Formula Booklet Physics Xi

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Formula Booklet – Physics XI

Dear students Most students tend to take it easy after the board examinations of Class X. The summer vacations immediately after Class X are a great opportunity for the students to race ahead of other students in the competitive world of IITJEE, where less than 2% students get selected every year for the prestigious institutes. Some students get governed completely by the emphasis laid by the teachers of the school in which they are studying. Since, the objective of the teachers in the schools rarely is to equip the student with the techniques reqired to crack IITJEE, most of the students tend to take it easy in Class XI. Class XI does not even have the pressure of board examinations. So, while the teachers and the school environment is often not oriented towards the serious preparation of IITJEE, the curriculum of Class XI is extremely important to achieve success in IITJEE or any other competitive examination like AIEEE. The successful students identify these points early in their Class XI and race ahead of rest of the 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 your velocity of preparations to the maximum. The formulae will help you revise your chapters in a very 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 Tyagi Gaurav Mittal Prasoon Kumar

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Formula Booklet – Physics XI

CONTENTS Description

Page Number

1.

Units, Dimensions & Measurements

03

2.

Motion in one Dimension & Newton’s Laws of Motion

04

3.

Vectors

06

4.

Circular Motion, Relative Motion, and Projectile Motion

07

5.

Friction & Dynamics of Rigid Body

09

6.

Conservation Laws & Collisions

12

7.

Simple Harmonic Motion & Lissajous Figures

14

8.

Gravitation

18

9.

Properties of Matter

20

10. Heat & Thermodynamics

25

11. Waves

30

12. Study Tips

35

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Formula Booklet – Physics XI

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 system a

n1u 2 = n 2 u 2

or

b

M  L  T  n 2 = n1  1  x  1  x  1   M2   L2   T2 

c

X1 − X 2 + ... + X N N

(iii)

Mean or average value: X =

(iv)

Absolute error in each measurement: |∆Xi| = | X –Xi|

(v)

Mean absolute error: ∆Xm=

(vi) (vii)

Σ | ∆ Xi | N

∆X X ∆X x 100 Percentage error = X

Fractional error =

(viii) Combination of error: If ƒ =

Xa Y b Zc

, then maximum fractional error in ƒ is:

∆ƒ ∆X ∆Y ∆Ζ =| a | +| b| +|c| ƒ X Y Ζ

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Formula Booklet – Physics XI

MOTION IN ONE DIMENSION & NEWTON’S LAWS OF MOTION (i)

Displacement: | displacement | ≤ distance covered

(ii)

Average speed: v =

(iii) (iv)

s1 + s 2 s +s = 1 2 s s t1 + t 2 1 + 2 v1 v 2

(a)

If s1 = s2 = d, then v =

(b)

If t1 = t2, then v =

2 v1 v 2 = Harmonic mean v1 + v 2

v1 + v 2 =arithmetic mean 2 →

→

Average velocity: (a) v av = →

→

→ r2 − r1 ; (b) | v av | ≤ v t 2 − t1 →

→ dr Instantaneous velocity: v = and | v | = v = instantaneous speed dt →

→

→

v −v = 2 1 t 2 − t1

(v)

Average acceleration: a av

(vi)

Instantaneous acceleration: a = d v / dt

→

→

 dv  v  dx 

In one – dimension, a = (dv/dt) =  (vii)

Equations of motion in one dimension: (a) v = u + at; 1 2 at ; 2

(b)

x = ut +

(c)

v2 u2 + 2ax;

(d)

x = vt –

(e)

1 2 at ; 2 v+u x=  t;  2 

1 2 at ; 2

(f)

s = x − x 0 = ut +

(g)

2 2 v = u + 2a (x–x0)

a (2n–1) 2

(viii) Distance travelled in nth second: dn = u + (ix)

Motion of a ball: (a) when thrown up: h = (u2/2g) and t = (u/g) (b) when dropped: v = √(2gh) and t = √(2h/g)

(x)

Resultant force: F = √(F12 + F22 + 2F1F2 cos θ)

(xi)

Condition for equilibrium: (a) F3 = − (F1 + F2 ) ; (b) F1 + F2 ≥ F3 ≥ |F1 – F2|

(xii)

Lami’s Theorem:

→

→

→

P Q R = = sin (π − α ) sin (π − β ) sin (π − γ ) →

→

→



→



(xiii) Newton’s second law: F = m a ; F =  d p / dt  

→

→

(xiv) Impulse: ∆ p = F ∆t and p 2 − p1 =∫12 (xv)



Fdt

Newton’s third law:

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Formula Booklet – Physics XI (1)

→

→

M

(a)

F12 = − F12

(b)

m F = F21 M+m F Acceleration: a = M+m

(c)

(2)

Contact force: F12 =

F

F21 Fig. 1

(xvi) Inertial mass: mI = F/a (xvii) Gravitational mass: mG =

F12

m

F FR 2 = ; m I = mG g GM

→

→

→

(xviii) Non inertial frame: If a 0 be the acceleration of frame, then pseudo force F = − m a 0 Example: Centrifugal force =

mv 2 = m ω2 r r

(xix) Lift problems: Apparent weight = M(g ± a0) (+ sign is used when lift is moving up while – sign when lift is moving down) m1

(xx)

T

Pulley Problems: (a) For figure (2): Tension in the string, T =

m1m 2 g m1 + m 2

Acceleration of the system, a = The force on the pulley, F = (b)

m2 g m1 + m 2

Frictionless surface

T m2

Fig. 2

m 2g

2 m1m 2 g m1 + m 2

For figure (3): Tension in the string, T =

2m1m 2 g m1 + m 2

Acceleration of the system, a =

The force on the pulley, F =

a

m 2 − m1 g m 2 + m1

•

T T

4m1m 2 g m1 + m 2

T T a

m1 Fig. 3

m2

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Formula Booklet – Physics XI

VECTORS →

→

→

→

→

→

→

→

→

(i)

Vector addition: R = A + B = B + A and A − B = A + ( − B )

(ii)

Unit vector: A = ( A/ A)

(iii)

Magnitude: A = √ (A 2x + A 2y + A 2z )

(iv)

Direction cosines: cos α = (Ax/A), cos β = (Ay/A), cos γ = (Az/A)

(v)

Projection:

→

^

→

(b)

Component

→

→

^ →

of B along A = A . B

^

^

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

→ →

A . B = AB cos θ,

(a) (vii)

→ ^

Component of A along B = A . B →

(vi)

→

(a)

(b)

→ →

A . B = A x B x + A y B y + A z Bz

Cross product: (a)

→

→

→

→

A x B = AB

(b)

^

sin θ n ;

A x A = 0; →

→

Ax B=

(c)

^

^

^

i Ax

j Ay

k Az

Bx

By

Bz

(viii) Examples: → →

(a)

W= F. r; →

→

→

(e) v = w x r ;

→ →

(b) P = F . v ; →

→

→

(f) τ = τ x F ;

→ →

(c) φ Ε = E . A; 

→



→

Area of a parallelogram: Area = | A x B |

(x)

Area of a triangle: Area =

(xii)

→

(g) F m = q  v x B 

(ix)

(xi)

→

→

→ →

(d) φ Β = B . A;

1 → → |Ax B | 2 ^ ∂ ^ ∂ ^ ∂ Gradient operator: V = i +j +k ∂x ∂y ∂z →

→

→

Volume of a parallelopiped: V = A .  B x C  



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Formula Booklet – Physics XI

CIRCULAR MOTION, RELATIVE MOTION & PROJECTILE MOTION (i)

Uniform Circular Motion: (a) v = ωr; 2 2 (b) a = (v /r) = ω r ; 2 (c) F = (mv /r); (d) (e)

→ →

r . v = 0;

→ →

v.a =0

(ii)

2 Cyclist taking a turn: tan θ = (v /rg)

(iii)

Car taking a turn on level road: v = √(µsrg)

(iv)

2 Banking of Roads: tan θ = v /rg

(v)

Air plane taking a turn: tan θ = v /r g

(vi)

Overloaded truck: (a) Rinner wheel < Router wheel (b) maximum safe velocity on turn, v = √(gdr/2h)

(vii)

Non–uniform Circular Motion: (a) Centripetal acceleration ar = (v2/r); (b) Tangential acceleration at = (dv/dt); (c) Resultant acceleration a=√ (a 2r + a 2t )

2

(viii) Motion in a vertical Circle: (a) For lowest point A and highest point B, TA – TB = 6 mg; v2A = v2B + 4gl ; vA ≥ √(5gl); and vB≥ √ (gl) (b) Condition for Oscillation: vA < √(2gl) (c) Condition for leaving Circular path: √(2gl) < vA < √(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 2w + v 2b ) ; (2) θ = tan–1 ( v w / v b ); (3) t=(x/vb) (it is minimum) (4) Drift on opposite bank = (v w/vb)x (b) Boat to reach directly opposite to starting point:

→

→

→

→

(1) sin θ = (vw/vb); (2) vresultant = vb cos θ ; (3) t= (xii)

x v b cos θ

Projectile thrown from the ground: g x2

(a)

equation of trajectory: y = x tan θ –

(b)

time of flight: T =

(c) (d)

Horizontal range, R = (u2 sin 2θ/g) 2 2 Maximum height attained, H = (u sin θ/2g)

2 u sin θ g

2u 2 cos 2 θ

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(e) (f) (g) (h) (i)

Formula Booklet – Physics XI

Range is maximum when θ = 450 0 Ranges are same for projection angles θ and (90 –θ) Velocity at the top most point is = u cos θ 2 tan θ = gT /2R 2 (H/T ) = (g/8)

(xiii) Projectile thrown from a height h in horizontal direction: (a) T = √(2h/g); (b) R = v√(2h/g); 2 2 (c) y = h – (gx /2u ) (d)

2 Magnitude of velocity at the ground = √(u + 2gh)

(e)

Angle at which projectiles strikes the ground, θ = tan

–1

2gh u

(xiv) Projectile on an inclined plane:

2u sin (θ − θ 0 ) g cos θ 0

(a)

Time of flight, T =

(b)

Horizontal range, R =

2 u 2 sin (θ − θ 0 ) cos θ g cos 2 θ 0

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Formula Booklet – Physics XI

FRICTION (i)

Force of friction: (a) ƒs ≤ µsN (self adjusting); (ƒs)max = µsN (b) µk = µkN (µk = coefficient of kinetic friction) (c) µk < µs

(ii)

Acceleration on a horizontal plane: a = (F – µkN)/M

(iii)

2 Acceleration of a body sliding on an inclined plane: a = g sin θ (1– µk cot t )

(iv)

Force required to balance an object against wall: F = (Mg/µs)

(v)

Angle of friction: tan θ = µs (µs = coefficient of static friction) DYNAMICS OF RIGID BODIES θ 2 − θ1 ∆θ = t 2 − t1 ∆t

(i)

Average angular velocity: ω =

(ii)

Instantaneous angular velocity: ω = (dθ/dt)

(iii)

Relation between v, ω and r : v=ωr; In vector form v = ω x r ; In general form, v = ωr sin θ

(iv)

Average angular acceleration: α =

(v)

2 2 Instantaneous angular acceleration: α = (dω/dt) = (d θ/dt )

(vi)

Relation between linear and angular acceleration: (a) aT = αr and aR = (v2/r) = ω2R (b) Resultant acceleration, a = √ (a 2T + a 2R ) (c) In vector form,

→

→

→

ω 2 − ω1 ∆ω = t 2 − t1 ∆t

→ → → → → → → → → → → a = a T + a R , where a T = α x r and a R = ω x u = ω x  ω x r   

→

(vii)

Equations for rotational motion: (a) ω = ω0 + αt; 1 (b) θ = ω0t + αt2; 2 (c) ω2 – ω02 = 2αθ

(viii) Centre of mass: For two particle system: (a)

x CM =

m 1x 1 + m 2 x 2 ; m1 + m 2

(b)

v CM =

m1 v1 + m 2 v 2 m1 + m 2

(c)

a CM =

m 1a 1 + m 2 a 2 m1 + m 2

Also v CM = (ix)

dx CM dv d 2 x CM and a CM = CM = dt dt dt 2

Centre of mass: For many particle system:

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(a) (b) (c) (d) (e) (f) (g)

X CM =

Formula Booklet – Physics XI

Σ mi x i ; M →

→

Σm i r i ; M

→

d r CM ; dt

→

d v CM ; dt

r CM =

→

v CM =

→

a CM =

→

→

→

P CM = M v CM = Σ m i v i ;

→

→

→

→

→

→

→

F ext = M a CM = Σm i a i = Σ Fi . If F ext = 0, a CM = 0, V CM = constant ; →

Also, moment of masses about CM is zero, i.e., Σm i r i = 0 or m1r1 = m 2 r2 2

(x)

Moment of Inertia: (a) I = Σ mi ri 2 (b) I = µr , where µ = m1m2/(m1 + m2)

(xi)

2 2 2 Radius of gyration: (a) K = √(I/M) ; (b) K = √[(r1 + r2 + … + rn )/n] = root mean square distance.

(xii)

Kinetic energy of rotation: K = →

→

1 Iω2 2

or I = (2K/ω2)

→

(xiii) Angular momentum: (a ) L = r x p ; (b) L = rp sin θ ; (c ) m v d →

→

→

(xiv) Torque: (a ) τ = r x F ; (b ) τ = r F sin θ (xv)

→

→







Relation between τ and L: τ =  dL/ dt  ;

(xvi) Relation between L and I: (a) L = Iω; (b) K =

1 2 Iω = L2/2I 2

(xvii) Relation between τ and α: (a) τ = Iα, (b) If τ = 0, then (dL/dt)=0 or L=constant or, Iω=constant i.e., I1ω1= I2ω2 (Laws of conservation of angular momentum) →

→

(xviii) Angular impulse: ∆ L = τ ∆t ∫ = (xix) Rotational work done: W

(xx)

τ d θ = τ av θ

→ →

Rotational Power: P = τ . ω

(xxi) (a) (b)

Perpendicular axes theorem: Iz = Ix + Iy 2 Parallel axes theorem: I = Ic + Md

(xxii) Moment of Inertia of some objects (a)

Ring:

I = MR2 (axis); I = 2

1 MR2 (Diameter); 2

I = 2 MR (tangential to rim, perpendicular to plane); 2 I = (3/2) MR (tangential to rim and parallel to diameter)

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(b) (c)

Formula Booklet – Physics XI

1 1 MR2 (axis); I = MR2 (diameter) 2 4 1 Cylinder: I = MR 2 (axis ) 2

Disc: I =

2

2

(d) (e) (f)

Thin rod: I = (ML /12) (about centre); I = (ML /3) (about one end) Hollow sphere : Idia = (2/3) MR2; Itangential = (5/3) MR2 Solid sphere: Idia = (2/5) MR2 ; Itangential = (7/5) MR2

(g)

Rectangular: I C =

(h)

Cube: I = (1/6) Ma

(i)

Annular disc: I = (1/2) M ( R12 + R 22 )

(j) (k) (l)

Right circular cone: I = (3/10) MR2 Triangular lamina: I = (1/6) Mh2 (about base axis) Elliptical lamina: I = (1/4) Ma2 (about minor axis) and I = (1/4) Mb2 (about major axis)

(

)

M l2 + b2 (centre) 12 2

(xxiii) Rolling without slipping on a horizontal surface: K=

 K2  1 1 1 MV 2 + I ω 2 = MV 2 1 +  2 2 2 2  R 

( Q V = Rω and I = MK2)

For inclined plane (a)

Velocity at the bottom, v =

(b)

Acceleration, a = g sin θ

(c)

Time taken to reach the bottom, t =

2gh

 K2  1 +  2   R  

 K2  1 +   r 2    K2  2s  1 + 2   R  

g sin θ

(xxiv) Simple pendulum: = T = 2π√ (L/g) 2 2 (xxv) Compound Pendulum: T = 2π√ (I/Mg l), where l = M (K + l ) Minimum time period, T0 = 2π√ (2K/g)

(xxvi) Time period for disc: T = 2π √(3R/2g) Minimum time period for disc, T = 2π√ (1.414R/g) (xxvii) Time period for a rod of length L pivoted at one end: T = 2π√(2L/3g

The 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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Formula Booklet – Physics XI

CONSERVATION LAWS AND COLLISIONS → →

(i)

Work done: (a) W = F . d ; (b) W = Fd cos θ; (c) W =∫xx2 1

(ii)

Conservation forces: ∫a(ba )

→

→

→

F . d r ∫ab(b)

Path 1

→

→

F . d r ;∫(c)

→

F .d r = 0

closed path

Path 2

→

For conservative forces, one must have: V x F = 0 Potential energy: (a) VU = − W; (b) F = − (dU/dX ) ;

(iv)

Gravitational potential energy: (a) U = mgh ; (b) U = −

(v)

Spring potential energy: (a ) 2U =

(vi)

Kinetic energy: (a) ∆K = W 2 =

(vii)

Total mechanical energy: = E = K + U

1

1

Kx 2 ; (b ) ∆2U = 1

1

2 mv 2f −

(viii) Conservation of energy: ∆K = – ∆U

mv i2 ;

(c)

→

(iii)

F = − VU GMm

(R + h )

(

K x 2 2 − x1 2

(b ) 12 K =

)

mv 2

or, Kƒ + Uƒ = Ki + Ui

In 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 2 2 (c) For unstable equilibrium: U(x) = maximum, (dU/dx) = 0 and (d U/dx ) is negative 2 2 (d) For neutral equilibrium: U(x) = constant, (dU/dx) = 0 and (d U/dx ) is zero

(xii)

Velocity of a particle in terms of U(x): v = ±

2 [E − U(x )] m

(xiii) Momentum: (a)

→

→

(b)

p = m v;

 →  F =  d p / dt  ,  

→

→

→

→

(b)

Conservation of momentum: If F net = 0, then p f = p i ,

(c)

Recoil speed of gun, v G = →

mB x vB mG

→

(xiv) Impulse: ∆ p = F av ∆t (xv)

Collision in one dimension: (a) Momentum conservation : m1u1 + m2u2 = m1v1 + m2v2 (b) For elastic collision,1 e = 1 =1 coefficient of restitution 1 2 1 2 2 2 2 1v 1 + m 2v 2 2 (c) Energy conservation: m1u21 + m22u2 = m (d)

st

nd

Velocities of 1 and 2 body after collision are:

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Formula Booklet – Physics XI

 m − m2 v1 =  1  m1 + m 2

(e) (f) (g)

 2 m1  2m 2      u1 +   m + m  u 2 ; v2 =  m + m  1 2  1  2 

 m − m1     u1 +  2  m + m u2  1  2 

If m1 = m2 = m, then v1 = u2 and v2 = u1 Coefficient of restitution, e = (v2–v1/u1 = u2) e = 1 for perfectly elastic collision and e=0 for perfectly inelastic collision. For inelastic collision 0 < e < 1

(xvi) Inelastic collision of a ball dropped from height h0 2n (a) Height attained after nth impact, hn = e h0 2 2 (b) Total distance traveled when the ball finally comes to rest, s = h0 (1+e )/(1–e ) (c)

Total time taken, t =

2h 0  1 + e    g 1− e 

(xvii) Loss of KE in elastic collision: For the first incident particle  m − m2 = 1 K i  m1 + m 2

Kƒ

   

2

and

∆K lost ∆K lost 4m1m 2 = = 100% ; If m1 = m 2 , Ki Ki (m1 + m 2 )2

(xviii) Loss of KE in inelastic collision: ∆ Klost = Ki – Kƒ=

1 m1 m 2 (u1 – u2)2 (1–e2) 2 m1 + m 2

Velocity after inelastic collision (with target at rest)  m − em 2 v1 =  1  m1 + m 2

 m (1 + e )  u 1 and v 2 = 1 u1  m1 + m 2 

(xix) Oblique Collision (target at rest): m1u1 = m1v1 cos θ1 + m2v2 cos θ2 and m1v1 sin θ1 = m2v 2 sin θ2 Solving, we get: m1u12 = m1v12 + m2v22 (xx)

dV dM = − v el dt dt  M0 − mb   [M0 = original mass of rocket plus fuel and mb = mass of fuel burnt] V = – vrel loge    M0 

Rocket equation: (a) M (b) (c)

If we write M = M0 – mb = mass of the rocket and full at any time, than velocity of rocks at that time is: V = vrel loge (M0/M)

(xxi) Conservation of angular momentum: (a) If τext = 0, then Lƒ = Li v max rmax = v min rmin

(b)

For planets,

(c)

I Spinning skater, I1ω1 = I2W 2 or ωƒ = ωi  i   





 Iƒ 

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Formula Booklet – Physics XI

SIMPLE HARMONIC MOTION AND LISSAJOUS FIGURES (i)

Simple Harmonic Motion: (a) F = – Kx ; K x or a = – ω2x, where ω = √(K/m); m

(b)

a=–

(c)

Fmax = ± KA and amax = ±ω2A d2x

+ ω2 x = 0

(ii)

Equation of motion:

(iii)

Displacement: x = A sin (ωt + φ) (a) If φ = 0, x = A sin ωt ; (b) If φ = π/2, x = A cos ωt 2 2 –1 (c) If x = C sin ωt + D cos ωt, then x = A sin (ωt + φ) with A= √(C +D ) and φ = tan (D/C)

(iv)

Velocity: (a) v = A ω cos (ω+ φ); (b) If φ=0, v = A ω cos ωt; (c) vmax =±ωA 2 2 (d) v = ± ω√(A – x ); (e)

x2 A

+

2

v2

ω2 A 2

dt 2

=1

(v)

Acceleration: (a) a = –ω2 x = – ω2A sin (ωt+φ) ; 2 (b) If φ=0, a=– ω A sin ωt 2 (c) |amax| = ω A; 2 (d) Fmax = ± m ω A

(vi)

Frequency and Time period: (a) ω = √(K/m) ;

(vii)

1 2π

(K / m );

(b)

ƒ=

(c)

T = 2π

m K

Energy in 1SHM: Potential Energy: (a) U = 2 Kx2 ; dU ; dx

(b)

F=–

(c)

Umax = 2 mω A ;

(d)

U =2 mω A sin ωt

1

1

2

2

2

2

2

(viii) Energy in SHM: Kinetic energy: 1

(a)

K = 2 mv2;

(b)

K= 2 mω2 (A2–x2);

(c)

2 2 2 K =2 mω A cos ωt ;

(d)

Kmax =2 mω2A2

1

1

1

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(ix)

Total energy: (a) E = K + U = conserved; 2 2 (b) E = (1/2) mω A ; (c) E = Kmax = Umax

(x)

Average PE and KE: (a) < U > = (1/4) mω2A2 ; 2 2 (b) < K > = (1/4) mω A ; (c) (E/2) = < U > = < K >

(xi)

Some relations: (a)

v12 − v 22

ω=

x 22 − x 12

(b) T = 2π

;

Formula Booklet – Physics XI

x 22 − x 12 v12 − v 22

; (c) A =

(v1x 2 )2 − (v 2 x1 )2 v12 − v 22

(xii) Spring– mass system: (a) mg = Kx0; (b)

x0 m = 2π K g

T = 2π

m + (m s / 3) K

(xiii) Massive spring: T = 2π (xiv) Cutting a spring: (a) K’ = nK ; (b) T’ = T0/√(n) ; (c) ƒ’ = √(n) ƒ0 (d)

 n +1  K, K2 =  n 

If spring is cut into two pieces of lengths l1 and l2 such that l1 = nl2, then K1 =  (n +1) K and K1l1 = K2l2

(xv)

Springs in parallel: (a) K = K1 + K2 ; (b) T = 2π √[m/(K1 + K2)] (c) If T1 = 2π√ (m/K1) and T2 = 2π√(m/K2), then for the parallel combination: 1 T2

=

1 T12

+

1

or T =

T22

T1T2 T12

+ T22

and ω 2 = ω12 + ω22

(xvi) Springs in series: (a) K1x1 = K2x2 = Kx = F applied (b) (c) (d)

1 1 1 = + K K1 K 2 1

ω

2

=

1

+

ω12

T = 2π

K=

or

1

ω22

K 1K 2 K1 + K 2

or T 2 − T12 + T22

m(K1 + K 2 ) K 1K 2

or

ƒ=

1 2π

K 1K 2 m (K 1 + K 2 )

(xvii) Torsional pendulum: (a)

Iα=τ–Cθ

or

(b) (c)

θ=θ0 sin (ωt+φ); ω = √(C/I) ;

d 2θ dt

2

+

C θ=0; I

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1 2π

Formula Booklet – Physics XI

C ; I

(d)

ƒ=

(e)

4 T = 2π√(I/C), where C = πηr /2l

(xviii) Simple pendulum: (a)

Iα = τ =– mgl sin θ or

(b)

ω = √(g/l) ;

(c)

ƒ=

(d)

T = 2π √(l/g)

1 2π

d 2θ

d 2θ g g +   sin θ = 0 or + θ=0; dt l dt 2 l 2

(g/l ) ;

(xix) Second pendulum: (a) T = 2 sec ; (b) l = 99.3 cm (xx)

Infinite length pendulum: (a)

T = 2π

(b)

T=2π

1 ; 1 1    g +   l Re 

Re (when l→∞) g

 θ2   A 2  (xxi) Anharmonic pendulum: T ≅ T0 1 + 0  ≅ T0 1 + 2  

16 

 

16 l 

(xxii) Tension in string of a simple pendulum: T = (3 mg cos θ – 2 mg cos θ0) (xxiii) Conical Pendulum: (a) v = √(gR tan θ) ; (b) T = 2π√ (L cos θ/g) (xxiv) Compound pendulum: T = 2π (a) (b)

(l + K / l) 2

2

For a bar: T = 2π√(2L/3g) ; For a disc : T = 2π√ (3R/2g)

(xxv) Floating cylinder: (a) K = Aρg ; (b) T = 2π√(m/Aρg) = 2π√(Ld/ρg) (xxvi) Liquid in U–tube: (a) K = 2A ρg and m = ALρ ; (b) T = 2π√(L/2g) = 2π√(h/g) (xxvii) Ball in bowl: T = 2π√[(R – r)/g] (xxviii) Piston in a gas cylinder: A2E ; V

(a)

K=

(b)

T = 2π

mV A2E

;

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(c)

T = 2π

(d)

T = 2π

Vm

Formula Booklet – Physics XI

(E–P for Isothermal process);

A 2P Vm

(E = γ P for adiabatic process)

Α2γ P

(xxix) Elastic wire: AY ; l

(a)

K=

(b)

T = 2π

lm AY

(xxx) Tunnel across earth: T = 2π√(Re/g) (xxxi) Magnetic dipole in magnetic field: T = 2π√(I/MB) (xxxii) Electrical LC circuit: T = 2π LC or (xxxiii) Lissajous figures – Case (a): ω1 = ω2 = ω

ƒ=

1 2π LC

or ω1 : ω2 = 1 : 1 x2

General equation:

a

2

+

y2 b2

−

2 xy cos φ = sin 2 φ ab

For φ = 0 : y = (b/a) x ; straight line with positive slope For φ = π/4 : For φ = π/2 :

x2 a

2

x2 a2

+ +

y2 b

2

y2 b2

−

2 xy 1 = ; oblique ellipse ab 2

= 1 ; symmetrical ellipse

For φ = π : y = –(b/a) x ; straight line with negative slope. Case (b): For ω1 : ω2 = 2:1 with x = a sin (2ωt + φ) and y = b sin ωt For φ = 0, π: Figure of eight

π 3π , : Double parabola 4 4 π 3π For φ = , : Single parabola 2 2

For φ =

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Formula Booklet – Physics XI

GRAVITATION (i)

Newton’s law of gravitation: (a)

2

–11

F = G m1m2/r ; (b) a = 6.67 x 10

2

2

K.m /(kg) ; (c)

(ii)

Acceleration due to gravity (a) g = GM/R2 ;

(iii)

Variation of g: (a) due to shape ; gequator < gpole 2 (b) due to rotation of earth: (i) gpole = GM/R (No effect) (ii) gequator =

GM R2

dF 2 dr =− F r

(b) Weight W = mg

− ω2 R

(iii) gequator < gpole 2 2 (iv) ω R = 0.034 m/s (v) If ω ≅ 17 ω0 or T = (T0/17) = (24/17)h = 1.4 h, then object would float on equator (c)

(iv)



At a height h above earth’s surface g’ = g 1 −

2h  , if h < < R g 

 d  (d) At a depth of below earth’s surface: g’ = g 1 −   R GM m 1 ≅ g earth Acceleration on moon: gm = 6 R 2m →

GM ^ r (outside) ; r2

(b)

→

^

r r (inside)

Gravitational field: (a) g = −

(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

(vii)

g =−

GM

(v)

R

3

U = mgh (if origin of potential energy is shifted to the surface of earth)

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)

(x)

(xi)

Three particles: U = –

Gm1m 2 Gm1m 3 Gm 2 m 3 − − r12 r13 r23

Escape velocity: 2GM or ve = √(2gR) = √(gD) R

(a)

ve =

(b)

ve = R

8πGρ 3

Maximum height attained by a projectile: h=

R

(v e / v)

2

−1

or v = v e

h h ≅ ve R+h R

(if h < < R)

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(xii)

Formula Booklet – Physics XI

Orbital velocity of satellite:

(a ) v 0 =

GM ; r

(b) v 0 = v e

R ; 2 (R + h )

(xiii) Time period of satellite: (a) T = 2π

(c) v0 ≅ve/√2

(R + h )3 GM

(xiv) Energy of satellite: (a) Kinetic energy K = (b) (c) (d) (xv)

(if h<
; (b ) T = 2 π

(if h << R )

1 1 GMm mv 20 = 2 2 r

GMm =– 2K ; r 1 GMm ; Total energy E=K + U=– r 2 1 GMm E = U/2 = – K ; (e) BE = –E = 2 r

Potential energy U =–

Geosynchronous satellite: (a) T = 24 hours ; (b) T2 = (c)

R g

 GMT 2 h =  4π 2 

4π 2 (R + h )3 ; GM

1/ 3

   

–R ;

(d) h ≅ 36,000 km.

(xvi) Kepler’s law: (a) Law of orbits: Orbits are elliptical (b) Law of areas: Equal area is swept in equal time 2 3 2 2 3 (c) Law of period: T ∝ r ; T = (4π /GM)r

The most powerful weapon on earth is human soul on fire!

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Formula Booklet – Physics XI

SURFACE TENSION Force F = ; Length l

(a)

(ii)

Combination of n drops into one big drop: (a) R = n1/3r

T=

(b) T =

Surface energy W = Surface area A

(i)

(b)

2 2 –1/3 Ei = n(4πr T), Eƒ = 4πR T, (Eƒ/Ei) = n ,

(c)

∆E = 4πR2T (n1/3 –1) = 4πR3T  −

1 r

(iii)

Increase in temperature: ∆θ =

(iv)

Shape of liquid surface:

(v)

∆E  1  = 1 −  1/3 E i  n 

1  R

3T  1 1   −  or ρs  r R 

3T ρsJ

1 1   −  r R Fcohesive

(a)

Plane surface (as for water – silver) if Fadhesive >

(b)

Concave surface (as for water – glass) if Fadhesive >

(c)

Convex surface (as for mercury–glass) if Fadhesive <

2 Fcohesive 2 Fcohesive 2

Angle of contact: (a) Acute: If Fa> Fc/√2 ; (b) obtuse: if Fa
cos θc =

Tsa − Tsl , (where Tsa, Tsl and Tla represent solid-air, solid- liquid and liquid-air Tla

surface tensions respectively). Here θc is acute if Tsl < Tsa while θc is obtuse if Tsl > Tsa (vi)

Excess pressure: (a) (b) (c) (d) (e)

(vii)

 1

1 

 General formula: Pexcess = T +   R1 R 2  For a liquid drop: Pexcess = 2T/R For an air bubble in liquid: Pexcess = 2T/R For a soap bubble: Pexcess = 4T/R Pressure inside an air bubble at a depth h in a liquid: Pin = Patm + hdg + (2T/R)

Forces between two plates with thin water film separating them: (a) (b) (c)

1 r

∆P = T  −

1 ; R

1 1  F = AT  −  ; r R

If separation between plates is d, then ∆P = 2T/d and F = 2AT/d

(viii) Double bubble: Radius of Curvature of common film Rcommon = (ix)

Capillary rise: (a)

h=

2T cos θ ; rdg

(b)

h=

2T (For water θ = 00) rdg

rR R−r

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(c) (d) (x)

Formula Booklet – Physics XI

r  rdg  h +  3   If weight of water in meniseus is taken into account then T = 2 cos θ 2T cos (π − θ) Capillary depression, h − rdg

Combination of two soap bubbles: (a) If ∆V is the increase in volume and ∆S is the increase in surface area, then 3P0∆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 or R3 = √ R 12 + R 22

(

)

ELASTICITY (i)

Stress: (a) Stress = [Deforming force/cross–sectional area]; 2 (b) Tensile or longitudinal stress = (F/π r ); (c) Tangential or shearing stress = (F/A); (d) Hydrostatic stress = P

(ii)

Strain: (a) Tensile or longitudinal strain = (∆L/L); (b) Shearing strain = φ; (c) Volume strain = (∆V/V)

(iii)

Hook’s law: FL A(∆L)

(a)

For stretching: Stress = Y x Strain or Y =

(b)

For shear: Stress = η x Strain or η = F/Aφ

(c)

For volume elasticity: Stress = B x Strain or B = –

P

(∆V/V )

(iv)

Compressibility: K = (1/B)

(v)

Elongation of a wire due to its own weight: ∆L =

(vi)

Bulk modulus of an idea gas: Bisothermal = P and Badiabatic = γP (where γ = Cp/Cv)

(vii)

Stress due to heating or cooling of a clamped rod Thermal stress = Yα (∆t) and force = YA α (∆t)

1 MgL 1 L2 ρ g = 2 YA 2 Y

(viii) Torsion of a cylinder: (a) r θ = lφ (where θ = angle of twist and φ = angle of shear); (b) restoring torque τ = cθ (c) restoring Couple per unit twist, c = πηr4/2l (for solid cylinder) and C = πη (r24 – r14)/2l (for hollow cylinder) (ix)

Work done in stretching:

1 (stress ) 1 1 x stress x strain x volume = Y (strain)2 x volume = x volume 2 2 Y 2 1 Potential energy stored, U = W = x stress x strain x volume 2 1 x stress x strain Potential energy stored per unit volume, u = 2 2

(a) (b) (c) (x)

W=

Loaded beam:

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(xi)

(xii)

Wl 3

(a)

depression, δ =

(b)

Depression, δ =

Formula Booklet – Physics XI

(rectangular )

4Ybd 3 Wl 3

12Yπr 2

(cylindrical )

Position’s ratio: ∆D − ∆r = D r

(a)

Lateral strain = −

(b)

Longitudinal strain = (∆L/L)

(c)

Poisson’s ratio σ =

(d)

Theoretically, –1 < σ < 0.5 but experimentally σ ≅ 0.2 – 0.4

lateral strain − ∆r/r = longitudinal strain ∆L / L

Relations between Y, η, B and σ: (a) Y = 3B (1–2σ) ; (b) Y = 2η (1+ σ); 1 1 1 = + Y 9 B 3η

(c)

(xiii) Interatomic force constant: k = Yr0 (r0 = equilibrium inter atomic separation) KINETIC THEORY OF GASES or P1V1 = P2V2

(i)

Boyle’s law: PV = constant

(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, µ Where µ = N/Na [N = number of molecules in the sample and NA = Avogadro’s number = 6.02 x 1023/mole] =

(iv)

M sample

The pressure on the wall : P =

mN mN 1 mN 2 1 < v 2x > = < v2 > = v rms = ρv 2rms V 3V 3 V 3

RMS speed: 2 2 (a) νrms = √(v1 + v2 + … + v 2N /N); (b) νrms = √(3P/ρ) ; (c) νrms = √(3KT/m); (d)

(vi)

[Msample = mass of gas sample and M = molecular weight]

Kinetic Theory: (a) Momentum delivered to the wall perpendicular to the x–axis, ∆P = 2m v x (b) Time taken between two successive collisions on the same wall by the same molecule: ∆t = (2L/v x) (c) The frequency of collision: νcoll. = (νx/2L) (d) Total force exerted on the wall by collision of various molecules: F = (MN/L) (e)

(v)

M

νrms = √(3RT/M) ; (e)

(ν rms )1 (ν rms )2

=

m2 = m1

M2 M1

Kinetic interpretation of temperature:

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(ix)

(x)

(a) (1/2) Mv2 rms = (3/2) RT ; 2 (b) (1/2) mv 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) Maxwell molecular speed distribution: 

m   2 π  KT 

3/2

v 2 e - mv

2

/ 2 KT

(a)

n (v) = 4πN 

(b)

The average speed: v =

(c)

The rms speed: v rms =

(d)

The most probable speed: νp =

(e)

Speed relations: (I) vp < v < vrms (II) v p : v : vrms = √(2) : √(8/π) : √(3) = 1.41 : 1.60 : 1.73

8KT 8 RT RT = = 1.60 πm πM M 3kT = m

3RT RT = 1.73 M M 2 KT = m

2RT RT = 1.41 M M

Internal energy: (a) Einternal = (3/2)RT (for one mole) (b) Einternal = (3/2 µRT (for µ mole) (c)

(xi)

Formula Booklet – Physics XI

Pressure exerted by a gas P =

2 E 2 = E 3 V 3

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 two rotational plus two vibrational) (e) Polyatomic gas (non–linear molecule, e.g., NH3, H2O etc): 6 (three translational plus three rotational) (f) Internal energy of a gas: Einternal = (f/2) µRT. (where f = number of degrees of freedom)

(xii) Dalton’s law: The pressure exerted by a mixture of perfect gases is the sum of the pressures exerted by the individual gases occupying the same volume alone i.e., P = P1 + P2 + …. (xiii) Van der Wall’s gas equation: (a)

2    P + a µ  (V - µb ) = µRΤ  V 2  

(b)

2    P + a µ  (V − b ) = RT (where Vm = V/µ = volume per mole); m  Vm2  

(c)

b = 30 cm3/mole

(d)

Critical values: Pc =

(e)

a 27 b 2

, VC = 3b, TC =

8a ; 27 Rb

PC VC 3 = = 0.375 RTC 8

(xiv) Mean free path: λ =

1 2 πd 2 ρn

,

Where ρn = (N/V) = number of gas molecules per unit volume and d = diameter of molecules of the gas

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Formula Booklet – Physics XI

FLUID MECHANICS (i) (i) (ii) (iii) (iv)

The viscous force between two layers of area A having velocity gradient (dv/dx) is given by: F = – ηA (dv/dx), where η is called coefficient of viscosity In SI system, η is measured I Poiseiulle (Pl) 1Pl = 1Nsm–2 = 1 decapoise. In egs system, the unit of η is g/cm/sec and is called POISE When a spherical body is allowed to fall through viscous medium, its velocity increases, till the sum of viscous drag and upthrust becomes equal to the weight of the body. After that the body moves with a constant velocity called terminal velocity. According to STOKE’s Law, the viscous drag on a spherical body moving in a fluid is given by: F = 6πηr v, where r is the radius and v is the velocity of the body. The terminal velocity is given by: v T =

2 r 2 (ρ − σ ) g 9 η

where ρ is the density of the material of the body and σ is the density of liquid (v)

Rate of flow of liquid through a capillary tube of radius r and length l V=

π pr 4 p p = = 4 8ηl R 8ηl/πr

where p is the pressure difference between two ends of the capillary and R is the fluid resistance (=8 ηl/πr4) (vi) The matter which possess the property of flowing is called as FLUID (For example, gases and liquids) (vii) Pressure exerted by a column of liquid of height h is : P = hρg (ρ = density of the liquid) (viii) Pressure at a point within the liquid, P = P0 + hρg, where P0 is atmospheric pressure and h is the depth of point w.r.t. free surface of liquid (ix) Apparent weight of the body immersed in a liquid Mg’ = Mg – Vρg (x) If W be the weight of a body and U be the upthrust force of the liquid on the 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) (xii)

Volume of immersed part of a solid density of solid = total volume of solid density of solid

Equation of Continuity: a1v1 = a2v2

1 2 v = constant 2 a (xiv) Accelerated fluid containers : tan θ = x g

(xiii) Bernouilli’s theorem: (P/ρ) + gh +

(xv)

ax θ ρ Fig. 4

Volume of liquid flowing per second through a tube: R=a1v 1 = a2v2

(

2gh

a 12

− a 22

)

(xvi) Velocity of efflux of liquid from a hole: v = √(2gh), where h is the depth of a hole from the free surface of liquid

I 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 Burkett

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Formula Booklet – Physics XI

HEAT AND THERMODYNAMICS (i)

L2 – L1 = L1α(T2 – T1); A2 – A1 = A2 β(T2 – T1); V2 – V1 = V1γ(T2 – T1) where, L1, A1, V1 are the length, area and volume at temperature T1; and L2, A2, V2 are that at temperature T2.α represents the coefficient of linear expansion, β the coefficient of superficial expansion and γ the coefficient of cubical expansion.

(ii) (iii) (iv)

If dt be the density at t0C and d0 be that at 00C, then: dt = d0 (1–γ∆T) α : β: γ = 1 : 2 : 3 If γr, γa be the coefficients of real and apparent expansions of a liquid and γg be the coefficient of the cubical expansion for the containing vessel (say glass), then γr = γa + γg

(v) (vi)

The pressure of the gases varies with temperature as : Pt = P0 (1+ γ∆T), where γ = (1/273) per 0C If temperature on Celsius scale is C, that on Fahrenheit scale is F, on Kelvin scale is K, and on Reaumer scale is R, then (a) (c)

(vii)

C F − 32 K − 273 R = = = 5 9 5 4 5 C = (F − 32) 9

(b)

F=

9 C + 32 5

(e)

K=

5 (F + 459.4) 9

(d)

K = C + 273

(a) (b)

Triple point of water = 273.16 K 0 Absolute zero = 0 K = –273.15 C

(c)

For a gas thermometer, T = (273.15)

(d)

For a resistance thermometer, Rθ = R0 [1+ αθ]

P (Kelvin ) Ptriple

(viii) If mechanical work W produces the same temperature change as heat H, then we 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 mass and specific heat i.e., W = mc (xii) When the state of matter changes, the heat absorbed or evolved is given by: Q = mL, where L is called LATENT HEAT (xiii) In case of gases, there are two types of specific heats i.e., cp and cv [cp = specific heat at constant pressure and Cv = specific heat at constant volume]. Molar specific heats of a gas are: Cp = Mcp 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 = µRT V =Constant (a) For ISOBARIC process: P = Constant ; T P (b) For ISOCHORIC (Isometric) process: V = Constant; =Constant T (c) For ISOTHERMAL process T = Constant ; PV= Constant (d) For ADIABATIC process: PVγ = Constant ; TVγ–1=Constant 1–γ and P( ) Tγ = Constant

(xvi) Slope on PV diagram

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(a) (b) (c) (d) (e)

Formula Booklet – Physics XI

For isobaric process: zero For isochoric process: infinite For isothermal process: slope = –(P/V) For adiabatic process: slope = –γ(P/V) 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=µRT loge (V2/V1) µRT x 2.303 x log10 (V2/V1) P1V1 x 2.303 x log10 (V2/V1) µRT x 2.303 x log10 (P1/P2)

µR (T1 − T2 ) (P1V1 − P2 V2 ) = (γ − 1) (γ − 1)

(d)

For adiabatic process:

(e)

In expansion from same initial state to same final volume

W=

W adiabatic < W isothermal < W isobaric (f)

In compression from same initial state to same final volume: W adiabatic < W isothermal < W isobaric

(xviii) Heat added or removed: (a) For isobaric process: Q = µCp∆T (b) For isochoric process = Q = µCv∆T (c) For isothermal process = Q = W = µRt loge (V2/V1) (d) For adiabatic process: Q = 0 (xix) Change in internal energy (a) For isobaric process = ∆U = µCv∆T (b) For isochoric process = ∆U = µCv∆T (c) For isothermal process = ∆U = 0 µ R (T2 − T1 ) (d) For adiabatic process: ∆U = –W = (γ − 1) (xx) Elasticities of gases (a) Isothermal bulk modulus = BI = P (b) Adiabatic bulk modulus BA = γP (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) (b) (c) (d)

3 RT (for one mole); 2 3 U = µRT (for µ moles) 2 3 ∆U = µR∆T (for µ moles); 2 1  ∆U  3 Cv=  = R µ  ∆Τ  2

U=

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(e)

Cp = Cv + R =

(f)

γ= 

 Cp C  v

 5 = R  2 

Formula Booklet – Physics XI

3 5 R+R= R 2 2 3  5 R  = = 1.67 2  3

(xxiii) Internal energy and specific heats of a diatomic gas 5 µRT (for µ moles); 2 5 ∆U = µR∆T (for µ moles) 2 1  ∆U  5 Cv =   = R; µ  ∆T  2

(a)

U=

(b) (c) (d)

Cp = Cv + R =

(e)

 C p   7R = γ =    Cv   2

5 7 R+R= R 2 2 5R  7  = = 1.4 2  5

(xxiv) Mixture of gases: µ = µ1 + µ2 M=

µ1M1 + µ 2 + M 2 N1m1 + N 2 m 2 = µ1 + µ 2 N1 + N 2

Cv =

µ1C v1 + µ 2 C v 2 µ1 + µ 2

and C p =

µ1C p1 + µ 2 C p 2 µ1 + µ 2

(xxv) First law of thermodynamics (a) (b) (c)

∆Q = ∆U + ∆W or ∆U = ∆Q – ∆W Both ∆Q, ∆W depends on path, but ∆U does not depend on the path For isothermal process: ∆Q = ∆W = µRT log | V2/V1|, ∆U = 0, T = Constant, PV = Constant and Ciso = ± ∞

(d)

For adiabatic process: ∆W =

µR (T2 − T1 ) , ∆Q = 0, ∆U = µCv (T2–T1), Q = 0, (1 − γ ) Cp 2 PVγ = constant, Cad = 0 and γ = = 1+ Cv ƒ

(where ƒ is the degree of freedom) (e) (f) (g) (h) (i)

For isochoric process: ∆W = 0, ∆Q = ∆U = µCv∆T, V = constant, and Cv = (R/γ–1) For isobaric process: ∆Q = µCp∆T, ∆U = µCv∆T., ∆W = µR∆T, P = constant and Cp = (γR/γ–1) For cyclic process: ∆U = 0, ∆Q = ∆W For free expansion: ∆U = 0, ∆Q = 0, ∆W = 0 n For polytropic process: ∆W = [µR(T2–T1)/1–n], ∆Q = µ C (T2–T1), PV = constant and C=

R R + γ −1 1− n

(xxvi) Second law of thermodynamics (a) There are no perfect engines (b) There are no perfect refrigerators

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Formula Booklet – Physics XI

Q 2 Q 2 T2 , = Q1 Q1 T1

(c)

Efficiency of carnot engine: η = 1–

(d)

Coefficient of performance of a refrigerator: β=

Q2 T2 Heat absorbed from cold reservoir Q 2 = = = Work done on refrigerator W Q1 − Q 2 T1 − T2

For a perfect refrigerator, W = 0

or

Q1 = Q2

β=∞

or

∆θ t , where K is coefficient of thermal ∆x

(xxvii) The amount of heat transmitted is given by: Q = –KA

conductivity, A is the area of cross section, ∆θ is the difference in temperature, t is the time of heat flow and ∆x is separation between two ends (xxviii)

Thermal resistance of a conductor of length d = RTh =

d KA

(xxix) Flow of heat through a composite conductor:

(K1θ1 / d1 ) + (K 2 θ2 / d 2 ) (K1 / d1 ) + (K 2 / d 2 )

(a)

Temperature of interface, θ =

(b)

Rate of flow of heat through the composite conductor: H =

(c)

Thermal resistance of the composite conductor R TH =

(d) (xxx) (a) (b) (c) (d)

Q A(θ1 − θ 2 ) = t (d1 / K1 ) + (d 2 / K 2 )

d1 d + 2 = (R Th )1 + (R Th )2 K 1A K 2 A

Equivalent thermal conductivity, K =

d1 + d 2

(d1 / K1 ) + (d 2 / K 2 )

Radiation absorption coefficient: a = Q0/Q0 Reflection coefficient: r = Qr/Q0 Transmission coefficient: t = Qt/Q0 Emissive power: e or E = Q/A .t

[t = time] eλeλ∫0∞

(e)

Spectral emissive power:

Q and e = eλ = At (dλ )

(f) (g) (h) (i)

Emissivity: Absorptive power: Kirchhoffs law: Stefan’s law: (a) For a black body: For a body:

(j)

Rate of loss of heat:

ε = e/E ; 0 ≤ ε ≤ 1 a = Qa/Q0 (eλ/aλ)1 = (eλ/aλ)B = ………= Eλ 4 –8 –2 –4 E=σT (where σ=5.67x10 Wm K ) 4 4 E = σ (T –T0 ) 4 4 e = εσ (T –T0 ) dQ – = εAσ (θ 4 − θ 0 4 ) dt

(k)

For spherical objects:

(dQ / dt )1 (dQ / dt )2

Rate of fall of temperature:

dθ εAσ 4 εAσ 4 = θ − θ 04 = θ − θ 04 dt ms V ρs

∴

=

(

(dθ/dt )1 (dθ/dt )2

=

r12 r22

)

(

)

A1 V2 r2 x = A 2 V1 r1

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Formula Booklet – Physics XI

(For spherical bodies) (l)

Newton’s law of cooling:

(m)

Wein’s displacement law:

(n)

Wein’s radiation law:

(o)

Solar Constant:

dθ = –K (θ–θ0) or (θ–θ0) α e–KT dt –3 λmT = b (where b = 2.9 x 10 m – K) A  A  –a/λT dλ Eλdλ=  5  ƒ (λT) dλ=  5  e λ  λ  2

1/ 4  RS  R 4 S  σT or T =    ES   σ   R S  R ES 

S = 

   

1/ 2

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Formula Booklet – Physics XI

WAVES 1.

Velocity: v = nλ and n = (1/T)

2.

Velocity of transverse waves in a string: v =

3.

Velocity of longitudinal waves: (a) In rods: v = √(Y/ρ) (Y – Young’s modulus, ρ = density) (b) In liquids: v = √(B/ρ) (B = Bulk modulus) (c) In gases: v = √(γP/ρ) (Laplace formula)

4.

Effect of temperature: (a) v = v0√ (T/273) or (b) (vsound/vrms) = √(γ/3)

5.

T = m

T πr 2 d

v = v0 + 0.61t

2π (vt–x) λ  t x (b) y = a sin 2π  −  T λ

Wave equation: (a) y = a sin

(c) y = a sin (ωt – kx), where wave velocity v =

ω = nλ k

(a) vparticle = (∂y/∂t) (b) maximum particle velocity, (vparticle)max = ω a

6.

Particle velocity:

7.

Strain in medium (a) strain = – (∂y/∂x) = ka cos (ωt – kx) (b) Maximum strain = (∂y/dx)max = ka (c) (vparticle/strain) = (ω/k) = wave velocity i.e., vparticle = wave velocity x strain in the medium

8.

Wave equation:

9.

Intensity of sound waves: (a) I = (E/At) (b) If ρ is the density of the medium; v the velocity of the wave; n the frequency and a the 2 2 2 i.e. I ∝ n2a2 amplitude then I = 2π ρ v n a –12 2 (c) Intensity level is decibel: β 10 log (I/I0). Where, I0 =Threshold of hearing = 10 Watt/m

10.

Principle of superposition: y = y1 + y2

11.

2 2 Resultant amplitude: a = √(a1 + a2 + 2a1a cos φ)

12.

Resultant intensity: I = I1 + I2 + 2√(I1I2 cos φ) 2 (a) For constructive interference: φ = 2nπ, amax = a1 + a2 and Imax = (√I1 + √I2) 2 (b) For destructive interference: φ = (2n–1) π, amin = a2 –a2 and Imin = (√I1=√I 2)

13.

(a) (b)

14.

Stationary waves: The equation of stationary wave, (a) When the wave is reflected from a free boundary, is:

∂ 2y ∂t

2

 ∂2y   = v2   ∂x 2   

2

Beat frequency = n1 – n2 and beat period T = (T1T2/T2–T1) If there are N forks in successive order each giving x beat/sec with nearest neighbour, then nlast = nfirst + (N–1)x

y = + 2a cos

2πx 2πt sin = 2a cos kx sin ωt λ T

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(b)

Formula Booklet – Physics XI

When the wave is reflected from a rigid boundary, is: Y + –2a sin

15.

2πx 2πt cos =–2a sin kx cos ωt λ T

Vibrations of a stretched string: 1 λ

T m

(a)

For fundamental tone: n1 =

(b)

For p th harmonic : np =

(c)

The ratio of successive harmonic frequencies: n1 : n2 : n3 :…….. = 1 : 2 : 3 : ……

(d)

Sonometer: n =

(e)

Melde’s experiment:

l 2l

p λ

T m

T m

(m = π r2 d) (i) Transverse mode: n =

(ii) Longitudinal mode: n = 16.

2p 2l

p 2l

T m

T m

Vibrations of closed organ pipe  v    4L 

(a)

For fundamental tone: n1 = 

(b) (c)

For first overtone (third harmonic): n2 = 3n1 Only odd harmonics are found in the vibrations of a closed organ pipe and 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 and n1 : n2 : n3 : ……=1 : 2 : 3 : …….

18.

End correction: (a) Closed pipe : L = Lpipe + 0.3d (b) Open pipe: L = Lpipe + 0.6 d where d = diameter = 2r

19.

Resonance column:

20.

Kundt’s tube:

21.

Longitudinal vibration of rods (a) Both ends open and clamped in middle: (i) Fundamental frequency, n1 = (v/2l) (ii) Frequency of first overtone, n2 = 3n1 (iii)Ratio of frequencies, n1 : n2 : n3 : …… = 1 3: 5 : ….. (b)

λ 3λ ; (b) l2 + e = 4 4 v l 2 − 3l1 ; (d) n = or λ = 2 (l 2 − l 1 ) (c) e = 2 (l 2 − l 1 ) 2

(a) l1 +e =

v air λ = air v rod λ rod

One end clamped (i) Fundamental frequency, n1 = (v/4l) (ii) Frequency of first overtone, n2 = 3n1

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Formula Booklet – Physics XI

(iii) Ratio of frequencies, n1 : n2 : n3 : ……= 1 : 3 : 5 : ……… 22.

Frequency of a turning fork: n α

t l2

E ρ

Where t = thickness, l = length of prong, E = Elastic constant and ρ = density 23.

Doppler Effect for Sound (a) Observer stationary and source moving: (i) Source approaching: n’ =

(ii) Source receding: n’ = (b)

v x n v − vs

v − vs xλ v

v + vs v x n and λ’ = xλ v + vs v

Source stationary and observer moving: (i) Observer approaching the source: n’ =

v + v0 xn v

(ii) Observer receding away from source: n’ = (c)

λ’ =

and

and

λ’ = λ

v − v0 x n and v

λ’ = λ

Source and observer both moving: (i) S and O moving towards each other: n’ =

v + v0 xn v − vs

(ii) S and O moving away from each other: n’ =

v − v0 xn v + vs

(iii) S and O in same direction, S behind O : n’ =

v − v0 xn v − vs

(iv)S and O in same direction, S ahead of O: n’=

v + v0 xn v + vs

v ± v m ± v0 v ± vm ± vs

(d)

Effect of motion of medium: n' =

(e)

Change in frequency: (i) Moving source passes a stationary observer: ∆n = For vs <
xn

2vs xn v

(ii) Moving observer passes a stationary source: ∆ n= (f)

2 vv s v 2 − v s2

2v 0 xn v

Source moving towards or away from hill or wall (i) Source moving towards wall (a) Observer between source and wall n’ =

v xn v − vs

n’ =

v x n (for reflected waves) v − vs

(for direct waves)

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Formula Booklet – Physics XI

(b) Source between observer and wall n’ =

v xn v + vs

n’ =

v x n (for reflected waves) v − vs

(for direct waves)

(ii) Source moving away from wall (a) Observer between source and wall n’ =

v xn v + vs

n’ =

v xn v + vs

(for direct waves) (for reflected waves)

(b) Source between observer and wall

(g)

n’ =

v xn v − vs

n’ =

v xn v + vs

(for direct waves) (for reflected waves)

Moving Target: (i) S and O stationary at the same place and target approaching with speed u v+u xn v−u

n’ = 

or



n’ = 1 + 

2u  xn v 

(for u <
(ii) S and O stationary at the same place and target receding with speed u v−u xn v+u

n’ = 

(h)

SONAR: n’ =

or



n’ = 1 − 

2u  xn v 

(for u <
v ± v sub 2 v sub   x n ≅ 1 ± xn v ± v sub v  

(upper sign for approaching submarine while lower sign for receding submarine) (i)

Transverse Doppler effect: There is no transverse Doppler effect in sound. For velocity component vs cos θ n’=

24.

v xn v ± v s cos θ

(– sign for approaching and + sign for receding)

Doppler Effect for light (a) Red shift (when light source is moving away): n’ =

1− v / c xn 1+ v / c

or

λ’ =

v c

For v << c, ∆ n = –   x n or (b)

1+ v / c xλ 1− v / c v c

∆λ’ =   x λ

Blue shift (when light source is approaching) n’ =

1+ v / c xn 1− v / c

or v c

For v << c, ∆ n =   n

λ’ = or

1− v / c xλ 1+ v / c v c

∆λ’ =–   λ

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Formula Booklet – Physics XI

v c

(c)

Doppler Broadening = 2∆λ = 2   λ

(d)

Transverse Doppler effect: For light, n’ = 1 −

(e)

v2 c

2

 1 v2  xn x n = 1 −  2 c 2  

(for v << c)

 2v  n  c 

RADAR: ∆n = 

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Formula Booklet – Physics XI

STUDY TIPS •

Combination of Subjects Study a combination of subjects during a day i. e. after studying 2–3 hrs of mathematics shift to any theoretical subject for 2 horrs. When we study a subject like math, a particular part of the brain is working more than rest of the brain. When we shift to a theoretical subject, practically the other part of the brain would become active and the part studying maths will go for rest.

•

Revision Always refresh your memory by revising the matter learned. At the end of the day you must revise whatever you’ve learnt during that day (or revise the previous days work before starting studies the next day). On an average brain is able to retain the newly learned information 80% only for 12 hours, after that the forgetting cycle begins. After this revision, now the brain is able to hold the matter for 7 days. So next revision should be after 7 days (sundays could be kept for just revision). This ways you will get rid of the problem of forgetting what you study and save a lot of time in restudying that topic.

•

Use All Your Senses Whatever you read, try to convert that into picture and visualize it. Our eye memory is many times stronger than our ear memory since the nerves connecting brain to eye are many times stronger than nerves connecting brain to ear. So instead of trying to mug up by repeating it loudly try to see it while reapeating (loudly or in your mind). This is applicable in theoritical subjects. Try to use all your senses while learning a subject matter. 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 Relaxation Take special care of your breathing. Deep breaths are very important for relaxing your mind and hence in your concentration. Pranayam can do wonders to your concentration, relaxation and sharpening your mined (by supplying oxygen to it). Aerobic exercises like skipping, jogging, swimming and cycling are also very helpful.

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