Motion PROJECTILE
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Volume 25 No. 8 August 2017
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PHYSICS FOR YOU
 
AUGUST ‘17
Rank
Cracking the
NEET
EXAM
Navdeep Singh
• MTG :
Why did you choose medical entrance?
Navdeep :
I studied for 810 hours daily.
Navdeep Singh :
From the starting it was my dream to get admission in AIIMS.
• MTG :
On which topic and chapters you laid more
exam.
stress in each subject?
• MTG :
What exams you have appeared for and what is your rank in these exams?
Navdeep :
I also appeared for AIIMS examination and my rank is 46.
• MTG :
Any other achievements?
(Please mention the name of exams and rank)
Navdeep :
I got 56
th
rank in NTSE
Navdeep :
If you have set your target for Medical you have to study for each and every topic.
• MTG :
How much time does one require for serious
All the medical books are of PMT preparation for this exam?
MTG books were very helpful.
Navdeep :
It depends on many things but I think minimum 68 hours
entrance level and MTG Magazines
are required.
also have unique and up to date
• MTG :
Any extra coaching?
• MTG :
How did you prepare questions. I would recommend MTG
Navdeep :
No.
for NEET and other medical exams?
Books to all the PMT aspirants.
Navdeep :
I have focused on my weak points and tried to
• MTG :
Which subjects/topics were you strong/weak at?
study for 8 hours regularly.
Navdeep :
Physics is a weak point of every medical student
• MTG :
What basic difference you found in various
but I took Physics as a challenge and Biology was my
papers you cleared?
favourite from the beginning.
Navdeep :
As far as NEET preparation is concerned it is
• MTG :
Which Books/Magazines/Tutorial/Coaching
more focused on NCERT. While one is preparing for AIIMS
classes you followed?
there has to be clarity of concepts.
Navdeep :
MTG Objective NCERT at Your Fingertips Biology,
• MTG :
How many hours in a day did you study to prepare for the examination?
MTG Magazines  Physics For You, Chemistry Today, Biology
Today, Tutorial  Helix Institute.
8 PHYSICS FOR YOU

AUGUST ‘17
• MTG : In your words what are the components of an ideal preparation plan?
Navdeep :
The components are :
(1) Hardwork (2) Dedication
(3) Regularity (4) Focus on target
• MTG :
What role did the following play in your success:
(a) parents
(b) teachers
(c) school?
Navdeep :
Parents  They always motivated and supported me.
Teachers  It was not possible for me to achieve my goals without the guidance of my teachers.
School  I did my 12
th
from Shivalik Public School and all my school teachers were supportive and were always ready to help me out.
• MTG :
Your family background?
Navdeep :
My father is physics lecturer and Principal at
Government Senior Secondary School and my mother is employee of LIC.
• MTG :
What mistake you think you shouldn’t have made?
Navdeep :
I think I should have studied more in 11
th
class.
• MTG :
How did you destress yourself during the preparation? Share your hobbies and how often could you pursue them?
Navdeep :
Listening to Gurbani and watching TV.
• MTG :
How have various MTG products like
Explorer, Books and Magazines helped you in your preparation?
Navdeep :
MTG books were very helpful. All the medical books are of PMT entrance level and MTG Magazines also have unique and up to date questions. I would recommend
MTG Books to all the PMT aspirants.
• MTG :
Was this your first attempt?
Navdeep :
Yes
• MTG :
Had you not been selected then what would have been your future plan?
Navdeep :
I would have preferred research field.
• MTG :
What do you think is the secret of your success?
Navdeep :
Hardwork
• MTG :
What advice would you like to give our readers who are PMT aspirants?
Navdeep :
Make a strategy. Work on your weak points.
Be focused. Maintain balance between your hobbies and studies.
All the Best!
10 PHYSICS FOR YOU

AUGUST ‘17
2
Kinematics is the branch of mechanics which deals with the study of motion of a body without taking into account the factors which cause motion.
MOTION
•
A body is said to be in motion if it changes its position with time, with respect to its surroundings.
e.g.
a bird flying in air.
Motion in one dimension
•
The motion of a body is said to be one dimensional motion if only one out of the three coordinates specifying the position of the body changes with respect to time. In such a motion, the body moves along a straight line.
Motion in two dimensions
•
The motion of a body is said to be two dimensional motion if two out of three coordinates specifying the position of the body change with respect to time. In such a motion, the body moves in a plane.
e.g
. motion of a body in a plane.
•
•
Distance and Displacement
•
Distance : The length of the actual path traversed by a body during motion in a given interval of time is called distance travelled by that body.
Distance is a scalar quantity.
Distance covered by a moving body can not be zero or negative.
K I N E M AT I C S
•
•
•
Displacement : The displacement of a body in a given interval of time is defined as the shortest distance between the two positions of the body in a particular direction during that time and is given by the vector drawn from the initial position to its final position.
Displacement is independent of the path.
The value of displacement can never be greater than the distance travelled.
Speed and Velocity
•
Average speed is defined for a time interval. Average
•
• speed of a trip
v av
=
Total distance travelled
Total time taken
If a particular travels distances
s
speeds
v
1
,
v
2
,
v
3 travelled
s
=
s
1 2
+
s
Total time taken =
s v
1
1
3
+ ...... +
s s s v s
1
s v
3
3
+
2
v s
2
2
1
etc. respectively, then total distance
+
s
+
Average speed of a trip =
s v
2
2
+
n s
,
s
....
3
+
2
,
s v
.....
s n n
3
etc. with
.....
+
+
s v n s n
1
n
The speed at a particular instant is defined as
• instantaneous speed (or speed).
If D
t
approaches zero, average speed becomes instantaneous speed.
v
= lim
∆
t
→ 0
∆
∆
t s
=
ds dt
12 PHYSICS FOR YOU

AUGUST ‘17
i.e.
, instantaneous speed is the time derivative of distance.
Velocity
Average velocity : The ratio of net displacement to the time intervals.
Instantaneous velocity :
The rate of change of position with time at any instant is called instantaneous velocity.
v v av
=
=
∆
∆
x
lim
∆
t
→
t
0
=
∆
x
∆
t x t f f
=
−
−
t dt x dx i i
•
If a particle moves 1 st
half of a distance with velocity
v
1
and 2 nd
half with
v
• velocity
v
third with
v
3
, then
v
avg
=
3
v v v v v
+
v v
+
v v
2
, then
v
avg
=
2
v v v
+
v
1 2
If a particle covers one third distance with
1
, next one third with
v
2
and last one
Acceleration
•
Average acceleration : The ratio of net change in velocity to the time taken for change is called the average acceleration.
Instantaneous acceleration :
The rate of change of velocity of an object is called acceleration of the object.
a
avg
a
=
=
v f
−
v i t f
−
t i
∆ lim
0
∆
v
∆
t
=
=
∆
v
∆
t dv dt
If a particle is accelerated for time
t
acceleration
a
1
and for time
t
2
1 with
with acceleration
a
2
, then
a
avg
=
a t
+
a t t
1
+
t
2
.
•
t
If a particle starts from rest and moves with uniform acceleration then distance covered by the body in
seconds is proportional to is 1 : 4 : 9 .......... .
t
2
.
\
Ratio of distances covered after 1 s, 2 s, 3 s ..........
•
If a body starts from rest and moves with uniform acceleration then distance covered by the body in
n
th
sec is proportional to (2
n
– 1) (
i
.
e
.,
s
\
Ratio of distance covered in 1 second is 1 : 3 : 5.
st
n
, 2
∝ nd
(2
n
– 1)
and 3 rd
Illustration 1 : A car moves from
A
to
B
with a speed of
30 km h
–1
and from
B
to
A
with a speed of 20 km h
–1
.
What is the average speed and average velocity of the car.
Sol. : Let
s
be distance between
A
and
B
.
Let
t
1
be time taken by the car to move from
A
to
B
with speed
v
1
and
t
2
be time taken by the car to move from to
A
with speed
v
2
. Then
B t
1
=
s
and
t v
2
=
v s
2
Average speed of the car
v av
=
Total distance travelled
Total time taken
=
t
1
2
+
s t
2
=
s v
1
2
+
s s v
2
=
v
2
2
v v
+
v
1
Here,
v
1
= 30 km h
–1
\
v
av
=
× ×
20
,
v
2
= 20 km h
–1
= 24 km h
–1
Since the total displacement of the body through out the journey is zero.
As D
x
=
x f
–
x i
= 0
\ average velocity
v av
=
∆
∆
t x
=
0
KINEMATIC EQUATIONS
OneDimensional Approach
•
•
v v
=
2
Uniformly
Accelerated Motion
•
•
u s ut
=
u
+
2
at
1
2
at
+ 2
as
2
S n
= +
2
( 2 1 )
Motion Under Gravity
Vertically downward Vertically upward
•
Free fall
u
= 0,
a
=
g
•
•
v
= 0,
a
= –
g
•
•
•
v
=
gt h
= (1/2)
gt
2
v
2
= 2
gh
•
•
u
=
gt h
=
ut
– (1/2)
gt
2
u
2
= 2
gh
•
•
Variable Acceleration
a
=
f
(
t
)
= +
t v u f t dt
0
t
∫
( )
0
∫
( ( ) )
•
v a
=
f
(
x
)
2
−
u
2
=
2
∫
( )
PHYSICS FOR YOU

AUGUST ‘17
13
•
If accelerationtime equation is given, we can get velocitytime equation by integration. From velocitytime equation, we can get displacementtime equation by integration.
Diff erentiation
Slope of (
x
–
t
curve)
Are a under (
v
–
t
curve)
Integration
Diff erentiation
Slope of (
v
–
t
curve)
Are a under (
a
–
t
curve)
Integration
Graphical Analysis
Case I : Motion of a freely falling body under garvity
Variation of distance with time
Variation of acceleration with time
Variation of velocity with time
Case II : Motion of a body projected vertically upward
Case III : Variable Acceleration
v
2
0
t
Velocity time graph for increasing acceleration
a a
θ
=
kt
0
t
Acceleration time graph for constant increasing acceleration
Case IV : Constant Retardation
y
Variation of displacement with time
v
•
For a vertically upward projected body with initial speed
v i
,
Maximum height,
H
max
=
Time of flight,
T
= 2
t
=
2
v i
2
g g v i
2
•
A ball is dropped from a height and it reaches on the earth after
t
seconds. From the same height, if two balls are thrown one upwards and other downwards, with the same speed
v
the earth surface after
t
1 then
t
=
t t
.
and
t
2
i
and they reach
seconds respectively,
Illustration 2 : Two balls are projected simultaneously with the same speed from the top of a tower, one vertically upwards and the other vertically downwards.
They reach the ground in 9 s and 4 s respectively. Find the height of the tower.
(Take
g
= 10 m s
–2
)
Variation of acceleration with time
v
A v
= –
at
B
O t
Velocity time graph for constant retardation
a
–
0
a a
= constant
t
Acceleration time graph for constant negative acceleration
Sol.: Let the downward direction be positive. Initial speeds with which the balls are thrown be
v
0
(say).
For the ball thrown upward,
u
= –
v
0
,
t
= 9 s,
s
=
h
,
a
= 10 m s
–2
Using
s
=
ut
+ 1
at
2
h
= (–
v
0
⇒
h
= 405 – 9
v
2
0
× 10 × 9
2
For the ball thrown downward,
u
= +
v t
= 4 s,
s
=
h
,
a
= + 10 m s
–2
,
0
Using
s
=
ut
+ 1
at
2
h
= (
v
0
) ⋅
⋅ 4
2
⇒
h
= 4
v
0
+ 80
Solving (i) and (ii), we get,
v
0
= 25 m s
–1
, and
h
= 180 m
... (i)
... (ii)
14 PHYSICS FOR YOU

AUGUST ‘17
Two Dimensional Approach
Position vector :
r xi y j
Displacement vector :
Suppose a body displaced from
P
to
P
′
and
r
′ be the position vector of
Net displacement =
∆
r
P
′
.
=
r
′
–
r
\
∆
r
(
x i y j
) (
xi
+
y j
) or
∆
r
= ∆
xi
+ ∆
y j r
P
•
•
•
The relative velocities of
A
and
B
with respect to each other depends on their respective displacements.
The displacement of
B
relative to
A
(
i.e.
displacement of
B
as measured from
A
) = (
x
B
–
x
A
).
The rate of change of this displacement is called the
y x r
P
velocity of
B
relative to
A
=
d
(
x
B
–
x
A
).
dt
\ The velocity of
B
relative to
A
=
=
v
B dt
–
v
−
•
dx
B dx
A dt
A
.
This is the velocity of
B
appears to have, when seen from
A
.
Average velocity :
RELATIVE MOTION
v av
=
Instantaneous Velocity :
Average acceleration
a av
=
∆
∆
t v
∆
t
lim
→
0
=
∆
t
u at
∆
∆
v v i
+
∆
r
= +
0
+
1
2
t
∆
=
dv
=
dt v j dv x
=
v
∆
∆
∆
=
∆
r v t t
∆
t x
Instantaneous acceleration
=
i
∆
∆ lim
→
0
+
t x
∆
i
∆
∆
∆
v t r t
+
y
=
∆
∆
t y d r dt
=
v x
=
u v y
=
u
=
=
dx dt
+
i
+
x
+
a x t y
+
a y t
+
dy dt j x
=
x
0
+
u x t
+ 1/2
a x t
2
y
=
y
0
+
u y t
+ 1/2
a y t
2
•
•
Let
r
and
r
BO
be the position vectors at time
t
, of two moving particles with respect to fixed origin velocities
dr
v
then given by,
v
AO
=
AO dt
AO
and and
v
BO v
O
BO
=
dr
. The
are
BO dt r
AO
A
O r
BA
By the triangle law of vectors,
OA AB OB r
BO
Boatriver problem
B a
•
•
=
Kinematic Equations
2
dt i
+
dv dt y
along
x
axis along
y
axis along
x
axis along
y
axis
= +
•
\ = − =
BO
−
r
AO
=
r
BA
AB
is the displacement of
A
relative to
B
and is the position vector of
B
relative to
A
.
•
•
The velocity of
=
\
d dt v
(
r
BO
BA
=
−
v
r
AO
BO
)
−
B
=
v
relative to
dr
AO
BO dt
−
dr
AO dt
A
is
v
BA
=
r
BA dr
BA dt
Relative velocity of a body
A
with respect to body
B
when the two bodies moving at an angle q
is given by
2
along
x
axis along
y
axis
One Dimensional Approach
•
•
Let two particles
A
and
B
move along the same straight line and at time
t
, their displacements measured from some fixed origin
O
on the line be
x
and
B
A
are,
and
x v
A
=
B dx dt
respectively. The velocities of
A
A
and
v
B
O
=
x
A dx
Relative Velocity
A v
A dt v v
B x
2
y
2
=
u
=
u x
B x
2
y
2
+ 2
a x
+ 2
a y
B x y v
B v
AB
tan β
=
=
If
v
AB
then,
=
v v v
makes an angle
A v
+
+
+
B v v v
− sin(
B
+
2
2
v v
180 cos(
° −
180 b
θ )
° − cos( cos
with the direction of
θ )
θ shortest path, the boat should be rowed upstream making an angle q
with
AB
such that
AB
gives the direction of resultant velocity.
C
180
=
v
° −
A v
−
B
θ
v
) sin
B
θ
B
A
θ cos θ
v
A
(a) To cross the river in the shortest path : For the
d
PHYSICS FOR YOU

AUGUST ‘17
15
So, sin θ = and
v
2
=
v v v
2
1
–
v
Also
t v d v
1
2
−
v
2
2
v
1
= velocity of boat in still water
v
2
= velocity of flow of water in river
d
= width of river
(b) To cross the river in the shortest time : For the boat to cross the river in
B x
C
shortest time, the boat should be directed along
AB
. Let
v
be the resultant velocity making an angle q with
AB
. Then tan
θ =
v v
2
=
v
+
v
2
2
.
d
A
θ
\
v
2
1 and
1
2
Time of crossing,
t
=
d
/
v
Now the boat reaches the point
C
rather than reaching point
B
due to velocity of river. If
BC
=
x
, then
1
.
tan θ =
v v
2
1
=
x d
or
v v
2
1
(c) If a man travels downstream in a river, time taken by the man to cover a distance by him to cover a distance
t t
1
2
=
2
2
.
d
is
t d
2
is
t
1
=
−
2
d
+
If a man swims upstream in a river, time taken
=
d v v
2
.
. So
Illustration 3 : Two ships are 10 2 km apart on a line running south to north. The one further north is moving west with a speed of 25 km h
–1 towards north with a speed of 25 km h
, while the other
–1
. What is their distance of closest approach and how long do they take to reach it?
Sol.: Let
A
and
B
be the initial positions of the two ships such that
AB
= 10 2 km.
The ship at
A
is moving with velocity
u
westward and the ship
B
northward with velocity
u
−
2
(25 km h given by
1
respectively.
Then
QR v
–1
BA
In figure (b)
).
=
v
B
−
v
A
=
and
PR
2
1
2
(
1
(25 km h
1
)
represent velocities
u
2
–1
)
The relative velocity of ship
B
with respect to ship
A
is
and
represents the relative velocity of
B
with respect to
A
.
QP
=
+
PROJECTILE MOTION
2
= and tan q
=
PR
QP
=
25
25
=
1
\ q
= 45°.
Now, we can suppose that the ship
A
is at rest at
P
and the ship
B
is moving along
QR
with a velocity of
25 2 km h
–1
.
The closest distance between the ships will be
PN
, which is the perpendicular distance of
P
from
QR
.
Now,
PN
PQ
= sin45°;
PN
=
PQ
sin45° or
PN
×
2
= 10 km
2
QN
=
PQ
×
2
= 10 km
= 25 2 km h
–1
\ Time required to reach the closest distance
=
QN
relative velocity
× 60 min. = 17 min
Oblique Projectile Motion
Equation of Trajectory
Maximum Height
Time of Flight
Horizontal Range
=
10
25 2
y x
tan
h
θ −
H u
2 sin
2
g
2
θ
T
=
u
θ
g
R
=
u
2 sin 2
θ
g
1
2
u
2
gx
2 cos
2
θ
16 PHYSICS FOR YOU

AUGUST ‘17
Horizontal Projectile Motion
Equation of Trajectory
y
=
1
2
Time of Descent
Horizontal Range
Instantaneous Velocity
T
=
2
gx
2
u
2
g h v
=
u
2
+ 2
gy
=
u
2
+
g t
tan
φ =
v v x y
;
φ = tan
− 1
gt u
Projectile Motion on an Inclined Plane of
Inclination ( )
Time of Flight
Maximum Height
Horizontal Range
Maximum range when thrown upward
Maximum range when thrown downward
T
=
g
θ cos α
=
2
g
2 sin
2 cos
θ
α
R
=
R
max
2
u g
2
=
g
θ cos
2
α
+
u
2
R
max
=
g u
2
)
Some important facts of angular projection of projectile :
Item
•
•
Acceleration of projectile
Velocity of projectile
Description
It is constant throughout the motion of projectile and it acts vertically downwards.
It is different at different instants. It is maximum at the starting point
O i.e.
u
and is minimum at the highest point
i.e.
u
cos q .
•
Linear momentum at the highest point
p
H
=
mu
cos q
•
Linear momentum at the lowest point
p
0
=
mu
•
•
•
•
Maximum horizontal range
Horizontal range will be same
Kinetic energy of projectile
Angular momentum of projectile at
H
R
max
=
u
2
/
g
. It is so when q = 45°
(i) if angle of projection is q or 90° – q
(ii) if angle of projection is (45° + q ) or (45° – q )
It is maximum at the starting point
O
and is minimum at the highest point
H.
L
= (
mu
cos q
) ×
H
Illustration 4 : A projectile
A
is thrown with velocity
v
1 at an angle of 30° to the horizontal from point
P
. At the same time, another projectile
B
is thrown with velocity
v
2
upwards from the point
Q
vertically below the highest point. For
(a) 1
B
to collide with
(b) 2
A
,
v v
2
1
should be
2
(d) 4
Sol. (c) : The time of ascent for the projectile
A
is
⇒
t
A
=
v
1 sin 30 °
g
Maximum height of the projectile or
v
1
2 sin
g
2
30 °
=
v v g
30 °
⇒
A
=
v v
2
1
v
1
2 sin
2
g
2
30 °
Height attained by projectile
B
=
v
2
t
– 1
gt
2
.
For collision, both heights must be same at the time
t
A
,
⇒
v
1
2 sin
2
g
2
30 °
=
v
2
v
1 sin
g
30 °
−
1
2
g
v
1
2 sin
g
2
2
30 °
PHYSICS FOR YOU

AUGUST ‘17
17
Illustration 5 : On an inclined plane of inclination
30°, a ball is thrown at an angle of 60° with the horizontal from the foot of the incline with a velocity of 10 3 m s
–1
. If
g
= 10 m s the inclined plane in
(a) 1 s
–2
, then the ball will hit
(b) 2 s
(c) 2 3 s (d) 4 3 s
Sol. (b) : Taking
x
axis parallel to inclined plane and
y
axis perpendicular to inclined plane,
u y
= 10 3 sin (60° – 30°) = 5 3 m s
–1
g
3
a y
= –
g
cos 30° = −
2
When the ball strikes back,
s y
= 0
Using,
s y
=
u y t
+ 1
a y t
2
0 = ( 5 3 )
t
−
1
2
g
2
3
t
2
⇒
t
= 0 or
t g
= 2 s.
CIRCULAR MOTION
•
When a particle moves in a plane such that its distance, from fixed (or moving) point remains constant then its motion is called as circular motion with respect to that fixed (or moving) point.
Angular displacement :
Angle position vector of a particle moving w.r.t. some fixed point.
∆θ
=
=
Arc
Radius
Arc
r
Angular velocity : It is defined as the rate of change of angular displacement.
ω =
=
Angle traced
Time taken
∆ lim
0
∆
∆
t
θ
=
d dt
θ
Relation between linear and angular velocity
Average angular velocity (
w
av
)
v
= ω
r
In vector form
v r
ω
av
=
θ
t
2
2
−
−
t
θ
1
1
=
∆
∆
t
θ
Instantaneous angular velocity
Angular
acceleration : Rate of change of angular velocity.
ω =
∆ lim
t
→
0
∆
∆
t
θ
=
d dt
θ
α =
∆ lim
0
∆ ω
∆
t
=
d dt
d dt
•
•
•
Uniform Circular Motion
•
When a body moves along a circular path with uniform speed, its motion is said to be uniform circular motion.
¾
Radius vector velocity vector
r
. .
0
•
¾
Velocity vector is always perpendicular to the
i.e
.
0
For circular motion, force towards centre (centripetal force) must act so that direction of changing.
v
keeps on
The work done by centripetal force is always zero.
Kinetic energy = constant
Since   = constant, so tangential acceleration
a t
= 0
NonUniform Circular Motion
•
•
•
The magnitude of the velocity of the particle in horizontal circular motion changes with respect to time.
The acceleration of particle is called tangential acceleration. It acts along the tangent to the circle at a point. It changes the magnitude of linear velocity of the particle.
Centripetal acceleration acceleration
a t a c
and tangential
act at right angles to each other.
\
a
2
=
a c
2
+
a t
2
=
v r
+
a t
2
.
tan φ =
a a c t
=
r
2
α
=
r v
2
2
α
.
Illustration 6 : A motor car is moving with a speed of 20 m s
–1
on a circular track of radius 100 m. If its speed is increasing at the rate of 3 m s
–2
, its resultant acceleration is
(a) 3 m s
Here,
a t a c
–2
(c) 2.5 m s
–2
Sol. (b) : Here,
v
= 20 m s
Centripetal acceleration,
a c
(b) 5 m s
–2
(d) 3.5 m s
–2
–1
,
r
= 100 m,
a
=
v
2
=
t
= 3 m s
–2
r
100
= 4 m s
− 2
is acting tangential to the circular path and is acting along the radius towards the centre of the circular path.
Resultant acceleration =
=
a c
2
+
a t
2
( ) ( )
2
= m s
− 2
18 PHYSICS FOR YOU

AUGUST ‘17
1.
Figure gives speedtime graph of a particle in motion along a constant direction.
Three equal intervals of time are shown. Among the three intervals, the incorrect statement is
Time
(a) The average acceleration in the second interval is greatest
(b) The average speed is greatest in the third interval
(c) The acceleration at
A
,
B
,
C
,
D
is zero.
(d) The distance travelled in the second interval is greatest.
2.
An engineer works at a factory out of town.
A car is sent for him from the factory every day and arrives at the railway station at the same time as the train. One day the engineer arrived at the station one hour before his usual time and without waiting for the car, started walking towards factory. On his way he met the car and reached his factory 10 min before the usual time. For how much time did the engineer walk before he met the car? The car moves with the same speed everyday.
(a) 105 min
(c) 75 min
(b) 55 min
(d) 45 min
3.
A projectile is projected at an angle with an initial velocity
u
. The time
t
at which its horizontal component will equal the vertical component is
(a)
(c)
u g u g
(cos
(sin
α
α
−
−
α
α
(b)
(d)
u g u g
(sin
(sin
2
α
α
+
−
α
4.
In the given figure, a particle starting from point
A
travelling upto
B
with speed
v
, then upto point
C
with speed 2
v
and to
A
with speed 3
v
. The average speed during the entire path is (Take radius =
R
.)
(a) 1.8
v
(b) 3
v
(c) 1.2
v
(d) 2
v
5.
An armoured car 2 m long and 3 m wide is moving at 13 m s
–1
when a bullet hits it in a direction making an angle tan
− 1
3
4
with the car as seen from the street. The bullet enters one edge of the car at the corner and passes out at the diagonally opposite corner. Neglecting any interaction between bullet and the car, the time for the bullet to cross the car is
(a) 0.15 s (b) 0.23 s (c) 0.4 s (d) 0.12 s
6.
A man can swim with a speed of 4 km h
–1
in still water. He crosses a river 1 km wide that flows steadily at 3 km h
–1
. If he makes his strokes normal to the river current, how far down the river does he go when he reaches the other bank?
(a) 500 m (b) 600 m (c) 750 m (d) 850 m
7.
A particle has an initial velocity of 9 m s and a constant acceleration of 2 m s
–2
–1
due east due west. The distance covered by the particle in the fifth second of its motion is
(a) zero (b) 0.5 m (c) 2 m (d) none
8.
Two boys are standing at the ends
A
and
B
of a ground where
AB
=
a
. The boy at
B
starts running in a direction perpendicular to
(c)
v
2
+
v
1
2
a
( −
1
)
(d)
AB
with velocity
a
2
(
v
2
−
v
1
2
a
( +
1
)
)
v
1 velocity
v
and catches the other boy in a time
t
, where
t
is
(a)
a
(b)
.
The boy at
A
starts running simultaneously with
9.
A gun is mounted on the top of a tower of height
h
.
When a bullet is fired from the gun at angle q
with the horizontal, bullet moves with speed
u
to cover maximum range
x m
. Then
(a) tan
(c) tan
θ =
θ =
u
2
2
u u
2
+ 2
gh u
+
2
(b)
(d)
x m x m
=
=
u u
2
u
2
g g
+
+
2
2
gh gh
PHYSICS FOR YOU

AUGUST ‘17
19
10.
A train is moving at constant speed
V
. Its driver observes another train in front of him on the same track and moving in the same direction with a constant speed
v
. If the distance between the trains be
x
, then the minimum retardation of the train so as to avoid a collision is
(a) (
V
+
v
)
2
(c) (
V
+
v
)
2
/
x
/2
x
(b) (
V
–
v
(d) (
V
–
v
)
)
2
2
/
/2
x x
11.
Two particles
A
and
B
get 4 m closer each second while travelling in opposite direction. They get
0.4 m closer every second while travelling in same direction. The speeds of
A
and
B
are respectively
(a) 2.2 m s
–1
and 0.4 m s
–1
(b) 2.2 m s
–1
and 1.8 m s
–1
(c) 4 m s
–1
and 0.4 m s
(d) none of these
–1
12.
A person is walking at the rate of 3 km h
–1
, the rain appears to fall vertically. When he increases his speed to 6 km h
–1
, it appears to meet him at angle of 45° with the vertical. The speed of rain is
(a) 3 2 km h
− 1
(c) 6 2 km h
−
1
(b) 3
2
(d) 2 3 km h
− 1 km h
− 1
13.
A balloon starts rising from ground with an acceleration of 1.25 m s
–2
. After 8 s a stone is released from the balloon. The maximum height from the ground reached by stone will be
(a) 40 m (b) 50 m (c) 45 m (d) 55 m
14.
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time
t
1
. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time
t
2
. The time taken by her to walk up on the moving escalator will be
(a)
t
2
t t
−
t
1
(c)
t
1
–
t
2
(b)
(d)
t
2
+
t
1
+
2
2
[NEET 2017]
15.
The
x
and
y
coordinates of the particle at any time are
x
= 5
t
– 2
t
2
and
y
= 10
t
respectively, where
x
and
y
are in m and
t
in s. The acceleration of the particle at
t
= 2 s is
(a) 5 m s
–2
(c) –8 m s
–2
(b) –4 m s
(d) 0
–2
[NEET 2017]
16.
A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity versus time?
(a)
(c)
(d)
[JEE Main Offline 2017]
17.
Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity?
(a) (b)
(c)
(b)
(d)
[JEE Main Online 2017]
18.
A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m s
4 m s
–2 time of
–2
and the car has acceleration
. The car will catch up with the bus after a
(a) 120 s (b) 15 s (c) 10 2 s (d) 110 s
[JEE Main Online 2017]
19.
Two cars
P
and
Q
start from a point at the same time in a straight line and their positions are represented by
x
P
(
t
) = (
at
+
bt
2
) and
x
Q
(
t
) = (
ft
–
t
2
At what time do the cars have the same velocity ?
).
(a)
(c)
1
+
+
b a f
2 1 +
b
)
(b)
(d)
2 (
b
− 1
2 1 +
b
)
)
[NEET PhaseII 2016]
20.
In the given figure,
a
= 15 m s
–2
represents the total acceleration of a particle moving in the clockwise direction in a circle of radius
R
= 2.5 m at a given instant of time. The speed of the particle is
20 PHYSICS FOR YOU

AUGUST ‘17
(a) 4.5 m s
–1
(c) 5.7 m s
–1
R
O
30°
a
(b) 5.0 m s
–1
(d) 6.2 m s
–1
[NEET PhaseII 2016]
SOLUTIONS
1. (d) :
As area under the curve in the third interval is largest, so distance travelled in the third interval is greatest. Hence statement (d) is incorrect.
2. (b)
3. (c) :
Horizontal component at any time
t
is
v x
Vertical component at any time
t
is
v y
=
u
cos
=
u
sin –
gt
From given problem,
v x
= v y
7. (b) :
Acceleration,
a
= –2 m s
\
Velocity after any time
t
,
v
=
u
+
at
or
t
Initial velocity,
=
=
=
v
∫
∫
 
a
cos θ
=
=
∫
4
+

∫
a
2
v
2
−
v
5
−
(
t dt t

−
(due east)
(due east)
\ Distance travelled in fifth seconds
(due east)
d t
2
v dt
5
= (9 –2
t
) m s
–1
4
t
1
4 5
(
u
= 9 m s
–1
)
dt
=
8. (b) :
The two boys meet at
C
after time
t
.
m
Taking horizontal motion of boy
A
from
A
to
C
,
a
=
v
cos q
t
9. (b) :
The equation of trajectory is tan tan
θ
θ
−
−
2
u
2
2
u gx
2
2 cos
gx
2
2 cos
2
θ
θ
…(i)
\
u
cos =
u
sin –
gt
or = (sin α − α
4. (a) :
t
AB
=
π
2
R v
,
t
BC
π
R v
=
π
3
R v
For
x
to be maximum,
dx d
θ
=
0 and
Differentiating eqn. (i) w.r.t. to q
d
θ
2
= negative
=
t v
CA
av
3 km h
=
=
t
5
12
AB
×
×
1
4
+
2
2
t
3
π
π
v
R
R
BC
4
=
+
t
5 π
18
CA
=
R v
=
1
2
2
1
v
+
3
v
+
750 m
5
18
v
= .
v
0 =
tan θ
2
dx g d
θ
u
2
+
x
sec
2 sec
2
θ
−
2
−
1
θ
+
⋅ 2
d
θ
u
4
g x m
+
At
x
=
x m
Put sec
2
,
dx d
θ
=
0 q
= 1 + tan
–
h
=
x
tan q −
2
gx u
2
2
2
\ tan
θ =
1 q
in eqn. (i)
2
u gx
2
x m
2
2
5. (a) :
Let the speed of the bullet be
v
. Velocity of bullet relative to car along
x
axis = (
v
cos q – 13) and along
y
axis =
v
sin q
.
Since bullet appears from diagonally opposite corner, its displacement relative to car along
x
axis and
y
axis are 2 m and 3 m respectively,
i.e.,
6. (c) :
To cross the river, the time taken by man
2 = (
=
t
=
4
v
13
1
1 cos
tan km km h q
– 13)
On eliminating
3
− 1
v
θ
=
−
1
4
2
t
and 3 = h
=
2
13
≈
v
sin q s
t
from above eqns., we get
During this time, the river takes him down stream by
(1 km = 1000 m)
x m
=
10. (d) :
Here relative velocity of the train w.r.t. other train =
gx u u
2
V – v
Hence, 0 – (
\
11. (b) :
If
u m g
+
V
and
2
gh
–
v gx
2
u m
2
v
)
2
and tan θ =
= 2
ax
or
a
Minimum retardation
= −
are speeds of then
u + v
= 4 and
u – v
= 0.4.
\ 2
u
= 4.4 or
u
= 2.2 m s
A
–1
(
u
= −
2
2
x
[using eqn. (ii)]
and
u
+ 2
gh
(
V v
)
2
)
2
2
B x
…(ii)
respectively,
;
v
= 1.8 m s
–1
PHYSICS FOR YOU

AUGUST ‘17
21
12. (a) :
Let velocity of rain be
Velocity of rain with respect to man,
v v
r xi y j
Velocity of man
v
v m
= 3
i rm
(
v
m
) = −
Rain seem to be vertical
3 ) +
\
x
– 3 = 0 or
x
= 3
In second case,
v m
= 6
i
\
v rm
(
x
\

x
 = 
y
 = 3
6 )
i y j
= − +
It seems to be at an angle 45° with vertical.
(
x
= 3)
\
v r
=
x
2
+
y
2
=
2
+
2
= 3 2 km h
− 1
13. (c) :
When a particle separates from a moving body it retains the velocity of the body but not its acceleration. The distance travelled by the balloon when stone is released is
s
0 and
v
= 0. Using
v
2
–
u
2 m above the ground and
.
For the motion of stone from the balloon to the highest point,
u
= 10 m s
= 2
a
(
,
a
= –
g
= –10 m s
s – s
0
–2
), where
s
is the displacement from the ground, we get
0 – (10)
2
= –2 (10) (
s
– 40) or
s
= 45 m
\ Maximum height attained by stone from the ground = 45 m.
14. (b) :
Let
v
1
is the velocity of Preeti on stationary escalator and
d
is the distance travelled by her,
\
v
1
=
d
/
t
Again, let
1
v
2
is the velocity of escalator
\
v
2
=
d
/
t
2
\ Net velocity of Preeti on moving escalator with respect to the ground is
v t
=
=
v d v
1
+
v
The time taken by her to walk up on the moving escalator will be
=
d
15. (b) :
x
= 5
t
– 2
t
2
=
d d t
1
+
t t
2
t
2
,
y
= 10
t d
2
=
=
d
+
2
2
16. (c) :
Velocity of the body going upwards is given by
v
=
v
0
–
gt
(
v
0
= initial velocity)
Hence, the graph between velocity and time should be a straight line with negative slope
g
and intercept
v
0
.
Also, during the whole motion, acceleration of the body is constant
i.e.,
slope should be constant.
17. (a) :
Here, acceleration is given by,
a
= –
c dv dt
= −
c
or
dx dt
or
vdv
= –
cdx
⋅
dv dx
…(i)
On integrating eqn. (i) we get
v
2
2 where
k
is a constant of integration.
∴
18. (c) :
x
= −
v
2
2
c
+
k c
, i.e.,
graph (a) is correct.
Acceleration of car,
a c dx dt
Acceleration of bus,
a
B
C
= 4 m s
= 2 m s
–2
Initial separation between the bus and car,
s
CB
= 200 m
Acceleration of car with respect to bus,
a
CB
=
a
C
–
a u
B
CB
= 2 m s
= 0,
t
= ?
= −
=
v
As,
s
CB
=
u
CB
×
t
+ 1
a
CB t
2
\ 200 = 0 ×
t
+ 1
19. (d) :
x v
P
P
(
t
(
t
) =
at
+
bt
) =
P dt
( )
Similarly, for car
x
Q
(
t
) =
ft
–
t
2
Q
,
× 2 ×
Position of the car
P t
2
⇒
t
= 10
The car will catch the bus after 10 2 s.
at any time
t
, is
2
s
v
Q
(
t
) =
v
P
(
t
) =
v
Q
(
t dt f
2
t
)
\
a
+ 2
bt
=
f
– 2
t
or 2
t
(
b
+ 1) =
f
–
a
\
t
=
−
( +
b
)
20. (c) :
Here,
a
= 15 m s
–2
R
= 2.5 m
From figure,
R a c
O
(Given)
30°
a a t
=
5 – 4
t
,
dy
=10 ⇒
v x
= 5 – 4
t
,
v y
= 10
dv dt x
=– 4,
dv dt y
= 0
⇒
a x
= –4,
a y
= 0
Acceleration,
a a i a j
= −
i i.e
.,
a
= –4 m s
–2
a c
=
a
cos 30° = 15 × 3
As we know,
a c
=
v
R
2
m s
⇒ =
–2
a R
\
v
=
15
×
2
3
× =
− 1
22 PHYSICS FOR YOU

AUGUST ‘17
EXAM
CL
AS
XI
S
PREP 2018
Useful for Medical/Engg. Entrance Exams
CHAPTERWISE MCQs FOR PRACTICE
WORK, ENERGY AND POWER
1.
When a body moves in a circle, the work done by the centripetal force is always
(a) > 0
(c) Zero
(b) < 0
(d) None of these
2.
Power applied to a particle varies with time as
P
= (3
t
2
– 2
t
+ 1) W, where
t
is in second. Find the change in its kinetic energy between
t
= 2 s and
t
= 4 s
(a) 32 J (b) 46 J (c) 61 J (d) 100 J
3.
A bomb moving with velocity
( 40
^
i
+
50
^
j
−
25 ) m s
−
1 explode into two pieces of mass ratio 1 : 4. After explosion the smaller piece moves away with
k
) velocity
(
(a)
(c)
45
45
^
j
−
−
200
^
i
+
70
^
j
+
15
35
35
m s . The velocity of larger piece after explosion is
^
j
(b)
(d)
45
−
i
^
35
^
i
−
+
35
^
j
45
4.
A body of mass 5 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to
1
10
th
of its original velocity. Then the mass of the second body is
(a) 4.09 kg (b) 0.5 kg (c) 5 kg (d) 5.09 kg
5.
A body of mass
m
strikes another body at rest of mass
m
9
.
Assuming the impact to be inelastic, the fraction of the initial kinetic energy transformed into heat during the contact is
(a) 0.1 (b) 0.2 (c) 0.5 (d) 0.64
6.
Which of the following is true for any collision?
(a) Both linear momentum and kinetic energy are conserved.
(b) Neither linear momentum nor kinetic energy may be conserved.
(c) Linear momentum is always conserved, however, kinetic energy may or may not be conserved.
(d) Kinetic energy is always conserved, but linear momentum may or may not be conserved.
7.
The kinetic energy of particle moving along a circle of radius
R
depends upon the distance covered
S
and is given by
K
=
aS
, where
a
is a constant. Then the force acting on the particle is
(a)
aS
R
(b)
2
aS
2
R
(c)
aS
R
2
2
(d)
2
aS
R
8.
A body of mass
M
is dropped from a height
h
on a sand floor. If the body penetrates
x
cm into the sand, the average resistance offered by the sand to the body is
(a)
(c)
Mg
Mgh
h x
+
Mgx
(b)
(d)
Mg
Mg
1 +
1 −
h x h x
9.
The displacement
x
and time
t
for a particle are related to each other as
t
=
x
+
3
. What is work done in first six seconds of its motion.
(a) 6 J (b) zero (c) 4 J (d) 2 J
10.
A small block of mass 0.1 kg
Y
is pressed against a horizontal spring fixed at one end to compress the
m
= 0.1 kg
2 m spring through 5.0 cm as
O
shown in figure. The spring constant is 100 N m
X
–1
.
When released the block moves horizontally till it
PHYSICS FOR YOU

AUGUST ‘17
23
leaves the spring, it will hit the ground 2 m below the spring
(a) At a horizontal distance of 1 m from free end of the spring.
(b) At a horizontal distance of 2 m from free end of the spring.
(c) Vertically below the edge on which the mass is resting.
(d) At a horizontal distance of 2 m from free end of the spring.
11.
A proton is kept at rest. A positively charged particle is released from rest at a distance
d
in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time
t
, the work done on the two moving charged particles is
(a) Same as the force law is involved in the two experiments.
(b) Less for the case of a positron, as the positron moves away more rapidly and the force on it weakens.
(c) More in the case of positron, as the positron moves away a larger distance.
(d) Same as the work done by charged particle on the stationary proton.
12.
A bullet losses 19% of its kinetic energy when passes through an obstacle. The percentage change in its speed is
(a) Reduced by 10% (b) Reduced by 19%
(c) Reduced by 9.5% (d) Reduced by 11%
13.
With what minimum speed
v
must a small ball should be pushed inside a smooth vertical tube from a height
h,
so that it may reach the
h v
R
top of the tube? Radius of the tube is
R
.
(Assume radius of crosssection of tube is negligible in comparison to
R
.)
(a) 2
+ )
R
(c)
5 − 2 ) )
14.
A ball is thrown from a height of
h
metre with an initial downward velocity
v
0
. It hits the ground, loses half of its kinetic energy and bounces back to the same height. The value of
v
(a) 2
gh
(b)
0
is
gh
(c) 3
gh
(d) 2 5
gh
−
15.
A girl in a swing is 2.5 m above ground at the maximum height and at 1.5 m above the ground at the lowest point. Her maximum velocity in the swing is (
g
= 10 m s
(a)
(c)
5 2
2 3 m s
−
1 m s
−
1
–2
)
(b)
(d)
2 5
3 2 m s
−
1 m s
−
1
SYSTEM OF PARTICLES AND ROTATIONAL MOTION
16.
A diver makes 2.5 revolution on the way from a
10 m high platform to the water. Assuming zero initial vertical velocity, the average angular velocity during the dive is
(a) 3
π
2 rad s
− 1
(b) 5
π
2 rad s
− 1
(c) 5
π
3 rad s
−
1
(d)
π
2 rad s
− 1
17.
The angular velocity of a rigid body about any point of that body is the same
(a) only in magnitude
(b) only in direction
(c) both in magnitude and direction necessarily
(d) both in magnitude and direction about some points but not about all points.
18.
An ice skater starts a spin with her arms stretched out to the sides. She balance on the tip of one skate to turn without friction. She then pulls her arms in so that her moment of inertia decreases by a factor of 2. In the process of her doing so, what happens to her kinetic energy?
(a) It increases by a factor of 4.
(b) It increases by a factor of 2.
(c) It remains constant.
(d) It decreases by a factor of 2.
19.
Two particles of equal mass have velocities
v
1
= 4
i
^ an acceleration
v
a
2
1
=
=
4
(
^
j i
+
− 1
j
) m s while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of
(a) straight line
(c) circle
(b) parabola
(d) ellipse
20.
Two particles of mass 1 kg and 3 kg have position vectors
(a)
i
^
3
2
j i
^
+
2
3
k
^
j
+ 4
k
The centre of mass has a position vector
+
^
− and − 2
i
^
+ 3
^
j
− 4
k
(b)
− − 3
^
j
−
respectively.
2
k
(c)
− + 3
^
j
+ 2
k
(d)
− + 3
^
j
− 2
k
24
PHYSICS FOR YOU

AUGUST ‘17
21.
Moment of inertia of ring about its diameter is
I
.
Then, moment of inertia about an axis passing through centre perpendicular to its plane is
(a) 2
I
(b)
I
2
(c)
3
2
I
(d)
I
22.
A flywheel rotates with a uniform angular acceleration. Its angular velocity increases from
20 rad s to 40 rad s in 10 s. How many rotations did it make in this period?
(a) 80 (b) 100 (c) 120 (d) 150
23.
Circular disc of mass 2 kg and radius 1 m is rotating about an axis perpendicular to its plane and passing through its centre of mass with a rotational kinetic energy of 8 J. The angular momentum in (J s) is
(a) 8 (b) 4 (c) 2 (d) 1
24.
From a circular ring of mass
M
and radius
R
, an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is
k
times
MR
2
. Then the value of
k
is
(a) 3/4 (b) 7/8 (c) 1/4 (d) 1
25.
Three identical spheres, each of mass
M
are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 2 m each.
Taking their point of intersection as the origin, the position vector of centre of mass is
(a)
(c)
1
3
2
3
(
(
−
+
)
)
(b)
(d)
2
3
1
3
(
−
( + )
)
26.
A thin circular ring of mass
M
and radius
R
is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity w . If two objects each of mass
m
be gently attached to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity.
(a)
ω
M
(b)
ω (
M
M
+
−
2
2
m m
)
(c)
M
ω
M
+ 2
m
(d)
ω
(
M
M
+
27.
A cylinder of mass
m
is rotated about its axis by an angular velocity w and lowered gently on an inclined plane as shown in figure. Then:
(a) it will start going upward
2
m
)
1/ 3
30°
(b) it will first going upward and then downward
(c) it will go downward just after it is lowered
(d) it can never go upward
28.
A thin uniform rod of length
L
is initially at rest with respect to an inertial frame of reference. The rod is given impulse at one end perpendicular to its length. How far does the centre of the mass translate while the rod completes one revolution about its centre of mass?
(a)
L
3
π
(b)
L
π
2
(c)
L
π
4
(d)
L
π
5
29.
A body rolls without slipping. The radius of gyration of the body about an axis passing through its centre of mass is
K
. The radius of the body is
R
. The ratio of rotational kinetic energy to translational kinetic energy is
(a)
K
R
2
2
(b)
K
2
R
+
2
R
2
(c)
K
2
K
+
2
R
2
(d)
K
2
+
R
2
30.
A uniform rod of length
L
(in between the supports) and mass
m
is placed on two supports
A
and
B
. The rod breaks suddenly at length
L
/10 from the support
B
. Find the reaction at support
A
immediately after the rod breaks.
(a) 9
40
mg
(b) 19
mg
(c)
mg
2
(d) 9
mg
SOLUTIONS
1. (c)
2. (b) :
P
= 3
t
2
– 2
t
+ 1 =
\
dE
= ( 3
t
2
t
= 4 s
E
=
∫
t
=
2
s
( 3
t
2
– 2
t
− + )
dt dE
+ 1)
dt dt
=
3
t
3
3
−
2
t
2
2
+
t
t t
= 4 s
= 2 s
= [(4
3
– 2
3
) – (4
2
– 2
2
E
= 56 – 12 + 2 = 46 J
) + (4 – 2)]
3. (a) :
Let the mass of the unexploded bomb be 5
m
.
It explodes into the two pieces of masses
m
and 4
m
respectively.
Initial momentum of the unexploded bomb
m
(
^
i
+
50
^
j
−
25 )
=
After explosion, momentum of the smaller piece
=
mv
1
=
m
( 200
i
^
+
70
^
j
+
15 ) and momentum of the larger piece
= 4
mv
2
PHYSICS FOR YOU

AUGUST ‘17
25
where
v
1 and respectively.
v
2
are the velocities of the two pieces
According to the law of conservation of momentum, we get
(
i
+ 50 − 25
4
mv
2
=
(
i
+
k
)
=
m
50
−
(
25
200
i
+
k m
(
70 + 15
)
+ 4
mv
200
i
+
70
+
15
k
^
)
2
v
2
4. (a) :
=
v
1
1
4
(
180
j
− 140
Here,
m
1
k
)
= 5 kg,
m
=
2
45
^
j
−
35
= ?,
u
2
= 0,
v
1
=
u
1
10
Velocity of the first body after collision is given by
=
2
∴
⇒
u
1
10
m
=
(
2
=
(
5
5
m
−
+
1
45
11
−
1
m
+
m
2
2
)
2 1
m m u
1
2
+
0
= 4.09 kg
+
m
1
+
m
or
2
1
10
=
5
5
−
+
m
2
m
2
5. (a)
6. (c) :
Linear momentum is conserved in all types of collision but kinetic energy is not conserved in all types of collision. Kinetic energy is conserved in elastic collision but not conserved in inelastic collision.
7. (d) :
Force on a particle moving on a circular path
2
= Centripetal force =
mv
=
1
2
mv
2
2 2
R R
K
=
2
R aS
R
8. (b)
9. (b) :
Given;
t
=
x
+ 3
⇒
x
= (
t
– 3)
2
Now,
v dx
= 2 (
t
− 3 )
At
t
= 0,
v
1
At
t
= 6,
v
2
= 2 (– 3) = – 6
= 2 (6 – 3) = 6
Work done = Change in kinetic energy
=
1
2
(
2
2
−
v
1
2
=
10. (a) :
Here;
m
= 0.1 kg,
h
= 2m,
k
= 100 N m
–1
,
x
= 5 cm, 1
1
2
( 100 )
5
100
kx
2
2
=
=
mv
1
2
mv
2
2
v
5
2
Then, a horizontal distance from free and of spring,
=
5
2 10
= 1m
11. (c) :
Force between two protons is same as that of between proton and a positron.
As positron is much lighter than proton, it moves away through much larger distance compared to a proton.
We know that, work done = force × distance.
As forces are same in case of proton and positron but distance moved by positron is larger, hence, work done will be more.
12. (a) :
The kinetic energy of a bullet is given by,
K
=
1
mv
2 kinetic energy,
K
2
′ =
K
2
m
When it losses 19% of its kinetic energy, then
81
100
K
\
v
′ =
2
K m
′ = × .
m
K
=
\ Percentage decrease in speed
v
= −
1
.
× 100 point will be zero.
% = 10 %
13. (d) :
For minimum
v
, velocity of ball at the topmost
mV
2
=
( 2
−
)
V
=
g R h
)
14. (a)
15. (b) :
At the highest point,
v
= 0,
h
\
Total energy,
E
1
At the lowest point,
=
mgh
1
v
=
v
max
+ 0 =
= ?,
h
2
1
= 2.5 m
mgh
1
= 1.5 m
∴
Total energy,
E
2
=
mgh
2
+
1
mv
2
2
According to the law of conservation of mechanical energy,
E
1
=
E
2
\
mgh
1
=
1
2
mv
2
+
mgh
2
or
v
2
=
2 (
1
−
h
2
) or
v
=
=
2 (
1
−
h
2
× × =
)
=
20
2
× m s
−
1
10
=
×
( .
2 5
− m s
−
1
. )
16. (b) :
Let, the freefall time be
t
.
∆ =
0
+
1
2
gt
2
⇒ =
2 ∆
g y
=
2 10 m )
10 m s
− 2
= 2 s
Thus, the magnitude of the average angular velocity is
ω avg
=
2 s
− 1
=
5 π
2 rad s
− 1
17. (c)
18. (b) :
As
L
=
I
w
L
is constant
Inertia decreases by a factor of 2
\
ω ∝
I
1
26
PHYSICS FOR YOU

AUGUST ‘17
\
Angular velocity increases by a factor of 2.
M oment of inertia of a ring about an axi s passing through
\ Her kinetic energy with be 1 double.
19. (a)
I
w
2
i.e.
, become
20. (d) :
r r
1
Here,
i
ˆ
j m
1
= 1 kg,
4
ˆ,
ˆ
i
+
3
ˆ
j
+
4
k
ˆ
2
m r
m r m m
2 2
2
m
2
i
ˆ
= 3 kg
j
4
k
ˆ
The position vector of the center of mass is
=
CM
=
+
) + (3)(2
ˆ
i
+ −
4
k
ˆ
) the centre and perpendicular to the plane of the ring i s
I
=
MR
2
M ass of the remaining portion of the ring as show n in figure (ii) is
M
−
M oment of inertia of the remaining portion of the ring about the gi ven axi s is
I
But
I
′ =
kMR
2
′ =
3
MR
2
4
(given) \
k
= 3/4
25. (c) :
y
A
(0,2)
M
4
=
3
M
4
=
= − +
21. (a)
i
ˆ
i
ˆ
+ +
12
4
j
ˆ
4
k
−
ˆ
8
− + −
1
k
ˆ
i
ˆ
ˆ
j k
ˆ
k
ˆ
22. (d) :
Here, w
1
= 20 rad s , w
t
= 10 s
As w
2
= w
1
+
t
2
= 40 rad s
\
40 = 20 + × 10 or = 2 rad s
From,
ω
2
2
− ω
1
2
=
2
αθ
– 1
(40 ) – (20 ) = 2 × 2 q or
θ =
1200
4 π
π
2
= 300 π
Number of rotations completed
=
θ
2 π
=
300
2 π
π
= 150
,
x
CM
2 m
=
O
Σ
Σ
Σ
m i
2 m
=
M
(2,0)
0
2
B x
M
M
0
0
M
M
×
×
3
y
CM
=
Σ
m i
=
M
\ Position vector of centre of mass is
26. (c) :
As no torque is being applied,
L
= constant,
i.e
,
I
ω
2
=
I
I
2
=
(
M
+
1
, w
1
=
I
MR
2
ω
2 )
2
2
= w
(
2
M
ω
+
M
2
m
)
3
2
3
(
ˆ
i j
ˆ
)
23. (b) :
Here, Mass of the disc,
M
= 2 kg
Radius of the disc,
R
= 1 m
Moment of inertia of the disc about an axis perpendicular to, its plane and passing through its centre of mass is
1
( 2 kg kg m
2
I
=
2
MR
2
1
2
= 1
Kinetic energy of rotation,
K
R
=
L
2
2
I
where
L
is the angular momentum
τ = or
L
= 2
K I
= × ×
= 4 J s
24. (a) :
9
10
27. (d) :
Since net force along the incline is zero, so cylinder will remain in position till it stops rotating.
After that it will start moving downwards.
28. (a)
29. (a) :
Rotational K.E.
Translational K.E.
=
2
1
1
2
I
ω
2
mv
2
=
mv
2
mK v
2
R
2
=
K
R
2
2
30. (a) :
Torque,
τ =
mg
9
20
L I
α
m
3
9
10
mg
9
10
L
2
20
α
9
L
=
A
I
α
m
(9 /20)
L
3
C
9
10
L
2
9
L
10
α
D
\
α =
Acceleration,
−
a
3
2
g
L
CM
A
= (
=
ma
AC
CM
)
=
9
10
mg
3
g
2
L
27
40
g
9
L
20 or
=
N
27
40
g
A
=
9
40
mg
27
g
40 or
N
A
=
9
40
mg
PHYSICS FOR YOU

AUGUST ‘17
27
Series 2
CHAPTERWISE PRACTICE PAPER
MOTION IN A PLANE LAWS OF MOTION
Time Allowed : 3 hours
GENERAL INSTRUCTIONS
( i ) A l l q u es t i o n s ar e c o m p u l s o r y .
( i i ) Q . n o . 1 t o 5 ar e v er y s h o r t an s w er q u es t i o n s an d c ar r y 1 m ar k ea c h .
( i i i ) Q . n o . 6 t o 10 ar
( i v ) Q . n o . 11 t o 22 ar e s h o r t an s w er q u es t i o n s an d c ar r y 2 m ar k s eac e al s o s h o r t an s w er q u es t i o n s an d c ar r y 3 m h .
ar k s eac h .
( v ) Q . n o . 23 i s a v al u e b as ed q u es t i o n an d c ar r i es 4 m ar k s .
( v i ) Q . n o . 24 t o 26 ar e l o n g an s w er q u es t i o n s an d c ar r y 5 m ar k s eac h .
( v i i ) U s e l o g t ab l es i f n ec es s ar y , u s e o f c al c u l at o r s i s n o t al l o w ed .
Maximum Marks : 70
SECTION  A
1.
Air is thrown on a sail attached to a boat from an electric fan placed on the boat. Will the boat start moving?
2.
Sand is thrown on tracks covered with snow. Why?
3.
It is easy to catch a table tennis ball than a cricket ball even both are moving with same velocity. Why?
4.
Can the magnitude of the resultant vector of the two given vectors be less than the magnitude of any of the given vectors?
5.
Why are porcelain objects wrapped in paper or straw before packing for transportation?
SECTION  B
6.
A skilled gunman always keeps his gun tilted above the line of sight while shooting, why?
7.
When a ball is thrown upward from the surface of the earth, its momentum first decreases and then increases. Is the conservation of momentum violated in this process?
8.
The driver of a threewheeler moving with a speed of 36 km h
–1
sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just to save the child. What is the average retarding force on the vehicle? The mass of the threewheeler is
400 kg and the mass of the driver is 65 kg.
OR
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must move in opposite directions.
9.
A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath.
Give explanation to support your diagram.
10.
A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of 9 m s
–2
, what would be the reading of the weighing scale? (
g
= 10 m s
–2
)
SECTION  C
11.
Define projectile. Show that a path of projectile projected with initial velocity
u
and making an angle q
with the horizontal is parabola.
28 PHYSICS FOR YOU

AUGUST ‘17
12.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev. per min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
13.
Derive the relation between linear velocity and angular velocity.
14.
State and prove the polygon law of vector addition.
15.
Express Newton’s second law of motion in component form. Give its significance.
16.
A cyclist starts from the centre
O
of a circular park of radius 1 km, reaches the edge
P
of the park, then cycles along the circumference, and returns to the centre along
QO
as shown in figure. If the round trip takes 10 min, what is the
(a) net displacement, (b) average velocity, and
(c) average speed of the cyclist?
17.
For ordinary terrestrial experiments, which of the following observers are inertial and which are noninertial?
(a) A child rounding in a giant wheel.
(b) A driver in sportscar moving with a constant speed of 200 km h
–1
on a straight road.
(c) The pilot of an aeroplane which is taking off.
(d) A cyclist negotiating a sharp turn.
(e) The guard of a train which is slowing down to stop at a station.
18.
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
OR
The position of a particle is given by
r
= .
ti
− .
t j
+ where
t
is in seconds and the coefficients have the proper units for
r
to be in metres.
(a) Find the
v
and of the particle?
(b) What is the magnitude and direction of velocity of the particle at
t
= 2.0 s?
19
Give some important implications of the third law of motion.
20.
What are the laws of the limiting friction? Is the coefficient of kinetic friction less than or greater than the coefficient of static friction?
21.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s
–2
. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force on the helicopter due to the surrounding air.
22.
Define angle of repose. Deduce its relation with coefficient of static friction?
SECTION  D
23.
From a school a group of boys went for a picnic in a village. They went through fields and enjoyed the beauty of nature. While walking, they saw a well which they had never seen in the city. They were very excited and started drawing water from well. They planned to have a competition in which they decided that the who would draw more water would become winner. A villager who was listening to them, went to them and told them about the importance of water.
He also explained that they use the water of this for irrigating their fields and also for drinking.
(i) What values were possessed by the villager?
(ii) If the two boys raising the bucket, pull it an angle
q
to each other and each exerts a force of 20 N then their effective pull is 30 N.
What is the angle between their arms?
(iii) What are the other means of irrigation in villages?
SECTION  E
24.
Figure shows the positiontime graph of a particle of mass 4 kg.
What is the (a) force on the particle for
t
< 0,
t
> 4 s, 0 <
t
< 4 s?
(b) Impulse at
t
= 0 and
t
= 4 s?
OR
Read each statement carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that
(a) is conserved in a process
(b) can never take negative values
(c) must be dimensionless
(d) does not vary from one point to another in space
(e) has the same value for observers with different orientations of axes.
PHYSICS FOR YOU

AUGUST ‘17
29
25.
Give four examples from daily life which illustrate the law of conservation of linear momentum.
OR
(a) State and prove Lami’s theorem.
(b) There are four forces acting at a point
P
produced by strings as shown in figure, which is at rest. Find the forces
F
1
and
F
2
.
26.
(a) Show that for a projectile the angle between the velocity and the
x
axis as a function of time is given by
θ =
− 1
v oy
−
v ox gt
(b) Show that the projection angle q
0
for a projectile launched from the origin is given by
θ
0
= tan
−
1
4
h
R m
where the symbols have their usual meanings.
OR
A racing car travels (without banking) on a track
ABCDEFA
(see figure).
ABC
is a circular arc of radius
2
R
.
CD
and
FA
are straight paths of length
R
and
DEF
is a circular arc of radius
R
= 100 m.
The coeffecient of friction on the road is = 0.1. The maximum speed of the car is
50 m s
–1
. Find the minimum time for completing one round.
SOLUTIONS
1.
According to Newton’s second law motion, a body moves only if an external force is applied to it. Since the fan is a part of the boat, the force exerted by the air thrown by it on the sail is an internal force.
As such, the boat cannot be put into motion.
2.
When tracks are covered with snow, there is considerable reduction of frictional force. So, the driving is not safe. When sand is thrown on the snowcovered tracks, the frictional force increases.
So, safe driving is possible.
3.
Due to its small mass, the momentum of the table tennis ball is much smaller than that of the cricket ball of same velocity. Less force is required to stop the table tennis ball than the cricket ball. Hence it is easy to catch the table tennis ball than the cricket ball.
4.
This is possible if the angle between the two vectors is obtuse (> 90°).
5.
When porcelain objects are wrapped in paper or straw, the time of impact between themselves is very much increased during jerk in transportation.
F
= ∆
P
∆
t
As D
t
increased,
F
decreased.
Hence, force on the porcelains is reduced during transportation, and saves them from breakage.
6.
As soon as a bullet is fired from a gun, it also starts falling on account of acceleration due to gravity.
Therefore, if the gun is along the line of sight, it will hit below the target. To avoid this, the gun is tilted towards the line of sight so that the bullet, after travelling along parabolic path, hits the target as shown in below figure.
7.
No. The combined momentum of the ball and the earth is conserved. The ball attracts the earth by the same force as the earth attracts the ball. When the ball moves upward, its momentum decreases in the upward direction but simultaneously the momentum of the earth increases in the upward direction at the same rate. Similarly, when the ball falls down, its momentum increases in the downward direction but simultaneously the momentum of the earth increases in the upward direction at the same rate.
8.
Here,
u
= 36 km h
–1
= m s
− 1
v
= 0;
t
= 4 s
Now,
v
=
u
+
at
\
0 = 10 +
a
× 4 or
a
= – 2.5 m s
–2
=
10 m s
− 1
m
= 65 + 400 = 465 kg
Therefore, retarding force (in magnitude) on the vehicle,
F
= 465 × 2.5 = 1162.5 N ~_ 1.2 × 10
3
N
30 PHYSICS FOR YOU

AUGUST ‘17
OR
Let
m
= mass of the nucleus at rest.
u
= its initial velocity = 0 as it is at rest.
Let
m
,
m
be the masses of the two smaller nuclei also called product nuclei and
v
1
,
v
2 be their respective velocities.
If
p
i
the nucleus and the two nuclei respectively, then and
p
f
be the initial and final momentum of
…(i) and
9.
i p f
=
=
=
1 1
0
+
2 2 or
1 1
v
2
2 2
…(ii)
According to the law of conservation of linear momentum,
i
=
2 2
f
= − or 0
=
or
m v
= −
m
2
…(iii)
The negative sign in equation (iii) shows that
v
1
v
i.e
., the two smaller nuclei are moved in opposite directions.
With respect to the ground observer, ball is thrown with velocity
u
at angle q with the horizontal. Hence it seems as a projectile. So path of the ball will be parabola.
u
= car speed,
u
= +
u
,
= initial vertical speed of ball.
θ = tan
−
1
u
2
–2
u
1
10.
Given
m
= 50 kg,
g
= 10 m s
Acceleration of lift,
a
= 9 m s
–2
(downward)
N
= Reading of weighing scale = Normal on the man
Applying Newton’s second law of motion,
mg
–
N
=
ma
⇒
N
=
m
(
g
–
a
) = 50(10 – 9)
⇒
N
= 50 N
Reading of weighing scale,
′ =
50
= 5 kg
m
10
11.
Any object launched in an arbitrary direction in space with an initial velocity and then allowed to move under the influence of gravity alone is called a projectile. Suppose a body is projected with initial velocity
u
, making an angle q with the horizontal.
The velocity
u
has two rectangular components :
(i) The horizontal component
u
cos q , which remains constant throughout the motion.
(ii) The vertical component
u
sin q
, which changes with time under the effect of gravity.
Equation of trajectory of a projectile : Suppose the body reaches the point
P
(
x
,
y
) after time
t
.
\ The horizontal distance covered by the body in time
t
,
x
= Horizontal velocity × time =
u
cos q
t
or
t
=
u x
cos θ
For vertical motion,
u
=
u
sin q ,
a
= –
g
, so the vertical distance covered in time
t
is given by or
s ut y u
sin
1
2
θ
at
⋅
u
2
x
cos
2
u
2
θ
g
−
1
2
g
.
u
2
x
2 cos
2
θ or
y x
tan
θ − cos
2
θ
x
2
Thus
y
is a quadratic function of
x
. Hence the trajectory of a projectile is a parabola.
12.
Frequency of revolution, = 40 rev/min = mass of stone,
m
= 0.25 kg; radius of circle,
r
= 1.5 m; angular speed of the stone, w
= 2
×
40
60
=
4 π
3
40
60 rev/s rad s
− 1
T
= tension in the string = ?
= 2 π
T v
m ax m ax
= maximum tension in the string = 200 N
= maximum speed of the stone = ?
The centripetal force is provided by the tension (
T
) in the string,
=
=
2
=
mr
ω
2
×
4
3
π
2
N
(...
v
=
r
w )
= 6.58 N = 6.6 N
PHYSICS FOR YOU

AUGUST ‘17
31
As the string can withstand a maximum tension of
200 N,
∴
T
max
=
mv
.
r
⇒ or
v
max
=
0 25
= 34.64 m s
=
v
max
=
1200
rT
max
m
–1
35.0 m s
–1
13.
Consider a particle moving along a circular path of radius
r
. As shown in figure, suppose the particle moves from
A
to
B
in time D
t
covering distance D
s
along the arc
AB
. Hence the angular displacement of the particle is
∆ θ =
∆
r s
Dividing both sides by D
t
, we get
∆ lim
t
0
∆
∆
θ
1
t r
∆ lim
t
0
∆
∆
t s
∆
∆
θ
Taking the limit D
t
0 on both sides,
=
1
∆
∆
t s
But lim
∆
t
→ 0
∆
∆
t
θ
=
d dt
= is the instantaneous angular velocity
∆
x
=
ds
=
, and lim
0
∆
t dt v
is the instantaneous linear velocity.
∴ ω
1
r v
or
v
= w
r
Linear velocity
=
Angular velocity
×
radius,
In vector notation,
v
For a given angular velocity, the linear velocity of particle is directly proportional to its distance from the centre.
14.
If a number of vectors are represented both in magnitude and direction by the sides of an open polygon taken in the same order, then their resultant is represented both in magnitude and direction by the closing side of the polygon taken in opposite order. Suppose we wish to add four vectors
Draw vector the head of
and the tail of
, ,
OP A
D
. Move vectors
, the tail of
,
touch the head of
touches the head of
B C
parallel to themselves so that the tail of touches
, as shown in figure (b). According to the polygon law, the closing side
OT
of the polygon taken in the reverse
. Thus
We apply triangle law of vector addition to different triangles of the polygon shown in figure (b),
In D
OPQ
,
OQ OP PQ A B
In D
OQS
,
OS OQ QS A B C
In D
OST
, or = + + +
This proves the polygon law of vector addition.
15.
In terms of their rectangular components, the force, momentum and acceleration vectors can be expressed as
F F i F j F k
;
=
x
+ + and
a a i a j a k
=
=
ma
z
The vector form of Newton's second law is
∴
x
+
y
+
z
=
dt
(
x
+
y
+
z
)
=
(
x y
2
)
Equating the components along the three coordinate axes, we get
+ +
F x
=
dp dt x
=
ma x
;
F y
=
dp dt y
=
ma y
;
F z
=
dp dt z
=
ma z
The above equations express Newton’s second law in component form.
The component form of Newton's second law indicates that if the applied force makes some angle with the velocity of the body, it changes the component along the direction of the force.
The component of velocity normal to the force remains unchanged. For example, in the motion of a projectile under the vertical gravitational force, the horizontal component of velocity remains unchanged.
32 PHYSICS FOR YOU

AUGUST ‘17
16.
(a) Net displacement is zero as both initial and final positions are same.
(b) Average velocity
=
Displacement
Time taken
As displacement is zero, average velocity of the cyclist is also zero.
(c) Average speed
=
Distance
Time taken
Now,
OP
=
QO
= 1 km;
PQ
=
1
4
( 2 π
r
) =
1
4
( 2 π ) .
= km
+ + and
t t
\
= 10 min =
10
60 h
Average speed
=
1
6 h
= +
.
/
+
= 3.571 × 6 = 21.43 km h
–1
17.
(a) A revolving wheel has acceleration so a child in it is noninertial observer.
(b) A sportscar moving with constant speed on a straight road has no acceleration so its driver is an inertial observer.
(c) When the plane takes off, it has accelerated motion hence its pilot is a noninertial observer.
(d) During sharp turn, a cyclist has centripetal acceleration. So, the cyclist is in noninertial frame.
(e) A slowing down train has decelerated motion so its guard is a noninertial observer.
18.
Here,
R
max
Now,
R
max
= 100 m
=
u g
2 or
u
=
R
max
×
g
For vertical motion, or
u
= 100
g
initial velocity,
u
= acceleration,
a
= –
g g v
= 0
Now,
v
2 2
= 2
aS
If
H
is the maximum height attained, then
2
–
u
− 100
g
)
2 or 2
gH
= 100
g
or
H g H
=
100
2
g
OR
g
= 50 m
(a) Velocity,
v dr
=
( .
=
i d dt
−
.
t j t i
−
) m s
− 1
+
Acceleration,
a dv
=
d dt
( .
= − .
(b) At time
t
= 2 s,
v
=
j
= −
−
j i
−
−
j j
.
t j
m s
)
− 2
\
v
= 3
2
+ − 8
2
= = .
If q
is the angle which
v
makes with
x
axis, then tan
θ =
v v y
8
\ q
= tan
–1
x
3
(–2.667) ~_ 70° with
x
axis
19.
Some important implications about the third law of motion :
(i) Newton’s third law of motion is applicable irrespective of the nature of the forces.
The forces of action and reaction may be mechanical, gravitational, electric or of any other nature.
(ii) Action and reaction always act on different bodies. If they act on the same body, the resultant force would be zero and there could never be accelerated motion.
(iii) The forces of action and reaction cannot cancel each other. This is because action and reaction, though equal and opposite, always act on different bodies and so cannot balance each other.
(iv) No reaction can occur in the an absence of an action. For example, in a tugofwar, one team can pull the rope only if the other team is pulling the other end of the rope. No force can be exerted if the other end is free. One team exerts the force of action and the other team provides the force of reaction.
20.
Laws of limiting friction are :
(i) The limiting friction depends on the nature of the surfaces in contact.
(ii) The limiting friction acts tangential to the two surfaces in contact and in a direction opposite to the direction of motion of the body.
(iii) The value of limiting friction is independent of the area of the surface in contact so long as the normal reaction remains the same.
(iv)The limiting friction (
f s
)
i.e.,
(
f s
µ
s
=
f
) max
∝
R
or (
f s
)
( )
R
max
= max max
is directly proportional to the normal reaction
R
between the two surfaces.
=
Limiting friction
Normal reaction
s
R
PHYSICS FOR YOU

AUGUST ‘17
33
21.
Here, mass of the helicopter,
M
= 1000 kg
Mass of the crew and the passengers,
m
= 300 kg
Acceleration,
a
= 15 m s
–2
g
= 10 m s
–2
(vertically upwards)
(a) Force on the floor by the crew and the passengers will be equal to their apparent weight. If the helicopter is rising up with an acceleration
a
, then the apparent weight is
m
(
g
+
a
) and hence the required force,
F
=
m
(
g
+
a
) = 300(10 + 15)
= 7500 N (vertically downwards)
(b) Action of the rotor of the helicopter on the
22.
The proportionality constant
s
is called coefficient of static friction. It is defined as the ratio of limiting friction to the normal reaction.
As
f k
<(
f s
) max
or
k
R
<
s
R
\ than the coefficient of static friction.
k
<
s
Thus the coefficient of kinetic friction is less surrounding air,
F
= (
M
+
m
force is given by
)(
F g
+
a
) = (1000 + 300)(10 + 15)
= 32500 N (vertically downwards)
(c) Force on the helicopter due to surrounding air will be equal and opposite to the action of the rotor of the helicopter on the surrounding air
(third law of motion). Therefore, the required
= 32500 N (vertically upwards).
Angle of repose is the minimum angle that an inclined plane makes with the horizontal when a body placed on it just begins to slide down.
As shown in figure, consider a body of mass of repose.
m
placed on an inclined plane. The angle of inclination of the inclined plane is so adjusted that a body placed on it just begins to slide down. Thus is the angle
Various forces acting on the body are :
(i) Weight
mg
of the body acting vertically downwards.
(ii) The limiting friction (
f s
) max
in upward direction along the inclined plane. It balances the component
mg
sin of the weight
mg
along the inclined plane. Thus
(
f s
) max
=
mg
sin
(iii) The normal reaction
R
perpendicular to the inclined plane. It balances the component
mg
cos of the weight
mg
perpendicular to the inclined plane. Thus
R
=
mg
cos
Dividing equation (i) and (ii), we get
f s
R
max
s
= tan
=
mg
sin φ
…(i)
…(ii) or
mg
cos φ
Thus the coefficient of static friction is equal to the tangent of the angle of repose.
23.
(i) The villager was nature loving and he knows the importance of natural resources.
(ii) Given,
A
= 20 N,
B
= 20 N,
R
= 30 N, q = ?
R
=
A
2
+
B
2
+ 2
AB
cos θ
30 =
=
20 20 2 20 20
2
−
2
2 20
−
2
2
= cos θ cos θ 82 49
(iii) Canals, lakes and ponds are other means of irrigation in villages.
24.
(a) (i) For
t
< 0, the position time graph is
AO
which means displacement of the particle is zero,
i.e
., particle is at rest at the origin. Hence force on the particle must be zero.
(ii) For
t
> 4 s, the position time graph
BC
is parallel to time axis. Therefore, the particle remains at a distance of 3 m from the origin,
i.e
., it is at rest.
Hence force on the particle is zero.
(iii) For 0 <
t
< 4 s, the position time graph
OB
has a constant slope. Therefore, velocity of the particle is constant in this interval
i.e
., particle has zero acceleration. Hence force on the particle must be zero.
(b) Impulse at
t
= 0
Impulse = change in linear momentum.
Before
t
= 0, particle is at rest,
i.e
.,
u
= 0.
After
t
= 0, particle has a constant velocity.
\
Impulse =
m
(
v
–
u
) = 4(0.75 – 0) = 3 kg m s
–1
Impulse at
t
= 4 s
Before
t
= 4 s, particle has a constant velocity
u
= 0.75 m s
After
t
= 4 s, particle is at rest,
i.e., v
= 0
\
Impulse =
m
(
v
–
u
) = 4 (0 – 0.75) = – 3 kg m s
–1
34 PHYSICS FOR YOU

AUGUST ‘17
OR
(a) The statement is false, as several scalar quantities are not conserved in a process.
For example energy being a scalar quantity is not conserved during inelastic collisions.
(b) The statement is false, because there are some scalar quantities which can be negative in a process.
For example, temperature being scalar quantity can be negative (–30 °C,–4 °C), charge being scalar can also be negative.
(c) The statement is false, there are large number of scalar quantities which may not be dimensionless.
For example, mass, density, charge etc. being scalar quantities have dimensions.
(d) The statement is false as there are some scalar quantities which vary from one point to another in space.
For example, temperature, gravitational potential, density of a fluid or anisotropic medium, charge density vary from point to point.
(e) The statement is true, orientation of axes does not change the value of a scalar quantity.
For example, mass is independent of the coordinate axes.
25.
Practical applications based on the law of conservation of linear momentum :
(i) Recoil of a gun : Let
M
be the mass of the gun and
m
be the mass of the bullet. Before firing, both the gun and the bullet are at rest. After firing, the bullet moves with velocity
v
and the gun moves with velocity
. As no external force acts on the system, so according to the principle of conservation of momentum, total momentum before firing
= total momentum after firing or 0 or
=
MV
= −
+
= −
m
M mv v
or
The negative sign shows that
v
opposite directions,
i.e
., the gun gives a kick in the backward direction or the gun recoils with velocity
. Further, as
M
>>
m
,
V
<<
v i.e.
, the recoil velocity of the gun is much smaller than the forward velocity of the bullet.
(ii) When a man jumps out of the boat to the shore, the boat slightly moves away from the shore :
Initially, the total momentum of the boat and the man is zero. As the man jumps from the boat to the shore, he gains a momentum in the forward direction. To conserve the momentum, the boat also gains an equal momentum in the opposite direction. So the boat slightly moves backwards.
(iii) An astronaut in open space, who wants to return to the spaceship, throws some object in a direction opposite to the direction of motion of the spaceship : By doing so, he gains a momentum equal and opposite to that of the thrown object and so he moves towards the spaceship.
(iv) Rocket and jet planes work on the principle of conservation of momentum : Initially, both the rocket and its fuel at rest. Their total momentum is zero. For rocket propulsion, the fuel is exploded. The burnt gases are allowed to escape through a nozzle with a very high downward velocity. The gases carry a large momentum in the downward direction. To conserve momentum, the rocket also acquires an equal momentum in the upward direction and hence starts moving upwards.
OR
(a) Given figure (a) shows a particle
O
under the equilibrium of three concurrent force
2 and between
F
3
F
1
F
3
.
b
Let
F
2
be angle between
between
F
3
F
As shown in figure (b), the forces can be represented by the sides of
D
ABC
F
, taken in the same order.
Applying law of sines of D
ABC
, we get sin(
F
1
π α
=
F
2
−
=
F
3
) or
F
sin
1
α
=
F
sin
2
β
=
F
sin
3
γ
PHYSICS FOR YOU

AUGUST ‘17
35
This is Lami’s theorem which states that if three forces acting on a particle keep it in equilibrium, then each force is proportional to the sine of the angle between the other two forces.
(b) As
P
is in equilibrium, so, net force along horizontal direction = 0
⇒
F
+ sin45° = 2sin45°
⇒
F
= sin45° = 1
2
N
Horizontal range,
So,
R h m
R
= or tan
=
θ
u
0
2 sin
g
tan
θ
2 θ
0
0
=
4
4
h
R m
= or
u g
2
θ
0
2
=
OR
= 0.1,
R
= 100 m,
v
= 50 m s
–1
θ tan
−
1
θ
0
4
h
R m
®
Net force along vertical direction = 0
F
= cos45° + 2cos45° = 3cos45°=
26.
(a) Let
v ox
and
v oy
3
2
N
be the initial component velocities of the projectile at
O
along
OX
direction and
OY
direction respectively, where
OX
is horizontal and the
OY
is vertical. Let the projectile go from
O
to
P
in time
t
and
v x
,
v y
be the component velocities of projectile at
P
along horizontal and vertical directions respectively.
Then,
v y
=
v oy
–
gt
and
v x
=
v ox
If q
is the angle which the resultant velocity makes with horizontal direction, then
v
tan or
θ =
θ =
v v y x
tan
=
− 1
v oy
−
gt
v ox v oy
−
gt v ox
(b) In angular projection, maximum vertical height,
h m
=
u
2 sin
2
g
2
θ
0
(a)
t
= time taken to cover the distance on arc
DEF
.
(
v m
= Maximum possible speed on track
DEF
)
Centripetal force is provided by friction force.
mv
R
2
m
= µ
mg
⇒
v m
= µ
gR
=
= 10 m s
–1
Length of arc
DEF
=
R
( /2)= 50 m (... q =
l
/
r
)
∴
t
1
=
50
10
π
=
5
π s
= s
(b)
t
= time taken by the car to cover the distance on circular track
ABC.
Again, centripetal force = friction force
(
v
′
⇒
v mv
2
R
′
′ =
= µ
mg m
= maximum possible speed on track
ABC
)
2 µ
gR
= 200
= 14.14 m s
–1
Length of arc =
ABC R
2
π
= 3
t
2
=
300 π
=
(c)
CD
=
FA
=
R
= 100 m,
v
= 50 m s
–1
R
= 300 m
t
= time taken to cover the path
CD
and
FA t
3
=
2
v
R
2 100
=
4 s
\ Total time taken,
t
=
t
+
t
+
t
50
= 15.7 + 66.6 + 4 = 86.3 s
36 PHYSICS FOR YOU

AUGUST ‘17
Class XI
T
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer.
Self check table given at the end will help you to check your readiness.
Work, Energy and Power
Total Marks : 120
NEET / AIIMS
Only One Option Correct Type
1.
Two spheres
A
and
B
of masses
m
1
and
m
2 respectively collide.
A
is at rest initially and
B
is moving with velocity
v
along
x
axis. After collision
B
has a velocity
v
2
in a direction perpendicular to the original direction. The mass
A
moves after collision in the direction
(a) same as that of
B
(b) opposite to that of
B
(c)
(d)
θ =
θ = tan tan
−
1
− 1
1
2
−
1
2
2.
A girl in a swing is 2.5 m above ground at the maximum height and at 1.5 m above the ground at the lowest point. Her maximum velocity in the swing is (
g
= 10 m s
(a) 5 2 m s
−
1
–2
)
(b) 2 5 m s
−
1
(c) 2 3 m s
− 1
(d) 3 2 m s
− 1
3.
An open watertight railway wagon of mass
5 × 10
3
kg is moving with an initial velocity of
1.2 m s
–1
without friction on a railway track. Rain falls vertically downwards into the wagon. What change will occur in the kinetic energy of wagon, when it has collected 10
3
kg of water?
(a) 1200 J (b) 300 J (c) 600 J (d) 900 J
4.
A body of mass 4 kg is moving with momentum of
8 kg m s
–1
. A force of 0.2 N acts on it in the direction
5.
(a) 10 (b) 8.5 (c) 4.5
Time Taken : 60 min
of motion of the body for 10 s. The increase in kinetic energy in joules is
(d) 4
One man takes 1 min to raise a box to a height of
1 m and another man takes 1 energy of the two men is
min to do so. The
(a) different (b) same
(c) energy of the first is more
(d) energy of the second is more
6.
When a long spring is stretched by 2 cm, its potential energy is
U
. If the spring is stretched by 10 cm, the potential energy in it will be
U
(a) 10
U
(b) 25
U
(c)
5
(d) 5
U
7.
The displacement of a body of mass 2 kg varies with time
t
as
S
=
t
2
+ 2
t
, where
S
is in m and
t
is in s.
The work done by all the forces acting on the body during the time interval
t
= 2 s to
t
= 4 s is
(a) 36 J (b) 64 J (c) 100 J (d) 120 J
8.
The work done by external agent in stretching a
9.
spring of force constant
k
from length
l
(a)
(c)
k k
(
(
l l
2
2
2
–
l
–
l
1
2
1
)
) (b)
(d)
1
2
1
2
k k
(
(
l l
2
2
2
–
l
+
l
1
2
1
)
1
)
to
l
2 is attached to the ceiling and an object attached to the other end is slowly lowered to its equilibrium position. If
S
be gain in spring energy and
G
be loss in gravitational potential energy in the process, then
(a)
S
=
G
(c)
G
= 2
S
(b)
S
= 2
G
(d) None of these
PHYSICS FOR YOU

AUGUST ‘17
37
10.
Concrete blocks of masses
m
A
and
m
on two identical vertical springs.
m
B
A
are balanced
= 2
m
B
. The gravitational potential energy of each system is zero at the equilibrium position of the springs. Which statement is true for the total mechanical energy of the systems when the blocks are balanced on the springs?
(a)
E
A
(c)
E
A
=
E
B
= 4
E
B
(b)
E
A
= 2
E
B
(d)
E
A
= – 2
E
B
–1
11.
A pendulum bob has a speed of 3 m s at its lowest position. The pendulum is 0.5 m long. The speed of the bob, when the length makes an angle of 60° to the verticle, will be
(a) 3 m s
–1
(b) 1/3 m s
–1
(c) 1/2 m s
–1
(d) 2 m s
–1
12.
A bead slides without friction around a loop the loop as shown in the figure. The bead is released from rest at a height
h
= 3.50
R
. How large is the normal force on the bead at point (
A
) if its mass is 50 g?
A
R
(a) 0.10 N downward (b) 0.10 N upward
(c) 1.0 N downward (d) 1.0 N upward
Assertion & Reason Type
Directions : In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as :
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If both assertion and reason are false.
13. Assertion :
A quick collision between two bodies is more violent than a slow collision, even when initial and final velocities are identical.
Reason :
The rate of change of momentum which determines the force is greater in the first case.
14. Assertion :
Potential energy of a stretched or compressed spring is proportional to square of extension or compression.
Reason :
Graph between potential energy of a spring versus the extension or compression of the spring is a straight line.
15. Assertion :
The mass equivalent of 1000 kW h
c
energy is 40 g.
Reason :
= 3 × 10
This follows from
8 m s
–1
.
E = mc
JEE MAIN / JEE ADVANCED
Only One Option Correct Type
2
,
where
16.
A block of mass
m
= 0.1 kg is released from a height of 4 m on a curved smooth surface. On the horizontal surface, path
AB
is smooth and path
BC
offers coefficient of friction = 0.1. If the impact of block with the vertical wall at
C
be perfectly elastic, the total distance covered by the block on the horizontal surface before coming to rest will be
(take
g
= 10 m s
–2
)
A B
C
(a) 29 m
(c) 59 m
1 m 2 m
(b) 49 m
(d) 109 m
17.
A locomotive of mass
m
starts moving so that its velocity varies as
v
=
s
2/3
, where is a constant and
s
is the distance traversed. The total work done by all the forces acting on the locomotive during the first
t
s after the starts of motion is
α
4 2
(b)
m t
162
(c)
81
(d)
2
18.
A uniform chain has a mass
m
and length
L
. It is placed on a frictionless table with length
l
0
hanging over the edge. The chain begins to slide down. The speed
v
with which the chain slides away from the edge is given by
(a)
(c)
v v
=
=
g
L gl
L
0
(
0
)
(b)
(d)
v v
=
= 2
gl
L
0
(
(
0
0
)
)
19.
A block of mass
m
is pulled by a constant power
P
placed on a rough horizontal plane. The friction coefficient between the block and surface varies with its speed
v
as
µ =
1
. The acceleration of
1 +
v
the block when its speed is 3 m s
–1
will be
38 PHYSICS FOR YOU

AUGUST ‘17
(a)
(c)
3
3
P m
P m
−
g
2
(b)
(d)
3
P m
+
g
2
g
2
More than One Options Correct Type
20.
One of the forces acting on a particle is conservative then which of the following statement(s) are true about this conservative force?
(a) Its work is zero when the particle moves exactly once around any closed path.
(b) Its work equals the change in the kinetic energy of the particle.
(c) Then that particular force must be constant.
(d) Its work depends on the end points of the motion, not on the path between.
21.
A man standing on the edge of the terrace of a high rise building throws a stone, vertically up with a speed of 20 m s of 20 m s
–1
–1
. After 2 s, an identical stone is thrown vertically downwards with the same speed
. Then
(a) the relative velocity between the two stones remains constant till one hits the ground
(b) both will have the same kinetic energy, when they hit the ground
(c) the time interval between their hitting the ground is 2 s
(d) if the collisions on the ground are perfectly elastic, both will rise to the same height above the ground.
22.
A box of mass
m
is released from rest at position 1 on the frictionless curved track as shown in the figure. It slides a distance
d
along the track in time
t
to reach position 2, dropping a vertical height
h
.
Let
v
and
a
be the instantaneous speed and instantaneous acceleration respectively, of the box at position 2. Which of the following equations is not valid for this situation?
(a)
h = vt
(b)
h
= (1/2)
gt
2
(c)
d
= (1/2)
at
2
(d)
mgh
= (1/2)
mv
2
23.
One end of a light spring of spring constant
k
is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. For a displacement
kx
2
. The possible cases are
(a) the spring was initially compressed by a distance
x
and was finally in its natural length.
(b) it was initially stretched by a distance finally was in its natural length.
x
and
(c) it was initially in its natural length and finally in a compressed position.
(d) it was initially in its natural length and finally in a stretched position.
Integer Answer Type
24.
Two identical beads of
m
= 100 g are connected by an inextensible massless string that can slide along the two arms
AC
and
BC
of a rigid smooth wireframe in a vertical plane. If the system is released from rest, the kinetic energy of the first particle when they have moved by a distance of 0.1 m is
16
x
× 10
–3
J. Find the value of
x
. (
g
= 10 m s
–2
)
25.
A block of mass
m
is released from rest at point
A
. The compression in spring (force constant
k
) when the speed of block is maximum is found to be
nmg
cos
θ
4
k
. What should be the value of
n
?
m
A
26.
An insects jumps from ball
A
onto ball
B
, which are suspended from inextensible light strings each of length
L
= 8 cm. The
L L
mass of each ball and insect is same.
A B
What should be the minimum relative velocity
(in m s
–1
) of jump of insect with respect to ball
A
, if both the balls manage to complete the full circle?
PHYSICS FOR YOU

AUGUST ‘17
39
Comprehension Type
We generally ignore the kinetic energy of the moving coil of a spring but consider a spring of mass
m
, equilibrium length
L
and spring constant
k
. Consider a spring, as described above, that has one end fixed and the other end moving with speed
v
. Assume that the speed of points along the length of the spring varies linearly with distance
L
from the fixed end. Assume also that the mass
m
of the spring is distributed uniformly along the length of the spring. Assume further that the force applied by the spring is spring constant times its deformation.
27.
The speed of small length (
dx
) at a distance
x
from fixed end is
(a)
x
L v
(b)
v
(c)
L x v
(d)
xv
28.
Kinetic energy of the spring
mv
2
mv
2
(C) The initial tangential acceleration
(R)
l
sin
l
R
(D) The radial acceleration
(S)
l
sin
−
sin
l
R
θ
+
+ sin
l
R
θ
A
(a) Q
(b) R
(c) Q
(d) P
B
R
S
P
Q
C
S
P
S
S
D
P
Q
R
R
30.
As shown in figure,
m
1
k
= 100 N m
–1
, = 0.2
= 8 kg,
m
2
= 16 kg,
1 2
(c)
mv
2
(d)
1
4
mv
2
Matrix Match Type
29.
A chain of length
l
and mass
m
lies on the surface of a smooth sphere of radius
R
>
l
with one end tied to the top of the sphere.
Column I
(A) Gravitational potential energy
(P) with respect to centre of the sphere
Column II
Rg l
1 − cos
R l
(B) The chain is released and slides
(Q) slide by q
2
l
Rg
down, its kinetic energy when it has sin
sin
θ −
l
R
sin
θ
+
+
l
R
Column I
(A) The minimum value of
F
Column II
(P) 5.12
(in N) in order to shift the block of mass
m
1
(B) Negative of work done by (Q) 40 friction (in J) on block
B
till this moment
(C) Work done by
F
till this moment
2
if it is applied on
A
(R) Zero
(D) The minimum value of
F
in (S) 32 order to shift the block of mass
m
A
(a) Q
(b) P
B
T
T
C
R, S
S
D
P
Q
(T) 6.4
(c) Q P
(d) S, R Q
T
P
S
T
Keys are published in this issue. Search now
!
Check your score! If your score is
> 90%
EXCELLENT WORK !
You are well prepared to take the challenge of final exam.
No. of questions attempted ……
No. of questions correct ……
Marks scored in percentage ……
9075% GOOD WORK !
You can score good in the final exam.
7460%
SATISFACTORY ! You need to score more next time .
< 60%
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40 PHYSICS FOR YOU

AUGUST ‘17
Class
XII
NEET JEE
ESSENTIALS
Maximize your chance of success, and high rank in NEET, JEE (Main and Advanced) by reading this column.
This specially designed column is updated year after year by a panel of highly qualified teaching experts welltuned to the requirements of these EntranceTests.
U n i t
CURRENT ELECTRICITY
2
ELECTRIC CURRENT
•
•
Electric Current
•
The rate of flow of charge through a crosssection of
• some region of a metallic wire (or an electrolyte) is called the current through that region.
Current at any instant is given by
I
=
dq
/
dt
. If a charge
q
flows through the circuit for time
t
, then
I
=
q
/
t
.
SI unit of current is ampere (A) or (C s
–1
).
In metallic conductors, the current is due to the motion of free electrons whereas in electrolytes and ionized gases, both positive and negative ions are responsible for current flow.
•
•
Drift Velocity
•
When an electric field is applied across a conductor, the free electrons are accelerated opposite to the direction of the field and therefore they have a net drift in that direction. Due to frequent collisions of electrons with the atoms, their average velocity is very small. This average velocity with which the electrons move in a conductor under an applied potential difference is called the drift velocity.
The acceleration of the electron is
a = e E/m
The drift velocity must be
•
v d
=
eE m
τ
E
= applied electric field
m
= mass of electron
= relaxation time
Current :
I
=
neAv d
{
n
= Number density of electrons}
•
Mobility : The drift velocity per unit electric field.
=
v d
/
E
=
e
/
m
Current Density
•
Current density at a point is defined as the amount of current following per unit area around that point, provided the area is normal to the direction of current.
J
=
I
/
A
, where
I
is current and
A
is area.
J
Current is perpendicular
to area : The average current density is given by
av
=
D
I
/
D
A
.
Current not perpendicular
to area : The average current density is given by
J av
=
∆
A
∆
I
cos θ
P
P
A
A
I
I
J av
A
OHM’S LAW AND RESISTIVITY
Ohm’s Law
•
The current flowing through a conductor is proportional to the potential difference across its ends, only when physical conditions (such as temperature) remain unchanged,
i
.
e
.,
V
∝
I
or
V
=
RI
.
The constant of proportionality
R
is called the resistance of the conductor.
PHYSICS FOR YOU

AUGUST ‘17
41
•
•
Resistance : The resistance of a conductor is a measure of the opposition offered by the conductor to the flow of current.
The resistance of a conductor is directly proportional to its length
l
and inversely proportional to the area of crosssection
A
,
i.e.
,
R
= ρ
l
A
where the constant depends on the nature of the material, is called the resistivity.
In vector form
= =
V
RA
=
( ρ /
El
ρ
σ
E
Sol. : As D
R
=
R
D
T
and in this case, D
R
=
R
\ D
T
= 1/ = 267 °C
T
=
T
+ D
T
= 0 °C + 267 °C = 267 °C
ELECTRICAL RESISTANCES
Combination of Resistance
each reasistance is same.
• where = 1
J
= σ
is called conductivity.
Resistivity : The resistivity of material of a conductor is given by
m
ρ =
ne
2
τ where
n
is number of free electrons per unit volume and is the relaxation time of the free electron.
Temperature Dependence of Resistivity
•
If
0
and
T
are the values of resistivity at 0° C and
T
° C respectively then over a temperature range that is not too large, we have approximately,
R
T
T
=
=
R
0
0
(1 +
T
)
Consequently, for resistance we have, approximately
(1 +
T
).
Colour Code of Resistors
Colour Figure
black 0
=
1 2
R S S
× 10
m
±
e
brown red orange yellow green
3
4
1
2
5 first two significant figures ( )
1 2
S S
tolerence of resistance ( )
e
blue violet
6
7
Colour
gold
Tolerence
5% grey 8 silver 10% white 9 no colour 20%
Illustration 1. The resistance of a silver wire at 0 °C is
1.25 . Upto what temperature it must be heated so that its resistance is doubled? [ silver
= 0.00375 °C
–1
]
Current and Potential Divider
•
Voltage divider : In a series circuit, current through each resistor is the same.
V
=
V
I
1
+
V
V
2
=
IR
1
+
IR
2
=
Voltage is divided as
V
1
=
R R
2
1
VR
+
1
R R
2
V
V
1
V
2
I
R
1
R
2
•
V
2
=
VR
1
+
2
2
Current divider : In a parallel circuit, potential drop across each resistance is same.
I
I
1
I
2
R
1
I
=
I
1
+
I
2
.. (i)
R
2
I
1
R
1
=
I
2
R
2
... (ii)
On solving eqns. (i) and (ii), we get
I
1
=
1
R
+
2
2
2
=
1
R
+
1
R R
2
I
V
V
42 PHYSICS FOR YOU

AUGUST ‘17
Note : Current through any resistance is total
×
Resistance of opposite branch
Total resistance
Resistances in special cases
Geometry of object
Object description Resistance of object
A solid cylinder
l
R
= ρ
π
b l
2
A hollow cylinder
l
R
=
π
(
ρ
l
2 2
)
Solid cylinder with different densities and
2
1
l
R
R
1
2
=
=
π
ρ
π
(
a
Net
R
=
2
2
l
ρ
1
−
l
2
R R
)
2
Rectangular slab
R
R
AB
XY
= ρ
b a
2
=
=
ρ
ρ
b a b a
A cubical frame with equal resistance
r
on edges
R
R
12
13
12
4
r r
R
17
6
r
Illustration 2. Let there be
n
resistors
R
R
max
= max {
R
1
...
R n
} and
R
min
= min {
R
1
1
...
R
...
R n n
with
}. Show that when they are connected in parallel, the resultant resistance
R
P
<
R
m i n and when they are connected in series, the resultant resistance
P
is
1
R
max
. Interpret the result physically.
Sol. : When
n
resistors are connected in parallel, the effective resistance
R
S
>
R
R
1
P
R
min
R
P
=
=
R
1
1
+
R
min
R
1
R
1
2
+
...
R
min
R
2
R n
R
min
R n
Since
∴
R
min
R n
R
min
R
P
≤ 1; for
> 1or
R
m i n
n
= 1, 2, 3, ...,
>
R
P n
Due to one resistor of resistance
R
m i n one route for the current in circuit.
, there is only
When
n
resistors are in parallel, there are additional
(
n
– 1) routes for the current in circuit.
When
n
resistors are in series, then
R
S
=
R
1
+
R n
\
R
S
2
+ .... +
R
>
R
m a x
Due to one resistor of resistance
R
series of effective resistance
R
S
.
m a x
, the current in circuit is more than the current due to
n
resistors in
CELLS, EMF AND INTERNAL RESISTANCES
EMF (ElectroMotive Force)
•
The potential difference across the terminals of a cell when it is not producing any current is called emf of the cell.
•
•
Emf depends on
f nature of electrolyte f metal of electrodes
Emf does not depend on
• f f f area of plates distance between the electrodes quantity of eletrolyte size of cell f
Terminal voltage : When current drawn through the cell or current is supplied to cell, the potential difference across its terminals is its terminal voltage.
Situation Potential difference (
V
) across the terminals of cell
A
A r r
I
I
B
B
Battery is open circuited or
V
A
– +
Ir
=
V
B
V
V
A
AB
V
A
–
V
B
or
V
A
–
V
B
= –
Ir
= –
Ir
or
V
– –
Ir
=
V
B
AB
<
V
AB
= +
Ir
= +
Ir
or
V
AB
>
V
AB
= as
I
= 0
r
A B
Battery is short circuited
I
=
r
ε
or
ε =
Ir
\
–
Ir
= 0 or
V
AB
= 0
A
ε
r
B
PHYSICS FOR YOU

AUGUST ‘17
43
Combination of Cells (Battery)
Illustration 3. 20 cells, each of 2 V emf and 0.5 internal resistance are given. How will you arrange them in a mixed grouping so as to send maximum current through a 2.5 external resistance? Calculate the value of the current.
Sol. : Since the external and the internal resistances are of the same order, the cells are to be placed in mixed grouping.
With usual notation,
N
= 20, = 2 V,
r
= 0.5 ,
R
= 2.5
Let there be
m
rows of
n
cells in parallel. Clearly,
N
=
nm
= 20 ... (i)
For maximum current,
R nr
⇒
n m n
= 5
m
... (ii)
From eqns. (i) and (ii) we get
m
× 5
m
= 20 or
m
2
= 4 or
m
= 2
From eqn. (ii),
n
= 5 × 2 or
n
= 10
I
=
mn
ε
=
V
(
×
.
+ ×
. )
Ω
=
40
10
V
Ω
= 4 A
KIRCHHOFF’S LAWS FOR ELECTRICAL NETWORK
Junction Rule
•
The algebraic sum of currents meeting at any junction in a circuit is zero,
i
.
e
.,
∑
I
= 0.
Conventionally the incoming currents at the point are taken as positive while those outgoing are taken as negative.
In figure for junction
O,
∑
I
=
I
1
+
I
2
–
I
3
+
I
4
–
I
5
= 0
I
1
I
2
O
This law is based on conservation of charge.
I
5
I
3
I
4
Loop Rule
•
According to this law in any closed part of an electrical circuit, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistance and current flowing through them.
⇒
∑
ε =
∑
IR
This law follows from the law of conservation of energy.
Wheatstone’s Bridge
•
In a balanced condition even though current flows in the rest of the circuit, galvanometer will not show any deflection (
i.e.
I g
= 0).
Also,
R
R
1
2
=
R
R
3
4
44 PHYSICS FOR YOU

AUGUST ‘17
Illustration 4 : In the given circuit,
1
R
1
= 3
= 2
R
2
= 2
4
= 6
3
= 6 V
R
3
= 2
R
2
= 4
C
= 5 F
C
2
R
1
R
2
1
R
3
R
4
3
Find the current in
R
3
and the energy stored in the capacitor.
Sol. : In steady state condition, no current flows through the capacitor
C
(5 F). Obviously no current flows through
R
1
(6 ).
6 V
6
B
C
A
1
5 F
I
2
I
2
2 4
I
1
1.5 A
D F
2 V
E
R
3
I
I
1
I
1 or
U
G
= 1
Illustration 5 : Five equal resistances, each of resistance
R
, are connected as shown in figure below. A battery of
V
volt is connected between
A
and
B
.
Find the current flowing in
FC
.
D
H
3 V
3
Current through
R
3
=
=
3
2
= 1.5 A
1
/
R
3
∴
I
=
6
4
Assume this
I
(1.5 A) divides as
I
1
and
I
2
as shown.
Apply Kirchhoff’s law to closed loop
DGHFED
,
3 – 3
I
or
I
1
– (4 1.5) – 2
I
1
= −
1
1
+ 2 = 0 or 5
I
1
= – 1
5
The negative sign of
I
1
shows that its direction is reverse of the direction shown in the figure.
V
A
– 2
I
or
V
A
1
+ 2 –
V
D
–
V
D
–
V
A
= 2
I
= 0 along passage
ABED
1
– 2 = 2 (– 0.2) – 2 = – 2.4 V or
V
D
= 2.4 V
\
Potential difference across capacitor = 2.4 V
∴ Energy stored in capacitor = 1 or
U
= 1 × (5 × 10 ) × (2.4)
2
CV
2
Sol. : The equivalent circuit diagram is as shown in figure.
I t i s a b a l a n c e d W h e a t s t o n e bridge, hence no current flows i n a r m
CD
. T h e r e f o r e r e s i s t a n c e o f a r m
CD
b e c o m e s i n e f f e c t i v e .
R e s i s t a nc e of up r a r m
FCE
=
R + R
= 2
R
Resistance of lower arm
FDE
=
R + R
= 2
R
The equivalent resistance between
A
a nd
B
i s
R
eq
=
2
2
R
R
+
2
2
R
R
=
R
V
C ur r e n t i n t he c i r c ui t i s
I
=
R
∴ Current in
FC
2 2
V
R
HEATING EFFECT OF ELECTRIC CURRENT
Joule’s Law of Heating
•
The amount of heat produced
H
in a conductor of resistance
R
, carrying current
I
for time
t
is
H
=
I Rt
(in joule) where
J
is Joule’s or
=
2
(in calorie) mechanical equivalent of heat (= 4.2 J cal
–1
).
Electric Power
•
Electric power,
P
= electric work done time taken
= =
�
=
R
�
•
•
•
•
The SI unit of power is watt (W).
The practical unit of power is kilowatt (kW) and horse power (hp).
1 kilowatt = 1000 watt.
Power loss in transmission cable,
=
2
c
=
P R
V c
2
c
P
R c
=
=
power delivered
Resistance of cable
Illustration 6 : A copper wire having crosssectional area of 0.5 mm
2
and a length of 0.1 m is initially at
25 °C and is thermally insulated from the surrounding.
If a current of 10 A is set up in this wire, (i) find the time in which the wire will start melting. The change of resistance with the temperature of the wire may be neglected. (ii) What will this time be, if the length of the wire is doubled ?
Melting point of copper = 1075 °C.
Specific resistance of copper = 1.6 10
–8
m
PHYSICS FOR YOU

AUGUST ‘17
45
PROJECTILE MOTION
CLASS XI
y
O t
= 0
u
Horizontal
Projectile
Motion
r
= 0
a x
=
a g y v y x x
( )
P x, y v x v
P
ROJECTILE
Motion
A body which is in flight through the atmosphere under the effect of gravity alone and is not being propelled by any fuel is called projectile and its motion is called projectile motion.
Projectile
Motion on an Inclined
Plane
y u y u g
sin
u x g g
cos
R
A t
= 0
v x
B v
T v y a = –g x
sin
a = –g y
cos
x h
Oblique
Projectile
Motion
A
R
B
·
Time of Flight
T
=
2 sin
u
cos
g
·
Maximum Height
H
2
=
u
2 cos
g
sin
2
·
Equation of Trajectory
y
=
1
2
gx
2
u
2
· Equation of Trajectory
y
=
x
tan –
1
2
2
u
2
gx
cos
2 is represents the parabolic path.
·
Horizontal Range
R
=
2
u
2
g
sin cos ( + ) cos
2
·
Time of Descent
T
=
2
h g
·
Horizontal Range
=
R u
2
h g y v y v a = x
0
v y
= 0
v = v x a y
= –
g v x
Maximum range occurs when
=
4
+
2
·
Instantaneous Velocity
v
=
u
tan =
2
+ 2
gy
=
v y v x
= tan
–1
u
2
+
g t gt u x u v x
( )
x, y y u = u x
cos
R
H v y v x
Maximum range along the incline when projectile is thrown upwards
2
R
max
=
u g
(1 + sin )
t
P r o j e c t i l e p a s s i n g through two different points on same height at time and
t
1
t
2
y
=
1 2
gt t
2
2
=
u
sin
g
1 + 1 –
2
gy u
sin
2
· Maximum Height
H
=
u
2 sin
2
2
g y t = t
1
t = t
2
u
2
t
1
=
u
sin
g
1 – 1 –
2
gy u
sin
O y y
·
Time of Flight
T
=
2 sin
u g
·
Horizontal Range
R
=
u
2 sin 2
g
Maximum range along incline when the projectile thrown downwards
2
R
max
=
u g
(1 – sin )
For complimentary angles and
(90 – ) range remains unchanged
Ratio of time of ights for projectiles at complimentary angles and 90 –
x
T
T
= tan
Range is times
R n
the maximum height
H
–1
= ; = tan [4/ ]
R nH n
Relation between horizontal range and maximum height
= 4 cot
R H
–1
If = then = tan (4)
R H
or = 76°
–1
If = 4 then = tan (1)
R H
or = 45°
ELECTRIC FLUX AND
GAUSS'S LAW
ELECTRIC FLUX
Electric flux is a measure of flow of electric field through a surface. It is equal to the dot product of an area vector and electric field f
Flux of electric field
E
through any area :
A
= cos or =
EA E A
f
For variable electric field or curved area
=
E dA dA
A
f
For a closed surface outward flux is taken to be positive while inward flux is taken to be negative.
E dA
cos
CLASS XII
A
E
Positive flux
E
90°
A
E
A
Negative flux
E
GAUSS’S LAW
s
E dA
1
0
Q enc
Problem Solving Strateg ies
Select a sy mmetric ga ussian su rface as p er the cha rge distrib ution.
surface.
Calculate
total elec tric charge i nside the gaussian
For unifo rm charge dens ity, simply
multiply i t by length, area and volume o f surface.
For non u niform ch arge densi ty integra te it over the region
enclosed the surfac e.
per the giv en uniform gaussians surface as charge dis tribution.
electric fie ld on the
Calculate
Flux across some definite symmetric closed surfaces
Field of a line charge
f
Hemispherical body
In uniform electric field
circular curved
;
circular
2
R E
E
In non uniform electric field
circular curved
;
curved
2
2
R E
f
Gaussian Cube
Charge kept at corner
cube
Charge kept at centre of face
8
Q
0
cube face
Q
24
0
2
Q
0
5
faces total total
Q
0
Q
0
E = E dA net
2
1
0
r
E
= 0
+ + + + + + + + + + + + + + + + + + + +
r dA
Field due to a long uniformly charged solid cylinder
Field of an infinite
Field of a uniformly charged solid sphere and charged conducting sphere plane sheet of charge
R
E in r
2
0
E s
R
2
0
E out
R
2
2
0
r
E
2
0
+
+
++
+
r
P
+
+
+
++ + ++
R
( )
E R
E
4
1
0
Q
2
R
4
1
0
Qr
3
R
E
E
4
1
0
r
Q
2
2
R
E
= Volume charge density
O R
E = E
EA
E l
(
E E dA
)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ +
E
= 0
+
EA
E = E
Result in field due to a sheet depends only on total charge of the sheet and independent of distribution of charge.
( )
E R
4
1
0
R q
2
E
( )/4
E R
( )/9
E R
0
E
= 0
R
2
R
3
R
E=
4
1
0
r q
2
r r
R
Density of copper = 9 10
3
kg m
–3
Specific heat of copper = 9 10
–2
cal kg
–8
°C
–1
Sol. : Mass of wire (
m
) = area length density
= 0.5 10
–6
0.1 9 10
3
= 45 10
–5
kg
Rise in temperature (
D
T
) = 1075 °C – 25 °C = 1050 °C
(i) If a current
I
flows in a wire of resistance
R
for a time
t
, the heat energy generated is
2
Q I Rt
cal
If
s
is the specific heat of copper, the wire will melt in time
t
given by
I Rt ms T
.....(i)
The resistance
R
is given by
R
= ρ
A
L
where is the specific resistance. Using this in the equation (i), we have
A
= or
t
=
t
× ×
− 5 − 2
) ×
10
2
×
( .
−
8
)
×
0 1
× ×
− 6
)
= 0.59 s
(ii) When the length of the wire is doubled, its resistance
R
as well as its mass
m
are doubled. From equation (i)
t
=
As
t
is proportional to
m
/
R
and as both
m
and
R
are doubled, it follows that
t
will remain the same,
i
.
e
., the wire will start melting in the same time.
CHARGING AND DISCHARGING OF CAPACITORS
Basic parameters
Charge on capacitor at time
t
Charging a capacitor
q
=
CV
(1–
e
–
t
/
RC
)
q
max
=
q
0
=
CV
C
R
V
Current in the circuit at time
Total voltage
t
I
=
V
R e
−
t RC
I
max
=
I
0
=
V/R
V
C
+
V
R
=
V
Voltage and current dependency over charge
As
q
increases
⇒
V
C
increases
\
V
R
decreases and
I
decreases
Illustration 7 : An uncharged capacitor is connected to a 15 V battery through a resistance of 10 . It is found that in a time of 2 s, potential difference across the capacitor becomes 5 V. Find the capacitance of the capacitor. Take ln (1.5) = 0.4.
Sol. : We know that charge on the capacitor at any time is given by
q
=
q
0
(1 –
e
–
t
/
) where
q
0
=
C
= 15
C
.
Here charge
q
at any time is given by
q
=
VC
where
Discharging of capacitor
q
=
q
0
e
–t/RC q
0
=
CV
= initial charge
C
R
I
I
= −
I e
V
0
0
−
/
R
initial current at
t
= 0
V
C
+
V
R
= 0; 
V
C
 = 
V
R

As
q
decreases
⇒
V
C
decreases
\
V
R
decreases and
I
decreases
V
is potential difference across capacitor at that time.
Here
V
= 5 V, so
q
= 5
C
. Putting the values, we get
5
C
= 15
C
(1 –
e
–
t
/
) or
e
–
t
/
= 2/3 or or
τ
t
C
=
R
3
2
t
/ or
=
t
RC
10
− 6
3
2
/
= µ F
48 PHYSICS FOR YOU

AUGUST ‘17
ELECTRICAL DEVICES
Devices and their principle Application
Meter Bridge
Principle : It is based on
Wheatstone bridge.
(i) To measure unknown resistance,
X R
( 100 −
l
)
l
.
(ii) To compare two unknown resistances
R
R
1
2
=
l l
2
1
(
( 100
100
−
−
l l
1
2
)
)
(iii) To measure the unknown temperature
θ =
R
R R
100
−
−
0
R
θ
× 100
Potentiometer
Principle :
Potential difference across any segments
∝
Length of the segm eent
(i) Comparision of emf of two cells
\
ε
ε
1
2
=
l l
1
2
(ii) Internal resistance of given primary cell
r
=
1
l
−
2
2
R
It is a safety device that protects the
Fuse
Principle : Heat produced by electric current in wire
H
=
I
2
Rt
appliance from getting damaged, by melting and opening the circuit.
~
r
Illustration 8 : Cells
A
and
B
and a galvanometer
G
are connected to a slide wire
OS
by two sliding contacts
C
and
D
as shown in figure. The slide wire is 100 cm long and has a resistance of 12 . With
OD
= 75 cm, the galvanometer gives no deflection when
OC
is 50 cm. If
D
is moved to touch the end of wire
S
, the value of
OC
for which the galvanometer shows no deflection is
62.5 cm. The emf of cell
B
is 1.0 V. Calculate
(i) the potential difference across
O
and
D
when
D
is at 75 cm mark from
O
(ii) the potential difference
O
across
OS
when
D
touches S
(iii) internal resistance and emf of cell
A
A
C
B
G
D
S
R
Fuse
Circuit diagram
l
Electric appliance
100 –
l
gradient of wire is 1/50 Vcm
–1
. Therefore, voltage drop across wire
OD
of length 75 cm is (1/50) × 75 = 1.5 V.
(ii) Now potential gradient of wire is 1/62.5 V cm is (1/62.5) 100 = 1.6 V.
–1
Therefore, voltage drop across wire
OS
of length 100 cm
.
(iii) For first case,
9
ε
+
r
For second case,
... (i)
12
ε
+
r
... (ii)
On solving eqns. (i) and (ii), we get
r
= 3 and = 2 V.
Sol. : Resistance of wire
OD
is 12
Let and
r
be the emf and internal resistance of cell
A
, respectively.
(i) Since 1.0 V is balanced across 50 cm, so potential
MPP4 CLASS XI
ANSWER
KEY
1.
(d)
2.
(b)
3.
(c)
4.
(c)
5.
(b)
6.
(b)
7.
(b)
8.
(b)
9.
(c)
10.
(c)
11.
(d)
12.
(c)
13.
(a)
14.
(c)
15.
(a)
16.
(c)
17.
(b)
18.
(c)
19.
(a)
20.
(a, d)
21.
(a, b, c, d)
22.
(a, b, c)
23.
(a, b)
24.
(4)
25.
(4)
26.
(8)
27.
(a)
28.
(b)
29.
(b)
30.
(c)
PHYSICS FOR YOU

AUGUST ‘17
49
1.
The number density of free electrons in a copper conductor is 8.5 × 10
28
m section of the wire is 2.0 × 10
.The area of cross
–6
m
2
and it is carrying a current of 3.0 A. How long does an electron take to drift from one end of a wire 3.0 m long to its other end?
(a) 8.1 × 10
(c) 9 × 10
4
s
s (b) 2.7 × 10
(d) 3 × 10
3
s
4
s
2.
If voltage across a bulb rated 220 V100 W drops by
2.5% of its rated value, the percentage of the rated value by which the power would decrease is
(a) 20% (b) 2.5% (c) 5% (d) 10%
3.
The equivalent resistance between the points
A
and
B
will be (each resistance is 15 )
(a) 10
(b) 40
(c) 30
(d) 8
4.
The resistance of the wire in the platinum resistance thermometer at ice point is 5 and at steam point is 5.25 . When the thermometer is inserted in an unknown hot bath its resistance is found to be
5.5 . The temperature of the hot bath is
(a) 100°C (b) 200°C (c) 300°C (d) 350°C
5.
A resistance of 2 is connected across the gap of a metrebridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2 , is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any correction, the unknown resistance is
(a) 3 (b) 4 (c) 5 (d) 6
6.
In a potentiometer experiment, when three cells
A
,
B
and
C
are connected in series the balancing length is found to be 740 cm. If
A
and
B
are connected in series balancing length is 440 cm and for
B
and
C
connected in series that is 540 cm. Then the emf of
A
,
B
and
C
are respectively (in V)
(a) 1, 1.2 and 1.5 (b) 1, 2 and 3
(c) 1.5, 2 and 3 (d) 1.5, 2.5 and 3.5
7.
In the circuit shown, if point
O
is earthed, the potential of point
X
is equal to
(a) 10 V
(b) 15 V
(c) 25 V
(d) 12.5 V
8.
Two cells of equal emf having internal resistances
r
1 and
r
2
(
r
1
>
r
2
) are connected in series. On connecting the combination to an external resistance
R
, it is observed that the potential difference across the first cell becomes zero. The value of
R
will be
(a) (
r
1
–
r
2
) (b) (
r
1
+
r
2
)
(c)
+
2
2
(d)
−
2
2
9.
A voltmeter of resistance resistance
R
A
R
V
and an ammeter of
are connected in series across a battery of emf and of negligible internal resistance. When a resistance
R
is connected in parallel to voltmeter, reading of ammeter increases to three times while that of voltmeter reduces to onethird. Then,
(a)
R
(c)
R
A
A
=
R
V
3
= 2
R
(b)
(d)
R
R
A
V
=
R
= 6
V
R
10.
The crosssection area and length of a cylindrical conductor are
A
and
l
, respectively. The specific conductivity varies as (
x
) =
0
l x
, where
x
is the distance along the axis of the cylinder from one of its ends.
(a) The resistance of the system along the cylindrical axis is
l A
0 is
0
V
0
.
σ
0
.
(b) The current density if the potential drop along the cylinder is
V
(c) The electric field at each point in the cylinder is
3
V
0
2
(d) None of these.
.
11.
Two wires
A
and
B
of the same material have their lengths in the ratio 1 : 2 and radii in the ratio
2 : 1. The two wires are connected in parallel across a battery. The ratio of the heat produced in
A
to the heat produced in
B
for the same time is
(a) 1 : 2 (b) 2 : 1 (c) 1 : 8 (d) 8 : 1
50 PHYSICS FOR YOU

AUGUST ‘17
12.
In the Wheatstone’s bridge shown in the figure, in order to balance the bridge we must have
(a)
(b)
(d)
R
R
(c)
R
= 1.5 ,
R
finite value
R
1
= 3 ,
R
1
= 6 ,
R
1
1
= 3 ,
R
2
2
= 3
= 1.5
2
= any
2
= any finite value
13.
The resistance of a wire is
R
. If it is melted and stretched to
n
times its original length, its new resistance will be
(a)
R
/
n
(b)
n
2
R
(c)
R
/
n
2
(d)
nR
[NEET 2017]
14.
A potentiometer is an accurate and versatile device to make electrical measurements of emf because the method involves
(a) potential gradients
(b) a condition of no current flow through the galvanometer
(c) a combination of cells, galvanometer and resistances
(d) cells
[NEET 2017]
15.
In the above circuit the current in each resistance is
(a) 1 A (b) 0.25 A (c) 0.5 A (d) 0 A
[JEE Main Offline 2017]
16.
Which of the following statements is false ?
(a) Wheatstone’s bridge is the most sensitive when all the four resistances are of the same order of magnitude.
(b) In a balanced Wheatstone’s bridge, if the cell and the galvanometer are exchanged, the null point is disturbed.
(c) A rheostat can be used as a potential divider.
(d) Kirchhoff’s second law represents energy conservation.
[JEE Main Offline 2017]
17.
1 1 1
A
1
A
2
A
3
4
B
9 V
0.5
V
4
4
1
1 1
A 9 V battery with internal resistance of 0.5 is connected across an infinite network as shown in the figure. All ammeters
A
1
,
A
2
,
A
3 are ideal. Choose correct statement.
and voltmeter
V
(a) Reading of
V
is 9 V.
(b) Reading of
A
1
is 2 A.
(c) Reading of
V
is 7 V.
(d) Reading of
A
1
is 18 A.
[JEE Main Online 2017]
18.
In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance
P
= 4 and the neutral point
N
is at 60 cm from
A
.
Now an unknown resistance
R
is connected in series to
P
and the new position of the neutral point is at
80 cm from
A
. The value of unknown resistance
R
is
(a)
20
3
Ω
(b)
33
5
Ω
A
P
Q
G
N
B
(c) 6 (d) 7
E
R h
K
[JEE Main Online 2017]
19.
A potentiometer
PQ
is set up to compare two resistances as shown in the figure. The ammeter
A
in the circuit reads 1.0 A when two way key open. The balance point is at a length
l
1
K
3
is
cm from
P
when two way key
K
3
is plugged in between 2 and 1, while the balance points is at a length
l
2
cm from
P
when key
K
3
is plugged in between 3 and 1. The ratio of two resistances
R
R
1
2
, is found to be
(a)
(b)
(c)
l
2
l
2
−
l
1
l
2
l
1
−
l
1
1
l
1
l l
2
(d)
1
l
1
l l
2
[JEE Main Online 2017]
20.
A uniform wire of length
l
and radius
r
has a resistance of 100 . It is recast into a wire of radius
r
.
2
The resistance of new wire will be
(a) 400
(c) 200
(b) 100
(d) 1600
[JEE Main Online 2017]
PHYSICS FOR YOU

AUGUST ‘17
51
SOLUTIONS
1. (b) :
Here,
Number density of free electrons,
n
= 8.5 10 m
Area of crosssection of a wire,
A
= 2.0 10 m
Length of the wire,
l
= 3.0 m
Current,
I
= 3.0 A
I
...(i)
The drift velocity of an electron is
v d
=
neA
The time taken by the electron to drift from one end
l lneA
I
(Using (i)) to other end of the wire is
∆
P t
=
v
=
=
3 0 m )( .
×
= 2.7 10 s
28 m
− 3
)( .
( .
A )
d
− 19
C )( .
2. (c) :
Power, =
R
2
As the resistance of the bulb is constant
2
− m )
∴
∆
V
Percentage decrease in power,
3. (d) :
P
=
V
∆
P
P
×
100
=
2 ∆
V
V
× 100
= 2 × 2.5% = 5%
By symmetry,
R
effective
of arm ( are in parallel
AE
+
EB
)
and arm
AB
= 3 0 and 15
+
1
15
=
1
10
or
R
1
= 10 .
Same for
CFD
and
CD
. Therefore 15 , 10 , 15 are in series and their resultant are in parallel with
AB
effective,
R
1
i e
1
40
+
1
10
=
5
40
⇒
R
total
4. (b) :
A s
R
Here,
t
R
= 5 ,
R
100
=
R
( 1 +
t
)
\
R
100
= 5.25
R
=
R
( 1 +
t
= 5.5
α =
R
R
100
0
×
−
R
0
100
Let the temperature of hot bath be
t
°C
...(i)
R t
=
R
(1 +
t
) or
α =
Equating equations (i) and (ii), we get
R
100
R
0
×
−
R
0
100
=
R t
−
R
0 or
t t
R
0
×
t
0
=
R
R R
100
−
0
R
0
× 100
...(ii)
= 200 °C
5. (a) :
Let
x
be the unknown resistance.
In first case, when Wheatstone’s bridge is balanced, we get
2
x
=
( 100
l
In second case, when Wheatstone’s bridge is balanced, then
−
2
l
)
+
80
20
−
l
Solving equations (i) and (ii), we get
x
= 3
...(i)
...(ii)
6. (a) :
Let
A
,
B
and
C
be the emf of three cells
A
,
B
and
C
respectively. As per question,
A
+
A
+
B
+ from (iii) into (i), we get
\
⇒
B
+
B
=
kl
C
=
kl
1
=
k
× 740
2
=
k
× 440
C
=
kl
=
k
× 540
Inserting the value of
Inserting this value of
A
A
:
B
:
= 1 V,
B
A
= 1.2 V,
+
B
from (ii) and (
C
= 300
k
and
B
...(i)
...(ii)
...(iii)
+
A
= 200
k
C
)
A
into (ii), we get
B
= 240
k
C
= 200
k
: 240
k
: 300
k
= 1 : 1.2 : 1.5
C
= 1.5 V
7. (b) :
In open loop
XABO
,
V
X
– 10 + 5 – 10 = 0 ⇒
V
8. (a) :
X
= 15 V
I
=
Across 1 st
2
ε
1 2
cell,
V
= –
Ir
1
⇒
V
= –
⇒ 2
r
1
=
R
+
r
1
2
+
r
ε
1 2
2
⇒
⋅
R r
1 or 0 = –
= (
r
1
–
r
2
)
(
2
ε
r
1
R r r
2
)
9. (a) :
Let initially a current
I
flow through the circuit as shown in figure.
Applying Kirchhoff’s voltage law on the circuit,
IR
A
+
IR
V
– = 0 or =
IR
A
+
IR
V
+ –
...(i)
Initial reading of ammeter is
I
and that of voltmeter is
IR
V
.
When resistance
R
is
I
I
connected in parallel with
R
A
I
R
V
voltmeter, reading of ammeter increases to three times, it means current 3
I
flows through ammeter but reading of voltmeter decreases to one third, it means current
I
/3 flows through the voltmeter.
Hence, remaining
3
− = through
R
as shown in figure.
First, applying KVL on mesh 2.
8
I
3
current passes
3
V
−
8
3
I R
= 0 or
R
V
= 8
R
52 PHYSICS FOR YOU

AUGUST ‘17
Now applying KVL on mesh 1
3
V
or
ε =
1
−
3
IR
IR
V
A
+
=
0
3
IR
A
...(ii)
1
3
IR
V
+ 3
IR
A
=
IR
A
+
IR
V
or
R
A
=
10. (c) :
Consider the cylinder as composed of thin discs of width
dx
connected in series. The resistance of a disc at a distance
x
away from the cylinder end is
R
V
3
A
;
∴
dx x
R
A
=
8
3
R dR
= where
σ
A
1
x dx
A
=
Al xdx
σ
is the crosssection area of the disc and
dx
is its width. Since the discs are connected in series, the total resistance is
R
=
∫
0
l dR
=
Al
across the cylinder is given by
1
I
σ
The current density is therefore,
∫
0 0
=
V
=
l
R
0
A xdx
=
=
3
3
2
=
σ
3
11. (d) :
Given
Since,
⇒
1
=
l l
ρ
l l
1
2
l
1
2
A
1
1
A
A
2
1
2
= and and
\ Ratio of heat =
12. (d) :
R
R
R
1
2
For
10
=
R
1
;
ABCD
\
R
1
l l
1
R
H
= 3
2
H
2
r r
1
2
1
2
r r
2
1
=
=
=
ρ
2
1
l
A
2
2
=
/
/
1
2
1
2
=
1
2
R
R
2
1
2
=
=
8
1
1
8 network, balance condition is
l
2
V l
A
σ
l
0
From Ohm’s law, we deduce that the current flowing
2
The electric field in the cylinder may be found by using Ohm’s law,
=
1
E x
=
σ
J
( )
2 or
=
r r
2
1
σ
0
=
J
1
2
l x
=
3
V x
2
l
The resistance of the inner network should have some finite value for having the bridge balanced.
Hence
R
2
should have any finite value.
13. (b) :
The resistance of a wire of length
l
and area
A
and resistivity is given as
R
=
l
/
A, l
′
= nl (Given)
As the volume of the wire remains constant,
\
A
′
l
′
=
Al
⇒
A
′
=
l
Al
′
=
Al nl
=
A n
54 PHYSICS FOR YOU

AUGUST ‘17
\
14. (b) :
R
′
=
ρ
A l
′
′
=
2
A
ρ
= =
n
2
R
A potentiometer is an accurate and versatile device to make electrical measurements of emf because the method involves a condition of no current flow through the galvanometer, the device can be used to measure potential difference, internal resistance of a cell and compare emfs of two sources.
15. (d) :
The potential difference across each loop is zero. Therefore no current will flow in the circuit.
16. (b) :
In a balanced Wheatstone bridge if the cell and the galvanometer are interchanged, the null point remains unchanged.
17. (b) :
Let equivalent resistance of the infinite network be
x
. Equivalent resistance between points
A
and
B
,
x
=
4
4
+
x x
+
2 or
x
2
−
2
x
8 0
x
= 4 (
∵
–
ve
value not accepted) ;
\
I
1
=
9
= 2
A
⇒
Reading of
A
1
is 2 A.
18. (a) :
∴
For
P
Q
=
P
= 4 ,
100
l
1
−
Q
= 4
⇒
+
8
3
R
∴
R
=
32
3
− =
3
Ω
19. (b) :
When key is plugged between 2 and 1,
2
l
1
=
Now,
P
′ =
P
+
R
,
l
′
1
l
1
= 60 cm
60
40
= 80 cm;
4
=
2
3
or
Q
=
2
P
8
=
=
\
P
Q
4
;
4
′
3
=
3
100
l
1
′
− ′
32
3
=
80
20
=
20. (d) :
Resistance of a wire of length
l
and radius
r
is
i
V
When key is plugged between 3 and 1,
V
On dividing eqn. (ii) by eqn. (i), given by
.
V
V e
1
2
1
2
=
iR
=
i
(
R
=
Here,
∴
.,
R
R
1
1
∝
1
=
Xl
1
1
R
+
+
R
1
R r
1
4
R
2
=
R
1
=
ρ
A l
=
ρ
A l
×
2
=
∴
) =
Xl l l
1
2 or
R
R
1
2
= 100 ,
r r r
1
2
4
2
=
R
R
1
2
A
A
r r
2
1
1
=
r, r
= 16
R
1
2
l
1
=
ρ
V
2
=
4
A
2
=
r
2
= 1600
1
ρ
V
π
r
(
,
R
2
= ?
V
=
Al
4
...(i)
...(ii)
)
EXAM
s
PREP 2018
ELECTRIC CHARGES AND FIELDS
1.
Charge
q
1
= + 6.0 nC is on
y
axis at
y
= + 3 cm and charge
q
2
= – 6.0 nC is on
y
axis at
y
= – 3 cm.
Calculate force on a test charge
q
0
x
axis at
x
= 4 cm.
= 2 nC placed on
(a) − ⋅
j
µ
N
(b)
51 8
j
µ
N
(c) − ⋅
j
µ
N
(d) 5 18
j
µ N
2.
Three charges
Q
1
,
Q
2
and
q
are placed on a straight line such that
q
is somewhere in between
Q
1
and
Q
2
. If this system of charges is in equilibrium, what should be the magnitude and sign of charge
q
?
(a)
(b)
(
Q
1
+
+
2
2
Q
2
CHAPTERWISE MCQs FOR PRACTICE
)
2
, positive
, positive
(a) 2.2 × 10
–4
(c) 6.6 × 10
–4
N
N
(b) 4.4 × 10
–4
(d) 8.9 × 10
–4
N
N
5.
Charges and
q
3
q
1
= 1.5 mC,
q
2
= 0.2 mC
= – 0.5 mC are placed at the
r
points
A
,
B
and
C
respectively, as shown in figure. If
r
1
= 1.2 m and
2
= 0.6 m, calculate the magnitude of resultant force on
q
(a) 3.125 kN
(c) 4.231 kN
2
.
A q
1
(b) 2.475 kN
(d) 6.541 kN
r
1
r
2
C q
3
B q
2
6.
Charge on an originally uncharged conductor is separated by holding a positively charged rod very close nearby, as shown in figure. Assume that the induced negative charge on the conductor is equal to the positive charge
q
on the rod. Then the flux through surface
S
1
is
(c)
(d)
(
Q
1
+
+
2
2
Q
2
)
2
, negative
, negative
3.
The electric field in a region is given by
E
= α
xi
, where is a constant of proper dimension. Find the total flux passing through a cube bounded by surfaces,
x
=
l
,
x
= 2
l
,
y
= 0,
y
=
l
,
z
= 0,
z
=
l
.
(a)
l
(c)
2
3
l
(b)
l
(d)
2
l
2
2
4.
A pendulum bob of mass 80 g carrying a charge of
2 × 10
–8
C is at rest in a uniform horizontal electric field
E
= 20,000 V m
–1 thread of the pendulum.
. Find the tension in the
(a) zero
(c) –
q
/
0
(b)
q
/
0
(d) none of these.
7.
Find the electric field vector at
P
(
a
,
a
,
a
) due to three infinitely long lines of charges along the
x
,
y
and
z
axes, respectively. The charge density,
i
.
e
., charge per unit length
2
y
of each wire is .
(a)
(b)
3
λ
πε
λ
2
πε
0
0
a a
(
(
i j k
)
)
3
z
P
( ,
,
a
)
1
x
PHYSICS FOR YOU

AUGUST ‘17
55
(c)
2 2
λ
πε
0
a
( )
(d) 2
0
λ
πε
a
( )
8.
A large sheet carries uniform surface charge density .
A rod of length 2
l
has a linear charge density on one half and – on the other half. The rod is hinged at midpoint
O
and makes angle q with the normal to the sheet. The torque experienced by the rod is
(a)
σλ
2
σλ
ε
l
2
l
0
2
(b)
σλ
ε
0
l
cos
2
θ
(c)
2 ε
0
(d)
σλ
ε
0
l
sin
2
θ
9.
Two spheres carrying charges +6 C and +9 C, separated by a distance
d
, experiences a force of repulsion
F
. When a charge of –3 C is given to both the sphere and kept at the same distance as before, the new force of repulsion is
(a) 3
F
(b)
F
9
(c)
F
(d)
F
3
10.
Three identical spheres, each having a charge
q
and radius
R
, are kept in such a way that each touches the other two. The magnitude of the electric force on any sphere due to the other two is
πε
0
q
R
2
(b)
4
3
πε
0
q
R
2
(c)
16
3
πε
0
q
R
2
(d)
16
5
πε
0
q
R
2
11.
A particle of mass
m
and charge
q
is released from rest in a uniform electric field of intensity
E
. Find the kinetic energy it attains after moving a distance
x
in the field.
(a)
qEx
2
(b)
qEx
3
(c)
qEx
(d)
qEm
12.
Two charges, each equal to
q
are kept at
x
= –
a
and
x
=
a
on the
x
axis. A particle of mass
m
and charge
q
0
=
q
2
is placed at the origin. If charge
q
0
is given a small displacement (
y
<<
a
) along the
y
axis, the net force acting on the particle is proportional to
(a)
y
(b) –
y
(c)
1
y
(d)
−
1
y
13.
An electron moves a distance of 6 cm when accelerated from rest by an electric field of strength
2 × 10
4
N C
–1
. Calculate the time of travel. The mass and charge of electron are 9 × 10
–31
1.6 × 10
–19
C respectively.
(a) 4 ns
(c) 5.85 ns
(b) 2.2 ns
(d) 7.2 ns
kg and
14.
A spherical shell of radius
R
= 1.5 cm has a charge
q
= 20 C uniformly distributed over it. The force exerted by one half over the other half is
(a) zero
(c) 500 N
(b) 10
–2
N
(d) 2000 N
15.
A large charged metal sheet is placed in a uniform electric field, perpendicularly to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet is
E
1
E
1
5
V m
–1
. The sheet experiences
= 5 × 10 right it is
E
2
5
V m
–1
= 3 × 10
and on the
m
2
(b) 0.9 × 10
–2
(c) 1.8 × 10
–2
m
2
(d) none of these
E
2
a net electric force of 0.08 N. Find the area of one face of the sheet. Assume the external field to remain constant after introducing the large sheet.
(a) 3.6 × 10
–2
m
2
ELECTROSTATIC POTENTIAL AND CAPACITANCE
16.
A charge
Q
is distributed over two concentric hollow spheres of radii
r
and
R
>
r
such that their surface charge densities are equal. Find the potential at the common centre.
(a)
4
πε
0
(
QR r
2
+
R
2
)
(b)
4 πε
0
(
Qr r
2
+
R
2
)
(c)
4 πε
Q r R
0
(
r
2
+
)
R
2
)
(d)
4 πε
0
(
QRr r
2
+
R
2
)
17.
A parallel plate capacitor of capacitance
C
is connected to a battery and is charged to a potential difference
V
. Another capacitor of capacitance
2
C
is similarly charged to a potential difference 2
V
.
The charging battery is then disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
(a) zero (b) 3
CV
2
CV
2
(d) 92
CV
2
56 PHYSICS FOR YOU

AUGUST ‘17
18.
Three capacitors
A
,
B
and
C
are connected in a circuit as shown in figure. What is the charge in C on the capacitor
B
?
22.
Find the equivalent capacitance across
A
and
B
.
(a) 1/3 (b) 2/3 (c) 1 (d) 4/3
19.
Consider a system of three charges
q q
3 3
−
2
3
q
placed at points
A
,
B
and
C
, respectively, as shown in the figure. The centre of the circle is
O
and radius is
R
and angle
CAB
= 60°. Then
(a) the electric field at point
O
is
8 πε
q
0
R
2
directed along the negative
x
axis
(b) the potential energy of the system is zero
(c) the magnitude of the force between the charges at
C
and
B
is
q
2
54 πε
0
R
2
(d) the potential at point
O
is
q
12
πε
0
R
20.
A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is
(a) zero everywhere
(b) nonzero and uniform
(c) nonuniform
(d) zero only at its center
21.
A network of four capacitors of capacitances equal to
C
1
=
C
,
C
2
= 2
C
,
C
3
= 3
C
and
C
2
and
C
4
is
4
= 4
C
are connected to a battery as shown in the figure. The ratio of the charges on
C
C
2
C
3
C
1
C
4
(b) 3
V
(d) 22
(a)
35
6
µ F (b) 25
(c) 15 F
µ F
(d) none of these
23.
A parallel plate capacitor has area of each plate
A
, the separation between the plates is
d
. It is charged to a potential
V
and then disconnected from the battery. How much work will be done in filling the capacitor completely with a dielectric of constant
K
?
ε
0
AV
V
2
d
ε
2
0
2
A
−
K
2
V
ε
0
2
ε
Kd d
0
AV
A
2
24.
Two isolated metallic solid spheres of radii
R
and
2
R
are charged such that both of these have same charge density . The spheres are located far away from each other and connected by a thin wire. Find the new charge density on the bigger sphere.
(a)
3
6
σ
(b)
5
6
σ
(c) 7
6
σ
(d)
5
7
σ
25.
Two thin wire rings each having a radius
R
are placed at a distance
d
apart with their axes coinciding. The charges on the two rings are +
q
and –
q
. The potential difference between the centres of the two rings is
(a)
4
q R
πε
0
d
2
(b)
2
q
πε
0
1
R
−
R
2
1
+
d
2
(c) zero (d)
q
4 πε
0
1
R
−
R
2
1
+
d
2
26.
Mark the correct statement:
(a) An electron and a proton when released from rest in a uniform electric field experience the same force and the same acceleration.
(b) Two equipotential surfaces may intersect.
(c) A solid conducting sphere holds more charge than a hollow conducting sphere of the same radius.
(d) No work is done in taking a positive charge from one point to another inside a negatively charged metallic sphere.
PHYSICS FOR YOU

AUGUST ‘17
57
27.
A uniformly charged solid sphere of radius
R
has potential
V
is measured with respect to infinity on its surface. For this sphere the equipotential surfaces with potentials 3 have radius
R
1
,
R
V
2
0
,
5
V
4
0
,
3
V
4
0 and
2
,
R
and
R
respectively. Then
V
4
0
(a)
R
(b) 8
R
(c)
R
(d)
R
1
1
1
= 0 and
R
1
<
R
4 and 2
= 0 and
R
0 and (
R
2
< (
R
–
R
)
2
R
1
>
R
> (
R
–
R
)
2
–
R
1
3
) > (
R
–
R
)
28.
A parallel plate air capacitor has a capacitance
C
.
When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be
(a) 400%
(c) 33.3%
(b) 66.6%
(d) 200%
29.
Two identical charged spheres, each of radius
r
having charge
q
are held at a distance
R
between their centres as shown in figure. The potential energy of the system is
r
= 4 cm (on equatorial line) = 4 × 10
–2
m
q
0
= 2 nC = 2 × 10
–9
C,
F
= ?
\
F = q
0
E
=
− 9
( .
[(
= 51.8 × 10
–6
N
q q
4 πε
0
(
r
2
×
+
2
a
×
−
)
− 9
(
a
)
)
−
− 2
)
) ]
/
9 10
9
F
= 51.8 N, along negative direction of
y
axis
∴
F
= − ⋅
j
µ N
2. (c) :
Since, the system is in equilibrium, net force on each charge is zero so
q
should be negative.
(a)
q
2
4 πε
0
( )
(b)
4
q
2
πε
0
(
R
+ 2
r
)
(c)
q
2
4 πε
0
2 1
(d) none of these
30.
Figure shows two identical parallel plate capacitors connected to a battery with the switch
S
closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
S
(a) 3 : 4
(b) 2 : 3
(c) 3 : 5
(d) 5 : 3
V
A
C
B
C
1. (a) :
Here,
q
= 6.0 nC = 6.0 × 10
–9
2
a
SOLUTIONS
= 6 cm = 6 × 10
–2
m
C
kQ Q
R
2
=
kqQ x
2
2 or
x
R
=
Q q
1
\ Net force on charge
q
is also zero.
\ or
(
R x
=
− )
Q
2
1
=
+
Q
2
kqQ x
Q
2
2
2 or or
x
Q q
1
=
Q
1
Q
2
Q
1
+
Q
2
Q
2 or
3. (a) :
q
=
(
Q
1
+
Q
2
)
2
is perpendicular to the faces
ABCD
and
EFGH
, whereas is tangential to all other faces.
y
E xi
^
z
0
A x = l
D
B
1
C x =
2
l
E
H
F
G
2
x
Flux entering the cube at face
ABCD
,
i
.
e
.,
1
=
E
1
S
= (
l
)(
l
2
) =
l
3
Flux leaving the cube at the face
EFGH
,
i
.
e
.,
2
=
E
2
S
= [ (2
l
)](
l
2
) = 2
l
3
[as
E
1
=
l
]
[as
E
2
= (2
l
)]
Total flux passing through the cube,
i
.
e
.,
c
=
2
–
1
= 2
l
3
–
l
3
=
l
3
58 PHYSICS FOR YOU

AUGUST ‘17
4. (d) :
Here ;
m
= 80 g,
E
= 20,000 V m
–1
q
= 2 × 10
At
B
, three forces
mg
,
T
and
F
=
qE
are in equilibrium as shown in figure.
Applying Lami’s theorem,
\
∴
F
OB
tan cos
T
=
θ
=
mg
θ =
=
mg
OA
×
= ×
AB
F
OA mg
=
=
F
AB
=
mg
tan q
qE mg
A
O
θ
–8
C
1 +
1 tan
OA
=
2
θ
mg
OA
=
= ×
1
×
− 8
=
×
− 6
20000
× 10
=
T
B mg
F
T
=
.
5. (a) :
Force on
q
F
1
= 1.875 × 10
Force on
q
F
2
As
=
= on
q
F
1
2
4
4 is
1
πε
0
is
1
πε
0
q q r
1
2
2
3
2
due to
= ×
9
q
×
1
1 5 10
− 3
1 2
2
due to
q q q r
2
2
3
− 6
×
= cos
OB
10
= .
×
N, along
AB
3
= ×
N, along perpendicular to
F
=
F
1
2
+
F
2
2
=
9
BC
F
×
3
2
mg
θ
− 4
0 2 10
N
×
.
− 3
0 6
2
×
.
×
− 3
− 3
, so the resultant force
N=3.125 kN
6. (b) :
Net charge on the conductor will be zero. So, net charge inside
S
will be due to the charge on the positively charged rod. Hence, flux through
S
1
is
q
/
0
.
7. (b) :
Let us consider the electric field due to wire 3 only.
E
3
=
Eu
ˆ
=
=
=
E
E
4
1
πε
= net
λ
2 2 πε
λ
=
a
4
2
λ
πε
πε
0
a
0
0
a
1
( )
(
2
E E a
2
λ
+
+
( )
( )
2
E a
3
) and
=
2
(
E i
λ
πε
2 cos
0
O a
=
(
3
45
4
y
° +
λ
πε
0
a
P
(
j
) cos
u
+
E
3
i k
45
°
)
)
x
8. (c) :
Consider a pair of small elements of rod, each of length
dx
, at a distance
x
from
O
.
Charge on each element,
dq
=
dx
Force on each element =
dq
(
E
) =
dx
σ
0
Perpendicular distance between the forces = 2
x
sin q
\
d
=
dx
2
σ
ε
0
2
x
sin q
2 ε
τ =
∫
l
0
σλ
ε sin
0
θ
xdx
=
λσ
2 sin
ε
0
θ
l
2
9. (d) :
Let us assume spheres as
A
and
B
.
+6 C +9 C
A B d
According to Coulomb’s law, the force of repulsion between
A
and
B
separated by a distance
d
is
F
= 1
0
(+ 6 µ C )(
d
2
+ 9 µ C )
...(i)
When a charge of –3 C is given to both the spheres, then charge on
A
= +6 C –3 C = +3 C and on
B
= +9 C –3 C = + 6 C
Again by Coulomb’s law, the new force of repulsion between
A
and
B
separated by the same distance
d
is
F
′
πε
0
(+ 3 µ C
d
)(
2
+ 6 µ
µ
C
C
)
Dividing eqn. (ii) by eqn. (i), we get
1
( + 3
+
µ C )( + 6
µ
C
d
2
)(
+
9
µ C
)
)
=
1
3
…(ii)
F
F
′ =
4
4
πε
1
πε
0
0
( 6
d
2
\
10. (c) :
F
′ =
F
3 charged sphere behaves as a point charge at its centre.
Since both
B
and
C
carry equal charge and are at equal distance from
A
\
F
AB
=
F
AC
=
F
PHYSICS FOR YOU

AUGUST ‘17
59
where,
F
=
4
1
πε
0
2
q
R
2
2
=
4
1
πε
0
4
q
R
2
2
Now as angle between
BA
and
CA
is 60°
∴
F
A
=
F
2
+
F
2
+ 2
FF
cos 60 ° =
16
3
πε
0
q
R
2
11. (c) :
Force acting on the charge
q
in the electric field of intensity
E
,
i
.
e
.,
F
=
qE
If
v
is the velocity acquired by the particle in moving through a distance then from
v
2
–
v x
= 2
after starting from rest
ax
, we get
v
2
= 2
ax v
0
= 0, or
v
2
=
qE m
x
a F
=
qE m
Thus, kinetic energy of the particle,
i
.
e
,
12. (a)
KE =
1
2
mv
2
=
1
2
=
qEx
13. (c) :
Force exerted on the electron by the electric field,
F
=
eE
\ Acceleration of the electron,
a F
=
eE m
=
.
− 19
2 10
−
31
4
= 0.35 × 10
16
m s
–2
Now
u
= 0,
s
= 6.0 cm = 0.06 m,
a
= 0.35 × 10
16
m s
–2
As
s
=
ut
+ 1
at
2
\
0.06 = 0 + 1 × 0.35 ×10
16
×
t
2 or
t
=
.
.
0 35 10
16
= 5.85 × 10
–9
s = 5.85 ns
14. (d)
15. (a) :
Here,
E
1
= 5 × 10
5
V m
F
= 0.08 N.
Let external field be
E
0
and surface charge density of sheet be (including both
–1
,
E
2
= 3 × 10
5
V m
–1
, surfaces). So,
Electric field due to sheet
E
1
=
E
0
–
E
P
E
P
ε
0
...(i)
From eqns. (i) and (ii),
–
E
2
=
E
0
+
E
P
E
0
=
E
1
−
2
E
2
= ×
5
− ×
2
3 10
5
=
10
5
...(ii)
V m
−
1 and
E
E
P
P
=
=
E
1
2
= –4 × 10
2
σ
ε
0
Force on sheet,
E
2
= − ×
5
V m
4 10
5
–1
5
2
− ×
∴ = − 8 ε
0
×
5
10
5 or
A
16. (c) :
If
q
radii
r qE
0 or 0.08 = 8
=
1
10
ε
−

0 or
12
0
and
and
R q
.
× 10
= 10
2
5
=
–12

σ
AE
0
×
A
× 10
5

× 36 × 10
[ ∵
are the charges on the spheres of
, respectively, then as per the given conditions,
q
1
+
q
2
=
Q
9
σ =
= 3.6 × 10
–2
m
2
...(i)
Since their surface charge densities are equal
∴
4
q
π
1
r
2
=
4
q
π
2
R
2
or
From eqns. (i) and (ii),
q
1
=
2
Qr
+
2
r R
2 and
Potential at the centre,
Substituting values of
q
V
(
r
17. (b) :
Here
Q
)
r q
1
2
1
q
=
q
R
2
2
2
=
(
=
and
q
r
2
+
R
2
r
QR
2
1
in eqn. (iii),
)
+
r
2
+
R
2
q
2
R
2
...(ii)
...(iii)
As the two capacitors are connected with opposite polarity, the common potential is
V
′ =
Q
1
2
−
Q
2
1
=
4
C
+ 2
C
=
Equivalent capacitance,
C
′ =
C
+ 2
C
= 3
C
Final energy of the configuration is
U
C
′ =
eq
=
1
2
2
1
(
+
(
R
=
CV
,
Q
=
2
2
14
9
)
=
)
×
1
2
2
4 πε
0
= 2
3
(
µ
C
× 2
C V
2
V
=
V
)
= 4
3
2
=
CV
9
CV
2
18. (b) :
Capacitors
B
and
C
are in parallel, then
A
is in series
=
( )
=
14
9
F =
14
9
×
10 7 6 ) C
10
14
− 6
µ C
F
Charge will be divided between the capacitors
B
and
C
. So charge on the capacitor
B
is
q
=
3
×
14
9
=
2
3
µ
C
60 PHYSICS FOR YOU

AUGUST ‘17
19. (c) :
The electric field at
O
due to the two charges
q
/3 will get cancelled. Electric field due to
−
3
2
q
towards
C
.
will be directed
E
4
1
πε
0
( / )
R
2
=
6
−
πε
0
q
R
2
Potential energy of the system
V
=
4
1
πε
0
q q
3 3
R
2
3
q
0
60°
=
4
1
πε
0
q
2
3
R
2
+
q
3
2
R
− sin
2
q
3
60 °
+
q
3
2
R
− cos
2
q
3
60
The magnitude of the force between
B
and
C
,
°°
F
=
4
1
πε
0
2
3
q
q
3
( sin 60
°
)
=
1
54
πε
q
0
2
R
2
The electric potential at
O
,
2
=
2
q
2
πε
0
R
22
≠
=
20. (b)
0
21. (b) :
The series combination of
C
parallel with
C
4
1
,
C
of the series combination is given by
2
and
C
3
is in
. The equivalent capacitance
C
123
C
1
123
=
C
1
1
+
C
1
2
+
C
1
3
1
C
2
1
C
+
3
1
C
=
11
6
C
or
C
123
Let
q
1
=
,
q
3
6
C
11
and
q
capacitor. Since in a series combination, charge on all the capacitors is same,
q
,
q
1
2
=
q
2
=
q
3
4
be the charges on the respective
=
C
123
×
V
= 6
CV
11
Also, the charge on the capacitor
C
4
,
q
4
=
C
4
V
= 4
CV
\ The ratio of the charges on
C
2
and
C
4
is,
q q
2
4
=
6
CV
4
11
CV
=
3
22
22. (a) :
Circuit can be redrawn as follows:
Now,
25
5
µ
µ F
F
=
10
2
µ
µ F
F
\
V x
=
V y
C eq
=
+
=
35
6
µ F
23. (d) :
Before filling with dielectric, energy stored is
U
1
=
1
2
C V
2
=
1
2
ε
0
d
A
V
2
On filling with dielectric, capacity will be
C
\
2
=
KC
Energy,
1
=
U
ε
2
d
=
=
1
2
1
2 and
C V
ε
V
0
d
2
2
=
V
K
V
K
1
2
2
Work done = decrease in energy
=
=
V
K
2
1
K
=
U
1
−
U
2
=
1
2
ε
0
d
A
V
2
ε
0
AV
K d
2
24. (b) :
Q
1
= (4
R
2
) = 4
R
Q
2
= [4 (2
R
)
2
2
] = 16
R
2
Total charge of both the spheres,
Q
=
Q
1
= 20
R
Total capacitance of both the spheres,
C
=
C
1
+
C
2
= 4
0
R
+ 4
0
+
(2
R
)=12
Q
2
0
2
R
Common potential,
= =
20
12
πε
0
R
=
5
σ
R
3
ε
0
If
Q
′
2
is charge on the bigger sphere after both the spheres are connected by a thin wire,
Q
′
2
=
C
2
V
=
( 8
πε
0
R
)
5
σ
R
3
ε
0
=
40
3
π
R
2
New charge density on bigger sphere,
i
.
e
.,
σ
σ ′ =
Q
′
2
2
=
40
3
π
R
2
16 π
R
2
σ
=
5
6
σ
PHYSICS FOR YOU

AUGUST ‘17
61
25. (b) :
The two charged rings are shown in the figure.
R
2
+
d
2
R
R
+q
A d B
–q
∴
V
A
=
4
1
πε
V
B
=
4
1
πε
0
V
A
−
V
B
0
q
R
R
+
R
2
−
R q
2
+
q d
+
2
d
2
=
4
1
πε
0
q
R
−
R
2
q
+
d
R
2
+
d
2
R
2
q
+
d
2
=
2
q
πε
0
1
R
−
1
2
26. (d) :
Option (a) is wrong because magnitude of force will be the same, but acceleration will be different because masses of electrons and protons are different.
Option (b) is wrong because at a point there can be only one potential.
Option (c) is wrong because charge always lies on the outer surface of a conductor.
Option (d) is correct because the whole conductor will be an equipotential body.
27. (a) :
Spherical surface of radius
r
inside sphere will be equipotential surface with potential (>
V
0
V
= (3
R
(3
R
2
–
r
2
)
V
0
=
)
kq
R
For
For
V
V
2
kq
R
3
=
=
3
5
V
2
V
4
0
0
;
,
2
–
3
5
V
2
V
4
r
0
2
0
) =
=
=
2
2
2
V
R
0
2
V
R
V
R
0
2
0
2
(3
R
2
–
R
(3
R
2
–
R
)
⇒
R
1
) ⇒
R
= 0
2
= be equipotential surface with potential
V
′ (<
V
0
)
R
2
Spherical surface of radius
r
′ outside this sphere will
\
For
V
V
Here
R
′
′
1
=
For
=
V
′
kq r
′
4
4 4
= 0,
R
=
= 3
2
V
0
0
r
′
;
=
< (
R
4
3
V
4
0
V R
R
4
–
R
=
V R
⇒
R
0
R
3
4
⇒
= 4
R
R
3
), and (
R
2
–
R
3
=
1
4
3
R
) < (
R
4
–
R
3
)
28. (b) :
Capacitance of the given capacitor,
C
=
ε
0
d
A
.
When the capacitor is halffilled with dielectric constant
K
= 5, capacitance is given by
C
′ =
d
2
ε
0
A
+
2
d
K
=
5
3
ε
0
d
A
5
=
5
3
C
Increase in capacitance
29. (c) :
=
3
C
its own charge
U
1
=
U
2
=
1
2
q
2
C
electric field of other
U
1
′ = ′ =
q
2
=
8
q
2
πε
0
r
× = . %
Mutual potential energy of each sphere due to
R
\
Total potential energy of the system of two spheres,
4 πε
0
U
U
= (
=
U
4
q
1
2
+
πε
0
U
2
) + (
U
′
1
+
U
′
2
)
=
8
2
q
2
πε
0
r
+
4
2
q
2
πε
0
R
30. (c) :
Initially when the switch
S
is closed, both the capacitors have same potential difference
V
across them. Therefore, initial energy stored in both the capacitors is
U i
=
U
A
+
U
B
=
1
2
CV
2
+
1
2
CV
2
=
CV
2
When the dielectric of dielectric constant,
K
= 3 is inserted, the capacitance of each capacitor becomes
3
C
. The potential difference across
A
is still
V
as it is still connected across the battery. With switch
S
open, the potential difference on
B
attains a new value
V
′ but charge
q
=
CV
does not change.
V
\
CV
= 3
C
×
V
′
or
V
′ =
3
The final total energy stored in both capacitors is
U f
U U
B
1
2
3
C V
2
1
2
3
\ The required ratio will be,
U
U f i
=
5
3
CV
CV
2
2
C
×
V
3
2
=
5
3
CV
2
62 PHYSICS FOR YOU

AUGUST ‘17
Series 3
CHAPTERWISE PRACTICE PAPER
Moving Charges and Magnetism  Magnetism and Matter
Time Allowed : 3 hours Maximum Marks : 70
GENERAL INSTRUCTIONS
( i ) A l l q u e s t i o n s ar e c o m p u l s o r y .
( i i ) Q . n o . 1 t o 5 ar e v e r y s h o r t an s w er q u es t i o n s an d c ar r y 1 m ar k eac h .
( i i i ) Q . n o . 6 t o 10 ar
( i v ) Q . n o . 1 1 t o 22 ar e s h o r t an s w er q u es t i o n s an d c ar r y 2 m ar k s eac e al s o s h o r t an s w er q u es t i o n s an d c ar r y 3 m h .
ar k s eac h .
( v ) Q . n o . 2 3 i s a v al u e b as ed q u es t i o n an d c ar r i es 4 m ar k s .
( v i ) Q . n o . 2 4 t o 26 ar e l o n g an s w er q u es t i o n s an d c ar r y 5 m ar k s eac h .
( v i i ) U s e l o g t ab l es i f n ec es s ar y , u s e o f c al c u l at o r s i s n o t al l o w ed .
SECTION  A
1.
Show that a force that does no work must be a velocity dependent force.
2.
If the ratio of the horizontal component of the
Earth's magnetic field to the resultant magnetic field at a place is 1/
2
, what is the angle of dip at that place?
3.
A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials?
4.
How does the (a) pole strength and (b) magnetic moment of each part of a bar magnet change if it is cut into two equal pieces transverse to length?
M
2
l
–m +m
5.
Soft iron is used to make electromagnets. Why?
SECTION  B
6.
A current of 2.00 A exists in a square loop of edge
10.0 cm. Find the magnetic field at the centre of the square loop.
7.
In a chamber, a uniform magnetic field of
6.5 G (1 G = 10
–4
T) is maintained. An electron is shot into the field with a speed of 4.8 × 10
6
m s
–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (
e
= 1.6 × 10
–19
C,
m e
= 9.1 × 10
–31
kg)
OR
What torque acts on a 40 turn coil of a 100 cm
2 area carrying a current of 10 A held : (a) parallel to and
(b) at right angles to a magnetic field of flux density
0.2 T?
8.
What is meant by magnetic screening or shielding?
9.
Find magnetic induction at centre
O
due to current
I
through the circuit shown in figure.
E
I
I
F
D C r
2 r
1
O
PHYSICS FOR YOU

AUGUST ‘17
63
10.
Draw diagrams to depict the behaviour of magnetic field lines near a bar of
(a) Copper
(b) Aluminium
(c) Mercury, cooled to a very low temperature
(4.2 K)
SECTION  C
11.
Assume the dipole model for earth’s magnetic field
B
V
which is given by
= vertical component of magnetic field
µ
2
m
cos
θ
B
q
=
4
π
r
H
= Horizontal component of magnetic field
=
µ
4
0
0
π
m r
sin
3
3
θ
= (90° – latitude measured from magnetic equator).
Find loci of points for which (i)
 B 
is minimum;
(ii) dip angle is zero; and (iii) dip angle is 45°.
12.
Verify the Gauss’s law for magnetic field of a point dipole of dipole moment at the origin for the surface which is a sphere of radius
R
.
13.
A straight wire of length ( /2) m is bent into a circular shape. If the wire were to carry a current of 5 A, calculate the magnetic field due to it, before bending, at a point distant 0.01 times the radius of the circle formed from it. Also, calculate the magnetic field at the centre of the circular loop formed, for the same value of current.
14.
Two identical circular coils of radius 0.1 m, each having 20 turns are mounted coaxially 0.1 m apart. A current of 0.5 A is passed through both of them (i) in the same direction, (ii) in the opposite directions. Find the magnetic field at the centre of each coil.
OR
A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3 A. A thin coil having
10 turns of wire and of radius 0.01 m carries a current of 0.4 A. Calculate the torque required to hold the coil in the middle of the solenoid with
( its axis perpendicular to the axis of the solenoid
0
= 4 10 Vs A
1 m
1
).
15.
A bar magnet of magnetic moment
M
and moment of inertia
I
(about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let
T
be the period of oscillations of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field .
What would be the similar period
T
′
for each piece?
16.
A moving coil galvanometer when shunted with a resistance of 5 gives a full scale deflection for
250 mA and when a resistance of 1050 is connected in series, it gives a full scale deflection for 25 V. Find the resistance of the galvanometer and the current required to produce a full scale deflection when it is used alone.
17.
(a) Why should a solenoid tend to contract when a current passes through it?
(b) A solenoid of length 1.0 m and 3.0 cm diameter has five layers of windings of 850 turns each and carries a current of 5 A. What is the magnetic field at the centre of the solenoid? Also calculate the magnetic flux for a crosssection of the solenoid at the center of the solenoid.
18.
Distinguish between Biot Savart’s law and Ampere’s circuital law.
19.
A combination of electric and magnetic fields is used as velocity selector for charged particles. Obtain the conditions required and also explain what happens to the particles of higher and lower velocities than one selected.
20.
A long horizontal wire
P
carries a current of 50 A.
It is rigidly fixed. Another fine wire
Q
is placed directly above and parallel to
P
. The weight of the wire
Q
is 0.075 N m
–1
and it carries a current of 25 A.
Find the position of the wire
Q
from the wire
P
so that
Q
remains suspended due to the magnetic repulsion. Also indicate the direction of current in
Q
with respect to
P
.
21.
A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian when the current in the coil is 0.35 A. the needle points west to east.
(a) Determine the horizontal component of earth's magnetic field at the location.
(b) The current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above.
Predict the direction of the needle. Take the magnetic declination at the places to be zero.
64 PHYSICS FOR YOU

AUGUST ‘17
22.
A beam of protons with a velocity of 4 10
5 m s
–1 enters a uniform magnetic field of 0.3 T. The velocity makes an angle of 60° with the magnetic field. Find the radius of the helical path taken by the proton beam and the pitch of the helix.
SECTION  D
23.
Manak was doing the activities in Physics lab , while using galvanometer he requested his
Physics teacher to help him performing a new activity of conversion of galvanometer into voltmeter and ammeter using standard resistances.
He also observed that the reading of ammeter was less than actual current in circuit and reading of voltmeter was less than actual value of potential difference between the two points where voltmeter was attached.
(a) What values are shown by Manak and his
Physics teacher?
(b) How could Manak convert galvanometer into ammeter and voltmeter?
(c) Why were the measured readings less than the actual value of current and potential difference?
SECTION  E
24.
Derive an expression for the force per unit length between two long straight parallel current carrying conductors. Hence define one ampere.
OR
(a) With the help of a diagram. Explain the principle and working of a moving coil galvanometer.
(b) Two moving coil galvanometers
M
1
and
M
2
have the following particulars :
R
1
= 10 ,
N
1
B
1
= 0.25 T
= 30,
A
1
= 3.6 × 10
–3
m
2
,
R
2
B
2
= 14 ,
N
2
= 42,
A
2
= 1.8 × 10
–3
m
2
,
= 0.50 T.
The spring constant are identical for two springs.
Determine ratio of (i) current sensitivity
(ii) voltage sensitivity of
M
2
and
M
1
.
25.
Answer the following questions :
(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece.
If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?
(c) A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?
Explain the meaning of this statement.
OR
(a) Define ‘intensity of magnetisation’ of a magnetic material. How does it vary with temperature for paramagnetic material?
I
B
(b) The given figure shows the variation of intensity of magnetisation
I
versus the applied magnetic field intensity
A
H
, for two magnetic materials
A
and
B
:
H
(i) Identify the materials
A
and
B
.
(ii) Why does the material
B
, have a larger susceptibility than
A
, for a given field at constant temperature?
26.
(a) What is the relationship between the current and the magnetic moment of a current carrying circular loop? Use the expression to derive the relation between the magnetic moment of an electron moving in a circle and its related angular momentum?
(b) A muon is a particle that has the same charge as an electron but is 200 times heavier than it. If we had an atom in which the muon revolves around a proton instead of an electron, what would be the magnetic moment of the muon in the ground state of such an atom?
OR
(a) Magnetic field lines show the direction (at every point) which a small magnetised needle takes up (at that point). Do the magnetic field lines also represent the lines of force of a moving charged particle at every point?
(b) If magnetic monopoles existed, how would
Gauss’s law of magnetism be modified?
(c) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current carrying wire exert a force on another element of the same wire?
(d) Magnetic field arises due to charges in motion.
Can a system have magnetic moment even though its net charge is zero?
(e) Magnetic force is always normal to the velocity of a charge and therefore does no work. An iron nail held near a magnet, when released, increases its kinetic energy as it moves to cling to the magnet.
What agency is responsible for this increase in kinetic energy if not the magnetic field?
PHYSICS FOR YOU

AUGUST ‘17
65
1.
When a force does no work
.
SOLUTIONS
We know that, work done,
)
=
0
i .e
.,
.
= 0
=
.(
)
∴
is perpendicular to
v
.
If direction of
v
changes then direction of also change so that condition
F v
should
.
=
0 is satisfied.
2.
As
B
H
=
B
cos or cos =
B
H
/
B
= 1/ 2
= 45°.
3.
Magnetic moment of proton,
µ
p
Magnetic moment of electron,
µ
∴ or
µ
µ
e p
=
m m e e
> >
p p
≈ 1837 > > 1
e
≈
≈
2
e m p
2
e m e
4.
(
When a bar magnet of magnetic moment
M m l
to length,
)
l l
–m +m –m +m
M/
2
M/
2
(a) the pole strength remains unchanged, since pole strength depends on number of atoms in crosssectional area.
(b) the magnetic moment is reduced to half (since
M
∝ length and here length is halved).
5.
An electromagnet should be a strong but temporary magnet and for that its material should have:
(i) high permeability and (ii) low retentivity. Soft iron possesses both the properties.
6.
The magnetic fields at the centre due to the four sides will be equal in magnitude and direction. The field due to one side will be
B
1
=
µ
0
Ia
I
2
π
d a
2
+
4
d
2
Here,
a
= 10 cm and
a d
O d
=
a
/2 = 5 cm
Thus,
B
1
=
µ
π
0
A
cm
–7
T m A
10 cm
10
)
2 cm
+ 4 5
–1
× 2 A ×
–6
T = 22.6 10
–6
T q cm )
2
1
= 2 × 10
= 5.66 10
–6
T m
Hence, the net field at the centre of the loop will be
4 5.66 10
−
2
7.
The magnetic force
F
=
qvB
acts normal to the direction of motion, thus provide the necessary centripetal force to follow the circular path.
qvB
=
mv
2
r mv
=
r
.
.
or
r
= 4.2 × 10
–2
− 31
− 19
×
× .
×
6
.
×
= 42 mm
− 4
OR
(a) With usual notation, we are given that
N
= 40,
A
= 100 cm
2
= 0.01 m
2
,
I
= 10 A,
B
= 0.2 T, q = 90°
Thus, =
NIAB
sin q
= [40 10 0.01 0.2 (sin 90 )]N m = 0.8 N m
(b) In this case, q = 0° and as such = 0.
8.
Magnetic screening or shielding is the phenomenon of protection of a region against any external magnetic effects. For example, when a soft iron ring is placed in a magnetic field, most of the lines are found to pass through the ring and no lines pass through the space inside the ring. Thus space inside the ring is shielded as shown in figure.
Superconductors also provide perfect magnetic screening as no magnetic lines of force pass through the superconductor as shown in figure.
9.
Magnetic field induction at
O
due to current through an arc
CD
is
B
1
=
4
0
π
r
1
upwards.
Magnetic field induction due to current through an arc
EF
is
B
2
=
4
0
π
r
2 downwards.
The current though
DE
and
FC
will not contribute the magnetic field induction at
O
. Therefore, total magnetic field induction at
O
due to current through the entire circuit
CDEF
will be
1
−
2
=
0
4 π
1 1
1 2
upwards.
66 PHYSICS FOR YOU

AUGUST ‘17
10.
(a) Copper is diamagnetic.
(b) Aluminium is paramagnetic.
(c) Mercury cooled to low temperature (4.2 K) is a superconductor. It behaves as a perfect diamagnet, so no lines of force pass through it.
11.
(i)
B
=
B
+
B
=
µ
4
0
π
r m
3
4 cos
2
θ +
sin
2
θ
=
k
3 cos
2
θ + 1
where
µ
4
0
π
r m
3
=
k
and cos
2
θ +
sin
2
θ =
Now
B
is minimum when cos q
= minimum = 0
1
or
2 which corresponds to magnetic equator. Thus, magnetic equator is the locus of points for which
is minimum.
(ii) If denotes dip angle, then tan δ =
B
B
V
H
=
2 cos sin θ
θ
=
2 tan θ
Now = 0° if tan q
= or θ π
Thus, it is again the magnetic equator which is the locus of points with zero dip.
(iii) When
2
δ = ± °
,
B
B
V
H
or tan
=
θ
1
=
2 or tan
θ
= or q
= tan
1
–1
(2) which is the required locus of points for which = ± 45°.
12.
Consider a Gaussian surface of radius around a point dipole of dipole moment , pointing along axis.
\
Let
m mk ds
be the surface area of an element around .
Magnetic field at along
∧ due to point dipole,
i .e
.,
B r
=
µ
4
0
π
Also,
Thus,
=
∫
ds r
µ
4
0
π
∫
B ds
3
m r
cos
3
∫
θ
d d r
B .ds
r r
.
θ θ φ )
r
3
θ
=
3 as
4
µ
0
π
∫
0
r m
π
∫
0
π
θ sin cos
θ θ
θ θ φ
= 0
= 0
13.
Since 2
r
=
π
2
m,
r
= 0.25 m
Magnetic field due to straight wire at a distance
a
= 0.01
r
= 0.01 0.25 m = 2.5 × 10
–3
m,
i.e
.,
B
=
k m
2
a
I
= 4 10
=
10
− 7
–4
T
T m
A
A
− 3 m
Magnetic field at the centre of the circular loop,
B
=
k m
2
r
π
I
= 1.256 10
=
–5
10
T
− 7
T m
A
.
.
m
A
14.
Here;
a
= 0.1 m,
N
= 20,
r
= 0.1 m,
I
= 0.5 A
Magnetic field at the center of each coil due to its own current is
B
1
=
µ
0
NI
2
a
=
4
π ×
10
−
7
×
.
=
− 5
T
Magnetic field at the centre of one coil due to the current in the other coil is
B
2
=
2 (
a
µ
2
0
NI r
+
r
)
2
=
4
π ×
10
−
7
× ×
2 0 1 + 0 1
×
2
= 2.22 × 10
–5
T
(i) When the currents are in the same direction, the resultant field at the centre of each coil is
–5
+ 2.22 10
–5
B
=
B
1
+
B
2
= 6.28 10
= 8.50 10
–5
T
(ii) When the currents are in opposite directions, the resultant field is
–5
– 2.22 10
–5
B
=
B
1
–
B
2
= 6.28 10
= 4.06 10
–5
T
OR
For solenoid,
l
= 0.4 m,
N
For coil,
N
2
= 10,
r
1
= 500,
I
= 0.01 m,
Field inside the solenoid,
l
2
1
= 3 A
= 0.4 A
PHYSICS FOR YOU

AUGUST ‘17
67
B m
=
=
µ
N
other.
2
I l
2
along the axis of solenoid.
Magnetic moment of coil, of solenoid,
and
Required torque,
= 10 0.4
A
=
N
2
I
2 r
(0.01)
2
2
, along the axis of coil.
will be perpendicular to each
=
m B
sin q =
N
2
I
2 r
2
.
µ
4
N I l
π × 10
. sin 90°
− 7
× ×
= 6
2
× 10
–7
= 6 9.87 = 5.92 10
–6
N m
15.
The moment of inertia of a bar magnet of mass
m
, length
l
about an axis passing through its centre and
2 perpendicular to its length is
I ml
. Let
M
be the magnetic moment of the magnet,
B
is the uniform magnetic field in which the magnet is oscillating, then time period of oscillation is
T
= 2 π
I
MB
When magnet is cut into two equal pieces, perpendicular to length, then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre is
I
′ =
m l
2 12
2
=
ml
12
2
I
8 8
Magnetic dipole moment,
M
′ =
M
/2
Its time period of oscillation is
T
′ = 2
I
′
= 2 π
M
I
/ 8
B
=
2
2
π
I
MB
=
T
2
16.
With shunt, current required to produce full scale deflection is given by
I g
\
=
I g
R
+
s
R R g s
⋅
R
I g
×
+ 5
−
3
When a resistance
R
is connected in series,
I g
=
R g
V
+
=
g
25
+ 1050
R g
.
+ 5
=
R g
25
+ 1050
...(i)
...(ii) or 1.25
R g
+ 1312.5 = 25
R g
+ 125 or 23.75
R g
= 1187.5
or
R g
= 50
17.
(a) We know that two parallel conductors carrying currents, in the same direction attract each other and in opposite directions repel each other.
Therefore, when current is passed through the coil of a solenoid, the parallel currents in the various turns of solenoid flow in the same direction. As a result, the various turns start attracting one another and solenoid tends to contract.
(b) Number of turns,
N
= 850 5
Area of cross section,
A
= π
r
7 2
l
= 1 m,
I
= 5 A
Magnetic field induction at the centre of solenoid is
B
=
2
=
22
3
× 10
− 2
2
0
NI
/
l
= 4 10
–7
(850 5) 5/1 = 2.671×10
Magnetic flux =
BA
= 2.671
× 10
–2
22
= 1.89
× 10
–5
7
Wb
×
3
2
× 10
− 2
–2
T
2
18.
S. No. Biot Savart’s law Ampere’s circuital law
1.
This law is based on the principle of magnetism.
This law is based on the principle of electromagnetism.
2.
This law is valid for asymmetrical
3.
This law is the
19.
Let current distribution.
differential form of magnetic induction
i.e
.,
 
and
=
µ
0
Idl
4 π
r
sin
2
θ
∫
denote the electric field and magnetic field respectively. If
B
, distributions.
This law is the integral form of
i e
⋅ = µ
0
I
,
is the velocity of the particles to be selected, force acting on a charged particle due to electric field, = and the force
F
m
For the particles with velocity or or or
(
=
F
B v
e q v B qE
),
= −
= −
F m
(
).
e
+
)
) (
m
=
0
)
Thus : (i) The direction of
v
to remain
should be that of
×
q
is being taken as positive.
68 PHYSICS FOR YOU

AUGUST ‘17
(ii)
.
These are the required conditions.
For the particles with velocities higher than
v v B
) dominates and hence the particles get deflected towards the direction of with velocities less than direction of +
.
− . But particles
v
, get deflected in the
20.
The magnetic force per unit length on the wire
Q
due to the current in wire
P
is
F
The currents in
P
and
Q
must have opposite directions, only then
Q
Q
will experience a repulsive force which would balance
P
the weight of
Q
.
\
F
=
I I
=
µ
2
0 1 2
π
F
W r
25 A
50 A
r
µ
µ
0 1 2
2 π
r
.
⇒
W
=
µ
0 1 2
2
− 7
π
r
× × or
r
=
2
π
W
= 3.33 10
–3
m = 3.33 mm
21.
Here,
N
= 30,
r
= 12 cm = 12 10
–2
m
I
= 0.35 A,
H
= ?
It is clear from figure the needle can point west to east only when
H
=
B
sin 45°, where,
B
= magnetic field strength due to current in coil
B
=
µ
4
0
π
\
H
=
2
µ
r
4
π
0
π
NI
2
= ×
π
r
NI
− 5 sin
−
7
W
×
2
π
H
N
S
× ×
45°
45°
− 2
.
Coil
B
⋅
1
E
2
T
(b) When current in coil is reversed and coil is turned through 90° anticlockwise, the direction of needle will reverse (
i.e
., it will point from east to west).
22.
The components of the proton's velocity along and perpendicular to the magnetic field are
v

= (4 10 and
v
⊥
5
m s
–1
) cos60° = 2 10
5
= (4 10
5
m s
–1
) sin60° =
2 3
As the force
qv B
is perpendicular to the magnetic field, the component
v

will remain constant. In the plane perpendicular to the field, the proton will describe a circle whose radius is obtained from the equation
m s
–1
10
5
B
m s
–1 or
r
=
=
mv mv r
⊥
qB
2
⊥
=
( .
0.012 m = 1.2 cm
( .
− 27 kg )(
− 19
2 3 10
5
C T ms
− 1
)
The time taken in one complete revolution in the plane perpendicular to
B
is
T
=
2 π
r v
⊥
= ×
.
×
×
5
.
m ms
−
1
The distance moved along the field during this period,
i.e
., the pitch
=
( 2 10
5 ms
− 1
) .
×
.
m
5 ms
−
1
= 0.044 m = 4.4 cm
The qualitative nature of the path of the protons is shown in figure.
23.
(a) Manak had shown curiosity, research oriented mind, ambitious, maturity. His Physics teacher has also shown helping nature, ability to implement theory into practical , involvement with student and readiness to motivate student.
(b) Galvanometer was converted into ammeter by connecting suitable small resistance in parallel and into a voltmeter by connecting suitable high resistance in series.
(c) An ammeter shows the reading less than actual as some current passes through shunt.
Similarly, a voltmeter shows less reading as some potential drops across high resistance connected in series.
24.
Refer to point 3.3 (9), page no. 174 (MTG Excel in
Physics)
OR
(a) Refer to point 3.4 (2), page no. 175 (MTG Excel in Physics)
(b) Current sensitivity of a moving coil galvanometer is defined as voltage sensitivity,
V.S
.=
NAB kR
(i) Ratio of current sensitivity,
C.S.
=
φ
=
NAB k
and
. .
. .
1
2
=
. .
. .
1
2
=
−
3
− 3
×
k
×
k
=
5
7
PHYSICS FOR YOU

AUGUST ‘17
69
(ii) Ratio of voltage sensitivity,
. .
. .
1
2
=
C S
C S
1
2
×
R
R
2
1
×
14
10
=
1
1
25.
(a) In a specimen of a ferromagnets, the atomic dipoles are grouped together in domains. All the dipoles of a domain are aligned in the same direction and have net magnetic moment. In an unmagnetised substance these domains are randomly distributed so that the resultant magnetisation is zero. When the substance is placed in an external magnetic field, these domains align themselves in the direction of the field. Some energy is spent in the process of alignment. When the external field is removed, these domains do not come back into their random positions completely.
The substance retains some magnetisation. The energy spent in the process of magnetisation is not fully recovered. The balance of energy is lost as heat. This is the basic cause for irreversibility of the magnetisation curve of a ferromagnetic substance.
(b) Carbon steel piece because the heat produced in complete cycle of magnetisation is directly proportional to the area under the hysteresis loop.
(c) Magnetisation of a ferromagnet is not a single valued function of the magnetising field. Its value for a particular field depends both on the magnetising field and on the history of its magnetisation
i.e
., how many cycles of magnetisation it has gone through etc. So, the value of magnetisation is a record or memory of its cycles of magnetisation.
If information bits can be made to correspond to these cycles, the system displaying such a hysteresis loop can act as a device for storing information.
OR
(a) Refer to point 3.8 (2) page no. 180 (MTG Excel in Physics)
In paramagnetic, intensity of magnetisation
I
varies inversely with temperature of the given
I
sample.
(b) (i)
A
is paramagnetic and
B
is
T
ferromagnetic.
(ii) Material
B
has larger susceptibility than
A
, because in ferromagnetic the domains easily orient themselves along the direction of magnetising field and hence, it can be easily magnetised.
26.
(a) Magnetic moment of current carrying loop is
M = IA
. For an electron revolving in a circular orbit of radius
r
with speed
v
, effective current is
\
I
= υ
2 π
r
Associated magnetic moment of electron revolving in orbit is
= =
2 π
r
× π
r
2
Related angular momentum is
L
=
rp
sin 90° =
rmv
× 1 or
L = mvr
⇒ = ×
⇒
L
=
2
m e
2
M
M e
or
2
vr
=
.
2
M e
(b) By Bohr’s condition of quantisation,
For ground state
n
= 1, so
2
π or
evr
2
For muon,
=
M
or
4
=
eh
π
vr m
.
=
2
h
π
m
or
− 19
×
M
=
C
4 π
m
× .
×
− 34
×
eh
.
×
− 31
J s kg
2
π or
M
= 4.6 × 10
–26
A m
2
(a) Force on a charge
\
q
OR
moving with a velocity uniform magnetic field of strength is
Magnetic force is always normal to .
= (
)
Magnetic field lines of
B
cannot represent the lines of force of moving charged particle.
(b) According to Gauss’s law in magnetism, magnetic flux over any closed surface is always zero.
∫
B dS
=
0
If monopoles existed, the magnetic flux would no longer be zero, but equal to strength enclosed by the surface.
0
times the pole
(c) No, there is no force or torque on an element due to the field produced by that element itself.
But there is a force (or torque) on an element of the same wire. However, for the special case of a straight wire, this force is zero.
(d) Yes, a system can have magnetic moment even if its net charge is zero. For example, every atom of paramagnetic and ferromagnetic materials has a magnetic moment, though every atom is electrically neutral. Again, a neutron has no charge, but it does have some magnetic moment.
(e) An iron nail is made up of a large number of atoms, in which so many electronic charges are in motion. All these charges in motion experience a magnetic force when held near a magnet. The magnetic forces do not change speed of the charges, but they do change their velocity. The velocity of centre of mass may increase at the expense of nail’s internal energy. Thus, internal energy of the nail is responsible for increase in kinetic energy of the nail, as a whole.
70 PHYSICS FOR YOU

AUGUST ‘17
Class XII
T
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer.
Self check table given at the end will help you to check your readiness.
Magnetic Effects of Current and Magnetism
Total Marks : 120
NEET / AIIMS
Only One Option Correct Type
1.
The charge on a particle
Y
is double the charge on another particle
X
. These two particles
X
and
Y
, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii
2
R
R
1
2
2
2
R
1
and
R
2 respectively. The ratio of the mass of
X
to that of
Y
is
(a)
2
R
R
2
1
R
1
R
2
2
2
(b)
(d) 2
R
1
R
2
2.
A triangular loop of side
l
carries a current
I
. It is placed in a magnetic field
B
such that the plane of the loop is in the direction of
B
. The torque on the loop is
(a)
IBl
(b)
I
2
Bl
4
BIl
2
(d) infinity
3.
Temperature above which a ferromagnetic substance becomes paramagnetic is called
(a) Critical temperature (b) Boyle temperature
(c) Debye's temperature (d) Curie temperature.
4.
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of
2 A. If the number of turns is 1000 per metre, the magnetisation at centre of solenoid is
(a) 4 × 10
(c) 7 × 10
5
A m
–1
5
A m
–1
(b) 6 × 10
(d) 8 × 10
5
5
A m
A m
5.
A short magnet of moment 6.75 A m
2
–1
–1
produces a neutral point on its axis. If horizontal component of earth’s magnetic field is 5 × 10
–5
Wb m distance of the neutral point should be
–2
, then the
Time Taken : 60 min
(a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm
6.
An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current
I
and the
O r
I
radius of the circular loop is
r
. Then the magnetic induction at its centre will be
(a)
µ
(
π +
1 )
2
r
I
I
4
0
π
(c) zero
2
r
I
(b)
µ
4
0
π
(d) infinite
(
π −
1 )
7.
A frog can be levitated in a magnetic field produced by a current in a vertical solenoid placed below the frog. This is possible because the body of the frog behaves as
(a) paramagnetic
(c) ferromagnetic
(b) diamagnetic
(d) antiferromagnetic.
8.
An electron experiences a force
F
=
( .
+ ×
N in a uniform magnetic field when its velocity is ( .
− 1
( .
i
−
i
10
× 10
7 m s
− 1
,
2 5 10
7
)
k
m s
the magnetic force on the electron is zero. The magnetic field vector is
.
When the velocity is redirected and becomes
− .
j
) +
−
.
.
j k
+
)
)
.
9.
A particle of mass
m
and charge
q
is projected from origin with initial velocity
( ).
Uniform electric and magnetic field exist in the region along
+
y
direction of magnitudes
E
and
B
, respectively.
Then particle will definitively return to the origin if
(a)
2
vB
π
E
is an integer
(b) (
+ )
is an integer
PHYSICS FOR YOU

AUGUST ‘17
72
(c)
vB
π
E
is an integer (d)
vB
3
π
E
10.
A proton of mass 1.67 × 10
–27
1.6 × 10
–19 is an integer
kg and charge
C is projected with a speed of 2 × 10
6
m s
–1 at an angle of 60° to the
x
axis. If a uniform magnetic field of 0.104 T is applied along
y
axis. The path of proton is
(a) a circle of radius = 0.2 m and time period × 10
–7
s
(b) a circle of radius = 0.1 m and time period 2 × 10
–7
(c) a helix of radius = 0.1 m and time period 2 × 10
(d) a helix of radius = 0.2 m and time period 4 × 10
–7
–7
s
s
s
11.
A circular current carrying loop of a radius
R
, carries a current the axis of coil is
I
1
(a)
R
2
R
times the value of magnetic
8 field at the centre. Distance of point from centre is
(b)
3
(c)
R
2 (d)
R
12.
A cyclotron’s oscillator frequency is 10 MHz.
If the radius of the dees is 60 cm, the kinetic energy of the proton beam produced by the accelerator is
(
e
= 1.6 × 10
–19
C,
m
P
= 1.6 × 10
–27
kg)
(a) 5 MeV (b) 6 MeV (c) 7 MeV (d) 8 MeV
Assertion & Reason Type
Directions :
In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as :
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If both assertion and reason are false.
13.
Assertion : A rectangular current loop is in an arbitrary orientation in an external uniform magnetic field. No work is required to rotate the loop about an axis perpendicular to its plane.
Reason : All positions represent the same level of energy.
14.
Assertion : When radius of a circular loop carrying current is doubled, its magnetic moment becomes four times.
Reason : Magnetic moment depends on the area of the loop.
15.
Assertion : The magnetic force on the closed loop in given figure is
× ×
×
B
× ×
2
B
× zero.
Reason : Force (magnetic) on the wire is
∫
dF
=
∫
B dl
× ×
×
× ×
I
×
×
× × × × ×
JEE MAIN / JEE ADVANCED
Only One Option Correct Type
16.
A particle of charge per unit mass is released from origin with a velocity magnetic
(0,
(a)
y
, 0), then
−
B
2
v
0
B
0
α
= −
y
(b)
B k
is equal to
B v
0
0
. If the particle passes through
α
(c)
B v v i
2
v
0
0
α
in a uniform
(d) −
v
B
0
0
α
17.
A uniform electric field
E
is present horizontally along the paper throughout the region but uniform magnetic field
B
0
is present horizontally
(perpendicular to plane of paper in inward direction) right to the line
PQ
as shown. A charge particle having charge
q
and mass
m
is projected vertically upward and crossed the line
PQ
after time
t
0
. Find the speed of projection, if particle moves with constant velocity after
t
0
(given
qE
=
mg
).
(a)
gt
0
(b) 2
gt
0
(c)
gt
0
2
18.
(d) particle can't move with constant velocity after crossing
PQ
A particle of specific charge with a velocity of 10 m s
–1 field
B
time
t
=
= −
k
1
12 s will be (in m s
–1
)
q m
= π
C kg
x
− 1
is
axis
v
of particle after
(a)
5 (
i
(
+
3
(c) 5 3
i j j
)
)
(b) 5 3
(d) 5
(
i j
( )
)
19.
The magnetic field
B
at the centre of a circular coil of radius
r
is times that due to a long straight wire at a distance
r
from it, for equal currents. Figure here shows three cases. In all case the circular part has radius
r
and straight ones are infinitely long.
For same current, the magnetic field
B
at the centre
P
in cases I, II, III have the ratio
(a)
(b)
−
2
2
2
1
:
3 π
π
4
2
+
−
1
1
2
:
3 π
4
+
1
2
PHYSICS FOR YOU

AUGUST ‘17
73
(c)
(d)
2 2
π
2
1
4
:
+
π
2
−
1
4
:
3
4
π
+
1
2
.
More than One Options Correct Type
20.
A charged particle with velocity
v xi y j
moves in a magnetic field
B yi x j
. The force acting on the particle has magnitude
F
. Which one of the following statements is/are correct?
(a) No force will act on charged particle if
x
=
y
(b) If
x
>
y
,
F
∝ (
x
2
–
y
2
)
(c) If
x
>
y
, the force will act along
z
axis
(d) If
y
>
x
, the force will act along
y
axis
21.
A wire of mass
m
and length
l
is placed on a smooth incline making an angle q
with the horizontal, whose front view is shown in figure. When a finite amount of charge is passed through it for an infinitesimal time, the wire immediately acquires some velocity and then ascends the incline by a distance
s
. For this small duration, we can neglect the gravitational force because the current can be considered very large due to small time duration. The amount of charge passed through the wire is
(a)
(c)
m gs
Bl
Bl
2 sin cos θ
θ
θ
(b)
mv
Bl
(d) information insufficient
22.
A charged particle of specific charge moves with a
B
0 velocity
v v i
in a magnetic field
B
=
2
(
Then (specific charge = charge per unit mass)
).
(a) path of the particle is a helix
(b) path of the particle is circle
(c) distance moved by the particle in time
t
= π
0
α is
π
B
0
v
0
α
(d) velocity of the particle after time
t
= π
0
α is
v
2
0
i
+
v
2
0
j
23.
A charged particle is fired at an angle q to a uniform magnetic field directed along the
x
axis. During its motion along a helical path, the particle will
(a) never move parallel to the
x
axis
(b) move parallel to the
x
axis once during every rotation for all value of q
(c) move parallel to the
x
axis at least once during every rotation if q
= 45°
(d) never move perpendicular to the
x
direction
Integer Answer Type
24.
A 10 cm length of wire with a mass of 20 g is attached frictionlessly to the vertical segments of a wire in which a current
I
flows. The surrounding has uniform horizontal field
B
= 10
4
G and the direction is shown in figure. What must be the current
I
(in A) to maintain the 10 cm wire in an equilibrium position?
I
B
I
10 cm
Frictionless slider
25.
A nonconducting nonmagnetic rod having circular crosssection of radius
R
is suspended from a rigid support as shown in figure. A light and small coil of 300 turns is wrapped tightly at the left end of the rod where uniform magnetic field
B
exists in vertically downward direction. Air of density hits one half of the right part of the rod with velocity
v
as shown in figure. The current must be in clockwise direction (as seen from
O
) in the coil so that rod remains horizontal. The magnitude of the current will be
Given
x
,
× 10
2
Lv
–2
x
A. Value of
x
will be
π
RB
ρ
=
1 A
− 1 2
26.
Three infinitely long thin wires, each carrying current
I
in the same direction, are in the
xy
plane of a gravity free space. The central wire is along the
y
axis while the other two are along
x
= ±
d
. The locus of the points for which the magnetic field
B
is zero is
x
= ±
d
α
. The value of is
Comprehension Type
In a certain region of space, there exists a uniform and constant electric field of magnitude
E
along the positive
y
axis of a coordinate system. A charged particle of
74 PHYSICS FOR YOU

AUGUST ‘17
mass
m
and charge –
q
(
q
> 0) is projected from the origin with speed 2
v
at an angle of 60° with the positive
x
axis in
x

y
plane. When the
x
coordinate of particle becomes
3
mv qE
2
a uniform and constant magnetic field of strength
B
is also switched on along positive
y
axis.
27.
Velocity of the particle just before the magnetic field is switched on is
(a)
vi
(b)
vi
+
2
3
v j
(c)
vi
−
2
3
v j
(d) 2
vi
−
3
2
v j
28.
z
coordinate of the particle as a function of time after the magnetic field is switched on is
(a)
mv qB
1 − cos
qB m t
(b)
−
mv qB
qB m t
(c)
−
mv qB
qB m t
(d)
mv qB
1
+ cos
qB m t
Matrix Match Type
29.
A charged particle with some initial velocity is projected in a region where nonzero electric and/or magnetic fields are present. In Column I, information about the existence of electric and/or magnetic field and direction of initial velocity of charged particle are given, while in Column II the probable path of the charged particle is mentioned.
Match Column I with Column II.
(A)
= 0 ,
B
≠ 0 , and initial velocity may be at any
Column II
(P) Straight line
(B)
B
velocity may be at any angle with
0 and initial
(Q) Parabola
(C)
(D)
0,  and initial velocity is
B
to both perpendicular to nonuniform pitch
and
v
(R) Circular
(S) Helical path
E
A
and
(a) P, R
(b) P, Q
(c) P, S
(d) P, S
B
P, Q
P, R
Q
R
C
S
S
R
Q
R
S
D
P
S
30.
A charged particle passes through a region that could have electric field only or magnetic field only or both electric and magnetic fields or none of the field. Match Column I with Column II.
Column I
(A) Kinetic energy of the particle remains constant
(B) Acceleration of the particle is zero
(C) Kinetic energy of the particle changes and it also suffers deflection
(D) Kinetic energy of the particle changes but it suffer no deflection
A
(a) P
B
Q
C
R, S
(b) P, R, S P, S P, Q
(c) P, R, S Q, R S
(d) Q, R P S
Column II
(P) Under special conditions, this is possible when both electric and magnetic fields are present
(Q) The region has electric field only
(R) The region has magnetic field only
(S) The region contains no field
D
P, Q
P, Q
P
R
Keys are published in this issue. Search now
!
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No. of questions attempted ……
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Marks scored in percentage ……
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PHYSICS FOR YOU

AUGUST ‘17
75
PHYSICS
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In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / NEET. The detailed solutions of these problems will be published in next issue of Physics For You.
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49
INTEGER TYPE
1.
In a slow reaction, heat is being evolved at a rate about 10 mW in a liquid. If the heat were being generated by the decay of
32
P, a radioactive isotope of phosphorus that has halflife of 14 days and emits only betaparticles with a mean energy of
700 keV, estimate the number of
32
P atoms in the liquid. Express your answer in the form of
A
× 26 × 10
15
and find
A
to nearest integer.
[Take : ln 2 = 0.7]
2.
A convex lens of focal length
20 cm and another plano convex
O d
lens of focal length 40 cm are
10 cm
f
= +20 cm
f
= 4 0 cm placed coaxially as shown in figure. The plano convex lens is silvered on plane surface. What should be the distance
d
(in dm) so that final image of the object
O
is formed on
O
itself.
3.
A point object is placed at the centre of curvature of a concave mirror
91°
y
O x
(taken as origin). A plane mirror is also placed at a distance of 10 cm from
20 cm the object as shown in figure. Consider two reflection first at plane mirror and then at concave mirror. The coordinates of the image thus formed are (
x
0
,
y
0
). Then find
π
x
20
y
0
0
76
4.
The principal section of glass prism is an isosceles
D
PQR
with
PQ
=
PR
. The face
PR
is silvered. A ray is incident perpendicularly on
P
face
PQ
and after two reflections it emerges from base
QR
, normal to it as shown in figure. The angle (in degree) of the prism is 9
x
. Find
x
.
Q
A
R
5.
The intensity received at the focus of the lens is
I
when no glass slab has been placed in front of the slit. Both the slits are of the same dimension and the plane wavefront incident perpendicularly on them, has wavelength . On placing the glass slab, the intensity reduces to 3
I
/4 at the focus. The minimum thickness of the glass slab is (2310 +
x
)Å if its refractive index is 3/2.
Given = 6933Å.
6.
A sound source is located somewhere along the
x
axis. Experiments show that the same wavefront simultaneously reaches listeners at
x
= –8 m and
x
= +2.0 m. A third listener is positioned along the positive
y
axis. What is
y
coordinate (in m) of third listener if the same wavefront reaches her at the same instant as it does the first two listeners ?
Solution Senders of Physics Musing
SET48
1. Nikita Verma, Kanpur (U.P.)
2. Prashant Pandey, Nagpur (Maharashtra)
3. Saumya Vyas, Surat (Gujarat)
By Akhil Tewari,
Author Foundation of Physics for JEE Main & Advanced, Professor, IITians PACE, Mumbai.
PHYSICS FOR YOU

AUGUST ‘17
COMPREHENSION TYPE
For questions 7, 8 and 9 :
The Doppler flow meter is a particularly interesting medical application of the
Doppler's effect. This device measures the speed of blood flow, using transmitting and receiving elements that are
Transmitter Receiver
Incident sound
Skin
Reected sound
Red blood cell
v
R
placed directly on the skin, as shown in figure. The transmitter emits a continuous sound whose frequency is typically about 5 MHz. When the sound is reflected from the red blood cells, its frequency is changed in a kind of Doppler effect because the cells are moving with the same velocity as the blood. The receiving element detects the reflected sound and an electronic counter measures its frequency, which is Dopplershifted relative to the transmitter frequency. From the change in frequency the speed of the blood flow can be determined. Typically, the change in frequency is around 600 Hz for flow speeds of about 0.1 m s
–1
.
7.
Assume that the red blood cell is directly moving away from the source and the receiver. What is the
(approx.) speed of the sound wave in the blood?
(a) 1700 m s
(c) 5000 m s
–1
–1
(b) 330 m s
–1
(d) 3000 m s
–1
8.
An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to onefourth the normal crosssectional area. What will be the change in frequency (approx.) due to reflection from the red blood cell in that region?
(a) 150 Hz (b) 300 Hz (c) 600 Hz (d) 2400 Hz
9.
At what extra rate does the heart have to work due to this narrowing down of the artery? Assume the density to be 1.5 g cm
–3
and the area of crosssection of the normal artery to be 0.1 cm
2
.
(a) 1.125 × 10
(c) 6.25 × 10
–4
W
–5
W
(b) 2.5 × 10
–4
(d) 5.625 × 10
W
–5
W
MATRIX MATCH
10.
Light from source
S
(
u
 < 
f
) falls on lens and screen is placed on the other side. The lens is formed by cutting it along principal axis into two equal parts and are joined as indicated in column II. Match the column I with column II.
78 PHYSICS FOR YOU

AUGUST ‘17
Column I
(A) Plane of image move towards screen if 
f
is increased
Column II
(P) Small portion of each part near pole is removed. The remaining parts are joined.
S
Screen
(B) Images formed will be virtual
(Q) The two parts are separated slightly.
The gap is filled by opaque material.
S
Screen
(C) Separation between images increase if

u
 decreases
(R) The two parts are separated slightly.
The gap is filled by opaque material.
S
Screen
(D) Interference pattern can be obtained if screen is suitably positioned.
(S) Small portion of each part near pole is removed. The remaining parts are joined.
S
Screen
A B
(a) P, Q R, S
C
R, S
(b) P, Q P, Q, R, S R, S
(c) Q, S P, R, S Q
(d) P Q, R R
D
P, Q
P
P
P, S
MPP4 CLASS XII
ANSWER
KEY
1.
(c)
2.
(c)
3.
(d)
4.
(d)
5.
(c)
6.
(b)
7.
(b)
8.
(c)
9.
(c)
10.
(c)
11.
(d)
12.
(c)
13.
(a)
14.
(a)
15.
(d)
16.
(c)
17.
(b)
18.
(b)
19.
(a)
20.
(a,b,c)
21.
(a,b)
22.
(b,c)
23.
(a,d)
24.
(2)
25.
(1)
26.
(3)
27.
(a)
28.
(c)
29.
(a)
30.
(b)
Scientists teleport firstever object to space Star Trekstyle
A team of Chinese scientists has teleported a photon over
480 km from the Earth into a special receiver orbiting the planet.
We’re many years away from sending people across galaxies Star
Trekstyle, but it still marks a huge leap in quantum mechanics.
The researchers brought together several disciplines, including advanced physics and rocket science, for the experiment.
Using a process called quantum entanglement, they teleported the photon to a satellite Micius, which was launched last year. On board the satellite is a sensitive photon receiver capable of picking up the elementary particle when it is sent from Earth.
Quantum entanglement occurs when two objects are formed at the same instant and point in space — existing in two separate places at once. Although they are separated, the two objects are immediately influenced by each other regardless of the distance between them. This is known as “wave function“. Thus, one photon in space and one photon on Earth can be created at the same time with a common quantum link. Scientists said this method could be used to transmit data associated with one object over to the other instantaneously . By altering the state of one they can alter the state of the other.
“Longdistance teleportation has been recognised as a fundamental element in protocols such as largescale quantum networks and distributed quantum computation,“ the team said.
Saturn
(Gas Gaint)
Robot lawyer can save you from parking fines
After helping overturn parking tickets worth
$10 million in the UK and the US, the world’s first robot lawyer is ready to offer free legal aid in 1,000 different areas, including tackling rogue landlords, harassment at work and credit card fraud, and getting a refund for malfunctioning electronics.
The online tool was developed by 20yearold British student Joshua Browder, currently enrolled at Stanford University, tw parking tickets and had to fight them. Acc he said the episodes helped him witness o years ago after he received several ording to tech website `Mashable’, misery“ for profit, especially impacting low how many lawyers “exploit human
income people.
Explaining the tool’s expansion, Browder said, “There’s so much exploitation properly. I’m really excited about going on where landlords aren’t behaving how it can help people.“
To seek the robot lawyer’s counsel and assistance, users need to log into
`DoNotPay.com’, where they will be asked to type in their problem. The website, which is currently only available in the US and the UK, then directs users to a chat bot that can solve their legal issue.
on problems ranging from credit
The bot can draft letters and offer advice card fraud to airline compensation, British daily `The Telegraph’ reported.
harassment at work it provides
With problems around parental leave and options with different levels of formality. It can send a casual letter, an official one, and file a complaint to the regulator.
At present the robot lawyer can solve problems that involve a single document. Browder is planning to develop the tool so it can handle more complicated processes, but insists it will always be free.
Jupiter
(Gas Gaint)
TRAPPISTI
Low mass star
EBLM J055557Ab
Low mass star
Smallestever star discovered
star in the universe — slightly larger than Saturn in may possibly have Earthsized planets with liquid water in its orbit.
Researchers from University of Cambridge in the UK years away, identified the star located about 600 light called EBLM J055557Ab as it passed in front of its much larger companion.
The star is likely as small as stars can possibly it has just enough mass to enable the fusion become, as of hydrogen nuclei into helium. The gravitational pull surface is about 300 times stronger than at its stellar what humans feel on Earth.
Courtesy : The Times of India
PHYSICS FOR YOU

AUGUST ‘17
79
ON
VELOCITY OF THE IMAGE
FOR SPHERICAL MIRROR
Er. Sandip Prasad*
When the position of the object changes with time on the principle axis relative to the mirror, the image position also changes with time relative to the mirror.
Hence to know the relation between object and image speed we use the mirror equation.
v
O
y
(c)
f
<
u
<
R
; 
m
 >1 \
v
I
(d)
u
<
f
;
>
v
O

m
 >1 \
v
I
>
v
O
Relation between object and image velocity is also valid for convex mirror. In convex mirror speed of image is slower than that of the object whatever the position of the object may be. This relation is not true in terms of acceleration of object and image.
(
O
,
x y
O
)
O
(
I
,
x y
I
)
I
O
(0, 0)
x
Let us consider an object is moving with speed
v
O
in front of a concave mirror the position of the object at this instant is (
x
O
,
y
O
). An image is formed at location
I
whose coordinate is (
x
I
,
y
I
) at this instant. As usual, we take
x
axis to be the principle axis.
Case I : Component of velocity of image along
x
axis
As,
v u f
or
x
1
I
+
x
1
O
=
f
1
Case II: Component of velocity of image along
y
axis
As we know that transverse magnification. size of the image size of the object
=
y y
I
O
y
Differentiating both sides w.r.t. time,
∴
I
or
=
dy dt
I
dy dt
I
f
=
d dt
=
y f
O
f
dy
y
O
dt
O
+
(
O
−
)
2
du dt
Differentiating both sides w.r.t. time,
−
x
1
I
2
dx dt
I
−
x
1
2
O or
v
Im
= –
m
2
.
v
Om
dx dt
O
= 0 or
dx dt
I
= −
x
⋅
x
I
2
2
O
dx dt
O where,
m
= −
x
I
x
O
v u
magnification
Negative sign indicates the object and image are always moving opposite to each other.
In concave mirror, depending on the position of the object, image speed may be greater or lesser or equal to the object speed.
(a)
R
<
u
< ; 
m
 < 1 \
v
I
(b)
u
=
R
; 
m
 = 1 \
v
I
<
v
O
=
v
O or
v
Im
y
=
f
−
v
Om
y
+
(
O
− )
2
v
Om
x
Where,
v v
Im
y
Om
y
= Velocity of image w.r.t. mirror along
y
axis
= Velocity of object w.r.t. mirror along
y
axis
y
at that instant
v
O
=
y
coordinate or height of the image from
y
axis
Om
x
= velocity of object w.r.t. mirror along
x
axis
Note :
In kinematics, we have studied that the velocity of the objects in
x
and
y
directions are independent of each other. But here, the velocity of the image in
y
direction will depend on its
x
coordinate and velocity in
x
direction also. Remember that these velocities are with sign convention. So the velocity along positive
x
direction will be positive and so on.
*Author is Director of Sandip Physics Classes and motivational speaker
80 PHYSICS FOR YOU

AUGUST ‘17
Some special cases regarding motion of object
Case I : When the object is on the principle axis and moving along principle axis
Component of velocity along
x
axis
(
v
Im
x v
I
= –
m
2
–
v
m
)
x v
Om
x
= –
m
2
(
v
O
–
v
m
)
x
Component of velocity along
y
axis
v y v
Im
Here,
Om
y
O
\
v
Im
y y
=
f
v
Om
y
+
(
f u
)
2
v
Om
x
= 0, since object is only moving along
x
axis
= 0, since object is lie on the principle axis
= 0
Case II : When the object is above the principle axis and moving parallel to principle axis
Component of velocity along
x
axis
v
Im
x
(
v
I
= –
m
–
v
m
)
x
2
v
Om
x
= –
m
2
(
v
O
–
v
m
)
x
Component of velocity along
y
axis
v y
Om
y
O
v
Here,
Im
y
=
f
v
Om
y
+
(
f u
)
2
v
Om
x
= 0, since object is only moving along
x
axis
0, since object is lie above the principle axis
∴
v
Im
y
=
(
y f
−
)
2
v
Om
x
Case III : When the object is on the principle axis and starts moving perpendicular to the principle axis
Component of velocity along
x
axis
v
Im
x
(
v
I
x
= –
–
v
=0 m
)
m x
2
v
Om
x
= –
m
2
(
v
O
–
v
m
)
x
= –
m
\
v
I
Component of velocity along
y
axis
2
(0 – 0)
v y v
Im
y
Here,
Om
y
O
=
v
O
=
f
v
Om
y
+
( − )
2
v
Om
x
, since object is only moving along
x
axis
= 0, since object is lie on the principle axis
∴
v
Im
y
=
f
⋅
v
Om
y
+ 0
=
f
⋅
v
Om
y
Case IV : When the object is on the principle axis and moving with a velocity which makes an angle
q
with the principle axis
82 PHYSICS FOR YOU

AUGUST ‘17
v
O sin
θ
v
O
θ
v
O cos
θ
Component of velocity along
x
axis
v
⇒ (
Im
x
= –
m
I m
)
x
2
v
Om
x
= −
f
2
⋅ (
v
O
−
v
m
)
x
Here, (
v
Here,
v
Om
y
Im
v
O
y
=
v
)
=
O
x
=
v
O
cos q
and (
v f f u
sin q ,
y
v
O
Om
y
+
( m
)
−
x
= 0
Component of velocity along
y
axis:
)
2
v
Om
x
= 0, since object is lie on the
∴
v
Im
y
=
f
⋅
v
Om
y
− 0
=
f
v
Om
y
Case V : When the object is above the principle axis and moving at an angle
q
with the horizontal
v
O cos
y x
Component of velocity along
x
axis
v
(
Im
x
I
Here, (
v
= –
m
−
O
)
x
m
)
x
2
=
v v
Om
x
= −
f
−
2
(
O
cos q
and (
v v
O m
)
−
v
m
)
x
Component of velocity along
y
axis
v
Im
Here,
v y
=
Om
y
f
=
v
O
v
Om
y
+ sin q
and
y
O
(
x
= 0
f u
)
2
v
Om
= constant
x
EXAMPLE 1 :
The velocity of image w.r.t. ground in the given figure is
f
= 30 cm
5 m s
–1
O
(Mirror is at rest given velocity with respect to ground)
20 cm
(a) 45 m s
(b) 45 m s
(c) 60 m s
(d) 60 m s
–1
–1
–1
–1
and approaches the mirror
and moves away from the mirror
and approaches the mirror
and moves away from the mirror
Soln.: (a) StepI:
First of all we need to draw the situation.
y f
= 30 cm
Soln.: StepI:
First of all we need to draw the situation.
StepII:
Place the origin of a coordinate system at the pole.
StepIII:
Find the component of the velocity of the object along
x
and
y
axis.
y
5 m s
–1
O x
20 cm
StepII:
Place the origin of a coordinate system at the pole .
StepIII:
Find the component of the velocity of the object along
x
and
y
axis.
Here, (
v
O
)
x
= 5 m s
–1
, (
v
O
)
y
= 0
StepIV:
Finally we have to apply the formula of velocity of the image along both the axes.
Component of velocity along
x
axis relative to mirror
v
Im
x
= –
m
Where,
m
2
v
Om
x u
.... (i)
f
Eqn. (i) can be written as
⇒
(
I m
)
x
= −
f f u
2
(
v
O
−
v
m
)
x
⇒ (
v
I
− 0 )
x
= −
− 30
30 20 )
2
( 5 0 )
x v
I
x
= –45 m s
–1
v
Here,
v y v
Om
O
Im
y
=
f
v
Om
y
+
(
y
−
axis
)
2
v
Om
x
= 0, since object is lie on the principle axis
Im
y
= 0 , since object is only moving along
x
axis
= 0 + 0; (
v
I
–
v
m
)
y
= 0
\
v
Im
y
= 0
EXAMPLE 2 :
Find the velocity of image w.r.t. ground in situation as shown in figure.
y
15 m s
–1
53°
O
2 m s
–1
30 cm
f
= 20 cm
x
15 m s
–1
O
53°
9 m s
–1
2 m s
–1
x
Here, (
v
O
(
v
O
)
)
x
=15 cos 53° = 9 m s
–1
y
= 15 sin 53° = 12 m s
–1
StepIV:
Finally we have to apply the formula of velocity of the image along both the axes.
Component of velocity along
x
axis relative to mirror
v
Im
x
= –
Where,
m m
2
v
Om
x u f
...(i)
Eqn. (i) can be written as
(
v v
m
)
x
= −
f
2
(
v
O
−
v
m
)
x
⇒
{
v
I
− −
2
x
= −
−
20
20 30 )
2
{ 9
v
I
= –(–2)
2
× 11 = – 44
= –46 m s
–1
Component of velocity along
y
axis
( )}
x v
Im
Here,
y
=
O
y
= 12 m s
–1
f
v
Om
y
+
( − )
2
v
Om
x v y
O
, since object is only moving along
x
axis
= 0, since object is lie on the principle axis
v
Im
y
=
20
− 20
( 30 )
(
v
O
−
v
m
)
y
− 0
(
(
v v
Ig
v
I
y
=
I
I
− m
– 0)
( )
y
)
^
y
+
=
= –24 m s
= –24
–1
( )
j
^
− 20
20 ( 30 )
= −
46 )
i
^
( + −
= Velocity of image w.r.t.ground
( 24 )
)
^
j y
PHYSICS FOR YOU

AUGUST ‘17
83
(
2
−
1
)
× =
2
2
α
B
SOLUTION SET48
1. (c) :
The angular velocity of the rod about the pivot when it passes through the horizontal is given by sin 30 ° =
mL
2
3
×
ω
2
2
ω =
3
2
g
L
Radial acceleration of the centre of mass (as centre of mass is moving in a circle of radius
L
/2) is given by
a r
2
= ω
2
L
2
=
3
4
g
. Torque about pivot, in the horizontal position, is
α =
τ = α
Tangential acceleration of the centre of mass,
2
α
mgL mL
=
2
3
g
4
/
/
2
3
=
3
2
g
L
. From Newton's second law of equation.
=
r
=
3
mg
4
mg R m a t
3
mg
4
or
R
2
=
mg
4
So, reaction force by the pivot on the rod,
R
=
R
[= tan
1
+
–1
R
= 10
mg
4
(1/3)]with the horizontal.
–1
R
2
/
R
1
2. (b) :
In figure (a) (
T
2
–
T
1
) × =
2
2
α
A
T
T a
1
2
–
Mg
=
Ma
A
A
= 2
Mg
A
=
R
α
A
=
2
3
g
R
In figure (b)
T
1
α
A
T
2
(a)
a
A
T
1
Mg
T
1
–
Mg
=
Ma
B
2
Mg
–
T
2
= 2
Ma
B a
B
=
So,
R
B
A
>
, α
B
B
=
2
g
7
R
3. (c) :
1 1 1
b a f
;
f
=
ab
AC
(
a
2
2
+
BC
+
c
2
2
) + (
b
⇒
c
2
=
ab
=
AB
2
2
+
c
2
) = (
a
+
b
)
2
∴
f
=
c
2
4. (d) :
Let acceleration of tube is
a
′ and radius of cylinder is
R
. There is no slipping between tube and cylinders. So, points
A
and
A
′
should have same acceleration.
\
a
′ =
a
+
R
... (i)
Similarly, there is no slipping between cylinder and the ground.
So, points
B
and
B
′
should have same acceleration.
B
B a
Ground
\ 0 =
a
–
R
From eqn. (i) and (ii), we get
a
′
= 2
a
Pressure at
A
is
p
A
=
p
atm
+
gH
Also, pressure at
A
is
H
... (ii)
(from vertical limb) narrow tube
A
L p
A
′ =
p
atm
=
p
atm
ρ
(from horizontal limb)
+ (2
a
)
L
84 PHYSICS FOR YOU

AUGUST ‘17
For no spilling from any opening,
p
A
=
p
\
p
atm
+
gH
=
p
atm
+ (2
a
)
L
=
2
5. (b) :
Time period for half part
T
=
2
π
l g
= π
g
=
π
π
=
2 s
So 2° part will be covered in a time
For the left 1° part : q = q
0 sin
2
T
π
×
t
⇒ sin( w
t
)
1
2
= sin
2 π
2
×
t
1s
⇒
6
t t
1
6 s
Total time
2
2
t
= + × = + =
4
3 s
6. (b) :
From the noninertial frame of tube, force on the liquid element is zero.
From eqn. (i) and (ii), we get
1
2
ρ
v
2
=
ρω
2
2
h
2
2
h
L
−
1
v
= ω
2
− 1
7. (d) :
Here,
T
= 75 dyn cm
= 75 × 10
–3
–1
N m
–1
mg
= 0.1 N, 2
R
= 20 cm
= 0.2 m
F
req
=
mg
+ 2[
T
(2
R
)]
= 0.1 + 2[75 × 10
–3
(0.2)] = 0.130 N
8. (b) :
With respect to the belt, pseudo force
ma
acts on cylinder at COM as shown in figure.
Here,
v
=
a
+
bt
2
dv dt
=
2
bt
Cylinder will be about to topple if pseudo force
=
mg
1
dx
2
v x h
L – h
(
dp
)
A
+ (
A dx
) w
dp
= – w
2
2
L x
= 0 (
A dx
= mass of element)
xdx
⇒
∫
p p
1
2
dp
= − ρω
2
∫
L
L h xdx p
2
p
= −
ρω
2
2
2
ρω
2
[
L
2
L h
(
2
− − +
2
Lh
)
−
2
=
ρω
2
(
2 −
2
)
... (i)
2
On applying Bernoulli’s theorem between points
(1) and (2) we have
p
1
+
1
2
ρ
0
2
=
p
2
+
1
2
ρ
v
2
p p
2
1
2
v
2
...(ii)
mg
9. (a) :
For small angle tan q = sin q q
α =
x
OC
,
β =
x
OC
′
⇒
α
β
=
OC
OC
′ =
1
µ
O
C r x
β µα =
4
3
Bending angle = b
– = 2°
C
10. (b) :
F
F
6°
B h a
A
I
II
Pure rotation about instantaneous point of contact.
Note:
If line of action passes through point of contact, it does not rotate or translate.
PHYSICS FOR YOU

AUGUST ‘17
85
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