IIT JEE-Advanced 2014 - Question paper with Solution - Exam Preparation

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

= 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)

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
-

= 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)

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
-

= 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
-
×  ×

```

## FAQs on IIT JEE-Advanced 2014 - Question paper with Solution - Exam Preparation

 1. What is the IIT JEE-Advanced 2014 exam?
Ans. The IIT JEE-Advanced 2014 exam is a competitive entrance examination for admission into the Indian Institutes of Technology (IITs) in India. It is considered one of the most prestigious engineering entrance exams in the country.
 2. How can I prepare for the IIT JEE-Advanced 2014 exam?
Ans. To prepare for the IIT JEE-Advanced 2014 exam, candidates should start by thoroughly understanding the exam pattern and syllabus. They should then create a study schedule, gather relevant study materials, and practice solving previous years' question papers and mock tests. It is also recommended to seek guidance from experienced mentors or join coaching institutes for additional support.
 3. Can I get the IIT JEE-Advanced 2014 question paper with solutions?
Ans. Yes, you can find the IIT JEE-Advanced 2014 question paper with solutions online. Many educational websites and coaching institutes provide free or paid access to these question papers and their solutions. Additionally, you can refer to books or study materials specifically designed for the IIT JEE-Advanced exam, which often include previous years' question papers with detailed solutions.
 4. What is the difficulty level of the IIT JEE-Advanced 2014 exam?
Ans. The IIT JEE-Advanced 2014 exam is known for its high difficulty level. The questions are designed to test the candidates' conceptual understanding, analytical skills, and problem-solving abilities. The exam often includes complex and challenging problems, requiring a strong foundation in physics, chemistry, and mathematics.
 5. How important is the IIT JEE-Advanced 2014 exam for admissions to the IITs?
Ans. The IIT JEE-Advanced 2014 exam is the second and final stage of the selection process for admissions to the IITs. It holds significant importance as it solely determines the candidates' eligibility for admission into the prestigious IITs. The final ranks obtained in the JEE-Advanced exam are used for seat allocation in various undergraduate engineering programs offered by the IITs.
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