Physics Formula Handbook by Competishun

PHYSICS FORMULA BOOKLET w.jeebooks INDEX S.No. Topic Page No. 1. Unit and Dimension 2 2. Rectilinear Motion 3–4 3. Projectile Motion & Vector 5–5 4. Relavitve Motion 5–7 5. Newton’s Laws of Motion 7–9 6. Friction 9–9 7. Work, Power & Energy 10 – 11 8. Circular Motion 11 – 14 9. Centre of Mass 14 – 18 10. Rigid Body Dynamics 18 – 25 11. Simple Harmonic Motion 26 – 28 12. Sting Wave 29 – 31 13. Heat & Thermodynamics 31 – 37 14. Electrostatics 37 – 40 15. Current Electricity 41 – 47 16. Capacitance 47 – 51 17. Alternating Current 52 – 54 18. Magnetic Effect of Current & Magnetic force on charge 54 – 56 19. Electromagnetic Induction 56 – 59 20. Geometrical Optics 59 – 66 21. Modern Physics 67 – 70 22. Wave Optics 70 – 73 23. Gravitation 73 – 75 24. Fluid Mechanics & Properties of Matter 75 – 77 25. Sound Wave 77 – 79 26. Electro Magnetic Waves 79 – 80 27. Error and Measurement 80 – 81 28. Principle of Communication 82 – 83 29. Semiconductor 84 – 85 WWW.JEEBOOKS.IN Page # 1

PHYSICS FORMULA BOOKLET w.jeebooks UNIT AND DIMENSIONS Unit : Measurement of any physical quantity is expressed in terms of an internationally accepted certain basic standard called unit. * Fundamental Units. S.No. Physical Quantity SI Unit Symbol Metre m 1 Length Kilogram Kg Second S 2 Mass Ampere A Kelvin K 3 Time Candela Cd Mole mol 4 Electric Current 5 Temperature 6 Luminous Intensity 7 Amount of Substance Supplementary Units : S.No. Physical Quantity SI Unit Symbol 1 Plane Angle radian r 2 Solid Angle Steradian Sr Metric Prefixes : S .N o . Prefix S ym b o l V a lu e C e n ti c 10–2 3 M ili m 4 M icro µ 3 5 Nano n 6 P ic o p 6 7 K ilo K M 9 ega 10–12 103 6 WWW.JEEBOOKS.IN Page # 2

RECTILINEAR MOTION Average Velocity (in an interval) : w.jeebooks vav = v = <v> = Total displacement = rf  ri Total time taken t Average Speed (in an interval) Total distance travelled Average Speed = Total time taken Instantaneous Velocity (at an instant) :    r  v inst = lim  t    t  0 Average acceleration (in an interval):   v vf  vi aav = t = t Instantaneous Acceleration (at an instant):  lim  v  dv t0  t  = dt = Graphs in Uniformly Accelerated Motion along a straight line (a  0)  x is a quadratic polynomial in terms of t. Hence x  t graph is a parabola. xx xi a<0 a>0 xi 0t 0t x-t graph v is a linear polynomial in terms of t. Hence vt graph is a straight line of slope a. v v slope = a u slope = a u a is negative a is positive 0t 0t WWW.JEEBOOKS.IN Page # 3

v-t graph  at graph is a horizontal line because a is constant. a a positive 0 a acceleration negative 0 t a acceleration a-t graph w.jeebooks Maxima & Minima dy =0 d  dy  < 0 at maximum dx & dx  dx  dy d  dy  and = 0 & dx  > 0 at minima. dx  dx  Equations of Motion (for constant acceleration) (a) v = u + at 11 1 (b) s = ut + at2 s = vt  at2 x = x + ut + at2 22 fi 2 (c) v2 = u2 + 2as (d) s = (u  v) t (e) a sn = u + 2 (2n  1) 2 For freely falling bodies : (u = 0) (taking upward direction as positive) (a) v = – gt 11 1 (b) s = – gt2 s = vt  gt2 h = h – gt2 22 f i2 (c) v2 = – 2gs g (d) sn = – 2 (2n  1) WWW.JEEBOOKS.IN Page # 4

PROJECTILE MOTION & VECTORS Time of flight : 2u sin  T= g w.jeebooks Horizontal range : R = u2 sin 2 g Maximum height : u2 sin2  H = 2g Trajectory equation (equation of path) : gx2 x y = x tan  – = x tan  (1 – ) 2u2 cos2  R Projection on an inclined plane y  x Up the Incline Down the Incline 2u2 sin  cos(  ) 2u2 sin  cos(  ) Range g cos2  gcos2  Time of flight 2u sin  2u sin  g cos  gcos  Angle of projection with incline plane for maximum   42 42 range u2 u2 Maximum Range g(1 sin) g(1 sin ) RELATIVE MOTION Page # 5   v AB (velocity of A with respect to B)  v A  vB   a AB (acceleration of A with respect to B)  aA  aB   Relative motion along straight line - xBA  xB  x A WWW.JEEBOOKS.IN

CROSSING RIVER A boat or man in a river always moves in the direction of resultant velocity of velocity of boat (or man) and velocity of river flow. 1. Shortest Time : w.jeebooks Velocity along the river, vx = vR. Velocity perpendicular to the river, vf = vmR The net speed is given by vm = v 2  v 2 mR R 2. Shortest Path : velocity along the river, vx = 0 and velocity perpendicular to river v = v 2  v 2 y mR R The net speed is given by vm = v 2  vR2 mR at an angle of 90º with the river direction. Page # 6 velocity vy is used only to cross the river, WWW.JEEBOOKS.IN

dd therefore time to cross the river, t = v y = v 2  v 2 mR R and velocity vx is zero, therefore, in this case the drift should be zero. w.jeebooks  v – v sin  = 0 or v = v sin  R mR R mR or  = sin–1  vR  v mR RAIN PROBLEMS   or vRm = v R2  v 2 vRm = vR – vm m NEWTON'S LAWS OF MOTION 1. From third law of motion FAB  FBA FAB = Force on A due to B FBA = Force on B due to A 2. From second law of motion Fx = dPx = ma F = dPy = ma F= dPz = ma dt x y dt y z dt z 5. WEIGHING MACHINE : A weighing machine does not measure the weight but measures the force exerted by object on its upper surface.  6. SPRING FORCE F  kx x is displacement of the free end from its natural length or deformation of the spring where K = spring constant. 7. SPRING PROPERTY K ×  = constant = Natural length of spring. 8. If spring is cut into two in the ratio m : n then spring constant is given by  m  = n. k = k11 = k22 1 = mn ; 2 mn WWW.JEEBOOKS.IN Page # 7

For series combination of springs 1  1  1  ....... For parallel combination of spring keq k1 k 2 k = k + k + k ............ eq 1 2 3 w.jeebooks 9. SPRING BALANCE: It does not measure the weight. t measures the force exerted by the object at the hook. Remember : Vp = V1  V2 2 aP = a1  a2 2 11. a  (m2  m1)g m1  m2 T  2m1m2g m1  m2 12. WEDGE CONSTRAINT: Components of velocity along perpendicular direction to the contact plane of the two objects is always equal if there is no deformations and they remain in contact. WWW.JEEBOOKS.IN Page # 8

13. NEWTON’S LAW FOR A SYSTEM    Fext  m1a1  m2a2  m3a3  ......  Fext  Net external force on the system. Frictionw.jeebooks static frictionm1,m2,m3arethemassesoftheobjectsofthesystemand  a1,a2,a3 are the acceleration of the objects respectively. 14. NEWTON’S LAW FOR NON INERTIAL FRAME :   FReal F Pseudo  ma Net sum of real and pseudo force is taken in the resultant force.  a = Acceleration of the particle in the non inertial frame  FPseudo =m  aFrame (a) Inertial reference frame: Frame of reference moving with con- stant velocity. (b) Non-inertial reference frame: A frame of reference moving with non-zero acceleration. FRICTION Friction force is of two types. (a) Kinetic (b) Static KINETIC FRICTION : fk = k N The proportionality constant k is called the coefficient of kinetic friction and its value depends on the nature of the two surfaces in contact. STATIC FRICTION : It exists between the two surfaces when there is tendency of relative mo- tion but no relative motion along the two contact surfaces. This means static friction is a variable and self adjusting force. However it has a maximum value called limiting friction. fmax = sN 0  fs  fsmax fstatic maximum sN kN WWW.JEEBOOKS.IN Applied Force Page # 9

WORK, POWER & ENERGY WORK DONE BY CONSTANT FORCE :  W= F . S w.jeebooks WORK DONE BY MULTIPLE FORCES    F = F1 + F + F3 + .....  2 W = [ F ] . S ...(i)      W = F 1 . S + F 2 . S + F 3 . S + ..... or W = W + W + W + .......... 123 WORK DONE BY A VARIABLE FORCE  dW = F.ds RELATION BETWEEN MOMENTUM AND KINETIC ENERGY p2 K = 2m and P = 2 m K ; P = linear momentum POTENTIAL ENERGY  U2 dU   r2   i.e., r2   W F  d r U2  U1   F dr  U1 r1 r1 r  U   F  d r  W  CONSERVATIVE FORCES U F= – r WORK-ENERGY THEOREM W +W +W = K C NC PS Modified Form of Work-Energy Theorem W = U C WNC + WPS = K + U WNC + WPS = E WWW.JEEBOOKS.IN Page # 10

POWER The average power ( P or pav) delivered by an agent is given by P or W pav = t w.jeebooks  =    =  .  P  F  dS F dS F v dt dt CIRCULAR MOTION 1. Average angular velocity 2  1   av = t2  t1 = t  d  2. Instantaneous angular velocity   = dt 3. Average angular acceleration  av = 2  1 =  t2  t1 t d d 4. Instantaneous angular acceleration   = dt =  d 5. Relation between speed and angular velocity  v = r and     v  r 7. Tangential acceleration (rate of change of speed) dV d dr  at = dt = r dt =  dt 8. Radial or normal or centripetal acceleration  a= v2 = 2r r r 9. Total acceleration a at   O v  a  at  ar  a = (at2 + ar2)1/2 P ar or a c Where      and      at  r ar  v WWW.JEEBOOKS.IN Page # 11

10. Angular acceleration   d    = dt (Non-uniform circular motion) ARCoWtation v2 mv2w.jeebooks 2  3 / 2 12. Radius of curvature R = = 1    dy  a F  If y is a function of x. i.e. y = f(x) R =   dx    d2y dx 2 13. Normal reaction of road on a concave bridge mv2  N = mg cos  + r O  N  V concave mgcos bridge mg 14. Normal reaction on a convex bridge  N = mg cos  – mv2 r NV  mgcos convex bridge mg  O 15. Skidding of vehicle on a level road  vsafe  gr 16. Skidding of an object on a rotating platform  max = g / r WWW.JEEBOOKS.IN Page # 12

17. Bending of cyclist v2  tan  = rg v2 18. Banking of road without friction  tan  = rg w.jeebooks 19. Banking of road with friction  v 2    tan rg 1  tan  20. Maximum also minimum safe speed on a banked frictional road  rg(  tan ) 1/ 2  rg (tan   ) 1/ 2 Vmax  Vmin    (1  tan )   (1  tan)      21. Centrifugal force (pseudo force)  f = m2 r, acts outwards when the particle itself is taken as a frame. 22. Effect of earths rotation on apparent weight  N = mg – mR2 cos2  ; where  latitude at a place 23. Various quantities for a critical condition in a vertical loop at different positions C DO B  P AN × (1) (2) (3) Vmin  4gL Vmin  4gL Vmin  4gL (for completing the circle) (for completing the circle) (for completing the circle) WWW.JEEBOOKS.IN Page # 13

24. Conical pendulum : fixed pointor suspension //////O/////// w.jeebooks T cos  T  L  h r mg T cos  = mg T sin  = m2 r  Time period = 2 L cos  g 25. Relations amoung angular variables : 0  Initial ang. velocity  = 0 + t d,  or  rO (Perpendicular  ar to plane of paper directed outwards at or V for ACW rotation)   Find angular velocity 1  = 0t + 2 t2   Const. angular acceleration 2 = 02 + 2    Angular displacement CENTRE OF MASS  Mass Moment : M = m r CENTRE OF MASS OF A SYSTEM OF 'N' DISCRETE PARTICLES    m1r1  m2r2  ........  mnrn  rcm = m1  m2  ........  mn ; rcm WWW.JEEBOOKS.IN Page # 14

n   1 n mi ri mi ri rcm = M i 1 i 1 =n mi i1 w.jeebooks CENTRE OF MASS OF A CONTINUOUS MASS DISTRIBUTION x=  x dm ,y =  y dm ,z =  z dm   cm dm cm dm cm dm  dm = M (mass of the body) CENTRE OF MASS OF SOME COMMON SYSTEMS  A system of two point masses m1 r1 = m2 r2 The centre of mass lies closer to the heavier mass.  Rectangular plate (By symmetry) b L x= y= c2 c2 WWW.JEEBOOKS.IN Page # 15

 A triangular plate (By qualitative argument) h at the centroid : yc = 3 w.jeebooks  A semi-circular ring 2R y = x =O c c  A semi-circular disc 4R yc = 3 xc = O R  A hemispherical shell y = x =O  A solid hemisphere c2 c  A circular cone (solid) y= 3R x =O c c 8 h yc = 4  A circular cone (hollow) h yc = 3 WWW.JEEBOOKS.IN Page # 16

MOTION OF CENTRE OF MASSAND CONSERVATION OF MOMENTUM: Velocity of centre of mass of system m1 dr1  m2 dr2  m3 dr3 .......... ....  mn drn w.jeebooks dt dt dt dt vcm = M    = m1 v1  m2 v 2  m3 v3 .......... mn vn M P System = M v cm Acceleration of centre of mass of system  m1 dv1  m2 dv 2  m3 dv 3 ..............  mn dv n acm dt dt dt dt = M    = m1a1  m2a2  m3a3..........  mnan M Net forceon system Net External Force  Net internal Force =M= M Net External Force = M  Fext = M acm IMPULSE Impulse of a force F action on a body is defined as :-  tf Fdt  (impulse - momentum theorem) J  ΔP J= ti Important points : 1. Gravitational force and spring force are always non-impulsive. 2. An impulsive force can only be balanced by another impulsive force. COEFFICIENT OF RESTITUTION (e) Impulse of reformation Fr dt e = Impulse of deformation = Fd dt Velocity of separation along line of impact = s Velocity of approach along line of impact WWW.JEEBOOKS.IN Page # 17

(a) e = 1  Impulse of Reformation = Impulse of Deformation (b) e = 0  Velocity of separation = Velocity of approach (c) 0 < e < 1  Kinetic Energy may be conserved  Elastic collision.  Impulse of Reformation = 0  Velocity of separation = 0  Kinetic Energy is not conserved  Perfectly Inelastic collision.  Impulse of Reformation < Impulse of Deformation  Velocity of separation < Velocity of approach w.jeebooks  Kinetic Energy is not conserved  Inelastic collision. VARIABLE MASS SYSTEM : If a mass is added or ejected from a system, at rate  kg/s and relative  velocity vrel (w.r.t. the system), then the force exerted by this mass  on the system has magnitude  vrel .  Thrust Force ( Ft )     dm  Ft v rel  dt  Rocket propulsion : If gravity is ignored and initial velocity of the rocket u = 0; v = v ln  m0  . r m RIGID BODY DYNAMICS 1. RIGID BODY : A 1 VA VAsin1 B A VB 2 VAcos1 VBsin2 B VBcos2 WWW.JEEBOOKS.IN Page # 18

If the above body is rigid VA cos  = VB cos  1 2 V = relative velocity of point B with respect to point A. BA w.jeebooksA B VBA Types of Motion of rigid body Pure Translational Pure Rotational Combined Translational and Motion Motion Rotational Motion 2. MOMENT OF INERTIA (I) : Definition : Moment of Inertia is defined as the capability of system to oppose the change produced in the rotational motion of a body. Moment of Inertia is a scalar positive quantity.  = mr 2 + m r 2 +......................... 1 22 =  +  +  +......................... S units of Moment of Inertia is Kgm2. Moment of Inertia of : 2.1 A single particle :  = mr2 where m = mass of the particle r = perpendicular distance of the particle from the axis about which moment of Inertia is to be calculated 2.2 For many particles (system of particles) : n  = miri2 i1 2.3 For a continuous object :  = dmr2 where dm = mass of a small element r = perpendicular distance of the particle from the axis WWW.JEEBOOKS.IN Page # 19

2.4 For a larger object :  = delement where d = moment of inertia of a small element w.jeebooks 3. TWO IMPORTANT THEOREMS ON MOMENT OF INERTIA : 3.1 Perpendicular Axis Theorem [Onlyapplicableto plane lamina(thatmeansfor2-Dobjectsonly)]. z = x + y (when object is in x-y plane). 3.2 Parallel Axis Theorem (Applicable to any type of object):  =  + Md2  cm List of some useful formula : Object Moment of Inertia 2 MR2 (Uniform) 5 Solid Sphere 2 MR 2 (Uniform) 3 Hollow Sphere MR2 (Uniform or Non Uniform) WWW.JEEBOOKS.IN Page # 20

w.jeebooksRing. MR2 (Uniform) Disc 2 MR2 (Uniform orNonUniform) Hollow cylinder Solid cylinder MR2 (Uniform) 2 ML2 (Uniform) 3 ML2 (Uniform) 12 WWW.JEEBOOKS.IN Page # 21

2m2 (Uniform) 3 w.jeebooks AB = CD = EF = Ma2 (Uniform) 12 Square Plate Ma2 (Uniform) Square Plate 6 Rectangular Plate  = M(a2  b2 ) (Uniform) 12 M(a2  b2 ) (Uniform) 12 Cuboid WWW.JEEBOOKS.IN Page # 22

4. RADIUS OF GYRATION :  = MK2 5. TORQUE :     rF w.jeebooks 5.5 Relation between '' & '' (for hinged object or pure rotation)  Hinge = Hinge  ext  Where  = net external torque acting on the body about Hinge ext Hinge point Hinge = moment of Inertia of body about Hinge point F1t F1c r1 F2t x F2c r2 F1t = M1a1t = M1r1 F2t = M2a2t = M2r2 resultant = F r + F r + ........ 1t 1 2t 2 = M1  r12 + M2  r22 + ............ ) =  resultant external Rotational Kinetic Energy = 1 ..2 2   P  MvCM  Fexternal  MaCM Net external force acting on the body has two parts tangential and centripetal.  FC = maC = m v2 = m2 rCM  Ft = mat = m rCM rCM WWW.JEEBOOKS.IN Page # 23

6. ROTATIONAL EQUILIBRIUM : For translational equilibrium. Fx  0 ............. (i) w.jeebooksand Fy  0 ............. (ii) The condition of rotational equilibrium is z  0  7. ANGULAR MOMENTUM ( L ) 7.1 Angular momentum of a particle about a point.  =   L = rpsin L r P  L =r ×P   L = P× r 7.3 Angular momentum of a rigid body rotating about fixed axis :  = H LH LH = angular momentum of object about axis H. IH = Moment of Inertia of rigid object about axis H.  = angular velocity of the object. 7.4 Conservation of Angular Momentum Angular momentum of a particle or a system remains constant if  ext = 0 about that point or axis of rotation. 7.5 Relation between Torque and Angular Momentum  dL  = dt  Torque is change in angular momentum WWW.JEEBOOKS.IN Page # 24

7.6 Impulse of Torque :  dt  J J  Change in angular momentum. w.jeebooksFor a rigid body, the distance between the particles remain unchanged during its motion i.e. rP/Q = constant For velocities with respect to Q with respect to ground P P VQ r r r Q wr Q  VQ VP  VQ2  r2  2 VQ r cos For acceleration : , ,  are same about every point of the body (or any other point outside which is rigidly attached to the body). Dynamics :  Fext    cm  cm  ,  Macm  Psystem  Mvcm , Total K.E. = 1 Mvcm2 + 1 cm 2 2 2  Angular momentum axis AB = L about C.M. + L of C.M. about AB    L AB  cm   rcm  Mvcm WWW.JEEBOOKS.IN Page # 25

SIMPLE HARMONIC MOTION S.H.M. F = – kx General equation of S.H.M. is x = A sin (t + ); (t + ) is phase of the motion and  is initial phase of the motion. w.jeebooks Angular Frequency () : 2  = T = 2f Time period (T) : T= 2 m  = 2 k k m Speed : v   A2  x2 Acceleration : a = 2x 11 1 Kinetic Energy (KE) : 2 mv2 = 2 m2 (A2 – x2) = 2 k (A2 – x2) Potential Energy (PE) : 1 Kx2 2 Total Mechanical Energy (TME) 1 11 = K.E. + P.E. = k (A2 – x2) + Kx2 = KA2 (which is constant) 2 22 SPRING-MASS SYSTEM km (1)   T = 2 k smooth surface m (2) T = 2  m1m2 K where  = (m1  m2 ) known as reduced mass WWW.JEEBOOKS.IN Page # 26

COMBINATION OF SPRINGS 1/keq = 1/k1 + 1/k2 Series Combination : k =k +k Parallel combination : eq 1 2 w.jeebooks  SIMPLE PENDULUM T = 2 g = 2 geff. (in accelerating Refer- ence Frame); geff is net acceleration due to pseudo force and gravitational force. COMPOUND PENDULUM / PHYSICAL PENDULUM Time period (T) :  T = 2 mg where,  = CM + m2 ;  is distance between point of suspension and centre of mass. TORSIONAL PENDULUM Time period (T) :  where, C = Torsional constant T = 2 C Superposition of SHM’s along the same direction x = A sin t & x = A sin (t + ) 11 22 A2 A A1 If equation of resultant SHM is taken as x = A sin (t + ) A= A12  A 2  2A1A2 cos & tan  = A 2 sin  2 A1  A 2 cos  1. Damped Oscillation  Damping force  F  –bv  equation of motion is mdv = –kx – bv dt  b2 - 4mK > 0 over damping WWW.JEEBOOKS.IN Page # 27

 b - 4mK = 0 critical damping  b2 - 4mK < 0 under damping  For small damping the solution is of the form. w.jeebooks x = A0e–bt/ 2m sin [1t +  ], where '   k  –  b 2  m   2m  For small b  angular frequency '  k / m,  0 –bt  Amplitude A  A0e 2m l  Energy E (t) = 1 KA2 e –bt /m 2  Quality factor or Q value , Q = 2 E = ' | E | 2Y where , '  k . b2 b m 4m2 , Y  2m 2. Forced Oscillations And Resonance External Force F(t) = F0 cos d t x(t) = A cos (dt + ) F0  A  tan   v0  m2 2  d2 2  d2 b2  and d x0     A F0 m 2  2d  (a) Small Damping (b) Driving Frequency Close to Natural Frequency A  F0 db WWW.JEEBOOKS.IN Page # 28

STRING WAVES GENERAL EQUATION OF WAVE MOTION : w.jeebooks2y 2y t2 = v2 x2 x y(x,t) = f (t ± ) v where, y (x, t) should be finite everywhere.  f  t  x  represents wave travelling in – ve x-axis.  v  f  t  x  represents wave travelling in + ve x-axis.  v y = A sin (t ± kx + ) TERMS RELATED TO WAVE MOTION ( FOR 1-D PROGRESSIVE SINE WAVE ) (e) Wave number (or propagation constant) (k) :  k = 2/ = (rad m–1) v (f) Phase of wave : The argument of harmonic function (t ± kx + ) is called phase of the wave. Phase difference () : difference in phases of two particles at any time t. 2 2  = x Also.  T t  SPEED OF TRANSVERSE WAVE ALONG A STRING/WIRE. T T  Tension v =  where   mass per unit length POWER TRANSMITTED ALONG THE STRING BY A SINE WAVE Average Power P = 22 f2 A2 v Intensity I = P = 22 f2 A2 v s REFLECTION AND REFRACTION OF WAVES yi = Ai sin (t – k1x) y t  A t sin (t  k2 x)  yr   Ar sin (t  k1x) if incident from rarer to denser medium (v2 < v1) WWW.JEEBOOKS.IN Page # 29

y t  A t sin (t – k 2x) if incident from denser to rarer medium. (v > v )  21 yr  Ar sin (t  k1x)  (d) Amplitude of reflected & transmitted waves. w.jeebooks A= k1  k2 Ai &A = 2k1 Ai r k1  k2 t k1  k2 STANDING/STATIONARY WAVES :- (b) y1 = A sin (t – kx + 1) y= A sin (t + kx + 2) 2 y1 + y2 = 2 A cos  kx  2  1  sin t  1  2    2   2 The quantity 2A cos kx  2  1  represents resultant amplitude at  2 x. At some position resultant amplitude is zero these are called nodes. At some positions resultant amplitude is 2A, these are called antin- odes. (c) Distance between successive nodes or antinodes =  . 2 (d) Distance between successive nodes and antinodes = /4. (e) All the particles in same segment (portion between two successive nodes) vibrate in same phase. (f) The particles in two consecutive segments vibrate in opposite phase. (g) Since nodes are permanently at rest so energy can not be trans- mitted across these. VIBRATIONS OF STRINGS ( STANDING WAVE) (a) Fixed at both ends : 1. Fixed ends will be nodes. So waves for which L=  2 L = 3 2 L= 2 2 are possible giving 2L n or  = n where n = 1, 2, 3, .... L= 2 T nT as v =  fn = 2L  , n = no. of loops WWW.JEEBOOKS.IN Page # 30

(b) String free at one end :  1. for fundamental mode L = 4 = or  = 4L w.jeebooks fundamental mode First overtone L = 3 Hence  = 4L 4 3  first overtone so f = 3T 1  (First overtone) 4L 5T Second overtone f2 = 4L  n  1   2 T  (2n  1) T so fn = 2L  4L  HEAT & THERMODYNAMICS 1 33 Total translational K.E. of gas = M < V2 > = PV = nRT 2 22 3P V= 3P 3RT 3KT < V2 > = rms = Mmol = m  Important Points : – Vrms  T V 8KT KT KT m = 1.59 m V = 1.73 rms m 2KT KT Most probable speed V = = 1.41  V > >V p m m rms V mp Degree of freedom : Mono atomic f = 3 Diatomic f = 5 polyatomic f = 6 WWW.JEEBOOKS.IN Page # 31

Maxwell’s law of equipartition of energy : Total K.E. of the molecule = 1/2 f KT For an ideal gas : f Internal energy U = nRT 2 w.jeebooks Workdone in isothermal process : W = [2.303 nRT log10 Vf ] Vi Internal energy in isothermal process : U = 0 Work done in isochoric process : dW = 0 Change in int. energy in isochoric process : f U = n 2 R T = heat given Isobaric process : Work done W = nR(T – T ) fi change in int. energy U = nCv T heat given Q = U + W Specific heat : f Cp =  f  1 R Cv = 2 R  2  Molar heat capacity of ideal gas in terms of R : (i) for monoatomic gas : Cp Cv = 1.67 (ii) for diatomic gas : Cp Cv = 1.4 (iii) for triatomic gas : Cp Cv = 1.33 In general : = Cp = 1  2 Cv f  Mayer’s eq.  Cp – Cv = R for ideal gas only Adiabatic process : Work done W = nR (Ti  Tf )  1 WWW.JEEBOOKS.IN Page # 32

In cyclic process : Q = W In a mixture of non-reacting gases : w.jeebooksMol. wt. =n1M1  n2M2 n1  n2 C= n1Cv 1  n2Cv 2 v n1  n2 = Cp (mix) = n1Cp  n2Cp  ..... Cv (mix) 12 n1Cv1  n2Cv2  .... Heat Engines Efficiency ,   work done by the engine heat sup pliedtoit = W  QH – QL  1– QL QH QH QH Second law of Thermodynamics  Kelvin- Planck Statement It is impossible to construct an engine, operating in a cycle, which will produce no effect other than extracting heat from a reservoir and perform- ing an equivalent amount of work. WWW.JEEBOOKS.IN Page # 33

 Rudlope Classius Statement It is impossible to make heat flow from a body at a lower temperature to a body at a higher temperature without doing external work on the working substance w.jeebooks Entropy Q f Q T i T  change in Sf entropy of the system is S = – Si   In an adiabatic reversible process, entropy of the system remains con- stant. Efficiency of Carnot Engine (1) Operation I (Isothermal Expansion) (2) Operation II (Adiabatic Expansion) (3) Operation III (Isothermal Compression) (4) Operation IV (Adiabatic Compression) Thermal Efficiency of a Carnot engine V2  V3  Q2  T2   1– T2 V1 V4 Q1 T1 T1 WWW.JEEBOOKS.IN Page # 34

Refrigerator (Heat Pump) Refrigerator w.jeebooksHot (T2) Hot (T1) Q2 Q1 W  Coefficient of performance,   Q2 =  1 =  1 W –1 T1 – 1 T1 T2 T2 Calorimetry and thermal expansion Types of thermometers : (a) Liquid Thermometer : T=    0  × 100 (b) Gas Thermometer :  100     0  Constant volume : T=  P  P0  × 100 ; P = P + g h  P100  P0  0   V Constant Pressure : T =  V  V T0 (c) Electrical Resistance Thermometer : T=  Rt  R0  × 100  R100  R   0  Thermal Expansion : (a) Linear : L  = L0T or L = L (1 + T) 0 WWW.JEEBOOKS.IN Page # 35

(b) Area/superficial : A or A = A (1 + T)  = A0T 0 (c) volume/ cubical : w.jeebooks V or V = V0 (1 +  T) r = V0T   23 Thermal stress of a material : F  Y  A Energy stored per unit volume : 1 or E  1 AY (L)2 E = K(L)2 2L 2 Variation of time period of pendulum clocks : 1 T = 2 T T’ < T - clock-fast : time-gain T’ > T - clock slow : time-loss CALORIMETRY : dQ dT Q dt = – KA dx Specific heat S =  m.T R= Q KA Molar specific heat C = n.T Water equivalent = mWSW HEAT TRANSFER Thermal Conduction : Thermal Resistance : WWW.JEEBOOKS.IN Page # 36

Series and parallel combination of rod : (i) Series :  eq = 1   2  ....... (when A1 = A2 = A3 = .........) (ii) Parallel : K eq K1 K2 w.jeebooks Keq Aeq = K1 A1 + K2 A2 + ...... (when  1 =  =  = .........) 2 3 for absorption, reflection and transmission r+t+a=1 Emissive power : U E = A t Spectral emissive power : dE E=  d Emissivity : E of a body at T temp. e = E of a black body at T temp. E (body ) Kirchoff’s law : a(body ) = E (black body) Wein’s Displacement law : m . T = b. b = 0.282 cm-k Stefan Boltzmann law : u =  T4 s = 5.67 × 10–8 W/m2 k4 u = u – u = e A (T4 – T 4) 0 0 d Newton’s law of cooling : dt = k ( – 0) ;  = 0 + (i – 0) e–k t ELECTROSTATICS Coulomb force between two point charges   1 |qr1q|32  =  1 |qr1q|22 rˆ F 4 0  r r 40r  The electric field intensity at any point is the force experienced by unit positive charge, given by   E F q0  Electric force on a charge 'q' at the position of electric field   intensity E produced by some source charges is F  qE  Electric Potential WWW.JEEBOOKS.IN Page # 37

If (W  )P ext is the work required in moving a point charge q from infinity to a point P, the electric potential of the point P is Vp  (Wp )ext  q  acc 0 w.jeebooks  Potential Difference between two points A and B is VA – VB   Formulae of E and potential V (i) Point charge E= Kq  rˆ = Kq  V = Kq  r3 r, r | r |2 (ii) Infinitely long line charge  rˆ = 2Krˆ 2 0r r V = not defined, v B – v = –2K ln (r / r ) A BA (iii) Infinite nonconducting thin sheet  nˆ , 20 V = not defined, vB  vA  rB  rA   20 (iv) Uniformly charged ring KQx  E = R2 3/2 , E =0 axis centre  x2 Vaxis = KQ KQ R2  x2 , Vcentre = R x is the distance from centre along axis. (v) Infinitely large charged conducting sheet  nˆ 0 V = not defined, vB  vA   rB  rA  0 (vi) Uniformly charged hollow conducting/ nonconducting /solid conducting sphere (a) for   kQ rˆ , r R, V= KQ E   r | r |2 (b)  for r < R, V= KQ E0 R WWW.JEEBOOKS.IN Page # 38

(vii) Uniformly charged solid nonconducting sphere (insulating material)  kQ rˆ KQ E  r (a) r |2 for r  R,V= w.jeebooks | (b)      for r  R,  E KQ r r V = 60 (3R2–r2) R3 30 (viii) thin uniformly charged disc (surface charge density is )   x  R2  x2  x 1   20  Eaxis = 20  Vaxis = R2  x2   Work done by external agent in taking a charge q from A to B is (W )= q (V – V) or (W ) AB = q (V – V) . B A el A B ext AB  The electrostatic potential energy of a point charge U = qV  U = PE of the system = U1  U2  ... = (U12 + U13 + ..... + U1n) + (U23 + U24 + ...... + U2n) 2 + (U34 + U35 + ..... + U3n) .... 1  Energy Density = E2 2  Self Energy of a uniformly charged shell = Uself  KQ2 2R  Self Energy of a uniformly charged solid non-conducting sphere = Uself  3KQ2 5R  Electric Field Intensity Due to Dipole   2KP (i) on the axis E = r3   KP (ii) on the equatorial position : E = – r3 (iii) Total electric field at general point O (r,) is E= KP 1 3 cos 2  res r3 WWW.JEEBOOKS.IN Page # 39

 Potential =En- epr.gEy of an Electric Dipole in External Electric Field: U  Electric Dipole in Uniform Electric Field :   w.jeebookstorque   p x E; F =0   Electric Dipole in Nonuniform Electric Field: torque   p x   Net force |F| = p E  E; r U =  p E ,  Electric Potential Due to Dipole at General Point (r, ) : p . r Pcos V = 40r2  40r3  The electric flux over the whole area is given by   E = E.dS = SEndS S  Flux using Gauss's law, Flux through a closed surface E =  qin . E  dS = 0  Electric field intensity near the conducting surface =  nˆ 0  Electric pressure : Electric pressure at the surface of a conductor is given by formula 2 P = 20 where  is the local surface charge density.  Potential difference between points A and B B  VB – VA = – E.dr A  =  ˆi  V  ˆj  V  kˆ  V  = –  ˆi   ˆj   kˆ   V E x x z x x z  = – V = –grad V WWW.JEEBOOKS.IN Page # 40

CURRENT ELECTRICITY 1. ELECTRIC CURRENT q Iav = t and instantaneous current w.jeebooks i =. Lim q  dq t dt t0 2. ELECTRIC CURRENT IN A CONDUCTOR I = nAeV. vd   , 1  eE 2 2 m  = 1 eE  , vd   2m I = neAVd 3. CURRENT DENSITY  dI  J n ds 4. ELECTRICAL RESISTANCE I = neAVd = neA  eE  =  ne2  AE  2m   2m  V I =  ne2   A  V =  A  V = V/R  V = IR E=  so  2m       is called resistivity (it is also called specific resistance) and 2m 1 = ne2 =  ,  is called conductivity. Therefore current in conductors is proportional to potential difference applied across its ends. This is Ohm's Law. Units: R  ohm(),   ohm  meter(  m) also called siemens,   1m1. WWW.JEEBOOKS.IN Page # 41

Dependence of Resistance on Temperature : R = R (1 + ). o Electric current in resistance I = V2  V1 R 5. ELECTRICAL POWER P = V Energy = pdt V2 P = I2R = V = . R H = Vt = 2Rt = V 2 t R H = 2 RT Joule = 2RT Calorie 4.2 9. KIRCHHOFF'S LAWS 9.1 Kirchhoff’s Current Law (Junction law)   =  in out 9.2 Kirchhoff’s Voltage Law (Loop law) IR + EMF =0”. w.jeebooks 10. COMBINATION OF RESISTANCES : Resistances in Series: R = R + R + R +................ + Rn (this means Req is greater then any 123 resistor) ) and V = V1 + V2 + V3 +................ + Vn . V1 = R1  R2 R1 V ; V2 = R2 V ;  ......... Rn R1  R2  ......... Rn 2. Resistances in Parallel : WWW.JEEBOOKS.IN Page # 42

11. WHEATSTONE NETWORK : (4 TERMINAL NETWORK) w.jeebooksWhen current through the galvanometer is zero (null point or balance PR point) Q = S , then PS = QR 13. GROUPING OF CELLS 13.1 Cells in Series :  Equivalent EMFEeq = E1  E2  .......  En [write EMF's with polarity] Equivalent internal resistance req = r1  r2  r3  r4  ....  rn 13.2 Cells in Parallel: Eeq  1  2 r2  ....  n [Use emf with polarity] r1  rn 1 1  .....  r1 r2 1 rn 1  1  1  ....  1 req r1 r2 rn 15. AMMETER A shunt (small resistance) is connected in parallel with galvanometer to convert it into ammeter. An ideal ammeter has zero resistance WWW.JEEBOOKS.IN Page # 43

Ammeter is represented as follows - w.jeebooksIf maximum value of current to be measured by ammeter is  then IG . RG = (I – IG)S  G .R G S = G RG when  >> G. S =   G  where  = Maximum current that can be measured using the given ammeter. 16. VOLTMETER A high resistance is put in series with galvanometer. It is used to measure potential difference across a resistor in a circuit. For maximum potential difference V = G . RS + G RG V If V R= G –R R << R  R  S G GS S G 17. POTENTIOMETER  = rR V – V =  .R A B Rr Potential gradient (x)  Potential difference per unit length of wire x = VA  VB =  R . L Rr L WWW.JEEBOOKS.IN Page # 44

Application of potentiometer (a) To find emf of unknown cell and compare emf of two cells. In case , In figure (1) is joint to (2) then balance length =  in case , w.jeebooks  = x 1 11 ....(1) In figure (3) is joint to (2) then balance length = 2  = x ....(2) 22 1  1 2 2 If any one of 1 or 2 is known the other can be found. If x is known then both  and  can be found 12 (b) To find current if resistance is known VA – VC = x 1 x1 IR = 1 x 1  = R1 Similarly, we can find the value of R2 also. Potentiometer is ideal voltmeter because it does not draw any current from circuit, at the balance point. (c) To find the internal resistance of cell. Ist arrangement 2nd arrangement WWW.JEEBOOKS.IN Page # 45

by first arrangement ’ = x1 ...(1) by second arrangement IR = x 2  = x2 , ' R also  = r'R w.jeebooks  ' = x 2  x1 = x 2 r'R R r'R R r’ =  1   2  R  2    (d)Ammeter and voltmeter can be graduated by potentiometer. (e)Ammeter and voltmeter can be calibrated by potentiometer. 18. METRE BRIDGE (USE TO MEASURE UNKNOWN RESISTANCE) If AB =  cm, then BC = (100 – ) cm. Resistance of the wire between A and B , R   [  Specific resistance  and cross-sectional area A are same for whole of the wire ] or R =  ...(1) where  is resistance per cm of wire. If P is the resistance of wire between A and B then P  P = () Similarly, if Q is resistance of the wire between B and C, then Q  100 –   Q = (100 – ) ....(2) Dividing (1) by (2), P Q = 100   WWW.JEEBOOKS.IN Page # 46

Applying the condition for balanced Wheatstone bridge, we get R Q = P X Q 100    x=R or X =  R P w.jeebooksSince R and  are known, therefore, the value of X can be calculated. CAPACITANCE 1. (i) q  V  q = CV q : Charge on positive plate of the capacitor C : Capacitance of capacitor. V : Potential difference between positive and negative plates. (ii) Representation of capacitor : ,( (iii) Energy stored in the capacitor : U = 1 Q2 = QV CV2 = 2 2C 2 (iv) Energy density = 1 r E2 = 1 2 2 K E2 r = Relative permittivity of the medium. K= r : Dielectric Constant For vacuum, energy density = 1 E2 2 (v) Types of Capacitors : (a) Parallel plate capacitor C = 0r A = K 0A d d A : Area of plates d : distance between the plates( << size of plate ) (b) Spherical Capacitor :  Capacitance of an isolated spherical Conductor (hollow or solid ) C= 4 r R R = Radius of the spherical conductor  Capacitance of spherical capacitor ab b 12 C= 4 (b  a)  C = 40K2ab a (b  a) b aK1 K2 K3 WWW.JEEBOOKS.IN Page # 47

(c) Cylindrical Capacitor :  >> {a,b} w.jeebooksCapacitance per unit length =2 F/m  b n(b / a)  (vi) Capacitance of capacitor depends on (a) Area of plates (b) Distance between the plates (c) Dielectric medium between the plates. (vii) Electric field intensity between the plates of capacitor V E = 0  d Surface change density q2 (viii) Force experienced by any plate of capacitor : F = 2A0 2. DISTRIBUTION OF CHARGES ON CONNECTING TWO CHARGED CAPACITORS: When two capacitors are C1 and C2 are connected as shown in figure (a) Common potential :  V= C1V1  C2V2 = Total charge C1  C2 Total capacitance (b) Q1' = C1V = C1 (Q1 + Q2) C1  C2 C2 Q2' = C2 V = C1  C2 (Q1 +Q2) WWW.JEEBOOKS.IN Page # 48

(c) Heat loss during redistribution : H = U – U = 1 C1C2 (V – V )2 i f C1  C2 12 2 w.jeebooks The loss of energy is in the form of Joule heating in the wire. 3. Combination of capacitor : (i) Series Combination 1 111 V1 : V2 : V3  1 : 1 : 1  C1 C2 C3 Ceq C1 C2 C3 +Q –Q +Q –Q +Q –Q C1 C2 C3 V1 V2 V3 (ii) Parallel Combination : Q+ –Q C1 Q+ –Q C2 Q+ –Q C3 V Q1: Q2 :Q3 = C1 : C2 : C3 Ceq = C1 + C2 + C3 4. Charging and Discharging of a capacitor : (i) Charging of Capacitor ( Capacitor initially uncharged ): q = q0 ( 1 – e– t /) R VC q0 = Charge on the capacitor at steady state Page # 49 q0 = CV WWW.JEEBOOKS.IN

Time constant = CReq. I = q0 V e – t /  e– t /  R w.jeebooks (ii) Discharging of Capacitor : q = q e –t/ 0 q0 = Initial charge on the capacitor I = q0 e – t /   q R q0 C 0.37v0 t 5. Capacitor with dielectric : (i) Capacitance in the presence of dielectric : C= K0A = KC0 d + + – – – – – – – – – – – – b V 0 b0 + + + + + + + + + + + + + + +b – – C = Capacitance in the absence of dielectric. 0 WWW.JEEBOOKS.IN Page # 50