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8/3/2019 Giai Chi Tiet de Ly Khoi a 2010
1/10
LUYN THI I HC - THY HIT: 01682 338 222
A CH MI KHU TP TH TRNGTHPT NGUYN HU - NG L DUN
TP VINH - NGH AN
GII THI I HC NM 2010-2011MN VT L
Thi gian lm bi: 90 pht;
(50 cu trc nghim)M thi 485
I. PHN CHUNG CHO TT C TH SINH (40 cu, tcu 1 n cu 40)
Cu 1: t in p xoay chiu c gi tr hiu dng khng i, tn s 50Hz vo hai u on mch mc ni tipgm in trthun R, cun cm thun c t cm L v tin c in dung C thay i c. iu chnh in
dung C n gi tr410
4F
hoc410
2F
th cng sut tiu th trn on mch u c gi tr bng nhau. Gi tr
ca L bng
A.1
.2
H
B.2
.H
C.1
.3
H
D.3
.H
Gii: Du hiu nhn dng bi ton C bin thin c P1 = P2 c quyn p dng:
=+
=+
== 3002
200400
221
0CC
CL
ZZZZ )(
3HL
= .
Cu 2: Mt mch dao ng in t l tng ang c dao ng in t t do. Tithi im t = 0, in tch trn mt bn tin cc i. Sau khong thi gian ngnnht t th in tch trn bn t ny bng mt na gi tr cc i. Chu k daong ring ca mch dao ng ny l
A. 6t. B. 12t. C. 3t. D. 4t.
Gii: t hnh v tTtT
t ==== 63
.2
3.
p n A.
Cu 3: Ti ni c gia tc trng trng g, mt con lc n dao ng iu ha vibin gc 0 nh. Ly mc th nng v tr cn bng. Khicon lc chuyn ng nhanh dn theo chiu dng n v tr cng nng bng th nng th li gc ca con lc bng
A. 0 .3
B. 0 .2
C. 0 .2
D. 0 .3
Gii: Trn vng trn LG c 4 im tng ng c Wd = WtC 2 im pha di ( M1, M2) chuyn ng theo chiu dng,v c 1 im M1 chuyn ng nhanh dn p n CCu 4:t in p u = 2cosU t vo hai u on mch ABgm hai on mch AN v NB mc ni tip. on AN gmbin trR mc ni tip vi cun cm thun c t cm L,
on NB ch c tin vi in dung C. t 11
2 LC = .
in p hiu dng gia hai u on mch AN khng ph thucR th tn s gc bng
A. 1 .2 2
B. 1 2. C.
1 .2
D. 21.
Gii: Ta c22
22
RL)(R
R.
CL
L
AN
ZZ
ZUZIU
+
+== URL khng ph thuc vo R th mu s c
22)( LCL ZZZ =
LCZZ 2= Hay L
C
2
1=
LC2
1= . Mt khc
LC2
11 = 12 = p n B.
Cu 5: Cho ba ht nhn X, Y v Z c s nucln tng ng l AX, AY, AZ vi AX = 2AY = 0,5AZ. Bit nnglng lin kt ca tng ht nhn tng ng l EX, EY, EZ vi EZ < EX < EY. Sp xp cc ht nhn ny
theo th t tnh bn vng gim dn l:
8/3/2019 Giai Chi Tiet de Ly Khoi a 2010
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A. X, Y, Z. B. Z, X, Y. C. Y, Z, X. D. Y, X, Z.
Gii: coi AY = 1, 4,2X == ZAA . Mt khc EZ < EX < EY ZXYA
E >>
=, p n D.
Cu 6: mt thong ca mt cht lng c hai ngun sng kt hp A v B cchnhau 20cm, dao ng theo phng thng ng vi phng trnh uA = 2cos40t vuB = 2cos(40t + ) (uA v uB tnh bng mm, t tnh bng s). Bit tc truynsng trn mt cht lng l 30 cm/s. Xt hnh vung AMNB thuc mt thong chtlng. Sim dao ng vi bin cc i trn on BM l
A. 19. B. 18. C. 20. D. 17.
Gii: Ta c cmf
v5,1
2030 === . Lp t s
=== 13,12,.....12,13333,135,1
20|| ctk
ABk
( kc =-13, -11, .... 11, 12)
Cu 7: Theo tin ca Bo, khi lectron trong nguyn t hir chuyn t quo L sang quo K th nguynt pht ra phtn c bc sng 21, khi lectron chuyn t quo M sang quo L th nguyn t pht raphtn c bc sng 32 v khi lectron chuyn t quo M sang quo K th nguyn t pht ra phtn cbc sng 31. Biu thc xc nh 31 l :
A. 31 = 32 - 21. B. 32 213132 21
=
+. C. 31 = 32 + 21. D. 32 2131
21 32
=
.
Gii: Theo ra ta c
=
=
)2(..........
)1(..........
3223
2112
hcEE
hcEE
Bt tm31
13
hcEE = (3). Ly (1) +(2)
213213
hchcEE += (4)
T (3) v (4) +=+=213231213231
111
hchchc 32 2131
32 21
=
+p n B.
Cu 8: lectron l ht scp thuc loiA. hipron B. nucln. C. mzn. D. leptn.
Gii: p n D. leptn.
Cu 9: Ti thi im t, in p 200 2 cos(100 )2
u t
= (trong u tnh bng V, t tnh bng s) c gi tr
100 2V v ang gim. Sau thi im 1
300s , in p ny c gi tr l
A. 100V. B. 100 3 .V C. 100 2 .V D. 200 V.
Gii: Ta c == 210021 0Uu c 2 v tr nm trn vng trn tho
mn3
= . Nhng v u ang gim nn chn im nm trn ( M1).
Gc qut3
100.300
1 === t . T vng trn lng gic ta suy ra v tr M2
cn tm nh hnh v v c hnh chiu ln trc u = -100 )(2 V p n C.Cu 10: Mt kim loi c cng thot lectron l 7,2.10-19J. Chiu ln lt vokim loi ny cc bc x c bc sng 1 = 0,18 m; 2 = 0,21 m, 3 = 0,32 m v 4 = 0,35 m. Nhng bcx c th gy ra hin tng quang in kim loi ny c bc sng l
A. 1, 2 v 3. B. 1 v 2. C. 3 v 4. D. 2, 3 v 4.
Gi s M d im dao ng cc i )12(202
)12(12 =+=
kdd
==+ kk )12(202
5,1)12( 5,022 M khng phi l im dao ng cc ai, nhng M nm gia
ng k =5 v k = 6.V tn sn dao n cc i trn on BM l( 5,4,3...-11, -12, -13). C 19 im. p n A.
8/3/2019 Giai Chi Tiet de Ly Khoi a 2010
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Gii: Ta c bc sng gii hn quang in )(10.276,010.2,7
10.3.10.625,6 619
834
0 mA
hc
=== .
Mt khc iu kin xy ra hin tng quang in ngoi l 0 ch c 21, tho mn p n B.Cu 11: Tia t ngoi c dng
A. tm khuyt tt bn trong sn phm bng kim loi.B. tm vt nt trn b mt sn phm bng kim loi.C. trong y t chp in, chiu in.D. chp nh b mt Tri t t v tinh.
Gii: Chn p n B. Tia t ngoi c dng tm vt nt trn b mt sn phm bng kim loi.Cu 12: Ni hai cc ca mt my pht in xoay chiu mt pha vo hai u on mch AB gm in trthunR mc ni tip vi cun cm thun. B qua in trcc cun dy ca my pht. Khi rto ca my quay u vitc n vng/pht th cng dng in hiu dng trong on mch l 1 A. Khi rto ca my quay u vi
tc 3n vng/pht th cng dng in hiu dng trong on mch l 3 A. Nu rto ca my quay uvi tc 2n vng/pht th cm khng ca on mch AB l
A. 2 3R . B.2
3
R. C. 3R . D.
3
R.
Gii: Ta c:
=+==
=+
==
3)(2
1)(2
23
2
03
2
23
21
2
01
1
11
LR
NZEI
LR
N
Z
EI
Mt khc === 13 32,
60
fnp
f )(
311 ==
RLZ
L .
V )(3
2222 121212 ====
RZZnn
LL . p n B.
Cu 13:t vo hai u cun scp ca mt my bin p l tng (b qua hao ph) mt in p xoay chiu cgi tr hiu dng khng i th in p hiu dng gia hai u cun th cp hl 100V. cun th cp, nugim bt n vng dy th in p hiu dng gia hai u hca n l U, nu tng thm n vng dy th in p l 2U. Nu tng thm 3n vng dy cun th cp th in p hiu dng gia hai u hca cun nybng
A. 100V. B. 200V. C. 220V. D. 110V.
Gii: Ta c: nN
U
U
N
nN
U
U
N
nN
UNN
3
2
100
2
11
2
11
2
11
2
=
=+
=
=
Vy khi s vng tng 3n vng th t s1
2
11
2
1
2
1
2 20023
U
U
UU
U
N
N
N
nN===
+
Cu 14: Mt cht im dao ng iu ha vi chu k T. Trong khong thi gian ngn
nht khi i t v tr bin c li x = A n v tr x =2
A, cht im c tc trung
bnh lA.
6.
A
TB.
9.
2
A
TC.
3.
2
A
TD.
4.
A
T
Gii: T vng trn lng gicT
A
t
SV
AAAS
Ttt
Tt
2
3
2
3
2
3
2
3
2
==
=+=
====
p n B.
Cu 15: Khi lectron quo dng th n th nng lng ca nguyn t hir c tnh theo cng thc
2
13,6nE
n= (eV) (n = 1, 2, 3,). Khi lectron trong nguyn t hir chuyn t quo dng n = 3 sang qu
o dng n = 2 th nguyn t hir pht ra phtn ng vi bc x c bc sng bng
U2 = 200(V) p n B.
8/3/2019 Giai Chi Tiet de Ly Khoi a 2010
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A. 0,4861 m. B. 0,4102 m. C. 0,4350 m. D. 0,6576m.
Gii: mE
hchcE
EEEE
656,0
10.6,1.6,13.5
10.3.10.625,6.36
5
36
36
5))
4(
9(
19
834
032
320
0023 ======
p n D.
Cu ny thc cht biu khng, khng cn gii cng nhp n D.Cu 16: Theo mu nguyn t Bo, bn knh quo K ca lectron trong nguyn t hir l r0. Khi lectronchuyn t quo N v quo L th bn knh quo gim bt
A. 12r0. B. 4 r0. C. 9 r0. D. 16 r0.Gii: Quo th L( ng n = 2 c bn knh r = 22r0 = 4r0 ), th N( ng n = 4 c bn knh r = 4
2r0= 16r0 ).
Vy khi (e) dch chuyn t qu do N v L th bn knh gim mt lng 16r0 - 4r0 = 12r0. p n A.Cu 17: Trong th nghim Y-ng v giao thoa nh sng, hai khe c chiu bng nh sng trng c bc sngt 380 nm n 760 nm. Khong cch gia hai khe l 0,8 mm, khong cch t mt phng cha hai khe n mnquan st l 2 m. Trn mn, ti v tr cch vn trung tm 3 mm c vn sng ca cc bc x vi bc sng
A. 0,48 m v 0,56 m. B. 0,40 m v 0,60 m. C. 0,40 m v 0,64 m. D. 0,45 m v 0,60 m.
Gii: 66 10.76,0.
10.38,0. ==
kD
xa
kD
xa
a
Dkx
Thay s .3,209,357,1 = kk Vy c 2 bc
sng tho mn. ng k =2 m 6,02 = . ng k = 3 m 4,02 = p n B.
Cu 18: Mt con lc l xo dao ng iu ha vi chu k T v bin 5 cm. Bit trong mt chu k, khong thi
gian vt nh ca con lc c ln gia tc khng vt qu 100 cm/s2 l3
T. Ly 2=10. Tn s dao ng ca
vt lA. 4 Hz. B. 3 Hz. C. 2 Hz. D. 1 Hz.
Gii: Theo ta suy 2/100|| scma v ng3
Tt
3
2= . V ta c th
biu din c qu trnh vt dh tho mn yu cu bi nh hnh v:
T hnh v suy c 22 /10021
scmAa ==
Thay A = 5cm ).(1)/(2 Hzfsrad == p n D.
Cu 19:t in p xoay chiu c gi tr hiu dng 200 V v tn s khng i vo hai u A v B ca on
mch mc ni tip theo th t gm bin trR, cun cm thun c t cm L v tin c in dung C thayi. Gi N l im ni gia cun cm thun v tin. Cc gi tr R, L, C hu hn v khc khng. Vi C = C1th in p hiu dng gia hai u bin trR c gi tr khng i v khc khng khi
thay i gi tr R ca bin tr. Vi C = 12
Cth in p hiu dng gia A v N bng
A. 200 V. B. 100 2 V. C. 100 V. D. 200 2 V.
Gii: V constZZ
IU
CL
=+
==22R )(R
U.RR
1
ng vi mi R( hu hn v khc khng ) th mu s c 1CL ZZ = .
KhiLCC ZZZ
CCC 22
2 121
2 ==== .
Vy khi )(200R)(
R.22
22
22
2
22
VUZR
ZU
ZZR
ZUZIU
L
L
CL
L
ANAN ==+
+=++== p n A.
Cu 20: Mt mch dao ng l tng gm cun cm thun c t cm 4 H v mt tin c in dung bini t 10 pF n 640 pF. Ly 2=10. Chu k dao ng ring ca mch ny c gi tr
A. t 2.10-8s n 3.10-7s. B. t 4.10-8s n 3,2.10-7s.C. t 2.10-8s n 3,6.10-7s. D. t 4.10-8s n 2,4.10-7s.
Gii: Ta c maxmin 222 LCcLCcLCc = Thay s c 4.10-8s 3,2.10-7s. p n B.
Cu 21: Mt si dy AB di 100 cm cng ngang, u B cnh, u A gn vi mt nhnh ca m thoa daong iu ha vi tn s 40 Hz. Trn dy AB c mt sng dng n nh, A c coi l nt sng. Tc truynsng trn dy l 20 m/s. K c A v B, trn dy c
A. 3 nt v 2 bng. B. 7 nt v 6 bng. C. 9 nt v 8 bng. D. 5 nt v 4 bng.
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Gii: Nhn dng: Xy ra TH1: 2 u l 2 nt ...)3,2,1(2
== kkl
. Thayf
v= vo biu thc trn ta suy ra
42000
40.100.22===
v
lfk . Vy s nt = k+1 = 5, s bng sng = k = 4. p n D.
Cu 22: Ba im O, A, B cng nm trn mt na ng thng xut pht t O. Ti O t mt ngun im phtsng m ng hng ra khng gian, mi trng khng hp th m. Mc cng m ti A l 60 dB, ti B l 20dB. Mc cng m ti trung im M ca on AB l
A. 26 dB. B. 17 dB. C. 34 dB. D. 40 dB.
Gii: Ta c 4060
1040lg10,10)(60lg10 ======B
A
B
ABAA
AA
II
IILLIIdB
IIL .
Mt khc ta li c 22
10=
=
A
B
A
B
B
A
d
d
d
d
I
I. V M l trung im AB nn
2
101
2ABA
M
dddd =
+=
Mt khc cng theo h thc22
2
101
=
=
M
A
A
M
d
d
I
IV LA LM = 10lg 06,34)
2
101lg(10lg10 2
2
==
=
A
B
B
A
d
d
I
I
)(2606,3460 dBLM = p n A.Cu 23:t in p u = U0cost vo hai u on mch gm in trthun R, cun cm thun c t cm Lv tin c in dung C mc ni tip. Gi i l cng dng in tc thi trong on mch; u1, u2 v u3 lnlt l in p tc thi gia hai u in tr, gia hai u cun cm v gia hai u tin. H thc ng l
A.2 21( )
ui
R LC
=+
. B. 3 .i u C= C.1 .
ui
R= D. 2
ui
L= .
Gii: d nhn thy i cng pha u1 = uR , v ng cho mi trng hp. p n C.
Cu 24: Dao ng tng hp ca hai dao ng iu ha cng phng, cng tn s c phng trnh li 5
3cos( )6
x t
= (cm). Bit dao ng th nht c phng trnh li 1 5cos( )6x t
= + (cm). Dao ng th
hai c phng trnh li l
A. 2 8cos( )6x t
= + (cm). B. 2 2cos( )6
x t
= + (cm).
C. 25
2cos( )6
x t
= (cm). D. 25
8cos( )6
x t
= (cm).
Gii: Ta bm my6
58A 1122
== AA p n D.
Cu 25: Phng x v phn hch ht nhnA. u c s hp th ntron chm. B. u l phn ng ht nhn thu nng lng.C. u l phn ng ht nhn ta nng lng. D. u khng phi l phn ng ht nhn.
Gii: Phng x v phn hch ht nhn u l phn ng ht nhn ta nng lng. p n C.Cu 26: t in p xoay chiu c gi tr hiu dng v tn s khng i vo hai u on mch gm bin trRmc ni tip vi tin c in dung C. Gi in p hiu dng gia hai u tu in, gia hai u bin trv hs cng sut ca on mch khi bin trc gi tr R1 ln lt l UC1, UR1 v cos1; khi bin trc gi tr R2 th
cc gi tr tng ng ni trn l UC2, UR2 v cos2. Bit UC1 = 2UC2, UR2 = 2UR1. Gi tr ca cos1 v cos2 l:
A. 1 21 2
cos ,cos3 5
= = . B. 1 21 1
cos ,cos5 3
= = .
C. 1 21 2
cos ,cos5 5
= = . D. 1 21 1
cos ,cos2 2 2
= = .
T h thc trn R12R21 , UUUU CC == vy khi 21 2 CC UU = R1R11 5,2 UUUUC ==
Vy5
2U
2
Ucos,
5
1U
cos R1R22R1
1 =====UUU
. p n C.
Gii: Ta c
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Cu 27: Trong th nghim Y-ng v giao thoa nh sng, hai khe c chiu bng nh sng n sc c bcsng 0,6 m. Khong cch gia hai khe l 1 mm, khong cch t mt phng cha hai khe n mn quan st l2,5m, b rng min giao thoa l 1,25cm. Tng s vn sng v vn ti c trong min giao thoa l
A. 19 vn. B. 17 vn. C. 15 vn. D. 21 vn.
Gii: Ta c mma
Di 57==
. Lp t s 166,4
5,1.2
5,12
2==
i
L
S vn sng l N = 2.4+1 = 9, S vn ti N = 2.4 = 8. Vy tng s vn sng v ti l 17 p n B.
Cu 28: Mt con lc l xo gm vt nh khi lng 0,02 kg v l xo c cng 1 N/m. Vt nhc t trn
gi cnh nm ngang dc theo trc l xo. H s ma st trt gia gi v vt nh l 0,1. Ban u gi vtv tr l xo b nn 10 cm ri bung nh con lc dao ng tt dn. Ly g = 10 m/s2. Tc ln nht vt nht c trong qu trnh dao ng l
A. 10 30 cm/s. B. 20 6 cm/s. C. 40 2 cm/s. D. 40 3 cm/s.Gii: V tr M m vt t c tc ln nht: )(02,0 mxmgkxFF msdh === p dng nh l bin thin cnng khi vt i t v tr ban u ( VT bin) n VT M:
)(.21
21
21
WWW )(222
max)(BienM xAmgSFkAkxmvA mzcmsc ==+==
)/(240)/(5
22)()(2 max
222max scmsmvxA
m
kxAgv ==+= p n C.
Cu 29: Xt hai mch dao ng in t l tng. Chu k dao ng ring ca mch th nht l T1, ca mch thhai l T2 = 2T1. Ban u in tch trn mi bn tin c ln cc i Q0. Sau mi tin phng in quacun cm ca mch. Khi in tch trn mi bn t ca hai mch u c ln bng q (0 < q < Q0) th t sln cng dng in trong mch th nht v ln cng dng in trong mch th hai l
A.1
4. B.
1
2. C. 4. D. 2.
Gii: T h thc c lp thi gian 1,12
02
2
2
0
2
01
1
2
0
=
+
=
+
I
i
Q
q
I
i
Q
q( lu cho Q0 = Q1= Q02, q1 = q2 = q.
T h thc trn ta suy ra2
02
2
2
01
1
=
I
i
I
ihay .2
1
2
02
01
02
01
2
1 ====T
T
Q
Q
I
I
i
i
p n D.
Cu 30: Lc ko v tc dng ln mt cht im dao ng iu ha c lnA. t l vi ln ca li v lun hng v v tr cn bng.B. t l vi bnh phng bin .C. khng i nhng hng thay i.D. v hng khng i.
Gii: Lc ko v ( Fhp = - kx) t l vi ln ca li v lun hng v v tr cn bng. p n A.Cu 31: Mt cht c kh nng pht ra nh sng pht quang vi tn s 6.1014 Hz. Khi dng nh sng c bcsng no di y kch thch th cht ny khng th pht quang?
A. 0,40 m. B. 0,45 m. C. 0,38 m. D. 0,55 m.
Gii: Theo thuyt lng t ta suy ra bc sng nh sng ca cht pht ra l: mf
c 614
8
0 10.5,010.6
10.3 ===
V iu kin kch thch cht ny khng pht quang l 0 >kt p n D.
Cu 32: Ht nhn 21084 Po ang ng yn th phng x, ngay sau phng x, ng nng ca ht A. ln hn ng nng ca ht nhn con.B. ch c th nh hn hoc bng ng nng ca ht nhn con.C. bng ng nng ca ht nhn con.D. nh hn ng nng ca ht nhn con.
Gii: T h thc dconddcon
d WW)(,W
W>>=
mm
m
mcon
con
con p n A.
Cu 33: Quang ph vch pht x
A. ca cc nguyn t khc nhau, cng mt nhit th nh nhau v sng ti ca cc vch.
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B. do cc cht rn, cht lng hoc cht kh c p sut ln pht ra khi b nung nng.C. l mt di c mu tn tm ni lin nhau mt cch lin tc.D. l mt h thng nhng vch sng (vch mu) ring l, ngn cch nhau bi nhng khong ti.
Gii: Quang ph vch pht x l l mt h thng nhng vch sng (vch mu) ring l, ngn cch nhau binhng khong ti. p n D.
Cu 34: Mt on mch AB gm hai on mch AM v MB mc ni tip. on mch AM c in trthun
50 mc ni tip vi cun cm thun c t cm1
H, on mch MB ch c t in vi in dung thay i
c. t in p u = U0cos100t (V) vo hai u on mch AB. iu chnh in dung ca t in n gi trC1 sao cho in p hai u on mch AB lch pha
2
so vi in p hai u on mch AM. Gi tr ca C1
bng A.54.10
F
B.
58.10F
C.
52.10F
D.
510F
Gii: Ta c ==+
=+
= CZ
ZRZUUU
L
LCCAB 125100
10050 2222maxRL
58.10F
p n B.
Cch 2: dng 1. =RLtgtg .Cu 35: Mt mch dao ng l tng gm cun cm thun c t cm L khng i v t in c in dung Cthay i c. iu chnh in dung ca t in n gi tr C1 th tn s dao ng ring ca mch l f1. tn
s dao ng ring ca mch l 15f th phi iu chnh in dung ca t in n gi trA. 1
C
5B. 1
C
5C. 5C1 D. 15C
Gii: Ta c5
52
1 12
2
1
2
1
2 CCC
C
f
f
LCf ===
=
p n A.
Cu 36: iu kin hai sng ckhi gp nhau, giao thoa c vi nhau l hai sngphi xutpht t hai ngundao ng
A. cng bin v c hiu s pha khng i theo thi gianB. cng tn s, cng phngC. c cng pha ban u v cng bin
D. cng tn s, cng phng v c hiu s pha khng i theo thi gianGii: iu kin 2 sng c th giao thoa vi nhau c l: cng tn s, cng phng v c hiu s pha khng itheo thi gian p n D.Cu 37: Mt ht c khi lng ngh m0. Theo thuyt tng i, ng nng ca ht ny khi chuyn ng vi tc 0,6c (c l tc nh sng trong chn khng) l
A. 0,36m0c2 B. 1,25 m0c
2 C. 0,225m0c2 D. 0,25m0c
2Gii: p dng cng thc:
20
202dd
202
20
d2
0 25,0)16,01
1(WW
1
W cmcmcm
c
v
cmcmE =
=+=
+= p n D.
Cu 38: Trong th nghim Y-ng v giao thoa nh sng, ngun sngpht ng thi hai bc x n sc, trong
bc x mu c bc sng 720 nm v bc x mu lc c bc sng (c gi tr trong khong t 500 mm n575 mm). Trn mn quan st, gia hai vn sng gn nhau nht v cng mu vi vn sng trung tm c 8 vnsng mu lc. Gi tr ca l
A. 500 nm B. 520 nm C. 540 nm D. 560 nm
Gii: Ta c to 2 vn sng trng nhau khi.k
l
a
Dl
a
Dkxx dluc
dlucdluc
=== SS
k = 9, l (nguyn dng), V 718,725,6575,05,0 = llmmluc
Vy nmmluc 56056,09
72,0.7=== p n D.
Cu 39: Mt vt dao ng tt dn c cc i lng gim lin tc theo thi gian lA. bin v gia tc B. li v tc
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C. bin v nng lng D. bin v tc Gii: Trong dao ng tt dn th c cc i lng cc i ( A, Vmax, Wtmax...) s gim dn theo thi gian p n C.Cu 40: Dng mt prtn c ng nng 5,45 MeV bn vo ht nhn 94 Be ang ng yn. Phn ng to ra ht
nhn X v ht . Ht bay ra theo phng vung gc vi phng ti ca prtn v c ng nng 4 MeV. Khitnh ng nng ca cc ht, ly khi lng cc ht tnh theo n v khi lng nguyn t bng s khi cachng. Nng lng ta ra trong ccphn ng ny bng
A. 4,225 MeV B. 1,145 MeV C. 2,125 MeV D. 3,125 MeVGii: NL to ra hay thu vo c tnh bi cng thc
dpddX
WWW +=
E .
Mt khc theo nh lut bo ton ng lng ta c:
ddPdX222 WWW mmmPPP pxpX +=+=
575,36
4.445,5WWW ddPdX =
+=
+=
X
p
m
mm MeV
Vy MeVE 125,245,54575,3WWW dpddX =+=+= p n C.
PHN RING [10 cu]Thsinh chc lm mt trong hai phn (phn A hoc B)A. Theo chng trnh Chun (10 cu, tcu 41 n cu 50)Cu 41: Trong th nghim Y-ng v giao thoa nh sng, hai khe c chiu bng nh sng n sc c bcsng . Nu ti im M trn mn quan st c vn ti th ba (tnh t vn sng trung tm) th hiu ng i canh sng t hai khe S1 , S2 n M c ln bng
A. 2,5 B. 3 C. 1,5 D. 2
Gii: M l vn ti2
)12(12
+== kdd V vn ti th 3 2= k
5,22
512 === dd p n A.
Cu 42: t in p u = U0cost vo hai u cun cm thun c t cm L th cng dng in qua cuncm l
A. 0U
i cos( t )L 2
= +
B. 0
Ui cos( t )
2L 2
= +
C. 0
Ui cos( t )
L 2
=
D. 0
Ui cos( t )
2L 2
=
Gii: Ta c bt i c dng: ))(cos(0 AtIi i += Vi 22,000
==== uLi
L L
U
Z
UI p n C.
Cu 43: Mt con lc n c chiu di dy treo 50 cm v vt nh c khi lng 0,01 kg mang in tch q =+5.10-6C c coi l in tch im. Con lc dao ng iu ho trong in trng u m vectcng intrng c ln E = 104V/m v hng thng ng xung di. Ly g = 10m/s2, = 3,14. Chu k dao ng iu ho ca con lc l
A. 0,58 s B. 1,40 s C. 1,15 s D. 1,99 s
Gii: V q > 0 nn Eqf = cng chiu E hng xung, v f suy ng hchy nhanh. M chy nhanh T < T suy ra g > g vy g = g+f/m
)(15,1
01,0
10.10.510
5,022'
46s
m
qEg
lT
+
=+
=
. p n C.
Cu 44: Khi chiu chm tia t ngoi vo mt ng nghim ng dung dch fluorexin th thy dung dch nypht ra nh sng mu lc. l hin tngA. tn sc nh sng B.phn x nh sng C. ho - pht quang D. quang - pht quang
Gii: quang - pht quang p n D.Cu 45: Vt nh ca mt con lc l xo dao ng iu ho theo phng ngang, mc th nng ti v tr cn bng.Khi gia tc ca vt c ln bng mt na ln gia tc cc i th t s gia ng nng v th nng ca vt l
A. 3 B.1
3C.
1
2D. 2
Gii: Ta c .31411W
W2
2
22
t
d ==
=
=x
A
x
A
p n A.
ppr
pr
Xpr
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Cu 46: Trong gi hc thc hnh, hc sinh mc ni tip mt qut in xoay chiu vi in trR ri mc haiu on mch ny vo in p xoay chiu c gi tr hiu dng 380V. Bit qut ny c cc gi tr nh mc :220V - 88W v khi hot ng ng cng sut nh mc th lch pha gia in p hai u qut v cng dng in qua n l , vi cos = 0,8. qut in ny chy ng cng sut nh mc th R bng
A. 180 B. 354 C. 361 D. 267Gii: Nhn xt ta c th coi qut ging nh cun dy c in trr 0 .
Khi ta c .35288
8,0.220coscos 2222
=====NB
NBNBquatNB
P
Ur
r
UPP
Mt khc
+) === ).(5,0.2 Ar
PIrIP NBNB === 7605,0
380I
UZ AB
AB.
+) =
== 264cos
cos1.
2
NB
NB
LL
NBrZ
r
Ztg
.
+) ++=++= 361264)352(760)( 222222 RRZrRZ LAB p n C.
Cu 47: Ban u c N0 ht nhn ca mt mu chtphng x nguyn cht c chu k bn r T. Sau khong thigian t = 0,5T, k t thi im ban u, s ht nhn chab phn r ca mu chtphng x ny l
A.0N
2 B.0N
4 C. N0 2 D.0N
2
Gii: p dng cng thc22200
/0
21
NNNN
Ttt=== p n A.
Cu 48: Trong thng tin lin lc bng sng v tuyn, ngi ta s dng cch bin iu bin , tc l lm chobin ca sng in t cao tn (gi l sng mang) bin thin theo thi gian vi tn s bng tn s ca daong m tn. Cho tn s sng mang l 800 kHz. Khi dao ng m tn c tn s 1000 Hz thc hin mt daong ton phn th dao ng cao tn thc hin c s dao ng ton phn l
A. 1600 B. 625 C. 800 D. 1000Gii: Tn s dao ng cng bc = tn s ngoi lc tc dng( tn s sng in t hay c gi l sng mang) p n C.Cu 49: Ti mt im trn mt cht lng c mt ngun dao ng vi tn s 120 Hz, to ra sng n nh trnmt cht lng. Xt 5 gn li lin tip trn mt phng truyn sng, v mtpha so vi ngun, gn th nhtcch gn th nm 0,5 m. Tc truyn sng l
A. 12 m/s B. 15 m/s C. 30 m/s D. 25 m/sGii: Nhn xt khong cch 5 gn li c chiu di 4 mm 125,05,0 == .Vy smfv /15120.125,0 === p n B.
Cu 50: Cho khi lng ca prtn; ntron; 40 618 3Ar ; Li ln lt l : 1,0073 u; 1,0087u; 39,9525u; 6,0145u v 1
u = 931,5 MeV/c2. So vi nng lng lin kt ring ca ht nhn 63 Li th nng lng lin kt ring ca ht nhn4018 Ar
A. nh hn mt lng l 3,42 MeV B. ln hn mt lng l 5,20 MeVC. ln hn mt lng l 3,42 MeV D. nh hn mt lng l 5,20 MeV
Gii: Ta c
=+
=
=+
=
+=
=
MeV
MeV
A
cmmPAmP
A
E
Li
Arhnnp
200,56
5,931).0145,60087,1.30073,1.3(
623,840
5,931).9525,390087,1.220073,1.18()).(.( 2
MeVLiAr 42,3== p n C.
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B. Theo chng trnh Nng cao (10 cu, tcu 51 n cu 60)Cu 51: kim chng hiu ng p-ple, ngi ta b tr trn mt ng ray thng mt ngun m chuynng u vi tc t 30 m/s,pht ra m vi tn s xc nh v mt my thu m ng yn. Bit m truyn trongkhng kh vi tc 340 m/s. Khi ngun m li gn th my thu o c tn s m l 740 Hz. Khi ngun m raxa th my thu o c tn s m l
A. 820 Hz B. 560 Hz C. 620 Hz D. 780 HzCu 52: Mt ng cin xoay chiu khi hot ngbnh thng vi in p hiu dng 220 V th sinh ra cngsut chc l 170 W. Bit ng cc h s cng sut 0,85 v cng sut to nhit trn dy qun ng cl 17
W. B qua cc haoph khc, cng dng in cc i qua ng clA. 2 A B. 1 A C. 2 A D. 3A
Cu 53: Mtbnh c momen qun tnh i vi trc quay cnh ca n l 0,4 kg.m2. bnh tng tc ttrng thi ng yn n tc gc phi tn cng 2000 J. B qua ma st. Gi tr ca l
A. 100 rad/s B. 50 rad/s C. 200 rad/s D. 10 rad/sCu 54: Mch dao ng dng chn sng ca mt my thu v tuyn in gm tin c in dung C0 vcun cm thun c t cm L. My ny thu c sng in t c bc sng 20 m. thu c sng in tc bc sng 60 m,phi mc song song vi tin C0 ca mch dao ng vi mt t in c in dung
A. C = 2C0 B. C = C0 C. C = 8C0 D. C = 4C0Cu 55: Chm tia Xpht ra t mt ng tia X (ng Cu-lt-gi) c tn s ln nht l 6,4.1018 Hz. B qua ngnng cc lectron khi bc ra khi catt. Hiu in th gia ant v catt ca ng tia X l
A. 2,65 kV B. 26,50 kV C. 5,30 kV D. 13,25 kVCu 56: Trong chuyn ng quay ca vt rn quanh mt trc c nh, momen qun tnh ca vt i vi trcquay
A. t l momen lc tc dng vo vt B. t l vi gia tc gc ca vtC.ph thuc tc gc ca vt D.ph thuc v tr ca vt i vi trc quay
Cu 57: Mt mch dao ng l tng gm cun cm thun c t cm L v t in c in dung C ang cdao ng in t t do. thi im t = 0, hiu in th gia haibn t c gi tr cc i l U0. Pht biu nosau y l sai ?
A. Nng lng t trng cc i trong cun cm l20CU
2
B. Nng lng t trng ca mch thi im t LC
2
= l
20CU
4
C. Hiu in th gia haibn t in bng 0 ln th nht thi im t LC2
=
D. Cng dng in trong mch c gi tr cc i l 0L
UC
Cu 58: Mt cht im khi lng m, quay xung quanh trc c nh theo qu o trong tm O,bn knh r.Trc qua tm O v vung gc vi mt phng qu o. Ti thi im t, cht im c tc di, tc gc, giatc hng tm v ng lng ln lt v, , an v p. Momen ng lng ca cht im i vi trc cxc nh bi
A. L = pr B. L = mvr2 C. L = man D. L = mrCu 59: Mt vt rn ang quay u quanh trc c nh vi tc gc 30 rad/s th chu tc dng ca mt
momen hm c ln khng i nn quay chm dn u v dng li sau 2pht. Bit momen ca vt rn nyi vi trc l 10 kg.m2. Momen hm c ln bng
A. 2,0 N.m B. 2,5 N.m C. 3,0 N.m D. 3,5 N.mCu 60: Bit ng v phng x 146 C c chu k bn r 5730 nm. Gi s mt mu g c c phng x 200 phnr/pht v mt mu g khc cng loi, cng khi lng ca mu g c , ly t cy mi cht, c phngx 1600 phn r/pht. Tui ca mu g c cho l
A. 17190 nm B. 2865 nm C. 11460 nm D. 1910 nm