OL-FT-I-JEEA-PAPER-2.pdf

AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

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1

ANSWERS, HINTS & SOLUTIONS

FULL TEST –I (Paper-2)

Q. No. PHYSICS CHEMISTRY MATHEMATICS 1. A B C

2. B B B

3. D B A

4. D B C

5. A, D

A, C A, C

6. A, B, C, D

A, B, C A, C

7. B, C A, B, D A, C

8. A, C A, B, D C, D

9. A, B, C B, C A, C, D

1.

(A) → (p, q, r, s ) (B) → (q, r) (C) → (q, r) (D) → (p, q, r, s)

(A) → (r, s) (B) → (p, r) (C) → (r, t) (D) → (q)

(A) → (s, t) (B) → (p, q) (C) → (r) (D)→ (p, q, r, s, t)

2.

(A) → (r) (B) → (p) (C) → (q) (D) → (s)

(A) → (q, r, t) (B) → (s, t) (C) → (s, t) (D) → (p, r, t)

(A) → (q) (B) → (r) (C) → (q) (D)→ (p, q, r, s, t)

1. 8 5 4

2. 4 2 8

3. 9 2 3

4. 5 5 5

5. 3 8 3

6. 1 8 5

7. 5 6 3 8. 8 7 3

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2

PPhhyyssiiccss PART – I

SECTION – A 1. By conservation of momentum J = mV + 2 mV V= J/3m. 2. 0 + 2 × 1 + 3 × 1 = VB ∴ VB = 5 volt

4. S < 21 gt2

⇒ S < 2g. (due to Lenz’s Law). 6. The molar heat capacity has the general definition

1 QCn T

∆= ⋅

∆

where n = number of moles, ∆Q = heat absorbed by the gas and ∆T = rise in temperature of gas. It is possible to obtain almost any set of values for ∆Q and ∆T by proper selection of a process.

7.

12 102usin 2Tg 10

× ×θ= = = 1 sec.

30º 30º

30º30º

8. M'g T M'a− = …(i) T = Ma …(ii) M'g a(M M')= +

M'ga(M M')

=+

masin mgcosθ = θ a gcot= θ

T

R

Mg

M

M'ggcot(M M')

θ =+

cot M cot M' M'θ + θ =

McotM'(1 cot )

θ=

− θ

T = M a = M. g cot θ

MgTtan

=θ

.

masinθ

ma

mgsinθmgcosθ

θ

+ macosθ



3

9. RF 2 2 iBa= .

2iBa

2iBa

SECTION –C

1. P = β v2

2mdv .v vdt

= β

0

0

2v t

v 0

dv dtv m

β=∫ ∫

ln2 .tmβ

=

mln2t =β

4 2t 0.6930.693×

= ×

t = 8 seconds. 2. After 5 s, speed of detector = 50 – 10 × 5 = 0 and that of source = 0 + 10 × 5 = 50 m/s

and the source has fallen a distance 21 10 (5) 1252

= × × = m

and the detector has rises a distance 2150 5 10(5) 125

2= × − × = m

So, 300 0f ' 130300 50

− = − = 156 Hz.

Ground

O v=0

50 m/sS

3. 2GMgR

=

2G.2mg'4R

=

gg'2

=

F mg'sin= − ∝

mgF .sin22

= − θ

αθmg’

2R

mgsinαθ2R

mgcosα

As θ is much small

mgma .22

= − θ

a g.= − θ

xa g.2R

= −



4

g2R

∴ ω = f

2RT 2g

= π

32 6400 10T 2 3.14

10× ×

= ×

= 2 × 3 × 800 × 1.41 T = 6768 s. 4. AB = v dt sin θ …(i) (along the circular arc)

OA = ANsinθ

ds = r dθ given vdt sin θ = OA dθ

2a 3

dθθO

B

N

A

ANvdt sin .dsin

θ = θθ

avdt sin dsin

θ = θθ

2d v sindt aθ= θ

d 5dtθ= rad/s.

5. We have, 10.2 = W + Kmax, 1 …(i) and 10.2 Z2 = W + Kmax, 2 …(ii)

Also de Broglieh h

2mK−λ = =ρ

1 22 1

2 1

K2.3 K 5.25 K

Kλ

∴ = = ⇒ =λ

…(iii)

Also 10.2 Z2 = energy corresponding to longest wavelength of the Lyman series = 3 × 13.6 ⇒ Z = 2. ∴ From equations (i), (ii) and (iii) W = 3 eV. 6. Velocity of efflux v 2gy=

Range 2hx 2gyg

= ×

The velocity of the block must be dxdt

.

V

A

x

ay

h

bdx 2h 1 dyV 2gdt g dt2 y

∴ = = × ×

bh dyV .

dty= …(i)



5

Using equation of continuity

Ady a 2gydt

= …(ii)

equation (i) and (ii)

bh aV 2gyy A

= ×

baV 2ghA

= ×

12020

= × = 1 ms–1.

7. If there were no hole

0 2GMgR

=

decrease in g due to hole (absent mass)

( )

1 2 2GM / 8 GMg

2RR2

= =

effective g = g0 – g1

2GM2R

=

= 5.5 m/s2.

8. q qdq .Rd .d2 R 2

= θ = θπ π

dq qddiT 2 2

θω= =

π π

2qdi .d4ω

= θπ

θ

dθRdθ

2

03

di(Rsin )dB

2Rµ θ

=

2

02

0

sin qdB d2R 4

π µ θ ω = θ π ∫ ∫

0qB

16 Rµ ω

=π

φ = Bπa2

2 0qa .

16 Rµ ω

φ = ππ

2

0q a16R

µ ωφ =

d| |dtφ

ε =

2

0qa| |

16Rµ

ε = α .

= 8 volt

8i1

= = 8 A.



6

CChheemmiissttrryy PART – II

SECTION – A

1. Wsolute = 3 gms, Wsolvent = 50 gm = 75 – 25

OH

3 xW40 56− = + .

( )+= = ×

3 x40 56m 5 1000

47

X = 9 2. Major product is formed via E2. H anti to halogen will be eliminated. 3. nHx = nNaOH = 0.25 × VNaOH, pH = pKa = 5 (acid buffer)

a1 1pH 7 pK logC 92 2

= + + = 1logC 1 C 10−⇒ = − ⇒ =

NaOHNaOH

0.25VC V 40 ml60 V

= ⇒ =+

nHx = 10 mm = 0.82 gm

0.82% 100 80%

1.025= × =

4. Time required for equifractional change in a first order reaction is constant.

SECTION – C 1.

C

H

N H CH

O

P Q

2.

Co

Cl

NH3H N3

Cl–

(Fac) (Mer)

+

Cl

NH3

H N3

Co

Cl

Cl

NH3

H N3

NH3 Cl

Co

Cl

Cl

H N3

Cl

NH3H N3

3. ( )

( )

2 4 7 2 3 34

5 3 4

Na B O H O 2Na B OH 2H BO

1 mol 2 mol

NaOH H BO NaB OH

2 mol 2 mol

+ → +

+ →



7

4. A = 2y1/3 nm. M = 6.023 × y × 4 amu

Density = m/a3 = 27

27

6.023 4 1.66 10 kg / m³8y 10

−

−

× × ××

40 5 kg / m³8

= =

5. Two double bonds capable exhibiting geometrical and one chiral carbon molecule asymmetric. 23 = 8

6. Pvmw 58.33 gmRT

= =

58.33% 100 77%

75= × =

7.

B – O – BO

O

O

O



8

MMaatthheemmaattiiccss PART – III

SECTION – A

1. PQ is a focal chord 21 a 2aQ , t tt

− = −

Normal drawn at P(t) meets the curve at Q 2t, t− −

∴ Normal drawn at 1Qt

−

meets the parabola again at R 12tt

+

21 1R a 2t , 2a 2t

t t

+ +

Equation of tangent at R is 21 1y 2t x a 2t

t t + = + +

….. (1)

Equation of focal chord PQ is ( ) ( )2t 1 y 2t x a− = − ….. (2) T is the point of intersection of line (1) and (2)

⇒ 22

a 1 2a 1T 4t , 4t3 3 tt − +

∴ Area of ∆ TQR = 328a 1t

3 t +

2. Let a point (3λ + 1, λ + 2, 2λ + 3) of the first line also lies on the second line. Then

3 1 3 2 1 2 3 21 2 3

λ + − λ + − λ + −= = ⇒ λ = 1

Hence the point of intersection P of the two lines is (4, 3, 5). Equation of plane perpendicular to OP where OP is (0, 0, 0) and passing through P is 4x + 3y + 5z = 50

3. For non triangle solution a b cb c a 0c a b

∆ = =

⇒ a + b + c = 0 ⇒ ‘1’ is a root of ax2 + bx + c = 0 4. Clearly we have

xy 2xdy ydx xdy ydx

e x+ −

=

⇒ xy yd(xy)e dx

− = ∫ ∫ ⇒ xyy e c

x−+ =

5. Let the roots of az3 + bz2 + cz + d = 0, z1 = x1, z2, z3 = x2 ± iy2

⇒ z1 + z2 + z3 = ba

−

⇒ ab > 0

Also z1 z2 z3 = ( )2 21 2 2

dx x ya

+ = −

⇒ ad > 0



9

6. On solving we have 12

α = and 2 34

β = ⇒ 32

β = ±

∴ New roots are ω and ω2 which are roots of x3 – 1 = 0 or x2 + x + 1 = 0

7. Redefining the function, we get

2x 4x 8 ; x 15 5 5f(x)

2 x ; x [1, 2)x 2 ; x 2

− + <= − ∈

− ≥

We have R.H.L at x = 1 is = h 0lim 2 (1 h) 1→

− + =

L.H.L at x = 1 is = ( ) ( )2

h 0

1 h 4 1 h 8lim 15 5 5→

− −− + =

∴ f(x) is continuous at x = 1

Now

2x 4 ; x 15 5

f '(x) 1 ; 1 x 21 ; x 2

− <= − ≤ < ≥

R.H.D at x = 1 ⇒ –1

L.H.D at x = 1 ⇒ 25

−

∴ Not differentiable at x = 1 ⇒ Not differentiable at x = 2 8. Let a , b , c lie in the xy plane

Let ˆa i= , 1 3ˆ ˆb i j2 2

= − + and 1 3ˆ ˆc i j2 2

= − −

∴ 1 3 1 3ˆ ˆ ˆ ˆ ˆp q r a b c i i j i j2 2 2 2

+ + = λ + µ + ν = λ + µ − + + ν − −

= ( )2

232 2 4µ ν λ − − + µ − ν

= 2 2 2λ + µ + ν − λµ − λν − µν

= ( ) ( ) ( )2 2 21 1 1 1 4 32 2

λ − µ + µ − ν + ν − λ ≥ + + =

∴ p q r+ + can take a value equal to 3 and 2. 9. z2 + az + a2 = 0 ⇒ z = aω, aω2 (where ‘ω’ is non real root of unity) ⇒ locus of z is a pair of straight lines and arg (z) = arg(a) + arg(ω) or arg(a) + arg(ω2)

⇒ arg(z) = ± 32π also, |z| = |a||ω| or |a| |ω2| ⇒ |z| = |a|.



10

SECTION – B

1. Clearly lines are 2x + y + 2 = 0 and x + 2y + 1 = 0 and third line is bx + y + 5 = 0 For four circles three lines must not be concurrent ⇒ b ≠ 5

For two circles exactly two lines should be parallel ⇒ 1b2

= and 2

For no circle three lines must be concurrent ⇒ b = 5 For infinite circles two lines must be identical ⇒ b ∈ φ

2. (A) Solving 22K Ky 03−

= > and 24K 2Kx 0

3−

= > ⇒ 1K , 22

∈

(B) Given ( )1g(x) ln cos x−= ; As 10 cos x x [ 1, 1]−≤ ≤ π ∀ − So domain of g(x) = [–1, 1]. Hence number of integers are two (i.e. –1 and 0) (C) Clearly, domain of expression = {–1, 1}. As x > 0 given So x = 1

Hence value of expression ( )201011 sin x 1−+ =

(D) Here x af(x)x b+

=+

….. (1)

and 1 a bxf (x)x 1

− −=

− ….. (2)

Given 1f(x) f (x)−= ⇒ x a a bxx b x 1+ −

=+ −

⇒ ( ) ( ) ( )2 21 b x b 1 x x 1 b 0+ + − − + = ∀ x ∈ Df Hence b = –1 and a ∈ R

SECTION – C

1. Using AM – GM inequality

(a + b)2 + (a + b + 4c)2 = (a + b)2 + (a + 2c + b + 2c)2 ≥ ( ) ( )2 22 ab 2 2ac 2 2bc+ +

= 4ab 8ac 8bc 16c ab+ + +

∴ ( ) ( ) ( ) ( )

2 2a b a b 4c 4ab 8ac 8bc 16c aba b c a b cabc abc

+ + + + + + ++ + ≥ ⋅ + +

= ( )4 8 8 16 a a b ba b c cc b a 2 2 2 2ab + + + + + + + + +

≥ 2 2

552 2 41 a b c8 5 5 100

2a b c 2

× =

2. Putting X = x + 1 and Z = z + 1 ∴ f(X) + f(Z) = f(X + Z)

∴ ( )h 0 h h 0

f(X h) f(X) f(X) f(h) f(X) f(h) f(0)f ' X lim lim limh h h→ →∞ →

+ − + − −= = =

= f′(0) = 4 ⇒ f′(X) = 4 ⇒ f(X) = 4X + c ∴ f(x) = 4x ⇒ f(2) = 8 3. Let f(θ) = ∏(tan θ – tan αi) – ( )( )itan tanθ − α∑

We have ( ) / 2f θ→−πθ → −∞ Again ( )1f 0α > , ( )3f 0α < and ( ) / 2f θ→πθ → ∞ ⇒ There will be 3 such values of ‘θ’



11

4. Let the equation be K(x – α)(x – β) = 0 ∴ Sum of coefficient = K(α – 1)(β – 1) is prime ⇒ K is 1 and α – 1 is 1 ⇒ One root is 2 and β – 1 is 2 Hence roots are 2 and 3. 5. Let 3t x= in I2

We have 1 1

2 t 1 t0 0

1 1 1 1I dt dt3 3e (2 t) e (1 t)−

= =− +∫ ∫

⇒ 1 t

12 1

20

I1 e 1I dt I 33e (1 t) 3e eI

= = ⇒ =+∫

6. Let 3, 4 and r be radii of the circles inscribed into the ∆’s ACD, BCD and ∆ABC respectively, we

get r AB c3 AC b= = ⇒ 3cb

r=

Similarly ∆ABC and ∆BCD are similar, we get r c4 a= ⇒ 4ca

r=

Now as 2 2 2c a b= +

⇒ 2 2

22 2

9c 16ccr r

= + ⇒ r = 5

7. Putting x = 2, 12

and – 1 successively

1f(2) f 32

+ =

….. (1)

( )1 3f f 12 2

+ − =

….. (2)

and f(–1) + f(2) = 0 ..... (3)

Solving, we get 3f(2)4

=

8. On solving we have 4 2A cos 1 cos 1= θ + − θ + = 2

2 1 7cos2 4

θ − +

⇒ 7A , 24 ∈

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