A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx =...
Transcript of A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx =...
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, INDRA VIHAR, KOTA-324005PHONE : +91 - 744 - 2436001, Fax : +91-744-2435003, E-mail: [email protected] Website: www.allen.ac.in KOTA / HS - 1/12
SOLUTION
PART-1 : MATHEMATICS ANSWER KEYPAPER-1
PAPER CODETM
Path to success KOTA (RAJASTHAN)0 1 C T 3 1 3 0 8 8
PATTERN : JEE (Advanced)LEADER & ENTHUSIAST COURSE
Date : 09 - 05 - 2014TARGET : JEE 2014
SCORE-II : TEST # 06
SECTION-I1. Ans. (C)
Total number of desired paths =(A ® C, C ® B) – (A ® C, C ® D, D ® B)Þ 6C3.9C4 – 6C3.4C2.5C2 = 1320
2. Ans. (B)2012 2
2014
x x 1x 1
+ ++
is always positive
at x = 02012 2
2014
x x 11
x 1+ +
=+
as x ® ¥ 2012 2
2014
x x 10
x 1+ +
®+
Þ (0,1]
hence range is 0,2pæ ù
ç úè ûintegral element is 1.
3. Ans. (C)
–1 – 12
12
1
1 1x 1, ,0, ,1
2 2= - - does not lies in the domain
of ƒ'(x)4. Ans. (D)
Py = x–22
x = y–22
Q. 1 2 3 4 5 6 7 8 9 10A. C B C D C B C A B BQ. 11 12 13 14 15 16A. A C D D C CQ. 1 2 3 4 5A. 3 5 2 0 3
SECTION-IV
SECTION-I
Let point be P(a2 + 2,a)
p
dy 1 9 11 P ,
dx 2 4 2æ ö æ ö= Þ a = Þç ÷ ç ÷è ø è ø
5. Ans. (C)
April has 30 days. For minimum probability,Saturday & Sundays should be maximum = 5Sundays & 5 Saturdays
Þ Sunday's dates Þ 2,9,16,23,30
Þ Saturday's dates Þ 1,8,15,22,29
Þ Prime dates Þ 2,3,5,7,11,13,17,19,23,29
Maximum number of days he does'nt hunt
Þ 5 + 5 + 10 – 3 = 17
hence minimum probability is 17 13
130 30
- =
6. Ans. (B)
Domain of g(x)
8x – x2 – 15 > 0 Þ [3,5]
ƒ'(x) = 3x– x2 – 4
ƒ'(x) < 0 " x Î [3,5]
ƒ(x)max = ƒ(3), ƒ(x)min = ƒ(5)
( )52 3
0
3z z 145ƒ 5 4z
2 3 6= - - = -
ƒ(3) = 456
-
Þ ƒ(3) – ƒ(5) = 503
01CT313088KOTA / HS - 2/12
09-05-2014TARGET : JEE 2014TM
Path to success KOTA (RAJASTHAN)
PAPER – 1
7. Ans. (C)
( )42
23
1
1 x 1x x x 1
x x
æ ö+æ ö+ + +ç ÷ç ÷è øè ø
( ) ( )5 42 4
13
x 1 x 1
x
+ +
Þ cofficient of x22 in (x2 + 1)5(x4+1)4
Þ 5Cr.x2r.4Crx4k
r = 3 & k = 4 r = 5 & k = 35C3.4C4
5C5.4C3
Þ 10 + 4 = 14
8. Ans. (A)ƒ(x) – x2 = (x – 1) (x – 2) (x – 3) (x – a)
ƒ(5) = 25 + 4.3.2 (5 – a)
ƒ(–1) = 1 + 2.3.4(1 + a)
ƒ(5) + ƒ(–1) = 26 + 24.6 = 170
9. Ans. (B)
9 2 8 3 3 3dy dy3x y 9x y y x 0
dx dx+ + - =
3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 03(x3y)2d(x3y) + y3dy – x3dx = 0
( )4 433 y x
x y C4 4
+ - =
y(0) = 0 Þ C = 0
4(x3y)3 + y4 – x4 = 0
10. Ans. (B)For infinite solutions D = Dx = Dy = Dz = 0
& No two planes are parallel
1 2 31
D 2 3 2sin 0 sin2
3 5 2
= - q = Þ q =
two possibilities q = 30º, 150º
So secq is either 2
3 or
2
3-
For 2
sec3
q = -
x
2 2 3
D 9 3 1
7 5 2
= - (Dx ¹ 0)
for secq = 2
3, Dx = Dy = Dz = 0
& No two planes are parallelhence only 1 value.
Paragraph for Question 11 & 12
Let ˆ ˆ ˆr xi yj zk= + +r
( )ˆ ˆ ˆr. 8i 10 j 6k r 8- + - - =r r
Þ x2 + y2 + z2 + 8x – 10y + 6z + 8 = 0
Þ Sphere having centre (–4,5,–3)
and radius 42
11. Ans. (A)
(3,4,–5)(1,1,–2)r
r
a 54 42, b 54 42= - = +
Þ (a2 + b2) = 19212. Ans. (C)
Normal vectors of plane
ˆ ˆ ˆ(5i 4 j k)- + ; through (1,1,–2)
Þ (x – 1)5 + (y – 1)(–4) + (z + 2)1 = 05x – 4y + z + 1 = 0
Paragraph for Question 13 & 14
13. Ans. (D)Number of H > No. of TÞ H are Þ 1008, 1009, ....... 2014
2014 2014 20141008 1009 2014
2014 2014 2014
C C CP ......
2 2 2= + +
2014 2014 20141008 1009 2014
2014
C C ..... CP
2+ +
=
2014 20141007
2014
2 CP
2.2-
=
0141007
2015
C1P
2 2
2
= -
KOTA / HS - 3/1201CT313088
LEADER & ENTHUSIAST COURSE 09-05-2014TM
Path to success KOTA (RAJASTHAN)
PAPER – 1
14. Ans. (D)
990 1024Þ 990C0.1024C0 +
990C1.1024C1+ ......+ ....... 990C990.1024C990
—————————————— 22014
Þ2014
9902014
C2
Paragraph for Question 15 & 1615. Ans. (C)
O
P
Q R(h,k)
Equation of PQ : xh 9yk1
9+
=
Homoginise C2 with PQ2
2 2 xh 9ykx 9y 9 0
9+æ ö+ - =ç ÷
è ø2
2 2 2hx 1 y (9 9k ) 18hk 0
9
æ ö- + - - =ç ÷
è øabove equation is same as C1
2 29 h 9(1 k )9 9
- -= Þ x2 = 9y2
Þ (x + 3y)(x – 3y) = 016. Ans. (C)
If OPRQ is a cyclic quadrilateralÞ ÐPOQ = 45°
Þ
2
12 9
4tan 4510
-
l-
= ° Þ 2 136l =
SECTION-IV1. Ans. 3
aeee = ah, aheh = ae
2 2e he e+ =
2 2e h2 2h e
a a2
a a+ ³
but can't be 2(ae ¹ ah) Þ 32. Ans. 5 A
4A5
A3
A2
A1
A12
A11A
10
A9
A8
A7
A6
both the diagonals of rectangle are equal.
There are six diameters of polygon all are equal
So to form rectangle we take any of the two.
Þ 6C2 = 15 = k
k5
3=
3. Ans. 2Let circle be (x – a)2 + (y – a)2 = 2(a – 1)2
eq. of common chord
Þ S1 – S2 = 0
(4x – 6y + 7) + a(2x + 2y – 2) = 0
that fixed point lies on 4x – 6y + 7 = 0
x + y = 2 Þ l + m = 2
4. Ans. 0
( ) ( ) ( )3 3 31 1 2 2 3 3p , p , p , p , p , p
O, G, H are collinear
Þ
31 1
32 2
33 3
1 p p
1 p p 0
1 p p
=
Þ (p1 – p2)(p2 – p3)(p3 – p1)(p1 + p2 + p3)
Þ p1 + p2 + p3 = 0 (p1 ¹ p2 ¹ p3)
m1 = 21p
3k 1
k 1 2 3k 1
( 1) m p | p | p-
=
- = - +å = 0
5. Ans. 3
O
y=ƒ(x)
y=xy=x2
1 2
1/2
3/2
1
1ƒ(x) {x}
2= +
1 22
0 1
1A x dx x dx
2æ ö= + +ç ÷è øò ò
1 22 3
0 1
x x x 8 1 101
2 2 3 3 3 3= + + = + - =
[A] = 3
01CT313088KOTA / HS - 4/12
09-05-2014TARGET : JEE 2014TM
Path to success KOTA (RAJASTHAN)
PAPER – 1
SOLUTION
PART-2 : PHYSICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10A. B A C D B A D A C AQ. 11 12 13 14 15 16A. A A A D A CQ. 1 2 3 4 5A. 5 8 1 8 8
SECTION-I
SECTION-IV
SECTION–I
1. Ans. (B)
Sol. 00
2E
2 dl
=pÎ
0 0del = pÎ
So ep = 0 2
dE
4rl
2. Ans. (A)
3. Ans. (C)
Sol. At terminal stage, torque applied on smaller discby rope = mga
current to the disc = 2B r
2Rw
(where w is terminal
angular velcity)
torque applied by magnetic field = 2 4B r4Rw
So, 2 4B r
mga4Rw
=
w = 100 rad/sec
4. Ans. (D)
Sol. Induced current in ring will create such amagnetic field which will repel bar magnet.
5. Ans. (B)
Sol. Electric field lines originate from positive chargeand terminate at negative charge
6. Ans. (A)
Sol. For the electric field, 2 2
0 0E x E yV K
2 2= - +
l l
from energy conservation2 2
20 0QE QE1k Mv 2 K
2 2 8- + + = - ´ +
l l
l l
v = 2 m/s
7. Ans. (D)
Sol.dm
2a v 5 1dt
= = ´
8. Ans. (A)
9. Ans. (C)
Sol.t
RCq Qe-
=
where C = 4pÎ0r
10. Ans. (A)
Sol. J – J' = mv
J' = 2mv
Jv
3mÞ =
11. Ans. (A)
12. Ans. (A)
Sol. (11 and 12)
In the case of water, (5 × 103) × 10 × 15 × 3600
= V × rw × S
w × (100 – 20)
In case of oil, (5 × 103) × 10 × 3.15 × 3600
= V × rO × S
O × (60 – 20)
After solving SO/S
w = 0.6 and
V × r0 = 5.7 × 103 kg
13. Ans. (A)
Sol. Using calorimetery principal
150 × 540 + 150 × 1 × (100 – T)
= (250 + 50) × 1 × (T – (–10))
+ 800 × 12
× (0– (–10)) + 800 × 80
+ 800 × 1 × (T – 0°)
KOTA / HS - 5/1201CT313088
LEADER & ENTHUSIAST COURSE 09-05-2014TM
Path to success KOTA (RAJASTHAN)
PAPER – 1
By solving it t = 20°C hence steam and ice bothbecome water so amount of water = 150 + 800
14. Ans. (D)
Sol. rB (20°C) = r
B (0°C)
11
10æ ö-ç ÷è ø
Hence density is decreased to 1
10 of its density
at 0°C
15. Ans. (A)16. Ans. (C)
SECTION–IV
1. Ans. 5
Sol. 2 21 2
1 1 1R
n n
é ù= -ê úl ë û
n1 = 3 for Paschen series
2. Ans. 8
Sol. 42 1Q T T 2Tµ Þ =
l1T
1 = l
2T
2
12 3000Å
2l
Þ l = =
K.Emax
= 2
hcW-
l
where K.Emax
= 13.6 2 2
1 12 4
é ù-ê úë û eV
3. Ans. 1
Sol. p(0.02)2 (1) = ( )2 4x0.01 2g
5æ öp ç ÷è ø
4. Ans. 85. Ans. 8
Sol. P = 700 × 103 × 1.6 × 10–19 × dtdN
= 10 × 10–3;
dtdN
= 167
11010
14
2
´´-
-
= 2.11
1012 = l N
0
l = 86400142n
´l
Þ N0 = 2n2.11
108640014 12
l´´
= 160 × 1015
SOLUTION
PART-3 : CHEMISTRY ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10A. B A C C C A B A C DQ. 11 12 13 14 15 16A. B A C B C CQ. 1 2 3 4 5A. 3 7 4 5 3
SECTION-I
SECTION-IV
SECTION-I
1. Ans. (B)2. Ans. (A)3. Ans. (C)
4. Ans. (C)
5. Ans. (C)
6. Ans. (A)
7. Ans. (B)
8. Ans. (A)
9. Ans. (C)
10. Ans. (D)
11. Ans. (B)
12. Ans. (A)
13. Ans. (C)
14. Ans. (B)
15. Ans. (C)
16. Ans. (C)
SECTION-IV1. Ans. 31. Ans. (25) [OMR Ans. 7]
3. Ans. 4
4. Ans. 5
5. Ans. 3
PAPER-2
PAPER CODETM
Path to success KOTA (RAJASTHAN)0 1 C T 3 1 3 0 8 9
PATTERN : JEE (Advanced)LEADER & ENTHUSIAST COURSE
Date : 09 - 05 - 2014TARGET : JEE 2014
SCORE-II : TEST # 06
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, INDRA VIHAR, KOTA-324005PHONE : +91 - 744 - 2436001, Fax : +91-744-2435003, E-mail: [email protected] Website: www.allen.ac.inKOTA / HS - 6/12
SOLUTION
PART-1 : MATHEMATICS ANSWER KEY
SECTION-I1. Ans. (A,B)
–ae = –4& 2b = 4A
(–4,0)
O
Focus of parabolais (–4 + A,0)Þ A = 4Þ b = 2A = 8 & ae = 4
b2 = a2(1 – e2) Þ a2 = 80 & 1
e5
=
Directrix of parabola is x + 4 + A = 0or x = –8
2. Ans. (A,B,C)
30º30º
60º
A
B CDÐB = 60ºA + C = 120º Þ A = 90º & C = 30º
( ) A A B Cr s a tan 4Rsin sin sin
2 s 2 2 2D
= - = =
Þ b c a bc 3 1
r R2 a b c 2
æ ö+ - -= = = ç ÷ç ÷+ + è ø
3. Ans. (B,C)Apply L'Hôpitals rule on L.H.S.
Þ 2
a x
ƒ(x) xƒ '(a)lim x
2a®
-=
-Þ ƒ(x) – xƒ'(x) = –2x3
Þ ƒ'(x) – 2ƒ(x)
2xx
=
Q. 1 2 3 4 5 6 7 8 9 10A. A,B A,B,C B,C A,B,C A,C A,B,C,D A,B A,C A,B,C,D A,B,D
A B C D A B C DP,Q,R,S,T R,S P,Q,R,T P P Q Q,T R,S
Q. 1 2 3 4A. 006 000 032 003
SECTION-I
SECTION-II Q.1
SECTION-III
Q.2
I.F = 1
dxx 1
ex
-ò =
Þ 21 1ƒ(x). 2x . dx
x x= ò
ƒ(x) = x3 + cxQ ƒ(1) = 0 Þ ƒ(x) = x3 – x
4. Ans. (A,B,C)1 1
1 1
sin x cos x ; 1 x 0y
sin x cos x ; 0 x 1
- -
- -
ì- + - £ £ï= í+ £ £ïî
12sin x ; 1 x 02y
; 0 x 12
-pì - - £ £ïï= ípï £ £ïî
3y ,
2 2p pé ùÎ ê úë û
5. Ans. (A,C)Possible when two circles touch each other &having same radii
r1 = r2 Þ 22 2a b= - ...(i)c1c2 = r1 + r2
Þ ( ) ( )2 2a 1 a 1 2 2+ + + = ...(ii)
Þ a = –3 & b = 16
(2,2)(3,3)(1,1)
L
Line L is x + y = 4
KOTA / HS - 7/1201CT313089
LEADER & ENTHUSIAST COURSE 09-05-2014TM
Path to success KOTA (RAJASTHAN)
PAPER – 2
6. Ans. (A,B,C,D)
y | x 1 |= - and y | x 1 |= -
0 1 2
1
y =(x–1)12
y = x–12 Ö
y =(x–1)32
y = –x4 Ö1
both curves intersect at (0, 1), (1, 0) and (2, 1)
required area =2
2 1
1
2 (y y )dx-ò
Þ A = 2
2
1
2 [ x 1 (x 1) ]dx- - -ò .......(1)
233/ 2
1
2 (x 1) 22 (x 1)
3 3 3
é ù-= - - =ê ú
ë û
Apply b b
a a
ƒ(x)dx ƒ(a b x)dx= + -ò ò in (1)
Þ2
2
1
A 2 [ 2 x (2 x) ]dx= - - -òRequired area can also be calculated as
1
4 3
0
2 (y y )dx-ò1
2
0
2 1 x (x 1) dxé ù= - - -ë ûò
Apply b b
a a
ƒ(x)dx ƒ(a b x)dx= + -ò ò
Þ1
2
0
A 2 x x dxé ù= -ë ûò7. Ans. (A,B)
Q opp
dym
dx=
Þ4 3
3 2h h h4h 3 h
h+ a + b
= + a + b
Þ h4 + ah3 + bh = 4h4 + 3ah3 + bh3h4 + 2ah3 = 0
Þ h3(3h + 2a) = 0 Þ h = 0, 2
h3a
= -
Þ a ¹ 0 and b Î Rfor two different values of h.
8. Ans. (A,C)
( ) 4
3 3 3 1P A
3 3´ ´
= =
( ) 4
3 1P B
3 27= =
A Ç B is 0 0
0 0
é ùê úë û
only
Þ P(A Ç B) = 4
13
9. Ans. (A,B,C,D)
|z|max = OA 2 1= + p/4p/4
q
(1,2)A(z )1
B(z )2
(1,0)O
(1,1)P
|z|min = OB 2 1 3= + =maximum value of amp(z)
is corresponding to point A i.e. 4p
minimum value of amp(z) is corresponding to
point B i.e. 1 1tan
4 4 2-p p
- q = -
(Q 1
tan2
q = in triangle OPB)
10. Ans. (A,B,D)
1ˆ ˆˆ ˆ ˆ ˆa.b b.c c.a2
= = =
Let ˆ ˆˆ ˆ ˆ ˆa b x, b c y,c a z+ = + = + =r r r
v (x y) z (x.z)y (y.z)x= ´ ´ = -r r r r r r r r r r
{ }{ } { }{ }ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ(a b).(c a) b c (b c).(c a) a b= + + + - + + +
{ } { }5 5ˆ ˆˆ ˆb c b a2 2
= + - + { }5 ˆ ˆc a2
= -
Þ vr
is coplanar with a and cÞ D is correct.
(A)5 ˆ ˆv c a2
= -r 5 11 1 2.
2 2= + -
52
=
(B)5 1 5ˆv.a 12 2 4
æ ö= - = -ç ÷è ø
r,
5 1 1ˆv.b 02 2 2
æ ö= - =ç ÷è ø
r,
5 1 5ˆv.c 12 2 4
æ ö= - =ç ÷è ø
r
01CT313089KOTA / HS - 8/12
09-05-2014TARGET : JEE 2014TM
Path to success KOTA (RAJASTHAN)
PAPER – 2
Þ ˆˆ ˆv.(a b c) 0+ + =r
Þ vr is perpendicular to ˆˆ ˆa b c+ +
(C)2
1 11
2 21 1ˆˆ ˆa b c 1 02 21 1
12 2
é ù = ¹ë û
SECTION – II
1. Ans. (A)®(P,Q,R,S,T); (B)®(R,S);
(C)®(P,Q,R,T); (D)®(P)
(A) 2x2 + mx + 6 = 0
H.M. of a & b =2 12
0m
ab= - <
a + b
(B) 72n+1 = 7(50 – 1)n
= 7 (nC050n – nC
150n+1+......+(–1)n nC
n)
( )7 50 1 : n even. N
7(50 1) : n odd
ì a += a Îí
a -î
Þ Remainder 7 : n even
18 : n odd
ì= í
î
(C) –10 < 2[x] – 5 < 10
5 15[x]
2 2- £ £ Þ n Î [–2,8)
(D) 2
1 1/ xI dx
x+
= ò put
22
1 11 t dx 2tdt
x x+ = Þ = -
Þ 3/ 2
2 2 1I 2 t dt 1 c
3 xæ ö= - = - + +ç ÷è øò
Þ a = –2; b = 3 : g = –1
2. Ans. (A)®(P); (B)®(Q); (C)®(Q,T);(D)®(R,S)
(A) Apply L o pital rule
( )( )ax
axx 0
ae blim 0
e bx c.coscx®
-=
-
Þ a b
0c-
= Þ a = b
(B) TEETWENTY
Þ TETET, W,N,E,Y
Þ Number of reqd. words =5! = 120
(C)2 |sin x|
|sinx| |cos |0
eI dx
e e
p
a=+ò
|sin x|
|sin x| |cosx|0
e2 dx
e e
p
=+ò
/ 2 sin x
sin x cosx0
e2.2 dx
e e
p
=+ò
Þ / 2 sinx cosx
sinx cosx0
e e2I 4 dx
e e
p +=
+òÞ I = p
(D) sinq(3 – 2cos2q) + 1 = 6sin2qÞ sinq(4 sin2q + 1) + 1 – 6sin2q = 0Þ 4sin3q – 6sin2q + sinq + 1 = 0
(sinq – 1) (4 sin2q – 2sinq – 1) = 0
Þ 2
sin 1 cos2 sin 0
14sin 2sin 1 0 cos2 sin
2
q = Þ q + q =ìïí
q - q - = Þ q + q =ïî
SECTION-III1. Ans. 006
AAT = IÞ a2 + b2 + c2 = 1 & ab + bc + ca = 0possible when a = ±1, b = 0, c = 0 ora = 0, b = ±1, c = 0 ora = 0, b = 0, c = ±1
2. Ans. 000y = –xey/x
y / x y / x2
xy ' y.1y ' x e e
x
é ù-ì ü= - -í ýê úî þë û
y / x xy ' y
e 1x-é ù= - +ê úë û
KOTA / HS - 9/1201CT313089
LEADER & ENTHUSIAST COURSE 09-05-2014TM
Path to success KOTA (RAJASTHAN)
PAPER – 2
y yy ' y ' 1
x xé ù= - +ê úë û
Þy y y
y ' 1 1x x x
æ ö æ ö- = -ç ÷ ç ÷è ø è ø
Þy
y 'x
=y
1x
ì ü¹í ýî þQ
Þ 2 2
yx. yxy ' y xy '' 0
x x
--= = =
3. Ans. 032
A(1,–1,2) C
B(0,0,0)D
x–y–2=0
AB 6=uuur
, ˆ ˆ ˆAB i j 2k= - + -uuur
C is foot of perpendicular from (0, 0, 0)
upon y = x – 2
Let C(a, b, g)
Þ0 0 0 0 0 2
1 –1 0 2a - b - g - - -æ ö= = = -ç ÷
è ø
Þ coordinates of C are (1, –1, 0)
Þ | AC | 2 | AD |= =uuur uuur
Þ AC and AD are sides of square which isbase of parallelopiped.
q
B
A DC
i – j = n
V = Base area × height
= (AC)(AD) ˆAB.nuuur
= ˆ ˆ(i j)ˆ ˆ ˆ2 2 ( i j 2k).
2
-´ ´ - + -
= 4 | 1 1 |
2
- -4 2=
4. Ans. 003
x.ƒ(x) – 0.1 = sinx
ƒ(b)b1
a ƒ(a)
ƒ(x)dx ƒ (x)dx bƒ(b) aƒ(a)-ì üï ï+ = -í ýï ïî þ
ò òQ
Þsin x
ƒ(x)x
=
.1 / 2ƒ
6 / 6p pæ öp =ç ÷ pè ø
= 3
Alter :
Apply Newton - Leibnitz theorem
Þ ƒ(x) + g(ƒ(x)).ƒ'(x) = cosx
Þ ƒ(x) + x.ƒ'(x) = cosx
(Q g(x) is inverse of ƒ(x))
Þ ( )dxƒ(x) cosx
dx=
Þ xƒ(x) = sinx + C
Q ƒ(0) = 1 Þsin x
ƒ(x)x
=
01CT313089KOTA / HS - 10/12
09-05-2014TARGET : JEE 2014TM
Path to success KOTA (RAJASTHAN)
PAPER – 2
SECTION–I1. Ans. (C, D)Sol. Stress is the property of material.
load = stress × cross-sectional area2. Ans. (A, B)
Sol. ( )mg
am M
=+
and Mg – N = Ma
3. Ans. (A, C)4. Ans. (A, B, C)Sol. Fm = 2 (Bi) (AC)
VB = VO = VD5. Ans. (A, D)6. Ans. (A, B, C, D)Sol. For convex lens
1 1 1f v u
= - m1 – m
2 =
( )( )2 2 u v v uv uuv uv
- +-= ...(ii)
1 u vf uv
-= ...(i) x = v – u ...(iii)
Using (i) & (ii) & (iii)
1 2
xf
m m=
-7. Ans. (A, D)8. Ans. (A, C)
Sol. The force exerted = I
AC
´
= 3
98
0.84 10100 0.28 10 N
3 10
--´
´ = ´´
The no. of photons per unit volume = 6I
7 10C h
= ´´ n
9. Ans. (B)
Sol. En = – 13.6 2
2
Zn
10. Ans. (A, B, C)
Sol.( )( )
2
1 2max2
min 1 2
a aII a a
+=
-
SOLUTION
PART-2 : PHYSICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10A. C,D A,B A,C A,B,C A,D A,B,C,D A,D A,C B A,B,C
A B C D A B C DR P Q S R,T Q,S P R
Q. 1 2 3 4A. 004 003 002 002
SECTION-III
Q.2
SECTION-I
SECTION-II Q.1
KOTA / HS - 11/1201CT313089
LEADER & ENTHUSIAST COURSE 09-05-2014TM
Path to success KOTA (RAJASTHAN)
PAPER – 2
1 1
2 2
a I 4 2a I 9 3
= = =
max
min
I25
I=
If idential paper is pasted acros second slit shifted C.B.F. will back in central point shift = (µ – 1) tDd
fringe width = n D
dl
n = ( )1 tshift
15fringe width
m -= =
lSECTION–II
1. Ans. (A) ® (R); (B) ® (P); (C) ® (Q); (D) ® (S)2. Ans. (A) ® (R, T); (B) ® (Q, S); (C) ® (P); (D) ® (R)Sol. F
net on the strip r
0vg – rvg = r
0vg – r
0vg (1 – 4y) = –4r
0vgy
\ a = 0
0
4 vgy4gy
vr
- = -r
Since a µ –y Þ SHM
\ Time period = 2 2
1sec4g
p p= =
wWhen the strip is in front of lower slit, central maxima shifts below point C & vice-versa.
SECTION–III1. Ans. 004Sol. mA = mB
rAvB = rBvB2
A B2
B A
rr
r=
rSo, mAsADq = mBsBDqri × vA sA = ri vBsB, depth of pentration of ice
A
B
h 4h 1
=
2. Ans. 003Sol. We have 10.2 = w + Kmax, 1 and ...(i)
10.2 Z2 = W + Kmax, 2 ...(ii)
Also lde-Broglie = h h
2mK=
r
\ 1 2
2 12 1
K2.3 K 5.25K
Kl
= = Þ =l ...(iii)
and 10.2 Z2 = ebergy corresponding to longest wavelength of the Lyman series = 3 × 13.6Þ Z = 2\ From equations (i), (ii) and (iii)
W = 3 eV.
01CT313089KOTA / HS - 12/12
09-05-2014TARGET : JEE 2014TM
Path to success KOTA (RAJASTHAN)
PAPER – 2
3. Ans. 002
Sol.i
r1r2
e
sin i = 21 21 sin r cos rm - -
since light strike grazingly so i = 90° r1 and r2 = 60°
so, µ = 2
4. Ans. 002
Sol. 22
113.6 Z 1 52.224
né ù- =ê úë û
( )2
2 2
1 113.6 Z 1.224
nn 1
é ùê ú- =ê ú-ë û
SOLUTION
PART-3 : CHEMISTRY ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10A. A, D A, B, D C, D A, B, C A,B,C,D B, C, D B, D B, C A, B, D B, D
A B C D A B C DR P, S Q, T P,Q,R,S Q, S P, R, S T Q, T
Q. 1 2 3 4A. 146 007 014 006
SECTION-III
Q.2
SECTION-I
SECTION-II Q.1
SECTION-I
1. Ans. (A,D)
2. Ans. (A, B, D)
3. Ans. (C, D)
4. Ans. (A,B,C)
5. Ans. (A,B,C,D)
6. Ans. (B, C, D)
7. Ans. (B, D)
8. Ans. (B,C)
9. Ans. (A,B,D)
10. Ans. (B,D)
SECTION-II
1. Ans. (A) ® (R); (B) ® (P, S); (C) ® (Q, T);(D) ® (P, Q, R, S)
2. Ans. (A) ® (Q, S); (B) ® (P, R, S);(C) ® (T); (D) ® (Q, T)
SECTION-III
1. Ans. 146
2. Ans. (6.972 kJ) [OMR Ans. 007]
3. Ans. 014
4. Ans. 006