IIT JEE - Sample Mock Test, Engineering, Exam Prepapration Notes | EduRev

Created by: Vinod Sharma

: IIT JEE - Sample Mock Test, Engineering, Exam Prepapration Notes | EduRev

 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 
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