1AITS-PT -III (Paper-2)-PCM(Sol)-JEE(Advanced)/17

FIITJEE

PART TEST - III

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......................~~M!~m¥..............................M~Img.;n~~....•.••...C A

PAPER-2

JEE(Advanced)-2017

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ANSWERS, HINTS & SOLUTIONS

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E~CIlCIl'"<3E~c,,"c .•.•0°..I'"e,;0'".::gCIl'"->c"~<t::lWu;~:; .=~"2~~<t~OCT~~c.",O.c~CIl.: eM'"",-.•..8'ri~C(=o<t

~gc._0-~CIlcc.- '""'''•... "ciu><t~g:f.•.. ~8'e~'"co,.- 0

"'-MQ.~o.c,gCIl ()cCil

'"'tl

"iilWW

~u:ui.•.•o

'""'"()c'">"<tWW..,.:

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2AITS-PT -III(Paper _2)_PCM(Sol)-JEE(Advanced)f17

Physics PART-I

SECTION-A

1. After refraction from water-glass-plane interface, the image must be formed at distance of 130 cmfrom point A (at centre of curvature of concave mirror), soV1.= image distance = -(130 - 42) = -88 emU1 = object distance

.&c.._llg = ~=O~ u =_1l9V, =_~x~=_9x88 =-99cm-v 1 U, 00 'Ilw 2 4/3 8This object will act as the image of refraction at glass air interface, soV2 = imagedistallce = -(99 - 9) = -90 emIl- 1 1 '_g __ = _ ~u = -30 em ~y = 39 cm-90 u 60

3. Frequency of vibrations of closed organ pipe = 6 kHz

f = Y.- = ~ ~YRT4f 4e Mo

focff

f =f;= (1+ 3~0rf' 1-=1+-f 600

or M = f' - f = _f_ = 10Hz600

4. J (T,dbsin8 - T2dbsin8) = ma

sin 8 (To+a.f - To).b = f .b.h.cr.a

a.sin8a=--hcr

5. t, + r, = 1T = t,t2 + t,r,r,t2 + t,rl, rl,t2 oo

t,t2 {1H,r, + (r,r2t + (r,r2)3 oo}3 1 3

=~ "4x"6 _ 24 _ 31-r,r2 1-~x~ - 19 -19

4 6 24

6. N = Noe-At

No _ N e-).t,3- 0

2No = Noe->J,3

2e-'.t, = e->J,

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3AITS-PT -III (Paper -2)-PCM(Sol)-JEE(Advanced)/17

~ A.(t2 - 11) = In2

~ (t2 - t1) = £n2 - 20 minutes(£n2)/20

7.

8. Concept Involved: Dimensional analysis.

E VB2'" E

oE2

V [B '" ~]o c2 C

'" ~ V '" Energy = mass, so2c

2(speedl

a1 = 1, b1 = 0, Cl = 0

9. BC = RI3

Since point B and point C are very dose so, e is approximately equal to ~-(u+13)2

Using sine rule for !lABC, we haveAC BC AC BC. AC BC RI3 R-=--~ --- (since 13-t 0 so)~ --=-=-=-sine sin213 COS(Ct+I3) sin213 'cosu 213 213 2

B

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4AITS-PT _III (Paper_2)_PCM(Sol).JEE(AdVanced)/17

11-12. Let N be the point of contact with the floor at any time t.0 = LNCG.If the floor be perfectly rough, N is the instantaneous centre ofrotation; hence, taking torque about it. LG is perpendicular to CN

I 2R . 0NU =mgx-slO (1)It

IN= IG+m(NG)2 (2)

and Ie =IG+m(CG)2 =mR2 (3)

With the help of equations ( 2) and ( 3), we can write

IN= mR2 _m(CG)2 +m(NG)2 = mR2 -m[ (CG)2 -(NGn .... ( 4)

With the help of triangle NCGR2 + (CG)2 _(NG)2

cosO=2R(CG)

R2 +(CG)2 _(NG)2~ 1= [ Since 0 ~O]

2R(CG)

~(CG)2 _(NG)2 = 4R2_R2

It

Putting this value in equation (4), we have

2mR2IN= --[It-21

1t

Putting this value in equation (1), we have2mR2 2R d

20 g

__ [1t_21u=mgx-sino~R[1t-21u=gsinO~--2 = [ 10

It 1t dt R1t-2

d20 g~ -2 + [ 10 = 0 [Equation of SHM]dt Rlt-2

Hence the required time is 2lt ~(lt -g2)R

Now, let the plane be perfectly smooth. Since there are no horizontal forces acting on the body, Gmoves In a vertical straight line and point of contact N will slide along the horizontal surface, so Lwill be the instantaneous centre of rotation. Taking torque about L, we have

I~. (5LU=mgx-sIOO .... )

It

IL=I

G+m(LG)2 (6)

and Ie = IG+ m(CG)2 = mR2 (7)

With the help of equations (6) and ( 7), we can write

IL= mR2 -m(CG/ +m(LG)2 = mR2 -m[ (CG)2 _(LG)2]

IL= mR2 _m(CG)2 [1-sin2 oJ = mR2 -m 4~2 cos

20

It

Since 0 ~O, so

IL= m~2 (lt2 _ 4) ( 8)

11:Putting this value in equation ( 5), we have

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5AITS-PT _III (P aper_2)_PCM(SOI)-JEE(AdVanced)/17

(n2 - 4)RHence the required time = 2n 2ng

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2

17. ilE = loss in energy due to collision = mvo4

1.67 X 10-21 x(6.25xl04)'= = 10.2 eV4x 1.6xlO-19

F=n,vA F =n2vA, h' 2 h, 2

F=F +F =val~+ n2],2 h h

1 2

':h=h,+h2

F = vArn. +_~~_Jlh, h-h,J

For minimum force F, ~ = 0 ~ -n, + n2

= 0dh, h; (h _h,)2

:::;. ~=1+~2 =3 ~ h =~=1m:::;. h =2mh n 1 3 2, ,F. =[~+0.4JX10X4=0.6X10X4=12Nm,n 1 2 2

18. (P) Loss in kinetic energy = gain in elastic potential energy1 1 . 1 °~ -mv~ = __x stress x stram x volume = -om" x ~ x At2 2 2 E

~Emv2~ ° = __ 0 = 2 unitmax A£

15-16. <v>=~8kT =475m/snmdN = 8../2 (~)"2 e-41'dv= 1.9 x 10-3 = 0.19%N n nkT

13-14.

AITS-PT -/II (Paper-2)-PCM(Sol)-JEE(Advanced)/17

(Q) al~T =~f. ltDdE

~ F = ltaDdE~ T4

~ am,x = 4F = DEa~T = 4 unitltd2 d

6

F- F

(R) 30 10-2 1 t2 _ 1 (40 x 102J 1 40 x 10-2 X 40 x 10-2

X -g --g -=~ =-xgx--- _-2 2 ~2gh 2 2xgxh

40x 40 40h- x10-2m=-cm ~ n=1

4x30 3

(S) Taking torque about 0 F2x ~ - F x ~ _ F x 4R = 03 1 3 B 3lt

R2e 1 2 f. 1 ltR2e 4

~ p,gTX3"-P2gR 2"x3"--4-P,gx3lt

=0

~ 12=3P,

R/3

19. Before ~ decay, neutron is at rest. Hence En = mnc2,Pn= 0

After ~ decay, from conservation of momentum:Pn=Pp+Pe

Or Pp + Pe = 0 ~ Ippl= IPel = P

1 ,

Ee = (m;c' + p;c2)2 = (m;c' + Ppc2)2

From conservation of energy:1 1

(m~c' + p2C2)2 + (m;c' + p2C2)2 = mnc2

mpc2 = 936MeV,mnc

2 = 938MeV,mec2 = 0.51MeV

Since the energy difference between nand p is small, pc will be small, pc« mpc2, while pc maybe greater than mec2.

2 p2C2 2~ mpc +--2 -, =mnc -pc

2mpc

To first order pc = mnc2- mpc

2 = 938MeV - 936MeV = 2MeVThis gives the momentum.Then,

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1

Ep= (m~c4+ p2C2)2= .,/9362+ 22 = 936MeV

E. = (m~c4+ p2C2)~= ~(0.51)2 +22 = 2.06MeV

AITS-PT _III (Paper_2)_PCM(Sol)-JEE(Advanced)/17

20. Since system is performing simple harmonic motion its centre of mass with time period

T=27t~

:. speed of block will same for the time t = !-= ~ [3m4 2V3kAlso at this time length of the spring is east.

L th f.' . t 3T 37t~meng 0 spnng ISmaximum a t = - = - -4 2 3k

Acceleration of both blocks is zero simultaneously for the first time at t = !.= 7t [2rTI .2 V3k

Second MethodTaking block of mass is reference then the reduced mass J.1 of block of mass 2m = 2m x m = 2m3m 3

Time period of oscillation = 27t [2rTIV3k

Speed of blocks will be same and length will be least their relative velocity is zero.

. 1 [2rTI 7t [2rTII.e., at t = 4"x 27tV3k ="2 V3k

length will be maximum at other extreme position

. 3 [2rTI 37t [2rTII.e., at t = 4"x 27tV3k = zV3k

acceleration of blocks will be zero when spring is unstretched

. 1~m~mI.e. at t = -2lt - = It -, 2 3k 3k

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8AllS-Pl -III(Paper -2)-PCM(Sol)-JEE(Advanced)/17

Chemistry PART -II

SECTION-A

1.

2.

d = a~ = ~ x 5.2 = 4.5A

PV 10-12 X 1.00-=n n=-------RT ' 82.058 x 760 x 298n = 5.38 x 10-20 mol

VVmotar = n 1.00 19 3

-S-.3-8-x-1-0--2-O'=- 1.86 x 10 em / mol

1

(ap)(~y(P-aP)

3. MgC03 ~MgO + CO2

MgSO, ~MgO +S02

Mg(N03)2 ~MgO+2N02 +'!022

MgSO, dilH,SO, lMgSO. +S02 +H20

S02 +K2Crp7 ~S03 +Cr3+ (green)

6. Facts.

7. Facts

( ---->. 110. Hp V)~H2+-02P-aP Pa 2

"P2

P 1I2pK = 0, H,

P PH °,Kp = 2 X 10-9

e.G" = -RTlnKp = -2.303 x 1.98 x 1200 x log2 x 10.9

= 47600 cale.Go = 47.6 kcal

Solution for the Q. No. 11 to 12.KI+CuS04 ~CU212 +1212+KHS03~KHS04 +1-

12+KCI03~103 +C12

1'+103~12

'2 +N2H4~N2 i +HI

Solution for the Q. No. 13 to 14.

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9AITS-PT-III (Paper-2)-PCM(Sol)-JEE(Advanced)/17

K _ MRTf2 _ (78x10-3k9)(8.314)(278.4t _ -1

f - a'USionH- 10.042x103 -5.0 K kgmol

Molality of solution = ~ = 0.2 mol kg-f

5.0WC"COOH

Mass fraction of acetic acid ---'~"--- = 0.015WCH,COOH+Wsenzene

Weenzene= 1000 gm

W ~OOO = 0.015

w= 15.22

Mole of CH3COOH = 15.22 = 0.25460i=~=0.788

0.254:. a =0.424

it

III~!l\I

ii

15.

16.

17.

18.

A~20 = A~+ + A~O-

o Kx1000I.m= at-. m = ---

C

Cell constant = K,R, = K2R2K, = KHA= 0.25 x 254 - 63.5 x 10-5

105

, _ K, 63.5x10-5/'~m- -=C 8 x 10-2

a. = Am= 7.93 x10-3

- 0 19A~ 420 x 10-4 .

Hg2(NO')2 +K2Cr04 ~Hg2Cr04 -l- (Red)+KNO,

S +NaOH ~Na2S +Na2Sp3

NH4CI+NaN02 ~N2 +Hp +NaCI

Na2Sp, + AgN0

3~Ag2SP, Na,S,O,)Na,[Ag(SP')2J

Peptisation: Phenomenon in which a precipitate is passed into colloidal form by adsorption of

common ions on its surface.Delta formation: When river water (sol) meets sea water (electrolyte) coagulation occurs and

delta is formed.Tyndall effect: Scattering of light rays by suspended particles in a colloid is called Tyndall effect

and sky looks blue due Tyndall effect.Protection of colloids: Protective power of colloids is inversely related to Gold number.

Hence (0).

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10Alrs-pr -III (Paper -2)-PCM(Sol)-JEE(Advanced)/17

(Q)~E=O- 0.059Iog~=_0.0591 0.01

1 -(R) ~ BrO; +3Hp+5e- ~-Br2 +OH

2

BrO; +2Hp+4e- ~BrO- +40H. 1-BrO- +H20+e- ~-Br2 +20H

2Eq. (1) = Eq. (2) + Eq. (3):. -5FEo = -4F(0.54) -1F(0.4S)

:. EO = 2.61 = 0.52 V5

RT(S)~EO =-lnKwFKw = 10-14:. EO = -0.828

20. Cu2++p. ~Cu

Cu2++H3P02~CU2H2

Cu2++PH3~CU3P2 J..

Bi(OH), +SnO~- ~Bi J.. +SnO;-

... (1)

... (2)

... (3)

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11AllS-Pl -III (Paper-2)-PCM( Sol)-JEE(Advanced)/17

Mathematics PART -III

SECTION-A

1.

2.

3.

4.

5.

6.

. . X5+ 1+ 1+ 1+ 32 'ir::s:;;;Let Xl be posItive then 1 >~Xl .32 = 2x1

5

::::> X{ -10Xt +35 ~ 0

Equality occurs if x{ = 1= 35 (not possible)

Alternate:fIx) = x5-10x + 35flO) = 35, f'(x) = 50 -10 ~ -10y'+10~0(y + 1Ox)~ (y + 1OX)x= 0y + 10x ~ 35fIx) ~ -10x + 35 'if X E R::::> fIx) ~ 0 X E (0, 3)Also, f(3) > 0 and f'(x) > 0 'if x > 3Hence, fIx) > 0 'if x E R+

(i), (i)2satisfy Z5+ Z + 1 = 0Dividing by Z2+ Z + 1 we get Z3- Z2+ 1

For Z3+ 1 = Z2, (e'8)3 +1 = (ei8)2cos 38 + 1 = cos 28 and sin 38 = sin 28No solution for 8

10P:x+2y+z=-

35 1 1x-- y-- z--

L: __ 4 = __ 2 = __ 42 1 -2

For point of intersection (2t +%) +2( t +~) +( -2t + ~) = 1~

2t+10=104 3.' .' (25 11 -7)Hence, pOint of mtersectlon IS -, -,-12 12 12

We have (I + A)(I + B)(I + C) = I::::> (I + B)(I + C)(1 + A) = I::::> BCA + BC + CA + BA + A + B + C = 0::::> ABC - BCA = (B + C)A - A(B + C)1 log. b log. C10g

ba 1 10g

bc = 0 so no unique solution

logea logeb 1

Let Xl, X2, ..... , Xn be the roots, then "L>1 = n

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12AITs-pr -III (Paper-2)-PCM(Sol)-JEE(Advanced)/17

I~X2 ="C2

I(Xl-1)2 = IX~ -2Ix1 +n = (IX1)2 -2IX1X2 -2I~ +n = 0

Hence Xl = X2 ..... X" = 1 one solution

7. 1523+ 2323 is divisible by 19 as (19 + 4)23 + (19 - 4)23 divides 19So it is 57k or 57k + 19 or 57k + 38, 1523+ 2323 when divided by 3 gives remainder 2Hence, 1523 + 2323 = 57k + 38

8. Let A = [ : :] we get four equations and four variables

a - b =-1c-d=2

Also A[ ~1] = [~]

=:;. -a + 2b = 1, -c + 2d = 0a = -1, b = 0, c = 4, d = 2

9. At least one of the cherry berry fruit in east is of type A. Hence chances of survival in east are 3-3

which are ! in west.2

,

10. , , 21t ' -Angle between a and b is 3where c is along the angle bisector, a - b is along another angle

bisector and hence perpendicular to cI;; - 61=.t3 hence I;; - 6 + cl = 2

14. We have x3 + y3 + Z3 = 3xyz hence xyz = a3Also xy + yz + zx = p hence X, y, z are roots of u3 + pu - a3 = 0Which gives us a2 = -p and x4 + y4 + Z4= _p(x2 + y2 + Z2) + a3(x + y + z) = 2p2x5 + y5 + Z5= _p(x3 + y3 + Z3) + a3(x2 + y2 + Z2) = -5pa3x7 + y7 + Z7= _p(x5 + y5 + Z5) + a3(x4 + y4 + Z4) = 7p2a3

15.-16. b" = (n + 2)2" + 2n + 2, Cn= (n + 4)2n + 2, dn = (n + 6)2nen = (n + 8)2n and so on

17.

18.

19.

f(n) = number of perfect squares in the list1,2,3, ..... n

Applying Rl --7 Rl + R2, R2 --7 R2 + R3 ..... and so onThen C2 --7 C2 - C1, C3 --7 C3 - C2 ..... and so onWe get IAnl = n + 1

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13

=> Izl4 -lzj2(a2 + Z) + 1 = -(z+ Z)2 :s:0

IZIE[-a+~,a+~]

Which is [ma, Mal

ZO. (P) 52C13.39!(Q) 52C26.Z6!'Z'Z'Z ..... 13 times(R) Cards of same denominator are together(S) 52C13.39C13.26C13.13C13

AilS-PI-III (paper_2)_PCM(SOI)-JEE(Advanced)/17

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