2010 – Lý- ĐH - kA- đ1,Đề Thi Và Gợi Ý bài giải của mônLý, chiều nay 4.7_ M136 ạậớạếờề ạườ !"#$#%!&'Ả(#ướầ ÐỀ THI TUYỂN SINH ĐẠI HỌC KHỐI A NĂM 2010 Môn thi : VẬT LÝ – Mã đề 136 (Thời gian làm bài : 90 phút) &)*+,#-./0/12%34 5 6%789:;<,=.30/%34 3> &7,8 )?.5%34 @ A% I. PHẦN CHUNG CHO TẤT CẢ THÍ SINH (40 câu, từ câu 1 đến câu 40) Câu 1: '8BC,DEF 4 %=);*DG,08-HB;;I 89,840/J,8)?K L%3012 4 1 M%405/ 4 1 &%4012 4 1 N%40112 4 1 Câu 2:MIO0L0MP+<8QDRSTUVO%BOW8X IC?SD90DRUY?%'ZDR8?B L/4M0BM14M%'ZDR8?BI'H)BLM L%4M% M%5M% &%1/M% N%3M% Câu 3:):[?\))0=DE*+G] CD9C40/µ%$)^_=30)^VWSZ= *`1020\8\))3012%a,??,C) \)) L%13?% M%32?% &%3?% N%3>?% Câu 4:'8B)8DbX8^cC8d^µ!8Y:C: *aV34e*/4e%fU;π 1 .34%&g)8<HB;Ch L%V1%34 @ *50/%34 % M%V%34 @ *10%34 % &%V%34@*501%34% N%V1%34 @ *5%34 % Câu 5:$<=)b`iB)VZg-DEH;<QDE=) Z 1 350/ n E n = − J=KJ.301050jK%$<=));<Q;IV`iB)V .5`iB)V.1g;<QZ9ZTBCD9C + L%40524µ% M%40@/3µ% &%40/2/µ% N%40341µ% Câu 6:&)B?k0[lC,DGZL k 0L [ 0L l 9L k .1L [ .402L l % M*-DE<*HVB?DGZ∆m k 0∆m [ 0∆m l 9∆m l n∆m k n∆m [ %"] T*B?;=)Zd\_^c( L%[0k0l% M%[0l0k% &%k0[0l% N%l0k0[% Câu 7:!B? 134 @ Po Z;<gCTBα0;CTBC08-HBα L%9G8-HB?)% M%FCIoG)W+8-HB?)% &%+8-HB?)% N%oG8-HB?)% Câu 8:'8UI)8\9g%))^R]UV h<C8T.L*hT. 1 A− 0UIC,8g L% 5 % 1 A T M% / % A T &% % A T N% > % 1 A T Câu 9:BGC,pDR08)]G)8\9<8Cα 4 o%fU;,*-bh?+%$)];I8c=)\DG* hC8-+*-g8CαH)]+ L% 4 % 5 α − M% 4 % 1 α − &% 4 % 1 α N% 4 % 5 α Câu 10:q=)BGU8)B L%=% M%<) &%<r% N%% Câu 11:Q)BDEP L%Ig*Z<\W^s+)B% M%);*IY:0*:% &%IY^\W UV:% N%Ig;*t<)^s+)B% Câu 12: W)c8GUH8;*DbJo`)K8: T);\Ch:Yag::Y_c8ZUIb 344%u8ZU0*^9?;g::Y_cIbHC v0*-<?;g:C1v%w*-<5?;b8ZUg ::Y_cIbH8;+ L%344% M%144% &%114% N%334% Câu 13: ):[?\))0XXRZTBG ]0)CZTBoCD9C14ZTBYCD9CλJCh ))^V244*22K%<`0_?cU P9??C@?Y%xhHλ L%244 M%214 &%24 N%2/4 Câu 14: NP8C8-202'=])B? > M= Z;<%#^Z B)B?kBα%!Bα;=)DGC9DG9HC 8-'=%$8-HB0U;,DEB=)Gh, DE;<Q+,,H%w-DEo)^Z;+ L%50312'= M%0112'= &%3032'= N%10312'= Câu 15: '8B)8DbX8^cC8d^faY:C :&;aDE% \F:HY:*h& 3 gc,)8< HBy 3 % Ic,)8<HB 3 2y g^\F:HY: *h L%2& 3 M% 3 & 2 &% 3 2& N% 3 & 2 Câu 16:#CTB?BB? L%\CdUYG)t% M%\^ZB?-DE% &%\^^ZB?% N%\^ZB?o-DE% Câu 17: W:. 1 )U t ω )c)BBLMX)BBLwwM ],*% )BLwX*bz],*98^cC8d^f0)BwM FCY:9:&% W 3 3 1 LC ω = % I::Y_c)BBLw Y8zgc,Cω+ L% 3 % 1 ω M% 3 % 1 1 ω &%1ω 3 N% 3 1% ω % Câu 18:'8UC^-`9c,/%34 3 !r%$P CD9C)D9?;IgU;khôngI`{ L%4022µ% M%402µ% &%405@µ% N%404µ% Câu 19:'8E?;LM344-0cM,h0cL]98H? ))8\9c,4!r%<?;LMC8CVah0LDE) C%,8;\C<?;14A%$I^LM0<?;C L%2Y% M%51Y% &%>@Y% N%/Y% Câu 20:'8B)8:VDbC)8:Vd)%BRI.40 :<8^Y:dB%")^R]U∆g:<^Y ;+8QhdB%&g)8<HB)8; L%∆% M%/∆% &%5∆% N%31∆% Câu 21: W:T);\Ch:Yc,a)c)BB X*bz],*9Y:C:&%xp::Y_c:0 _c*b:,UH)BB*bChz 3 cDEv &3 0v z3 )ϕ 3 7*bChz 1 ghDGZC<v &1 0v z1 )ϕ 1 %M*v &3 . 1v &1 0v z1 .1v z3 %xhH)ϕ 3 )ϕ 1 ( L% 3 1 3 3 ) 0) 2 5 ϕ ϕ = = % M% 3 1 3 1 ) 0) 5 2 ϕ ϕ = = % &% 3 1 3 1 ) 0) 2 2 ϕ ϕ = = % N% 3 1 3 3 ) 0) 1 1 1 ϕ ϕ = = % Câu 22: ):[?\))0=DE*+]C D9CV5@4*/4%$)^_=40@0)^VWS Z=*`1%<0Bh??5C?H ZTB9D9C L%40@µ402/µ% M%404µ40/4µ% &%402µ40/4µ% N%404µ40/µ% Câu 23: W:T);\Ch:Y144c,a)cLM H)BB],*=)ZdX*bz08^cC8d^fY: C:&;a%xpwI,_8^cY:%&hz0f0&_ B%9&.& 3 g::Y_c*bzCha ;ahzH*b%9&. 3 1 C g::Y_Lw+ L% 144 1 % M%344% &%144% N% 344 1 % Câu 24:BRI0: 144 1 )J344 K 1 u t π π = − J)C+0+K Ch 344 1V ^%"RIC 3 544 s 0:;Ch L% 344 1 %V− M%−344% &% 344 5 %V N%144% Câu 25:k|B)8:VDb%&g)8<HBZU 3 0 HBZ 1 .1 3 %Mc:<}^Y:C89dB~ 4 %"C }Y:C:`8^HB%$:<}^YHB\C 89+`J4n`n~ 4 KgF,89DR8:)BZU89DR 8:)BZ L%1% M%% &% 3 1 % N% 3 % Câu 26:=)<\HM)0<=));<Q;IV`iB)f`iB) $g;<QCD9Cλ 13 0<=);IV`iB)'`iB)f g;<QCD9Cλ 51 <=);IV`iB)'`iB)$ g;<QCD9Cλ 53 %MIZThλ 53 ( L% 51 13 53 13 51 λ λ λ λ λ = − % M%λ 53 .λ 51 λ 13 % &%λ 53 .λ 51 •λ 13 % N% 51 13 53 51 13 λ λ λ λ λ = + % Câu 27: '8)BBLMX)BBL''M],*% )BBL'C :bc24Ω],*98^cC8d^ 3 π !0)BB'MFCY :9:;aDE% W:.v 4 )344πJK)c)BBLM% \ F:HY:*h& 3 )):c)BBLM: 1 π )9 :c)BBL'%xhH& 3 + L% 2 @%34 e − π M% 2 34 e − π &% 2 %34 e − π N% 2 1%34 e − π Câu 28:=)€;<QM)0`iB)$H<=));<Q 4 %$ <=);IV`iB)w\`iB)fg`iB)^9 L%31 4 % M% 4 % &%> 4 % N%3/ 4 % Câu 29: \:ICGW0))DE9C^TU VX)8 L%P<8C:,a=)R M%Pc,0PDG &%CPcP<8 N%Pc,0PDGC:,a=)R Câu 30:w,dH8;:T);\8)c)BBLMX :bcz],*98^c%Mo`:b8?;H;%$ )H;`;\9,8AgDR8::Y))BB 3L%$)H;`;\9,85AgDR8::Y) )BB 5 L%w*)H;`;\9,81Ag^H)B BLM L% 5 R % M% 5R % &% 1 5 R % N% 1 5R % Câu 31:uW)H8UoCXC*ELM140) 8=)DGSZ9DGg L .1)4π M .1)J4π•πKJ L M +0+K%M*,8;\C<WUo54A%k|g L'wM8W)Uo%",I)89<8dB<)BM' L%3>% M%3@% &%3% N%14% Câu 32: W:T);\Ch:Ya0c,24!r)c)BB ],*X:bcz08^cC8d^fY:C:&; aDE% \F:&*h 34 F π − )W 34 1 F π − gU<Y<)B B\Ch+%xhHf+ L% 3 % 5 H π M% 3 % 1 H π &% 5 %H π N% 1 %H π Câu 33:'8)]T)Xto,DE4041T)C8Z3wA%to DEW<•,h+p=)YT)%!:,DE_•t o403%Mc_tbhT)h|34X‚I)])8]c% fU;.34A 1 %,89UtoBDE)`g)8 L% 4 5 A% M% 14 / A% &% 34 54 A% N% 4 1 A% Câu 34:N)8aEH)8\PDG0Pc,CDGg 8 2 5)J K / x t π π = − JK%M*)8ZUCDGg8 3 2)J K / x t π π = + JK% N)8ZCDGg8 L% 1 @)J K / x t π π = + JK% M% 1 1)J K / x t π π = + JK% &% 1 2 1)J K / x t π π = − JK% N% 1 2 @)J K / x t π π = − JK% Câu 35:fd|)\Y<8UI)8\C89 L%D9a% M%F:989H8D9\h?+% &%F:9gDG<8% N%aDD9;a% Câu 36:~aBTB L%H;<,0bP8:8gD\8F,HB% M%8:,_BJBK<ƒ0-b_)^,% &%)U]0Uo)WUCU9hC% N%8^CVo*,\8<Y% Câu 37: W:.v 4 )ω)c)BBX:bcz08^cC 8d^fY:C:&],*%xpDR8:ZR) )BB7 3 0 1 5 cDE:ZR_c:b0_c8^ _cY:%!:Z L% 1 u i L ω = % M% 3 % u i R = &% 5 %i u C ω = N% 1 1 3 J K u i R L C ω ω = + − % Câu 38: '8t)8]cCBDE^<Y=)R L%<8-DE M%8,8 &%<8,8 N%<8, Câu 39:'8)BC)<=)01%34 3> 6%&*cDE))B;Z TBCD9Cλ 3 .403@µ7λ 1 .4013µ0λ 5 .4051µλ .4052µ%w_ZTBCI ?;:DE`:b)B;CD9C L%λ 3 0λ 1 λ 5 % M%λ3λ1% &%λ 1 0λ 5 λ % N%λ 5 λ % Câu 40:'8)]T))8\9g<82%M*)8g0 )^RItoH)]C89,DE`344A 1 5 T %fU; π 1 .34%c,)8Ht L%!r% M%5!r% &%3!r% N%1!r% PHẦN RIÊNG [10 câu] Th sinh ch đưc lm mt trong hai phn (phn A hoc B) A. Theo chương trWnh ChuXn (10 câu, t" câu 41 đ$n câu 50) Câu 41: )Rpd0p],*8`B:T);\9:bzX ]c)BB;):T);\Ch:Y5@4%M*`B;C hhZ(114@@„)B8UhZg8:_: bc`BDR8:`Cϕ09)ϕ.40@% I`B:;B; UhZgz+ L%52Ω M%5/3Ω &%1/Ω N%3@4Ω Câu 42: &),DEH7G)7 4 / 3@ 5 L7 f cDE(3044573044@75>0>2127 /04323.>5302'=A 1 %")9-DE<*<HB? / 5 f g-DE <*<HB? 4 3@ L L%9G8DE2014'= M%9G8DE501'= &%G8DE501'= N%oG8DE2014'= Câu 43: ):[?\))0=DE*+G] CD9Cλ%w*BI'<`C?,ZJV??K g:DRHV=" 3 0" 1 *'C89+ L%1λ M%302λ &%5λ N%102λ Câu 44: McCw 4 B?H8€UCTB;<UCg…%" )^R.4020IVRIc0,B?Dh?…H€UC TB; L% 4 w 1 M% 4 w 1 &% 4 w N%w 4 1 Câu 45: B8I<WUoC8X)89c,314!r0B)Ca h<WUo%k|2EX<*<8DG;\C0b\8)9 X0EZUEZ-402%,8;\C L%54A M%32A &%31A N%12A Câu 46: W:.v 4 )ω)c8^cC8d^fgDR8 :`8^ L% 4 v )J K f 1 π = ω + ω M% 4 v )J K 1 f 1 π = ω + ω &% 4 v )J K f 1 π = ω − ω N% 4 v )J K 1 f 1 π = ω − ω Câu 47: $*PQ)B)8,:dhy)=T<gU; h;Y% C:DE L%^TB M%`` &%)` N%] Câu 48: toH8)]T))8\)=)DG0,*-Bh ?+%$,HtC89+8Q89,dBgF,_8 -*-Ht L% 3 1 M%5 &%1 N% 3 5 Câu 49: '8)]GC\?;=)24toC,DE4043: `.•2%34 / &DE):I%&)])8\)):DR\ =GDR8:DRC89m.34 AD9SZT,D9%fU;.34 A 1 0π.503%&g)8\)H)] L%402@ M%30>> &%304 N%3032 Câu 50: )<B+C;*0DRQY*:<80Z )<8HC:V)cJpCK*<=)R9c, +c,H)8?c%&)c,C@44!r%$)8?cCc, 3444!rd:8)8)cg)8)cd:DE,)8) c L%@44 M%3444 &%/12 N%3/44 B. Theo chương trWnh Nâng cao (10 câu, t" câu 51 đ$n câu 60) Câu 51: '8C)=`,9Y`;,hHC40% 1 % I -,VBZ;<*,8Cω^,14446%Mo`%xhH ω L%34A M%144A &%344A N%24A Câu 52: IIZ:Z ,=0DR,<8DR;S8X? ;I8\9,854A0?9c,Th8;?Z;<% M*?;\)9,854A%$X?Bcg;)DEc ,?4!r%$X?Tg;)DEc,? L%/14!r M%@14!r &%@4!r N%2/4!r Câu 53: &PkV8,kJ,&GKCc,9U/0%34 3@ !r%Mo `8-<=)Za%!::*_H,k L%35012 M%2054 &%10/2 N%1/024 Câu 54: 'B)8PIpCH8;;*:XY:C: & 4 8^cC8d^f%';;DEC:VCD9C14% I DEC:VCD9C/40^]))9Y:& 4 HB)8 98Y:C: L%&.& 4 M%&.1& 4 &%&.@& 4 N%&.& 4 Câu 55: '8UI,DE0`;T`Y,h∆=)`iB))?O0 %Y∆`?OC9WS`iB)%BRI0UIC ,80,8C0,D9?8DEcDEν0ω0 %')=8 DEHUI,9d∆DEThb L%f. M%f.ω &%f. 1 N%f. Câu 56: '8B)8DbX8^cC8d^fY:C: &C)8:Vd)%uRI.40::*_^YChdB v 4 %#I)?;{ L%w-DEVDRdB)8^ 1 4 &v 1 M%&DR8:)BChdB 4 f v & &%!::*_^Y:+4cZUbRI f& 1 π = N%w-DEVDRHBbRI f& 1 π = 1 4 &v Câu 57: '88G:T);\)B8gDR9::Y114g UGp34„%M*8GC:,U40@2U)^:< ?;`U8G3„%Mo`)0DR8:dB`8G L%1L M% 5L &%3L N% 1 L Câu 58: );I8`;Ht]`8Y,h0)=`Ht, 9Y`; L%Y8,8CHt M%F:9,CHt &%Y8hHt,9Y`; N%F:)=dY)t Câu 59: '8t]`;\`Y,h∆9,8C54AghY H8)=…C89a<`;tc\VB1%M* )=Ht];,9Y∆34% 1 %')=…C89+ L%502w% M%504w% &%102w% N%104w% Câu 60: M*XhCTB 3 / & Cg…254-%x^Q8€}aC8C TB144?…A8€}P)B0P,DEH€}aC0U;V?; 9W0C8CTB3/44?…A%aH€}a…) L%3>34- M%1@/2- &%33/4- N%33>4- ****************@@@@@@@@@@**************** . 2010 – L - ĐH - kA- đ1,Đề Thi Và Gợi Ý bài giải của môn Lý, chiều nay 4. 7_ M136 ạậớạếờề ạườ. W:T);Ch:Y 144 c,a)cLM H)BB],*=)ZdX*bz08^cC8d^fY: C:&;a%xpwI,_8^cY:%&hz0f0&_ B%9&.& 3 g::Y_c*bzCha ;ahzH*b%9&. 3 1 C g::Y_Lw+ L% 144 1 % M% 344 % &% 144 % N% 344 1 % Câu 24: BRI0: 144 1 )J 344 K 1 u t π π = − J)C+0+K Ch 344 1V ^%"RIC 3 544 s 0:;Ch L% 344 . &%1/Ω N%3 @4 Ω Câu 42 : &),DEH 7 G) 7 4 / 3@ 5 L 7 f cDE(3 044 5 7 3 044 @ 7 5>0>212 7 / 043 23.>5302'=A 1 %")9 - DE<*<HB? / 5 f g - DE <*<HB? 4 3@ L L%9G8DE20 14 '=