start 2015 sample test paper clsss 11 maths

8/10/2019 STaRT 2015 Sample Test Paper Clsss 11 Maths

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(SPACE FOR ROUGH WORK)

STaRT2015 XI_(Maths) - 3


PART-I (PHYSICS)

Straight Objective Type This section contains ( 1-15) questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

lh/ks oLrqfu"B izdkjbl [k.M esa(1-15) cgq&fodYih iz'u gSaA izR;sd iz'u ds4 fodYi (A), (B), (C) rFkk(D) gSa] ftuesa lsflQZ ,d lghgSA

1. A projectile is launched at an angle ! with the horizontal from point A, and finally falls at point B. A and B areon a horizontal line. During the course of the motion from point A to point B, for the projectile :,d iz{ksI; fcUnqA ls {kSfrt ds lkFk ! dks.k ij NksM+k tkrk gS] var esa ;g /kjkry ds fdlh fcUnqB ij igqprk gSAA o B ,d gh{ksfrt js[kk ij fLFkr gSaA fcUnqA ls B rd xfr ds nkSjku iz{ksI; ds fy;s:

!

A B

(A) magnitude of normal acceleration first increases then decreases.(B) magnitude of tangential acceleration first increases then decreases.(C) radius of curvature first increases then decreases.(D) external force acting on it first increases then decreases.(A) vfHkyEcor~ vfHkdsUnz h; Roj.k dk ifjek.k igys c


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5. Value of ! is increased gradually from ! = 0. At ! = tan 1 " # $%

& '

21

both the blocks just start slipping. Then value of ( 2 is:

(g = 10 m/s 2)

! = 0 ls ! dk eku yxkrkj c


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9. An insect of mass m = 3 kg is inside a vertical drum of radius 2 m that is rotating with anangular velocity of 5 rad/sec. If the insect does not fall off , then the minimum co-efficientof friction required is (g = 10m/s 2) :,d 3 kg nzO;eku dk dhM+k2 eh- f=kT;k ds m/okZ/kj Me ds vUnj gS rFkk Me5 rad/sec ds dks.kh; osxls ?kwe jgk gSA ;fn dhM+k uhps ugh fxjrk rks U;wure vko';d ?k"kZ.k xq.kkad gS(g = 10m/s 2) :

(A) 0.5 (B) 0.4 (C) 0.2 (D) none of these buesa ls dksbZ ugha

10. A mass M hangs stationary at the end of a light string that passes through a smoothfixed tube to a small mass m that moves around in a horizontal circular path. If " isthe length of the string from m to the top end of the tube and ! is angle between thispart and vertical part of the string as shown in the figure, then time taken by m tocomplete one circle is equal to,d nzO;ekuM gYdh jLlh ls fLFkj :i ls yVd jgk gS rFkk jLlh dk nwljk fljk fpduh tM+or~V~;w c ls xqtkjus ds ckn NksVs nzO;ekum ls ca/kk gSA tks {kSfrt o kh; iFk esa ?kwe jgk gSA ;fn m ls V~;w cds ijh fljs rd jLlh dh yEckbZ " vkSj jLlh ds bl Hkkx rFkk jLlh ds /okZ/kj Hkkx ds chp dks.k

! gS rksm }kjk ,d pDdj yxkus esa o k iwjk djus esa fy;s x;s le; dk eku gksxk M

m

!

(A) !)

sing2

"(B) !

)cosg

2 "

(C) !)

sinMgm

2 "

(D) Mgm

2 ")

11. A uniform thin rod of mass m and length L is held horizontally by two vertical strings attached to the two ends. Oneof the string is cut. Find the angular acceleration soon after it is cut :,d m nzO;eku rFkkL yEckbZ dh NM+ dks nks yEcor~ Mksjh;ks a }kjk {kSfrt yVdk;k x;k gS Mksjh fdukjks a ls ca/kh gSA ;fn ,d Mksjh dksrksM+ fn;k tk;s rks Mksjh VwVus ds rqjUr ckn dks.kh; Roj.k gksxk :

(A) L2

g

(B) L

g

(C) L2

g3

(D) L

g2

12. A uniform cylinder of mass m can rotate freely about its own axis which is horizontal. A particle of mass m 0hangs from the end of a light string wound round the cylinder which does not slip over it. When the system isallowed to move, the acceleration of the descending mass m 0 will be:m nzO;eku dk ,d le:i csyu mldh Loa; v{k tks {kSfrt gS ds lkis{k ?kw .kZu ds fy, LorU=k gSAm o nzO;eku dk ,d d.k ,dMksjh ds fljs ls yVdk gqvk gS rFkk Mksjh csyu ij fyiVh gqbZ gS tks bl ij fQlyrh ugha gSA tc fudk; dks xfr ds fy, NksM+k tkrkgS] nzO;ekum 0 dk uhps dh vksj Roj.k gksxk &

(A)2

2m g

m mo

o*(B)

m gm m

o

o*(C)

2m gm m

o

o*(D)

m gm m

o

o2 *


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13. In the figure shown the initial velocity of boat (30 kg) + person (15 kg )is 2 m/s. Find velocity v of person with respect to boat so that velocity ofboat will be 1 m/s towards right (Neglect friction between boat and water)fp=kkuqlkj "uko+ euq";" dk izkjfEHkd osx 2 eh0@lS0 gSA euq"; dk osx v uko ds lkis{kKkr djks ftlls uko dk osx 1 eh0@ls0 nk;ha vksj gks tk;s uko vkSj ikuh ds chp ?k"kZ.kux.; gS

(A) 3 m/s towards right nkW;h vkSj 3 m/s ls(B) 3 m/s towards left ck;h vkSj 3 m/s ls(C) 4 m/s towards right nk;h vkSj 4 m/s ls(D) 4 m/s towards left ck;h vkSj 4 m/s ls

14. The spacecraft of mass M moving with a velocity v in free space explodes and breaks up into two pieces. If afterexplosion a piece of mass m comes to rest, the other piece of spacecraft will have velocity.M nzO;eku dk vUrfj{k;ku eqDr {ks=k esa v osx ls xfr'khy gS rFkk foLQksV ds dkj.k nks Hkkxksa es a VwV tkrk gSA ;fn foLQksV ds i'pkr~m nzO;eku dk ,d VqdM+k fojkekoLFkk esa vk tkrk gS] vUrfj{k;ku ds vU; VqdM+ s dk osx gksxk &

(A) - .mMMv

+ (B) - .mMMv

* (C) mv / (M

m) (D) - .mMmv

*

15. Arc AB is half section of ellipse (with AB as the major axis). An ideal gasis taken from A to B as shown in diagram. The work done by gas for A toB process is : [use ) = 22/7]pki AB nh?kZo k dk v/kZ&Hkkx(AB nh?kZ v{k) gS A ,d vkn'kZ xSl dks fp=kkuqlkjAls B rd ys tk;k x;k gSA bl f;k esa xSl }kjkA ls B rd fd;k x;k dk;Z gS:[) = 22/7] A B

210 3 910 3

310 3

510 3

V(m )3

P(N/m )20

(A) 11 J (B) 22 J (C) 11 J (D) 22 J

PART-II (CHEMISTRY)


lh/ks oLrqfu"B izdkjbl [k.M esa(16-25) cgq&fodYih iz'u gSaA izR;sd iz'u ds4 fodYi (A), (B), (C) rFkk(D) gS a] ftuesa lsflQZ ,d lghgSA


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16. The most stable anion is :

lcls vf/kd LFkk;h ,uk;u _.kk;u gS %

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

17. The correct set of four quantum numbers for the valence electron of Rubidium (Z = 37) is ::fcfM;e (Z = 37) ds la;ksth bysDVkWu ds fy, pkj Dok.Ve la[;k dk lgh leqPp; crkvks

(A) n = 5, " = 0, m = 0, s = + 21

(B) n = 5, " = 1, m = 0, s = + 21

(C) n = 5, " = 1, m = 1, s = + 21

(D) n = 6, " = 0, m = 0, s = + 21

18. Fluorine has the highest electronegativity among the ns 2 np 5 group on the Pauling scale, but the electron affinity offluorine is less than that of chlorine because :(A) the atomic number of fluorine is less than that of chlorine.(B) fluorine being the first member of the family behaves in an unusual manner.(C) chlorine can accommodate an electron better than fluorine by utilising its vacant 3d orbital.(D) small size, high electron density and an increased electron repulsion makes addition of an electron to fluorine

less favourable than that in the case of chlorine in isolated stage.ikWfyx ekie ds vuqlkj ns 2 np 5 oxZ esa Ykksjhu lcls T;knk fo|qr_.kh gSaA ysfdu Ykksjhu dh bysDVkWu ca/kqrk] Dyksjhu ls de gksrh gSa] D;ksafd %(A) yksjhu dk ijek.kq ekad Dyksjhu ls de gksrk gSaA(B) yksjhu oxZ dk igyk lnL; gSa] blfy;s ;s vyx O;ogkj djrk gSaA(C) Dyksjhu fjDr 3d d{kd dk mi;ksx yksjhu ls vPNh rjg djds ,d bysDVkWu j[k ldrk gSaA(D) NksVk vkdkj mPp bysDVkWu ?kuRo vkSj bysDVkWu frd"kZ.k esa o f) ds dkj.k foyfxr voLFkk esa Dyksjhu dh rqyuk es yksjh

esa ,de tksM+uk de vuqdqy gksrk gSA



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19. S 1 : Always cyclic part is main chain and acyclic is substituent in alicyclic molecules.S 1 : ,fylkbfDyd v.kq esa ges'kk ph; Hkkx eq[; J a[kyk gksrk gS rFkk vph; Hkkx izfrLFkkihAS 2 : Correct IUPAC name of the following is 3-Bromo-2-Chlorohepta-2,4-diene.S 2 : fuEu dk lghIUPAC uke3-czkseks-2-DyksjksgsIVk-2,4- MkbZbZu

S 3 : DU of the following compounds is 6.S 3 : fuEu ;kSfxd dkDU = 6 gSA

S 4 : Di, tri., tetra of complex radical are considered for alphabetization.S 4 : ladqy ewyd dk MkbZ] VkbZ] VsVk vxzsth o.kZekyk esa lfEefyr fd;k tkrk gSA(A) F T T F (B) F T F T (C) F F T T (D) T T F F

20. A 5.2 molal aqueous solution of methyl alcohol (CH 3OH), is supplied. What is the mole fraction of methyl alcohol inthe solution ?esfFky ,sYdksgkWy] CH3OH dk ,d 5.2 eksyy tyh; foy;u fn;k tkrk gSA bl foy;u esa esfFky ,sYdksgkWy dk eksy izHkka'k D;kgS\(A) 0.100 (B) 0.190 (C) 0.086 (D) 0.050

21. 2K, + ,2 + 22 HNO 3 / 2H,O3 + 2K ,O3 + 22NO 2 + 10H 2OIf 3 moles of K , & 2 moles ,2 are reacted with excess of HNO 3. Volume of NO 2 gas evolved at NTP is;fn K, ds3 eksy rFkk,2 ds2 eksyHNO 3 ds vkf/kD; ls f;k djrs gSA ekud rki o nkc ijNO 2 dk izkIr vk;ru gS(A) 739.2 Lt (B) 1075.2 Lt (C) 44.8 Lt (D) 67.2 Lt

22. Identify the incorrect statement(A) Formal charge do not indicate real charge separation with in the molecule.(B) A pair of +1 and 1 formal charges on the adjacent atoms is considered as coordinate bond in lewis dot

structures.(C) The lowest energy lewis dot structure is the one with minimum formal charge on the atoms.(D) Formal charge is the factor based upon electronegativity of bonded atom.vlR; dFku dks igpkfu;sA(A) ,d v.kq esa vkSipkfjd vkos'k(formal charge) okLrfod vkos'k iFkDdj.k dks ugh n'kkZrkA(B) ysfol fcUnq lajpuk esa fudVorhZ (adjacent) ijek.kq ij +1 rFkk 1 vkSipkfjd vkos'k dks milgla;kstd ca /k ds :i es a ekurs

gS aA(C) U;wure tkZ okyh ysfol fcUnq lajpuk esa ijek.kqvksa ij U;wure vkSipkfjd vkos'k gksrk gSA(D) vkSipkfjd vkos'k ,d ,slk dkjd gS tks cfU/kr ijek.kqvksa dh fo|qr _.krk ij vk/kkfjr gSA


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23. Decreasing I effect of given groups is :fn;s x;s lewgks ds I izHkko dk ?kVrk gqvk e gSA(i) CN (ii) NO 2 (iii) NH2 (iv) F(A) iii > ii > i > iv (B) ii > iii > iv > i (C) iii > ii > iv > i (D) ii > i > iv > iii

24. Which of the following compound will not give positive iodoform test.

fuEu esa ls dkSulk ;kSfxd /kukRed vkW;ksMksQkWeZ ijh{k.k ugha nsrk gSA

(A) (B) (C) (D) CH 3 CHO

25. 7.5 ml of a gaseous hydrocarbon was exploded with 36 ml of oxygen. The volume of gases on cooling was found tobe 28.5 ml, 15 ml of which was absorbed by KOH and the rest was absorbed in a solution of alkaline pyrogallol. Ifall volumes are measured under same conditions, the formula of hydrocarbon is,d xSlh; gkbMksdkcZu ds 7.5 ml dks vkWDlhtu ds36 ml ds lkFk foLQksfVr fd;k x;kA B.Mk djus ijxSl dk vk;ru 28.5 ml ik;k x;k] ftles a ls 15 ml dks KOH ds }kjk vo'kksf"kr fd;k x;k rFkk cps gq, vk;ru dks {kkjh; ik;jksxSyksyds foy;u es a vo'kksf"kr fd;k x;kA ;fn lHkh vk;ru leku ifjfLFkfr ij ekis x;s rks gkbMksdkcZu dk lw=k gS\(A) C 3H4 (B) C 2H4 (C) C 2H6 (D) C 3H6

PART - III (MATHS)


lh/ks oLrqfu"B izdkj

bl [k.M esa(26-40) cgq&fodYih iz'u gSaA izR;sd iz'u ds4 fodYi (A), (B), (C) rFkk(D) gS a] ftuesa lsflQZ ,d lghgSA

26. Modulus of non zero complex number z, satisfying 0zz 0* and |z| 2 4zi = z 2 is0zz 0* rFkk|z| 2 4zi = z 2 dks larq"V djus okyh v'kwU; lfEeJ la[;kz dk ekikad gkssxk&

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

27. If ax2 + bx + 8 = 0 does not have two distinct real roots and the minimum value of 2a + b is 1, then the value of | 1|is;fn ax 2 + bx + 8 = 0 nks fHkUu okLrfod ewy ugha j[krh gS vkSj 2a + b dk U;wure eku1 gS] rks|1| dk eku gksxk&(A) 4 (B) 2 (C) 3 (D) 1


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35. The coefficient of x 48 in the product (x 2) (x 4)........(x 98) isxq.kuQy (x 2) (x 4)........(x 98) esax48 dk xq.kkad gksxk&(A) 99 2 (B) 2450 (C) 2500 (D) none of these bues a ls dksbZ ugha

36. If log10 (x3 + y3) log 10 (x

2 + y2 xy) < 2. Where x, y are positive real numbers. Then the maximum value of xy is;fn log 10 (x3 + y3) log 10 (x2 + y2 xy) < 2 tcfd x, y /kukRed okLrfod la[;k,a gS] rcxy dk vf/kdre eku gksxk&

(A) 2500 (B) 3000 (C) 1200 (D) 3500

37. The interior angles of a polygon are in A.P. If the smallest angle is 120 and the common difference is 5, thenthe number of sides in the polygon is:fdlh cgqHkqt ds vUr%dks.k lekUrj Js


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40. If ) < ! < 2 ) , then the equation cosec ! = log 1/3 " # $%

& ' +

xx32

is possible for the values of x in the interval

;fn ) < ! < 2 ) rc lehdj.k cosec ! = log 1/3 " # $%

& ' +

xx32

ds lEHko gksus ds fy,x ds eku fdl vUrjky esa gksxs a&

(A) " # $

%& '

32

,0 (B) 456

%& '

31

,0 (C) ( 7 , 0) U " # $

%& '

32

,31

(D) none of these buesa ls dksbZ ugha

PART-IV (MENTAL ABILITY)

Straight Objective Type

This section contains ( 41-50) questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

lh/ks oLrqfu"B izdkjbl [k.M esa(41-50) cgq&fodYih iz'u gSaA izR;sd iz'u ds4 fodYi (A), (B), (C) rFkk(D) gS a] ftuesa lsflQZ ,d lghgSA

Direction (41 to 43) : Find the missing terms

funsZ'k 41 ls 43 vKkr in Kkr djksA

41. 94, 166, 258, ?, 4912(A) 3610 (B) 1644 (C) 1026 (D) 516

42. AD 17 39 CFBP 258 108 HJGN ? ? LM(A) 203, 181 (B) 179, 239 (C) 18, 210 (D) 9, 29



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43.

(A) 38 (B) 64 (C) 4 (D) 16

44. If AJAY is written as 1117 , then in same code NAMA would be written as :;fn AJAY dks 1117 fy[kk tk;s rks] mlh dksM+ esaNAMA dks D;k fy[kk tk;sxk \(A) 5114 (B) 5411 (C) 5141 (D) 4511

45. Arrange the following group such that when arranged in a specific order, meaningful word is formed.

fuEu v{kjks ds lewg dks bl izdkj O;ofLFkr djks rkfd buls ,d vFkZ iw.kZ 'kCn cu tk;s \V A H Y E1 2 3 4 5(A) 2, 3, 4, 5, 1 (B) 3, 2, 5, 1, 4 (C) 3, 5, 2, 1, 4 (D) 1, 5, 2, 3, 4

46. If + is x, is +, x is 8 and 8 is , then what is the value of given equation;fn + dk eryc ] dk eryc +, dk eryc rFkk dk eryc ekus rks bl lwpuk ds vk/kkj ij fuEu lehdj.k dkeku gksxkA21 8 + 2 12 3 = ?(A) 14 (B) 9 (C) 13.5 (D) 11

47. What is the product of all the numbers in the dial of a telephone ?

VsfyQksu ds Mk;y ij fLFkr vadksa dk xq.kuQy D;k gksxk ?(A) 1,58,480 (B) 1,59,450 (C) 1,59,480 (D) None of these

Directions (48) : In the following questions, there is a diagram marked (X), with one or more dots placed in it. The diagram is followed by four other figures, marked (A), (B), (C) and (D) only one of which is such as to make possible the placement of the alternative in each case.

funsZ'k(48) : fuEufyf[kr iz'u esa] ,d vkd fr(X) nh xbZ gS ftlls ,d ;k vf/kd fcUnq yxs gq, gSA ;g vkd fr vU; pkj vkd fr;ksa dk vuqlj.k djrh gS ftudks fd (A), (B), (C) rFkk(D) esa fpfUgr fd;k x;k gSA bues a ls flQZ ,d vkd fr esa bu fcUnqvksa ds lgh fpfUgr dh lEHkkouk gSA ml fodYi dks pqfu;s?


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48.

(X)

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

Directions : (49) Figures 1 and 2 are related in a particular manner. Establish the same relationship between figures 3 and4 by choosing a figure from amongst the four alternatives, which would replace the question mark in figure (4).

funsZ'k: (49) iz'u vkd fr ds izFke Hkkx esa fn;s x;s nwljs fp=k dk ftl izdkj igys fp=k ls lEcU/k gS ogh lEcU/k nwljs Hkkx ds rhljfp=k dk mkj vkd fr esa fn;s x;s pkj fodYiksa esa ls fdlls lEcU/k gksxkA

49. Problem Figures iz'u vkfr

........... ......... ....... ........... . ............ .......... .... .... .. ........

..... ?........... ......... ........ ........... . ...

.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .: :: :Answer Figures mkj vkfr

. .

. . . . . .

. . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

. . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.. . . . . . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . .

. . . .

. . . . . . . . . . . . . . . . . . . . . . .

. . . .

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

Directions : (50) Out of the four figures (A), (B), (C) and (D), given in each problem, three are similar in a certain way.However, one figure is not like the other four. Choose the figure which is different from the rest.

funsZ'k: (50) uhps fn;s x;s4 fp=kksa(A), (B), (C) rFkk(D) esa rhu fdlh rjhds ls leku gS vkSj ,d mu rhuksa ls vyx gS vFkkZr~lewg esa ls rhu fp=k leku izd fr ij vk/kkfjr gS ml fp=k dks crkb, ftldh izd fr ckfd rhuksa ls vyx gSaA

50.

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

ANSWER KEYCLASS : XI | STREAM : SCIENCE-MATHS

1. (A) 2. (C) 3. (A) 4. (B) 5. (C)6. (A) 7. (B) 8. (C) 9. (C) 10. (D)11. (C) 12. (A) 13. (A) 14. (A) 15. (C)16. (A) 17. (A) 18. (D) 19. (B) 20. (C)21. (A) 22. (D) 23. (D) 24. (C) 25. (B)

26. (B) 27. (A) 28. (C) 29. (D) 30. (C)

31. (B) 32. (B) 33. (A) 34. (A) 35. (B)36. (A) 37. (B) 38. (A) 39. (B) 40. (B)41. (A) 42. (A) 43. (C) 44. (C) 45. (C)

46. (B) 47. (D) 48. (A) 49. (B) 50. (B)

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