FIITJEE JEE(Advanced)-2013 - FIITJEE LTD.-An … Key REE...

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

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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

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2

PPhhyyssiiccss 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

V3V

+ 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 = 2g

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 = 12

CV2 .

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) 66 6+ ×

+ + 7 = 3 + 7 = 10 Ω



3

⇒ i1 = 1010

= 1 A.

at t = ∞

⇒ 7 + 4 66 4

×+

= 9.4

⇒ i2 = 109.4

so 1

2

i 1i (10 / 9.4)

= = 0.94

10. Power delivered by mg is converted to the heat dissipated in R1 and R2

⇒ mgv =

1 2R RP P+

solving, we get 2RP = 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 ∝ 1T

⇒ T1 > T2 > T3 as λ1 < λ2 < λ3 (Wien's law)

Now as the area under the curve Eλ and λ gives the intensity; so



4

σ Ae3T34 > σ Ae2T2

4 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 = = = ∵

2h p T K.E.2m2mT

⇒ λλ

A

B = B

A

TT

Since λA = λB12

⇒ 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 T3

⇒ V4 ∝ T

(B) Relation between P and T

PV = nRT ⇒ P 1/ 3P

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. =µTv

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

2. f1 = = − 340 10f f

340 34 9



5

and f2 = = − 340 20f f

340 17 19

and 1

2

ff

= =

10199

20 1819

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 lengthof waves wavelength

For B and C :

Total number of waves =

+λ λ

B1 2tn (1.6)3 3 ....(2)

Equating (1) and (2) ⇒ nB = 1.3



6

CChheemmiissttrryy 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 CH3 CH

OH

group.

C2H5CONH2 2Br KOH+→ C2H5NH2 2HNO→2HNO2 3 2N CH CH OH→ +

C2H5N C S

(Y)CS2 2I NaOH+

3HCOONa CHI+ 2. OH

NO2

2PhSO Cl→

O

NO2

SO2Ph

(A)

NaF / DMSO→

F

NO2(B)

F– is very reactive unsolvated which can displace −3PhSO which being a good leaving group in

presence of strong withdrawing group —NO2. 3.

3 2 3

Limiting reagent

1PCl O POCl2

11 1.34137.3 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

2PV 1 1n 4.46 10RT 0.0821 273

−×= = = ×

×

mass of this one litre sample is known from density

molar mass of the gas, M = 20.714 16g/mole

4.46 10−=

×



7

5. rms3RT 3 8.314 300U 432.78 433M 0.03995

× ×= = = ≈

As molar mass of Ar = 39.95 g/mol = 0.03995 kg/mol

6. 22 21 2

1 1 1Z Rhn n

= −

λ

2 72 21 16 1.097 101 3

= × × −

7 1836 1.097 10 m9

− = × ×

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. 3 2 4NCl 4H O NH OH 3HOCl+ → +

10.

21

22

Cu 2e Cu, G 2F 0.34

Cu Cu e G F 0.16

Cu e Cu

+ −

+ + −

+ −

+ → ∆ = − ×

→ + ∆ = ×

+ →

3 1 2G G G 0.52 F∆ = ∆ + ∆ = − ⇒ cu / CuE 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

VV

= – (40 – 20) + 0 + 2.303 × 1 × 0.082 × log C

A

VV

= – 6.13 litre-atmosphere = – 620.77 J



8

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

0

2 2

120 C heated4 2 4 2 4 2 23H O H O strongly

Plaster of Paris2 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

dNN

↑ Fraction

V → Velocity

SECTION – C

1.

C10H203O

reductive→ CH3 CH2 CH

CH3

CHO (C5H10O)

Structure of A⇒ CH3 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 n1

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

8w6

b

K 10 10K 10

−−

−= =



9

MMaatthheemmaattiiccss PART – III

SECTION – A

1. 3x3 + bx2 + bx + 3 = 0 ⇒ 3 (x3 + 1) + bx (x + 1) = 0 has roots −1, 2, 12

.

So, ( )f x 2 2

1 1x x2 2

e 1 3x bx bx 3lim lim2x 1 2x 1→ →

− + + +=

− −

( )( )1x2

13 x 1 x 2 x2lim

12 x2

→

+ − − =

−

= 3 3 3 272 2 2 8

− −× × =

7. f′′(x) = 1 x1 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 c15 8 17

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

Note that a2 + b2 = c2 Triangle ABC is right angled

( )1R 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

⇒ 1 2 3r r r r 34 2 4R 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 c2 = a2m2 – b2) ∴ k = ± 2√2 ∴ T1 and T2 are x + y + 2√2 = 0 and x + y − 2√2 = 0

∴ Distance between them = 4 2 42

=

11. x 3y 0x 1

+= >

+ x 3 or x 1⇒ < − > −

or x 3 xy 0 y 1

→ − → ∞→ →

( ) ( )0,1 1,∪ ∞ 12. ( )P x is an even function.

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

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

So, at ( )bx ,P x2a−

= ± has minima.



10

b2 2 b 2a2a−

= ⇒ = −

∴ Maximum at (0, 1).

Also, ( )( )b

a4 2

0

8 2 2 1 ax bx 1 dx15

−

= − + +∫ 1b 1 a2

⇒ = − =

Now, ( ) ( ) ( )( )

2x 0

P x g x g xlim

x→

− + −is finite.

⇒ 1C ,B 12

= = −

Also, y = 1 is tangent to ( )2 1Ax x f x2

− + =

⇒ 2 1Ax x 12

− + = has equal roots

⇒ 1A2

= −

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 ⇒ 52 2 1+

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

2 2

S N PS 40 2 2S 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 3 2abc abc 2 4cm.

R8R 2R∆

= = =

2. A = ( )4 3 4 3 2 21 2

3A 1, 2, 3,2 , B , , m 3 c 2 c 36 2 34 2 4π π π = = = − − = =

hence number of

divisors of m is 9

4. ( ) ( )20 2020 20 2 20 4 20 20

0 2 4 201 x 1 x

c c x c x c x2

+ + −= + + + − − − − −

20 20 20 18 20 16 200 2 4 20c x c x c x c= + + + − − − − −

So, 9

20 202r 2r 2

r 0

c c +=

=∑ coeff of x22 in ( ) ( )2n 2n1 x 1 x

4

+ + −

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

FIITJEE JEE(Advanced)-2013 - FIITJEE LTD.-An … Key REE...

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