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Years IIT-JEE
CHAPTERWISE SOLVED PAPERS
PHYSICS
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Preface
Whenever a student decides to prepare for any examination, her/his first and foremost curiosity
about the type of questions that he/she has to face. This becomes more important in the context of
competitive examinations where there is neck-to-neck race.
We feel great pleasure to present before you this book. We have made an attempt to provide
chapter wise questions asked in IIT-JEE / JEE Advanced from 1978 to 2016 along with solutions.
Solutions to the questions are not just sketch rather have been written in such a manner that the
students will be able to under the application of concept and can answer some other related
questions too.
We firmly believe that the book in this form will definitely help a genuine, hardworking student.
We have tried our best to keep errors out of this book. Comment and criticism from readers will be
highly appreciated and incorporated in the subsequent edition.
We wish to utilize the opportunity to place on record our special thanks to all team members of
Content Development for their efforts to make this wonderful book.
Career Point Ltd.
CONTENTS
Chapter
Page No.
1.
Unit, Dimension & Error
♦ Answers
♦ Solutions
1‐16
9
10
2.
Kinematics
♦ Answers
♦ Solutions
17‐32
23
24
3.
Laws of Motion & Friction
♦ Answers
♦ Solutions
33‐54
41
42
4.
Work, Power and Energy
♦ Answers
♦ Solutions
55‐68
61
62
5
Conservation Law
♦ Answers
♦ Solutions
69‐88
77
78
6.
Rotational Motion
♦ Answers
♦ Solutions
89‐130
106
108
7.
Gravitation
♦ Answers
♦ Solutions
131‐142
135
136
8.
Simple Harmonic Motion
♦ Answers
♦ Solutions
143‐162
151
152
9.
Properties of Matter & Fluid Mechanics
♦ Answers
♦ Solutions
163‐188
10.
Wave Motion
♦ Answers
♦ Solutions
189‐222
201
203
11.
Heat and Thermodynamics
♦ Answers
♦ Solutions
223‐282
246
248
173
174
Cont....
12.
Electrostatics
♦ Answers
♦ Solutions
283‐332
303
305
13
Current Electricity
♦ Answers
♦ Solutions
333‐360
345
346
14.
Magnetic Effect of Current
♦ Answers
♦ Solutions
361‐400
378
380
15.
Electromagnetic Induction and Alternating Current
♦ Answers
♦ Solutions
401‐434
415
417
16.
Optics
♦ Answers
♦ Solutions
435‐504
461
463
17.
Modern Physics
♦ Answers
♦ Solutions
505‐564
529
531
18.
Model Test Papers
♦ Practice Test‐1 [Paper‐1]
♦ Practice Test‐1 [Paper‐2]
♦ Practice Test‐2 [Paper‐1]
♦ Practice Test‐2 [Paper‐2]
565‐598
565
573
581
589
Chapter
1
Unit, Dimension & Error
ONLY ONE CORRECT ANSWER
1.
4.
(B) [M–3 L–2 T4 Q4]
(D) [M–3 L–2 T4 Q]
AB
(B) [ML2 T–2]
(D) [ML2 T–1]
∆V
, where ε0 is
∆t
the permittivity of free space, L is a length, ∆V
is a potential difference and ∆t is a time
interval. The dimensional formula for X is the
[2001, 2M]
same as that of
(A) resistance
(B) charge
(C) voltage
(D) current
A cube has a side of length 1.2 × 10–2m.
Calculate its volume
[2003, 2M]
–6
3
(A) 1.7 × 10 m
(B) 1.73 × 10–6 m3
–6
3
(C) 1.70 × 10 m
(D) 1.732 × 10–6 m3
α – kθ
In the relation p =
e
β
(A) [M0 L2 T0]
(C) [ML0 T–1]
(B) [ML2 T]
(D) [M0 L2 T–1]
A wire has a mass (0.3 ± 0.003) g, radius
(0.5 ± 0.005) mm and length (6 ± 0.06) cm.
The maximum percentage error in the
[2004, 2M]
measurement of its density is
(A) 1
(B) 2
(C) 3
(D) 4
S N H
:: : :: : : ′0
E
5
K
R
A quantity X is given by ε0 L
p is pressure, Z is distance, k is Boltzmann
constant and θ is the temperature. The dimensional
formula of β will be [2004, 2M]
6.
The circular scale of a screw gauge has
50 divisions and pitch of 0.5 mm. Find the
diameter of sphere. Main scale reading is 2.
[2006, 3M]
1
ε0E2 (ε0 : permittivity of
2
free space; E : electric field) is
[2000, 2M]
αZ
5.
8.
The dimensions of
(A) [MLT–1]
(C) [MLT–2]
3.
Which of the following sets have different
dimensions?
[2005, 2M]
(A) Pressure, Young's modulus, Stress
(B) Emf, Potential difference, Electric potential
(C) Heat, Work done, Energy
(D) Dipole moment, Electric flux, Electric field
In the formula X = 3Y Z2, X and Z have
dimensions of capacitance and magnetic
induction respectively. What are the
dimensions of Y is MKSQ system? [1995, 2M]
(A) [M–3 L–1 T3 Q4]
(C) [M–2 L–2 T4 Q4]
2.
7.
M
AB
S N H
: : ::: : : ′′2
E
25
K
R
M
(A) 1.2 mm
(C) 2.20 mm
9.
(B) 1.25 mm
(D) 2.25 mm
A student performs an experiment to determine
the Young's modulus of a wire, exactly 2 m
long, by Searle's method. In a particular
reading, the student measures the extension in
the length of the wire to be 0.8 mm with an
uncertainty of ±0.05 mm at a load of exactly
1.0 kg. The student also measures the diameter
of the wire to be 0.4 mm with an uncertainty of
±0.01 mm. Take g = 9.8 m/s2 (exact). The
Young's modulus obtained from the reading is
close to
[2007, 3M]
11
2
(A) (2.0 ± 0.3) × 10 N/m
(B) (2.0 ± 0.2) × 1011 N/m2
(C) (2.0 ± 0.1) × 1011 N/m2
(D) (2.0 ± 0.5) × 1011 N/m2
2
39 YEARS TOPIC- WISE JEE Advanced Questions with Solutions
In the experiment to determine the speed of
[2007, 3M]
sound using a resonance column
(A) prongs of the tuning fork are kept in a
vertical plane
(B) prongs of the tuning fork are kept in a
horizontal plane
(C) in one of the two resonances observed, the
length of the resonating air column is close to
the wavelength of sound in air
(D) in one of the two resonances observed, the
length of the resonating air column is close
to half of the wavelength of sound in air
10.
The density of a solid ball is to be determined
in an experiment. The diameter of the ball is
measured with a screw gauge, whose pitch is
0.5 mm and there are 50 divisions on the
circular scale. The reading on the main scale is
2.5 mm and that on the circular scale is
20 divisions. If the measured mass of the ball
has a relative error of 2%, the relative
[2011]
percentage error in the density is
(A) 0.9%
(B) 2.4%
(C) 3.1%
(D) 4.2%
14.
In the determination of Young's modulus
4MLg 

Y =
 by using Searle's method, a wire
πld 2 

of length L = 2 m and diameter d = 0.5 mm is
used. For a load M = 2.5 Kg, an extension
l = 0.25 mm in the length of the wire is
observed. Quantities d and l are measured
using a screw gauge and a micrometer,
respectively. They have the same pitch of
0.5 mm. The number of division on their
circular scale is 100. The contribution to the
maximum probable error of the Y
[2012]
measurement.
(A) due to the errors in the measurements of d
and l are the same.
(B) due to the error in the measurement of d is
twice that due to the error in the
measurement of l
(C) due to the error in the measurement of l is
twice that due to the error in the
measurement of d
(D) due to the error in the measurement of d is
four times that due to the error in the
measurement of l
15.
The diameter of a cylinder is measured using a
Vernier callipers with no zero error. It is found
that the zero of the Vernier scale lies between
5.10 cm and 5.15 cm of the main scale. The
Vernier scale has 50 divisions equivalent to
2.45 cm. The 24th division of the Vernier scale
exactly coincides with one of the main scale
divisions. The diameter of the cylinder is
[2013]
(A) 5.112 cm
(B) 5.124 cm
(C) 5.136 cm
(D) 5.148 cm
Students I, II and III perform an experiment for
measuring the acceleration due to gravity (g)
using a simple pendulum.
They use different lengths of the pendulum
and/or record time for different number of
oscillations. The observations are shown table.
Least count for length = 0.1 cm, Least count for
time = 0.1 s
Student
11.
Length
of the
pendulum
(cm)
Number of
oscillations
(n)
Total time
for (n)
oscillations
(s)
Time
period
(s)
I
64.0
8
128.0
16.0
II
64.0
4
64.0
16.0
III
20.0
4
36.0
9.0
If EI, EII and EIII are the percentage errors in g,
 ∆g

× 100  for students I, II and III,
i.e., 
 g

respectively
[2008, 3M]
(A) EI = 0
(B) EI is minimum
(C) EI = EII
(D) EII is maximum
12.
13.
A vernier calipers has 1 mm marks on the main
scale. It has 20 equal divisions on the vernier
scale which match with 16 main scale
divisions. For this vernier calipers, the least
[2010]
count is
(A) 0.02 mm
(B) 0.05 mm
(C) 0.1 mm
(D) 0.2 mm
3
UNIT, DIMENSION & ERROR
some simple dynamical systems in onedimension. For such systems, phase space is a
plane in which position is plotted along
horizontal axis and momentum is plotted along
vertical axis. The phase space diagram is x(t) vs
p(t) curve in this plane. The arrow on the curve
indicates the time flow. For example, the phase
space diagram for a particle moving with
constant velocity is a straight line as shown in
the figure. We use the sign convention in which
position or momentum upwards (or to right) is
positive and downwards (or to left) is negative.
[2011]
LINKED COMPREHENSION TYPE
Passage # 1 (1 & 2)
A dense collection of equal number of electrons
and positive ions is called neutral plasma.
Certain solids containing fixed positive ions
surrounded by free electrons can be treated as
neutral plasma. Let N be the number density of
free electrons, each of mass m. When the
electrons are subjected to an electric field, they
are displaced relatively away from the heavy
positive ions. If the electric field becomes zero,
the electrons begin to oscillate about the
positive ions with a natural angular frequency
ωp, which is called the plasma frequency. To
sustain the oscillations, a time varying electric
field needs to be applied that has an angular
frequency ω, where a part of the energy is
absorbed and a part of it is reflected. As ω
approaches ωp, all the free electrons are set to
resonance together and all the energy is
reflected. This is the explanation of high
[2011]
reflectivity of metals.
1.
2.
3.
Momentum
(A)
(A)
(B)
mε0
Ne
(C)
Ne2
mε 0
(D)
mε0
Ne2
Estimate the wavelength at which plasma
reflection will occur for a metal having the
density of electrons N = 4 × 1027 m–3. Take
ε0 ≈ 10–11 and m ≈ 10–30, where these quantities
are in proper SI units.
(A) 800 nm
(B) 600 nm
(C) 300 nm
(D) 200 nm
Passage # 2 (3 to 5)
Phase space diagrams are useful tools in
analyzing all kinds of dynamical problems.
They are especially useful in studying the
changes in motion as initial position and
momentum are changed. Here we consider
Momentum
(B)
Position
Position
Momentum
Taking the electronic charge as e and the
permittivity as ε0, use dimensional analysis to
determine the correct expression for ωp.
Ne
mε 0
The phase space diagram for a ball thrown
vertically up from ground is
(C)
4.
Momentum
(D)
Position
Position
The phase space diagram for simple harmonic
motion is a circle centered at the origin. In the
figure, the two circles represent the same
oscillator but for different initial conditions,
and E1 and E2 are the total mechanical energies
respectively. Then
Momentum
E2
a
(A) E1 = 2 E2
(C) E1 = 4 E2
E1
2a
E1
Position
(B) E1 = 2E2
(D) E1 = 16 E2
4
5.
39 YEARS TOPIC- WISE JEE Advanced Questions with Solutions
Consider the spring-mass system, with the mass
submerged in water, as shown in the figure.
The phase space diagram for one cycle of this
system is
Momentum
(A)
Momentum
(A) Reynolds number and coefficient of friction
(B) Curies and frequency of a light wave
(C) Latent heat and gravitational potential
(D) Planck's constant and torque
4.
Let [ε0] denote the dimensional formula of the
permittivity of the vacuum and [µ0] that of the
permeability of the vacuum. If M = mass,
L = length, T = time and I = electric current.
[1998, 2M]
–1 –3 2
(A) [ε0] = [M L T I]
(B) [ε0] = [M–1 L–3 T4 I2]
(C) [µ0] = [ML T–2 I–2]
(D) [µ 0] = [ML2 T–1 I]
5.
The SI unit of the inductance, the henry can by
[1998, 2M]
written as
(A) weber/ampere
(B) volt-second/ampere
(D) ohm-second
(C) joule/(ampere)2
6.
A student uses a simple pendulum of exactly
1 m length to determine g, the acceleration due
to gravity. He uses a stop watch with the least
count of 1 s for this and records 40 s for
20 oscillations. For this observation, which of
the following statement(s) is/are true?
(A) Error ∆T in measuring T, the time period,
is 0.05 s
(B) Error ∆T in measuring T, the time period,
is 1 s
(C) Percentage error in the determination of g
is 5%
(D) Percentage error in the determination of g
is 2.5%
7.
Planck's constant h, speed of light c and
gravitational constant G are used to form a
unit of length L and a unit of mass M. Then
[2015]
the correct option(s) is(are)
(B)
Position
Position
Momentum
(C)
Momentum
(D)
Position
Position
ONE OR MORE THAN ONE CORRECT ANSWERS
1.
L, C and R represent the physical quantities
inductance,
capacitance
and
resistance
respectively. The combinations which have the
[1984, 2M]
dimensions of frequency are
(A) 1 / RC
(B) R / L
(C) 1 /
2.
3.
LC
(D) C / L
The dimensions of the quantities in one
(or more) of the following pairs are the same.
[1986, 2M]
Identify the pair (s)
(A) torque and work
(B) angular momentum and work
(C) energy and Young's modulus
(D) light year and wavelength
The pairs of physical quantities that have the
[1995, 2M]
same dimensions is (are)
(A) M ∝
(C) L ∝
8.
c
(B) M ∝
G
h
(D) L ∝
G
In terms of potential difference V,
current I, permittivity ε0, permeability
speed of light c, the dimensionally
equation (s) is (are)
(A) µ0I2 = ε0V2
(B) ε0I = µ0V
(C) I = ε0cV
(D) µ0cI = ε0V
electric
µ0, and
correct
[2015]
5
UNIT, DIMENSION & ERROR
9.
A length-scale (l) depends on the permittivity
(ε) of a dielectric material, Boltzmann constant
(kB), the absolute temperature (T), the number
per unit volume (n) of certain charged particles,
and the charge (q) carried by each of the
particles. Which of the following expressions
(s) for l is (are) dimensionally correct ? [2016]
 nq 2 

(A) l = 

ε
k
T
 B 
(C) l =
10.
 εk T 
(B) l =  B2 
 nq 


q2


 εn 2 / 3 k T  (D) l=
B 



q2


 εn 1 / 3 k T 
B 

In an experiment to determine the
acceleration due to gravity g, the formula
used for the time period of a periodic motion
7( R – r )
is T = 2π
. The values of R and r
5g
are measured to be (60 ± 1) mm and
(10 ± 1) mm, respectively. In five successive
measurements, the time period is found to be
0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The
least count of the watch used for the
measurement of time period is 0.01 s. Which
of the following statement(s) is(are) true ?
[2016]
(A) The error in the measurement of r is 10%
(B) The error in the measurement of T is
3.57%
(C) The error in the measurement of T is 2%
(D) The error in the determined value of g is
11%
MATRIX MATCH TYPE
1.
2.
Column I gives three physical quantities. Select
the appropriate units for the choices given in
Column II. Some of the physical quantities may
have more than one choice.
[1990, 3M]
Column I
Column II
Capacitance
ohm-second
Inductance
coulomb2 – joule–1
Magnetic induction coulomb (volt)–1,
newton (ampere metre)–1,
volt-second (ampere)–1
3.
Match the physical quantities given in Column I
with dimensions expressed in terms of mass
(M), length (L), time (T), and charge (Q) given
in Column II and write the correct answer
against the matched quantity in a tabular form
in your answer book.
[1993, 6M]
Column I
Column II
Angular momentum [M L2 T–2]
Latent heat
[M L2 Q–2]
Torque
[M L2 T–1]
Capacitance
[M L3 T–1 Q–2]
Inductance
[M–1 L–2 T2 Q2]
Resistivity
[L2 T–2]
Some physical quantities are given in Column I
and some possible SI units in which these
quantities may be expressed are given in
Column II. Match the physical quantities in
Column I with the units in Column II.
[2007, 6M]
Column I
Column II
(p) (volt)
(A) GMeMs
(coulomb)
G — universal
(metre)
gravitational constant,
Me — mass of the earth,
Ms — mass of the sun.
(q) (kilogram)
3RT
(B)
(metre)3
M
(second)–2
R — universal gas constant,
T — absolute temperature,
M — molar mass.
(C)
F2
q 2 B2
F — force,
q — charge,
B — magnetic field.
GM e
(D)
Re
G — universal
gravitational constant,
Me — mass of the earth,
Re — radius of the earth.
(r) (metre)2
(second)–2
(s) (farad)
(volt)2
(kg)–1
6
4.
5.
39 YEARS TOPIC- WISE JEE Advanced Questions with Solutions
Column II gives certain systems undergoing a
process. Column I suggests changes in some of
the parameters related to the system Match the
statements in Column I to the appropriate
process (es) from Column II.
[2009]
Column I
Column II
(A) The energy of the (p) System : A capacitor,
system is increased.
initially uncharged.
Process : It is connected
to a battery.
(B) Mechanical energy (q) System : A gas in an
adiabatic container
is provided to the
fitted
with
an
system, which is
adiabatic piston.
converted into energy
Process : The gas is
of random motion of
compressed
by
its parts.
pushing the piston.
(C) Internal energy of (r) System : A gas in a
the
system
is
rigid container.
converted into its
Process : The gas
mechanical energy.
gets cooled due to
colder atmosphere
surrounding it.
(D) Mass of the system (s) System : A heavy
nucleus, initially at
is decreased.
rest.
Process : The nucleus
fissions into two
fragments of nearly
equal masses and
some neutrons are
emitted.
(t) System : A resistive
wire loop.
Process : The loop
is placed in a time
varying magnetic
field perpendicular
to its plane.
Column II shows five systems in which two
objects are labelled as X and Y. Also in each
case a point P is shown. Column I given some
statements about X and/ or Y. Match these
statements to the appropriate system(s) from
Column II.
[2009]
Column I
(A) The force exerted
by X on Y has a (p)
magnitude Mg.
Column II
Y
X
P
Block Y of mass M left
on a fixed inclined
plane X, slides on it
with
a
constant
velocity.
(B) The gravitational
potential energy of
X is continuously (q)
increasing.
P
Z
Y
X
Two ring magnets Y
and Z, each of mass M,
are kept in frictionless
vertical plastic stand so
that they repel each
other. Y rests on the
base X and Z hangs in
air in equilibrium. P is
the topmost point of
the stand on the
common axis of the
two rings. The whole
system is in a lift that
is going up with a
constant velocity.
(C) Mechanical energy
of the system
(r)
X + Y is
continuously
decreasing.
Y
P
X
A pulley Y of mass m0
is fixed to a table
through a clamp X. A
block of mass M hangs
from a string that goes
over the pulley and is
fixed at point P of the
table.
The
whole
system is kept in a lift
that is going down
with
a
constant
velocity.
7
UNIT, DIMENSION & ERROR
(D) The torque of the
weight of Y about
(s)
point P is zero.
Y
X
P
A sphere Y of mass M
is put in a non-viscous
liquid X kept in a
container at rest. The
sphere is released and
it moves down in the
liquid.
(t)
The dimensions of electrical conductivity
is ……..
[1997,1M]
4.
The equation of state of a real gas is given by
a 

 p + 2  (V – b) = RT
V 

where p, V and T are pressure, volume and
temperature respectively and R is the universal
gas constant. The dimensions of the constant a
in the above equation is …..
[1997, 2M]
ANALYTICAL & DESCRIPTIVE QUESTIONS
1.
Give the MKS units for each of the following
quantities
[1980, 3M]
(a) Young's modulus,
(b) Magnetic induction,
(c) Power of a lens.
2.
A gas bubble, from an explosion under water,
oscillates with a period T proportional to pa db
Ec, where p is the static pressure, d is the
density of water and E is the total energy of the
explosion. Find the values of a, b and c.
[1981, 3M]
3.
Write the dimensions of the following in terms
of mass, time, length and charge
[1982, 2M]
(a) Magnetic flux,
(b) Rigidity modulus.
4.
N divisions on the main scale of a vernier
callipers coincide with (N + 1) divisions on the
vernier scale. If each division on the main scale
is of a units, determine the least count of
instrument.
[2003, 2M]
5.
In a Searle's experiment, the diameter of the
wire as measured by a screw gauge of least
count 0.001 cm is 0.050 cm. The length,
measured by a scale of least count 0.1 cm, is
110.0 cm. When a weight of 50 N is suspended
from the wire, the extension is measured to be
0.125 cm by a micrometer of least count
0.001 cm. Find the maximum error in the
measurement of Young's modulus of the
material of the wire from these data.
[2004, 2M]
Y
P
X
A sphere Y of mass M
is falling with its
terminal velocity in a
viscous liquid X kept
in a container.
6.
3.
Match List I with List II and select the correct
answer using the codes given below the lists :
[2013]
List I
List II
(1) ML2T–1
(P) Boltzmann constant
(Q) Coefficient of viscosity (2) ML–1T–1
(R) Planck constant
(3) MLT–3K–1
(S) Thermal conductivity
(4) ML2T–2K–1
Codes :
P
Q
R
S
(A) 3
1
2
4
(B) 3
2
1
4
(C) 4
2
1
3
(D) 4
1
2
3
FILL IN THE BLANKS
1.
Planck's constant has dimensions ……..
[1985, 2M]
2.
In the formula X = 3Y Z2, X and Z have
dimensions of capacitance and magnetic
induction respectively. The dimensions of Y in
MKSQ system are ……..
[1988, 2M]
8
6.
7.
39 YEARS TOPIC- WISE JEE Advanced Questions with Solutions
The pitch of a screw gauge is 1 mm and there
are 100 divisions on the circular scale. While
measuring the diameter of a wire, the linear
scale reads 1 mm and 47th division on the
circular scale coincides with the reference line.
The length of the wire is 5.6 cm. Find the
curved surface area (in cm2) of the wire in
appropriate number of significant figures.
[2004, 2M]
8.
The edge of a cube is measured using a vernier
calliper. (9 divisions of the main scale is equal
to 10 divisions of vernier scale and 1 main
scale division is 1 mm). The main scale
division reading is 10 and 1 division of vernier
scale was found to be coinciding with the main
scale. The mass of the cube is 2.736 g. Calculate
the density in g/cm3 upto correct significant
figures.
[2005, 2M]
9.
To find the distance d over which a signal
can be seen clearly in foggy conditions, a
railways engineer uses dimensional analysis
and assumes that the distance depends on the
mass density ρ of the fog, intensity
(power/area) S of the light from the signal
and its frequency ƒ. The engineer finds that d
is proportional to S1/n. The value of n is.
[2013]
The energy of a system as a function of time t is
given as E(t) = A2 exp(–αt), where α = 0.2 s–1.
The measurement of A has an error of 1.25 %.
If the error in the measurement of time is 1.50 %,
the percentage error in the value of E(t) at
t = 5 s is
[2015]
9
UNIT, DIMENSION & ERROR
ANSWERS
¾ Only One Correct Answer
1. (B)
2. None of the four choices
3. (D)
4. (A)
5. (A)
6. (D)
7. (D)
8.(A)
9.(B)
10. (A)
11. (B)
12. (D)
13. (C)
14. (A)
15. (B)
3. (D)
4. (C)
5. (B)
5. (A,B,C,D)
¾ Linked Comprehension Type
1. (C)
2. (B)
¾ One or More than One Correct Answers
1. (A,B,C)
2. (A,D)
3. (A,B,C)
4. (B,C)
7. (A,C,D)
8. (A,C)
9. (B,D)
10. (A,B,D)
6.(A,C)
¾ Matrix Match Type
1. See the solution
2. See the solution
4. A→ p,q,s,t; B→ q; C→ s; D → s
3. A → p,q ; B→ r, s; C→ r,s; D→ r, s
5. A → p, t; B → q,s,t; C → p,r,t; D → q
6. (C)
¾ Fill in the Blanks
1. [ML2T–1]
2. [M–3L–2T4Q4]
3. [M–1L–3T3A2]
4. [ML5T–2]
¾ Analytical & Descriptive Questions
1. (a) N/m2 (b) Tesla (c) m–1
4.
a
N +1
2. a =
5. 1.09 × 1010 N / m2
–5
1
1
,b= ,c=
6
2
3
6. 2.6 cm2
3. (a) [ML2T–1Q–1]; (b)[ML–1T–2]
7. 2.66 g/cm3
8. 3
9. 4
10
39 YEARS TOPIC- WISE JEE Advanced Questions with Solutions
SOLUTIONS
α 
 kθ 
∴ [β] =   =  
p
 Zp 
Dimensions of kθ are that to energy. Hence,
 ML2 T –2 
= [M0L2T0]
[β] = 
–1 – 2 
 LML T 
Therefore, the correct option is (A).
Only One Correct Answer
1.
X
[Y] =  2 
Z 


Capaci tan ce
= 
2
 (Magnetic induction ) 
 M –1L–2Q 2 T 2 
=  2 – 2 – 2  = [M–3L–2T4Q4]
 M Q T 
∴ Correct answer is (B).
2.
6.
1
ε0 E2 is the
2
expression of energy density (Energy per unit
volume)
(None of the four choices)
 ML2 T –2 
1
–1 –2
2
=
E
ε
 2 0   L3  = [ML T ]

 

3.
∆q
∆V
A
∆q
or ε0
=
L
∆V
(∆q ) L
or ε0 =
A.(∆V)
C=
X = ε0 L
Density ρ =
πr 2 L
∆r ∆L 
∆ρ
 ∆m
+2
+
∴
× 100 = 
 × 100
r
L 
ρ
 m
After substituting the values, we get the
maximum percentage error in density = 4%
Hence, the correct option is (D).
7.
Dipole moment = (charge) × (distance)
Electric flux = (electric field) × (area)
Hence, the correct option is (D).
8.
Least count (LC)
Pitch
Number of divisions on circular scale
0 .5
= 0.01 mm
=
50
Now,
Diameter of ball = (2 × 0.5 mm) + (25 – 5) (0.01)
= 1.2 mm
=
∆V
∆V
(∆q )L
L
=
∆t
∆t
A (∆V)
but [A] = [L2]
∆q
∴ X=
= current
∆t
9.
Y =
=
4.
5.
3
m
–2
3
–6
3
V = l = (1.2 × 10 m) = 1.728 × 10 m
Q Length (l) has two significant figures, the
volume (V) will also have two significant
figures. Therefore, the correct answer is
V = 1.7 × 10–6 m3
 αZ 
0 0 0
 kθ  = [M L T ]
 
 kθ 
[α] =  
Z
α 
Further [p] =  
β
4FL
FL
=
Al
πd 2 l
(4)(1.0 × 9.8)(2)
π(0.4 × 10 – 3 ) 2 (0.8 × 10 – 3 )
= 2.0 × 1011 N / m2
∆Y
 ∆d   ∆l 
= 2
Further
 + 
Y
 d   l 
  ∆d   ∆l 
∴ ∆Y = 2
 +   Y
  d   l 
 0.01 0.05 
11
= 2 ×
+
 × 2.0 × 10
0 .4
0.8 

= 0.225 × 1011 N/m2
= 0.2 × 1011 N/m2 (By rounding off)
or (Y + ∆Y) = (2 + 0.2) × 1011 N/m2
∴ Correct option is (B).
11
UNIT, DIMENSION & ERROR
10.
11.
Length of air column in resonance is odd integer
λ
multiple of .
4
∴ Correct option is (A).
Now density d =
Here, r is the diameter.
∴
l
T = 2π
g
or
∆m
 ∆r 
× 100 + 3 ×   × 100
m
 r 
1
= 2% + 3 ×
= 3.11%
2 .7
=
∆g
 ∆l 2∆t 
× 100 =  +
 × 100
g
t 
 l
 0.1 2 × 0.1 
+
EI = 
 × 100 = 0.3125%
128 
 64
14.
 0.1 2 × 0.1 
+
EII = 
 × 100 = 0.46875%
64 
 64
 0.1 2 × 0.1 
EIII = 
+
 × 100 = 1.055%
36 
 20
Hence EI is minimum.
∴ Correct option is (B).
12.
Least count of vernier calipers
LC = 1 MSD – 1 VSD
Smallest division on main scale
=
Number of divisions on vernier scale
20 divisions of vernier scale = 16 divisions of
main scale
16
∴ 1 VSD =
mm = 0.8 mm
20
∴ LC = 1 MSD – 1 VSD
= 1 mm – 0.8 mm
∴
= 0.2 mm
∴ Correct option is (D).
13.
Least count of screw gauge =
0.5
50
= 0.01 mm = ∆r
Diameter r = 2.5 mm + 20 ×
or
 ∆m
∆d
 ∆r 
+ 3  × 100
× 100 = 
d
m
 r 

(4π 2 )(n 2 )l
l
t
∴g =
= 2π
n
g
t2
% error in g =
∆r
0.01
=
r
2.70
∆r
1
× 100 =
r
2.7
0 .5
= 2.70 mm
50
m
m
=
3
V
4 r
π 
3 2
Y=
4MLg
& % ymax
πld 2
= %M + %L + %l + 2%d
Least count of both instrument,
0.5
∆l = ∆d =
= 5 × 10−3
100
%l =
5 × 10−3
∆l
× 100 =
= 2%
0.25
l
%d =
5 × 10−3
∆d
× 100 =
× 100 = 1%
d
0.5
here we see that, contribution of l, = 2%
contribution of d = 2% d = 2 × 1 = 2%
hence both terms l and d contribute equally.
15.
Here
1 MSD = 0.05 cm
2.45
= 0.049 cm
1 VSD =
50
Least count = 1 MSD – 1VSD = 0.001 cm
diameter = 5.10 + (0.001) × 24
diameter = 5.124 cm
Linked Comprehension Type
1.
N = Number of electrons per unit volume
∴ [N] = [L–3], [e] = [q] = [It] = [AT]
[ε0] = [M–1L–1T4A2]
Substituting the dimensions we can see that,
 Ne 2 

 = [T–1]
ε
m

0 

Angular frequency has also the dimension [T–1]
∴ Correct option is (C).
12
2.
39 YEARS TOPIC- WISE JEE Advanced Questions with Solutions
ω = 2πf =
∴λ =
2πc
λ
2πc
=
ω
3.
Reynold's number and coefficient of friction
are dimensionless quantities.
Curie is the number of atoms decaying per unit
time and frequency is the number of
oscillations per unit time.
Latent heat and gravitational potential both
have the same dimension corresponding to
energy per unit mass.
4.
F=
2πc
Ne 2 / mε 0
Substituting the values, we get
– 600 nm.
λ ~
∴ Correct option is (B).
3.
4.
Momentum is first positive but decreasing.
Displacement (or say position) is initially zero.
It will first increase. At highest point,
momentum is zero and displacement is
maximum. After that momentum is downwards
(negative) and increasing but displacement is
decreasing. Only (D) option satisfies these
conditions.
[ε0] =
=
1
mω2 A2
2
or E ∝ A2,
In all the given four figures, at mean position
the position coordinate is zero.
At the same time mass is starting from the
extreme position in all four cases. In figures (C)
and (D), extreme position is more than the
initial extreme position. But due to viscosity
opposite should be the case.
Hence, the answer should be either option (A)
or option (B) . Correct answer is (B), because
mass starts from positive extreme position
(from uppermost position). Then, it will move
downwards or, momentum should be negative.
[F][r 2 ]
[IT] 2
[MLT – 2 ][L2 ]
∴ [µ0] =
2
A 
E2
 a 
=  2  =  
E1
 2a 
 A1 
or E1 = 4E2
∴ Correct option is (C).
5.
[q 1 ][q 2 ]
= [M–1 L–3 T4I2]
Speed of light,
1
c=
ε0µ 0
E=
2
q q
1
⋅ 12 2
4πε 0 r
=
1
[ε 0 ][c] 2
1
[M
–1
L T I 2 ][LT –1 ] 2
5.
(A) L =
φ
weber
or henry =
i
ampere
 di 
(B) e = – L  
 dt 
e
∴L = –
(di / dt )
or henry =
(C) U =
One or More than One Correct Answers
(A) Torque and work both have the
dimensions [ML2T–2].
(D) Light year and wavelength both have the
dimensions of length ie, [L].
∴ Answer is (A) and (D).
4
= [MLT–2I–2]
1 2
Li
2
or henry =
2.
–3
volt − sec ond
ampere
∴ L=
2U
i2
joule
(ampere) 2
1 2 2
Li = i Rt
2
∴ L = Rt or henry = ohm-second
(D) U =
13
UNIT, DIMENSION & ERROR
6.
40 s
= 2 s.
20
Further, t = nT = 20T or ∆t = 20 ∆T
∆t
∆T
∴
=
t
T
T
or ∆T = ⋅ ∆t
t
∴ L=
T=
L=
∴ L∝
8.
 2 
=   (1)
 40 
= 0.05 s
Further,
h
(hc)
1/ 2
L∝
∆g
∆T
1
∴
× 100 = –
× 100
×
T
2
g
ε0 =
q2
Fr 2
or % error in determination of g is
∆g
∆T
× 100 = – 200 ×
T
g
ε0 =
A 2T 2
MLT − 2 × L2
ε0 = M–1L–3T4A2
c = LT–1
Similarly µ0I2 = ε0V2
I = ε0cV
A = M0L0T0A1
200 × 0.05
2
= – 5%
∴ Correct options are (A) and (C).
=–
9.
hc
E=
λ
– Gm 1 m 2
E=
r
M∝
c
E = mc2
hc
E=
λ
hc
= mc2
λ
h
λ=
m
l α εa kb Tc nd qe
(A) l =
hc
GM 2
=
λ
r
unit of λ and r is L
 hc 
So M =  
G
l
l
=
2
L
L
(B) l =
ε kB T
nq 2
…(1)
1
G
Ans. (A)
L–3 × A 2 T 2
M –1A 2 T 4 L– 3M1L2 T – 2 θ –1θ
l=
1/ 2
M∝
h
W = qV
ML2T–2 = ATV
V = M1L2T–3A–1
I⇒A
q2
1
4πε 0 r 2
l
or T ∝ g–1/2
g
G1/2 from (1)
G
F=
T = 2π
7.
h
M
=
(M –1A 2 T 4 L–3 )M1L2 T –2 θ –1θ
L– 3A1T 2
=
L2 = L
(C) l =
A 2T 2
M –1A 2T 4 L– 3L– 2 M1L2T – 2θ –1θ
(D) l =
A 2T 2
M A T L L M +1L2T – 2θ –1θ
=
–1
2
L2 = L
4 – 3 –1
14
10.
39 YEARS TOPIC- WISE JEE Advanced Questions with Solutions
Time period
T = 2π
I
Capacitance
Inductance
Magnetic
induction
7( R – r )
5g
Time period
I 0.52 sec
II 0.56 sec
III 0.57 sec
IV 0.54 sec
V 0.59 sec
0.52 + 0.56 + 0.57 + 0.54 + 0.59
Av 〈 T 〉 =
5
2.78
〈T〉=
5
〈 T 〉 = 0.556 sec
0.03
0.01
absolute error = 0.1/5 = 0.02
0.02
0.01
0.03
2
T g (R – r)–1 = const
∆ (R – r )
2∆T
∆g
+
–
=0
T
g
R–r
2.
6.
Error =
II
coulomb-volt coulomb2 joule–1
ohm-sec, volt second ampere–1
newton (ampere-metre)–1
The correct table is as under
I
Angular momentum
Latent heat
Torque
Capacitance
Inductance
Resistivity
II
[M L2T–1]
[L2T–2]
[M L2T–2]
[M–1 L–2T2Q2]
[M L2Q–2]
[M L3T–1 Q–2]
Boltzmann constant
3
Q E = KT
2
2 –2
ML T = k[K]
k = [ML2T–2K–1]
Coff. of viscocity
F = 6πηRV
MLT −2
L × LT –1
η = [ML–1T–1]
Planck constant
E = hν
ML2T–2 = h.T–1
h = [ML2T–1]
Thermal conductivity (K)
K = [MLT–3K–1]
η=
2∆T
∆g
1
+
–
(∆R – ∆r) = 0
T
g
(R – r )
∆g
 2 × 3.57 2 
=
+ 
g
50 
 100
 2 × 3.57 2 
% error = 
+  × 100 ≈ 11%
50 
 100
Matrix Match Type
1.
t≡
U≡
L
R
∴ L ≡ tR ≡ ohm-second
2
q
2C
q ≡ CV
Fill in the Blanks
1.
2
q
≡ coulomb2 / joule
U
q
∴ C≡
≡ coulomb/volt
V
∴ C≡
–e
di / dt
e (dt )
∴ L≡
≡ volt-second/ampere
(di)
L≡
F = ilB
F
∴ B≡
≡ Newton / ampere-metre
il
2.
E = hν
∴ h=
E
ν
or [h] =
[ML2 T –2 ]
[E ]
=
= [ML2T–1]
–1
[ν ]
[T ]
[X] = [C] = [M–1L–2T2Q2]
[Z] = [B] = [M T–1 Q–1]
∴ [Y] =
[M –1 L–2 T 2 Q 2 ]
[MT –1Q –1 ] 2
= [M–3L–2T4Q4]
15
UNIT, DIMENSION & ERROR
3.
Electrical conductivity σ =
=
qi
i/A
=
F/q
FA
=
(it )(i)
i2t
=
FA
FA
σ =
J
E
5.
[A 2 ][T]
[MLT – 2 ][L2 ]
= [M–1 L–3 T3 A2]
4.
= 2.24 × 1011 N/m2
Now,
∆d
∆Y
∆L
∆l
=
+
+2
Y
L
l
d
 a 
 2  = [p]
V 
∴[a] = [pV2]
= [ML–1T–2] [L6]
= [ML5T–2]
 0.1   0.001 
 0.001 
=
 +
 + 2

 110   0.125 
 0.05 
= 0.0489
∆Y = (0.0489)Y
= (0.0489) × (2.24 × 1011) N/m2
= 1.09 × 1010 N/m2
Analytical & Descriptive Question
2.
4.
[T] = [padbEc] = [ML–1T–2]a [ML–3]b [ML2T–2]c
Equating the powers both sides, we have
a+b+c=0
… (i)
– a – 3b + 2c = 0 … (ii)
– 2a – 2c = 1
… (iii)
Solving these three equations, we have
5
1
1
and c =
a=– ,b=
6
2
3
(N + 1) divisions on the vernier scale
= N divisions on main scale
∴ 1 division on vernier scale
N
divisions on main scale
=
N +1
Each division on the main scale is of a units.
 N 
∴ 1 division on vernier scale = 
a
 N + 1
units = a′(say)
Least count = 1 main scale division
– 1 vernier scale division
 N 
= a – a′ = a – 
 a
 N + 1
=
a
N +1
Young's modulus of elasticity is given by
stress
Y =
strain
FL
F/A
FL
=
=
=
l/L
lA
 πd 2 

l 

 4 
Substituting the values, we get
50 × 1.1 × 4
Y=
–3
(1.25 × 10 ) × π × (5.0 × 10 – 4 ) 2
6.
1mm
100
= 0.01 mm
Diameter of wire = (1 + 47 × 0.01) mm
= 1.47 mm
d
Curved surface area (in cm2) = (2π)   (L)
2
Least count of screw gauge =
S = πdL
= (π) (1.47 × 10–1) (5.6) cm2
= 2.5848 cm2
Rounding off to two significant digits
S = 2.6 cm2
or
7.
1 MSD = 1 mm
9 MSD = 10 VSD
∴ Least count,
LC = 1 MSD – 1 VSD
9
mm
= 1 mm –
10
1
mm
=
10
Measure reading of edge
= MSR + VSR (LC)
16
39 YEARS TOPIC- WISE JEE Advanced Questions with Solutions
= 10 + 1 ×
1
10
9.
= 10.1 mm
Volume of cube V = (1.01)3 cm3 = 1.03 cm3
[After rounding off upto 3 significant digits, as
edge length measured upto 3 significant digits]
2.736
∴ Density of cube =
1.03
= 2.6563 g/cm3
= 2.66 g/cm3
(After rounding off to 3 significant digits)
8.
d = ρa Sb ƒc
b
 M1L1T − 2 LT −1  –1 c
 (T )
d′ = (ML ) 

L2


0 1 0
a + b –3a –3b – c
MLT =M L T
1
a + b = 0; – 3a = 1, a = – ; – 3b – c = 0
3
b=–a
1
⇒ b=
3
⇒n=3
–3 a
E = A2e–αt
E = A2e–0.2t
ln E = 2lnA – 0.2t
dE
dA
=2
– 0.2 dt
E
A
dE
dA
=2
– 0.2 dt
E
A
dt
= 1.5 %
t
∴ dt = 1.5 × 5 %
= 1.5 × 5 = 7.5
= 2 × 1.25 + 0.2 × 7.5
= 2.5 + 1.5
=4%