Physics Handbook PDF - JEE Main & Advanced & NEET Formulae
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This is a physics handbook designed for the JEE Main, Advanced, and NEET examinations. It provides a comprehensive overview of various physics topics and their formulae. The handbook contains sections on topics including mechanics, SHM, waves, gravitation, fluid dynamics, heat, optics, electromagnetism, and modern physics.
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PHYSICS HANDBOOK For [JEE Main & JEE Advanced & NEET] Examination MOTION CORPORATE OFFICE Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) : 1800-212-1799, 8003899588 | url : www.motion.ac.in | : [email protected] CONTENTS S.No....
PHYSICS HANDBOOK For [JEE Main & JEE Advanced & NEET] Examination MOTION CORPORATE OFFICE Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) : 1800-212-1799, 8003899588 | url : www.motion.ac.in | : [email protected] CONTENTS S.No. Topic Page No. 1. Mechanics 05 – 63 2. SHM Waves 64 – 76 3. Gravitation, Fluid, Heat 77 - 100 4. Optics 101– 127 5. Electromagnetism 128 – 176 6. Modern Physics 177 – 186 IIT - JEE SYLLABUS GENERAL PHYSICS Units and dimensions, dimensional analysis; least count, significant figures; Methods of measurement and error analysis for physical quantities pertaining to the following experiments: Experiments based on using vernier calipers and screw gauge (micrometer), Determination of g using simple pendulum, Young's modulus by Searle's method, Specific heat of a liquid using calorimeter, focal length of a concave mirror and a convex lens using u-v method, Speed of sound using resonance column, Verification of Ohm's law using voltmeter and ammeter, and specific resistance of the material of a wire using meter bridge and post office box. MECHANICS Kinematics in one and two dimensions (Cartesian coordinates only), projectiles; Circular motion (uniform and non-uniform); Relative velocity. Newton's laws of motion; Inertial and uniformly accelerated frames of reference; Static and dynamic friction; Kinetic and potential energy; Work and power; Conservation of linear momentum and mechanical energy. Systems of particles; Centre of mass and its motion; Impulse; Elastic and inelastic collisions. Law of gravitation; Gravitational potential and field; Acceleration due to gravity; Motion of planets and satellites in circular orbits. Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of uniform bodies with simple geometrical shapes; Angular momentum; Torque; Conservation of angular momentum; Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres; Equilibrium of rigid bodies; Collision of point masses with rigid bodies. SHM , WAVES Linear and angular simple harmonic motions. Wave motion (plane waves only), longitudinal and transverse waves, Superposition of waves; progressive and stationary waves; Vibration of strings and air columns. Resonance; Beats; Speed of sound in gases; Doppler effect (in sound). GRAVITATION , FLUID , HEAT Pressure in a fluid; Pascal's law; Buoyancy; Surface energy and surface tension, capillary rise; Viscosity (Poiseuille's equation excluded), Stoke's law; Terminal velocity, Streamline flow, Equation of continuity, Bernoulli's theorem and its applications. Thermal expansion of solids, liquids and gases; Calorimetry, latent heat; Heat conduction in one dimension; Elementary concepts of convection and radiation; Newton's law of cooling; Ideal gas laws; Specific heats (Cv and Cp for monatomic and diatomic gases); Isothermal and adiabatic processes, bulk modulus of gases; Equivalence of heat and work; First law of thermodynamics and its applications (only for ideal gases). Blackbody radiation: absorptive and emissive powers; Kirchhoff's law, Wien's displacement law, Stefan's law. Hooke's law, Young's modulus. OPTICS Rectilinear propagation of light; Reflection and refraction at plane and spherical surfaces; Total internal reflection; Deviation and dispersion of light by a prism; Thin lenses; Combinations of mirrors and thin lenses; Magnification. Wave nature of light: Huygen's principle, interference limited to Young's double-slit experiment. ELECTROMAGNETISM Coulomb's law; Electric field and potential; Electrical Potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field, Electric field lines; Flux of electric field; Gauss's law and its application in simple cases, such as, to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Capacitance; Parallel plate capacitor with and without dielectrics; Capacitors in series and parallel; Energy stored in a capacitor. Electric current: Ohm's law; Series and parallel arrangements of resistances and cells; Kirchhoff's laws and simple applications; Heating effect of current. Biot-Savart law and Ampere's law, magnetic field near a current-carrying straight wire, along the axis of a circular coil and inside a long straight solenoid; Force on a moving charge and on a current-carrying wire in a uniform magnetic field. Magnetic moment of a current loop; Effect of a uniform magnetic field on a current loop; Moving coil galvanometer, voltmeter, ammeter and their conversions. Electromagnetic induction: Faraday's law, Lenz's law; Self and mutual inductance; RC, LR and LC circuits with d.c. and a.c. sources. MODERN PHYSICS Atomic nucleus; Alpha, beta and gamma radiations; Law of radioactive decay; Decay constant; Half-life and mean life; Binding energy and its calculation; Fission and fusion processes; Energy calculation in these processes. Photoelectric effect; Bohr's theory of hydrogen-like atoms; Characteristic and continuous X-rays, Moseley's law; de Broglie wavelength of matter waves. PART-I MECHANICS 1. PHYSICAL QUANTITY The quantities which can be measured by an instrument and by means of which we can describe the laws of physics are called physical quantities. Types of physical quantities : Fundamental Derived Supplementry Seven physical fundamental quantities (1) length (2) mass (3) time (4) electric current, (5) thermodynamic temperature (6) amount of substance (7) luminous intensity 2. DEFINITIONS OF SOME IMPORTANT SI UNITS (i) Metre : 1 m = 1,650, 763.73 wavelengths in vaccum, of radiation corresponding to organ-red light of krypton-86. (ii) Second : 1 s = 9,192, 631,770 time periods of a particular form Ceasium - 133 atom. (iii) Kilogram : 1kg = mass of 1 litre volume of water at 4°C (iv) Ampere : It is the current which when flows through two infinitely long straight conductors of negligible cross-section placed at a distance of one metre in vacuum produces a force of 2 × 10–7 N/m between them. (v) Kelvin : 1 K = 1/273.16 part of the thermodynamic temperature of triple point of water. (vi) Mole : It is the amount of substance of a system which contains as many elementary particles (atoms, molecules, ions etc.) as there are atoms in 12g of carbon - 12. (vii) Candela : It is luminous intensity in a perpendicular direction of a surface of 1 2 m of a black body at the temperature of freezing point under a pressure 600000 of 1.013 × 105 N/m 2. : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 5 (viii) Radian : It is the plane angle between two radii of a circle which cut-off on the circumference, an arc equal in length to the radius. (ix) Steradian : The steradian is the solid angle which having its vertex at the centre of the sphere, cut-off an area of the surface of sphere equal to that of a square with sides of length equal to the radius of the sphere 3. CHARACTERISTICS OF BASE UNITS OR STANDARDS : Some special types of units : 1. 1 Micron (1) = 10–4 cm = 10–6m (length) 2. 1 Angstrom (1A) = 10–8 cm = 10–10m (length) 3. 1 fermi (1 f) = 10–13 cm = 10–15m (length) 4. 1 inch = 2.54 cm (length) 5. 1m = 39.37 inch = 3.281feet (length) 6. 1 mile = 5280 feet = 1.609 km (length) 7. 1 atmosphere = 105 N/m 2 = 76 torr = 76 mm of Hg pressure (pressure) 8. 1 litre = 10–3 m 3 = 1000 cm3 (volume) 9. 1 carat = 0.0002 kg (weight) 10. 1 pound (lb) = 0.4536 kg (weight) Different quantities with units. symbol and dimensional formula. Quantity Symbol Formula S.I. Unit D.F. Disp. s Metre or m M0LT0 Area A ×b (Metre)2 or m2 M0L2T0 Volume V ×b×h (Metre)3 or m3 M0L3T0 s Velocity v v m/s M0LT–1 t Momentum p p = mv kgm/s MLT–1 v Acceleration a a m/s2 M0LT–2 t Force F F = ma Newton or N MLT–2 Impulse - F×t N.sec MLT–1 Work W F.d N.m ML2T–2 1 Energy KE K.E. mv 2 Joule or J ML2T–2 2 or U P.E. = mgh Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 6 W Power P P watt or W ML2T–3 t Density d d = m/v kg/m3 ML–3T0 dv Coefficient F = A kg/ms(poise ML–1T–1 dx of viscosity in C.G.S.) Gm1 m2 N m2 Gravitation G F M–1L3T–2 r2 kg 2 4. APPLICATION OF DIMENSIONS (a) To find out the unit of a physical quantity (b) To derive the dimensions of a physical constant (c) To check the dimensional correctness of a given physical equation : Every physical equation should be dimensionally balanced. This is called the 'Principle of Homogeneity'. The dimensions of each term on both sides of an equation must be the same. (d) Conversion of units : This is based on the fact that the numerical value (n) and its corresponding unit (u) is a constant, i.e., n[u] = constant or n1[u1] = n2 [u2] (e) To establish the relation among various physical quantities : If we know the factors on which a given physical quantity may depend, we can find a formula relating the quantity with those factors 5. LIMITATION OF DIMENSION ANALYSIS (i) The method works only if the dependence is the product type. 1 2 e.g. s = ut + at 2 (ii) The numerical constants having no dimensions cannot be deduced by the method of dimensions. e.g. = 2 (iii) The methods works only if there are as many equation available as there are unknown. : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 7 1. LOGARITHMS : (i) e 2.7183 (ii) If ex = y, then x = loge y = ln y (iii) If 10x = y, then x = log10y (iv) log10y = 0.4343 loge y = 2.303 log10y (v) log (ab) = log (a) + log (b) a (vi) log = log (a) – log (b) b (vii) log an = n log (a) 2. ANGLE CONVERSION FORMULAS 1 degree = ( 0.02) radian Degrees to radians : multiply by 180 180 1 radian 57 degrees 180 Radians to degrees : multiply by Measurement of positive & Negative Angles : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 8 3. IMPORTANT FORMULAS (i) sin2 + cos2 = 1 (ii) 1 + tan2 = sec2 (iii) 1 + cot2 = cosec2 sin A sin B sin C Sine Rule a b c Cosine rule a2 = b2 + c2 – 2bc cos A 4. SMALL ANGLE APPROXIMATION It is a useful simplification which is only approximately true for finite angles. It involves linearrization of the trigonometric functions so that, when the angle is measured in radians. sin ~ 2 cos ~ 1 or cos ~ 1 – 2 for the second - order approximation tan ~ Range 1° to 10° are small angle 5. BINOMIAL THEOREM : n(n – 1)x 2 (1 ± x)n = 1 ± nx +........... 2! : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 9 n(n 1) 2 (1 ± x)–n = 1 nx + x......... 2! If x vbr ; v v smin = w r at = sin –1 br v br vr (c) Flag Problems : If flag will flutter in direction of vwc then vwc = vw – vc : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 25 (d) Aircraft Wind Problems v a = v aw v w v aw = velocity of aircraft with respect to wind or velocity of aircraft in still air v w = velocity of wind v a = absolute velocity of aircraft (e) Rain Problems In these type of problems we again come across three terms vr , vm and vrm , Here, vr = velocity of rain vm = velocity of man (it may be velocity of cyclist or velocity of motorist also) and vrm = velocity of rain with respect to man. 14. CONDITION FOR COLLISION OF TWO PROJECTILES : For collision to take place, in following figure h2 h1 tan = tan x Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 26 1. STRING CONSTRAINT : Ist format : - (when string is fixed) IInd format (when pulley is also moving) vAP = – vBP (–ve sign indicate the direction of each block is opposite with respect to Pulley) v A vB vA – vp = – vB + vP vP = 2 III format : 1. First choose the longest string in the given problem which contains the point of which velocity/acceleration to be find out. 2. Now mark a point on the string wherever if comes in contact or leaves the contact of real bodies. 3. If due to motion of a point, length of the part of a string with point is related, increases then its speed will be taken +ve otherwise –ve. 2. WEDGE CONSTRAINED : Components of velocity and acceleration perpendicular to the contact surface of the two objects is always equal if there is no deformation and they remain in contact. : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 27 1. FORCE A pull or push which changes or tends to change the state of rest or of uniform motion or direction of motion of any object is called force. Force is the interaction between the object and the source (providing the pull or push). It is a vector quantity. Effect of resultant force : may change only speed may change only direction of motion. may change both the speed and direction of motion. may change size and shape of a body kg. m unit of force : newton and (MKS System) s2 g. cm dyne and (CGS System) s2 1 newton = 105 dyne Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 28 2. NORMAL REACTION Normal reaction always act perpendicular to contact surface in such a direction that it always tries to compress the body. 3. TENSILE FORCE in following figure T = mg (i) Tension always acts along the string and in such a direction that it tries to reduce the length of string (ii) If the string is massless then the tension will be same along the string but if the string have some mass then the tension will continuously change along the string. 4. NEWTON’S FIRST LAW OF MOTION : According to this law “A system will remain in its state of rest or of uniform motion unless a net external force act on it. 1st law can also be stated as “If the net external force acting on a body is zero, only then the body remains at rest.” 5. NEWTON’S SECOND LAW OF MOTION : Newton’s second law states, “The rate of change of a momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts” dp dp i.e., F or F k dt dt where k is a constant of proportionality. p mv, So F k (dmv ) dt dv For a body having constant mass, F km k ma dt From experiments, the value of k is found to be 1. So, Fnet ma : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 29 6. APPLICATIONS OF NEWTON'S LAW : 6.1 Motion of a Block on a Horizontal Smooth Surface. Case (i) : When subjected to a horizontal pull: F = ma or a = F/m Case (ii) : When subjected to a pull acting at an angle ( ) to the horizontal : 6.2 MOTION OF BODIES IN CONTACT. Case (i) : Two body system : F m 2F a = m m and f = m m 1 2 1 2 Case (ii) : Three body system : F a = m m m 1 2 3 6.3 Motion of a body on a smooth inclined plane : a = g sin , N = mg cos Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 30 6.4 PULLEY BLOCK SYSTEM If m 1 > m 2 (m1 m 2 )g a (m1 m2 ) T = m 2(a + g) 7. NEWTONS’ 3RD LAW OF MOTION : Statement : “To every action there is equal and opposite reaction”. According to Newton third law two forces are equal in magnitude and opposite in direction FAB –FBA 7.1 Climbing on the Rope : T > mg man accelerates in upward direction, T < mg man accelerates in downward direction 7.2 SPRING FORCE : 1. When spring is in its natural length, spring force is zero. F x Graph between spring force v/s x : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 31 2. Cutting of the spring (i) when the block A is released then it take some finite time to reach at B. i.e., spring force doesn’t change instantaneously. (ii) When point A of the spring is released in the above situation then the spring forces. changes instantaneously become zero because one end of the spring is free. (iii) In string tension may change instantaneously. 8. PSEUDO FORCE : To apply newton's first law and second law in non inertial frame we introduce a new type of force that is pseudo force. Pseudo force = ma2 N – ma2 = m(a1 – a2) N = ma1 Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 32 1. FRICTION Friction is a contact force that opposes the relative motion or tendency of relative motion of two bodies. 1.1 Static frictional force When there is no relative motion between the contact surfaces, frictional force is called static frictional force. It is a self-adjusting force The direction of static friction on a body is such that the total force acting on it keeps it at rest with respect to the body in contact. : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 33 1.2 Limiting Frictional Force This frictional force acts when body is about to move. This is the maximum frictional force that can exist at the contact surface. (i) The magnitude of limiting frictional force is proportional to the normal force at the contact surface. flim N flim = sN Here s is a constant the value of which depends on nature of surfaces in contact and is called as ‘coefficient of static friction’. 1.3 Kinetic Frictional Force Once relative motion starts between the surface in contact, the frictional force is called as kinetic frictional force. The magnitude of kinetic frictional force is also proportional to normal force. fk = kN From the previous observation we can say that k < s Although the coefficient of kinetic friction varies with speed, we shall neglect any variation i.e., when relative motion starts a constant frictional force starts opposing its motion. 2. MINIMUM FORCE REQUIRED TO MOVE THE PARTICLE : mg Fmin F = mg / (cos + sin) 1 2 3. FRICTION AS THE COMPONENT OF CONTACT FORCE : The perpendicular component is called the normal contact force or normal force and parallel component is called friction. Fc max = 2N2 N2 {when fmax = N} Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 34 4. MOTION ON A ROUGH INCLINED PLANE a = g sin – g cos ANGLE OF REPOSE : when, mg sin c = mg cosc tan c = where C is called angle of repose : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 35 1. Angular Velocity (i) Average Angular Velocity Total Angle of Rotation av = ; Total time taken 2 – 1 av = t – t = 2 1 t where 1 and 2 are angular position of the particle at time t1 and t2 respectively. (ii) Instantaneous Angular Velocity The rate at which the position vector of a particle with respect to the centre rotates, is called as instantaneous angular velocity with respect to the centre. lim = d = t0 t dt Relation between speed and angular velocity : v v = r is a scalar quantity ( ) r 2. Relative Angular Velocity ( VAB ) AB rAB here VAB Relative velocity to position vector AB 3. Angular Acceleration : (i) Average Angular Acceleration : Let 1 and 2 be the instantaneous angular speeds at times t1 and t2 respectively, then the average angular acceleration av is defined as 2 – 1 av = t – t = 2 1 t Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 36 (ii) Instantaneous Angular Acceleration : It is the limit of average angular acceleration as t approaches zero, i.e., lim = d = d = t0 t dt d Important points : It is also an axial vector with dimension [T–2] and unit rad/s2 If = 0, circular motion is said to be uniform. d d d2 As = , = = , dt dt dt 2 i.e., second derivative of angular displacement w.r.t time gives angular acceleration. is a axial vector and direction of is along if increases and opposite to if decreases 4. Radial and tangential acceleration dv at = rate of change of speed dt 2 2 v and ar 2r r v r r Following three points are important regarding the above discussion : dv (i) In uniform circular motion, speed (v) of the particle is constant, i.e., 0. dt Thus, at = 0 and a = ar = r2 dv (ii) In accelerated circular motion, = positive, i.e., at is along e t or tangential dt dv acceleration of particle is parallel to velocity v because v r e t and at ˆ e dt t : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 37 dv (iii) In decelerated circular motion, = negative and hence, tangential acceleration is dt anti-parallel to velocity v. Relation between angular acceleration and tangential acceleration dv d at = =r or at = r dt dt 5. Relations among Angular Variables These relations are also referred as equations of rotational motion and are - = 0 + t...(1) 1 2 = 0t + t...(2) 2 2 = 02 + 2...(3) Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 38 6. Centripetal Force : Concepts : This is necessary resultant force towards the centre called the centripetal force. mv 2 F= = m2r r (i) A body moving with constant speed in a circle is not in equilibrium. (ii) It should be remembered that in the absence of the centripetal force the body will move in a straight line with constant speed. (iii) It is not a new kind of force which acts on bodies. In fact, any force which is directed towards the centre may provide the necessary centripetal force. 7. Centrifugal Force : Centrifugal force is a fictitious force which has to be applied as a concept only in a rotating frame of reference to apply Newton’s law of motion (in that frame) FBD of ball w.r.t non inertial frame rotating with the ball. 8. SIMPLE PENDULUM : T – mgcos = mv2/L or, T = m(gcos + v2/L) 2 mv 2 |Fnet | = (mg sin )2 L 2 2 v4 = m g sin L2 : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 39 9. CONICAL PENDULUM : r g speed v = and (L2 – r 2 )1/ 4 mgL Tension T = (L – r 2 )1/ 2 2 10. CIRCULAR TURNING ON ROADS : Centripital force are provided by following ways. 10.1 By Friction Only: For a safe turn without sliding safe speed v rg 10.2 By Banking of Roads Only v = rg tan 10.3 By Friction and Banking of Road Both rg( tan ) Maximum safe speed vmax = , (1 – tan ) rg( – tan ) Minimum safe speed vmin = (1 tan ) 11. DEATH WELL : mv 2 mv 2 N= , f = mg , fmax = R R Cyclist does not drop down when mv 2 gR fmax mg mg , v R Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 40 1. Work Done By a constant force W F. s F. ( r f r i ) Fs cos = Force × displacement in the direction of force. In cgs system, the unit of work is erg. In mks system, the unit of work is Joule. 1 erg = 10–7 joule 2. WORK DONE BY A VARIABLE FORCE : F Fx i Fy j Fzk ds dx i dy j dzk xB yB zB W= Fxdx + Fy dy + F dz z xA yA zA 3. AREA UNDER FORCE DISPLACEMENT CURVE : Graphically area under the force-displacement is the work done The work done can be positive or negative as per the area above the x-axis or below the x-axis respectively. : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 41 4. DIFFERENCEBETWEEN CONSERVATIVEAND NON COSERVATIVE FORCES S. Conse rvative forces Non-Conse rvative forces No. 1 W ork done does not depend upon path W ork done depends on path. 2 W ork done in a round trip is zero. W ork done in a round trip is not zero. 3 Central in nature. Forces are velocity-dependent and retarding in nature. 4 W hen only a conservative force acts W ork done against a non- within a system, the kinetic energy and conservative force may be potential energy can change. However disipiated as heat energy. their sum, the mechanical energy of the system, does not change. 5 W ork done is completely recoverable. W ork done in not completely recoverable. 5. WORK DONE BY CONSERVATIVE FORCES Ist format : (When constant force is given) dw = F.d r ( dr dx i dyj dzk ) II format : (When F is given as a function of x, y, z) dw = (Fx i Fy j Fzk ).(dx i dyj dzk ) dw = Fxdx + Fydy + FZdz IIIrd format (perfect differential format) dw = ( y i xj ).(dx i dyj ) dw = ydx + xdy 6. ENERGY A body is said to possess energy if it has the capacity to do work. When a body possessing energy does some work, part of its energy is used up. Conversely if some work is done upon an object, the object will be given some energy. Energy and work are mutually convertiable. 6.1 Kinetic Energy 1 K.E. = mv 2 2 Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 42 6.2 Potential Energy W.D = – U where U is change in potential energy 6.2.1 Gravitational Potential Energy : It is possessed by virtue of height. GPE = ± mgh 6.2.2 Elastic Potential Energy : It is a property of stretched or compressed springs. 1 2 Elastic Potential Energy = kx 2 7. Relation Between Potential Energy (U) and Conservation Force (F) dU (i) If U is a function of only one variable, then F slope of U-r graph dr (ii) If U is a function of three coordinate variables x, y and z, then U U U F î ĵ k̂ x y z : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 43 8. EQUILIBRIUM Physical Stable Unstable Neutral situation equilibrium equilibrium equilibrium Net force Zero Zero Zero Potential energy Minimun Maximum Constant When displaced A restoring nature A force will act Force is again mean (equilibrium) of force will act on which moves the zero position. the body, which body away from brings the body mean position. back towards mean position. In U-r graph At point B At point A At point C U B A C r In F-r graph At point A At point B At point C F C A B r 9. WORK ENERGY THEOREM : W net = K Work done by net force Fnet in displacing a particle equals to the change in kinetic energy of the particle i.e. (W.D)c + (W.D)N.C + (W.D)ext. + (W.D)pseudo = K where (W.D)c = work done by conservative force 10. Power of a Force (i) Average power Total work done WTotal : Pav Total time taken t Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 44 (ii) Instantaneous power dW Pins. = rate of doing work done F. v F v cos dt 11. Vertical Circular Motion A bob of mass m is suspended from a light string of length R as shown. if velocity at bottommost point of bob is u, then depending on the value of u following three cases are possible (i) If u 5gR , bob will complete the circle. (ii) If 2gR < u < 5 gR , string will slack between (iii) If u 2gR , bob will oscillate between CAB. In this case v = 0 but T 0. If u 5gR ,bob will just complete the circle, In this case, velocity at topmost point is v gR , Tension in this critical case is zero at topmost point and 6 mg at bottommost point. (iv) Condition for the body to reach B : if u 2gR then the body will oscillate about A. (v) At height h from bottom velocity of bob will be, v u 2 – 2gh u2 – 2gR (vi) 2gR < u < 5 gR then cos = v 3gR It is the angle from the vertical at which tension in the string vanishes to zero. And after that its motion is projectile. (vii) When u 5 gR Tension at A : TA = 6mg Tension at B : TB = 3mg : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 45 1. Centre of Mass of a System of ‘N’ Discrete Particles : m r m 2 r2 ...........mn rn rcm 1 1 m1 m 2 .........mn where, mi ri is called the moment of mass of particle with respect to origin. n xCOM = m x i i i 1 M n n Similarly, yCOM = i 1 mi y i and m z i i i1 z COM M M 2. Position of COM of two particles : - 0 m2l m2l r1 = m m = m m...(1) 1 2 1 2 m2l m1l r2 = l – = m m...(2) m1 m 2 1 2 m 1r1 = m 2r2...(3) Centre of mass of two particle system lie on the line joining the centre of mass of two particle system. Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 46 3. Centre of Mass of a Continuous Mass Distribution For continuous mass distribution the centre of mass can be located by replacing summation sign with an integral sign. Proper limits for the integral are chosen according to the situation xcm xdm , y cm y dm , z cm zdm dm dm dm dm = M (mass of the body) here x,y,z in the numerator is the coordinate of the centre of mass of the dm mass. 1 rcm = M rdm 2R (a) C.O.M of a semicircular Ring ycm = 4R (b) C.O.M of a semicircular Disc ycm = 3 : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 47 R (c) C.O.M. of a Hollow Hemisphere ycm = 2 3H (d) C.O.M. of mass of a solid cone = 4 3R (e) C.O.M. of a solid Hemisphere ycm = 8 Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 48 2H (f) C.O.M. of mass of triangular plate ycm = 3 4. CAVITY PROBLEMS : If some mass or area is removed from a rigid body then the position of centre of mass of the remaining portion is obtained by assuming that in a remaining part +m & – m mass is there. 5. Velocity of C.O.M of system : To find the velocity of centre of mass we differentiate equation (1) with respect to time dr1 dr2 dr3 m1 m2 m3 ....... drcom dt dt dt dt m1 m 2 m3 ......... d r1 dr dr m1 m 2 2 m 3 3 ....... Vcom dt dt dt m1 m2 m3 ......... m v m2 v 2 m3 v 3 ...... Vcom 1 1...(2) m1 m2 m3 ... : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 49 6. Acceleration of centre of mass of the system : - To find the acceleration of C.O.M we differentiate equation (2) dv dv dv m1 1 m 2 2 m 3 3 ...... dVcom dt dt dt dt m1 m2 m 3 ........ m1a1 m 2 a 2 m 3 a 3 ...... a com ...(3) m1 m 2 m 3 ........ Now (m 1 + m 2 + m 3) a com = m1a1 m 2 a 2 m 3 a 3 ...... Fnet( system) F1net F2 net F3 net ........ The internal forces which the particles exert on one another play absolutely no role in the motion of the centre of mass. Case I : If Fnet = 0 then we conclude : (a) The acceleration of centre of mass is zero ( a com 0 ) If a1, a2, a3.... is acceleration of m 1, m 2, m 3 mass in the system then a1, a2, a3 may or may not be zero. (b) K.E. of the system is not constant it may change due to internal force. (c) Velocity of centre of mass is constant ( v com cons tan t) but v1, v2, v3 may or may not constant. It may be change due to internal force. from eq (2) m1v 1 m 2 v 2 m 3 v 3 .......... cons tan t This is called momentum conservation. "If resultant external force is zero on the system, then the net momentum of the system must remain constant". Case II : When centre of mass is at rest. Vcom 0 then Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 50 drcom 0 rcom = constant. dt i.e. r1, r2 , r3......... may or may not change 7. SPRING BLOCK SYSTEM : 2 Maximum extension x0 = v 0 m 3k 2mF xmax = x1 + x2 = k (m M) 8. IMPULSE : Impulse of a force F acting on a body for the time interval t = t1 to t = t2 is defined as t2 I= t1 F dt dv I = F dt m dt md v dt I = m(v 2 v1 ) P change in momentum due to force F t2 Also (impulse - momentum theorem) IRe = t1 FRe s dt P : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 51 9. Impulsive force : A force, of relatively higher magnitude and acting for relatively shorter time, is called impulsive force. An impulsive force can change the momentum of a body in a finite magnitude in a very short time interval. Impulsive force is a relative term. There is no clear boundary between an impulsive and Non-Impulsive force. N1, N3 = Impulsive; N2 = non-impulsive Both normals are Impulsive 10. COEFFICIENT OF RESTITUTION (e) The coefficient of restitution is defined as the ratio of the impulses of reformation and deformation of either body. Im pulse of reformation F dt r e= = Im pulse of deformation F dt d Velocity of separation of point of contact e Velocity of approach of point of contact 11. Line of Motion The line passing through the centre of the body along the direction of resultant velocity. Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 52 12. Line of Impact The line passing through the common normal to the surfaces in contact during impact is called line of impact. The force during collision acts along this line on both the bodies. Direction of Line of impact can be determined by : (a) Geometry of colliding objects like spheres, discs, wedge etc. (b) Direction of change of momentum. If one particle is stationary before the collision then the line of impact will be along its motion after collision. velocity of seperation along line of impact e= velocity of approach along line of impact 13. Conservation of Linear Momentum (i) For a single mass or single body If net force acting on the body is zero. Then, p = constant or v = constant (if mass = constant) (ii) For a system of particles or system of rigid bodies If net external force acting on a system of particles or system of rigid bodies is zero, then, PCM cons tan t or v CM = constant 14. COLLISION In every type of collision only linear momentum remains constant (i) Head on elastic collision In this case linear momentum and kinetic energy both are conserved. After solving two conservation equations. we get : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 53 m – m2 2m 2 v 1' 1 v 1 v 2 m1 m2 m1 m 2 m – m1 2m1 and v ' 2 2 v 2 v 1 m 1 m 2 m1 m 2 (ii) Head on inelastic collision (a) In an inelastic collision, the colliding particles do not regain their shape and size completely after collision. (b) Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus, the kinetic energy of the particles no longer remains conserved. (c) However, in the absence of external forces, law of conservation of linear momentum still holds good. (d) (Energy loss)Perfectly Inelastic > (Energy loss)Partial Inelastic (e) 0 R, the motion is said to be forward slipping and if vp < vQ < R, the motion is said to be backward slipping. Now, suppose an external force is applied to the rigid body, the motion will no longer remain uniform. The condition of pure rolling on a stationary ground is, a = R Thus, v = R, a = R is the condition of pure rolling on a stationary ground. Sometime it is simply said rolling. Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 62 12.1 Pure rolling when force F act on a body : a = linear acceleration, = angular acceleration from linear motion : F + f = Ma from rotational motion : Fx – f R = I F(R x) F(x – RC) a= , f= MR(C 1) R(C 1) 12.2 Pure Rolling on an Inclined Plane: g sin a= 1 c So body which have low value of C have greater acceleration. Note : We can represent the moment of inertia of a different rigid body in a following way. I = CMR2 value of C = 1 for circular ring (R), 1 C= for circular disc (D) and solid cylinder (S.C.) 2 2 2 C= for Hollow sphere (H.S) , C= for solid sphere (S.S) 3 5 13. TOPPLING Torque about E Fb = (mg) a Fb or a= mg F increases Fb + N (a – x) = mg a if x = a Fmax b = mga mga Fmax = b : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 63 PART-II SHM, WAVES 1. Different Equations In SHM (i) F = – kx F k (ii) a x 2 x m m k (iii) angular frequency of SHM m d2 x (iv) 2 x dt 2 (v) F x or a x is the sufficient and necessary condition for a periodic motion to be simple harmonic. dx dv (vi) If x = A sin t then v A cos t and a 2 A sin t dt dt From these three equations we can see that x-t, v-t and a-t all three functions oscillate simple harmonically with same angular frequency , Here x oscillates between +A and –A, v between +A and –A and a between +2A and –2A. 2. x A sin t , x A sin t x A cos t , x A cos t Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 64 3. If x A sin t. Then, v A cos t and a 2 A sin t. Corresponding x-t, v-t and a-t graphs are shown. x v A +A T/2 3T/4 T/2 T t t T/4 T T/4 3T/4 –A 2 A T , a 2 A T/4 T t T/2 3T/4 2 A 4. Physical At mean At extreme At general quantity position position point Speed A zero A2 x2 Acceleration zero ±2A –2x Force zero ±kA – kx 1 2 1 Kinetic energy kA zero k (A2 x2 ) 2 2 1 m 2 A 2 2 1 1 2 Potential energy U0 U0 k A2 U0 kx 2 2 1 1 1 Total mechanical U0 k A2 U0 k A2 U0 k A2 2 2 2 energy U0 or potential energy at mean position can be zero. : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 65 5. Potential energy versus x or kinetic energy versus x graph is parabola. While total energy versus x graph is a straight line as it remains constant. 6. Spring Block System k 2 m 1 1 k (i) ,T 2 ,f m k T 2 m k (ii) or or m h oot sm m In all three cases, T 2 k (iii) Parallel combination k1 k2 or In both cases ke = k1 + k2 Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 66 (iv) Series combination k1 k 2 1 1 1 ke or k1 k2 k1 k 2 k e k1 k 2 (v) In case of two body oscillation, T 2 where, k m1 m 2 = Reduced mass of two blocks m1 m 2 (vi) A plank of mass m and area of cross section A is floating in a liquid of density . When depressed, it starts oscillating like a spring-block system. Effective value of k in this case is k Ag m T 2 Ag ms m (vii) If mass of spring m s is also given, then T 2 3 k YA (viii) Every wire is also like a spring of force constant given by k . l From here itself we conclude that force constant of a spring is inversely proportional to its length. If length of spring is halved its force constant will become two times. 7. Pendulum (i) Only small oscillations of a pendulum are simple harmonic in nature : l T 2. g : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 67 (ii) Second’s pendulum is one whose time period is 2 s and length is 1 m. x A sin t , x A sin t x A cos t , x A cos t (iii) Time period of a pendulum of length of the order of radius of earth is 1 T 2 1 1 g l R From here we can see that R T 2 or 84.6 min if l . g R Hence time period of a pendulum of infinite length is 2 g or 84.6 min, Further, l T 2 if l 1 Fmed < Fair r 2. ELECTRIC FIELD : If a charge q0 placed at a point in electric field, experiences a net force F on it, then electric field strength at that point can be F or E....(1) q0 [q0 test charge] (a) Electric Field Strength due to Point Charge : Kq E x2 (b) Vector Form of Electric field due to a Point Charge : Kq E 3.x x 2.1 Graph of electric field due to binary charge configuration 1. 2. 3. 4. : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 129 2.2 Electric field Strength at a General Point due to a Uniformly Charged Rod : KQ in ||-direction Ex = (cos 2 – cos 1 ) Lr k = (cos 2 – cos 1 ) r KQ in -direction Ey = (sin 1 sin 2 ) Lr k = (sin 1 sin 2 ) r r is the perpendicular distance of the point from the wire 1 and 2 should be taken in opposite sense 2.3 Electric field due to infinite wire ( >> r) k 2k Enet at P (1 1) r r 2.4 Electric field due to semi infinite wire 2 k Enet at P r Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 130 2.5 Electric field due to Uniformly Charged Ring : KQx EP = (along the axis) (R x 2 ) 3 / 2 2 2.6 Electric field Strength due to a Uniformly Surface Charged Disc : x E = 2 1 – 2 [along the axis] 2 0 x R Case (i) : If x >> R KQ E= x2 Case (ii) : If x f1) with the same intensity. The stopping potential increases linearly with frequency. Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 178 (iv) The figure shows that the maximum 1 2 kinetic energy mv max of ejected 2 electron is directly proportional to the frequency f of the incident light if it is greater than the threshold frequency. 1 2 mv max eV0 h ( v v th ) 2 FORCE DUE TO RADIATION (PHOTON) (A) Case (I) : a = 1, r=0 F IA I pressure = = = A cA c Case : (II) when r = 1, a = 0 F 2IA 2I pressure P A cA c Case : (III) When 0 < r < 1 a+r=1 IA 1 I Now pressure P (1 r ) = (1 r ) c A c (B) Case - I a = 1, r = 0 F cos IA cos2 I 2 Pressure = P= = cos A cA c Case II When r = 1, a = 0 2IA cos2 2I cos2 Pressure , P cA c : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 179 Case III 0 < r < 1, a + r = 1 Pressure I cos 2 I cos2 I cos 2 (1 – r ) 2r (1 r ) c c c r 2I (C) Pressure = c Note that integration is done only for the hemisphere that faces the incident beam ATOMIC LINE SPECTRA In 1885, Balmer discovered a formula for the wavelengths of certain spectral lines observed coming from excited hydrogen. l 1 1 R 2 2 for n = 3, 4 , 5....... 2 n where, R = 1.097 × 107 m –1 is called the Rydberg constant. Later on, other similar formulae were also discovered for hydrogen atoms. These are Lyman Series, l 1 1 R 2 2 for n = 2, 3 , 4........ 1 n Paschen Series l 1 1 R 2 2 for n = 4, 5, 6........ 3 n Brackett Series l 1 1 R 2 2 for n = 5, 6, 7....... 4 n Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 180 The Bohr Model (a) The electrons move in circular orbits and the centripetal force is provided by the Coulomb force of attraction between then nucleus and electron. (b) Only certain orbits are allowed in which the angular momentum of the electron h is an integral multiple of 2 nh that is, mvr = where n = 1, 2, 3,.......... 2 (c) Energy is emitted or absorbed when the electron jumps from one level to another. For a transition from a higher orbit to a lower orbit, the photon is emitted and the energy difference is hf, where f is the frequency of the photon. (d) Photon absorption causes a transition of the electron to a higher orbit but only if the enegy and frequency are just appropriate. Bohr's Formulae (a) The radius of the nth orbit is defined as n2 rn 0.53 Å Z (b) The speed of electron in n th orbit is Z c vn n 137 Where c = 3 × 108 ms–1 is the speed of light. 2.19 10 6 Z or vn ms 1 n The speed of electron is independent of its mass. (c) The total energy of an atom is the sum total of kinetic and potential. If the electron is in the n th shell, then Z2 U E –13.6 ( eV ) or E K U K n2 2 and the total energy is directly proportional to the mass of the electron. : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 181 If the potential energy is assumed to be zero at n = 1, then the total energy of the atom will be 13.6 Z 2 1 E– 27.2 Z 2 or E 2 2 13.6 Z 2 n2 n (d) The wavelenth of photon emitted when electron jumps from n = n2 to n = n1 is given by 1 E E 2 E1 13.6 Z 2 1 1 2 2 hc hc hc n1 n2 1 1 1 or R Z2 2 2 n1 n 2 If the nucleus of atom is not considered to be stationary, then equations for radius energy and wavelength may be modified as n2 m rn 0.53 1 Å Z M Z2 M En – 13.6 eV n2 m M 1 R 1 1 Z 2 2 2 m 1 m n1 n2 1 M where m is the mass of electron, and M is the mass of nucleus. Note : (i) Total number of emission lines from some excited state n1 to another energy state (n1 n2 ) (n1 n2 1) n2 ( < n1) is given by 2 n(n 1) For example total number of lines from n1 = n to n2 = 1 are 2 Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 182 (ii) As the principal quantum number n is increased in hydrogen and hydrogen like atoms, some quantities are decreased and some are increased. The table given below show which quantities are increased and which are decreased. Increased Decrease d Radius Speed Potential ene rgy Kinetic ene rgy Total energy Angualr speed Time period Angular momentum X-Rays X-rays are highly penetrating electromagnetic radiations of wavelength of the order of 1 Å hc hc That is, min 1/ 2 mv 2 eV The cut-off wavelength is independent of the target material. It depends only on the kinetic energy of the electron beam. The characteristic spectrum depends only on the target material, and it is independent of the kinetic energy of the electron beam. The figure shows the variation of intensity I of X-rays as a function of their wavelength . The graphs shows a minimum wavelength min ( called out-off wavelength) but there is no maximum wavelength. The graph shows two basic features. (a) a continuous curve with min shows continuous spectrum of X-rays (b) two sharp peaks of high intensity show characteristic spectrum. : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 183 (c) Variation of l versus the kinetic energy of the electron beam is increased from K1 to K2 keeping the same target. It is observed that cutoff wavelength of continuous spectrum decreases (2 < 1) but the characteristic spectrum remains unchanged. (d) Variation of I versus The target material is changed (Z2 > Z1) but the kinetic energy of the electron beam remains unchanged. It is observed that the cut-off wavelength remains same but the characteristic spectrum is the unique property of tha target material. Moseley's Law f a( Z b ) 1 1 R ( Z 1)2 1 2 n n = 2, 3, 4.............. Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) Page : 184 Radioactivity (a) After n half lives, n 1 (i) number of nuclei left = N0 2 n 1 (ii) fraction of nuclei left = and 2 n 1 (iii) percentage of nuclei left = 100 2 (b) Number of nuclei decayed after time t, = N0 – N = N0 – N0 e–t = N0 (1 – e–t) The corresponding graphs is as shwn in figure. (c) Probability of a nucleus for survival after time t, : 1800-212-1799 | url : www.motion.ac.in | : [email protected] Page : 185 N N0 e t P (survival) = e t N0 N0 The corresponding graph is shown in figure. (d) Probability of a nucleus to disintegrate in time t is, P (disintegration) = 1 – P (survival) = 1 – e–t The corresponding graph is as shown. (e) Half life and mean life are related to each other by the relation. t1/2 = 0.693 tav or tav = 1.44 tt/2 (f) As we said in point number (2), Number of nuclei decayed in time t are N0 (1 – e– t ) So, to avoid it we can use. N N = t where, N are the number of nuclei decayed in time t, at the instant when total number of nuclei are N. But this can be applied only when t
