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PART-1 : PHYSICSANSWER KEY : PAPER-1
PART-2 : CHEMISTRY
PART-3 : MATHEMATICS
Path to Success
ALLENCAREER INSTITUTEKOTA (RAJASTHAN)
CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2014-2015)
PAPER CODE 0 0 C T 2 1 4 0 0 5
PATTERN : JEE (Advanced)TEST TYPE : MAJOR
ENTHUSIAST & LEADER COURSE
Date : 10 - 05 - 2015TARGET : JEE (Advanced) 2015
ALL INDIA OPEN TEST #02
T M
Q. 1 2 3 4 5 6 7 8 9 10A. C B B B C B C D D CQ. 11 12A. C A
A B C D A B C DP S,T R,T Q P,R Q Q S,T
Q. 1 2 3 4 5 6 7 8A. 6 4 9 9 6 6 8 4
SECTION-I
SECTION-IV
SECTION-II Q.1 Q.2
Q. 1 2 3 4 5 6 7 8 9 10A. B D A A D D C C B DQ. 11 12A. D A
A B C D A B C DQ,S R,T S P Q,R,S,T P,Q,R,S,T P Q,S
Q. 1 2 3 4 5 6 7 8A. 2 2 4 9 3 7 2 5
Q.1 Q.2
SECTION-I
SECTION-IV
SECTION-II
Q. 1 2 3 4 5 6 7 8 9 10A. B D D C A D D B A DQ. 11 12A. B C
A B C D A B C DQ P S R P,Q,S P,Q,S P,Q,S R,T
Q. 1 2 3 4 5 6 7 8A. 2 7 1 7 6 1 2 2
Q.1 Q.2
SECTION-I
SECTION-IV
SECTION-II
PART-1 : PHYSICSANSWER KEY : PAPER-2
PART-2 : CHEMISTRY
PART-3 : MATHEMATICS
Path to Success
ALLENCAREER INSTITUTEKOTA (RAJASTHAN)
CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2014-2015)
PAPER CODE 0 0 C T 2 1 4 0 0 6
PATTERN : JEE (Advanced)TEST TYPE : MAJOR
ENTHUSIAST & LEADER COURSE
Date : 10 - 05 - 2015TARGET : JEE (Advanced) 2015
ALL INDIA OPEN TEST #02
T M
Q. 1 2 3 4 5 6 7 8 9 10A. A,B,C B,D A,C,D B,D B,C,D A,B,C,D B,C B,C,D D AQ. 11 12 13 14 15 16A. C A A B D CQ. 1 2 3 4A. 7 5 1 9
SECTION-IV
SECTION-I
Q. 1 2 3 4 5 6 7 8 9 10A. A,D B,C B,C,D A,B,D A,B,D A,C A,B,C,D B,C C CQ. 11 12 13 14 15 16A. D B C A D AQ. 1 2 3 4A. 6 2 1 6
SECTION-I
SECTION-IV
Q. 1 2 3 4 5 6 7 8 9 10A. A,C,D A,B,C A,C B A,C, A,B A,B B,C D AQ. 11 12 13 14 15 16A. B D A C B AQ. 1 2 3 4A. 6 2 6 1
SECTION-I
SECTION-IV
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-2436001 [email protected] HS-1/14
PART-1 : PHYSICS SOLUTIONPAPER-1
Path to Success
ALLENCAREER INSTITUTEKOTA (RAJASTHAN)
T M
CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2014-2015)
PAPER CODE 0 0 C T 2 1 4 0 0 5
PATTERN : JEE (Advanced)TEST TYPE : MAJOR
ENTHUSIAST & LEADER COURSE
Date : 10 - 05 - 2015TARGET : JEE (Advanced) 2015
ALL INDIA OPEN TEST #02
SECTION-I1. Ans. (B)Sol. As msinq = constant so 1 × sin 60°
= ( )max2 3 y sin90- °
60°
y
x
max
3 3y
2Þ =
2. Ans. (D)Sol.
6C 3C
3C 6C
V
S
net charge zero
–q +q
O
Initial condition
6C 3C
3C 6C
V
–(3CV/2) +3CV
Þ
6C 3C
3C 6C
V
Final conditionNet charge in initial condition on the boundedsystem is zero.
Final charge on the same system
is +3CV –3CV 3CV
2 2=
Therefore 3CV
2charge will flow through the
switch.
3. Ans. (D)
Sol.y
y60cmd
x
d
Deviation produced by prism
d = (m–1)A = (1.5–1)0
4 2 1rad
90æ ö æ ö= =ç ÷ ç ÷è ø è øp p
y = fd = (60) 190
æ öç ÷è ø
=23
cm
4. Ans. (C)
Sol. x
y
BB
BABA
5. Ans. (A)Sol. T
1 = 60g
\\\\\\\\\\\\\\\\\
T1
T
T – T1 – 50g
= 50 × 2Þ T = 110 g + 100
= 1200 N6. Ans. (D)
Sol.1 1F q(v B) aq j q(ai B)= ´ Þ - = ´r r rr $ $
$0B B kÞ =
r
Now $1 2v v (ai) (ai b j) abk´ = ´ + =r r $ $ $
Therefore 1 2F q (v v ) B 0é ù= ´ ´ =ë ûr rr r
HS-2/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-1
Kota/00CT214005
7. Ans. (D)Sol. Let frequencies of A and B are n1 and n2 then
n1 = 1
v v v4 4 15 60
= =´l
and 21
v v vn
2 2 30.5 61= = =
´l
But n1 – n2 = 6 so v v
60 61- = 6
Þ v = 21960 cm/s
1
21960n
60Þ = = 366 hz, n2
21960
66= = 360 Hz
8. Ans. (B)
Sol. Least count = pitch 0.5
No.of divisions 50=
= 0.01 mm ;Diameter of wire = 6 × 0.5 + 46 × 0.01= 3.46 mm
9. Ans. (A)
Sol.
Rsin /np
p/n
I = 2
2 2 2m.(2R sin )
nn m.R cos MR12 n
pé ùê úp
+ +ê úë û
Þ I =
2
2 2 2sin
nnmR cos MR3 n
pé ùê úp
+ +ê úë û10. Ans. (D)
Sol.
where 2
2 2 2sin
nI nmR cos MR3 n
pé ùê úp
= + +ê úë û
(M + nm) g sin q – f = (M + nm) a ........(i)fR = Ia/R ........(ii)
form (i) & (ii) f = 2
(M nm)gsin I[I (M nm)R ]
+ q+ +
Þ µ(M + nm) g cosq = 2
(M nm)gsin I[I (M nm)R ]
+ q+ +
µ = 2
I tanI (M nm)R
q+ +
11. Ans. (B)
12. Ans. (C)
Sol. For Q. 11 & 12
For resonance condition w of electric field must
be equal to the angular frequency of electrons
rotation
\ w = qB
*mAt low temperature the lattice vibration
decreases and increasing w will increase the path
length between two collisions.
Holes are positively charged thus direction of
electric field must be reversed
SECTION-II
1. Ans. (A)®(Q); (B)®(P); (C)®(S);(D)®(R)
2. Ans. (A) ® (P,Q,S) ; (B) ® (P, Q, S) ;(C) ® (P,Q, S) ; (D) ® (R, T)
Sol. (A) It represents induce electric field for time
varying magnetic field upto some
distance R.
For magnetic field it may be field of a
cylinder with varing current density
(B) Induce electric field lines or magnetic field
lines of a current carring wire.
(C) Induce electric field lines or magnetic field
lines of a solenoid.
(D) Only electrostatics field of a dipole
SECTION-IV
1. Ans. 2
Sol. Assumeing initial angular velocity of the disc
to be zero, we can assume it to be performing
the pure translational motion as the net torque is
also zero.
We can treat this as a smiple pendulum.
T 2g
Þ = pl
= 0.993
29.8
p
T ; 2 sec
HS-3/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-1
Kota/00CT214005
2. Ans. 7
+ x
P(x,y)
(3,0)( 3,0)-
y
q -2q
Potential at
P = 0 = ( )
( )( )2 22 2
K 2qKq0
x 3 y ) 3 x y- =
+ + - +
Þ ( ) ( )2 22 23 x y 4 x 3 4y- + = + +
Þ 9-6x + x2 + y2 = 4x2 + 24 x + 36 + 4y2
Þ 3x2 + 30x + 27 + 3y2 =0
Þ (x+5)2 + y2 = 42 Þ a =5, b=2 Þ a + b =7
3. Ans. 1
Sol. p = charge × (distance between mass centres of
both the portions) = 2
2 2 4l læ ö´ =ç ÷è øl l l
= 1 mC–m
4. Ans. 7
Sol.34 8
min 19 3
hc 6.63 10 3 100.62Å
eV 1.6 10 20 10
-
-´ ´ ´
l = = =´ ´ ´
minhc 12400
OR 0.62ÅeV 20000
æ öl = = =ç ÷è ø
Also, ( )2 2 2K 1 2
1 1 1R Z 1
n n
é ù= - -ê ú
l ê úë û
Þ ( )27K
K
1 11.09 10 41 1 1 0.76Å
4é ù= ´ - - Þ l =ê úl ë û
Now, l - l = - =K min 0.76 0.62 0.14Å
- -= ´ = ´10 80.14 10 m 14 10 m
5. Ans. 6
Sol. From PV = NkT
Þ (10–5 × 13.6 × 103 × 9.81) × 7.4
= Nk × 293 \ Nk = 0.0338
Energy discharged through capacitor
= 2 51CV 12.0 10 J
2= ´
Energy transformed to plasma K.E. = 24 × 104 J
As average kinetic energy associated with the
gas molecules is 32
NkT, but each deuterium
molecule produces two ions and two electrons,
have 4 × 32
NkT = 24 × 104
Þ T = 1.18 × 106 Þ So a = 6
6. Ans. 1
Sol. By symmetry circuit can be reduced to
5W
10V
I 5W I
10
I 1A5 5
= =+
7. Ans. 2
Sol. Electric field at
P : 2 2
2 2 5/2 2 2 5/2
kp(2x y ) 3kpxyˆ ˆE i j(x y ) (x y )
-= +
+ +
r.
At P xE 0=r
Þ y x 2 2m= =
8. Ans. 2
Sol. For power to be minimum current should be
zero Þ n n 2
n 2n 1 n 4
+= Þ =
+ +
HS-4/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-1
Kota/00CT214005
PART–2 : CHEMISTRY SOLUTIONSECTION - I
1. Ans. (B)
Kh = w
a
KK
1410
1410 1010
--
- =
2W
Ha
K CKK (1 )
a= =
- aa << 1, hence10–10 = 0.01 a2
a = 10–4
% a = 10–2
2. Ans. (D)3. Ans. (A)
Specific conductance , K = 2 11
. 10 ScmR A
- -=l
\ molar conductance eqL = ´K 1000
M
= 2
31
1010
10
-
- ´
= 102 S cm2 mol–1
4. Ans. (A)5. Ans. (D)6. Ans. (D)7. Ans. (C)8. Ans. (C)
CH COONa2Electrolysis CH2
CH2CH COONa2
Br2
CCl4CH –Br2
CH2
Alc K
OH
CH CH
(A) (B)
(C)
Br–
9. Ans. (B)
C(graphite) + 2H2(g) ® CH4(g)
DrHº = {– 400 + 2 (–300)} – {–900}
DrHº = – 100
DrHº = – 100 = 700 + 880 – 4EC–H
E(C–H) = 420kJ
mole
10. Ans. (D)11. Ans. (D)12. Ans. (A)
SECTION - II
1. (A)®(Q, S) ; (B)®(R, T) ; (C)®(S) ; (D)®(P)
2. (A)®(Q,R,S,T); (B) ® (P,Q,R,S,T);
(C) ® (P); (D)®(Q,S)
SECTION - IV1. Ans. 2
2 = hc
- fl
........(i)
6 = 2hc
- fl
........(ii)
Solving : f = 2eV Ans.
2. Ans. 2
2 NO (g) + O2 (g) ® 2NO
2 (g)
t = 0 moles 2 0.5 0
t = t moles 2–2 × 0.5 0.5 – 0.5 2 × 0.5
= 1 = 0 = 1
nf = 1 + 1 = 2
Dn = 2.5 - 2 = 0.5 moles
\ change in pressure
nRT 1 300P 0.5 2atmV 12 6.25
DD = = ´ ´ =
3. Ans. 4
3A
3 3 2 M 3 3N 64r2
´= ´
´
3A
3A
1 2M64 r N2
4 M 2 2dN 64r
=´ ´
´ ´=
´
1 22 d 4 2 2
=´ ´
d = 4
HS-5/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-1
Kota/00CT214005
4. Ans. 9
5. Ans. 36. Ans. 7
Ph–C + NaHCO 3 ® Ph–C + H O + CO2 2
||O
OH
||O
ONa(A) PhOH + NaNH2 ® PhONa + NH3
(B) CH3OH + NaH ® CH3ONa + H2
(C) CH3SH + ELLi ® CH3SLi + CH3–CH3
(D) 2PhSO3H + 2Na ® 2PhSO3Na + H2
(E) (y + z + w + v) – (x)
(17 + 2 + 30 + 2) – ( 44) = 7
7. Ans. 2
COOEtCOOEt
(i) NaOEt/D(ii) H O(iii) (iv) Zn-Hg/HCl
3+
D
NBS/hv
Br
(two)product are possible
8. Ans. 5
PART-3 : MATHEMATICS SOLUTIONSECTION-I
1. Ans. (C)
Let ( )5x
ƒ x ......... x 15!
= + + +
( )4 3x x
ƒ ' x ......... x 14! 3!
= + + + +
( )3 2x x
ƒ" x x 13! 2!
= + + +
( )2x
ƒ"' x x 1 0 x R2!
= + + > " Î
Þ ƒ" is increasing with only 1 real root
say a so 3 2
1 03! 2!
a a+ + a + =
Þ ƒ'(x) have only one point of minima x = aÞ ƒ'(a) = +ve Þ ƒ'(x) > 0 " xÎR
Þ ƒ'(x) > 0 Þ ƒ(x) = 0 have only one real root
2. Ans. (B)
Let M = cosx cos2x........cos999x
( ) ( )( )
999
999
2 sin xsin 2x.........sin 999x cosx cos2x....cos999x
2 sin x sin 2x......sin 999x=
999
sin 2x sin 4x.......sin1998x
2 sin xsin 2x.....sin 999x=
( )999
sin1000xsin 1002x .......sin1998x
2 sin xsin 3x.........sin 999x=
999
1
2=
Q sinx = sin1998x sin3x = –sin1996x and so on.
3. Ans. (B)z1 + z3 = z2 + z4 Þ PQRS is a parallelogramand set A contain points on the following circle
P31
R C A Q B(–4,0) (–2,0) (–1,0) (2,0)
(–5/2,0)
parallelogram inscribed in a circle is rectangleand area of rectangle will be maximum when it
is a square so area ( )21 93
2 2= =
4. Ans. (B)
x ycos sin 1
a bq + q = , OP
bm tan
a= q
aP(a cos ,bsin )q q
O
T
bm cot
a= - q
HS-6/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-1
Kota/00CT214005
2
2
b btan cot
a atanb
1a
q + qa =
-2 2
abtan cot
a b= q + q
-
)( )
( )22 2least
2 2 32abtan
a b 2 3 1
+a = =
- + -
least
1
63
p= Þ a =
5. Ans. (C)
1 2x x 90S =x1x2.....x5 = 243
as 1 2x x9
10
S= & ( )1/10
1 2x x 9P =
Þ x1x2 = x2x3=...... = 9Þ x1 = x2 =...... = x5 = 3so a = –5C13b = –5C33
3
c = 5C434
a > b, a + b < 0, b + c > 0, c > 06. Ans. (B)
( )/ 2
13
1
0
I cosx cosxdxp
= ò
( ) ) ( ) ( )/ 2/2
13 13 1
0 0
cosx sin x 13 cosx sin x sin xdxpp
-= - -ò
( ) ( )/ 2
13 1 2
0
13 cos x 1 cos x dxp
-= -ò
( )1 2 1I 13 I I= -
2 21 2
1 1
13 I 1 II I 1 1
I I13 1 13
é ù= Þ = + Þ =ê ú+ ë û
7. Ans. (C) ƒ(x,y) > x2 + 2y " x,yÎR
and ƒ(x,y) > y2 + 4x " x,yÎR
2ƒ(x,y) > x2 + 4x + y2 + 2y
ƒ(x,y) > ( ) ( )2 21x 2 y 1 5
2é ù+ + + -ë û
5
2³ -
so least value of ƒ(x,y) is 5
2æ ö-ç ÷è ø
obtained when
x =–2 & y = –1
8. Ans. (D)
x2 = 1002 + 1002 – 2(100)2cosq
= 2.1002[1 –cosq]
4dx d2x 2.10 sin
dt dt
q= q ...(1)
Qx
S100
100
Also s =100q Þ ds d d 1
100dt dt dt 10
q q= Þ = ...(2)
(2) & 2
pq = in (1)
Þ 4dx 2.10 .1 1
.dt 102.100 2
= 5 2=
Paragraph for Question 9 & 10
x : Judge see a plus sign
P(x) =æ ö æ ö æ ö æ ö+ + =ç ÷ ç ÷ ç ÷ ç ÷è ø è ø è ø è ø
4 2 2 44
2
1 1 2 2 41C
3 3 3 3 81
y : A originally wrote a plus sign
4 21 1 2 1
3 .P(x y) 133 3 3 3
P(y / x)41P(x) 4181
æ ö æ ö+ç ÷ ç ÷Ç è ø è ø= = =
9. Ans. (D)
10. Ans. (C)
11. Ans. (C)
12. Ans. (A)
Q Reflection of (x, y) in
L is 3x 4y 4x 3y,
5 5- + +æ ö
ç ÷è ø
\ 2 23x 4y 3x 4y 4x 3y 4x 3y12 7 12 25 0
5 5 5 5- + - + + +æ ö æ ö æ ö æ ö- - + =ç ÷ ç ÷ ç ÷ ç ÷
è ø è ø è ø è ø
Þ 12(9x2 + 16y2 – 24xy) – 7(–12x2 + 12y2 +7xy) – 12(16x2 + 9y2 + 24xy) + 625 = 0
Þ xy = 1
T.A. of xy = 1 is y = x and its reflection iny = 2x gives T.A. of S1 which is
4x 3y 3x 4y5 5+ - +
= Þ y = 7x
HS-7/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-1
Kota/00CT214005
SECTION – II1. A ns. (A)®(P); (B)®(S,T); (C)®(R,T);
(D)®(Q)
( )
sin x , x 0,2
ƒ x 1 , x ,2
cosx , x ( ,2 ]
ì pé ùÎï ê úë ûïï pï æ ù= Î pí ç úè ûïï- Î p pïïî
which is differentiable in [0,2p]
(A) ( )( ) ( )10
0
I x 1 x 2 .... x 9 dx= - - -òI = – I using king.Þ I = 0
(B) ( )( )1
1/ 44 3 5 4
0
x x 4x 5x dx+ +ò4x5 + 5x4 = tdt = 20(x4 + x3)dx
9 5/ 41/ 4
0
dt tt c
520 20.4
= +ò 19 3
25=
(C) ƒ(t) = 16t3 – 4t2 – 16t + 8, t Î [–1,1]ƒ'(t) = 48t2 – 8t – 16= 8[6t2 – t – 2]= 8(6t2 – 4t + 3t – 2)= 8[2t(3t – 2) + (3t – 2)]
= 8(2t + 1) (3t – 2) = 0 Þ 1 2
t ,2 3
= -
ƒ(–1) = 4, ƒ(1) = 4, 1
ƒ 132
æ ö- =ç ÷è ø
,
2 8ƒ
3 27æ ö =ç ÷è ø
Þ ƒmax = 13
(S) sin–1x = x + c(1, /2)p
(–1,– /2)p
y = x + c Þ c = y – x
c 1, 12 2
p pé ùÎ - + -ê úë ûc = 0 only value
2. Ans. (A) ® (P,R); (B) ® (Q); (C) ® (Q); (D) ® (S,T)
(A) ( ) 22tan
7a + b =
B D C3 17
h
A
a b
2
3 1722h h
51 71h
+=
-
2
2
20 h 22.
h h 51 7=
- Þ 70h = 11(h2 – 51)
Þ h = 11
(B) ah 1 2 s 12
r r a a
S D= = D
Då å
= (a + b + c) 1 1 1
9a b c
æ ö+ + ³ç ÷è ø
(by AM > GM)
(C) Let ( ) ( ) ( ) ( )D 0 . A a , B b ,C cr rr r
Þ a 2b b 2c 2a c
E F G3 3 3
æ ö æ ö+ + +æ öç ÷ ç ÷ ç ÷
è øè ø è ø
r rr r r r
D
AG
CF
BE
1V DE DF DG
6é ù= ë ûuuur uuur uuur
3
1a 2b b 2c c 2a
6.3é ù= + + +ë û
r rr r r r
1 2 01
0 1 2 a b c 96.27
2 0 1
é ù= =ë ûrr r
(D)
1 2 3
2 3 1 a b c
3 1 2
é ùë ûrr r
( )18 a b .c= - ´rr r
218c= -r
so max. value of 218c-r
is 18
{as c a b£rr r
}
HS-8/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-1
Kota/00CT214005
SECTION-IV1. Ans. 6
Multiply gh in(1), ƒh in(2) and gh in (3)and add(ƒgh)' = 6(ƒgh)2 + 6
Þ 2
dt6dx
t 1=
+Þ tan–1t = 6x + c
tan–1(ƒ(x) g(x) h(x)) = 6x + c
by x = 0 Þ c4
p=
Þ ( ) ( ) ( )ƒ x g x h x tan 6x4
pæ ö= +ç ÷è ø
2. Ans. 41
2 x 1 x2
± - aa + a = ± Þ =
a
put in 2 1x x x
24= a + a +
Þ3 2
,2 3
a = , 13 601
12
- ±-
3. Ans. 9by x ® 3 – x & x ® 8 – x
ƒ(6 – x) = ƒ(x) = ƒ(16 – x)
Þ period = 10 ...(1)
by x = 3, ƒ(6) = ƒ(0) ...(2)
also ƒ'(3 + x) = –ƒ'(3 – x) Þ ƒ'(3) = 0 by x = 0& ƒ'(6) = 0 by x = 3
and ƒ'(8 + x) = –ƒ'(8 – x) Þ ƒ'(8) = 0 by x = 0
so roots are 0,3,6,8,10,–10,–7,–4,–2
4. Ans. 9The curve is(x – 2) + 4(y + 1)2 = 4
P A B(0,–1)
(2,–1)(–2,–1) (4,–1) so
2
2
Max PB
Min PA=
5. Ans. 6(0,40)
(40,0)
Let z = x + iy Þ [ ]x y, 0,1
40 40Î
Þ x,y Î [0,40]
also [ ]2 2 2 2 2
40 40z 40x 40y, 0,1
z | z | x y x y= Þ Î
+ +Þ 40x < x2 + y2 & 40y < x2 + y2
Þ (x – 20)2 + y2 > 202 & x2 + (y – 20)2 > 202
( ) ( )2 22 1 1A 40 40 20
4 2= - - p
( )A2 6
100= - p Þ Least integer greater then
A
100 is equal to 6.
6. Ans. 6
x+y=p
x+y=0
x+y=p
x + y Î ......[–p,0]
Area 1 12
62
æ öp =ç ÷pè ø
7. Ans. 8
D1 = D2 = D3 = 0 Þ a + c = 2b
Þ max (b) = 8
8. Ans. 4
b c b cM N
2 2
æ ö æ öl + + mç ÷ ç ÷è ø è ø
r rr r
( ) ( )1 1 1AMN b c b c
2 2 2D = l + ´ + m
r rr r
1b c b c
8é ù= lm ´ - ´ë û
r rr r
D
C
N
M
A O E B bl b
mC
c
( )1BCDE CE BD
2D = ´
uuur uuur
( ) ( )1b c c b
2= l - ´ m -
r rr r
1b c b c
2= lm ´ - ´
r rr r
BCDE4
AMN
D=
D
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-2436001 [email protected] HS-9/14
PART-1 : PHYSICS SOLUTIONPAPER-2
Path to Success
ALLENCAREER INSTITUTEKOTA (RAJASTHAN)
T M
CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2014-2015)
PAPER CODE 0 0 C T 2 1 4 0 0 6
PATTERN : JEE (Advanced)TEST TYPE : MAJOR
ENTHUSIAST & LEADER COURSE
Date : 10 - 05 - 2015TARGET : JEE (Advanced) 2015
ALL INDIA OPEN TEST #02
SECTION-I1. Ans. (A,C,D)
Sol. For image of B BB
1 1 1v 60cm
v 30 20- = Þ =
- ;
For image of A AA
1 1 1v 40cm
v 40 20- = Þ =
-
length of image = vB – vA = 20 cm2. Ans. (A,B,C)Sol. Let velocity of dog be
1 2v i v j+$ $ such that2 21 2v v 89+ = .
To catch the rabbit
1 2 0(v i v j)t (5i 10j) (3i 4j)+ = + + +$ $ $ $ $ $ t0Þ v1t0 = 5 + 3 to & v2 t0 = 10 + 4 t0Þ 2v1 = v2 + 2 Þ 5v1
2 – 8v1 – 85 = 0
1
8 64 1700 8 1764 8 42v 5
10 10 10± + ± ±
Þ = = = =
Þ v2 = 8ms–1 and 01
5 5 5t s
v 3 5 3 2= = =
- -3. Ans. (A,C)Sol. For (A) : Power consumed P = I2R
But 0
q= f
Î. so q = aÎ0t2
0
dqI 2 t
dtÞ = = a Î
Þ P = a2Î02t4R
For (B) , Assuming initial charge in resevoir beq0 then electric flux through a closed spherical
surface around S2 will be 2
20 0
s0
q t- a Îf =
Î
For (C) 2sd
2 tdt
f= - a
4. Ans. (B)
Sol. vsound = Er Here E is same and rsea > rfresh
5. Ans. (A, C)We add a resistance R in series to achieve avoltmeter of range 50V.
Þ 50 × 10–6 = 50
100 R+ Þ R = 106 W
Now, when a current of 10 mA is passed throughthe ammeter, 50 mA should go through the coil.We add a resistance S in parallel to the ammeter,50 mA should go through the galvanometer is :
50 × 10–6 = (10 × 10–3) S
S 100+Þ S = 0.5 W
6. Ans. (A,B)
Sol. Average rate of rise of temperature( ) ( ) ( )
( ) ( )-1 1
P 0.99 50kV 20mA2 C/ sec
t ms 1kg 495 J kg C-
Dq= = = °
D °
The minimum wavelength of the X-rays emitted
min 3
hc 124000.248Å
eV 50 10l = = =
´
7. Ans. (A,B)Try yourself by using BE/nucleon curve
8. Ans. (B, C)9. Ans. (D)Sol. The block will move updown periodically
\ centre of mass moves periodically in vertical.(In horizontal both will have same vertical andacceleration in magnitive but opposite indistance)
10. Ans. (A)Sol. When smaller block will be at point C it will
has zero xE but max potential energy \ byconservation principle of mechanical energy wecan say that velocity of bigger block will bezero.
11. Ans. (B)
Sol. Emax = hcl
+ evA = f (work function)
12400124
= ev + 10,000 ev – 10 ev = 10,090 ev
Þ min
hcl
= 10,090 ev Þ min
1240010,090
l = =1.24 Å
HS-10/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-2
Kota/00CT214006
SECTION - I1. Ans. (A, D)2. Ans. (B,C)
Since 'A' is more volatile than 'B' & initial molesare equal : Y
A2 > Y
B2 , Y
B2 < X
B2but on vapourisation, composition changes andvapour pressure decreases P
T2 > P
T33. Ans. (B,C,D)4. Ans. (A,B,D)
3I2 + 6NaOH ¾¾®NaIO
3 + 5NaI + 3H
2O
5. Ans. (A, B, D)6. Ans. (A, C)7. Ans. (A,B,C,D)
(A)I show poor rate of SN2 due
to steric hindrane
(B) Ph—CH—CH—Me
OH D
Optically pure
HIS 1N
CH—Me + Ph CH—Me
H
I
Ph
I
H
D D
Diastercomer Not Racemic mixture
(C) Me—Cl Me—OH Not Me—O—MeMoistAg O2
(It is obtained in dry Ag2O)
(D) MeI + NH 3 Me NI4dÅ
(Excess)8. Ans. (B,C)9. Ans. (C)10. Ans. (C)11. Ans. (D)12. Ans. (B)13. Ans. (C)14. Ans. (A)15. Ans. (D)16. Ans. (A)
SECTION - IV1. Ans. 62. Ans. 23. Ans. 1
O O ¾¾¾®
2
KOHI
CHI3 + 2CH3 – COOK
4. Ans. 6
12. Ans. (D)
Sol.Power outputPower input
=ò3 6
0.5 100 100= =
´Þ 6%
13. Ans. (A)Sol. n = a (z–b) (Mosley's law)
b is less for K so compared to L due to lessshielding effect in K.Energykb > Energy ka Þ nkb > nka \ ab > aa
14. Ans. (C)Sol. For (A) By symmetry E = 0
For (B) By superposition principle
E = 20
q4 ap Î
For (C)0
6qV
4 a=
p Î
For (D)0
5qV
4 a=
p Î15. Ans. (B)
Sol. dF Idl B; B® ® ® ® ®
= ´ t = m´ò ò
16. Ans. (A)
Sol.P RQ S
= \ S = RQP
In option (a), P = 100, Q = 10 \ S = R10
Since correct R lies between 400 and 500 W,correct value of S will lie between 40 and 50.Similarly for other options.
SECTION-IV1. Ans. 6
Number of electric field lines are drawn in pro-portion to charge magnitudes.
2. Ans. 2
Sol. n0 =
1 T
2 ml
T = y A a D T (D T = 80)µ = r An0 = 200
3. Ans. 6
( )2
2 d I RI R mgy mgv v
dt mg= = Þ =
4. Ans. 1Sol. 30VS = 29MS and 1 MS = (1/2)°; least count
= 1MS – 1VS = 1'
PART–2 : CHEMISTRY SOLUTION
HS-11/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-2
Kota/00CT214006
PART-3 : MATHEMATICS SOLUTIONSECTION-I
1. Ans. (A,B,C)
cos2x = –1 Þ x = (2n + 1)2
p
2. Ans. (B,D)
For min value a = 0
–10 10
ƒ(0) = 50
and for maximum value a = 11 –10 10 11
( ) 1ƒ 11 22.20 220
2= =
3. Ans. (A,C,D)
1234
1235
43214322
Shaded area = Q
1234
1235
43214322
Shaded area = P
P < R < Q
Q – R < R – P Þ P + Q < 2R
4. Ans. (B,D)
ƒ2(x) = x Þ ƒ4(x) = ƒ6(x)........ = x
Þ fn(x) = nx
(n + 1)x > nx Þ x > 0 so A false, B true.
for (D) 4x = 5 sinx,
Þ 3 solutions
5. Ans. (B,C,D)
as ƒ(x) is a continuous, odd function in closed
interval therefore its range will be [-a,a], where
a Î R+ so a = –a & b = a
Now check options
6. Ans. (A,B,C,D)
Let tn is k, 122333........ k k.......k
k times
tn
Þ ( ) ( )k k 1 k k 1
1 n2 2
- ++ £ £
Þ 4k2 – 4k + 8 < 8n < 4k2 + 4k
Þ (2k – 1)2 + 7 < 8n < (2k + 1)2 – 1
Þ (2k – 1)2 < 8n – 7 < (2k + 1)2 – 8 <(2k + 1)2
Þ 2k 1 8n 7 2k 1- £ - < +
Þ8n 7 1
k k 12
- +£ < +
Þ n
8n 7 1t
2
é ù- += ê ú
ë û
7. Ans. (B,C)
by x = 7 Þ ƒ(6) = 0
by x = 4 Þ ƒ(4) = 0
Þ ƒ(x) = (x – 4) (x – 6) g(x) ...(1)
(1) in given relation
(x – 4) (x – 5) (x – 7) g(x – 1)
= (x – 7)(x – 4) (x – 6) g(x)
Þ (x – 5) g(x – 1) = (x – 6) g(x) ...(2)
by x = 5 Þ g(5) = 0 Þ g(x) = (x – 5) h(x) ..(3)
(3) in (2) gives (x – 5) (x – 6) h(x – 1)
= (x – 6) (x – 5) h(x)
Þ h(x) = h(x – 1) Þ h(x) = l ...(4)
by (4), (3) in (1) ƒ(x) = (x – 4) (x – 6) (x – 5)lby ƒ(7) = 6, l = 1
8. Ans. (B,C,D)
( ) ( )2
2ƒ ' x x 2
x= -
y
2 xx y® ±¥ Þ ® ¥
x 0 y-® Þ ® -¥
x 0 y+® Þ ® ¥
HS-12/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-2
Kota/00CT214006
Paragraph for Question 9 to 10Locus of Q is a circle
C A F D
B(2,3,4)
Let BC is maximum value of
QB and BD is minimum value
Line x 2 y 3 z 4
BF :1 1 1
- - -= = = l
F(l + 2, l + 3, l + 4) Þ 3l = –6 Þ l = –2
Þ F(0,1,2)
so BC2 = BF2 + FC2
= 12 + (FA + AC)2
= 12 + ( )2
2 2 2 30+ =
BC 30= so 'D' is correct.
Now F is mid point of
( )1 1 1AD , , 0,1,2
2 2 2
a + b + g +æ öÞ ºç ÷è ø
Þ D(–1,1,3)
9. Ans. (D)10. Ans. (A)
Paragraph for Question 11 to 12
( )1
n
0
ƒ n 1 x sin x dx2
pæ ö+ = ç ÷è øò
1 1n n 1
00
2 2x cos x nx cos x dx
2 2-ùp pæ ö æ ö= - +ç ÷ ç ÷úp pè ø è øû
ò
Þ ( ) ( )2nƒ n 1 g n+ =
p...(1)
Now ( )1
n 1
0
nƒ n nx sin x dx2
- pæ ö= ç ÷è øò
1 1n n
0 0
x sin x x cos dx2 2 2
p p pæ ö= -ç ÷è ø ò
( )1 g n 12
p= - +
Þ ( )nlim nƒ n 1
®¥=
Q ( ) ( )nlim ƒ n g n 0
®¥= = ...(2)
Þ ( ) ( )nlim n 1 ƒ n 1
®¥+ =
Þ ( ) ( )2
n nlim n g n lim nƒ n 1
2®¥ ®¥
p= +
( ) ( ) ( )nlim n 1 ƒ n 1 ƒ n 1
2®¥
pé ù= + + - +ë û 2
p= ...(3)
by (1) (2) & (3)
( ) ( )( ) ( )2n
3n 1 ƒ n 3lim
24n 4n 1 g n®¥
+=
p+ +
11. Ans. (C)
12. Ans. (A)
13. Ans. (A)
( )2cos t 1 0
A t 1 2cos t 1
0 1 2 cos t
é ùê ú= ê úê úë û
|A(t)| = 8cos3t – 4cost
= 4 cost cos2t sin 4t
sin t=
|A(4t)| = 4cos4t cos8t = sin16t
sin 4t
(P) t 0
sin 4t sin16tlim 16
sin t sin 4t®=
(Q) |A(t)| |A(3t)| = 4cost cos2t.
4cos3t cos6t < 16
which is maximum at t = 0
(R)t
17
4 sin16tA A 1
17 17 sin t p=
p pæ ö æ ö ö= =ç ÷ ç ÷ ÷è ø è ø ø
(S)0
sin16tdt 0
sin t
p
=ò
HS-13/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-2
Kota/00CT214006
14. Ans. (B)
by x = i
(1 + i)62 = (C0 – C
2 + C
4–.......–C
62)
+ i(C1 – C
3 + C
5–....+ C
61)
Þ ( )62i / 42e p = (C
0 – C
2....... –C
62)
+ i(C1 – C
3 + C
5..... +C
61)
Þ comparing real parts
0 2 4 62
610 2 4 62
600 4 4 60
602 6 62
C C C ...... C 0
C C C ....... C 2
C C C ..........C 2
C C ..............C 2
- + - =ìïí
+ + + =ïîÞ + + =
Þ + + =
comparing imaginary parts
311 3 5 61
611 3 61
C C C ......... C 2
C C ................. C 2
- + - + = -
+ + + =
by adding C1 + C
5 + C
9.... + C
61 = 230(230–1)
by subtraction C3 + C
7+.....+C
59 = 230(230+1)
15. Ans. (D)2 4 34
i i i18 18 18A 1,e ,e , .....e
p p pì üï ï= í ýï ïî þ
2 4 94i i i18 18 48B 1,e ,e , .....e
p p pì üï ï= í ýï ïî þ
LCM of 2 2
,18 48 3
p p pæ ö =ç ÷è ø
so
2 4 5i i i i
i3 3 3 3A B 1,e ,e ,e ,e ,ep p p p
pì üï ïÇ = í ýï ïî þ
( )1 22k 2k 2
i i i18 48 144
1 2zw e e e 8k 3kp p p
= = +
by k1= 0 & k
2Î{0,1,.....47} ® 48 values of zw
by k1= 1 & k
2Î{0,1,.....47} ® 48 values of zw
by k1= 2 & k
2Î{0,1,.....47} ® 48 values of zw
for rest of the values of k1 Î {3,........17} the
values of zw repeats. \ 144 differentiablevalues will be there so n(C) = 144
n(A Ç B) = 4. {Plot 18, sided polygon insidecircle of unit radius}
n(B Ç D) = 9. {Plot 48, sided polygon insidecircle of unit radius}
16. Ans. (C)
( )n
210
k 1
cos k log k 3k 2 1=
p + + =å
( ) ( )( )n
k 1
cos k log k 1 log k 2 1=
p + + + =å
I case : n even
Þ |– log2 – log3 + log3 + log4 – .....
.....+ log(n + 1) + log(n + 2)| = 1
Þ |log (n + 2) – log2| = 1
Þ n 2 1
10, n 182 10
+= Þ =
II case : n odd
|– log2 – log3 + log3 + log4.... – log(n+1)
– log(n+2)| =1
|– log(n + 2) – log2| = 1 Þ 2(n + 2) = 10, 1
10Þ n = 3.
for equation (2)
I case : n even
|log (n + 2) – log2| = 2 Þ 2
2
n 2 110 ,
2 10
+=
n = 98 one solution.
II case : n odd.
|– log(n + 2) – log2| = 2 Þ 2(n + 2) = 102, 2
1
10n = 48 rejected as n is odd.
SECTION-IV
1. Ans. 7
( ) ( ) ( )3 3 3
1 1 1
ƒ x ƒ x ƒ xdx dx dx
x x 2= -ò ò ò
( )3
1
1 1ƒ x dx
x 2æ ö= -ç ÷è øò
( )3
1
1 1ƒ x dx
x 2æ ö£ -ç ÷è øò
3
1
1 1dx
x 2£ -ò
HS-14/14
ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-2
Kota/00CT214006
2 3
1 2
1 1 1 1dx dx
x 2 2 xæ ö æ ö= - + -ç ÷ ç ÷è ø è øò ò
1 1 3n2 n
2 2 2= - + -l l
4n
3= l
Note : Maximum can be achieved by letting
( ) [ ]1, x 1,2ƒ x
1, x (2,3]
ì Î= í
- Îî
2. Ans. 5
Power of R with respect to 2nd circle gives x.2x= 10.26
Þ x2 = 130
2PQ 526
=
3. Ans. 1
ex(1 + yex)(dy + ydx) = 2x dx
Þ (1 + yex) (exdy + yexdx) = 2xdx
Þ ( )2x
21 ye
x c2
+= + Þ x 2ye 1 2x 4+ = +
\ ( ) ( )ƒ 1 6 1 e- = -
4. Ans. 9
R'' R'
R Q
C
B Q' Q'' A
P
R 'Q PQCA AB
= Þ102 dPQ .85
102-
=
R'' R PRBC AB
= Þ90 dPR .85
90-
=
QR = d =102 d 90 d.85 .85
102 90- -
+
Þ85 85d 1 85 85
102 90æ ö+ + = +ç ÷è ø
Þ5 17d 1 1706 18
æ ö+ + =ç ÷è ø
Þ 170.18d
50=
Þ5d 934
=