Tips Tricks Shortcut Methods for Jee Physics
Short Description
Tips Tricks Shortcut Methods for Jee Physics...
Description
SYNOPSIS AND SHORTCUTS FOR JEE
Sr.No.
Subjects
1.
Physics
2.
Chemistry
3.
Mathematics
Page No. 1 - 79 80 - 155 156 - 210
1
SPECIAL MODULE (PHYSICS)
TIPS, TRICKS & SHORTCUT METHODS
PHYSICS Mechanics Vectors o o
1.
Au B
o
o o
2.
o
o
o
C and then C A A as well as B . o o
A B when T
Au B
450 .
3.
The rectangular components can’t have magnitude greater than vector
4.
itself ' cos T d 1, T JJG JJJG JJJG JJJG If A1 A 2 A 3 ....... A n A1
A2
A3
.......
each other at angle
0.
A n , then the adjacent vectors can be inclined to 2S . 9
5.
If any two vectors are parallel or equal, then the scalar triple product is zero.
6.
The magnitude of P u Q can vary from maximum value
o o
o
o
o
P Q
to
o
minimum value P Q . 7.
While finding the angle between two vectors one should check that the two vectors are directed towards the point or away from point.
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2 8.
SPECIAL MODULE (PHYSICS)
For resultant of two vectors of equal magnitude. T
Magnitude of resultant
600q
3a
900
2a
1200
a
where ‘a’ stands for magnitude of each vector. o
9.
o
o
o
o
o
a O b , (a) a is parallel to b , if O > 0 ; (b) a is antiparallel to b , if O < 0.
Significant figures and error analysis 1.
The limit of accuracy of a measuring instrument is equal to the least count of the instrument.
2.
When a quantity is squared, the number of significant digits is not squared. Algebraic operations with significant figures General ruleFinal result shall have significant figure corresponding to the number of significant digits in the least accurate variable. (i)
Addition and subtraction Suppose in the measured values to be added or subtracted the least number of significant digits after the decimal is n. Then in the sum or difference also, the number of significant digits after the decimal should be n. Example:
1.9 + 2.77 + 3.456 = 8.126 | 8.1 (to correct significant digits)
Here the least number of significant digits after the decimal is one. Hence, the result will be 8.1 (when rounded off to smallest number of decimal places.) Example:
17.36 - 11.4 = 5.96 | 6.0 (to correct significant digits) Lakshya Educare
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SPECIAL MODULE (PHYSICS)
(ii)
Multiplication or division Suppose in the measured values to be multiplied or divided the least number of significant digits be n. Then in the product or quotient, the number of significant digits should also be n. Example:
2.5 u 13.41 = 33.525 | 34 (to correct significant digits)
The least number of significant digits in the measured values are two. Hence the result when rounded off to two significant digits become 34. Therefore the answer is 34. Example:
3570 = 313.157895 | 313 11.4
3.
When two quantities are multiplied, the maximum relative error in the result is the sum of maximum relative errors in those two quantities.
4.
When we are considering result involving quotient of two quantities, the maximum relative error in the result is the sum of maximum relative errors in those quantities.
Projectile Motion H
-
Maximum Height
T
-
Time of Flight
R
-
Range
ux
-
Initial Velocity along x-direction
uy
-
initial Velocity along y-direction
uy = u sin T , ux = u cos T , ax = 0 and ay = -g
?
H=
u2 sin2 T 2g
T=
2usin T g
uy
gT 2
u2y 2g
…(1)
2uy g …(2) Lakshya Educare
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SPECIAL MODULE (PHYSICS)
Sub. (2) in (1) we get, H=
g 2 T2 8g
R=
u2 sin 2T g
= 2
gT 2 8
2u2 sin T cos T g
2u2 sin2 T cot T 2g
R = 4 H cot T
[Relation between H and T]
§ u2 sin2 T · 4¨ ¸¸ cot T ¨ 2g © ¹ [Relation between R and H]
…(3)
[Relation between R and H]
…(4)
From (3) and (2) R= 4
R=
gT 2 cot T 8
gT 2 cot T 2
tan D
H R 2
tan D
2H R
2
u2 sin2 T g u 2 2g u sin 2T
tan D
sin2 T 2sin T cos T
tan T 2
For body projected from height H with horizontal velocity – Ux = u t=
uy = 0
2H g
sx = u
2H g Lakshya Educare
5
SPECIAL MODULE (PHYSICS)
1.
If a body is dropped from the Aeroplane moving with horizontal velocity, then the body will also have horizontal velocity because of the horizontal velocity of the Aeroplane and problem reduces to body projected from height H with horizontal velocity.
2.
In case of throwing a particle in a moving train. The particle has horizontal velocity because of the motion of the train and particle will have projectile motion. If one is interested in motion respect to the train, then it will not be projectile motion [observer is in ground frame].
Projectile motion on inclined planeRange of the projectile on the inclined plane. R=
2u2 sin D E cos D g cos 2 E
Time of flight T=
2usin D E . g cos E
R=
g T 2 cos D . 2sin D E
[Relation between ‘R’ and ‘T’ on inclined plan] 3.
When tan D
2 tan E particle strikes the plane horizontally.
4.
When cot E
2 tan D E particle strikes the plane at right angles.
Relative Velocity 1.
Identify the observed body and two observers.
2.
Find out what velocities/displacements are given in problem.
3.
Draw the velocity/displacement vectors with suitable co-ordinate system.
4.
Use the relative velocity equation : v AB
o
o
o
v AC v CB
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SPECIAL MODULE (PHYSICS)
Newton’s Law 1.
If a body is in equilibrium, then it does not mean that no force acts on the body but it simply means that the net force (resultant of any number of forces) acting on the body is zero.
2.
F
d § o· ¨m v ¸ dt © ¹
If m is constant, then F
m
dv dt
3.
Mass of a body is a measure of the resistance offered by the body to the change in velocity of the body. In other words, it is a measure of inertia of the body.
4.
Whenever a body loses a contact with the surface, the normal force becomes zero. In problems where a body loses contact, this concept should be used.
5.
Area under the force time graph gives magnitude of impulse of the given force in given time.
6.
Frictional force varies depending on whether the body is in motion or not -
If the body is at rest with respect to the surface then f < P s N
-
If the body is just in motion with respect to the surface f
-
If the body is in motion with respect to the surface f
P sN
Pk N
7.
If the maximum force of friction is greater than the applied force, then the force of friction will be equal to the applied force.
8.
Acceleration of a body sliding down an inclined planea = g(sin D - P k cos D )
If sin D ! P s cos D P s tan D
a=0
If
P s t tan D
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7 9.
SPECIAL MODULE (PHYSICS)
Work done in moving a body up an inclined plane through a distance s along the incline W = mg (sin T + P k cos T )x x = distance
10.
If the body is moved down an inclined plane with constant speed then work done is given by W = mg P k cos T sin T x
Tension 1.
Tension force always pulls a body.
2.
Tension across massless pulley or frictionless pulley remains constant.
Application of Newton’s law to circular motion tion type of problems 1.
Draw free body diagram.
2.
Identify the direction of acceleration
3.
Write equation of motion.
To find out the direction of normal force Normal force will perpendicular to the surface of the body
or If perpendicular to the surface of contact can’t be drawn, the normal force will act perpendicular to the body. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
or If neither can be done, normal force has to be drawn as two components one in the x-direction and one in the y-direction. Example: What minimum velocity v must be given to the solid cylinder of radius r, so that continues its motion without a jump down the incline.
Solution: In the given position, Let Z be angular velocity of the cylinder about the point O. ?
mg(r – r cos T ) =
mgr(1 - cos T ) =
g(1 - cos T ) =
1 I Z2f Z2i 2 1 § mr 2 v2 · 2 ·§ 2 mr ¨ ¸¸ ¨¨ Z ¸ 2 ¨© 2 r 2 ¸¹ ¹©
1 § 3r · § 2 v 2 · ¨ ¸ ¨ Z 2 ¸¸ 2 © 2 ¹ ©¨ r ¹
4 v2 = Z2 r g (1 - cos T ) + 3 r
…(1)
Writing force equation, -N + mg cos T = m Z2 r
…(2)
(Note: Normal reaction passes through the centre of the cylinder.) Using equation (1) and (2), ª 4g v2 º N = mg cos T - m « 1 cos T » r »¼ «¬ 3 Lakshya Educare
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SPECIAL MODULE (PHYSICS)
N=
7mg cos T 4mg mv 2 3 3 r
N t 0 v2 t g
7 cos T 4 r 3
At critical point T ? v t
D
gr 7 cos D 4 3
;
? Minimum velocity required =
gr 7 cos D 4 ; 3
[Ans.]
For two bodies in contact Force at contact (between two bodies) =
Mass on which force is not directly applied u Applied force Total mass of system
Note: Not applicable for more than two bodies
1.
f=
m2 F m1 m2
f=
m1F m1 m2
Bodies, which move together, can be considered as one system. If bodies have different motions, they should be considered as separate bodies. To find internal forces for bodies moving together, treat them as single system to find acceleration then to find internal forces consider one of the bodies as system
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10
SPECIAL MODULE (PHYSICS)
Example: Two blocks of masses 4 kg and 2 kg are placed in contact with each other on a smooth surface. A horizontal force of 18 N is applied on 2kg block such that both blocks move together calculate the value of contact force between the two blocks. Solution: Let the acceleration = a a = 3m/s2
? 18 = (4 + 2)a
4 kg block will move only because of contact force ? f = 4 u 3 = 12 N 2.
Tension in string connecting two bodies =
Mass on which force is not directly applied u Applied force Total mass of system T=
m1F m1 m2
Collision 1.
When a massive body suffers one – dimensional elastic collision with a stationary light body, the velocity of massive body remains practically unchanged but light body begins to move with a velocity which is double the velocity of massive body.
2.
Momentum and total energy are conserved during elastic collisions.
3.
The coefficient of restitution gives you an idea of the degree to which kinetic energy is conserved.
4.
When kinetic energy is conserved one can either use kinetic energy equation or the coefficient of restitution formula.
5.
In the perfectly inelastic collision, the relative velocity of the bodies after the collision is zero.
6.
If a ball is dropped from a height h. On the ground and the coefficient of restitution be e, then after striking the ground n times, it rises to a height H = e2nh. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
7.
In head on collision (also called as one dimension collision). Bodies move along same straight line before and after collision.
Work, Power, Energy 1.
For conservative forces work done along a closed path is Zero.
2.
For non – conservative forces work done along a closed path is not equal to zero.
3.
Work done against friction depends on the path followed. Viscosity and friction are non-conservative forces.
4.
Work done by electric force and gravitational force does not depends upon path followed. They are called as conservative forces.
5.
Work done depends on the frame of reference.
6.
Conservative laws can be used to describe the behavior as mechanical system even when the exact nature of forces is not known.
7.
Kinetic energy is always +ve
8.
Kinetic energy of a body cannot change if the force acting on a body is perpendicular to the instantaneous velocity.
9.
In case of conservative force (power is not dissipated), work does not depend upon the path followed. It depends upon initial and final position of the body
10.
In case of friction, power depends on the path followed.
11.
Kinetic energy of the particle performing SHM is given by K=
1 mZ2 A2 x 2 2
Where
x – displacement
m – mass
Z – Angular frequency
A – Amplitude
And potential energy is given by U = ? Total energy = k + V =
1 MZ2 x 2 2
1 mZ2 A 2 2
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12
SPECIAL MODULE (PHYSICS)
12.
When momentum increases by factor n kinetic energy increases by factor n2 (if mass is constant).
13.
A body starting from rest moves along a smooth inclined plane of length A , height h and having angle of inclination T . (i)
It’s acceleration down the plane is g sin T
(ii)
It’s velocity at the bottom of the inclined plane
(iii)
Time taken to reach the bottom T=
=
2A 2 = gA sin T
2A = g sin T 2h2
=
2
ghsin T
2gh =
2gA sin T
2A 2 gh
2h g sin2 T
1/2
l § 2h · t= ¨ ¸ sin T © g ¹ (iv)
Angle of inclination is changed keeping length constant. t1 t2
(v)
1/2
§ sin T2 · ¨ ¸ © sin T1 ¹
Angle of inclination changed keeping height constant. t1 t2
sin T2 sin T1
Centre of mass 1.
Centre of mass coincides with the geometrical centre of the body for all symmetrical bodies with uniform distribution of mass. Sr.No.
Body
1
Plane triangular lamina
Centroid of lamina
Cone or pyramid
Line joining apex and centre of base, 1/4th of length from base.
2 2.
Position of centre of mass
When a freely falling body explodes in to pieces the centre of mass as the whole system still lies on the same vertical line. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
3.
For a shell moving on a parabolic path explodes in to pieces, the centre of mass of the system shall lie along the same parabolic path.
4.
If centre of mass is chosen as the origin then some of moments of the masses of the system about the centre of mass is zero.
5.
Acceleration of centre of mass multiplied by total mass of the system gives total resultant force on the system.
To find out centre of mass of combination of bodies 1.
Find the system where whether it is one dimensional two dimensional or three dimensional.
2.
If the system is one dimensional use length instead of mass
3.
If the system is two dimensional use area instead of mass
4.
If the system is three dimensional use volume instead of mass
5.
Locate the axis and locate the centre of mass of everybody.
6.
Use following formula to locate c.m. of entire system § m x m2 x 2 ... · Xcm = ¨ 1 1 ¸ © m1 m2 ... ¹ For two dimension find Xcm and Ycm For three dimension find Xcm, Ycm and Zcm e.g. Xcm =
Mu 0 mu 0 Mm Mu
Ycm = Zcm =
h § R· mu¨ ¸ 4 © 2¹ Mm
Mu 0 mu 0 Mm
(Note: Use coordinates of the centre of mass of components 7.
Some complex COM problems can be quickly and easily solved using concept of negative mass (hypothetical concept). Lakshya Educare
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SPECIAL MODULE (PHYSICS)
e.g.
Find the center of mass? Solution: Consider the system as
2 ª §r· º §r· U u Sr u 0 « U u S u ¨ ¸ » u ¨ ¸ © 2 ¹ »¼ © 2 ¹ «¬ ,Ycm = 0, Zcm = 0 2 ª §r· º 2 USr « U u S u ¨ ¸ » © 2 ¹ »¼ «¬ 2
X cm
8.
Use following formula to calculate velocity of c.m. Vcm x
mV1x m2 V2x ... m1 m2 ...
For two dimension find X and Y component of velocity and for three dimension find X, Y and Z component of velocity 9.
Do the same to find out acceleration.
Gravitation 1.
Areal velocity of a planet is constant (Keplar’s second law) and is given by Lakshya Educare
15
SPECIAL MODULE (PHYSICS)
dA dt
L 2m
Here, L is the angular momentum of the planet about sun. 2.
Most of the problems of gravitation are solved by two conservation laws: (i)
Conservation of angular momentum about Sun and
(ii)
Conservation of mechanical (potential + kinetic) energy [If drag is negligible]
Hence, the following two equations are used in most of the cases. mvr sin T = constant
…(i)
1 GMm mv2 = constant 2 r
…(ii)
At aphelion and perihelion position T = 900 Hence, equation (i) can be written as, mvr sin 90o = constant or
mvr = constant
…(iii)
Further, since mass of the planet (m) also remains constant, equation (i) can also be written as vr sin T = constant
3.
or
v1r1 = v2r2
?
r1 ! r2
v1 v2
…(iv) ( T = 900)
Applying the above mentioned conservation laws in aphelion and perihelion positions with r1 = a(1 + e) and r2 = a(1 – e) Lakshya Educare
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SPECIAL MODULE (PHYSICS)
We can show that ° v min ° ® ° ° v max ¯ 4.
GM § 1 e · ¨ ¸ a ©1 e ¹
v1
and total energy of the planet E = –
GM § 1 e · ¨ ¸ a ©1 e ¹
v2
If
F v rn
then
T2 v (r)1–n
GMm 2a
and if U v rm then 5.
T2 v (r)2–n
If masses of sun and planet are comparable and motion of sun is also to be considered, then both of them revolve around their centre of mass with same angular velocity but different angular velocities in the circles of different radii. The centre of mass remains stationary T=
2Sr3/2 , Z = GM
GM r3
L = Pr 2 Z k=
1 2 2 Pr Z 2
Moment of Inertia
I = Pr 2
We can see in above formulae (except period and angular velocity) all ª have same form only ‘m’ is replaced by «P ¬
m1m2 º » call reduce mass. m1 m2 ¼
6.
The orbital velocity of a satellite is independent of the mass of the satellite but depends upon the mass and radius of the planet around which the rotation is taking place.
7.
If a body is at height ‘h’ from the surface of earth ves =
2g h R h Lakshya Educare
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SPECIAL MODULE (PHYSICS)
2 u orbital velocity
8.
Escape velocity =
9.
If the radius of a planet decreases by n% , keeping the mass constant the acceleration due to gravity on its surface decreases by 2n% .
10.
When a body falls from a height ‘h’ to the surface of the earth, its velocity on reaching +the surface of the earth is given by 1/2
v 11.
ª § h ·º « 2gR ¨ R h ¸ » © ¹¼ ¬
A body rises to a height nR (where R = radius of the earth), if thrown upwards with a velocity 1/2
v 12.
ª h ·º § « 2gR ¨ 1 1 n ¸ » © ¹¼ ¬
1/2
ª § n ·º «2gR ¨ 1 n ¸ » © ¹¼ ¬
Trajectory of a body projected from point A in the direction AB with different initial velocities: Let a body be projected from point A with velocity v in the direction AB. For different values of v the paths are different. Here are the possible cases (i)
If v = 0, path is straight line from A to O
(ii)
If 0 v v 0 , path is an ellipse with centre o of the earth as a focus
(iii)
If v = v0, path is a circle with O as the centre
(iv)
If
v 0 v v e , path is again an ellipse with O as a focus
A o v 0 v v e (v)
If v = ve, body escapes from the gravitational pull of the earth and path is a parabola Lakshya Educare
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SPECIAL MODULE (PHYSICS)
If v ! v e , body again escapes but now the path is hyperbola. Here,
(vi)
§ GM · v0 = orbital speed ¨ ¸ at A and ve = escape velocity at A ¨ r ¸¹ © Note:
13.
1.
From case (i) to (iv) total energy of the body is negative. Hence these are the closed orbits. For case (v) total energy is zero and for case (vi) total energy is positive. In these two cases orbits are open.
2.
If v is not very large the elliptical orbit will intersect the earth and the body will fall back to earth.
If the rate of rotation of the earth increases, the value of acceleration due to gravity decreases at all places on the surfaces of the earth except of poles.
Optics 1.
When an object is placed with its length along the principal axis, then the magnification is known as longitudinal magnification and is denoted by mL. In this case mL =
we know
ª v 2 v1 I = « O «¬ u2 u1
1 1 v u
º dv » = du »¼
(for small objects)
1 f Lakshya Educare
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SPECIAL MODULE (PHYSICS)
?
dv v2
mL
du u2
dv du
0 §v· ¨ ¸ © u¹
or
dv du
2
§v· ¨ ¸ © u¹
2
m2 .
where ‘m’ is known as transverse magnification (m) which is defined as m= 2.
I §v· = ¨ ¸ O © u¹
When a two dimensional object is placed with its plane perpendicular to principal axis, then it’s magnification is known as superficial magnification (ms). ms =
3.
4.
5.
1 1 v u
ma mb area of image = = m2 area of object aub
1 f
If the object is placed between ‘F’ and ‘P’, then concave mirrors give enlarged, erect and virtual image. Due to their converging property, they used as reflectors in automobile head lights and search lights. 1 v P 1 . The f refractive index will be different for different colours. Hence the focal length of a lens is different for different wavelengths. For red colour it is maximum and for violet colour it is minimum irrespective of the nature of lens. The focal length of a lens depends upon ‘ P ’. Actually
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20 6.
SPECIAL MODULE (PHYSICS)
If a lens is made of a number of layers of different refractive indices, then for a given wavelength of light, the lens will have as many focal lengths as the number of P 's .
7.
When an equi-convex lens of focal length ‘f’ is cut into two equal parts by a horizontal plane as shown in figure below, then focal length of each part remains the same but intensity of image formed by each part is reduced to half.
8.
When a equi-convex lens is cut into the two equal parts by a vertical plane CD (see figure). Then the focal length of each part (f’) becomes twice. i.e., f’ = 2f
9.
Limitations of the lens maker’s formula: (A)
The lens should be thin so that the separation between the two refracting surface should be small.
(b)
The medium on either side of the lens should be same.
If any of the limitation is violated then we have to use the refraction at the curved surface formula for both the surfaces.
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SPECIAL MODULE (PHYSICS)
10.
If should be remembered that both the focal lengths f1 and f2 of a thin lens are not always f P1 . equal. Actually it is 1 f2 P 2
11.
When two sides of a given equi-convex lens have different medium then we can write P 2 P1 v u
1 f0
ª 2P 0 P1 P2 º » « «¬ 2 P 0 1 »¼
where P 0 is the refractive index of material and f0 is the focal length of lens in axis. Example: An equiconvex lens of glass ( P 0 = 1.5) of focal length f0 = 40 cm is placed such § that on left side of it is air ( P1 = 1) and that on the right side is water ¨ P 2 ©
4· . 3 ¸¹
Determine the focal length of the lens. Due to unsymmetrical condition of medium around the lens, the first and second focal lengths of the lens are unequal For an equiconvex lens made of glass the equation may be simplified as P 2 P1 v u
1 f0
3 P1 P 2
and the equation
f1
ª 2P P 0 1 º P2 P f0 « 1 » and f2 f «¬ 2P P1 P 2 »¼
1 f0
ª 2P 0 P1 P 2 º » « «¬ 2 P 0 1 »¼
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SPECIAL MODULE (PHYSICS)
or f2
ª 2P 2 P 0 1 º f0 « » ¬« 2P 0 P1 P 2 ¼»
are simplified as f1
P1f0 and f2 3 P1 P 2
Here 12.
P1 = 1 and P 2
P 2 f0 3 P1 P 2 4 ; f0 = 40 cm and f2 = 80 cm 3
When a lens is kept in medium other than air, then 1 fm
§ P ·§ 1 1 · 1¸ ¨ ¸ ¨ © Pm ¹ © R1 R 2 ¹
where P m = refractive index of the medium in which the lens is placed. ?
fm f
P 1 § P · 1¸ ¨ P © m ¹
where fm = focal length in the medium and f = focal length in air. (A)
We can write the above expression in this way: fm f
Pm P 1 P Pm
fm ! 1 . So focal length increases and power f decreases. But nature of the lens remain unchanged. Now if P m P then
(B)
(C)
fm is infinite. So focal length will become infinite f and power becomes zero. So the lens behaves like a plane glass plate. If P
P m , then
When the lens is placed in a medium for which ‘ P ’ is greater then that of the lens, the nature of the lens changes. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
Example If an object is placed on left at a distance a from the lens (see figure) and its real image is formed at a distance ‘b’ from the length of the given lens.
Solution: From Newton’s formula x1x2 = f1f2 or
(a – f1)(b – f2) = f1f2
or
ab – af2 – bf1 + f1f2 = f1f2
or
af2 + bf1 = ab
?
f1 f2 a b
1
Thermodynamics For a closed curve (a)
Work done in clockwise direction is +ve
(b)
Work done in anticlockwise direction is –ve
(c)
Internal energy is a state function. So 'U for a closed path is zero.
(d)
The adiabatic exponent of a gaseous mixture is given by n1 n2 v 1
(e) (f)
n2 n2 v1 1 v 2 1
dU = nCvdT (always) dQ = nCpdT (constant pressure process) If a cyclic process is represent by a circle on the P-V diagram, then the work done is given by
W=
S (P2 – P1)(V2 - V1) 4 Lakshya Educare
24 (g)
SPECIAL MODULE (PHYSICS)
For transfer of heat when conductors are in parallel combination, 1 Rp
1 1 1 A ... where R = . R1 R 2 R 3 kA
when they are in series then Rp = R1 + R2 (Thermal moisture) (h)
In P – V diagram, for a closed path, work done is always area of that closed path.
(i)
For the following cases:
(1)
(2)
(j)
In questions, identify the system properly and identify correctly the type of processes taking place. (a)
e.g. If gas is in thermally insulated vessel undergoing volume change then processes are adiabatic.
(b)
If gas is in diathermic conducting ( 'Q = 0) undergoing reversible changes then it is isothermic.
(c) 1.
Number of moles for closed system is constant in absence of any chemical reaction.
The work done by a gas in irreversible cycle cannot be calculated from p – v diagrm Lakshya Educare
25 2.
SPECIAL MODULE (PHYSICS)
For specific heat of a as, we may use the following formulae Cv
3.
R r 1
and
Cp
rR r 1
The function on heat energy used to increase internal energy of a gas is dU dQ
Cv Cp
1 r
Fluid Mechanics 1.
At same point on a fluid pressure is same in all directions. In the figure p1 = p2 = p3 = p4
2.
Forces acting on fluid in equilibrium have to be perpendicular to its surface.
3.
In the same liquid pressure will be same at all points at same level. For example, in the figure p1 z p2 p3 = p4 and p5 = p6 P3 = p4 P0 + U1gh1 = p0 + U2gh2
4.
Work done in breaking a drop of radius ‘R’ into. ‘n’ drops of equal size = 4SR 2 V n1/3 1 .
5.
Angle of contact increases with rise in temperature. addition of soluble impurities.
6.
Detergents decrease both the angle of contact as well as surface tension.
7.
A liquid does not wet the containing vessel if its angle of contact is obtuse.
8.
The liquid rises in a capillary tube, when the angle of contact is acute.
9.
The equation 2SrV
It decreases on
Sr2 hUg is to be applied only for vertical cylindrical
tubes. This equation should not be used for capillary tubes of other shapes. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
10.
According to Stoke’s law, F = 6SKrv
11.
According to poiseuille’s equation Q=
dV dt
S P2 P1 8 KA
V t
S Pr 4 8 KA
where the letters have usual meanings. 12.
When a body of mass ‘m’ is floating in a liquid, then the excess mass ‘m0’ to be kept on the body so that it just sinks in liquid is given by (m + m0)g = VUg where ‘V’ is the volume of the body and ‘ U ’ is density of the liquid
13.
When a body of volume ‘V’ and density ‘ U ’ is dropped into a liquid of density ‘ V ’, then the effective downward acceleration of the body in liquid is, a=
14.
V UV g VU
§UV· ¸g ¨ © U ¹
The density of liquid of bulk modulus ‘k’ at a depth ‘h’ is given by Uh
Ugh º ª U0 «1 k »¼ ¬
Where ‘ U ’ is the average density of liquid and U0 is the density of liquid on its surface. 15.
The total pressure inside an air bubble of radius ‘r’ at a depth ‘h’ below the surface of liquid of density ‘ U ’ is p = p0 + hUg
2S r
where p0 is the atmospheric pressure and ‘S’ is the surface tension of liquid. 16.
Poiseuille’s equation can also be written as Q=
p1 p2 § 8 KL · ¨ ¸ © SR 4 ¹
'U X Lakshya Educare
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SPECIAL MODULE (PHYSICS)
8KL
X
SR 4
This equation can be compared with the current equation through a resistance i.e. 'V R
i
Here, 'V = potential difference and R = electrical resistance For current flow through a resistance, potential difference is a requirement similarly for flow of liquid through a pipe pressure difference is must. 17.
Problems of series an parallel combination of pipes can be solved in the similar manner as is done in case of an electrical circuit. The only difference is., (i)
Potential difference
'V
is replaced by the pressure difference
'P (ii)
The electrical resistance is replaced by X
(iii)
§ 8K L · ¨ ¸ and © SR 4 ¹
The electrical current is replaced by volume flow rate ‘Q’ or
dv . dt
The following example illustrates the theory. Example A liquid is flowing through horizontal pipes as shown in figure. Length of different pipes has the following ratio L AB
L CD
L EF 2
L GH 2
Similarly, radii of different pipes has the ratio, R AB
R EF
R CD
R GH 2 Lakshya Educare
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SPECIAL MODULE (PHYSICS)
Pressure at ‘A’ is 2P0 and pressure at ‘D’ is P0. The volume flow rate through the pipe AB is ‘Q’ Find, (A) (B)
volume flow rates through EF and GH pressure at ‘E’ and ‘F’.
Solution: The equivalent electrical circuit can be drawn as under, Xv
?
§ ¨ as X ©
L R4
X AB : X CD : XEF
8 KL · ¸ SR 4 ¹
§1· 1 ¨ ¸ 1 1 2 : X GH = 2 4 : © ¹4 : : 4 4 1 §1· §1· §1· ¨ ¸ ¨ ¸ ¨ ¸ © 2¹ © 2¹ © 2¹ = 8 : 8 : 16 : 1
(A)
As the current is distributed in the inverse ratio of the resistance (in parallel). The ‘Q’ will be distributed in the inverse ratio of ‘X’. Thus, volume flow rate through EF will be will be
(B)
16 Q., 17
ª 16X X Xnet = 8X + « ¬ 16X X ?
Q and that from GH 17
Q
º 288 X » + 8X = 17 ¼
'P X net
2P0 P0 288 X 17
§ ¨ as i ©
'P · R ¸¹
17P0 288X
Now, let P1 be the pressure at ‘E’, then Lakshya Educare
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SPECIAL MODULE (PHYSICS)
2P0 – P1 = 8QX =
8 u 17P0 288
17 u 8 · § P1 = ¨ 2 P = 1.53P0 288 ¸¹ 0 ©
?
Similarly, if P2, be the pressure at ‘F’, then P2 – P0 = 8QX 8 u 17 P0 288
?
P2 = P0 +
or
P2 = 1.47 P0
Sound and Waves 1.
Any function of ‘t’, say y = y(t) oscillates simple harmonically if d2 y dt 2
v y
or we can say if above condition is satisfied ‘y’ will oscillate simple harmonically. 2.
All sine and cosine functions of ‘t’ are simple harmonic in nature i.e., for the function y = a sin Zt r I y = a cos Zt r I d2 y
is directly proportional to –y. Hence, they are simple harmonic in dt2 nature. 3.
A function f(t) is said to be periodic of time period ‘T’ if f(t + T) = f(t) All sine or cosine functions of time are periodic. Thus Y = A sin Zt or or
period T =
A cos Zt is periodic
2S Z Lakshya Educare
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SPECIAL MODULE (PHYSICS)
4.
If a wire of length ‘ A ’, area of cross-section ‘a’, Young’s modulus ‘Y’ is stretched by suspending a mass ‘m’, then the period of vibration of suspended mass is T = 2S
5.
If two masses of m1 and m2 are connected at the two ends of the spring of spring constant ‘k’ then time period of their oscillation is T = 2S
6.
mA Ya
P , where P k
m1m2 , called reduced mass m1 m2
When the springs are connected in series between the two masses, 1 ks
1 1 1 1 ..... k1 k 2 k 3 k 4
When springs connected in parallel Kp = k1 + k2 + k3 + …… 7.
In disturbances that can be represented as a group of waves, the energy may be transported with a velocity of individual wave. This is called the group velocity.
8.
Two identical waves moving in opposite directions along the string will still produce standing waves even if their amplitudes are unequal. This is the case when an incident travelling wave is only partially reflected from a boundary, the resulting superposition of two waves having different amplitudes and travelling in opposite directions gives a standing wave pattern of waves whose envelop is shown in figure.
The standing wave ratio (SWR) is defined as 9.
A max A min
Ai Ar Ai Ar
If the two emitted waves from sources S1 and S2 already have a phase difference of ‘ S ’ the conditions for maximas and minimas are interchanged i.e., path difference. Lakshya Educare
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10.
SPECIAL MODULE (PHYSICS)
'x
O 3O , …….. 2 2
(constructive interference)
'x
0, O,2O ………
(for destructive interference)
Most of the problems of interference can be solved by calculating the path difference 'x and then by putting 'x
0, O,2O ……..
(constructive interference)
'x
O 3O , ……… 2 2
(destructive interference)
Provided waves from S1 and S2 are in phase. 11.
Sound waves can be refracted, diffracted but cannot be polarized. Due to large wavelength, sound waves required large reflecting surfaces.
12.
Consider the superposition of two sinusoidal waves of same frequency at a point. Let us assume that the two waves are travelling in the same direction with same velocity. The equation of the two waves reaching at a point can be written as y1
A1 sin kx Zt
y2
A 2 sin kx Zt I
The resultant displacement of the point where the waves meet is, Y = Y1 + Y 2 Solving we get Y Here
A sin kx Zt T
A1 A 2 cos I A 2 sin I
Or
and
A cos T
A sin T 2
A2
A1 A 2 cos I
A
A12 A 22 2A1A 2 cos I
tan T
A sin T A cos T
A2 sin I
2
A 2 sin I A1 A 2 cos I Lakshya Educare
32 13.
SPECIAL MODULE (PHYSICS)
If wind blows at a speed vw from the source to observer. v o v + vw. If in opposite direction v o v – vw. fc
14.
§ v r vw r vs · ¨ ¸f © v vw B vs ¹
When a source is revolving, with velocity vs around a stationary listener vn and nmin v vs
nmax
vn v vs
Beat frequency = nmax – nmin Similarly when a listener is revolving around source with velocity vL nmax
v vL n v
and nmin
v vL n v
(n is frequency of the source)
Beat frequency = nmax – nmin J u r.m.s. speed of molecules of air 3
15.
Speed of sound =
16.
The frequency of note produced by the organ pipe varies
17.
(i)
directly as
r , where ‘r’ is the thermodynamic constant.
(ii)
directly as
T , where ‘T’ is absolute temperature of the gas.
(iii)
inversely as
(iv)
inversely as length of the tube.
U , where ‘ U ’ is the density of the gas.
When a wave is reflected from (i)
rarer medium no change of phase
(ii)
denser medium there is phase change of ‘ S ’
Rotational Motion 1.
Theorem of parallel axes is application for any type of rigid body whether it is a two dimensional or three dimensional, while the theorem of Lakshya Educare
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SPECIAL MODULE (PHYSICS)
perpendicular axes is applicable for laminar type or two dimensional bodies only 2.
The point of intersection of three (X, Y and Z) axes, in theorem of perpendicular axes, may be any point on the plane of body (it may even lie outside the body). This point may or may not be the centre of mass.
3.
Moment of inertia of a symmetrically cut part of a rigid body has same form as that of the whole body. Example in figure (a) moment of inertia of a sector of circular disc shown about an axis perpendicular to its axis plane and passing through point 1 ‘O’ is M1R 2 as the moment of inertia of inertia of the complete disc is 2 1 also M2R 2 (where M1 is mass of the sector and M2 is mass of the 2 complete disc). This can be shown as in figure. If ‘M’ is the mass of
1 th n
part of the disc then mass of the disc = nM Idisc Isector
4.
1 nM R 2 2 1 Idisc n
1 MR 2 2
Angular velocity of a rigid body Z is
dT . Here ‘ T ’ is the angle between dt
the line joint any two points (say ‘A’ and ‘B’) on the rigid body and any reference line (dotted) as shown in figure. For example AB is a rod of length 4m. End ‘A’ is resting against a vertical wall OY and ‘B’ is moving towards right with constant speed VB = 10 m/s. to find the angular speed of rod at T proceed as under
300 , we can
Lakshya Educare
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SPECIAL MODULE (PHYSICS)
OB = x = AB cos T x = 4 cos T dx § dT · = 4sin T ¨ ¸ dt © dt ¹ dx /dt dT = dt 4sin T ZT =
10 = -5 rad/sec 4sin 30
§ dT · Here –ve sign implies that ‘ T ’ decreases as ‘t’ increases ¨ 0¸ . © dt ¹ 5.
In law of motion for a single particle acted on by a torque o
o
dL dt
W
o
o
Holds only if W and L are measured with respect to any point ‘O’ fixed in an inertial frame. o
o
dL dt
6.
W ext
7.
Suppose a rod is suspended from a support at ‘O’ and particle strikes the rod at any point. The angular momentum of the ‘rod + particle’ system remains conserved only about point of suspension or point ‘O’. Because in this case W ext on the system is zero only about ‘O’.
8.
The path of a point on circumference of a tyre is a cycloid and the distance moved by this point in on full rotation is 8R if it’s rotating without slipping.
9.
The equation W ext
ID does not hold good in a non-inertial frame.
However, there exists a very special case when W ext
ID does hold even if
the angular acceleration ‘ D ’ is measured from a non-inertial frame. That special case is , when the axis of rotation passes through the centre of Lakshya Educare
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SPECIAL MODULE (PHYSICS)
mass; otherwise the pseudo forces produce a pseudo torque about the axis. 10.
Work done by friction in pure rolling on a stationary ground is zero as the point of application of the force is at rest.
11.
The body rolls down the inclined plane without slipping only when the coefficient of limiting friction P bears following reaction. ª K2 º Pt« 2 tan T 2» ¬« K R ¼» In case of a rolling body, all points of the rigid body have same angular speed but different linear speed. The linear speed is maximum (= 2V) for the point ‘T’ and minimum ( = zero) for point ‘P’.
12.
In cases where pulley is having some mass and friction is sufficient enough to prevent slipping, the tension on two sides of the pulley will be different and rotational motion of the pulley is also to be considered.
13.
A = g sin T
: If surface is smooth
A = g sin T - Pg cos T
: If surface is rough but friction is insufficient to prevent slipping forward slipping will take place.
A=
g sin T I 1 MR 2
: If pure rolling is taking place, i.e., friction is sufficient to prevent slipping.
Electrostatics 1.
If ‘q’ be the inducing charge, the charge induced on a body having dielectric constant ‘K’ is given by q'
=
1· § q ¨ 1 ¸ k¹ © Lakshya Educare
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SPECIAL MODULE (PHYSICS)
2.
The magnitude of charge is not affected by its motion like mass i.e. change is invariant.
3.
A charge at rest produces only electric field around itself, a charge having unaccelerated motion produces electric field as well as magnetic field around itself while a charge having accelerated motion emits electromagnetic radiation also in addition to producing electric and magnetic fields.
4.
The dielectric constant of a metal is infinity.
5.
Permittivity of the medium is the property of the medium that determines the number of electric lines of force passing through it.
6.
Two identical pith balls each of mass ‘m’ are charged with a charge ‘q’. If the two balls are suspended by a silk thread of length ‘ A ’ from the same hook as shown
then
F mg
tan T
=
7.
The absolute electric potential at any point in the electric field is defined as the work done per unit positive charge required to move the test charge from infinity to that point.
8.
Potential difference between two points ‘A’ and ‘B’ in an electric field is defined as the work required to move a unit positive charge from the point ‘A’ to the point ‘B’ against the intensity of the electric field VB – VA
=
WAB q0
(q0 is the test charge)
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SPECIAL MODULE (PHYSICS)
9.
The work done in moving a charge between two points in a electric field is independent of the path followed between these two points since the electric field is a conservative field.
10.
The electric potential at a distance ‘r’ from a point charge ‘q’ is given by V
=
1 q 4SH0 r
11.
The locus of all points which are at the same potential is known as equipotential surface.
12.
The electrical capacitance of the earth is 7 × 10–4 farad.
13.
Force of attraction between oppositely charged plates of a capacitor is given as : F =
Q2 2H 0 A
or
Force per unit area i.e. 14.
F A
=
V2 2H 0
1 H 0 E2 2
(where ‘E’ is the field between the plates)
H0 A ªt t t t º d (t1 t2 t3 ..... t n ) « 1 2 3 ....... n » kn ¼ ¬ k2 k2 k 3
IN series combination 1 Ceff
17.
1 H 0 E2 A 2
If a number of slabs of thickness t1, t2, t3, …….., tn and dielectric constants k1, k2, ………, kn are inserted in between the plates the capacity is given by C =
16.
=
Energy stored per unit volume in the electric field is called energy density (u) u =
15.
F
=
1 1 1 1 ...... C1 C2 C3 Cn
In parallel combination Ceff
=
C1 + C2 + C3 + …….+ Cn Lakshya Educare
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SPECIAL MODULE (PHYSICS)
18.
Electric field due to a point charge at its own location is zero.
19.
Coulomb’s force between any two charges does not depend upon the presence or absence of any other charge.
20.
When a charge ‘q’ is brought from infinity to a point, where potential due to any source is ‘V’, work done = qV, which is also the potential energy of charge at this point.
21.
However, when a condenser is given charge ‘q’ so as to raise its potential 1 to ‘V’, energy stored in the condenser is qV. This energy resides in the 2 dielectric medium between the plates of the condenser.
22.
In case of a charged conductor (i)
Charge resides only on the outer surface of conductor.
(ii)
Electric field at any point inside the conductor is zero.
(iii)
Electric potential at any point inside the conductor is constant and equal to potential on the surface of the conductor, wherever be the shape and size of the conductor.
(iv)
Electric field at any point on the surface of charged conductor is directly proportional to the surface density of charge at that point, but electric potential does not depend upon the surface density of charge. 1
Due to a point charge, E v
24.
Due to an electric dipole, E v
25.
Due to an infinite uniform line of charge, E v
26.
Due to an infinite plane sheet of charge, E v r 0 and V v r
27.
On the surface of an irregularly shaped charged conductor, electric field intensity may be different at different points, but electric potential is same at every point.
28.
At the centre of a charged ring of radius ‘r’, E = 0.
r
2
and V v
1 . r
23.
1 r
3
and V v
1 r2 1 and V v loge r r
Lakshya Educare
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SPECIAL MODULE (PHYSICS)
And at a distance
r 2
on the axis of charged ring, E = max.
29.
When a cavity is made in a conductor, electric field in the cavity is zero.
30.
In a region, when E = 0; ‘V’ may be zero or ‘V’ may be constant.
31.
If V = 0, at a point, then ‘E’ at that point may be zero or may not be zero.
32.
Electric potential energy of any number of point charges is U =
33.
1 4SH 0
¦ all pairs
q jq k rjk
Electric field intensity due to a point charge ‘q’, at a distance (t1 + t2) where t1 is the thickness of medium of dielectric constant K1 and t2 is thickness of medium of dielectric constant K 2 is luq
E =
4SH 0 T1 K1 t2 K 2
2
Electric potential at the same point in the case described above is V = 34.
1 q 4SH 0 t1 K1 t2 K 2
For maintaining equilibrium of a charged soap bubble, V =
8H 0 T r
and q = 8Sr 2H0 rT
where ‘T’ is surface tension of soap solution and ‘r’ is radius of the bubble. 35.
When a number of dielectrics of same thickness ‘d’ having different areas of cross section A1, A2, A3, ……… with dielectric constants K1, K2, K3, ………. respectively are placed between the plates of a parallel plate capacitor, its capacitance is given by C =
36.
H 0 (K1A1 K 2 A2 K 3 A3 ......) d
Total energy stored in any grouping of capacitors is equal to sum of the energies stored in individual capacitors. Lakshya Educare
40 37.
38.
SPECIAL MODULE (PHYSICS)
Suppose there are ‘n’ charged drops, each of capacity ‘C’, charged to potential ‘V’ with charge ‘q’, surface density ‘ V ’ and potential energy ‘U’ coalesce to form a single drop. For such a drop, Total charge
=
nq
Total capacity
=
n1/3 C
Potential
=
n2/3 V
Surface density of charge
=
n1/3 V , and
Total potential energy
=
n2/3 U
When an insulating slab of dielectric constant ‘K’ is introduced between the plates of a parallel plate capacitor. (a) and the charging battery is on : (i)
potential difference ‘V’ remains constant.
(ii)
electric field intensity ‘E’ remains constant.
(iii) capacity ‘C’ becomes ‘K’ times. (iv)
potential energy ‘U’ becomes ‘K’ times.
(v)
charge ‘q’ becomes ‘K’ times.
(vi)
surface density of charge ‘ V ’ becomes ‘K’ times.
(b) and the charging battery is disconnected (i)
capacity ‘C’ becomes ‘K’ times
(ii)
charge ‘q’ remains constant
(iii) ‘ V ’ remains constant (iv)
‘V’ becomes
1 times K
(v)
‘E’ becomes
1 times K
(vi)
Potential energy becomes
1 times K
Lakshya Educare
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SPECIAL MODULE (PHYSICS)
Current Electricity 1.
The flow of electrons from ‘A’ to ‘B’, will make the conventional current from ‘B’ to ‘A’.
2.
The slope of the graph showing the variation of potential difference (V) on X-axis and current (I) on Y-axis gives the conductance of the conductor carrying current.
3.
The reciprocal of slope of V-I graph gives the resistance.
4.
The resistance as well as conductance of a conductor depend upon (i) the length (ii) the area of cross-section, (iii) nature of material of conductor and (iv) temperature of the conductor.
5.
The resistance (R) of a conductor is directly proportional to its length ( A ), when the area of cross-section of the conductor is constant. Therefore RvA
or
R A
a constant or
R1 A1
R2 A2
6.
Resistance of a conductor increases with decrease in density or when it is subjected to mechanical stress.
7.
Resistance of pure metals and metallic alloy increases with increase in temperature but the resistance of semiconductor decreases with increases in temperature.
8.
The value of temperature coefficient of resistance (D ) of a conductor is different at different temperature. The temperature coefficient of resistance averaged over the temperature range t10C to t20 C is given by D =
R 2 R1 R1 (t 2 t1 )
where R1, R2 = resistance of conductor at t10C to t20 C respectively. 9.
The conductor behaves as superconductors at a very low temperature.
10.
If the value of temperature coefficient of resistance of a material is zero (i.e. D = 0), there will be no change in its resistance with temperature. The temperature coefficient of resistance for superconductor is zero.
Lakshya Educare
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SPECIAL MODULE (PHYSICS)
11.
The resistivity of antimony, bismuth and semiconductor decreases with increase of temperature.
12.
Magnetic field increases the resistivity of all metals except iron, cobalt and nickel (which are ferromagnetic materials). The magnetic field decreases the resistivity of ferromagnetic materials like iron, cobalt and nickel.
13.
If ‘n’ identical resistances are first connected in series and then in parallel, the ratio of the equivalent resistance is given by Rs : Rp
14.
=
n2 : 1
or
Rp : Rs
=
1 : n2
If equivalent resistance of R1 and R2 in series and parallel by Rs and Rp respectively, then R1 =
1ª R s R 2s 4R s R p º »¼ 2 «¬
R2 =
1ª R s R 2s 4R s R p º « ¬ ¼» 2
15.
Approximate percentage change in resistance = 2 × percentage change (for small changes) in length by stretching.
16.
If three identical resistors each of the resistance ‘R’ are connected in the form of a triangle, the equivalent resistance between the ends of a side is § 2R · equal to ¨ ¸. © 3 ¹
17.
If an equilateral triangle is made of a uniform wire of resistance ‘R’ ; the § 2R · equivalent resistance between the ends of a side is ¨ ¸. © 9 ¹
18.
If a wire of resistance ‘R’ is bent in the form of a circle, the effective R . resistance between the ends of a diameter is 4
19.
If a square is made of a uniform wire of resistance ‘R’, the equivalent R resistance between the ends of a diagonal is . 4
Lakshya Educare
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SPECIAL MODULE (PHYSICS)
20.
If four identical resistors each of resistance ‘R’ are connected in the form of a square, the effective resistance between ends of a diagonal is ‘R’.
21.
If we have ‘n’ identical conductors, each of equal resistance, then the number of combination, we can have, using all at a time is 2 n-1.
22.
If we have ‘n’ different conductors, then the number of possible combination are 2n.
23.
If 12 wires; each of resistance ‘r’ ohm, are connected to form a skelton cube, then the total resistance between two diagonally opposite corners 5r of the cube = ohm. 6
24.
If 12 wires; each of resistance ‘r’ ohm, are connected to form a skelton cube, then the total resistance between the corners of the same edge of 7r cube = ohm. 12
25.
If 11 wires; each of resistance ‘r’ ohm, are connected to form a skelton cube, then the total resistance from one vacant edge to the other end is = 7r ohm. 5
26.
When current is drawn from the cell (i.e. during discharging of a cell), e.m.f. of a cell = terminal potential difference + voltage drop across the internal resistance of a cell. The direction of current inside the cell is from negative terminal to positive terminal. During charging of a cell, terminal potential difference = emf of a cell + voltage drop across internal resistance of a cell i.e. terminal potential difference becomes greater than the e.m.f. of the cell. The direction of current inside the cell is from positive terminal to negative terminal.
27.
Due to internal resistance of a cell (a) the energy is consumed inside the cell. (b) There is a potential drop inside the cell.
28.
In series grouping of cells, the effective e.m.f of ‘n’ cells, each of e.m.f. ‘E’ becomes nE. In parallel grouping of ‘n’ cells each of e.m.f. ‘E’, the effective e.m.f. is ‘E’ because in parallel combination of cells, the sizes of the electrodes will increase without affecting the e.m.f. of the cell. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
29.
The current in the external resistor will be maximum (i) in series grouping of cells, provided the value of internal resistance of a cell is very very small as compared to external resistance, (ii) in parallel grouping of cells, provided the value of internal resistance of a cell is very very large as compared to external resistance, (iii) in mixed grouping of cells, provided the value of external resistance is equal to total internal resistance of all the cells.
30.
When ‘m’ cells are wrongly connected in series of cells combination, the total e.m.f. of all the cells decreases by e.m.f. of 2 cells. For example, if ‘n’ cells each of e.m.f. ‘E’ and internal resistance ‘r’ are connected in series and by mistake ‘m’ cells are wrongly connected, to an external resistance ‘R’, then effective e.m.f. of all the cells = (n – 2m) E and effective resistance of circuit = (nr + R).
31.
Using Kirchoff’s law, while dividing the current having a junction through different arms of a network it will be same through different arms of same resistance if the end points of those arms are equilocated w.r.t. exit point for current in network and will be different through different arms if the end points of those arms are not equilocated w.r.t. exit point for current of the network. For illustration, in figure ‘A’, the currents going in arms AB, AD, and AL will be same because the locations of end points, ‘B’, ‘D’ and ‘L’ of these arms are symmetrically located w.r.t. exit point ‘N’ of the network.
In Fig. ‘B’, the currents, in arm AB and AL will be same as the end points ‘B’ and ‘L’ of these arms are located symmetrically w.r.t. exit-point ‘D’ of the network. Since the end point ‘B’ and ‘D’ of arms AB and AD are not Lakshya Educare
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SPECIAL MODULE (PHYSICS)
symmetrically located w.r.t. exit point ‘D’, hence current in arm AD is different from that of arm AB of the network.
32.
The potentiometer is equivalent to an ideal voltmeter of infinite resistance because while measuring the e.m.f. of a cell by a potentiometer, at the position of null point, no current flows in the cell circuit i.e. the cell is in the open circuit. Hence we obtain actual value of e.m.f. of the cell.
33.
Thermistors are mostly prepared from metal oxides. Their temperature coefficient of resistivity is large. The thermistor is used to measure the temperature due to the fact that its resistivity changes with temperature. A thermistor can measure a change in temperature of the order of 10-1 0C.
34.
Wheatstone bridge is most sensitive if all the arms of bridge have equal resistances i.e. P = Q = R = S.
35.
If the temperature of the conductor placed in the right gap of metre bridge is increased, then the balancing length decreases and the jockey moves towards left on bridge wire.
36.
If the temperature of a semiconductor placed in the right gap of metre bridge is increased, then the balancing length increases and the jockey moves towards right on bridge wire.
37.
If a resistor is connected in series to the right gap resistor in the metre bridge, then the balancing length decreases and hence jockey moves towards left. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
38.
If a resistor is connected in parallel to the right gap resistor in the metre bridge, then the balancing length increases and hence jockey moves towards right.
39.
The balanced position of Wheatstone bridge is not interchanging the positions of battery and galvanometer.
40.
The resistance between ‘A’ and ‘C’ of Wheatstone bridge in balanced position is RAC =
41.
affected
to
(P Q)(R S) (P Q) (R S)
Potentiometer is an ideal voltmeter.
Thermal and Chemical Effects of Current 1.
For a given voltage V, if resistance is changed from R to (R/n), power consumed changes from P to nP. [' P = V2/R When R’ = R/n, then P’ = V2/(R/n) = nV2/R = nP]
2.
In parallel combination, the effective power of various electrical appliances is more than power of individual appliance and is given by P = P1 + P2 + P3
3.
In parallel groupings of bulbs across a given source of voltage, the bulb of greater wattage will give more brightness and will allow more current for it, but will have lesser resistance and same potential difference across it.
4.
In series combination, the effective power of the various appliances becomes less than the power of individual appliance and is given by 1 1 1 1 . P P1 P2 P3
5.
The resistance of a bulb fitted in a house varies inversely as its power i.e. R v 1/P.
6.
In series grouping of bulbs, across a given source of voltage, the bulb of greater wattage will give less bright light and will have lesser resistance and potential difference across it but same current. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
7.
Filament of lower wattage bulb is thinner than that of higher wattage bulb i.e. the filament of 60 watt bulb is thinner than that of 100 watt bulb.
8.
The resistance of lower wattage bulb is higher than that of higher wattage bulb. It means, the resistance of 60 watt bulb is higher than that of 100 watt bulb.
9.
The material of heating element of electric heater or iron rod should have high resistivity and high melting point.
10.
The material of fuse wire should have high resistance and low melting point. The length of fuse wire is immaterial.
11.
The heat generated in a wire is doubled when both the radius and length of the wire are doubled.
12.
If one heater boils a certain mass of water in time t1 and another heater boils the same mass of water in time t2, then connecting both the heaters in series, the same water will boil in time (t1 + t2); then connecting both t ut the heaters in parallel the same water will boil in time t = 1 2 . t1 t 2
13.
If Hp and Hs are heat produced by a equal resistors, when they are connected in parallel and in series respectively, the H p = n2Hs.
14.
The ratio of chemical equivalent (E) and electrochemical equivalent (z) is same for all elements and is called Faraday’s constant F i.e. E/z = F = Faraday constant.
15.
A change of 96500 C is required to liberate 1.008 g of hydrogen.
16.
E.C.E. of substance = E.C.E. of hydrogen × chemical equivalent of the substance.
17.
V-I curve for a voltameter is a straight line beyond the voltage of back e.m.f. of electrolyte, hence Ohm’s law is obeyed there.
18.
The energy is consumed inside the cell due to its internal resistance and is given by = I2 r t, where r is the internal resistance of the cell.
19.
The direction of flow of conventional current inside the cell is from negative electrode to the positive electrode, while outside the cell, the Lakshya Educare
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SPECIAL MODULE (PHYSICS)
direction of this current is from positive electrode to the negative electrode. 20.
E.M.F. (E) is the characteristic of each cell and its value remains constant for a given cell, while terminal potential difference (V) goes on decreasing when more and more current is drawn from the cell.\
21.
The maximum current that can be drawn from the cell is Imax = E/r, where E is the E.M.F. of the cell and r is the internal resistance of a cell.
22.
The power dissipated in the external resistance (R) in a circuit is maximum when the internal resistance (r) of the cell is equal of the external resistance i.e. R = r
23.
R E where R is Rr V the external resistance and r is the internal resistance of a cell. In general, the efficiency of a cell is given by
K =
24.
When the power dissipated in the external circuit is maximum, the efficiency of the cell is 50%.
25.
If the cell is short circuited, its terminal potential difference and efficiency are zero.
26.
The output power of a cell is given by P0 =
E2 (R r)2
R.
Output power P0 is zero when external resistance R is zero. 27.
The maximum value of P0 is E2/4r.
28.
While charging a secondary cell, we send a current in the cell by some external electric source (i.e. a battery or battery charger) by connecting the positive terminal of battery charger to the positive electrode of the cell and the negative terminal of battery charger to the negative electrode of the cell. While charging, the direction of current inside the cell will be from positive electrode to negative electrode. In this case, the potential difference between the two electrodes of the cell well be greater than the e.m.f. of the cell V = E + Ir.
29.
Charging current for a secondary cell =
e.m.f. of ch arg e e.m.f. of cell . total resis tan ce of the circuit Lakshya Educare
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SPECIAL MODULE (PHYSICS)
30.
When two cells are connected in series so as to support each other, the effective e.m.f. becomes = E1 + E2.
31.
When two cells are connected in series so as to oppose each other, the effective e.m.f. becomes = E1 – E2 or E2 – E1 as E1 > E2 or E2 > E1.
32.
When identical cells are connected in parallel, their effective e.m.f. will be equal to e.m.f. of one cell. It means in parallel combination of cells, the electric capacity of the cell increases without changing its e.m.f.
33.
The chemical equivalent expressed in gram is called gram equivalent.
34.
The charge required to liberate one gram equivalent of a substance at an electrode during electrolysis is called Faraday. If is equal 96500 C/gram mole.
35.
The variation of thermo e.m.f. (E) and temperature of hot junction (T) is a 1 parabolic curve which can be given by the equation. E = DT E T 2 . 2
36.
§ The graph between thermo electric power, S ¨ © straight line.
dE · ¸ and temp T is a dT ¹
heat = S I t
37.
For Peltier effect,
38.
For Thomson effect, heat = V I t.
39.
Temperature of inversion of a thermocouple depends upon the temperature of cold junction but neutral temperature is independent of temperature of cold junction.
40.
Thermocouple should not be used to measure the temperature above the neutral temperature.
41.
Thomson’s coefficient for lead is zero.
42.
Seebeck effect is the resultant of Peltier effect and Thomson’s effect.
43.
For Peltier effect and Thomson’s effect, the heat evolved or absorbed is directly proportional to current. But for Joule’s law of heating effect, the heat produced is directly proportional to the square of the current flowing through conductor.
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SPECIAL MODULE (PHYSICS)
44.
Neutral temperature is independent of temperature of cold junction whereas temperature of inversion depends upon the temperature of cold junction.
45.
(i) The thermoelectric power is positive if the temperature of hot junction lies in between temperature of cold junction and neutral temperature. (ii) The thermoelectric power is negative if temperature of hot junction lies in between neutral temperature and temperature of inversion.
46.
See beck effect, Peltier effect and Thomson’s effect are reversible effects but Joule’s heating effect is irreversible effect.
Magnetic Effect of Current 1.
An electric charge in motion, in free space produces both electric and magnetic field, whereas a static electric charge produces only electric field.
2.
When an electric current is passed through a conductor, only magnetic field is associated with the conductor as the conductor is electrically neutral.
3.
The magnetic field produced due to current through a conductor is always in a place perpendicular to the plane of conductor.
4.
A moving charge through a conductor produces only magnetic field around the conductor but there is no electric field around the conductor.
5.
The magnetic field interacts with moving charge only as a moving charge produces magnetic field. Therefore, there is no interaction of magnetic field with stationary charge.
6.
When a charged particle is moving parallel to the direction of magnetic field, no force acts on the particle due to magnetic field. The path of charged particle is st. line. The values of momentum and kinetic energy of charged particle remains constant.
7.
When charged particle is moving perpendicular to the direction of magnetic field, the force on particle F = q v B, which is maximum. The
Lakshya Educare
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SPECIAL MODULE (PHYSICS)
path of particle in magnetic field will be circular, whose radius r = mv/qB. The momentum of the particle (p) = mv = q B r, will remain constant in magnitude but its direction will be changing. The K.E. of the particle in magnetic field is constant. 8.
A solenoid carrying current behaves as a bar magnet and the magnetic field inside the solenoid carrying current is uniform but the magnetic field just outside the solenoid on its curved face is zero.
9.
The magnetic field induction is constant at all points equidistant from the linear conductor.
10.
When solenoids are in parallel and connected to a source of electric current, the ratio of their magnetic field induction is independent of number of turns of solenoids but varies inversely as their radii.
11.
The magnetic field induction at the centre of O of the circular metal path of radius r subtending an angle D (in radians) at the centre O, carrying current i is (XL – XC).
12.
When a cell is connected between any two points of a uniform circular conductor, the magnetic field induction at the centre will be zero.
13.
The S.I. unit of magnetic field induction is tesla (denoted by T) and its C.G.S. unit is gauss (denoted by G), where l T = 104 G.
14.
Biot-Savart law for a test charge q0 moving with velocity v is S. I. units
o
o
is;
o
P vu r B = 0 q0 2 . 4S r Lakshya Educare
52 15.
SPECIAL MODULE (PHYSICS)
The magnitude of magnetic field induction at a point well inside the current carrying straight solenoid = P 0 nI and at one edge P 0 nI /2 where n is the no. of turns per unit length of solenoid. o
16.
The Lorentz magnetic force, on a charge q, moving with velocity v in a o
magnetic field B is given by; o
o o
F m = q(v u B) .
17.
The direction of magnetic Lorentz force is perpendicular to the direction of motion as well as to the magnetic field. Therefore, the Lorentz force does not perform any work on the charged particle. Due to this, the magnitude of velocity of charged particle does not change, only direction of motion changes.
18.
Lorentz magnetic force is maximum, when the charged particle moves at right angles to the magnetic field.
19.
Lorentz magnetic force is zero if (i)
The particle is uncharged (i.e. q = 0, ? F = qv B sin T = 0).
(ii)
The charged particle is at rest (i.e. v = 0).
(iii)
The charged particle is moving parallel or antiparallel to the magnetic field i.e. T = 00 or 1800. o
20.
Since the magnetic Lorentz force is always perpendicular to velocity v , therefore, the charged particle describes a circular path.
21.
The speed and kinetic energy of the charged particle moving perpendicular to the magnetic field do not change because force is always perpendicular to velocity but momentum of the particle changes due to change in direction of motion in the magnetic field.
22.
If a charged particle is moving in a magnetic field at some angle T , its direction of velocity continuously changes. The charged particle describes a helical path of radius = mv sin T /q B. It moves along the direction of magnetic field with speed v cos T .
23.
The momentum of the charged particle, moving along the direction of magnetic field does not change. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
24.
The centripetal force one the charged particle in magnetic field moving at some angle T with the direction of magnetic field is qv B and T . It is independent of the mass of the particle.
25.
Total Lorentz force acting on charged particle moving in electric and magnetic field is, o
o
o
o
F = (E v u B) .
26.
Lorentz force is important in detecting the presence of electric and magnetic fields.
27.
Lorentz force between two charges q1 and q2, moving with velocity v1, v2 separated by distance r is Fm =
P 0 (q1v1 )(q 2 v2 ) . 4S r2
28.
If the charges move, the electric as well as magnetic field are produced. In case, the charges move with speeds comparable to the speed of light, magnetic and electric forces between them would become comparable.
29.
Force on a conductor of length l, carrying current I2 and placed at a distance r parallel to another infinitely along conductor carrying current P 2I I l I1 is; F = 0 1 2 4S r
30.
The force on the long conductor is above case is equal and opposite to the force on the short conductor.
31.
A current loop carrying current I and area vector A behaves as a
o o
o
magnetic dipole, whose magnetic dipole moment, P m = I A o
o
32.
A current carrying coil is in stable equilibrium when P m is parallel to B .
33.
When magnetic dipole of moment P m moves from unstable equilibrium
o o
to stable equilibrium in a magnetic field B , the kinetic energy gained by it will be 2pmB.
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SPECIAL MODULE (PHYSICS)
34.
In order to increase the range of voltmeter n times, its total resistance should also be increased by n times.
35.
In order to reduce the range of a voltmeter of resistance G from V to V/n, a shunt equal to nG should be connected in parallel with it.
36.
If a charged particle of charge q is moving with a velocity v parallel to a straight current carrying conductor at a perpendicular distance r from it, the force on the particle due to magnetic field is F = Bqv =
P 0 2i u u qv . 4S r
37.
The coil in a moving coil galvanometer is suspended in a uniform magnetic field by phosphor bronze fibre as it has high Young’s modulus of elasticity and low value of modulus of Rigidity. It makes the galvanometer sensitive.
38.
A radial magnetic field is linked with the coil in moving coil galvanometer, in order to make the scale of galvanometer linear.
39.
In order to increase the range of an ammeter n times, the value of shunt resistance to be connected in parallel is S = G/(n-1).
40.
In order to increase the range of voltmeter n times the value of resistance to be connected in series with galvanometer is R = (n-1)G.
Magnetostatics 1.
In a moving coil galvanometer the current ‘I’ is directly proportional to deflection ‘ I ’ I D
I
or
I = kI
where ‘k’ is a constant of the galvanometer and is known as galvanometer constant. K =
C NAB
where C = elastic torsional constant of the suspension wire N = Number of turns in the coil, A = area per turn of the coil and Lakshya Educare
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SPECIAL MODULE (PHYSICS)
B = magnetic induction of radial magnetic field. The current sensitivity,
S =
I I
=
l K
=
NAB C
2.
In dead beat galvanometer, the frame on which the coil is wound is metallic. Due to strong induced current, the coil is not allowed to oscillate.
3.
In ballistic galvanometer, the frame is non-metallic. The coil oscillates for a long time due to absence of induced currents.
4.
An ideal ammeter has a zero resistance.
5.
The resistance of a milliammeter is more than that of an ammeter.
6.
To connect a galvanometer which gives full scale deflection for a current Ig so that it may be used to read a current ‘I’, the value of the shunt required is given by S =
Ig G (I Ig )
where G = galvanometer resistance.
7.
An ideal voltmeter has an infinite resistance so that it may not change the current in the element.
8.
The work done in deflecting the magnet through an angle ‘ T ’ from equilibrium position is given by W = MB (I – cos T )
9.
If two magnets of moments M1 and M2 are arranged with like poles touching and their axes are inclined making an angle ‘ T ’, the magnetic moment of the system M =
1/2
ªM12 M22 2M1M2 cos Tº ¬ ¼
10.
In general, atoms with paired electrons exhibit diamagnetism.
11.
In general, atoms with finite magnetic moment are paramagnetic. Such materials have unpaired electrons in their valance field.
12.
Ferromagnetism is due to the existence of magnetic domains.
13.
Ferromagnetic materials exhibit a phenomenon called hysteresis. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
14.
The lines joining the places of equal declination are called isogonal lines.
15.
The lines joining the places of zero declination are called agonal lines.
16.
The lines joining the places of equal dip are called isoclinal lines.
17.
The lines joining the places of zero dip are called aclinal lines.
18.
The lines joining the places of equal horizontal component of earth’s magnetic field are called isodynamical lines.
19.
When a bar magnet is cut into two equal pieces along its length, we get two magnets each of half the pole strength and half the dipole moment.
20.
When a bar magnet is cut into two equal pieces perpendicular to its length, each piece is a magnet with same pole strength, but half the dipole moment.
21.
Magnetic lines of force are endless, i.e., they run in closed loops inside and outside the bar magnet continuously.
22.
Unlike poles of equal strength form a magnetic dipole. As the poles cannot be separated, therefore, magnetic flux over a closed surface is always zero.
23.
When a magnetic dipole of moment M moves in a magnetic field of intensity B from position of unstable equilibrium to position of stable equilibrium, the kinetic energy gained by it = 2 MB.
24.
At magnetic equator, angle of dip is 00 and at magnetic poles, angle of dip is 900.
25.
Angle of dip and declination change from place to place and also from time to time.
26.
At the poles, horizontal component of earth’s field is zero, i.e., H = 0.
27.
At equator, vertical component of earth’s magnetic field is zero, i.e. V = 0.
28.
Magnetic field does not interact with stationary charges, i.e., charges at rest.
29.
i)
Due to a single pole, magnetic potential at a distance d from the pole is V =
P0 m u 4S d Lakshya Educare
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SPECIAL MODULE (PHYSICS)
ii)
Due to a bar magnet, potential at a point on axial line is V =
P0 M u 2 4S (d A2 )
and at a point on equatorial line, V = 0. iii)
Also, B =
dV . dr
Magnetic potential is a scalar and its SI units are JA -1 m-1. 30.
S.I. unit of magnetic induction is tesla and cgs unit of the same is gauss. l T = 104 gauss
31.
S.I. unit of magnetic intensity is A m-1 and its cgs unit is oersted.
32.
S.I. unit of magnetic moment is joule/tesla or A m2, and S.I. unit of magnetic pole strength is A – m.
33.
All substance exhibit diamagnetism. In paramagnetic and ferromagnetic substance, diamagnetism is neutralized by the large intrinsic dipole moment of spinning electrons.
34.
The origin of diamagnetism is traced to the orbital motion of electrons.
35.
The origin of paramagnetism and ferromagnetism is due to intrinsic magnetic moment of spinning of electrons.
36.
Formation of domains is responsible for enhancing the magnetism of ferromagnetic materials.
37.
Relative magnetic permeability (P r ) is less than l in case of diamagnetic Pr ! l
materials,
for
paramagnetic
materials
and
P r !! l
for
ferromagnetic materials. 38.
Fm is independent of temperature in case of diamagnetic substances.
39.
Some of the important relations in magnetism are : P
=
Pr =
B I ; Fm = H H P ; B = P0 (H + I), P r = 1 + Fm P0 Lakshya Educare
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SPECIAL MODULE (PHYSICS)
I
=
magnetic moment pole strength = volume area
40.
Electromagnets are made of soft iron because it has high retentivity and low coercivity.
41.
In a tangent galvanometer, the plane of the coil has to be set in magnetic meridian.
42.
In a vibration magnetometer, the reference line about which suspended magnet vibrates must be in magnetic meridian.
43.
In a vibration magnetometer, time period of a combination of magnets in difference position is more than the time period of the combination of magnets in sum position.
44.
Force between two magnetic dipoles in the end on position is F =
P 0 6M1M2 4S r4
45.
If O is latitude at a place, the dip G at this place is given by tan G = 2 tan O .
46.
In the determination of dip by a dip circle, in vertical plane in magnetic meridian, R is effective. And in vertical plane perpendicular to magnetic meridian, V is effective.
Electromagnetic Induction and Alternating Currents 1.
When a magnet is moved with respect to a coil, an e.m.f. is induced in the coil, which opposes the motion of the magnet towards or away from the coil. For example, when N pole of a bar magnet is moved towards a coil, the upper face of the coil behaves as North so as to oppose the incoming N pole, the current induced in the coil is anticlockwise. The reverse is also true.
2.
When a current carrying coil is moved towards a stationary coil, the direction of current induced in stationary coil is opposite to that in the moving coil. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
3.
When a current carrying coil is moved away from a stationary coil, the direction of current induced in stationary coil is same as that in the moving coil.
4.
When strength of current in a coil is increased, induced current in the coil is in a direction opposite to that of main current.
5.
When strength of current in a coil is decreased, induced current in the coil is in the direction of main current.
6.
When two coils A and B are arranged as shown in figure then on pressing K, current in A increases in clockwise direction. Therefore, induced current in B is anticlockwise, shown in figure.
However, when key K is released, current in A decreases in clockwise direction. Therefore, induced current in B is clockwise. 7.
When current in a straight conductor AB is increased, induced current in loop is clockwise. Shown in figure, so as to oppose the increase in magnetic flux in the loop. If current in AB is decreasing, the induced current in the loop is anticlockwise, so as to oppose the decrease in magnetic flux.
8.
When a magnet is dropped vertically in a long solenoid of copper, acceleration of magnet is less than acceleration due to gravity g. However, when the magnet is dropped freely in a very long vertical copper tube, the acceleration of magnet may become even zero. This is because resistance of copper tube is negligible and eddy currents may become too strong.
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60 9.
SPECIAL MODULE (PHYSICS)
The direction of induced e.m.f. depends on : i)
direction of magnetic flux
ii)
rate of change of magnetic flux, i.e.,
dI dt
is increasing or
decreasing. 10.
11.
Note that e.m.f. is induced in the following cases : i)
When a train moves horizontally in any direction.
ii)
When an aeroplane flies horizontally.
iii)
When a conductor falls freely in east west direction.
iv)
Landing or taking off an aeroplane with its wings in east west direction.
v)
The plane of orbit of a metallic satellite is inclined to the equatorial plane at any angle.
Further, e.m.f. is not induced in the following cases : i)
When a coil and magnet move, but there is not relative motion between the two.
ii)
When a conductor falls freely in N-S direction.
iii)
Landing or taking off an aeroplane with its wings in north south direction.
iv)
When plane of orbit of a metallic satellite coincides with equatorial plane of earth.
12.
The dimensions of L/R. RC and LC are the dimensions of time. Their reciprocals have the dimensions of frequency.
13.
In RLC circuit, impedance is infinite for Z = 0, and also for Z f . The 1 1 impedance is minimum Z = R, when ZL or Z . It is ZC LC corresponding to this angular frequency that resonance occurs.
14.
In case of a half wave rectifier, Im
I0 / S and I v
15.
In case of a full wave rectifier, Im
2I0 / S and I v
I0 / 2 . I0 / 2 . Lakshya Educare
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SPECIAL MODULE (PHYSICS)
Iv Im
0.707 I0 0.637 I0
16.
From factor of a.c. is K
17.
Coefficient of mutual inductance of two long co-axial solenoids with air P 0 N1N2 A , where N1, N2 are total number of turns of the two core, M l solenoids, each of length l and area of cross section A. If two solenoids have different areas of cross-section, then A is taken as area of cross section of inner solenoid. For two coupled coils, M
1.11 .
K L1L 2 where K is coefficient of coupling
between the two coils. When two coils are tightly coupled, so that magnetic flux produced in primary is fully linked with secondary, K = 1. When two coils are lying close, K < 1. For uncoupled coils, K = 0. Infact, K= 18.
magnetic flux linked with sec ondary magnetic flux linked with primary
Behavior of inductor and capacitor in a.c. and d.c. circuit is different XL = Z L = 2 S nL, XC =
l l = . For d.c., n = 0, ZC 2SnC
? XL = 0, XC = f Total reactance is (XL – XC) and not (XL + XC), because through L and C, alternating current has a relative phase difference of 1800. Electrons and Photons 1.
The value of sparking potential is 30,000 V for air at normal pressure.
2.
The value of sparking potential depends upon
3.
i)
nature of the gas
ii)
pressure of the gas
iii)
size of electrodes
iv)
distance between the electrodes
The value of sparking potential is less for pointed electrodes under the similar conditions.
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SPECIAL MODULE (PHYSICS)
4.
The value of sparking potential (V) is proportional to the product of gas pressure (p) and spark length (l) of the gas, i.e., V v p l. It is called Paschen’s law.
5.
The kinetic energy gained by electron of charge e, accelerated through a potential difference V is eV. If electron is accelerated from rest, the velocity acquired by electron is v = (2e V/m)1/2, where m is the mass of the electron.
6.
When an electron is subjected to a perpendicular uniform electric field, it moves on a parabolic path.
7.
If an electron moving with velocity v is subjected to a perpendicular uniform magnetic field B, it will describe a circular path of radius r, given by B e v = mr2/r or e/m = v/(rB),
8.
If an electron moving with velocity v, subjected to the combined effect of electric field E and magnetic field B, goes undeviated, then Ee = mv2/r = Bev or
v = E/B
And e/m = v2/rE = v/rB 9.
The deflection of electron beam perpendicular to its initial direction of motion due to perpendicular electric field is y =
1 1 eE l2 at2 = 2 2 m v2
……… (i)
When, under the combined effect of electric and magnetic fields, the cathode rays meet the screen at undeflected position, then eE = evB
or
v = E/B
Putting this value in (i), we have
10.
y =
1 eE l2 2 m E /B
=
1 § e · l2B2 ¨ ¸ 2©m¹ E
2
or
=
e 2y E = 2 2 m l B
When a charged particle of charge q, mass m, moving with a velocity v, is subjected to a perpendicular uniform magnetic field B, it will describe a circular path of radius r, given by Lakshya Educare
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SPECIAL MODULE (PHYSICS)
r = mv//qB
11.
i)
For the given value of q and B, r v mv, i.e., radius v (momentum).
ii)
For the given value of v and B, r v l/(q/m), i.e., radius v 1/(sp. Change). Since the specific charge of D -particle is less than that of photon, hence, radius of circular path in a given magnetic field for a particle moving with a given velocity will be smaller for proton that for D -particle.
iii)
The centripetal force on the charged particle, F = qvB, is independent of the mass of particle.
iv)
The centripetal acceleration a = qvB/m, is smaller for heavier particles.
v)
The angular frequency of charged particle in the magnetic field is, Z = qB/m, which is less for heavier particle than the lighter particle having same charge.
If charged droplet is stationary under the applied electric field E, then neE = mg =
4 3 Sr Ug 3
or
ne =
4Sr3Ug , where n is an integer. 3E
12.
Gases at normal pressure act as bad conductors and when subjected to low pressure and high voltage, they act as conductors.
13.
The properties of cathode rays are independent of (i) nature of gas in the discharge tube and (ii) nature of material of cathode.
14.
The value of specific charge (e/m) of cathode rays or of electrons is 1.76 × 1011 C kg-1. This value is constant and is independent of nature of gas and nature of material of cathode.
15.
A charged particle at rest if moving parallel to the direction of magnetic field experiences no force. Total force on the moving charge q due to both o
o
the electric field E and magnetic field B is called Lorentz force and is given by o
F
§o o o· q ¨E V u B¸ © ¹ Lakshya Educare
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SPECIAL MODULE (PHYSICS)
16.
A charged particle experiences a force whether it is at rest or in motion in an electric field, which is independent of the velocity of the charged particle. This force acts in the direction of field for positive charged particle and opposite to the direction of electric field for negatively charged particle.
17.
If two charged particles of masses m1 and m2 having equal K.E. and charge are subjected to the same magnetic field applied perpendicular to their direction of motion, then radii of their circular paths r 1 and r2 are given by r1 r2
=
m1 m2
18.
The value of specific change is same for all the particles of the cathode rays but they are different for different particles of positive rays.
19.
The value of specific change for hydrogen = 1.6 × 10-19 C/1.67 × 10-27 kg = 9.6 × 107 Ckg-1.
20.
Photons are packets of energy which are emitted by source of radiations. i)
Each photon is of energy E = hv = hc/ O .
ii)
All photons travel in straight line with the speed of light in vacuum.
iii)
Photons are electrically neutral.
iv)
Photons have zero rest mass.
v)
Photons are not deflected by electric and magnetic fields.
vi)
The equivalent mass of a photon while moving is given by m =
E
hy
2
2
c
c
hc 2
c O
h . cO E c
hv c
h . O
vii)
Momentum of the photon = mc =
viii)
No. of photons of wavelength O emitted in t second from a lamp of Pt O power P is, in . hc Lakshya Educare
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SPECIAL MODULE (PHYSICS)
21.
Photoelectric effect has established the particle nature of light.
22.
Work
function
of
a
metal
is
W0
hv 0
hc O0
1243.1 u 10-9 eV O0
1240 u 140-9 eV . O0 23.
Work function (W0) varies from metal to metal.
24.
That material is better for photoelectric emission whose work function least. As caesium has least work function, hence it is best metal for photoelectric emission.
25.
When atomic number of elements increases, the work function will decrease.
26.
When the temperature of metal increases, the work function will decrease.
27.
The maximum kinetic energy of the emitted photoelectron is equal to the product of charge of electron and stopping potential, i.e., Maximum K.E. = eVs.
28.
Stopping potential (i) is directly proportional to the frequency of the incident light and (ii) is inversely proportional to the work function of metal.
29.
Stopping potential is independent of i)
intensity of incident light,
ii)
illuminating power of the source and
iii)
the distance of the source from the metal surface.
30.
Photoelectric effect is based on law of conservation of energy.
31.
Work function of a metal depends upon
32.
(a)
nature of metal,
(b)
the impurities present on metal surface.
Photoelectric current is independent of (a)
frequency of the incident light (v)
(b)
energy of the incident light (E) Lakshya Educare
66 33.
SPECIAL MODULE (PHYSICS)
The variation between energy (E) of the incident light and wavelength O is a curve as shown in figure (a)
34.
The variation between the frequency (v) of the incident light and photoelectric current (I) is a curve as shown in figure (b).
35.
The graph between stopping potential and frequency of the incident light is a straight line whose slope gives the ratio of Planck’s constant to electronic charge.
36.
Einstein’s photoelectric equation was experimentally verified by Milliken for radiations of lower frequencies and De-Broglie for higher frequency radiations.
37.
Photo electric effect was discovered by Hall-watch
38.
Kinetic energy of emitted photoelectrons varies from zero to h (v – v0) where v is the frequency of the incident radiation and v0 is the threshold frequency of photosensitive surface.
39.
Infra red radiations can not eject photoelectrons from a metal surface, whereas X-rays will always do it.
40.
Photoelectric effect was explained by Einstein following quantum theory of light predicted by Maximum Planck.
41.
The kinetic energy of photoelectrons does not depend upon the intensity of incident radiations.
42.
The photoelectrons emitted from a metal surface have different kinetic energies even through the incident photons have the same energy. This is so because all the electrons are not emitted from the outer surface of metal. These electrons emitted from below the surface of metal required more energy for making them free. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
43.
In photoelectric effect, the incident photon is completely absorbed by the electron of the photosensitive surface.
44.
The de-Broglie hypothesis established the wave nature of moving material particles.
45.
The de-Broglie wavelength O = h/mv is independent of the nature and charge of the material particle.
46.
The de-Broglie waves are not electromagnetic waves.
47.
Out of photon and electron having same de-Broglie wavelength, the total energy of electron is more than that of photon.
48.
Out of photon and electron having same de-Broglie wavelength, the kinetic energy of photon is greater than that of electron.
Atoms, Molecules and Nuclei 1.
2.
Problem solving technique (for nuclear physics) (a)
Balance atomic number and mass number on both the sides.
(b)
Calculate the total energy of the reactants and products individually and equate them.
(c)
Finally equate the momenta of reactants and products.
If a particle of mass ‘m’ and charge ‘q’ is accelerated through a potential different ‘v’ then wavelength associated with is given O =
3.
The de-Broglie wavelength of a gas molecule of mass ‘m’ at temperature ‘T’ (in Kelvin) is given by O =
4.
h 1 u 2mq v
h 3mkT
where h = Boltzmann constant.
Mass defect is given by 'm = [Zmp + (A – Z) mn mZ A ] where mp, pn and mZ A be the masses of proton, neutron and nucleus respectively, ‘Z’ is number of proton, (A – Z) is number of neutrons. Lakshya Educare
68 5.
6.
SPECIAL MODULE (PHYSICS)
Packing fraction = p =
'm A
MA A
Mass defect mass number
n2 where n is mZ principal quantum number, ‘Z’ is charge number of element, ‘m’ is mass of particle revolving around the nucleus. As per Bohr’s theory, radius of nth Bohr orbit is rn v
Z n
c §Z· ¨ ¸ 137 © n ¹
§Z· 2.2 u 106 ¨ ¸ m /s . ©n¹
7.
Velocity of electron in nth orbit Vn v
8.
Angular frequency of electron Zn v
9.
Frequency of electron f v
10.
Period of revolution of electron Tn v
11.
Electric current due to electron motion In
12.
Magnetic induction produced at the nucleus due to electron motion Bn =
13.
Z2m
4.159 u 106 Z2
n3
n3
Z 2m
6.62 u 1015 Z 2
n3
n3
P0 I0 Z3 m2 v 2rn n5
n3 Z2 m
12.58Z 2 n3
rad./s
c /s .
1.5 u 10-16 efn v
n3 Z2
sec.
Z2 m
1.06Z2
n3
n3
mA
tesla.
Magnetic moment produced due to electron motion Mn = A n In
Srn2 In
neh . 4Sm
14.
Potential energy of electron in nth orbit P.E. =
15.
K.E. of electron =
16.
Total energy of electron (En) = P.E. + K.E. =
Ke2 2rn
Ke2 rn
Un
1 (P.E.) 2 KZe2 2rn
13.6Z 2 n2
eV .
Lakshya Educare
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SPECIAL MODULE (PHYSICS)
17.
Note that only Lyman series is obtained in both, emission as well as absorption spectrum. All other spectra are obtained only in the emission spectrum.
18.
Spectrum of hydrogen has fine structure i.e. each spectral line consists of a large number of fine lines. This is explained on the basis of spin quantum number and uncertainty principle.
19.
Small letters are used to represent the quantum numbers of single electron atoms, whereas capital letters are used to represent the quantum numbers of multielectron atoms.
20.
According to Pauli’s exclusion principle, no two electrons in atoms can occupy the same quantum state i.e. in a neutral atom, all the four quantum numbers can never be identical for two or more than two electrons.
21.
Electrons occupy the orbitals in increasing order of energy i.e. Es < Ep < Ed < Ef.
22.
Most stable state of the atom is that of minimum potential energy.
23.
Total number of elements for a given value (n) of quantum number. i)
For n = 2, N = 2 (12 + 22) = 10
ii)
For n = 3, N = 2 (12 + 22 + 32) = 28
iii)
For n = 4, N = 2 (12 + 22 + 32 + 42) = 60 En (V) e
13.6Z2
24.
Ionisation potential of electron =
25.
ª 1 1 º Excitation potential of electron = 13.6Z 2 « 2 » (n l)2 ¼» ¬« n
en2
(V) .
26.
Wavelength of spectral line increases with the increase in order of the series, OPfund ! OBrackett ! OPaschen ! O Balmer ! O Lyman
27.
In any series, wavelength of spectral line decreases with increase of its order.
Lakshya Educare
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SPECIAL MODULE (PHYSICS)
28.
Maximum number of spectral lines obtained on account of transition of n(n 1) electron present in nth orbit to various lower orbits = . 2
29.
The maximum number of electrons in a shell can be 2 n2.
30.
Maximum number of electrons in a subshell can be 2 (2i + 1).
31.
Using standard symbols, i)
r1 r2
2
§ n1 · § Z 2 · ¨ ¸ ¨ ¸ © n2 ¹ © Z1 ¹
ii)
Vn1 Vn2
§ n2 · § Z1 · ¸ ¨ ¸¨ © n1 ¹ © Z 2 ¹
32.
Note that neutrons are required for fission and protons are required for fusion.
33.
Whereas radioactive disintegration is a spontaneous process, fission is not. Energy obtained in radioactive disintegration is much smaller than that released in fission.
34.
The decay constant of a stable element is zero.
35.
The value of decay constant is also equal to negative of the slope of N-t curve.
36.
Uranium in which fraction of U233 is increased from 0.7% to 2.3% is called enriched uranium.
37.
In uranium ore, the ratio of U235 to U238 is 1 : 40.
38.
Critical mass for uranium fuel is 10 kg.
39.
Heavy water is the best moderator. Characteristics of moderator are : i)
its atomic weight must be low,
ii)
it should not absorb neutrons,
iii)
it should undergo elastic collisions with neutrons and reduce their velocity.
40.
Fusion reactors are better than fission reactors as no unwanted radioactive substances are produced in them.
41.
Geiger Nuttal law log e O = A + B loge R Lakshya Educare
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SPECIAL MODULE (PHYSICS)
where A and B are constant. For radioactive series, ‘B’ is same and ‘A’ different. ‘R’ is range. 42.
Relation between range ‘R’ of ' D ' particles and their energy E is R = 0.318 E3/2.
43.
Whenever ' D ' or ' E ' particle is emitted by a nucleus, then the daughter nucleus is left in the excited state. It suddenly transfers to ground state by emitting J -rays.
44.
J -rays interact with matter, as a consequence of which the phenomena of photoelectric effect, Compton effect and pair production occur. At low energies photoelectric effect occurs and at high energies pair production occurs. Z2
45.
Coefficient of photoemission Pphoto v
46.
Coefficient of Compton effect P compton v Z
47.
Coefficient of pair production Ppp v Z 2
48.
Intensity of J -rays at a depth ‘x’ inside the matter is I
(hJ )7/2
I0 e-Px where ‘ P ’
is coefficient of absorption of that material. It is the reciprocal of the l distance inside matter, at which the intensity I reduces to or 37% of e dI /I its max. value (I0). Also, P dx P depends on wavelength of J -rays and nature of absorbing material (P v O3 ).
Lakshya Educare
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SPECIAL MODULE (PHYSICS)
49.
Specific activity is the activity of 1 gram of material.
50.
Four radioactive series are 92U
i)
Uranium series
ii)
Actinum series
92U
iii)
Thorium series
90Th
iv)
Neptunium series
–
298
235
–
232
94Pu
206 82Pb 82Pb
–
207
82Pb
241
–
208
81Tl
204
The first three are natural radioactive series and the last one is artificial. Decay constant of end product of every radioactive series is zero. 51.
Geiger Muller Counter is used for detecting ' D ' and ' E ' particles.
52.
Cloud chamber is used for detecting radioactive radiations and for determining their paths, range and energy.
53.
The quantities which are conserved in disintegration are : Angular momentum; linear momentum; charge, mass number or Baryon number, Lepton number, energy and mass.
54.
Baryon number B = 1, for a neutron and also for a proton.
55.
Lepton number (L) L = 1 for electron; and also for neutrino. L = -1 for positron and also for antineutrino.
56.
Activity of a radioactive element A
dN dt
ON, A 0
A
A 0 e-Ot , where A0 is maximum initial activity.
57.
O
O D OE , W
58.
Ox y X o Y oZ
O N0
WD WE WD WE O
O
In this case, a)
dNx dt
O x N x
Lakshya Educare
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SPECIAL MODULE (PHYSICS)
b)
dN y dt
O x Nx O y Ny
c)
dNz dt
O y Ny
d)
Ny (t)
N0 O Ox Oy
ªe-O y t e-O x t º where N0 is initial number of nuclei «¬ »¼
present. 59.
When a point light source is located at a distance ‘d’ from a given surface, then effective energy reaching to that given surface will be E'
E 4Sd 2
× (vertical maximum cross sectional area of given surface)
Here ‘d’ is distance between source and centre of the given surface.
a)
b)
If surface sphere then
If cylinder,
or
c)
60.
E'
If cuboid, then
Time of disintegration t
E'
E'
E'
E 4Sd
2
E 4Sd 2 E 4Sd 2
E 4Sd 2
u Sr 2
u 2rA s
u Sr 2
u (ac)
T log10 (N0 / N) log10 2 Lakshya Educare
74 61.
SPECIAL MODULE (PHYSICS)
Note that D -particles and J -rays have line spectra, but ‘ E ’ particles have a continuous spectrum.
62.
1 milli Curie = 37 Rutherford.
63.
The age of earth is 5 billion years.
64.
A-N curve is as shown in the figure. We find that A v N .
65.
Penetrating power varies inversely as mass and ionizing power is proportional to energy.
66.
Radioactivity is a nuclear process and not an atomic process. It is not associated with electron configuration in atoms.
67.
Number of atoms left after n half lives, N = N0 / 2n or
68.
1 º ª Number of atoms decayed in a half lives = N0 – N = N0 «1 n » ¬ 2 ¼
69.
Percentage of radioactive material left at time ‘t’ is where n
70.
71.
N N0
§1· ¨ ¸ ©2¹
N u 100 N0
n
1 2n
u 100
t . T
Nuclei of radioactive element ‘A’ are being produced at a constant rate ‘ D ’. The element has a decay constant O . At time t = 0, there are N0 nuclei of the element. (A)
Calculate the number ‘N’ of nuclei of ‘A’ at time ‘t’.
(B)
If D = 2 N0 O , calculate the number of nuclei of ‘A’ after one half-life of ‘A’ and also, the limiting value of ‘N’ as t o f .
In U238 ore containing Uranium, the ratio of U234 to Pb206 nuclei is 3. Calculate the age of the are, assuming that all the lead present in the ore Lakshya Educare
75
SPECIAL MODULE (PHYSICS)
is the final stable product of U238. Take half life of U238 to be 4.5 × 109 years. 72.
The element Curium 96Cm248 has a mean life of 1013 seconds. Its primary decay modes are spontaneous fission and D -decay, the former with a probability of 8% and the latter with a probability of 92%. Each fission releases 200 MeV of energy. The masses involved in D -decay are as follows: 96Cm248
= 248.072220u
94Pu244
= 244.064100u and 2He4 = 4.002603u
Calculate the power output from a sample of 1020 Cm atoms. § ¨1u ©
931
MeV · ¸ C2 ¹
Solids and Semiconductors 1.
According to child Langmuir law, the plate current Ip in diode valve is directly proportional to three half power of plate potential (Vp) i.e. Ip v Vp3/2 or Ip = k Vp3/2 . where ‘K’ is a constant whose value depends on the area and temperature of cathode and also on its distance from the plate.
2.
In an energy band diagram there are about 1029 energy levels per cubic metre with energy separation of 10-23 eV.
3.
If there is no forbidden gap (i.e. energy gap) in the energy band diagram, the solids behaves as a good conductor.
4.
If there is a large energy gap or forbidden gap in the energy bond diagram, the solid behaves as insulator or bad conductor.
5.
If there is a small energy gap or forbidden gap in the energy band diagram, the solid behaves as a semiconductor.
6.
At absolute zero, a pure semiconductor has filled valence band and empty conductor band. Hence it behaves as an insulator.
7.
The forbidden gap in pure germanium is 0.72 and for silicon is 1.1 eV.
8.
At ordinary temperature, some of the electrons acquire kinetic energy due to thermal vibrations and move from valence band to conduction Lakshya Educare
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SPECIAL MODULE (PHYSICS)
band after crossing over the energy gap. Then the number of electrons in the conduction band is equal to the number of holes in valence band. 9.
The number of electrons or holes in a semiconductor at temperature TK is given by ne
nh
ni
AT 3/2e
-E g /kT
It means on increasing temperature, the number of current carriers increases. This increases the conductivity of the semiconductor with increase in temperature. 10.
It is extremely difficult to obtain a completely pure semiconductor.
11.
A pure semiconductor (called intrinsic semiconductor) has negative temperature coefficient of resistance.
12.
The electrical conductivity of intrinsic semiconductor is low. It depends upon the number of thermal charge carriers generated at a given temperature.
13.
The electrical conductivity of a semiconductor increase with increase in temperature.
14.
A very small doping can semiconductor too much.
15.
The mobility of electron is greater than that of hole in a semiconductor.
16.
The mobility of electrons as well as of holes in a semiconductor decrease with increase in temperature but it is independent of the number density of the electrons or holes.
17.
In a semiconductor, the conventional current is contributed by the movement of electrons towards the positive end of semiconductor and movement of holes towards the negative end of the semiconductor.
18.
In a doped semiconductor, the number density of electrons and holes is not equal. It can be shown that
change
the
conductivity
of
intrinsic
nenh = n2i
where ne, nh are the number density of electrons and holes respectively d ni is number density of intrinsic carriers (i.e. electrons and holes.) 19.
In n-type semiconductor :
ne | Nd !! nh Lakshya Educare
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SPECIAL MODULE (PHYSICS)
where Nd is the density of donor atoms doped in a semiconductor. In p-type semiconductor :
nh | Na !! ne
where Na is the density of acceptor atoms doped in a semiconductor. 20.
A thin layer formed on both the sides of the p-n junction, which is devoid of the free charge carriers but has immobile ions is called depletion layer. The thickness of depletion layer at p-n junction is about 10-6 m.
21.
The value of potential barrier of germanium p-n junction is 0.3 V and for silicon p-n junction diode is 0.7 V. The value of potential barrier depends upon the amount of doping of the semiconductor crystals forming p-n junction.
22.
The electric field set up across the barrier is of the order of 7 × 105 Vm-1 for a Si diode and 3 × 105 Vm-1 for a Ge diode.
23.
A p-n junction diode can be considered to be equivalent to a capacitor with ‘p’ and ‘n’ regions acting as the plates of a capacitor and depletion layer as the dielectric medium.
24.
During forward biasing of p-n junction the width of depletion layer becomes thin and the p-n junction offers a low resistance to the motion of change carriers. Infact an ideal junction diode when forward biased offers zero resistance. Voltage drop across such a junction diode is zero.
25.
During the reverse biasing of the p-n junction diode the width of depletion layer becomes thick and the p-n junction offers a high resistance to the motion of majority charge carriers. Infact an ideal junction diode when reverse biased offers infinite resistance and acts as an open circuit. Voltage drop across such a junction diode is equal to voltage applied.
26.
Potential barrier at p-n junction opposes the forward current but aids the reverse current.
27.
The diffusion current through p-n junction during forward biasing is given by I = I0 (eeV/kT – l) where Io is the value of diffusion current when the junction diode is unbiased; ‘k’ is Boltzmann constant.
Lakshya Educare
78 28.
SPECIAL MODULE (PHYSICS)
The diffusion current through p-n junction during revers biasing is given by I = I0 (e-eV/kT – l) = -I0 where ‘V’ is large. where I0 is the value of diffusion current when the junction diode is unbiased.
29.
The p-n junction diode is a one way device. It offers low resistance when forward biased and high resistance when reverse biased, hence it can be used as a rectifier.
30.
The ratio of change in junction voltage
'V
to the change in junction
current 'I is called as dynamic resistance i.e. R d
'V . 'I
31.
Knee voltage of the p-n junction for Ge is 0.3 V and for Si 0.7 V.
32.
An ideal diode offers zero resistance in forward biasing and infinite resistance in reverse biasing.
33.
An ideal diode offers zero resistance in forward biasing the infinite resistance in reverse biasing.
34.
In a transistor the length of collector is maximum, less of emitter and least of base.
35.
In reverse biasing of p-n junction the reverse current is, Ir = -I0 i.e. saturated reverse current.
36.
In a transistor, emitter is heavily doped and its main function is to supply majority charge carriers when forward biased. The collector is moderately doped as compared to emitter. Its function is to collect majority charge carriers. The base is very lightly doped as compared to emitter and collector. Its function is to bring an interaction between collector and emitter.
37.
Transistor can be used as amplifier and oscillator but not as a rectifier.
38.
In a common base transistor amplifier (i) the input and output signals are in the same phase (ii) there is no amplification in current of a given signal (iii) there is an amplification in voltage and power of the given signal. Lakshya Educare
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SPECIAL MODULE (PHYSICS)
39.
In a common emitter transistor amplifier, (i) the input and output signals are out of phase by ‘ S ’ or 1800 (ii) there is amplification in the current, voltage and power of the given signal.
40.
Common emitter amplifier is preferred over common base transistor D . amplifier due to larger current gain of that E 1 D
41.
An ideal diode works as an open switch in reverse biasing.
42.
The Boolean expressions obey commutative law, associative law as well as distributive law, i.e.
43.
(i)
A+B=B+A
(ii)
(iii)
A + (B + C) = (A + B) + C
a.B=B.A
In digital electronics; two types of logic are used (i)
Positive logic : In this logic, ‘0’ stands for lower level (say 0 V) and 1 stands for higher level (say 5 V).
(ii)
Negative logic : In this logic, ‘0’ stands for higher level (say 5 V) and 1 stands for level higher (say 0 V).
44.
The NAND gate and NOR gate are the building blocks of digital system.
45.
i)
AB
AB
ii)
AB
AB
AB
ii)
AB
A B
iv)
AB
AB
A B
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