mock test-02 (jee) solutions test-02 (jee) solutions p… · mock test-02 (jee) solutions 1. (a)...
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(1)
SAFE HANDS & IIT-ian's PACE
MOCK TEST-02 (JEE) SOLUTIONS
1. (a) Sol. Let the retardation produced by instantaneous opposition
= v (where is a constant)
Net instantaneous acceleration = g - v
i.e., dv/dt = (g - v)
Integrating,
t
0
v
0
dtvg
dv
teg
vgeit
g
vg
.,.ln
i.e., v =
g (1 – e-t)
i.e., te1g
dt
dS
i.e., dS = te1g
dt
Integrating,
t
0
t
s
0
dte1g
dS
=
t
0
t
t
0
dteg
dtg
or S = 2
t
2
ge
gt
g
or S = 2
g
(e-t – 1 ) +
gt
2. (b)
Sol. Velocity before strike u = 2gh Component of acceleration along the inclined plane = g sin
and the perpendicular component = g cos Using S = ut + 1/2 at2 For vertical direction we get,
0 = v cos t – 1/2 g cost2 and For horizontal direction
x = u sint + 1/2 g sin t2
= u sin
g
u2Qt
g
u2sing
2
1
g
u22
= g
sinu4
g
sinu2
g
sinu2 222
= 4 × g
singh2 = 8h sin
3. (c) Sol. Due to buoyant force on the aluminium block the reading of spring balance A will be less than 2 kg but it increase the reading of balance B. 4. (c)
Sol. Thrust on the rocket
5. (c)
Sol. Moment of inertia of rod AB about its centre and
perpendicular to the length = = I
Now moment of inertia of the rod about the axis which is passing through O and perpendicular to the plane of
hexagon Irod= [From the theorem of parallel
axes]
Now the moment of inertia of system Isystem
=
Isystem = 5 (12 I) = 60 I [As ] 6. (a)
Sol. Young's modulus (As Y, L and F are
constant) From the graph it is clear that for same load elongation is minimum for graph OD. As elongation (l) is minimum therefore area of cross-section (A) is maximum. So thickest wire is represented by OD. 7. (c)
Sol. Specific gravity of alloy of water Density
alloyof Density
of water densityalloyof Volume
alloyof Mass
wp
mm
mm
2
2
1
1
21
2
2
1
1
21
2
2
1
1
21
// s
m
s
m
mm
mm
mm
ww
of water density
substanceof density substanceof gravityspecific As
8. (b)
Sol. At V = V0, P =2
P0
Ti =nR
PV=
R
)V(2
P0
0
R2
VP 00 (n = 1)
and at V = 2V0, P =5
P4 0 dt
udmF )40(105 4
N6102
12
2ml
Iml 122
22
12mx
ml
6
5
2
3
12
22
2 mllm
ml
6
566
2
rod
mlI
25ml
Iml 122
Al
FLY
Al
1
l l
l
x
O
B A
(2)
Tf =nR
PV=
R
5
P4)V2( 0
0
=R5
VP8 00
T = Tf – Ti =
2
1
5
8
R
VP 00 =R10
VP11 00
9. (a) Sol. Temp. of source T1 = 100 + 273 = 373 K. Temp of sink T2 = 0 + 273 = 273 K.
Efficiency of carnot engine = 1 –1
2
T
T= 1 –
373
273 =
373
100
To increase =373
100+
373
100×
5
1=
373
120
= 1 –373
T2 =373
120
or T2 =373
120373× 373 = 253 K.
.K253tempksinNew
10. (b) Sol.
6° 3°
A
B
C
Time taken by pendulum in going from A to B
= 4
T where T = 2
g
Time taken by pendulum in going from B to C
= 12
T
Time period of pendulum
=
12
T
4
T2
= 3
T2=
3
2.
5
=
15
2sec
Altier :
30°
T = 360
240. T
= T3
2
11. (d)
Sol. 2r4
P
= I for an isotropic point sound source.
P = I.4pr2
= (0.008w/m2)(4..10
2)
= 10 watt. 12. (b)
Potential at the centre of first ring
220
2
0
1
44 RR
Q
R
QVA
Potential at the centre of second ring
220
1
0
2
44 RR
Q
R
QVB
Potential difference between the
two centres 24
))(12(
0
21
R
QQVV BA
Workdone24
))(12(
0
21
R
QQqW
13. (c) Initially the potential at centre of sphere is
VC =
After the sphere grounded, potential at centre becomes zero. Let the net charge on sphere finally be q.
+ = 0 or q =
The charge flowing out of sphere is
14. (c)
Here angle between and is 0º
U = – PE cos 0 = – P = –
15. (d) Sol. Magnetic field at P
Due to wire 1, d
B)8(2
.4
01
and due to wire 2, d
B)16(2
.4
02
2
0
2
022
21
12.
4
16.
4
ddBBBnet
dd
00 510
2
4
16. (b)
x
Q3
4
1
x
Q2
4
1
x
Q
4
1
000
04
1
r
q
x
Q3
4
1
0 x
Qr3
x
Qr3
P
E
+ –
–q
P
E
q
0 0
P
8 A
6 A
(0, 0, d)
X
Y
B2
B1
P
(3)
Sol. By using
2/3
2
2
1
r
x
B
B
axis
centre ,
given r = R and centreaxis BB8
1
2/3
2
2
18
R
x
32/1
2
22 1)2(
R
x
2/1
2
2
12
R
x
2
2
14R
x Rx 3
17. (b) Sol. When the slider moves towards B, the resistance of the circuit (bigger loop) decreases. Therefore, the current in the bigger loop increases. The increasing current results in
increasing flux ( i) linked by the smaller coil. Consequently, induced emf will be generated in the smaller loop causing an induced current so as to oppose the increase in flux. Therefore the current flows anticlockwise in the inner loop. Hence (B) is correct. 18. (b) Sol.
1 sin 600 = sin 300
= 19. (c)
Sol. Y = BA
111
001
010
000
YBA
= AND
20. (b)
Sol. eVs =
= 3.1 – 1.9 eVs = 1.2 eV Vs = 1.2 V
21. Ans. 5 Sol. m ur = m (v0 + v) a
u = as v0 =
also mu2 – = m (v0 +v)
2 –
22. Ans. 6 Sol.
We can assume that 9V battery with 1Ω internal resistance is
in parallel with 0V battery with internal resistance 2Ω.
Eeq =
2
1
1
12
0
1
9
= 9 2
3= 6 volt
4 1 1 3 6V B A
3V
P.d. across 2 = 6 V
23. Ans. 2 Sol.
IL
IC INet = IL – IC
= 0.8 – 0.6 = 0.2 Amp.
24. Ans. 2 Sol. Shortest wavelength of Brackett corresponds to n = 4 and
n = and shortest wavelength of Balmer series corresponds to n
= 2 and n =
(Z2)
16
6.13=
4
6.13 Z = 2
25. Ans. 5
Sol. d = f1 ~ f2
26. (c)
Sol. It is 2, 4, 6-trinitrophenol
27. (d)
Sol.
+ CHCl3 + 3KOH + 3KCl + 3H2O
28. (c)
Sol. In aryl halides the C–X bond has partial double bond
character due to resonance so it will not give SN reaction
29. (b)
Sol.
intra-molecular H-bonding
30. (d)
Sol.
31. (a)
Sol.
30°
30° 60°
=1
3
9.14000
12400
2
e
r4
aGM5
a
GM e
2
1
r
mGMe
2
1
a
mGMe
NH2
CH3
N C
CH3
=
(4)
+
32. (c)
Sol.
+ CH2 = CHCH2Cl
33. (d)
Sol. Follow conditions of geometrical isomerism.
34. (b)
Sol.
(I) +
(II) + H2NOH
(III)
(IV)
+
35. (a)
Sol. (i) 4p, (ii) 4s, (iii) 3d, (iv) 3p
The order of increasing energy,
(iv)< (ii)< (iii)< (i)
36. (a)
Sol. The vapour pressure increase with increase in
intermolecular forces. When the forces are weak, the liquid
has high volatility and maximum vapour pressure diethyl
ether has higher vapoour pressure while water has lowest
vapour pressure.
37. (a)
Sol. Electrical work obtained FnEGW cell0
max
JJ 212300)1.1(965002 kJ3.212 38. (a)
Sol. Adding the first two reactions, we get the third equation
and using the free energy concept, we have 03
02
01 GGG ; FEnFEnFEn 0
33022
011
(n = number of electron involved)
Vn
EnEnE 741.0
3
)900.0(2)424.0(1
3
022
0110
3
39. (b)
Sol. s = )AgCl(
10
m
3
=
)6763(
106.210 6–3
= 2 × 10
–5 (M)
Ksp = ( 2 × 10–5
)2 = 4 × 10
–10.
40. (b)
S o l . 1 M H2SO4 = 2g eq. of H2SO4
hence, y = 2x or x = 2
1 y.
41. (a)
S o l . Wnet = area enclosed
V = P
nRT
VA = 1
R600
1
300R2
VB = 1
R300
2
300R2
; Vc =
1
R400
2
400R2
VD = 1
R800
WAB = –nRT TAln
A
B
V
V= –2R (300)
2
1ln= 600 Rln2
WBC = –2(400 –300) R= – 200 R
WCD = – 2 R(400)ln
C
D
V
V= – 800 Rln2
WAD = –1(600 R – 800 R) = 200 R
W Total dqy = WAB + WBC + WCD + WAD =– 200 Rln2
= – 100 Rln4
42. (c)
S o l . STHG 156.30 molkJH ; 11066.0 KkJmolS ; 0G at
equilibrium; ?T STH or 066.056.30 T
KT 463
43. (a)
Sol. We know that Antipyretics are those compounds which
are applied to reduce the body temperature in fever.
Ex. Aspinic (acyl salicylic acid) paracetamol, phenacetin,
novalgin, & analgin
44. (c)
Sol. Both Be and Al are rendered passive due to formation of
inert, insoluble and impervious oxide layer on their surface.
45. (a)
Fe/Br
2
3AlCl
(5)
Sol. The process is truly adsorption as gangue particles are
wetted with water and sulphide ore particles are wetted with
pine oil.
INEGER TYPE QUESTIONS
46. 6 47. 7 48. 8 49. 6 50. 2
51. (c)
Sol. }4,,|),{( 22 yxZyxyxR
)2,0(),2,0(),1,0)(1,0(),1,1(),0,1(),0,2{( R )}0,2(),1,1(),0,1(
Hence, Domain of }2,1,0,1,2{ R .
52. (d)
Sol. Because, inequality is not applicable for a complex
number.
53. (c)
Sol. tr = r
1 (1 + 2 + 3 ….. + r) =
2
1.
r
1r (r + 1)
S = tr =
n
1r
)1r(2
1 = 115
54. (a)
Sol. Let f(x) = ax2 – bx + 1.
Now f(0) = 1 and roots are imaginary, f(x) > 0 x R
f(–1) = a + b + 1 > 0
55. (c)
Sol. Each group will have m
mn= n students.
Also, there is no distinction between groups.
Required no. of ways = !m),timesm......!n!n(
!mn
= )!m()!n(
!mnm
56. (c)
Sol. Let P the image of P in the line x + 3y = 7.
Then x + 3y = 7 will be perpendicular bisector of PP.
Equation of PP is y – 8 = 3(x-3)
3x –y – 1 = 0.
On sol v ing 3x-y-1 = 0 and x + 3y = 7,we get the
coordinates of M as (1, 2). Since M is mid point of PP’,
coordinates of P are (-1, -4)
57. (a)
Sol. centre(3, –4), radius = 5
Shortest distance = OM – radius
= – 5 = – 5 =
58. (a)
Sol. Centre of conic is
22
,hab
afgh
hab
bghf
Here, ,14a 2h , 11b , 22g , 29f , 71c Centre
22 )2()11)(14(
)29)(14()2)(22(,
)2()11)(14(
)22)(11()29)(2(Centre )3,2(
59. (d)
S o l .
3
2 5dx
xx
xI
Put tx 32 dtdx
dxxx
xdt
tt
tI
2
3
3
2 5
5)(
5
5 and
3
2
3
2
15
52 dxdx
xx
xxI
1][2 22 xI 2/1I
60. (a)
Sol. = x
= dx
= dx
put t = sin x – cos x
dt = (cos x + sin x) dx
also t2 = 1 –sin 2x
I =
= sin–1
t
= sin–1
(sin x –cos x) + c
61. (b)
Sol. We have , I. F. = sin x = sin x = log
sin x P = (log sin x) = cot x.
Hence (B) is the correct answer
62. (a)
Sol.
.
63. (b)
Sol. P(B) = 1 – P ( ) = 1 -
and P(AUB) = P(A) + P(B) – P(AB)
or,
P(A) =
P(A) P(B) =
Hence, A and B are independent.
64. (d)
Sol. The given expression is equal to
= 1+ 4 +1 9 = 15
x+3y = 7
P
M
P(3, 8)
5
25169
5
32
5
7
O(3, –4)
M
xcos
xsin
xsin
xcos
xcosxsin
xsinxcos
2
x2sin
xsinxcos
2 2t1
dt
2
2
dxPe dxP
dx
d
2
4
npq
np
8,2
1,
2
1 npq
256
28
2
1.28
2
1
2
1)2(
8
62
28
CXp
B2
1
2
1
3
1
2
1)A(P
6
5
3
2
)BA(P3
1
2
1x
3
2
2121 3cotcot12tantan1
(6)
65. (c)
Sol.
– sin–1
x
sin–1
x + sin–1
y + sin– 1
z =
sin–1
x = sin–1
y = sin– 1
z =
x = y = z = 1
Also f(p + q) = f(p). f(q) p, q R …(1)
Given f(1) = 1
from (1),
F(1 + 1) = f(1). F(1) f(2) =12 = 1
from (2), f(2 + 1) = f(2) . f(1)
f(3) = 12. 1 = 1
3 = 1
Now given expression = 3 – = 2
66. (d)
Sol. We know that . We have i.e.
and
67. (c)
Sol. The given expression is equal to
68. (b)
Sol. Required distance =
metre
69. (b)
Sol. )34).(34(|34| bababa
baba .24||9||16 22
.122
14324144144
70. (b)
Sol. Angle between two diagonals of a cube = cos–1
3
1
71. Ans. 1
Sol. Tr = tan–1
22
22
)1r(r1
r)1r(
= tan
–1 (r + 1)
2 – tan
–1 r
2
S = Tr = tan–1
(n +1)2 –
4
tan
4)1n(tanlim 21
n= tan
42 = (1)
72. Ans. 1
Sol. 1
73. Ans. 2
Sol.
– +
2
1
f (x) = n2x + 1 – 1 = n2x
f2
1
2
1
, f
2
e
2
e
ne
2
e = 0
Range
0,
2
1 a + b = 2
74. Ans. 9
Sol. Equation of plane is a
z
a
y
a
x = 1
It passes through the point (1,1,1).
So a
1
a
1
a
1 = 1 a = 3
hence A(3, 0, 0), B(0, 3, 0), C(0, 0, 3)
volume of tetrahedron = 6
1
2
9
300
030
003
= 2
9 2 = 9
75. Ans. 2
Sol. z = 8
16
2
122
161
81
62
62
81
)z(|z|
|z||z|
z
z
z
z
arg(z) = – 8 arg )z( 1 – 6arg )z( 2 + 2k, kI
= 8 arg (z1) + 6 arg (z2) + 2k
= 8.6
+ 6.
4
+ 2k =
2
3
3
4
– 2 =
6
5
4 sin = 4 sin 6
5= 2
2
2
2
3
2
3
3
||
)(1
A
AadjA
)(11 AadjK
A
|| AK
110
121
423
K )1(4)1(2)3(3 11429
222
222
22
22
acbbc2
bc2acb
cba
acb
2
Acot
Acos1
Acos1
Acosbc2bc2
bc2Acosb2 2
o15cot60
13
1360
15°
15°
60
x