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BO˜ GIAO DUˇC VA AO TAˇO E THI TUYE¯N SINH AˇI HOˇC NA˚M 2013 ---------- Mon: TOAN; KhoÆi A val khoÆi A1 E CH˝NH THC Thli gian lalm bali: 180 phuøt, khong ke thli gian phaøt æe ------------------- I. PHAN CHUNG CHO TA`T CA TH˝ SINH (7,0 æiem) Cau 1 (2,0 æiem). Cho halm soÆ y = -x 3 +3x 2 +3mx - 1 (1), vøi m lal tham soÆ thc. a) Khaßo saøt s bieÆn thien val veı æo th cußa halm soÆ (1) khi m =0. b) Tm m æe halm soÆ (1) nghch bieÆn tren khoaßng (0; + ). Cau 2 (1,0 æiem). Giaßi phng trnh 1 + tan x =2 2 sin x + π 4 . Cau 3 (1,0 æiem). Giaßi he phng trnh x +1+ 4 x - 1 - y 4 +2= y x 2 +2x(y - 1) + y 2 - 6y +1=0 (x, y R). Cau 4 (1,0 æiem). Tnh tch phan I = 2 1 x 2 - 1 x 2 ln x dx. Cau 5 (1,0 æiem). Cho hnh choøp S.ABC coø æaøy lal tam giaøc vuong tai A, ABC = 30 , SBC lal tam giaøc æeu canh a val maºt ben SBC vuong goøc vøi æaøy. Tnh theo a the tch cußa khoÆi choøp S.ABC val khoaßng caøch tl æiem C æeÆn maºt phaœng (SAB). Cau 6 (1,0 æiem). Cho caøc soÆ thc dng a, b, c thoßa maın æieu kien (a + c)(b + c)=4c 2 . Tm giaø tr nhoß nhaÆt cußa bieu thøc P = 32a 3 (b +3c) 3 + 32b 3 (a +3c) 3 - a 2 + b 2 c . II. PHAN RIE´NG (3,0 æiem): Th sinh ch æc lalm mot trong hai phan (phan A hoaºc phan B) A. Theo chng trnh Chuan Cau 7.a (1,0 æiem). Trong maºt phaœng vøi he toa æo Oxy , cho hnh chı nhat ABCD coø æiem C thuoc ælng thaœng d :2x + y +5=0 val A(-4; 8). Goi M lal æiem æoÆi xøng cußa B qua C , N lal hnh chieÆu vuong goøc cußa B tren ælng thaœng MD. Tm toa æo caøc æiem B val C , bieÆt raLng N (5; -4). Cau 8.a (1,0 æiem). Trong khong gian vøi he toa æo Oxyz , cho ælng thaœng Δ: x - 6 -3 = y +1 -2 = z +2 1 val æiem A(1; 7; 3). VieÆt phng trnh maºt phaœng (P ) æi qua A val vuong goøc vøi Δ. Tm toa æo æiem M thuoc Δ sao cho AM =2 30. Cau 9.a (1,0 æiem). Goi S lal tap hp taÆt caß caøc soÆ t nhien gom ba chı soÆ phan biet æc chon tl caøc chı soÆ 1; 2; 3; 4; 5; 6; 7. Xaøc ænh soÆ phan tß cußa S . Chon ngaªu nhien mot soÆ tl S , tnh xaøc suaÆt æe soÆ æc chon lal soÆ chaün. B. Theo chng trnh Nang cao Cau 7.b (1,0 æiem). Trong maºt phaœng vøi he toa æo Oxy , cho ælng thaœng Δ: x - y =0. lng troln (C ) coø baøn knh R = 10 caØt Δ tai hai æiem A val B sao cho AB =4 2. TieÆp tuyeÆn cußa (C ) tai A val B caØt nhau tai mot æiem thuoc tia Oy . VieÆt phng trnh ælng troln (C ). Cau 8.b (1,0 æiem). Trong khong gian vøi he toa æo Oxyz , cho maºt phaœng (P ):2x +3y + z - 11 = 0 val maºt cau (S ): x 2 + y 2 + z 2 - 2x +4y - 2z - 8=0. Chøng minh (P ) tieÆp xuøc vøi (S ). Tm toa æo tieÆp æiem cußa (P ) val (S ). Cau 9.b (1,0 æiem). Cho soÆ phøc z =1+ 3 i. VieÆt dang lng giaøc cußa z . Tm phan thc val phan aßo cußa soÆ phøc w = (1 + i)z 5 . ------HeÆt------ Th sinh khong æc sß dung tali lieu. Caøn bo coi thi khong giaßi thch g them. Ho val ten th sinh: ..................... ..................... ; SoÆ baøo danh: .....................
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Khoi A-A1 2013

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  • BO GIAO DUC VA AO TAO E THI TUYEN SINH AI HOC NAM 2013 Mon: TOAN; Khoi A va khoi A1

    E CHNH THC Thi gian lam bai: 180 phut, khong ke thi gian phat e

    I. PHAN CHUNG CHO TAT CA TH SINH (7,0 iem)

    Cau 1 (2,0 iem). Cho ham so y = x3 + 3x2 + 3mx 1 (1), vi m la tham so thc.a) Khao sat s bien thien va ve o th cua ham so (1) khi m = 0.

    b) Tm m e ham so (1) nghch bien tren khoang (0; +).

    Cau 2 (1,0 iem). Giai phng trnh 1 + tanx = 22 sin

    (x+

    pi

    4

    ).

    Cau 3 (1,0 iem). Giai he phng trnh

    { x+ 1 + 4

    x 1

    y4 + 2 = y

    x2 + 2x(y 1) + y2 6y + 1 = 0(x, y R).

    Cau 4 (1,0 iem). Tnh tch phan I =2

    1

    x2 1x2

    lnx dx.

    Cau 5 (1,0 iem). Cho hnh chop S.ABC co ay la tam giac vuong tai A, ABC = 30, SBC latam giac eu canh a va mat ben SBC vuong goc vi ay. Tnh theo a the tch cua khoi chopS.ABC va khoang cach t iem C en mat phang (SAB).

    Cau 6 (1,0 iem). Cho cac so thc dng a, b, c thoa man ieu kien (a + c)(b+ c) = 4c2. Tm gia tr

    nho nhat cua bieu thc P =32a3

    (b+ 3c)3+

    32b3

    (a+ 3c)3a2 + b2

    c.

    II. PHAN RIENG (3,0 iem): Th sinh ch c lam mot trong hai phan (phan A hoac phan B)

    A. Theo chng trnh Chuan

    Cau 7.a (1,0 iem). Trong mat phang vi he toa o Oxy, cho hnh ch nhat ABCD co iem C thuocng thang d : 2x+ y + 5 = 0 va A(4; 8). Goi M la iem oi xng cua B qua C , N la hnh chieuvuong goc cua B tren ng thang MD. Tm toa o cac iem B va C , biet rang N(5;4).

    Cau 8.a (1,0 iem). Trong khong gian vi he toa o Oxyz, cho ng thang :x 63 =

    y + 1

    2 =z + 2

    1va iem A(1; 7; 3). Viet phng trnh mat phang (P ) i qua A va vuong goc vi . Tm toa o iemM thuoc sao cho AM = 2

    30.

    Cau 9.a (1,0 iem). Goi S la tap hp tat ca cac so t nhien gom ba ch so phan biet c chon tcac ch so 1; 2; 3; 4; 5; 6; 7. Xac nh so phan t cua S. Chon ngau nhien mot so t S, tnh xac suate so c chon la so chan.

    B. Theo chng trnh Nang cao

    Cau 7.b (1,0 iem). Trong mat phang vi he toa o Oxy, cho ng thang : x y = 0. ngtron (C) co ban knh R =

    10 cat tai hai iem A va B sao cho AB = 4

    2. Tiep tuyen cua (C)

    tai A va B cat nhau tai mot iem thuoc tia Oy. Viet phng trnh ng tron (C).

    Cau 8.b (1,0 iem). Trong khong gian vi he toa o Oxyz, cho mat phang (P ) : 2x+ 3y+ z 11 = 0va mat cau (S) : x2 + y2 + z2 2x+ 4y 2z 8 = 0. Chng minh (P ) tiep xuc vi (S). Tm toa otiep iem cua (P ) va (S).

    Cau 9.b (1,0 iem). Cho so phc z = 1+3 i. Viet dang lng giac cua z. Tm phan thc va phan ao

    cua so phc w = (1 + i)z5.Het

    Th sinh khong c s dung tai lieu. Can bo coi thi khong giai thch g them.

    Ho va ten th sinh: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ; So bao danh: . . . . . . . . . . . . . . . . . . . . .

  • B GIO DC V O TO CHNH THC

    P N THANG IM THI TUYN SINH I HC NM 2013

    Mn: TON; Khi A v khi A1 (p n - thang im gm 04 trang)

    Cu p n ima. (1,0 im) Khi m = 0 ta c 3 23 1y x x .= + Tp xc nh: .D = \ S bin thin:

    - Chiu bin thin: hoc 2' 3 6 ; ' 0y x x y x= + = = 0 2.x = 0,25

    Khong ng bin: (0; 2); cc khong nghch bin: ( ; 0) v (2; ).+ - Cc tr: Hm s t cc tiu ti x = 0, yCT = 1; t cc i ti x = 2, yC = 3. - Gii hn: lim ; lim .

    x xy y += + =

    0,25

    - Bng bin thin:

    Trang 1/4

    0,25

    th:

    0,25

    b. (1,0 im)

    Ta c 2' 3 6 3y x x= + + .mHm s (1) nghch bin trn khong (0; )+ khi v ch khi ' 0, 0y x > 0,25

    2 2 , 0.m x x x > Xt 2( ) 2f x x x= vi Ta c 0.x > '( ) 2 2; '( ) 0 1.f x x f x x= = =

    0,25

    Bng bin thin:

    0,25

    1 (2,0 im)

    Da vo bng bin thin ta c gi tr m tha mn yu cu ca bi ton l m 1.

    x 'y

    y

    + 0 2 0 0 +

    +

    1

    3

    2 O

    y

    x

    3

    1

    x

    ( )f x

    0 + 1 0

    0 +

    1

    + '( )f x

    0,25

  • Trang 2/4

    Cu p n imiu kin: Phng trnh cho tng ng vi cos 0.x sin1 2(sin co

    cosx s )x xx

    + = + 0,25 (sin cos )(2cos 1) 0.x x x + = 0,25

    sin cos 0 ( )4

    x x x k k + = = + ] . 0,25

    2 (1,0 im)

    2cos 1 0 2 ( )3

    x x k k = = + ] .

    i chiu iu kin ta c nghim: 4

    x k= + hoc 2 ( )3

    x k k= + ] . 0,25

    44

    2 2

    1 1 2

    2 ( 1) 6 1 0 (2)

    x x y y

    x x y y y

    + + + = + + + =(1)

    ,

    iu kin: T (2) ta c suy ra 1.x 24 ( 1)y x y= + 0.y 0,25

    3 (1,0 im)

    t 4 1,u x= suy ra u Phng trnh (1) tr thnh: 0. 4 42 2 (3).u u y y+ + = + + Xt 4( ) 2 ,f t t= + + t vi Ta c 0.t

    3

    4

    2'( ) 1 0, 0.2

    tf t tt

    = + > +

    Do phng trnh (3) tng ng vi ,y u= ngha l 4 1.x y= +

    0,25

    Thay vo phng trnh (2) ta c 7 4( 2 4) 0 (4).y y y y+ + =Hm c 7 4( ) 2 4g y y y y= + + 6 3'( ) 7 8 1 0g y y y= + + > vi mi 0.y 0,25

    M nn (4) c hai nghim khng m l (1) 0,g = 0y = v 1.y = Vi ta c nghim ( ; vi 0y = ) (1; 0);x y = 1y = ta c nghim ( ; ) (2; 1).x y = Vy nghim ( ; )x y ca h cho l v (1; 0) (2; 1).

    0,25

    t 2

    21 dln , d d d , .x xu x v x u v x 1

    x xx= = = = + 0,25

    Ta c 22

    1 1

    1 1ln dI x x x 1 xx x x

    = + + 0,25

    2 2

    1 1

    1 1lnx x xx x

    = + 0,25

    4 (1,0 im)

    5 3ln 2 .2 2

    = 0,25 Gi H l trung im ca BC, suy ra SH BC. M (SBC) vung gc vi (ABC) theo giao tuyn BC, nn SH (ABC). 0,25

    Ta c BC = a, suy ra 3 ;2

    aSH = osin30 ;2aAC BC= =

    o 3cos30 .2

    aAB BC= =

    Do 3

    .1 . .6 1S ABC

    a .6

    H AB AC= =V S

    0,25

    Tam gic ABC vung ti A v H l trung im ca BC nn HA = HB. M SH (ABC), suy ra SA = SB = a. Gi I l trung im ca AB, suy ra SI AB.

    0,25

    5 (1,0 im)

    Do 2

    2 13 .4 4

    AB aSI SB= =

    Suy ra . .3 6 39( ,( )) .

    . 1S ABC S ABC

    SAB

    V V ad C SABS SI AB

    = = =3

    0,25

    S

    A B

    C

    I

    H

  • Trang 3/4

    Cu p n imt ,ax y

    c c= = .b Ta c iu kin ca bi ton tr thnh 0, 0.x y> > 3.xy x y+ + =

    Khi 33 2 2

    3 33232 .

    ( 3) ( 3)yxP x

    y x= + ++ + y

    v> >

    Vi mi u ta c 0, 03

    3 3 3 3 3 ( )3( ) 3 ( ) ( ) ( )4 4u v .v u v uv u v u v u v ++ = + + + + =u

    Do 333 23

    3 332 ( ) 2 3 332 8 83 3 3 3 9( 3) ( 3)

    y y x y xy xx xy x xy x yy x

    + + + + + = + + + + + + + .y

    0,25

    Thay 3xy x= y vo biu thc trn ta c 333 3

    3 332 ( 1)( 6)32 8 (2( 6)( 3) ( 3)

    y x y x yx x yx yy x+ + + + = + + + + + 1) . Do

    3 2 2 3 2 3 2( 1) ( 1) ( ) 2 ( 1) ( ) 2( ) 6.P x y x y x y x y xy x y x y x y + + = + + = + + + +

    0,25

    t t x Suy ra t v .y= + > 0 3 2( 1) 2 6.P t t t + Ta c

    2 2( )3 ( )4 4

    x y tx y xy x y t+= + + + + = + . nn ( 2)( 6) 0t t + Do 2.t

    Xt 3 2( ) ( 1) 2 6,f t t t t= + vi t Ta c 2. 22

    1'( ) 3( 1) .2 6

    tf t tt t

    += +

    Vi mi t ta c v 2 23( 1) 3t 221 7 71 1 2 2( 1) 72 6

    ttt t

    + = + + =+ + 3 2 , nn

    3 2'( ) 3 0.2

    f t > Suy ra ( ) (2) 1 2.f t f = Do 1 2P .

    0,25

    6 (1,0 im)

    Khi a th b c= = 1 2P = . Do gi tr nh nht ca P l 1 2 . 0,25 Do C d nn ( ; 2 5).C t t Gi I l tm ca hnh ch nht ABCD, suy ra I l trung im ca AC.

    Do ( )4 2 3; .2 2t tI + 0,25 Tam gic BDN vung ti N nn IN = IB. Suy ra IN = IA. Do ta c phng trnh

    ( ) ( )2 22 24 22 3 45 4 4 82 2 2t tt t + + = +

    7.a (1,0 im)

    32+

    1.t = Suy ra C(1; 7).

    0,25

    Do M i xng vi B qua C nn CM = CB. M CB = AD v CM||AD nn t gic ACMD l hnh bnh hnh. Suy ra AC||DM. Theo gi thit, BN DM, suy ra BN AC v CB = CN. Vy B l im i xng ca N qua AC.

    0,25

    ng thng AC c phng trnh: 3 4 0.

    .

    x y+ + =ng thng BN qua N v vung gc vi AC nn c phng trnh 3 17 0x y = Do (3 17; ).B a a+ Trung im ca BN thuc AC nn

    3 17 5 43 4 0 7.2 2

    a a a+ + + + = = ( 4; 7).B Vy 0,25

    c vct ch phng l ( 3; 2;1).u = JG 0,25 (P) qua A v nhn u

    JG lm vct php tuyn, nn (P) c phng trnh

    3( 1) 2( 7) ( 3) 0 3 2 14 0.x y z x y z + = + = 0,25 M thuc nn (6 3 ; 1 2 ; 2 ).M t t + t 0,25

    8.a (1,0 im)

    2 2 2 22 30 (6 3 1) ( 1 2 7) ( 2 3) 120 7 4 3 0AM t t t t t= + + + = = 1t = hoc 3 .7t Suy ra M= (3; 3; 1) hoc ( )51 1 17; ;7 7 7M . 0,25

    A D

    B C M

    N I

  • Trang 4/4

    Cu p n imS phn t ca S l 37A 0,25 = 210. 0,25 S cch chn mt s chn t S l 3.6.5 90= (cch). 0,25

    9.a (1,0 im)

    Xc sut cn tnh bng 90 3 .210 7

    = 0,25 Gi M l giao im ca tip tuyn ti A v B ca (C), H l giao im ca AB v IM. Khi (0; ),M t vi H l trung im

    ca AB. Suy ra

    0;t 2 2.

    2ABAH = =

    0,25

    2 21 1 1 ,

    AH AM AI= + 2 suy ra 2 10.AM =

    Do 2 2 4 2.MH AM AH= = M | |( , ) ,

    2tMH d M= = nn 8.t = Do (0; 8).M

    0,25

    ng thng IM qua M v vung gc vi nn c phng trnh 8 0.x y+ = Do ta im H tha mn h

    .0

    (4;4)8 0

    x yH

    x y = + =

    0,25

    7.b (1,0 im)

    A

    I

    B

    H

    M

    Ta c 2 2 12 ,4

    IH IA AH HM= = = nn 1 .4

    IH HM=JJJG JJJJG Do (5;3).I

    Vy ng trn (C) c phng trnh 2 2( 5) ( 3) 10x y + = .0,25

    (S) c tm v bn knh (1; 2;1)I 14.R = 0,25

    2 2 2

    | 2.1 3( 2) 1.1 11| 14( ,( )) .142 3 1

    d I P R+ + =+ +

    = = Do (P) tip xc vi (S). 0,25

    8.b (1,0 im)

    Gi M l tip im ca (P) v (S). Suy ra M thuc ng thng qua I v vung gc vi (P). 0,25 (1 2 ; 2 3 ;1 ).M t t+ + + t Do

    Do M thuc (P) nn Vy 2(1 2 ) 3( 2 3 ) (1 ) 11 0 1.t t t+ + + + + = =t (3;1;2).M 0,25 1 31 3 22 2

    z i i = + = +

    0,25 9.b

    (1,0 im) 2 cos sin .3 3

    i = + 0,25 5 5 5 52 cos sin 16(1 3 ).

    3 3z i = + = iSuy ra 0,25

    16( 3 1) 16(1 3) .w i= + + Do 0,25

    Vy w c phn thc l 16( v phn o l 3 1)+ 16 (1 3).

    ------------- Ht -------------

  • B GIO DC V O TO

    CHNH THC ( thi c 6 trang)

    THI TUYN SINH I HC NM 2013

    Mn: VT L; Khi A v Khi A1 Thi gian lm bi: 90 pht, khng k thi gian pht

    M thi 859

    H, tn th sinh:.......................................................................... S bo danh:............................................................................

    Cho bit: hng s Plng h = 6,625.1034 J.s; ln in tch nguyn t e = 1,6.1019 C; tc nh sng trong chn khng c = 3.108 m/s; gia tc trng trng g = 10 m/s2. I. PHN CHUNG CHO TT C TH SINH (40 cu, t cu 1 n cu 40)

    Cu 1: t in p u = 120 2cos2ft (V) (f thay i c) vo hai u on mch mc ni tip gm cun cm thun c t cm L, in tr R v t in c in dung C, vi CR2 < 2L. Khi f = f1 th in p hiu dng gia hai u t in t cc i. Khi 2 1f = f = f 2 th in p hiu dng gia hai u in tr t cc i. Khi f = f3 th in p hiu dng gia hai u cun cm t cc i ULmax. Gi tr ca ULmax gn gi tr no nht sau y?

    A. 85 V. B. 173 V. C. 57 V. D. 145 V. Cu 2: Mt vt nh dao ng iu ha theo phng trnh x A cos 4 t= (t tnh bng s). Tnh t t = 0, khong thi gian ngn nht gia tc ca vt c ln bng mt na ln gia tc cc i l

    A. 0,104 s. B. 0,125 s. C. 0,083 s. D. 0,167 s. u(cm)

    t2

    t1x(cm)60

    5

    -5 30

    N 0

    Cu 3: Mt sng hnh sin ang truyn trn mt si dy theo chiu dng ca trc Ox. Hnh v m t hnh dng ca si dy ti thi im t1 (ng nt t) v t2 = t1 + 0,3 (s) (ng lin nt). Ti thi im t2, vn tc ca im N trn dy l

    A. 39,3 cm/s. B. 65,4 cm/s. C. 65,4 cm/s. D. 39,3 cm/s. Cu 4: Trong mt th nghim Y-ng v giao thoa nh sng, bc sng nh sng n sc l 600 nm, khong cch gia hai khe hp l 1 mm, khong cch t mt phng cha hai khe n mn quan st l 2 m. Khong vn quan st c trn mn c gi tr bng

    A. 1,2 mm. B. 0,3 mm. C. 0,9 mm. D. 1,5 mm. Cu 5: t in p (V) (vi U0u U cos t= 0 v khng i) vo hai u on mch gm cun dy khng thun cm mc ni tip vi t in c in dung C (thay i c). Khi C = C0 th cng dng in trong

    mch sm pha hn u l (1 <

  • C ca mt kim loi l 0,75 i kim loi ny bng

    C i u cu y bin p xoay chi dng 200 V.

    D. 8. C i l nng ln htn nh sng nng lng ca phtn nh sng lc, V l nng

    V > . D. L> > V. C c truy n mt kh ng dy u sut truyn

    B. 92,8%. C. 87,7%. D. 85,8%. C on lc l xo g khi lng 1 o c cng

    u 9: Gii hn quang in m. Cng thot lectron ra khA. 26,5.1032 J. B. 2,65.1019 J. C. 2,65.1032 J. D. 26,5.1019 J. u 10: t vo ha n s cp ca m M1 mt in p u c gi tr hiu

    Khi ni hai u cun s cp ca my bin p M2 vo hai u cun th cp ca M1 th in p hiu dng hai u cun th cp ca M2 h bng 12,5 V. Khi ni hai u cun th cp ca M2 vi hai u cun th cp ca M1 th in p hiu dng hai u cun s cp ca M2 h bng 50 V. B qua mi hao ph. M1 c t s gia s vng dy cun s cp v s vng dy cun th cp bng

    A. 4. B. 15. C. 6. u 11: G g ca p ; L l

    lng ca phtn nh sng vng. Sp xp no sau y ng? A. > V > L. B. V > L > . C. L>u 12: in nng n t ni pht u dn c bng mt pha vi hi

    ti l 90%. Coi hao ph in nng ch do ta nhit trn ng dy v khng vt qu 20%. Nu cng sut s dng in ca khu dn c ny tng 20% v gi nguyn in p ni pht th hiu sut truyn ti in nng trn chnh ng dy l

    A. 89,2%. u 13: Mt c m vt nh c 00 g v l x 40 N/m

    c t trn mt phng ngang khng ma st. Vt nh ang nm yn v tr cn bng, ti t = 0, tc dng lc F = 2 N ln vt nh (hnh v) cho con lc dao ng iu ha n

    thi im t = 3 s th ngng tc dng lc F. Dao ng iu ha ca con lc sau khi khng

    c gi tr b n gn gi tr no nht sau y? A. 11 cm. B. 5 cm.

    F

    cn lc F tc dng

    iC. 9 cm. D. 7 cm.

    C mch ni tip m thun, o v t in (hnhu 14: on gm cun c n mch X v). Khi t vo hai u A, B in p uAB = U0cos(t+) (V) (U0, v khng i) th:

    LC2 = 1, UAN = 25 2 V v UMB = 50 2 V, ng thi uAN sm pha 3 so vi uMB.

    Gi tr ca U0 l

    A.

    A M N

    L X C

    B

    B. 12,525 7 V. 7 V. C. 12,5 14 V. D. 25 14 V. C : M ng dy dn t, ch nht ch 6 2, quay anh c i xng

    b. C. 0,6.10 Wb. D. 1,2.10 Wb. C t v tinh dng trong truyn thng so vi m o xc nh

    n kinh 85 20T.

    C

    C o eo trc O m, chu k m t = 0, vt

    u 15 t khu phng, d hnh c din t 0 cm u qu mt tr(thuc mt phng ca khung) trong t trng u c vect cm ng t vung gc vi trc quay v c ln 0,4 T. T thng cc i qua khung dy l

    A. 2,4.103 Wb. B. 4,8.103 W 3 3

    u 16: Gi s m ang ng yn t t mt catrong mt phng Xch o Tri t; ng thng ni v tinh vi tm Tri t i qua kinh s 0. Coi Tri t nh mt qu cu, bn knh l 6370 km, khi lng l 6.1024 kg v chu k quay quanh trc ca n l 24 gi; hng s hp dn G = 6,67.1011 N.m2/kg2. Sng cc ngn (f > 30 MHz) pht t v tinh truyn thng n cc im nm trn Xch o Tri t trong khong kinh no nu di y?

    A. T kinh 79o20 n kinh 79o20T. B. T kinh 85o20 o

    C. T kinh 81o20 n kinh 81o20T. D. T kinh 83o20T n kinh 83o20. u 17: Bit bn knh Bo l r0 = 5,3.1011 m. Bn knh qu o dng M trong nguyn t hir bng A. 132,5.1011 m. B. 84,8.1011 m. C. 47,7.1011 m. D. 21,2.1011 m. u 18: Mt vt nh da ng iu ha dc th x vi bin 5 c 2 s. Ti thi i

    i qua v tr cn bng O theo chiu dng. Phng trnh dao ng ca vt l

    A. x 5cos( t )= (cm). B. x = 5cos(t + 2 )

    2(cm).

    C. x = 5cos(2t + 2 ) (cm). D. x 5cos(2 t )

    2= (cm).

    Trang 2/6 - M thi 859

  • C ao ng iu ha to ra sng trn ng tm O t nc vi bc sn t nc, nm trn hai phng truyn sng m cc phn t nc ang dao

    l xo nh, c treo thng ng im O c nh. Khi l xo c chiu di t nhin th = MN = NI = 10 n vt nh vo i I ca l xo v kch

    g n sc: , vng, lam, tm l vng. . tm. lam.

    Ht prtn bay ra theo ph t . Cho

    Bit 1u = 931,5 MeV

    ao ng in t t do. in tch ca t in trong mch da nht v th hai q1 v q2 vi: 4 3.10 , q tnh thi im t, in

    iu mt pha vo hai u on mch A, B mc ni ti 69,1 , thun c t t in c i 76,8 F. B qua

    ng ng (khi lng tng i tnh) c i chuyn ng l tc rong chn khn

    p gm in tr R, t i ng C, cun c t cm L th Khi L = L1 v n p hiu dng

    cng gi tr;

    cc vt nh lc ang v ng thi tru g cc vn tc sao cho hai con

    u 19: Mt ngun pht sng d truyn trn mg . Hai im M v N thuc m

    ng. Bit OM = 8, ON = 12 v OM vung gc vi ON. Trn on MN, s im m phn t nc dao ng ngc pha vi dao ng ca ngun O l

    A. 6. B. 4. C. 5. D. 7. Cu 20: Gi M, N, I l cc im trn mt

    OM cm. G u dthch vt dao ng iu ha theo phng thng ng. Trong qu trnh dao ng, t s ln lc ko ln nht v ln lc ko nh nht tc dng ln O bng 3; l xo gin u; khong cch ln nht gia hai im M v N l 12 cm. Ly 2 = 10. Vt dao ng vi tn s l

    A. 3,5 Hz. B. 2,9 Hz. C. 1,7 Hz. D. 2,5 Hz. Cu 21: Trong chn khng, nh sng c bc sng ln nht trong s cc nh sn

    A. nh sng B. nh sng C. nh sng D. nh sngCu 22: Dng mt ht c ng nng 7,7 MeV bn vo ht nhn 147 N ang ng yn gy ra phn ng

    14 1 177 1 8N p O. + ng vung gc v ng bay ti ca h+ i ph

    khi lng cc ht nhn: m = 4,0015u; mp = 1,0073u; mN14 = 13,9992u; mO17 = 16,9947u. /c2. ng nng ca ht nhn 178O l

    A. 2,214 MeV. B. 6,145 MeV. C. 2,075 MeV. D. 1,345 MeV. Cu 23: Tia no sau y khng phi l tia phng x?

    A. Tia . B. Tia X. C. Tia +. D. Tia . Cu 24: Hai mch dao ng in t l tng ang c d

    o ng th ln lt l 2 21 2q q 1,+ = bng C. 17tch ca t in v cng dng in trong mch dao ng th nht ln lt l 109 C v 6 mA, cng dng in trong mch dao ng th hai c ln bng

    A. 6 mA. B. 4 mA. C. 10 mA. D. 8 mA. Cu 25: Ni hai cc ca mt my pht in xoay ch

    p gm in tr cun cm cm L v n dung 1in tr thun ca cc cun dy ca my pht. Bit rto my pht c hai cp cc. Khi rto quay u vi tc n1 = 1350 vng/pht hoc n2 = 1800 vng/pht th cng sut tiu th ca on mch AB l nh nhau. t cm L c gi tr gn gi tr no nht sau y?

    A. 0,6 H. B. 0,8 H. C. 0,2 H. D. 0,7 H. Cu 26: Mt ht c khi lng ngh m0. Theo thuyt tng i, khi l

    a ht ny kh vi tc 0,6c (c nh sng t g) l A. 1,75m0. B. 1,25m0. C. 0,36m0. D. 0,25m0.

    Cu 27: t in p 0u = U cos t (U0 v khng i) vo hai u on mch mc ni tin c in du m thun c ay i c. L = L2: i

    hai u cun cm c lch pha ca in p hai u on mch so vi cng dng in ln lt l 0,52 rad v 1,05 rad. Khi L = L0: in p hiu dng gia hai u cun cm t cc i; lch pha ca in p hai u on mch so vi cng dng in l . Gi tr ca gn gi tr no nht sau y?

    A. 1,57 rad. B. 0,26 rad. C. 0,83 rad. D. 0,41 rad. Cu 28: Hai con lc n c chiu di ln lt l 81 cm v 64 cm c treo trn mt cn phng. Khi ca hai con tr cn bng, yn cho chn cng hng

    lc dao ng iu ha vi cng bin gc, trong hai mt phng song song vi nhau. Gi t l khong thi gian ngn nht k t lc truyn vn tc n lc hai dy treo song song nhau. Gi tr t gn gi tr no nht sau y?

    A. 8,12 s. B. 2,36 s. C. 0,45 s. D. 7,20 s. Cu 29: Hai dao ng iu ha cng phng, cng tn s c bin ln lt l A1 = 8 cm, A2 = 15 cm v lch

    pha nhau 2

    A. 11 c . B. 17 cm. C. 7 cm.

    . D p ca hai c bin bao ng tng h dao ng ny ng m D. 23 cm.

    Trang 3/6 - M thi 859

  • C g mt th nghi thoa sng nc n sng kt hp O o ng cng pha, u 30: Tron m v giao , hai ngu 1 v O2 dacng bin . Chn h ta vung gc xOy (thuc mt nc) vi gc ta l v tr t ngun O1 cn ngun O2 nm trn trc Oy. Hai im P v Q nm trn Ox c OP = 4,5 cm v OQ = 8 cm. Dch chuyn ngun O2 trn trc Oy n v tr sao cho gc n2PO Q c gi tr ln nht th phn t nc ti P khng dao ng cn phn t nc ti Q dao ng vi bin c . Bit gia P v Q khng cn cc i no khc. Trn on OP, im gn P nht m cc phn t nc dao ng vi bin cc i cch P mt on l

    A. 3,4 cm. B. 2,0 cm. C. 1,1 cm.

    c i

    D. 2,5 cm. C phn ng ph ng sut 200 M ng ton b nn l phn ng ny

    61,6 g. C. 461,6 kg. D. 230,8 kg. C t si dy , hai u c sng dng v (k c hai u

    C. 0,5 m. D. 2 m.

    C n a hai ng v ph v ht v s ht

    u 31: Mt l n hch c c W. Cho r g lng msinh ra u do s phn hch ca 235 U v ng v ny ch b tiu hao bi qu trnh phn hch. Coi mi nm c 365 ngy; mi phn hch sinh ra 200 MeV; s A-v-ga-r NA = 6,02.1023 mol1. Khi lng 235 U m l phn ng tiu th trong 3 nm l

    A. 230,8 g. B. 4u 32: Trn m n hi di 1 m nh, ang c i 5 nt sng

    dy). Bc sng ca sng truyn trn dy l

    A. 1,5 m. B. 1 m.

    u 33: Hin nay urani t hin ch ng x 235 i t l sU 238 U, v 235 U238 U l 7

    1000. Bit chu k bn r ca 235 U v 238 U ln lt l .10 7,00 m Cc

    m, ur

    8 n v 4,50.109 nm. h y bao

    n ani t nhin c t l s h v t 238 U l nhiu t 235 U s h 3 ?

    A. 3,15 t nm. B. 1,74 t n C. m.

    100

    m. 2,22 t n D. 2,74 t nm. C khi l ng iu 2 s v c c th nng u 34: Mt vt nh ng 100 g dao ha vi chu k 0, nng l 0,18 J (mti v tr cn bng); ly 2 = 10. Ti li 3 2 cm, t s ng nng v th nng l

    A. 1. B. 3. C. 2. D. 4.

    C t in p u 35: u = 220 2co i u p gm s100t (V) vo ha on mch mc ni ti in tr 20 , cun m cm thun c t c 0,8 H

    v t in c in dung

    310 F

    . Khi in p tc thi gia hai in tr

    bng 6

    u

    110 3 V th in p i gia hai u cun cm c n l

    A.

    tc th l

    440 3 V. B. 330 3 V. C. 330 V. D. 440 V.

    Cu 36: Cc ng l ca c ng thi dn n t hir h bng biu thc mc n ng c tr g ca nguy c xc n

    n 2

    13,6En

    = (eV) (n = 1, 2, 3,...). Nu nguyn t hir hp th mt phtn c nng lng 2,55 eV th bc t ca bc x m nguyn t hir c th pht ra l

    A. 4,87.10sng nh nh

    8 m. D. 1,46.108 m. C t c tn n trong ch c sng l

    . 30 m. C t vt nh dao n a theo mt qu di 12 cm. Dao bin l

    C nhn c ht k n th c B. nng lng lin kt ring cng nh.

    ln.

    8 m. B. 9,74.108 m. C. 1,22.10u 37: Sng in s 10 MHz truy n khng vi b A. 3 m. B. 60 m. C. 6 m. Du 38: M g iu h o thng ng ny cA. 3 cm. B. 24 cm. C. 12 cm. D. 6 cm. u 39: Ht hi cng lA. nng lng lin kt cng nh. C. nng lng lin kt ring cng D. nng lng lin kt cng ln.

    Trang 4/6 - M thi 859

  • Cu 40: t in p u = 220 2cos100t (V) vo hai u on mch mc ni tip gm in tr R = 100 ,

    t in c C = 410 F

    2

    v cun cm thun c L = 1 H

    . Biu thc cng dng in trong on mch l

    A. i 2, 2cos(100 t )4

    = + (A). B. i 2, 2cos(100 t )4

    = (A).

    C. i 2, 2 2 cos(100 t )4

    = (A). D. i 2, 2 2 cos(100 t )4

    = + (A). II. PHN RING (10 cu) Th sinh ch c lm mt trong hai phn (Phn A hoc Phn B) A. Theo chng trnh Chun (10 cu, t cu 41 n cu 50) Cu 41: Mt con lc n c chiu di 121 cm, dao ng iu ha ti ni c gia tc trng trng g. Ly 2 = 10. Chu k dao ng ca con lc l

    A. 2 s. B. 2,2 s. C. 1 s. D. 0,5 s. Cu 42: Cho khi lng ca ht prtn, ntron v ht nhn teri ln lt l 1,0073u; 1,0087u v 2,0136u. Bit 1u = 931,5 MeV/c

    21D

    2. Nng lng lin kt ca ht nhn l 21DA. 3,06 MeV. B. 1,12 MeV. C. 2,24 MeV. D. 4,48 MeV.

    Cu 43: Mt mch LC l tng ang thc hin dao ng in t t do. Bit in tch cc i ca t in l q0 v cng dng in cc i trong mch l I0. Ti thi im cng dng in trong mch bng 0,5I0 th in tch ca t in c ln l

    A. 0q2

    . B. 0q 22

    . C. 0q 52

    . D. 0q 32

    .

    Cu 44: Thc hin th nghim Y-ng v giao thoa vi nh sng n sc c bc sng . Khong cch gia hai khe hp l 1 mm. Trn mn quan st, ti im M cch vn trung tm 4,2 mm c vn sng bc 5. Gi c nh cc iu kin khc, di chuyn dn mn quan st dc theo ng thng vung gc vi mt phng cha hai khe ra xa cho n khi vn giao thoa ti M chuyn thnh vn ti ln th hai th khong dch mn l 0,6 m. Bc sng bng

    A. 0,7 m. B. 0,6 m. C. 0,5 m. D. 0,4 m. Cu 45: Trong mt th nghim v giao thoa sng nc, hai ngun sng kt hp dao ng cng pha t ti hai im A v B cch nhau 16 cm. Sng truyn trn mt nc vi bc sng 3 cm. Trn on AB, s im m ti phn t nc dao ng vi bin cc i l

    A. 11. B. 10. C. 9. D. 12. Cu 46: Khi ni v quang ph vch pht x, pht biu no sau y l sai?

    A. Quang ph vch pht x ca cc nguyn t ha hc khc nhau th khc nhau. B. Trong quang ph vch pht x ca nguyn t hir, vng nh sng nhn thy c bn vch c trng l

    vch , vch lam, vch chm v vch tm. C. Quang ph vch pht x ca mt nguyn t l mt h thng nhng vch sng ring l, ngn cch nhau

    bi nhng khong ti. D. Quang ph vch pht x do cht rn hoc cht lng pht ra khi b nung nng.

    Cu 47: Mt vt nh dao ng iu ha vi bin 4 cm v chu k 2 s. Qung ng vt i c trong 4 s l A. 32 cm. B. 16 cm. C. 8 cm. D. 64 cm.

    Cu 48: t in p 0u U cos(100 t )

    12= (V) vo hai u on mch mc ni tip gm in tr, cun

    cm v t in th cng dng in qua mch l 0i I cos(100 t )

    12= + (A). H s cng sut ca on

    mch bng A. 1,00. B. 0,87. C. 0,50. D. 0,71.

    Cu 49: Gi s mt ngun sng ch pht ra nh sng n sc c tn s 7,5.1014 Hz. Cng sut pht x ca ngun l 10 W. S phtn m ngun sng pht ra trong mt giy xp x bng

    A. 2,01.1019. B. 2,01.1020. C. 0,33.1019. D. 0,33.1020.

    Trang 5/6 - M thi 859

  • Cu 50: t in p xoay chiu u = U 2cost (V) vo hai u mt in tr thu th cng dng in qua in tr c gi tr hiu dng bng 2 A. Gi tr ca U bng

    n R = 110

    A. 110 2 V. B. 220 V. C. 220 2 V. D. 110 V. B. Theo chng trnh Nng cao (10 cu, t cu 51 n cu 60) Cu 51: Mt bnh xe ang quay u quanh trc c nh vi ng nng l 225 J. Bit momen qun tnh ca bnh xe i vi trc l 2 kg.m2. Tc gc ca bnh xe l

    A. 15 rad/s. B. 56,5 rad/s. C. 112,5 rad/s. D. 30 rad/s. Cu 52: Mt vt rn quay quanh mt trc c nh vi tc gc 30 rad/s. Momen qun tnh ca vt rn i vi trc l 6 kg.m2. Momen ng lng ca vt rn i vi trc l

    A. 20 kg.m2/s. B. 180 kg.m2/s. C. 500 kg.m2/s. D. 27000 kg.m2/s. Cu 53: lectron l ht s cp thuc loi

    A. mzn. B. hipron. C. leptn. D. nucln. Cu 54: Ban u mt mu cht phng x nguyn cht c N0 ht nhn. Bit chu k bn r ca cht phng x ny l T. Sau thi gian 4T, k t thi im ban u, s ht nhn cha phn r ca mu cht phng x ny l

    A. 01 N .4

    B. 015 N .16

    C. 01 N .

    16 D. 0

    1 N .8

    Cu 55: t mt in p xoay chiu c gi tr hiu dng khng i v tn s f thay i c vo hai u mt cun cm thun. Khi f = 50 Hz th cng dng in qua cun cm c gi tr hiu dng bng 3 A. Khi f = 60 Hz th cng dng in qua cun cm c gi tr hiu dng bng

    A. 3,6 A. B. 2,0 A. C. 4,5 A. D. 2,5 A. Cu 56: Mch dao ng LC l tng ang hot ng, in tch cc i ca t in l q0 = 106 C v cng dng in cc i trong mch l I0 = 3 mA. Tnh t thi im in tch trn t l q0, khong thi gian ngn nht cng dng in trong mch c ln bng I0 l

    A. 1 ms.2

    B. 1 ms.6

    C. 10 ms.3

    D. 1 s.6

    Cu 57: Mt con lc l xo c khi lng vt nh l m1 = 300 g dao ng iu ha vi chu k 1 s. Nu thay vt nh c khi lng m1 bng vt nh c khi lng m2 th con lc dao ng vi chu k 0,5 s. Gi tr m2 bng

    A. 75 g. B. 100 g. C. 25 g. D. 150 g. Cu 58: Trn mt ng ray thng c mt ngun m S ng yn pht ra m vi tn s f v mt my thu M chuyn ng ra xa S vi tc u. Bit tc truyn m l v (v > u). Tn s ca m m my thu nhn c l

    A. fvv u . B.

    f (v u)v . C. fv

    v u+ . D. f (v u)

    v+ .

    Cu 59: Mt a trn, phng, ng cht c momen qun tnh 8 kg.m2 i vi trc c nh i qua tm a v vung gc vi b mt a. a quay quanh vi gia tc gc bng 3 rad/s2. Momen lc tc dng ln a i vi trc c ln l

    A. 24 N.m. B. 83

    N.m. C. 12 N.m. D. 38

    N.m.

    Cu 60: Hai qu cu nh khi lng ln lt l 2,4 kg v 0,6 kg gn hai u mt thanh cng v nh. Momen qun tnh ca h i vi trc quay i qua trung im ca thanh v vung gc vi thanh l 0,12 kg.m2. Chiu di ca thanh l

    A. 0,6 m. B. 0,3 m. C. 0,8 m. D. 0,4 m.

    ---------------------------------------------------------- HT ----------

    Trang 6/6 - M thi 859

  • B GIO DC V O TO

    THI CHNH THC

    P N

    THI TUYN SINH I HC NM 2013 Mn thi: VT L; Khi A v Khi A1

    (p n c 02 trang)

    M - p n Cu 318 426 528 681 794 859

    1 D A C A A D 2 D A D B A C 3 A B A C C D 4 C C D A A A 5 B C C D D D 6 A A D D C A 7 D C A B C D 8 D C D A A C 9 A A C A C B

    10 A A B B C D 11 D C D C A C 12 C D D A B C 13 C C B B B C 14 D D A C B A 15 B B B C B A 16 B C C C B C 17 C C C C D C 18 A B D B B A 19 A C D A D A 20 D C D C B D 21 A D B B B B 22 A A A A D C 23 C A A B C B 24 C B D D B D 25 D D C B C A 26 C D B D C B 27 A D B C A C 28 D C C A B C 29 A A A B A B 30 C B A B C B 31 B D C D C D 32 D D D B B C 33 B A C C D B 34 A C C D D A 35 C D D D C D 36 B C C A D B 37 A B B A A D 38 C B B D D D 39 B B B A D D 40 B D B D A A 41 C B C A D B 42 A D A C A C 43 C A B A C D

  • M - p n Cu 318 426 528 681 794 859

    44 A D A D B B 45 B C C B C A 46 C B A C D D 47 B B C D B A 48 C A B A D B 49 B B D D D A 50 D A B B A B 51 B B C C D A 52 D B C D B B 53 D D A D C C 54 B D A C C C 55 C B A B B D 56 A D B D A B 57 B B A C D A 58 A A A C B B 59 D C D B A A 60 D A D D A D

  • B GIO DC V O TO

    CHNH THC ( thi c 6 trang)

    THI TUYN SINH I HC NM 2013

    Mn: HA HC; Khi A Thi gian lm bi: 90 pht, khng k thi gian pht

    M thi 617

    H, tn th sinh:.......................................................................... S bo danh:............................................................................

    Cho bit nguyn t khi ca cc nguyn t: H = 1; C = 12; N = 14; O = 16; Na = 23; Mg = 24; Al = 27; P = 31; S = 32; Cl = 35,5; Ca = 40; Cr = 52; Fe = 56; Cu = 64; Zn = 65; Br = 80; Ag = 108; Ba = 137.

    I. PHN CHUNG CHO TT C TH SINH (40 cu, t cu 1 n cu 40) Cu 1: Khi c chiu sng, hirocacbon no sau y tham gia phn ng th vi clo theo t l mol 1 : 1, thu c ba dn xut monoclo l ng phn cu to ca nhau?

    A. isopentan. B. neopentan. C. butan. D. pentan. Cu 2: Khi lng Ag thu c khi cho 0,1 mol CH3CHO phn ng hon ton vi lng d dung dch AgNO3 trong NH3, un nng l

    A. 16,2 gam. B. 21,6 gam. C. 10,8 gam. D. 43,2 gam. Cu 3: Phenol phn ng c vi dung dch no sau y?

    A. NaHCO3. B. NaCl. C. HCl. D. KOH. Cu 4: Trong iu kin thch hp, xy ra cc phn ng sau:

    (a) 2H2SO4 + C 2SO 2 + CO2 + 2H2O. (b) H2SO4 + Fe(OH)2 FeSO 4 + 2H2O. (c) 4H2SO4 + 2FeO Fe 2(SO4)3 + SO2 + 4H2O. (d) 6H2SO4 + 2Fe Fe 2(SO4)3 + 3SO2 + 6H2O. Trong cc phn ng trn, phn ng xy ra vi dung dch H2SO4 long l A. (b). B. (a). C. (d). D. (c).

    Cu 5: Cho m gam Fe vo bnh cha dung dch gm H2SO4 v HNO3, thu c dung dch X v 1,12 lt kh NO. Thm tip dung dch H2SO4 d vo bnh thu c 0,448 lt kh NO v dung dch Y. Bit trong c hai trng hp NO l sn phm kh duy nht, o iu kin tiu chun. Dung dch Y ha tan va ht 2,08 gam Cu (khng to thnh sn phm kh ca N+5). Bit cc phn ng u xy ra hon ton. Gi tr ca m l

    A. 4,06. B. 2,40. C. 4,20. D. 3,92. Cu 6: Oxi ha hon ton 3,1 gam photpho trong kh oxi d. Cho ton b sn phm vo 200 ml dung dch NaOH 1M n khi phn ng xy ra hon ton, thu c dung dch X. Khi lng mui trong X l

    A. 12,0 gam. B. 14,2 gam. C. 11,1 gam. D. 16,4 gam. Cu 7: Cho bt Fe vo dung dch gm AgNO3 v Cu(NO3)2. Sau khi cc phn ng xy ra hon ton, thu c dung dch X gm hai mui v cht rn Y gm hai kim loi. Hai mui trong X v hai kim loi trong Y ln lt l:

    A. Fe(NO3)2; Fe(NO3)3 v Cu; Ag. B. Cu(NO3)2; Fe(NO3)2 v Cu; Fe. C. Cu(NO3)2; Fe(NO3)2 v Ag; Cu. D. Cu(NO3)2; AgNO3 v Cu; Ag.

    Cu 8: Tn thay th (theo IUPAC) ca (CH3)3CCH2CH(CH3)2 l A. 2,2,4,4-tetrametylbutan. B. 2,4,4-trimetylpentan. C. 2,2,4-trimetylpentan. D. 2,4,4,4-tetrametylbutan.

    Cu 9: Tin hnh in phn dung dch cha m gam hn hp CuSO4 v NaCl (hiu sut 100%, in cc tr, mng ngn xp), n khi nc bt u b in phn c hai in cc th ngng in phn, thu c dung dch X v 6,72 lt kh (ktc) anot. Dung dch X ha tan ti a 20,4 gam Al2O3. Gi tr ca m l

    A. 25,6. B. 50,4. C. 51,1. D. 23,5.

    Trang 1/6 - M thi 617

  • Cu 10: iu kin thch hp xy ra cc phn ng sau: (a) 2C + Ca CaC 2. (b) C + 2H2 CH 4. (c) C + CO2 2CO. (d) 3C + 4Al Al 4C3. Trong cc phn ng trn, tnh kh ca cacbon th hin phn ng A. (c). B. (b). C. (d). D. (a).

    Cu 11: trng thi c bn, cu hnh electron ca nguyn t Na (Z = 11) l A. 1s22s22p63s2. B. 1s22s22p53s2. C. 1s22s22p43s1. D. 1s22s22p63s1.

    Cu 12: Cho 1,37 gam Ba vo 1 lt dung dch CuSO4 0,01M. Sau khi cc phn ng xy ra hon ton, khi lng kt ta thu c l

    A. 2,33 gam. B. 1,71 gam. C. 3,31 gam. D. 0,98 gam. Cu 13: Hp cht X c thnh phn gm C, H, O, cha vng benzen. Cho 6,9 gam X vo 360 ml dung dch NaOH 0,5M (d 20% so vi lng cn phn ng) n phn ng hon ton, thu c dung dch Y. C cn Y thu c m gam cht rn khan. Mt khc, t chy hon ton 6,9 gam X cn va 7,84 lt O2 (ktc), thu c 15,4 gam CO2. Bit X c cng thc phn t trng vi cng thc n gin nht. Gi tr ca m l

    A. 12,3. B. 11,1. C. 11,4. D. 13,2. Cu 14: Dung dch no sau y lm phenolphtalein i mu?

    A. axit axetic. B. glyxin. C. alanin. D. metylamin. Cu 15: Trong mt bnh kn cha 0,35 mol C2H2; 0,65 mol H2 v mt t bt Ni. Nung nng bnh mt thi gian, thu c hn hp kh X c t khi so vi H2 bng 8. Sc X vo lng d dung dch AgNO3 trong NH3 n phn ng hon ton, thu c hn hp kh Y v 24 gam kt ta. Hn hp kh Y phn ng va vi bao nhiu mol Br2 trong dung dch?

    A. 0,20 mol. B. 0,25 mol. C. 0,10 mol. D. 0,15 mol. Cu 16: Cc cht trong dy no sau y u to kt ta khi cho tc dng vi dung dch AgNO3 trong NH3 d, un nng?

    A. vinylaxetilen, glucoz, imetylaxetilen. B. vinylaxetilen, glucoz, anehit axetic. C. glucoz, imetylaxetilen, anehit axetic. D. vinylaxetilen, glucoz, axit propionic.

    Cu 17: Cho s cc phn ng:

    X + NaOH (dung dch) Y + Z; Y + NaOH (rn) T + P; ot ot ,CaOT Q + Ho1500 C 2 ; Q + H2O Z. ot ,xtTrong s trn, X v Z ln lt l: A. CH3COOCH=CH2 v CH3CHO. B. CH3COOCH=CH2 v HCHO. C. HCOOCH=CH2 v HCHO. D. CH3COOC2H5 v CH3CHO.

    Cu 18: Ln men m gam glucoz to thnh ancol etylic (hiu sut phn ng bng 90%). Hp th hon ton lng kh CO2 sinh ra vo dung dch Ca(OH)2 d, thu c 15 gam kt ta. Gi tr ca m l

    A. 18,5. B. 7,5. C. 45,0. D. 15,0. Cu 19: Cho cc cn bng ha hc sau:

    (a) H2 (k) + I2 (k) 2HI (k). (b) 2NO 2 (k) N2O4 (k). (c) 3H2 (k) + N2 (k) 2NH 3 (k). (d) 2SO2 (k) + O2 (k) 2SO3 (k). nhit khng i, khi thay i p sut chung ca mi h cn bng, cn bng ha hc no

    trn khng b chuyn dch? A. (c). B. (b). C. (a). D. (d).

    Cu 20: Cht no sau y khi un nng vi dung dch NaOH thu c sn phm c anehit? A. CH2=CHCOOCH2CH3. B. CH3COOC(CH3)=CH2. C. CH3COOCH2CH=CH2. D. CH3COOCH=CHCH3.

    Cu 21: Cho X l hexapeptit AlaGlyAlaValGlyVal v Y l tetrapeptit GlyAlaGlyGlu. Thy phn hon ton m gam hn hp gm X v Y thu c 4 amino axit, trong c 30 gam glyxin v 28,48 gam alanin. Gi tr ca m l

    A. 77,6. B. 73,4. C. 83,2. D. 87,4. Trang 2/6 - M thi 617

  • Cu 22: Kim loi st tc dng vi dung dch no sau y to ra mui st(II)? A. HNO3 c, nng, d. B. MgSO4. C. CuSO4. D. H2SO4 c, nng, d.

    Cu 23: Hn hp X gm Na, Ba, Na2O v BaO. Ha tan hon ton 21,9 gam X vo nc, thu c 1,12 lt kh H2 (ktc) v dung dch Y, trong c 20,52 gam Ba(OH)2. Hp th hon ton 6,72 lt kh CO2 (ktc) vo Y, thu c m gam kt ta. Gi tr ca m l

    A. 21,92. B. 23,64. C. 39,40. D. 15,76. Cu 24: Hn hp X gm 3,92 gam Fe, 16 gam Fe2O3 v m gam Al. Nung X nhit cao trong iu kin khng c khng kh, thu c hn hp cht rn Y. Chia Y thnh hai phn bng nhau. Phn mt tc dng vi dung dch H2SO4 long (d), thu c 4a mol kh H2. Phn hai phn ng vi dung dch NaOH d, thu c a mol kh H2. Bit cc phn ng u xy ra hon ton. Gi tr ca m l

    A. 5,40. B. 7,02. C. 3,51. D. 4,05. Cu 25: Cho 100 ml dung dch amino axit X nng 0,4M tc dng va vi 80 ml dung dch NaOH 0,5M, thu c dung dch cha 5 gam mui. Cng thc ca X l

    A. NH2C3H5(COOH)2. B. (NH2)2C4H7COOH. C. NH2C3H6COOH. D. NH2C2H4COOH.

    Cu 26: Cho 0,1 mol tristearin ((C17H35COO)3C3H5) tc dng hon ton vi dung dch NaOH d, un nng, thu c m gam glixerol. Gi tr ca m l

    A. 27,6. B. 4,6. C. 9,2. D. 14,4. Cu 27: Tin hnh cc th nghim sau:

    (a) Sc kh etilen vo dung dch KMnO4 long. (b) Cho hi ancol etylic i qua bt CuO nung nng. (c) Sc kh etilen vo dung dch Br2 trong CCl4. (d) Cho dung dch glucoz vo dung dch AgNO3 trong NH3 d, un nng. (e) Cho Fe2O3 vo dung dch H2SO4 c, nng. Trong cc th nghim trn, s th nghim c xy ra phn ng oxi ha - kh l A. 5. B. 2. C. 4. D. 3.

    Cu 28: Dung dch axit axetic phn ng c vi tt c cc cht trong dy no sau y? A. Na, CuO, HCl. B. NaOH, Cu, NaCl. C. Na, NaCl, CuO. D. NaOH, Na, CaCO3.

    Cu 29: Bit X l axit cacboxylic n chc, Y l ancol no, c hai cht u mch h, c cng s nguyn t cacbon. t chy hon ton 0,4 mol hn hp gm X v Y (trong s mol ca X ln hn s mol ca Y) cn va 30,24 lt kh O2, thu c 26,88 lt kh CO2 v 19,8 gam H2O. Bit th tch cc kh o iu kin tiu chun. Khi lng ca Y trong 0,4 mol hn hp trn l

    A. 11,4 gam. B. 19,0 gam. C. 17,7 gam. D. 9,0 gam. Cu 30: Cht no sau y khng to kt ta khi cho vo dung dch AgNO3?

    A. HNO3. B. HCl. C. K3PO4. D. KBr. Cu 31: Thc hin cc th nghim sau:

    (a) Cho dung dch HCl vo dung dch Fe(NO3)2. (b) Cho FeS vo dung dch HCl. (c) Cho Si vo dung dch NaOH c. (d) Cho dung dch AgNO3 vo dung dch NaF. (e) Cho Si vo bnh cha kh F2. (f) Sc kh SO2 vo dung dch H2S. Trong cc th nghim trn, s th nghim c xy ra phn ng l A. 6. B. 5. C. 3. D. 4.

    Cu 32: Lin kt ha hc gia cc nguyn t trong phn t HCl thuc loi lin kt A. hiro. B. cng ha tr khng cc. C. cng ha tr c cc. D. ion.

    Cu 33: Ha tan hon ton 1,805 gam hn hp gm Fe v kim loi X bng dung dch HCl, thu c 1,064 lt kh H2. Mt khc, ha tan hon ton 1,805 gam hn hp trn bng dung dch HNO3 long (d), thu c 0,896 lt kh NO (sn phm kh duy nht). Bit cc th tch kh u o iu kin tiu chun. Kim loi X l

    A. Zn. B. Al. C. Cr. D. Mg.

    Trang 3/6 - M thi 617

  • Cu 34: Ha tan hon ton m gam Al bng dung dch HNO3 long, thu c 5,376 lt (ktc) hn hp kh X gm N2, N2O v dung dch cha 8m gam mui. T khi ca X so vi H2 bng 18. Gi tr ca m l

    A. 17,28. B. 21,60. C. 19,44. D. 18,90. Cu 35: Cho hn hp X gm 0,01 mol Al v a mol Fe vo dung dch AgNO3 n khi phn ng hon ton, thu c m gam cht rn Y v dung dch Z cha 3 cation kim loi. Cho Z phn ng vi dung dch NaOH d trong iu kin khng c khng kh, thu c 1,97 gam kt ta T. Nung T trong khng kh n khi lng khng i, thu c 1,6 gam cht rn ch cha mt cht duy nht. Gi tr ca m l

    A. 9,72. B. 3,24. C. 6,48. D. 8,64. Cu 36: ng vi cng thc phn t C4H10O c bao nhiu ancol l ng phn cu to ca nhau?

    A. 4. B. 3. C. 2. D. 5. Cu 37: Dy cc cht u tc dng c vi dung dch Ba(HCO3)2 l:

    A. HNO3, NaCl v Na2SO4. B. HNO3, Ca(OH)2 v KNO3. C. NaCl, Na2SO4 v Ca(OH)2. D. HNO3, Ca(OH)2 v Na2SO4.

    Cu 38: T nilon-6,6 l sn phm trng ngng ca A. axit aipic v etylen glicol. B. axit aipic v hexametyleniamin. C. etylen glicol v hexametyleniamin. D. axit aipic v glixerol.

    Cu 39: Hn hp X gm Ba v Al. Cho m gam X vo nc d, sau khi cc phn ng xy ra hon ton, thu c 8,96 lt kh H2 (ktc). Mt khc, ha tan hon ton m gam X bng dung dch NaOH, thu c 15,68 lt kh H2 (ktc). Gi tr ca m l

    A. 19,1. B. 24,5. C. 16,4. D. 29,9. Cu 40: Hn hp X cha ba axit cacboxylic u n chc, mch h, gm mt axit no v hai axit khng no u c mt lin kt i (C=C). Cho m gam X tc dng va vi 150 ml dung dch NaOH 2M, thu c 25,56 gam hn hp mui. t chy hon ton m gam X, hp th ton b sn phm chy bng dung dch NaOH d, khi lng dung dch tng thm 40,08 gam. Tng khi lng ca hai axit cacboxylic khng no trong m gam X l

    A. 18,96 gam. B. 9,96 gam. C. 12,06 gam. D. 15,36 gam.

    II. PHN RING (10 cu) Th sinh ch c lm mt trong hai phn (Phn A hoc Phn B)

    A. Theo chng trnh Chun (10 cu, t cu 41 n cu 50) Cu 41: Dy cc cht u c kh nng tham gia phn ng thy phn trong dung dch H2SO4 un nng l:

    A. glucoz, tinh bt v xenluloz. B. glucoz, saccaroz v fructoz. C. fructoz, saccaroz v tinh bt. D. saccaroz, tinh bt v xenluloz.

    Cu 42: Th nghim vi dung dch HNO3 thng sinh ra kh c NO2. hn ch kh NO2 thot ra t ng nghim, ngi ta nt ng nghim bng:

    (a) bng kh. (b) bng c tm nc. (c) bng c tm nc vi. (d) bng c tm gim n. Trong 4 bin php trn, bin php c hiu qu nht l A. (b). B. (d). C. (c). D. (a).

    Cu 43: t chy hon ton hn hp X gm 0,07 mol mt ancol a chc v 0,03 mol mt ancol khng no, c mt lin kt i, mch h, thu c 0,23 mol kh CO2 v m gam H2O. Gi tr ca m l

    A. 2,70. B. 5,40. C. 8,40. D. 2,34. Cu 44: Cho cc cp oxi ha - kh c sp xp theo th t tng dn tnh oxi ha ca cc ion kim loi: Al3+/Al; Fe2+/Fe; Sn2+/Sn; Cu2+/Cu. Tin hnh cc th nghim sau:

    (a) Cho st vo dung dch ng(II) sunfat. (b) Cho ng vo dung dch nhm sunfat. (c) Cho thic vo dung dch ng(II) sunfat. (d) Cho thic vo dung dch st(II) sunfat.

    Trang 4/6 - M thi 617

  • Trong cc th nghim trn, nhng th nghim c xy ra phn ng l: A. (b) v (c). B. (b) v (d). C. (a) v (c). D. (a) v (b).

    Cu 45: Trong cc dung dch: CH3CH2NH2, H2NCH2COOH, H2NCH2CH(NH2)COOH, HOOCCH2CH2CH(NH2)COOH, s dung dch lm xanh qu tm l

    A. 4. B. 3. C. 1. D. 2. Cu 46: Cho 25,5 gam hn hp X gm CuO v Al2O3 tan hon ton trong dung dch H2SO4 long, thu c dung dch cha 57,9 gam mui. Phn trm khi lng ca Al2O3 trong X l

    A. 80%. B. 60%. C. 20%. D. 40%. Cu 47: Cho X v Y l hai axit cacboxylic mch h, c cng s nguyn t cacbon, trong X n chc, Y hai chc. Chia hn hp gm X v Y thnh hai phn bng nhau. Phn mt tc dng ht vi Na, thu c 4,48 lt kh H2 (ktc). t chy hon ton phn hai, thu c 13,44 lt kh CO2 (ktc). Phn trm khi lng ca Y trong hn hp l

    A. 57,14%. B. 42,86 %. C. 28,57%. D. 85,71%. Cu 48: Hn hp X gm H2, C2H4 v C3H6 c t khi so vi H2 l 9,25. Cho 22,4 lt X (ktc) vo bnh kn c sn mt t bt Ni. un nng bnh mt thi gian, thu c hn hp kh Y c t khi so vi H2 bng 10. Tng s mol H2 phn ng l

    A. 0,075 mol. B. 0,070 mol. C. 0,050 mol. D. 0,015 mol. Cu 49: Cho cc pht biu sau:

    (a) Trong bng tun hon cc nguyn t ha hc, crom thuc chu k 4, nhm VIB. (b) Cc oxit ca crom u l oxit baz. (c) Trong cc hp cht, s oxi ha cao nht ca crom l +6. (d) Trong cc phn ng ha hc, hp cht crom(III) ch ng vai tr cht oxi ha. (e) Khi phn ng vi kh Cl2 d, crom to ra hp cht crom(III). Trong cc pht biu trn, nhng pht biu ng l: A. (a), (c) v (e). B. (b), (c) v (e). C. (a), (b) v (e). D. (b), (d) v (e).

    Cu 50: Cho phng trnh phn ng aAl + bHNO3 cAl(NO 3)3 + dNO + eH2O. T l a : b l A. 1 : 3. B. 1 : 4. C. 2 : 3. D. 2 : 5.

    B. Theo chng trnh Nng cao (10 cu, t cu 51 n cu 60) Cu 51: Trng hp no sau y khng xy ra phn ng?

    (a) CH2=CHCH2Cl + H2O ot

    (b) CH3CH2CH2Cl + H2O (c) C6H5Cl + NaOH (c)

    ot cao, pcao ; (vi C6H5 l gc phenyl)

    (d) C2H5Cl + NaOH ot

    A. (a). B. (b). C. (c). D. (d). Cu 52: Cho cc pht biu sau:

    (a) Glucoz c kh nng tham gia phn ng trng bc. (b) S chuyn ha tinh bt trong c th ngi c sinh ra mantoz. (c) Mantoz c kh nng tham gia phn ng trng bc. (d) Saccaroz c cu to t hai gc -glucoz v -fructoz. Trong cc pht biu trn, s pht biu ng l A. 1. B. 2. C. 4. D. 3.

    Cu 53: Cho 12 gam hp kim ca bc vo dung dch HNO3 long (d), un nng n phn ng hon ton, thu c dung dch c 8,5 gam AgNO3. Phn trm khi lng ca bc trong mu hp kim l

    A. 65%. B. 30%. C. 55%. D. 45%. Cu 54: Cho cc pht biu sau:

    (a) x l thy ngn ri vi, ngi ta c th dng bt lu hunh. (b) Khi thot vo kh quyn, freon ph hy tng ozon.

    Trang 5/6 - M thi 617

  • (c) Trong kh quyn, nng CO2 vt qu tiu chun cho php gy ra hiu ng nh knh. (d) Trong kh quyn, nng NO2 v SO2 vt qu tiu chun cho php gy ra hin tng ma axit. Trong cc pht biu trn, s pht biu ng l A. 1. B. 4. C. 3. D. 2.

    Cu 55: Trng hp no sau y, kim loi b n mn in ha hc? A. Thp cacbon trong khng kh m. B. t dy st trong kh oxi kh. C. Kim loi km trong dung dch HCl. D. Kim loi st trong dung dch HNO3 long.

    Cu 56: Cho phng trnh phn ng aFeSO4 + bK2Cr2O7 + cH2SO4 dFe 2(SO4)3 + eK2SO4 + fCr2(SO4)3 + gH2O. T l a : b l A. 6 : 1. B. 2 : 3. C. 1 : 6. D. 3 : 2.

    Cu 57: Cho s phn ng 2o+Cl , + NaOH,tCr X Y. d dungdch dCht Y trong s trn l A. Na2Cr2O7. B. Na[Cr(OH)4]. C. Cr(OH)3. D. Cr(OH)2.

    Cu 58: Peptit X b thy phn theo phng trnh phn ng X + 2H2O 2Y + Z (trong Y v Z l cc amino axit). Thy phn hon ton 4,06 gam X thu c m gam Z. t chy hon ton m

    gam Z

    cn va 1,68 lt kh O2 (ktc), thu c 2,64 gam CO2; 1,26 gam H2O v 224 ml kh N2 (ktc). Bit Z c cng thc phn t trng vi cng thc n gin nht. Tn gi ca Y l

    A. glyxin. B. alanin. C. axit glutamic. D. lysin. Cu 59: Hn hp X gm ancol metylic, ancol etylic v glixerol. t chy hon ton m gam X, thu c 15,68 lt kh CO2 (ktc) v 18 gam H2O. Mt khc, 80 gam X ha tan c ti a 29,4 gam Cu(OH)2. Phn trm khi lng ca ancol etylic trong X l

    A. 23%. B. 46%. C. 16%. D. 8%. Cu 60: Cho 13,6 gam mt cht hu c X (c thnh phn nguyn t C, H, O) tc dng va vi dung dch cha 0,6 mol AgNO3 trong NH3, un nng, thu c 43,2 gam Ag. Cng thc cu to ca X l

    A. CHCCH2CHO. B. CH3CCCHO. C. CHC[CH2]2CHO. D. CH2=C=CHCHO.

    ---------------------------------------------------------- HT ----------

    Trang 6/6 - M thi 617

  • B GIO DC V O TO

    THI CHNH THC

    P N

    THI TUYN SINH I HC NM 2013 Mn thi: HA HC; Khi A

    (p n c 02 trang)

    M - p n Cu 617 531 463 374 286 193

    1 D D B C B A 2 B C B C C D 3 D B C D D B 4 A B D A B B 5 A B D A B B 6 B D D B D B 7 C D D D B B 8 C C B B C B 9 C A D A A D

    10 A B C A C D 11 D C C B B A 12 C D C A B D 13 A C C B C D 14 D A A D C B 15 D A B D D C 16 B B A C A C 17 A D B A B C 18 D B D D A A 19 C B B B D C 20 D D A B A A 21 C B B C B B 22 C A C D A D 23 D B D C B C 24 B A C C C D 25 C A A C C A 26 C C D C B C 27 C C B B D A 28 D A C D A D 29 A D B A A A 30 A D C D C C 31 B A D B B D 32 C D A A C B 33 B C C A D D 34 B C D A D A 35 D D B A B C 36 A D B C C B 37 D C A B D B 38 B B A B D B 39 B A C D C C 40 C C A A B A 41 D A A D A D 42 C C A A D B 43 B D C B C B

  • M - p n Cu 617 531 463 374 286 193

    44 C C D B A C 45 D A B B C A 46 C B B B A D 47 B D A C A D 48 A C C C C D 49 A B A D D C 50 B B B C D D 51 B D C D A A 52 D B A C B C 53 D A D A D A 54 B A D C D C 55 A B A A B C 56 A C C D A A 57 B A B B A A 58 A D D D C A 59 A C D D A C 60 A A A C D B

  • B GIO DC V O TO

    CHNH THC ( c 6 trang)

    THI TUYN SINH I HC NM 2013

    Mn: TING ANH; Khi A1 Thi gian lm bi: 90 pht, khng k thi gian pht

    M thi 951

    H, tn th sinh:.......................................................................... S bo danh:............................................................................

    THI GM 80 CU (T QUESTION 1 N QUESTION 80) Mark the letter A, B, C, or D on your answer sheet to indicate the word(s) OPPOSITE in meaning to the underlined word(s) in each of the following questions. Question 1: We are now a 24/7 society where shops and services must be available all hours.

    A. an active society B. an inactive society C. a physical society D. a working society Question 2: We'd better speed up if we want to get there in time.

    A. lie down B. turn down C. slow down D. put down Question 3: Her thoughtless comments made him very angry.

    A. thoughtful B. honest C. pleasant D. kind Question 4: A chronic lack of sleep may make us irritable and reduces our motivation to work.

    A. uncomfortable B. miserable C. calm D. responsive Question 5: She is a very generous old woman. She has given most of her wealth to a charity organization.

    A. kind B. hospitable C. mean D. amicable Mark the letter A, B, C, or D on your answer sheet to indicate the word whose underlined part differs from the other three in pronunciation in each of the following questions. Question 6: A. reign B. vein C. reindeer D. protein Question 7: A. work B. form C. stork D. force Question 8: A. eleven B. elephant C. examine D. exact Question 9: A. overboard B. cupboard C. aboard D. keyboard Question 10: A. assure B. pressure C. possession D. assist Mark the letter A, B, C, or D on your answer sheet to indicate the sentence that is CLOSEST in meaning to the sentence given in each of the following questions. Question 11: How brave you are! he said to the firemen.

    A. He criticized the firemen for their discouragement. B. He asked how brave the firemen were. C. He blamed the firemen for their discouragement. D. He praised the firemen for their courage.

    Question 12: I could not get the job because I did not speak English well. A. I would have spoken English well if I could get that job. B. I wish I had got the job so that I could speak English well. C. Despite my poor English, I was successful in the job. D. I failed to get the job because of my poor English.

    Question 13: A small hotel was the only choice of place to stay at during my trip to London. A. I had different choices of where to stay during my trip to London. B. I had no alternative but to stay at a small hotel during my trip to London. C. There were a lot of hotels for me to choose from during my trip to London. D. I was talked into staying at a small hotel during my trip to London.

    Question 14: I would have worn the right shoes if I had known I was going to do all this climbing. A. I did not go climbing because I did not have the right shoes. B. As I did not know I was going to do so much climbing, I did not wear suitable shoes. C. I would have gone on the climb if I had been wearing the right shoes. D. I would love to go climbing, but I do not have any shoes that would be suitable.

    Trang 1/6 - M thi 951

  • Trang 2/6 - M thi 951

    Question 15: I wish I hadnt accepted the invitation to her birthday party. A. I insisted on having been invited to her birthday party. B. I regret accepting the invitation to her birthday party. C. If only I had come to her birthday party. D. I was very glad to accept the invitation to her birthday party.

    Question 16: Although the teacher explained the theory clearly, the students found it hard to understand it.

    A. Though explained clearly, the theory of teaching was difficult to the students. B. Despite the teachers clear explanation of the theory, the students had difficulty understanding it. C. In spite of explaining the theory clearly, the students themselves found it hard to understand it. D. Although the teaching theory was clear, it was a real challenge to the students.

    Question 17: Thanks to the efforts of environmentalists, people are becoming better aware of the problems of endangered species.

    A. Environmentalists are doing their best to make people aware of the problems of endangered species. B. People owe their growing awareness of the problems of endangered species to the efforts of environmentalists. C. People have no idea about the problems of endangered species in spite of the efforts of environmentalists. D. Environmentalists are expressing their gratitude towards people who are better aware of the problems of endangered species.

    Question 18: The man in that painting reminds me of my uncle. A. Whenever my uncle sees the man in that painting, he misses me. B. Whenever I see the man in that painting, I remember to meet my uncle. C. Memories of my uncle come back whenever I see the man in that painting. D. I am recalling my uncle whenever I look at the man in that painting.

    Question 19: No one but Jane succeeded in giving the correct answer. A. Jane was among those who failed to guess the answer. B. Only Jane failed to answer the question correctly. C. Everyone but Jane failed to provide the correct answer. D. All but Jane managed to produce the correct answer.

    Question 20: Peters main subject at university is electronics. A. Peter thinks electronics is a special subject. B. Peter majors in electronics at university. C. The university lets Peter major in electronics. D. Electronics is among the subjects that Peter likes.

    Mark the letter A, B, C, or D on your answer sheet to indicate the correct answer in each of the following questions. Question 21: It is ______ work of art that everyone wants to have a look at it.

    A. a so unusual B. such unusual a C. such an unusual D. so an unusual Question 22: What ______ if the earth stopped moving?

    A. happened B. would happen C. will happen D. happens Question 23: The children ran away as if they ______ a ghost.

    A. have seen B. had seen C. see D. would see Question 24: I would join that running competition ______.

    A. when I was 5 years older B. if it would happen 5 years ago C. if I were 5 years younger D. unless I were 5 years younger

    Question 25: I'll give this dictionary to ______ wants to have it. A. anyone B. everyone C. whatever D. whoever

    Question 26: He's sometimes bad-tempered but he's a good fellow ______. A. in heart B. with heart C. at heart D. by heart

    Question 27: The water supply of our home city has failed to ______ average purity requirements. A. meet B. hold C. see D. own

    Question 28: No one can avoid ______ by advertisements. A. influencing B. being influenced C. to be influenced D. having influenced

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    Question 29: Books and magazines ______ around made his room very untidy. A. lying B. which lied C. laying D. that lie

    Question 30: My brother tried to learn Japanese at a night school, ______ he gave up after 2 months. A. until B. therefore C. when D. but

    Question 31: Having traveled to different parts of our country, ______. A. many interesting lifestyles and customs have been learned by usB. we have learned a lot about interesting lifestyles and customsC. much has been learned about interesting lifestyles and customsD. we are seeing a lot of interesting lifestyles and customs

    Question 32: He wasn't attending the lecture properly and missed most of ______. A. what the teacher said B. which the teacher said C. things said by the teacher D. that the teacher said

    Question 33: Connecticut was the fifth of the original thirteen states ______ the Constitution of the United States.

    A. have ratified B. to ratify C. ratified D. ratify Question 34: You can use my car ______ you drive carefully.

    A. though B. as long as C. as though D. lest Question 35: Her mother, ______ has been working for thirty years, is retiring next month.

    A. that B. whose C. whom D. who Question 36: The Moon is much closer to Earth ______, and thus it had greater influence on the tides.

    A. but the Sun is B. where the Sun is C. than is the Sun D. unlike the Sun Question 37: A large number of workmen ______ because of the economic recession.

    A. has laid aside B. has been laid out C. have laid down D. have been laid off Question 38: We expressed ______ the missing child would be found alive.

    A. the hope that B. as we hoped C. the hope which D. the hope for Question 39: The government was finally ______ by a minor scandal.

    A. pulled down B. put back C. brought down D. taken down Question 40: His honesty is ______; nobody can doubt it.

    A. without question B. out the question C. beside the question D. in question Question 41: In our hospital, patients ______ every morning.

    A. are examined B. can examine C. have examined D. were examining Question 42: We decided to take a late flight ______ we could spend more time with our family.

    A. in order to B. so as to C. in order D. so that Question 43: Thanh: Lans the best singer in our school.

    Nadia: ______ A. Yes, please. B. Thats OK! C. I cant agree with you more! D. Yes, tell me about it!

    Question 44: Scarcely had he stepped out of the room ______ he heard a loud laughter within. A. until B. then C. when D. than

    Question 45: Mai: Do you want another serving of chicken soup? Scott: ______. A. No longer B. No way C. No comment D. No thanks

    Read the following passage and mark the letter A, B, C, or D on your answer sheet to indicate the correct answer to each of the questions from 46 to 55.

    Archimedes' Principle is a law of physics that states that when an object is totally or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced. The principle is most frequently applied to the behaviour of objects in water, and helps to explain floating and sinking, and why objects seem lighter in water. It also applies to balloons.

    The key word in the principle is upthrust, which refers to the force acting upward to reduce the apparent weight of the object when it is under water. If, for example, a metal block with a volume of 100 cm3 is dipped in water, it displaces an equal volume of water, which has a weight of approximately 1 N (3.5 oz). The block therefore seems to weigh about 1 N less.

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    An object will float if its average density is less than that of water. If it is totally submerged, the weight of the water it displaces (and hence the upthrust on it) is greater than its own weight, and it is forced upward and out of the water, until the weight of the water displaced by the submerged part is exactly equal to the weight of the floating object. Thus a block of wood with a density six tenths that of water will float with six tenths of its volume under water, since at that point the weight of fluid displaced is the same as the blocks own weight. If a dense material is made into a suitable shape, it will float because of Archimedes principle. A ship floats, whereas a block of iron of the same mass sinks.

    It is also because of Archimedes principle that ships float lower in the water when they are heavily loaded (more water must be displaced to give the necessary upthrust). In addition, they cannot be so heavily loaded if they are to sail in fresh water as they can if they are to sail in the sea, since fresh water is less dense than sea water, and so more water must be displaced to give the necessary upthrust. This means the ship is lower in the water, which can be dangerous in rough weather.

    From "Archimedes' Principle", Microsoft Student 2008 [DVD]. Microsoft Corporation, 2007.

    Question 46: What happens when something is immersed in a fluid? A. It receives a downward force, equal to the weight of the fluid displaced. B. The fluid will expand the object and overflow to the floor. C. It will be pushed further down with a force, equal to the weight of the fluid displaced. D. It receives an upward force, equal to the weight of the fluid displaced.

    Question 47: The word volume in the passage refers to ______. A. quantity B. frequency C. loudness D. length

    Question 48: The word displaces in the passage almost means ______. A. puts in position B. takes the place of C. takes place D. replaces with a new one

    Question 49: If an objects average density is less than that of water, the object will ______. A. float B. sink C. drift D. inflate

    Question 50: A block of wood with a density seven tenths that of water will ______. A. float with an equal volume of its volume under water B. float with a half of its volume under water C. go up and down then sink D. sink immediately when submerged

    Question 51: A ship floats, whereas a block of iron of the same mass sinks because the ship ______. A. has buoys B. is made of wood C. has a special shape D. is lighter

    Question 52: The phrase six tenths in the passage means ______. A. 10/6 B. 6/10 C. 6 and 10 D. 10 of 6

    Question 53: The word upthrust in the passage refers to the ______. A. upward push B. upside-down turn C. upper side of an object D. upturned force

    Question 54: Ships cannot be so heavily loaded if they want to sail in fresh water as they sail in the sea, because ______.

    A. fresh water is lighter than sea water B. fresh water is more polluted C. sea water is saltier than fresh water D. theres too much salt in sea water

    Question 55: Archimedes' Principle explains why ______. A. Archimedes became famous B. objects seem lighter in water C. humans can swim D. all objects will float

    Read the following passage and mark the letter A, B, C, or D on your answer sheet to choose the word or phrase that best fits each of the numbered blanks from 56 to 65.

    In a world where 2 billion people live in homes that don't have light bulbs, technology holds the key (56)______ banishing poverty. Even the simplest technologies can transform lives and save money. Vaccines, crops, computers and sources of solar energy can all reduce poverty in developing countries. For example, cheap oral-rehydration therapy developed in Bangladesh has dramatically cut the death (57)______ from childhood diarrhoea.

    But even when such technologies exist, the depressing fact is that we cant make them (58)______ for those who most need them. Solar panels, batteries and light bulbs are still beyond the purse of

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    many, but where they have been installed they change lives. A decent light in the evening gives children more time for homework and extends the productive day for adults.

    Kenya has a thriving solar industry and six years ago Kenyan pioneers also (59)______ connecting schools to the Internet via radio links. These people were fortunate (60)______ being able to afford solar panels, radios and old computers. How much bigger would the impact be if these things (61)______ and priced specifically for poor people?

    Multinationals must become part of the solution, because (62)______ they own around 60 per cent of the world's technology, they seldom make products for poor customers. Of 1,223 new drugs marketed worldwide from 1975 to 1996, for example, just 13 were for tropical diseases.

    People think those enterprises should do more to provide vital products such as medicines (63)______ different prices around the world to suit (64)______ people can afford. Alternatively, they could pay a percentage of their profit towards research and development for (65)______.

    Adapted from The Price is Wrong in Focus on IELTS Foundations by Sue OConnell, Pearson Longman, 2006

    Question 56: A. for B. with C. at D. to Question 57: A. amount B. penalty C. toll D. number Question 58: A. cheaply enough B. enough cheaply C. enough cheap D. cheap enough Question 59: A. were starting B. had started C. started D. have been starting Question 60: A. in B. at C. on D. by Question 61: A. are made B. made C. were made D. have been made Question 62: A. while B. however C. when D. unless Question 63: A. with B. to C. on D. at Question 64: A. what B. that C. where D. which Question 65: A. the poor B. the wealthy C. the better-off D. the rich Read the following passage and mark the letter A, B, C, or D on your answer sheet to indicate the correct answer to each of the questions from 66 to 75.

    In 1826, a Frenchman named Nipce needed pictures for his business. He was not a good artist, so he invented a very simple camera. He put it in a window of his house and took a picture of his yard. That was the first photograph.

    The next important date in the history of photography was 1837. That year, Daguerre, another Frenchman, took a picture of his studio. He used a new kind of camera and a different process. In his pictures, you could see everything clearly, even the smallest details. This kind of photograph was called a daguerreotype.

    Soon, other people began to use Daguerre's process. Travelers brought back daguerreotypes from all around the world. People photographed famous buildings, cities, and mountains.

    In about 1840, the process was improved. Then photographers could take pictures of people and moving things. The process was not simple and photographers had to carry lots of film and processing equipment. However, this did not stop photographers, especially in the United States. After 1840, daguerreotype artists were popular in most cities.

    Matthew Brady was one well-known American photographer. He took many portraits of famous people. The portraits were unusual because they were lifelike and full of personality. Brady was also the first person to take pictures of a war. His 1862 Civil War pictures showed dead soldiers and ruined cities. They made the war seem more real and more terrible.

    In the 1880s, new inventions began to change photography. Photographers could buy film ready-made in rolls, instead of having to make the film themselves. Also, they did not have to process the film immediately. They could bring it back to their studios and develop it later. They did not have to carry lots of equipment. And finally, the invention of the small handheld camera made photography less expensive.

    With a small camera, anyone could be a photographer. People began to use cameras just for fun. They took pictures of their families, friends, and favorite places. They called these pictures "snapshots".

    Documentary photographs became popular in newspapers in the 1890s. Soon magazines and books also used them. These pictures showed true events and people. They were much more real than drawings.

    Some people began to think of photography as a form of art. They thought that photography

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    could do more than show the real world. It could also show ideas and feelings, like other art forms. From Reading Power by Beatrice S. Mikulecky and Linda Jeffries

    Question 66: The first photograph was taken with ______. A. a daguerreotype B. new types of film C. a small handheld camera D. a very simple camera

    Question 67: Daguerre took a picture of his studio with ______. A. a very simple camera B. special equipment C. a new kind of camera D. an electronic camera

    Question 68: The word this in the passage refers to the ______. A. carrying of lots of film and processing equipment B. taking of pictures of people and moving things C. stopping of photographers from taking photos D. fact that daguerreotype artists were popular in most cities

    Question 69: The word ruined in the passage is closest in meaning to ______. A. terribly spoiled B. poorly-painted C. badly damaged D. heavily-polluted

    Question 70: The word lifelike in the passage is closest in meaning to ______. A. realistic B. manlike C. touching D. moving

    Question 71: The latest invention mentioned in the passage is the invention of ______. A. rolls of film B. handheld cameras C. daguerreotypes D. processing equipment

    Question 72: The word handheld in the passage is closest in meaning to ______. A. handling manually B. controlling hands C. operated by hand D. held by hand

    Question 73: Matthew Brady was well-known for ______. A. inventing daguerreotypes B. taking pictures of French cities C. the small handheld camera D. portraits and war photographs

    Question 74: As mentioned in the passage, photography can ______. A. show the underworld B. convey ideas and feelings C. replace drawings D. print old pictures

    Question 75: Which of the following could best serve as the title of the passage? A. Story of Famous Photographers B. Photography and Painting C. Different Steps in Film Processing D. Story of Photography

    Mark the letter A, B, C, or D on your answer sheet to indicate the underlined part that needs correction in each of the following questions. Question 76: In the early 1900's, Pennsylvania's industries grew rapidly, a growth sometimes

    A B accompanied by disputes labor. C D

    Question 77: Looking from afar, the village resembles a small green spot dotted with tiny fireballs. A B C D

    Question 78: Some people often say that using cars is not as convenient than using motorbikes. A B C D

    Question 79: It was not until the end of prehistoric times that the first wheeled vehicles appearing. A B C D

    Question 80: I like the fresh air and green trees of the village which I spent my vacation last year. A B C D

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    44 B D D B C C 45 A A D B D D 46 B C B D B D 47 B B C C A A 48 A B A C D B 49 B A B A D A 50 C A A D A A 51 D B A B C C 52 D A B C B B 53 A C C C B A 54 A C C A B A 55 C C D D A B 56 B C A C C D 57 C B C A B C 58 A B B D D D 59 A B C B B C 60 D C D B C A 61 B A C C A C 62 A D C D A A 63 A C A D A D 64 D C D A B A 65 B A D B B A 66 A C D D C D 67 B A D A A C 68 D B B C D A 69 B A C C A C 70 B C B A C A 71 D D D C B B 72 A B C A B D 73 C A D D D D 74 C D B B C B 75 C C C C D D 76 C C B D A D 77 C C C B B A 78 D D C A D C 79 D D D A B D 80 A A D C C C