53198602 formula booklet physics xi 0001

7/31/2019 53198602 Formula Booklet Physics XI 0001

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Dear studentsMost students tend to take it easy after the board examinations of Class X. Thesummervacations immediately after Class X are a great opportunity for the students torace ahead ofother students in the competitive world of IITJEE, where less than 2% students get selectedevery year for the prestigious institutes.Some students get governed completely by the emphasis laid by the teachers of the school inwhich they are studying. Since, the objective of the teachers in the schools rarely is to equip thestudent with the techniques reqired to crack IITJEE, most of the students tend to take it easy inClass XI. Class XI does not even have the pressure of board examinations.So, while the teachers and the school environment is often not oriented towardsthe seriouspreparation of IITJEE, the curriculum of Class XI is extremely important to achieve success inIITJEE or any other competitive examination like AIEEE.The successful students identify these points early in their Class XI and race ahead of rest ofthe competition. We suggest that you start as soon as possible.

In this booklet we have made a sincere attempt to bring your focus to Class XI and keep yourvelocity of preparations to the maximum. The formulae will help you revise yourchapters in avery quick time and the motivational quotes will help you move in the right direction.Hope you.ll benefit from this book and all the best for your examinations.Praveen TyagiGaurav MittalPrasoon Kumar


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Call 1600-111-533 (toll-free) for info. Formula Booklet . Physics XI


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CONTENTSDescription Page Number1. Units, Dimensions & Measurements 032. Motion in one Dimension & Newton.s Laws of Motion 043. Vectors 064. Circular Motion, Relative Motion, and Projectile Motion 075. Friction & Dynamics of Rigid Body 096. Conservation Laws & Collisions 127. Simple Harmonic Motion & Lissajous Figures 148. Gravitation 189. Properties of Matter 2010. Heat & Thermodynamics 2511. Waves 3012. Study Tips 35


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UNITS, DIMENSIONS AND MEASUREMENTS(i) SI Units:(a) Time-second(s);(b) Length-metre (m);(c) Mass-kilogram (kg);(d) Amount of substance.mole (mol); (e) Temperature-Kelvin (K);(f) Electric Current . ampere (A);(g) Luminous Intensity . Candela (Cd)(ii) Uses of dimensional analysis(a) To check the accuracy of a given relation(b) To derive a relative between different physical quantities(c) To convert a physical quantity from one system to another systemc21b21a211 2 2 2 2 1 TT

xLLxMMn n or u n u n ..

.

..

.

..

.

..

...

.

..

.= =(iii) Mean or average value:NX X ... XX 1 - 2 + + N=(iv) Absolute error in each measurement: |.Xi| = |X .Xi|(v) Mean absolute error: .Xm=

NS | .Xi |(vi) Fractional error =X.X(vii) Percentage error = x 100X.X(viii) Combination of error: If = ca b


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ZX Y , then maximum fractional error in is:.= . + . + ... | c |Y

| b | YX| a | X


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MOTION IN ONE DIMENSION & NEWTON.S LAWS OF MOTION(i) Displacement: | displacement | = distance covered(ii) Average speed:22111 21 21 2vsvss st ts sv++=++

=(a) If s1 = s2 = d, then1 21 2v v2 v vv+= = Harmonic mean(b) If t1 = t2, then2v = v1 + v2 =arithmetic mean(iii) Average velocity: (a)

2 1av 2 1 t tr rv--=. ..; (b) | v av |= v.(iv) Instantaneous velocity: and | v |dt

v d r...= = v = instantaneous speed(v) Average acceleration:2 1av 2 1 t tv va


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--=. ..(vi) Instantaneous acceleration: a d v/ dt. .=In one . dimension, a = (dv/dt) = vdxdv ......(vii) Equations of motion in one dimension:(a) v = u + at;(b) x = ut +21 at2 ;(c) v2 u2 + 2ax;(d) x = vt .21 at2;(e) ..

...= . +2x v u t;(f) at ;2s x x ut 1 2= - 0 = +(g) v2 = u2 + 2a (x.x0)(viii) Distance travelled in nth second: dn = u +2a (2n.1)

(ix) Motion of a ball: (a) when thrown up: h = (u2/2g) and t = (u/g)(b) when dropped: v = v(2gh) and t = v(2h/g)(x) Resultant force: F = v(F12 + F22 + 2F1F2 cos .)(xi) Condition for equilibrium: (a) F3 (F1 F2 ) ;. . .= - + (b) F1 + F2 = F3 = |F1 . F2|(xii) Lami.s Theorem: ( ) ( ) (p - .)=p - =p - a sin

RsinQsinP(xiii) Newton.s second law: . .... ...= =


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. . . .F m a ; F d p/ dt(xiv) Impulse: . = . - =. .p F t and p2 p1 Fdt(xv) Newton.s third law:.21


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(a). .F12 = -F12(b) Contact force: 12 F F21M mF m =+=(c) Acceleration: a =M mF+(xvi) Inertial mass: mI = F/a(xvii) Gravitational mass: mG = I G2; m mGMFRgF = =(xviii) Non inertial frame: If.a0 be the acceleration of frame, then pseudo force

. .F = -ma 0Example: Centrifugal force = m rrmv2= .2(xix) Lift problems: Apparent weight = M(g a0)(+ sign is used when lift is moving up while . sign when lift is moving down)(xx) Pulley Problems:(a) For figure (2):Tension in the string, T = gm mm m

1 21 2+Acceleration of the system, a = gm mm1 22+The force on the pulley, F = gm m2m m1 2

1 2+(b) For figure (3):Tension in the string, gm m2m mT1 22 1+=


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Acceleration of the system, gm mm ma2 12 1+-=The force on the pulley, gm m4m mF1 22 1+=a TT TTam2m1Fig. 3

m2m1 TFrictionlesssurfacem2gFig. 2TF(1)(2)MF21m F12

Fig. 1


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VECTORS(i) Vector addition: R A B B A and A B A ( B). . . . . . . . .= + = + - = + -(ii) Unit vector: ) A / A ( A ^.=(iii) Magnitude: ) A A (A A 2z2y2x= v + +(iv) Direction cosines: cos a = (Ax/A), cos = (Ay/A), cos . = (Az/A)(v) Projection:(a) Component of.Aalong.B=^ B . A .(b) Component of.Balong

.A=

.B. A ^

(c) If.A= A i A j,^y^x + then its angle with the x.axis is . = tan.1 (Ay/Ax)(vi) Dot product:(a)

. .A . B = AB cos ., (b) A . B = AxBx + AyBy + AzBz

. .(vii) Cross product:(a) =. .A x B AB sin .^n;(b) A x A = 0;. .(c) . .A x B =

x y zx y z^ ^ ^B B BA A Ai j k(viii) Examples:(a) W F . r ;. .= (b) P F . v ;


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. .= (c) E . A;. .f. = (d) B . A;. .f. =(e) v w x r ;. . .= (f) x F;. . .t = t (g) . .... .. .=. . .F m q v x B(ix) Area of a parallelogram: Area = | A x B |. .(x) Area of a triangle: Area = | A x B |21 . .(xi) Gradient operator:z

k

y jxV i^ ^ ^.+ ..+ ..= .(xii) Volume of a parallelopiped: . ...

. ...=. . .V A. B x C


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CIRCULAR MOTION, RELATIVE MOTION & PROJECTILE MOTION(i) Uniform Circular Motion:(a) v = .r;(b) a = (v2/r) = .2r ;(c) F = (mv2/r);(d) r . v = 0;. .(e) v . a = 0. .(ii) Cyclist taking a turn: tan . = (v2/rg)(iii) Car taking a turn on level road: v = v(srg)(iv) Banking of Roads: tan . = v2/rg(v) Air plane taking a turn: tan . = v2/r g(vi) Overloaded truck:(a) Rinner wheel < Router wheel(b) maximum safe velocity on turn, v = v(gdr/2h)(vii) Non.uniform Circular Motion:(a) Centripetal acceleration ar = (v2/r);(b) Tangential acceleration at = (dv/dt);(c) Resultant acceleration a=v ) a a ( 2t2r+(viii) Motion in a vertical Circle:

(a) For lowest point A and highest point B, TA . TB = 6 mg; v2A = v2B + 4gl ; vA = v(5gl); and vB=v (gl)(b) Condition for Oscillation: vA < v(2gl)(c) Condition for leaving Circular path: v(2gl) < vA < v(5gl)(ix) Relative velocity: v BA v B v A. . .= -(x) Condition for Collision of ships: ( r A - v B) x ( v A - v B) = 0. . . .(xi) Crossing a River:(a) Beat Keeps its direction perpendicular to water current

(1) vR =v( ; ) v v ( 2b2w + (2) . = tan.1 (vw / vb );(3) t=(x/vb) (it is minimum) (4) Drift on opposite bank = (vw/vb)x(b) Boat to reach directly opposite to starting point:(1) sin . = (vw/vb); (2) vresultant = vb cos . ; (3) t=v cos .xb(xii) Projectile thrown from the ground:(a) equation of trajectory: y = x tan . .2 2 .2

2u cosg x(b) time of flight:gT = 2 u sin .(c) Horizontal range, R = (u2 sin 2./g)(d) Maximum height attained, H = (u2 sin2 ./2g)


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(e) Range is maximum when . = 450(f) Ranges are same for projection angles . and (900..)(g) Velocity at the top most point is = u cos .(h) tan . = gT2/2R(i) (H/T2) = (g/8)(xiii) Projectile thrown from a height h in horizontal direction:(a) T = v(2h/g);(b) R = vv(2h/g);(c) y = h . (gx2/2u2)(d) Magnitude of velocity at the ground = v(u2 + 2gh)(e) Angle at which projectiles strikes the ground, . = tan.1u2gh(xiv) Projectile on an inclined plane:(a) Time of flight, T = ( )00.. - .g cos2u sin(b) Horizontal range, ( )0

0.. - . .=g cos2u sin cosR 22


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FRICTION(i) Force of friction:(a) s = sN (self adjusting); (s)max = sN(b) k = kN (k = coefficient of kinetic friction)(c) k < s(ii) Acceleration on a horizontal plane: a = (F . kN)/M(iii) Acceleration of a body sliding on an inclined plane: a = g sin . (1. k cott2)(iv) Force required to balance an object against wall: F = (Mg/s)(v) Angle of friction: tan . = s (s = coefficient of static friction)DYNAMICS OF RIGID BODIES(i) Average angular velocity:t t t

1 .= ..-. - .. =22 1(ii) Instantaneous angular velocity: . = (d./dt)(iii) Relation between v, . and r : v=.r; In vector form

. . .v = . x r ; In general form, v = .r sin .(iv) Average angular acceleration:t2 t1 .t= ..-. - .a = 2 1(v) Instantaneous angular acceleration: a = (d./dt) = (d2./dt2)(vi) Relation between linear and angular acceleration:(a) aT = ar and aR = (v2/r) = .2R(b) Resultant acceleration, a = v (a a 2 )R

2T +(c) In vector form,. .... ...= + = a = . = . .. . . . . . . . . . . .a aT aR , where aT x r and aR x u x x r(vii) Equations for rotational motion:(a) . = .0 + at;(b) . = .0t +

21 at2;(c) .2 . .02 = 2a.(viii) Centre of mass: For two particle system:(a) ;m mx m x m x1 21 1 2 2


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(a) ;MX mixiCM= S(b) ;Mr CM mi r i.. = S(c) ;dtv d r CMCM..=(d) ;dta d v CMCM..

=(e). . .P CM = M v CM = S mi vi;(f) F ext M a CM mi ai Fi .. . . .= = S = S If F ext = 0, a CM = 0, VCM = constant ;. . .(g) Also, moment of masses about CM is zero, i.e., Smi r i = 0 or m1r1 = m2r2.(x) Moment of Inertia: (a) I = S mi ri2(b) I = r2, where = m1m2/(m1 + m2)

(xi) Radius of gyration: (a) K = v(I/M) ; (b) K = v[(r12 + r22 + . + rn2)/n] = root mean square distance.(xii) Kinetic energy of rotation: K =21 I.2 or I = (2K/.2)(xiii) Angular momentum: (a) L = r x p ; (b) L = rp sin . ; (c) m v d. . .(xiv) Torque: ( ) t = ( ) t = .. . .a r x F; b r F sin(xv) Relation between t and L: . ..

.

. ..

.t =. .dL/ dt ;(xvi) Relation between L and I: (a) L = I.; (b) K =21 I.2 = L2/2I(xvii) Relation between t and a:


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(a) t = Ia,(b) If t = 0, then (dL/dt)=0 or L=constant or, I.=constant i.e., I1.1= I2.2(Laws of conservation of angular momentum)(xviii) Angular impulse: . L = t .t. .(xix) Rotational work done: W = t d . = tav.(xx) Rotational Power:. .P = t ..(xxi) (a) Perpendicular axes theorem: Iz = Ix + Iy(b) Parallel axes theorem: I = Ic + Md2(xxii) Moment of Inertia of some objects(a) Ring: I = MR2 (axis); I =21 MR2 (Diameter);I = 2 MR2 (tangential to rim, perpendicular to plane);I = (3/2) MR2 (tangential to rim and parallel to diameter).


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(b) Disc: I =21 MR2 (axis); I =41 MR2 (diameter)(c) Cylinder: I = MR (axis)21 2(d) Thin rod: I = (ML2/12) (about centre); I = (ML2/3) (about one end)(e) Hollow sphere : Idia = (2/3) MR2; Itangential = (5/3) MR2(f) Solid sphere: Idia = (2/5) MR2 ; Itangential = (7/5) MR2(g) Rectangular: ( )12I M b2 2C= + l (centre)(h) Cube: I = (1/6) Ma2(i) Annular disc: I = (1/2) M ( 222R1 + R )(j) Right circular cone: I = (3/10) MR2(k) Triangular lamina: I = (1/6) Mh2 (about base axis)

(l) Elliptical lamina: I = (1/4) Ma2 (about minor axis) and I = (1/4) Mb2 (aboutmajor axis)(xxiii) Rolling without slipping on a horizontal surface:. .... ...= + .2 = +222 2RMV 1 K

2I 12MV 12K 1 (Q V = R. and I = MK2)For inclined plane(a) Velocity at the bottom, v = . . ... ...+ 2

2R2gh 1 K(b) Acceleration, a = g sin . . .... ...+ 2


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2r1 K(c) Time taken to reach the bottom, t = . . .... ...+ g sinR2s 1 K22(xxiv) Simple pendulum: = T = 2pv (L/g)(xxv) Compound Pendulum: T = 2pv (I/Mg l), where l = M (K2 + l2)Minimum time period, T0 = 2pv (2K/g)(xxvi) Time period for disc: T = 2p v(3R/2g)Minimum time period for disc, T = 2pv (1.414R/g)(xxvii) Time period for a rod of length L pivoted at one end: T = 2pv(2L/3gThe heights by great men reached and kept..were not attained by sudden flight,but they, while their companions slept..were toiling upwards in the night.


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CONSERVATION LAWS AND COLLISIONS(i) Work done: (a) W F . d ;. .= (b) W = Fd cos .; (c) W =(ii) Conservation forces: (a) F . d r (b) F . d r ; (c) F . d r = 0. . . . . .For conservative forces, one must have:.FV x = 0(iii) Potential energy: (a) VU = - W; (b) F = - (dU/dX) ; (c)F = -VU.(iv) Gravitational potential energy: (a) U = mgh ; (b) (R h)U GMm+= -(v) Spring potential energy: ( ) ( ) ( 2 )122a U = Kx2 ; b .U = K x - x(vi) Kinetic energy: (a) .K = W = 2 ( ) 2i

2 mvf - mv ; b K = mv(vii) Total mechanical energy: = E = K + U(viii) Conservation of energy: .K = . .U or, K + U = Ki + UiIn an isolated system, Etotal = constant(ix) Power: (a) P = (dw/dt) ; (b) P = (dw/dt) ; (c) P =. .F . v(x) Tractive force: F = (P/v)(xi) Equilibrium Conditions:(a) For equilibrium, (dU/dx) = 0(b) For stable equilibrium: U(x) = minimum, (dU/dx) = 0 and (d2U/dx2) is positive

(c) For unstable equilibrium: U(x) = maximum, (dU/dx) = 0 and (d2U/dx2) is negative(d) For neutral equilibrium: U(x) = constant, (dU/dx) = 0 and (d2U/dx2) is zero(xii) Velocity of a particle in terms of U(x): v = [E U(x)]m2 -(xiii) Momentum:(a) ( ) . .... ...= =. . . .

p m v; b F d p/ dt ,(b) Conservation of momentum: If Fnet 0, then p f p i,. . .= =(c) Recoil speed of gun, BGBG x vmm


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v =(xiv) Impulse: . p = Fav .t. .(xv) Collision in one dimension:(a) Momentum conservation : m1u1 + m2u2 = m1v1 + m2v2(b) For elastic collision, e = 1 = coefficient of restitution(c) Energy conservation: m1u12 + m2u22 = m1v12 + m2v22(d) Velocities of 1st and 2nd body after collision are:. 21xxPath 1 Path 2 closedpath.ba .ba .121

2121212121212

12


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22 12 111 212 22 1211 21 21 um mu m mm mu ; v 2mm mu 2mm mm m v . .... ..

.+-+ . .... ...+= . .... ...+

+ . .... ...+-=(e) If m1 = m2 = m, then v1 = u2 and v2 = u1(f) Coefficient of restitution, e = (v2.v1/u1 = u2)(g) e = 1 for perfectly elastic collision and e=0 for perfectly inelastic collision. For inelasticcollision 0 < e < 1(xvi) Inelastic collision of a ball dropped from height h0

(a) Height attained after nth impact, hn = e2nh0(b) Total distance traveled when the ball finally comes to rest, s = h0 (1+e2)/(1.e2)(c) Total time taken, t = ......-+1 e


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1 eg2h0(xvii) Loss of KE in elastic collision: For the first incident particle( ) 100%K; If m m , Km m

4m mKand K

m mm mKKilost2 1 21 21 2ilost21 2

1 2i= . =+. =. .... ...+ = -(xviii) Loss of KE in inelastic collision: . Klost = Ki . K=1 2

1 2m mm m21+(u1 . u2)2 (1.e2)Velocity after inelastic collision (with target at rest)( )11 211 2

1 21 21 um mm 1 eu and vm mm emv+


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+= . .... ...+-=(xix) Oblique Collision (target at rest):m1u1 = m1v1 cos .1 + m2v2 cos .2 and m1v1 sin .1 = m2v2 sin .2Solving, we get: m1u12 = m1v12 + m2v22(xx) Rocket equation: (a)dtv dMdtM dV = - el(b) V = . vrel loge . .... ... -0

0 bMM m [M0 = original mass of rocket plus fuel and mb = mass of fuel burnt](c) If we write M = M0 . mb = mass of the rocket and full at any time, than velocity of rocks atthat time is:V = vrel loge (M0/M)(xxi) Conservation of angular momentum:(a) If text = 0, then L = Li(b) For planets,minmaxmin

maxrrvv =(c) Spinning skater, I1.1 = I2W2 or . = .i . .... ...IIi


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SIMPLE HARMONIC MOTION AND LISSAJOUS FIGURES(i) Simple Harmonic Motion:(a) F = . Kx ;(b) a = .mK x or a = . .2x, where . = v(K/m);(c) Fmax = KA and amax = .2A(ii) Equation of motion: x 0dtd x22+ .2 =(iii) Displacement: x = A sin (.t + f)(a) If f = 0, x = A sin .t ;(b) If f = p/2, x = A cos .t(c) If x = C sin .t + D cos .t, then x = A sin (.t + f) with A= v(C2+D2) and f =tan.1 (D/C)(iv) Velocity:(a) v = A . cos (.+ f);(b) If f=0, v = A . cos .t;(c) vmax =.A(d) v = .v(A2 . x2);

(e) 1AvAx2 2222=.+(v) Acceleration:

(a) a = ..2 x = . .2A sin (.t+f) ;(b) If f=0, a=. .2A sin .t(c) |amax| = .2A;(d) Fmax = m .2A(vi) Frequency and Time period:(a) . = v(K/m) ;(b) = (K /m);21p(c) T = 2pKm

(vii) Energy in SHM: Potential Energy:(a) U = Kx2 ;(b) F = .dxdU ;(c) Umax = m.2A2;(d) U = m.2 A2 sin2 .t(viii) Energy in SHM: Kinetic energy:(a) K = mv2;(b) K= m.2 (A2.x2);


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(c) K = m.2A2 cos2 .t ;(d) Kmax = m.2A221212121212121


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(ix) Total energy:(a) E = K + U = conserved;(b) E = (1/2) m.2 A2 ;(c) E = Kmax = Umax(x) Average PE and KE:(a) < U > = (1/4) m.2A2 ;(b) < K > = (1/4) m.2A2;(c) (E/2) = < U > = < K >(xi) Some relations:(a) . = ;x xv v21222221-- (b) T = 2p 22212

122v vx x-- ; (c) A = ( ) ( )22212 121 2v v

v x v x-- 2(xii) Spring. mass system:(a) mg = Kx0;(b) T = 2pgx2Km = p 0(xiii) Massive spring: T = 2p ( )K

m + ms / 3(xiv) Cutting a spring:(a) K. = nK ;(b) T. = T0/v(n) ;(c) . = v(n) 0(d) If spring is cut into two pieces of lengths l1 and l2 such that l1 = nl2, then K1 = K,nn 1...


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..

. + K2 =(n +1) K and K1l1 = K2l2(xv) Springs in parallel:(a) K = K1 + K2 ;(b) T = 2p v[m/(K1 + K2)](c) If T1 = 2pv (m/K1) and T2 = 2pv(m/K2), then for the parallel combination:2221222211 222212 andT T

or T T TT

1T1T1 . = . + .+= + =(xvi) Springs in series:(a) K1x1 = K2x2 = Kx = F applied(b)1 21 21 2 K K

K K or KK1K1K1+= + =(c) 222

12211 1 1 or T - T + T.+.=. 22


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2(d) T = 2p ( )(1 2)1 21 21 2m K KK K

2or 1

K Km K Kp ++ =(xvii) Torsional pendulum:(a) Ia=t.C. or 0ICdtd22. + . = ;(b) .=.0 sin (.t+f);

(c) . = v(C/I) ;


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(d) =IC

21p;(e) T = 2pv(I/C), where C = p.r4/2l(xviii) Simple pendulum:(a) Ia = t =. mgl sin . or ...... . +2 lgdtd2sin . = 0 or g 0dtd22. + . =

l;(b) . = v(g/l) ;(c) = (g/ ) ;21lp(d) T = 2p v(l/g)(xix) Second pendulum:(a) T = 2 sec ;(b) l = 99.3 cm(xx) Infinite length pendulum:

(a) ;Rg 1 1T 2 1e. .... ...+= pl(b) T=2p

gRe (when l.8)(xxi) Anharmonic pendulum: T . T0 . .... ...+ . . ..


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.

. .

.

. .+ 2202016

T 1 A161l(xxii) Tension in string of a simple pendulum: T = (3 mg cos . . 2 mg cos .0)(xxiii) Conical Pendulum:(a) v = v(gR tan .) ;(b) T = 2pv (L cos ./g)(xxiv) Compound pendulum: T = 2p ( )2l + K2 / l(a) For a bar: T = 2pv(2L/3g) ;(b) For a disc : T = 2pv (3R/2g)(xxv) Floating cylinder:(a) K = A.g ;

(b) T = 2pv(m/A.g) = 2pv(Ld/.g)(xxvi) Liquid in U.tube:(a) K = 2A .g and m = AL. ;(b) T = 2pv(L/2g) = 2pv(h/g)(xxvii) Ball in bowl: T = 2pv[(R . r)/g](xxviii) Piston in a gas cylinder:(a) ;VK A E2=(b) ;A E

T 2 mV2 = p


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(c)A PVT 2 2= p m (E.P for Isothermal process);(d)PT 2 Vm. .= p 2 (E = . P for adiabatic process)(xxix) Elastic wire:(a) K = AY ;l(b) T =AY2 m l p(xxx) Tunnel across earth: T = 2pv(Re/g)(xxxi) Magnetic dipole in magnetic field: T = 2pv(I/MB)(xxxii) Electrical LC circuit: T = 2p LC or2 LC1p =

(xxxiii) Lissajous figures .Case (a): .1 = .2 = . or .1 : .2 = 1 : 1General equation: + - cos f = sin fab2xybyax 22222

For f = 0 : y = (b/a) x ; straight line with positive slopeFor f = p/4 : ; oblique ellipse21ab2xybyax222

2+ - =For f = p/2 : 1 ; symmetrical ellipsebyax222


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2+ =For f = p : y = .(b/a) x ; straight line with negative slope.Case (b): For .1 : .2 = 2:1 with x = a sin (2.t + f) and y = b sin .tFor f = 0, p: Figure of eightFor f = , 3 :4p4p Double parabolaFor f = , 3 :2p2p Single parabola


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GRAVITATION(i) Newton.s law of gravitation:(a) F = G m1m2/r2 ; (b) a = 6.67 x 10.11 K.m2/(kg)2 ; (c)r2 drFdF = -(ii) Acceleration due to gravity (a) g = GM/R2 ; (b) Weight W = mg(iii) Variation of g:(a) due to shape ; gequator < gpole(b) due to rotation of earth: (i) gpole = GM/R2 (No effect)(ii) gequator = RRGM 22 -.(iii) gequator < gpole(iv) .2R = 0.034 m/s2(v) If . . 17 .0 or T = (T0/17) = (24/17)h = 1.4 h, then object wouldfloat on equator(c) At a height h above earth.s surface g. = g , if h Rgh 2 1 < < . ...

. ...-(d) At a depth of below earth.s surface: g. = g ...... -R1 d(iv) Acceleration on moon: gm = 2 earthmm g6

1RGM .(v) Gravitational field: (a) ( ) () r r (inside)Rr outside ; b g GMrg GM^3^2 = - = -. .

(vi) Gravitational potential energy of mass m:(a) At a distance r : U(r) = . GMm/r(b) At the surface of the earth: U0 = . GMm/R(c) At any height h above earth.s surface: U . U0 = mgh (for h < < R)or U = mgh (if origin of potential energy is shifted to the surface of earth)(vii) Potential energy and gravitational force: F = . (dU/dR)(viii) Gravitational potential: V(r) = .GM/r(ix) Gravitational potential energy of system of masses:(a) Two particles: U = . Gm1m2/r(b) Three particles: U = .


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232 3131 3121 2rGm mrGm mrGm m - -(x) Escape velocity:(a) ve =R2GM or ve = v(2gR) = v(gD)(b) ve =3R 8pG.(xi) Maximum height attained by a projectile:( ) Rv hR h

or v v h

v / v 1h R 2 e ee.+=-= (if h < < R)


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(xii) Orbital velocity of satellite:( ) ( ) ( );2 R h; b v v Rra v0 GM 0 e += = (c) v0 .ve/v2 (if h


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SURFACE TENSION(i) (a)lFLengthT = Force = ; (b)AWSurface areaT = Surface energy =(ii) Combination of n drops into one big drop: (a) R = n1/3r(b) Ei = n(4pr2T), E = 4pR2T, (E/Ei) = n.1/3, . .... .... = -1/3i n

1 1EE(c) .E = 4pR2T (n1/3 .1) = 4pR3T ...

... -R1r1(iii) Increase in temperature: .. =s3T......

. -R1r1 or ...... -. R1r1

sJ

3T(iv) Shape of liquid surface:(a) Plane surface (as for water . silver) if Fadhesive >2Fcohesive(b) Concave surface (as for water . glass) if Fadhesive >2Fcohesive(c) Convex surface (as for mercury.glass) if Fadhesive Fc/v2 ;(b) obtuse: if Fa Tsa(vi) Excess pressure:(a) General formula: Pexcess = . .. .. ...+1 R21RT 1(b) For a liquid drop: Pexcess = 2T/R

(c) For an air bubble in liquid: Pexcess = 2T/R(d) For a soap bubble: Pexcess = 4T/R(e) Pressure inside an air bubble at a depth h in a liquid: Pin = Patm + hdg + (2T/R)(vii) Forces between two plates with thin water film separating them:(a) .P = T ;R1r1 ...... -(b) ;

R1r1 AT F ...... = -(c) If separation between plates is d, then .P = 2T/d and F = 2AT/d(viii) Double bubble: Radius of Curvature of common film Rcommon =R rrR-(ix) Capillary rise:(a) ;

rdgh = 2T cos .(b)rdgh = 2T (For water . = 00)


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(c) If weight of water in meniseus is taken into account then T =....... +2 cos3rdg h r(d) Capillary depression, ( )rdgh - 2T cos p - .(x) Combination of two soap bubbles:(a) If .V is the increase in volume and .S is the increase in surface area, then3P0.V + 4T.S =0 where P0 is the atmospheric pressure(b) If the bubbles combine in environment of zero outside pressure isothermally,then .S = 0 orR3 = v ( ) 222R1 + RELASTICITY(i) Stress: (a) Stress = [Deforming force/cross.sectional area];

(b) Tensile or longitudinal stress = (F/p r2);(c) Tangential or shearing stress = (F/A);(d) Hydrostatic stress = P(ii) Strain: (a) Tensile or longitudinal strain = (.L/L);(b) Shearing strain = f;(c) Volume strain = (.V/V)(iii) Hook.s law:(a) For stretching: Stress = Y x Strain or A( L)Y FL.=(b) For shear: Stress = . x Strain or . = F/Af(c) For volume elasticity: Stress = B x Strain or B = . (V/V)

P.(iv) Compressibility: K = (1/B)(v) Elongation of a wire due to its own weight: .L =YL g2

1YAMgL21 = 2.(vi) Bulk modulus of an idea gas: Bisothermal = P and Badiabatic = .P (where . =

Cp/Cv)(vii) Stress due to heating or cooling of a clamped rodThermal stress = Ya (.t) and force = YA a (.t)(viii) Torsion of a cylinder:(a) r . = lf (where . = angle of twist and f = angle of shear);(b) restoring torque t = c.(c) restoring Couple per unit twist, c = p.r4/2l (for solid cylinder)and C = p. (r24 . r14)/2l (for hollow cylinder)


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(ix) Work done in stretching:(a) W =21 x stress x strain x volume = Y21 (strain)2 x volume = ( ) x volumeYstress

21 2(b) Potential energy stored, U = W =21 x stress x strain x volume(c) Potential energy stored per unit volume, u =21 x stress x strain(x) Loaded beam:


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(a) depression, d = (rectangular)4YbdW3l3(b) Depression, d = (cylindrical)12Y rW23pl(xi) Position.s ratio:(a) Lateral strain =rrD- .D = -.(b) Longitudinal strain = (.L/L)(c) Poisson.s ratio s =L / L

r rlongitudinal strain

lateral strain.= -. /(d) Theoretically, .1 < s < 0.5 but experimentally s . 0.2 . 0.4(xii) Relations between Y, ., B and s:(a) Y = 3B (1.2s) ;(b) Y = 2. (1+ s);(c).= +319B

1Y1(xiii) Interatomic force constant: k = Yr0 (r0 = equilibrium inter atomic separation)KINETIC THEORY OF GASES(i) Boyle.s law: PV = constant or P1V1 = P2V2(i) Chare.s law: (V/T) = constant or (V1/T1) = (V2/T2)(ii) Pressure . temperature law: (P1/T1) = (P2/T2)(iii) Avogadro.s principle: At constant temperature and pressure, Volume of gas,V . number of moles, Where = N/Na [N = number of molecules in the sample

and NA = Avogadro.s number = 6.02 x 1023/mole]M= Msample [Msample = mass of gas sample and M = molecular weight](iv) Kinetic Theory:(a) Momentum delivered to the wall perpendicular to the x.axis, .P = 2m vx(b) Time taken between two successive collisions on the same wall by the same molecule: .t =(2L/vx)(c) The frequency of collision: .coll. = (.x/2L)(d) Total force exerted on the wall by collision of various molecules: F = (MN/L


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) (e) The pressure on the wall : P = 2rms2rms2 2x v3v 1VmN3v 13Vv mNVmN < > = < > = = .(v) RMS speed:(a) .rms = v(v12 + v22 + . + v 2N /N);(b) .rms = v(3P/.) ;

(c) .rms = v(3KT/m);(d) .rms = v(3RT/M) ; (e) ( )( ) 1212rms 2rms 1MMmm= =

..(vi) Kinetic interpretation of temperature:


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(a) (1/2) Mv2rms = (3/2) RT ;(b) (1/2) mv2 rms = (3/2) KT(c) Kinetic energy of one molecule = (3/2) KT ;(d) kinetic energy of one mole of gas = (3/2) RT(e) Kinetic energy of one gram of gas (3/2) (RT/M)(ix) Maxwell molecular speed distribution:(a) n (v) = 4pN 2 -mv / 2KT3/2 2 v e2 KTm ......p(b) The average speed:M1.60 RTM

8 RTmv 8KT =p

=p=(c) The rms speed: vrms =M1.73 RTM3RTm3kT = =(d) The most probable speed: .p =M1.41 RT

M2RTm2KT = =(e) Speed relations: (I) vp < v < vrms(II) vp : v : vrms = v(2) : v(8/p) : v(3) = 1.41 : 1.60 : 1.73(x) Internal energy:(a) Einternal = (3/2)RT (for one mole)(b) Einternal = (3/2 RT (for mole)(c) Pressure exerted by a gas P = E32V

E32 =(xi) Degrees of freedom:(a) Ideal gas: 3 (all translational)(b) Monoatomic gas : 3 (all translational)(c) Diatomic gas: 5 (three translational plus two rotational)(d) Polyatomic gas (linear molecule e.g. CO2) : 7 (three translational plus tworotational plus twovibrational)


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(e) Polyatomic gas (non.linear molecule, e.g., NH3, H2O etc): 6 (three translational plus threerotational)(f) Internal energy of a gas: Einternal = (f/2) RT. (where f = number of degreesof freedom)(xii) Dalton.s law: The pressure exerted by a mixture of perfect gases is the sum of the pressuresexerted by the individual gases occupying the same volume alone i.e., P = P1 + P2 + ..(xiii) Van der Wall.s gas equation:(a) ( ) . = . .... ... + V - b RVP a 22(b) (V b) RTVa P m 2m 2= - . ..

.. .

.

. + (where Vm = V/ = volume per mole);(c) b = 30 cm3/mole(d) Critical values: Pc = ;27 Rb, V 3b, T 8a27 ba2 C C = =(e) 0.3758

3RTP VCC C = =(xiv) Mean free path: . =2 d n1p 2.,Where .n = (N/V) = number of gas molecules per unit volume andd = diameter of molecules of the gas


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FLUID MECHANICS(i) The viscous force between two layers of area A having velocity gradient (dv/dx) is given by: F = ..A (dv/dx), where . is called coefficient of viscosity(i) In SI system, . is measured I Poiseiulle (Pl) 1Pl = 1Nsm.2 = 1 decapoise. Inegs system, the unitof . is g/cm/sec and is called POISE(ii) When a spherical body is allowed to fall through viscous medium, its velocity increases, till the sumof viscous drag and upthrust becomes equal to the weight of the body. After thatthe body moveswith a constant velocity called terminal velocity.(iii) According to STOKE.s Law, the viscous drag on a spherical body moving in afluid is given by: F =6p.r v, where r is the radius and v is the velocity of the body.(iv) The terminal velocity is given by: vT = ( ).r . - s g92 2where . is the density of the material of the body and s is the density of liquid(v) Rate of flow of liquid through a capillary tube of radius r and length l

Rp8 / rp8V pr 44=. p=.= pl l

where p is the pressure difference between two ends of the capillary and R is the fluid resistance(=8 .l/pr4)(vi) The matter which possess the property of flowing is called as FLUID (For example, gases andliquids)(vii) Pressure exerted by a column of liquid of height h is : P = h.g (. = density of the liquid)(viii) Pressure at a point within the liquid, P = P0 + h.g, where P0 is atmospheric pressure and h is thedepth of point w.r.t. free surface of liquid(ix) Apparent weight of the body immersed in a liquid Mg. = Mg . V.g(x) If W be the weight of a body and U be the upthrust force of the liquid on th

e body then(a) the body sinks in the liquid of W > U(b) the body floats just completely immersed if W = U(c) the body floats with a part immersed in the liquid if W < U(xi)density of soliddensity of solidtotal volume of solidVolume of immersed part of a solid =(xii) Equation of Continuity: a1v1 = a2v2


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(xiii) Bernouilli.s theorem: (P/.) + gh +21 v2 = constant(xiv) Accelerated fluid containers : tan . =ga x(xv) Volume of liquid flowing per second through a tube: R=a1v1 = a2v2 ( )222a1 a2gh-(xvi) Velocity of efflux of liquid from a hole:v = v(2gh), where h is the depth of a hole from the free surface of liquidI do not ask to walk smooth paths, nor bear an easy load.I pray for strength and fortitude to climb rock-strewn road.Give me such courage I can scale the hardest peaks alone,And transform every stumbling block into a stepping-stone.. Gail Brook Burkettax..Fig. 4


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HEAT AND THERMODYNAMICS(i) L2 . L1 = L1a(T2 . T1); A2 . A1 = A2 (T2 . T1); V2 . V1 = V1.(T2 . T1)where, L1, A1, V1 are the length, area and volume at temperature T1; and L2, A2,V2 are that attemperature T2.a represents the coefficient of linear expansion, the coefficientof superficialexpansion and . the coefficient of cubical expansion.(ii) If dt be the density at t0C and d0 be that at 00C, then: dt = d0 (1...T)(iii) a : : . = 1 : 2 : 3(iv) If .r, .a be the coefficients of real and apparent expansions of a liquid and .g be the coefficient of thecubical expansion for the containing vessel (say glass), then.r = .a + .g(v) The pressure of the gases varies with temperature as : Pt = P0 (1+ ..T), where . = (1/273) per 0C(vi) If temperature on Celsius scale is C, that on Fahrenheit scale is F, on Kelvin scale is K, and onReaumer scale is R, then(a)4R5

K 273

9F 325C = - = - = (b) C 325F = 9 +(c) (F 32)9C = 5 -(d) K = C + 273 (e) (F 459.4)9K = 5 +(vii) (a) Triple point of water = 273.16 K

(b) Absolute zero = 0 K = .273.150C(c) For a gas thermometer, T = (273.15) (Kelvin)PPtriple(d) For a resistance thermometer, R. = R0 [1+ a.](viii) If mechanical work W produces the same temperature change as heat H, thenwe can write:W = JH, where J is called mechanical equivalent of heat(ix) The heat absorbed or given out by a body of mass m, when the temperature changes by .T is: .Q= mc.T, where c is a constant for a substance, called as SPECIFIC HEAT.(x) HEAT CAPACITY of a body of mass m is defined as : .Q = mc

(xi) WATER EQUIVALENT of a body is numerically equal to the product of its massand specific heati.e., W = mc(xii) When the state of matter changes, the heat absorbed or evolved is given by: Q = mL, where L iscalled LATENT HEAT(xiii) In case of gases, there are two types of specific heats i.e., cp and cv [cp = specific heat at constantpressure and Cv = specific heat at constant volume]. Molar specific heats of a gas are: Cp = Mcp


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and Cv = Mcv, where M = molecular weight of the gas.(xiv) Cp > Cv and according to Mayer.s formula Cp . Cv = R(xv) For all thermodynamic processes, equation of state for an ideal gas: PV = RT(a) For ISOBARIC process: P = Constant ;TV =Constant(b) For ISOCHORIC (Isometric) process: V = Constant;TP =Constant(c) For ISOTHERMAL process T = Constant ; PV= Constant(d) For ADIABATIC process: PV. = Constant ; TV..1=Constantand P(1..) T. = Constant(xvi) Slope on PV diagram


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(a) For isobaric process: zero(b) For isochoric process: infinite(c) For isothermal process: slope = .(P/V)(d) For adiabatic process: slope = ..(P/V)(e) Slope of adiabatic curve > slope of isothermal curve.(xvii) Work done(a) For isobaric process: W = P (V2 . V1)(b) For isochoric process: W = 0(c) For isothermal process: W=RT loge (V2/V1)RT x 2.303 x log10 (V2/V1)P1V1 x 2.303 x log10 (V2/V1)RT x 2.303 x log10 (P1/P2)(d) For adiabatic process: ( )( )( )(. -1)= -. -1W = R T1 - T2 P1V1 P2V2(e) In expansion from same initial state to same final volumeWadiabatic < Wisothermal < Wisobaric(f) In compression from same initial state to same final volume:Wadiabatic < Wisothermal < Wisobaric

(xviii) Heat added or removed:(a) For isobaric process: Q = Cp.T(b) For isochoric process = Q = Cv.T(c) For isothermal process = Q = W = Rt loge (V2/V1)(d) For adiabatic process: Q = 0(xix) Change in internal energy(a) For isobaric process = .U = Cv.T(b) For isochoric process = .U = Cv.T(c) For isothermal process = .U = 0(d) For adiabatic process: .U = .W = ( )( )1T T R 1 2-.

-(xx) Elasticities of gases(a) Isothermal bulk modulus = BI = P(b) Adiabatic bulk modulus BA = .P(xxi) For a CYCLIC process, work done .W = area enclosed in the cycle on PV diagram.Further, .U = 0 (as state of the system remains unchanged)So, .Q = .W(xxii) Internal energy and specific heats of an ideal gas (Monoatomic gas)(a) U =23 RT (for one mole);(b)

2U = 3 RT (for moles)(c) .U =23 R.T (for moles);(d) Cv= R23.U 1 = ..


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.

..

.

.

.


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(e) Cp = Cv + R =23 R + R =25R(f) . = 1.673R 52R 325CCvp = = ...... = . .... ..

.(xxiii) Internal energy and specific heats of a diatomic gas(a)2U = 5 RT (for moles);(b) .U =25 R.T (for moles)(c) Cv = R;25TU 1 = ..

...

.

.

.(d) Cp = Cv + R =25 R + R =27 R(e) 1.45

725R27RCCvp = = ...


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..

. = . .

.

.

. .

.

.

. =(xxiv) Mixture of gases: = 1 + 21 21 2 1 1 2 2N NM M N m N mM++= + + +=1 21 21 21 2

1 21 2 + + = + + = 1 2 p1 p2pv vvC C

and C

C CC(xxv) First law of thermodynamics(a) .Q = .U + .W or .U = .Q . .W(b) Both .Q, .W depends on path, but .U does not depend on the path(c) For isothermal process: .Q = .W = RT log | V2/V1|, .U = 0, T = Constant, PV =Constantand Ciso = 8(d) For adiabatic process: .W = ( )( ) ,1R T2 T1- .

- .Q = 0, .U = Cv (T2.T1), Q = 0,PV. = constant, Cad = 0 and. = = 1+ 2CCvp(where is the degree of freedom)(e) For isochoric process: .W = 0, .Q = .U = Cv.T, V = constant, and Cv = (R/..1)


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(f) For isobaric process: .Q = Cp.T, .U = Cv.T., .W = R.T, P = constant andCp = (.R/..1)(g) For cyclic process: .U = 0, .Q = .W(h) For free expansion: .U = 0, .Q = 0, .W = 0(i) For polytropic process: .W = [R(T2.T1)/1.n], .Q = C (T2.T1), PVn = constant and1 nC R R-+. -1=(xxvi) Second law of thermodynamics(a) There are no perfect engines(b) There are no perfect refrigerators


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(c) Efficiency of carnot engine: . = 1.121212TTQQ,QQ=(d) Coefficient of performance of a refrigerator:=1 221 22 2T TT

Q QQWQWork done on refrigeratorHeat absorbed from cold reservoir-=-= =For a perfect refrigerator, W = 0 or Q1 = Q2 or = 8(xxvii) The amount of heat transmitted is given by: Q = .KA t.x

.. , where K is coefficient of thermalconductivity, A is the area of cross section, .. is the difference in temperature, t is the time of heatflow and .x is separation between two ends(xxviii) Thermal resistance of a conductor of length d = RTh =K Ad(xxix) Flow of heat through a composite conductor:(a) Temperature of interface, . = ( )( )(1 1) ( 2 2)1 1 2 2K / d K / dK / d K /d

+.1 + .2(b) Rate of flow of heat through the composite conductor: H = ( )(d1 /K1) (d2 /K2)AtQ+= .1 - .2(c) Thermal resistance of the composite conductor


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( ) ( )2Th 1 Th2211TH R RK AdK AdR = + = +(d) Equivalent thermal conductivity, K = (1 1) (2 2)1 2d /K d /Kd d++(xxx) (a) Radiation absorption coefficient: a = Q0/Q0(b) Reflection coefficient: r = Qr/Q0(c) Transmission coefficient: t = Qt/Q0(d) Emissive power: e or E = Q/A .t [t = time](e) Spectral emissive power: e. = At (d.)Q and e = e.e.

(f) Emissivity: e = e/E ; 0 = e = 1(g) Absorptive power: a = Qa/Q0(h) Kirchhoffs law: (e./a.)1 = (e./a.)B = ...= E.(i) Stefan.s law: (a) E=sT4 (where s=5.67x10.8 Wm.2 K.4)For a black body: E = s (T4.T04)For a body: e = es (T4.T04)(j) Rate of loss of heat: . A ( )dtdQ 40= e s .4 - .

For spherical objects: ( )( ) 222121rr

dQ/ dtdQ/ dt=(k) Rate of fall of temperature: ( ) ( 4)

04 404 . - ... = e s . - . = e sV s

AmsA


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dtd. ( )( ) 12122121rrVx VAAd dtd dt= =././.80


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(For spherical bodies)(l) Newton.s law of cooling:dtd. = .K (...0) or (...0) a e.KT(m) Wein.s displacement law: .mT = b (where b = 2.9 x 10.3 m . K)(n) Wein.s radiation law: E.d.= A .......5 (.T) d.= .......5A e.a/.Td.(o) Solar Constant: S =2ESSRR. ...

. ... sT4 or T =1/ 2SES1/ 4RRS. .... ...

...

..

.s


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WAVES1. Velocity: v = n. and n = (1/T)2. Velocity of transverse waves in a string: v =r dTmTp 2=3. Velocity of longitudinal waves:(a) In rods: v = v(Y/.) (Y . Young.s modulus, . = density)(b) In liquids: v = v(B/.) (B = Bulk modulus)(c) In gases: v = v(.P/.) (Laplace formula)4. Effect of temperature:(a) v = v0v (T/273) or v = v0 + 0.61t(b) (vsound/vrms) = v(./3)5. Wave equation: (a) y = a sin.2p (vt.x)(b) y = a sin 2p ......

.- xTt(c) y = a sin (.t . kx), where wave velocity v = . = n.k6. Particle velocity: (a) vparticle = (.y/.t)(b) maximum particle velocity, (vparticle)max = . a7. Strain in medium (a) strain = . (.y/.x) = ka cos (.t . kx)(b) Maximum strain = (.y/dx)max = ka(c) (vparticle/strain) = (./k) = wave velocityi.e., vparticle = wave velocity x strain in the medium8. Wave equation: . .

..

. .

.

.

.= ...2222

2xv yty9. Intensity of sound waves:(a) I = (E/At)(b) If . is the density of the medium; v the velocity of the wave; n the frequency and a theamplitude then I = 2p2 . v n2 a2 i.e. I . n2a2


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(c) Intensity level is decibel: 10 log (I/I0). Where, I0 =Threshold of hearing =10.12 Watt/m210. Principle of superposition: y = y1 + y211. Resultant amplitude: a = v(a12 + a22 + 2a1a2 cos f)12. Resultant intensity: I = I1 + I2 + 2v(I1I2 cos f)(a) For constructive interference: f = 2np, amax = a1 + a2 and Imax = (vI1 + vI2)2(b) For destructive interference: f = (2n.1) p, amin = a2 .a2 and Imin = (vI1=vI2)213. (a) Beat frequency = n1 . n2 and beat period T = (T1T2/T2.T1)(b) If there are N forks in successive order each giving x beat/sec with nearestneighbour, thennlast = nfirst + (N.1)x14. Stationary waves: The equation of stationary wave,(a) When the wave is reflected from a free boundary, is:y = + 2a cos 2a cos kx sin tT2 x sin 2pt = ..p


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(b) When the wave is reflected from a rigid boundary, is:Y + .2a sinT2 x cos 2pt.p =.2a sin kx cos .t15. Vibrations of a stretched string:(a) For fundamental tone: n1 =m1 T.(b) For p th harmonic : np =mp T.(c) The ratio of successive harmonic frequencies: n1 : n2 : n3 :.... = 1 : 2 : 3: ..(d) Sonometer:mT

2n ll

= (m = p r2 d)(e) Melde.s experiment: (i) Transverse mode: n =mT

2pl(ii) Longitudinal mode: n =mT22pl

16. Vibrations of closed organ pipe(a) For fundamental tone: n1 = ......4Lv(b) For first overtone (third harmonic): n2 = 3n1(c) Only odd harmonics are found in the vibrations of a closed organ pipeand n1 : n2 : n3 : ...=1 : 3 : 5 : ..17. Vibrations of open organ pipe:(a) For fundamental tone: n1 = (v/2L)(b) For first overtone (second harmonic) : n2 = 2n1

(c) Both even and odd harmonics are found in the vibrations of an open organ pipe andn1 : n2 : n3 : ..=1 : 2 : 3 : ...18. End correction: (a) Closed pipe : L = Lpipe + 0.3d(b) Open pipe: L = Lpipe + 0.6 dwhere d = diameter = 2r19. Resonance column: (a) l1 +e =4. ; (b) l2 + e =4


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(iii) Ratio of frequencies, n1 : n2 : n3 : ..= 1 : 3 : 5 : ...22. Frequency of a turning fork: n a.t El 2Where t = thickness, l = length of prong, E = Elastic constant and . = density23. Doppler Effect for Sound(a) Observer stationary and source moving:(i) Source approaching: n. =v vv- sx n and .. =vv - vs x .(ii) Source receding: n. =v vsv+x n and .. =vv + vs x .(b) Source stationary and observer moving:

(i) Observer approaching the source: n. =vv + v0 x n and .. = .(ii) Observer receding away from source: n. =vv - v0 x n and .. = .(c) Source and observer both moving:(i) S and O moving towards each other: n. =s0v vv v-

+ x n(ii) S and O moving away from each other: n. =s0v vv v+- x n(iii) S and O in same direction, S behind O : n. =s0v vv v

-- x n(iv) S and O in same direction, S ahead of O: n.=s0v vv v++ x n(d) Effect of motion of medium:


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m sm 0v v vv v vn' =(e) Change in frequency: (i) Moving source passes a stationary observer: .n = 2s2sv v2vv-x nFor vs


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(b) Source between observer and walln. = x nv vv+ s(for direct waves)n. =v vsv-x n (for reflected waves)(ii) Source moving away from wall(a) Observer between source and walln. = x nv vv+ s(for direct waves)n. =v vsv+x n (for reflected waves)

(b) Source between observer and walln. = x nv vv- s(for direct waves)n. =v vsv+x n (for reflected waves)(g) Moving Target:(i) S and O stationary at the same place and target approaching with speed u

n. = x nv uv u ......-+ or n. = x nvu 2 1 ...... + (for u


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u 2 1 ...... - (for u


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cv ...... or ... =. ......cv .


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STUDY TIPS Combination of SubjectsStudy a combination of subjects during a day i. e. after studying 2.3 hrs of mathematicsshift to any theoretical subject for 2 horrs. When we study a subject like math,aparticular part of the brain is working more than rest of the brain. When we shift to atheoretical subject, practically the other part of the brain would become activeand thepart studying maths will go for rest. RevisionAlways refresh your memory by revising the matter learned. At the end of the dayyoumust revise whatever you.ve learnt during that day (or revise the previous daysworkbefore starting studies the next day). On an average brain is able to retain thenewlylearned information 80% only for 12 hours, after that the forgetting cycle begins. Afterthis revision, now the brain is able to hold the matter for 7 days. So next revision shouldbe after 7 days (sundays could be kept for just revision). This ways you will ge

t rid of theproblem of forgetting what you study and save a lot of time in restudying that topic. Use All Your SensesWhatever you read, try to convert that into picture and visualize it. Our eye memory ismany times stronger than our ear memory since the nerves connecting brain to eyearemany times stronger than nerves connecting brain to ear. So instead of trying tomug upby repeating it loudly try to see it while reapeating (loudly or in your mind).This isapplicable in theoritical subjects. Try to use all your senses while learning a

subjectmatter. On an average we remember 25% of what we read, 35% of what we hear, 50%of what we say, 75% of what we see, 95% of what we read, hear, say and see. Breathing and RelaxationTake special care of your breathing. Deep breaths are very important for relaxing yourmind and hence in your concentration. Pranayam can do wonders to your concentration,relaxation and sharpening your mined (by supplying oxygen to it). Aerobic exercises likeskipping, jogging, swimming and cycling are also very helpful.


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53198602 formula booklet physics xi 0001

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