FIITJEE INTERNAL TESTHOME PAPER

PAPER- 2

DATE: 14/04/2018

ANSWER KEY

PART.1 : PHYSICSQ. 1 2 3 4 5 10A. B,C A,B,C,O B A,B,C A,C A,B,C,DQ. 11 12 13 14A. C A 0 0q. 1 2 3 4 5A. 4 5 5 3 8

PART.2 : CHEMISTRYQ. 1 2 3 4 5 10A. C,O A,O C,O A,C B,C BQ. 11 12 13 14A. A A B BQ. 1 2 3 4 5A. 4 4 6 8 4

PART.3 : MATHEMATICSQ. 1 2 3 4 5 10A. A,B,C B,C 0 A,B A,B B,CQ. 11 12 13 14A. A B A CQ. 1 2 3 4 5A. 8 1 6 5 8

PART-1 : PHYSICSSECTION-!

I. Ans. (B, C)

SOLUTION20rnls

~

,v = \\ 20 = 5n Il1/S

Sol. A Is~---------:c

Ffor complete rod: ma = F => a = - ...( I)

III

at any distance x from edge let there be a crosssection for which FBD is given:

I

I

\' ,

4()\\'

d

11Ig

E .\,) ((

v,= 10 (2 + ll):. d = VI . 2 = 40 + 20llx" = 20 II

Ans. (B)

In frame of ring work energy theorem

(just before collision)

At t = 2 sec => v = 20 - lOx 2 = 0,=> collision takes place at maximum height ofprojectile

202Maximum height => H = -- = 20m

2x 10:. After collision velocity of particle = 20 + lOll

-flgl (I - cos 8) + ma sin 8 = 0 - 0=> 8 = 60°

4. AilS. (A,H,C)

Sol. ~mv~ = mgh = 50 => Vu = 10 m/s

y+-(A)'~ ~,

3.Sol.

IB,

x

T---ia-C

( IIlx)for portion BC : mile a = T => T = L a;

( IIlx) F F F,T = - -;::; - x ::::> stress = ::::::>-

L III L AL

FxdC sTress LA Fx

(C) - =--=-=-dr Y Y LAY

'fFX M F=> /',C = ----fix => - = --"LAY L 2AY

L I F'(D) Energy stored = f ;;- -- x' dx A

(I L. L2A2y

IIand angle rotated by plate in 2 sec is 0 =W1=4"'

II2= -

2

.!. F' A (L' 1 F'L

= ~'A' Y l3) = 6 AY

2. Ans. (A,H,C,D)Sol. For ball x = 40 => 20 t = 40 => I = 2 sec

HS.l/B

... just after second collisionAns. (A.C)Ans. (A,II.CD)Ans. (A,II,CmAns. (C)

Momentum conservationv,=6Iv,

I , I ( )'-x60xv;+-xlx v, -v, =mgh2 2

=>v=~IOO, 3660

v = 61 v, ,10 {60V61

8.9.Ill.II.

v.-m Dv,-m

Time taken to travel from shown position is 55days

12. Ans. (A)13. Ans. (I))

...(I)

a cos U1

a = -~g--(ml sinal -1112sina2)Int +m2

U,m in horizontal direction.. _ I11lacosal -Ill:! cosu2aon -"'1 ..' (2)

IV' + III I + III .2

From cquation (I) and (2)

Sol.Impulse m (v, - v,) = mv"=> v, - v, = v" = 10 m/s

v~

(C)'~ ~,

v, = 10 m/s10

v, = 61 v, => v, = 61... v = v, _ v, = I()_.J.Q = 600

61 61

mv:!fA =mgsin8+ R

... If car does not slip at A. it will not slipanywhere

mv:!so avoide slipping => fA- mg sin 8 =R

mv:!=> IJmg cos 8 - mg sin 0 =R=> v = ~gR(flCos8-sin8)

Ans. (A,II.C)7.

5. Ans. (A,C)6. Ans. (B, DjSol. At A, requirment of friction is max (for circular Sol.

motion)

Sol.

Just after first collision... just after first collision V,,,, = 0

"'"(M+Ill] +1112)x(ml +fll2) x[m~ sin al cos u1- Ill:! Ill] sin U2COS <Xl+ In;!

1111 sin at cos <Xl - m~ sin uzcos all

D

Just before second collision

gal'm= (M+ITI] +1112)x(m] +1112) x [m,sinu]Imlcas al + 1112 cos u21- m2sin u21mt cos al +m, cos u, IJ

HS.2/7

..

a = O. so<;111

g

~ x [m,sin a, [m,cos a, + m, cos a,j-ml+m~ - -

l11,sin a, 1m, cos a, + m,cos a,1I= 0m,sin U2 - m2 sin 0.2 = 0

14. Ans. (D)

I , g <Jt ~-gt" =-t t +.2.....2' 2" 4

I, 1+13~-=--l2 :2

3. Ans.S4. Ans. 3

Sol. t, =phg

Along the incline.

~2gh I g,h=--t,+-x-t;

2 "2 2"

Ans.4

Nisin 0.1 = N2 sin 0.2

mig cos 0.1 sin 0.1= m2g cos 0.2 sin 0..2

ml sin 2 Cl1 = 1112 sin 2(x2SECTION-IV

1.

Sol.

S I ~l:=0.26o. 2S. Ans.8Sol. Total time taken on the circular track.

~0.26 x4xl0

2.

a, = 2.6 mis'

a, = 5.2

W,= ~(mg)(x, -x,)

= - 0.26 x 4 x 10 (1O-~X2.6X :.~)

=-0.26x4x IOx5W, =- 52 JAns. 5

6.

Total time. t = 2x (_2_) + t,0.25

t = 48Ans.3

man

x

Iv,Hv,I+Iv,1 = 10+9+20 =l13 13"

apple

thus.

2v~ T = 4 = --' ~ v = 20,

"c

~v-IO=O~v=IO, ,Now, range = 60 = (v, + 6) x 4~ v =9,

v"ho,,"=IOi +6jv, ,"ppl,' = (v, + 10) + (v, + 6)

60

- ( )- -v =v = v -10 i+ v +6 '+v k~rplc wr_l. m;m am (\ ) y J ,

Sol.

2x

(nut': dongatllJn)I P{lSI!IOn 2,,F i"'7l--------r-l I'J

I

r--'

(ma, cUlllprcsslI.JIl) '\ = UpoSitIOn 3 I posllion I

i

Apply WET between 2 & 3

3 ,--kX" =-3Fx2

2F 2xl0 5x=-=--:;::;-m

k 16 4Now. applying WET between I & 2

I ' I , ( )-k(2x)" --111\'" = F 2,2 2~ v = 5 I11ls

Sol.

HS.3/7

PART-2 : CHEMISTRYSECTION. I

SOLUTIONI I. Ans. (A)

I. Ans. (C,D)

2. Ans. (A,D)

Sol: SN' reaction does not invol ve carbocation

3.

formation and takes place with inversion of

configuration.

Ans. (C,D)

12. Ans. (A)

Sol. ~OH B,,,, yO(e) CH,- C11.- CH = C - CfI, - Oil ~ CII,- C11.- CH :: C - ('HO. tH,' . ttl.

IH/N,v

CH~CII,- ell, -CH - ClL - ou.. b, .

13. Ans. (B)

14. Ans. (H)

SECTION - IV

I. Ans.4

Sol. (I)

ID) CH,-CH:-7'=CH-ClI:-OI{ /dnO.} CH,-CHI-1'=CH-CHOell CII, 2.

k~iv

CH,- CII,- ~H - ell, - CH,- OHCII,

Ans. 4

is better leaving group than

HS.4/7

PART-3 : MATHEMATICS SOLUTION1. Ans. (A,B,C)

- - -axb+b=cxa+bxc ..... ( i)

4. Ans. (A,B)

5z' - 2z (ky + x) - (x' + 2y' + 4xy) = 0

taking dot product with a and b respectively

a.b =[abc] & b' =[abc]

2(ky+x):tJ4(ky+x)' +20(x' +2y' +4xy)z=----~~---------

10

=:> 4( k + 10)' = 24(k' + 10)

discriminant of quadratic inside square rootmust be zero

2.

Let a = b = k

=:> [abc] = k'cose & [abc] = k'

=:> cose = I where e = a f\ b =:>

putting in (i)

a=cxa+axc =:> a=O =:> b=O

and c can be any general vector

Ans. (B,C)

=ky + x :t ~(k' + IO)y' + 2( k + lO)xy + 6x '

5

sin-Ix = cos-Iy

Letu = [\=:> 0 S; a S; 1(/2 & 0 S; [) S; 1(/2

=:> sina = sin[) =:> sin'a = sin'[)

=:> x' + y' = Iwhere 0 S; x. y S; I=:> 5, represent quadrant of a circle similarly 5,tan-'x = cot-'y

let y = I) where x = tany=:> tany = tanl). and y = cot I)

I I=:> x=- =:> y=-

y x

s.

=:> 5k' - 20k - 40 = 0

=:> k' - 4k - 8 = 0

k =amAns. (A,B)

LetP(2/ .. -3i~./~)

Q(3~l + 2. I -5~l. 2~1+ 2)

3.

wherer x > O. y > 0which represent exactly one branch ofhyperbola

Ans. (D)

If n, =n, +i~n, then P,. P,. P,

=:> (A) is incorrect (B) is incorrect

if n, = 2n, +3n, PQ.b = 0 and PQ.c = 0

then :3 no i. E R for which n, = n, + /~n,

(e) is incorrect

if n, = 2n, +3n,

then :3 no A E R such that n, = n, + An,

(D) is correct

to get A and ~l

hence equation of line PQ will be

I 2x--=y-2=z--3 3

HS.5/7

~ (n + ~) 11< a < (n + 1)11

Ans. (B,D)

a.k <0~ sin(2a) < 0~ 2nll + 11< 2a < 2nll + 211

(2n+l) (2n+2)---11 < a < ---11

2 2

..

6.

~

11~ -<a<lI or

2b.c =0

311-<a <2112

9.

~ f(a) = 7, f([3)= 9

~ ordered (a, [3)= 2 x 4 = 8

similarly, if f(a) = 9, f([3)= 7

~ ordered (a, [3)= 2 x 4 = 8

and if f(a) = f([3)= 8

~ ordered (a, [3)= 4 x 4 = 16

Ans. (B,D)~ Towl ordered pairs (a, [3)= 32

('n" )"'"''f(x)= x' ""2 + x' .

10. Ans. (B,C)r+rxa=axb

7.

8.

~ tan'a - tana - 6 = 0 and sin'::' > 02

~ tan'a = 3 or-2~ tan'a = -2~ a = 11- tan"2Ans. (A,R,C)

.[x +~ + ,...J. b[l + x+~. ~+ ...J.[l- x .• .:..:._~ + ..J3 2~ 3! 2! 3!

lim ,I.~o x:. a = 6. b = -3 & c = 3Ans. (A,B,D)

t'() • • 5x=x+x+-,x'

-I

t" 6' 4' 10(x)= x' + x'--,x

put f(x) = 0~ 3x' + 2x' - 5 = 0Letx'=t

~ 3t' +2t' -5 =0

~ (t-I)(3('+5t'+5t+5)=0~t=l... t<Oif 3t' + 5t' + 51+ 5 = 0f(a) + f([3) = 16f =7

nlln

11.

(1 )' ILt - =-

:t-+I 3 9

~2,,'"1~Lt e(2 .•.,,:!I1I~1 = e-4

x:-.""

Taking dot with a, b and a x b. we get

Ta = 0

Til+[rail]=o

and [r a: il] = Ia: x ill'~ (8). (e). (D) are correctParagraph for Question No. 11 and 12Ans. (A)

L rxb=axb

~ (r-a)xb=O

~ L: r=a +"-b

A.-J

-,B

'- -1',: Tn, =d,

1',: Tn, = d,HS,6/7

- -~ b.n, = b.n, =0

12. Ails. (II)If Lis II toP,&P,

(- d, - a.n,) S' '1 I ( d, - a.n,)=> A a + _ I Inl ar y B a + . _ .

b.n, b.nz

Solve L & P,

~ (a+Ab).n,=d,1 d, - a.n,=> fI.=~-~

b.n,2.

~ vectors must be of the form {d.IllJ. nk I2 x 2 x 2 = 8 unordered triplets of non zerovectors<-: e. m, n E ( I, 2))AilS. IAt x ~ 0'. (.) = x

X" cos(x,in,,+I)lim --------- = k,-,,- cos(tan x - x) cos( x - x)

k=l.as lilllx'=1\ ....•0.

13.

and a.n, otd, and a.n, ot d,

Paragraph for Question No. 13 and 14AilS. (A)

f(x) J-~2 ::~

lre/2 x>O4.

Ails. 6Lc( f(x) = allxll + :.l]xn-I + a~xn-2+ anPutting in given equation and comparingcoefficients of powers of x.we get f(x) = x + b. where b is constant:. 1'(20) - 1'(14) = 6Ails. 5The domain of the function

SECTION-IV

10 . xotkre+~

f(x) =re

I x = kre+-1

and a,; + b,J + c, k are mutually perpendicular

:::::> ala! + b,b1 + CIC! = 0~ a,a, + b,b, + c,c, = 0 and a,a, + b,b, + c,c, = 0

.: a" b" c, i {O, I, 2, I

Range of g(f(x)) is 0 or 1Now 2 sinh (xl = cos(reg(x)) + cos(2f(x))

3re ?' h reLet x = -+ _SII1( (x)) = 1+-2 2

(3re) . "( I re)~ h 2+ =SII1 2'+'4

3re 2' h reAt X=-. SII1( (x))=-I+-2 2

h(3re) . _,(re I)~ 2 =SII1 '4-2'

3re~ h(x) is discontinous at x = 2

AilS. (C)h

_,(x' -4x+5)Sin"(I-X)).y = cos --------- ISX= 2cot-'(cos" (3 - x))

consider point (2. re). and circle(x - 8)' + (y - re)' = I

@...(2.IT).•••• B

C..... ";;...,

(8.n)

re~ 1'(0') = f(0-) = f(0) ~ 1-- = b

2

_Sl_'n_-_'h_-h_h_C_O_S'_'_h.- ( I - 2:2)

~ f'«n= limh_O

~ a = I ~ 6a + 2b + re = 8Ans .••

I tan( rex)f(a) = I+sin(l + x)-cos(rex)

-tan(rex) 0 < x' < I

lim f(x) = 0, and g( I') = g(l-)H'

All is the minimum distance = BC - AC=6-1=5

Ails. 8at x = O. f(x) is continuous

~ 6=2+bb=4

5.

6.

- - -a,i + b,j +c,k

Ans.8- - -Leta,i+b,j+c,k.

I.

14.

HS.7/7