HINTS & SOLUTIONS PAPER-1 PART-I : PHYSICS€¦ · Sol. Mass of solution = 180 g Since the salt is sparingly soluble in water, neglecting mass of ions dissolved in water, Mole of

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DATE : 13-05-2018 COURSE : ALL INDIA TEST SERIES (VIKALP) | CLASS: XII/XIII

HINTS & SOLUTIONSPAPER-1

PART-I : PHYSICS1. In the circuit .......................Sol:

C1 = C C2 = C+( Q0 – q)

–( Q0 – q)

+q

–q

dqi

dt

R

0Q q

c

– iR –

q0

c

0Q 2q Rdq

c dt

q t

00 0

dq 1dt

Q 2q Rc

2t /Rc0Qq 1 e

2

2t /Rc0Qdqi e

dt Rc

2t /Rc00

QQ q 1 e

2

20Q

H4c

2. The electric .......................Sol. V = A(x2 – 2y2 + z2)

Plane parallel to x – z plane y = constant x2 + z2 = constant its a circlePlane parallel to x-y plane z = constantx2 – 2y2 = constant its a hyperbola

3. The figure .......................

Sol. E = 00

QQ AE

A

QABCD = 0AE1 QQ 1 1

k k k

Udielectric =

2 20

0

E At1 Ek At

2 k 2k

4. A neutron .......................Sol.

n v H

21mv k

2 (k)max =

21 m kv

2 2 2

E2 – E1 = 10.2 eV(k)max > 10.2 eV k > 20.4 eV

E4 – E1 = 12.75 eV

k12.75eV

2 K > 25.5 eV

5. Which of the .......................Sol. (A) [0E] = [Mº L–2 T1 A1]

(B) [] = [M–1 L-3 T3 A2](C) [force] = [M1 L1 T–2]

(D)E

B

= [Mº L1 T–1]

6. A solenoid of .......................Sol.

L R/20 = 2E/R

R

L R/20

R

0 =2E

R, U =

2 2 2

2 2

1 2E L 4E 2LEL

2 R 2 R R

sol

R

H 1

H 2 ,

2

S 2

1 2 LEH U

3 3 R

7. Force on a .......................Sol.

r

(x, y)

y

x

r ˆF A(r B) A(r B)rr

F

is conservative

fr

y

x

ir

dw = F.dr A(r B)dr




w =f

c

r

r

A (r B)dr

2 2f i

f i

r rA B r r

2

(A) ri =B

2, rf =

3B

2

w =

2 29B B3B B4 4A B

2 2 2

= –A (B2 – B2) = 0(B) ri = rf = Bw = 0(C) ri = B rf = 2B

W = 2 24B B

A B 2B B2

=2 2

23B ABA B

2 2

8. Consider the .......................

Sol. (1) 10 kg100

(f12)max = 20

(2) 10 kg 100

(f23)max = 40

(3) 10 kg15

(f34)max = 60friction between (1) & (2) = 20 Nfriction between (2) & (3) = 40 Nfriction between (3) & ground = 25 N

9. A L shaped .......................Sol.

3V0

3V0

V0

V0B

C

3L0L0

AL0

2L0

y

x

VA - BB =0

0

3L

00 0

0L

B3V (x L )dx

L 0 0 018V B L

VC – VB =0

0

2L

00

0L

BV

L (3L0 + y) dy = 0 0 09V B L

2

VA – VC = (VA – VB) – (VC – VB)

= 18V0B0L0 – 0 0 00 0 0

9V B L 27V B L

2 2

10. Some amount .......................

Sol.

BA

FBD of A part (horizontal force balance)

dg (d )

2

= gd

d2

2d

g

FBD of A + B (Horizontal force balance)

(1 + sin) = 2gt

2

= 90 –

tg t

2

(1 + sin) = 2gt

2

2 (1 sin )t

g

11. The ends of .......................

Sol. x 0 x Hot end cool end

dTincreasing

dx

–dT

KAdx

= Constant A =Cons tant

dTK

dx

SincedT

dx = increasing

A decreasing




12. A person riding .......................Sol.

Observer

S

10 m/s310 m/s

1240 m/s

S1

20 m/s

D

Frequency observed by S1 =310

310 10

(48) = 49.6 Hz

Frequency reflected & transmitted by S1= 49.6 HzFrequency of reflected sound received by observer =

310 10

310

310

310 10

48 = 51.2 Hz

Beat frequency observed by observer = (51.2 – 48) = 3.2 Hz

Frequency received by detector =1240 20

1240

31048 50.4

300

hz

wavelength observed =1260

25m50.4

13. There is a .......................Sol.

+Q +2 Q

+Q

A B C D

S1 S2

A D

0

B C

+q1 –q1 +q2 + Q – q1 + q2) 0

-q2

–( Q – q1 + q2)

2 1 2

0 0

q Q q qd d 0

A A

q2 + Q – q1 + q2 = 0 q1 – 2q2 = Q .......(1)

1 2

0 0

q qd d 0

A A

q1 + q2 = 0 ........(2)

from (1) & (2)

q1 =Q

3

q2 =Q

3

(qB)i = +Q (qB)f = – q1 + q2 =2Q

3

charge flown from S1 =5Q

3

(qD)i = +2Q (qD)f = – Q + q1 – q2

Q

3

change flown through S2 =7Q

3

14. A person A .......................Sol.

6 6

A1 m/s

6 3

BA

121

6

VAB = 1 m/s

VBA =

A

B1 m/s

16 3

6

= 2 m/s

dr

dt = – 1 m/s = m

m = 1, n = 4, P = 3

15. A wooden cubical .......................Sol.

N

x

mg

M = a2B0

for no toppling

mgx = a2B0 x =2

0a B a

mg 2

0a B 1

mg 2

0

mg

2aB

x =2

0a Ba

4 mg

=

0

mg

4aB




16. In a radioactive .......................Sol. NA = N0 e

–t

NB = N0 (1 – e–t)

1t00

NN e

5 t1 = n5 1

n5t

600

NN e

8 6 = n = n8 =

n8

6

09N

10 = N0 (1 – e–t) t9

1 e10

2t 1e

10 2

n10t

t2 – t1 =n10 n5

n2n8

6

= 6n2

3 n2

= 2 sec

t1 =6 n5

n8

=6 n5

3 n2

= 2 log25

t2 = 2 2

n106 2log 10 log 100

3 n2

17. A non-uniform .......................

Sol.

F

x = 0 x = L

dm = dx =

LL 2

0 0 0

0 0

x x 31 dx x L

L 2L 2

a0 =0

0

F 2F3 3 LL2

Mx =2

0

x x

2L

T =2

00

x 2Fx

2L 3 L

T =22F x

x3L 2L

LL 2 3 2 2

0 0

2F x x 2F L L 4FLTdx

3L 2 6L 3L 2 6 9

18. A triangular .......................Sol. t = 0

x=+L

x=0

x = –L

Vy

h

0 x –L

y(x,0) =hx

hL

–L × 0

hxh –

L0 ×+ L

0 x +Lat any time t .

x = Vt + Lx = Vt – L

y

h

x = Vt

0 x Vt – L ;y

0x

y

0t

y(x, t) =hx

hL

Vt – L × Vt ;y h

x L

y hV

t L

hx

h –L

Vt × Vt + L ;y h

x L

y hV

t L

0 x Vt + L ;y

0x

y

0t

Instantaneous power

P = –y y

fx t

0 x Vt – L

P =

2h

FVL

Vt – L × Vt

2h

FVL

Vt × Vt + L

0 x Vt + L

19. Choose the correct .......................

Sol.

01

Esin t

2R 3

02

Esin t

2R

12

=00E

2 cos30 sin t2R 6




03Esin t

2R 6

= 0Esin t

Z 6

Z =2R

3, Phase difference =

6

, Power factor = cos =

3

2, tan =

Reac tance x 1

Resis tance R 3

2 2R x Z R = R, x =R

3

20. An electron .......................

Sol. L =nh

2

L = nh mh hn m

2 2 2

n – m = integer, so integer multiple ofh

2 are allowed.

PART-II : CHEMISTRY21. A 180 mL saturated ………….Sol. Mass of solution = 180 g

Since the salt is sparingly soluble in water, neglecting massof ions dissolved in water,Mole of water in solution = 180 / 18 = 10 mol(Pº – Ps) / P

º = (mol of ions / mol of water)Moles of ions dissolved in water = 0.0054 molMoles of MCl2 dissolved in water = 0.0018 mol[M2+] = 1.8 mmol / 180 mL = 0.01 M[Cl–] = 3.6 mmol / 180 mL = 0.02 MKsp = [M2+][Cl–]2 = 4 × 10–6

22. Which step(s) is/are ………….Sol. Bond breaking and ionisation energy are always

endothermic. Lattice making is always exothermic. Electrongain enthalpy of halogens is exothermic.

23. Which of the following ………….Sol. For exothermic reaction, equilibrium constant decreases

with increasing temperature.But, rate constant of any chemical reaction increases withincreasing temperature.

24. A solution of an unknown ………….Sol. White ppt with dilute HCl implies presence of group 1

cations. AgCl dissolves in ammonia solution. White ppt ofPbCl2 dissolves in hot water, PbI2 is yellow ppt. Possibility ofHg2Cl2 is ruled out because white ppt of Hg2Cl2 does notdissolve in ammonia solution..

25. Given for H2CO3 ………….Sol. [H+] for H2CO3

= M1021.0104CK 4701

[H+] for NaHCO3

= M104104104KK 911721

26. When solid NH4SCN………….Sol. Since gas is produced from solids entropy has increased. ∆ng >

0.In closed container, P-V work is zero. w = 0. Temperature dropimplies heat is absorbed by the system during the process. q >0. Thus, as per first law ∆E = q + w > 0.∆H = ∆E + RT∆ng.Thus, ∆H > 0.∆G < 0, because the process is spontaneous.

27. 1 mol red lead………….Sol. Pb3O4 + 8HCl 3PbCl2 + Cl2 + 4H2O

Pb3O4 + 4HNO3 2Pb(NO3)2 + PbO2 + 2H2O

28. Following is/are ………….Sol. Two oxygen atoms per tetrahedron are shared forming rings.

Hybridisation of each Si is sp3.

29. The following diagram ………….Sol. Low accuracy because measurements are far off from correctly

value. But, they are highly precise, because they are very closeto each other.

30. Two mole of electrons………….Sol. In first cell, at anode: 2Cl–(I) Cl2(g) + 2e–

at cathode, Na+ (l) + e⎯ Na(s)

In second cell,

at anode : 2 Cl⎯ (l) Cl2 (g) + 2 e⎯at cathode, 2H2O(I) + 2e– H2(g) + 2OH–

31. Which statement(s) ………….Sol. Soap and detergent both have hydrophillic as well as

hydrophobic groups, hence both can form micellles.Carboxylates are more basic than alkyl sulphates, becuasecarboxylic acids are weaker acids.In acidic pH, stearate gets protonated to form correspondingfatty acid, which is not much soluble in water due to longhydrophobic chain. Soap precipitates in hard water becuase itscalcium salt is insoluble.

32. Which diagram represents ………….Sol. In option C, Copper cannot reduce ferric ion to iron.

33. Which of the following ………….Sol. Conjugate base of Q is aromatic, hence it is the most acidic

compound among given hydrocarbons.

–HC

–C

34. Xconc. NaOH Y ………….

Sol.CHO|CHO

conc. NaOH

int ramolecularcannizaro reaction

2CH OH|COO

H 2CH OH|COOH




35. The commonly used ………….Sol. Nitration of phenol with dilute HNO3, gives mixture of o-

nitrophenol and p-nitrophenol; of which o-nitrophenol (X) issteam-volatile. Y on reduction gives p-amino phenol.

36. Two ketones, P and Q………….Sol. Racemization of an optically active ketone in basic medium

is possible due to tautomerisation only if the alpha-acidic His attached to a chirality centre.In option A, both P and Q will undergo racemization. Inoption B, Q is optically inactive. In option C, Q will undergoracemisation.

37. C – NH2

||O

52OP

………….

Sol.

C – NH2

||O

52OP

+

(i) CH Mg3 I

(ii) H O3

C||O

+ 2CHI3

38. Which of the following ………….Sol. 2-bromopropane + Mg/ether followed by CO2; followed by

hydrolysis would give isobutyric acid.

39. Observed the following ………….Sol. Cyclobutanecarbonitrile is

CN

LAH/ether reduces it to primary aminecyclobutylmethylamine (A). A undergoes exhaustivemethylation by 3 eq. of CH3I to give quaternary ammoniumsalt (B). B is formed by substitution.

B is iodide salt ofN+

B further undergoes hoffman elimination on heating withAgOH.

C is CH2

40.O O O

AlCl3Benzene P Q

Zn(Hg)

SOCl2SOCl2

R

AlCl3

SNaBH4T

H2SO4U

NBS(1eq.)V

CCl4,hv

W

NaOEt/EtOH, heat

HCl

………….

Sol.

+

O

O

O

AlCl3

O

P OHO

Zn(Hg)HCl

Q OHO

SOCl2

R OCl

AlCl3

OS

NaBH4

OHT

H2SO4

U

NBS/CCl4 Br

VEtONa/EtOH

W

PART-III : MATHEMATICS

41. Let ‘m’ be .............................Sol. Suppose that both r and r + 1 are solutions to the equation

3 2x 3x 34x m

Then3 2r 3r 34r m , and also3 2(r 1) 3(r 1) 34(r 1) m

Subtracting the first of these equations from the second gives2(3r 1) 3(r 1) 34 0,

or23r 9r 30 0 or

2r 3r 10 0 r = –5, 2 and put in given equation

42. If a triangle .............................

Sol. AB BC AC

2u u v u vBC .

u u v u v

u v u v ˆ ˆ ˆ ˆAB . BC u v . u v 1 1 0.u v u v

B 90 1 cos2A cos2B cos2C 0

4cos A cosBcosC 0




43. Let 1 1 1 2 2 2Z x iy , Z x iy ...................

Sol.1 2i i

1 2 1 2

1 2

Z e ,Z e ,Re(Z Z ) 0

/ 2,

1 1i i3 4 5 1Z e ,Z ie ,Z cos (1 i)

6 1Z sin ( 1 i)

44.n 2n n 2n 2

0 m 1 m.................C C C .C ... .Sol. The given series can be written as

nn 2n 2r r

r mr 0

S C C ( 1)

n

n rr

r 0

C ( 1)

× coefficient of xm in (1 + x)2n–2r

= coefficient of xm in

n n rn r 2r

r 0

C ( 1) (1 x)

= coefficient of xm in [(1 + x)2 – 1]n

= coefficient of xm in (x2 + 2x)n

= coefficient of xm in xn(x + 2)n

= coefficient of xm–n in (x + 2)n

n n (m n) 2n mm n

nC 2 2

m n

if m n and 0 if m < n.

45. If A is non –.............................

Sol.1 13ABA A 2A BA

1 13ABA A 2A 2A BA 2A 1 13A(BA ) 2(A B )A

1 13A(B A)A 2A (B A )A 1 13A(B A)A 2A (B A)A

Let B + A = X

1 13AXA 2A XA n 1 n 13 | A || X || A | 2 | A || X || A | n n3 | x | 2 | X | (as | A | 0) (i)

| X | 0 or | A B | 0

Let 1 1M ABA A BA 2 1 2 1AM A BA BA BA A BA AM

1 1

1 2 1

1

3ABA A 2A BA

2A (A BA AM)

2ABA 2M

1

1

1

ABA A 2M

A(BA ) 2M

A(A B)A 2M

46. 5 players of .............................

Sol.

5 6110

C 2 5P(A) ,

162

5 6110

C 2 5P(B) ,

162

5 32

10

C 2 2 5P(A B) ,

322

15P(A B)

32

47. The area bounded .............................

Sol. The desired area is shown as shaded region

x1O

1 /2 22 2

0 1

4A 4 (x x )dx 4 (x x)dx (2 3)

3 6

48. Suppose that .............................

Sol. f(1000)f f(1000) 1

f(1000)f(999) 1

999f(999) 1

1f(999)

999

The numbers 999 and1

999 are in the range of f.

Hence, by intermediate value property of continuous function,

function takes all values between 999 and1

999, then there

exists.

1,999

999

such that f( ) 500

Then 1f( ) f f( ) 1 f(500)

500

Similarly,1

199 ,999999

, thus 1f 199

199

But there is nothing to show that 1999 lies in the range of f. Thus

(d) is not correct and (c) is also incorrect.




49. A function f : R R .............................

Sol.2 nx

nxn

ax bx c ef(x) lim

1 c.e

n2 x

nn x

n2 x

nn x

2

nx

n

nx

ax bx c elim ; x 0

1 c. e

ax bx c elim ; x 0

1 c. e

ax bx c1

elim ; x 0

1c

e

2ax bx c; x 0

1; x 0

1; x 0

c

since f(x) is continuous function x R

x 0 x 0lim f(x) lim f(x) f(0)

2

x 0 x 0

1lim lim (ax bx c) 1

c

11 c 1

c

c 1, a, b R

50. Let f(x) be .............................

Sol. 34

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

f(x) 1 f (x)

Integrating on the interval (a, b), we getb b

1 2 ba4

a a

f(x)f '(x) 1dx dx tan f (x) b a

1 f (x) 2

1 2 1 2

x b x a

1b a lim tan f (x) lim tan f (x)

2 24

51. A straight line.............................Sol. Let equation of line passes through A(–2, –3) is

x 2 y 3r

cos sin

Point on it (rcos –2, rsin –3)If this point is B lies on x + 3y = 9

AB =20

cos 3sin If this point is C lies on x + y + 1 = 0

AC =4

sin cos as AB.AC = 20

3sin2 + cos2 + 4sin cos= 4 3cos2 – 4sin cos+ sin2= 0 (cos – sin)(3cossin) = 0 tan = 1,3Equation of line y + 3 = x + 2 or y + 3 = 3(x + 2) y = x – 1 and y = 3x + 3

52. If graph of .............................

Sol. Let image of point A( , ) about y 2x is B(a, b).

Mid point of AB

i.e.a b

M ,2 2

lie on y 2x 0

b a

2 b 2 2a2 2

(i)

and slope of AB1

2

b 1 1 a

ba 2 2 2

(ii)

Subtracting (i) and (ii),

5 3 4b 3a

2b a2 2 5

8b 6a 3b 4a

2a b5 5

( , ) lies on xy = 1

4b 3a 3b 4a

15 5

2 212b 7ab 12a 25

2 212x 7xy 12y 25 0

53. A circle touches.............................Sol. Family of circles touching x + y – 2 = 0 at (1, 1) is

2 2(x 1) (y 1) (x y 2) 0 2 2x y ( 2)x ( 2)y 2 2 0 (1)

Given circle is:2 2x y 4x 5y 6 0 (2)

Equation of common chord PQ is S – S’ = 0.

( 6)x ( 7)y 8 2 0 (3)

(a) PQ || line: x + y – 2 = 0

61 6 7;

7

which is impossible

(b) PQ line : x y 2 0

6 131 ;

7 2

which is impossible

But when13

2 , we can see that the circles (1) and (2) are

not intersecting each other and their radical axis is perpendicular

to the given line x y 2 0 Eq. (3) can be written as

6x 7y 8 (x y 2) 0

which is in the form 1 2L L 0

Solving 1L and 2L , we get the point of intersection (6, –4)




54. Let A, B and C .............................

Sol. Let 2 2 21 1 2 2 3 3A (2t ,4t ),B (2t ,4t ) and C (2t ,4t )

Slope of AB = 2 t1 + t2 = 1 and t1 + t2 + t3 = 0So, t3 = – 1

Also,

2 2 22 21 2 3

1 2

2(t t t ) 4t t 1

3 3

1 2t 1, t 0 A (2,4),B (0,0) andC (2, 4) HenceP (6,0)

55. The volume .............................

Sol. Volume = | 2b c 3c a 4a b | 18

224 a b c 18

3| a b c |

2

Now ,

(1 sin ) cos sin 2

2 2 4a b c sin cos sin 2

3 3 3

2 2 4sin cos sin 2

3 3 3

Applying 1 1 2 3R R R R and expanding

3a b c 3 | cos3 |

2

1 2cos 3 3 ,

2 3 3

2,

9 9

56. Identify the correct .............................

Sol. (A)

–2 – O 2

No of solutions are 5

(B)

8

38

8

5

2

8

7


(C) 2


(D) |12–x| 2

2 = |x|2 |x2–12| = |x|


57. Which of the .............................Sol. If f(x) is not constant on [–1, 1]

f '(x) 0 at some point in –1 < x < 1

f is increasing at that point and we can’t have f(–1) = f(1)(b) Not trueIf f is bounded curve changes its concavityIf f is unbounded then curve crosses x-axis(c) Not truee.g. f(x) = x4

(d) f '(x) =2x2 + 2x – 12 = 2(x + 3)(x – 2) < 0 for x ( 3, 2) where f(x) decreases.

Also f "(x) 4x 2 0 for1

x2

58. If A(x + y) .............................Sol. A(x + y) = A(x)A(y)

A(0 + 0) = A(0)A(0) A(0) = 1Put y = –x, we getA(0) = A(x)A(–x) …..(i)

2

A( x)B( x)

1 (A( x))

2

1

A(x)1

1(A(x))

2

A(x)

1 (A(x))

= B(x)

Thus, B(x) is even.2011 2010 2011

2010 2010 2010

B(x)dx B(x)dx B(x)dx

2010 2011

0 2010

2 B(x)dx B(x)dx 2010 2011

0 0

B(x)dx B(x)dx

59. Letdy

y f(x)dx .............................

Sol.dy

y f(x)dx is linear differential equation.

I.F. = ex

Solution is yex =xe f(x)dx C

now if 0 x 2 thenx x xye e e dx C

yex = x + Cy(0) = 1, C = 1 yex = x + 1




x

x 1y ;

e

2

y(1)e

Alsox x

2x

e (x 1)ey '

e

2 2

e 2e e 1y '(1)

ee e

If x > 2

x x 2ye e dx yex = ex – 2 + C y = e–2 + Ce–x

As y is continuous

2 xxx 2 x 2

x 1lim lim(e Ce )

e

2 2 23e e Ce C 2 for x > 2y = e–2 + 2e–x

y(3) = 2e–3 + e–2 = e–2 (2e–1 + 1)

xy ' 2e

3y '(3) 2e

60. If

n

0k 3k

1Ck .............................

Sol.

n

0k 3k

1Ck

=

n

0k )3n)(2n)(1n(

)2k)(1k( n+3Ck+3

=)3n)(2n)(1n(

1

n

0k

2 )2)6k2(–6k5k( n+3Ck+3

=)3n)(2n)(1n(

1

n

0k

n

0k3k

3n3k

3n3k

3nn

0k

2 C2C)3k(2–C)6k5k(

=)3n)(2n)(1n(

1

n

0k

n

0k3k

3n2k

2nn

0k1k

1n C2C)3n(2–C)2n)(3n(

=)3n)(2n)(1n(

1

2

)3n)(2n(–)4n(–22)]3n(–2)[3n(2–)1–2)(3n)(2n( 3n2n1n

on solving we get

=)3n)(2n)(1n(

2–2)2nn( 1n2

a = 1 p = 1b = 1 q = 2c = 2 r = 3d = – 2

PAPER-2PART-I : PHYSICS

1. In the figure ...........................Sol.

k2

massless rod

(1) k

/2 /2

m

/2

Let the block is displaced x.Elongation in spring (2) = x –

m

k(x–)

/2k/2

k(x–)2

k k(x )2

22x

4

x =5

4

Restoring force = k (x – )

= k24 kx

x x m x5 5

=k

5m, T =

5m2

k

2. A particle A ...........................Sol.

y

xA B




BAˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆV 2i j k 2i 2 j k 4i 3 j

37ºA

4

3

B

dmin = 5 sin 37 = 3m

3. Two small ...........................

Sol.

m m v0

v0

0v

2

0v

2

Vcm

In COM frame

0v

2m

m0v

2

0v

2

0v

2

0V

2

m

k1 45º

0V

2

m45º

K1 +x

v

v

m

COME

20mV

2 = mV2 +

21kx

2COAM

0V x2 m 2 mv

2 2 2

0mV2

mV ( + x) v = 0V

2( x)

20mV

2 =

2 220

2

mV 1kx

24( x)

after substituting the values we get v0 = 1.3 m/s

4. A non-conducting ...........................

Sol.

45º53º 37º

53º

37º

E

= 45º

Electric field will be along angle bisector.

5. Two uniform ...........................

Sol.

22

zero

mg2 2

2 21 m 1 mcos mg cos2 (2 )

2 2 2 3 2 3

=

constant

2 22 21 m

mg 1 1 2 (1 4)2 2 2 3

= constant

2 2 25 3m mg

6 4 = constant

22m 5 3g

2 3 2

= constant

2 25 3g

3 2 = constant

5 d 32 g2 0

3 d 2

103g 0

3

2 9g(angularfrequency)

10

T =10

29g




6. There are ...........................Sol. Suppose we have connected battery across points A and B.

All points other than A and B will be at same potential soequivalent circuit will look like as given below

r

r

18 Points

So equivalent resistance will be 6

7. Moment of ...........................

Sol. = 2 2Ma b

2

= M (aa + bb) + m2 2a b

2

A0 = 74B0 = 222.

8. A glass sphere ...........................Sol. A = 2(1 – cos2C)R2

2C C

C

= 2 (2sin2C) R2

= 4R229 36 R

25 25

9. A uniform ...........................Sol.

A

Acosg

mg

N

fs

2A cos

C

A cos sin

C

N =2A cos

mgC

s

A cos sinf

C

=A

mg2C

s

Af

2C

fs N

A Amg

2C 2C

A

A 2mgC

10. Four identical ...........................

Sol.

q

q

R

Rq

4

3

q2

R

R

2R

k1

0

12 3

3q R. 2Rsin

4 ˆB k4 2R

= 0q k8 2 R

0

13 3

q. R.2Rsin2 ˆB k

4 2R

= 0q k16 R

0

14 3

q. R. 2Rsin4 ˆB k

4 2R

= 0q k8 2 R

01

q 1 ˆB 2 k8 R 2

1 1 1F q v B

F1 =

2 20q 1

28 2

12. Find the minimum ...........................Sol.

x

y2, 2y1, 1

1 1 2 2

1 2

y A( T x) y A( T x)

(1 T) (1 T)

y1 (1T – x)(1–1T) = y2 (2T + x)(1–2T)




x = 1 1 2 2

2 2 1 1

(y y ) T0

y (1 T) y (1 T)

Tension = 1 1

1

y A T

(1 T)

= y1A1T (1 – 1T)

= y1A1T = (109)(10–5) (6 × 10–4) (20)

= 120 Newton

Fundamental frequency of 1st wire alone

=1 120 20

10Hz2(1) 0.3 2

Fundamental frequency of 2nd wire alone

=1 120 40

20Hz2(1) 0.075 2

Frequency for which both the wires will vibrate (joint on

node)

= LCM of individual frequency = 20 Hz

14. Which of the ...........................

Sol. E =KQq 1 KqQ

– ma 2 ma

=KQq KQq KQq

– –a 2a 2a

If length of semi major axis is A

E =KQq

–2a

=KQq

–2A

A = a

+Q

a

u0 m

a

60°

3

2

Length of semi minor axis =3

2a

2 2

KQq GMm

a a

M =KQq

Gm

T2 =2

0

41 Qq

G4 Gm

a3 =

3 3016 ma

Qq

T =

3 3016 ma

Qq

16. Which of the ...........................Sol.

T

5T0

2T0

A C

B

5P02P0

A – B T = constant T

V = constant T = KV

PV = nRT P = constant V TB – C = constant V = constantC – A = constantP

C

B AV

V

CB

A

TP

AB

C

T

A – B P = constant , U0 = 0

3nRT

2WAB = nRT = nR (–3T0) = –3nRT0 = –2U0

BC V = constantWBC = 0

C – A WCA = nR(5T0) n i

f

= 5nRT0 n5

2

= 0

10 5U n

3 2

Process Q U W

AB –5U0 –3U0 –2U0

BC +3U0 +3U0 0

CA0

10 5U n

3 2

0 10

3v 0

10 5U n

3 2

0

n(2.5)

Net heat = 0

10 5U n

3 2

– 2U0




18. Radius of the ...........................

Sol. +e

–e

m

Pr

2

2

ke( e) mv

r r

22ke

mvr

(m) vr = n

nr

mv

22ke

mvn

mv

2

2kemv mv

n

2kev

n

2 2

2 2

nr

ke m

2ke( e) 1E mv

r 2

=2

2 2ke 1 1mv mv

r 2 2

2 4 3

2 2

k e mE

2 n

For hydrogen atom = 1

V0 =2Ke

r0 =

2

2Ke m

E0 =2 4

2

K e m–

2For this atom = 3

V =n

V3 0 , for n = 1 V = 3V0

r = 0

2

r9

n for n = 2 r = 0r9

4

E = 02E

n

27

PC = E3 – E1 =9

27– E0 –

0E

1

27–

= –3E0 + 27E0 = 24E0P=C

E24 0

20. The velocity ...........................

Sol.

J

a

COM

VCM

3

a2

J = 6aVCM VCM =a6

J

CM = 3(2a) 12

a2 2

+ (2a )

2

3

a2

+ 2 (2a)

9

aa

22

= 2a3 + 3a9

8 + 3a

9

40

= 3a9

66 = 3a

3

22

AMAIT w.r.t com

Ja = 3a3

22

2a22

J3

Velocity of P

J/6a

2

3J 2a

322 a

= ia11

JJ

a6

J

=

6

J

11

i

a

J

PART-II : CHEMISTRY

21. If P = total number of ………….

Sol. 6 GI with following trans pairs are possible: (aa)(bb)(cd),

(ab)(ab)(cd), (ac)(bd)(ab), (ad)(bc)(ab), (aa)(bc)(bd),

(bb)(ac)(ad). Out of which 3rd and 4th are chiral. So there are 2

pairs of enantiomers and total 4 chiral stereoisomers.

22. The following wave………….

Sol. The wave function is independant of angle which implies it

belongs to spherically symmetric s orbital. Further, it has only

one radial node at r = Z/2 which indicates 2s orbital. 2s orbital is

first excited state.




23. A first order reaction………….

Sol. K =2.303

2 log

100

80 K =

2.303 5log

2 4min–1

K =2.303

2 ×0.1 min–1

t½ =K

693.0 = 6 min

24. 5.6 L of an unknown………….

Sol. Mol of gas = n = 5.6 / 22.4 =1

4E = nCvT

Cv = 3 cal.mol–1.K–1

Cp = 5 cal.mol–1.K–1

=5

3

ideal gas is monoatomic.

25. A crystal is made up ………….

Sol. Effective no. of atoms of X = 4 – 4×(1/8) = 3.5

Effective no. of atoms of Y = 4 – 1 = 3

Effective no. of atoms of Z = 8 – 4 = 4

Hence, formula is X7Y6Z8

26. When the following ………….

Sol. Bromine disproportionates in basic solution of carbonate.

3Br2 + 3Na2CO3 5NaBr + NaBrO3 + 3CO2

27 Consider the ………….

Sol. x = 6, y = 3.

Final product is benzene.

H+/vkf/kD;leku iqujko`fr

28. How many of the following………….

Sol. Reactivity order towards Grignard reagent is as follows

C

O

R H > C

O

R R > C

O

R OR > C

O

R NR2

Only (i), (iv), (vi) and (vii) undergoes nucleophilic addition

reaction.

29. How many hyperconjugable………….

Sol.

CH–CH(CH3)2

+

CH2CH3

CH3

1

2

3

Total hyperconjugable H-atoms = 6

30. How many of the following………….

Sol. (S1, S3, S5 & S7 are correctly matched)

31. Metal M ………….

Sol. K2Cr2O7 is a very common lab reagent whose aqueous solution

is orange coloured but turns yellow in alkaline medium due to

formation of chromate.

K2Cr2O7 is obtained from Chromite ore FeO.Cr2O3 which can be

written as Fe(CrO2)2. Thus, M is Cr.

32. The gas C ………….

Sol. 2NH4Cl + K2Cr2O7 N2 + Cr2O3 + 4H2O + 2KCl

33. Suppose two hypothetical………….

Sol.

*2Px *2Py *2Px *2Py

One state has both electrons in different π* orbitals but with

opposite spins (violating Hund's rule).

Other state has 2 outermost electrons paired in a single *

orbital (again violating Hund's rule).Total spin S in both the

cases is zero, hence both are singlet states.

34. Which of the following………….

Sol. Acid catalysed dehydration of alcohol (E1 elimination) proceeds

through carbocation intermediate, which are singlet species as

they contain no upaired electron. Thus, total spin is zero.

Reimer-Tiemann formylation and Carbylamine formation involve

formation of singlet dichlorocarbene by alpha elimination of

chloroform by an alkali.




35. Which of the following………….

Sol. Y is aromatic cyclopentadienyl anion. Z is ferrocene, in

which ferrous ion is central atom and two cyclopentadienyl

anion (Y) act as ligand each donating 6 pi electrons to

ferrous ion.

EAN = 26 – 2 + 2 × (6) = 36.

38. Choose the correct ………….

Sol. (37) & (38)

H3OOO

COOH(X)

HOH

HOH

CH3–CHOH

OH+

OH OHCOOH

–H2O

CH3–CH=O

(Z)

(–H2O)

(Y)

O

OH

O

40. Compound P and R ………….

Sol.

O O

H /OH

OH

(P)

KOHCHCl3

OH

(Q)CHO

Ac2O / AcO,

(T)

OO

42SOH

OH

(S)COOH

Ni/H2

OH

(R)CH=CH–COOH


41. If a, b, c ..............................

Sol. c a b, b c a, a b c are all positive

Using A.M. G.M.

1/3a b c abc

3 (i)c a b a b c b c a (c a b)(a b c)(b c a)

Also,2 2 2a a (b c)

2a (a b c)(a b c)

Similarly2b (b c a)(b c a)

2c (c a b)(c a b)

2 2 2 2 2 2a b c (a b c )(b c a) (c a b)

Thus abc

(a b c)(b c a)(c a b)

abc1

(c a b)(a b c)(b c a)

hence from (i)abc

1(c a b)(a b c)(b c a)

42. If a, b, c, d > 0..............................

Sol. We have

2 2 2 2 2 2 2(a b c )x 2(ab bx cd)x b c d 0

2 2 2(ax b) (bx c) (cx d) 0

2 2 2(ax b) (bx c) (cx d) 0

b c dx

a b c

2b ac or 2 log b log a log c,

Now 1 1 3 2

33 14 log a

65 27 log b [Apply R R R 2R ]

97 40 log c

0 0 0

65 27 log b

97 40 log c

43. Suppose that ..............................

Sol. h'(x) f(x)

h''(x) f'(x)

h(1) 0,f(1) f'(1) h'(1) h''(1) 1 g(1)

f g(x) x

'g' is inverse of 'f'

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

f ' g(1) .g'(1) 1

f '(1).g'(1) 1

g'(1) 1

44. A be a square..............................

Sol. Letm n q n

A , adj(a)p q p m

Let | A | = d = mq – np




m qd n(1 d)A d adjA 0

p(1 d) q md

2 2 2 2

2 2 2 2

3 2 2 2

2 2

mq m d q d mqd np 2npd npd 0

(mq np) (mq np)d m d q d 2mqd 2d 0

(d d 2d ) d(m q 2mq) 0

d[(d 1) (m q) ] 0 d 1,m q 0

2| A d adj | A | (m q) 4(mq np) 4d 4

45. Two persons ..............................

Sol.4 2 2 4 3

2 3C p (1 p) C p (1 p)

6(1 p) 4p

3p 0.6

5

46. If the radius..............................

Sol.

22 h ab 4 2tan

a b 7

1tan

2 2 2

1 2(8 )sin

2 3 2

6

Hence, equation of circle is2 2(x 6) (y 6) 8

47. Possible value..............................

Sol. x2 – 5|x| + 1 = (3 – k)x

draw the graph of x2 – 5|x| + 1 and (3 – k)x

y = x2 – 5|x| + 1y = (3–k)x

There are two ways for line (3 – k)x to have 3 distinct

solutions with x2 – 5|x| + 1 as shown in figure

Case-I

so y = (3 – k)x, line touches y = x2 – 5|x| + 1 at some x > 0

and cuts at two points for x < 0

x2 – 5x + 1 = (3 – k)x

x2 + (k – 8)x + 1 = 0 D = 0 (k – 8)2

k = 8 ± 2, k = 6 is only possible value for which x > 0

k = 6 or 10

Similarly for x < 0

x2 + 5x + 1 = (3 – k)x

x2 + (k + 2)x + 1 = 0

(k + 2)2 = 4 or k = – 4

x > 0 so rejected

k + 2 = ±2

k = – 4 or 0 k = 0

x < 0 is possible value

so total possible values of k are 0 and 6

48. If the eccentric..............................

Sol.

If P is the point (acos , b sin ), the coordinates of Q are

a cos , bsin2 2

i.e., ( asin , b cos ) where2 2a 28, b 7

The area of 1CPQ

2

2 2| ab cos ab sin |

1 2 2 21

area (x y x y )2

1ab

2

1( 28)( 7 ) 7 sq. units

2

49. The straight line..............................

Sol.




Let the perpendicular distance of P from the line be h,

1h 5 6( 2 1)

2

12h ( 2 1)

5

Also tangent parallel to the given line is 4y 3x 12 2

Its distance from the line 4y 3x 12 IS

12H ( 2 1)

5

Hence there are three points as shown in the figure.

50. Number of the ..............................

Sol. 2 sin 11x cos 3x 3 sin 3x 0

1 3cos 3x sin 3x sin 11x 0

2 2

osin(30 3x) sin 11x 0

2sin 7 x cos 4 x 012 12

sin 7 x 0 or cos 4 x 012 12

7 x n or 4 x n ,n Z12 12 2

n n 7x or x ,n Z

7 84 4 48

51. [ a b b c c a

]..............................

Sol. [ a b b c c a

] = [ a b c

]2

52. If xa yb zc md

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

Sol. (a b) ( c d) (b c) (a d) (c a) (b d)

= [a b d] c [b c d]a [c a d]b – 6d

Now d a b rc

, where , , r are scalars

Take dot product with a b

(a b) d r [a b c]

= 2r

r =(a b) d

2

Similarly find and

So 2 d

= [a b d]c [b c d]a [c a d]b

m = 2

53. The value of ..............................

54. The minimum ..............................

Sol.x yf(x y) 2 f(y) 4 f(x) .........(i)

Interchanging x and y, we get

y xf(x y) 2 f(x) 4 f(y) .........(ii)

x y y x

x x y y

x x

2 f(y) 4 f(x) 2 f(x) 4 f(y)

f(x) f(y)k

4 2 4 2

f(x) k(4 2 )

f(1) k(4 2) 2

k 1

x x

4 4

x 2 x x 2

f(x) 4 2

f(4) 4 2 240

f(x) (2 ) 2 (2 1/ 2) 1/ 4

Thus, f(x) has least value as 1/ 4

55. The value of ..............................

56. The value ..............................

Sol.2 2n 2

2nn

(x ax 1) x (2x x b)f(x) lim

1 x

2

2

x ax 1, | x | 1

2x x b, | x | 1

a b 3, x 1

2a b 5

, x 12

x 1lim

f(x) exists if

x 1lim

f(x) =

x 1lim

f(x)

2 2

x 1 x 1lim (2x x b) lim (x ax 1)

2 1 b 1 a 1

a b 1 ........ (i)

x 1lim

f(x) exists if

x 1lim

f(x) =

x 1lim

f(x)

2 2

x 1 x 1lim (x x 1) lim (2x x b)

1 a 1 2 1 b

a b 1 ...... (ii)

Solving Eqs. (i) and (ii), we get a = 1 and b = 0.




57. Let f be differentiable....................

Sol. f is differentiable at x = 1 and g(x) is continuous at x = 3.

i.e. a = 2b and 2a = b

Also 3a + 3 – c = 10 + b

a b 0 and c 7

2x 7 x 49k (1 ) 0 has real and distinct roots

for R

2

2

2

49 4(49k)(1 ) 0 R

4k 4k 0 R

16k 16k 0

k(k 1) 0

k ( 1, 0)

58.x 2

f(x)lim

| g(x) | 1 exists ..............................

Sol. h 0 h 0

f(2 h) b(4 h)LHL lim lim

| g(2 h) | 1 | c(2 h) d | 1

4b

| 2c d | 1

h 0 h 0

f(2 h) b(4 h)RHL lim lim

| g(2 h) | 1 | a(2 h) 3 c | 1

4b

| 2a 3 c | 1

f is differentiable at x = 1 i.e., a = b = 0

So, LHL =RHL = 0 (it is given that limit exists)

59. ABC ..............................

60. The inradius ..............................

Sol. LetA B C

tan x, tan y, tan z2 2 2

CP

B

A

R

Q

r

r

r

I

r r rx , y , z

AR BP CQ

put the values of AR, BP and CQ in given relation, we get

2x 5y 5z 6

r r r r

2x 5y 5z 6 (1)

Also in the triangle, xy + yz + zx = 1 (2)

If we interchange y and z, then both equations (1) and (2)

remain unchanged

ABC is isosceles with B C y z

So from (1) and (2), we get

x = 3 – 5y (3)

and 2xy + y2 = 1 (4)

Solving we get,4 1

x , y z3 3

3rAR , BP 3r, CQ 3r

4

Now perimeter 2s = 2 . AR + 2 . BP + 2 .27r

CQ2

Given perimeter is smallest integer r 2

27s

2

27rs 2 27 sq. unit

2




JEE PREPARATORY TEST-2 (JPT-2)(JEE ADVANCED PATTERN)

TARGET : JEE (MAIN+ADVANCED) 2018

DATE : 13-05-2018 COURSE : ALL INDIA TEST SERIES (VIKALP) | CLASS: XII/XIII

ANSWER KEYCODE-0


1. (ABD) 2. (BD) 3. (BC) 4. (AC) 5. (ACD) 6. (BC) 7. (ABCD)

8. (AB) 9. (AD) 10. (AC) 11. (AD) 12. (BCD) 13. (C) 14. (ABCD)

15. (BD) 16. (BCD) 17. (ABCD) 18. (ABC) 19. (ABCD) 20. (BC)

PART-II : CHEMISTRY

21. (BD) 22. (AD) 23. (AC) 24. (AC) 25. (AD) 26. (ACD) 27. (AD)

28. (ABCD) 29. (BC) 30. (ABD) 31. (ABD) 32. (ABD) 33. (ABC) 34. (BD)

35. (ABD) 36. (D) 37. (BD) 38. (ACD) 39. (ABC) 40. (AD)


41. (AC) 42. (AC) 43. (ABCD) 44. (AB) 45. (BC) 46. (ACD) 47. (AD)

48. (AB) 49. (BCD) 50. (AC) 51. (AC) 52. (BCD) 53. (ABC) 54. (ABC)

55. (AB) 56. (BCD) 57. (AD) 58. (BD) 59. (ABD) 60. (ABC)


1. (5) 2. (3) 3. (2) 4. (9) 5. (9) 6. (6) 7. (3)

8. (1) 9. (4) 10. (2) 11. (A) 12. (C) 13. (B) 14. (A)

15. (B) 16. (D) 17. (D) 18. (A) 19. (A) 20. (D)

PART-II : CHEMISTRY

21. (6, 4) 22. (1) 23. (1, 2, 3, 4) 24. (1) 25. (7, 6, 8) 26. (5, 1, 3)

27 (9) 28. (4) 29. (6) 30. (4) 31. (D) 32. (B)

33. (C) 34. (C) 35. (C) 36. (B) 37. (A) 38. (D)

39. (A) 40. (C)


41. 0,1,2,3 42. 1,7 43. 2,3 44. 1,2,4 45. 2,3,5 46. 1,2,4,8 47. 0, 6

48. 0,1,2,3,4,5,6 49. 3 50. 2,5,8 51. (A) 52. (A) 53. (B)

54. (C) 55. (B) 56. (C) 57. (A) 58. (C) 59. (B) 60. (C)

Documents

HINTS & SOLUTIONS PAPER-1 PART-I : PHYSICS€¦ · Sol. Mass of solution = 180 g Since the salt is sparingly soluble in water, neglecting mass of ions dissolved in water, Mole of