(JEE ADVANCED PATTERN) TARGET : JEE (MAIN+ADVANCED) … · 2019-11-13 · Website : | E-mail :...

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005

Website : www.resonance.ac.in | E-mail : [email protected]

Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029SOL02JPAIOT250218-1

ALL INDIA OPEN TEST (AIOT)(JEE ADVANCED PATTERN)

TARGET : JEE (MAIN+ADVANCED) 2018DATE : 25-02-2018 COURSE:VVIIJJEETTAA ((0011JJPP,, 0022JJPP)),,

VVIISSHHWWAAAASS ((0011JJFF)),, VVIIJJAAYY ((0011JJRR,, 0077JJRR))

HINTS & SOLUTIONSPAPER-1

MATHEMATICS1. If a, b, c, d are .......................

Sol. Let b = a + p, c = a + 2p, d = a + 3p

2 2

2

1 11 1a a 3pa d

1 1 1 1

b c a p a 2p

(a p)(a 2p) a 3ap 2p1

a(a 3p) a 3ap

1 1 1 1

a d b c

1 1(a d)

b c

1 1(a a 3p)

a p a 2p

2

2 2

2

2 2

(2a 3p)

a 3ap 2p

p4 4.

a 3ap 2p

2. The value of the.......................

Sol.

1 x 1 xa x

2x1 a 1 x

sin e cos e eI dx

e 1tan e tan e

=

a x

2x1 a 1 x

1 edx

2 e 1tan e tan e

= ae

1 a

dt

2 t tan e

as tan–1 (ex) = t

= ln22

3. Let f(x) = sin(x) .......................

Sol. f(x) = sin(x) – 4x(1 – x)

f(x) = cosx – 4 + 8x, f(x) = –2 sin x + 8

3 1f '"(x) – cos x 0 x 0,

2

and1

f "'(x) 0 x , 12

Hence f(x) 0 x (0, 1)

4. Two parabola .......................

Sol. Given parabolas are confocal and their axes in opposite

direction hence they will cut each other orthogonally. Therefore

MPN 90 , also common focus is (5, 2).

5. Let f(x) = x3 + x + 1.......................

Sol. Let x = 2, a x , put in x3 + x + 1 = 0

We get p(x) = x3 + 2x2 + x – 1

6. Let two curves.......................

Sol. y – f(x) = f’(x) (y – x) and y – g(x) = g’(x) (y – x)

x/4 x/4

f '(x) g'(x) 1f(x) g(x)

f(x) g(x) 4

f(x) e ,g(x) 4e

1 2

x / 4–1

x 4 13I dx

124 4e

Area

1x/4 x/4 1/4

0

(4e – e )dx 12(e – 1)

7. Let –1–1

32 log (cos x)log (sin x)f(x) 2 3 .............

Sol. f(x) = sin–1x + cos–1x =2

x (0, 1) and f(x) is not define

for x < 0




8. L1 is a tangent .......................

Sol.2 2x y

116 4

1

Tangent A is 5x – 6y = 16

Tangent at B is 5x – 6y = –16

Normal at A is 6x + 5y =75

2

Normal at B is 6x + 5y =75

2

Radius of circumcircle of rectangle is

9 10925

4 2

9. Let A and B .......................

Sol. (ABAT)T = ABTAT which is not symmetric.

(AB – BA)T = (AB)T – (BA)T = BTAT – ATBT

adjAB | A | adjA

| A |

(adj AT – B)T = ((adj A)T – adj A)T = (adj A) – adj (AT)

B15 = A15, S is skew-symmetric.

10. The coordinates .......................

Sol. Let for required pointx 1 y 1

z2 3

1 2 , 1 3 ,

Now distance between this point with (1, –1,0) is 4 14

2 2 22 3 16.14

214 16.14

4.

(9, –13,4) and (–7,11, –4) Ans.

11. The volume of .......................

Sol.1 1 1

AD BC AC sin45 ,3 2 6

soAC

AD BC 1.2

apply AM GM

13. Let for .......................

Sol. apply lebnitz we get

2

2

xf '(x) – f(x) 2x

x x 1

On solving differential equation2f(x) x n(x 1)

14. Let the end .......................

Sol. Given circle will intersect major axis y = x at the foci of ellipse.

So foci of ellipse are (2 2 , 2 2 ) & (–2 2 , –2 2 )

length of major axis of ellipse is 10

15. Let the unit .......................

Sol. | a | | b | 1

a.b 0 ; (as a b)

c a b (a b) ......(i)

Taking dot product by a,

2a.c | a | (a.b) [aab]

| a | . | c | cos .1 0 0

1. | c | .cos

As | c | 1 ; cos

Taking dot product of (i) by b

2b.c b.a | b | [b a b]

| b | | c | cos 0 .1 0

1.1.cos cos

1|| 2c 2 2 2 1

2 2 2cos cos 1

2 21 2cos cos2

Hence,2cos , cos2




16. The figure shows .......................Sol. Let radius of circles with centres A, B, C are a, b and c

respectively then

XZ = 22 )c–a(–AC a – c = 9 ....... (1)

Area of quadrilateral ACZX =2

1(a+ c)(40) = 300

a + c = 15 ......... (2)hence a = 12 and c = 3Now for finding the radius of circle touching both circles withcentres A and CXZ = XY + YZ

40 =22 )12–b(–)12b( +

22 )3–b(–)3b(

40 = b48 + b12 b =27

400

Now area of ABC

x

A C

B

y zABC = Area of quadrilateral ABYX + Area of quadrilateralBCZY – Area of quadrilateral ACZX

D =2

1

27

40012

3

80+

2

1

27

4003

3

40– 300

D = 476 +27

8cm2 [] = 476 cm2

19. Angle between .......................Sol. Focus of parabola is (3, 5) and focal chord is

mx – y + 5 – 3m = 0

2

5 3m3

1 m

8

m15 or

8

tan15

2

1 18 15cot tan

15 8

20. Let2

1f(x)

6 x 2

.......................

Sol. f(x) is defined when26 x 2 > 0

2x 8 2x 9 3 x 3

PHYSICS21. A uniform solid ...................

Sol. AL

= 2

cm

R mRˆ ˆmV k k 02 2

2

B

mR ˆL k2

2C

3 ˆL mR k2

2D

3 ˆL mR k2

2

2O cm

R mR 3ˆ ˆ ˆL mV k k mR k2 2 2

22. Consider a ...................Sol.

0 5 10 15 20 25

0 4 8 12 16 20d

2d3d

d2d

3d

Main scale

Vernier scale

d = 0.25 mm

LC = 0.25 mm

zero error = – 0.75 mm

while taking the reading zero of vernier will be between 9th and

10th of main scale division.

Reading = 9mm + 0.50 mm + 0.75 m = 10.25 mm

23. Consider an ...................

Sol. Equivalent Circuit

1 2 2 3 3 4 4 5 5 6

C1 C2 C3 C4 C5

PQ

Q

R P S

V

S

C4

C5C2

R

C3

P

Q

C1

C1 = C2 = C3 = C4 = C5 = C= 0A

d

Balanced wheat stone bridge




C

CC

C

Q= CV

V

WB = CV2 = 20AV

d

Electric field between 3 & 4

=

0

33

0 0

A VQ Vd 2EA A 2d

Potential difference between 2 and 5 = VQ – VS =V

2

Potential difference between 1 and 4 = VP – VS =V

224. Consider the .........................Sol.

mg

F =2R cos sin a

C

T

F = 2 2R cos

1 rC

T =2R cos sin a

C

; mg =

2 2R cos1 r

C

1T a 13tan tan tan

2mg 1 r 513

mgtanT

5

25. Two blocks .........................Sol.

1 2 1 2 1 2max

1 2 1 2 1 2

k k k k m ma A, , T 2

m m m m k k

fmax + K2A = m21 2

1 2

K KA

m m

m2g = 1 22 2

1 2

K Km K A

m m

A = 2 1 2

1 2 2 1

m g(m m )

k m k m

26. Two cars A .........................

Sol. (180)2 + (255 – 15t)2 = (300t)2

32400 +65025 + 225 t2 – 7650 t = 90000 t2

89775 t2 + 7650 t – 97425 = 0

t = 1 sec

f =300 15cos37

300 10cos53

f0

=300 12

300 6

f0

=312

294× 2.94 khz

f = 3.12 khz

27. In the diagram.........................

Sol. In right ring emf induced in ABC part and CDA part will be same.

Simplified diagram may be

R2

R1

C

A

R1 R2

So there will be current in each ring and so magnetic force on

them will be non-zero.




28. Consider a .........................

Sol.

xo

y

P

V

Q

V

+B = 0

1B

r

B = 0B =0r

For a point inside pipe, pipe will act as solenoid

For a point outside pipe, it will act as infinite long wire

29. Current in .........................

Sol. 0t

i q2

0

0i

i t it

i0 =2q

t 0t

i i 1t

dH = i2 R dt =20i R

2t

1 dtt

s

H =20i R

T 2

20

t 2t1

tt

dt = 20

ti R

3

=

24 q R

3 t

30. A cone-shaped .........................

Sol.

Fc sin

Fc cos mg sin

mg cos

N

For equilibrium

N = mg sin + Fc cos

Fc sin = mg cos

m2r sin = mg cos

r =2

gcot

31. Consider .........................Sol. For inner sphere initially

P = 40T 4R2 = 4 4 2

0T RFor outer shell

P = T444R2

4 40T R2 = T44.4R2

40T = 4T4

T = 0T

2After covering for inner sphere temperature of inner surface willbe greater than T0

32. Consider the .........................

Sol.

XL = R

R

~

XC = 2R

E = E0 sin(t +4

)

3

2

1

1 =0E

Rsin(t +

4

)

2 =0E

Rsin(t +

4

–

2

)

3 =0E

2Rsin(t +

4

+

2

)

= 1+ 2+ 3

03

E

2R

01

E

R

(t +4

)

2 = 0E

R

0E

R

0E

2R

05E

2R

t4

tan =1

2

= 05Esin t

2R 4

z =12 2R x =

2R

5

212 2 4R

R x5

tan =1

x 1

2R R1 = 2x

2 × 12 + x2 = 5x2 =24R

5 x2 =

24R

25

2Rx

5




33. Three point .........................

Sol. Assuming cube with ABCD and CDEF as two adjacent

faces.

through CDEF = flux due to q at A only

=0

1 q

3 8

Also, VD > Vc since q at A is closer to D then C.

34. Choose the .........................

Sol. (A) eV0 =min.

hc

, min. =

12400

30000 = 0.4Å

(B) Energy of photon=12400

5000 = 2.48 eV

(C) In hydrogen atom speed V 1

nHere in ground state n = 1 and in second exided state

n = 3

(D) Since balmer series lie in visible region so only balmer

will be seen by observer.

35. Four point .........................

Sol. U = – 62 2Gm 3 2Gm

–aa 2

2

3

–GmF on A a i ak– a j ak a i– a j

(a 2)

|F| =2

2 2 2

3

Gm4a 4a 4a

a 2 2

=2

2

3 Gm

2 a

A (a, 0, a)Z

(0, a, a)

(a, a, 0)y

x

36. A particle is .........................

Sol. 2 = 7t +1

2(–6) t2 t =

1

3sec, 2 sec.

5t sec.

3

width d =5 35 7

7 3 m.3 3 3

37. An incompressible .........................Sol.

P0 +1

2V0

2 = P +1

2V2

P0 +1

2V0

2 = P +1

216V0

2 ; P = P0 –15

2V0

2

38. One mole of .........................

Sol. 3 1 2 31 2

1 2 3 eq

n (n n n )n n

1 1 1 1

1 2 35 7 8

1 1 13 5 6

=eq

6

r 1eq =

43

31

39. A soap bubble .........................Sol.

Atmosphere

A B

Pressure at A = P0 +4S

2R

Pressure at B = P0 +4S 4S

2R R = P0 +

6S

R40. A long string .........................Sol. 1 = 0.2 Kg/m = 0

10

TV

2 = 0.8 Kg/m = 40

12

0

VTV

4 2

At =

1

2i i

11 21

V22V 22A A A

VV V 3V2

Ei =1

202A2V1

Et =1

2(40) 2 2

t 2A V = 2 2 10

V1 44 A

2 9 2

2 20 1

1 16A V

2 18

=2 2

0 1 i i

8 1 8 KA V E E

9 2 9 9

K = 8




CHEMISTRY

41. The hexagonal unit cell of ...........

Sol. (A)

C3

(B) There are total 12 tetrahedral voids in 1 hexagonal unit

cell. Hence in each equal unit cell, tetrahedral voids

=12

43 .

(C) Packing fraction remains same, i.e. 74%.

(D) In original hcp also, no atom lie completely in unit cell.

So here also, it will be the same case.

43. At 25ºC, in one litre saturated ...........

Sol. Ag2C2O4 (s)

22 4

2s 0.1 M

2Ag aq C O aq

2 2 1sp 2 4K [Ag ] [C O ]

1.6×10–8 = (2s)2 (0.1)1

s = 2×10–4 M

So Ksolution = 22 4Ag C O Na

K K K

= 460 4 10 140 0.1 50 0.2

1000 1000 1000

–~ 24×10–3 S cm–1

Also M M MAgCl Ag Cl

44. Aqueous solution of boric ...........

Sol. B(OH)3 + H2O [B(OH)4]– + H+

OH2BO 22

+ 2BO +

+ 2H2O

Optically resolvable due to asymmetric structure

45. Identify the correct ...........

Sol. NaOH can co-exist with Na2HPO3 in aqueous solution as the

latter compound does not contains acidic H.

47. Amongs the following...........

Sol. Fe/Fe2+, Cr/Cr3+ shows positive value of standard electrode

potential.

48. What is/are true for hydrogen ...........

Sol. (A) In alkaline solution, its reducing character is more than in

acidic medium.

H2O2 O2 + 2H+ + 2e– ;

2OH– + H2O2 O2 + 2H2O + 2e–

(B) Correct statement.

(C) Due to polar nature, it has high dielectric constant.

(D) K2MnO4 + O3 + H2O KMnO4 + 2KOH + O2

49. Which property given in b...........

Sol. (A)23CO

Non polar,23SO

polar

(B*) CO2 and SO2 both are planar

(C*) COCl2 and SOCl2 both are polar

(D) CF4 and SF4 both are non planar

50. Which of the following ...........

Sol. [Pt(ox)3]2– is optically active due to 3 didentate ligands. [PdCl4]

2–

and [Pt(ox)3] 2– do not show G.I. as all ligands are same.

Hybridization of Ni2+ is dsp2 in [Ni(CN)4] 2– and that of Hg2+ is sp3

in [Hg(SCN)4] 2–

Both CN– and SCN– ligands are ambidentate. Hence both

[Ni(CN)4] 2– and [Hg(SCN)4]

2– show linkage isomerism.

K2[Ni(CN)4] is diamagnetic but Co[Hg(SCN)4] is not as Co2+ is d7

– paramagnetic.




51.X

(Hydrocarbon)

(i) O3

(ii) Zn/H2O O...........

Sol. (A)

Me

CH = CMe

Me

(i) O3

(ii) Zn/H2O

Me

H

OH

OO

OO2CH3 – C – CH3 +

(C)Me

Me

Me (i) O3

(ii) Zn/H2O

H

OH

OO

OOCH3 – CHO + CH3 – C – CH3 +

(D) (i) O3

(ii) Zn/H2O

Me Me

MeO

+

H

O

OH

OOO +

H

52. The possible products in ...........

Sol. CH3–CH=O + HCHO OH

OH

CH2–CH2–CH=O + CH3–CH–CH2–H=O

OH

OH)ii(

H/KCN)i(

3

OH

CH2–CH2–CH–COOH + CH3–CH–CH2–CH–COOH

OHOH OH

OH

CH2–CH2–CH–COOH + CH3–CH–CH2–CH–COOH

OHOH OH


Sol. (A)CH3–C–Cl

O

H

)excess(MgBrCH3

CH3–C–OH

CH3

CH3

(3º R–OH)

(B) H

OH2

OH (3º R–OH)

(C)O 2S

LiAlH

N

4 H

OH

(3º R–OH)

(D) O

Cl Ether

MgO

MgCl

H

OH

54.

CONH2

...........

Sol. [Moderate]




55. In which of the following ...........

Sol. (A)

(B)

(C)

(D)(1) CO2

(3) CH3COCl

(2) H3O+

O – C – CH3

OHOOC

56. A magnesium ribbon...........

Sol. Mg + air MgO + Mg3N2

0.25 mol 0.1 0.05

MgO + 2HCl MgCl2 + H2O.

0.1 0.2mol

Mg3N2 + 8HCl 3MgCl2 + 2NH4Cl.

0.05 mol 0.4 mol 0.1 mol

NaOH

NH3 0.1 mol

57. For a first order reaction ...........

Sol. rate =t

dx 1tan K[A ] K 0.1

dt 3 ; K = 5.77 s–1

58. In He+, electron is initially ...........

Sol. Initially, electron is in an orbital with no radial nodes,

So n– – 1 = 0; and wavefunction changes sign once across a

plane i.e. there is one angular node so = 1

n = 2

Hence initially electron is in 2p.

Now this electron absorbs75

100× 13.6 eV or

3

4× 13.6 eV

energy to move to higher orbit (shell)

In He+, E1 = – 4 × 13.6 eV

E2 =–4 13.6

4

eV = – 13.6 eV

E3 = –4 13.6

9

eV

E4 = –4 13.6

16

eV = –

13.6

4eV

Hence we can see that E4–E2 =3

4× 13.6 eV.

Hence electron goes in 4th shell.

So, nintial + nfinal = 2 + 4 = 6

59. C4H8 (unsaturated hydrocarbons) ...........

Sol. CH2=CH–C2H5

Meso

CH3–CH=CH–CH3

trans

60. Total number of -Keto ...........

Sol. ,




PAPER-2MATHEMATICS

1. P(x) = 0 is a ........................

Sol. P(x) – 5 = (x – 2)Q(x)

P(x) – 3 = (x – 4)R(x)

Let P(x) has one integral root n such than P(n) = 0

P(n) – 5 = (n – 2)Q(n) .......... (1)

–2–n

5 = Q(n)

as Q(n) is also an integer

so n – 2 can be ±1 and ±5 only

n – 2 = 1 n = 3

n – 2 = – 1 n = 1

n – 2 = 5 n = 7

n – 2 = – 5 n = – 3

are possible values of n

P(n) – 3 = (n – 4)r(n)

–4–n

3 = r(n)

r(n) is also on integer

hence n – 4 can be ± 1 and ± 3

n – 4 = 1 n = 5

n – 4 = – 1 n = 3

n – 4 = 3 n = 7

n – 4 = – 3 n = 1

So possible value of n should satisfy both equation hence

possible values of n are 1, 3, 7

So, (C) is incorrect option

So, (C) is incorrect option

2. A is a 3 × 3 ........................

Sol. Let A =

11 12 13

21 22 23

31 32 33

a a a

a a a

a a a

u

=

x

y

z

A u

=

11 12 13

21 22 23

31 32 33

a x a y a y

a x a y a y

a x a y a z

A u

. u

= 0

a11x2 + a27y

2 + a33z2 + (a12 + a21)

xy + (a23 + a32) yz + (a31 + a13) zx = 0 x,y,z R

Hence a11 = a22 = a3 = 0 and aij + aji = 0 i j

3. Let f(x) ........................

Sol. R.H.D. =

2 h

0

h 0

(5 1 t )dt 11

f '(2 ) limh

1 2 h

0 1

h 0

(6 t)dt (4 t)dt 11

limh

0 2 h2 2

1 1

h 0

(6 t) (t 4)11

2 2lim

h

h 0

h(12 h)lim 6

2h

L.H.D. = 5 Continuous but non differentiable at x = 2

4. Letrn

r 0

(–1)X

r 1

........................

Sol.n n 1

2 3 n 1 (–1) x1– x x – x ..... (–x)

1 x

12 3 n

0

1 1 n 1n

0 0

(1– x x – x .... (–x) )dx

1 xdx (–1) dx

1 x 1 1

1 n 1

0

x| X – In2 | dx,

1 x

where

1 n 1

0

1 x 1dx .

2(n 2) 1 x 2(n 1)

5. The value of ........................

Sol. zn – 1 = (z – 1)(z – z2)(z – z3)……(z – zn)

n

2 3 n

z – 1(z – z )(z z )....(z z )

z – 1

n–1 n 22 3 nz z .... 1 (z z )(z z )....(z z )

Putting z = z1 = 1

|1 + 1 + …… + 1| = |(1 – z2)(1 – z3) ….(1 – zn)|

2 3 n| 1– z || 1– z | .... | 1– z | n.




6. The value of |A1A2|2 ........................

Sol.i2 / n i2 /n2

2 1 11

ze | z – z | | z || (e – 1) |

z

2 2 2 21 2 2 1| A A | | z – z | 2 sin

n

2 2 21 2 1 3 1 n

2 2 2

| A A | | A A | ...... | A A |

2 (n – 1)2 sin 2 .... sin

n n n

2 4 2(n – 1)2 (n – 1) – cos cos .... cos

n n n

= 2(n – 1 – (– 1)) = 2n.

7. The value ........................

8. The value of ........................

Sol. D =

1 1 1

2 1 3

1 2 p

= –p,

D1 =

4 1 1

6 1 3

q 2 p

= –2(p – q + 6)

D2 =

1 4 1

2 6 3

1 q p

= –2p – q + 6,

D3 =

1 1 4

2 1 6

1 2 q

= 6 – q

x = 1D

D =

2(p q 6)

p

,

y = 2D 2p q 6

D p

, z = 3D q 6

D p

New for unique solution p 0, q R

total no. of order pairs (p, q) = 20 × 21 = 420

L = 420

For no solution p = 0, q 6

total no. of ordered pairs of (p, q) = 1 × 20 = 20

M = 20

for infinite many solution p = 0, q = 6

total no. of ordered pairs (p, q) = 1 × 1 = 1N = 1

L M N 19 420 20 1 19

420 420

= 1

9. The length ........................

10. Area of the ........................

Sol. Let DC = x

BQ BP 2x – 2; CQ CR

x – 2 CB 3x – 4

2 2(3x – 4) 16 x x 3 AB 6

PBQ is isosceles and PB = BQ = 4. Since BC = 5,

4 1 4 32sinB area of PBQ 4 4

5 2 5 5

4sinB

5 1 4 32

PBQ 4 42 5 5

11. If the expansion ........................

Sol. Since (1 + x + x2)n = a0 + a1x + a2x2 + ….. + a2nx

2n ….. (1)

Substituting x = , 2 and 1 and then adding them together

a0 + a3 +a6 + …. = 3n–1

Again multiplying (1) by x and then substituting x = , 2 and 1

and then adding the three expansions thus obtained

a2 + a5 + a8 + …… = 3n–1

0 3 6 1 4 7

2 5 8

a a a ..... a a a .....

a a a ....

The required ratio isn–1

n–1

2.32.

3

12. If number of ........................

Sol. 2m – 2 = 10 m = 6

13. If solution of ........................

Sol. Dividing throughout by2 2sin y cos x , we get

2 2

2 2

3 tan x sec x dx 7sec x dx

7cot y cosec y dy 5cosec y dy

2 24(cot y sec x dx tan x cosec y) dy 0

Integrating,

2 2tan x cot y3 7 tanx 7 5 cot y 4 (cot y tanx) c

2 2 .




14. If sin , sin , ........................

Sol. From the given conditions we have

2 sin = sin + sin .....(i)

cos2 = cos cos .....(ii)

Squaring (i), 4 sin2 = sin2 + sin2 + 2 sin sin

Using (ii), 4(1 – cos cos )

= 1 – cos2 + 1 – cos2 + 2 sin sin

cos2 + cos2 – 4 cos cos = 2 (sin sin – 1)

sinsin1

coscos4coscos 22 = – 2

15. A given quantity........................

Sol. Let r be radius and h be height of the cylinder

volume (v) =1

2r2 h(given)

Total surface area

S =1

2 (2rh) + 2

21r

2

+ 2rh

= (+ 2)2v

rrh + r2

= (+ 2)+ r2

ds

dr =

2

–2v( 2)

r

+ 2r ....... (1)

2

2

d s

dr=

3

4v( 2)

r

+ 2 ....... (2)

For maximum or minimumds

dr= 0 r3 =

v( 2)

Hence S is minimum when – r2h (+ 2) + 22 r3 = 0

h

2r=

2

k = 2.

16. The distance ........................

Sol. 1 2 1 24 1 2 – 2 –3 ……. (1)

1 2 1 2– 3 – 4 –1– 3 4 – 3 –2 ……. (1)

From (1) and (2),

– 5 x 2 = – 10

2 12, 1

point of intersection (4 + 1, – 3 – 4, – 1 + 7)

i.e. (5, – 7, 6).

Distance of (5, – 7, 6) from (1, – 4, 7)

= 16 9 1 26 .

17. If point (a, b), ........................

Sol. a2 = 4 – b2 [0, 4]

f(–1) = 1 + 4 – a2 = 5 – a2 > 0

f(0) = –a2 < 0

f(5) = 5 – a2 > 0

f(4) = –a2 < 0

-1 0 4 5

two roots lie in (–1, 5)

18. If the equation ........................

Sol. 4

2

xcos3xsin

2

xsinxcos3

= a2 + 3 sin 2x – cos 2x

2cos 2x + 2 = a2

– 1 2

2–a2

1

0 a2 4

– 2 a 2

a can be (1) and (2)

20. Let f : R 0,2

........................

Sol. The equation 3x2 + 6x + a = 0 must have equal roots.

So D 0 a 3




PHYSICS21. A small ball .........................

Sol. By work energy theorem: MgR cos = 21Mv

2

T = mg

Mg

v = 2gRcos

Radial force equation : T = Mg cos +2Mv

Rv2/R = 2gcos

90° –

g sin

T = Mg cos +M

R2gR cos

Mg = 3Mg cos cos =1

3

tan =2gcos

2cotgsin

= tan–1 222. A bird in air .........................Sol.

x

y

z

For direct image:

happ = x +3

4y

appdh

dt =

dx

dt + 3

4

dy

dt

= – 5 + 2 +3

4 (–2 – 2) = – 6

After reflection from mirror happ = x + 3

4 (y + 2z)

appdh

dt =

dx

dt+

3

4

dy

dt

+3 dz

2 dt

= – 5 + 2 +3

4 (–2 – 2) +

3

2(2) = – 3

23. Figure shows .........................

Sol.

f T

F

T cos = f sin but T mgT cos = mg sin F = T sin + f cos

= mg

cos

sin sin + mg cos

= mg

cos

)cos(

For Fmax = < 45° Fmax = mg sec But tension in the string can not be more than mg

T =mg sin

cos

< mg should be less than 45°

Now if > 45° T = mg = 90° – F = mg sin + mg cos = 2mg sin

If < 45° then Fmax = mg sec If > 45° then Fmax = 2mg sin

24. Three balls .........................

Sol.m mm mv1

u v2

Just before collision Just after collision

applying conservation of linear momentummu = mv2 + mv1coefficient of restitutionv2 – v1 = eu

v2 =u

2 (1 + e)

v1 =u

2 (1 – e)

for collision between A and B for collision betweenB and C

vB =u 1

12 2

=3u

4

vC =1 3u 1 u

12 4 3 2

vA =u 1

12 2

=u

4

vB =1 3u 1 u

12 4 3 4

Total number of collisions = 2Total energy loss E

=

221 m 1 1 m 3u 1

u 1 12 2 4 2 2 4 9

=2 21 3 1 9 8

mu mu4 4 4 16 9

=2

23mu 1mu

16 8 = 25

mu16




25. Magnitude of .........................26. Maximum magnitude .........................Sol. Let z-coordinate of the loop at any time t is z and current

through the loop is .Magnetic flux associated with the loop t = (z + B0) a

2 + Lt = 0 = B0a

2

since t = t = 0 =2z

aL

due to the magnetic fieldsforce on the frame at any z.

F (d B)

R

Q

S P

y

x

Bz = z + B0

Bx = –x

FPQ = FRS = 0

QR xˆF aB ( k)

=2z a ˆa a k

L 2

=2 4a ˆzk2L

SP xˆF aB ( k)

=

2 4a ˆzk2L

2 4a z ˆF kL

equation of motion2

2

d zm

dt =

2 4az mg

L

2 2 4

2

d z az g

mLdt

z = A sin(t + 0) + z0

where =2a

mL

at z = z0

2

2

d z

dt = 0

z0 = 2 4

mgL

x a =

2

g

and amplitude A = –Z0 = 2 4

mgL

a =

2

g

z = z0 (1 – sin (t + 0)

at t = 0 ; z = 0 0 =2

z = z0 (1 – t) =2

g

(cost – 1)

z = 2

2

g t2sin

2

=2

2g

2 tsin

2

27. The instant .........................

28. The instant at .........................

Sol. Path difference

y

mean

S2

S1

0.25 mm

4m

x1 = d sin = 0.5 10–3 ×4

1025.0 3

=32

10 6

metre

x2 = d sin = 0.5 × 10–3 ×1

10y 3

=2

10y 6metre

For maximum intensity x1 + x2 = n

2

10y

32

10 66 = 4000 × 10–10

y =80

59

sin t =80

59

t =

80

59sin

1 1

For minimum intensity

2

10y

32

10 66 = 2000 × 10–10

y =80

27

t =

80

27sin

1 1

For central maxima path difference is zero

d sin = – d sin

1

y

4

y0

Position of central maximum is y0 = –4 sin t




29. The trajectory .........................

30. The radius .........................

Sol. B =2

)PCJ(

2

)POJ( 00

=2

0 )]PCPO(J[

=2

)COJ(0

=4

JR0

OP

C

00

JRB

4

(Uniform)

J

For positive charge r =0

0

qB

mV = 0

0

4mv

q RJ

31. The given figure .........................

Sol.

N

extreme position

mg

A

f

f = m2A ; N = mg

3a af N.

2 2

m2A 3 = mg A =2

g

3

32. A uniform .........................

Sol. Terminal speedmg

2 = 6rnv

de-broglie wavelength =h h

p mv =

h

mgm

12 r

=2

12 h r

m g

33. Two radio .........................

Sol. N = N0 e-t

At t =n2

; NA = 0N

2 , NB = 0N

4

CdN

dt = NA + 2NB = N0

34. Consider an .........................

Sol. If we connect positive terminal of a battery to A and negative to

infinity then the current through branches will be as shown. If we

connect negative to B and positive to infinity current will be in

opposite direction by super-position of two situation net current

in AB branch will be2

3

. Potential difference between A and B

will be2

R3

. Effective resistance between A and B will be

2R

3.

A

/3

/6/6

/3 /3/6/6

35. A uniform heavy .........................

Sol. compression in rod A =mg

3young modulus = YA

compression in another rod =2mg

3 young modulus = YB

Since compression in both rod is same

A B

mg 2mg

3S 3SY Y B

A

Y

Y 2

Given

YA A T = 3YB B T




36. A glass hemisphere .........................

Sol. The image of object O by refraction at plane surface is

formed at I such that

AI =4

3d

I acts as object for curved surface. The curved

surface makes image of I at I

1

v–

4

34

R d3

=

41

3R

or

1

v

=1

3R–

4

3R 4d

I acts as object for mirror. Mirror makes i ts image

at I distant v above B.

I acts as virtual object for the curved surface which makes

its image at infinity

4

3

–1 4

3R 3R 4d

=

1

3R

solving we get d =3

4 R =

3

4 × 4 = 3 cm Ans.

37. The chain of .........................

Sol. i = f

mgL

2sin + 0=m1g

(L x)

2

sin – m2g

x

2+

1

2mv2

mgL

2sin =

g

L (L2 + x2 – 2Lx)sin –

2x

L + v2

gL sin =g

L (L2 + x2 – 2Lx) sin –

2x

Lg + v2

v =5

g8

38. A vertical hollow .........................

Sol. Suppose the radius of the cylinder is R and length of rod is 2.

Consider the case when the end A has sliding tendency up.

Forces acting on the rod are shown in the figure.

N1

N1W

BG

N2

N2

A

Resolving forces horizontally and vertically, we have

N2 = N1 cos + N1 sin (i)

and N1 sin = N2 + N1 cos + W (ii)

Taking moment about A,

N1(2R cosec ) = W ( sin ) (iii)

From equations (i), (ii) and (iii), we get

2R = [(1 – 2) sin – 2 cos ] sin2 (iv)

Similarly, when the rod makes least angle , we get, A will have

sliding tendency downward.

2R = [(1 – 2) sin + 2 cos ] sin2 (v)

From equations (iv) and (v), we get

= tan

cossincossin

sinsintan

2

122

331

after substituting the values

= tan11 37

tan2 84




39. Two point masses .........................

Sol.2X0 X0

2a2mcomm

d

F =2 20

Gm(2m) GM(2m)

9X 4a

GM =2

20

4 Gma

9 X

T2 =

2 2 2 32 23 3 0 0

2

20

9 X a X4 4 9a a

GM Gm 2 Gm4 Gma

9 X

T =3 / 20

3X

2Gm

< V > =2 2

0 0 0 02 2 3 2

0 0

X 2X 4X 4X 8GmT T T X 9 X92 2 Gm

40. An ideal diatomic .........................

Sol. Equation of straight line AB P = 00

0

2PV 5P

V

Slope of the straight line AB = 0

0

2P

V

When process changes from endo to exo. slope should be

equal to adiabatic P-V curve slope =P 7 P

V 5 V

0

0

2P 7P

V 5V

P = 0

0

P10V

7 V

10

70

0

PV

V

= 00

0

2PV 5P

V

0

35V V

24

0

0

P10P

7 V

35

24 V0 = 0

25P

123P0V0 = nRT0

0

25P

12

0

35V

24

=825

288 P0 V0 = nRT

0

T 875

T 288(3) =

875

864

T =875

864 T0

CHEMISTRY

41. Select the correct ...........

Sol. (A) Van Arkel method

2 4(pure)high temperature(vapour) (vapour)

Ti 2 Ti Ti + 22

(B) Fools gold FeS2 Fe2+ S —S

(C)


Sol. (A) Both on heating form an oxosalt (K2CrO4 & K2MnO4), a

metal oxide (Cr2O3 & MnO2) & oxygen gas.

(B) Both have their colour (orange and purple respectively)

because of charge transfer spectrum (since no unpaired

electrons).

(C) Only K2Cr2O7 behave as primary titrant while KMnO4

behave as self-indicator in redox titrations.

K2Cr2O7 requires diphenyl amine indicator for end point

detection while KMnO4 behave as secondary titrant.

(D) Both form a slightly coloured precipitate (S) with hydrogen

sulphide in acidic medium.


Sol. Correct product for (C) option.

2H O


Sol.

CH3

HCl

CH3

H Cl

is chiral (no POS and COS)




45. Select correct formula ...........

Sol.4 2 2 7 3 2 4(NH ) Cr O NaOH NH (g) Na CrO

(orange–red) [Y] (yellow)[X] [Z]

Na2CrO4 + BaCl2 4BaCrO NaCl

(yellow)

46. Coloured solution 'R' ...........

Sol. (NH4)2Cr2O7 (s) 2 2 3 2N (g) Cr O (s) 4H O(g)'P' 'Q'

Cr2O3 dil HCl3CrCl aq(green)R' '

2NaOH Br3 2 4CrCl (aq) Na CrO

(yellow)

3 3NH sol excess NH sol 33 3 3 6CrCl (aq) Cr(OH) [Cr(NH ) ]

pgreen ale pink( ) ( )

2 4dil H SO3 2 4 3CrCl (aq) Cr (SO )

.green sol

NaOH excess NaOH –3 3 4CrCl (aq) Cr(OH) [Cr(OH) ]

.green solgreen( ) ( )

47. pH of solution of cathode ...........

Sol. At Cathode:

Saturated with H2S, all CuSO4 is precipitate as CuS.

CuSO4 + H2S CuS + H2SO4

0.1 M – – 0.1 M

Finally cathode contain 0.1 M H2SO40.2 M H+ (aq.)

pH(cathode) Finally = –(log 2×10–1) = 1–log2 = 0.7 Ans.

48. Select the correct ...........

Sol. H2S 2H+ (aq.) + S2–(aq.) Kaq = K1.K2.

0.1 M 0.2 M

Kaq. =2 2–

2

[H ] [S ]

[H S]

= K1.K2

[S2–] = 1 2 22

K K [H S]

[H ]

=–21 –1

2

10 10

[0.2]

= 2.5 × 10–21

Again KSP(CuS) = [Cu2+][S2–] = 10–44

[Cu2+]2.5×10–21= 10–44

[Cu2+] = 4 × 10–24 M Ans.

At Anode :

Since KSPof ZnS is not very small

Again[Zn2+] = 0.1 M

H2S H+ + HS– K1

0.1 0 0

(0.1 – x) x + y (x – y)

~ 0.1 ~ x ~ x

HS– H+ + S2– K2

x – y (x + y) y

~ x ~ x

K2 = y = [S2–] = 10–14 Ans.

50. When X and W are ...........

Sol.(X)

O3

Zn/H2O/

O

O

OH

(Y)

O

(Z)

N2H4

OH/(W)

Pd/C/

(V)

H2/Pd




51. To 500 mL of xM NaOH, ...........Sol. 2NaOH + CuSO4 Cu(OH)2

+ Na2SO4.mmoles 500x 500 × 0.6

mmoles left 500x–600 300 Initial pOH = – log x

Final pOH = – log500x 600

1000

pOHf – pOHi = –log1000

600–x500 ×

x

1

1 500x 600

10 1000x

1000 x = 5000 x – 6000

4000 x = 6000

x =6 3

4 2

52. The graph of compressibility ...........

Sol. Z = 1 +Pb

RT (high pressure)

dZ

dP =

b

RT=

1

2.8

b =RT

2.8 =

22.4

2.8= 4(NA ×

4

3r3)

(NA ×4

3r3) =Volume of 1 mol gas molecules =

5.6

2.8= 2

53. How many of the following ...........

Sol. C60 &–2

72OCr

54. It is observed that ...........

Sol. C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g)

G = H – TS

(useful energy) = – 3000 – 300 × 0.18

= – 3054 KJ/mol

Energy required per hour =128

9× 60 × 60 J

Glucose needed per hour =

128 12860 60 60 60

9 9mol3054 1000 3054 1000

× 180 g 3 g Ans.

56. In how many of the following...........

Sol. (A) Gold sol is negatively charged.

(B) Sulphur sol is negatively charged.

(C) TiO2 sol is positively charged.

(D) Sb2S3 sol is negatively charged.

(E) Al(OH)3 sol is positively charged.

(F) Congo red is acidic dye. It is negatively charged.

(G) Methylene blue is basic dye. It is positively charged.

(H) Soap micelle is negatively charged.

57. Calculate of pI of glutamic ...........

Sol. pI =2.19 4.25

2

= 3.22

58. How many moles of ...........

Sol.O O O O O O O O OH

H3O

CH3–OH + 4CH3–CH=O + 4HCHO

1234

59. How many positional ...........

Sol.

Cl

Cl,

Cl

Cl

,

Cl

Cl

,

Cl

Cl

, Cl

Cl,

Cl

Cl

60. How many of the following...........

Sol. Only (iv), (vi) and (ix) are less basic than aniline. (iv) Pyrole is

least basic on lone pair is involved in aromaticity. (vi) has ortho

substituent thus reducing the solvation and basicity. (ix) has –I

group at meta position hence lesser basic than aniline.





TARGET : JEE (MAIN+ADVANCED) 2018DATE : 25-02-2018 COURSE: VVIIJJEETTAA ((0011JJPP,, 0022JJPP)),,


ANSWER KEYCODE-0

PAPER-1

MATHEMATICS

1. (CD) 2. (AC) 3. (ABC) 4. (ACD) 5. (ABCD) 6. (BC) 7. (ABC)

8. (BD) 9. (CD) 10. (AC) 11. (ABCD) 12. (ABD) 13. (BD) 14. (ABD)

15. (AD) 16. (6) 17. (5) 18. (4) 19. (7) 20. (5)

PHYSICS

21. (ABD) 22. (AC) 23. (ABCD) 24. (AD) 25. (C) 26. (AC) 27. (AD)

28. (AD) 29. (AC) 30. (AD) 31. (AC) 32. (BD) 33. (BC) 34. (ACD)

35. (AD) 36. (5) 37. (5) 38. (6) 39. (6) 40. (8)

CHEMISTRY

41. (ABD) 42. (CD) 43. (BCD) 44. (ACD) 45. (ABD) 46. (ACD) 47. (AB)

48. (ABCD)49. (BCD) 50. (ACD) 51. (CD) 52. (BD) 53. (ABCD)54. (BCD)

55. (AC) 56. (6) 57. (6) 58. (6) 59. (6) 60. (6)

PAPER-2

MATHEMATICS

1. (ABD) 2. (AD) 3. (AD) 4. (BC) 5. (B) 6. (B) 7. (A)

8. (C) 9. (D) 10. (B) 11. (2) 12. (6) 13. (6) 14. (2)

15. (2) 16. (5) 17. (2) 18. (5) 19. (9) 20. (3)

PHYSICS

21. (ABC) 22. (AC) 23. (ABD) 24. (ABCD) 25. (A) 26. (B) 27. (C)

28. (B) 29. (C) 30. (B) 31. (3) 32. (6) 33. (1) 34. (5)

35. (6) 36. (3) 37. (1) 38. (6) 39. (1) 40. (5)

CHEMISTRY

41. (BC) 42. (ABD) 43. (ABD) 44. (ABC) 45. (C) 46. (C) 47. (D)

48. (D) 49. (C) 50. (B) 51. (3) 52. (2) 53. (2) 54. (3)

55. (2) 56. (5) 57. (3) 58. (4) 59. (6) 60. (3)







ANSWER KEYCODE-1

PAPER-1

MATHEMATICS

1. (BD) 2. (AD) 3. (ABD) 4. (BCD) 5. (ABCD) 6. (BD) 7. (ABD)

8. (AD) 9. (BD) 10. (BC) 11. (ABCD) 12. (ABC) 13. (BC) 14. (ACD)

15. (AB) 16. (6) 17. (5) 18. (4) 19. (7) 20. (5)

PHYSICS

21. (BCD) 22. (AC) 23. (ABCD) 24. (BC) 25. (A) 26. (AC) 27. (BC)

28. (BC) 29. (AC) 30. (BC) 31. (AC) 32. (BD) 33. (AD) 34. (ABC)

35. (BC) 36. (5) 37. (5) 38. (6) 39. (6) 40. (8)

CHEMISTRY

41. (ACD) 42. (BC) 43. (ABD) 44. (BCD) 45. (ABD) 46. (ACD) 47. (AB)

48. (ABCD)49. (ABC) 50. (ACD) 51. (BC) 52. (BC) 53. (ABCD)54. (BCD)

55. (AC) 56. (6) 57. (6) 58. (6) 59. (6) 60. (6)

PAPER-2

MATHEMATICS

1. (ACD) 2. (AB) 3. (AB) 4. (CD) 5. (C) 6. (C) 7. (B)

8. (D) 9. (C) 10. (D) 11. (2) 12. (6) 13. (6) 14. (2)

15. (2) 16. (5) 17. (2) 18. (5) 19. (9) 20. (3)

PHYSICS

21. (ACD) 22. (AC) 23. (BCD) 24. (ABCD) 25. (C) 26. (D) 27. (A)

28. (D) 29. (A) 30. (D) 31. (3) 32. (6) 33. (1) 34. (5)

35. (6) 36. (3) 37. (1) 38. (6) 39. (1) 40. (5)

CHEMISTRY

41. (AC) 42. (ABC) 43. (ABC) 44. (ABC) 45. (A) 46. (A) 47. (B)

48. (B) 49. (D) 50. (B) 51. (3) 52. (2) 53. (2) 54. (3)

55. (2) 56. (5) 57. (3) 58. (4) 59. (6) 60. (3)







ANSWER KEYCODE-2

PAPER-1

MATHEMATICS

1. (CD) 2. (AC) 3. (ABC) 4. (ACD) 5. (ABCD) 6. (BC) 7. (ABC)

8. (BD) 9. (CD) 10. (AC) 11. (ABCD) 12. (ABD) 13. (BD) 14. (ABD)

15. (AD) 16. (6) 17. (5) 18. (4) 19. (7) 20. (5)

PHYSICS

21. (ABD) 22. (AC) 23. (ABCD) 24. (AD) 25. (C) 26. (AC) 27. (AD)

28. (AD) 29. (AC) 30. (AD) 31. (AC) 32. (BD) 33. (BC) 34. (ACD)

35. (AD) 36. (5) 37. (5) 38. (6) 39. (6) 40. (8)

CHEMISTRY

41. (ABD) 42. (CD) 43. (BCD) 44. (ACD) 45. (ABD) 46. (ACD) 47. (AB)

48. (ABCD)49. (BCD) 50. (ACD) 51. (CD) 52. (BD) 53. (ABCD)54. (BCD)

55. (AC) 56. (6) 57. (6) 58. (6) 59. (6) 60. (6)

PAPER-2

MATHEMATICS

1. (ABD) 2. (AD) 3. (AD) 4. (BC) 5. (B) 6. (B) 7. (A)

8. (C) 9. (D) 10. (B) 11. (2) 12. (6) 13. (6) 14. (2)

15. (2) 16. (5) 17. (2) 18. (5) 19. (9) 20. (3)

PHYSICS

21. (ABC) 22. (AC) 23. (ABD) 24. (ABCD) 25. (A) 26. (B) 27. (C)

28. (B) 29. (C) 30. (B) 31. (3) 32. (6) 33. (1) 34. (5)

35. (6) 36. (3) 37. (1) 38. (6) 39. (1) 40. (5)

CHEMISTRY

41. (BC) 42. (ABD) 43. (ABD) 44. (ABC) 45. (C) 46. (C) 47. (D)

48. (D) 49. (C) 50. (B) 51. (3) 52. (2) 53. (2) 54. (3)

55. (2) 56. (5) 57. (3) 58. (4) 59. (6) 60. (3)







ANSWER KEYCODE-3

PAPER-1

MATHEMATICS

1. (BD) 2. (AD) 3. (ABD) 4. (BCD) 5. (ABCD) 6. (BD) 7. (ABD)

8. (AD) 9. (BD) 10. (BC) 11. (ABCD) 12. (ABC) 13. (BC) 14. (ACD)

15. (AB) 16. (6) 17. (5) 18. (4) 19. (7) 20. (5)

PHYSICS

21. (BCD) 22. (AC) 23. (ABCD) 24. (BC) 25. (A) 26. (AC) 27. (BC)

28. (BC) 29. (AC) 30. (BC) 31. (AC) 32. (BD) 33. (AD) 34. (ABC)

35. (BC) 36. (5) 37. (5) 38. (6) 39. (6) 40. (8)

CHEMISTRY

41. (ACD) 42. (BC) 43. (ABD) 44. (BCD) 45. (ABD) 46. (ACD) 47. (AB)

48. (ABCD)49. (ABC) 50. (ACD) 51. (BC) 52. (BC) 53. (ABCD)54. (BCD)

55. (AC) 56. (6) 57. (6) 58. (6) 59. (6) 60. (6)

PAPER-2

MATHEMATICS

1. (ACD) 2. (AB) 3. (AB) 4. (CD) 5. (C) 6. (C) 7. (B)

8. (D) 9. (C) 10. (D) 11. (2) 12. (6) 13. (6) 14. (2)

15. (2) 16. (5) 17. (2) 18. (5) 19. (9) 20. (3)

PHYSICS

21. (ACD) 22. (AC) 23. (BCD) 24. (ABCD) 25. (C) 26. (D) 27. (A)

28. (D) 29. (A) 30. (D) 31. (3) 32. (6) 33. (1) 34. (5)

35. (6) 36. (3) 37. (1) 38. (6) 39. (1) 40. (5)

CHEMISTRY

41. (AC) 42. (ABC) 43. (ABC) 44. (ABC) 45. (A) 46. (A) 47. (B)

48. (B) 49. (D) 50. (B) 51. (3) 52. (2) 53. (2) 54. (3)

55. (2) 56. (5) 57. (3) 58. (4) 59. (6) 60. (3)

(JEE ADVANCED PATTERN) TARGET : JEE (MAIN+ADVANCED) … · 2019-11-13 · Website : | E-mail :...

Documents

Transcript of (JEE ADVANCED PATTERN) TARGET : JEE (MAIN+ADVANCED) … · 2019-11-13 · Website : | E-mail :...