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

AITS-FT-IV-(Paper-1)-PCM(S)-JEE(Advanced)/13

JEE(Advanced)-2013

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ANSWERS, HINTS & SOLUTIONS FULL TEST –IV (Paper-1)

ALL INDIA TEST SERIES

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1

Q. No. 1.

PHYSICS D

CHEMISTRY A

MATHEMATICS 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

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AITS-FT-IV-(Paper-1)-PCM(S)-JEE(Advanced)/13

2

Physics

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}  V  ⇒ WAB + 0 + nRT0 ln  0  + nR (2T0 – T0) = 4nRT0 {∵ DA is isobaric process}  3V0  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 = 2π

g

.

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. V

7.

The work done by cell =

1

∫ V dq = ∫ CV dV = 2 CV . 2

0

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 ⇒ Req. =

(4 + 2) × 6 + 7 = 3 + 7 = 10 Ω 6+6

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3

AITS-FT-IV-(Paper-1)-PCM(S)-JEE(Advanced)/13

10 = 1 A. 10 at t = ∞ 4×6 ⇒ 7 + = 9.4 6+4

⇒ i1 =

⇒ i2 =

10.

10 9.4

so

i1 1 = = 0.94 i2 (10 / 9.4)

Power delivered by mg is converted to the heat dissipated in R1 and R2 ⇒ mgv = PR1 + PR2 solving, we get PR2 = 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 VP = VR

All electrons accumulate at R VP > VR

electrons accumulate at P VP < VR

electrons accumulate at R VP > VR

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 ⇒ T1 > T2 > T3 as λ1 < λ2 < λ3 (Wien's law) T Now as the area under the curve Eλ and λ gives the intensity; so

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AITS-FT-IV-(Paper-1)-PCM(S)-JEE(Advanced)/13

14.

4

σ Ae3T34 > σ Ae2T24 Now as T3 < T2 ⇒ e3 > e2. 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  p2  h ∵ = T = K.E. ⇒ λ=  2mT  2m ⇒

λA = λB

{Areas of the bodies are same, given}

TB TA

1 λB 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

Since λA =

15.

P = V3 , for ideal gas PV = nRT (A) relation between V and T V × (3V3) = nRT  nR  4 ⇒ V4 =  T ⇒ V ∝ T  3  (B) Relation between P and T 1/ 3

P PV = nRT ⇒ P   3 ⇒ P4/3 ∝ T

= nRT

(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.

v=

T µ

T can be calculated by using Hooke’s Law and on stretching µ also changes. 2.

f1

10  340  =  f f = 9  340 − 34 

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5

AITS-FT-IV-(Paper-1)-PCM(S)-JEE(Advanced)/13

20  340  =  f f = 19  340 − 17  10 f1 19 and = 9 = 20 18 f2 19

and f2

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 : (1.5) t λ  Totalnumber   optical path length  ∵  =  wavelength  of waves    For B and C :  1  2t  nB   (1.6)   3 + 3 Total number of waves = λ λ Equating (1) and (2) ⇒ nB = 1.3

Total number of waves =

....(1)

....(2)

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6

AITS-FT-IV-(Paper-1)-PCM(S)-JEE(Advanced)/13

Chemistry

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 H C CH group. 3

OH Br2 +KOH

HNO2 C2H5CONH2  → C2H5NH2  → N2 + CH3 CH2OH (Y) I2 + NaOH CS2

C2H5N 2.

OH

O

PhSO Cl

2   →

C

S HCOONa + CHI3

SO 2Ph

F

NaF / DMSO   →

NO2

NO2 NO2 (A) (B) F– is very reactive unsolvated which can displace PhSO3− which being a good leaving group in presence of strong withdrawing group —NO2. 3.

PCl3

+

11 137.3 = 0.0801

1 O2 → POCl3 2 1.34 32 = 0.0419 0.0801

Limiting reagent

Moles of POCl3 formed = 0.0801 Mass of POCl3 = 0.0801 × 153.3 = 12.3 g 11.2 Percentage yield = × 100 = 91% 12.3 4.

d = 0.714 g/L at STP ⇒ m = 0.714 g T = 273 K P = 1 atm V=1L PV 1× 1 n= = = 4.46 × 10 −2 RT 0.0821× 273 mass of this one litre sample is known from density 0.714 = 16g / mole molar mass of the gas, M = 4.46 × 10−2

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7

5.

6.

AITS-FT-IV-(Paper-1)-PCM(S)-JEE(Advanced)/13

3RT 3 × 8.314 × 300 = = 432.78 ≈ 433 M 0.03995 As molar mass of Ar = 39.95 g/mol = 0.03995 kg/mol 1 1 1 = Z2Rh  2 − 2  λ  n1 n2  1 1 = 62 × 1.097 × 107  2 − 2  1 3  8  = 36 × 1.097 × 107   m−1 9  λ = 2.85 nm Urms =

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.

10.

NCl3 + 4H2O → NH4 OH + 3HOCl Cu+2 + 2e− → Cu,

∆G1 = −2F × 0.34

Cu+

∆G2 = F × 0.16

→ Cu+2 + e−

Cu+ + e− → Cu ∆G3 = ∆G1 + ∆G2 = −0.52 F

⇒ Ecu+ / 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 V ∴ w = –P (VB – VA) + 0 + 2.303 nRT log C VA = – (40 – 20) + 0 + 2.303 × 1 × 0.082 × log

VC VA

= – 6.13 litre-atmosphere = – 620.77 J

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8

AITS-FT-IV-(Paper-1)-PCM(S)-JEE(Advanced)/13

12.

Milli equivalent of Ba(MnO4)2 = Meq of Fe+2 = Meq of FeCrO4 = Meq of K2Cr2O7

13.

120 C ∆ heated 2 [CaSO4 .2H2 O]  → 2CaSO4 .H2 O → 2CaSO 4  → 2CaO + 2SO2 + O2 −3H O −H O strongly 0

2

2

Plaster of Paris

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 CH3

1. O3 C10H20   → reductive

H3C

CH2

CH

CHO (C5H10O)

CH3 ⇒ Structure of A

H3C

CH2

CH

CH3 CH

CH

CH

CH2

CH3

No. of stereoisomers of A = 6 3.

x×1+y×1=4×1 x × 1 + 2y = 5 + 1 From Eq. (1) and Eq. (2) we get y=1 x=3 ⇒x/y=3

4.

t1/2 ∝

5.

Eq. (1) Eq. (2)

1 1−n

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 K 10−14 = 10−8 Kh = w = −6 Kb 10

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9

Mathematics

AITS-FT-IV-(Paper-1)-PCM(S)-JEE(Advanced)/13

PART – III SECTION – A

1.

1 . 2 1  3 ( x + 1)( x − 2 )  x −  2 2 3x + bx + bx + 3 3 3 −3 −27 2  = lim = × × = 1 1 2 2 2 8 2x − 1   x→ 2 x −  2 2  

3x3 + bx2 + bx + 3 = 0 ⇒ 3 (x3 + 1) + bx (x + 1) = 0 has roots −1, 2, e ( ) −1 = lim 2x − 1 x → 1 f x

So, lim x→

1 2

2

1+ x . Also f′′(x) > 0 ⇒ f′(x) is increasing ⇒ f′(x) > f′(0) = 0 ∀ x > 0 1 + f(x)

7.

f′′(x) =

8.

When P < –3, F(x+p) will have 4 positive roots

9.

a b c ⇒ a = 15λ, b = 8λ, c = 17λ (λ being a positive constant) = = 15 8 17 2 2 2 Note that a + b = c Triangle ABC is right angled 1 R = (17λ ) , ∆ = area of ∆ABC = 60λ2 2 s = 20λ, s − a = 5λ, s − b = 12λ, s − c = 3λ ∆ ∆ ∆ ∆ = 12λ + 5λ + 20λ − 3λ = 34λ r1 + r 2 + r 3 − r = + + − s−a s−b s−c s r1 + r 2 + r3 − r 34λ = ×2 = 4 ⇒ R 17λ

We have

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 c2 = a2m2 – b2) ∴ k = ± 2√2 ∴ T1 and T2 are x + y + 2√2 = 0 and x + y − 2√2 = 0 4 2 ∴ Distance between them = =4 2

11.

y=

or

x+3 > 0 ⇒ x < −3 or x > −1 x +1 x → −3 x→∞

y→0

y →1

( 0,1) ∪ (1, ∞ ) 12.

P ( x ) is an even function. ∴

(

P ( x ) = ax 4 + bx 2 + 1 and P′ ( x ) = 4ax 3 + 2bx = 2x 2ax 2 + b

)

It has two minima. Hence, a > and b < 0. −b , P ( x ) has minima. So, at x = ± 2a

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AITS-FT-IV-(Paper-1)-PCM(S)-JEE(Advanced)/13

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

Also,

8 2 =2 15

Now, lim

−b a

∫ (1 − (ax

))

+ bx 2 + 1 dx ⇒ b = −1 a =

0

P ( x ) − ( g ( x ) + g ( −x )) x2

x →0

⇒ C=

4

1 2

is finite.

1 ,B = −1 2

Also, y = 1 is tangent to Ax 2 − x + ⇒ Ax 2 − x + ⇒ A=−

1 = f (x) 2

1 = 1 has equal roots 2

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 ⇒ Foci are S1(1, 1) and S2(4, 5)

5 2 2 +1

S1N PS1 = = S2N PS2

40 5

=2 2

SECTION – C

1.

2.

4.

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] abc abc 2∆ = = = 4cm. = 4R R 8R3 2R2

{

}

 π π 3π  A = A = 1, 2, 3,2 , B =  , ,  4 2 4  divisors of m is 9

(1 + x )20 + (1 − x )20 =

20

c0 9

So,

∑ r =0

2 x 20 +

20

20

20

c 2r

=

20

c0 +

c 2 x18 +

20

20

c2 x2 +

m = 34 −

20

c 2r + 2 = coeff of x

3

20

)

c1 24 − 3 c 2 = 36 = 22 32 hence number of

c4 x4 + − − − − −

c 4 x16 + − − − − − 22

(

20

c 20 x 20

c 20

(1 + x )2n + (1 − x )2n    in  4

⇒ a = 10 ⇒ xy = 40 has total order pair (x, y) solution = 4 × 2 = 8

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