Download pdf - BUET Admission Question

Transcript
Page 1: BUET Admission Question

j Admission Tech

BUET

Admission Question Year : 2013

Page 2: BUET Admission Question

02.

" Admission Tech

BUET Admission Test 2012-2013

atbematics (Written)

� l!l<l� 10% IBM <1>fu.>l�iirn <1J<1�lll <1>c1,; 1 IBM,;� l!l�,; 01w1Ti<1 <1>P1�iil* �r<��h .,.-�«· �� ¢,!,?

Solve: P (L) = 95%; P (I)= 10% P (L n I) = P (L) x P(I)

f.!tcfu � = P(L)-P(L) x P(I)P(L)

= 1-·P{I) = 1- .10=0.90=90%ta�cos-1 x)= sin(tan-1 2) l!lil ,fll)ilfil � 1

Solve: tan(cos-1 x)=sin(tan-1 2)

( -1 '�) • ( • -I 2 ) [f fi ]=> tan tan -

x- = sm sm Js rom 1g.

I..nl

I

� 2 1-x 2 4 2 2 2 , 5 .fs =>-x-=Ts => �=

5 =>5-5x =4x =>9x =5 => x- =

9 => x = ±

3

Js Js � x;t-- "lfl��'l�ffl-lt :.x=-(Ans)3 3

03. lil� � � i'!"m l!l<!i 2ffC" 10 kg '(JlSw!il l!l<!ilu <I& � "WI, � .'.l!P' � I m � lil� � '@o,'i'!

. -oil1"1_fll41�1C<l ��I � '@9111 ITTl:9111 ffl'l 30 kg wt W'! � C1vlJ '() '()ts?'! f.lcfu � I Solve: <:lll<lRI, � 'Wfi'! 'l_il! �.

Im

---

-----+1r.!:._1� Im

I \2 .J 2 .. "' ,, I ''

w R=-30 kg-wt

P=JO kg-wt

w P R

-=i_1=e 2 2

2I

I

20 2x30 R=P+w =>30=10+w :::>w =20kg-wt =>-=-- =>f.=3m(Ans)I f.

6 � ffi '() 4 � 9f,fit � $11 C� 6 mil l!llfi1t � $, <fliW1i WI I�� lil� $; <f.i'!1 �"1ml cir,,�� ,i�,m� �?Solve:

Mat�6)6

5

Physics(4 )0-(i)

I -(ii)

(i)<.!lil � �,i,,m -6C6 x4 C0 -1

(ii)l!lHfiTI�'l'l�'ffi =6C

5 x4 C

1 =24

4 2-(iii) (iii)l!lHFTI�'l'l�">ffi =6C4 x4 C

2 =90

:. t'l!1U � 'l'!�'IJT°

= I+ 24 + 90 = 1 15 (Ans.)

l Admission Tech

Page 3: BUET Admission Question

" Admission Tech

05. �l<lC<l, f(x)=x 11'<!!1l'll'R�m'llN x=e �1

Solve: f(x) = x • ; f'(x) =dx

!. d ( x • !J

!. ! dx d =x' [.J..-+lnx- - -(u )=u (--+lnu.-x dx dx x dx u dx dr

(')1[d ,. ,vdu dv

)l

f"(x)=-d(

x' !.I' ---, J +x' --dx x x'

lnx

![- 2 x1 3

<!l<lt f"(x)<!l x=e WI

x·.;-�

x.2x' 1 j

x

= +

dx_c!_

()XJ...-�J

x2 x1 x x-;[_23

x-2xlnx x4

= :r[-) J[:2 _ �: ]+)[-:3

_ 1-�i'nx]

_x x

��.�'WU, f'(x) = 0

:.x�,- �- ln

:J=o

!._2 1,-2 =>x 1 (1-lnx]=O:.�x· =0��

1 - ln x = 0 '®. 1 - �1 x = 0 => lnx =1= lne => x =e

!.)-d(f"(c)= x'·1· ) [I In-c],- +e• 2 -----l 1 ] =O+e'!. -d( x=eWI x' =0 dx e· e1 e e e

!.[ 3

-23

ne [-2-1

3

+2]dx

(x=c) _I_

(-

� 1)= e' x ll1 �� :. x = c Wi f(x) � � 1 (Showed) (i51'<!<11 � "i�R>l\!J fn RC!l <ll!ll W<: :

06. <!l<!iffl '51f� @� �iITTJ � 16, @� Ji -!171� ..!Fl sll'� � �I-ill!::� � � I Glf� y �'1R'rn�I

Solve: 1S"fW5 � � = 2ae l!J�, 2ae = 16 => 2a{F2)= 16 => a =4J2° .... (i)

' b2 b2 bla a a

�. e· =1+ 2 =>(F2f =1+

2 =>2=1+2

x'�-+--+--+---+--+-4---+--X

(-ac, o) (ae, o)

y' xi 1

=>--L = 1 => x2 -y2 =32 (Ans.) 32 32

o- i\BC <!!<!$���qi � 1 �Fl� AB <!!<It AC l!J� C'1'1l 4 q I A, B <!l<l� C ri"l.,� m

W'1l '5W<iS <l� 8, 8 <!l<!� 16 kg-m qcf� WI '8 'ifR>"i� f.icfu 'if.<! I olve: !3C = J A82 + AC2 = J42 + 42 = 4-hm·�.:Cf A <l 8�'1TC�<!l'l1ll�,fl!f.l� A<l B�m<!l'l1ll�oj"ij�,flll'!m��

4./i � �!,s '.ff AB��-;, z.TP-� � I AG..LGH<l BH..LGH � I C ::: :: - . �-AG= BH = x ('®)

�,;=_: F(x+4)=16[:.CG..LGH] =>Fx+4F=l6 =>8+4F=l6 = : =: wt (Ans.) i..g-

� :'\=) =>2x=8 =>x=4m ��<"" --� .<s- . .\8£i!�<l AB� 4m� A� 8<!!1l�� 1 x

: Admission Tech

Page 4: BUET Admission Question

� Admission Tech

08. X ,!!<! �?fV:I' C<rl�'1 <!>@ X = y2 ,!!<!t y = X - 2 � 'lffi � � � f.;,fu <ll'<! I

Solve: (0, 1)

y2 = x . . . . (i)

y=x-2 . . . . (ii) :::>y=/-2 :::> y2 -y-2=0

. _1±.Ji+8.. y- 2

= 1±3 =2 -l2 '

y2 = x ..... (i) y = x-2 ..... (ii) (i)'8 (ii)�t� y= y2 -2:::>y2 -y-2=0:. y=-1,2

(4,2)

y=x-2

(9, 0)

---y2=x

y=2 WI (i)l.!I, x=4 y=-IW!, (i).!! x=I �iffe0l'm� R'rn�W!I

{t1: 2x2 4 . =-[1-0]=-sq.urut

3 3 4

A1 = f[Fx-(x-2)}ix

-1

=J(Fx-x+2ftx

(x +a).!!<!� 21� � 7it°<ffi <J� 729, 7290, <!l<lt 30375 WJ a <!l<! °lW1 f.:cf,J <ll'<! I

Solve: (x +a)"= x" +�x"-1a + n(n -l) x"-2a1 + ....... +a" [N91� <s"i"il� �I)! 2!

"=729 (")· � n-i =7290 (") (iii)·:::> n(n -l)x".�=30375X .... I , X a .. , . II • 21 2 I . x

n(n2�

I) x"-1a1 = 30375 .... (iii) :::> 112

;n (729{': J = 30375

(ii) I!!, nx" .! = 7290 x

:::> n(729).� = 7290 [·: x" = 729]

a 10 . :::>-=- .... (1v)x n

n 2 -n 5 :::>--=-n 2 6

=>6n1 -6n =5n2

:::::> n2 =6n

=> n = 6[n *O]

Admission Tech

(i) :::> x6 = 729 = 36

:::>x=3 .) a IO(v1; :::>3

=6

:::>a= 5 (Ailr'

Page 5: BUET Admission Question

� Admission Tech

Ph sics Written)

10. �cm300 m ��-:l<l'>1U����?;f'IJ 500 m ���-:l<l'>1U�c�;;�� � 1 ����Mm���� <'1fT.71! � � �-sm �i'l' � 'ffl' �f.$ '5111�f.1 • .,:.;,m'<I � C<m'f f.w=l."1 � m?

Solve:�. t�"fil�� :. ucosat=500 . . . . . (i) -:l<!� h=!gt 2

2

:. 300 - h = u sin at - ! gt 2 ::::, 300 = u sin at ... (ii)

2 .. . u sin at 300 3 _1 3 (11)-:-(1): --=-. -::::, 1ana =

-:::::,a= tan - (Ans.)

ucosat :,00 :, 5

v�om!s

h300m

1 1. .,i<l'ffl �11'l 200 m � �_''l <'lfl'.<'l � r,,.,(j 500 kg '>',f.; � � 1 � 20% � '51"rnl �-.: V �����?

, mgb , (500x9.8x 2QO'\ Solve: P =-; 20% ':;;'l!i5lq"i�Ui�� 11=!00%-20%=80% = "'

t 60 J p• 16.33 16 33 :.:ii�� p� 11=-=16.33kw ::::,0.8=--kw ::::, P =-'-kw =20.416kwP P 0.8

�. 11P= mgb :::c,P= mgb =(500x9.8x200)w =20 416w =20. 416kw(Ans.)

t 111 0.8x60 12. ff�� l l.4gcm-3 <.!i<1t�'ll'ill�i!l 0.8x!010Nm·2

� 2x l<f Nm-2 m?1W��m1

Solve: B = ddp => dV = dp V; V'= V -dV = V -dp V m constant.v 8 B

v

£. = V' . , V V 11.4 -3 69 3

p' v .. p = p x v, = px

v(,-!)=

.975 gmcm = ll. grncm-

13. -:l<l'>1U � � -s <.!i<l'ffl ctfi'lRi•rn "frlfil � m.m 1i'lfliif,tll 80 dB -:l<it 78 dB. ,!)zyrn � ��'lff.l!t�?

Solve: Pvc=80dB; Prv=78dB Now, 80=l 0log10(1

1:c) ::::,Ivc=l08Io

Again, 78 = 10 log10

(1rv J ::::, Irv = 107

'8 [Now, I= lw + Ive = I 01

·3 Io +108 1 o = (101

·8 + IO'V1

lo lo

I { (

1078 +J08) }

P=l0log10 �dB = !Olog

10 Jo 1° dB =82.124dB

!.; <-:,J•.M <lJ<!m � � -:l<l'>1U �� 60 W GSL ef-ms -:l<l'>1U 13 W CFL ffi ffmJ � � � vJ 1i� Tk. 30 -:l<!� Tk. 250 zj% � � � Tk. 4 WI <.!ii!l � 'llt� d � 4 ';2 .t :::��f.-� � � �'ut CFL mm� m?

1, c: \foney required to rep lace the pre vious one = (250 -30)tk = 220 tk.:: ' .::.::o = {0.047)kw x t hour x 365 x 4 [Basically how much e nergy you will save, 60-13=4 .i

220 _; = = 3.206 hours . .! ,O(l.!-x365

1')

Admission Tech

Page 6: BUET Admission Question

Admission Tech

15. � mt"il:"l' � 'It� �'l 42°<!!<1t � mt� �m 'ltW �'1 48" WI fl mt�� 'It� �'lf.lcfu�

1 1 Solve: µ = -- = 1 .494; ,µ ... = -- = 1.3456 '

1 sin 42° " sin 48°

fl mt�� 'It�� 0 WI, . 0 1 1 ,µw 0 . -1(1.3456) 64 20 CA )sm =-=--=- :::::;, =sm -- = . · ns.

�· µ, • µ& • µ, 1.494 ,µw

16. ,!I� 5 McV ,� � b m ,!I� "l� W ��'Jl?,i Cl!'WI � C'il'l,1!8 «iit'l.ef.:<> � � .. c�m � � 1 ii .a� lTR 1.s T WI ,� t9fil �'Jl?ot <1oi f.l,rn � 1 ,� � <.!l<!t �� "ll�

l.7xl0- 27Kg�t I.6xI0-19C I

I [ 2 x 8 x I 0-1 3 l . Solve: EK =5xl06 xl.6xl0· 19J =>-mv 2 =8x!O-n:::::;, v = 27

ms I

2 l.7xW-

= 30.678x 106m/s: F = qvBsin 90° = (!.6x 10·19 x30.678x 106 x i.5)N = 7.3628x 10·12 N (Ans.)17. <!!���� 3x!O.., m2 1!l<lt'5!"fllliJ!! 727'C 1�f.!sffl"!'T<!�'l�?�'5!��

��'lWl'?(�� 5.?x!O-s Wm"2 K""]Solve: E = oA T' ... ... (i) -K5.7x IO"'}x (3x 10_, }x(727 + 273)4 �s-1 = 1.71 x 1·0-1 J IS (Ans.)lffil, T

2����fw,i�.i�1 :.E

2=3E; E

2=crAr; .. .. . (ii)

(ii)+( i)· �= T24 :::::;, 3E = T24 =>T =l316.07K(Ans.) I E T4 E (IOOOK)4 l

S. � ems� f.l� � 2.3 eV •�1 680 nm �C"i"'lll f.lf-rs � ill:(S'!l � � �

����mr�"t'f"R�?Solve: cj> = 2.3cV

E= he =(6.636x!0-34x3xl08 )

J =2_927x10_191 =(2.927xl0·

19

)ev =l.8297eV A. 680 x 10·9 1.6 x 10·19

cj>>E :.�.���mii�f:i�.J11(Ans.)

Chemist ritten

l.O g l!l� � � ?if<l'l 25.0 mL 1.0 M HCI !i<!t'l � <lllll � 1 �� '1'l'lr<l5 ""'i.ffl<>f 2l"M �50.0 mL � � � !i<l'l <'lltll 1 � � 15?1 '<l 'It� f.lcfu � ISolve: 11rn, l:f1W<l <ll1ff � M C03

MC03 +2HCl---tMq +Hp+C0

2

HCI + NaOI-1----t NaCl+ Hp I --C.-.. 25 _3

50 I _3 ""co =-[M="'1«1<>.,,'Sil); n11c1 =-3 x1=25x!O mole; nN,m1 =-3 x-=5x!O mole.. , ' M 10 ' 10 10 - - ....,:;, 2 -l -3 00 ( ��--i.,,2nMC0,+nN,ou = n

11c1 =>M+5xlO =25xl0 =>M=l Ans.)

-:-.fl::. n � � � CaC0 3 (Ans.)

Admission Tech

Page 7: BUET Admission Question

� Admission Tech

20. 4.60p!Nll .n<I$ <!l<l'!ll 5<1'1 cw::il � m 1 .illW(j <!)� � O.OIM� .!)� 10.0mL!i<l'l vrni� 1 .!l� 0.0 IM� '51Jlf>ic'otE4 � � (mL) 5'!'1 C'il'l � WI? (pl<, = 4.75]

S 6 K., 10 0 0 -4 ? olve: pH=4. : p =4.75: nacid =-1 x . I =lx!O mole: n,,11 =. IO

<ffii, <!!<l'l<l � � = V L: �JT�WR �'<! eqn � "l�. K (Salt] - n v -O!S n pH= p '+ log-- => 4.6 = 4.7) + log-.!!L_ => 10 = � :. n,,i, = 7.079x 10-smole

(acid] n� V Ix 10·

Again, ns,i, = V5,i, xMs.:, => 7.079xl0-� = v ... , x0.01 => V,.i, =7.079x!0-3 L = 7.079mL (ans.) 21. (a)� CW.'1 ·•i'�tJltt0,, e � �<::o ;;"ffffe .!l� � �'fil 'IT,li �IWTil'l\'i: c<l5R � �

5<1'1 � � �? �'1 :11mfu � 5<1'1 � � � �? �'1 � }l�f!j'2 �iir,m,r.«, �'1 � ,

Solve: (a) � 5<1'1 FcCl3 .!!Hr� 5<1'1

� B<l'1 �= � � clll� .!l<I$ '1!'!'1 � Na��� -.m ™ � a4i�\:, <m � -.m �HC)'ITT!T���If.!fJRTT: NaCNS+ FcCl3 �Fc(CNS)]Cl2 + NaCl

N "11<15<1'1 (b) ��'I"!! ,!I'll'! <!l� a-� <!lP.t'-"ll "111 {J � }l�� �I��� -II ��'!lTil �qji-;)Jl � I

NH2 == I

..A.-. Solve: (b) ; ;,,,,_,,,.,: H - C - COOH � '!Ii� ,,,..,.� 'I"!! <liBf'l <.'l"C\!i C<liT'! "'ffi.lWI � ¢ 1

H 22. (a) t'1iS }ll!ln! c<IIITT <ill <ill�€ <ill �r . .,dJ,.,(�E � �? (<liWl f.liv� �� f.i� �{J'ffi1l � fu

Solve: (a)�' � � (Pb)�t"1<1>d;1e,1"®, H2SO 4 � fo;�51�1, 1�1"1'm: Pb02 (s)+Pb(s)+2H2S04 (aq) � 2PbS04 (s)+2H20(1) �

(b) 27°C�"ilITT!rn<.'l��ffi 0.02molcbn-1 HO .!!llB<!'i�<lllll� 1����f.lcrn�

Solve:(b)i5!11l'1�: _!_H2 -e·�2W E\,11 =0V, n=l, T=300K2

Ecca = E0"11 - RT ln[W] = 0- &.3 !4 x 300

ln[0.02] = 0.JOI IV (Ans.)nF I x96500 _: �r,;,j�����=

a) CaCO 3 i:<li HCI -£if-ws � ffl � �['51 B<!<l fir<!Wi'-1' e: CaC01 +2HCl----tCaCl2 +Hp+C02 CaCl2, H20-t2i"f'll

C02��l:f':rt - C �. �� �<q �<!<" "l� Kl �'1 ffl Na 2S203:5<!'1 f.ml � �

t; :C::3), -4K.1�2K2S04 +Cu2I2 + I2

. - :,:!.:5=0, � Na,S,06 + 2 al

--------� Admission Tech

Page 8: BUET Admission Question

- Admission Tech

(c) NaHC03� 180"Cl:51���

180°C Solve: 2NaHC03

Na2C0

3 + H2

0 + C02

(d) � 'lr.�Na 2C03 ffl m � � i�9i� �

Solve: Na2C0

3 +Si0

2 �Na

2Si03 +C0

2

24. (a)� 9f1f.@ � � � o1?'!'!J m GTf.R "fNHll lTl'!l I Ill� "l'<l�C� � t.s ':.., "ll<l'!:T �? llPlf!lf.Rs�fii',j1Solve: (a) .ii�� 91'1-om � <!i1lR � � � � �� � -..�t � -..� � � 1������.l!l�<>i-im1���'Wl�� .. 11Pl-:�"IITT2NaBr+Cl

2 -->2NaCI + Br

2

(b) dsp2, sp3d <!!<ft sp3d2 � ��tim � R-IWi � R6?

Solve:(b) dsp2 �� <1�; sp3d ��·rr�liHl� �; sp3d2 ��25. (a)�� 'l!llfJ'C'I! Zn(Hg)i!l11 "llt� .<flra(fum) llllfJ'C'I! c!?1di�•-1i\C<t> 1<1��1rn� � �� r.<-,;'0., &

�ll1'!11�-ifl1�'6��">11Solve:

��I Solve:

H2C=CH2 HC .. CH+I-12(�� .. ) &li�'flllJI!

HgCJi HC .. CH+HCI CH

2 =CIICl('lt��)

150-200){:26. (a) <li"ffil � m !l<ft'l l'f"El \cf C1!1'1 � � � 'IW?

Solve: Zn� CuSO 4 � Cu� �"Am IZn(s) + CuSO 4 (aq) � ZnS0

4 (aq) + Cu(s)(b)·��H2004 1!lil��:�<>rrnC<ll"i!?����'Solve: 2PbS0

4(s)+2HP c1wg;ng >Pb02

-Pb+2H2S0

4

(s) (I) (s) (s) (aq) .ii���� C11 "f1R � � \ll<ft .ii� 'iR� H

2S04

� � 'll'm H2SO 4 .ii11 � � 9flll I

(C) � <f1tiI 1lNf 9fTf.rn <11""1� � '1f?5 � .l!l� � crnl � i l!lll l!l� ;r]lffil'12flrl'l 1il'G ISolve: � � 9fTf.rn �8 <lfP'l� � '['8 <IT'P9f �"A .l!l'RF!!rn 'lt'l' � lTR! 1 � � �<lm <ITt""fll"tml"l "llt'll 'ITlll vt'"'f � "l'R-ll �11 "lfu � .2ira,;1rrn � � � � cm �ii I

:1. (a) fu�ll.lllf<r'l -o �1w1w�·· � � 'lli!"!j"i,cf 0t'1 , � �� l!lv>RI 91'1<H!l 'l!Tll ,!Fl� \ll�!<fl RI� �m Solve: �llll.ll�i\S ll1C!llt.1l�'IS �g��� �g� �8 02� *-8Ci11<1�il

Admission Tech

Page 9: BUET Admission Question

Admission Tech

(b) CaC2 +1lp�CaO+IICsCH, HC:O-f�H

2C=CTI

2, n(H

2C=CH

2)�-(01

2 -CH2 -)0

�"Im ��toml 'l!llfl"Cll ��(�"Im�� iR!llrn I 80kg� � (� � W�"lmR��ISolve:��. 64grn CaC

2 � � � 28 grn

28x80x!0 3 :.80xl0 3

grn CaC2

� �� = gm =35xl0 3 grn =35kg (Ans.) 64

d!bMmittti\ Read the following passage and answer the question numbers 28-29 that on the basis of the information provided in the passage.: Galileo was born 1564 at Pisa Italy. He was the eldest of seven children. His father was a musician and a scholar. Galileo himself played the organ and the flute, but it was his contributions to science that made him famous. Al first young Galileo had a tutor at home in Pisa. Then he went to school in a monastery at Florence. His father later sent him to the University of Pisa to study medicine, but Galileo was more interested in mathematics and physics. He left Pisa without finishing medicine course. But in 1589 at the age of 25 he became a Professor of mathematics. In 1609 Galileo made a small telescope. When he tuned his telescope on the sky, he gradually discovered four moons circling the planet Jupiter, crater on the moon, spots on the sun and rings round Saturn. He also observed that the planet Venus has phases like the moon. This could only mean that Venus travels round the sun. Galileo became convinced that the Earth and all other planets orbit the sun. At that time the Christian church thought that any idea that the earth was not the centre of our universe went against the Bible. A book published by the astronomer Copernicus in 1543 setting out such a theory was officially banned by the church. Galileo's views on the subject and the books he wrote got him into serious troubles with the church. As the church was very powerful in those days, Galileo was forced to say publicly that he did not agree with the ideas of Copernicus in order to avoid torture and even death. Although he made this declaration, he neve,: changed his real belief. Galileo died in 1642 at the age of seventy seven.

28 . (a) Answer whether the following statements arc true or false. If false, provide the correct information. (i) Although Galileo was the son of a musician, science was of grater attraction for him than music.(ii) Copernicus was the first man to know that all the planets in our solar system move round the

sun(iii) According to ecclesiastical belief of Gal ileo's time the sun was the centre of our

universe.(iv) Galileo's father was an ordinary man having little education. (b) Fill in the gaps with correct form of verbs provided in the brackets.(a) Galileo __ (know) as the inventor of telescope.{b) Though his father wanted him to be a doctor, he __ (finish) his medical course. (c) He __ (choose) teaching as his profession.d) He knew that the Earth with other planets __ (move) round the sun.! !1e __ (threaten) with torture and even death by the church.

: he: (a): (i) True (ii) True ?i.se Correct Ans.: The earth was the centre. : ' � Co:-rect Ans.: His father was a scholar.

(ii) did not finish(iv) move (v) w�s fhr��IMPn

Admission Tech

Page 10: BUET Admission Question

Admission Tech

29. Answer each of the following questions in one complete sentence.(a) How did Galileo contribute to science?(b) What did Galileo do with his telescope'!(c) Why did Galileo have trouble with the church?(d) What notion do you get about the church of Galileo's time?(c) What do you know about the schooling of Galileo?Solve:(a) Galileo invented telescope and observed planetary motions and underwent scientific experiments.(b) Galileo observed the planetary motions and discovered that actually sun was the centre of oursolar system.(c} Galileo had trouble with church because his discoveries were against the words of Bible.(d) I think that the church of his time was of ill culture, they did not understand the meaning ofthere religion rather they gave wrong explanation.(e) Galileo had a tutor at home in Pisa, then went to a school in a monastery at Florence, laterwent to university of Pisa to study medicines.

30. (a) Fill in the blanks with correct form ofvcrbs:(i) I never thought I __ (sec) you again.(ii) He __ (do) everything he could to help me.(iii) Most of the earth's surface __ (cover) by water.(iv) I tried to pacify him but he went on __ (grumble).(v) The e-mail __ (send) a week ago.(b) Fill in the bl.anks with appropriate prepositions.(i) He has a reputation __ honesty.(ii) She said nothing __ reply.(iii) You must conform ___ the regulations.(iv) I shall look the matter. (v) We went to the airport to see ___ our uncle.

:\lath (MCQ)

� f(x) = x2 -�xi <!l�t g(x) = x2 + I � � (fog )(2) .!Iii 'llR m-(a) 0 (b) 15 (c) 25 (d) 5

Solve:(b) (fog) (2} = f(g(2)); f (x) = x2 - �xi

g(x)=x2 +I =52-21� =f(22 +l)=f(S) =25-10=15

� x = .!. {-1+.J-J) <!l�t y= .!.(-1-f-J) � � (1-x-y+xy) <!Iii 'llR m-2 2

(a) 0 (b) I (c) 2 (d) 3Solve:(d} x = .!.(-1 + f-3) = ro; y = .!.(-1- f-3)= ci

2 2

1-x-y+ xy = l-ro-ro2 +ro.ro2 = l-ro-ro2 +cil = 1-(ro+ci>2)+1[ro3 = 1]= l-{-l)+t[.· 1+w+ro2 =0] =l+l+l =3

e: Ad - . -

m1ss1on Tech

[Ans: would sec] [Ans: did] f Ans: is covered] [Ans: grumbling] (Ans: was sent]

[Ans: for] [Ans: in] [Ans: to] (Ans: into] [Ans: oil]

Page 11: BUET Admission Question

" Admission Tech

03. k <!I� 'llf.l � W'! (k + !)x2 +2(k +3)x +2k + 3 � l!l� "l.<f-1,f WI?(a) 3,2 (b) 3,-2 (c) -3.2 (d) -3,-2Solve: ilffe!l'u 1.f<l<fwi (k+l)x2 +2(k+3)x+2k+3=0 �� "B mD = [2(k +3)]2-4(k + IX2k +3) � 0 =4(k +3)2 -4(2kl + 2k+3k +3)=>0=4[(k2 +6k+9)-(2k2 +5k+3)] =>0= -k2 +k+6 =>k2 - k-6 =0 .=>(k-3Xk+2)=0 :. k =-2,3 04. k <!I� t<1SR � � A= [k -3 - 2 ] $_.,..,,1f&'<<1 m 'll?-2 k-2

(a) 3 {b) 2 (c) s±J0 2 ad"(A) Solve: A -1 =-�-; A �m'llwi !A!=O�!Ai(d) �2

IAl=jk_-i3 k--221 �0=(k-3Xk- 2) -(-2X-2) => 0= k2 -3k-2k+6 -4=>0=k2 -5k+2 =>k= 5±� = S±� = 5±.Ju2 2 2 05. � A� Jx3�<!Fl� IA!=-7� j(2AJ-'[.!iiPll'lWl-

(a) -� (b) -� (c) _!14 56 7Solve: A is a 3 x 3 matrix

So Let A = [� � r] ; 2A = rn� }! Ir l IA[ = ,� � f I= -7We know, IA-1 1 = !�I

[Prove yourself]1

[2a 2b 2c

] Now, 1(2A) 1 1 = -I( \I l(2A) = 2d 2e 2f 2A11 ' 2g 2h 2i

2(d) --7

=2x2x21� � rl[c: =�1 J[c>�2 ][c>i] = 23(-7) =-56 :.1(2Af'l=l(Z�(-516 Shortcut: n � A� wlJ l(mA� = m"JA[ :� 8(�) I'& W:l � � � � <!l<llfG � offi'Wl <!l� m-� � CW3 "!ml? 2) -! (b) 8! (c) 7.!. (d) !

2 2

- � � ; "'--- C...t.. 7! Sehe: -:;��o.R!10ul'1�«JM =(8-1)!=7!; <ll�'Sft'4J�'f<l'is4w1�0f :.1•1r.01�m-.,R"/Jl =-2 - • . - : '. - ! : l - .......... � 100 'i5l'I ?f'l W C<ll� WI fb) IO(lo100_1)_ IOO (c) .!.Q(10100_1)_ 100 (d) .!.Q(i0100_1)-ioI 9 9 9 9

f Admission Tech

Page 12: BUET Admission Question

Admission Tech

Solve: S0 =I+ 11 + 111 + .......... up ton => 9S

0 = 9 + 99 + 999 + ..... up ton

=> 9S0 = (10-1 )+ (102 -1)+(103 -1)+ ... up ton

::>9S0 ={10+ 102 + 103 + ... + 10• )-(I +1+ 1+ ... up ton)

10° -1 ·,.· -1 1-r" ::>9S0= l0---nf��tr!m S =a--;r>l=a--;r<I]

10-1 " r-1 l-r

=> 9S = IO (10• -1)- n => S = 10(10• -1)-�n 9 n 81 9 . , l O ( 100 ) 100 .. 100"fli"flf"C1JI�� n=lOO :.S

100=-10 -1 --

81 9 08 . x <!Ill� "l!1m '5RJ (!,-xi (1, x) <!l<I� (x2 ,-1) ��<!I�� Cll"<lrn � �?

(a) -1, 0, I (b) 2, 3, 4 (c) -3, 2, 3 (d) -4, 3, 4Solve:� <!I� 'lil<'!Cil*-1 � �.

l l -/ L o

x2

-1 �

=> 0-0+ l(0-(-2x)(l-x2 n = O[ C3 � �)

::>2x(I-xXl+x)=O :.x=O

Or I - x = 0 => x = I

=>l-x2 x+l 0=0[r;=r1 -r2 ,r12=r2 -r3] x=-1,0,1

0 -2x 01

Or I + x = 0 => x = -1

x2

-1 I

09. x2 +y2 +2x+c=0 <!l<I� x2 +y2 +2y+c=orern�""'!"f� c<!lil1f!,I�-

(a) 0 {c) I

Solve: x2+y2+2x+c=O; c1(-l,O ); r,=�

x2+ y2+2y+c=O; c2( 0,-I); r2 =�

c,c2= J(-1-of+(o+1f =ful=-n

(d) 2

From fig.�� c1c

2 =r.+r

2 =>.J2= Jl-c+� =>.f2=2Jl-c

=>.J2=2·JI-c =>2=4(1-c) =:>..!..=1-C=>c=l-� =� 2 2 2

Note: <!I� � �if 'fllR � � � <f!"'IVlT 'lll""rooM � � ,rn [� Iii� W"1 <fiilJ ' iffef y= x1"' �WI ;(!)m-

(a) 2lnx (b) l(21nx}x xSolve: y = x'"' dy = x'"'[lnx .l+tnx..!.Jdx x x

[-: ! (uv )=uv

(�: +lnx.:)J

(c) 2lnx

= 2lnx

[ref. LutfuzzammSir] [�� � � en f.lal <fiill llWI rl

=xm•rln

xx + �

x l= xm•r2

�x l

Admission Tech

(d) 2 ylnx

Page 13: BUET Admission Question

11.

Admission Tech

x xl!l'i!'lffil�Wl�t""I-I f(x)=- l!lil'llfol�m?ln x

(a)_! (b) c (c) !c c

(d) -e

In x.J-x..!.x Solve: f(x)=ln x; f'(x)=

(lnx) 2

x(lnx)2 (.!.-o)-(1n x -1(21n x..!.)

=rlnxx)ll; f"(x)= x

(lnx)4x

For ma x imum or minimum f'(x)=O :. lnx -,I =0 =>ln x -1 =0 =>lnx =l=lnc =>x=c(In x)"

12(!-0)-0-1{2. !) f"(x)l!l x=c W'I f"(c)= e ( c -!>o :. x=c w,�m

1 c 12. 12 m !'l'i1 l!l7l5iu � ¥1J 1'ftG'i! llllll � 9 kg��� 1 � � ,� 5.25 m � 1ffei 1!17l5iu ifu'i! ��

�����WC<l"i'�!ITTl��m-(a) 65kg (b) 6lkg (c) 63kg (d) 47.25 kgSolve: ______ iz_m ___ _

.75m 5.25m

,� I , \II R 9kg

6m

�Wfil�-9 W R -=-=-

0.75 5.25 6 9 W

:.-=- =>W=63kg 0.75 5.25

1 3 . l!l7l5iu � 8 km b- 1 Ci!C'l � I '1Tfi C� 16 km h "1 Ci!C'l I!!� <!"' Cffi m �"! � � � C<U'R 'lllt� '!1ITT51t<f �? (a) 120' (b) 150'Solve: ,d� � Scenario�-��� 1�<f�v���'"I

(c) 135"

v 8 v+ucosa=O =>cosa=-- =>cosa=--=>a=l20°

u 16

(d) 90"

14. 1!1���3cm��l�lml��C<l'1�1���¥11W'I�����?(a) O.Scm (b) l.Ocm (c) !.5cmSolve: <fiRli .. � t"l"f111J �'1 � � � Short � 1

�- f=<IJ�q"f

(d) 2 .0cm

F(3)= .!_mu 2 -.!_m(�)2

•. • (i); BC ��twt FS=.!_m(�)2

• • • (ii) 2 2"\2 2"\2

=1-(i)

u2

I

�=_L_= 4 =>�=.!_ =>S=lcm3

U2_U

2

1-.!_ 3 3 4 4

A�C u

Lo

lcm scm

_ - 1 s(n-1)2 . . ��T::, ��� -<:it'"IC<l'1'$WI� x��� x=--[forobJCCt1veJn (2n -!) ·

Admission Tech

Page 14: BUET Admission Question

15.* f Admission Tech

lP)'I � iorn fllnr � l!l<fi!u � l!l<ll � � 9 kg i5ffi .!)� � 6 kg om s� � ��"l"Cil �I�� "fllffil'f �'f �-(a) l.662ms-2

( c) 1.862ms-2 Solve: ,fllffil'l �'IF=9-6x9.819+6 = l.962m/s2

,�r

6kg 9kg

(b) l.762ms-2

(d) 1.962ms-2

16. ax+y+l=O, x+ y+a=O � x+ay+l=O �R�� a l!liPlR�-(a) 1,-2 (b) -1,2 (c) 1, 2 (d) -1,-2Solve: �-.slilrn � "WI 1� } !I= 0 :=> la O 1 1 � a1 a 1 I- a a-I

211 0 II c c =>(a-I) 0 -1 a =0 re;=-( I )' c� =-( 2 )] -I I 1 a-1 a-I ==> {a-lY[l(-1-a)-o+ 1(0-1)]= 0 [r1 � f.rnl] :=> (a -1)2 [-1-a-0-1] =0 �(a-1)2 {-a- 2 )= 0 :. a-1=0:=>a=I; -a-2 =0:=>a=-2 :.a=l,-2 17. 1!1<15� 1 2 'Sr'l<l'f ��-irctiJ 8 �� ,AA�� 7 wi�m'1� 9f!Wl�-irc"IJ 5���?(a) 360 (b) 390 (c) 336 (d) 480Solve: �=8 � :."fliffi!il<!1j=l 2-8=4 � :.��"f!ifC<lil =8c x4c =336

, l

I (c) -.Ji

(d) ../32 Solve: sin[ cos-1(-i)+ tan-1( � )] = sin[120° + 30°] =sinl50° = iShortcut : Use Calculator

3. �� [x y][� :J[:Jl!l9��'1TF!�(a) [x2a+xyh xyh+y2b] (b) [x2a+2xyh+y2b](c) [x2a+xyh] (d) [ 2x2a+xyh+2 y2b]xyh+y2bSolve: [x y][ � : ][:] = [ax+hy hx +by][; J = [ax2 + hxy + hxy + by2 ]= [x2a + 2xyh + y2b J Shortcut: a, b, h, x, y 1!19 � 'lTF! m � m calculator -!! calculate � 1

f Admission Tech

Page 15: BUET Admission Question

Admission Tech

20. <!!� � � � f.R;u � 'to �<( �<( ff<!1 � ffl I � � �'1"flif""'fm � 1 � A l!lq� B � � � � ffl, � A �'<1<11 B '-'111 � � :!!Gl���?(a) 0.50 (b) 0.72 (c) 0.91 (d) 0.93Solve: Herc the events arc iudcpcndcnt :. P(A) = 0.7; P(B) = 0.7P(AuB) = P(A) + P(B)-P(AnB) = P(A) + P(B)-P(A)P(B) = 0.7 +0.7-0.7x0.7 = 0.91

21. a '-'111 � 'IITC'!il � (1 +axf l!lil R•·,��·.-1. x i £� x' .!il �'! � 'i1!1<l �?s (a) -4

Solve: T,. 1 =8c 18.'(ax)'r

(c) � s

(d) �16

8! 8! =>-=-a [a"#O�mfilfuf.i��;iij3!5! 4!4!

I a x'<!!i11fl<.'1 =Sc a3 �--= l 3!(5}4! 4(3!}4!

4 .. 8 4 4 x <!111�·, = Ca =>a=-4 5

22. tanx + tan2x +tan3x = tanx tan2x tan3x�'1 x <!111 l!R m-

(a) mt12 (b) mt

4 (c) nn3 (d) mt

5

Solve: tanx +tan2x +tan3x = tanx tan2x tan3x => tanx +tan2x =-tan3x(l-tanx tan2x) tanx+tan2x ru: => -tan3x => tan3x =-tan3,c => 2 tan3x =0 => tan3x = 0 => 3x = rm: => x = -

1-tanx tan2x 3 Shortcut: Use Calculator

23. C<lSR � �"l"il <!I�� raF!IT'� � C<l<I 11� 7ms-1 ,Sms- 1 <!1q� 13ms·1� � � 't.::i::.C<l"1'm!il �� C<li!'l m-( a) 30° (b) 60° (c) 45° (d) 900Solve:�<!1$�� 7m/se 8m/s<!l!l�Wf 13m/s<!!!l,fl!Ref.l��:.(13)2=(7t+{8}7 +2.7.8coso. [�1fl]l 7m/

;r(., 13m/s.

lf-72-82 I . => cos a = a,; a 2x7x8 2 13m/s

=> a = 600 8m/s 24.* 36kg'5r.ffil <!l��t'>filFli"ITTflll'l<lof�ll� <!1<!5� '-'l11C<l<I 'il� 15km� "ffe<I?

(a) 4 N (b) 6 N (c) 2.S N (d) 1.6 N 15x10J

Solve: 6V=15km/hr =--mis =4.1667m/s :.6t=60sec 3600

F=rm =ml\v·= 36x4.1667 =Z.5N

' -

6.t 60 -•·!

;::; -_-. '-_-_ .- .!:11 ':;Fl �-

(b) -5I

(c) -7 � Admission Tech

(d) -7

Page 16: BUET Admission Question

Solve: lim 5n+l + 7"+!n �oo 5" -7"

26_ Jex (l+sinx)dx<!!il"l!R�-1 + cosx

f Admission Tech

(a) ex cos(i)+c (b) e' sin (%}-c (c) e• tan(f )+c (d) e' coi(i)+c

Solve: I= Je'( l+smx)dx = Je' 2 2 x = f e' -2 +tan� dx. [l + 2sin �cos�} [sec2 �

;

1 + cos x 2cos2 � 2 2 .-Po x . , ( ) 2 x 1 q1�, f(x)=tan- .. f x =sec-.-2 2 2 :. I= J e'[f'(x)+ f(x)]dx = c'f(x) +c [·: J e'[f(x) + f'(x)]dx = e'f(x)+ c] = ex ( tani)+ c

27. tan-I . <!Jil��$J�-(cos x -sin x) COSX +sm x (a) I (b) -I

1 (c) -2 (d) 2_1(cosx-sinx) -i[ ::: <!:] _1(1-tanx) Solve: y = tan . = tan . = tan --cos x + sm x cosx + smx l+tanx

cosx cos x

[ tan�-tanx J ( ) =tan -1 4 rt =tan- 1 tan �-x =�-x :. dy =-Il+tan-tanx dx

:S el��C"������l�'l�'iJ166WJ�l"IFll,fSlllt��? (a) 33 (b) 11 (c) 24 (d) 12Solve: ,fSlll n wl � � � ,i�"'iJI � "C

2 '.:'l�'llcili,"C2=66 => n(n-l)=66 =>n2-n =l32 =>n2 -n-132=0

2 :.n=l±.Ji+s28 =1±23 =12,-11 :.n=l2·: n :;c-ll2 2 y2 = 9x � t� P � C<!>tfu 12 WI <!I� t� � Wl-2) 9.50 (b) 18.25 (c) IO.SO (d) 20.25

f Admission Tech

Page 17: BUET Admission Question

Admission Tech

Solve: m, SP = PM = x + a

Ill�, y2 =9x => y 2 ={{)x :.a =�(y2 =4ax l!lil ��<mi I]

144 Ill�, p�� y=l2 .'.}2 2=9x =>X=-=16 9

:.SP=x+a =16 +2 =18.2 5 4

30. � y=px 2 +qx.""2�.� 2x 2y'-xy'�-

(a) 2y (b) 0 (c) Y (d) 2y2

( 3 2) ( 1 ...!.) 3 _.!. 1 _.!.2x2y"-xy' =2x2 2p+

4qx 2 -x 2px-

2qx 2 =4px2 +

2qx 2 -2px2 +

2qx 2

=2px2+2 qx-i ={px2 +qx-i)=2y

3 1. � A= 8i-1-3klll<lt B= 3i+4]-sk�. '5TWI s �isorn X ��"Im

(a) 3M (b) � (c) � (d) 35'/2IO IM lM IO

Solve: B lllit �<rn A lllit �"f =AcosO = A.B = (8)(3)+(4)(-1)+(-3)(-5) 3 5 35.fi.

181 �3 2+4 2 +(-5)2 =

5.fi. =

10

32. � 3scc4 0+8=I0scc2 0�,� tanOlllit ��-

(b) ±I

Solve: 3 scc4 0+8 = 10 scc2 0 => 3x2 +8 -lOx =0 [ x =sec2 Oimi]

=>X=IO±.Jioo:96 =10±2=2i

6 6 '3

(d)±1,±1

scc2 0=2 -==>l+tan2 0=2 -::>tan0=±1; scc2 0=� -::>l +tan2 0=i-::>tan0=± � 3 3 v3

33. ��'illMIPl�""'l"ffi�<f 2x-4y-9=0 lll<lt 6x-12y+7=0���11�?

(a) fi (b) 17 () 17

(d) ...!2._5 3.Js

c 5fj 6..J5

Solve: "'1"M>•mll � = 2R

�: 2x-4y-9=0 ... (i); 6x-12y+7=0 -::>2x-4y+2=o ... (ii) 3

I 7 34

-- (-9) -:R = 3 => 2R = _3_ => R = .22._ 2R

,_=-(-4)2 2.Js 6.J5

Admission Tech

Page 18: BUET Admission Question

" Admission Tech

34. <!l<!ifu �"Jl (-1,3) <!l<l� (4,-2)�f?ml C'>W-i � 1�f1iil 'll� m �r·w ?'I:" D-(a) 2J5 (b) 3,12 . . (c) 2 4 2 _ (d) :,� 1 _:_: Solve: .:+ 2'. = 1; (-1,3)-)-=I + i =l ... (1), ( 4,-2)-)----- l ... (u)�2 -b ab a b ab 2 Solving (i) and (ii ) b a=2, b=2 ·.� =·h2 + 22 =2.Ji � \!11<! � mm �911'-' WI (a)'2Ji:3 (b)3J2:2 (c)2:JF3 (d)2h:4Solve: From the fig. ( p ) = _Q_ = Q/3 :. _Q = ...JL. ::::;, sin a=!sin a-90° sin90° sina I 3sina 3

. �')2 - l 2.Ji Agam, cosa=-�I-l3J [�a�<r =-3 <!l">R, P � "Rf� .. r P + Q cos a = 0 ::::;,cosa=-!_ ::::;,- 2.Ji =-� ::::;,P:Q=2h:3

Q 3 Q _;6. 200 m <!l<i� 300 111 � � @1 <!l<llfu rz•,.; cm <!l� � � � C!fo'!"R� <1� 40 km 11· 1 l!l<i� 30kmh-10C"l <lPl!'<ffil I <Ti'1J "lm <!l'ill "1� � �?(a) 2 minutes (b) 3 minutes (c) 4 minutes Solve: �'!! � (<tf<ITT m � � � 40km I h <!lil � � 1 t z.rni � � [ t hour -!!<ml] 40t = 30t + 0.2 + 0.3 ::::;, t = 0.05 hour ::::;, t = 3 min utcs(Ans.)

(d) 3.5 minutes .2km c=J--- v=40km/h�u=30kwJh.3km -!!<fill <ll'ffi! @"ill F = (si -6] + 3k)N ¥! -211m"l � � <!i'f!W<l d = (31 + dy] + 5k)m Jfll'1 � 1 dy <!l'll 'l!V; <I"-' W'! � � "1ffi1'1 "['U WI? (a)O (b)5 (c)6 (d) -6 Solve: W =F.d ::>0=5.3+(-6}dY +3.5 ::>0=15 -6dY +15 ::::;,6dY =30 ::::;,dY =5� -2i"<!i!P'fli> <!l� m � x(t)=16t-3t3Cll�IR ,JmJ t c,rc-<m � 1 � � 1Wli � �Tc<' <I� t <!lil 'lfR-(a) 0.75s (b) l.30s (c) 5.30s (d) 7.30s Solve: x=l6t-3t3 ::::;,v=l6-9t2[v= �;]

,:;.,.,, 2 /16 Now, 1�ll� v=O :.16-9t =0 ::::;,t=f9 =l.3333seco::1.30sec � � � � C�ITT' -!l<!ifu "W!<!i '$ <!l� ff 7l""'iC<li C<l'<'T1 � I "I� �'1 WI- [Ans: b] (a) ?it� <i� Cffil wrr (b) h <!00 ,fl'!J,-l (c) h <!00 m'1! � (d) 98ms·2 Solve: <ff["fU � 'i<I "i1r m �'1 ;p;rr,, 1

f Admission Tech

Page 19: BUET Admission Question

" Admission Tech

40. I OOOkgmil <!l<!iW 'ii(ll,1<$11�.ler � C<lc<1 c>T%i1 "it'4 Gi&S�l-t � 1 � <l<l<! � I 800N. l&Cl\,liSTl�ICG\il ;s9f"il-21���Wl-(a) ON (b) l 1800N (c) 1800N (d) 9800NSolve: (a);!q<JP0��<l<'l IF=O

41. 16 kg \5m1 <!l<!ifu f0 � �"iil 4 s <m"tt 8 N «1 � � 1 � � C<lc<lil o;Ri,rs:i'-1 m-

(a) O.Sms·1 (b) 2.0rns·1 (c) 4.0ms" 1 (d) 8.0ms· 1

,W Fi'>t 8x4 Solve:(b); F=ma =>F=rn- =nV=-=>i'>V=-=2m/snr m 16

42. � ('ITT� i�'*-t''· 800 � (;<>'.�� .c� ����"I� 200 N � � �� 1 �

�'1-;m�rn �-(a) 0.25 (b) 0.125 (c) 0.50

F 200 Solve: (a); FK = 200N; R = 800N ; �i = i = - = 0.25R 800

(d) 4.00

43. ��"I� 'lTCWl <!l<!IW � <!l� l� � 5.2xl0"11m<!Wllt� <!l<!IW � "':t2.18xl06m s·1rn<1�mcrm 1�� 9.lxl0-11kg�����m? [Ans: BJ(a) 3.8lx]0-6 N (b) 8.32xl0"8N (c) 2.17x10...:7N (d) l.2Sxl026N

m _ 9.lx10"31

31 Solve: m=--0-[·:v,c<!l"il1-t�·w) = =9.10024x10- kg� 1_ (2.18x106 }2f-2"

(3x108}2

F = mV2

= 9.10024x 10"31 x (2. I 8x I 06

)2

= 8_3169x 10_8NC [ 5.2xlo·II

[DlW!. �'{3 ���mm m im � � 'lTrn 13•1-1 � �.,,� �WfGfJ m -11 C<l5'1 W"1 � 1] 44. f.!oo � 111� ��wit? [Ans: CJ

(a) 'Sll 3M <!l<!Z l<l<1 V (b) \S"il 3M <!l<!Z C<IS'f 2V (c) � 2M <!l<!� m 3V (d) 'Sll M <!l<!Z C<!<14V I Solve: (a) <!Ji! wn EK = x 3M x V2 = 1.5MV22

(b) <!l"il � EK = .!. x 3M x (2V)2 = 6M Y2

2

( c) <!l'il iSRJ EK

= _!. x 2M x (3V)2 = 9MV:

2

(d) <!l'il� EK = .!.x Mx(4V)2 =8MV2

45. "(ffl � M <!l<!� <lW!l� R � � � � <!lil �"fll!i m­G

[Ans: BJ

R2 M (a)M (b) R2 (c) MR2

GM g M Solve: (b)· g=-· =>-=-' R2 G R 2

(d) MR

.!.� .::<::"'; � x = 2cos(50t)� � � <1� �, � x <!lil •"i � <!l<!Z t <!l"il '>frn"l1!"i v,.<==, !"ts�".<''' ms-1<!l�m- [Ans:C]-= :OOsin(50t) (b) 100 cos(50t) (c) 100

t: ;; : A = 2 = w = 50 ; v run = Aw = 2 x 50 = I OOms -i

f Admission Tech

(d) 200

Page 20: BUET Admission Question

f Admission Tech

',{Y".JUVIJ V. \,)U!YI.;

47. <!"12ic!IK'iil � cl'llfu � � 1% "lliM..£<1 WI cl"il 'lW! � � Ei'l "'�[£� D? ""� �"iN 0.2.(a) 1% (b) 2% (c) 0.2% (d) 5�o

e I dL dx!OO 0.2 . Solve: cr=0.2 · -= -· cl"<Fl cr=- => 0.2 =-- => d =-D =>d = D .::� 0.2° 0

' L 100' ' De D 100 48. cl<rii: m,i � 250 mL "IWl 1.25 L "llR � ei'llfu <>iWll wn � 1 m,i ,s "Im "I� T<>-;-z- -� t•.,

80°Cei<1� 20°CW! ��"!"ii "IW!<I "i51"1"ITTll � WI?(a) 25°C (b) 30°C (c) 35°C Solve: (b); �. )11"� "i519f"ilf.lfl = 0

(d) 40°C

:. rn 1s

1(80-0) = rn

2s i (0-20) => v

1ps

1(80-0) = v

2ps i (0-20) => v

1(80-0)= v

2(0-20)

=> 0.25(80-0) = 1.25(0-20); By s olving, 0 = 30°C-l9. cl<rii: � �- [Ans:AJ

(a) lf� "'["IT R'"I� ������<AT WI(c)����

50. � <!Y � � 0Flfi; 15ll� i?f!<llstT?

(b)���(d)��OO!<!>ll�����

(a) ffl <f1IW! (b) ffl "llRSolve: (b ); � fl �-,1f.1·<1> 15191 C"<!P'l I

( c) <RTti'

2 Mach '1� .. � ie_m � i:JWWl cl'llfu � � � -.iro <f.""llt<!> � WI? (a)<!!� (b)�<f (c)�<f (d)ol'l�<f

. v + 2v 3v Solve: (c); Mach means the velocity of sound. f = -'--' f. = -' f. :. f = 3f. Vs Vs

200 � �rz <!l<filu � � � � � f.l� � �t� IOHz�? (a) 100 m (b) 50 m (c) 180 m (d) 200 mSolve: rn=200, f=IOHz, n=?

f=rrm =>10=200n =>n=0.05scc- 1 = 0·�5 Hour-1 =>n=l80Hour-1

3600 � 5Trci' � � CW!<li M ""� ,� N cli:l 1!�7'9fr'f � � I l5T"il "I.Jl'C<l­(a) ffl C'l'fl'rlS � m'Sl offi(b) t<t1<1'fi M � mcsf � cl<lt c� �� N � �(c) t� M � m bl�� ei<i� c� N mi' rffi"l'>;'f>(d) t� M mi' rffi"f':1' ..!!<It C'1T'1<1> N � mcsf &1R;{,:,Solve: (a); Both spheres are possitivcly charged.�"fl� 1!1f!! 30 Cm l.OxJ08 v� "fl� lit� 2.ox10-2s;pmi f.rn• �I,!/� 1!R'lj� .. ffi ffl'l �-la) l.5xl011J (b) 3.0xl09J (c) 6.0xl07J (d) 3.3xl06JSolve: (b ); E = qv = 30 x Ix I 08 = 3 x 109 J< <{'jcf 1il1RJol "!Ni � � I pF 1 "IITTKl '11� � ��'1 � "I�� "l!� � �'m9f Q!J1l�� Nell 91.cf <l>lli � 1 lW'I � 2 p FBT I C'l11V@ '5l�l<1<flrri<!> � �-a) 0.25 (b) 0.50 (c) 2.0 (d) 4.0

-olvc:(d)· C =E, A . C =ke. A . C =!pf· C =2pF> I d'2 2d'l • 2

. C _ 2C k E. A _ 2 E, A k _ 4- =>-----=> -. z I 2d d f Admission Tech

Page 21: BUET Admission Question

f Admission Tech

56.* 110 v� ;rfC'<j ,j�� <!1<!$ mtWl. "J.""0 � 0.4 cm 2 \!)re 3000K 's!il"i"[email protected] i5lfC'11 � 1 �

f<qi� 2FlITT:i'f 1"!F! �? [cr = 5.7 x 10-s W m-2 K -.; ] (a) IA (b) I .SA (c) l.68A (d) 2A

Solve:(c)· E=AcrT4==Vxl _._AcrT' =I =>l=0.4xl0-•x5.7x10-sx(3000)

4

=l.6789A' \' 110 57. 10 "fITTiil '-'l<l>fi, �f ,,fq •. rnt"il � .. �<�" 3.5 mH. ,,;;.,tm.l!oil l!c<!J 2 Af<i,J� -21<ffi� Wl �� "I�

iw!J � i]'f-;; Wl-(a) 0 Wb (b) 3.5xl0' \\b (c) 7.0xl0-4 Wb Solve: (c); \"9=LI =>10o=3.5xi0-'x2 =>�=7xl0-\vb

58. BJt I!!?. '>i� ..... � .!:� �- � � '<""Z-s <!':<"' �-

(a) m�1 <,-.; ,.�

(d) 7.0x\0-i 'Nb

(c) "1,'<R 1.<r.�< "ll'<R'i � F-!i (d) .£-efG ITT E � � '-'l<i� �"l"l!� B � 'SR]

Solve: (c); To create a path difference59. !Ohcnry ffl� "1<!$� "ll<!J � 12 /\rqi�� 1 '-'l� 50ms-'-'l 600V��wr,:; -w=.

� Wl �� -21<!lm 1!R �? (a)6 A (b)9 A (c)12 A (d)l5A

Solve: (b); i::=-L � =:,600=-10 2 _3

::::>3=12-12 ::::>12 =9A (I -I

) (I -12)

t 50x\O 60. 4 m0i�w@ JJ'efi!l�-'f!""h'I "lfGJ!l��"ll�� 36 cm . 01'1l1f�1ITTI C<!'T<WI�<!>��·

(a) 7.2 cm, 28 8 cm (b) 5 cm, 20 cm (c) 45 cm, 180 cm (d) 27 cm, 108 cmf 36 Solve: (a); -!.=4::::>f =4f ::::>fe =-=7.2cmf, • ' 5

f. +f, =36 ::::>4f,+f, ==36; ( =4f, =4x7.2 =28.8cm61. ..!!<!$ 40 WV3 '-'l'efi!l 60 w� �q1 � � wi C<f5R � c<r''lt @"'"''1 � m1

(a) 40 W lamp (b) 60 W lamp(c)lf��,mr;i (d)-rr��i"lil�<f.i!"C<lSolve: (a); G� � <ll"ll � <ft� @�l"l'-'�"11.<l �

- • 8 &'' • � 62. miffil wr.i c� � !f!"l 3. x 10· mWl � ,i�� oWf<lJl(jm ·l(j CIRRi �PfKI � a:ic.�1·w•1� ,im:-

(a) l.3s (b) 2.5s (c) 8.0s (d) 8.0m in Solve: (b); d = 2x 3.8x 108ml"I�� ..!l<lR!lil � 'm<!Rl � 11

v = 3x 108111 / s[Radar signal is an E-M wave]; t = � = 2x 3·8x l08

2.Ssccv 3x 108

63. JJ'efi!l f.iWl � � � V'rnl w; <!lftw?. G"c� �� � � 1 � �� � C<l'R <\!'11 �ZTI(a)�� (b) c� (c) � f.iWl i5ffilr! (d) � RW1 � Solve: (c); Posit ively charged neon ions proceed towards cathode.

i;..: 450 W <!lil IPS '-'lil "l!l'm1! � 00 � mR 1flJq � � fi "l1'iA 90 W '-'l<I� �ml; <!fRj 45 W � a)-+ fans and 3 lights (b) 3 fans and 4 lights (c) 3 fans and 5 l ights (d) 4 fans and 4 ligh3 olH•: (a) -!!i:"""11 =4x90+3x45=495w [Ans:: _ !<� 3x90+4x45=450w [SotheAnsis(B)] : .£?;""'" =3x90"'-5x45=495w (d)'-'1°9� =4x90+4x45=540w

f Admission Tech

Page 22: BUET Admission Question

f Admission Tech

(a) EC (b) EC2 (c) JFE (d) ic

[:\ns: d]

Solve: We know that General Fonnula, E2 = p2c2 + m:c·1; m. = O :. E2 = p2c2 => p = �

66. <!!� 'l;l� 'l;l�"lm p - �� n -'Wl Fl'TI! <Vil!�-(a)���� ?fm-� c;.rc91 (b) \� @n:� "R1 r,.rr.91(c) �Rl '519f � ffi?f (d) �-� "l�l'".;SW, �Solve: Hall-effect experiment determines the maJority career of current now.

67. �<!!R,��-

[Ans: d]

(a) C:!lW<l (b) 1<1�"68. � "ITTP'lfoil 'l•1'511CW11�1�@f.mr•R �-

(b) IDffl

(c) "lfu'wi (d) -£if.u c� [Ans: c][Ans: d]

(d)�!!P'B

(a) C<I�� (b)� (c)� ( d) '11<170. �'t!i 91'11'<1�-

( a) f«!i firni �<f'1'5fr<l' � �ll(c) � fu-!i >:f1� � �ll

(b) F,«ll firnl 21� �� �

(d)�fu-!i����71. <!!<;{-@ <!!!! :::;;'IJ � ,;wis <ffi?

(a)<!!�� oTefi �(c) .!lfu � NCTJ f4�,\:) � .r1

(b) <!!fv ��r-,i;;ii m � � '11(d) .!lfu .!l'<!>ffl � .!lf<11,,{� �r <rm

72. � c<W1 0Pi2J <!l<1 C111J C<J �wn <!I� �:i'l-� � �"l'D "ff@? (a)� (b)�w,rnq (c)�Solve: Polarisation is a unique property of transverse wave.

(d)�

(Ans: a] [Ans: d]

[Ans: d) [Ans: d]

73. liWi'hl '3l<l",if"!! f.l!IN>:ji � .!l� � C«li1 <r-<rC'1 C<l "ffi "IIB<ll!l•t � �fl<'I <!'11 �- [Ans: d] (a)�1� (b)•rirll<f� (c)�'°11.>JN?, (d)�<f"IWSolve: According to the definition of electron affinity.

74_ ;w.oll �c'lf?. l!1:l:fJ ���cw�! <11� � �? (a) Alp3 (b) Si0 2 (c)Li02

·s. 9f,fui-rriqri sif91B11Wil IIAm"i i� �FITTl 'w!$ � �(a) ls2 2s22p6 3s23p64s2

(c) ls22s22p63s2 3p6

(b) ls22s22p63s23p64s2 3d2

(d) None of the above

(Ans: d]

[Ans: a]

"6. Co(NHJ4 Cl3 <!!!! � ff<!t"f 'lit� "ITI!m'I Agi'-103 ff<l<f ClJ1°1 � ��g0l'f'"f<i> AgCI .!l!! cm,, ,r�<IJ1 �(a) I (b) 2 (c) 3 (d) 0 [Ans: a] Solve: Co(NH3)4 Cl3 = [Co(NHJ,(C!}i ]' er [Co3

' .!lll ,if:iw1 "!�"'in 6]

[Co(NHJi CI)2 ]cl+ Agi'-103 --> [co(NI-1,},i( Cl)i ]N03 + AgCI J. So I mole AgCI is precipitated [Ans: c] ·7, � ,fffl%CcJrn �\1J C<J>r,,ju �'1-r<l�Cf f<ifu.m?

(a) Crp;- + 201-1�--> 2Crp!- + HP(c)K+02 -->K02 (d) ·ca(HC03)2 --> CaC0 3 + C02 + Hp

Admission Tech

Page 23: BUET Admission Question

Solve: I

(0)

K +

Admission Tech

f<fG!r,�

78. .!)<1$ CW'rn � f.rnt"f'=l' "l111Tl'fl w,,; b � � ,@ll?(a) � 1f'l1Fl 11t� C.2fW,l, � "<3 � �

(b) � 1!'llt-1 '1t� � (l (2l"W-l �( c) � ,jllf,I 'it�� (l c�'1U-l �(d) � 'i'l1R 11t'l!J<!' � "<3 � �

Solve: They have the same number of electrons and protons.79. "flfm 'ilil1ml � �9!�m � "<3 ,s� ;;.sd: �?

(a) IA, IIA, IIIA, IV A (b) llA, IlIA, IV A, VA (c) !IIA, !VA, VA, VIA (d) IIA, !VA, VA, Vl1A

80. 'I'� �� 5<1'liu �-(a) <!l� <!]� 1%��'1 (b)�l!lmil 1%�5<1'1(c) l!lPIITT> l!lPrm 5% n �'l (d) � ,!!Pfm 5% � �'l

81. ��'1ilmm��ffl��-

(Ans: c]

(Ans: c]

[Ans: c]

<a>�� (b)�� (c)�R� (d) b '<l cm [Ans: b]82. '5Wl im -a'!WPj �'1 � 'a' 'llR! 'ITI:'lf ��-

(a) �&-�<fml�<f (b)���fffl<t (c)��t?'ffil��(d)b"<3cffl [Ans: a][Ans: a]

(d)fu'W1�'183. ����2i�f1i�?

(a) '5Jfil Ftil <J� �'1 (b) � <!Wl (c) � <!lvi84. f.lm���zjire��'l (O-N-0)���?

(a) N02 <NO;< NO; (b) NO;< N0

2 < NO;

(c) NO; <NO; <N02

(d) NO; <N02 <NO;

[Ans: d]

Solve: N "rnll!'J:il � mi m!l � m � �ii �'1 m ( VSEPR �) ll't'1" � �'l � I :. i!F'II: No; < N0

2 < No;

85. f.lm � DNA '<l RNA <!l� ,ii�91.cf � t"fl'iFl? [Ans: d] (a) Mangnesium(b) Sodium (c) Sulphur (d) PhosphorousSolve: Phosphorous (P)

86. f.lm � lllm-� >rn ll� c,roi � mRl ,m;'l! � '<fITT'i? [ Ans: a J (a) -C

6H

5

(b) -N02 (c) -C = N (d) -COzI-I

[Ans: b]

�Q

CI Br

jal� Cl

Cl

(b)6'�Cl

(d} Equal amount ofb & ;

Br Admission Tech

Page 24: BUET Admission Question

" Admission Tech

Solve: - Cl is an ortho-para indicator.

Lsr+ � Cl

88. f.rWl <!l�m C<ISR ��� � �?2 0 3 0 4 5

II II CH3 CH2 - C-CH

1- C -OCH1 - CH

3

(less)

(Stcric hindercnce)

(a) I (b)2 (c) 3Solve:

1 2 0 3 0 4 5 II II CH

3 Cll

1-C-CI-1

1-C-OCH

2- CH

3

no. 3 is most acidic. 89. C<'TiS(II)�����mmq><fj�?

(a) HN03

(b) H2S0 4

(c) HCISolve: Only PbN03 is soluble.

(d) 4

(d) HotHp

90. � xrirn <(rn'ffil � <lJWTI.Wf!!i1 <!l<llfu 1!5/mqn: � I 'ilJW!T-�'ffil �(<lW!) �-

(a)-10-9 m (b) 1-IOOnm (c) 0.59xl0"8cmn mm C<!iR .?lfiFllfu '019f '@<1.9f�?

(a)� (b)<11""1� (c)�"fw-1.-iz. � C<ISR cifts$ C'<ITT> �c- � <l'>il1 � -11?

OH

(•)6

Solve:@--CH,

NH,

(b)6

CH= CH, CH,

(,)6 (d)6

(d) O.IA

(d)�

[Ans: c]

(Ans: a)

[Ans: a]

[Ans: a] [Ans: d]

:3. !1<rrrs�fw.i<.!l�����ci:r·������- [Ans:b] R

(a) CH3Mg8r (b) HCHO (c)RCHO (d) ""'-c=O

-! <t>l-1�1$:®d,ll fii� tl1'll � � �� m ,rrr;� � � � � �-RI.......-

(a) HCHO (b) HCHO+ RMgX (c) NaN02 + HX (d) HCN [Ans: d)

-� <.!l<llfum��"fTCill�w,ci:ri519i'lf1il!l'!l<!l""i<.!1<1���������- [Ans:b](a) "<t>'l!' �"fllm (b) 'V'!J�<!> �'>fllTcm (c) &.n�� �"f'l!r.ill (d) tir C<llFl �9f:1r3ll

'5 �����<!'l��? [Ans:a] (a) rr�� 91lf.! (b) � 9fRrn (c) �� '51Jl'1l<>l�'1 (d) �

• - 27°C i5f9flWilfl! <.!l<llfu 'i1'ifro <I� � <Im'� J 3Q4cms -I , � �-

(a) 02

(b) C02

(c) NH3

(d) S01

[Ans: Blank) f Admission Tech

Page 25: BUET Admission Question

� Admission Tech

Solve: C= /3RT �C 2=3RT: ill=

3R: =3x8.316xl 07x300 =44_015xlO'gm/molc

\ M M . c- (1304)2 There is no answer.

98. 40°C 'i5J''i'l1'f!ml N 20$ l!l?. l'lC!!l§il 'zz.n �� 45 min . �ffl C'lS1 � �-

(a) 0.105min-1 (b) O.OlSmin (c) 0.02s-1 (d) o.01ss· 1 f/\ns: b]

Solve: K= 0·693 =0.015min-'45

99. �lTT'l �?fllf.m <3 �.,.. � �, , '"i< � - \.,!,� � CG"ri? [Ans: d](a) I-!, (b)He (c) :\'itrogcn (d)�-m

100. � -;;'h- ""T .!�­

(a) °J's" <c

.l•l.

[Ans: d](d) m'l-�'l 1<1fhm

[Ans: b](a)�-!.<�-: 01:'!: � (b) � �W. (57:!) ".'P'1(c) � � ,;r.pj (d) ;;r�I?. iS"i!: � m r.,1 C<'iR� m ffl

102. lWITTffl f<traRrrn Cl! �"f'M "I��� @�,i � � f/\ns: d](a) "lfw•rfu; (b) � (c) rirw �rfu; (d) �-::r·\i'ioi--1fus

103. ���f.Rsc!l<1��,i�rn�<t? [Ans:b](a) NH0Cl(s) � NH3 (g)+HCl(g) (b) Nll4Cl(aq) �NH,'.(aq)+Cr(aq)

(c) C 12H2p11 (s) � c1i1-12p11 (aq) (d) CaC0 1 (s).,.--=!Ca 2 '(g)+Coi-(g)

Solve: NH4Cl(aq)� Ntl;(aq)+cr(aq)104. 1Wl �R!lfll (i5ll 238 <B<!t "iHllll'IP!<!i 'it'<lJI 92 ) l� l!l<f.W a - <f.'fl f..*'1; �I. 1S�"llf"f\5 � � -

"l"J]l!l'ff.f<ls 11�<!TI 'll'<IK."Vl [ Ans: al(a) 234 and 90 (b) 234 and 92 (c) 238 and 90 (d) 238 and 92

2)8 · 234

Solve: 92 · U - a ---t90 Th105. ���1\.llprn'i"!l? [Ans:c]

(a) Benzidine (b) Glycine (c) Adenine (d) HistidineSolve: Adenine is a base

106. 01-i cm•r C= 18.5%, H = 1.55%, Cl =55.04% \Jl<l� 0=2181%©"1� � 1 C<Tl"ITITT111Z� RIO"IWl?(a) CHCIO

Solve: C : H : Cl: O = �: 1.55: 5

5.04: 23.8112 1 35.5 16

= 1.54: I.SS: 1.55: 1.48 "'I: I: I: 1 :.1o-J 'iZ� CHCIO,-- ��DNACl:3�?

a, Adenine (b) Guanine (c) Uracil � ::g C.2C0 3 ��ffl���� CaC0 3 �r-fil��?

� o 550 g (b) 9.966 g (c) 9.881 g- he: 6.0nx102ir,�'1.CaC0 3=100g

(d) Thymine

(d) 9.662 g

2 ::: ��:: CaC O, =0.0332g :.wfi =(I0-0.0332)g=9.966g

f Admission Tech

(/\ns: a]

[Ans: c]

[Ans: b]

Page 26: BUET Admission Question

� Admission Tech

For question numbers I 09-112, choose the correct oprion that ,1i1J complete the corresponding sentences.

I 09. She was very pleased by the appreciation she received like everyone els.:. ,he .:-:::o�i.'d h.:mg _. (a)entcrtained (b)praiscd (c)playful (d),indicatcc '.\u, B,

110. Pioneer men and women endured terrible hardships, and ' . .\!:.;; :::>; (a) so do their children (b) neither did the children(c) also the child (d) so did their children

111. 111c changes in this city have occurred __ . :.\.ns: DJ (a) with swillness (b) rapidly (c) fastly (d) in rapid ways

112. IOOO species of linch have been identified. [Ans::\; (a) As many as (b) As many (c) As much as (d) Much as For question numbers 113-116, each sentence has four underlined words or phrases. The fourunderlined parts of the sentence arc marked (A), (B), (C) and (D). ldcnli fy the one undcrlmedworld or phrase that must be changed in order to make the sentence correct.

113. Visitors were not pcrmilted ente ring the park after dark bccuasc ofthc lack Qf scctuity and Lighting. I AllS: BJ A B C D

114. Writers like William Shakespeare and Edgar Allan Poe arc not only prolific but too interesting. f Ans: DJA B C D

115. I need both fine brown sugar as well ill? powdered stigar to bake a Hawaiian cake. [Ans: A]A B C D

116. Evm a professional psychologist may have difficultly talking calm and logically about his ow11 problems.A B C D

(Ans: BJFor questfon numbrs 117-120, read the following passge and choose the best answer fromthe four options marked (A), (B), (C) and (D) that following each question.Although speech is the most advanced form of communication, there arc many ways ofcommuncating without using speech. Signals, sings, symbols, and gestures may be found inevery kwon culture. The basic function of a signal is to impinge upon the environment in such away that ii attracts allcnlion, as, for example, t.hc dots and dashes of a telegraph circuit. Symbolsarc more d ifficult to discribe than either signals or signs because of their intricate relationshipwith the receiver's cultural perceptions, ln some culttires, applauding in a theatre providesperformers with an auditory symbol of approval. Gestures such as waiving and hand shaking alsocommunicate certain cultureal messages.

117. What docs the author say about the speech? [Ans: DJ (a) II is the only t1ire form of communication(b) It is dependent upon the advances made by inventors(c) II is necessary for communication lo occur(d) II is the most advanced form of communication

118. According to the passage, what is a signal? [Ans: CJ (a) The most difficult form of communication to describe(b) A form of communication which may be used across long distances(c) A fonn of communication that interrupts the environment( d) The fonn of communication most related to cultural perceptions

119. The word "intricate" could best be replaced by which of the following? [Ans: BJ (a) inefficient (b) complicated (c) historical (d) uncertain

120. Applauding was cited as an example of [ Ans: q (a) a signal (b) a sign (c) a symbol (d) a gesture

f Admission Tech


Recommended