FIITJEE Ltd., FIITJ EE House, 29-A, Kalu Sarai, S arvapriya Vihar, New Delhi -110016, Ph 461 06000, 26569493, Fax 265139 42 web site: www.fi itjee.co mANSWE RS, HINTS & SOLUTIONSFULL TEST – II (Main) Q. No. PHYSICS CHEMISTRY MATHEMATICS 1. A A B2. B C A3. C B A4. B D A5. D B B6. C C B7. A C B8. C C C9. A. B B10. C B B11. D D D12. C B B13. C C A 14. A B A15. B C B16. D B A17. A C A18. C B A19. A C A20. C B C21. A B D22. D B C23. A D D24. C B B25. D D C26. B B C27. B D C28. D D A29. C C A30. D D AA L L I N D I A T E S T S E R I E S FIITJEE JEE (Main)-2014F r o m C l a s s r o o m / I n t e g r a t e d S c h o o l P r o g r a m s 7 i n T o p 2 0 , 2 3 i n T o p 1 0 0 , 5 4 i n T o p 3 0 0 , 1 0 6 i n T o p 5 0 0 A l l I n d i a R a n k s & 2 3 1 4 S t u d e n t s f r o m C l a s s r o o m / I n t e g r a t e d S c h o o l P r o g r a m s & 3 7 2 3 S t u d e n t s f r o m A l l P r o g r a m s h a v e b e e n A w a r d e d a R a n k i n J E E ( A d v a n c e d ) , 2 0 1 3
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ANSWERS, HINTS & SOLUTIONS
FULL TEST – II (Main)
Q. No. PHYSICS CHEMISTRY MATHEMATICS
1. A A B
2. B C A
3. C B A
4. B D A
5. D B B 6. C C B
7. A C B
8. C C C
9. A. B B
10. C B B
11. D D D
12. C B B
13. C C A
14. A B A
15. B C B 16. D B A
17. A C A
18. C B A
19. A C A
20. C B C
21. A B D
22. D B C
23. A D D
24.C B B
25. D D C
26. B B C
27. B D C
28. D D A
29. C C A
30. D D A
A
L L
I N D I A
T E S T
S E
R I E S
FIITJEE JEE (Main)-2014
F r o m C
l a s s r o o m / I n t e g r a t e d S c h o o l P r o g r a m s 7 i n T o p
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P P h h y y s s i i c c s s PART – I
1. Avg. speed2 2
r mr (v t) (v t)3
t
2
r v 5 9 r v 2m/ s
2. avg
1 t 1v v 3v2 2 2vt 4
3.m m
g T a2 2
…(i)
T cos 60 =ma
cos 60
…(ii)
Solving (i) and (ii) acceleration of ring =2g9
4. Work done by all the forces on the block equal to change in kinetic energy.
6. No effect of ‘a’ and ‘g’ on time period of spring pendulum.
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11. Potential across capacitor is zero, hence energy stored is zero.
12. vA – vB = E. dx = 10 (1)= 10 volt.
13. Z = 2 2R X = 29 X
but cos =R 3Z 5
X = 4 .
14. =0
vA – vB =0 0
(b) (a)
W = Q(vA – vB) =
0
Q(b a)
15. = 0 + 1 10 = 10 rad/sec2 v = r = 1 10 = 10 m/s
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0 0
1 2Q 1 Q4 PA 4 PB
= 0
2 1PA PB
4PB2 = PA2
(x 5a)2 + y2 = (4a)2
19. i = 5A2 2
2mq q 1Li
2C 2C 2 qmax = 6 C
21. f = ma2
=2f
I mR
a2 = F/4m, f = F/4
F
a1
f
f
a2
22. The process is equivalent to TV1/2 = C
Compare with TVx1 = C x = 3/2
C =R R
1 1 X
=
R R2 3 1 3 2
=3 1
R 2R R2 2
23. Intensity will be highest at the nearest point.
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C C h h e e m m i i s s t t r r y y PART – II
14. 19hcE 2.9 10 J
Total energy of 10 quanta = 10 × 2.9 × 10-19J
Energy stored for process3
1923
112 4.18 107.8 10 J
6 10
% efficiency19
19
7.8 10100
29 10
= 26.9%
15. a b
b a
r M2r 1 M
rms
TV
M (As
rms
3RTV
M )
rms a a b
b arms b
V T M 2 2 2 2
V T M 1 1 1
16. 1 1
1 1 2
k E 1 1n
k R T T
... (i)
2 2
2 1 2
k E 1 1n
k R T T
... (ii)
Eqn. (ii) – Eqn.(i) 2 12 1
2 1 1 2
E Ek k 1 1n
k k R T T
(For equimolar formation of B and C, 2 1k k )
1 2
2 2k T 3008314n k 8.314 300 T
n2 2
2
T 30083148.314 300 T
T2 = 329.77 K.
17.
3 2 22SO g 2SO g O g
at equilibrium 1 2x 2x x
Only SO2 will oxidized.Equivalent of SO2 = Equivalent of KMnO4 2x × 2 = 0.2 × 52x = 0.5
2
c 2
0.5 0.252 2
K 0.1250.52
18. b bT i k m 2.08 = 0.52 × 1 × ii = 4It means salt on dissociation gives 4 ions. Thus the salt that gives 4 ions is K3[Fe(CN)6].
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19. 2 o o1 1 1Cu e Cu ; E X V G FX ... (i)
3 1 o o2 2 2In 2e In E X V G 2FX ...(ii)
2 o o3 3 3In e In E X V G FX ... (iii)
From Eqn. (i) + (iii) – (ii)2 2 3 o o
In Cu In Cu G FE
o o
1 3 2G F X X 2X FE
o1 3 2E X X 2X
20. Due to Fajan’s rule.23. Green ppt. is – Cr(OH)3
25.
CH3 C
O
OH2
HVZBr /P C OH
O
2BrCH3
KCN
H O
Y
||
2
O
OH C CH COOH
X Malonic acid
26. 1o aromatic amine on diazotisation followed by coupling with -napthol gives azo dye test.
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M M a a t t h h e e m m a a t t i i c c s s PART – III
1. Given equation can be written as |x – 1|2 + |x – 1| – 6 = 0t2 + t – 6 = 0 t = 2 or –3 |x – 1| = 2
x – 1 = 2 or x = 3, –1
2. |z1 – z2|2 + |z2 – z3|
2 + |z3 – z4|2 + |z4 – z1|
2 = 2(|z1|
2 + |z2|2 + |z3|
2 + |z4|2) + |z1 + z3|
2 + |z2 + z4|2 8 + |z1 + z3|
2 + |z2 + z4|2
3.
r 4 22 2
r r T
r 4 r 2 4r
=
2 2
r
r 2r 2 r 2r 2
=2 2
1 1 14 r 2r 2 r 2r 2
= 2 2
1 1 14 r 1 1 r 1 1
n 2 2
1 1 1 1S 1
4 2 n 1 n 1 1
51 1 1 1 1 38 37 26 1
S 14 2 26 37 4 26 37
=
1 1380 3454 26 37 26 37
57 338
37S 1326 26
4. Number of terms = Number of non–negative integral solution of the equation p + q + r + s = 50= 50 + 4 – 1C50 =
53C50 =53C3
5. If a function is symmetric about two mutually perpendicular lines, it must be symmetric about theirpoint of intersection.
6.5 5
2 26 6x y dy dya x ydx dx1 x 1 y
5 5
2 2
6 6
y dy xay ax
dx1 y 1 x
6 2 6 3
6 2 6 3
1 ydy x a 1 x xdx 1 x y a 1 y y
..... (1)
Also,
6 6
3 3
6 6
1 x 1 ya x y
1 x 1 y
or 3 3 6 6x y a 1 y 1 x
6 3 6 3a 1 y y a 1 x x
Hence, from equation (1),6 2
26
1 ydy xdx y1 x
7. For a triangle formed by joining any three co–normal points, orthocentre consides with the fourthpoint.
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9. If a circle cuts two circles orthogonally, radical axis of the two circles pass through the centre ofthe first circleHere radical axis of given circles is 6x – 4y + 4 = 0 or 3x – 2y + 2 = 0
Hence 3 – 2 + 2 = 0 3 14 2 2
10. sin =21
21
r r r r
= 2 sin –1
21
21
r r r r
O1 O2
r 2
r 1
11. Let jbiar a2 + b2 = 1 …(1)Also,
cos 45° =| ji||r |
) ji(r
a + b = 1 …(2)
cos 60° =| j4i3||r |
) j4i3(r
3a – 4b =
25
…(3)
There exists no real values of a and b satisfying (1), (2) and (3)Hence no such unit vector exists
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15. x (2, 3) –1 < x2 – 4x + 3 < 0, so f(x) is increasing in (–1, 0)
f(sin x) is increasing onn I
(4n 1) ,2n2
16. Let point P on the straight line x + y = 4 be (m, 4 – m), this will be nearest to the parabola if at
this point to the straight line becomes normal to the parabola.Let it is normal at x – 10 = t2, y = 2tPerpendicular to x + y = 4 at (m, 4 – m) is y – (4 – m) = (x – m) …..(1)Normal at parabola at (t2 + 10, 2t) is y + t(x –10) = 12t + t3 …..(2)
(1) and (2) are same t = –1, m =172
so required point is17 9
,2 2
17. a a c 3b 3 b
a . a c sin 3.12
3 = a . a c sin
3 = 3.2 sin
sin =
1
2
cos
2
=
3
4
18. (x) =
x 2 2
2 x 2
e 2cos2x 2xsec x
ln 1 x cosx sinx
cosx e 1 sinx
+
x 2
2 x 2
e sin2x tanx
1sin x cosx
1 x
cosx e 1 sinx
+
x 2
2 x 2
e sin2x tanx
ln 1 x cosx sinx
2xsinx e 2xcosx
= B + 2Cx + ….
Put x = 0, B =
1 2 0 1 0 0 1 0 0
0 1 0 1 0 1 0 1 0
1 0 0 1 0 0 0 1 0
= 0
19. Let g(x) be the inverse of f, then f(g(x)) = x 3g (x)(g(x)–2) = x (g(x))2 – 2g(x) – log3 x = 0
g(x) = 33
2 4 4log x1 1 log x
2
Since g : [1, ] [2, ]So g(x) = 31 1 log x
20. Since 2x 3x 2 0 0 1 2tan x 3x 2 <2
Since 24x x 3 0 0 < 1 2cos 4x x 3 2
0 < L.H.S < The given equation has no solution
21. Let f = (5 2 – 7)19 x – f = an integer [x] + {x} – f = an integer {x} – f = an integer, but – 1 < {x} – f < 1 {x} = fSo, x{x} = x.f = 119 = 1
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23.1x
= 2e1 2 3
...3 ! 5 ! 7 !
= 2eS
S =r 1 r 1 r 1
r 1 (2 r 1) 1 1 1 1(2 r 1)! 2 (2 r 1)! 2 (2 r)! (2 r 1)!
= 1 1 1 1 1 ...2 2 ! 3 ! 4 ! 5 !
=12
e1 =1
2 e
1x
= 2eS =2 e2 e
= 1
So,1
0
f (y) logy x dy = 0
24. Let any point on second line be (, 2, 3)
cos = 642
sin = 642
OAB =12
(OA).OB sin =12
3 . 14 6
42 = 6
= 2So, B is (2, 4, 6) B(, 2, 3)O
A (1, 1, 1)
25. Centroid of triangle will bea b c
3
Now line joining the orthocentre and the circumcentre is divided by centroid in 2 : 1 ratio
internally, so orthocentre will be a b c .26. If we draw the graph of tan x and cot x, we
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28. f(x) =x 4
|t x| |t x|
0 x
e dt e dt =x 4
x t t x
0 x
e dt e dt =x 4x t t x x 4 x
0 xe e e e 2
f (x) = ex – e4–x = 0 x = 4 – x x = 2f(0) = f(4) = e4 – 1, f(2) = 2(e2 – 1), so maximum value of f(x) is e4 – 1.
29. Given inequation can be rewritten as2sec x 2 2y 23 y
3 9 1
2
2sec x 1 1
3 y3 9
1
Now,2
2sec x 1 1 1
3 3 and y3 9 3
So, we should have sec2 x = 1, y =13
x = 0, , 2, 3
30. arc(AC) = 3, arc(AB) = 4, arc(BC) = 5Let radius of the circle be ‘r’ r = 3 = 3 /r, = 4/r, = 5/r