IIT JEE - Sample Mock Test, Engineering, Exam Prepapration
IIT JEE - Sample Mock Test, Engineering, Exam Prepapration
| Download, print and study this document offline |
Page 1
Vidyamandir Classes
VMC/2013/Solutions 1 Mock IIT Advanced/Test - 2/Paper-1
Solutions to Mock IIT Advanced/Test - 2[Paper-1]/2013
[CHEMISTRY]
1.(D) n = 2
3
r
2
=
Change in entropy for reversible adiabatic process is zero. (Always).
2.(B) V = 1Litre
T = 300 K
P = 190/760
? Since at t=8 , 390 mm of Hg pressure is observed
It means, out of the initial 200 mm, only 190 mm should be the pure reactant.
? A
0
= 190 mm Hg ? 1/4 atm.
? PV = nRT
?
1 1
n 10 millimoles
4R 100
= ˜ =
A
0
= 10.
After 10 minutes 0.2 gm of NaOH required
? 5 millions of HBr formed.
? After 10 minutes, 5 millimoles of C
2
H
5
Br has been consumed out of 10.
? Half life = 10 minutes.
3.(A)
1
1 2
SN
1
SN SN
r
% of SN
r r
=
×
=
( ) ( ) ( )
6 6
5 6 5 6
0.8 10 [S] 0.8 10
100 100
3.12 10 [S][Nu ] 0.8 10 [S] 3.12 10 0.8 10
- -
- - - - -
× ×
× = ×
× + × × + ×
= 2.5%
4.(B) Beryllium compounds are covalent (more polarising power of Be
+2
)
Mg, Ca, Sr, Ba form divalent ionic compounds.
5.(B)
2 3
CaO SiO CaSiO (slag) + ?? ?
6.(A)
7.(B)
2 2 8
H S O
O O
|| ||
HO S O O S OH
|| ||
O O
- - - - - Peroxy
2 2 7
H S O
O O
|| ||
HO S O S OH
|| ||
O O
- - - - Non-Peroxy
Page 2
Vidyamandir Classes
VMC/2013/Solutions 1 Mock IIT Advanced/Test - 2/Paper-1
Solutions to Mock IIT Advanced/Test - 2[Paper-1]/2013
[CHEMISTRY]
1.(D) n = 2
3
r
2
=
Change in entropy for reversible adiabatic process is zero. (Always).
2.(B) V = 1Litre
T = 300 K
P = 190/760
? Since at t=8 , 390 mm of Hg pressure is observed
It means, out of the initial 200 mm, only 190 mm should be the pure reactant.
? A
0
= 190 mm Hg ? 1/4 atm.
? PV = nRT
?
1 1
n 10 millimoles
4R 100
= ˜ =
A
0
= 10.
After 10 minutes 0.2 gm of NaOH required
? 5 millions of HBr formed.
? After 10 minutes, 5 millimoles of C
2
H
5
Br has been consumed out of 10.
? Half life = 10 minutes.
3.(A)
1
1 2
SN
1
SN SN
r
% of SN
r r
=
×
=
( ) ( ) ( )
6 6
5 6 5 6
0.8 10 [S] 0.8 10
100 100
3.12 10 [S][Nu ] 0.8 10 [S] 3.12 10 0.8 10
- -
- - - - -
× ×
× = ×
× + × × + ×
= 2.5%
4.(B) Beryllium compounds are covalent (more polarising power of Be
+2
)
Mg, Ca, Sr, Ba form divalent ionic compounds.
5.(B)
2 3
CaO SiO CaSiO (slag) + ?? ?
6.(A)
7.(B)
2 2 8
H S O
O O
|| ||
HO S O O S OH
|| ||
O O
- - - - - Peroxy
2 2 7
H S O
O O
|| ||
HO S O S OH
|| ||
O O
- - - - Non-Peroxy
Vidyamandir Classes
VMC/2013/Solutions 2 Mock IIT Advanced/Test - 2/Paper-1
2 5
H SO
O
||
HO S O O H
||
O
- - - - Peroxy
5
CrO
Peroxy
8.(D)
2 3
HCN H O H COOH NH
(X)
+ ?? ? +
2 4
H SO
2
HCOOH CO H O
(Y)
?
???? ? +
473K
Bond order 3
CO NaOH HCOO Na
+
=
+ ????
T
2
CO FeO Fe CO + ?? ? +
9.(B)
3 3
AgNO KI AgI K NO
+ -
?
+ ?? ? + +
AgI due to the field of Ag+, further I
-
are attached to it
? it becomes (AgI) I
-
negative sol.
10.(D) For a saturated solution of CaF
2
.
2
2 (aq) (aq)
CaF (s) Ca 2f
s 2s
+ -
+
2 3
sp
k (s) (2s) 4s = ?
1
3
sp
2
k
S [Ca ]
4
+
? ?
= =
? ?
? ?
2
[Ca ] [F ]
+ -
=
11.(BC)
B.E. of F
2
is lowest and Cl
2
is highest
Acidic character : Increases down the group and left to right across the period.
I.E. of Be > B.
12.(ABCD) Write the cell reaction : Anode : Ag(s) Cl (aq) AgCl(s) 1e
- -
+ ?? ? +
Cathode : Ag (aq) 1e Ag(s)
+ -
+ ?? ?
Ag (aq) Cl (aq) AgCl(s)
+ -
+ ?? ?
?
0 0
cell
Ag Ag Cl AgCl Ag
0.059 1
E E E log
1 [Ag ][Cl ]
+ -
+ -
= - -
Observe that as cell works, the precipitation of AgCl increases in anodic compartment.
When cell stops working, I.P = K
sp
which means [Ag
+
] is now simply =
sp
K = constant.
When E
cell
= 0, it means I.P = K
sp
. [Note that cell is in equilibrium]
Let [Ag
+
]
Cathode
= concentration of Ag
+
ion in cathodic department.
And [Ag
+
]
anode
= concentration of Ag
+
ion in anodic department.
sp c a
1 1
Q (at equilibrium)
K [Ag ] [Cl ]
+ -
= =
eqm.
a
1
Q
[Cl ]
-
=
+
a
[Ag ]
?
Cathode Anode
[Ag ] [Ag ]
+ +
=
*Note that [Ag
+
]
anode
is not visible during the working of cell. It comes to picture only at equilibrium.
Page 3
Vidyamandir Classes
VMC/2013/Solutions 1 Mock IIT Advanced/Test - 2/Paper-1
Solutions to Mock IIT Advanced/Test - 2[Paper-1]/2013
[CHEMISTRY]
1.(D) n = 2
3
r
2
=
Change in entropy for reversible adiabatic process is zero. (Always).
2.(B) V = 1Litre
T = 300 K
P = 190/760
? Since at t=8 , 390 mm of Hg pressure is observed
It means, out of the initial 200 mm, only 190 mm should be the pure reactant.
? A
0
= 190 mm Hg ? 1/4 atm.
? PV = nRT
?
1 1
n 10 millimoles
4R 100
= ˜ =
A
0
= 10.
After 10 minutes 0.2 gm of NaOH required
? 5 millions of HBr formed.
? After 10 minutes, 5 millimoles of C
2
H
5
Br has been consumed out of 10.
? Half life = 10 minutes.
3.(A)
1
1 2
SN
1
SN SN
r
% of SN
r r
=
×
=
( ) ( ) ( )
6 6
5 6 5 6
0.8 10 [S] 0.8 10
100 100
3.12 10 [S][Nu ] 0.8 10 [S] 3.12 10 0.8 10
- -
- - - - -
× ×
× = ×
× + × × + ×
= 2.5%
4.(B) Beryllium compounds are covalent (more polarising power of Be
+2
)
Mg, Ca, Sr, Ba form divalent ionic compounds.
5.(B)
2 3
CaO SiO CaSiO (slag) + ?? ?
6.(A)
7.(B)
2 2 8
H S O
O O
|| ||
HO S O O S OH
|| ||
O O
- - - - - Peroxy
2 2 7
H S O
O O
|| ||
HO S O S OH
|| ||
O O
- - - - Non-Peroxy
Vidyamandir Classes
VMC/2013/Solutions 2 Mock IIT Advanced/Test - 2/Paper-1
2 5
H SO
O
||
HO S O O H
||
O
- - - - Peroxy
5
CrO
Peroxy
8.(D)
2 3
HCN H O H COOH NH
(X)
+ ?? ? +
2 4
H SO
2
HCOOH CO H O
(Y)
?
???? ? +
473K
Bond order 3
CO NaOH HCOO Na
+
=
+ ????
T
2
CO FeO Fe CO + ?? ? +
9.(B)
3 3
AgNO KI AgI K NO
+ -
?
+ ?? ? + +
AgI due to the field of Ag+, further I
-
are attached to it
? it becomes (AgI) I
-
negative sol.
10.(D) For a saturated solution of CaF
2
.
2
2 (aq) (aq)
CaF (s) Ca 2f
s 2s
+ -
+
2 3
sp
k (s) (2s) 4s = ?
1
3
sp
2
k
S [Ca ]
4
+
? ?
= =
? ?
? ?
2
[Ca ] [F ]
+ -
=
11.(BC)
B.E. of F
2
is lowest and Cl
2
is highest
Acidic character : Increases down the group and left to right across the period.
I.E. of Be > B.
12.(ABCD) Write the cell reaction : Anode : Ag(s) Cl (aq) AgCl(s) 1e
- -
+ ?? ? +
Cathode : Ag (aq) 1e Ag(s)
+ -
+ ?? ?
Ag (aq) Cl (aq) AgCl(s)
+ -
+ ?? ?
?
0 0
cell
Ag Ag Cl AgCl Ag
0.059 1
E E E log
1 [Ag ][Cl ]
+ -
+ -
= - -
Observe that as cell works, the precipitation of AgCl increases in anodic compartment.
When cell stops working, I.P = K
sp
which means [Ag
+
] is now simply =
sp
K = constant.
When E
cell
= 0, it means I.P = K
sp
. [Note that cell is in equilibrium]
Let [Ag
+
]
Cathode
= concentration of Ag
+
ion in cathodic department.
And [Ag
+
]
anode
= concentration of Ag
+
ion in anodic department.
sp c a
1 1
Q (at equilibrium)
K [Ag ] [Cl ]
+ -
= =
eqm.
a
1
Q
[Cl ]
-
=
+
a
[Ag ]
?
Cathode Anode
[Ag ] [Ag ]
+ +
=
*Note that [Ag
+
]
anode
is not visible during the working of cell. It comes to picture only at equilibrium.
Vidyamandir Classes
VMC/2013/Solutions 3 Mock IIT Advanced/Test - 2/Paper-1
13.(ABD)
14.(BC) 0.1 m urea : i = 1 ; 0.04m Al
2
(SO
4
)
3
: i = 5 ; 0.05 m CaCl
2
: i = 3 ; 0.005 m NaCl : i = 2.
Freezing point is highest for solution IV.
Boiling point is highest for solution II.
15.(ABCD) As we know
rms
3RT
C
M
=
For heavier gas, M will be high.
C
rms
will have a lower value.
So graph A correspond to either a heavier gas or same gas at lower temperature.
Similarly graph B correspond to either a lighter gas or same gas at higher temperature.
16.(4) Use :
cell 1 1
P
dE 1.70 1.2
S nF 2 100000 5000 J mol 5 kJ mol
dT 45 25
- -
- ? ? ? ?
? = = × × = =
? ? ? ?
-
? ? ? ?
17.(8) Positronium consists of positron and electron
For positron, the mass of electron will be considered as reduced mass since the nucleus is formed by positron only
and hence the nucleus is movable
e p e
e reduced
e p
(m )(m ) m
(m )
m m 2
= =
+
? The radius of 2
nd
orbit of positronium =
2
o 0
n
(2R ) 8R
Z
× =
?
Due to reduced mass concept
18.(4)
2 2
1
CO O CO
2
x x 2
+ ?? ?
2 2
CO CO
1 x
?? ?
-
Volume of O
2
used = 2 1.6 0.4 - =
?
x
0.4
2
= Hence x = 0.8 Now mole ratio =
0.8
4
0.2
= .
19.(3) x millimoles : Salicylic acid ; y millimoles : Malonic acid in 1000 ml
25 ml extract will contain
x
m.moles
40
of Salicylic and
y
m.moles
40
of Malonic acid
NaOH used for titrating this extract = 2 m.moles
?
2x 2y
2
40 40
+ = ? x + y = 40 . . . .(i)
50 ml extract contains
x
20
m.moles salicylic and
y
20
m.moles malonic acid. On heating only malonic acid will
give CO
2
.
?
y 11.2
y 10 m.moles
20 22.4
= ? = ? x = 30 m.moles ? x/y = 3
20.(6)
Page 4
Vidyamandir Classes
VMC/2013/Solutions 1 Mock IIT Advanced/Test - 2/Paper-1
Solutions to Mock IIT Advanced/Test - 2[Paper-1]/2013
[CHEMISTRY]
1.(D) n = 2
3
r
2
=
Change in entropy for reversible adiabatic process is zero. (Always).
2.(B) V = 1Litre
T = 300 K
P = 190/760
? Since at t=8 , 390 mm of Hg pressure is observed
It means, out of the initial 200 mm, only 190 mm should be the pure reactant.
? A
0
= 190 mm Hg ? 1/4 atm.
? PV = nRT
?
1 1
n 10 millimoles
4R 100
= ˜ =
A
0
= 10.
After 10 minutes 0.2 gm of NaOH required
? 5 millions of HBr formed.
? After 10 minutes, 5 millimoles of C
2
H
5
Br has been consumed out of 10.
? Half life = 10 minutes.
3.(A)
1
1 2
SN
1
SN SN
r
% of SN
r r
=
×
=
( ) ( ) ( )
6 6
5 6 5 6
0.8 10 [S] 0.8 10
100 100
3.12 10 [S][Nu ] 0.8 10 [S] 3.12 10 0.8 10
- -
- - - - -
× ×
× = ×
× + × × + ×
= 2.5%
4.(B) Beryllium compounds are covalent (more polarising power of Be
+2
)
Mg, Ca, Sr, Ba form divalent ionic compounds.
5.(B)
2 3
CaO SiO CaSiO (slag) + ?? ?
6.(A)
7.(B)
2 2 8
H S O
O O
|| ||
HO S O O S OH
|| ||
O O
- - - - - Peroxy
2 2 7
H S O
O O
|| ||
HO S O S OH
|| ||
O O
- - - - Non-Peroxy
Vidyamandir Classes
VMC/2013/Solutions 2 Mock IIT Advanced/Test - 2/Paper-1
2 5
H SO
O
||
HO S O O H
||
O
- - - - Peroxy
5
CrO
Peroxy
8.(D)
2 3
HCN H O H COOH NH
(X)
+ ?? ? +
2 4
H SO
2
HCOOH CO H O
(Y)
?
???? ? +
473K
Bond order 3
CO NaOH HCOO Na
+
=
+ ????
T
2
CO FeO Fe CO + ?? ? +
9.(B)
3 3
AgNO KI AgI K NO
+ -
?
+ ?? ? + +
AgI due to the field of Ag+, further I
-
are attached to it
? it becomes (AgI) I
-
negative sol.
10.(D) For a saturated solution of CaF
2
.
2
2 (aq) (aq)
CaF (s) Ca 2f
s 2s
+ -
+
2 3
sp
k (s) (2s) 4s = ?
1
3
sp
2
k
S [Ca ]
4
+
? ?
= =
? ?
? ?
2
[Ca ] [F ]
+ -
=
11.(BC)
B.E. of F
2
is lowest and Cl
2
is highest
Acidic character : Increases down the group and left to right across the period.
I.E. of Be > B.
12.(ABCD) Write the cell reaction : Anode : Ag(s) Cl (aq) AgCl(s) 1e
- -
+ ?? ? +
Cathode : Ag (aq) 1e Ag(s)
+ -
+ ?? ?
Ag (aq) Cl (aq) AgCl(s)
+ -
+ ?? ?
?
0 0
cell
Ag Ag Cl AgCl Ag
0.059 1
E E E log
1 [Ag ][Cl ]
+ -
+ -
= - -
Observe that as cell works, the precipitation of AgCl increases in anodic compartment.
When cell stops working, I.P = K
sp
which means [Ag
+
] is now simply =
sp
K = constant.
When E
cell
= 0, it means I.P = K
sp
. [Note that cell is in equilibrium]
Let [Ag
+
]
Cathode
= concentration of Ag
+
ion in cathodic department.
And [Ag
+
]
anode
= concentration of Ag
+
ion in anodic department.
sp c a
1 1
Q (at equilibrium)
K [Ag ] [Cl ]
+ -
= =
eqm.
a
1
Q
[Cl ]
-
=
+
a
[Ag ]
?
Cathode Anode
[Ag ] [Ag ]
+ +
=
*Note that [Ag
+
]
anode
is not visible during the working of cell. It comes to picture only at equilibrium.
Vidyamandir Classes
VMC/2013/Solutions 3 Mock IIT Advanced/Test - 2/Paper-1
13.(ABD)
14.(BC) 0.1 m urea : i = 1 ; 0.04m Al
2
(SO
4
)
3
: i = 5 ; 0.05 m CaCl
2
: i = 3 ; 0.005 m NaCl : i = 2.
Freezing point is highest for solution IV.
Boiling point is highest for solution II.
15.(ABCD) As we know
rms
3RT
C
M
=
For heavier gas, M will be high.
C
rms
will have a lower value.
So graph A correspond to either a heavier gas or same gas at lower temperature.
Similarly graph B correspond to either a lighter gas or same gas at higher temperature.
16.(4) Use :
cell 1 1
P
dE 1.70 1.2
S nF 2 100000 5000 J mol 5 kJ mol
dT 45 25
- -
- ? ? ? ?
? = = × × = =
? ? ? ?
-
? ? ? ?
17.(8) Positronium consists of positron and electron
For positron, the mass of electron will be considered as reduced mass since the nucleus is formed by positron only
and hence the nucleus is movable
e p e
e reduced
e p
(m )(m ) m
(m )
m m 2
= =
+
? The radius of 2
nd
orbit of positronium =
2
o 0
n
(2R ) 8R
Z
× =
?
Due to reduced mass concept
18.(4)
2 2
1
CO O CO
2
x x 2
+ ?? ?
2 2
CO CO
1 x
?? ?
-
Volume of O
2
used = 2 1.6 0.4 - =
?
x
0.4
2
= Hence x = 0.8 Now mole ratio =
0.8
4
0.2
= .
19.(3) x millimoles : Salicylic acid ; y millimoles : Malonic acid in 1000 ml
25 ml extract will contain
x
m.moles
40
of Salicylic and
y
m.moles
40
of Malonic acid
NaOH used for titrating this extract = 2 m.moles
?
2x 2y
2
40 40
+ = ? x + y = 40 . . . .(i)
50 ml extract contains
x
20
m.moles salicylic and
y
20
m.moles malonic acid. On heating only malonic acid will
give CO
2
.
?
y 11.2
y 10 m.moles
20 22.4
= ? = ? x = 30 m.moles ? x/y = 3
20.(6)
Vidyamandir Classes
VMC/2013/Solutions 4 Mock IIT Advanced/Test - 2/Paper-1
Solutions to Mock IIT Advanced/Test - 2[Paper-1]/2013
[PHYSICS]
21.(C)
22.(B)
0
1
2 I 1
4 2
= × × B
a
µ p
p
(due to semicircular part)
0
2
2I
4
= × B
a
µ
p
(due to parallel part) there two field are at right angle to each other
Hence resultant field
2 2 2 0
1 2
I
4
4
= + = + B B B
a
µ
p
p
23.(C) ?
3
K
m
=
2 2 = = =
max
f mg cos tan mg cos mg sin µ ? ? ? ?
2 2
? ? f mg sin m A f m A mg sin ? ? - = ? = +
As
2
? = ? =
max
f f A g sin?
3mg sin
A
k
?
=
24.(D) Thermal Resistance of
4
0 1 1 3
336
336 10
-
= = = =
×
L . .
AC R
KA
Thermal Resistance of
4
0 2
2
336 10
L .
BC R
KA
-
= = =
×
Heat flow rates are
1 2
20 40 20
2
H H
R R R
= , = =
1 2
3
40 40 336
13 44
10
×
= + ? = = H H H . W
R
Rate of melting of ice =
( ) ( ) 13 44 4 18
80
=
f
. / .
H
L
= 40 mg/s
25.(D)
3 3
1 4 4
2
2 3 3
+ × = r r dg r g p s p p ?
( ) ( )
3 12
Diameter 2
2 2
= ? = =
- -
r r
g d g d
s s
? ?
26.(B) 12 6 = ? × = u cos u cos t ? ?
Hence 0 5 = t . sec
Now
2
1
3 75
2
= × - . m u sin t gt ?
10 = u sin m / sec ?
Hence
2 2
1
15 10 10 2 3 0
2
T T T T - = - ? - - = ? T = 3
27.(C) Let AP PB AB = = = when the truck at rest
? When truck moves (AP = x let)
? ( )
2
2 2 2
3
2 4 3
4
x x x x + = - ? = ? =
?
3 4
3 5 37
2 3 4
/
sin /
/
?= = ? ?=
-
? T cos ma ?=
T T sin mg + ?= ?
2
10 4 5
5
1 1 3 5
g cos /
a m / s
sin /
? ×
= = =
+ ? +
x
?
?
T
mg
Page 5
Vidyamandir Classes
VMC/2013/Solutions 1 Mock IIT Advanced/Test - 2/Paper-1
Solutions to Mock IIT Advanced/Test - 2[Paper-1]/2013
[CHEMISTRY]
1.(D) n = 2
3
r
2
=
Change in entropy for reversible adiabatic process is zero. (Always).
2.(B) V = 1Litre
T = 300 K
P = 190/760
? Since at t=8 , 390 mm of Hg pressure is observed
It means, out of the initial 200 mm, only 190 mm should be the pure reactant.
? A
0
= 190 mm Hg ? 1/4 atm.
? PV = nRT
?
1 1
n 10 millimoles
4R 100
= ˜ =
A
0
= 10.
After 10 minutes 0.2 gm of NaOH required
? 5 millions of HBr formed.
? After 10 minutes, 5 millimoles of C
2
H
5
Br has been consumed out of 10.
? Half life = 10 minutes.
3.(A)
1
1 2
SN
1
SN SN
r
% of SN
r r
=
×
=
( ) ( ) ( )
6 6
5 6 5 6
0.8 10 [S] 0.8 10
100 100
3.12 10 [S][Nu ] 0.8 10 [S] 3.12 10 0.8 10
- -
- - - - -
× ×
× = ×
× + × × + ×
= 2.5%
4.(B) Beryllium compounds are covalent (more polarising power of Be
+2
)
Mg, Ca, Sr, Ba form divalent ionic compounds.
5.(B)
2 3
CaO SiO CaSiO (slag) + ?? ?
6.(A)
7.(B)
2 2 8
H S O
O O
|| ||
HO S O O S OH
|| ||
O O
- - - - - Peroxy
2 2 7
H S O
O O
|| ||
HO S O S OH
|| ||
O O
- - - - Non-Peroxy
Vidyamandir Classes
VMC/2013/Solutions 2 Mock IIT Advanced/Test - 2/Paper-1
2 5
H SO
O
||
HO S O O H
||
O
- - - - Peroxy
5
CrO
Peroxy
8.(D)
2 3
HCN H O H COOH NH
(X)
+ ?? ? +
2 4
H SO
2
HCOOH CO H O
(Y)
?
???? ? +
473K
Bond order 3
CO NaOH HCOO Na
+
=
+ ????
T
2
CO FeO Fe CO + ?? ? +
9.(B)
3 3
AgNO KI AgI K NO
+ -
?
+ ?? ? + +
AgI due to the field of Ag+, further I
-
are attached to it
? it becomes (AgI) I
-
negative sol.
10.(D) For a saturated solution of CaF
2
.
2
2 (aq) (aq)
CaF (s) Ca 2f
s 2s
+ -
+
2 3
sp
k (s) (2s) 4s = ?
1
3
sp
2
k
S [Ca ]
4
+
? ?
= =
? ?
? ?
2
[Ca ] [F ]
+ -
=
11.(BC)
B.E. of F
2
is lowest and Cl
2
is highest
Acidic character : Increases down the group and left to right across the period.
I.E. of Be > B.
12.(ABCD) Write the cell reaction : Anode : Ag(s) Cl (aq) AgCl(s) 1e
- -
+ ?? ? +
Cathode : Ag (aq) 1e Ag(s)
+ -
+ ?? ?
Ag (aq) Cl (aq) AgCl(s)
+ -
+ ?? ?
?
0 0
cell
Ag Ag Cl AgCl Ag
0.059 1
E E E log
1 [Ag ][Cl ]
+ -
+ -
= - -
Observe that as cell works, the precipitation of AgCl increases in anodic compartment.
When cell stops working, I.P = K
sp
which means [Ag
+
] is now simply =
sp
K = constant.
When E
cell
= 0, it means I.P = K
sp
. [Note that cell is in equilibrium]
Let [Ag
+
]
Cathode
= concentration of Ag
+
ion in cathodic department.
And [Ag
+
]
anode
= concentration of Ag
+
ion in anodic department.
sp c a
1 1
Q (at equilibrium)
K [Ag ] [Cl ]
+ -
= =
eqm.
a
1
Q
[Cl ]
-
=
+
a
[Ag ]
?
Cathode Anode
[Ag ] [Ag ]
+ +
=
*Note that [Ag
+
]
anode
is not visible during the working of cell. It comes to picture only at equilibrium.
Vidyamandir Classes
VMC/2013/Solutions 3 Mock IIT Advanced/Test - 2/Paper-1
13.(ABD)
14.(BC) 0.1 m urea : i = 1 ; 0.04m Al
2
(SO
4
)
3
: i = 5 ; 0.05 m CaCl
2
: i = 3 ; 0.005 m NaCl : i = 2.
Freezing point is highest for solution IV.
Boiling point is highest for solution II.
15.(ABCD) As we know
rms
3RT
C
M
=
For heavier gas, M will be high.
C
rms
will have a lower value.
So graph A correspond to either a heavier gas or same gas at lower temperature.
Similarly graph B correspond to either a lighter gas or same gas at higher temperature.
16.(4) Use :
cell 1 1
P
dE 1.70 1.2
S nF 2 100000 5000 J mol 5 kJ mol
dT 45 25
- -
- ? ? ? ?
? = = × × = =
? ? ? ?
-
? ? ? ?
17.(8) Positronium consists of positron and electron
For positron, the mass of electron will be considered as reduced mass since the nucleus is formed by positron only
and hence the nucleus is movable
e p e
e reduced
e p
(m )(m ) m
(m )
m m 2
= =
+
? The radius of 2
nd
orbit of positronium =
2
o 0
n
(2R ) 8R
Z
× =
?
Due to reduced mass concept
18.(4)
2 2
1
CO O CO
2
x x 2
+ ?? ?
2 2
CO CO
1 x
?? ?
-
Volume of O
2
used = 2 1.6 0.4 - =
?
x
0.4
2
= Hence x = 0.8 Now mole ratio =
0.8
4
0.2
= .
19.(3) x millimoles : Salicylic acid ; y millimoles : Malonic acid in 1000 ml
25 ml extract will contain
x
m.moles
40
of Salicylic and
y
m.moles
40
of Malonic acid
NaOH used for titrating this extract = 2 m.moles
?
2x 2y
2
40 40
+ = ? x + y = 40 . . . .(i)
50 ml extract contains
x
20
m.moles salicylic and
y
20
m.moles malonic acid. On heating only malonic acid will
give CO
2
.
?
y 11.2
y 10 m.moles
20 22.4
= ? = ? x = 30 m.moles ? x/y = 3
20.(6)
Vidyamandir Classes
VMC/2013/Solutions 4 Mock IIT Advanced/Test - 2/Paper-1
Solutions to Mock IIT Advanced/Test - 2[Paper-1]/2013
[PHYSICS]
21.(C)
22.(B)
0
1
2 I 1
4 2
= × × B
a
µ p
p
(due to semicircular part)
0
2
2I
4
= × B
a
µ
p
(due to parallel part) there two field are at right angle to each other
Hence resultant field
2 2 2 0
1 2
I
4
4
= + = + B B B
a
µ
p
p
23.(C) ?
3
K
m
=
2 2 = = =
max
f mg cos tan mg cos mg sin µ ? ? ? ?
2 2
? ? f mg sin m A f m A mg sin ? ? - = ? = +
As
2
? = ? =
max
f f A g sin?
3mg sin
A
k
?
=
24.(D) Thermal Resistance of
4
0 1 1 3
336
336 10
-
= = = =
×
L . .
AC R
KA
Thermal Resistance of
4
0 2
2
336 10
L .
BC R
KA
-
= = =
×
Heat flow rates are
1 2
20 40 20
2
H H
R R R
= , = =
1 2
3
40 40 336
13 44
10
×
= + ? = = H H H . W
R
Rate of melting of ice =
( ) ( ) 13 44 4 18
80
=
f
. / .
H
L
= 40 mg/s
25.(D)
3 3
1 4 4
2
2 3 3
+ × = r r dg r g p s p p ?
( ) ( )
3 12
Diameter 2
2 2
= ? = =
- -
r r
g d g d
s s
? ?
26.(B) 12 6 = ? × = u cos u cos t ? ?
Hence 0 5 = t . sec
Now
2
1
3 75
2
= × - . m u sin t gt ?
10 = u sin m / sec ?
Hence
2 2
1
15 10 10 2 3 0
2
T T T T - = - ? - - = ? T = 3
27.(C) Let AP PB AB = = = when the truck at rest
? When truck moves (AP = x let)
? ( )
2
2 2 2
3
2 4 3
4
x x x x + = - ? = ? =
?
3 4
3 5 37
2 3 4
/
sin /
/
?= = ? ?=
-
? T cos ma ?=
T T sin mg + ?= ?
2
10 4 5
5
1 1 3 5
g cos /
a m / s
sin /
? ×
= = =
+ ? +
x
?
?
T
mg
Vidyamandir Classes
VMC/2013/Solutions 5 Mock IIT Advanced/Test - 2/Paper-1
28.(C)
( )
1 1
t / RC Rt / L t / LC t / LC
V V
e e e e
R R
- - - -
= - ? = -
?
1
2
t / LC
e
-
= ? 2
t
n
LC
= ? 2 2 t LC n RC n = =
29.(A)
44
10 1 44 700 308
4 100
cm C f ms
?
? ? = + ? = ? = = × =
Next resonance occurs for length =
3
1 32
4
cm
?
- =
1 cm = d/3 ? diameter = 3cm.
30.(D)
31.(CD) Velocity of the 1st particle when 2nd is projected = ( )( ) 100 10 1 + -
& it is at height of = ( )( )
2 1
100 1 10 1 90
2
m / s × + - = = 95 m.
After this till both are in motion, relative velocity which in zero, will not change as relative acc. is zero.
32.(ABD)
( ) ( )
1 4 1
Rt / L t
E
i e e
R
- -
= - = -
= 2 Ampere at 2 t n sec =
? rate of energy supplied by battery = E i = 16 J/S
rate of heat dissipated across R = i
2
R = 8 J/S
8 4
a b
V V iR V - = - =
33.(ACD) When S is open current through both resistance is zero in steady state ? potential of
a = 18V & potential of b is zero.
When S is closed then a current of 2A flows through each
resistor & the switch as shown, with change of
72 18 c & c µ µ on C
1
& C
2
respectively
? V
a
= V
b
= 6V.
1 2
for both 36 Q C & C C ? =- µ
34.(AC)
2 2
2
n z z
R , v , E
z n
n
? ? ?
35.(AC)
2
radius pitch
mv mV
&
qB qB
p
?
= =
?
27
27
2 2
19 19
2 4 003 1 6 10
4 003 1 6 10
5 10 7 5 10
2 1 6 10 5 2 1 6 10 5
. . V
. . V
& .
. .
p
p
-
-
- - ?
- -
× × ×
× ×
× = × =
× × × × × ×
?
6
25
10
2
V m / s
?
= ×
6
37 5
10
4
.
V m / s
?
? = ×
? Velocity of a particle =
2 2 6
62 5
10
4
.
V V m / s
?
+ = ×
7
1 56 10 . m / s ˜ ×
KE of a particle =
( )( )
2
2 27 7
1 1
4 003 1 6 10 1 56 10
2 2
m V . . . J
a
-
= × × × ×
=
( )
2
27 7
13
4 003 1 6 10 1 56 10
1
2
1 6 10
. . .
MeV
.
-
-
× × × ×
×
×
4 7 . MeV ˜
E
L
R
i
18V
+ +
– –
+ +
– –
6O
18 C µ
2A
2A
2A
72 C µ
18V
+ +
– –
+ +
– –
6O
3O
108 C µ
3 F µ
54 C µ
A
B
Read More
FAQs on IIT JEE - Sample Mock Test, Engineering, Exam Prepapration PDF Download
| 1. What is the IIT JEE exam? |  |
Ans. The IIT JEE exam, also known as the Indian Institutes of Technology Joint Entrance Examination, is a highly competitive engineering entrance exam conducted in India. It is the gateway to undergraduate engineering programs offered by the Indian Institutes of Technology (IITs) and other prestigious institutes in the country.
| 2. How can I prepare for the IIT JEE exam? | |
Ans. To prepare for the IIT JEE exam, it is important to have a structured study plan. Start by understanding the exam pattern and syllabus thoroughly. Then, refer to standard textbooks and study materials recommended by experts. Practice solving previous year question papers and take mock tests to assess your preparation level. It is also beneficial to join a coaching institute or online courses to get guidance from experienced faculty and to clarify doubts.
| 3. Are sample mock tests helpful for IIT JEE preparation? | |
Ans. Yes, sample mock tests are extremely helpful for IIT JEE preparation. They provide a simulated exam-like environment and help in understanding the exam pattern, time management, and identifying strengths and weaknesses. Regularly solving mock tests allows you to assess your performance and work on improving your problem-solving skills. It also boosts your confidence and reduces exam anxiety.
| 4. How can I manage my time effectively during the IIT JEE exam? | |
Ans. Time management is crucial during the IIT JEE exam. One effective strategy is to divide the time available for each section based on the number of questions and difficulty level. Start with easier questions and move on to the more challenging ones. Avoid spending too much time on a single question. If you are unsure about an answer, mark it for review and move ahead. Use the last few minutes to review your answers and ensure you haven't made any silly mistakes.
| 5. What are some important tips for success in the IIT JEE exam? | |
Ans. Here are some important tips for success in the IIT JEE exam:
- Start early and have a disciplined study routine.
- Focus on conceptual understanding rather than rote learning.
- Practice regularly and solve a variety of problems from different topics.
- Analyze your mistakes and work on improving them.
- Stay updated with the latest exam pattern and syllabus.
- Take care of your health and manage stress effectively.
- Stay motivated and believe in your abilities.
About this Document
|
Vinod Sharma |
Document Description: IIT JEE - Sample Mock Test, Engineering, Exam Prepapration for 2023 is part of for preparation.
The notes and questions for IIT JEE - Sample Mock Test, Engineering, Exam Prepapration have been prepared according to the exam syllabus.
Information about IIT JEE - Sample Mock Test, Engineering, Exam Prepapration covers topics like and IIT JEE - Sample Mock Test, Engineering, Exam Prepapration Example, for 2023
Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for IIT JEE - Sample Mock Test, Engineering, Exam Prepapration.
Introduction of IIT JEE - Sample Mock Test, Engineering, Exam Prepapration in English is available as part of our for & IIT JEE - Sample Mock Test, Engineering, Exam Prepapration in Hindi
for course. Download more important topics related with , notes, lectures and mock test
series for Exam by signing up for free. : IIT JEE - Sample Mock Test, Engineering, Exam Prepapration
Description
Full syllabus notes, lecture & questions for IIT JEE - Sample Mock Test, Engineering, Exam Prepapration | Plus excerises question with solution to help you revise complete syllabus | Best notes, free PDF download
Information about IIT JEE - Sample Mock Test, Engineering, Exam Prepapration
In this doc you can find the meaning of IIT JEE - Sample Mock Test, Engineering, Exam Prepapration defined & explained in the simplest way possible.
Besides explaining types of IIT JEE - Sample Mock Test, Engineering, Exam Prepapration theory, EduRev gives you an ample number of questions to practice IIT JEE - Sample Mock Test, Engineering, Exam Prepapration tests,
examples and also practice tests.
IIT JEE - Sample Mock Test, Engineering, Exam Prepapration is the part of for 2023 exam preparation. The content of IIT JEE - Sample Mock Test, Engineering, Exam Prepapration has been prepared
for learning according to the exam syllabus. IIT JEE - Sample Mock Test, Engineering, Exam Prepapration covers topics like for
2023 Exam.
Find important questions, notes, tests & features of IIT JEE - Sample Mock Test, Engineering, Exam Prepapration in this document.
How EduRev helps you in preparation?
EduRev provides you with complete coverage and for 2023 IIT JEE - Sample Mock Test, Engineering, Exam Prepapration as a part of our plan for
syllabus to prepare for exam. Our subject experts have curated special courses, tests & mock
test series. All previous year questions (PYQs) & topic wise tests like tests for IIT JEE - Sample Mock Test, Engineering, Exam Prepapration are provided with detailed
solutions. Download more important topics, notes, lectures and mock test series for Exam by signing up for free
Download as PDF
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Download free EduRev App
Track your progress, build streaks, highlight & save important lessons and more!
(Scan QR code)
|
|
Additional Information about IIT JEE - Sample Mock Test, Engineering, Exam Prepapration for Preparation
IIT JEE - Sample Mock Test, Engineering, Exam Prepapration Free PDF Download
The IIT JEE - Sample Mock Test, Engineering, Exam Prepapration is an invaluable resource that delves deep into the core of the exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the IIT JEE - Sample Mock Test, Engineering, Exam Prepapration now and kickstart your journey towards success in the exam.
Importance of IIT JEE - Sample Mock Test, Engineering, Exam Prepapration
The importance of IIT JEE - Sample Mock Test, Engineering, Exam Prepapration cannot be overstated, especially for aspirants.
This document holds the key to success in the exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
IIT JEE - Sample Mock Test, Engineering, Exam Prepapration Notes
IIT JEE - Sample Mock Test, Engineering, Exam Prepapration Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to IIT JEE - Sample Mock Test, Engineering, Exam Prepapration.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, IIT JEE - Sample Mock Test, Engineering, Exam Prepapration Notes on EduRev are your ultimate resource for success.
IIT JEE - Sample Mock Test, Engineering, Exam Prepapration Questions
The "IIT JEE - Sample Mock Test, Engineering, Exam Prepapration Questions" guide is a valuable resource for all aspiring students preparing for the
exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study IIT JEE - Sample Mock Test, Engineering, Exam Prepapration on the App
Students of can study IIT JEE - Sample Mock Test, Engineering, Exam Prepapration alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the IIT JEE - Sample Mock Test, Engineering, Exam Prepapration,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of IIT JEE - Sample Mock Test, Engineering, Exam Prepapration is prepared as per the latest syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup
