Physics For You - May 2016

73 Pages·2016·11.41 MB·English

Volume 24 No. 5 May 2016
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Physics Musing Problem Set 34 8
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JEE Advanced Practice Paper 10
T JEE Main Solved Paper 2016 22
PMT Practice Paper 31
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BITSAT Practice Paper 39
Brain Map 46
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Exam Prep 2016 51
T AIIMS Practice Paper 56
Olympiad Problems 65
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JIPMER Practice Paper 69
Core Concept 74
O
Live Physics 81
Physics Musing Solution Set 33 82
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You Ask We Answer 84
Crossword 85
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Physics For you | may ‘16 7
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Physics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment
the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material.
In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed
solutions of these problems will be published in next issue of Physics For You.
The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who
send atleast five correct solutions will be published in the next issue.
We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.
1. A particle is projected from ground in vertical A
direction at t = 0. At t = 0.8 s, it reaches h = 14 m. It
will again come to same height at t = (g = 10 m s–2) B

14 7 r
(a) 2 s (b) s (c) 3 s (d) s 
5 2
C
2. In the figure shown, the blocks A and C are pulled (a) 3 sin a = 2 cos b (b) 2 sin a = 3 cos b
down with constant velocities u . Acceleration of (c) 3 sin b = 2 cos a (d) 2 sin b = 3 cos a
block B is
5. The intensity of radiation emitted by the Sun has its
maximum value at a wavelength of 510 nm and that
b b
  emitted by the North star has the maximum value
at 350 nm. If these stars behave like black bodies,
then the ratio of the surface temperatures of the
A u C u Sun and the North star is
B (a) 1.46 (b) 0.69 (c) 1.21 (d) 0.83
(a) u2 tan2θsecθ (b) u2 tan3θ 6. N(< 100) molecules of a gas have velocities
b b 1, 2, 3........ N km s–1 respectively. Then
u2 (a) rms speed and average speed of molecules is
(c) sec2θtanθ (d) zero
same
b
(b) ratio of rms speed to average speed of molecules
3. Two particles A and B each of mass m are attached
is (2N+1)(N+1)/6N
by a light inextensible string of length 2l. The
(c) ratio of rms speed to average speed of molecules
whole system lies on a smooth horizontal table
with B initially at a distance l from A. The particle is (2N+1)(N+1)/6
at end B is projected across the table with speed u (d) ratio of rms speed to average speed of molecules
perpendicular to AB. Velocity of ball A just after the
(2N+1)
jerk is is 2
6(N+1)
u 3
(a) (b) u 3 7. A thermodynamic process of one mole ideal
4
gas is shown in the
V
u 3 u figure. The efficiency 4T
(c) (d) 2V 0C
2 2 of cyclic process ABCA 0
4. A particle initially at rest starts moving from point will be
A
V B
A on the surface of a fixed smooth hemisphere of (a) 25% 0 T0 2T0
radius r as shown in the figure. The particle looses (b) 12.5%
P
its contact with hemisphere at point B. C is centre of (c) 50% P0 2P0
the hemisphere. The equation relating a and b is (d) 7.7%
Contd. on page no. 80
By Akhil Tewari, author Foundation of Physics for JEE main & advanced, Senior Professor Physics, RaO IIT aCaDEmy, mumbai.
8 Physics For you | may ‘16
exam on
22nd May
JJEEEE 2016
JJEEEE
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PRACTICE PAPER 2016
PRACTICE PAPER 2016
PaPer-1
Section 1 (MaxiMuM MarkS : 32) total distance (nearest integer) in m through which
• This section contains EIGHT questions the block moves before it comes to rest completely.
• The answer to each question is a SINGLE DIGIT [g = 10 m s–2]
INTEGER ranging from 0 to 9, both inclusive
1. A current I flows in a rectangularly shaped wire
whose center lies at (x , 0, 0) and whose vertices
0
are located at the points A(x + d, –a, –b),
0
B(x – d, a, –b), C(x – d, a, b), and D(x + d, –a, b)
0 0 0
respectively. Assume that a, b, d << x . Find the A B D C
0
magnitude of magnetic dipole moment vector of the 5. An open organ pipe containing air resonates in
rectangular wire frame in J T–1. (Given: b = 10 m, fundamental mode due to a tuning fork. The
d = 4 m, a = 3 m, I = 0.01 A) measured values of length l (in cm) of the pipe
2. A very long, straight, thin wire carries –3.60 nC m–1 and radius r (in cm) of the pipe are l = 94 ± 0.1,
r = 5 ± 0.05. The velocity of the sound in air is
of fixed negative charge. The wire is to be surrounded
accurately known. The maximum percentage error
by a uniform cylinder of positive charge, radius
in the measurement of the frequency of that tuning
1.50 cm, coaxial with the wire. The volume charge
fork by this experiment is given by a2%. Find the
density r of the cylinder is to be selected so that
value of 10 a.
the net electric field outside the cylinder is zero.
Calculate the required positive charge density r 6. An initially uncharged capacitor C is fully charged
(in mC m–3). by a constant emf e in series with a resistor R. Rate
of energy dissipation in the resistor is equal to rate
3. A long coaxial cable consists of two thin-walled
of energy stored in capacitor at time CR ln k. Find
conducting cylinders with inner radius 2 cm and
the value of k.
outer radius 8 cm. The inner cylinder carries a steady
7. You are at a distance of R = 1.5 × 106 m from the
current 0.1 A, and the outer cylinder provides the
centre of an unknown planet. You notice that if you
return path for that current. The current produces a
throw a ball horizontally it goes completely around
magnetic field between the two cylinders. Find the
the planet hitting you in the back 90,000 s later with
energy stored in the magnetic field for length 5 m of
exactly the same speed that you originally threw it.
the cable. Express answer in nJ (use ln 2 = 0.7).
The length of semi major axis of the motion of ball
4. A 0.5 kg block slides from the point A on a horizontal is 2R. The mass of the planet in scientific notation is
track with an initial speed 3 m s–1 towards a weightless a × 1021 kg. Find a.
horizontal spring of length 1 m and force constant
8. In a certain polytropic process, the volume of
2 N m–1. The part AB of the track is frictionless and
argon was increased four times. Simultaneously,
the part BC has the coefficient of static and kinetic
the pressure decreased eight times. Find the molar
friction as 0.22 and 0.20 respectively. If the distance
heat capacity (in SI unit) of argon in this process,
AB and BD are 2 m and 2.14 m respectively, find the
assuming the gas to be ideal.
10 Physics For you | may ‘16
Section 2 (Maximum Marks : 40) (a) Minimum rate of fuel consumption to just lift
• This section contains TEN questions it off the launching pad is 2.45 kg s–1.
• Each question has FOUR options (a), (b), (c) and (d). (b) Minimum rate of fuel consumption to give it an
ONE OR MORE THAN ONE of these four option(s) acceleration of 20 m s–2 is 3.5 kg s–1.
is(are) correct (c) The speed of the rocket is 4.2 km s–1 when the
9. For a certain radioactive substance, it is observed rate of consumption of fuel is 10 kg s–1 after
that after 4 h, only 6.25% of the original sample is whole of the fuel is consumed.
left undecayed. It follows that (d) All are correct.
(a) the half-life of the sample is 1 h
13. One mole of a diatomic ideal gas (γ = 1.4) is taken
1
(b) the mean life of the sample is h through a cyclic process starting from point A.
ln2 The process A → B is an adiabatic compression,
(c) the decay constant of the sample is ln(2) h–1 B → C is isobaric expansion, C → D is an adiabatic
(d) after a further 4 h, the amount of the substance expansion, and D → A is isochoric. The volume
left over would by only 0.39% of the original ratios are V /V = 16 and V /V = 2 and the
A B C B
amount. temperature at A is T = 300 K.
A
10. A system consists of two identical m (a) Temperature of the gas at B is 909 K.
cubes, each of mass m, linked (b) Temperature of the gas at D is 791 K.
together by the compressed (c) The efficiency of the cycle is 61.4%.
weightless spring of stiffness k. The (d) The efficiency of the cycle is 38.6%.
cubes are also connected by a thread 14. A point charge q is located
which is burned through at a certain at centre O of a spherical
moment. m uncharged conducting
b a
layer provided with a
(a) The lower cube will bounce up after the thread O q
small orifice as shown
has been burned through when the initial
in the figure. The inside
2mg
compression of the spring is . and outside radii of the
k
layer are equal to a and b
(b) If the initial compression of the spring is
respectively. What amount of work has to be
7 mg/k, then centre of gravity of this system will
performed to slowly transfer the charge q from the
8mg point O through the orifice and into infinity ?
rise to height .
k q2 1 1 q2 1 1
(c) The lower cube will bounce up after the thread (a) 8pe a−b (b) 8pe b−a
has been burned through when the initial 0 0
compression of the spring is 5mg . (c) q2 1−1 (d) q2 1−1
4pe a b 4pe b a
k 0 0
(d) All are correct.
15. n drops of a liquid each with surface energy E join
11. It is desired to make a long cylindrical conductor to form a single drop. Then
whose temperature coefficient of resistivity at 20°C (a) some energy will be released in the process
will be close to zero. If such a conductor is made by (b) some energy will be absorbed in the process
assembling alternate disks of iron and carbon, find (c) the energy released will be E(n – n2/3)
the ratio of the thickness of a carbon disk to that an (d) the energy absorbed will be nE(22/3 – 1)
iron disk.
16. A 30 cm violin string with linear mass density
(For carbon, r = 3500 × 10–8 W m and 0.652 g m–1 is placed near a loudspeaker that is fed
a = –0.50 × 10–3 °C–1 for iron, r = 9.68 × 10–8 W m
by an audio oscillator variable frequency. It is found
and a = 6.5 × 10–3 °C–1)
that the string is set into oscillation only at the
(a) 0.36 (b) 0.036 (c) 1.0 (d) 2.0 frequencies 880 Hz and 1320 Hz as the frequency of
the oscillator is varied continuously over the range
12. A rocket set for vertical firing weighs 50 kg and
500-1500 Hz. What is the tension in the string ?
contains 450 kg of fuel. It can have a maximum
exhaust velocity of 2 km s–1. (a) 120 N (b) 60 N (c) 90.8 N (d) 45.4 N
12 Physics For you | may ‘16
17. A particle of mass m moves in a certain plane due
Column-I Column-II
to a force F whose vector rotates in that place with a
(A) Frequency of orbiting (P) is directly
constant angular velocity w. Assuming the particle
electron proportional
to be stationary at the moment t = 0, then
to Z2
(a) its velocity as a function of time is
 F  wt (B) Angular momentum (Q) is directly
 sin 
mw  2  of orbiting electron proportional
to n
(b) its velocity as a function of time is
 2F  wt (C) Magnetic moment of (R) is inversely
 sin 
mw  2  orbiting electron proportional
(c) the distance covered by the particle between to n3
8F (D) The average current (S) is independent
two successive stops is
mw2 due to orbiting of of Z
(d) the mean velocity over two successive stops electron
4F
is . 20. In Young's double-slit experiment, the point source
pmw
S is placed slightly off the central axis as shown in the
18. A massless rope is tossed over a wooden dowel of
figure. If l = 500 nm, then match the following.
radius r in order to lift a heavy object of weight W
off the floor, as shown in the figure. The coefficient
of sliding friction between the rope and the dowel S1 P
is m. Which of the following relation is correct S
y= 10 mm
2 mm
for minimum downward pull (F ) on the rope
min
m O
necessary to lift the object ? m
0
2
S
2
r
1 m 2 m
Column-I Column-II

(A) Nature and order of (P) Bright fringe
W F
interference at point P, of order 80
(a) Fmin = We–pµ (b) Fmin = We+pµ OP = 10 mm
(c) F = –Wepµ (d) F = We–p
min min (B) Nature and order of (Q) Bright fringe
Section 3 (Maximum Marks : 16) interference at point O of order 262
• This section contains TWO questions
(C) If a transparent paper (R) Bright fringe of
• Each question contains two columns, Column I and
(refractive index µ = order 62
Column II
1.45) of thickness t =
• Match the entries in Column I with the entries in
0.02 mm is pasted on S ,
Column II 1
i.e., one of the slits, the
• One or more entries in Column I may match with one
nature and order of the
or more entries in Column II
interference at P
19. In each situation of Column I, a physical quantity
related to orbiting electron in hydrogen-like atom (D) After inserting the (S) Bright fringe of
is given. The terms Z and n given in Column II transparent paper in order 280
have usual meaning in Bohr's theory. Match the front of slit S1, the nature
quantities in Column I with the terms they depend and order of interference
on in Column II. at O
Physics For you | may ‘16 13
PaPer-2
Section 1 (MaxiMuM MarkS : 32) 5. In two calorimeters, we poured 200 g of water each
• This section contains EIGHT questions at temperatures of +30 °C and +40 °C. From the
• The answer to each question is a SINGLE DIGIT hot calorimeter 50 g of water, is poured into cold
INTEGER ranging from 0 to 9, both inclusive
calorimeter and stirred. Then from cold calorimeter
1. For the given circuit in the steady state condition, 50 g of water is poured in hot and again stirred. How
charge on the capacitor is q0 = 16 µC. If now the many times do you have to pour the same portion
battery is removed and the nodes A and C are of water back and forth so that the temperature
shorted. The time during which charge on the
difference between water in the calorimeters
capacitor becomes 4 µC is t(µs) and emf of battery
becomes less than 3 °C ? Heat loss during the transfer
3t
is e(V). Find the value of . and head capacity of calorimeters is neglected.
eln2
6. The tap in the garden was
2 B 3
closed inappropriately
1 +–C= 4 µF 3 rfleoswulitning g infr etehlye waotuert
of it which forms a
D
A C downward narrowing
4 4
beam. The beam of water d = 6 mm
1 3 cm
d = 3 mm
has a circular cross- 2
24 V
+ – section, the diameter
of the circle is 6 mm at one point and 3 cm below it
2. A glass of refractive index 1.5 is coated with a is only 3 mm as shown in figure. If the rate of water
thin layer of thickness t of refractive index 1.8. wasted is 55.65 × 10–n L s–1 then find the value of n.
Light of wavelength l travelling in air is incident (Neglect the effect of viscosity and surface tension
normally on the layer. It is partly reflected at the of the flowing water.)
upper and the lower surfaces of the layer and the
7. Nuclei A and B convert into a stable nucleus
two reflected rays interfere. If l = 648 nm, obtain
the least value of t(in 10–8 m) for which the rays C. Nucleus A is converted into C by emitting 2
a-particles and 3 b-particles. Nucleus B is converted
interfere constructively.
into C by emitting one a-particle and 5 b-particles.
3. A charged particle enters a uniform magnetic field
At time t = 0, nuclei of A are 4N and nuclei of B
with velocity v = 4 m s–1 perpendicular to it, the 0
0 are N . Initially, number of nuclei of C are zero.
0
3
length of magnetic field is x = R, where R is the Half-life of A (into conversion of C) is 1 min and
2 that of B is 2 min. Find the time (in minutes) at
radius of the circular path of the particle in the field.
Find the magnitude of change in velocity (in m s–1) which rate of disintegration of A and B are equal.
of the particle when it comes out of the field. 8. A block of mass m is being pulled up the rough
4. Two tuning forks A and B each incline, inclined at an angle 37° with horizontal by
of natural frequency 85 Hz move A B an agent delivering constant power P. The coefficient
with velocity 10 m s–1 relative to of friction between the block and the incline is µ.
O
stationary observer O. Fork A Find the maximum speed (in m s–1) of the block
moves away from the observer during the course of ascent.
while the fork B moves towards [Take: P = 60 W, m = 1 kg, m = 0.5]
him as shown in the figure. A wind
with a speed 10 m s–1 is blowing in the direction of
µ
motion of fork A. Find the beat frequency measured
by the observer in Hz. [Take speed of sound in air
as 340 m s–1]
14 Physics For you | may ‘16
Section 2 (Maximum Marks : 32) 12. A conducting wire of length l and mass m is placed
• This section contains EIGHT questions on two inclined rails as shown in the figure. A
• Each question has FOUR options (a), (b), (c) and (d). current I is flowing in the wire in the direction
ONE OR MORE THAN ONE of these four option(s)
shown. When no magnetic field is present in the
is(are) correct
region, the wire is just on the verge of sliding. When
9. A thin uniform bar lies on a A a vertically upward magnetic field is switched on,
frictionless horizontal surface 10 m s–1 the wire starts moving up the incline. The distance
and is free to move in any
travelled by the wire as a function of time t will be
way on the surface. Its mass is B
6 m s–1
0.16 kg and length 3 m.
y
Two particles, each of mass 0.08 kg, are moving on C
the same surface and towards the bar in a direction l
I x
perpendicular to the bar, one with a velocity of
10 m s–1, and other with 6 m s–1 as shown in the
figure. The first particle strikes the bar at point A and 
z
the other at point B. Points A and B are at a distance
of 0.5 m from the centre of the bar. The particles 1IBl 
(a) −2g t2
strike the bar at the same instant of time and stick  
2 m 
to the bar on collision. The loss of the kinetic energy
1IBl 1 
of the system in the collision process is (b) × −2gsinq t2
 
2 m cosq 
(a) 2 J (b) 4 J
(c) 2.72 J (d) 5.44 J (c) 1IBl−2gsinqt2
 
2 m 
10. Consider the earth as a uniform sphere of mass
M and radius R. Imagine a straight smooth tunnel (d) 1IBl cos2q−2gsinqt2
 
made through the earth which connects any two 2 m cosq 
points on its surface. The time taken by a particle to
13. For a certain metal, the K absorption edge is at
go from one end to other through the tunnel is
0.172 Å. The wavelength of Ka, Kb and Kg lines of K
R3 R3 series are 0.210 Å, 0.192 Å, and 0.180 Å, respectively.
(a) 2p (b) p
GM GM The energies of K, L and M orbits are E , E and E ,
K L M
R3 R3 respectively. Then
(c) (d) (a) E = –13.07 keV (b) E = –7.52 keV
GM 2GM K L
(c) E = –3.21 keV (d) E = 13.04 keV
11. A linear object of size 1.5 cm is placed at 10 cm M K
from a lens of focal length 20 cm. The optic centre 14. Two light springs of force constants k1 and k2 and
of lens and the object are displaced a distance D. a block of mass m are in one line AB on a smooth
The magnification of the image formed is m. (Take horizontal table such that one ends of each spring
optic centre as origin). The coordinates of image of is fixed on rigid supports and the other end is free
A and B are (x , y ) and (x , y ) respectively. Then as shown in the figure.
1 1 2 2
60cm
k k
1 v 2
A C D B
The distance CD between the free ends of the
springs is 60 cm. If the block moves along AB with a
velocity 120 cm s–1 in between the springs, calculate
the period of oscillation of the block.
(a) (x , y ) = (–20 cm, –1 cm) (k = 1.8 N m–1, k = 3.2 N m–1, m = 200 g)
1 1 1 2
(b) (x , y ) = (–20 cm, 2 cm) (a) 3 s (b) 4 s
2 2
(c) m = 3 (d) m = 2 (c) 2.83 s (d) 4.35 s
Physics For you | may ‘16 15
15. PQR is an equilateral triangular frame of mass m The temperature difference across the unit is 13 K. It has
and side r. It is at rest under the action of horizontal a cross-sectional area of 1.3 m2 and the rate of heat flow
magnetic field B as shown in the figure and the through it is 65 W. Glass has a thermal conductivity of
gravitational field. 1 W m–1 K–1.
r 3
4
17. Select the correct statement.
(a) The frame remains at rest if the current in the (a) The unit is in steady state and in thermal
2mg equilibrium.
frame is .
rB (b) The unit is in steady state but not in thermal
(b) The frame remains at rest if the current in the equilibrium.
2mg (c) The unit is not in steady state but is in thermal
frame is .
equilibrium.
rB 3
(c) The frame is in simple harmonic motion (d) The unit neither in steady state nor in thermal
when frame is slightly displaced in its plane equilibrium.
perpendicular to PQ. The period of oscillation
18. The thermal conductivity of air is
1/2
is pr 3 (a) 1 W m–1K–1 (b) 1 W m–1K–1
 g  10 12
(d) For same as in above option, the period of 1 9
(c) W m–1K–1 (d) W m–1K–1
3r 1/2 14 130
oscillation is p
 
2g PARAGRAPH 2
16. Two identical cylindrical vessels with their bases at
In the given setup, the parallel plate capacitor AB
the same level each contain a liquid of density r.
has vertical plates with separation, d = 50 mm and
The height of the liquid in one vessel is h and in
1 capacitance C . From the plate A a small conducting
0
other vessel is h . The area of either base is A. What
2 ball hangs on a non-conducting silk thread of
is the work done by gravity in equalizing the levels
length L = 100 mm. The mass of the ball is m and
when the two vessels are connected ?
capacitance is C . It initially touches the plate A, as
1
(a) rAg (h – h )2 (b) rAg (h + h )2 shown in the figure. The plate B is grounded while plate
1 2 1 2
4 4 A is connected to a power supply of potential V for a
0
(c) rAg (h – h )2 (d) rAg (h + h )2 short time by closing the switch S and then opening
1 2 1 2
2 2 it again.
Section 3 (Maximum Marks : 16)
• This section contains TWO paragraphs
• Based on each paragraph, there will be TWO questions
• Each question has FOUR options (a), (b), (c) and (d).
ONE OR MORE THAN ONE of these four options(s)
is(are) correct The motion of the conducting ball was observed. It was
PARAGRAPH 1 found that due to the charge deposited on the plate
The figure shows a cross-section of a double glass unit of and the ball, the ball swings across, touches the plate
a window on a vertical wall. A graph of the temperatures B, swings back, touches A and finally swings out again
at different points within the unit is shown next to it. such that it almost touches plate B. Take g = 10 m s–2.
16 Physics For you | may ‘16
19. The ratio of the charge carried by the plates and the Energy in volume of element (length l)
ball finally is m i2 m i2l dr
(a) C1 : C0 (b) C12 : C02 dUB = uB dV = 8p02r2 (2prl)dr = 40p r
(c) C2 : C2 (d) C : C
0 1 0 1 m i2l bdr m i2l b
U = 0 ∫ = 0 ln
20. Based on the description given in the passage, the B
4p r 4p a
required power supply voltage (V ) is given by a
0 Using values, we get U = 7 nJ
B
 C  m
(a) V0 = 1+C1  2 3C 4. (4) : As the track AB is frictionless, the block moves
0 1 1
this distance without loss in its initial KE = mv2
(b) V0 = 1+C0  m 1 2
 C  2 3C = × 0.5 × 32 = 2.25 J.
1 1 2
 C  5m In the path BD as friction is present, so work done
(c) V0 = 1+ 0 
 C  3C against friction
1 1
= µ mg × (BD) = 0.2 × 0.5 × 10 × 2.14 = 2.14 J
k
(d) V0 = 1+C1  5m So, at D the KE of the block = 2.25 – 2.14 = 0.11 J.
 C  3C Now, if the spring is compressed by x
0 1
1
0.11 = × k × x2 + µ mgx
solutions 2 k
paper-1 1
or 0.11 = × 2 × x2 + 0.2 × 0.5 × 10x
1. (2) : Magnetic moment of a current carrying loop, 2
m=IS or x2 + x – 0.11 = 0
  
Area of the loop, S=AB×BC On solving, x = 0.1 m or –1.1 m
  x ≠ –1.1 m, so, x = 0.1 m
  
Here, AB=−2d i+2aj,BC=2bk
After moving the distance x = 0.1 m the block comes

\ S=(−2d i+2aj)×(2b k) =4bdj+4abi to rest. Now the compressed spring exerts a force:
  F = kx = 2 × 0.1 = 0.2 N
\ |m|=I|S|=4Ib a2+d2
on the block while limiting frictional force between
= 4 × 0.01 × 10 × 32+42 block and track is fL = µs mg = 0.22 × 0.5 × 10 = 1.1 N.
= 0.4 × 5 = 2 J T–1 Since, F < fL. The block will not move back. So, the
total distance moved by the block
2. (5) : We don't really need to write an integral, we
= AB + BD + 0.1 = 2 + 2.14 + 0.1 = 4.24 m
just need the charge per unit length in the cylinder
v
to be equal to zero. This means that the positive 5. (4) : u =
charge in cylinder must be +3.60 nC m–1. This 2(l+2e)
v v
positive charge is uniformly distributed in a circle \ u = =
of radius R = 1.50 cm, so 2(l+2×0.6r) 2(l+1.2r)
3.60nCm−1 3.60nCm−1 Du Dv D(l+1.2r) Dv Dl+1.2Dr
r = = ≈ 5 µC m–3. \ = − = −
pR2 p(0.015m)2 u v l+1.2r v l+1.2r
Dv Du Dl+1.2Dr
3. (7) : The magnetic field inside Here =0; ×100=− ×100%
v u l+1.2r
is only due to the current of the
For maximum % error: Dl = 0.1 cm, Dr = 0.05 cm
m i
inner cylinder. B = 0 r Du  0.1+1.2×0.05
2pr  ×100% = ×100%
Magnetic field energy density  u max 94+1.2×5
is not uniform in the space = 0.16% = a2%
between the cylinders. At a \ a = 0.4. Hence, 10a = 4.
distance r from the centre
6. (2) : The capacitor charge as a function of time is
B2 m i2
u = = 0 given by
B 2m 8p2r2 q = Ce(1 – e–t/RC),
0
Physics For you | may ‘16 17