Fulltest IV Advanced Paper 1 Answer Sol Aits 2013 Jeea Ft IV Paper 1

ANSWERS, HINTS & SOLUTIONS

FULL TEST –IV

(Paper-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. D A C 2. B B A 3. C D A 4. D B D 5. C C B 6. B A C 7. A A A 8. A C C 9. C C D 10. D D B 11. B, D D B, C, D 12. A, B A, B, C A, B, D 13. A, C A, C, D A, C, D 14. A, B, C B, C, D A, B, C, D 15. A, B, C, D B, C, D A, B, C 1. 2 6 4 2. 9 2 9 3. 6 3 3 4. 2 3 8 5. 3 8 9

ALL INDIA TEST SERIES

FIITJEE

JEE(Advanced)-2013

From Long Term Classroom Programs and Medium / Short Classroom

Program 4 in Top 10, 10 in Top 20, 43 in Top

100, 75 in Top 200, 159 in Top 500 Ranks & 3542 t

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PART – I

SECTION – A

1. On heating a metal sheet, distance between any two points increases. 2. Heat taken by ice to convert to water at 100°C fully :

5 ×(40) × (0.5) + 5 × 80 + 5 × (100) × (1) = 1000 cal. Heat given by steam to condense fully :

2 × 500 = 1000 cal. Hence everything will be water at 100°C. 3. WAB + WBC + WCD + WDA = ∆Q {∵ ∆U = 0 in cyclic process}

⇒ WAB + 0 + nRT0 ln 0 0 V 3V      + nR (2T0 – T0) = 4nRT0 {∵ DA is isobaric process} get WAB = nRT0 ln3 + 3nRT0

4. As α > β , no collision will occur with the wall and the ball is doing SHM with amplitude β only.

T = 2

g

π .

5. As the fluid is at rest; pressure at the same horizontal level in a connected fluid is same if it is at equilibrium.

6. The magnetic field will make electrons revolve around the direction of B. It may cause some electrons not to reach the collector plate. If it is very strong; it will not let electrons reach the collector. As the magnitude is not given, hence photo current may decrease.

7. The work done by cell =

∫

=

∫

V

0

V dq CV dV = 1

2CV

2 _.

8. As the magnet-1 falls into solenoid A, the magnetic flux associated with solenoid A increases. From Lenz's law, induced current in solenoid A will oppose this increase in magnetic flux. Hence direction of induced current in solenoids is as shown.

The nature of magnetic field produced by solenoid B is as shown. Therefore magnet 2 will be attracted by magnetic field due to solenoid B.

9. at t = 0

⇒ Re q . = (4 2) 6

6 6

+ ×

⇒ i₁ =10 10 = 1 A. at t = ∞ ⇒ 7 + 4 6 6 4 × + = 9.4 ⇒ i₂ = 10 9.4 so 1 2 i 1 i =(10 / 9.4) = 0.94

10. Power delivered by mg is converted to the heat dissipated in R₁ and R₂ ⇒ mgv = 1 2 R R P +P solving, we get 2 R P = 6W

11. The changing magnetic field inside the plane produces electric lines of forces in anticlockwise direction.

There is no direct connection in the shown conductors, so electrons, experiencing electric force, try to accumulate as shown.

All electrons accumulate at Q symmetrically V_P = V_R

All electrons accumulate at R V_P > V_R

electrons accumulate

at P V_P < V_R

electrons accumulate at R V_P > V_R

12. Fission of a nucleus is feasible only if the binding energy of daughter nuclei is more than the parent nucleus.

A = 55 will have more BE than 110.

A = 70 will have same BE as 110 but A = 40 will have more B.E. A = 100 will have same BE as 110 but A = 10 will have lesser B.E. A = 90 will have same BE as 110 but A = 20 will have lesser B.E.

13. λ ∝ 1

T ⇒ T1 > T2 > T3 as λ1 < λ2 < λ3 (Wien's law)

σAe₃T₃ > σAe₂T₂ {Areas of the bodies are same, given} Now as T₃ < T₂ ⇒ e₃ > e₂.

14. hv = K.E. (T) + work function (W) ⇒ hv = T + W ⇒ 4.25 eV = TA + WA (for Metal A) ⇒ 4.70 eV = TB + WB (for Metal B) Since TB = (TA – 1.5) eV Also λ = h/p ⇒ λ =  = =  ∵  2 h p _{T K.E.} 2m 2mT ⇒ λ λA_B = B_A T T Since λA = 1λ_B 2 ⇒ TA = 4TB ⇒ TB = TA – 1.50 gives TB = 4TB – 1.5 ⇒ TB = 0.5 eV ⇒ TA = 2 eV ⇒ WA = 2.25 eV ⇒ WB = 4.20 Ev

15. P = V3 _{, for ideal gas}

PV = nRT

(A) relation between V and T V × (3V3_{) = nRT} ⇒ V4 ₌ nR _T 3       ⇒ V 4 _{∝ T} (B) Relation between P and T

PV = nRT ⇒ P 1/ 3 P 3       = nRT ⇒ P4/3 ∝ T (C) For expansion V → increases work-done : positive internal energy : increases

Hence, heat will have to supplied to the gas. (D) As T ∝ V4

with increase in temperature, volume increases Hence work done is positive.

SECTION –C

1. =

µ T v

T can be calculated by using Hooke’s Law and on stretching µ also changes.

2. f1 =   = −   340 10 f f 340 34 9

and f2 =   = −   340 20 f f 340 17 19 and 1 2 f f = = 10 19 9 20 18 19

3. For 1st reading of oscillator fA = (514 ± 2)Hz

⇒ fA = 516 Hz or 512 Hz

For 2nd _{reading of oscillator}

fA = (510 ± 6) Hz

⇒ fA = 516 Hz or 504 Hz

⇒ A has a frequency of 516 Hz

4. Velocity of approach of man towards the bicycle = (u – v) Hence velocity of approach of image towards bicycle is 2(u – v). 5. For A :

Total number of waves =

λ

(1.5) t _....(1)

∵ _  _{ }= _

   

Totalnumber optical path length of waves wavelength

For B and C :

Total number of waves =

         ₊   λ λ B 1 2t n (1.6) 3 _{3 ....(2)} Equating (1) and (2) ⇒ nB = 1.3

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PART – II

SECTION – A

1. Reaction of X with Br2 + KOH suggests that X is an amide. Evolution of N2 and formation of

alcohol suggests that Y is a 1° aliphatic amine. Iodoform test of Z suggests that it is an alcohol containing _{H3C CH}

OH

group.

C2H5CONH2 Br KOH2+ → C₂H₅NH₂ HNOHNO22→

2 3 2 N CH CH OH → + C₂H₅N C S (Y) CS₂ I2+NaOH 3 HCOONa CHI+ 2. _OH NO2 2 PhSO Cl → O NO2 SO₂Ph (A) NaF / DMSO → F NO2 (B)

F– is very reactive unsolvated which can displace PhSO which being a good leaving group in −₃ presence of strong withdrawing group —NO2.

3. 3 2 3 Limiting reagent 1 PCl O POCl 2 11 1.34 137.3 32 0.0801 0.0419 0.0801 + → = =

Moles of POCl3 formed = 0.0801

Mass of POCl3 = 0.0801 × 153.3 = 12.3 g Percentage yield = 11.2 100 91% 12.3× = 4. d = 0.714 g/L at STP ⇒ m = 0.714 g T = 273 K P = 1 atm V = 1 L 2 PV 1 1 n 4.46 10 RT 0.0821 273 − × = = = × ×

mass of this one litre sample is known from density molar mass of the gas, M = 0.714 ₂ 16g / mole

5. U_rms 3RT 3 8.314 300 432.78 433

M 0.03995

× ×

= = = ≈

As molar mass of Ar = 39.95 g/mol = 0.03995 kg/mol 6. 2 _{2 2} 1 2 1 1 1 Z Rh n n   =  −  λ _{ } 2 7 2 2 1 1 6 1.097 10 1 3   = × × _ − _   7 8 1 36 1.097 10 m 9 −   = × × _{ }   λ = 2.85 nm 7. _N Cl O

Central N-atom is bonded to two other atoms and has one l.p., the electron pair arrangement is trigonal planar. The C − −N O bond angle is about 120° (we expect it to be slightly less than 120 because of greater lp-bp-repulsions) and the molecule is V-shaped.

8. NCl₃+4H O₂ →NH OH 3HOCl₄ + 10. 2 1 2 2 Cu 2e Cu, G 2F 0.34 Cu Cu e G F 0.16 Cu e Cu + − + + − + − + → ∆ = − × → + ∆ = × + → 3 1 2 G G G 0.52 F ∆ = ∆ + ∆ = − ⇒ E_{cu / Cu}+ =0.52 V 11. A B C

Thus it is a cyclic process. Hence, ∆E = 0, ∆H = 0, ∆S = 0

and ∆E = q + w (1st _law)

∴ 0 = q + w or q = – w

Total work done = WA →B + WB → C + WC → A

∴ w = –P (VB – VA) + 0 + 2.303 nRT log C A V V = – (40 – 20) + 0 + 2.303 × 1 × 0.082 × log C A V V = – 6.13 litre-atmosphere = – 620.77 J

12. Milli equivalent of Ba(MnO4)2 = Meq of Fe = Meq of FeCrO4 = Meq of K2Cr2O7 13.

[

]

0 2 2 120 C heated 4 2 3H O 4 2 H O 4 strongly 2 2 Plaster of Paris

2 CaSO .2H O ₋ →2CaSO .H O→₋∆ 2CaSO →2CaO 2SO+ +O

14. In case of B, C and D, the salt are of weak base and strong acid which undergo hydrolysis to give acidic solutions CH3COONa however, on hydrolysis gives basic solution.

15. On increasing temperature, the Maxwell curve of distribution of molecular velocity is flattened and maximum is shifted to higher velocity.

T1 T2 T2 > T1 dN N ↑ Fraction V → Velocity SECTION – C 1. C₁₀H₂₀ O3 reductive → H₃C CH₂ CH CH₃ CHO (C₅H₁₀O) Structure of A ⇒ H₃C CH₂ CH CH₃ CH CH CH CH₃ CH₂ CH₃ No. of stereoisomers of A = 6 3. x × 1 + y × 1 = 4 × 1 Eq. (1) x × 1 + 2y = 5 + 1 Eq. (2)

From Eq. (1) and Eq. (2) we get y = 1 x = 3 ⇒ x / y = 3 4. t1/2 ∝ _{1 n}1 a− ⇒ t1/2 = K a1–n K–1 rate constant

Log t1/2 = log K + (1 – n) log a

Y = C + mx Slope = (1– n) = –2 ⇒ n = 3 5. Kh = 14 8 w 6 b K 10 10 K 10 − − − = =

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PART – III

SECTION – A 1. 3x3 _{+ bx}2 _{+ bx + 3 = 0 ⇒ 3 (x}3 _{+ 1) + bx (x + 1) = 0 has roots −1, 2,} 1 2. So, ( ) f x 2 2 1 1 x x 2 2 e 1 3x bx bx 3 lim lim 2x 1 2x 1 → → − + + + = − −

(

)(

)

1 x 2 1 3 x 1 x 2 x 2 lim 1 2 x 2 →   + − _ − _   =  ₋      = 3 3 3 27 2 2 2 8 − − × × = 7. f′′(x) = 1 x 1 f(x) + + . Also f′′(x) > 0 ⇒ f′(x) is increasing ⇒ f′(x) > f′(0) = 0 ∀ x > 0 8. When P < –3, F(x+p) will have 4 positive roots

9. We have a b c

15 =8 =17 ⇒ a = 15λ, b = 8λ, c = 17λ (λ being a positive constant)

Note that a2 _{+ b}2 _{= c}2

Triangle ABC is right angled

(

)

1 R 17 , 2 = λ ∆= area of ∆ABC = 60λ2 s = 20λ, s − a = 5λ, s − b = 12λ, s − c = 3λ r1 + r2 + r3 − r s a s b s c s ∆ ∆ ∆ ∆ = + + − − − − = 12λ + 5λ + 20λ − 3λ = 34λ ⇒ r1 r2 r3 r 34 _{2 4} R 17 + + − _λ = × = λ

10. Any line perpendicular to x – y +10 = 0 is of the form x + y = k

If this line is a tangent to the hyperbola x2 – 2y2 = 16, k2 = 16 (−1)2 _{− 8 = 8 (using c}2 _{= a}2_m2 _{– b}2₎

∴ k = ± 2√2

∴ T1 and T2 are x + y + 2√2 = 0 and x + y − 2√2 = 0

∴ Distance between them = 4 2 4 2 = 11. y x 3 0 x 1 + = > + ⇒ < −x 3 or x> −1 or x 3 x y 0 y 1 → − → ∞ → →

( ) (

0,1 ∪ 1,∞

)

12. P x is an even function.

( )

∴ _{P x}

( )

_=ax4_+bx2_{+1and P x′}

_{( )}

_=4ax3_{+2bx=2x 2ax}

(

2_+b

)

It has two minima. Hence, a > and b < 0.

So, at x b,P x

( )

2a −

b 2 2 b 2a 2a − = ⇒ = − ∴ Maximum at (0, 1). Also,

(

(

)

)

b a 4 2 0 8 2 2 1 ax bx 1 dx 15 − =

∫

− + + b 1 a 1 2 ⇒ = − = Now,

( )

(

( ) ( )

₂

)

x 0 P x g x g x lim x → − + − is finite. ⇒ C 1,B 1 2 = = − Also, y = 1 is tangent to Ax2 x 1 f x

( )

2 − + = ⇒ _Ax2 _x 1 ₁ 2

− + = has equal roots

⇒ A 1

2 = −

13. z1 and z2 are end points of diameter.

14. S′ is radical circle of S1, S2 and S. S′′ is circle of centre = radical centre and radius = 8 and r1 = 4,

r = 8

15. 2ae = 5, 2a = (2√2 + 1) √5, e ⇒ 5

2 2 1+

Foci are S1(1, 1) and S2(4, 5) 1 1

2 2

S N PS 40 _{2 2}

S N=PS = 5 =

SECTION – C 1. Perimeter of ∆ DEF = a cos A + b cos B + c cos C

= R [sin 2A + sin 2B + sin 2C] = R [4 sin A + sin B + sin C]

= 4R abc₃ abc₂ 2 4cm. R 8R 2R ∆ = = = 2. A = A

{

1, 2, 3,2 , B

}

, ,3 m 34

(

3c 2₁ 4 3c₂

)

36 2 32 2 4 2 4 π π π   = =_{ } = − − = =   hence number of divisors of m is 9 4.

(

)

(

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20 20 20 20 2 20 4 20 20 0 2 4 20 1 x 1 x c c x c x c x 2 + + − = + + + − − − − − 20 20 20 18 20 16 20 0 2 4 20 c x c x c x c = + + + − − − − − So, 9 20 20 2r 2r 2 r 0 c c ₊ = =

∑

coeff of x22 in

(

)

(

)

2n 2n 1 x 1 x 4  _{+ + −}     