IIT JEE-Advanced 2014 - Question paper with Solution - Exam Preparation JEE Notes | EduRev

JEE : IIT JEE-Advanced 2014 - Question paper with Solution - Exam Preparation JEE Notes | EduRev

 Page 1


 
(1) 
 
IIT - JEE 2014 (Advanced) 
 
 
CODE : 1 
Page 2


 
(1) 
 
IIT - JEE 2014 (Advanced) 
 
 
CODE : 1 
(2) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution 
(2) 
 
Page 3


 
(1) 
 
IIT - JEE 2014 (Advanced) 
 
 
CODE : 1 
(2) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution 
(2) 
 
IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (3) 
(3) 
PART I : PHYSICS 
SECTION - 1 : (One or More Than One Options Correct Type) 
 
This section contains 10 multiple choice questions. Each question has four choices (A), 
(B), (C) and (D) out of which ONE or MORE THAN ONE are correct. 
 
 
 
1. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and 
an alternating current I(t) = I
0
 cos (?t), with I
0
 = 1A and ? = 500 rad s
-1
 starts flowing in 
it with the initial direction shown in the figure. At t = 
7
6
p
?
, the key is switched from B to 
D. Now onwards only A and D are connected. A total charge Q flows from the battery to 
charge the capacitor fully. If C = 20µF, R = 10 O and the battery is ideal with emf of 50V, 
identify the correct statement (s). 
 
 
 
 
 
 
 
 
(A) Magnitude of the maximum charge on the capacitor before t = 
7
6
p
?
 is 1 × 10
-3
 C. 
(B) The current in the left part of the circuit just before t = 
7
6
p
?
 is clockwise. 
(C) Immediately after A is connected to D, the current in R is 10A. 
(D) Q = 2 × 10
-3
 C. 
 
1. (C), (D) 
 If q represents the charge on capacitor’s upper plate: 
 I (t) = I
0
 cos(?t) = 
dq
dt
 ? q(t) = 
0
I
?
 sin(?t) 
 Max charge = 
1
1A
500 rad s
-
 
  
 = 2 × 10
-3 
C 
 Charge on upper plate at t = 
7
6
p
?
 = 
0
I
?
 sin 
7
6
p ??
??
??
 = 
0
I
2
-
?
  
 When capacitor is fully charged; charge on upper plate = 50 V × 20µF 
                = 1 × 10
-3
 C 
 ?  Q = 1 × 10
-2
 C - 
0
I
2
??
-
??
?
??
 = 2 × 10
-3
 C 
 Voltage across capacitor when A and D are connected = 
3
6
110 C
20 10 F
-
-
 × 
 × µ
 = 50V  
 Total voltage across resistor = 100V 
 ? current = 
100v
10O
 = 10A 
 Current is anti clockwise.  
 
Page 4


 
(1) 
 
IIT - JEE 2014 (Advanced) 
 
 
CODE : 1 
(2) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution 
(2) 
 
IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (3) 
(3) 
PART I : PHYSICS 
SECTION - 1 : (One or More Than One Options Correct Type) 
 
This section contains 10 multiple choice questions. Each question has four choices (A), 
(B), (C) and (D) out of which ONE or MORE THAN ONE are correct. 
 
 
 
1. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and 
an alternating current I(t) = I
0
 cos (?t), with I
0
 = 1A and ? = 500 rad s
-1
 starts flowing in 
it with the initial direction shown in the figure. At t = 
7
6
p
?
, the key is switched from B to 
D. Now onwards only A and D are connected. A total charge Q flows from the battery to 
charge the capacitor fully. If C = 20µF, R = 10 O and the battery is ideal with emf of 50V, 
identify the correct statement (s). 
 
 
 
 
 
 
 
 
(A) Magnitude of the maximum charge on the capacitor before t = 
7
6
p
?
 is 1 × 10
-3
 C. 
(B) The current in the left part of the circuit just before t = 
7
6
p
?
 is clockwise. 
(C) Immediately after A is connected to D, the current in R is 10A. 
(D) Q = 2 × 10
-3
 C. 
 
1. (C), (D) 
 If q represents the charge on capacitor’s upper plate: 
 I (t) = I
0
 cos(?t) = 
dq
dt
 ? q(t) = 
0
I
?
 sin(?t) 
 Max charge = 
1
1A
500 rad s
-
 
  
 = 2 × 10
-3 
C 
 Charge on upper plate at t = 
7
6
p
?
 = 
0
I
?
 sin 
7
6
p ??
??
??
 = 
0
I
2
-
?
  
 When capacitor is fully charged; charge on upper plate = 50 V × 20µF 
                = 1 × 10
-3
 C 
 ?  Q = 1 × 10
-2
 C - 
0
I
2
??
-
??
?
??
 = 2 × 10
-3
 C 
 Voltage across capacitor when A and D are connected = 
3
6
110 C
20 10 F
-
-
 × 
 × µ
 = 50V  
 Total voltage across resistor = 100V 
 ? current = 
100v
10O
 = 10A 
 Current is anti clockwise.  
 
(4) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution 
(4) 
2. A light source, which emits two wavelengths ?
1
 = 400 nm and ?
2
 = 600 nm, is used in a 
Young's double slit experiment. If recorded fringe widths for ?
1
 and ?
2
 are ß
1
 and ß
2
 and 
the number of fringes for them within a distance y on one side of the central maximum are 
m
1
 and m
2
, respectively, then 
(A) ß
2
 > ß
1
 
(B) m
1
 > m
2
 
(C) From the central maximum, 3
rd
 maximum of ?
2
 overlaps with 5
th
 minimum of ?
1
 
(D) The angular separation of fringes for ?
1
 is greater than ?
2
 
 
2. (A), (B), (C) 
 ß = ? 
D
d
  
 
1
22 1
nm
2
?
?= 
 n
2
600 = 
1
400
m
2
 
 3n
2
 = m
1
 
 For n
2
 = 3 
 m
1 
= 9 
 which is 5
th
 minima. 
 
3. One end of a taut string of length 3m along the x axis is fixed at x = 0. The speed of the 
waves in the string is 100 ms
-1
. The other end of the string is vibrating in the y direction 
so that stationary waves are set up in the string. The possible waveform(s) of these 
stationary waves is (are) 
(A) y(t) = 
x50t
Asin cos
63
pp
        (B) y(t) = 
x100t
Asin cos
33
pp
    
(C) y(t) = 
5 x 250 t
Asin cos
63
pp
       (D) y(t) = 
5x
Asin cos t
2
p
   250 p 
3. (A), (C), (D) 
 ? = 
k
?
  
 
1
AB C D
100ms
-
? = ? = ? = ? =   
 x = 3 is an antinode. This eliminates (B)
 
 
4. A parallel plate capacitor has a dielectric slab of dielectric constant K 
between its plates that covers 1/3 of the area of its plates, as shown in 
the figure. The total capacitance of the capacitor is C while that of the 
portion with dielectric in between is C
1
. When the capacitor is charged, 
the plate area covered by the dielectric gets charge Q
1
 and the rest of 
the area gets charge Q
2
. The electric field in the dielectric is E
1
 and that 
in the other portion is E
2
. Choose the correct option/options, ignoring 
edge effects. 
(A) 
1
2
E
E
 = 1         (B) 
1
2
E
E
 = 
1
K
 
(C) 
1
2
Q
Q
 = 
3
K
         (D) 
1
C
C
 = 
2
K
 + ?
 
Page 5


 
(1) 
 
IIT - JEE 2014 (Advanced) 
 
 
CODE : 1 
(2) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution 
(2) 
 
IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (3) 
(3) 
PART I : PHYSICS 
SECTION - 1 : (One or More Than One Options Correct Type) 
 
This section contains 10 multiple choice questions. Each question has four choices (A), 
(B), (C) and (D) out of which ONE or MORE THAN ONE are correct. 
 
 
 
1. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and 
an alternating current I(t) = I
0
 cos (?t), with I
0
 = 1A and ? = 500 rad s
-1
 starts flowing in 
it with the initial direction shown in the figure. At t = 
7
6
p
?
, the key is switched from B to 
D. Now onwards only A and D are connected. A total charge Q flows from the battery to 
charge the capacitor fully. If C = 20µF, R = 10 O and the battery is ideal with emf of 50V, 
identify the correct statement (s). 
 
 
 
 
 
 
 
 
(A) Magnitude of the maximum charge on the capacitor before t = 
7
6
p
?
 is 1 × 10
-3
 C. 
(B) The current in the left part of the circuit just before t = 
7
6
p
?
 is clockwise. 
(C) Immediately after A is connected to D, the current in R is 10A. 
(D) Q = 2 × 10
-3
 C. 
 
1. (C), (D) 
 If q represents the charge on capacitor’s upper plate: 
 I (t) = I
0
 cos(?t) = 
dq
dt
 ? q(t) = 
0
I
?
 sin(?t) 
 Max charge = 
1
1A
500 rad s
-
 
  
 = 2 × 10
-3 
C 
 Charge on upper plate at t = 
7
6
p
?
 = 
0
I
?
 sin 
7
6
p ??
??
??
 = 
0
I
2
-
?
  
 When capacitor is fully charged; charge on upper plate = 50 V × 20µF 
                = 1 × 10
-3
 C 
 ?  Q = 1 × 10
-2
 C - 
0
I
2
??
-
??
?
??
 = 2 × 10
-3
 C 
 Voltage across capacitor when A and D are connected = 
3
6
110 C
20 10 F
-
-
 × 
 × µ
 = 50V  
 Total voltage across resistor = 100V 
 ? current = 
100v
10O
 = 10A 
 Current is anti clockwise.  
 
(4) Vidyalankar : IIT JEE 2014 - Advanced : Question Paper & Solution 
(4) 
2. A light source, which emits two wavelengths ?
1
 = 400 nm and ?
2
 = 600 nm, is used in a 
Young's double slit experiment. If recorded fringe widths for ?
1
 and ?
2
 are ß
1
 and ß
2
 and 
the number of fringes for them within a distance y on one side of the central maximum are 
m
1
 and m
2
, respectively, then 
(A) ß
2
 > ß
1
 
(B) m
1
 > m
2
 
(C) From the central maximum, 3
rd
 maximum of ?
2
 overlaps with 5
th
 minimum of ?
1
 
(D) The angular separation of fringes for ?
1
 is greater than ?
2
 
 
2. (A), (B), (C) 
 ß = ? 
D
d
  
 
1
22 1
nm
2
?
?= 
 n
2
600 = 
1
400
m
2
 
 3n
2
 = m
1
 
 For n
2
 = 3 
 m
1 
= 9 
 which is 5
th
 minima. 
 
3. One end of a taut string of length 3m along the x axis is fixed at x = 0. The speed of the 
waves in the string is 100 ms
-1
. The other end of the string is vibrating in the y direction 
so that stationary waves are set up in the string. The possible waveform(s) of these 
stationary waves is (are) 
(A) y(t) = 
x50t
Asin cos
63
pp
        (B) y(t) = 
x100t
Asin cos
33
pp
    
(C) y(t) = 
5 x 250 t
Asin cos
63
pp
       (D) y(t) = 
5x
Asin cos t
2
p
   250 p 
3. (A), (C), (D) 
 ? = 
k
?
  
 
1
AB C D
100ms
-
? = ? = ? = ? =   
 x = 3 is an antinode. This eliminates (B)
 
 
4. A parallel plate capacitor has a dielectric slab of dielectric constant K 
between its plates that covers 1/3 of the area of its plates, as shown in 
the figure. The total capacitance of the capacitor is C while that of the 
portion with dielectric in between is C
1
. When the capacitor is charged, 
the plate area covered by the dielectric gets charge Q
1
 and the rest of 
the area gets charge Q
2
. The electric field in the dielectric is E
1
 and that 
in the other portion is E
2
. Choose the correct option/options, ignoring 
edge effects. 
(A) 
1
2
E
E
 = 1         (B) 
1
2
E
E
 = 
1
K
 
(C) 
1
2
Q
Q
 = 
3
K
         (D) 
1
C
C
 = 
2
K
 + ?
 
IIT JEE 2014 Advanced : Question Paper & Solution (Paper – I) (5) 
(5) 
4. (A), (D) 
 If c = 
o
A
3d
e
 then   c
1
 = kc   c
2
 = 2c 
 c
1
 + c
2
 = C ? (k + 2) c = C ? c = 
c
k2 + 
  
 ?  c
1
 = 
kc
k2 +
   c
2
 = 
2C
kz + 
  
If charging voltage is V: charges will be in the ratio of capacities and as potential 
difference is same. Electric field should be equal. 
 
5. Let E
1
(r), E
2
(r) and E
3
(r) be the respective electric fields at a distance r from a point 
charge Q, an infinitely long wire with constant linear charge density ?, and an infinite 
plane with uniform surface charge density s. If E
1
(r
0
) = E
2
(r
0
) = E
3
(r
0
) at a given distance 
r
0
, then 
(A) Q = 
2
0
4r sp         (B) r
0
 = 
2
?
ps
 
(C) E
1
(r
0
/2) = 2E
2
(r
0
/2)      (D) E
2
(r
0
/2) = 4E
3
(r
0
/2) 
5. (C) 
 
2
0
2
00
000
0
2
0
0 2
00
0
1q
2r q
2r 2 r
42r
r
1q
q2r
r
42
r
? ?  =  ? ? = 
? p s = ? ? pe pe
? 
? ? s ? s = 
?  =  ?  = p s
p
? pe e
?  
 
10
E(r /2)
4
 = 
0
E(r / 2)
2
  
 
6. A student is performing an experiment using a resonance column and a tuning fork of 
frequency 244 s
-1
. He is told that the air in the tube has been replaced by another gas 
(assume that the column remains filled with the gas). If the minimum height at which 
resonance occurs is (0.350 ± 0.005) m, the gas in the tube is 
(Useful information : 167RT = 640 j
1/2
 mole
-1/2
; 140RT = 590 J
1/2
 mole
-1/2
. THe 
molar masses M in grams are given in the options. Take the values of 
10
M
 for each gas 
as given there) 
(A) Neon
10 7
M20,
20 10
??
 =   = 
??
??
    (B) Nitrogen
10 3
M28,
28 5
??
 =   = 
??
??
 
(C) Oxygen
10 9
M32,
32 16
??
 =   = 
??
??
    (D) Argon
10 17
M36,
36 32
??
 =   = 
??
??
 
6. (D) 
 f = 244 Hz 
 
4
?
 = 0.356 ± 0.005 ? ? = 1.400 ± 0.020 ? ? = (341.6 ± 4.88) 
m
s
  
? = ?f = ? = 
rRT
M
 = 
100 rRT
100M
 
 = 
3
100RrT
100 M 10 kg/mol
-
 ×  ×  
  
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