Download 9th class bridge course 62-73

CLASS-IX
plate which is silvered at its one surface.
The other surface is then reflecting
surface of the plane mirror.
Laws of Reflection: The reflection at a
surface obeys the following two laws,
which are called the laws of reflection.
1. The angle of incidence ‘i’ is equal to the
angle of reflection r (i.e.  i   r) . In
figure  AON =  BON.
For a ray incident normally on a surface,
 i = 0 0, therefore  r = 0 0. Thus, a ray
of light incident normally on a surface is
reflected back along the same path.
2. The incident ray, the reflected ray and
the normal at the point of incidence, lie
in the same plane.(Plane of paper in
figure)
MPC BRIDGE COURSE
i.e the sum of angle of incidence, angle of
reflection and angle of deviation is 180 0
 d = 180 – (i + r) = 180 – (i + i) (  i=r)
 d  180 – 2i
Therefore, for an angle of incidence i, the
angle of deviation is equal to 180 – 2i =
  2i
Note : The deviation produced by n
reflections from two plane mirrors
inclined at an angle  is given by D =
n(180 –  ) = 360 - 2  , if n=2 where n is
even.
Image: When the rays of light, diverging
from an object point, after reflection or
refraction, either actually meet at some
other point, or appear to meet at some
other point, then that point is called
image of that object.
N
A
DAY-13 : WORKSHEET
In
ci
de
nt
Normal
B
ed
ct
le
ef
R
R
ay
i
M
1. For a ray incident normally on a surface,
angle of incidence is
ay
R
1) 0º
r
M1
O
2) 90º
3) 45º
4) 180º
2. Formula for the angle of deviation due
to reflection d is
Reflection at a plain surface
Formula for the angle of deviation due
to reflection:
In the figure angle of incidence = i; Angle
of deviation = d
N
A
B
i r
M
0
1) 90 – 2i
2) 90 – iº
3) 180 – 2i
4) 360 – 2i
3. The phenomenon of left appearing right
and right appearing left an reflection in
a plane mirror is called
1) magnification
2) lateral inversion
3) virtual
4) none of these
4. The least height of the mirror should be
M1
d
c
1) half of the observer
2) one fourth of the observer
3) same height of the observer
Consider the straight line AOC, i + r +d
= 1800
NARAYANA GROUP OF SCHOOLS
4) 2/3rd of the observer
62
CLASS-IX
MPC BRIDGE COURSE
5. If the mirror is displaced by a distance
x, towards or away from the object, the
image is displaced by
1) 4x
2) x
3) 3x
DAY-14 : SYNOPSIS
Mirro rs: A smooth, highly polished
reflecting surface is called a mirror. One
surface of the mirror is made opaque by
silvering followed by a thin coat of red
lead oxide paint. There are two types of
mirrors.
4) 2x
6. Number of images formed for an object
kept in between two mirrors placed
perpendicular to each other
1) 2
2) 3
3) 4
Plane mirrors:A highly polished plane
surface is called a plane mirror.
4) 5
7. If the object is displaced by a distance x
towards or away from the mirror, the
image suffers displacement
1) x
2) 2x
3) 3x
Spherical mirrors: A mirror in which
the reflecting surface is curved is called
a spherical mirror.
4) 4x
8. If a plane mirror is rotated through an
angle  , the reflected ray rotate
through an angle.
1)  /2
2) 2 
3) 4 
reflecting
surface
4) 
9. A light ray is made to incident on a glass
plate with an angle of incidence 30º.Find
the angle of reflection is
1) 60º
2) 30º
3) 90º
2) 300º
3) 90º
2) 120º
3) 145º
4) 60º

Aperture: The width (distance) of the
spherical mirror from which reflection
can take place is called its aperture. It is
denoted by MM

Pole: The centre of a spherical mirror
is called its pole. It is denoted by P.

Centre of curvature: The geometric
centre of the hollow sphere of which the
spherical mirror is a part is called the
centre of curvature of the spherical
mirror. It is denoted by C.

Radius of curvature: The radius of the
hollow sphere of which the spherical
mirror is a part is called the radius of
curvature of the spherical mirror. In
other words, the
12. A plane mirror reflects a beam of light
to form a real image. The incident beam
is
1) parallel
2) convergent
3) divergent
4) any one of the above
13. When an object is placed between two
parallel mirrors, then number of images
formed is
1) 2
2) 4
3) 8
NARAYANA GROUP OF SCHOOLS
b. Spherical mirror
In spherical mirrors the polished
reflecting surface is a part of a hollow
sphere of glass. Depending upon the
nature of the reflecting surface of the
mirror, spherical mirrors are of two
types.
4) 60º
11. A ray of light, after reflection from a
plane mirror, suffers a deviation of 60º.
Find the angle between the incident
and reflected rays.
1) 130º
a. Plane mirror
4) 0º
10. The two mirrors are inclined at an angle
30º. If a ray of light is obliquely incident
on the first mirror, the deviation after
two reflections is
1) 250º
reflecting
surface
4) infinite
63
CLASS-IX
MPC BRIDGE COURSE
M
X
C
M
P
F
X
C
Note: 1. By sign convention, for concave
mirror f  ve and for convex mirror
P
F
f   ve
2. The relation between focal length and
M'
M'
a.Concave mirror
radius of curvature is R =
b.Convex mirror
distance between the pole and centre of
curvature of the spherical mirror (PC) is
called its radius of curvature. It is
denoted by r.
3. mirror formula is
f
2
1 1 1
 
f v u
Principal axis: The straight line
DAY-14 : WORKSHEET
mages formed by a concave mirror for
different positions of the object:
1. A spherical mirror whose inner hollow
surface is the reflecting surface is called

Position of the object
Position of the image
Nature and size of
the image
At infinity
At the focus (F)
Real, inverted and very
small (highly diminished)
Beyond the centre of
curvature (C)
Between the focus(F)
Real, inverted and
and centre of curvature(C) diminished
At the centre of
curvature (C)
At the centre of
curvature (C)
Real, inverted, same size
as the object
Between the centre
of curvature (C) and
focus(F)
Beyond the centre
of curvature (C)
Real, inverted, bigger
than the object(magnified)
At the focus (F)
At infinity
Real, inverted and
enlarged (highly magnified)
Between the focus
(F) and Pole (P)
Behind the mirror
Virtual, erect and enlarged
(magnified)
mirror for different positions of the
object
Nature and size of
the image
Position of the object
Position of the image
Between the focus(F)
and infinity
Behind the mirror
between P and F
Virtual, erect and
diminished
Behind the mirror at a
distance of f/2 from the
pole
Virtual, erect and
diminished
Between the focus(F)
and the pole (P)
Behind the mirror
between P and F
Virtual, erect and
diminished
At infinity
Behind the mirror at
the focus (F)
At the focus(F)
Virtual, erect and highly
diminished (point size)
Focus: If a beam of light parallel to the
principal axis falls on a concave mirror,
all the rays after reflection meet at a
point. This point is called the focus (F) of
the concave mirror.
NARAYANA GROUP OF SCHOOLS
2) convex mirror
3) plane mirror
4) none of these
2. The centre of the speherical mirror is
called
1) focus
2) pole
3) centre of curvature 4) none of these
3. Any ray of light travelling parallel to the
principal axis of a concave mirror, after
reflection passes through
1) pole of mirror
2) focus of mirror
3) centre of curvature of mirror
4) in between pole and focus of mirror
4. Virtual effect and enlarged image will
be formed by a concave mirror when
object is placed in between
1) infinity and centre of curvature
2) focus and centre of curvature
3) pole and focus
passing through the centre of curvature
and the pole of a spherical mirror is
called its principal axis (PX).

1) concave mirror
4) at centre of curvature
5. When object is placed between centre of
curvature (c) and focus (f) of a concave
mirror the nature and site of the image
is
1) real, inverted and very small
2) real, inverted, same site of object
3) real, inverted and bigger than the
object
4) virtual, erect and enlarged
64
CLASS-IX
MPC BRIDGE COURSE
6. Real, inverted and same site as the
object, image will be formed by concave
mirror, when the object is at
1) centre of curvature
2) focus
3) between focus and pole
4) beyond centre of curvature
7. The image formed by a convex mirror of
real object is larger than the object.
1) When u  2f
2) When u  2f
3) for all values of u 4) for no value of u
8. The diameter of spherical mirror in which
reflection takes place is called 1
)
radius of curvature 2) centre of curvature
3) linear aperture. 4) focal length.
9. When object is placed between principle
focus and pole for a concave mirror the
image is formed at
1) pole
DAY-15 : SYNOPSIS
Principle of Calorimetry : (or Law of
Mixtures)
When
two
bodies
at
different
temperatures are brought in contact with
each other the heat flows from the body at
higher temperature to that at lower
temperature until the temperatures of
both bodies become equal. Then as there
is no change of state and no heat is lost/
gained to/ from the surroundings, the
heat lost by the hot body must be equal to
the heat gained by the cold body. Thus,
during transfer of heat from a hot body to
the cold body, heat energy must be
conserved. This is the principle of
calorimetry.
Mathematically,
Heat lost by the hot body = Heat gained by
the cold body. ( provided there is no loss of
heat to the external surroundings)
Note : Heat lost = heat gained, will hold
good only if there is no change of state
and no heat is lost/gained to / from the
surroundings.
Thermal Equilibrium :
2) principal focus
3) centre of curvature
Two bodies in contact will be in thermal
equilibrium only when they are at the
same temperature.
4) behind the mirror
Units of specific heat :
10. The rear-view mirror of a car is
1)Plane
2) Convex
3) Concave 4) either convex or concave
11. The focal length of spherical mirror is
S.I. unit of specific heat = J kg –1 K–1
CGS unit of specific heat is cal g –1 °C–1
Since 1cal = 4.18 J therefore 1 cal g –1
°C–1 = 4180 J kg –1 K–1
Note :Water is a liquid having highest
specific heat capacity i.e.,4180 J kg –1K–1.
1).Maximum for red light
Heat capacity or Thermal capacity :
2) Maximum for blue light
Heat capacity of a body is defined as the
amount of heat required to raise the
temperature of the (whole) body through
1°C or 1K.
3) Maximum for white light
4). Same for all lights
NARAYANA GROUP OF SCHOOLS
Heat capacity = heat required to rise the
temperature of the body through 1°C
65
CLASS-IX
MPC BRIDGE COURSE
Mathematically, heat capacity=
DAY-15 : WORKSHEET
amount of heat
Q
Q
=m×C.
=
H
Rise in temperature t
t
Hence Thermal Capacity = mass ×
specific heat.
Units of heat capacity :
S.I. unit of heat capacity is J K
CGS unit of heat capacity is cal °C –1
Note : Thermal capacity depends on mass
of the substance whereas specific heat
does not depend upon the mass of the
substance.
(OR)
PRINICPLE: If two liquids at different
temperatures are mixed together, the
heat loss by hot body is equal to the heat
gain by the cold body. This is called law
of mixture.
 When three substances of different
masses m1, m2 and m3 specific heats s1,
s2, s3 and at different temperatures 1 ,
2 , and 3 respectively are mixed, then
the
resultant
temperature
 If two substances of specific heats s1 , s2
having masses m1 , m2 are mixed at the
same temperature, effective specific heat
m1s1  m2 s2
m1  m2
When “x” gram of steam is mixed with
“y” gram of ice, the resultant temperature
80(8 x  y )
is t 
( x  y)
NARAYANA GROUP OF SCHOOLS
2) temperature
3) force
4) mass
1) latent heat
2) Specific heat
3) Heat
4) Heat capacity
3. The mathematical expression for specific
heat “S”
1)
Q
m
2)
Q
t
3)
Q
m.t
m
t
Q
4)
4. The amount of heat required to raise the
temperature of given mass of a substance
through 1°C is called
1) Heat capacity 2) Thermal capacity 3 )
Specific heat
4) both (1) and (2)
5. The relation between heat capacity ‘Q’
and specific heat ‘S’ is
is
m1s11  m2 s22  m3s33
 = m s m s m s
1 1
2 2
3 3
of the mixture is s 
1) heat
2. The amount of heat required to raise the
temperarue of unit mass of a substance
through 1°C is called
–1
LAW
OF
MIX TURES
CALORIMETRY
1. Calorimetry is the measurement of
1) Q = m.S
3) Q 
S
m
2) Q 
m
S
4) Q  m  S
6. The C.G.S unit of specific heat is
1) cal g °C–1
2) cal g–1 °C–1
3) cal g–1 °C
4) cal–1 g °C
7. S.I unit of thermal capacity is
1) JK
2) JK–1
3) cal K
4) cal °C
9. The substance with the highest specific
heat capacity is
1) Mercury
2)Gold
3)Tungsten
4)Water
66
CLASS-IX
MPC BRIDGE COURSE
10. The ratio of thermal capacity per unit
volume in terms of densities and specific
heats is
1) 2d1s1 = d2s2
2) d1s1 = 2d2s2
3) d2s1 = d1s2
4) d1s1 = d2s2
11. If two substances of masses m 1, m 2
specific heats s1, s2 at initial temperature
1
and
2
are
mixed
then
temperature of mixture is
losses)
1)
m1s11  m2s 22
m1s1  m2s 2
m1s11  m2s 22
3)
m1s1  m2s 2
2)
final
(no heat
m1s1  m2s2
m1  m2
12. If two liquids of specific heats s1,s2 having
masses m1,m 2 are mixed at the same
temperature, effective specific heat of the
mixture (s) is
m2s1  m1s2
2) m  m
1
2
m1s1  m2s2
3) m  m
1
2
m2s1  m1s2
4) m  m
1
2
2) 2 : 1
3) 1 : 4
4) 4 : 1
2) 30 cal/0C
3) 300 cal/0C
4) 0.3 cal/0C
1
2
3)
1
4
NARAYANA GROUP OF SCHOOLS
4)
mass
or
volume

m
v
In case of a liquid, sometimes an another
term relative density (R.D.) is
defined. It is the ratio of density of the
substance to the density of water 4°C.
Density of subs tance
Hence, R.D  Density of water at 4C
iii) Density of a mixture of two or more
liquids
Here, we have two cases.
Case 1: Suppose two liquids of densities
15. Two spheres of copper of diameters
10cm and 20 cm will have thermal
capacities in the ratio
2)
3. DENSITY OF LIQUID
ii) Relative Density (R.D)
1) 3 cal/0C
1
8
While dealing with fluids we are more
interested in properties that vary from
point to point in the extended substance
rather than properties of a small piece
of the substance. This is why we talk
about density and pressure rather than
mass and force in case of fluids.

14. The thermal capacity of 100g lead shot
is (specific heat of lead is 0.03 cal/g/ 0C)
1)
The substances which flow are called
fluids. Fluids include both liquids and
gases. The science of fluids at rest is
called fluid statics while that of moving
fluids is hydro-dynamics. Fluid statics
includes hydrostatic pressure, floatation,
Pascal’s law and Archimedes’ principle
while hydrodynamics includes continuity
equation Bernoullis principle and
Torricell’s theorem. This all the subject
of this chapter.
i) Density (  ) of any substance is defined
as the mass per unit volume or
13. The densities of two substances are
in the ratio 5 : 6 and their specific heats
are in the ratio 3 : 5. Then ratio of their
thermal capacities per unit volume is
1) 1 : 2
1. INTRODUCTION
2. DENSITY AND PRSSURE
m1s1  m2s2
4) m  m
1
2
m1s1  m2s2
1) m  m
1
2
DAY-16 : SYNOPSIS
1 and 2 having masses, m1 and m2 are
mixed together. Then the density of the
mixture will be
1
6
67
CLASS-IX

 m1  m2    m1  m2 
Total mass

Total volume
 V1  V2   m1  m2 


2 
 1
212
If m1 = m2, then     
1
2
5. Absolute Pressure and Gauge Pressure
Case 2: If two liquids of densities 1 and
2 having volumes V1 and V2 are mixed,
then the density of the mixture is,


If V1 = V2, then  
1  2
2
3. PRESSURE IN A FLUID
Consider a small surface of area dA
centered on a point on the fluid, the
normal force exerted by the fluid on each
side is dF^ . The pressure P is defined at
that point as the normal force per unit
P
area, i.e.,
dF
dA
If the pressure is the same at all points
of a finite plane surface with area A, then
P
The excess pressure above atmospheric
pressure is usually called gauge
pressure and the total pressure is called
absolute pressure. Thus,
Gauge pressure = absolute pressure –
atmospheric pressure
 m1  m2 
Total mass

Total volume
 V1  V2 
1V1  2V2
V1  V2
MPC BRIDGE COURSE
Note: Fluid pressure acts perpendicular
to any surface in the fluid no matter how
that surface is oriented. Hence, pressure
has no intrinsic direction of its own, its
a scalar. By contrast, force is a vector
with a definite direction.
F
A
DAY-16 : WORKSHEET
1. The pressure inside a liquid of density
d at a depth ‘h’ below its surface is
h
1) d g
d
3) h g
2) hdg
4)
hg
d
2. Pressure at any point inside a liquid is
1) directly proportional to density of the
liquid
2) inversely proportional to density of the
liquid
3) directly proportional to square root of
density of the liquid
Where F is the normal force on one side
of the surface. The SI unit of pressure is
pascal, wh er e 1 pascal=1 Pa=1.0 N/ m 2
One unit used principally in meterology
is the Bar which is equal to 105 Pa.
4) inversely proportional to square of density of liquid
3. Liquid pressure at a point in a liquid does
not depend on
1) density of liquid
5
1 Bar = 10 Pa
4. Atmospheric Pressure (P0)
It is pressure of the earth’s atmosphere.
This changes with weather and
elevation. Normal atmospheric pressure
at sea level (an average value) is
1.013 × 105 Pa. Thus, 1 atm = 1.013 × 105
Pa = 1.013 Bar
2) shape of the vessel in which the liquid
is kept
3) depth of the point from the surface
4) acceleration due to gravity
4. As the depth of a liquid increases, the
pressure of liquid
1) decreases
2) increases
3) remains same 4) cannot say
NARAYANA GROUP OF SCHOOLS
68
CLASS-IX
MPC BRIDGE COURSE
5. Pressure at a certain depth in river
water is P1, and at the same depth in
sea water is P2. Then [ here density of
sea water is greater than that of river
water)
1) P1 = P2
2) P1 > P2
3) P1 < P2
4) P1 - P2 = atmospheric pressure
6. If the total pressure inside a base of
liquid tank is 3 atm, then calculate the
pressure due to water column in tank.
[atmospheric pressure = 1 atm]
1) 1 atm
2) 2 atm
3) 3 atm
4) 0.5 atm
7. The height of mercury in a barometer
is 760 mm,then the atmospheric pressure is [g=9.8 m / s 2]
1) 0.101  105 Pa
2)10.1  105 Pa
3) 1.01  105 Pa
4) 1.01  104 Pa
8. Pressure is a scalar quantity because
1) pressure is always compressive in
nature.
2) a vector (force) divided by a scalar
(area) is a scalar quantity.
3) a vector (force) divided by a vector (area)
is a scalar quantity.
4) a scalar (force) divided by a vector
(area) is a scalar quantity.
9. If density of kerosene in C.G.S is 0.8
gm/cm3 then density in S.I is
1) 800 kg/m3
2) 1000 kg/m3
3) 600 kg/m3
4) 1200kg/m3
10. Mass of stone is 900 gm dropped in a
liquid and density of the stone is
3
g
cm3
. Then volume of the stone is
1) 200 cm3
2) 300 cm3
3) 400 cm3
4) 100 cm3
NARAYANA GROUP OF SCHOOLS
11. Water of density 4kg/m3 and ice of
density 2 kg/m3 are mixed together. If
their masses are equal then the density of mixture is
1)
8
kg m–3
3
2)
7
kg m–3
3
3)
5
kg m–3
3
4)
2
kg m–3
3
DAY-17 : SYNOPSIS
Pascal’s Law: It states that “pressure
applied to an enclosed fluid is transmitted
undiminished to every portion of the fluid
and the walls of the containing vessel”.
A well known application of Pascal’s law
is the hydraulic lift used to support or
lift heavy objects. It is schematically
illustrated in figure.
F1
A1
CAR
A2
A piston with small cross section area A1
exerts a force F1 on the surface of a liquid
F1
such as oil. The applied pressure P  A is
1
transmitted through the connecting pipe
to a larger piston of area A2. The applied
pressure is the same in both cylinders,
F1 F2
so P  A  A
1
2
A2
or F2  A .F1
1
Now, since A2 > A 1, therefore, F2 > F 1.
Thus, hydraulic lift is a force multiplying
device with a multiplication factor equal
to the ratio of the areas of the two pistons.
Dentist’s chairs, car lifts and jacks, many
elevators and hydraulic brakes all use
this principle.
69
CLASS-IX
MPC BRIDGE COURSE
2. PRESSURE DIFFERENCE IN
or
ACCELERATING FLUIDS
Consider a liquid kept at rest in a beaker
as shown in figure(a). In this case we
know that pressure do not change in
horizontal direction (x-direction), it
decreases upwards along y-direction. So,
we can write the equations,
dP
dP
 g
 0 and
dy
dx
y
  Ag dy    dP  A   A dy   a y 
dP
or dy    g  a y 
Similarly, if the beaker moves along
positive x-direction with acceleration ax,
the equation of motion for the fluid
element shown in figure is,
O
ax
P+dP
P
(P+dP)A
PA
A
dx
ax
x
 A dx  a x
or
ax
x
or
PA – (P + dP)A = (mass)(ax) or –(dP)A
ay
o
 A dy   a y 
(dP)A =
y
But suppose the beaker is accelerated
and it has components of acceleration ax
and ay in x and y directions respectively,
then the pressure decreases along both
x and y directions. The above equation
in that case reduces to,
y
–W –
x
o
=
dP
 a x
dx
DAY-17 : WORKSHEET
1. Oil can be extracted from the oil seeds
using a device called
(a)
dP
dP
 a x and
   g  a y 
dx
dy
1) Simple machine 2) Hydraulic machine
3) Both (1) and (2) 4) Neither (1) nor (2)
These eqations can be derived as
follows: Consider a beaker filled with
some liquid of density  accelerating
upwards with an acceleeration along
positive y-direction. Let us draw the free
body diagram of a small element of fluid
of area A and length dy as shown in
figure.
y
(P+dP)A
P+dP
A
P
dy
ay
A
x
PA
Equation of motion for this fluid element
is, PA – W – (P + dP)A = (mass)(ay)
NARAYANA GROUP OF SCHOOLS
2. ______________ is used for punching holes
in metals.
1) hydraulic brakes
3) Both (1) and (2)
2) hydraulic press
4) Neither (1) nor (2)
3. Hydraulic lift works on the principle of
1) Charle’s law
2) Boyle’s law
3) Newton’s law
4) Pascal’s law
4. Which of the following is used for
servicing automobils in service stations
1) Hydraulic press 2) Bramah press
3) hydraulic brake 4) simple machine
5. The ______________ exerted at any point
in an enclosed and incompressible liquid
is transmitted equally is all directions.
1) Pressure
2) Temperature
3) area
4) force
70
CLASS-IX
MPC BRIDGE COURSE
6. Pascal’s law is used to multiply force in
machines such as
1)hydraulic press2) Bramah press
3) hydraulic lift 4) all of these
7. According to Pascal’s law, pressure is
isotropic, which means
1) pressure is same at all the points.
2) pressure is same at all the points lying
in the same horizontal plane.
3) pressure at a point is same in all
directions.
DAY-18 : SYNOPSIS
Electric current
An isolated metallic conductor, say a
wire, contains a few electrons which are
moving at random with high speeds.
These are called conduction electrons.
The rate at which these electrons pass
from left to right through a point in a
wire is the same as the rate at which
they pass from right to left through the
same point, i.e., net rate is zero.
4) pressure can change only in specific
directions.
8. A force of 50 kgf is applied to the smaller
piston of a hydraulic machine. Neglecting
friction, then find the force exerted on
the large piston, the diameters of the
piston being 5 cm and 25 cm respectively.
1) 250 kgf
2) 750 kgf
3) 1250 kgf
4) 1000 kgf
1) 100 N 2) 10 N 3) 150 N 4) 1000 N
10.Two strings of different area of cross
sections filled with water are connected
with a tight rubber tube filled with water. Radius of the smaller piston and
larger piston are 2 cm and 4 cm.Find the
force exerted on the larger piston when
force of 100 N is applied to the smaller
piston
3) 300N
4) 400N
11. If a room has dimensions 3 m × 4 m × 5
m, what is the mass of air in the room if
density of air at NTP is 1.3 kg/m3 ?
1) 57 kg
2) 78 kg
3) 65 kg
4) 1 kg
12. What force does water exert on the
base of a house tank of base area 1.5 m2
when it is filled with water up to a height
of 1 m ? [Density of water is 103 kg/m3
and g = 10 m/s2]
1) 104 N
2) 1.5 × 104 N
3) 2 × 104 N
4) 2.5 × 104 N
NARAYANA GROUP OF SCHOOLS
+
+
A
+
I
9. The areas of the pistons in a hydraulic
machine are 5 cm2 and 625 cm2. What
force on the smaller piston will support
a load of 1250 N on the larger piston ?
1) 100 N 2) 200N
+
+
To define the current mathematically,
suppose charged particles are moving
perpandicular to a surface of area A as
in figure (this area could be the crosssectional area of a wire). The current is
defined as the rate at which electric
charge flows through this surface. If Q is
the amount of charge that passes
through this area in time interval t ,
then average current Iavg, over this time
invervel through this area is the ratio of
the charge to the time interval.
i.e.,
Iavg 
Q
_______________(1)
t
It is possible for the rate at which the
charge flows to vary with time. We define
the instantaneous current I as the limit
of the preceding expression as t goes to
zero;
I  Lt
t 0
Q
dQ

____________(2)
t
dt
When a wire is connected to a battery,
an electric field is set up at every point
within the wire. This field exerts a force
71
CLASS-IX
on each conduction electron. Although
the
electrons
are
continuously
accelerated by the field, but due to their
frequent collisions with atoms of the
wire, they on an average simply drift at
a small contact speed in the direction
opposite to the field. Thus, there isa net
flow of charge inthe wire at a small rate.
The total charge passing through any
cross-section per second is the electirc
current in the wire.
MPC BRIDGE COURSE
To consider the current density at a point
P of the conductor, we draw a small area
S through P perpendicular to the flow
of charges as shown in Figure. Let i be
the current through area S , then
average current density is given by
jav 
i
s
........(1)
+
In the steady state of current through
each section of the conducting loop would
be same-no matter what is the location
or orientation of area of that section see
figure. This is because of the fact that
charge is conserved.
P
i
+
J
S
So, the current density at point P is
i
di

s 0 S
dS
j  lim
I
Unit of Electric Current
It is ampere (A) in S.I system.
“The current is said to be one ampere
when one columb of charge flows past
any cross-section of a conductor every
second.”
Current Density
Current i is a characteristic of a
particular conductor. This is a
macroscopic quantity like mass, volume,
length, etc. The current is a scalar
quantity. A related microscopic quantity
is the current density J.
This is a vector quantity and is
characteristic of a point inside the
conductor rather than conductor as a
whole. The current density J at a point
is defined as a vector having magnitude
equal to current-per unit area
surrounding that point and normal to the
direction of charge flow, i.e., direction
in which current passes through that
point.
NARAYANA GROUP OF SCHOOLS
......(2)
If the current i is uniformly distributed
over an area S and is perpendicular to
it, then j 
i
S
...(3)
Now consider the situation in which the
cross-sectional area S is not normal to
direction of current flow. Let S is
oriented at an angle  to the direction
of current flow as shown in the figure.
The current density is given by
j
i
Scos 
or i  iScos 
Where i is the current through S .
+

i
S
j
+
Scos
72
CLASS-IX
If S be the area vector corresponding to
area S , then i  j  S
The total current through finite surface
 
S is i   j  dS
s
It is clear from above equation that
current is an scalar as the integral j.dS
is a scalar. The arrow associated with
current does not indicate that it is a
vector but this simply shows the sense
of charge flow. On the other hand, current
density is a vector quantity.
The SI unit of electric current density is
coulomb per second metre2 (C/s-m2).
DAY-18 : WORKSHEET
1. The conventional electric current flows
from
1) higher potential to lower potential
2) lower potential to higher potential
3) higher potential to higher potential
4) lower potential to lower potential
2. Electric current is a
1) Scalar
2) Vector
3) Tensor vector
4) Psedo vector
3. Electric current has
1) Both magnitude and direction
2) only magnitude and no direction
3) no magnitude but it has direction
4) no magnitude and direction
4. The direction of conventional electric
current is
1) is opposite to the direction of flow of
electrons
2) is same as that of direction of flow of
electrons
3) has no direction
4) can’t say
5. The direction of electronic current is
1) is opposite to the direction of flow of
electrons
2) is same as that of direction of flow of
electrons
3) has no direction
MPC BRIDGE COURSE
6. The Random motion of free electrons
1) doesnot contribute to their drift velocity
2) contribute to their drift velocity
3) Both (1) and (2)
4) neither (1) nor (2)
7. A large number of free electrons are
present in metals. Why is there no
current in the absence of electric field
across it ?
1) motion of free electrons is random
2) motion of free electrons is in a
perticular direction
3) both (1) and (2)
4) neither (1) nor (2)
8. A steady current is passing through a
linear conductor of nonuniform crosssection. The net quantity of charge
crossing any cross section per second is
1) independent of area of cross-section
2) directly proportional to the length of
the conductor
3) directly proportional to the area of
cross section.
4) inversely proportional to the area of
the conductor
9. What is the number of electron that
constitutes a current of one ampere
1) 6.25 × 10–18 s–1
2) 6.25 × 1018 s–1
3) 8.22 × 10–12 s–1
4) 8.12 × 1022 s–1
10. A wire carries a current of 2.0 A. What
is the charge that has flowed through
its cross – section in 1.0 sec.
1) 2C
2) 3C
3) 4C
4) 5C
11. In the above question, How many
electrons does this correspond to ?
1) 1.25 × 1019
2) 1.6 × 108
3) 2.3 × 109
4) 3.2 × 108
12. Two wires of the same material but of
different diameters carry the same
current I. If the ratio of their diameters
is 2 : 1, then the corresponding ratio of
their mean drift velocities will be
1) 4 : 1
2) 1 : 1
3) 1 : 2
4) 1 : 4
4) can’t say
NARAYANA GROUP OF SCHOOLS
73