8/2/2019 Iit Jee 2012 Pet4 Solns p2
1/22
32
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
BRILLIANTSPROGRESSIVE EVALUATION TEST
FOR STUDENTS OF
OUR ONE/TWO-YEAR POSTAL COURSES
TOWARDS
IIT-JOINT ENTRANCE EXAMINATION, 2012
SECTION I
1. (A)
= =
221 2
1 1 1R
n n
For first Lyman transition,
= = =
L 2 2
1 1 1 3R R
41 2
For first Paschen transition,
p2 2
1 1 1 7R R
1443 4
= = =
=PL3 7
: :4 144
73 :
36=
= 108 : 7
PART A: CHEMISTRY
IIT/ELITE 2012PET I/PET IV/CPM/P(II)/SOLNS
PAPER II SOLUTIONS
CHEMISTRY PHYSICS MATHEMATICS
8/2/2019 Iit Jee 2012 Pet4 Solns p2
2/22
33
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
2. (D) Species Hybridization
4NH+ 5 4 1 8 4
2 2+ = = sp3
3NO
5 0 1 6
32 2
+ += = sp
2
2NO+
5 0 1 4
22 2
+ = = sp
3. (A) In the reaction,
H Li LiH+
Li undergoes oxidation and H undergoes reduction
( )H e H+ . Hence
hydrogen acts as an oxidising agent.
4. (A) Density of the solution,
= +
1 mol. wtd M
m 1000
1 98d 11.07
21.91 1000
= +
= 11.07 [0.0456 + 0.098]
= 11.07 0.1436 = 1.589 g/mL
5. (B) 2 2 31 3
N H NH
2 2
+
Partial pressure of 324.91
NH 100 24.91 atm100
= =
Pressure of (N2
+ H2) = 100 24.91 = 75.09 atm
Partial pressure of 23
H 75.09 56.32 atm4
= =
Partial pressure of 21
N 75.09 18.77 atm4
= =
( ) ( )= =
3
2 2
NH
p 1/2 3/2 1/ 2 3/ 2HN
P 24.91
K P P 18.77 56.32
24.91
4.3 421=
= 0.0136 atm
1
8/2/2019 Iit Jee 2012 Pet4 Solns p2
3/22
34
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
6. (C) For gas A, A AA
A
d RTP
M=
For gas B, B BBB
d RTP
M=
A A A B
B B B A
P d T M
P d T M
=
[since M
A= M
B]
A
B
P 1 3 31
P 2 1 2
= =
or PA: PB = 3 : 2
7. (A) NO2
group exhibits I and M effect. These ve effects are greater for
NO2
group than for CN group. Also I effect is more when the group is
present in the ortho position. Thus
2CH
2NO
is the most stable carbanion
among them.
8. (C) I and M effects increase the strength of acid. Since NO2 group present in
para position exhibits M and I effect,
3NH
2NO
is most acidic.
SECTION II
9. (A), (C)
2 2 3 2 2 4 62Na S O I 2NaI Na S O+ +
(tetrathionate)
8/2/2019 Iit Jee 2012 Pet4 Solns p2
4/22
35
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
10. (A), (B), (D)
NaI < NaCl < BaO < CaO (Lattice energy)BaCO
3> SrCO
3> CaCO
3> MgCO
3(Thermal stability)
Dipole moment of NH3
is greater than that of NF3.
Nuclear spins are in opposite direction in para hydrogen.
11. (A), (B), (D)
G = H TS
When H is ve and S is +ve, G becomes ve. So exothermic reactions arespontaneous at all temperatures.
When H is +ve, and S is ve, G is + ve. Hence such endothermic reactions
are not spontaneous at all temperatures.
When H is +ve and S is +ve, then G is ve only when H < TS i.e., G is
ve and spontaneous when temperature is high.
12. (A), (C)
Reductive ozonolysis
3 2 3CH C C CH CH3O
3 2 3CH C C CH CH
O
O O
2H O / Z n 3 2 3CH C C CH CH
O O
ZnO+
2 4
4
dil H SO3 2 3 3 2 31% HgSO
CH C C CH CH CH C C CH CH =
H OH
Rearrange
3 2 2 3CH CH C CH CH
O
4
cold alkaline3 2 3 KMnO
CH C C CH CH 3 2 3CH C C CH CH
OH
OH
OH
OH
22H O3 2 3CH C C CH CH
O O
3
2
(i) O3 2 3 3(ii) H O
CH C C CH CH CH COOH + 3 2CH CH COOH
8/2/2019 Iit Jee 2012 Pet4 Solns p2
5/22
36
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
SECTION III
13. (5)
3CHOH
3CH
H
heat
+
3CH
2OH
3CH
2H OC H
3CH
3CH
Ring
expansion
3CH
3CH H
3CH
3CH+
14. (6)
All the six carbon atoms are in the sp2
hybrid state.
15. (2) + + CH CH 2 RMgX 2R H XMg C C MgX
Acetylene
16. (4) (g) (g) (g) (g)A 2B 2C D+ +
Initial 1 1.5 conc.
At (1 x) (1.5 2x) 2x xequilibriumconc.
At equilibrium since [A] = [D], (1 x) = x or x = 0.5
= =
2 2
c 2 2
[C] [D] (2 0.5) (0.5)K
[A] [B] (1 0.5) [1.5 1]
1 0.5 14
0.5 0.25 0.25
= = =
8/2/2019 Iit Jee 2012 Pet4 Solns p2
6/22
37
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
17. (4)2 2
2 24 3C O 2H O 2CO 4H 2e 3 + + + +
2 24MnO 4H 3e MnO 2H O 2 + + + +
Add2 2
2 2 24 4 33C O 2MnO 2H O 6CO 2MnO 4H +
+ + + +
Since the reaction takes place in the basic medium,2 2
2 4 2 2 24 33C O 2MnO 2H O 4OH 6CO 2MnO 4H O
+ + + + +
or22
2 4 2 24 33C O 2MnO 4[OH ] 6CO 2MnO 2H O
+ + + +
18. (1) rmsT
vM
= =2
2
rms (H at 5 0 K )
rms (O at 800 K )
v 50 321
v 2 800
SECTION IV
19. (A) (p), (q); (B) (p), (r), (t); (C) (p), (s), (t); (D) (p), (q), (t)
(A) 50 ml of 0.1 M HNO3
+ 50 ml of 0.1 M KOH gives 50 10 3
0.1 mole of KNO3.
Complete neutralisation takes place. KNO3
does not undergo hydrolysis. It
simply ionises. Hence the pH of solution is 7.
(B) +3CH COOH NaOH
(moles) 3
50 0.2 10 350 0.2 10
+3 2CH COONa H O
gives 350 0.2 10 3moles of CH COONa
which is present in 100 ml of solution.
3
350 0.2 10
[CH COONa] 1000 moles / lit100
=
= 10 10 3
10 = 0.1 M
CH3COONa undergoes hydrolysis in solution
[ ]= + +w a1
pH pK pK log [salt]2
[ ]1
14 4.8 log 0.1
2
= + +
[ ]= + =1 1
14 4.8 1 (17.8)2 2
= 8.9
8/2/2019 Iit Jee 2012 Pet4 Solns p2
7/22
38
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
(C) 4 4 23 3 3
HCl NH OH NH Cl H O
(moles) 50 0.1 10 50 0.1 10 50 0.1 10 in 100 ml
+ +
32
450 0.1 10
[NH Cl] 1000 5 10 M100
= =
NH4Cl undergoes salt hydrolysis.
[ ]= w b1
pH pK pK log (salt)2
( )21 14 4.8 log 5 102
=
= + 1
[14 4.8 2 0.6990]2
= 1
[11.2 0.6990]2
= 5.6 0.3495 = 5.25
(D) 5 10 3
moles of NH4OOC CH
3is formed in 100 ml.
Hence
= =
32
3 45 0.1 10 1000
[CH COONH ] 5 10 M10
This salt undergoes hydrolysis.
++3 4 3 4CH COONH CH COO NH
2H O3 3 44CH COO NH CH COOH NH OH
+++ +
= + w a b1
pH [pK pK pK ]2
1[14 4.8 4.8] 7
2= + =
20. (A) (p), (r), (t); (B) (p), (t); (C) (q); (D) (s)
(A) 34SF sp dii
(B)
3 3
6XeF sp d
ii
(C) CCl4 sp
3
(D) SF6 sp
3d
2
8/2/2019 Iit Jee 2012 Pet4 Solns p2
8/22
39
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
SECTION I
21. (D) Let the initial velocity be iv along the +x direction and the final velocity fv
along the direction making an angle with iv .
Now = iv vi
= + fv v cos i v sin j
Change in velocity = ifv v
i.e., v v cos i v sin j vi = +
v (cos 1) i v sin j= +
2 2 2v v (cos 1) (vsin ) = +
2 2v cos 2cos 1 sin= + +
v 2 2 cos=
2v 4 sin2
=
2vsin2
=
average acceleration2vsinv 2a
t t
= =
22. (C) Since the rod is in rotational equilibrium, torque on it is zero.
N mg = 0
N = 39.2 N
Taking moments about O,
0.5 0.560 m sin N cos m
2 2 =
sin N
cos 60
=
39.260
=
= 0.65
= tan 1
(0.65)
iv
x
fv
y
F.B.D. for the rod
P
mg
O
N
PART B: PHYSICS
8/2/2019 Iit Jee 2012 Pet4 Solns p2
9/22
40
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
23. (B) 1Elastic potential energyU stress strain
per unit volume 2
=
= 21
B (strain)2
, where B is bulk modulus
But strainy
(a coskx cos t)x x
= =
= ak sin kx cos t
= 21
U B ( ak sin kx cos t)2
2 2 2 21B a k sin kx cos t
2
=
The velocity of longitudinal wave is given by
Bc =
2B c=
Also ck
=
= 2 2 2 21
U a cos kx cos t2
= 2 2 21
U (x, 0) a cos kx2
In units of2 2a
2
U(x, 0) = cos2
kx
24. (A) For x-motion, x = vt (i)
But v 2gh= from Torricellis theorem
For y-motion, 21
(H h) gt2 = at the initial moment.
2(H h)t
g
=
8/2/2019 Iit Jee 2012 Pet4 Solns p2
10/22
41
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
From (1),
2(H h)x 2ghg=
2 h(H h)=
1 3 32 m
4 4 2
= =
25. (A) P = k Fa
b
c
[ML2
T 3
] = [MLT 2
]a
[L]b
[ML 3
]c
a + c = 1
a + b 3c = 2
2a = 3
Solving, we get,
3 1a , b 1 and c
2 2= = =
=
3 112 2P kF
Let the power be P1
when weight, length and density are all altered. Then,
( ) ( )
=
1 1
3/221P k 2F 2
2
( ) =
3/ 2 11P 12 (2)2
12 2 2 4
2= =
26. (A) Centre of mass lies along the x-axis.
cm
x dmx
dm =
xL
L0
0
xLL
0
0
x e dx
e dx
=
8/2/2019 Iit Jee 2012 Pet4 Solns p2
11/22
42
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
Let u = x and
x
Ldv e dx
=
x
Lv L e
=
L xx LLL
00
cm xLL
0
x L e L e dx
x
e dx
=
L
0
Lx
2 2L
0
x
L
L e L e
L e
=
2 2 2L e L e L
L (e 1)
+=
L L
(e 1) (2.718 1)= =
= 0.582 L
27. (C) Frequency of source = 1000 Hz
At t = 3 sec, velocity of detector v = a t
= 10 3 = 30 m/s
The frequency detected = 1100 Hz at t = 3 sec
c 301100 1000
c
+ =
, where c is the velocity of sound.
i.e., 11c = 10 c + 300
c = 300 m/s
28. (C) From conservation of momentum we have
m1
v1i
+ m2
v2i
= m1
v1f
+ m2
v2f
(1.60) (4.00) + (2.10) ( 2.50) = (1.60) (v1f
) + (2.10) ( 1.75)
8/2/2019 Iit Jee 2012 Pet4 Solns p2
12/22
43
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
v1f
= 3.00 m/s
Again by law of conservation of energy
2 2 2 2 21 2 1 22i1i 1f 2f
1 1 1 1 1m v m v m v m v kx
2 2 2 2 2+ = + +
On substituting the numerical values and simplifying, we get compression inthe spring as
x = 0.17 m
SECTION II
29. (A), (B), (D)
=dQ dQ
AKdt dx
in the steady state, where K is coefficient of thermal conductivity.
Since dQdx
is constant,
dQ
Adt
. Hence rate of flow of heat decreases from A to C but increases from
C to B. The section at P at a distance x from A and the section at Q are at thesame distance from C. Since temperature gradient is uniform, temperature at
C is 50. Since the rod is symmetrical about C, rate of flow of heat throughP and Q is the same.
30. (A), (C), (D)
Since blocks move with uniform velocity, acceleration a = 0 for the system.
The F.B.D. for B is
T = N = mB
g (1)
The spring is stretched by tension T.By considering the F.B.D. for A, we have
T = mA
g (2)
From (1) and (2) we get mB
= 10 kg
For the spring we have
T = k = mA
g
2 9.8 = 1960
=1
100= 10
2m = 1 cm
Energy stored 2 41 1
kx 1960 102 2
= = = 9.8 10 2
J
Bgm
TNa
8/2/2019 Iit Jee 2012 Pet4 Solns p2
13/22
44
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
31. (A), (B)
Potential energy for the system is
= + 2p1
U kx mgx sin 37 mg sin 372
For Up
to be a minimum
= >
2p p
2
U U0 , 0
x x
kx = mg sin 37
mg sin 37
x k
=
2 10 0.6
20
= = 0.6 m
Value of = + 2min1
V k (0.6) mg sin 37 mg (0.6) sin 372
21 20 (0.6) 2 10 0.4 0.62
= +
= 10 (0.6)2
+ 8 0.6 = 8.4 J
Maximum elastic potential energy occurs when
2 21 1mg sin 37 kx 20 x2 2
= =
2 10 (0.6) 1 = 10 x2
x = 1.2 m = 1.1 m
32. (A), (B), (C), (D)
Volume of the block = 125 cm3
Let h1 be the height of the block above mercury.
Then (5 h1) 25 13.6 g = 125 7.2 g
h = 5 2.65 = 2.35 cm
8/2/2019 Iit Jee 2012 Pet4 Solns p2
14/22
45
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
Let h2
be the height of water required to just submerge the block. Then
wt. of block = wt. of (mercury + water) displaced
i.e., 900 = h2 2.5 1 + (5 h
2) 25 13.6
h2
= 2.54 cm
After pouring water, depth of mercury = 5 2.54 = 2.46 cm
SECTION III
33. (8) The kinetic energy of the system about BD is given by = K.E. about BC
=21
I K.E.2
of A about BD
21 mv2
=
22 Iv
m
=
22m( sin 60 )
m
=
2 23
4=
= 2 23
K.E. m8
Aliter
About BD about BC
2 21 1I mv2 2
= (K.E. of A)
=
22 2 2m
vm 4
= 2 23
4
2 21 3K.E. m2 4
=
A
B D C60
8/2/2019 Iit Jee 2012 Pet4 Solns p2
15/22
46
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
34. (1) We have 1
0 1
q
4 r =
, where q is the charge on the metal sphere. The charge
induced on the shell is
q = 4 0
r1
1
Hence the potential acquired by the shell is2
2
q
r =
0 11
0 2
4 r
4 r
=
112
2
r
r
=
On comparison n = 1
35. (4) Average energy
T
0T
0
u dt
u
dt
=
T
0
1u dt
T=
But ( )2 2 21
u m A x2=
But x = A sin (t + )
( )T
2 2 2
0
1u mA cos t dt
2T = +
( )T
2 2 2
0
1mA cos t dt
2T= +
ButT
2
0
1cos ( t ) dt T
2 + =
2 21u mA4
=
n = 4
8/2/2019 Iit Jee 2012 Pet4 Solns p2
16/22
47
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
36. (2) By the law of conservation of energy,
21 mghmvh2
1R
=
+
, where h is measured from the surface of the earth.
Given ev kv k 2gR= =
=
+
21 mg (r R)mk (2gR)r R2
1R
, where r is the distance measured from the centre
of the earth.
2 r Rk R 1 (r R)
R
+ =
i.e., k2
r = (r R)
2
Rr
1 k =
On comparison r = 2
37. (6) Due to extension produced in the cord, energy stored in it is converted into
kinetic energy K, when the stone flies away. Assuming that there is no loss of
energy in this process, the K.E. of the stone is given by21
K mv 4 J2
= = . This
must be equal to the work done in stretching the cord.
Therefore = =1
W F 4J2
4 2F
0.2
= = 40 N
Stress 6 22 3 2
F 40 401.415 10 Nm
A r (3 10 )
= = =
Strain20
0.47642
= =
Youngs modulus
= = = 6
6 2stress 1.415 10Y 2.97 10 Nm
strain 0.476
On comparison n = 6
8/2/2019 Iit Jee 2012 Pet4 Solns p2
17/22
48
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
38. (4) The angular impulse about the centre of mass
J = Lf
Li
= I 0
But
=
J p
2
2mp
2 12
=
6p
m =
=
6p
2t m
n mt
48 p
=
On comparison n = 4
SECTION IV
39. (A) (p), (r), (t); (B) (p), (s), (t); (C) (p), (q), (t); (D) (p), (t)
Points x1
and x3
are points of zero displacement. Just to the left of x1, the
displacement is negative indicating that the gas molecules are displaced to the left,away from point x
1at some instant. Just to the right of x
1, the displacement is
positive, indicating that the molecules suffer displacement to the right which is
again away from x1. So at point x
1, the pressure is minimum.
At point x2, the pressure does not change because the gas molecules on both sides of
that point have equal displacements in the same direction. Hence pressure is
normal.At x3, the pressure is maximum, because the molecules on both sides of that point
are displaced towards point x3.
Since pressure p is proportional to density , pressure and density are in phase.
40. (A) (p), (q), (s); (B) (q), (r), (t); (C) (p), (q), (s); (D) (q), (r), (t)
Process AB is an isothermal process with T = constant. Therefore (P V) graph is
a hyperbola. (V T) graph is a straight line since T is constant. By ideal gas
equationPM
RT = . Therefore P. As T is a constant, ( T) graph is a straight
line.
Process BC is an isobaric process with P = constant. As P is constant, (P V) graph
is a straight line. In isobaric process, V T and hence (V T) graph is a straight
line.
When P is constant1
T . Hence ( T) graph is a hyperbola.
For isothermal process U = 0, but for isobaric process U 0. What applies toprocess AB, also applies to CD. Similarly what applies to BC, also applies to DA.
8/2/2019 Iit Jee 2012 Pet4 Solns p2
18/22
49
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
SECTION I
41. (A) ax2
+ bx + c a (x ) (x ) p 1 p 2
a x xp p 1
+ +
+
Putting x = 1,1 1 a
a b c ap p 1 p(p 1)
+ + = =
+ +
( )2
2 2
2
b 4aca a ( ) 4 a b 4ac
a
= = + = =
42. (B)a
31 r
=
and3
3
a 108
131 r=
2
2
1 r r 13
41 2r r
+ + =
+
1r or 3
3 = . Here
1r
3= and a = 2
Now( )na 1 r 7281 r 243
>
n
11
72831 243
3
>
> < = >n n 51 728 1 1 11 n 5
729 7293 3 3
Minimum value of n = 6
43. (D)( ) ( )x 7 7log 2 log x log x5 x 2 24 + =
( )7 7 7log x log x log x5 2 2 24 2 4 + = = log7 x = 2 x = 72
= 49
44. (C) x + y + z = 16 and x, y, z 3
Let u = x 2, v = y 2, w = z 2
Then u + v + w = 10,
The number of positive integral solutions =9C
2= 36
PART C: MATHEMATICS
8/2/2019 Iit Jee 2012 Pet4 Solns p2
19/22
50
Brilliant Tutorials Pvt. Ltd. IIT/ELITE/PET/CPM/P(II)/Solns
45. (D) xA
= 1, xM
= 2, AP = 2
MAP 60 =
Required area = area QRPMQ
12 2 3
3=
43
3
=
46. (D) f ( ) 5 cos 3 cos cos 3 sin sin 33 3
= + +
( )13 3 3
cos sin 3 7 cos 32 2
= + = + + where =13
cos14
Range of f() = [3 7, 3 + 7] = [ 4, 10]
47. (B) P(a, a) must be outside the circle 2a2 4a 6 > 0
either a > 3 or a < 1
= = 21 1PT S 2a 4a 6 and 1CT r 2 2= =
< < < < >
1tan
3 6 2 2 2 3
2
2
2 2 1a 2a 3 2 3
32a 4a 6 >