Concept Recap Test Mains 4 Sol

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ANSWERS, HINTS & SOLUTIONS CRT(Set–IV)

ANSWERS KEY PHYSICS CHEMISTRY MATHEMATICS

Q. No. ANSWER ANSWER ANSWER 1. C B D 2. C D C 3. A A B 4. C D A 5. D B D 6. A D B 7. A C D 8. D B B 9. C D C 10. A D C 11. A C C 12. C B B 13. A C A 14. A C A 15. D A A 16. B B B 17. A C A 18. B B A 19. C B D 20. D C A 21. B D C 22. B A B 23. B A D 24. B C A 25. D D D 26. D D B 27. B D B 28. B C D 29. B B D 30. A A D

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AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/13


2

PPhhyyssiiccss PART – I

SECTION – A

1. Since VL = VC, so it is resonance.

and ViR

=

2. Energy lost in the collisions rel

2 2r

1 m U (1 e )4

= − .

3. The equivalent circuit is

eq

V 2VIR R

= ⇒ and current through ammeter is halved

ammeterI VI2 R

= =

V A

R

R

4. 0

qV4 d

=πε

; 20

qE4 d

=πε

5. initially the current in the circuit is zero, and there is no potential drop across the internal

resistance.

1 21 2

di diL L ,dt dt

= = ε at t = 0.

At any time t,

1 21 2

di diL L irdt dt

= = ε −

6. 3 2 3 4(1.2 10 10 10 10 ) (10 10 ) F− −× × × × × × − = Buoyant force = weight ⇒ F = 20 N in down ward direction 7. % error in V → 6% % error in i → 1% maximum percentage error in value of R is 6%. ∴ R = (20 ± 1.2). 8. Kinetic energy is maximum at centre and potential energy is minimum.

10. 32

5 9C F −= .

11. dtdva −=

dtdvkv2 −= , ∫∫ −=

v

v2

t

0 0vdvkdt


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3

v1

v1kt

0

=+

tkv1

vv0

0

+=

12. When an incident ray I is reflected by a mirror whose normal is N, the reflected ray is given by

the following :

( I.N)NR 2 I(N.N)

= − +

Using this expression twice, we get the result,

ˆ ˆ ˆ2 j 2k iR3

+ −=

13. θ =60º Let A = acceleration of (5M) (horizontal rightward) FBD of (5M) along X-axis T 2T

Nsinθ Tcosθ

A T Tcos Nsin 5MA+ θ− θ = N 2Mg cos 2Mg cos 60 Mg= θ = ° =

1 3T T Mg 5MA2 2

⇒ + − =

3T 3Mg 10MA⇒ − = …(i)

y

x(horizontal)

FBD of (M) along X-axis Let a = acceleration of M T + MA = Mg …(ii) FBD of 2M along incline 2 Mg sin θ - 2 MA cos θ - T = 2 Ma

T

a

MA

3 12Mg 2MA T 2Ma

2 2− − =

3Mg MA T 2Ma− − = …(iii) (i), (ii) and (iii) ⇒ A = 0 14.

1 2

B2

Bres

x

y

B1

15. minλ for n = 2 → 1. 17. The relative speed between the two image is

zero

I2

→uu I1←

u



4

18. 21 1 .

2mgxdU AdxLA y

= × ×

2 2

22 2

0

12

lm g AU x dxL A Y

= ∫

⇒ 2 2

6m g lUAY

= .

19. The torque due to the normal reaction (N) and friction (f) about the C.M. topple the body in the

forward direction. 21. A node is a point where there is no vibration. There is vibration at other points where nodes are

not formed. 22. W F.dr= ∫

b dr .d r| d r | R

= ∫

b 2 Rb| dr |R R

π= =∫ = 2 π bJ.

23.

LC

LC

2L 2L

i i

25. This balanced wheat stone bridge

4 82 48 42 4

4

××++ =x

; 8 32

6 4 12.=

× x

⇒ x = 8Ω T

30° α

PG

4Ω

8Ω

x

4Ω

8Ω

2Ω 4Ω

27. Use Kirchoff’s current law. 28. Using A1v1 = A2v2 ……(i) and Bernoulli’s principle we get

2 2

1 1 2 2

1 1P v P v2 2

+ ρ = + ρ …..(ii)

From (i) and (ii)

1 21 2 2 2

1 2

2(P P )v A(A A )

−=

ρ −

Rate of flow = A1v11 2

1 2 2 21 2

2(P P )A A(A A )

−=

ρ −

But P1 – P2 = ∆P



5

∴ rate of flow of glycerine 1 2 2 21 2

2 PA A(A A )

∆=

ρ −

29. Surface integral E d S.→ →

= ∫

Edscos EdS= π = −∫ ∫

02Gm

dS m Gr

0 4= = − π∫

E→

dS→

m0

30. In adiabatic compression, the temperature always increases and since PV = nRT, the quantity PV

also increases.



6

CChheemmiissttrryy PART – II

SECTION – A

1. M C O

+

3. cell cellP

EG nFE ;n 2; and H nF E TT

∂ ∆ = − = ∆ = − + ∂

4. [ ] [ ][ ] [ ][ ][ ] [ ][ ]3

1 3 2 2 3 3

d OK M O K O O M K O O

dt= + −

On applying ssa for intermediate ‘o’ [ ][ ] [ ][ ][ ] [ ][ ]1 3 2 2 3 3K M O K O O M K O O= +

[ ] [ ][ ][ ][ ] [ ]

1 3

2 2 3 3

K M OO

K M O K O=

+

Then we get [ ] [ ][ ]

[ ][ ] [ ]

23 1 3 3

2 2 3 3

d O 2K K M Odt K M O K O

−=

+

And if K3[O3] >> K2[O2][M]

Then [ ] [ ][ ]

[ ] [ ][ ]2

3 1 3 31 3

3 3

d O 2K K M O2K M O

dt K O−

= =

Hence K1 = K (for overall process) So in the given condition.

1a aE E= (for overall process) 5.

S+

O

( )3 2CH CO ONaOAc→

S+

OC

CH3

O C

O

CH3

O

AcO−−→

S+

O

H

COCH3

AcO−AcOH−

S+

O COCH3

S

O COCH3

( )AcO

RDS−−

←

S+

S

AcO−

←

S OAc

6. (A) Bulky groups are stabilized at anti position.

(B) N Hθ60oθ ≈ hence lone pair resides in almost sp2 hybrid orbital, i.e. in an orbital

with ≈ 33% in S-character. While in

N H

lone pair of N-occupy sp3 hybrid orbital so its donation is more easy.

(C) At high pH : COOH COO H− +− − + , so two anionic groups stabilised at anti position.

7. B

H

H

HB

H

HB

H



7

8.

O

O

OH

H

O

O

O 10. More stabilised group or ion is better leaving group. 11.

O

S

S

O

O

S OOO

O

OO

12. Probability density 21s 3

0

1a

ψ =π

(at nucleous r = 0)

13. According to Werner’s theory, only those ions are precipitated which are attached to the metal

atoms with ionic bonds and are present outside the co-ordination sphere. 14. It is cyanide process of extraction of Ag and Au. 15. [Co(en)2Cl2]+ shows geometrical as well as optical isomerism. 16.

NH3Si SiH3

SiH3

Trigonal planar. Lone pair is no longer available to donate to any lewis acid.

N

CH3

CH3 CH3

Pyramidal, l.p. is localized on N.

18. 3A

4b 4 N r3

= × × π put the values of b, NA and π and get r = 0.2944 nm.

20. This is because acidity increases with increase in the size of borane. In larger borane, the charge

formed upon deprotonation can be better delocalized over a large anion with many boron atoms than over a small one.

21. ( )2V V

1

T 3S nC n ;C Ar RT 2

∆ = =



8

22.

Me

OH

Me

H+

→

1,2 methyl shift

( )o2Me

OH

Me

Me

OH

Me

H

( )o3

Me

OH

Me

( )more stable ( )less stable

23. Diels-Alder reaction is given by conjugated (S)-cis and (S)-trans conformations. 24.

H

o490 C→CH3

CH2

25. In (I) lone pair is delocalized to activate the ring while in (III) lone pair is not delocalized. Among I

and II, N is more basic than sulphur because the lone pair of N resides in a orbital of high p-character.

26. For E – Z priority order is:

OH

OH

O5

43

2 1

Priority for R – S configuration – OH > = bond > R-grp. > H

C OH

OH

OH4

3

2

1

R

But least prior group is attached to solid line so result is ‘S’. 28.

H

CH3( )2

2

Hg OAcH O,ether→

Hg(OAc)

CH3

OH

H

4

2

NaBHH O,EtOH→

H

CH3

OH

H1

23

5 6

1 methylcyclohexanol− 29. β-ketoacid decarboxylate at a faster rate. 30. The ‘H’ is abstracted from the axial position due to stereo electronic factors.



9

MMaatthheemmaattiiccss PART – III

SECTION – A 1. R ∩ R′ = (x, y); −3 ≤ x ≤ 3, 0 ≤ y ≤ 5 dom R ∩ R′ = [−3, 3] Range R ∩ R′ = [0, 5] ⊃ [0, 4] ∵ (0, 0) ∈ R ∩ R′ and (0, 5) ∈ R∩R′ ∴ 0 is related to 0 as well as 5. Hence R∩R′ does not define a

function.

(0, 5)

(3, 4)(–3, 4)

2. |z1 + z2| ≤ |z1| + |z2|

3. For real λ f(x) = x2 + 2bx + 2c2

⇒ (x + b)2 + 2c2 – b2 ≥(2c2 – b2) f(x)min = 2c2 – b2 g(x) ⇒ −x2 – 2cx + b2

⇒ −(x + c)2 + c2 + b2 ⇒ (b2 + c2) – (x + c)2 ≤ (b2 + c2)

g(x)max = b2 + c2 By the given condition 2c2 – b2 > b2 + c2 | c | 2| b |>

4. Since cos (n! πx) will be a proper fraction between –1 and +1 (excluding 0 and 1) and (It is) → 0 as m → ∞. 5. x2 – y2 = 8

⇒ dy2x 2y 0dx

− =

⇒ 1 ydy xdx

− −=

At point 5 3,2 2

−

⇒ 11 3 m

dy 5dx

−= =

Also 9x2 + 25y2 = 225

⇒ 18x + 50y dy 0dx

=

⇒ dx 25ydy 9x

− =

At point 5 3,2 2

−

2dx 5 mdy 3

− = − =

1 2 m m 1= −∵

so θ = 90º = 2π



10

6. ⇒

22

2

1x 1 dxx

1 1x x xx x

−

+ + α + + β

∫

Put 1x zx

+ =

∴ 211 dx dzx

− =

⇒ dz(z )(z )+ α + β∫

⇒ 2

dz

z ( )z+ α + β + αβ∫

⇒ 2 2

dz

z2 2

α + β α − β + −

∫

⇒ 2 2( )log z z c

2 2 2α + β α + β α −β + − + − +

⇒ ( )log z (z )(z ) c2

α + β+ − + α + β +

7. Since S. D. ≤ Range = (b – a) ∴ Variance ≤ (b – a)2 ∴ Var.(X) ≤ (b – a)2 or (b – a)2 ≥ Var.(X)

8. ∵ sin–1x + sin–1y + sin–1z = π ∴ sin–1x + sin–1y = π − sin–1z

⇒ 1 2 2 1sin x 1 y y 1 x sin z− − − + − = π −

⇒ 2 2x 1 y y 1 x z− + − =

9. ⇒ |z| = |z – 2| ⇒ |z|2 = |z – 2|2 ⇒ zz (z 2)(z 2)= − − ⇒ z z 2+ = …..(i) also |z| = |z + 2|

⇒ |z|2 = |z + 2|2 ⇒ zz (z 2)(z 2)= + +

⇒ z z 2+ = − ….(ii) By (i) and (ii) | z z | 2+ =

10. The system of linear equation will have a non zero solution if

3 3 3a (a 1) (a 2)a (a 1) (a 2) 01 1 1

+ ++ + =

Now operate c2 – c1 and c3 – c2, then expand. 11. B = CAC–1 B2 = (CAC–1) (CAC–1) = CA2C–1 B3 = B2B = (CA2C–1)(CAC–1) = CA2(C–1C)AC–1 = CA2.AC–1 = CA3C–1



11

12. The number of students answering at least r questions incorrectly = 2n – r ∴ The number of students answering exactly r questions (where 1 ≤ r ≤ (n – 1)) Incorrectly = 2n – r – 2n – (r + 1) The number of students answering all questions wrongly ⇒ 2º = 1 Thus the total number of wrong answer is

⇒ 1.(2n – 1 – 2n – 2) + 2(2n – 2 – 2n – 3) ⇒ 2n – 1 = 4095 ⇒ 2n = 4096 ⇒ n = 12

13. 2b = a + c …..(i)

2prqp r

=+

…..(ii)

and b2q2 = (ap) (cr) …..(iii) put b from (i) and q from (ii) in equation (iii) then on simplification we get required result. 14.

x 1 x 1lim f(x) lim (p[x 1] q[x 1])

− −→ →= + − − = p(1) – q(–1) = p + q

x 1 x 1lim f(x) lim p[x 1] q[x 1]

+ +→ →= + − − = 2p – q(0) = 2p

∵ f(x) is constant at x = 1 ∴ p + q = 2p ⇒ p = q ⇒ p – q = 0

15. Since f′(x) = g(x).(x – a)2 ∴ f′(x) > 0 if g(a) > 0 and < 0 if g(a) < 0 ∴ f is increasing in the nbd of a if g(a) > 0 and f is decreasing in the nbd of a if g(a) < 0 16. f(x) = x3 + bx2 + cx f(1) = 1 + b + c f(2) = 8 + 4b + 2c By Role’s Theorem f(1) = f(2) ⇒ 3b + c + 7 = 0 …..(i) f′(x) = 3x2 + 2bx + c

4f ' 03

=

By Rolle’s theorem ⇒ 8b + 3c + 16 = 0 …..(ii) By (i) and (ii)

b c 15 8 1= =

−

b = −5 c = 8. 17. Put a = 2, b = 3, c = 0

2 20

dx2(2 3)(3 0)(0 2) 60(x 4)(x 9)

∞π π

= =+ + ++ +∫



12

18. ⇒ Area of OABCO

⇒

1 12 2

2 2

0 0

2 c 1 x dx (1 1 x )dx

− − − −

∫ ∫

⇒

1 12 2

2

0 0

2 (c 1) 1 x dx 1dx

+ − −

∫ ∫

O X

ABC

(0, 1)

⇒ 2

12

2

0

12 1 x dx2

− −

∫

⇒ 2 24 2π− …..(i)

∴ Area inside the circle and outside the ellipse

⇒ 2 24 2

ππ − −

⇒ 1(4 2)4 2π

− +

19. Put y zx=

⇒ dy dzx zdx dx

= +

∴ dz 1x z zdx z

+ = + φ

⇒ dz 1xdx z

= φ

⇒ dz dx log | c |1 xz

= + φ

∫ ∫

Since x 1log | cx |y z

= =

⇒ 1 dz1zz

= φ

∫

⇒ 21 1

1zz

−=

φ

⇒ 2

22

1 yzz x

φ = − = −

⇒ 2

2x yy x

φ = −



13

20. In right angled ∆PAC

2 2

1

CA g f ctan2 PA Sθ + −= =

Now 2

2

1 tan2cos

1 tan2

θ−

θ =θ

+

θ/2θ/2

A

B

C (–g, –f)P(x , y )1 1

21. Equation of the tangent at the vertex is x – y + 1 = 0 …..(i) Equation of the axis of the parabola is x + y + k = 0 …..(ii) Since it passes through (0, 0) 0 + 0 + k = 0 ⇒ k = 0 ∴ Equation of axis is x + y = 0 …..(iii)

Z A S

Vertex Focus

Where z is a point on directrix

Solving (i) and (iii) we get 1 1A ,2 2−

∴ z is (−1, 1) Now directrix is x – y + c = 0 …..(iv) But equation (iv) is passes through z –1 – 1 + c = 0 ⇒ c = 2 So directrix is x – y + 2 = 0 …..(v) Using PS = PM PS2 = PM2

22. Equation of a tangent to the ellipse 2 2

2 2x y 1a b

+ = is

xcos y sin 1a b

θ θ+ =

a bA ,0 B 0,cos sin

→ → θ θ

Y

XOA

B

P

Let P is the mid point of AB, P → (h, k)

ah2cos

=θ

…..(i)

bk2sin

=θ

….(ii)

Eliminate θ by (i) and (ii)

23. PQ = 2b tan θ

2 2 2 2OQ OP a sec b tan= = θ + θ since OQ = OP = PQ

221sin

3(e 1)θ =

−

∵sin2θ < 1

21 1

3(e 1)<

−

O (0, 0)

P(asec , btan )θ θ

Q(asec , –btan )θ θ



14

2e3

>

24. Since a is perpendicular to b Let V xa yb z(a b)= + + × ….. (A)

Given a.b 0, V.a 0, V.b 1, [V ab] 1= = = = By equation (i) 2 2V.a x a x a 0 x 0= = = ⇒ = ….. (i)

Again 2 2

21V.b y b 1 yb y

b= ⇒ = ⇒ = ….. (ii)

Again 2V.(a b) z a b× = ×

22

11 z a b z| a b |

= × ⇒ =×

….. (iii)

Put (i)/(ii)/(iii) in equation (A).

25. Probability that problem is not solved by 1st = 1 112 2

− =

Probability that problem is not solved by 2nd = 1 213 3

− =

Probability that problem is not solved by 3rd = 1 314 4

− =

∴ Probability that problem is not solved by any one of the three = 1 2 3 1. .2 3 4 4

=

Hence the required Probability 1 314 4

− =

26. Mean = np = 4 Variance = npq = 3

∴ n = 16, 3 1q , p4 4

= =

So P(X ≥ 1) = 1 – P(X = 0) = 1631

4 −

27. sin (θ + φ) = sin θ cos φ + cos θ.sin φ 12 3 5 413 5 13 5

= − + −

5665

= −

28. cos 2θ = sin α

cos2 cos2π θ = − α

2 2n2π θ = π ± − α

29. h1 = x tan 60º h2 = x tan 30º.

30º60º

h1 h2

A B Cx x

30. Put b = ar, c = ar2, d = ar3, e = ar4, f = ar5.

Concept Recap Test Mains 4 Sol

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