Trac Nghiem Co Luu Chat

7/25/2019 Trac Nghiem Co Luu Chat

1/49

1

A - NGN HNG CU HI KIM TRA NH GI TRC NGHIM KHCH QUAN

Chng 1: Khi nim chung v cc tnh cht vt l c bn ca lu cht.

1. Ngn hng cu hi vp n chi tit chng 1

tt Cu hi v p n p n1 Cc nghin cu ca mn thu lc c thc hin cho:

a) Lu cht trong iu kin khng b nn.

b) Cht kh trong iu kin khng b nn.

c) Cht lng.

d) C 3 p n kia u ng.

D

2 Trong thu lc hc ngi ta p dng cc phng php nghin cu:

a) M hnh ho.

b) Dng cc i lng trung bnh.

c) Dng cc i lng v cng nh.

d) Cc p n kia u ng.

D

3 Cu no sau y sai:

a) Cht lng mang hnh dng bnh cha n

b) Cht lng b bin dng khi chu lc ko

c) Mun n hi th tch ca khng kh ln hn ca nc

d) H s nn ca khng kh ln hn ca nc

C

4 Trng lng ring ca cht lng l:

a) Trng lng ca mt n v khi lng cht lng.

b) Khi lng ca mt n v trng lng cht lng.

c) Trng lng ca mt n v th tch cht lng.

d) Khi lng ca mt n v th tch cht lng.

C

5 Khi lng ring ca cht lng l:

a) Khi lng ca mt n v th tch cht lng.

b) Khi lng ca mt n v khi lng cht lng.

c) Khi lng ca mt n v trng lng cht lng.

d) Trng lng ca mt n v th tch cht lng.

A

6 T trng () ca mt loi cht lng l:

a) T s gia trng lng ring v khi lng ring ca cht lng .

b) T s gia trng lng ring ca cht lng v trng lng ringca nc 40C

c) T s gia trng lng ring ca nc 40C v trng lng ring ca cht lng

d) Cha c p n chnh xc.

B

7 Mt loi du c t trng = 0,75 th khi lng ring bng:

a) 750 N/m3 b) 750 kg/m3 c) 750. 9,81 N/m3 d) 750. 9,81 kg/m3

B

8 M un n hi th tch E ca cht lng: D


2/49

2

x

z

u

D

C A

B

a) L nghch o ca h s nn. b) C tr s nh khi cht lng d nn.

c) C n v l N/m2 d) C 3 cu kia u ng

9 H s nn pca cht lng c tnh theo cng thc:

a)dpV

dV

0

1p b)

dp

1

V

dV

0

p c) dpdV

V

0

p

d)dp

1

dV

V

0

p

A

10 H s dn n Tca cht lng c tnh theo cng thc:

a)dTV

dV

0

1T

b)dT

1

V

dV

0

T

c) dTdV

V

0

T

d)dT

1

dV

V

0

T

B

11 H s nn ca mt cht lng th hin:

a) Tnh thay i th tch theo nhit ca cht lng.

b) Bin thin ca th tch tng i khi bin thin p sut bng 1.

c) Cng sinh ra khi bin thin tng i ca th tch bng 1.


B

12 Tnh dn n ca cht lng:

a) Tnh thay i th tch tng i ca cht lng.

b) Tnh thay i th tch ca cht lng khi nhit thay i.

c) c c trng bng h s nn p.


B

13 Hai tm phng AB v CD t song song v st nhau, gia l du bi trn. Tm CDc nh, tm AB chuyn ng vi vn tc u. Lc ma st gia hai tm phng c tnh

theo cng thcdydu.S.T vi y l phng:

a) Trng vi phng x, gc ta t trn tm CD

b) Trng vi phng x, gc ta t trn tm AB.

c) Theo chiu chuyn ng u.

d) Trng vi phng z.

A

14Trong cng thc

dy

duST , l:

a) H s nht ng lc ph thuc vo ch chy ca cht lng

b) H s nht ng lc vi th nguyn l Pa.s

c) H s nht ng hc ph thuc vo nhit ca loi cht lng


B


3/49

3

du/dy

1

2

3

15 Ghp cc ng cong di y cho ph hp vi loi cht lng:

a) 1: Cht lng Newton, 2: Cht lng l tng

b) 3: Cht lng l tng, 2: Cht lng phi Newton

c) 1: Cht lng phi Newton, 3: Cht lng l tng

d) 2: Chtlng phi Newton, 1: Cht lng Newton

C

16 Gi y l phng vung gc vi dng chy. Cht lng Newton l cht lng c:

a) H s nht ng lc khng ph thuc vo vn tc bin dng.

b) Quan h gia v du/dy l quan h tuyn tnh

c) C 3 p n kia u ng.

d) ng quan h v du/dy i qua gc ta

C

17 Cht lng l tng:

a) C nht bng 0. b) C tnh di ng tuyt i.

c) Hon ton khng nn c. d) C 3 p n kia u ng.

D

18 nh lut ma st trong ca Newton biu th mi quan h gia cc i lng sau:

a) ng sut php tuyn, vn tc, nhit .

b) ng sut tip tuyn, vn tc bin dng, nht.

c) ng sut tip tuyn, nhit , nht, p sut.

d) ng sut php tuyn, vn tc bin dng.

B

19 n v o nht ng lc l:

a) Poaz. b) N.s/m2 c) Pa.s. d) C 3 p n kia u ng.

D

20 n v o nht ng hc l:

a) m2 / s b) Pa.s c) N.s/m2 d) C 3 p n kia u sai.

A

21 Khi nhit tng:

a) nht ca cc cht th lng v th kh tng.

b) nht ca cc cht th lng v th kh gim.

c) nht ca cc cht th lng gim. d) nht ca cc cht th kh gim.

C

22 Khi p sut tng:a) nht ca cc cht th lng tng b) nht ca cc cht th lng gim

c) nht ca cc cht th lng v th kh tng

d) nht ca cc cht th lng v th kh gim

A

23 nht ng lc ca cht lng 1 l 1, cht lng 2 l 2. nht ng hc cacht lng 1 l 1, cht lng 2 l 2. Nu 1 > 2th:

a) 1 lun ln hn 2 b) 1lun nh hn 2

c) Khng ph thuc vo nhau d) Cn ph thuc vo loi cht lng

D


4/49

4

Chng 2: Tnh hc cht lng.1 Ngn hng cu hi vp n chi tit chng 2

tt Cu hi v p n p n1 Cc lc sau thuc loi lc khi :

a) Trng lc, lc ma st b) Lc ly tm, p lc

c) Ap lc d) Trng lc, lc qun tnh

D

2 Cc lc sau thuc loi lc b mt:

a) Trng lc b) Lc ly tm, p lc

c) Ap lc, lc ma st d) Trng lc, lc qun tnh

C

3 Cht lng l tng:

a) Mt gi thit cn thit khi nghin cu v ng hc cht lng

b) Mt gi thit hu ch trong bi ton thu tnh

c) Cht lng rt nht

d) Mtgithit cn thit khi nghin cu vng lc hc cht lng

D

4 i vi cht lng thc trng thi tnh:

a) ng sut tip t l tuyn tnh vi nhit

b) ng sut tip khng tn ti

c) nht bng khng

d) ng sut tip t l tuyn tnh vi trng lng cht lng

B

5 Mt at k thut bng:

a) 10 mH2O b) 736 mmHg c) 9,81.104Pa d) C 3 p n kia u ng

D

6 thit lp phng trnh vi phn cn bng ca cht lng tnh ngi ta xt:

a) Tc ng ca lc b mt ln mt vi phn th tch cht lng.

b) Tc ng ca lc khi ln mt vi phn th tch cht lng.

c) S cn bng ca lc b mt v lc khi tc ng ln mt vi phn th tch cht lng.

d) S cn bng ca lc b mt v lc khi tc ng ln mt th tch cht lng ln hu hn.

C

7 Phng trnh vi phn cn bng ca cht lng tnh tuyt i c th vit di dng sau:

a) dz = - dpb) C 3 cu kia u saic) dz = dp/ d) dp = - dz

B

8 Hai dng ca phng trnh c bn thu tnh l:

a) Dng 1: hpp o Dng 2: constg2

upz

2

b) Dng 1: constg2

upz

2

Dng 2: gzaxpp o

c) Dng 1: hpp o Dng 2: constp

z

d) Dng 1: hp Dng 2: constp

z

C


5/49

5

p1 p2

1

2

9 Gi p l p sut tc dng ln mt phng S ti im A:

a) p phi vung gc vi su h ca A.

b) p c gi tr khng i khi S quay quanh A.

c) p c gi tr thay i khi S quay quanh A.

d) C 3 p n kia u sai.

B

10p sut thy tnh ti mt im trong cht lng c tnh cht:a) Thng gc vi din tch chu lc. b) C n v l Pa.

c) L lc php tuyn ca cht lng tc dng ln mt n v din tch.

d) C 3 cu kia u ng.

D

11 Chn cu ng:

a) p sut thu tnh ti mt im theo cc phng khc nhau th khc nhau.

b) p sut thu tnh l i lng v hng.

c) p sut thu tnh l vc t nhng c tnh cht nh i lng v hng.

d) p sut thu tnh lun c gi tr khc khng.

C

12 p sut tuyt i ca cht lng:

a) Thng gc vi mt tc dng v nm ngang.

b) Thng gc v hng vo mt tc dng.

c) C tr s bng 0 ti mt tip xc vi kh tri.

d) Thng gc v hng theo phng thng ng.

B

13 Chn cu ng trong cc cu sau y:

a) p sut tuyt i c gi tr bng 1at ti im c p sut l p sut kh tri.b) p sut dti A c gi tr > 0, c ngha l p sut tuyt i ti A ln hn p suttuyt i ca kh tri.

c) Ap sut chn khng ti A c gi tr > 0, c ngha l p sut tuyt i ti A nh hnp sut tuyt i ca kh tri.


D

14 Hai bnh thng nhau cha hai loi cht lng. Mt thong ca haibnh c th ngang nhau khi:

a) p2< p1, 1 > 2. b) p2> p1, 1 > 2.

c) p1= p2, 1 < 2. d ) p1= p2, 1 > 2.

B

15 cao o p sut d ti mt im trong cht lng l hd= 15m ct nc. Ap sut dti im bng:

a) 1,5 at b) 14 at c) 1,3 at d) 2,5 at

A

16 Mt ng ch U cha cht lng n khong na ng. Khi xe chuyn ng v pha trcvi vn tc khng i, ta quan st thy:

a) Mc cht lng trong ng a dng cao hn

b) Mc cht lng trong ng b dng cao hn

c) Mc cht lng trong hai ng bng nhau

d) Cha xc nh c

C


6/49

6

B A

C D

v

DC

Z

a b

17 Mt bnh h cha nc chuyn ng ngang chm dn u vi gia tc a = -9,81m/s2. nghing ca mt thong (tg) bng:

a) 1/4 b) - c) - 1 d) 1

D

18 Trong bi ton tnh tng i, lc khi tc dng ln cht lng l:

a) Trng lc. b) Trng lc v lc qun tnh.

c) Trng lc v p lc. d) p lc v lc qun tnh.

B

19 Trong cc pht biu sau y, pht biu no sai:

a) p sut d l phn p sut ln hn p sut kh tri.

b) p sut tuyt i lun c gi tr dng.

c) p sut chn khng c th c gi tr m.

d) p sut chn khng lun l mt gi tr khng m.

D

20 Mt xe hnh hp ch nht kn nh hnh v cha y cht lng chuyn ng vi gia tcchmdn a = 9,81 m/s2. Mi quan h v p sut ti cc im gc xe l:

a) pA < pB< pC < pD.b) pB < pA< pC < pD.

c) pA > pB> pC > pD.

d) pB > pC> pA > pD.

B

21 Mt xe cha y xng nh hnh v:

a) p sut ti gc A s ln nht khi xe chuyn ng u

b) p sut ti gc B s nh nht khi xe chuyn ng chm dn

uc) p sut ti gc C s ln nht khi xe chuyn ng nhanh dn u

d) p sut ti gc D s ln nht khi xe chuyn ng nhanh dn u

C

22 Xe cha cht lng ln dc chm dn u vi gia tc chm dn u, so vi mt phngngang (ng nt lin) th mt thong cht lng (ng nt t) s nh hnh v:

Hnh 1 Hnh 2 Hnh 3

a) Hnh 1 b) Hnh 3 c) Hnh 2 d) Cha xc nh c

C

23 Mt ng ch U cha cht lng n khong na ng. Khi ng quay quanh trc thngng nh hnh v vi vn tc khng qu ln ( cht lng cha trn ra khi ng), taquan st thy:

a) Mc cht lng trong ng a dng cao hn

b) Mc cht lng trong ng b dng cao hn

c) Mc cht lng trong 2 ng khng i

d) Cha xc nh c nu khng tnh ton.

B


7/49


8/49

8

2

h2

A+

B +

h1

1 h3

3

a) nh lut Archimede

b) Lc tc dng ca cht lng lnthnh phng

c) S truyn nguyn vn p sut ti mi im trong lng cht lng tnh(pascal)

d) Lc nht ca Newton

33 n v o p sut chun l:

a) N/m2 b) at c) mH2O d) mmHg

A

34 Khi p sut kh quyn pa= 0,8at, p sut d pd= 3,8at th:

a) p sut tuyt i bng 4,8at b) p sut chn khng bng 2,8at

c) p sut tuyt i bng 46mH2O d) Cha c p n chnh xc

C

35 Hp lp phng kn cha y nc c t trong mt thang my chuyn ng. plc tc dng ln mt y sovi khi ng yn s thay i:

a) Tu thuc vo vn tc thang my b) Tng khi thang my i xung chm dn u

c) Gim khi thang my i xung chm dn u

d) Khng th xc nh c

B

36 Mt hnh tr trn khng np thng ng cao 1m, cha y chtlng. Cho bnh quayquanh trc ca n vi vn tc gc khng i sao cho sao cho th tch cht lng cn litrong bnh bng 2/3 th tch ban u. p sut ti mt im A trn thnh bnh so vilc bnh ng yn s:

a) Tng b) Khng i c) Gim d) Tu thuc v tr ca im A

B

37 Phng trnh p = po + h ng cho:

a) Ch trng hp cht lng tnh tuyt i.

b) Ch trng hp cht lng tnh tng ic) C cht lng tnh tuyt i v cht lng tnh tng i

d) Mi trng hp cht lng chuyn ng

C

38 Gia bnh A (cha cht lng c 1) v bnh B (cha cht lng c 3) l p k ch U(cha cht lng c 2) Hiu p sut gia hai im A v B c tnh theocng thc:

a) pA- pB= - h1 1+ h2 2 - h3 3

b) pA- pB= h1 1- h2 2 + h3 3

c) pA- pB= h2 2 + h3 3 - h1 1

d) pA- pB= - h1 1- h2 2 + h3 3

C

39Khi du c t trng = 0,8 quay vi vn tc gc 81,9 1/s. p sut trn mt

thong p = pa. im nm di mt thong 0,2 m s c p sut d bng:

a) 0,02 at b) Khng th xc nh c v khng bit bn knh R

c) 0,02 m ct du d) 0,16 m ct nc

D


9/49

9

aB

vL

0,5L

40 Xe hnh hp ch nht, di L ;cao 0,5L, cha y cht lng c trng lng ring .Gia np ca xe c mt l nh. Khi xe chuyn ng nhanh dnu vi gia tc a = 9,81m/s2, p sut d ti im B (gc trncng pha sau xe) bng:

a) 0,5 L b) 0

c) L d) - 0,5 L

A

41 So snh p lc thy tnh P tc dng ln y ca 3 b cha cht lng (b 1: nc, b 2:thy ngn, b 3: xng), c din tch y S v chiu cao ct cht lng H nh nhau. Tac:

H

S S S

P1 P2 P3

papa

pa

1 2 3

a) P3> P1> P2 b) P1= P2 = P3 c) C 3 cu kia u sai. d) P3< P1< P2

D

42 im t ca p lc thu tnh tc dng ln thnh bn phng ca b cha nc:

a) Lun trng vi trng tm ca thnh phng

b) Lun nm di trng tm ca thnh phng

c) Ph thuc vo hng t lc ln thnh phng

d) Lun nm trn trng tm ca thnh phng

B

43 Thnh phn nm ngang ca p lc tc dng ln mt cong l:

a) Trng lng khi cht lng nm trn b mt cong

b) Tch s p sut ti trng tm vi din tch b mt o

c) p lc tc dng ln hnh chiu ca b mt y ln mt phng nm ngang

d) p lc tc dng ln hnh chiu ca b mt y ln mt phng thng ng

D

44 Khi tnh p lc tc dng ln thnh cong, thnh phn tc dng theo phng ngang Py=pdcy.Sx vi pdcy l p sut d ti:

a) Trng tm ca thnh cong

b) im chiu ca trng tm ca thnh cong ln trc 0x

c) im chiu ca trng tm ca thnh cong ln mt phng vung gc vi trc 0x

d) Trng tm ca hnh chiu ca thnh cong ln mt phng vung gc vi trc 0y

D

45 Trong cng thc tnh p lc thy tnh tc dng ln thnh phng P = .hc.S, hcl:

a) Khong cch thng ng t mt ngn cch vi cht kh n trng tm b mt

b) Khong cch thng ng t mt mt chun n trng tm b mt

c) Khong cch thng ng t mt thong t do c p sut pan trng tm b mt

d) Khong cch thng ng t mt thong t do c p sut pan im t lc

C

46

Trong phng trnh Sz

J

zzC

C

CD , trc z l :

a) Trc thng ng hng t di ln

D


10/49

10

0H1

H2

b) Trc thng ng hng t trn xung

c) Mt trc bt k nm trong mt phng cha din tch chu lc

d) Giao tuyn ca mt phng thng ng vi mt phng cha din tch chu lc vhng t trn xung

47 Hai din tch phng hnh trn v hnh vung cng nm trong mt cht lng c trngtm ngang nhau v c din tch bng nhau. Ap lc cht lng tc dng ln hai din tch

phng c quan h nh sau:a) Ptrn= Pvung b) Ptrn< Pvung c) Ptrn> Pvung

d) Cha xc nh c v ph thuc vo hng t ca hai thnh phng

A

48 Thnh phn thng ng ca p lc tc dng ln mt cong bng:

a) Vi thnh phn nm ngang

b) p lc tc dng ln hnh chiu thng ng ca b mt

c) Tch tr s p sut ti trng tm vi din tch ca b mt

d) Trng lng khi cht lng nm trong vt th p lc

D

49 Mt ng b tng hnh tr trn ngn i b nc di L . Mc nc hai bn l H1, H2.Phn lc theo phng ngang Pxca p lc nc tc dng ln ng b tng l:

a) L)HH(4

3P 22

2

1x b) )HH(L4

3P 21

2

2x

c) L)HH(2

1P 22

2

1x d) )HH(L2

1P 22

2

1x

D

50 Khi xc nh vt th p lc tnh p lc ln thnh cong theo phng z, mt phng

chiu thnh cong ln l:a) Bt buc phi l mt thong c p sut l p sut kh quyn

b) Mt nm ngang c) Mt mt ng p no d) Mt nm nghing

A

51 Khi xc nh chiu dy ca thnh ng dn c kch thc ln v chu p sut cao, ngita c:

a) Vn dng phng trnh Bernoulli xt lc tc dng ln thnh ng

b) Xt n ng sut ko cho php ca vt liu lm ng

c) Vn dng phng trnh ng lng xt lc tc ng ti khuu

d) Khng c p n chnh xc

B

52 Lc y Archimede tc dng ln mt vt chm trong cht lng:

a) t ti trng tm ca khi cht lng m vt chim ch

b) Bng trng lng ca khi cht lng b vt chim ch

c) t ti trng tm ca vt khi vt ng cht

d) Cc p n kia u ng

D

53 Chn cu sai trong cc cu sau y.

Lc y Archimede tc dng ln mt vt ngp trong cht lng:a) t ti trng tm ca vt khi vt ng cht

b) Lun lun t ti trng tm ca vt

B


11/49

11

CD

V1 A

V2

V3

B

R

c) C gi tr bng trng lng ca vt khi vt v tr cn bng

d) C gi tr nh hn trng lng ca vt khi vt chm xung y bnh

54 Mt vt ng cht ni trong nc nh hnh v, ta c:

a) T trng ca vt < 1 b) T trng ca vt >1

c) T trng ca vt = 1 d) Cha xc nh c

A

55 Mt vt cn bng trong nc nh hnh v; C l trng tm ca vt; D l tm y:

a) Vt trng thi cn bng phim nh

b) Vt trng thi cn bng khng n nh

c) Vt trng thi cn bng n nh d) Cha xc nh c

B

56 Mt vt gm 2 phn A v B chm trong cht lng. Phn A c khi lng ring nh hnphn B. vt c cn bng n nh ta nn t:

a) Phn B nm di b) Phn A nm dic) Ph thuc vo th tch ca 2 phn A,B d) Khng th xc nh c

A

57 Vt th p lc cho mt cong AB l :

a) V3 b) V2

c) V1 d) V2 + V3

C

58 p lc theo phng thng ng (PZ) tc dng ln ng hnh tr trn c bn knh R v

chiu di L, mt bn ngp trong nc nh hnh v c tnh theo cng thc sau:

a) LR2

1 2 b) R2

1 2

c) LR2

1 2 d) R

4

3 2

A

59 Vt chm trong cht lng trng thi cn bng n nh khi:

a) Trng tm C nm cao hn tm y D b) Trng tm C nm ngang vi tm y D

c) Trng tm C nm thp hn tm y D d) Ty theo trng lng vt

C

60 Khi mt chic tu i t bin vo sng th:

a) Chic tu s hi ni ln so vi lc i ngoi bin.

b) Chic tu s hi chm xung so vi lc i ngoi bin.

c) Hi chm hay ni hn so vi lc i ngoi bin ph thuc vo tu lm bngg haybng st.

d) Khng thay i so vi lc i ngoi bin.

B


12/49


13/49

13

A

h

h=?

A

h

A

hv

L

c) pdA = 0,2at; pA = 12m H20; pdA = 147,2 mm Hg

d) pA = 1,2 at; pdA = 0,2 at; pckA = - 2,2 at

68 Mt xe cha du (t trng l 0,8) chuyn ng vi gia tc khng i nh hnh bn.im A nm su h = 0,6m so vi mt thong c:

a) p sut d l 0,06 at

b) p sut tuyt i l 1,6 at

c) p sut d l 0,048 at

d) p sut tuyt i l 1,48 at

C

69 Hp lp phng kn c cc cnh bng 2 m mt na cha nc v mt na cha duc t trng 0,75 c t trong mt thang my chuyn ng thng ng ln trn vi

gia tc nhanh dn a = 5,19 m/s2. Chnh lch gia p sut tc dng ln y v nhca hnh hp (KPa) l:

a) 12,88 b) 11 c) 26,25 d) 34,29

C

70 Mt bnh h hnh tr cha cht lng (c t trng 1,3) quay trn u quanh trc Z vi

vn tc gc . Mc Glycerin ln ti mp bnh. p sut d ti im A gia y bnh oc l 0,4at. Chiu cao h ca ct Glycerin nm trn im A bng:

a) 3,08m

b) 3,56m

c) 5,2m

d) 3,67m

A

71 Mt bnh cha du (t trng l 0,75) quay quanh trc thng ng vi vn tc gckhng i nh hnh bn. p sut tuyt i ti im A cch mt thong c p sut khtri mt khong h = 0,8m bng:

a) 1,08 at

b) 0,06at

c) 1,03 at

d) 1,06 at

D

72 Cht lng thy tnh tuyt i c t trng = 0,8. Mt thong c p sut chn khngpcko= 0,5at ; im c p sut d pd= 0,7at su :a) 10 m b) 15 m c) 12 m d) 6,4 m

B

73 ng ch U t trn xe chuyn ng chm dn u, ngi ta o c L = 15 cm, chnh cht lng trong hai nhnh ng h = 20cm. Gia tc ca xe c gi tr bng (m/s2):

a) 7,36

b) 13,08

c) 14,72

d) 6,54

B


14/49

14

R

h

pa

2m

4m60o

pa A

B

C

L

r

A

3m

A

74 Toa xe ch du c t trng 0,8 chuyn ng vi vn tc v = 36 km/htheo ng vng vi bn knh cong R = 300m. Gc nghing ca mtdu hp vi phng ngang (tg) bng:

a) 0,028 b) 0,034 c) 0,072 d) 0,068

B

75 Mt bnh h hnh tr trn c bn knh R = 2m, chiu cao H = 4m cha nc n 1/3

th tch bnh. Ngi ta quay bnh quanh trc thng ng vi vn tc gc ti a maxsao cho nc khng trn ra bn ngoi, khi lc tc dng vo y bnh l:

a) 128,42 kN b) 164,28 kN c) 146,78 kN d) 246,42 kN

B

76 Mt bnh lp phng h c cc cnh bng 0,8 m cha nc n 1/2 th tch cabnh. Ngi ta t bnh ln 1 chic xe chuyn ng nhanh dn u trn mt phngngang vi gia tc ti a sao cho nc khng trn ra ngoi, khi p lc tc dng lny bnh bng (N):

a) 2842 b) 2511 c) 5684 d) 5302

B

77

Mt ci vm bn cu kn hon ton nh hnh v c xy dng di y h nc su quan st. Cho R = 8m v h = 40m. Lc thu tnh PZ tc dng ln vm l:

a) 68342 kN

b) 98057 kN

c) 78342 kN

d) 88057 kN

A

78 Ca van ABC chn nc c kch thc nh hnh v. Van rng 2m. Thnh phn p lcnm ngang tc dng ln van ABC:

a) 298,42 kNb) 420,55 kN

c) 480,69 kN

d) 333,54 kN

D

79 Mt bnh hnh tr kn cha y xng c X= 0,7; bn knh r = 0,2m; di L = 1,5m;t nm ngang nh hnh v. Bit p sut d ti im A bng 1,4at. Lc ca xng tcdng ln np tri ca bnhl:

a) 17,08 kN

b) 15,85 kNc) 14,91 kN

d) 16,85 kN

A

40 Mt cnh ca hnh ch nht cao 3m rng 1m, t thng ng, ng b nc c ncva ngp n cnh trn, mt thong thng vi kh tri. M men i vi im A ycnh ca (Nm) l:

a) 4,5 b) 13,5 c) 18 d) 27

A

41 Thnh ca mt b cha xng c t trng x = 0,7 thng vi kh tri c chiu cao 3m,rng 5m, di 5mcha y xng. p lc P ca khi xng tc dng ln y b l:

a) 1030,05 kN b) 1545,075 kN c) 515,025 kN d) 735,75 kN

C


15/49

15

A

h

C

B

R

HPx

H

pa

R

A

B

0

pd

d

ba

D

42 Mt my p thy lc piston nh c ng knh d = 5cm; piston ln c ng knh D =25cm. B qua ca trng lc v lc ma st. nhn c lc tc dng ln piston ln l20kN, ta phi tc dng ln piston nh mt lc l:

a) 900 N b) 800 N c) 4000 N d) 1250 N

B

43 Mt mu g hnh lp phng c cc cnh bng 0,5m c th xung nc, khilng ring ca g l 200kg/m3. Th tch phn g chm di nc (m3):

a) 0,05 b) 0,025 c) 0,125 d) 0,075

B

44 B cha cht lng su h = 9m c mt ca thng ng hnh ch nht AC gm 2 tmphng chng ln nhau theo chiu cao. Mun cc tm chu p lc nh nhau th chiucao tm AB phi bng:

a) hAB = 4,5 m

b) hAB = 6 m

c) hAB = 5,14 m

d) hAB = 6,36 m

D

45 Mt ng dn nc c d = 0,25m chu p sut thy tnh pd= 1,4MPa; ng sut ko cho php cavt liu lm ng [ ] = 70MPa, chiu dy ca thnh ng (mm):

a) 1,6 b) 2,5 c) 4,2 d) 5,0

B

46 Mt ng b tng hnh tr trn ngp trong b nc di L =5m, ct nc H =2R = 6m.Phn lc theo phng ngang Pxca p lc nc tc dng ln ng b tng l:

a) 662,175 kN

b) 1386,15 kN

c) 882,9 kN

d) 220,725 kN

C

47 Cho 1/4 mt tr trn AB c bn knh R = 1m v chiu di L = 1 m. Chiu cao ct nctrn im A: H = 1,5 m,. Thnh phn thng ng ca p lc nc (PZ) tc dng lnmt AB bng:

a) 18073 N b) 15784 N

c) 16824 N d) 17275 N

C

48

Trong b ch ho kh xng c iu ho bng phao hnh cu gn vo cn quay quanhO. Gi s mc xng trong bnh khng i v khi l xng vo bt kn th phao chm mtna. Bit a = 50mm; b = 20mm; d = 7mm; trng lng phao G = 0,262N; trng lngvan kim f = 0,135N; p sut d ca xng tc dng ln van kim pd= 0,6at; t trng ca

xng x= 0,7. ng knh D ca phao bng:

a) 60mm

b) 75mm

c) 85mm

d) 80mm

C


16/49

16

pcko

khong kh

A

2,8m

B

2m

H

a

P

H

d

n

C

QD

b

a

d

Pe

H1

H2PL

v

a

bDL/2L/2

dau

49 Bnh cha nc c p sut chnkhng trn mt thong pcko= 0,1at. Bnh c ngnbi mt van AB hnh vung c cnh bng 2m quay quanh trc nm ngang qua im Acch mt thong 2,8m. van AB v tr thng ng nh hnh v th p sut tuyt ica khng kh trong ng phi bng:

a) 1,28 at

b) 1,13 at

c) 1,43 atd) 2,12 at

C

50 Trn thnh phng nghing 45o ca mt b cha nc c mt l hnh ch nht kchthc a = 0,2m; b = 0,3m. Np hnh bn tr ng kn l c gi vo b nh ccbuloong. cao H = 1m. Lc ko P tc dng ln cc buloong bng:

a) 152 N

b) 556 N

c) 1314 N

d) 2529 N

B

51 Ngi ta c xi lanh rng c chiu cao H = 380mm v ng knh trong ln nht d =400mm bng cch rt gang lng vo khun ri cho khun quay quanh trc thng ngca n vi s vng quay n = 300v/ph. B dy thnh xilanh di dy hn thnh trnl:

a) 2 cm

b) 3 cm

c) 4 cm

d) 5 cm

A

52 Vt C v piston tr nng 3kN; d = 6cm; D = 30cm; a = 30cm; b = 5cm; ma st lp ltkn bng 5% lc nn ca piston tr. to ra mt lc p ln vt C l P = 35 kN th lcQ tc dng vo cn ca my p thu lc bng:

a) 187 N

b) 267 N

c) 378 N

d) 488 N

B

53 Mt bc tng hnh ch nht c chiu rng l b = 4m chu ct nc tc dng t haipha l H1 = 6m; H2= 4m. Trs im t ca p lc nc tc dng ln tng l:

a) P = 392,4 kN; L = 2,33m

b) P = 392,4 kN; L = 2,54m

c) P = 595,958 kN; L = 2,33m

d) P = 595,958 kN; L = 2,54m

B

54 Xe ch du (t trng 0,8). Bit ng knh D = 1m; kch thcb = 1,5m;L= 2m; gia tc ca xe a = 2m/s2. Tr s ca p lc

du tc dng ln np trc (PT) xe bng:a) PT= 4904 N

b) PT= 5518 N

A


17/49

17

0H1

H2

c) PT= 6130 N

d) PT= 3130 N

55 Mt ng b tng hnh tr trn ngn i b di L =4m. Mc nc hai bn l H1=10m, H2= 5m. Gi tr ca p lc nc tc dng ln ng b tng bng:

a) 1470,88 kN

b) 2938,64 kN

c) 2739,08 kN

d) 2629,52 kN

C

Chng 3: ng hc lu cht.

1 Ngn hng cu hi vp n chi tit chng 3

tt Cu hi v p n p n1 ng dng trng vi qu o khi:

a) Chuyn ng khng ph thuc thi gian b) Chuyn ng c xoy

c) Chuyn ng ph thuc thi gian d) Chuyn ng c th

A

2 ng dng l :

a) ng biu din qu o chuyn ng ca mt phn t cht lng

b) ng bt k c t ra thun tin cho vic nghin cu

c) ng biu din vn tc trong dng chy

d) ng m vc t vn tc ca mi phn t cht lng trn n tip tuyn vi n

D

3 ng dng trong dng chy u:

a) Lun lun vung gc vi mt ct t i qua n

b) Lun lun song song vi nhau

c) Lun lun tip tuyn vi cc vect vn tc

d) Cc p n kia u ng

D

4 Dng chy u l:

a) Vn tc khng i trn mt ct bt k

b) Lu lng khng i dc theo dng chy

c) Phn b vn tctrn mt ct t khng i dc theo dng chy

d) Vn tc khng i trn mt mt ct t

C

5 Chuyn ng dng l chuyn ng m:

a) Cc thng s ca dng chy ti v tr quan st c nh lun ph thuc vo t

b) Vn tc ti v tr quan st c nh ph thuc vo t cn p sut ti v tr quan st cnh khng ph thuc vo t

c) Vn tc v p sut ti v tr quan st c nh ph thuc vo t, cn khi lng ringkhng ph thuc vo t

d) Vn tc, p sut v khi lng ring ti v tr quan st c nh khng ph thuc vothi gian t

D


18/49

18

6 Dng chy mt chiu l:

a) Dng chy b qua s thay i ca cc thng dng chy theo phng vung gc vidng chy

b) Dng chy c ng dng l nhng ng thng

c) Dng chy u n nh

d) Dng chy u

A

7 Dng nc c lu lng Q = 6 m3/s, lu lng M (kg/s):

a) 6000 b) 5000 c) 49050 d) 58860

A

8 Dng cht lng c lu lng Q = 4 m3/s, lu lng G (N/s):

a) Khng xc nh c b) 4000 c) 49050 d) 9810A

9 Bn knh thy lc Rh bng :

a) a/4 trong trng hp dng chy c p trong ng vung c cnh l a

b) d/4 trong trng hp dng chy c p trong ng trn

c) Din tch mt ct t chia chu vi t

d) Cc p n kia u ng

D

10 Dng chy c p trong ng trn c bn knh ca ng r0= 60mm, bn knh thy lc R hbng:

a) 60 mm b) 15 mm c) 30 mm d) Cha xc nh c

C

11 Cho dng cht lng khng nn c chuyn ng dng, ta c:

a) Q = const, vi Q l lu lng th tch

b) M = const, vi M l lu lng khilng

c) G = const, vi G l lu lng trng lng

d) Cc p n kia u ng

D

12 Lu lng th tch l mt i lng c tnh bng:

a) Lng cht lng i qua mt ct t ca dng chy

b) S

udSQ vi S l mt mt ct t ca dng chy

c) Lng cht lng i qua mt mt ct trong mt n v thi gian


B

13 Phng trnh lin tc c xy dng t:

a) nh lut bo ton nng lng cho khi cht lng chuyn ng

b) nh lut bo ton khi lng cho khi cht lng chuyn ng

c) nh lut bo ton ng lng cho khi cht lng chuyn ng

d) nh lut bo ton moment ng lng cho khi cht lng chuyn ng

B

14Trong phng trnh lin tc di dng vi phn 0udiv

t

nu cht lngchuyn

ng n nh (dng) th :

a) 0)u(div

b) 0)u(div

c) 0t

d) Cc p n kia u ng

D


19/49

19

h

b

15 Phng trnh vi phn lin tc ca dng cht kh chuyn ng n nh (dng):

a) const)u(div

b) 0)u(div

c) 0udivt

d) 0

t

B

16 Cho dng cht lng khng nn c, chuyn ng n nh, vi iu kin no ca a, b, cth c th tn ti dng chy c: uX= ax + by uy = cy + bz uZ = az + bx

a) c b a c) a = - b d) (a,b,c)

C

17 Chuyn ng ca cht lng c cho trc bi cc thnh phn vn tc:

uX= ax + bt; uy = -ay + b; uZ = 0

a) Chuyn ng trn khng th xy ra b) Chuyn ng trn l chuyn ng dng

c) Chuyn ng trn l chuyn ng khng dng d) Cha d liu xc nh

C

18 Trong chuyn ng n nh:

a) ng dng trng vi qu o

b) Dng ca cc ng dng thay i theo thi gian

c) Cc ng dng song song vi nhaud) ng dng khng trng vi qu o

A

19 Chuyn ng c xoy khi:

a) Cc phn t cht lng khng t quay quanh mt trc tc thi i qua bn thn n

b) grad(u) = 0 c) 02urot


C

20 Phng trnh lin tc ca cht lng chuyn ng dng chy c p trong ng trn cdng:

a) Q = vS b) 1S1= 2S2 c)

2

22

2

11

dvdv

d) u1dS1= u2dS2

C

21 Trong trng hp no sau y th ux, uy, uzc th l thnh phn vn tc ca mt dngchy khng nn c (a, b, c, d l cc hng s):

a) ux= -dx + b; uy= -ay + c; uz= d b) ux= a + bx; uy= cy + d; uz= cy + x

c) ux= -ax + b; uy= ay + c; uz= c d) ux= -ax + b; uy= -ay + c; uz= cz

C

22Mt chuyn ng c vec t vn tc kxyjxy4ix2u 2

, y l:

a) Chuyn ng cht lng khng xoy, n nh

b) Chuyn ng cht lng xoy, n nhc) Chuyn ng cht lng xoy, khng n nh

d) Khng phi l chuyn ng ca mt cht lng

D

23 Dng chy trong mt knh hnh ch nht c b rng y b v chiu su ct nc l h.Bn knh thy lc l R l:

a)h)(b

bhR

. b)

)h2(b

bhR

c)

2hb

bhR

. d) Khng s liu tnh.

C

24Dng chy c p trong ng trn, nu ng knh d1 = 2 d2; th vn tc v2 bng:

A


20/49

20

a) 4 v1 b) 2 v1 c) 1/4 v1 d) 1/2 v1

25 Xt mt dng chy c p n nh trong ng, lu lng khi lng trong ng:

a) C n v l kg/s

b) L khi lng cht lng i qua mt ct t ca dng chy trong mt n v thi gian

c) L khi lng cht lng i qua mt mt ct ngang bt k ca ng ng trong mtn v thi gian

d) C 3 cu kia u ng

D

Chng 4: ng lc hc lu cht.1 Ngn hng cu hi vp n chi tit chng 4

tt Cu hi v p n p n1 V nng n v l:

a) z b) z + p/ c) c n v l J d) C ba p n kia u saiA

2 cao chn khng:

a) pck b) pck / c) C n v l N/s d) Cha c p n chnh xcB

3 Th nng n v l:

a) z + p/ b) C n v l m

c) Th nng ca mt n v trng lng cht lng

d) Cc p n kia u ng

D

4 caovn tc l:

a) v b) u2/2g c) gh2 d) Khng c cu tr liB

5 Cng m mt n v trng lng cht lng c kh nng to ra do p sut l:

a) p b) p/ c) gh2 d) Khng c cu tr liB

6 H s hiu chnh ng nng bng :

a) 1 b) 2 c) Ty thuc loi cht lng d) Cha yu t xc nhD

7 Mt phn t cht lng cao z so vi mt chun v c p sut p. Th nng ca mt nv khi lng cht lng l:

a) gz b) gz + p/ c) z + p/ d) mgz

B

8 H s hiu chnh ng nng:

a) C gi tr bng 2 khi dng chy tng

b) L t s gia ng nng thc v ng nng tnh theo vn tc trung bnh

c) c a vo do s phn b vn tc khng u ca cc phn t cht lng trn mt mtct t

d) Cc p n kia u ng

D

9H s hiu chnh ng lng:a) C gi tr bng 4/3 khi dng chy ri

b) L t s gia ng lng thc v ng lng tnh theo vn tc trung bnh

B


21/49

21

c) c s dng trong phng trnh Bernoulli

d) Cc p n kia u ng

10 H s hiu chnh ng nng s dng trong phng trnh:

a) Lin tc b) ng lng

c) Bernoulli ca cht lng thc d) Phng trnh Euler

C

11 ng nng v ng o p:a) C th trng nhaub) Khng bao gi trng nhau c) Lun lun dc ln

d) Lun lun dc xung

A

12 ng o p (z+p/) dc theo mt ng ng trn nm ngang c ng knh khng i:

a) Lun lun dc ln theo chiu dng chy

b) Lun lun dc xung theo chiu dng chy

c) Lun lun trn ng nng

d) C th tng hoc gim ty thuc vo tn tht trn ng ng

B

13 iu no p dng c c cho cht lng thc v cht lng l tng:

a)g2

vpze

2

b) Phng trnh ng lng

c) Cng thc : T = .S.dy

du d) Cc p n kia u c

D

14 ng Ventury l dng c o:

a) Lu lng tc thi trong ng b) Lu lng trung bnh trong ng

c) Vn tc trung bnh trong ng d) Vn tc tc thi trong ng

B

15Trong phng trnh: FvvQ 1122

, F l :

a) Tng ngoi lc tc dng ln ton dng chy

b) Tng ngoi lc tc dng ln khi cht lng c xt

c) Tng ngoi lc tc dng, b qua trng lc

d) Lc do cht lng tc dng ln thnh rn

B

16Xt dng chy qua mt on ng m rng dn, b qua ma st th tng ngoi lc F

trong phng trnh ng lng p dng cho on ng s bao gm:

a) Trng lc ca th tch kim tra; phn lc t thnh ng ln th tch kim tra; lc do psut gy nn trong on ng

b) Lc do ng sut ct to ra xung quanh thnh ng; phn lc t thnh ng ln th tchkim tra

c) p lc ti hai mt ct vo v ra on ng; phn lc t thnh ng ln th tch kim tra;trng lc ca cht lng trong th tch kim tra

d) Cc p n kia u sai

C

17Trong dng chy c p, nu p sut ti mt ct trc l p1, ti mt ct sau l p2, ta cquan h gia p1 v p2:

a) Ph thuc vo iu kin c th ca dng chy b) p1 < p2

A


22/49

22

1

1

2

2

v2

v1p1

p2

c) p1 = p2 d) p1 > p2

18 Trong dng chy c p trong ng trn nm ngang c ng knh l d, p sut ti mt cttrc l p1, ti mt ct sau l p2, ta c quan h gia p1 v p2:

a) Ph thuc vo iu kin c th ca dng chy b) p1 < p2

c) p1 = p2 d) p1 > p2

D

19

Phng trnh th hin nguyn l D'Alambe tng qut nht l:a) Phng trnh Euler thy ng b) Phng trnh Euler thy tnh

c) Phng trnh Navier - Stoke d) Phng trnh ng lng

C

20 iu no sau y l iu kin cn p dng phng trnh:

21w

2

22

22

2

111 h

g2

vpz

g2

vpz

1. im 1 v 2 nm trn mt ng dng. 2. Tnh theo p sut d.

3. Cht lng chuyn ng dng, khng nn c, lc khi ch c trng lc.

4. Cht lng nn c. 5. Dng chy u hoc bin i dn.

a) 1 , 2 , 3 b) 3 , 4 , 5 c) 1 , 3 , 5 d) 2 , 3 , 4

C

21

Cc s hng trong phng trnh: constg2

upz

2

c n v l:

a) m.N/m3 b) m.N/kg c) m.N/N d) m.N/s

C

22 Nng lng n vca mt dng chy (e) l:

a) Nng lng ca mt n v khi lng cht lng b) C n v l J/kg

c) C n v l m d) Cc p n kia u ng

C

23 ngha ca cao vn tc:

a) Ch n thun l mt s c tnh theo v

b) L nng lng ca mt n v khi lng cht lng

c) L ng nng ca mt n v khi lng cht lng

d) L cao thng ng ti a m mt n v trng lng cht lng t c khi c vntc ban u l v

D

24 Phng trnh Bernoulli th hin:

a) nh lut bo tonnng lng cho khi cht lng chuyn ng

b) nh lut bo ton khi lng cho khi cht lng chuyn ng

c) nh lut bo ton ng lng cho khi cht lng chuyn ng

d) nh lut bo ton moment ng lng cho khi cht lng chuyn ng

A

25 Dng chy qua ct cong nm trn mt phng ngang, bit mt ct 1-1 c psut p1, vn tc v1, mt ct 2-2 c vn tc v2. B qua tn tht. p sut ti mtct 2-2:

a) p2 = pa

b) p2 > pac) p2 < pa

D


23/49

23

1

1

2

2

v2

v1p1

p2

1

1

2

2

v2

v1p1

p2

x

y

P

pa

A

B

d) Gi tr ca p2ph thuc vo v1,v2v p1

26 Dng chy ri qua ct cong nm trn mt phng ngang, bit mt ct 1 - 1 c din tch S1,p sut p1, vn tc v1, mt ct 2 - 2 c din tch S2, hW l tn tht nng lng ca dngchy trong ng, p sut kh quyn l pa. p sut ti mt ct 2 - 2:

a) p2 = pa

b)

W

2

2

2

1

12 h2

vv

pp

c)

W

2

2

2

112 h

2

vvpp

d)

W

2

2

2

1

12 h2

vvpp

B

27 Dng chy ri qua ct cong nm trn mt phng ngang, mt ct 1 - 1 c din tch S1, psut p1, vn tc v1, mt ct 2 - 2 c din tch S2, p sut p2, vntc v2. Pxl lc do thnhng tc dng ln cht lng theo phng x. B qua tn tht, ta c:

a) - Px+ p1S1 - p2S2 cos

= S1v1(v2cos

- v1)b) - Px- p1S1 - p2S2 cos = S1v1(v2cos - v1)

c) - Px+ p1S1 + p2S2 cos = S1v1(v2cos - v1)

d) - Px + p1S1 - p2S2 cos = S2v2(v2cos + v1)

A

28

Cc gi thit v dng chy dn dt n cng thc: const

dp

2

ugz

2

l:

a) L tng, dng, khng nn c, dc theo 1 ng dng

b) Dng, u, khng nn c, dc theo 1 ng dng

c) L tng, u, khi lng ring l hm ca p sut p, dc theo 1 ng dng.

d) L tng, dng, khi lng ring l hm ca p sut p, dc theo 1 ng dng

D

29 Dng chy t b qua ng nh hnh v, xt p = pA- pB. Ta c:

a) p > 0

b) p < 0

c) p = 0

d)

p dng hay m ph thuc vo vn tc dng chy qua ng.

A

30 i vi dng cht lng v kh chuyn ng dng trong ng ta lun p dng c phngtrnh:

a) Q = const b) .Q = const c) constg2

vpz

2

d) v.Q = const

B

31 Trong dng cht lng chuyn ng:

a) p sut phn b theo qui lut thu tnh trn mi mt ct t

b) p sut phn b theo qui lut thu tnh ch trn ng dng

c) p sut phn b theo qui lut thu tnh ch trn mt ct t ni dng chy u hocbin i chm

C


24/49

24

1 2

ng o ap

5m 3m

1 2

ng nang

5m 3m

1 2 3

d) Cc p n kia u sai

32 Chuyn ng trong ng trn nm ngang c ng o p nh hnh v. Gi tr 3m o ttm ng biu din:

a)2g

v

p 2111 b)

p1 c)2g

v

pZ

2

222

2

d) Cc p n kia u sai.

D

33 Chuyn ng trong ng trn nm ngang c ng nng nh hnh v. Gi tr 5m o t tmng biu din:

a)2g

v

p 2

111 b)

p1

c)

pz 22 d)

2g

v

p 2222

A

34 Cht lng m chuyn ng ca n c m t bi phng trnh Euler thu ng

dt

udgradp

1F

l cht lng:

a) Khng nht b) Nht c) Nn c d) Phi Newton

A

35

Phng trnh Bernoulli: 21w

2

22

22

2

11

11 h

g2

vpz

g2

vpz

c s dng tnh cho:

a) Dng chy ca cht lng nn c

b) Dng chy ca cht lng ch chu tc dng ca trng lcc) Dng chy n nh v khng n nh ca cht lng

d) Mi loi dng chy

B

36 Dng cht lng chy trong ng nm ngang nh hnh bn, ngi ta lp 3 ng o p ti 3 vtr. Mc cht lng dng ln trong cc ng ny s l:

a) Dng cao nh nhau trong 3 ng.

b) Dng cao nht trong ng 1, trong ng 2 v 3 cao bng nhau.

c) Dng cao nht trong ng 1, sau n ng 2 v thp nht trong

ng 3d) Thp nht trong ng 1, trong ng 2 v 3 cao bng nhau.

C

37 dc thu lc J = 0,03 c ngha l:

a) ng ng nghing so vi mt phng ngang mt gc sao cho tg = 0,03

b) ng ng nghing so vi mt phng thng ng mt gc sao cho tg = 0,03

c) Trung bnh c 1m chiu di dng chy th tn tht nng lng n v l 0,03 m

d) Trung bnh c 1m chiu di dng chy th tn tht nng lng n v l 0,03 J

C

38 Th tch kim tra dng ch cho:

a) Mt h thng c lp b) Mt h thng kn

c) Mt khi lng c nh d) Mt vng c nh trong khng gian

D


25/49

25

A

B

Z

Q, v

Pl-tv0, Q0

v2

v1

h

t

B

A

Z

FQ,V

h

D

d

39 ng ng nmtrong mt phng thng ng nh hnh v Z = 2 m; pA= 0,6 at; pB = 0,5at; 1 ; dA>dB; th chiu chy ca nc trong ng:

a) Ch xc nh c khi bit ng knh cc ng

b) Chc chn t B sang A

c) Chc chn t A sang B

d) Ch xc nh c khi bit lu lng.

B

40 Du (t trng d= 0,8) chy ra khi vi phun vi Q = 50lit/s v v=10m/s. B qua tntht v trng lc. gi cho van phng c cn bng th lc l xo phi bng N):

a) 800

b) 1250

c) 400

d) 625

C

41 Du (t trng d= 0,8) c vn tc v0 = 5m/s, lu lng Q0= 70lit/s. B qua tn thtnng lng v trng lc. Lc do tia du Pl-ttc dng ln mt cnh go ng yn c gi trl (N):

a) 560

b) 700

c) 350

d) 280

A

42 Cho h = 96cm v z = 0,2m. Cht lng trong p k l thy ngn c t trng bng 13,6.Chnh lch p sut gia hai im A v B ca mt ng dn kh (p = pA- pB)bng:

a) 109 kN/m2

b) 118 kN/m2

c) 128 kN/m2

d) 98 kN/m2

C

43 Tia du (c t trng 0,8) c vn tc V = 4,8m/s, lu lng Q = 21dm3/s bn vo gia gugn sau xe. Gu l mt cong i xng 180o. B qua ma st. Lc F tc dng vo xe theophng ngang xe ng yn c gi tr l (N):

a) F = 63

b) F = 161

c) F = 144

d) F = 156

B

44 Ngi ta mun thit k mt vi phun nc ln cao h = 12m. B qua tn tht. Vn tc ncva ra khi ming ng bng (m/s):

a) 9,9

b) 15,34

c) 12,52

d) 14

B

45 Ngi ta mun thit k mt vi phun nc ln cao 5m, vi vn tc ra khi vi phun v = D


26/49

26

h

D

d

ck

hd

Q

x

ck

h

v

H = 30m

dQ

Q,v

Q2,v

Q1,v

9,9m/s. ng knh ca ng dn D = 15cm, ca ming vi phun d = 3cm. Bqua tn tht v chnh lch cao gia vi phun v ng dn. p sut d canc trong ng l (kPa):

a) 49 b) 97,8

c) 24,5 d) 78,2

46 Dng cht lng c = 0,8 chuyn ng trong mt on ng trn nm ngang ng knh

khng i c chiu di L = 10m. u on ng c p sut p1=1,54at, cui ng c p sutp2 = 1,22 at. dc o p l:

a) - 0,2 b) - 0,3 c) - 0,25 d) - 0,4

D

47 ng xi phng ht nc t sng vo rung. Bit h = 3,5 m; pck= 0,75 at; = 1; hw= 0.Ming ra ca ng thp hn mc nc sng (x) l:

a) 3,5 m

b) 4,5 m

c) 5 m

d) 4 m

D

48 ng xi phng ht nc t sng vo rung. Bit h = 4m; p ck= 0,8at; = 1; hw= 0. Vn tcnc chy qua ng bng(m/s):

a) 9,9

b) 8,86

c) 12,5

d) 7,71

B

49 Nc t b c H = 30m, chy (ri) qua ng c ng knh d = 5cm ra ngoi. B qua tntht. Lu lng nc chy ra ngoi bng (lt/s):

a) 47,6

b) 52,7

c) 76,5

d) 60,5

A

C th gii phi s ngi M v ngi M ni l lm. Th nhng ngi M li s ngi

Nht v ngi Nht lm xong mi ni. Vy ngiNht s ai, xin tha, l ngi TrungQuc, v ngi Trung Quc khng ni cng lm. Ngi Nht lun phng ngi TrungQuc v Trung Quc l mt cng quc c tim lc qun s. Vy xin hi ngi Trung Qucs ai, l ngi Vit Nam, ngi Vit Nam ni mt ng lm mt no b ai bit u m ln-----> Viet Nam Pro50

Tia nc c Q = 50,7 /s phun ra theo phng ngang. Khi gp bn phng tvung gc vi n b phn thnh hai phn. B qua ma st v trng lc. Nu

phn chy xung c lu lng Ql= 21 /s, th phn kia s lchmt gc so vi ban u bng.

a) 600

b) 300

D


27/49

27

h

Hg

Q

d1

d2

Qd2

h

d1pd1

S2QS1

k

d, l ck

B

z

c) 570

d) 450

51 Bit ng knh d1= 300mm;d2= 100mm; chnh ct thu ngn p k h =600mm;= 1;hw=0. T trng ca thungn bng 13,6. Lu lng nc chy trong ngbng (lt/s):

a) 24

b) 96

c) 30

d) 78

B

52 ng dn nc nm ngang c d1= 75mm; d2= 25mm. T ch ng co hp ngi ta cmmt ng nh vo mt bnh cha nc. Bit p sut d pd1=0,09at, lu lngQ = 3,1 lt/s.

B qua tn tht nng lng. Nc chy ri. Chiu cao h nc c th ht t bnh diln ng dn l:

a) 0,5 mb) 0,6 m

c) 1,1 m

d) 0,7 m

C

53 u phun ca vi cha chy c tit din S2= 0,01m2c vn vo ng trn c tit din S1= 0,05m2. p sut d ti tit din S1 l pd1= 1,5at. B qua ma st v trng lc. Nc chyri. Lc tc dng ln cc buloong khi nc phun ra bng:

a) 14730 N

b) 7350 Nc) 110675 N

d) 4910 N

D

54 Bm B ht du t mt bnh cha qua ng ng di l = 1,8m vi Q = 4 l/s. Bit z =1,2m; ti kho k= 4,8. Du c = 1cm2/s; = 860kg/m3. p sut chn khng ti mtct vo bm pck= 0,55at. Du chy tng. ng knh ng dn du d bng:

a) 45 mm

b) 50 mm

c) 56 mm

d) 62 mm

C


28/49

28

k

d, L ak

B

z

F 900d

H

Qd2d1

h

Q

d

55 Bm B y du t mt bnh cha qua ng ng di L = 1,4m, ng knh d =0,03m vi Q = 6dm3/s. Bit z = 3m; k = 4. Du c nht=2cm2/s; = 8450 N/m3. Du chy tng. p sut y (c trn p k)ca bm bng:

a) 2,93 at

b) 1,95 atc) 1,61 at

d) 0,85 at

B

56 Nc ra khi vi phun c ng knh d = 2cm t b nc c ct nc H=12m. B qua mast v trng lc.Lc F gi cho vt cn hnh nn (t i din vi tia nc v c gc nh bng 90o) c cn bng c tr s bng:

a) 18,04 N

b) 21,7 N

c) 35,15 N

d) 47,8 N

B

57 Bit d1= 50mm, d2= 100mm, lu lng Q = 30m3/h. p k ch U cha thu ngn c ttrng = 13,6. Nc trong ng chy ri. B qua tn tht dc ng. chnh ct thungn h p k l:

a) 15mm

b) 22mm

c) 27mm

d) 33mm

C

58 Bit lu lng nc chy trong ng Q = 2,4m3/pht; ng knh ng d=120mm; p sutd ca nc trong ng pd= 1,8at. B qua lc ma st v trng lc.Lc do nc tc dngln on ng cong nm ngang ni hai on ng vung gc vi nhau bng:

a) 3022 N

b) 2137 N

c) 6198 N

d) 9692 N

A

Chng 5: Chuyn ng mt chiu ca cht lng.1 Ngn hng cu hi vp n chi tit chng 5

tt Cu hi v p n p n1 Mt dng cht lng chy c p trong ng trn c s Reynolds tnh theo cng thc

v.R

Reh

Rh = 2320, vi Rhl bn knh thy lc, th dng chy l:

a) Tng b) Ri c) Qu d) Khng th xc nh c

B


29/49

29

2 i vi dng chy c p trong ng trn, quan h gia tn tht dc ng hd v vn tc vtheo:

a) Bc 1 b) Bc 2 c) Bc trong khong t 1 n 2

d) Tu thuc ch chy

D

3 Tn tht cc b htti ch ng cohp t ngt t tit din 1 sang tit din 2 l:

a) 2gvv 2

21 b)g2v

SS1

2

2

2

1

2

c)2g

v

S

S10,5

2

2

1

2

d)

2g

vvSS

2

21

12

C

4 Tn tht cc b hmti ch ng m rng t ngt t tit din 1 sang tit din 2 l:

a)

g2

vv 2

21 b)

2g

v1

S

S 2

2

2

1

2

c)

2

1

2

S

S10,5

d)

2g

vvSS

2

21

12

B

5 Phn b vn tc dng chy tng c p trong ng trn c dng:

a) Parabol b) Logarit c) Hyperbol d) Bc nhtA

6 Phn b vn tc dng chy ri c p trong ng trn c dng:

a) Parabol b) Logarit c) Hyperbol d) Bc nhtB

7Dng chy tng c p trong ng trn dng cng thc tnh h s

Re

64 vi:

a) Re =

d.v b) Re =

hR.u

c) Re =

hR.v

d) Re =

hR.v

A

8Cng thc tnh hd =

g2

v

d

l 2 p dng c cho trng hp:

a) Dng chy trong knh b) Dng chy khng p trong ng trn

c) Dng chy c p trong ng trn d) C 3 trng hp kia u c

C

9Cng thc tnh hd =

g2

v

R4

l 2

h

p dng c cho trng hp:

a) Dng chy trong knh b) C 3 trng hp kia u c

c) Dng chy khng p trong ng trn d) Dng chy c p trong ng trn

B

10 Tn tht nng lng dc ng hd ca dng chy c p trong ng trn:

a) T l bc 2 vi ng knh ng

b) T l nghch vi ng knh ng bc 2

c) T l nghch vi ng knh ng bc 4 khi chuyn ng tng

d) T l nghch vi ng knh ng bc 1 khi chuyn ng ri

C

11 S Reynolds phn gii di ca cht lng chy c p trong ng trn:

a) C gi tr bng 2320 b) C 3 cu kia u ngB


30/49

30

(A) (B)

c) L c s phn bit trng thi chy ca dng cht lng

d) C ngha quan trng trong vic nghin cu dng chy trong ng

12 Khi vn tc nc chy trong ng hai trng hp nh nhau, tn tht cc b ca dngchy theo chiu A: hcA= 0,8m th khi chy theo chiu B tn tht cc b hcBs bng:

a) 0,4 m

b) 0,8 m

c) 1,6 m

d) 0,6 m

C

13 Quy lut phn b vn tc trn mt mt ct t ca dng chy tng trong khe hp gia haibn phng song song ng yn:

a) Thay i theo quy lut bc hai b) Thay i theo quy lut bc nht

c) L tng hp ca dng Poaz v dng Cuet d) Khng i

A

14 Quy lut phn b vn tc trn mt mt ct t ca dng chy tng trong khe hp gia haibn phng song song 1 ng yn, 1 chuyn ng vi vn tc khng i:

a) Thay i theo quy lut bc hai b) Thay i theo quy lut bc nht

c) L tng hp ca dng Poaz v dng Cuet. d) Khng i

C

15 Trng thi chy tng thng xut hin trong trng hp:

a) Dng chy trong cc khe rt hp b) Cht lng c nht rt nh

c) Dng chy rt nhanh d) Dng chy trong cc ng c ng knh rt ln

A

16Cng thc sau 3D

L

p

12

1Q

dng tnh lu lng ca dng chy:

a) Tng trong ng trnb) Tng trong khe hp gia 2 mt tr trn ng tm

c) Tng trong khe hp gia 2 bn phng song song ng yn

d) Tng qua bu lc du

B

17Cng thc sau 3

12

1Bh

L

pQ

dng tnh lu lng ca dng chy:

a) Tng trong ng trn b) Tng trong khe hp gia 2 mt tr trn


d) Tng qua bu lc du

C

18Cng thc Bh

2

uBh

L

p

12

1Q 03

dng tnh lu lng ca dng chy tng:

a) Trong ng trn b) Trong khe hp gia 2 mt tr trn

c) Trong khe hp gia 2 bn phng song song 1 ng yn,1 chuyn ng.

d) Trong khe hp gia 2 bn phng song song ng yn

C

19

Cng thc

2

23 e

2

31D

L

p

12

1Q dng tnh lu lng ca dng chy:

a) Tng trong ng trn b) Ri trong ng trn

D


31/49

31

H d


d) Tng trong khe hp gia 2 mt tr trn lch tm

Cng thc3Bh

LQ12p

dng tnh st p ca dng chy :

a) Tng trong ng trn b) Ri trong ng trn

c) Tng trong khe hp gia 2 mt tr trn lch tm

d) Tng trong khe hp gia 2 bn phng song song ng yn

D

20 Lu lng cht lng r r qua khe h gia piston v xilanh tr:

a) Tng khi dng cht lng c nht ln hn b) Tng khi lch tm tng

c) Tng khi chiu di piston tng

d) Tng khi lch tm gim v chiu di piston tng

B

21 Dng chy tng trong khe hp gia 2 bn phng song song ng yn nh c vn tc trungbnh v = 2 m/s. Ti tm khe hp vn tc bng:

a) 1,33m/s b) 1,24m/s c) 0,88m/s d) 3m/s

D

22Trong cng thc tnh st p qua bu lc

o

3 r

Rln

h

Q6p

, Q l:

a) Lu lng cht lng i vo bu lc b) Lu lng cht lng i qua mt khe h lc

c) Lu lng cht lng i ra khi bu lc d) Cha c p n chnh xc

B

23 nh lut Haghen-Poise xc nh chnh p ca dng chy tng c p trong ng trnbng cng thc:

a) o3 r

R

lnh

Q6

p

b) 4d

LQ128

p c) Bh

QL12

p 3

d) Bh

QL128

p 3

B

24 So snh tn tht dc ng ca dng chy trong ng vung v ng trn c h s ma st,din tch mt ct t, chiu di v lu lng bng nhau. Ta c t s gia tn tht dcng trong ng vung so vi trong ng trn (hdvung/ hdtrn) bng:

a) 1,128 b) 0,886 c) 1,333 d) 1,50

A

25 Cht lng c nht 10mm2/s, chy tng c p trong ng nm ngang L =500m, d =100mm vi Q = 10lit/s. Tn tht nng lng dc dng bng:

a) 1,56 m b) 2,08 m c) 3,12 m d) 4,24 m

B

26 Dng chy y trong ng c mt ct ngang l mt hnh vung c cnh 3m. Chiu ding l 981m. Vn tc chy trong ng v = 3m/s. H s ma st = 0,03. Tn tht nnglng dc ng hdca dng chy trong ng bng:

a) 6 m b) 3 m c) 4,5 m d) 9 m

C

27 Nc chy t b cha c H = 12m ra ngoi qua ng c ng knh d = 1m. Tng tn thtnng lng ca dng chy hW= 2m, nc chy ri. Lu lng nc chy ra bng:

a) 16 m3/s

b) 12 m3/s

c) 14 m3/s

d) 11 m3/s

D


32/49

32

d1d2

h

h

h

v F

(1) (2)

Van tiet lu

SQ

28 Dng chy vi lu lng Q = 0,02 m3/s trong ng ng c tit din thu hp t ngt tS1=0,05 m2sang S2= 0,005 m2. Tn tht nng lng t thu htbng:

a) 0,37 m b) 0,66 m c) 1,32 m d) 0,41 m

A

29 Dng chy vi lu lng Q = 0,02m3/s trong ng ng c tit din m rng t ngt tS1=0,005m2sang S2= 0,05m2. Tn tht nng lng t m hmbng:

a) 0,37 m b) 1,32 m c) 0,66 m d) 0,41 m

C

30 Nc c dn t A n B cch nhau 0,5km trong on ng trn nm ngang c ngknh d = 15cm, h s ma st= 0,025. Lu lng dng chy trong ng Q = 40l/s. B quatn tht cc b. dc thu lc l:

a) 0,052 b) 0,043 c) 0,035 d) 0,028

B

31 Nc chy (ri) qua ng nm ngang nh hnh v, ng knh d1= 100mm; d2= 50mm.Trong ng d1vn tc ca nc v1= 0,4m/s. B qua tn tht. Chnh lch cao h tronghai ng o p bng (mm):

a) 24,5

b) 152c) 122

d) 75,5

C

32 Dng chy tng c p trong khe hp gia 2 bn phng song song ng yn c sReynolds Rek= 480 th k bng:

a) 0,03 b) 0,133 c) 0,04 d) 0,05

D

33 Nc 80o C (ly = 0,484.10-6m2/s), chy qua ng ca b tn nhit c tit din hnhch nht c cnh 2mm X10mm. Vn tc nh nht dng lun chy ri (truyn nhit tt

hn chy tng) l:a) 12,5cm/s b) 18,7 cm/s c) 33,7 cm/s d) 22,7 cm/s

C

34 ng ng dn du nm ngang c ng knh d= 0,1m v di l = 10km,du chy tngv nht = 0,002Ns/m2. Mun du chy vi lu lng Q =13,9dm3/s th chnhlch p sut gia 2 u ng phi bng:

a) 1,16 at b) 3,78 at c) 11,3 at d) 2,38 at

A

35 Tm phng c din tch S = 0,6m2, c ko trt ngang gia, cch thnh c nh h =5mm trong du c = 0,163N.s/m2, vi vn tc v=1,0m/s. Lc F cn thit ko tm

phng l:

a) 13,28 N b) 39,12 N

c) 19,56 N d) 6,64 N

B

36 Lu lng du d r qua khe h gia xilanh v piston tr khi ng tm l 6cm3/s. Khi

gia chng c lch tm tng i 5,0e

th lu lng du d r s l(cm3/s):

a) 8,25 b) 9 c) 7,5 d) 12

A

37 Nc chy qua van tit lu c tit din l S = 0,25m2, h s lulng =0,6. Khi p sut p1= 5at, p2= 1at, th lu lng bng(m3/s):

a) 4,2

A


33/49

33

A

d, Ld, Ld, L

D, L D, L

d

kM

L

L

h

h

d

L

B

d,LZh

ck

u

r

B

d,LZh

ck

u

r

b) 3,6

c) 2,1

d) 1,2

38 Du bi trn n cc trc theo ng ng gm cc on ng c chiu di L = 500mm;d = 4mm, D = 10mm. Du chy tng v p sut ti mi trc nh nhau. B qua tn thtcc b. mi trc nhn c khng di 8 cm3/s th lu lng ti nt A phi bng:

a) QA= 24,82 cm3/s

b) QA= 34,52 cm3/s

c) QA= 38,75 cm3/s

d) QA= 44,15 cm3/s

A

39 Nc chy t bnh di ln bnh trn vi lu lng Q = 3,5 l/s. Bit d=50mm, L = 3,5m;h s cn ca kha k = 8; nhm: = 0,3mm, h s ma st:

2)74,1r

lg.2(

1

. Ch s trn p k M bng:

a) 1,1 at

b) 1,29 at

c) 1,4 at

d) 1,96 at

C

40 Trc trn ng knh d = 4cm quay trong mt lt di L = 6cm. Khe hp gia trc v lt rng = 0,02mm c bi trn bi du c =0,04Pa.s. Trc quay vi n = 180v/ph.Cng sut tiu hao cho ma st bng:

a) 0,136 W

b) 1,339 W

c) 2,140 W

d) 3,929 W

C

41 Chiu cao ht ca bm zh= 5m, ng ht c ng knh d = 120mm; L=10m; h s ma st = 0,035; h s tn tht cc b ti li chn rc r=8; ti ch un u= 0,15; Q = 6,6lt/s. Nc chy ri. p sut chn khng pckti mt ct vo bm:

a) pck= 0,48 at

b) pck= 0,52 at

c) pck= 0,63 at

d) pck= 0,37 at

D

42 Bit cao chn khng ti mt ct vo bm hck= 4,5m ct nc; ng ht c d = 150mm;L = 10m; h s ma st = 0,03; h s tn tht ti li chn rc r= 6,0 ; ch un u= 0,2 ; lu lng qua bm Q = 16dm3/s. Nc chy ri. cao t bm Zhbng:

a) 3,5 m

b) 5,5 m

c) 4,3 md) 4,1 m

D


34/49

34

B

Zh

d2d1

v

H

c

c

v

H

c

c

43 Mt bm nc c N = 60kW, = 80%, lm vic vi Q = 0,15m3/s v cao chn khngcho php [HCK] = 6m. Tng tn tht cc b trn ng ng ht 5mhwh , b qua tntht dc ng, d1= d2= 250mm. Nc chy ri. Chiu cao ht cho php [Zh] ca bmbng:

a) 6,5 m

b) 6,8 m

c) 3,2 m

d) 3,5 m

C

Chng 6: Dng chy qua l, vi.

1 Ngn hng cu hi vp n chi tit chng 6

tt Cu hi v p n p n1 Vi cng ct p H v tit din S, lu lng ca cht lng chy t do qua vi c chiu di l

= (34)d so vi lu lng qua l:

a) Ln hn b) Nh hn c) Bng d) Cha iu kin so snh

A

2Trong cng thc tnh lu lng dng chy t do qua l t mt b h: gH2SQ , H l:

a) Chnh lch cao gia mt thong v y bnh.

b) Chnh lch cao gia mt thong v tm l

c) Chnh lch cao ca mt thong ti cc thi im khc nhaud) Cha c p n chnh xc

B

3 Xt dng chy qua l tho, v l vn tc dng chy khi ra khi l ti mt ct co hp. Ta c:

a) Lun lun 2gHv

b) Lun lun gH2v

c) Lun lun gH2v

d) Cha c p n chnh xc

A

6 nghch l tn ti Vit Nam:1.Ai cng c vic lm nhng khng ai lm vic2.Ai cng khng lm vic nhng ai cng c lng3.Ai cng c lng nhng khng ai sng4.Ai cng khng sng nhng ai cng sng5.Ai cng sng nhng khng ai hi lng6.Ai cng khng hi lng nhng ai cng gi tay ng .4 Dng chy qua l mng nh hnh v, v l vn tc dng chy ra khi l ti mt ct co hp,

S l tit din l, Scl tit din dng chy ti mt ct co hp, l h s vntc v l h s lu lng. Ta c:

a) Q = .S.v

b) Q = .Sc.v

D


35/49

35

Hx

y

h = 2m

c) Q = S.v

d) Q = Sc.v

5 H s lu lng ca dng chy qua vi:

a) Lun lun nh hn 1 v co hp dng

b) Lun lun nh hn 1 v tn tht dc ng

c) C th ln hn hoc nh hn 1

d) Lun lun nh hn 1 v co hp dng v tn tht nng lng

D

6 H s lu lng trong cng thc tnh lu lng qua l s nh khi:

a) Tn tht cc b qua l nh

b) Khng c tn tht nng lng

c) Dng chy qua l b co hp nhiu


C

7 Ngi ta khoan mt l trn thnh b c ct nc bng H tho nc. B qua ma st.

Vi h trc gn nh hnh v th phng trnh qu o chuyn ng ca dng chy ra ll:

a) 2x.Hy

b) 2x.gH2y

c) .gH2

xy

2

d) .H4

xy

2

D

8 Mt b cha di 10m, rng 5m, cha cht lng cao h = 2m. B qua tn tht. Thi giantho cn b qua l c din tchS = 0,5dm2lp di y b bng (giy):

a) 3193

b) 6385

c) 7385

d) 9577

B

Chng 7: Tnh ton thy lc ng ng c p.1 Ngn hng cu hi vp n chi tit chng 7

tt Cu hi v p n p n1 Khi nim ng ng di trong tnh ton thy lc ng ng l loi ng ng:

a) Chiu di L >> ng knh d ca ng

b) nhm


36/49

36

A

B

C D

h1h2

Qra

1

2

l Q1

Q2

d,L

(1) (2) (5)

(4)(3)

a) Ct p cc nhnh c cng li tnh ct p ca h thng

b) Tn tht nng lng trong cc nhnh bng nhau

c) Vic xc nh nhnh c bn l cn thit

d) Lu lng trong cc nhnh bng nhau

3 Khi tnh ton thy lc ng ng phn nhnh h, nhnh c bn l nhnh:

a) C tn tht nng lng ln nht

b) C yu cu v nng lng ln nht

c) Cao nht

d) Di nht

B

4 Khi tnh ton thy lc h thng ng ng ni tip:

a) Lu lng trong cc on ng bng nhau

b) Tntht nng lng trong cc on ng bng nhau

c) Ct p ca h thng bng ct p ca tng on ng

d) Lu lng ca h thng bng tng lu lng ca cc on ng

A

5 Khi tnh ton thy lc h thng ng ng song song:

a) Tt c cc cu kia u ng

b) Tn tht nng lng trong cc nhnh bng nhau

c) Ct p ca h thng bng ct p ca cc nhnh

d) Lu lng ca h thng bng tng lu lng ca cc nhnh

A

6 Hai b cha A, B v mt h thng ng ng nh hnh v. cui h thng lu lngnc cung cp l Qra. Ta c mi quan h v lu lng nc chy trong cc nhnh:

a) QAC= QDB+ Qra

b) Qra= QBD+ QAC

c) QBD= QAC+ Qra

d) Cha c c s kt lun

D

7 Hai nhnh ng 1 v 2 c ng knh d v chiu di L nh nhau c lp song song nhhnh v. nhnh 1 ngi ta lp thm mt li lc c tn tht cc b l. Qi,hwi l lulng v tn tht nng lng qua nhnh i (i=1,2). Ta c:

a) Q1< Q2

b) hw1 > hw2

c) Q1> Q2

d) hW1< hW2

A

8 Dng chy qua mng ng nh hnh v. Qi, hil lu lng v tn tht nng lng dcng trong nhnh th i. Ta c:

a) Q1= Q2+ Q3+ Q4

b) hd2= hd3

c) Q1 = Q5

d) hd2= hd3+ hd4-hd5

C


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37

Bm

Q1

Q2

Q3

0

III

III

E

AB

C

D

F

G

V3, d3

V1,d1

Q2

1. Dau

DH

2.Nc

DH 3.Thu n an

DH

9 Mt h thng ng ng r nhnh c gn mt my bm nh hnh v. Lulng Q1, Q2,Q3s l:

a) Q1= Q2 - Q3

b) Q1= Q2= Q3

c) Q1= Q2 + Q3

d) Q1= Q3 - Q2

C

10 Ba bn cha ni vi nhau nh hnh v, mc nc trong bn I cao hn bn II v mc ncbn II cao hn bn III. Dng chy trong ng s l:

a) Bn I chy v 0, 0 chy v bn II

b) Bn I chy v 0, bn II chy v 0

c) Bn II chy v 0, 0 chy v bn III

d) Cha c s xc nh

D

11 Nc chy t b qua mng li ng dn nh hnh v, lu lng nc ly ra khi cc

im B, C, D, E, Fl q. Lu lng nc chy trong ng BD l:a) 2q

b) 5q

c) 3q

d) 4q

D

12 Nc chy trong h thng ng ng, vn tc v tnh bng cng thc:

a)

4

dQ

d

dvv

2

12

2

1

3

31

b)

2

1

3

32

1

21

d

dv

d

Q4v

c)2

1

2

2

1

3

31d

Q4

d

dvv

d)

2

1

3

321d

dvQv

C

13 Trng hp no sau y iu kin cho ta xc nh c vn tc trung bnh ca mtdng chy c p trong ng trn:

a) Bit lu lng v ng knh ng

b) Bit trng thi ca dng chy v vn tc ti tm ng

c) Bit h s nhm ca ng, ng knh ng v dc thu lcd) C 3 trng hp kia u c

D

14 Ba bnh hnh tr c kch thc bng nhau v cha cht lng vi cao H nh nhau(Bnh 1: du; 2: nc; 3: thy ngn), b qua ma st, so snh thi gian T tho htcht lng qua l nh c cng ng knh bng D di y bnh, ta c:

a) T1 < T2< T3

b) T1 > T2> T3

c) T1 = T2= T3

d) Cha c c s so snh

C


38/49

38

(1) (2)

Van tiet lu

d

1 2

34

5

6

A B

15 Khi tnh lu lng dng chy n nh t qua van tit lu (c chiu dy = 0,5d) sang, ta c th p dng cng thc tnh lu lng ca bi ton:

a) Dng chy t do qua l nh, thnh mng, ct p khng i

b) Dng chy ngp qua l, thnh mng, ct p khng i

c) Dng chy qua vi tr

d) C 3 cu kia u sai

B

16 Cc ng ng c chiu di v ng knh khc nhau ni vi nhau nh hnh v. Gi HA,HBl ct p thy tnh ti A v B; hWi l tn tht ct p trn ng th i. Ct p thy tnhti B l:

a) HB= HA- hW1- hW2 - hW3- hW4- hW5- hW6

b) HB= HA- hW1+ hW2

c) HB= HA- hW1- hW2 - hW3

d) HB= HA- hW4- hW5- hW6

D

17 iu kin p dng cng thc tnh ct p LK

QH

2

2

l:

1. dng chy tng

2. dng chy u c p

3. dng chy ri

4. dng chy ri thnh trn thy lc

5. dng chy ri thnh nhm thy lc

6. ng ng di

a) 1, 2, 6 b) 3, 5, 6 c) 2, 4, 6 d) 2, 5, 6

D

18Cng thc

4dg

LQ128H

trong bi ton thy lc ng ng n gin, c th dng

tnh:

a) Tn tht dc ng ca dng chy u

b) Tn tht dc ng ca dng chy tng

c) Ct p ca dng chy ri

d) Ct p v tn tht dc ng ca dng chy tng, ng ng di

D

19Cng thc tnh tn tht dc ng L

K

Qh

2

2

d c dng tnh cho:

a) Tt c cc trng hp chy tng hoc chy ri

b) Ch cho trng hp chy ri thnh hon ton nhm

c) Cho tt c cc trng hp chy ri


B

20Trong cng thc tnh lu lng JKQ , n v ca K l:

a) m3/s b) m/s c) J/N d) s/m3

A

21 Mt ng ng bng gang mi c chiu di L = 1000m, chnh ct p tnh H = 5m. B


39/49

39

H

d, L

d, L

A

B

C

D

V3, d3

V1,d

Q2

Ngi ta tra c h s c trng lu lng K = 340,8lit/s. Lu lng nc chy trongng bng (lit/s):

a) 19,4 b) 24,1 c) 23,2 d) 25,8

22 Mt ng ng bng gang mi c chiu di L = 500m, lu lng nc chy trong ng Q= 244 lit/s. Ngi ta tra c h s c trng lu lng K = 1726 lit/s. chnh ct ptnh H bng (m):

a) 20 b) 30 c) 40 d) 10

D

23 Mt ng ng bng gang mi c chiu di L = 2500m, chnh ct p tnh H = 30m.Lu lng nc chy trong ng Q= 250lit/s. H s c trng lu lng K (m3/s):

a) 3,245 b) 2,502 c) 2,282 d) 2,722

C

24 ng ng di 2L, ng knh d, ni hai bnh c chnh H. Nc chy tng, b quatn tht cc b. Nu ta ni t gia ng 4 nhnh song song c chiu di tng ng L,ng knh d th khi lu lng nc chy trong ng s tng ln:

a) 1,6 ln

b) 1,5 ln

c) 3 ln

d) 2,66 ln

A

25 Cho 3 on ng ghp r nhnh nh hnh v. Chiu di on ng AB l L1=20m, ngknh D1= 40mm, h s ma st 1= 0,025. chch lch ct p ti hai u on ng ABl 6m. Lu lng tho ra C l 2 lit/s. Lu lng tho ra D l (lit/s):

a) 1,46

b) 1,85

c) 0,73

d) 3,15

B

26 Nc chy trong ng nh hnh v, lu lng Q2= 1,16lit/s, ng knh d1= 11cm, d3=15cm. Khi V3= 18cm/s th V1c gi tr bng:

a) 11,8 cm/s

b) 45,7 cm/s

c) 31,3 cm/s

d) 58,3 cm/s

B

Cnh Chm Cy Chi Cht C, C Con Cua Canh Chng Chp Con C. Cht Con Cng Chp Con CCa Cua. -Cua C : Con C Ca Cua. -Cng Ci : Con C Ca Cng. Cua Cng C Ca~i C : CaCua - Ca Cng, Ca Cua - Ca Cng ... Cnh C Con Cng, Con C Cng Cng Coi Chng Ci C.-C : Con C Ca Cng , Cng C Cng chp. -Cng c : con c ca cua , cua c cng canh chng .Cng c cng ci c : ca cua - ca cng, ca cua - ca cng ... Cht c con cp, cp ci: con c cacp. Cua chy, cng chy, con cng con c cng ct Cnh. Cp chp con c....


40/49

40

A

B

C

1

(2)

(3)

Z = 15m

A

B

C

pdA

1

2

3

Z = 15m

pckB

dz

H H

dz

H2H

dz

H2H1

27 Nc chy t b A kn phn nhnh sang 2 b h B v C. Bit tn tht nng lng trongng ng 1 l: hW1 = 5m. p sut d trong b A l pdA=63,765kPa. Tn tht nnglng trong ng ng l:

a) 18,5 m

b) 16,5 m

c) 17,5 m

d) 15,5 m

B

28 Nc chy t b A kn phn nhnh sang 2 b B v C. Bit tn tht nng lng trongng ng 1: hW1= 3m, trong ng ng 2: hW2= 3m, p sut chn khng trong b Bbng 6,53kPa . p sut d trong b A l:

a) 52,33 kPa

b) 58,86 kPa

c) 49,85 kPa

d) 37,91 kPa

A

29 ng c ng knh d = 150mm. Ct nc Hl= 3,5m. Tn tht t b vo ng l hv=0,5m ct nc. B qua tn tht dc ng v cc ch un. Ct nc H2bng:

a) 1,5 m

b) 2 m

c) 2,5 m

d) 3 m

B

30ng c ng knh d= 150mm. Ct nc Hl= 3,5m. Tn tht t b vo ng l hv=0,5m ct nc. B qua tn tht dc ng v cc ch un. Lu lng Q bng:

a) 78,2 lt/s

b) 85,4 lt/s

c) 88,7 lt/s

d) 97,3 lt/s

A

31

Nc chy trong ng xi phng c cao vn tc m1

g2

v2

, ct nc H=3,5m; z = 8m;

tn tht t b vo ng hvo= 0,5m. B qua tn tht dc ng v cc ch un, nc chyri. p sut chn khng ti im A cao nht trong ng xi phng bng:

a) 0,4 at

b) 0,5 at

c) 0,6 at

d) 0,7 at

C


41/49

41

v

H

c

c

v

H

c

c

H

h

L

32 Dng chy qua l mng nh hnh v. Cho din tch co hp Sc= 3cm2; h s vn tc =0,8; H = 3m. Lu lng Q chy qua l l:

a) 0,43 lit/s

b) 0,81 lit/s

c) 1,84 lit/s

d) 2,54 lit/s

C

33 Dng chy qua l mng nh hnh v. Din tch l S = 5cm2; h s lu lng = 0,6; H= 4m. Lu lng chy qua l l:

a) 1,73 lit/s

b) 2,66 lit/s

c) 3,94 lit/s

d) 4,03 lit/s

B

34 B cha du c ct du cao 4m khng i. Vn tc l thuyt (b qua tn tht) ca du

chy qua l y l:

a) 6,61 m/s b) 8,86 m/s c) 14,34 m/s d) 11,45 m/s

B

35 Mt l khoan trn thnh ca b cch y h = 1,5m. Gi s cht lng khng c ma st. on tia nc phng ra xa nht L = 10m, th H phi bng :

a) 18,17 m

b) 16,67m

c) 8,50 m

d) 17,60 m

B

36 Dng chy t b cha c ct nc H = 300mm qua l thnh mng. H s vn tc =0,96 th vn tc ti mt ct co hp vcbng:

a) 2,33 m/s b) 3,38 m/s c) 4,55 m/s d) 5,34 m/s

A

Chng 9

Nhng khi nim c bn v my thy kh v my bm.

2

Ngn hng cu hi v

p n chi ti

t ch

ng 9

tt Cu hi v p n p n1 Cc bm sau thuc nhm bm th tch:

a) Ly tm, piston b) Bnh rng, hng trc

c) Piston, bnh rng d) Bnh rng, ly tm

C

2 Cc my thu lc sau thuc nhm my cnh dn:

a) Bm ly tm, bm piston b) Bm bnh rng, bm hng trc

c) Bm piston, bm bnh rng d) Bm ly tm, bm hng trc

D

3 So snh kh nng lm vic ca bm hng trc vi bm ly tm:

a) Bm hng trc to c ct p cao v lu lng cao hn

C


42/49

42

b) Bm hng trc to c ct p cao v lu lng thp hn

c) Bm hng trc to c ct p thp v lu lng cao hn

d) Bm hng trc to c ct p thp v lu lng thp hn

4 So snh kh nng lm vic ca bm pison vi bm ly tm:

a) Bm piston to c ct p cao v lu lng cao hn

b) Bm piston to c ct p cao v lu lng thp hn

c) Bm piston to c ct p thp v lu lng cao hn

d) Bm piston to c ct p thp v lu lng thp hn

B

5 My th tch c dng khi cn:

a) Lu lng nh, p sut ln b) Lu lng ln, p sut nh

c) Cng sut ln d) Hiu sut cao

A

6 Ct p ca my thu lc th tch c tnh bng chnh:

a) p sut ca dng chy trc v sau my

b) ng nng ca dng chy trc v sau my

c) V tr b ht v b y

d) Nng lng n v ca dng chy trc v sau my

D

7 tnh ct p ca bm ang lm vic, sau khi c ch s trn ng h o p sut ti mtct vo v ra ca bm, ta cn phi bit:

a) Chnh lch v tr ca 2 ng h; ng knh ng ht, ng y v lu lng ca bm

b) ng knh ng ht, ng y v cng sut ca bm

c) ng knh ng ht, ng y v lu lng ca bm

d) Chnh lch v tr ca 2 ng h v lu lng ca bm

A

8 Hin tng ni bt (xm thc) xy ra trong bm khi p sut ti thiu trong bm:

a) Thp hn p sut kh tri

b) Thp hn p sut hi bo ho ti nhit lm vic

c) Cao hn p sut hi bo ho ti nhit lm vic

d) Cao hn p sut ht ca bm

B

9 Hin tng ni bt (xm thc) xy ra trong bm khi c du hiu:

a) n, rung v ct p gimb) Ct p v hiu sut gim

c) Lu lng tng

d) n, rung v cng sut tng

A

10 Hin tng ni bt (xm thc) d xy ra khi b mt bnh cng tc:

a) Nhn, kim loi dn b) Nhn, kim loi do

c) Nhm, kim loi dn d) Nhm, kim loi do

C

11 Khi c hin tng ni bt (xm thc) xy ra trong bm th nn:

a) Gim chiu cao t bm b) Gim lu lng bm

c) Gim tc quay

D


43/49

43

d) Tu thuc tnh hnh thc t c th gim 1 trong 3 thng s trn

12 p sut hi bo ho ca cht lng:

a) Tng khi nhit gim b) Ph thuc vo p sut kh tri

c) Gim khi nhit gim d) Cha c p n chnh xc

C

13 Chiu cao ht cho php ca bm c xc nh t:

a) iu kin chng xm thc b) p sut chn khngc) iu kin thc t ca cng trnh d) Yu cu ni tiu th

A

14 Khi nhit lm vic ca nc l 100oC, p sut trn mt thong b ht l 1at, chiucao ht cho php ca bm:

a) Chc chn c gi tr m b) Chc chn c gi tr dng

c) Chc chn bng khng d) Ph thuc iu kin c th

A

15 Chiu cao ht ca bm nc:

a) Khng bao gi ti 10m

b) Lun nh hn 10m khi p sut trn mt thong b ht l p sut kh tri

c) C th ln nh bt k d) Khng bao gi c gi tr m

B

16 Nng lng trao i gia bnh cng tc v cht lng trong bm ly tm gm:

a) Ch yu l ng nng b) Chyu l p nng

c) Ch yu l th nng d) ng nng v p nng

D

17 Nng lng trao i gia my bm v cht lng trong bm th tch gm:

a) Ch yu l ng nng b) Ch yu l p nng

c) Ch yu l th nng d) ng nng v p nng

B

18 c tnh h thng (HC) l ng biu din:

a) Nng lng yu cu ca h thng

b) Quan h gia ct p yu cu ca h thng v lu lng

c) S thay i tn tht nng lng ca h thng theo lu lng

d) Quan h gia cng sut yu cu v lu lng

B

19 Quan h gia ct p yu cu (HC) v lu lng (Q) trong h thng l:

a) ng cong bc 2 b) Tuyn tnh

c) Khng th bit c d) Ty thuc vo ch chy ca dng trong h thng

A

20 i vi my thy lc, hiu sut thy lc tng khi:

a) nht tng b) Lu lng tng

c) Tn tht nng lng cc b v tn tht do ma st ca dng cht lng gim

d) chnh p sut bung y v bung ht tng

C

21 Bin php no sau y c th lm tng hiu sut c kh ca my bm:

a) Tng cng lm kn

b) Tng nhn bng b mt lm vic ca bnh cng tcc) Gim ct p

D


44/49

44

d) Gim ma st gia b phn chuyn ng v khng chuyn ng trong my

22 ng c tnh tng hp ca bm ly tm cho ta quan h gia:

a) Ct p v lu lng

b) Cc thng s lm vic ca my cc s vng quay khc nhau

c) Cc thng s lm vic ca my mt s vng quay

d) Ct p v lu lng ca my mt s vng quay

B

23 im lm vic ca my bm l:

a) Giao im ca c tnh my bm v c tnh h thngtrong cng h trc to

b) Giao im ca c tnh lm vic ca my bm v c tnh h thng trong cng h trcto

c) im c ct p v lu lng tnh ton

d) im c cng sut ln nht

B

24 Trong tam gic vn tc, khi < 90o:

a) cu< u b) cu= u c) cu> u d) cu< u v ph thuc c vo

D

25 Bm ly tm c 2> 90os c:

a) Ct p l thuyt tng khi lu lng tng

b) Ct p l thuyt tng khi lu lng gim

c) Ct p l thuyt tng khi cng sut tng

d) Ct p khng i khi lu lng thay i

A

26 Bm ly tm c 2= 90os c:

a) Ct p l thuyt tng khi lu lng tngb) Ct p l thuyt tng khi lu lng gim

c) Ct p l thuyt tng khi cng sut gim

d) Ct p l thuyt khng i khi lu lng thay i

D

27 iu chnh lu lng ca bm ly tm, ta thng dng cc phng php sau:

a) iu chnh van kha, thay i gc vo 1.

b) Thay i s vng quay, thay i gc ra 2

c) Thay i s vng quay, thay i gc ra 2v iu chnh van kha

d) Thay i s vng quay v iu chnh van kho

D

28 Khi khi ng bm ly tm, ngi ta:

a) ng van kho trn ng ng yv cng sut ch Q = 0 l nh

b) M van kho trn ng ng y v hiu sut ch Q = 0 l ln

c) ng van kho trn ng ng y v hiu sut ch Q = 0 l ln

d) M van kho trn ng ng y v cng sut ch Q = 0 l ln

A

29 Cc thng s sau nh hng n ct p l thuyt ca cc bm ly tm hin nay l:

a) Tc quay, ng knh trong ca bnh cng tc v gc t cnh li vo

b) Tc quay, ng knh ngoi ca bnh cng tc v gc t cnh li ra

B


45/49

45

c) ng knh bnh cng tc li vo v ra, gc t cnh li vo v ra

d) Tc quay, ng knh trong ca bnh cng tc v gc t cnh li vo v lira

30 Trong bm ly tm, gc t cnh l gc gia:

a) Vn tc tng i w v vn tc tuyt i c

b) Vn tc tng i w v vn tc vng u theo chiu ngc li

c) Vn tc tng i w v vn tc vng u

d) Tip tuyn ca cnh v phng ca vn tc vng

B

31 Ct p l thuyt v cng Hlca bm ly tm c xy dng:

a) Vi gi thit cht lng l tng

b) Vi gi thit s cnh dn nhiu v cng, mng v cng v cht lng l tng

c) Vi gi thit s cnh dn nhiu v cng, mng v cng

d) T ct p yu cu ca h thng

B

32 Ct p l thuyt v cng Hlca bm ly tm s tng khi:

a) Gc t cnh ti li ra 2tng

b) Gc t cnh ti li ra 2gim

c) Gc t cnh ti li vo 1tng

d) Gc t cnh ti li vo 1gim

A

33 Thnh phn ct p l thuyt ng ca bm ly tm Hll:

a)g2

ccH

2

1

2

2

tl

b)g2

wwH

2

1

2

2

tl

c)g2

uuH

2

1

2

2tl

d)g2

ww

g2

uuH

2

2

2

1

2

1

2

2tl

A

34 Ct p l thuyt tnh ca bm ly tm Hltc c nh cc thnh phn vn tc dngchy trong vng bnh cng tc nh sau:

a) Do thay i vn tc tng i w

b) Do thay i vn tc theo u

c) Do thay i vn tc w, u

d) Do thay i vn tc tuyt i c

C

35 Cc nguyn nhn lm gim hiu sut thy lc ca bm ly tm l:

a) Tntht nng lng do ma st c kh

b) Tn tht nng lng do ma st ca cht lng khi i qua my bm

c) Tn tht nng lng trong ng ng

d) Tn tht nng lng do ma st cht lng khi i qua my bm v ng ng

B

36 Cc nguyn nhn lm gim hiu sut lu lng ca bm ly tm l do:

a) Tht thot r r cht lng ra ngoi myb) Cc khe h cho php mt phn cht lng quay tr ngc v bnh cng tc ca bm

D


46/49

46

c) S cnh hu hn v tht thot r r cht lng ra ngoi my

d) Tht thot r r cht lng ra ngoi my v mt phn cht lng quay tr ngc v bnhcng tc ca bm

37 Khi iu chnh im lm vic bng van kho:

a) c tnh lm vic ca bm H - Q s thay i

b) ng c tnh li HC - Q ca h thng s b thay i

c) ng cong - Q ca bm s b thay i

d) ng cong N - Q ca bm s b thay i

B

38 Khi iu chnh im lm vic bng cch thay i s vng quay:

a) c tnh lm vic ca bm H - Q s thay i

b) ng c tnh HC - Q ca h thng s b thay i

c) ng cong - Q ca bm s b thay i

d) ng cong N - Q ca bm s b thay i

A

39 Trong cc phng php thay i lu lng ca bm ly tm di y, phng php no lkhng thc hin c:

a) Thay i ng knh ra ca bnh cng tc

b) Thay i s vng quay lm vic ca bm

c) Thay i bng van kho d) Thay i ng knh vo ca bnh cng tc

D

40 Bm ly tm gn vi h thng c van iu khin, lc u hot ng n nh lu lng Q,ct p H. Khi kho bt mt phn van tng tr lc ng th:

a) Lu lng cung cp tng

b) Lu lng cung cp gim v ct p gimc) Lu lng cung cp khng i

d) Lu lng cung cp gim

D

41 Thng i vi my ly tm, gc t cnh li ra 2cly:

a) Cng ln cng tt to c ct p cao

b) Cng ln cng tt to c lu lng cao

c) Trong khong gi tr no c cng sut to c cao

d) c xc nh do yu cu c th v ct p tnh v ct p ng

D

42 Bung xon c trong bm ly tm c tc dng:

a) Dn hng dng cht lng vo bm nhm tng hiu sut ca bm

b) Bin mt phn p nng ca dng cht lng thnh ng nng

c) Bin mt phn ng nng ca dng cht lng thnh p nng


C

43 Cc my thy lc no c dng chy qua my c coi l dng n nh:

a) Bm ly tm, bm hng trc, bm piston

b) Bm ly tm, bm roto, bm pistonc) Bm ly tm, bm hng trc, tua bin nc

d) Bm bnh rng, bm ly tm, bm piston

C


47/49

47

44 i vimy thy lc, tn tht lu lng tng khi:

a) Lu lng gim

b) Tng cng lm kn

c) nht cht lng bm tng

d) chnh p sut bung y v bung ht tng.

D

45Mt thng cha 2m

3

xng, trng lng ca khi xng l 14715N, khi lng ring caxng bng (kg/m3):

a) 735,75 b) 800 c) 750 d) 9810

C

46 Cht lng c th tch ban u l 10 m3c nn thm mt p sut p = 50at. M unn hi th tch ca cht lng E = 2.109N/m2. Th tch cht lng sau khi nn gim i mtlng V (m3) l:

a) 1,955 b) 0,1496 c) 0,0245 d) 0,245

C

47 nht ng hc ca du =15mm2/s, du c t trng = 0,8. nht ng lc (Ns/m2) l:

a) 12.103 b) 12.10-3 c) 18,75 d) 120

B

48 Du c nht ng lc = 0,001Ns/m2, = 9810N/m3, nht ng hc (cm2/s)bng:

a) 0,01 b) 1 c) 9,81.103 d) 0,1

A

49 Bm m hnh c ng knh bnh cng tc 0,6m lm vic vi lu lng l 0,04m3/s, svng quay 1000v/ph. Bm thc c ng knh bnh cng tc 0,9m lm vic vi s vngquay 1500v/ph, lu lng l:

a) 202,5 lt/s b) 405,5 lt/s c) 502,4 lt/s d) 558,4 lt/s

A

50 Bm m hnh c ng knh bnh cng tc 0,6m lm vic vi s vng quay 1000 v/ph, ctp l 10m. Bm thc c ng knh bnh cng tc 0,9m lm vic vi s vng quay1500v/ph th ct p l:

a) 50,6 m b) 74,5 m c) 114 m d) 33,8 m

A

51 Mt bm ly tm khi lm vic s vng quay nh mc nm= 1000 v/ph c ct p H =80m; s vng quay n = 600v/ph ct p ca bm s l:

a) 28,8 m b) 60,1 m c) 50,2 m d) 55,4 m

A

52 Mt bm ly tm khi lm vic s vng quay nh mc nm= 1000 v/f c cng sut N =5 kW, s vng quay n = 600 v/f cng sut ca bm s l (kW):

a) 1,08 b) 1,80 c) 3 d) 2,28

A

53 Mt my bm a nc t mt ging h ln mt b cha c p sut d 1,4at caodng nc Z = 22m, tn tht nng lng trn ng ng l 2m. Ct p ca bm tnhtheo m ct nc l:

a) 38 m b) 28 m c) 32 m d) 34 m

A

54 Mt my bm a nc t mt ging kn c p sut d trn mt thong 1,5 at ln mt bcha h; cao dng nc Z = 29m, tn tht nng lng trn ng ng l 2,5m. Ct

p ca bm l:a) 16,5 m b) 15,5 m c) 17,5 m d) 21,5 m

A

55 Trong h thng bm nc ang lm vic, chn khng k ti mt ct vo bm ch B


48/49

48

d

dak

ck

B

y

B

D,

D

C

A

368mmHg, p k ti mt ct ra bm ch 2,35at. ng knh ng ht v y bng nhau,p k t cao hn chn khng k 1,5m. Ct p H ca bm bng:

a) 40m b) 30m c) 20m d) 60m

56 Bm nc c lu lng Q = 20dm3/s, ct p H = 40m nc, cng sut N=10kW, hiusut (%) l:

a) 76,5 b) 89,2 c) 78,5 d) 66,7

C

57 Mt bm piston lm vic vi lu lng Q = 10 lt/s v p sut p = 3at, hiu sut =75%. Cng sut trn trc bm bng:

a) 18,75 kW b) 3,924 kW c) 7,365 kW d) 6,53 kW

B

58 Bm piston c cng sut trn trc N= 4kW, lm vic vi lu lng Q=8,1 l/s, hiu sut = 80% th p sut ca bm bng:

a) 3 at b) 3,5 at c) 4 at d) 4,5 at

C

59 Bm lm vic trong iu kin: mt thong b ht thng vi kh quyn, trn ng ht c

tn tht nng lng hwh= 0,5m, cao vn tc , 0,5m2g

v

2

22 , cao chn khng

cho php [Hck] = 8m. Chiu cao ht cho php ca bm [Zh] bng:

a) - 7 m b) - 9 m c) 7 m d) 9 m

C

60 Bm lm vic trong iu kin: bm nc 100oC, mt thong b ht thng vi khquyn, tn tht nng lng trn ng ht hwh= 0,8m . cao d tr chng xm thc lyh = 0,3 pbh/ . Chiu cao ht cho php ca bm [Zh] bng:

a) - 6,2m b) 6,2m c) - 3,8m d) 3,8m

C

61 Trong mt th nghim bm ly tm o c:

p sut chn khng ng ht pck= 0,3 at, ng knh ng ht d1= 50 mm. p sut dtrn ng y pAK= 5,5at, ng knh ng y d2= 40 mm. p k t cao hn chn khngk y = 1,3m. Bm chy 5 pht th nc y thng th tch V = 1,8m3. Cng

sut tiu th ca ng c in l 6 kW v hiu sut ng c in l0,9. Nc chy ri. Hiu sut ca my bm bng:

a) 61%

b) 71%

c) 73%

d) 65%

D

62 Mt my bm nc t b ln bn cha nc trn cao vi Q = 1,25dm3/s. Cao trnh A = 5m. Cao trnh C =15m. ng ht v ng y c tng chiu di l L = 17m; ng knh

D = 3cm, h s ma st = 0,02. Hiu sut = 0,7. B qua tn tht cc b. Cng suttrn trc bm bng:

a) 206,8 W

b) 171,8 W

c) 143,8 W

d) 258,2 W

A


49/49

d d

z pa

B

63 Mt bm nc c N = 60 kW, = 80%, lm vic vi Q = 0,15 m3/s v cao chnkhng cho php [HCK] = 6m. Tng tn tht cc b trn ng ng ht 5h h , b qua

tn tht dc ng, d1= d2= 250mm. Nc chy ri. Chiu cao ht cho php [zh] cabm bng:

a) 5,8 m

b) 3,2 m

c) 3,5 m

d) 5,5 m

B

TH CH TIN

Tin l ci chi chi!Tin l mnh giy s ghi r rngC tin ph qu giu sangKhng tin lm k c hn iu linhC tin lm k chung tnhKhng tin n cho mnh quay lC tin k i ngi chKhng tin bn h th chng nhnC tin y h hngKhng tin c bc bng hong chi vi!C tin tho thch n chi!

Khng tin lm tot m hi c ngyC tin sng xn chiu sayKhng tin bng i sut ngy nm phiC tin bc ph xe hiKhng tin nm ng chao i i lngC tin ci v g chngKhng tin ng t hng cng qun seC tin anh ni em ngheKhng tin anh ni em ch anh ngho!Khng tin kh lm ai i!Khng tin cam phn m ci mt mnh!

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.......,/../Tay........

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...../..../Tri..........

..//'...'/`,Hn/'/.../..../..../''\i('(........~/'...')V

.\...*Shock*..../i

...\'...\........_............\.............(........

.......\.............\......

B, khng chu hc gi l bc hc.

Hc cch lm sch c th khi tm ra gi l hc k. Hc nhng

iu v ch, lm ph phm thi gian tin ca gi l hc ph. Hc

nhng iu by b v b, khng ng ch gi l hc b. Hc

cch tr chuyn trong giao tip gi l hc tr. Hc nm nm thc

n, thc ung gi l hc v. Hc cch dy ng gi, chng ng g

ng gt gi l hc thc. Hc cch lm gi gi l hc gi.

Ngi gi i hc gi l hc c. By iu cn phi hc gi l ththc. Hc ang v, tip thu tt gi l v hc. Hc lu di gi l

trng hc. B, khng chu hc gi l bc hc.
http://binhdinhffc.com/diendan/blog.php?u=1338&b=320&sid=9873ba2531dcf0411439d58c10d8b0d6http://binhdinhffc.com/diendan/blog.php?u=1338&b=320&sid=9873ba2531dcf0411439d58c10d8b0d6http://galavui.luyenthianhvan.org/2008/05/b-khng-chu-hc-gi-l-bc-hc.htmlhttp://galavui.luyenthianhvan.org/2008/05/b-khng-chu-hc-gi-l-bc-hc.htmlhttp://galavui.luyenthianhvan.org/2008/05/b-khng-chu-hc-gi-l-bc-hc.htmlhttp://binhdinhffc.com/diendan/blog.php?u=1338&b=320&sid=9873ba2531dcf0411439d58c10d8b0d6

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