole cube. Then the work done by gravity in this process assuming that the complete water finally lies at the bottom of the cube is : a Hkqtk

rFkk V vk;ru dk ,d [kks[kyk ?ku fp=k esa iznf'kZr gSA fp=kkuqlkj blesa ,d NksVk

{kS=k gSA bl y?kq {kS=k dks

m fdxzk

V 4

vk;ru dk ?ku

nzO;eku ds ikuh ls Hkj fn;k tkrk gSA ikuh ,d Nsn }kjk fudydj lkjs ?ku

esa QSyrk gSA ;g ekuuk gS vUr esa lkjk ikuh ?ku ds ry esa vk tkrk gS] rks bl izfØ;k esa xq:Ro }kjk fd;k x;k dk;Z gksxk

Nnz

(A)

1 mg a 2

(B)

3 mg a 8


®

(C)

5 mga 8

(D)

1 mga 8

)



P

PHYSICS

13.

The force acting on the system is represented against distance r as shown. The potential energy of the system will be best represented by. (At x = 0, potential energy = 0)

nwjh r ds lkis{k fudk; ij dk;Zjr cy n'kkZ;s vuqlkj gSA fudk; dh fLFkfrt ÅtkZ dks lcls lgh çnf'kZr djrk gSA

(x = 0 ij, fLFkfrt

ÅtkZ

= 0)

(A)

(B)

(C)

(D)


®

)



S C I S Y H P

PHYSICS

14.

Figure shows a plot of potential energy function U(x) = kx 2 where x = displacement and S k = constant. Identify the correct conservative force function F(x) C I 2 fn;k x;k fp=k fLFkfrt ÅtkZ dk Qyu U(x) = kx tgk¡ x = foLFkkiu rFkk k = fu;rkad gSA la j{kh cy F(x) S Y H P

dk lgh Qyu gSA

15.

(A)

(B)

(C)

(D)

A block of mass 20 kg is acted upon by a force F = 30 N at an angle 53° with the horizontal in downward direction as shown. The coefficient of friction between the block and the horizontal surface is 0.2. The friction force acting on the block by the ground is (g = 10 m/s 2) 20 kg nzO;eku ds ,d CykWd ij ,d cy F = 30 N {kSfrt ls 53° ds dks.k ij uhps dh vks j yxk;k tkrk gS

tSlk fd fn[kk;k x;k gSA CykW d vkSj {kSfrt lrg ds chp ?k"kZ.k xq.kkad x;s ?k"kZ.k cy dk eku gS (g = 10 m/s2)

0.2 gSA

/kjkry }kjk CykWd ij yxk;s

F 53°

Space for Rough Work ( (A) 40.0 N

(B) 30.0 N

) (C) 18.0 N


®

(D) 44.8 N )



PHYSICS

S C I S Integer value correct Type ( ) Y This section contains 15 questions. The answer to each question is a single digit integer , ranging from H P

SECTION – II (

– II)

0 to 9 (both inclusive).

bl [k.M esa 15 16.

gSaA çR;sd ç'u dk mÙkj 0 ls 9 rd ¼nks uksa 'kkfey½ ds chp dk ,dy

gSA

A block of unknown mass is at rest on a rough horizontal surface. A force F is applied to the block. The graph in the figure shows the acceleration of the block with respect to the applied force. The mass of the block (in kg) is

,d vKkr nzO;eku dk CykWd {kS frt [kq jnjh lrg ij j[kk gS A CykWd ij ,d {kSfrt

F cy

xzkQ esa vkjks fir cy ds lkis{k CykW d dk Roj.k çnf'kZ r gSA CykWd dk nzO;eku

&


®

(kg esa) gS

yxk;k tkrk gSA

)



PHYSICS

17.

A point mass of 0.5 kg moving with a constant speed of 5 m/s on an elliptical track experience an S

C

inward force of 10 N when at either end of the major axis and a similar force of 1.25 N at each end I S Y of minor axis. Find the ratio of length of major axis to the length of minor axis : 0.5 kg nzO;eku

fdukjks ij

dk fcUnqor nzO;eku fu;r pky

10 N cy

5 m/s ls

fn?kZo`Ùkkdkj iFk ij xfr'khy gS A ;g nh?kZ v{k ds

vUnj dh rjQ rFkk y?kq v{k ds fdukjs ij

1.25 N cy

vUnj dh rjQ eglql djrk gSA

rks nh?kZ v{k rFkk y?kq v{k dh yEckbZ;ks dk vuqikr Kkr djks A

18.

A pendulum of length 1 m is released from  = 60°, then the rate of change of speed of the bob at  = 30° is x m/s 2 (g = 10m/s2). Find x

1 m yEckbZ

dk ,d yksyd

x m/s2 (g = 10m/s2) gks

 = 60° ls

NksM+k tkrk gSA rc

 = 30° ij

xksyd dh pky esa ifjorZu dh nj

rks x Kkr djksA


®

)



H P

PHYSICS

19.

A particle moves along a circle with constant speed such that magnitude of its instantaneous n acceleration is a. The magnitude of average acceleration during the time it covers half cycle is a.  Find the value of n. ,d d.k o`Ùk esa fu;r pky ls bl izdkj xfr djrk gS fd blds rkR{kf.kd Roj.k dk ifjek.k a gSA vk/kk pØ

r; djus ds i'pkr vkS lr Roj.k dk ifjek.k 20.

n a gSA n dk 

eku Kkr djksA

Two particles A and B are revolving with constant angular velocity on two concentric circles of radius 1m and 2m respectively as shown in figure. The positions of the particles at t = 0 are shown 



in figure. If m A = 2kg, m B = 1kg and P A and PB are linear momentum of the particles then what is 



the maximum value of P A  PB in kg-m/sec in subsequent motion of the two particles.

nks d.k A rFkk B fu;r dks.kh; osx ls Øe'k% 1m rFkk 2m f=kT;k ds nks ldsUnzh; o`Ùkks ij fp=kkuq lkj pDdj yxk jgs gSA t = 0 ij d.kksa dh fLFkfr;k¡ fp=k esa iznf'kZr gSA ;fn m A = 2kg, mB = 1kg rFkk P A o PB d.kksa ds js[kh; la osx gS rks nksuksa d.kksa dh xfr esa P A  PB dk vf/kdre eku ¼ kg-m/sec esa) Kkr djksA 







2m/s

A

B 3m/s


®

)



S C I S Y H P

PHYSICS

21.

A small block is projected up from the bottom of a fixed incline, inclined at 60º from the horizontal. S

C

Coefficient of friction between the block & the incline is

I 3 . If the ratio of descending time to S Y 2

H P

ascending time is  then find the value of 2 ?

{kSfrt

60º dks.k

okys ,d fLFkj urry ij isans ls ,d NksVk xqVdk Åij dh vksj iz{ksfir fd;k tkrk gSA xqVds

o urry ds e/; ?k"kZ.k xq.kad

3 gSA 2

;fn uhps vkus o Åij tkus ds le; esa vuqikr  gS rks

2 dk

eku

Kkr djksa

22.

Block B of mass 2 kg rests on block A of mass 10 kg. All surfaces are rough with the value of coefficient of friction as shown in the figure. If the minimum force required to cause the raltive motion between A & B is F(in Newton). Then find value of

F . (g = 10 m/s 2) 6

nzO;eku 2 kg dk CykW d B ,d vU; 10 kg nzO;eku ds CykWd A ij fLFkj j[kk gSA fp=k esa fn[kk;s x;s vuqlkj ?k"kZ.k xq.kkad ds eku lHkh [kqjnjh lrg ds fy, fp=k esa iznf'kZ r gSA ;fn yxk;s tkus okyk U;wure cy F gS tks F 6

fd A rFkk B ds chp lkis{k xfr mRiUu dj lds] rks dk eku Kkr dhft,A ¼ g = 10 [email protected] 2½


®

)



PHYSICS

23.

Two blocks of mass m and 2m are arranged on a wedge that is fixed on a horizontal surface. Friction coefficient between the block and wedge are shown in figure. Find the magnitude of acceleration of two blocks. m rFkk 2m nzO;eku ds nks CykWd ost ij O;ofLFkr gS ;g ost {kSfrt lrg ij fLFkj gSA CykWd rFkk ost ds e/;

?k"kZ.k xq.kka d fp=k esa iz nf'kZ r gSA nksuksa CykWdksa ds Roj.k dk ifjek.k Kkr djks A

µ=1

m Fixed 2m wedge 53°

24.

smooth

m µ=1

37°

LFkj ost

53°

2m

?k"kZ.kjfgr 37°

Two blocks of same mass are kept on a horizontal rough surface as shown in the figure. Coefficient of friction between the blocks and surface is 0.1. A force of 16 N is applied on one block as shown in figure. Find the tension (in Newton) in the string.

leku nzO;eku ds nks CykWd fp=kkuqlkj {kSfrt [kqjnjh lrg ij j[ks gSA CykW d rFkk lrg ds e/; ?k"kZ.kxq.kkad 0.1 gSA 16N dk ,d cy ,d CykWd ij fp=kkuqlkj yxk;k tkrk gSA Mksjh esa ruko U;w ( Vu esa) Kkr dhft,A

25.

A 200 gm block a pressed with a minimum force against a rough vertical wall so that the block does not fall. If coefficient of static friction between the wall & the block is 0.4. Find the minimum force (in Newton) (g = 10 m/s 2) ,d 200 gm nzO;eku dk xqVdk ,d Å/okZ/kj [kq jnjh nhokj ij U;wure cy ls bl izdkj nck;k tkrk gS rkfd

xqVdk uhps ugh fxjrk gS] ;fn fnokj o xqVds ds e/; LFkS frd ?k"kZ.k xq.kkad dhft,A (g=10 m/s2) Space for Rough Work (

®

0.4 gSA

( Vu es)a Kkr U;wure cy U;w

)



S C I S Y H P

PHYSICS

26.

Potential energy of a particle of mass m, depends on distance y from line AB according to given relation U =

y=

3 a towards line AB (perpendicular to it) then minimum velocity so that it cannot return to its

initial point is

O;eku m nz K

S C I K , where K is a positive constant. A particle of mass m is projected from S 2 2 y a Y H P

K , calculate N. aNm

ds d.k dh fLFkfrt ÅtkZ AB js [kk ls nw jh y ds lkFk

/kukRed fu;rka d gSA ;fn

m nzO;eku

ds d.k dks

y =

U =

K 2

y  a2

3 a ls AB js[kk

tkrk gS rks og U;wure~ osx ftlls ;g iqu% iz kjfEHkd fcUnq ij ykSVdj ugha vk;s


®

lEcU/k ds vuq lkj cnyrh gSA tgkW

dh vks j ¼blds yEcor~½ iz{ fir ks fd;k K ls fn;k aNm

tkrk gS rks N gksxkA

)



PHYSICS

27.

A block of mass m = 2kg is pulled along a rough horizontal surface by applying a constant force at an angle  = tan –1 2 with the horizontal as shown in the figure. The friction coefficient between the block and the surface is  = 0.5. If the block travels at a uniform velocity v = 5 m/s then calculate average power the (Watt) of the applied force. (Take acceleration due to gravity g = 10 m/s 2) 5 m = 2kg nzO;eku dk CykWd {kSfrt [kqjnjh lrg ds vuq fn'k CykWd ij {kS frt ls  = tan –1 2 dks.k ij fu;r

cy vkjks fir djds fp=kkuqlkj [khpka tkrk gSA CykWd rFkk lrg ds e/; ?k"kZ.k xq.kkad  = 0.5 gSA ;fn CykWd nh xbZ vkSlr 'kfDr fu;r osx v = 5 m/s ls xfr'khy gks rks vkjksfir cy }kjk dk eku Kkr djksa xq ( :Ro ds 5

dkj.k Roj.k g ekuks)

28.

Initially spring is in its natural length and blocks A & B are at rest. Find maximum value of constant force F that can be applied on B such that block A remains at rest. (g = 10 m/s 2) (give answer in Newton) izkjEHk esa fLiazx mldh iz kd`frd yEckbZ esa gS rFkk A o B CykW d fojkekoLFkk esa gSA B CykWd ij yxk;s tk ldus

okyk vf/kdre~ fu;r cy (mÙkj U;wVu es nhft,A)

F dk

eku Kkr dhft, rkfd CykWd


®

A fojkekoLFkk

esa jg lds A

(g = 10 m/s2)

)



S C I S Y H P

PHYSICS

29.

Work done by force F to move block of mass 2kg from A to C very slowly is (76 x)J. Force F is always acting tangential to path. Equation of path AB is x 2 = 8y and BC is straight line which is tangent on curve AB at point B (  between block and path ABC is 0.5). Then value of ’x’ is [g = 10 m/s2] : 2kg CykWd dks A ls C rd /khjs&/khjs ys tkus ij cy F }kjk fd;k x;k dk;Z (76 X)J gSA cy F ges'kk iFk dh

Li'kZ js[kk ds vuqfn'k dk;Zjr gSA AB iFk dh lehdj.k x2 = 8y vkS j BC ljy js[kk gS] tks oØ AB ds fcUnq ij Li'kZ js [kk gSA iFk ABC o CykWd ds e/; ?k"kZ.k xq.kkad  = 0.5 gSA rc] ’X’ dk eku gSS& [g = 10 m/s2] :

30.

B

A block of mass 60 kg is released from rest when compression in the spring is 2m (natural length of spring is 8m). Surface AB is smooth while surface BC is rough. Block travels x distance before coming to complete rest. Value of x is : [g = 10 m/s 2] 60 kg nzO;eku ds ,d CykWd dks fojkekoLFkk ls NksM+k tkrk gS] tc fLizax 2m laihfMr gSA fLiz ( ax dh izkÑfrd

yEckbZ 8m gSA) lrg AB fpduh gS] tcfd gSA rc x dk eku gS&

BC [kq jnjh


®

gSA CykWd iw.kZ :i ls :dus ls igys x nw jh r; djrk

)



S C I S Y H P

CHEMISTRY

PART-II : CHEMISTRY Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, Si = 28, Y P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75, R T Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207] SECTION – I (

S I M E H C

- I)

Straight Objective Type (

)

This section contains 15 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.

bl [k.M es a ( 31.

15

iz 'u gS aA iz R;s d iz'u ds 4 fodYi

a] (A), (B), (C) rFkk (D) gS

ftues a ls

gSA)

What volume of air at 1 atm and 273 K containing 21 % of oxygen by volume is required to completely burn sulphur(S 8) present in 200 g of sample, which contains 20 % inert material which does not burn. Sulphur burns according to the reaction

1 S8 (s) + O2 (g)   SO2 (g) 8 (A) 23.52 litre 1 atm rFkk 273 K ij

(B) 320 litre

(C) 112 litre

(D) 533.33 litre

vk;ru ds vuqlkj 21% vkWDlhtu ;q Dr ok;q dk vko';d vk;ru D;k gS] tks

izkn'kZ es mifLFkr lYQj (S8) ds iw. r% kZ tyus ds fy, vko';d gksrk gSA izkn'kZ esa

20 %

200 g

vfØ; inkFkZ mifLFkr

gS] tks tyrk ugh gSA lYQj fuEu vfHkfØ;k ds vuqlkj tyrk gS& 1 S8 (s) + O2 (g)   SO2 (g) 8 (A) 23.52 yhVj

(B) 320 yhVj

(C) 112 yhVj

(D) 533.33 yhVj

Space for Rough Work

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JP/JFMONT2050117C0-17

CHEMISTRY

32.

Some older emergency oxygen masks containing potassium superoxide, KO 2 which reacts with CO2 and water in exhaled air to produce oxygen according to the given equation. If a person

Y R T S 4KO2 + 2H2O + 4CO2   4KHCO3 + 3O2 I M dqN iqjkus vkikrdkyhu vkWDlhtu ekLdks es iksVsf'k;e lqijvkWDlkbM] KO2 gks rh gS] tks mPNoflr (exhaled) E H ok;q es CO2 rFkk ty ds lkFk fØ;k djds nh x;h lehdj.k ds vuqlkj vkWDlhtu mRikfnr djrh gSA ;fn C exhales 0.667 g of CO 2 per minute, how many grams of KO 2 are consumed in 5.0 minutes ?

,d O;fDr iz fr

feuV 0.667

g, CO2 dk

mPNolu ¼fu"dklu½ djrk gS] rks

5.0 feuV

es fdrus xzke KO2

iz;qDr ¼[kpZ½ gksrh gS\ 4KO2 + 2H2O + 4CO2   4KHCO3 + 3O2 (A) 10.7 33.

(B) 0.0757

(C) 1.07

(D) 5.38

Phosphoric acid (H3PO4) prepared in a two step process.  P O (1) P4 + 5O2  4 10

 4H PO (2) P4O10 + 6H2O  3 4

We allow 62 g of phosphorus to react with excess oxygen which form P 4O10 in 85 % yield. In the step (2) reaction 90% yield of H 3PO4 is obtained. Produced mass of H3PO4 is :

QkWLQksfjd vEy (H3PO4) dks nks in izØe esa cuk;k tkrk gS&  P O (1) P4 + 5O2  4 10

 4H PO (2) P4O10 + 6H2O  3 4

ge 62 xzke QkW LQks jl dh vkf/kD; vkWDlhtu ds lkFk fØ;k djrs gS] ftlls (2) es H3PO4 dh 90% yfC/k (A) 37.485 g

85 % yfC/k

esa

P4O10 curk

gSA in

izkIr gksrh gSA H3PO4 dk mRikfnr nzO;eku gS&

(B) 149.949 g

(C) 125.47 g

(D) 564.48 g




CHEMISTRY

34.

100 mL of H2SO4 solution having molarity 1 M and density 1.5 g/mL is mixed with 400 mL of water. Calculate final molarity of H 2SO4 solution, if final density is 1.25 g/mL:

eksyjrk

1 M rFkk

?kuRo

1.5 g/mL j[kus

35.

100 mL H2SO4 foy;u

(B) 0.145 M

dks

400 mL ty

Y

ds lkFk fefJr R T fd;k tkrk gSA H2SO4 foy;u dh vfUre eksyjrk ifjdfyr dhft,] ;fn vfUre ?kuRo 1.25 g/mL gS& S I (A) 4.4 M

okys

(C) 0.52 M

(D) 0.227 M

Which species are oxidized and reduced in the reaction ? FeC2O4 + KMnO4

 

3+

Fe + CO2 + Mn

2+

(A) Oxidised : Fe , C ;Reduced : Mn

(B) Oxidised : Fe ; Reduced : Mn

(C) Reduced : Fe , Mn ; Oxidised : C

(D) Reduced : C ; Oxidised : Mn , Fe

vfHkfØ;k es dkSulh Lih'kht vkWDlhÑr rFkk vipf;r gksrh gS\ FeC2O4 + KMnO4 (A)

 

vkWDlhÑr: Fe , C ; vipf;r

(C) vipf;r : Fe , Mn ; 36.

3+

Fe + CO2 + Mn

2+

: Mn

vkWDlhÑr : C

(B) vkWDlhÑr : Fe ; (D)

vipf;r : Mn

vipf;r : C ; vkWDlhÑr : Mn , Fe

Which of the following is incorrectly matched ? Hybridisation 3

Geometry

Orbitals used

Trigonal bipyramidal

s + p x + py + pz + dz2

Pentagonal bipyramidal

s + p x + py + pz + dx 2 -y 2 + d z2 +d xy

3 2

Capped octahedral

s + p x + py + pz + dx 2 -y 2 + d z 2

3

Tetrahedral

s + p x + p y + pz

(A) sp d 3

(B) sp d (C) sp d (D) sp

3




M E H C

CHEMISTRY

fuEu es ls dkSulk xyr lqe fyr s gS\ 3

f=kdks.kh; f}fijkfeMh;

s + p x + py + pz + dz2

iapdks.kh; f}fijkfeMh;

s + p x + py + pz + dx 2 -y 2 + d z2 +d xy

3 2

Vksihuqek v"VQydh;

s + p x + py + pz + dx 2 -y 2 + d z 2

3

prq"Qydh;

s + p x + p y + pz

(A) sp d 3

(B) sp d

3

(C) sp d (D) sp

37.

Species having maximum ‘Cl-O bond order is :

vf/kdre ‘Cl-O ca/k Øe j[kus okyh Lih'kht gS& –

(A) ClO 3 38.

–

–

(B) ClO4

(C) ClO

–

(D) ClO 2

Choose the correct code of characteristics for the given order of hybrid orbitals of same atom,

sp < sp2 < sp3 (i) Electronegativity (iii) Size (A) (i), (iii) and (iv)

(ii) Bond angle between same hybrid orbitals (iv) Energy level (B) (iii), (iv)

(C) (ii) and (iv)

(D) (i), (ii), (iii) & (iv)

leku ijek.kq ds ladj d{kdks ds fn, x;s Øe ds fy, vfHky{k.kks ds lgh dwV dk p;u dhft,& sp < sp2 < sp3

q (i) oS| r_.krk

(ii) leku

ladj d{kdks ds e/; ca/k dks.k (iv) ÅtkZ Lrj

(iii) vkdkj (A) (i), (iii) o (iv)

(B) (iii), (iv)

(C) (ii) o (iv)

(D) (i), (ii), (iii) o (iv)




Y R T S I M E H C

CHEMISTRY

39.

–

–

+

–

+

molecular orbital (MO)

Which of the following statement is correct for resulting molecular orbital ?

Y R T S I M E H C

(A) It is a gerade MO. (B) It has 2 nodal plane. (C) It is an antibonding MO and less stable than pure atomic orbitals. (D) It is a bonding MO. –

+

–

+

vkf.od d{kd (MO)

–

fuEu es a ls dkSulk dFku ifj.kkeh vkf.od d{kd ds fy, lgh gS \ (A) ;g

,d ftjs M+

(B) ;g 2 uksMy

40.

(gerade) vkf.od

d{kd gSA

ry j[krk gSA

(C) ;g

,d iz frcfU/kr vkf.od d{kd gS rFkk ;g 'kq ) ijek.oh; d{kdks ls de LFkk;h gksrk gSA

(D) ;g

,d cfU/kr vkf.od d{kd gSA

The product obtained from the following reaction is ( fuEu +

Cl

Anhy. AlCl3

utZy AlCl3

vfHkfØ;k ls iz kIr mRikn gS&)

+

(A)

(B)

(C)

(D) All of the above mijks ( Dr

lHkh)




CHEMISTRY

41.

How many monobrominated structural isomeric reaction?

products will be obtained in the following

fuEu vfHkfØ;k es fdrus eksu cz ks kseksuhÑr la jpukRed leko;oh mRikn izkIr gksxsa\ Br 2/hv

Products mRikn

(A) 6 42.

(B) 4

(C) 5

(D) 3

The product obtained during hydrogenation of following is shown as

fuEu ds gkbMªkstuhdj.k ds nkSjku cuus okyk mRikn gS% Me C=C H

CC–CC

(A)

H

H

H H

H

H C=C

H C=C

Me

Me C=C H

H H

H C=C Me

C=C

C=C H



C=C H

(C)

H H

HH

H

Me C=C

C=C

C=C

HH

BaSO4

H

C=C

C=C

Me (B)

H H H

H

H2-Pd

C=C

H Me

Me

H

(D) CH3 –(CH2)8 –CH3 Space for Rough Work



Y R T S I M E H C

CHEMISTRY

43.

Choose the correct option for the following molecule :

fuEu v.kq ds fy, lgh fodYi dk p;u dhft,A H

Cl C=C= C=C H

(A)

Cl

vleryh;

(A) Non-planar 44.

(B)

µ  0

(C) (A) o (B) nksuksa

(D)

µ  0

(B)

µ  0

(C) both (A) and (B)

(D)

µ  0

Alkene which give same product on reaction with hydroboration and oxymercurationdemercuration is

fuEu es ls dkSulh ,Ydhu gkbMª kscksjhdj.k o vkDlheD;wZjhdj.k&foeD;wZjhdj.k ds lkFk vfHkfØ;k ij leku mRikn nsrh gSA (A) CH3 – CH = CH2

(B) CH3 – C = CH 2

(C)

(D)

CH3 Cl2 / h

45.

    

P

P may be P gksxk% CH3

CH3

CH3

CH2 –Cl

Cl (A)

(B)

Cl

(C)

(D) Cl




Y R T S I M E H C

CHEMISTRY

SECTION – II (

– II)

Y R This section contains 15 questions. The answer to each question is a single digit integer, ranging from T S gSaA çR;sd ç'u dk mÙkj 0 ls 9 rd ¼nks uksa 'kkfey½ ds chp dk ,dy I 0 to 9 (both inclusive). ( bl [k.M esa 15 M E gSA) H C – 3– 2– + Integer value correct Type (

46.

xNO3 + yAs2S3 + zH2O   - - AsO4 + - - NO + - - -SO4 + - - -H .

In the redox reaction, What is the value of

)

x

?

z

3– 2– +  - - AsO4 + - - NO + - - -SO4 + - - -H . fuEu jsMkWDl vfHkfØ;k es] xNO –3 + yAs2S3 + zH2O  x z

47.

dk eku D;k gS\

2. 0 g of polybasic organic acid (Molecular wt. = 600) required 100 mL of a

M NaOH solution for 6

complete neutralisation . Find the basicity of acid. 2. 0 g cgq{kkjh;

dkcZ fud vEy ( v.kqHkkj = 600) dks iw.kZ mnklhuhdj.k ds fy, 100 mL,

M NaOH foy;u dh 6

vko';drk gksrh gSA vEy dh {kkjdrk Kkr dhft,A 48.

4.3 g of saturated hydrocarbon (C nH2n + 2) is burnt completely and all CO 2 were absorbed in aqueous NaOH solution. A 300 ml, 2M NaOH solution was required for complete absorption of CO2. Hence, find the value of n. 4.3 g ,d larÌr gkbMªksdkcZu (CnH2n + 2) dk iw.kZ ngu djk;k tkrk gS rFkk lEiw.kZ CO2, tyh; NaOH foy;u

es vo'kksf"kr djk;h tkrh gSA ;fn CO2 ds iw.kZ vo'kks"k.k ds fy,] vko';drk gksrh gS] rks n dk eku Kkr dhft,A

2 M NaOH foy;u

ds

300 ml

dh




CHEMISTRY

49.

200mL of 1 M HCl is mixed with 300 mL of 6M HCl and the final solution is diluted to 1000 mL. + Calculate molar concentration of [H ] ion. 200mL, 1 M HCl dks 300 mL, 6M HCl ds

lkFk fefJr fd;k tkrk gS rFkk vfUre foy;u dks ruq fd;k tkrk gSA [H+] vk;u dh eksyj lkUnzrk ifjdfyr dhft,A 50.

16 g of SO x gas occupies 5.6 L at 1 atm and 273 K. What will be the value of x ?

l 1 atm rFkk 273 K ij 5.6 L vk;ru 16 g SOx xS 51.

1000 mL rd Y

?ksjrh gSA x dk eku D;k gksxk\

Total number of molecules which cannot form H-bond amongst themselves.

bues ls] v.kqvks dh dqy la[;k crkb,] tks H- ca/k ugha cuk ldrs gSa& SiH3OH, HCN, B(OMe) 3, NHMe2, CH3CONH2, HCHO, HCOOH, NH 2OH, H4SiO4 52.

If Hund’s rule is violated, then find the total number of species among following which will become diamagnetic from paramagnetic :

;fn gq.M fu;e dk mYya?ku gksrk gS] rks fuEu es ls ,slh Lih'kht dh dqy la[;k crkb,] tks vuqpqEcdh; ls izfrpqEcdh; gks tk;sxh\ B2 ,O2 ,N –2 ,C2 ,NO,OF,N22– ,BN 53.

–

2–

–

3–

2–

2–

Consider the following species NO3 ,SO4 ,ClO3 ,SO3 ,PO4 , XeO3 ,CO3 ,SO3 Then calculate value of X, where X:Total number of species which have bond order 1.5 or greater than 1.5

– 3– 2– 2– fuEu Lih'kht dk voyksdu dhft, NO –3 ,SO2– 4 ,ClO3 ,SO3 ,PO4 , XeO3 ,CO3 ,SO3

rc X dk eku ifjdfyr dhft,] tgk¡ X: Lih'kht dh dqy la[;k tks 1.5 ;k 1.5 ls vf/kd ca/k Øe j[krh gSaA




R T S I M E H C

CHEMISTRY

54.

Amongst the following, total no. of planar species is :

fuEu es ls leryh; Lih'kht dh dqy la[;k gSa& (i) SF4

(ii) BrF3 –

(v) SbF4 55.

(iii) XeF2

–

+

(vi) SF5

(vii) CH3

(iv) IF5 +

(viii) PCl4

Total number of species which used all three p-orbitals in hybridization of central atoms and should be non-planar also.

,slh Lih'kht dh dqy la[;k crkb,] tks dsfUnz; ijek.kqvks ds ladj.k es leLr rhuks p- d{kdks dk iz;ksx djrh gS rFkk vleryh; Hkh gksuh pkfg,A +

+

XeO2F2, SnCl2, IF5, l3 , XeO4 ,SO2 , XeF7 ,SeF4 Br

56.

Br

  H – C – C – H  Br

Br

NaNH 2 Excess

How many halogen will the product have? Br

Br

H–C–C–H Br

Br

  NaNH 2

vkf/kD;

cuus okys mRikn es fdrus gsy tu ks gksxs\




Y R T S I M E H C

CHEMISTRY

57.

Total number of possible structurally isomeric alkenes in reaction (i),(ii) and (iii) is X,Y and Z respectively , then value of (X+Y+Z) is: Cl Cl (i)

Alc. KOH

Alc. KOH

(ii)

(iii)

Cl

vfHkfØ;k (i), (ii) o (iii) esa lEHko la jukRed leko;oh ,Ydhuksa dh dqy l[a;k Øe'k% dk eku gksxk&

Y R T Alc. KOH S I M X, Y o Z gS rc (X+Y+Z) E H C

Cl Cl (i)

{kkjh; KOH

{kkjh; KOH

(ii)

(iii)

{kkjh; KOH

Cl CH2 –CH3

58.

X HCCH + Y CH3 –CH2 –CCH  Write the sum of value of (x + y) ? (x + y) dk eku fyf[k, \

59.

How many alkenes, alkynes and alkadienes can be hydrogenated to form isopentane (Consider only structural isomers)

fdruh ,Ydhu] ,YdkbZu o ,YdsMkbZu gkbMª ksthuÑr gksdj vkblksisUVs u ¼dsoy lajpukRed leko;oh ekusa½ cukrs gSaA 60.

2 CHCl3 + 6Ag

 

6AgCl + X

How many  bonds are there in the product (X)? mRikn (X) es fdrus  cU/k gSa \




MATHEMATICS

MATHEMATICS

S C I T Straight Objective Type ( ) A M This section contains 15 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for E H its answer, out of which ONLY ONE is correct. T A bl [k.M es a 15 iz 'u gS aA iz R;s d iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gS a] ftues a ls gSA M SECTION – I (

61.

- I)

Sum of all the terms of two AP’s having 2n + 1 and 2m + 1 terms is equal to 4n + 2 and 6m + 3 th

th

respectively. Then sum of their (n + 1) and (m + 1) term respectively is 2n + 1

muds

62.

vkSj 2m + 1 inks okyh nks lekUrj Js.kh;ksa ds lHkh inksa dk ;ksxQy Øe'k%

(n + 1) os in

rFkk

(m + 1) os a

in dk ;ksxQy gS

4n + 2

vkSj

6m + 3 gS

-

(A) 5

(B) 4

(C) 3

(D) 6 2

If two of the roots of equation |x| – 10|x| + 16 = 0 are first and third term of a Increasing G.P. then their second term could be

;fn lehdj.k

2

|x| – |x| + 16 = 0 ds

in gks ldrk gS

nks ewy] o/kZeku xq.kksÙkj Js.kh ds izFke vkSj rhljs in gS rc mudk nqljk

-

(A) 4

(B) –2

(C) 32

(D) – Space for Rough Work (

®

1 2 )



MATHEMATICS

63.

x

y

y

z

2y

x z

If (p – q ) = (q – r ), q = p r and pqr = 1, where p, q, r  (0,  ) – {1} then value of

1 1 1 + + x y z

will be

;fn

x

y

y

z

2y

(A) p + q + r

64.

3

3

vkSj pqr = 1 tgk¡

(B) pqr

p, q, r  (0,  ) – {1} rc

(C)

1 1 1 + + p q r

1 1 1 + + dk x y z

eku gksxk

(D) 0

3

If x + y + z – 3xyz = 0, where x, y, z > 0 and x + y + z = 3 then value of log 3xyz is equal to

;fn

3

3

3

x + y + z – 3xyz = 0 tgk¡ x, y, z > 0

(A) 1

The value of

  (r

2

n

2

log3xyz dk

(C) 0

eku cjkcj gS (D)

-

1 3



 r – 1) r –1 is

r 1

  (r

vkSj x + y + z = 3 rc

(B) 3 n

65.

x z

(p – q ) = (q – r ), q = p r



 r –1) r –1 dk eku cjkcj gS -

r 1

(A) n n + 2 n – 2

(B) (n + 2) n – n

(C) n n + 2 n

(D) (n + 2) n + 2 Space for Rough Work (

®

)



S C I T A M E - H T A M

MATHEMATICS

66.

n

If sec x + tan x = p then value of sec x + tan x is

;fn

sec x + tan x = p gks n

n

n

(A) [ C0p + C2p (C) 2

n–1 n

n–4

n

n

 r 0

(1  x)r x

r 0

(A)

eku gS

n–8

n

Coefficient of x in



n

sec x + tan x dk + ..........]

2

n r

.x2n

(1  x)r x

nr

n

n

esa xk dk xq.kkad gS

2

n

4

2

n

4

-

n+1

(B) Ck+1

n

n

(D)

Ck+1

rks

Ck

(1  2x  3 x 2  4 x 3  ........ )

(1  x  x 2  x 3  ........) (1  2x  3 x  4 x  ........ ) 2

n+1

(1  x  x 2  x 3  ........ )

3

|x| < 1 gks

n

( C0 + C2p + C4p +..........)

.x2n is

If |x| < 1 then coefficient of x in

;fn

n–1 n

(D) C0 + C2p + C4p + ............

(C) Ck

68.

-

(B) 2

( C0 + C1p + C2p + ......)

k

n

n

rks

+ C4p

n

67.

n

3

is

esa x3 dk xq .kkad gS

-

3

(A) 0

(B) C1

3

(D) C1

4

(C) C3


®

)



S C I T A M E H T A M

MATHEMATICS

69.

2

2

2 40

(1  x 2 ) 40 x

(B) 42

n+2

n+3

n+k

n

n+1

n+2

n+3

n+k

(A)

C1 + C1 +

C2 + C2 +

C3 + ......... + C3 + ......... +

n+k+1

Cn+1

(B)

n+k–1

is equal to

-

(C) 43

n+1

C0 +

(D) 44

Ck is equal to Ck dk

eku cjkcj gS

Ck+1

(C)

n–k+1

-

Cn

(D)

n+k+1

Ck+1

2

Number of solution of equation cos4x = cos 3x in x  (0, 10] is

lehdj.k

2

cos4x = cos 3x ds

(A) 10

72.

ds foLrkj esa inks a dh la[;k cjkcj gS

n

C0 +

x

2 40

+ 2(1 + x )

2

2 40

+ 2(1 + x )

2

(A) 40

71.

2 40

Number of terms in the expansion of x (1 + x ) +

x (1 + x ) +

70.

(1  x )

2 40

vUrjky x  (0, 10] esa gyksa dh la[;k gS

(B) 50

(C) 25

2

Number of solution of the equation sin x . cos x = x +

lehdj.k

2

sin x . cos x = x +

(A) 0

1 ds x

gyksa dh la[;k gS

(B) 1


®

(D) 5

1 is x

(C) 3

(D) 4

)




MATHEMATICS

73.

4

2

Number of triplet of (x, y, z) satisfying the equation sin x + cos y = 1 + sec z (A) infinite

lehdj.k (A)

74.

4

(B) 0 4

4

(C) 1 2

sin x + cos y = 1 + sec z dks

vuUr

larq"B djus okys

(B) 0


(D) 2

(x, y, z) ds

f=kiys Vksa dh la[;k gS

(C) 1

-

(D) 2

2

Range of a for which there exist solution of following equation sin x sin y = a and cos x cos y = a is a ds

ekuksa dk ifjlj gksxk ftlds fy, fuEu lehdj.kksa   

(A)  – ,

 1  5  1– 5  ,      2 2   

 1– 5 1  5  ,  2 2  



(B)  1,





(D)  – , 

(C)  

75.

2

sin x sin y = a ,cos x cos y = a dk



gy fo|eku gS

1 5   2  1– 5   2 

 1– 5  ,    2  

Number of solution of equation cot x – 2sin x + cosec x = 0 in x  (0,  ) are x  (0, ) es a

lehdj.k

cot x – 2sin x + cosec x = 0 ds

(A) 1

(B) 2


®

gyks a dh la[;k gS

-

(C) 3

(D) 0

)



-

MATHEMATICS

SECTION – II (

– II)

S C I Integer value correct Type ( ) T A This section contains 15 questions. The answer to each question is a single digit integer , ranging from M E 0 to 9 (both inclusive). H T A bl [k.M esa 15 gSaA çR;sd ç'u dk mÙkj 0 ls 9 rd ¼nks uksa 'kkfey½ ds chp dk ,dy gSA M 76.

2

2

2

If sum of 97 term of the series 1 .2 + 2 .3 + 3 .4 + .......... is 97 × 98 × 99 ×

a where a and b are b

co-prime then 25b – a is equal to

;fn Js.kh gS rc

77.

If p =

;fn

2

2

2

1 .2 + 2 .3 + 3 .4 + ......... . ds 97 inks

25b – a cjkcj

4a 2  4a  1 2a

gS

dk ;ksx

97 × 98 × 99 ×

a gS b

tgk¡

a vkS j b lg

vHkkT;

-

, where a > 0 then number of solution of min(p) = 4 sin  in   [0, 2] will be

4a 2  4a  1 tgk¡ a > 0 gks p= 2a

rks   [0, 2] esa lehdj.k


®

min(p) = 4 sin ds

gyksa dh la[;k gS 

)



MATHEMATICS

78.

2

3

4

5

6

If (1 – p)(1 + 3x + 9x + 27x + 81x + 243x )= 1 – p , p  1 then integral value of

;fn (1 – p)(1 + 3x + 9x2 + 27x3 + 81x4 + 243x5)= 1 – p6 , p  1 gks rks

79.

p dk x

p is x

iw.kk±d eku gS -

Three number, third of which is 4 form a decreasing G.P. If last term is replaced by 3 then three number form an A.P. then first number of G.P. is

rhu la[;k,¡] ftlesa rhljk in 4 gS ákleku xq.kksaÙkj Js .kh cukrh gSA ;fn vfUre in dks 3 ls gVk;k tkrk gS rc rhuks la[;k,¡ lekUrj Js.kh cukrh gS] rc xq.kksÙkj Js.kh dh izFke la[;k gS

80.



x

x If 1, log81(3 + 48) and log9  3 –



;fn

x

1, log81(3 + 48)



-

8   are in A.P. then value of x is 3  8  3 

vkSj log9  3 x –  lekUrj Js .kh esa gS rc x dk eku gS 


®

-

)




MATHEMATICS

81.

17

6

2

3 4

(1 – x) (1 + x + x + x ) esa x

1005

+ (1007)

dk xq.kkad S ds cjkcj gS rc

1005

+ (1007)

1007

|S| =

is equal to

1007

ds ;ksxQy es a vfUre vad cjkcj gS

10


-

2 6

Coefficient of x in (1 + x – 2x ) is equal to a4 then number of prime factors of |a 4| is equal to 2 6

(1 + x – 2x ) esa x

84.

17

Last digit in the sum of (1007) (1007)

83.

3 4

Coefficient of x in (1 – x) (1 + x + x + x ) is equal to S then |S| = 6

82.

2

1 15!

+

1 1!14!

+

10

dk xq.kkad a4 gS rc

1 2!13!

+ ........ +

|a4| ds vHkkT;

1 2a is equal to then (b – a) is equal to 7! 8! b!

1 1 1 1 + + + ........ + dk 15! 1!14! 2!13! 7! 8!

eku

2a ds b!


®

xq .ku[k.Mksa dh la[;k gS&

cjkcj gS rc

(b – a) dk

eku gS

-

)



MATHEMATICS

85.

The remainder when 1! + 2! + 3! + .......... + p! is divided by 15. (p > 90)

'ks"kQy gksxk tc

86.

6

If tan

;fn

88.

6

fd;k tkrk gS

(p > 90)

2

Equation sin x + cos x = a has real solution if a  [p, q]  [r, s] then p + q + r + s is

lehdj.k

87.

1! + 2! + 3! + .......... + p! dks 15 ls foHkkftr


5

tan

If sin

;fn



6

6

2

sin x + cos x = a dk

+ 2tan

 5

okLrfod gy gksxk ;fn a  [p, q]  [r, s] rc

p + q + r + s gS -

2 4 a + 4cot = cot (a,b  N) then least value of a + b is equal to 5 5 b

+ 2tan

2 4 a + 4cot = cot (a,b  N) gks 5 5 b

rks a + b dk U;wure eku cjkcj gS

-

5x 10 y + cos = 2 then the value of [x + y] is where [] is G.I.F, where x, y  [0, ) 2 9

sin

5x 10y + cos = 2 gks 2 9

rks

[x + y] dk

eku gS tgk¡ [] egÙke iw .kkZ ad Qyu gS tgk¡


®

x, y  [0, )

)



MATHEMATICS

89.

2

[0, 2] esa

90.

7

2

Number of solution of the equation (sin  + tan ) +

lehdj.k

2

2

(sin  + tan ) +

7 12

=

2 3

12

=

tan – sin ds

2 3

tan – sin in [0, 2] is equal to

gyks dh la[;k cjkcj gS

2

-

2

If in  ABC, tan A and tan B are the roots of equation pq(x + 1) = r x. Then  C is equal to

a , b

then a + b is equal to (where a and b are coprime natrual number)

;fn  ABC esa a + b dk

2

2

tan A vkSj tan B lehdj.k pq(x + 1) = r x ds

eku gS tgk¡

a vkS j b lgvHkkT;

izkd`r la[;k gS

ewy gSA rc C dk eku

a b

ds cjkcj gS rc

-

(


®

)




Code

MAIN PATTERN ONLINE TEST-2

P25-16

0

(MONT-2) XI TARGET : JEE (MAIN+ADVANCED) 2017

DATE : 05-01-2017 | COURSE : VIJETA (JP), VISHWAAS (JF), ANOOP (EP), AKHIL (EF) 11.

Since it is not possible to erase and correct pen filled 11. pawfd bubble, you are advised to be extremely careful while

isu ls Hkjs x, xksys feVkuk vkSj lq/kkjuk laHko ugha gS

blfy, vki lko/kkuh iw oZ d vius mÙkj ds xksyksa dks HkjsaA

darken the bubble corresponding to your answer. 12.

Neither try to erase / rub / scratch the option nor 12. fodYi make the Cross (X) mark on the option once filled.

dks u feVk,a@u LØs p djsa vkSj u gh xyr

dks HkjsaA

Do not scribble, smudge, cut, tear, or wrinkle the

ORS dks

dkVs u gh QkMs u gh xUnk ugha djsa rFkk

dks bZ Hkh fu'kku ;k lQsnh

ORS. Do not put any stray marks or whitener

(X) fpUg

ORS ij

ugha yxk;sA

anywhere on the ORS. 13.

If there is any discrepancy between the written data

13.

and the bubbled data in your ORS, the bubbled data

ORS esa

fdlh izdkj dh fy[ks x, vka dMksa rFkk xksys

fd, vka dMksa esa fojks/kkHkkl gS] rks xksys fd, vkadMksa dks gh

will be taken as final. C.

;fn

lgh ekuk tkosxkA

Question Paper Format

C.

This question paper consists of three part. Each part

bl iz'u&i=k esa rhu Hkkx

consists are two section.

14. Section 1 contains 15 multiple choice questions.

14.

Each question has Four choices (A), (B), (C) and (D)

1 esa 15

gaSA gj iz'u esa pkj fodYi (A),

(B), (C) vkS j (D) gSa ftuesa ls dsoy

gSA

out of which only ONE is correct. 15. Section 2 contains 15 questions. The answer to

15.

each question is a single-digit integer , ranging from

2 esa 15

gSaA izR;sd iz'u dk mÙkj 0

'kkfey½ ds chp ,d

9 rd ¼nksuksa

gSA

0 to 9 (both inclusive). D.

Marking Scheme

D.

16. For each question in Section 1 , you will be awarded

16.

4 marks if you darken the bubble corresponding to

dks dkyk djus ij 4 vkS j dks bZ Hkh cq ycw yk dkyk ugha a djus ij a xsA vU; lHkh fLFkfr;ks a es a (0) va d iznku fd, tk;s

only the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark

(–1) va d nku iz fd;k tk;s xkA

will be awarded. 17. For each question in Section 2 , you will be awarded

1 es a gj iz 'u es a ds oy lgh mÙkj okys cq ycqy(s BUBBLES)

17.

2 esa gj iz'u esa dsoy lgh mÙkj okys cqycqys (BUBBLES)

4 marks if you darken the bubble corresponding to

dks dkyk djus ij 4

only the correct answer and zero mark if no bubbles

ij (0) vad iznku fd, tk;saxsA vU; lHkh fLFkfr;ks a es a (0) vad

are darkened. In all other cases, zero will be awarded.

Name of the Candidate (

):

I have read all the instructions and shall abide by them

eSaus lHkh funs Z'kksa dks i<+ fy;k gS vkS j eSa mudk vo'; ikyu d:¡xk@d:¡xhA ...................................... Signature of the Candidate

ijh{kkFkhZ ds gLrk{kj

vkS j dksbZ Hkh cqycwyk dkyk ugha djus

iz nku fd;k tk;sxkA Roll Number (

):

I have verified all the information filled by the candidate.

ijh{kkFkhZ }kjk Hkjh xbZ lkjh tkudkjh dks eSusa tk¡p fy;k gSA ...................................... Signature of the Invigilator

ijh{kd ds gLrk{kj

Resonance Eduventures Ltd.

P25-16

CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Ph.No. : +91-744-3012222, 6635555 | Toll Free : 1800 258 5555 Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222 Website : www.resonance.ac.in | E-mail : [email protected] | CIN: U80302RJ2007PLC024029

MAIN PATTERN ONLINE TEST-2 (MONT-2) XI TARGET : JEE (MAIN+ADVNACED) 2017 DATE : 05-01-2017

COURSE : VIJETA (JP), VISHWAAS (JF), ANOOP (EP), AKHIL (EF)

HINTS & SOLUTIONS T

PART-I PHYSICS 1.

a  32

 42  5 m/s

fricition force ?k"kZ.k cy f = m a = 2×5 = 10 N

w 2.

mg sin 30º = ma cos 30º  a = g tan 30º

a=

T–W =

g



3 ma

 7.

9. 3.

02 = V2 – 2gs

 4.

V2 s= . 2g

(A).

Body will be in equilibrium at both x 1 & x2 as at these points force will be zero. At x2 on increasing x force becomes -ve & on decreasing x force becomes + ve. so force & displacement have opposite signs. So it a pt. of stable eq. oLrq x1 rFkk x 2 ij lkE;koLFkk esa gksxhA D;ksafd bl txg cy 'kw U;

gksxkA x ij x c<+us ij cy _.kkRed gksxk] vksj x de gksu s ij cy /kukRed gksxkA ;kfu cy vks j foLFkkiu dh laKk ¼ signs½ foifjr gksxh] rks ;gk¡ ij LFkk;h foLFkkiu gksxkA P0 = Fv



2v

v

x

x

0

As tangential acceleration Li'kZ



T = 1.019 W

g R

At the highest point and lowest point the component of normal reaction between M and m in the horizontal is zero. But in between these points, normal reaction in the horizontal direction was non-zero. That means normal reaction in horizontal direction first increases then decreases and so the acceleration of mass ‘M. mPpre~ fcUnq rFkk fuEure~ fcUnq ij M rFkk m ds e/; vjksfir

vfHkyEc cy dk {kSfrt ?kVd 'kwU; gSA ijUrq bu nksu aks fcUnqvksa ds e/; vfHkyEc cy dk {kSfrt ?kVd v'kwU; gSA vFkkZr~ {kSfrt fn'kk esa vfHkyEc cy igys ?kVrk gS rRi'pkr~ ?kVrk gS] vksj blfy, ‘M dk Roj.k Hkh blh izdkj ifjofrZr gksrk gSA 10. For anti-clockwise motion, speed at the highest point should be

gR . Conserving energy at (1) & (2) :

okekorZ fn'kk es a xfr ds fy,] mPpre fcUnq ij pky gR gksuh pkfg,A (1) rFkk (2) ds e/; ÅtkZ laj{k.k ls:

7Mv3 3P0

js[kh; Roj.k

a = dV/dt = dr/dt

but  = 4 and dr/dt = 1.5 (reel is turned uniformly at the rate of 2 r.p.s.) ijUrq  = 4 rFkk dr/dt = 1.5 jhy dks 2 pDdj çfr lSd.M dh (

,d leku nj ls ?kqek;k tkrk gSA) 

ij

 dv   Mv dx  v  P0  

 M  v 2 dv  P0  dx  6.

T = W (1 + a/g) put a = 6  T = W (1 + a/g) a = 6  j[kus

mg = m2 R ,  =

2

5.

W a g

a = 6 ,

Now by the F.B.D. of the mass. nzO;eku dk F.B.D.

1 R 1 mv a2 = mg  m(gR) 2 2 2  v = gR + gR = 2gR  v = 2

a

a

2gR

For clock-wise motion, the bob must have atleast that much speed initially, so that the string must not become loose any where until it reaches the peg B.

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SOLJP/JFMONT2050117-1

nf{k.kkorZ xfr ds fy, ckWc ds ikl çkjEHk es a de ls de bruk osx gksuk pkfg, ftlls oks
mv c2

13. By F = 

. Graph A is correct.

dr du }kjkA vkjs[k A lgh gSA F=  dr

;

R

du

14. F = –

dU dx

U(x) = kx2



VC being the initial speed in clockwise direction. VC dh çkjfEHkd pky nf{k.korZ For VC min : Put T = 0 ; VC min : ds fy, T = 0 j[kh

 

VC =

gR 2

(Easy) Max.

f max = N

 V /V = C

frictional force

vf/kdre ?k"kZ.k cy

gSA

a

VC : Va = 1 : 2

= (mg + F sin53°)

gR 1 2 = 2 2gR

= 0.2 (20 × 10 + 30 ×

N

T = 2

Lcos  g

mg

L1 cos1 = L2 cos2

L1 L2 L1 L2

 

cos 2

=

cos 1

cos45 cos30

Fcos53°

53°

N

T1 = T2



4 ) 5

Ans.

11. since pwafd

 

15.

fn'kk esa gSA

F

Fsin53° = 44.8 N As applied horizontal force is Fcos53° = 18N < f max, friction force will also be 18 N.

2

pw af d yxk;k x;k {kSfrt cy Fcos53° gS = 18N < f , ?k"kZ.k cy Hkh 18 N gksxkA

3

max

12. Let h be the height of water surface, finally ekuk vUr esa ikuh ds lrg dh Åpk¡ bZ h gSA

16. From graph ; xzkQ ls ; when tc

a a a a2h = a . . ; h= 2 2 4

a a   4 8

C.M. gets lowered by a – 

=a–

3a 5a = 8 8

5a 8

= mg

v)Z nh?kZ v{k b = semi minor axis v)Z y?kq v{k Let ekuk x = a cost V = velocity osx, A = acceleration Roj.k y = b sint at t = 0 ij

A =

a a 4  8  

 vr% xq :Ro }kjk fd;k x;k dk;Z

= a –

3a 8

V2 R1

v = b

A = a2

,

v{k

a2 =

(b)2 , R1

R1 =

b2 a

a2 at minor axis y?kq v{k ij R2 = b F=

5a

F=6

a = semi major axis

major axis nh?kZ

5a  Work done by gravity = mg 8

=

F = 4 = fmax

6 – 4 = m(A) m = 2kg

17. Let ekuk

 nzO;eku dsUnz uhps vk;sxk a –

a = 0,

when tc a = 1,  F – f max = ma

 



Let m be mass of block

ekuk m CykW d dk nzO;eku gSA

mv 2 R

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If ;fn v = 5m/sec then gks

1

rks F 

ROC

22. FBD of B

(aB)max =

F1R1 = F2R2

f max mB

= µSg = 2.5 m/s2

a =2 b 18.

FBD of combined system FBD f k = 0.15 (2 + 10) g = 18 N

la;qDr fudk; dk 60° 30°

Fmax – f k = (m A + mB) (aB)max



Fmax = f k + 12 ×

2.5 = 48 N. Ans.8 23. Acceleration Roj.k = 0 Here ;gk¡

30°

mg sin 30°

2mg sin37° < mgsin53° + mgcos57°

mg cos 30° 24.

mg at =

dv dt

 gsin30 16 – 10 = 10a a = 0.6 T – 5 = 5a T – 5 = 5 × 0.6 = 3 T = 8.

2

= 5 m/s

19. In half cycle the direction of final velocity reverses i.e.

vf

  vi

vk/ks pØ esa vfUre osx dh fn'kk foijhr gks xh vFkkZr~ v f   v i

 a  =

V t

a

2Vi

=

t

=

=

Vf

 Vi t

 V    V  i

=

i

25.

(F = minimum force U;wure

F = mg F=

mg



t

2u (assuming ekfu, v = u) t

Also using distance = speed x time mi;ksx ls nwjh = pky x le;

R = u t t = 

a

=

R u 2u R u

2 u2 2 = = a  R 

20. Since angular velocities of the particles are different, after some time, two particles may move parallel. In such case 



 PB

is maximum. 



d.k lekUrj py jgs gks xs a A rc P A  PB vf/kdre gksxkA 



 PB

= (2 × 2 + 1 × 3) kg m/s = 7 kg m/s max

1 1 2 2 (g sin  + µ g cos ) t a = (g sin  – µ g cos ) t d 2 2 td sin   µcos   = = 3 =  ta sin   µcos 

21. S =



2 = 3

Ans.

K 1 K  mV02 = 2a 2 a mV02 2

pqafd d.kksa ds dks.kh; osx fHkUu gSa] vr% dqN le; i'pkr nksuksa

P A

5N

26. Ui + ki = Uf + kf

Hence vr% n = 2

P A

0.2kg  10m / s2 = = 0.4

(  = 60º & µ =

3 ) 2

=

K  V = 2a 0

K am

27. F cos  = N N =mg – F sin

 mg cos   sin  mgvcos  P = F v cos = cos    sin  mgv = = 25 1   tan  F=

W

Ans.5



cy)

28. Values of mg for A and B are 2N & 2N A o B ds fy, mg dk eku 2N o 2N gSA Work energy theorm on B gives B ij dk;Z ÅtkZ izes; ls Fx – 2x – (K/2) x2 = 0 F = 2 + kx/2 For equilibrium of A, kx = mg = 2 A dh lkE;koLFkk ds fy,, kx = mg = 2 so vr%

71

.......(i)

0.85 ×

3.335

 W = 5.38g

44 62 = mol P4O10 124

Now mole of P4O10 × 0.9 =

mol of H3PO4 4

 Mass of H3PO4 = 149.949g

dy 2x 2 4 = = = 1 dx 8 8



33. 0.85 × mol of P4 = mol of P4O10

F = 2 + 1 = 3N

29. Slope of line BC js ( [kk BC dk =

W



0.85 × P4 ds eks y = P4O10 ds eksy

 = 45º

0.85 ×

vc

62 = mol P4O10 124

P4O10 ds

eksy × 0.9 =

H3PO4 ds eksy 4

 H3PO4 ds eksy = 149.949g 34. Mol of H2SO4 = 1×0.1 = 0.1 mol using mass conservation 100 × 1.5 + 400× 1 = V × 1.25  V = 440 mL

vc

Now, by

Now conc. of H2SO4 =

Wnet = KE = 0

H2SO4 ds

eksy = 1×0.1 = 0.1 m ol nzO;eku laj{k.k dk iz ;ks x djus ij

W F – mg(10 + 2) – mg(10 + 4) = 0



W F = 380 = 76 × 5

30. By Wnet = K.E = 0

 

x = 5.

2

100 × 1.5 + 400× 1 = V × 1.25  V = 440 mL

1 k(x – x ) = mgx 2 2

2

vc H2SO4 dh lkUnzrk

0

1 1 × 200(2 – x ) = × 60 × 10x 2 2



x = 1m Also at this moment bl {k.k ij f max > kx So, block will not move so total distance travelled = 2 + 1 = 3m. vr% CykWd xfr ugha djsxk blfy, dqy r; nwjh = 2 + 1 = 3m

36. In sp3d2 Sol. sp3d2 es 37.

PART-II CHEMISTRY 31. Let volume of air = V  Vol. of O2 = 0.21 V



1 160 0.21V   32 22.4

1 160 32





ClO –3

B.O= 5/3

ClO –4

B.O= 7/4

ClO –

B.O= 1

ClO –2

B.O=3/2 nodal plane

V = 533.33L +

+

+

+

39. uksMy

22.4

ry

0.21V  V = 533.33L 22.4

3.335 44

geometry is octahedral

T;kfefr v"VQydh; gSA

22.4

+

32. Mass of CO2 consumed in 5 min = 0.667 × 5 = 3.335g mol of KO2 = mol of CO2

W 71

0.1 = 0.227 M 0.44

Vol. of O2

ekuk ok;q dk vk;ru = V  O2 dk vk;ru = 0.21 V S8 ds eksy  8 O2 dk vk;ru



=

2



molof S8  8

0.1 = 0.227 M 0.44

 W = 5.38g

 46. [3e – + 4H+ + NO3 [20H2O + As2S3

5 feuV es [kpZ gqbZ CO2 dk nzO;eku = 0.667 × 5 = 3.335g KO2 ds eksy = CO2 ds eksy

+

+

+

 NO + 2H2O] × 28

 2AsO34  3SO24 +40 H+ + 28e –]

×3 x = 28

and rFkk z = 4



2

47.

600

 x =

48. C H n

2n + 2

1 6

 0.1

PART-III MATHEMATICS

 x = 5

st

2n  1 (a + b) = 4n + 2 2 ab = 2 T’n 1 = 2

+ O2   nCO2 + (n + 1)H2O

CO2 + 2NaOH

S’2n 1 =

 Na CO + H O 2

3

2

moles of NaOH (NaOH ds eksy) = 0.3 × 2 = 0.6

1 × 0.6 = 0.3 2 1 moles of C H (C H ds eksy) = × 0.3 n 4.3 0.3 = So (vr%), n = 6 12n  (2n  2)  1 n

So (vr%), moles of CO2 (CO2 ds



n

2n + 2

n



5.6 22.4

st

eks y) =

2m  1 (c + d) = 6m + 3 2 c  d = 3 T"m1 = 2 T’n 1 + T"m 1 = 5

S" 2m1 =



x =2

ugha cuk ldrs gSaA

ekuk c izFke in gS rFkk d lekUrj Js<+h dk vfUre in gSA 2m  1 (c + d) = 6m + 3 2 c  d = 3 T"m1 = 2 T’n 1 + T"m 1 = 5

S" 2m1 =

53. Species

Bond order

Lih'kht

ca/k Øe

NO –3

1.33

SO2– 4

1.50

2

ClO –3

62. |x| – 10|x| + 16 = 0 roots : –2, – 8, 2, 8 st rd 1 and 3 term of an increasing G.P. could be 2, 8 or – 8, – 2 nd 2 term in respective cases is 4 or – 4 2 Hindi |x| – 10|x| + 16 = 0 ewy : –2, – 8, 2, 8

1.66

SO3

2.00

PO3– 4

1.25

XeO3

2.00

CO32–

1.33

SO2– 3

1.33

o/kZ eku xq.kksÙkj Js.kh ds izFke o rhljs in nksu aks inksa dh fLFkfr esa 4 ;k – 4 63.

 r z

px

2 y 2

y

x

z

 px, qy, r z in G.P. x

 p = q = r z = k  x = logpk , y = logqk, z = logr z 1 1

 54. BrF3, XeF2 CH3

55. XeO2F2 ,IF5, XeO4 , XeF7 , SeF4



56. Br–C C–Br 64.

y=

3

= xyz

x, y, z > 0 AM = GM  x = y = z and vkSj x + y + z = 3

,

,

 y3  z3

x3

y

1 + + = logk(pqr) = logk (1) = 0 x y z





2, 8 ;k – 8, – 2 gSA

 px, qy, r z in A.P.

= q

and (q ) = p r

x=4

57. x =

in gSA

2n  1 (a + b) = 4n + 2 2 ab = 2 T’n 1 = 2

B2 ,O2 ,N22–

52.

b vfUre

S’2n 1 =

51. B(OMe)2 and HCHO can not form H-bond. B(OMe)2 rFkk HCHO H- ca/k

ekuk lekUrj Js .kh dk a izFke in rFkk

Hindi

 M = 2 M

nd

Let c is 1 and d is last term of 2 A.P.

2n + 2

49. 200 × 1+ 300 × 6 = 1000 × M

16 50. 32  16x

st

61. Let a is 1 and b is last term of 1 A.P.

 x = y = z = 1 log3(1) = 0

z=

n

65.

  (r r 1

59.

CH3 CH3

C CH2

2

 r –1) r –1

Tr =

r  1 – r – 1

Sn =

n  1 + n – 1 – 0

=

n (n + 2) – 2 = n n + 2 n – 2

60. H–CC–H Corporate Office : 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 | CIN: U80302RJ2007PLC024029


1

66. sec x – tan x =

2

71. 2cos 2x – 1 =

p

2

1

n

p n

n

n

n

n

n–4

= 2( C0p + C2p

n

n

n

+ ..........)

n –1

n– 2

k

Coefficient of x in

2

n 1   n  1 x   x  – 1   x     1 x

x

– 1

  1  x n1  x  – 1   x     esa xk dk xq.kka d 1 x

n

n+1

n+1

Ck

68. coefficient of x in

(1  x  x 2

 x 3  ......... ) (1  2x  3 x 2  ......... ) (1 – X) –1

3

= coefficient of x in

(1 – X) – 2

3

= coefficient of x in (1 – X) = 0

=

(1  x  x 2

(1 – X)

2 40

69. (1 + x )

esa x3 dk xq.kkad

= (1 + x )

=

 1  x    x 

2

x  number of term = 43 70.

n

n+1

4

4

 a 

75. cot x + cosec x = 2sin x  cos x + 1 = 2sin 2 x  cos x = 2sin 2 x –1  cos x = – cos 2x  cos x + cos 2x = 0

3x x cos = 0 2 2 3x x  cos = 0 or cos = 0 2 2

 O

= 43

/3

2/3 O

2

(1  x 2 ) 42

inks dh la [;k

2

=0

1   2 x  2  2  x  

2 40

4

es a x3 dk xq.kka d

+1

= (1 – X)

4

73. sin x + cos y  2, 1 + sec z  2, sin x = 1, cos y = 1, 2 sec z = 1

 2cos

 x 3  ......... ) es a x3 dk xq .kkad 2 (1  2x  3 x  ......... ) – 2

1 1  2 or ;k x +  – 2 x x

 no solution dksbZ gy ugha

a  R

n+1

(1 – X) –1

2

72. –1 < sin x cos x < 1 and x +

 1– 5 1  5  ,   2  2   1– 5 1  5   1  5  , a     a   1,  2 2  2   

– 1

3

Hindi

gS

   

2

Coefficient of x in (1 + x) – x n+1 n+1 k (1 + x) – x esa x dk xq.kkad gS -



3 3 or cos2x = – 2 2 number of solution of equation in x  (0, 2] = 2 + 4 + 4 = 10 x  (0, 2] esa lehdj.k ds dqy gy = 2 + 4 + 4 = 10 total solution in (0, 10] = 50 (0, 10] esa dqy gy = 50

74. cos(x – y) = a + a 2 – 1  a + a  1 2 2 a + a + 1  0  a + a – 1  0

n

k

3 4

 cos 2x = 1 or cos2x =

n–8

+ C4p

k

+

 1   p –   p 

67. Coefficient of x in (x + x (1+x) + x (1 + x) +......+ (1 + x) ) n n –1 n–2 2 n k (x + x (1+x) + x (1 + x) +......+ (1 + x) ) esa x dk xq.kkad -

x

3

 cos 2x = 1 or cos22x =

 1  2 (sec x + tan x) =   p  p    n

2

4cos 2x – 2 = 1 + 4 cos (2x) – 3cos 2x 3 2 cos 2x = t  0 = 4t – 4t – 3t + 3 2  0 = (t – 1)(4t – 3)

1 2sec x = p + p 2 tan x = p –

1  cos 6 x

/2



2

76. Tr = r (r + 1) Tr = r(r + 1)(r + 2 – 2) Tr = r(r + 1)(r + 2) – 2r(r + 1) Sn =

1 2 n(n + 1)(n + 2)(n + 3) – n(n + 1)(n + 2) 4 3

n  3 2  –   3  4 2  S17 = 97 × 98 × 99 25 –  3  Sn = (n)(n + 1)(n + 2)

n+k

C0 + C1 + ........ + Ck n n+1 n+k = Cn + Cn + ......... + Cn n+1 n+1 n+k = Cn+1 + Cn + ........... + Cn n+2 n+2 n+k = Cn+1 + Cn + ........ + Cn n+k+1 n+k+1 = Cn+1 = Ck

a = 73 b=3 25b – a = 2 Corporate Office : 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 | CIN: U80302RJ2007PLC024029




1

77. p = 2a + 2 +

since pwfd 2a +

 4

2a

1 2a

214 2a so = 15 b!  a = 14, b = 15  b–a=1

 2

min(p) = 4 = 4 sin 

 sin  = 1   =

 2

85.

6

1 – (3 x ) 6 78. (1 – p) = 1 – p 1 – 3 x only dsoy

p = 3 x

;fn if p = 3x

6

2

2



4

r 2

=

es a gSA

and vkS j

ar = 4

r 2 4

and vkSj

3 2r – 1

a=

 sin2 2x =

 

8    3x –  are in G.P. xq .kksÙkj Js.kh esa gSA 3    x 8  x (3 + 48) = 9  3 –  3  

87. Let ekuk

x

2 4

2

81. (1 – x) (1 + x) (1 + x ) = (1 – x) (1 – x ) (1 +x ) = (1 – x) (1 4 4 – x )  Coefficient of x17 in (1– x4)4 + x2(1 – x4)4 – 2x(1 – x4)4 4 = 0 + 0 – 2 C4 4 4 2  (1– x ) + x (1 – x4)4 – 2x(1 – x4)4 = 0 + 0 – 24C4 esa x17 dk

 S = – 2  |S| = 2 4n + 2

4n+3

4n

82. 7 = 7 , 7 = 9, 7 = 3, 7 = 1 1005 4n+1 Now (1007) = (1007)  last digit vfUre 1007

4n + 3

vad = 7  last digit vfUre vad = 3

and (1007) = (1007)  last digit of sum 7 + 3 which is 0  ;ksx dk vfUre vad 7 + 3 tks fd 0 gSA 2 6

6

6

83. (1 + x – 2x ) = (1 – x) (1 + 2x) 10 6 6 Coefficient of x in (1 – x) (1 + 2x) 6 6 4 6 6 5 6 6 6 = C6 C4(2) + (– C5) C5(2) + C4 C6(2) = (1 × 15 × 16) – (6 × 6 × 32) + (15) × (1) × (64) = 240 – 1152 + 960 = 48 84.

1 1 1 1 1 + + + ........ + = 15! 1!14! 2!13 ! 7! 8! 15!

15! 15! 15!    .........    7!8!  15! 1!14! =

1 15 15 15 { C0 + C1 + ......... + C7} 15!

215 15 15 15 14 Now C0 + C1 + ............. + C7 = = 2 2



=   tan + 2tan 2 + 4cot 4

5

4 tan 4

4(1– tan2 2)  tan  + 2tan 2 + 2tan2  tan  + 2tan 2 + 2cot 2 – 2tan 2 2(1– tan2 )   cot  = cot 2tan  5

 tan  + 88. sin

xq.kkad gS

1  1  ,1  2   2 

 tan  + 2tan  +

8.3 = 72 x=2 2 4

2

(1– a )

 sum = 0

1/2

2

3

 a   – 1, –

9, (3x + 48) ,

2 4

4

4 2 (1 – a )  1 3

 0 

Js.kh esa gSA

4

1 2 (1– a ) 3

2ar = a + 3

1 8   log9(3x + 48), log9  3 x –  are in A.P lekUrj 80. log99, 2 3  

4n + 1

2

 sin2 x cos2 x =

3 2 , r = ,a=9 2r – 1 3

6

6

86. sin x +cos x = a  (sin2x + cos2x)3 – 3sin2 x cos2x(sin2x + cos2x) = a2  1 – 3sin2 x cos2 x = a2

79. Let a, ar, ar be in G.P. ekuk a, ar, ar 2 xq.kksÙkj Js.kh

a=

1! 2! 3! 4! 5! ............p! = 1  2  6  24 + 15k = 33 + 15k = 15 + 3  remainder 'ks"kQy = 3

5x

5x

= 1

=





x=

2 2 2 10y 10 y cos = 1 = 0y = 0 9 9

5

[x + y] = 0 2

2

2

89. sin  + tan  + sin –

1     sin    2    sin  = –

1 2

2

+

3

tan +

7 = 0 12

   tan  – 1  3  

and vkSj tan  =

2

= 0

1 3

7 11 7  , and vkS j  = , 6 6 6 6 7   = 6 pq 90. tanA . tanB = = 1 pq   =

 A + B =

 2



C=

 2



MONT 2

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