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SOLSCT140413 - 1
PAPER-1PART-I (Physics)
1. When a gas is ........................................Sol. (B)
When the particle collides with moving piston, they strike withless speed and bounce with greater speed.
2. A rigid container ........................................Sol. (C)
Initially mass of air is 200 100 = 100 gm, finally mass of air is150 100 = 50 gm. As there is a hole in the wall, pressureinside the container will remain constant = P
0
PV = nRT T n
1
as number of moles of gas is halved, the temperature should bedoubled (in K)
Ti= 300 K So T
f= 600 K T = 300 K = 300 C
3. A wire AB of ........................................Sol. (C)
at x = 0, A = maximum, so A(x) = A cos (kx)at t = 0, all the particles are passing through mean position so y(x, t) = (A coskx) sin (t)
4. Two particle A ............................... .........Sol. (B)
vrel
2 = urel
2 + 2 arel
Srel
0 = (15)2 + 2(3) Srel
Srel
=2
75m.
5. Four Students perform ........................................Sol. (C)
1+
0f4
V
2+
0f4
V3
3+ =
0f4
V5
2
= 31
+ 2 3
= 51
+ 4
so second resonance length is more than 31
and third
resonance length is more than 51.
6. In a cylindrical ........................................Sol. (D)
dp = gdh
PP
PP
10h
0h
2
0
gdh)h6100(dp
Pf P
0= 10h 0h3h20h1000
P
f= 105 + 0.3 105 = 1.3 105 P
a
7. For a cyclic process ........................................Sol. (A)
from B C T = constantfrom C A, P = constant
HINTS & SOLUTIONS
FULL SYLLABUS TEST (FST)-XITARGET : JEE (ADVANCED)-2013
V T T1 2
rmsC
1
8. A canon of mass ........................................Sol. (B)
Initially n
=m
k=
1
100= 10 sec1
and v = 22n xA = 10
22 )2/3(1
= 5 m/sec.Velocity of the cannon after firing the bullet can be foundusing momentum conservation
(m) (5) =
2m
(20) +
2m
V
v = 10m/sec.
v = 221n xA 10
=
2
21
2
3A
)2/1(
100
A
1=
2
5m.
9. A uniform solid ........................................
Sol. (C)
Whether the cylinder rolls up or rolls down, the static friction
will always act up the incline and its value will be fr=
2mR1
sinmg
which is less than mg sinSo about point C which is above cm. torque can be balanced.
So about point C, its angular momentum will be conserved.
We can also find the distance of point C from cm. suppose it is
x. Applying torque balance about point C.
(mg sin) (x) =
2
11
sinmg
(R + x)
Solving we get x = 2R
10. On a horizontal ........................................Sol. (A)
R = 22 Nfr R = 222 mgrm = 109 N
Date : 14-04-2013 COURSE : REVISION CLASSES
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SOLSCT140413 - 2
11. Two identical ........................................
Sol. (B)
For equilibrium of upper shpere 2N cos 30 = mg
N =3
mg
For the equilibrium of lower sphere
Breaking force welding strength (limiting friction)N sin 30 (
s) (mg + N cos 30)
Solving we get s
s
33
1
12. A man squatting ........................................
Sol. (D)
Initially he is sitting so Vi= 0 and finally be becomes stationery
so Vf= 0. So graph of his velocity v/s time graph will be roughly
as shown below
As slope of Vt graph will be initially positive and then negative,
so acceleration of the man will also be initially positive and then
negative. Initially when acceleration is +ve, N = m (g + a) so
reading will be greater than mg, then acceleration will become-ve, N = m (g a) so reading of weighing machine will be less
than mg. And finally man becomes stationery so reading of
weighing machine will be equal to mg.
13. In a container ........................................Sol. (B)
To convert 0 ice into 0 water heat required = 1 80 = 80 calTo convert 0 water into 100 water heat required= 1 1 100 = 100 calTo convert 100 water into 100 steam heat required= 1 540 = 540 calTo convert 100 steam into 190 steam heat required
= ncpT =
18
1(4R) 90 = 40 cal
Total heat given = 80 + 100 + 540 + 40 = 760 cal.
14. Velocity of particle ........................................Sol. (AD)
For t (2, 2.5) speed so motion will be acceleratingfor t (2.5, 3) speed so motion will be retarding
15. On a triangular ........................................
Sol. (ABD)
(A) If ma cos = mg sin a = g tan then no friction will act onthe block.
(B) If acceleration towards x direction is more than g cot,then the block will loose the contact with the wedge. So again
no friction will act on the block.
(D) If downward acceleration is greater than g then the block
will again loose the contact with the wedge. So again fr= 0
16. A wedge of mass ........................................
Sol. (BD)
When the particle reaches diametrically opposite point, sup-
pose displacement of the wedge is x till that time.
(Sx)cm
=21
2x21x1
mm
)S(m)S(m
0 =2/mm
)R2x()2/m()x()m(
x =3
2R which will be maximum displacement of the wedge
because at that instant both the particles and the wedge will
come at rest for a moment. In this way the wedge will perform
oscillatory motion.
As extreme to extreme displacement of wedge is3
R2so its
amplitude is3
R
17. A horizontal cylinder ........................................Sol. (AD)
N =R
mu2+ mg (3 cos 2), at = 120,
N = 0 N =R
mu2+ mg(3cos120 2) = 0
u= gR5.3
If the block is moving clockwise, then to cross the point P, theblock has to cross the highest point, so to cross the highest
point u = gR5
18. A Uniform verticle ........................................
Sol. (AD)
The ball will strike the rod with a velocity of
V = )cos1(gR2 V = g .
Applying angular momentum conservation about the hinge point
gm + 0 = 0 +
3
)2(m 2() get =
4
3
g
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SOLSCT140413 - 3
also e = collisionbeforeapproachofVelocity
collisiontheaftersaperationofVelocity
e =
g
0 where =
4
3
g
so e =4
3
19. An elastic rod ........................................Sol. ( A)
If we increase the temperature by T,Compressive force generated is F = YA TFor buckling F = YA T critical load of buckling
(Y) (r2) T 2
43
L
Yr T 2
22
L
r
.
20. An isolated part ........................................Sol. (B)
N1+ N
2= mg ..................(i)
(N1)
2
= (N2) 4
..................(ii)
Solving we get,
N1=
3
mg, N
2=
3
mg2
To avoid buckling,
N2= 2
43Yr
3
mg2
L/2 L/4
cm
mg
N2
N1
21. Velocity of water ........................................Sol. (B)
difference in water level
2
1=
2
=
0f2
V=
1602
320
= 1 m
Pressure difference (P1 P
2) =
wgy
= (103) (10) (1) = 104 Paapplying Bernoulli equation between section (1) and section(2)
P1+
2
1V2 = P
2+
2
1(2V)2
P1 P
2=
2
3V2 104 =
2
3(103)V2 V =
3
20
22. The difference ........................................Sol. ( A)
Balancing pressure at level (1) and (2)
P1
+ wg (h
0+ h) = P
2+
wgh
0+
gh
(P1 P
2) = (
+
w) gh
104 = (11 103 103) (10) h h = 10 cm
23. The horizontal ........................................Ans. (B)
24. The value of ........................................Ans. (A)
Sol.(23 to 24)
Horizontal range =g
h2u
R =g
)y1(2gy2
R = )y1(y2 R will be maximum when R2 will be maximum and R2 will be
maximum when 0dy
)R(d 2
0dy
)R(d 2 = 4(1 2y) = 0 y =
2
1
PART-II (Chemistry)
25. Consider the balanced...............
Sol. (D)
It follows directly form definition of stoichiometry.
26. Given : LiCl . 3NH3(s) LiCl . NH
3(s) + 2NH
3(g) ;
KP
= 9 atm2 ..............
Sol. (D)
Moles of NH3
in vessel =RT
PV=
3000821.0
1.82atm3
= 10 mol
These moles must be present in the vessel before the equilibrium
begins to move backwards and conversion of LiCl . NH3(s) to
LiCl . 3NH3
(s) even begins.
Moles of NH3required for conversion of LiCl . NH
3(s) to LiCl
. 3NH3(s) is 12.
27. The work done in adiabatic compression................
Sol. (B)
q = 0, U = WnC
V(T
2T
1) = P
ext(V
2V
1)
n2
3R (T
2 300) = 2
1
300nR
2
nRT2
2
3(T
2300) = (600T
2) T2 = 420 K
W =nCV(T
2T
1) = 2
2
3 2 (420 300) = 720 cal.
28. Which of the following is the best ...............
Sol. (D)
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SOLSCT140413 - 4
29. Which of the following molecule ...............
Sol. (A)
30. A new flurocarbon of molar mass 102 g mol1 .....................Sol. (A)
Molar mass = 102 gram/mole
P = 650 torr ; T = 77 + 273 = 350 K
Q = i t = 0.25 600 = 150 C
E = Q V = 150 12 = 1800 J
This heat is supplied to the system at constant pressure thats
why this is used for change in enthalpy For vaporisation of 1.8 gram, amount of heat required q =
1800 J
For vaporisation of 102 gram, amount of heat required q
=8.1
1800 102 J
= 102 103 J = 102 KJ/mole
H = U + PVH = U + n
gRT
For determination of U per mol (ng
= 1)
102 (KJ/mol) = U + (1 8.3 350) 103 U = 102 2.9 = 99.1 KJ/mole
31. The correct order of increasing ...............
Sol. (C)
32. Two mole of an ideal gas is expanded .............
Sol. (B)
Ssystem
= nR In1
2
V
V= 2 R In 2 = 11.52 J/K
Ssurrounding
= 310
100041.3 = 11 J/K
Stotal
= + 11.52 11 = + 0.52 J/K
33. ................
Sol. (B)
(A) 2NHC
||O
, when attached to a ring, is a named as
carboxamide.
(B) has higher priority than C N
(C) C N has prefix cyano
34. Which of the following pairs ..............................
Sol. (B)
35. Silane, SiH4, cannot be ..................
Sol. (C)
36. Which of the following pairs consists ................
Sol. (D)
37. The enthalpy of neutralisation of .................
Sol. (C)
Hionization = Hneutralization of WA + SB Hneutralizationof SA + SB
= 56.1 (57.3) = 1.2 KJ eq1
Enthalpy of ionization for making 100% ionization when there is
no ionization at all = 1.5 KJ eq1
% of ionization =5.1
2.15.1 100 = 20%
38. Point out the correct statement(s) ..................
Sol. (BC)
39. Among B2H
6, CH
4, C
2H
6and SiH
4, ...............
Sol. (AD)
40.
Consider an isothermal previously .........................
Sol. (BD)
41. 1 mole each of H2(g) and I2(g) are ...................
Sol. (BCD)
42. Which of the following statement (s) is/are true ..................
Sol. (A,B,C,D)
(A) Due to the formation of metal ion clusters
(B) M + (x + y) NH3 M+(NH3)X + e(NH3)y
(C) due to the formation of metal clusters.
(D) M (NH3)6 true statement
43. 266HB has ................
Sol. (A)
[6 3 + 6 + 2 = 26 e or 13 e pairs (total)] [6 2 = 12 e = 6e
pairs] = 7 e pairs
n B atoms have (n +1) e pairs for cluster bonding.closo structure (wades rules)
44. B4H
10has ............
Sol. (C)
[4 3 + 10 = 22 electron 11 electron pairs] [4] = 7 electronpairs for cluster bonding.
n = 4, (n + 3) electrons pairs are available for cluster bonding,hence arachno structure.
closo structure (wades rules)
45. Lycopene is expected to absorb ..................
Sol. (D)
46. Look at the structure of -carotene ................
Sol. (A)
47. When the bird is vertical, the ..............
Sol. (A)
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SOLSCT140413 - 5
Due to cooling, low pressure created in bird. So due to
capaliary action liquid flows upward.
48. A person purchases the toy and keeps ............
Sol. (A)
Again alcohol is more volatile and high vapour density. so this
process again repeat.
PART-III (Mathematics)
49. If equation sin4x + asin2x + 1 = 0....................Sol. (A) Let t = sin2x t [0, 1]
f(0).f(1) 0 2 + a 0 a (, 2]
50. Given Sn
=
n
0rr2
1, ....................
Sol. (C) Sn
=
2
11
2
11
1n= 2 n2
1, S =
1
11
2
= 2
S Sn
=n2
1 1000
now, 512 < 1000 < 1024 29 < 1000 < 210
if 2n > 1000, then least value of n is 10.
51. The remainder obtained ........................ .Sol. (A) 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33
and n!/15 for n 5 required remainder is same as the remainder obtained bydividing 33 with 15, i.e. equal to 3.
52. Total number of polynomials ........... ..............Sol. (A) We must have
i3 + ai2 + bi + c = 0
and (
i)
3
+ a(
i)
2
+ b(
i) + c = 0 c a + (b 1) i = 0 b = 1, a = cThus total number of such polynomials = 10C
1= 10.
53. If
113
010
121
z
y
x
=
1
2
0
.........................
Sol. (B)
zyx3
y
zy2x
=
1
2
0
y = 2, x + 2y + z = 0 x z = 4, 3x y + z = 1
3x + z = 3 x =4
7, z =
4
9
54.0x
lim [sinx], where [.] .........................
Sol. (C) LHL at x = 0 is 1RHL at x = 0 is 0
so limit doesn't exist
55. A variable line L is drawn.........................Sol. (D) x = rcos, y = rsin
putting in L1
we getL1
rsin= rcos+ 10
OA
1=
10
cossin
Similarly putting the general point of drawn line in the equation of
L2, we get
OB1 =
20cos
sin
rcos= h, rsin= k
we haver
2=
10
cossin +
20
cossin
40 = 3rsin 3rcos 3y 3x = 40
56. In triangle ABC equation ...................... ...Sol. (B) Reflection of P(5, 8) in BC will lie on circumcircle
Clearly P1 (8, 5)
Thus equation of circumcircle is (x 2)2 + (y 3)2
= (8 2)2 + (5 3)2
57. If two different tangents of ..................... ....
Sol. (B) Tangent to y2 = 4x in terms of m is y = mx +m
1
normals to x2 = 4by in terms of m is y = mx + 2b + 2m
b
If these are same lines, thenm
1= 2b + 2m
b
2bm2 m + b = 0For two different tangents, we must have
1 8b2 > 0 |b| a
Similarly3
2m
a+
3
2m
b> c,
3
2m
a+
3
2m
c> b
3
4(ma + mb + mc) > a + b + c ma + mb + mc > 43
(a + b + c)
62. In an arithmetic progression.........................
Sol. (BC) Let the successive terms of the A.P. be a1,a2 ...., a9 a10.
By hypothesis a2 + a4 + a6 + a8 + a10 = 15
i.e., (a + d) + (a + 3d) + ...... + (a + 9d) = 15
i.e. , 5a + 25d = 15 .... (i)
and a1 + a3 + a7 + a9 = 2
112
5a + 20d =2
112
From (i) and (ii)., we get 5d =
2
12 or d =
2
1and a =
2
1
Hence the A.P. is2
1, 1, 1
2
1, 2, .....
63. If a, b, c are in H.P., .........................Sol. (ABCD)
(A) ifacb
a
,
bac
b
,
cba
c
are in H.P..
a
acb ,
b
bca ,
c
cba are in A.P..
a
cba
, b
cba
, c
cba
are in A.P..
a
1,
b
1,
c
1are in A.P..
a, b, c are in H.P.
(B)b
2=
ab
1+
cb
1
2(b2 bc ab + ac) = 2b2 b(a + c) b(a + c) = 2ac
b =ca
ac2
a, b, c are in H.P.
(C) a 2
b,
2
b, c
2
bare in G.P..P.
2
bc
2
ba
4
b2
4
b2= ac
2
b(a + c) +
4
b2
ac =2
)ca(b
b =ca
ac2
a,b,c are in H.P.
(D)cb
a
,
ac
b
,
ba
c
are in H.P..
a
cb ,
b
ac ,
c
ba are in A.P..
a
cba ,
b
cba ,
c
cba are in A.P..
a
1
, b
1
, c
1
are in A.P.. a,b,c are in H.P.
64. Total number of ways of.........................Sol. (CD) Required number = number selections of one or more out
of three 25 paise coins and two 50 paise coins= 4 3 1 = 11
65. A(1, 2) and B(7, 10) .........................
Sol. (BC)
P must lie on the perpendicular bisector of AB.Equation of perpendicular of AB is AB
y 6 = 102
)71((x 4) 3x + 4y 36 = 0
also AB = 10 AL = 5
AH = ALcosec60 =3
10
66. If (acosi, asin
i) ; .........................
Sol. (AB) The given points lie on the circle x2 + y2 = a2 for all (i), sincethe points are the vertices of an equilateral triangle, thecircumcentre and the centroid of the triangle are same.
3
1a (sin
1+ sin
2+ sin
3) = 0
cos1+ cos
2+ cos
3= sin
1+ sin
2+ sin
3= 0
67. Latus rectum of .........................Sol. (C) If (x, y) is the centroid of triangle formed by A(0, 0), P(t)
and Q
t
1is
3x = 2a
2
2
t
1t
3y = 4a
t
1t
Eliminating t we get
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SOLSCT140413 - 7
9y2 = 8a(3x 4a)
or y2 =8a
3
4ax
3
\ length of latus rectum is3
a8
68. Vertex of parabola .........................
Sol. (B) Since y2 =3a4 4ax
3
is the equation of parabola P1
hence vertex is
0,
3
a4
69. (C) In the expansion of ...................... ...70. (B) The number of integral .........................Sol. 69 T
r + 1= 6561C
r(71/3)6561 r
(111/9)r
= 6561Cr.72187 r/3
11r/9
Tr + 1
is rational
If9
rand
3
rare integers
r is a multiple of 9.0 r 6561
9
r= 0,1,2,....., 729
Total terms = 730
70. Tr + 1
= 256Cr(31/2)256 r
(51/8)r
= 256Cr.2128 r/2
5r/8
for Tr + 1
is integer, then 128 2
r0 ,
8
r0
r = 0, 8, 16,...., 256
Number of integral terms = 33
71. If1p
1+
2p
1+
3p
1=
2
1.........................
72.
1p
Acos+
2p
Bcos+
3p
Ccos=
Sol.71 (D)1p
1+
2p
1+
3p
1=
2
1
AM G.M.
3
p
1
p
1
p
1
321
3/1
321 p1.
p1.
p1
or
3
6
1
321 ppp
1
p1
p2
p3 216
72. (B)1p
Acos+
2p
Bcos+
3p
Ccos
=
2
acosA +
2
bcosB +
2
ccosC
=
2
)C2sinB2sinA2(sinR
=
R2 a b c. .
2R 2R 2R
=
2R4
R4=
R
1
PAPER-2PART-I (Physics)
1. A Cylindrical rod ......................................Sol. (D)
for equilibrium Buoancy force (B) = mgNow lets displace the rod by an angle
net
= (mg)
4
sin
net
=
4
mg
net
= K K=4
mg
T = mg
4
2
cm
Where cm
+
2
4m
=
0
cm
= 0
16
m 2.
2. In a region, potential ..................................... .Sol (C)
The potential energy graph can be assumed to be equivalent toa mountain. A particle is released from x 10+ with forwardinitial velocity. The particle will move forward and then returnback. And when it will pass through x = 0, its speed will bemaximum. applying energy conservation between x = 10 to x =0 :
Ki+ U
i= K
f+ U
f
21 (2) (2)2 + 10 =
21 (2) v2 + 0
V = s/m14 .
3. A football player ......................................
Sol. (C)
t
VVa if
=t
)jV()jV(
a
= )i(t
V
+ j
t
V
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SOLSCT140413 - 8
As average acceleration is towards north west, the external
force also should be towards north west. This external forcemust be static friction.
4. A uniform rod of ......................................Sol. ( A)
Applying torque balance about the hinge point
(mg)
2
= (T) ()
T =2
mg= 40 N
So wave velocity will be
V = T
=1.0
40 = 20 m/sec
Time taken to reach the topmost point
t =speed
cetandis=
20
1= 5 102 seconds.
5. A river is 240 m ......................................Sol. (C)
Time to cross the river
t =y
y
V
S=
4
240= 60 seconds
Drifting during this timeS
x= V
xt
Sx
= (3 1) (60)S
x= 120 m
Time to walk along the shore
t =1
120= 120 seconds
Total time to go from A to B = 60 + 120 = 180 seconds = 3minutes.
6. 20 gm of water at ......................................
Sol. (C)
Heat available by water till 0C
= msT = 20 1 30 = 600 calHeat required by ice to reach 0C water
= mL + msT = 5 80 + 5 80 + 5 0.5 10 =425 calThus we have 25 gm of water at 0C and extra heat 600 425
= 175 cal
Hence, 175 = 25 1 TT = 7C (temp. of mixture)
7. A cuboidal vessel of ......................................Sol. (B)
dgha2 = ah2
dgh h = 2m.
8. Surface mass ......................................
Sol. (A)dm = 2 rdr () = (Kr2) 2 rdr
= 1
0
2
2
dmr= 1 kg.m2 .
9. A 1m long tube is ....................... ...............
Sol. (B)
Temperature of the gas remains constant so
PiV
i= P
fV
f
(P0) (0.25) = (P
0gh) (1 h)
solving h = 50 cm
10. Three blocks and ......................................Sol. (C)
a =
428
)g4)5.0(g2()g8(
a = 7
g2m/sec
2
to find the tension at the middle point of the rod, draw the FBDof the part :
7g T = (7) (a)
Solving T = 50 N = 5 kg-force
11. A uniform cubical ......................................Ans. (A) q, s ; (B) p, q, s ; (C) s, t ; (D) p, q, r
Sol.
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SOLSCT140413 - 9
If surface is rough, the breaking force in x direction is zero,
while welding strength (limiting friction) will be non-zero. So
sliding cannot occur. But the cube can topple about the corner
point A.
For topping :-
(Fa) > (mg)
2
a F >
2
mg
(A) If f = 3
mg
then the block will neither slide nor topple. But F
force will create an anticlockwise torque which will lift the right
half and press the left half. So at right half pressure is less and
at left half pressure will be more. So net pressing force (Normal
reaction) will be shifted to the left of the middle point.
(B) F =3
mg2: The block cannot slide, but as the force is
greater than2
mg, So it will start topple about the corner point
A.
(C) F =2mg3
, since F > mg, so the block will be lifted up. so it
cannot slide also.
(D) F =4
mg3, and the surface is smooth. As F >
2
mg, so the
block start toppling about the corner point A. To keep the centre
of mass stationary in x direction, the corner point A will have to
slide toward right.
12. In a container, an ........................ ..............Ans. (A) q,s,t ; (B) p,r,t ; (C) p,r ; (D) p,sSol. (A) Initially the gas is compressed slowly so process is
isothermal. Then the gas is expended, so the process will bealmost adiabatic. So, pv graph will be as shown :
So , pf< p
iand T
f< T
iand W
net= negative
(B) Initially the gas is expanded slowly against atmosphericpressure. So the process is isobaric expansion. Then it iscompressed adiabatic. So the pv graph will be as shown :
So, pf> p
i, T
f> T
iand W
net= negative
(C) As the piston moves very slowly its acceleration a 0, Fnet
0 upwards and downwards forces should be almostbalanced. So PA = P
0A + Kx
Pressure of gas will be P = P0
+A
Kx
As the gas is expanded, pressure will increase. As both P andV are increasing. So temperature will also increase. As volumeis increasing, W
gas= positive.
(D) dQ = 2dU
dQ =
RdT
23n2
R3n
dT/dQ
Molar heat capacity C = 3R = constant, So the process must be
a polytropic process. For a polytropic process C = Cv
+x1
R
= 3R x =3
1
C = Cv+
x1
R
= 3R x =
3
1
So, equation of the polytropic process will be
PV1/3 = constant 3/1v
1P So, V P
Now lets convert this equation into vT
3/1vV
nRT
= constant T V2/3
So, V T
and V So, Wgas will be positive.
13. An object is moving ......................................Ans. 9
Sol. V = 2t (12 6) =dt
ds
ss
0sds =
tt
0tt2 (12 t) dt S = 12t2
3
t2 3
Now =t
S
=t
3
t2t12
32
= 12t 3
t2 2
will be maximum whendt
vd = 0
12 3
t4= 0 t = 9 second
14. Helium gas is ......................................Ans. 8Sol. Assume pressure and volume at state B are P
Band V
Brespec-
tively.from A B process is isothermalso (2P
0) (2V
0) = (P
B) (V
B) .....(1)
from B C process is adiabatic
PBVBr
= P0V0r
.....(2)
Solving we get VB
=8
V0.
15. An astronaut is ......................................Ans. 5
Sol.g2
u2
= 10m
g
u2= 4 sec
Solving , g = 5 m/s2 .
16. Wind entering in .................... ..................
Ans. 2
Sol. Energy entering in the windmill
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SOLSCT140413 - 10
=2
1mv2
Pin
=dt
dE=
2v2
1
dt
dm
Pin
=
2v2
1(AV) =
2
1 A V3
Electrical power output
Pout
=
3AV
2
1
3
1
Pout
=3AV
6
1 =
3)10(102.16
1
Pout
= 2 kW.
17. An aeroplane of ......................................
Ans. 5
Sol.
Applying bernaullis equation for air above and below the wing
P1+
2
1V
12 = P
2+
2
1V
22
(P2 P
1) =
2
1(V
1+ V
2) (V
1 V
2)
Here V1
= V + VV
2= V V
So, (P2 P
1) =
2
1(2V) (2V) = 2V V
For level flight
(P2 P
1) = mg
(2V V)A = mgPutting the values we will get
V = 5 m/sec.
18. A wooden block ......................................
Ans. 2
Sol. Due to the perfectly inelastic collision,
Eloss
= )mm(2mm
21
21 (u1 u2)
2 (1 e2)
Eloss
=)m9m(2
)m9)(m(
(u 0)2 (1 02)
Eloss
=2u
20
m9
Heat produce =
2u20
m9
2
1=
2u40
m9
Due to this heat, the temperature of the bullet will reach its
melting point, and then the bullet will melt, so
Heat produced =2u
40
m9
= mST + mL
f
22. TlI3contains ..........
2u40
m9
= m(2000) (200) + m(5 105)
Solving u = 2 103 m/sec.
19. A small uniform ..................................... .
Ans. 2
Sol. Lets see relative to the truck
For perfect rolling relative to the truck
arel
= R .............(i)ma f
r= (m) (a
rel) .............(ii)
(fr) (R) = )(2
MR2
.............(iii)
Solving we get arel
=3
a2=
3
32
arel = 2 m/sec2
Srel
= urel
+2
1a
relt2
4 = 0 +2
1(2) t2 t = 2 seconds.
20. A particle is moved ......................................Ans. 2Sol. work done in path 2
w2
= )jdyidx(F
= )jdxidx(jxas the line : y = x
= 1
0dxx =
2
1J.
work done in path 1w
1= w
OB+ w
BA
= idxF
+
jdyF
= jdyjxidxjx
= 0 +
1
0y
jdy)j1(= 1 J
2
1
w
w= 2. Ans. 2
PART-II (Chemistry)
21. Which of the following is the hybridisation for the ................
Sol. (A)
P
Cl
||O
Cl Cl
sp3
I
F
||O
F
F
F
sp d3 2
F
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SOLSCT140413 - 11
Sol. (A)
23. Bubbling CO2 through which of the ..................
Sol. (A)
2NaAlO2 + CO2 +3H2O Na2CO3 + 2Al(OH)3
24. Which of the following is a water .................
Sol. (D)
25. Addition of water to which of the ...................
Sol. (D)
26. A certain mass of gas is expanded ..................
Sol. (B)
State 1 State 2
(1 L, 10 atm, 300 K) (4L, 5 atm, 600 K)
Heat given, q = 50 (600 300) = 15000 J
W = Pext
(V2 V
1) = 1 (41) = 3 L atm = 300 J
E = q + W = 15000 300 = 14700 JH = E + P
2V
2 P
1V
1= 14700 J + (20 10) 100 J = 15.7 KJ
27. Which of the compounds does not ..............
Sol. (D)
28.
Which alkyl group is .......................
Sol. (A)
29. Which of the following pairs consists of ................
Sol. (A)
30. The average OH bond energy ......................
Sol. (C)
(A) H2O() H
2O(g) H = 40.6 KJ/mole
(B) 2H(g) H2(g) H = 435.0 KJ/mole
(C) 2O(g) O2(g) H = 489.6 KJ/mole
(D) H = 571.6 KJ/mole
(A) Calculation of Hf
(H2O,)
2H2(g) + O
2(g) 2H
2O() H = 571.6 KJ/mole
Hr= 2H
F(H
2O, ) 2H
F{H
2,(g)} H
F(O
2, g)
Zero Zero571.6 = 2H
F(H
2O, ) so H
F(H
2O, ) = 285.8 KJ/mole
(B) Calculation of HF(H
2O, g)
H2O() H
2O(g) H = 40.6
Hr= H
F(H
2O, g) H (H
2O,)
HF
(H2O, g) = H
F(H
2O, ) + H
r
= 285.8 + 40 = 245.8 KJ/mole
(C) H = 245.8 KJ/mole
Hr = HH
+2
1
OO 2
OH
245.8 = + 435 + 2
1(489.6) 2
OH
2OH
= 435 + 244.8 + 245.8
2OH = 925.6
OH
= 462.8 KJ/mole
31. (A) Na3B
3O
6...................................
Sol. (A) p, r, t (B) p, s, t (C) p, q, r, s, t (D) p, s
(A) Na3B
3O
6
(B) NaCaAlSi3O
9
(C) Borax [Na2B
4O
7.10H
2O Na
2B
4(OH)
4O
5.8H
2O]
(D) P4O
10
32. (A) Perfectly crystalline solid .................... .....
Sol. (A q) ; (B p, r, s, t) ; (C p, s) ; (D r, t).
(A) IIIrd law states 0TLim
S 0 for perfect solid.
(B) In reversible cyclic process the every net energy change is zero
(i.e. Hsys = 0, Usurr = 0 and net heat exchange = 0) and Ssurr = 0.(C) In irreversible cyclic process Ssys
= 0 but Ssurrounding
> 0 by IInd
law and all state function are zero for it.
(D) In adiabatic process net heat change is zero so Ssurrounding
= 0.
33. Sum of oxidation numbers of all the ..............
Sol. (7)
KO2, K = + 1 On two 'O' atoms, Sum
PbO2, O = 2 2 2 = 4. BaO
2: Ba = + 2
On two 'O' atoms
34. Consider the equilibrium ..................
Sol. (4)
Ni(s) + 4CO(g) Ni(CO)4(g)
P P
For backward reactionQ
p K
p
p4K
P
P
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SOLSCT140413 - 12
3-
3atm125.0
P
1
P3 8 atm3
P 2 atmP
Total= 2 P = 4 atm.
35. ICl3 is an orange colored solid ....................
Sol. (08)
I2Cl6 is a planar molecule.
36. How many of these species are ..................
Sol. (4)
O2, O
2+, O
2, B
2
37. reacts with NH2OH to form ...................
Sol. (4)
38. 1 mole of how many of the following acids ......................
Sol. (4)
(HCl, HNO3, H3PO2, H3BO3) H2SO4, H2SO3, H3PO3, H4P2O5 are
diprotic. HCl, HNO3, H3PO2, H3BO3 are monoprotic.
39. How many of these are hydrolysed .............................
Sol. (6)
SOCl2 + H2O H2SO3 + 2HCl
SO3 + H2O H2SO4
XeO4 + 2H2O H2XeO4
POCl3 + H2O H3PO4 + 3HCl
PCl5 + H2O H3PO4 + 5HCl
SiCl4 + H2O H2SiO3 + 4HCl
40. Intramolecular H-Bonding is observed .................Sol. (2)
,
PART-III (Mathematics)
41. Number of integral .......................................
Sol. (B) 0)5x(.)1x(log
)8x()3x()2e()8x2x(22
2
x2
(x 3) (x 8) 0 and x 5 x [3,8] x2 2x + 8 > 0 xR
ex + 2 > 0 x R(x 5)2 > 0 x R {5}
and log2(x2 + 1) x R
Number of integral values of x is 5 (x = 3,4,6,7,8 x 5).
42. Sum of first 2n terms ...................................Sol. (D) (13+33+53+....+ (2n 1)3 + (1.12 + 2.32 + 3.52 +....+ n(2n 1)2)
=
n
1r
n
1r
22n
1r
3 )1r3()1r2()1r2(r)1r2(
=
n
1r
23 )1r7r16r12( =6
)1n(n (18n2 14n + 5) n
43.
n
0r
rn )rx(sinC is .......................................
Sol. (A) In
=
n
0r
xrir
n eC
= In
= ((1 + eix)n)
= In
=
n2
2
xcos.
2
xsini2
2
xcos2
=
2
nxsin.
2
xcos.2 nn
44. 15 identical balls have to ..............................Sol. (A) Let the balls put in the box are x
1, x
2, x
3, x
4and x
5.
we must havex
1+ x
2+ x
3+ x
4+ x
5= 15, x
i 2
(x1
2) + (x2
2) + (x3
2) + (x4
2) + (x5
2) = 55 + 5 1C5= 9C
5
45. Line ax + by + p = 0 .......................................Sol. (B) Lines xcos + ysin = p, xsin ycos = 0 are mutually
perpendicular. Thus, ax + by + p = 0 will be equally inclined tothese lines and would be the angle bisector of these lines.
Now equations of angle bisectors is
xsin ycos = ( xcos + ysin p) x(cos sin) + y(sin + cos) = por x(sin + cos) y(cos sin) = pcomparing these lines with ax + by + p = 0, we get
a
cos sin =
cossinb
= 1 a2 + b2 = 2
46. If the circles x2 + y2 + 2ax + 2by = 0 ............Sol. (C) Both circles pass through origin. If they touch each other
then tangents drawn to respective circles at origin must beidentical lines.i.e. ax + by = 0 and bx + cy = 0 should represents same line
c
b
b
a b2 = ac
47. Maximum number of.....................................Sol. (D) Normals to y2 = 4ax and x2 = 4by in terms of 'm' are
y = mx 2am am3 and y = mx + 2b + 2
b
mfor a common normal
2b + 2m
b+ 2am + am3 = 0
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SOLSCT140413 - 13
am5 + 2am3 + 2bm2 + b = 0That means there can be atmost 5 common normals
48. If the tangent to the.....................................
Sol. (A)4
y
16
x 22 = 1.The tangent at P() is
4
xcos +
2
ysin = 1.
The focus of the parabola x2 = 8(y 6),is (0, 8).
0 +
2
8sin = 1 sin=
4
1.
49. Locus of the point of intersection ..................Sol. (D) Let the normal drawn to the curve at P(t
1) cuts the curve
again at Q(t2), then
If R(h,k) be the point of intersection of tangents drawn to curveat P and Q, then
h =21
21
tt
tct2
, k = 21 ttc2
k
h= t
1t2= 2
1t
1
Also h
31
1 t
1
t = 2ct1.
31t
1
h(t14 1) = 2ct
1
h2(t14 1)2 = 4c2t
12
h2
2
2
2
1h
k
= 4c2
h
k
Thus locus is (x2 y2)2 + 4c2xy = 0
50. Ifa
xcos=
b
xsin,.......................................
Sol. (D)a
xcos =b
xsin tanx =ab
cos2x = 22
22
ba
ba
and sin2x = 22 ba
ab2
so acos2x + bsin2x = 3 22 2a aba b
= a
51. Ans. (A) (q,s), (B) (p,t), (C) (r,s,t), (D) (q)(A) If , are the .................................
Sol. (A) sinx = 2
1 x =
6
7,
6
11
and cosx = 2
3 x =
6
5,
6
7
so + = 3 and =
(B ) cotx = 3 x = 6
5,
6
11
and cosecx = 2 x =6
7,
6
11
so + = 2 and = (C) sinx =
21 x =
67 ,
611
and tanx =1
3 x =
6
,
6
7
so + = 3 and + = 2and = (D) = 0 =
52. Ans. (A) (r,t), (B) (p), (C) (q,s), (D) (q)(A ) The solution of ..............................
Sol. (A) |x + y| = 10 ........(1)
and log10
y log10
|x| = log100
4
y = 2|x| ........(2)
so from (1) and (2) we have x =3
10and y =
3
20
or x = 10 and y = 20
(B) Let log2x = A and log
2y = B
4A2 + 1 = 2B and 2A B
A =2
1
also
2
2
14
+ 1 = 2B B = 1
log2x =
2
1and log
2y = 1
x = 2 and y = 2
(C) log4x = log
2y
xlog 22 = log2y
x = y2 and x2 5y2 + 4 = 0 x2 5x + 4 = 0 x = 1, x = 4 y = 1, y = 2 Let y > 0 Solution set are {1, 1}, {4, 2}
(D) (x 1)2 + (y 1)2 = 0x = 1, y = 1
53. Ifxlog3log 44 3x2 = 27, .... ...................... ...
Sol. (4)xlog3log 44 3x2 = 27 log
4x = 2
x = 4
54. If 3 be the remainder, ................... ...........Sol. (5) 22005 = 2.22004 = 2.(24)501 = 2(17 1)501
= 2.{501C0. 17501501C
1. 17500 +..........}
= 34( 1) + 32= 34( 1) + 17 + 15
3 = 15 = 5
55. Among the 8! permutations .................... ......Sol. (7) Let the arrangement be x
1x
2x
3x
4x
5x
6x
7x
8.
Clearly 5 should occupy the position x4
or x5.
Thus, required number = 2(7!) i.e. = 7
56. If Sn
= 31
1+ 33 21
21
..............................
Sol. (1) Tr= 333 r...21
r.....321
= 2
r
1
1r
1
n
1r)r(T = 2
1n
1
1 = Sn
nlim [S
n] = 1
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57. Consider the point ................... ...........Sol. (2) We have |PA PB| AB Thus for |PA PB| to be maximum,
point A,B and P must be collinear. Equation of AB is x + 2y = 2.
Solving it with give line we get P 24 17
,5 5
so a + 2b =24 34
5
= 2
58. Locus of the mid point .................... ..........Sol. (2) Clearly the points A,B and P are concyclic, we have
PD2 = DA2 = BD2 = OA2 OD2
Let D (x, y) x2 + (y b)2 = a2 (x2 + y2) 2x2 + 2y22by + b2 a2 = 0 + + = 2
59. Consider an ellipse with .................... ..........Sol. (2) Equation of ellipse can to take as
2 2x y1
25 16 with e =
5
3
Thus one focus is (3, 0)Let the drawn circle touches the ellipse at point ' P'. Focal distance
of P = a ex1
= 5 5.5
3cos.
= 5 3cosIts minimum value is 2. Thus radius of largest circle is 2 units.
60. Ellipse 1by
ax
2
2
2
2 and ............................
Sol. (2) We have aee
= Aeh
and b = B
ee2 a2 = A2.e
h2
ee2 .
)e1(
b2e
2
= eh2 .
2
2h
B
(e 1)
2e
1
11
e
=
2he
11
1 2
h2e e
1
e
1 = 2