SYNOPSIS
AND
SHORTCUTS
FOR JEE
Sr.No.
Subjects
Page No.
1.
Physics
1 - 79
2.
Chemistry
80 - 155
TIPS, TRICKS & SHORTCUT METHODS PHYSICS Mechanics Vectors 1. o o o o oo o oo o o o u uu u A B C and then o o o o o o o o A A A A C A as well as o oo o B . 2. o o o o o o o o o o o o o o o o u uu u A B A B when T T T 45T 0.
3. The rectangular components can’t have magnitude greater than vector itself ''''cosT d TT d TT d TT d T1, 4. If JJG JJJG JJJG JJJG JJGJJG JJJGJJJG JJJGJJJG JJJGJJJG JJG JJJG JJJG JJJG A1 A2 A3 ... An 0.
A1 A2 A3 ... An, then the adjacent vectors can be inclined to
each other at angle 2SSSS 9 .
5. If any two vectors are parallel or equal, then the scalar triple product is zero. 6. The magnitude of o o o o o o o o u uu u
P Q can vary from maximum value
o o o o o o o o P Q to minimum value o o oo oo o o 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.
8. For resultant of two vectors of equal magnitude. TTTT Magnitude of resultant
600qqqq 3a
900 2a
1200 a
where ‘a’ stands for magnitude of each vector. 9. o o oo oo o o O O O O a b , (a) o oo o a is parallel to o o o o b , if OOOO> 0; (b) o oo o a is antiparallel to o o o o b , if < 0. O OO O
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 rule-
Final 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)
(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 uuuu 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
11.4 = 313.157895 |||| 313
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 TTTT , ux = u cos TTTT , ax = 0 and ay = -g
H = u sin TTTT u 2g 2g 2 2 2 y …(1) T = 2usinTTTT 2u g g y ? ?? ? u gT 2 y …(2)
Sub. (2) in (1) we get, H = g T gT
8g 8
2 2 2
[Relation between H and T]
R = u sin 2TTTT 2u sin cosTTTT TTTT
g g 2 2 = TTTT T T T T ¨¨¨¨¨¨§¨§§¨§ TTTT¸¸¸¸¸¸¸¸···· TTTT © ¹ © ¹ © ¹ © ¹ 2u sin u sin 2 cot 4 cot 2g 2g 2 2 2 2
R = 4 H cot TTTT [Relation between R and H] …(3) From (3) and (2) R = 4gT cotTTTT 8 2 R = gT cotTTTT 2 2
[Relation between R and H] …(4) D D D D H tan R 2 D D D D 2H tan R T T T T T T T T u u u u T T T T T T T T TTTT u sin g sin 2
2g u sin 2 2sin cos
2 2 2 2 TTTT D D D D tan tan 2
For body projected from height H with horizontal velocity – Ux = u uy = 0 t = 2H g sx = 2H u g
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 plane-
Range of the projectile on the inclined plane. R = D ED ED ED E DDDD E EE E 2u sin cos g cos 2 2 Time of flight T = D ED ED ED E E EE E 2usin g cos . R = DDDD D E D E D E D E g T cos 2sin 2 .
[Relation between ‘R’ and ‘T’ on inclined plan] 3. When tan D D D 2 tan EEEE particle strikes the plane horizontally. D
4. When cot E E E E 2tan D ED ED E particle strikes the plane at right angles. D E
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 :
o o o o o o o o o o o o vAB vAC vCB
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. o o o o § · § · § · § · ¨¨¨¨ ¸¸¸¸ © ¹ © ¹ © ¹ © ¹ d F m v dt If m is constant, then F mdv 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 <PPPPsN
- If the body is just in motion with respect to the surface Pf P P PsN - If the body is in motion with respect to the surface Pf P P PkN
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 plane-
a = g(sin DDD - PD PPPkcos DD ) DD If sin D ! PD ! PD ! PD ! Pscos DDDD
P P P P s tanDDDD a = 0 If P tP tP tP ts tanDDDD
9. Work done in moving a body up an inclined plane through a distance s along the incline
W = mg (sin TTTT + PPPPk cos TTTT )x x = distance
10. If the body is moved down an inclined plane with constant speed then work done is given by W = mg PPPPkcosT T T T sinTTTT 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.
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 ZZZZ be angular velocity of the cylinder about the point O. ? ?? ? mg(r – r cos TTTT ) = 1I ZZZZ Z Z Z Z 2 2 2 i f mgr(1 - cos TTTT ) = §¨¨§¨¨¨¨¨§§¨ ¸ ¨· §¸ ¨¸ ¨¸ ¨· §¸ ¨· §¸ ¨¸ ¨· §¸ ¨ZZZZ ¸···¸¸·¸¸¸¸¸ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹ 1 mr v mr 2 2 r 2 2 2 2 2 g(1 - cos TTTT ) = §¨¨¨¨§§§ ¸¸¸····¸ ¨¨§¨§¨§§¨¨¨¨ZZZZ ··¸·¸¸¸·¸¸¸¸ © ¹ © ¹ © ¹ © ¹ ©©©© ¹¹¹¹ 1 3r v 2 2 r 2 2 2 4g 3 (1 - cos TTTT ) + v r 2 = ZZZZ r2 …(1)
Writing force equation,
-N + mg cos TTTT = m ZZZ rZ2 …(2)
(Note: Normal reaction passes through the centre of the cylinder.) Using equation (1) and (2),
N = mg cos TTTT - m«ª«ª«ª«ª T T T T º»º»º»º» « » «« »» «¬ »¼ ¬¬ ¼¼ ¬ ¼ 4g v 1 cos 3 r 2
N = 7mg cosTTTT4mg mv 3 3 r 2 N tttt 0 v2 tttt g 7 cosT T T T 4 r 3 At critical point T DT DT DT D ? ?? ? v tttt gr 7 cosD D D D 4 ; 3 ? ??
? Minimum velocity required = gr 7 cosD D D D 4 ;
3 [Ans.]
For two bodies in contact
Force at contact (between two bodies)
= Mass on which force is not directly applieduuuuApplied force Total mass of system
Note: Not applicable for more than two bodies
f = m F m m 2 1 2 f = m F m m 1 1 2
1. 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
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 ?
??
? 18 = (4 + 2)a a = 3m/s 2
4 kg block will move only because of contact force ?
??
? f = 4 uuuu 3 = 12 N
2. Tension in string connecting two bodies
= Mass on which force is not directly applieduuuuApplied force Total mass of system
T = m F m m 1 1 2 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.
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 = 1mZZZZ A x 2
2 2 2
Where x – displacement m – mass
Z Z Z
Z – Angular frequency A – Amplitude And potential energy is given by U = 1MZZZZ x
2 2 2 ? ? ? ? Total energy = k + V = 1mZZZZ A 2 2 2
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
A A
A, height h and having angle of inclination TTTT . (i) It’s acceleration down the plane is g sin TTTT
(ii) It’s velocity at the bottom of the inclined plane 2gh = 2g sinAAAA TTTT (iii) Time taken to reach the bottom
T = TTTT A A A A 2 g sin = TTTT A A A A A A A A 2 g sin 2 = 2AAAA gh 2 = TTTT 2h ghsin 2 2 = TTTT 2h g sin2 t = ¨¨§§¨§§¨ ¸¸·¸···¸ TTTT ©©©© ¹¹¹¹ l 2h sin g 1/2
(iv) Angle of inclination is changed keeping length constant.
§ · § · § · § TTTT · ¨¨¨¨ ¸¸¸¸ TTTT © ¹ © ¹ © ¹ © ¹ sin t t sin 1/2 2 1 2 1
(v) Angle of inclination changed keeping height constant. TTTT TTTT sin t t sin 2 1 2 1 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 Position of centre of mass
1 Plane triangular lamina Centroid of lamina
2 Cone or pyramid Line joining apex and centre of base, 1/4th of length from base.
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
Xcm = § · §§ ·· § · ¨ ¸ ¨¨ ¸¸ ¨ ¸ © ¹ ©© ¹¹ © ¹ m x m x ... m m ... 1 1 2 2 1 2
For two dimension find Xcm and Ycm
For three dimension find Xcm, Ycm and Zcm
e.g. Xcm = u u u u u u u u M 0 m 0 M m Ycm = § · § · § · § · u u uu u u u u ¨¨¨¨ ¸¸¸¸ © ¹ © ¹ © ¹ © ¹ h R M m 4 2 M m Zcm = u u u u uu u u M 0 m 0 M m
(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).
e.g.
Find the center of mass? Solution:
Consider the system as
ª º ª º ª º ª § ·§ ·§ ·§ · º § ·§ ·§ ·§ · U u S u U u S u u U u SU u S uu U u S u U u S u uu U u S u U u S u«««« ¨ ¸¨ ¸¨ ¸¨ ¸ »»»»u¨ ¸¨ ¸¨ ¸¨ ¸ © ¹ © ¹ © ¹© ¹ © ¹© ¹ © ¹ © ¹ « » « » « » « » ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ª º ª º ª º ª § ·§ ·§ ·§ · º US U u S u US U u S u US U u S u US U u S u«««« ¨ ¸¨ ¸¨ ¸¨ ¸ »»»» © ¹ © ¹© ¹ © ¹ « » « » « » « » ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ r r r 0 2 2 X r r 2 2 2 cm 2 2 ,Ycm = 0, Zcm = 0
8. Use following formula to calculate velocity of c.m.
mV m V ... V m m ... 1x 2 2x cm x 1 2
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
dA L
dt 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 TTTT = constant …(i)
1 2mv
2 - GMm
r = constant …(ii)
At aphelion and perihelion position TTTT = 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 TTTT = constant …(iv)
or v1r1 = v2r2 ( TTTT = 900) ? ? ? ? r1 !!!! r2 v1 v2
3. Applying the above mentioned conservation laws in aphelion and perihelion positions with
We can show that §§§§ ···· °°°° ¨¨¨¨ ¸¸¸¸ © ¹ © ¹ © ¹ © ¹ °°°° ®®®® °°°° §§§§ ···· ¨ ¸ ¨¨ ¸¸ ¨ ¸ °°°° ©©©© ¹¹¹¹ ¯¯¯¯ GM 1 e v v a 1 e GM 1 e v v a 1 e min 1 max 2
and total energy of the planet E = –GMm 2a
4. If F vvvv rn
then T2 vv (r)vv 1–n
and if U vvv rv m
then T2vvvv (r)2–n
5. 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 = 2 rSSSS GM 3/2 , ZZZZ = GM r3 L = PP ZP ZP Zr2Z k = 1P ZP ZP ZP Zr 2 2 2 Moment of Inertia I = PPPPr2
We can see in above formulae (except period and angular velocity) all have same form only ‘m’ is replaced by ª«ª«ª«ª«P P P P º»º»º»º»
¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ m m m m 1 2 1 2
call reduce mass. 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 = 2gh Rh
8. Escape velocity = 2 orbital velocity uuuu
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
ª º ªª ºº ª §§§§ ····º «««« ¨¨¨¨ ¸¸¸¸»»»» © ¹ ©© ¹¹ © ¹ ¬ ¼ ¬¬ ¼¼ ¬ ¼ h v 2gR R h 1/2
11. A body rises to a height nR (where R = radius of the earth), if thrown upwards with a velocity
ª º ª º ªª ºº ªª ºº ª §§§§ ····º ª §§§§ ····º ¨ ¸ ¨ ¸ ¨¨ ¸¸ ¨¨ ¸¸ ¨ ¸ ¨ ¸ « » « » «« »» «« »» « » « » © ¹ © ¹ ©© ¹¹ ©© ¹¹ © ¹ © ¹ ¬ ¼ ¬ ¼ ¬¬ ¼¼ ¬¬ ¼¼ ¬ ¼ ¬ ¼ h n v 2gR 1 2gR 1 n 1 n 1/2 1/2
12. 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 v0, 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 v0 v ve, path is again an ellipse with O as a focus A oooov0 v ve
(v) If v = ve, body escapes from the gravitational pull of the earth and
(vi) If v!!!!ve, body again escapes but now the path is hyperbola. Here, v0 = orbital speed § · § · § · § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ © ¹ © ¹ © ¹ GM
r at A and ve = escape velocity at A Note:
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.
13. 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 = I O = ª º ª º ª º ª º «««« »»»» « » « » « » « » ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ v v u u 2 1 2 1 = dv
du (for small objects) we know 11 1
dvdu 0 v2 u2 or § · § ·§ · § · ¨ ¸¨ ¸¨ ¸¨ ¸ © ¹ © ¹© ¹ © ¹ dv v du u 2 ? ? ? ? § ·§ ·¨ ¸¨ ¸§ ·§ ·¨ ¸¨ ¸ © ¹ © ¹© ¹ © ¹ dv v m m du u 2 2 L .
where ‘m’ is known as transverse magnification (m) which is defined as m = I O = § · § · § · § · ¨ ¸¨ ¸¨ ¸¨ ¸ © ¹ © ¹ © ¹ © ¹ v u
2. 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 = area of image area of object = uuuu ma mb a b = m 2 3. 11 1 v u f
4. 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.
5. The focal length of a lens depends upon ‘ PPPP ’. Actually v P 1v P v P v P 1
f . The
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.
6. 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 PPPP '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.
10. If should be remembered that both the focal lengths f1 and f2 of a thin lens are not always
equal. Actually it is PPPP P PP P f f 1 1 2 2 .
11. When two sides of a given equi-convex lens have different medium then we can write
ª º ª º ª º ª º P P P P PP P P PP P P P PPPP P P P «««« »»»» P P P P « » « » « » « » ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ 2 1 v u f 2 1 2 1 0 1 2 0 0
where PPPP0 is the refractive index of material and f0 is the focal length of lens in axis.
Example:
An equiconvex lens of glass ( PPPP0 = 1.5) of focal length f0 = 40 cm is placed such
that on left side of it is air ( PPPP1 = 1) and that on the right side is water ¨¨§§¨§¨§P P P P ¸¸·¸···¸
© ¹ © ¹ © ¹ © ¹ 4 3 2 .
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 PP P PPPP P P P P P P 1 3 P P v u f 2 1 1 2 0
and the equation
ª º ªª ºº ª P P P P P P P P º «««« »»»» P P P P P PP P P P P P « » «« »» «¬ »¼ ¬¬ ¼¼ ¬ ¼ 2 1 f f 2 0 1 0 1 2 and PPPP PPPP ª««ª«ª«ª P P PP P PP P PP P P º»º»º»º» f P ff P P f ¬¬¬¬«««« P »»»¼»¼¼¼ 2 1 f f 2 1 2 1 0 1 2 2 0 0
or ª«ª«ª«ª« PPPP P P P P º»º»º»º» P P P P P P P P P P P P « » « » « » « » ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ 2 1 f f 2 2 0 2 0 0 1 2 are simplified as P PP P P P P P P P P P f f 3 1 0 1 1 2 and PPPP P P P P P P P P f f 3 2 0 2 1 2 Here PPPP1 = 1 and P P P P 4 3 2 ; f0 = 40 cm and f2 = 80 cm
12. When a lens is kept in medium other than air, then
§ · § · § · § · § · § · § · § PPPP · ¨¨¨¨ ¸¸¸¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨P ¸ PP P © ¹ © ¹ © ¹ © ¹ ©©©© ¹¹¹¹ 1 1 1 1 fm m R1 R2
where PPPPm = refractive index of the medium in which the lens is placed. ? ? ? ? P P P P § · §§ ·· § PPPP · ¨ ¸ ¨¨ ¸¸ ¨PPPP ¸ © ¹ ©© ¹¹ © ¹ f 1 f 1 m m
where fm = focal length in the medium
and f = focal length in air.
(A) We can write the above expression in this way:
P P PP P P P P P P P P P P P P 1 f f m m m Now if PPPPm P P P P then f !!!!1 f
m . So focal length increases and power
decreases. But nature of the lens remain unchanged. (B) If P PP PP PP Pm, then f
f
m is infinite. So focal length will become infinite
and power becomes zero. So the lens behaves like a plane glass plate.
(C) When the lens is placed in a medium for which ‘ PPPP ’ is greater then that of the lens, the nature of the lens changes.
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 ? ? ? ? f f 1 a b 2 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
n n n n v 1 v 1 v 1 1 2 2 2 1 2 (e) dU = nCvdT (always)
dQ = nCpdT (constant pressure process)
(f) If a cyclic process is represent by a circle on the P-V diagram, then the work done is given by
W = SSSS
(g) For transfer of heat when conductors are in parallel combination, 1 1 1 1 ... Rp R1 R2 R3 where R = A A A A 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) Number of moles for closed system is constant in absence of any chemical reaction.
1. The work done by a gas in irreversible cycle cannot be calculated from p – v diagrm
2. For specific heat of a as, we may use the following formulae R C r 1 v and rR C r 1 p
3. The function on heat energy used to increase internal energy of a gas is C dU 1 dQ C r v p 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 zzzz p2
p3 = p4 and p5 = p6
P3 = p4
P0 + UUUU gh1 1 = p0 + UUUU gh2 2
4. Work done in breaking a drop of radius ‘R’ into. ‘n’ drops of equal size = S4 RSSS 2VVVV n1/31 .
5. Angle of contact increases with rise in temperature. It decreases on 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 S V S2 rS V SS V SS V Sr h g2 UUUU is to be applied only for vertical cylindrical
tubes. This equation should not be used for capillary tubes of other shapes.
10. According to Stoke’s law, F = SK6SKSKSKrv 11. According to poiseuille’s equation
Q = SSSS K K K KAAAA P P dV dt 8 2 1 SSSS K K K KAAAA V Pr t 8 4
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 = UV gUUU
where ‘V’ is the volume of the body and ‘ UUUU ’ is density of the liquid
13. When a body of volume ‘V’ and density ‘ UUUU ’ is dropped into a liquid of density ‘ VVV ’, then the effective downward acceleration of the body in liquid V is, a = U VU VU VU V ¨ §¨§¨§¨§U VU VU VU V·¸·¸¸·¸· U U UU UU U ©©©© U ¹¹¹¹ V g g V
14. The density of liquid of bulk modulus ‘k’ at a depth ‘h’ is given by U UU U ª º ª º ª º ª º U U U UU U U U «««« »»»» ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ gh 1 k h 0
Where ‘ UUUU ’ is the average density of liquid and UUUU0 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 ‘ UUUU ’ is
p = p0 + U U U U
2S h g
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 = p p 8 L X R 'U'U'U'U K K K K § · § · § · § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ S SS S © ¹ © ¹ © ¹ © ¹ 1 2 4
K K K K S SS S 8 L X R4
This equation can be compared with the current equation through a resistance i.e. ' '' ' V i R
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 K KK K § · § · § · § · ¨¨¨¨ ¸¸¸¸ S SS S © ¹ © ¹ © ¹ © ¹ 8 L X R4 and
(iii) 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 L L L 2 2 GH EF AB CD
Similarly, radii of different pipes has the ratio,
R
R R R
2
GH AB EF CD
Pressure at ‘A’ is 2P0 and pressure at ‘D’ is P0. The volume flow rate
through the pipe AB is ‘Q’
Find, (A) volume flow rates through EF and GH (B) pressure at ‘E’ and ‘F’.
Solution:
The equivalent electrical circuit can be drawn as under, v vv v L X R4 K K K K § · § · § · § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ S SS S © ¹ © ¹ © ¹ © ¹ 8 L as X R4 ? ? ? ? XAB : XCD: XEF : XGH= § · § ·§ · § · ¨ ¸ ¨ ¸¨ ¸ ¨ ¸ © ¹ © ¹© ¹ © ¹ § · § · § · § · § · § · § · § · § · § · § · § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹ 1 1 1 1 2 2 : : : 1 1 1 1 2 2 2 4 4 4 4 = 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 Q
17 and that from GH will be 16Q., 17 (B) Xnet = 8X + ª º ª º ª º ª º « » « » « » « » ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ 16X X 16X X + 8X = 288 X 17 ? ?? ? Q ''''P Xnet ' ' ' ' § · §§ ·· § · ¨ ¸ ¨¨ ¸¸ ¨ ¸ © ¹ ©© ¹¹ © ¹ P as i R 2P P 17P 288X 288X 17 0 0 0
2P0 – P1 = 8QX = u uu u 8 17P 288 0 ? ?? ? P1 = u uu u § · §§ ·· § · ¨ ¸ ¨¨ ¸¸ ¨ ¸ © ¹ ©© ¹¹ © ¹ 17 8 2 P 288 0 = 1.53P0
Similarly, if P2, be the pressure at ‘F’, then
P2 – P0 = 8QX ? ?? ? P2 = P0 + u uu u 8 17 P 288 0 or P2 = 1.47 P0
Sound and Waves
1. Any function of ‘t’, say y = y(t) oscillates simple harmonically if v v v v d y y dt 2 2
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 Z r IZ r IZ r IZ r It y = a cos Z r IZ r IZ r IZ r It d y dt 2
2 is directly proportional to –y. Hence, they are simple harmonic in
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 ZZZZt or A cos ZZZZt is periodic or period T = SSSS Z Z Z Z 2
4. If a wire of length ‘ AAAA ’, 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 = S2SSS mAAAA Ya
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 = 2SSSS PPPP k , where P P P P m m m m 1 2 1 2
, called reduced mass
6. When the springs are connected in series between the two masses,
1 1 1 1 1 ... ks k1 k2 k3 k4
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 A A A A A A max i r min i r
9. If the two emitted waves from sources S1 and S2 already have a phase
O O O O O O O O ' ' ' ' x ,3 2 2 …….. (constructive interference) ' O O ' O O ' O O
' x 0, ,2 ……… O O (for destructive interference)
10. Most of the problems of interference can be solved by calculating the path difference ''''x and then by putting
' O O ' O O ' O O ' x 0, ,2 …….. O O (constructive interference) O O O O O O O O ' ' ' ' x ,3 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
Z Z Z Z y1 A sin kx1 t Z I Z I Z I Z I y2 A sin kx2 t
The resultant displacement of the point where the waves meet is, Y = Y1 + Y2 Solving we get Z T Z T Z T Z T Y A sin kx t
Here A1A cos2 I I I I A cosTTTT
I T I T I T I T A sin2 A sin Or A2 A1A cos2 IIII 2 A sin2 IIII 2 I I I I A A12 A22 2A A cos1 2 and T T T T TTTT IIII T I TT II T I A sin A sin tan A cos A A cos 2 1 2
13. If wind blows at a speed vw from the source to observer. v oooo v + vw. If in opposite direction v oooo v – vw. § · § · § · § rrrr rrrr · cccc ¨¨¨¨ ¸¸¸¸ © ¹ © ¹ © ¹ © BBBB ¹ v v v f f v v v w s w s
14. When a source is revolving, with velocity vs around a stationary listener
vn n v v max s and vn n v v min s
Beat frequency = nmax – nmin
Similarly when a listener is revolving around source with velocity vL
v v n n v L max and v v n n v L
min (n is frequency of the source)
Beat frequency = nmax – nmin
15. Speed of sound = JJJJ uuuu
3 r.m.s. speed of molecules of air 16. The frequency of note produced by the organ pipe varies
(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 UUUU , where ‘ UUUU ’ is the density of the gas.
(iv) inversely as length of the tube.
17. When a wave is reflected from
(i) rarer medium no change of phase
(ii) denser medium there is phase change of ‘ SSSS ’
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
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 ‘O’ is 1M R
2
2
1 as the moment of inertia of inertia of the complete disc is
also 1M R 2
2
2 (where M1 is mass of the sector and M2 is mass of the
complete disc). This can be shown as in figure. If ‘M’ is the mass of 1 n
th
part of the disc then mass of the disc = nM 1 I nM R 2 2 disc 1 1 I I MR n 2 2 sector disc 4.
Angular velocity of a rigid body ZZZZ is dTTTT
dt . Here ‘ TTTT ’ is the angle between 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 T T T 300, we can proceed as under
OB = x = AB cos TTTT x = 4 cos TTTT dx dt = TTTT § · § · § · § · T TT T ¨¨¨¨ ¸¸¸¸ © ¹ © ¹ © ¹ © ¹ d 4sin dt TTTT d dt = TTTT dx / dt 4sin ZT ZT ZT ZT = 10 4sin 30 = -5 rad/sec
Here –ve sign implies that ‘ TTTT ’ decreases as ‘t’ increases ¨§¨§§¨§¨ TTTT ·¸·¸·¸·¸
© ¹ ©© ¹¹ © ¹ d 0 dt .
5. In law of motion for a single particle acted on by a torque
o oo o o o o o W W W W d L dt Holds only if o o o o
WWWW and ooooL are measured with respect to any point ‘O’ fixed in an inertial frame. 6. o oo o o oo o W WW W d L dt 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 WWWWext 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 WWWWext D D D DI does not hold good in a non-inertial frame. However, there exists a very special case when WWWWext D D D DI does hold even if the angular acceleration ‘ DDDD ’ is measured from a non-inertial frame. That
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 PPPP bears following reaction.
ª º ª º ª º ª º P t T P tP t TT P t «««« »»»» T « » « » « » « » ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ K tan K R 2 2 2
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 TTTT : If surface is smooth
A = g sin TTTT - PPPPg cos TTTT : If surface is rough but friction is insufficient to prevent slipping forward slipping will take place. A = TTTT g sin I 1 MR2
: 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 ' = q 1 1
k
§ ·
¨ ¸
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 =
W q
AB 0
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 = Q 2H A 2 0 or F = 1 E A 2H 2 0
Force per unit area i.e. F
A = 2
V H
2
0
14. Energy stored per unit volume in the electric field is called energy density (u)
u = 1 E
2H
2
0 (where ‘E’ is the field between the plates)
15. 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 = A t t t t d (t t t ... t ) ... k k k k H ª º « » ¬ ¼ 0 2 3 1 n 1 2 3 n 2 2 3 n 16. IN series combination 1 Ceff = 1 1 1 1 ... C1 C2 C3 Cn 17. In parallel combination Ceff = C1 + C2 + C3 + …….+ Cn
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 to ‘V’, energy stored in the condenser is 1qV.
2 This energy resides in the 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.
23. Due to a point charge, E 1 r
v 2 and V 1 r v . 24. Due to an electric dipole, E 1
r v 3 and 1 V r v 2
25. Due to an infinite uniform line of charge, E 1 r
v and V v loge r
26. Due to an infinite plane sheet of charge, Evr0 and Vvr
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.
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 = 1 q q
4SH
¦
rj k 0 all pairs jk
33. 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 K2 is
E = l q
4 T K t K
u
SH0 1 1 2 2 2
Electric potential at the same point in the case described above is
V = 1 q
4SH0 t1 K1t2 K2
34. For maintaining equilibrium of a charged soap bubble, V = 8 T
r H0
and q = 8 r 2S H0rT
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 = (K A K A K A ...)
d
H0 1 1 2 2 3 3
36. Total energy stored in any grouping of capacitors is equal to sum of the energies stored in individual capacitors.
37. 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
38. 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
K times (v) ‘E’ becomes 1
K times (vi) Potential energy becomes 1
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
A R v or A R a constant or A A R R 1 2 1 2
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 t C10 to t20C is given by
D = R R R (t t ) 2 1 1 2 1
where R1, R2 = resistance of conductor at t C10 to t20C 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.
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 = n2 : 1 or Rp : Rs = 1 : n2
14. If equivalent resistance of R1 and R2 in series and parallel by Rs and Rp
respectively, then R1 = 1 R R 4R R 2 ª º « » ¬ ¼ 2 s s s p R2 = 1 R R 4R R 2 ª º « » ¬ ¼ 2 s s s p
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 equal to 2R
3
§ ·
¨ ¸
© ¹.
17. If an equilateral triangle is made of a uniform wire of resistance ‘R’ ; the equivalent resistance between the ends of a side is 2R
9
§ ·
¨ ¸
© ¹.
18. If a wire of resistance ‘R’ is bent in the form of a circle, the effective resistance between the ends of a diameter is R
4 .
19. If a square is made of a uniform wire of resistance ‘R’, the equivalent resistance between the ends of a diagonal is R
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 2n-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 of the cube = 5r
6 ohm.
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 cube = 7r
12 ohm.
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
5 ohm.
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.
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
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.
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 affected to interchanging the positions of battery and galvanometer.
40. The resistance between ‘A’ and ‘C’ of Wheatstone bridge in balanced position is
RAC = (P Q)(R S)
(P Q) (R S)
41. 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
. PP1P2 P3
5. The resistance of a bulb fitted in a house varies inversely as its power i.e. Rv1/ 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.
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
the heaters in parallel the same water will boil in time t = t t .
t t
u
1 2 1 2
13. If Hp and Hs are heat produced by a equal resistors, when they are
connected in parallel and in series respectively, the Hp = 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
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. In general, the efficiency of a cell is given by K = R E
Rr V where R is the external resistance and r is the internal resistance of a cell.
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 =
E R (Rr)
2 2 .
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
