Formula Booklet Physics Xii

Formula Booklet – Physics XII

Dear students It is the dream of every one of you to make a mark in this arena called life. The IITJEE is one of the most rigorous test that you have to clear to take firm steps towards achieving this. All of you want to do well in this examination or some other engineering entrance competition in order to realize your dreams and also to see those tears of satisfaction in the eyes of your parents who are proud of your success. To succeed in any endeavour what a person needs is velocity. Yes, the same velocity that you study in physics. Velocity is a combination of speed and direction. Speed is the ability to do our work with utmost efficiency and negligible wastage of resources such as time. At this stage of our career speed is very important for all of us. I would like to take the help of an example to explain the term ‘direction’. A young guy was lost on a road and he asked an elderly fellow, “Sir, could you tell me where this road will take me?” Without a moment’s hesitation the elderly chap said, “ Son, this road will take you anywhere in the world you want to go, if you are moving in the right direction.” So direction is the right path towards our aim. You can tread on that path only when you have your goal clearly in front of your eyes and you are working for that goal intensely with a strong desire and an unshakable determination level, always believing that you’ll do it. Your attitude and motivation are of utmost importance. Remember, this test that you are taking is not just a test of your knowledge but also of how strong are you, mentally. In this booklet we have made a sincere attempt to 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

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

1

Formula Booklet – Physics XII

CONTENTS Description

Page Number

1.

Elecrotstatics and Capacitance

03

2.

Current Electricity

12

3.

Magnetic Effects of Current

15

4.

Magnetic Properties of Current

18

5.

Electromagnetic Induction

20

6.

Alternating Current

22

7.

Light

26

8.

Modern Physics

35

9.

Some Study Tips

43

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

2

Formula Booklet – Physics XII

ELECTROSTATICS & CAPACITANCE ELECTROSTATICS 1. (a)

Coulomb’s Law 1 q 1q 2 , K = Dielectric constant or relative permittivity of the medium Fm = 4πε 0 K r 2

(b)

Fm =

(c) (d)

F0 [F0 – Force between point charges placed in vacuum] K –1 –3 4 2 [ε0] = [M L T A ] Fe =2.4 x 1039 [For electron–proton pair) Fg

=1.2 x 1036 (For proton–proton pair) 2. (a) (b)

Electric field →

→

E=

F q0 →

Electric field due to a point charge: (i) E = (ii)

1 q ^ . r (if charge q is placed at the origin) 4πε 0 r 2

→ →

→ 1 q( r − r 0) (if charge q is placed at some point having position vector E= r0) 4 πε0 → → 3 | r− r0|

→

(c)

[E] = [M1L1T–3A–1]

3. (a)

Electric field on the axis of a uniformly charged ring 1 qx Eaxial = (R = radius of the ring) 4πε 0 R 2 + x 2 3 / 2

(b)

Ecentre = 0

4. (a)

Electric dipole dipole moment p = q(2l) (where 2l = length of the dipole) 1 2pr Eaxial = (r = distance of axial point w.r.t. centre of dipole) 4πε 0 r 2 − l 2 2

(b)

≅

1 2p 4πε 0 r 3

(

)

[

]

(if r >>l)

p 1 2 4πε 0 r + l 2

≈

1 p 4πε 0 r 3

(c)

Eequat. =

(d) (e)

(Eaxial/Eequat.) = 2/1 Dipole field at an arbitrary point (r, θ) 1 2p cos θ 1 p sin θ (i) Er = ; (ii) Eθ = 3 4πε 0 4 πε r r3 0 1 p 1 + 3 cos 2 θ (iii) E = E 2r + E θ2 = 4πε 0 r 3

(f)

Dipole field component at (x, y, z) point 1 3xz p 1 3yz p ; (ii) Ey = ; (i) Ex = s 4πε0 r 5 4πε 0 r

[

]

3/ 2

(if r >>l)

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

3

Formula Booklet – Physics XII

(iii)

(g)

Ez =

(

1 p 3z 2 − r 2 4πε 0 r5 →

) →

→

(ii) τ = pE sin θ

Torque on a dipole : (i) τ = p x E ; → →

(h)

Potential energy of a dipole: (i) U = − p . E = − pE cos θ (ii) Work done in rotating a dipole from angle θ1 to angle θ2 W = U2 – U1 = pE (cos θ1 – cos θ2)

5.

Electric flux

(a)

dφ = E . dS

(b)

φ ∫ E . dS = EA cos θ

→ →

→ →

(If electric field is constant over the whole surface)

2

(c) (d)

Unit of φ = (Nm /Coulomb) = J.m/Couplomb 1 3 –3 –1 [φ]= [M L T A ]

6.

Gauss’s Law:

→ →

7.

q ε0 Electric field due to various systems of charges

(a)

Isolated Charge:

∫

E . dS =

r q

E⊥ (b)

Electric dipole: • –q

1 q 4 π ε0 r 2

E=

Fig. 1 P

y• •

• +q

2a

E||

P •

→

→

1 2p (i) E || = 4π ε 0 x 3

→

→

(ii) E ⊥ = −

x

1 p 4 π ε0 y 3

Fig. 2 • r • –q

θ

• +q

→

(iii)E =

p

1 p 1 + 3 cos 2 θ 4π ε 0 r 3

Fig. 3

(c)

+++ + + q + R + + + A ring of charge:+ • + x + + + + + + + ++ Fig. 4 +

•P

E=

1 qx 4π ε 0 R 2 + x 2

(

)

3/ 2

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

4

Formula Booklet – Physics XII

(d)

(e)

+ ++ + +q + R + + + A disc of charge: • + + x + + + + + + Fig. 5 ++ + +

•P

+ + + + σ + +

Infinite sheet of charge:

σ  x 1 − 2 2ε 0  x + R2 

E=

E=

  

σ 2ε 0

Fig. 6

(f)

+ + + + λ

Infinitely long line of charge:

E=

•P

x

λ 2π ε 0 x

Fig. 7

(g)

+ + + Finite line of charge: + + +

α β

E⊥=

P E||=

λ 4 π ε0

λ (sin α + sin β) 4 π ε0 x

(cos α+cos β)

x Fig. 8

(h)

(i)

+ + Charged spherical shell: + + +

Solid sphere of charge:

+ + + R

+ q + +

+ + + Fig. 9

+ + + R + ++ + + + + + ++ Fig. 10

+

(i) Inside: 0 ≤ r ≤ R E=0 (ii) Outside: r ≥ R E=

q 4π ε 0 r 2

(i) Inside: 0 ≤ r ≤ r E = ρr/3ε0 (ii) Outside : r ≥ R E=

ρR 3ε 0

R   r 

2

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

5

Formula Booklet – Physics XII

8.

Force on a charged conductor: The force per unit area or electric pressure Pelec. =

9.

10. (a) (b) (c) (d)

Charged soap bubble: (a) Pin – Pout =

4T σ 2 − r 2ε 0

(b)

If air pressure inside and outside are assumed equal then: Pin = Pout and

or

r=

or

r = [Q /128π ε0T]

8ε 0 T

or T =

σ2 2

2

σ2r 8ε 0

or σ = √(8ε0T/r)

→

→

E = −VV 1 q 4π ε0 r

(f)

Potential due to a group of charges, V =

(g)

Potential due to a dipole:

(k)

Q = 8πr√(2ε0rT)

Electric potential: V = (W/q) Unit of V = Volt 2 –3 –1 [V] = [ML T A ]

Potential due to a point charge, V =

(i) (j)

or

4T σ 2 = r 2ε 0

1/3

(e)

(h)

dF σ 2 = dA 2ε 0

1 p ; (ii) equatorial point, V = 0; 4π ε 0 r 2 p cos θ

(i)

Axial point, V =

(iii)

V (r, θ) =

1 4π ε 0

1  q1 q 2 q 3  +   + 4π ε 0  r1 r2 r3 

ρ2

Potential due to a charged spherical shell 1 q 1 q (ii) surface: V = ; 4π ε0 r 4π ε 0 R

(i)

outside: V =

(iii)

inside : V = Vsurface =

1 q 4π ε 0 R

Potential due to a charged spherical conductor is the same as that due to a charged spherical shell. Potential due to a uniformly charged nonconducting sphere (i)

outside: V=

(iii)

inside: V=

1 q ; 4π ε 0 r

(ii) surface: V =

(

)

1 q 4πε0 R

1 q 3 R 2 − r2 ; (iv) centre: V = 3 x 1 q = 1.5 Vsurface 3 4πε0 2 4π ε0 R 2R

Common potential (Two spheres joined by thin wire) (i)

common potential V=

(ii)

q1 =

1 4π ε 0

r1 (Q1 + Q 2 ) rQ = 1 ; (r1 + r2 ) r1 + r2

 Q1 + Q2     r +r   1 2 

q2 =

r2 Q r1 + r2

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

6

Formula Booklet – Physics XII

(iii) 11.

q1 r1 = q 2 r2

or

σ1 r2 = σ 2 r1

(a)

Potential energy 1 q1 q 2 = q1V1 U= 4πε 0 r

(b)

U=

(c)

U = − p .E

12. (a)

If n drops coalesce to form one drop, then 1/3 2/3 (c) V = n Vsmall ; Q = nq; (b) R=n r ; 1/3 1/3 (e) E=n Esmall (d) σ = n σsmall

13.

Energy density of electrostatic field: u =

(For a system of two charges)

1 q1q 2 1 q1q 3 1 q 2 q3 + + 4π ε 0 r12 4 π ε 0 r13 4π ε0 r23

(For a system of three charges)

→ →

(For an electric dipole)

1 ε0E 2 2

CAPACITANCE 14. (a) (b) (c)

Capacitance: C = (q/V) Unit of C = farad (F) –1 –2 4 2 Dimensions of C = [M L T A ]

15.

Energy stored in a charged capacitor U=

1 CV2 ; 2

17.

1 1 Q2 U= QV; (c) U = 2 2 C 1 1 σ2 2 Energy density: (a) u = ε0E ; (b) u = 2 ε0 2 Force of attraction between plates of a charged capacitor

(a)

F=

1 σ2A ε0E2A; (b) F = ; 2ε 0 2

18. (a)

Capacitance formulae Sphere: (i) Cair = 4π ε0R;

(b)

Spherical capacitor: (i) Cair =

(c)

Parallel plate capacitor: (i) Cair =

(d)

Cylindrical capacitor: (i) Cair =

(e)

Two long parallel wires: C=

(a) 16.

(b)

(c)

F=

Q2 2ε 0 A

(ii) Cmed = K (4π ε0R) 4π ε 0ra rb ; rb − ra

ε0A ; d

(ii) Cmed = (ii) Cmed =

4π ε 0 K ra rb (rb − ra )

Kε 0 A d

2πKε 0 l 2 π ε 0l ; (ii) Cmed = loge (rb / ra ) log e (rb / ra )

ε0 l where d is the separation between wires and a radius of log e (d/a )

each wire (d>>a) 19. (a) (b)

Series Combination of Capacitors (Charge remains same) q1 = q2 = q3 = q q q q V1 = (Potential difference is different) , V2 = , V3 = C1 C2 C3

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

7

Formula Booklet – Physics XII

(d) (e)

1 1 1 1 = + + C C1 C 2 C 3 For two capacitors in series: C = C1C2/(C1 + C2) Energy stored: U = U1 + U2 + U3

20. (a) (b) (c) (d)

Parallel Combination of Capacitors (Potential difference remains same) V1 = V2 = V3 = V (Charges are different) q1 = C1V, q2 = C2V, q3 = C3V C = C1 + C2 + C3 U = U1 + U2 + U3

21. (a) (b)

Effect of dielectric Field inside dielectric, Ed = E0/K Polarization charges on surface of dielectric: Qp 1 1   (ii) σp = =σ 1 −  (i) Qp = Q  1 −  ; A  K  K

(c)

Polarization vector: (i) | P | =Qp/A ;

22.

Capacitance formulae with dielectric

(a)

C−

(b)

C=

ε0 A d−t

(c)

C=

ε 0 A  K1 + K 2    d  2 

(c)

→

K ε0 A ε0 A = 1  K d − t (K − 1)  d − t 1 −   K

→

(ii) | P | =ε0χEd

(For a dielectric slab of thickness t)

(For a metallic slab of thickness t)

K1

K2

A/1

A/2

d

Fig. 11

(d)

2ε A  K K C = 0  1 2 d  K1 + K 2

d/2

K1

  

K2 A/2

d/2

A/2

Fig. 12

(e)

C=

2K 2 K 3  ε0 A   K1 +   4d  K 2 + K 3 

K1

K2

d

K3

d

QUEST - Power Coaching for IITJEE Fig. 13

1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

8

Formula Booklet – Physics XII

(f)

For n plates with alternate plates connected: C = (n–1) ε0A/d

(g)

C=

ε0A

K1 K2 K3

 t1 t2 t3    K + K + K  2 3   1

t1

t2

t3

Fig. 14 23.

Spherical capacitor with inner sphere grounded

(a)

C=

(b)

Charge on inner sphere = –q1, while charge on outer sphere = +q2

(c) 24. (a)

4π ε 0 r1 r2

(r2 − r1 )

+ 4 π ε0 r2

r  Magnitude of charge on inner sphere: q1 =  1  q2  r2  Insertion of dielectric slab Battery remains connected when slab is introduced (i) V’ = V; (ii) C’ = KC ; (iii) Q’ = KQ ; (iv) E’ = E; (v) U’ = KU

(b)

Battery is disconnected after charging the capacitor and slab is introduced (i) Q’ = Q; (ii) C’ = KC ; (iii) E’ = E/K; (iv) V’ = V/K; (v) U’ = U/K

25. (a)

Charge transfer, Common potential and energy loss when two capacitors are connected C V + C 2 V2 q1 + q 2 = Common potential: V = 1 1 C1 + C 2 C1 + C 2

(b)

Charge transfer: ∆q =

(c) 26. (a) (b)

C1C 2 (V1 − V2 ) C1 + C 2

1 C1C 2 (V1 − V2 )2 2 C1 + C 2 Charging and discharging of a capacitor –t/RC ) ; (ii) V = V0 (1–e–t/RC); (iii) I = I0e–t/RC; Charging: (i) q = q0 (1–e (iv) I0 = V0/R –t/RC –t/RC –t/RC ; (ii) V = V0e ; (iii) I = – I0e Discharge: (i) q = q0e (b) Time constant: τ = RC

Energy loss: ∆U =

Whether you think you can… …or you think you can’t… …either ways you are right! QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

9

Formula Booklet – Physics XII

CURRENT ELECTRICITY 1. (a)

Electric Current I = (q/t); (b) I = (dq/dt); (c) I = (ne/t); (d) q = ∫ IdT

2. (a)

Ohm’s law, Resistivity and Conductivity V = IR ; (b) R= ρ(l/A) ; (c) σ = (1/ρ) (d) vd = (eEτ/m); (e) I = neAvd; ne 2 τ m  m  l  R =  2    ; (g) ρ = ; (h) σ = m ne 2 τ  ne τ   A 

(f) 3. (a)

Current density J = (I/A); (b) J = nevd; (c) J = σE; (d) µ = (vd/E); (e) σ =neµ

4.

Temperature Coefficient of Resistance

(a)

R = R0[1 + α(T–T0)] ; (b) α =

(d)

α=

5.

Cell: (a) E =

6. (a) (c)

Series Combination of Resistances (b) V = V1 + V2 + V3; R = R1 + R2 + R3; V1 = IR1, V2 = IR2, V3 = IR3 I = constant = I1 = I2 = I3; (d)

7.

Parallel Combination of resistances 1 1 1 1 = + + ; R R1 R 2 R 3 I= I1 + I2 + I3; V = constant = V1 = V2 = V3; V V V I1 = , I2 = , I3 = R1 R2 R3 For a parallel combination of two resistances: R 1R 2 R2 R= ; (ii) I1 = I; (iii) (i) R1 + R 2 R1 + R 2 Heating effect of current W = VI t; P = VI ; 2 2 P = I R = V /R; I 2 Rt H = I2 Rt Joule = Calorie J

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

8. (a) (b) (c) (d)

R − R0 ; (c) ρ = ρ0 [1+α(T–T0)] ; R 0 (T − T0 )

ρ − ρ0 ρ0 (T − T0 )

W ; Q

(b) I =

E ; (c) V = E – Ir r+R

(where V = IR)

I2 =

R1 I R1 + R 2

9.

Electric bulb: (a) Resistance of filament R = V2/P; (b) Maximum current that can be allowed to pass through bulb, Imax = (P/V)

10. (a)

Total Power Consumed Parallel combination: P = P1 + P2 + P3 1 1 1 1 Series combination: = + + P P1 P2 P3

(b)

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

10

Formula Booklet – Physics XII

11.

Effect of stretching a resistance wire 2

2

r  R 2 l 1 A1  l 2   A1   =  1  = x =  = R 1 l 2 A 2  l 1   A 2   r2  E nE = nr + R R

12.

Cells in series: I =

13.

Cells in parallel: I =

E

(r / n ) + R

4

[ Q l1A1 = l2A2]

(if n r <
=

E (if r << R) R

nE (if r >>R) r Mixed Combination (m rows with each containing n cells in series) nE mnE I= = ; (nr / m) + R nr + m R I is maximum when n r = m R ; mnE Imax = 2 mnrR =

14. (a) (b) (c) 15. (a) (b)

Chemical effect of current: Faraday’s first law of electrolysis: m = Zq = ZIt Faraday’s second law of electrolysis: (i) m ∝ W (W = ECE) or m/W = constant (where W = atomic weight/valency) (ii)

(c) (d)

As

m1 Z1 = and m 2 Z2

m1 W1 Z W = ; so 1 = 1 m 2 W2 Z 2 W2

Faraday : 1 Faraday = 96,500 Coulomb W = F = Faraday’s constant Z βθ 2 2

16.

Thermo e.m.f. : e = αθ +

17.

α Neutral temperature: θN = –   β

18.

Temperature of inversion: θ Ν =

19.

Thermoelectric power or Seebeck Coefficient: S =

20. (i) (ii)

Peltier effect: Heat absorbed per second at a junction when a current I flows = πI (where π = Peltier coefficient) Peltier coefficient, π = SθH

21. (i)

Thomson Coefficient: θΗ Heat absorbed/ sec ∫= θI Χ

(where θ = θH = θC)  de     = 0    dθ  θΝ

θ1 + θ C [ Q θI – θN = θN – θC] 2 de =α + βθ dθ

σdθ

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

11

Formula Booklet – Physics XII

(ii)

Thomson coefficient, σ =

∆Q/time I ∆θ

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

12

Formula Booklet – Physics XII

MAGNETIC EFFECTS OF CURRENT µ 0 I dl sin θ 4π r2

22.

Biot–Savart law : dB =

23.

Field due to a long straight wire: B =

24.

Field due to a circular coil: µ NI at centre: B = 0 ; 2a µ0 N I a 2 at an axial point: B = 3/ 2 2 a2 + x2

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

(

on axis when x >> a : B =

µ0 I 2 πr

)

µ0 N I a 2

2 x3 point of inflexion: It occurs at x = a/2  4µ N I   = 0.716 Bcentre Field at the point of inflexion: B =  0  5 5a 

25.

Magnetic moment of circular coil: (a) M = NIA ; (b) Field: B=

26. (a) (b) (c)

Field due to an arc of current: B = (µ0Il/4πR2) ; B = (µ0Iθ/4πR) At the centre of a semicircular coil: B = (µ0I/4R)

27.

Field due to finite length of wire: B =

28.

µ 0 2M 4π x 3

µ0I (sin φ1 + sin φ2) 4πa  2 2 µ0I   Field at the centre of a square loop: B =   πl    →

→

→

→

29.

Ampere’s law: (a∫)

30.

Field due to a current in cylindrical rod:

(a)

outside: B = (µ0 I/2πr) ; (b) surface: B = (µ0 I/2πR); (c) inside: B=

31. (a)

Field due to a current carrying solenoid: inside: B = µ0n I ; (b) at one end : B = (µ0n I/2) µ nI at an axial point: B = 0 (cos α2 – cos α1) 2

(c)

B . d l = µ0 I ; (b) ∫

H. d l =1

µ0I r 2π R 2

32.

Field due to a toroid: (a) inside: B=µ0nI – µ0NI/2πR ; (b) outside: B = 0

33.

Force on electric current: F = H x B

34.

Force between two parallel conductors:

35.

Comparison of magnetic and electric forces between two moving charges: (Fmagnetic/Felectric) = 2 2 (v /c )

→

→

→

F µ 0 I1 I 2 = l 2πd

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

13

Formula Booklet – Physics XII

→

36.

Force on a current loop in a magnetic field: F = 0 (any shape)

37.

Torque on a current loop in a magnetic field: τ = M x B

38. (a) (b)

Moving coil galvanometer: τ–NIAB; τ =Kθ ;

(c)

I= 

(d)

Current sensitivity = (θ/I) = (NaB/K) ; (e) Voltage sensitivity = (θ/V) = (θ/IR) = (NAB/KR)

39. (a) (b) (c) (d)

Ammeter: Shunt resistance S = (IgG/ I– Ig); Length of shunt wire, l = S πr2/ρ; Effective resistance of ammeter, RA = GS/(G+S); For an ideal ammeter, RA = 0

40.

Voltmeter:

(a)

V High resistance in series, R =  − G  ;

(b) (c)

For converted Voltmeter, RV = R + G; For an ideal Voltmeter, RV = ∞

41.

Force on a moving charge: → → → F = q v x B  ; (b) F = q v B sin θ  

(a)

→

→

→

or τ M B sin θ

 K  θ ;  N AB 

  Ig 

  

42.

Path of a moving charge in a magnetic field

(a)

When v is ⊥ to B : (i) path = circular; (ii) r = (mv/qB) ; (iii) ν = (qB/2πm); (iv) T = (2πm/qB) ; (v) ω = qB/m)

(b)

When angle between v and B is θ: (i) path=helical ; (ii) r = (mv⊥/qB) = (mv sin θ/qB); 2πm (iii) ν = (qB/2πm); (iv) T = ; (v) ω = (qB/m); qB (vi) pitch p = 2πr/tan θ (where tan θ = (v⊥/v ||)

43. (i) (ii) (iii) (iv) (v) (vi)

Cyclotron: T = (2πm/qB) ; ν = (qB/2πm) ; ω = (θB/m) ; radius of particle acquiring energy E, r = [√(2mE)/qB]; velocity of particle at radius r, v = qBr/m; the maximum kinetic energy (with upper limit of radius = R)

→

→

→

K max =

→

1 q 2 B2 R 2 2 m

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

14

Formula Booklet – Physics XII

44. (a)

(b)

Magnetic field produced by a moving charge: →

→

→

µ q (v x r ) B= 0 ; 4π r3

B=

µ 0 q v sin θ 4π r2

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

15

Formula Booklet – Physics XII

MAGNETIC PROPERTIES OF CURRENT 45. (a) 46. (a) (c)

Magnetic field: F B = max ; qv

1  dF  B=   I  dl  max

(b)

Atomic magnetic moments: eL eS µL = – ; (b) µS = – ; 2m m eh eJ µJ = – g ; (d) µB = = 0.927 x 10–23 J/T 4πm 2m

47.

Intensity of magnetization: I = (M/V)

48.

Magnetizing field: B −I; H= µ0 B ; For vacuum, H = µ0

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

For medium, h = B/µ; H =nI (solenoid); H = I/2 πr (straight wire);

(f)

H=

(g)

∫

49.

Magnetic susceptibility: χ = (I/H)

50.

Magnetic permeability:

(a)

µ = (B/H) ; (b) µr = (µ/µ0) ; (c) µr = (B/B0)

51. (a) (c)

Other relations: µ = µ0 (1+χ) ; (b) (d) B = B0 (1+ χ) :

52. 53.

Pole strength: m = F/B Magnetic moment of dipole : M = m x 2l

54.

Field due to a pole: B =

→

I dl sin θ r2

(Biot-Savart law) ;

→

H . d l = Ifree

µr = 1 + χ or χ = µr – 1; B = µ0 (H + I)

55.

µ0  m    4π  r 2  Field due to a bar magnet:

(a)

Axial point: B =

(b)

Equatorial point: B =

µ0 M 4π r 2 + l 2

(c)

At arbitrary point: B =

µ0 M 1 + 3 cos 2 θ 3 4π r

µ 0 2 Mr 4π r 2 − l 2

(

)

2

(

=

µ 0  2M    4π  r 3 

)

3/ 2

=

(if r > > l)

µ0  M    4π  r3 

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

16

Formula Booklet – Physics XII

56.

Force and torque on a dipole in uniform magnetic field (a)

→

F = 0;

(b)

→

→

→

τ = M x B ; (c) τ MB sin θ → →

57.

Potential energy of a dipole in magnetic field: U = − M . B = – MB cos θ

58.

Tangent galvanometer:

(a)

B = BH tan θ;

(b)

I = K tan θ,

59.

Vibration magnetometer: T = 2π

where K =

2r B H µ0n I M BH

Nothing will happen until you generate the will to make it happen!

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

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

17

Formula Booklet – Physics XII

ELECTROMAGNETIC INDUCTION 60.

Magnetic flux:

(a)

dφ = B. d A = BdA cos θ;

(b) φ= ∫ B . d A ;

(c)

φ = BA cos θ;

(d) ∫ B . d S = 0 ;

61. (a)

(c)

Faraday’s laws of e.m. induction: Induced e.m.f., e = – (dφ/dt); e 1 dφ Induced current, I = =− ; R R dt Induced charge, q = (φ1 – φ2)/R

62.

Motion of a conducting rod:

(b)

(a) (b) (c)

→

→

→

→

→

→

→

→

(e)

→

V. B = 0

→

F = − e ( v x B) ;

Induced e.m.f., e = B/v For a rod rotating with angular frequency ω or rotating disc, induced e.m.f., e=

1 2 Bl ω = Βπfl 2 = Baƒ 2

63. (a) (c) (e) (f)

Motion of conducting loop in a magnetic field: Induced e.m.f. e = Blv ; (b) Induced current, I = (e/R) = (Blv/R) F = IlB = B2l2v/R ; (d) P = Fv = IlBv = B2l2v 2R; H = I2R = (B2l2v2/R); In non uniform magnetic field, e = (B1–B2) lv and I = (B1–B2)lv/R

64. (a) (b)

Rotating loop: φ = NAB cos ωt = φ0 cos ωt, with φ0 = NAB; e = e0 sin ωt, where e0 = NaBω; (c) I = (e0 sin ωt/R) = I0 sin ωt, with I 0 = e0/R

65.

Induced electric field: Induced e.m.f. =

66. (a) (b) (c) (d) (e)

Self Inductance: L = φ/ I ; e = – (LdI/dt); 2 2 L = µ0N A/l = µ0n Al L = µrµ0N2A/l 2 L = µ0N πR/2

67. (a)

Mutual inductance: M = (φ2/ I1) ; (b) e2 = – M(dI 1/dt); (c) M = µ0NsNp A/lp

68. (a) (b)

Series and parallel combination L = L1 + L2 (if inductors are kept far apart and joined in series) L = L1 + L2 ±2M (if inductors are connected in series and they have mutual inductance M) LL 1 1 1 L= 1 2 or = + L1 + L 2 L L1 L 2 (if two inductors are connected in parallel and are kept for apart)

(c)

∫

→ →

E . dl

(For a solenoid with air core); (For a solenoid with a material core); (For a plane circular coil)

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

18

Formula Booklet – Physics XII

(d) 69. (a) 70. (a) (b)

M = K√(L1L2)

(if two coils of self inductances, L1 and L2 are over each other)

Energy stored in an inductor: 1 2 U = LI 2 ; (b) uB = (B /2µ0) 2 Growth and decay of current in LR circuit –t/ (for growth), where τ = L/R I = I0 (1–e τ) –t/τ (for decay), where τ = L/R I = I0 e

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.

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

19

Formula Booklet – Physics XII

ALTERNATING CURRENT 71. (a) (b)

A.C. Currrent and e.m.f. : I = I0 sin (ωt ± φ) ; e= e0 sin (ωt ± φ);

(c)

< I > = 0, < I >1/2 =

(d) (e) (f)

< I 2 > I 02 / 2 ;

2I 0 =0.637 I0 ; π

Irms = (I0/√2) = 0.707 I0 ; form factor = π/2√2

72. (a)

A.C. response of R, L, C and their series combinations Resistance only: (i) e = e0 sin ωt; (ii) I = I0 sin ωt ; (iii) phase difference φ = 0; (iv) e0 = I0R; (v) erms = Irms R

(b)

Inductance only: (i) e = e0 sin ωt; (ii) I = I0 sin (ωt–π/2) ; (iii) current lags the voltage or voltage leads the current by a phase π/2;

I0XL;

(iv) (c)

erms = Irms XL ;

(iv)

e0 =

(vi) XL = ωL

Capacitance only: (i) e = e0 sin ωt ; (ii) I = I0 sin (ωt + π/2); (iii) current leads the voltage or voltage lags the current by a phase π/2 ; (iv) (v) erms = Irms XC ; (vi) XC = (1/ωC)

e0 = I0XC;

(d)

Series LR circuit: (i) e = e0 sin ωt ; (ii) I = I0 sin (ωt + φ); –1 (iii) the current lags the voltage or voltage leads the current by a phase φ = tan (XL/R); (iv) cos φ = (R/Z) and sin θ = (XL/Z); 2 2 (v) Impedance, Z = √ [R +(ωL) )] ; (vi) e0 = I0Z; (vii) erms = Irms Z

(e)

Series RC circuit: (i) e = e0 sin ωt ; (ii) I = I0 sin (ωt + φ); –1 (iii) The current leads the voltage or voltage lags behind the current by a phase φ = tan (XC/R) (iv) cos φ = (R/Z); 2 2 (v) Impedance, Z = √[R + (1+ωC) )]; (vi) e0 = I0Z ; (vii) erms = Irms Z

(f)

Series LCR circuit:

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

20

Formula Booklet – Physics XII

(i) (ii)

e = e0 sin ωt ; I = I0 sin (ωt – φ);

(iii)

φ=tan

(iv) (v) (vi)

current lags and circuit is inductive if XL < XC ; current leads and circuit is capacitive if XL < XC ; 2 2 Impedance, Z = √[R + (XL – XC) ];

(viii)

cos φ = (R/Z) and sin φ = 

–1

 XL − XC  R 

  , φ is positive for XL > XC, φ is negative for XL<XC; 

 XL − XC Z 

(vi) e0 = I0Z;

  

73.

Resonance

(a)

Resonance frequency, ƒr = 

(b)

At resonance, XL = XC, φ = 0, Z = R (minimum), cos φ = 1, sin φ = 0 and current is maximum (=E0/R)

74.

Half power frequencies

(a)

lower, ƒ1 = ƒr –

(b)



    2π LC 

R 4 πL R upper, ƒ2 = ƒr + 4 πL

or or

R 2 πL

1

R 2L R ω2 = ωr + 2L

ω1 = ωr –

∆ω =

R L

75.

Band width: ∆ƒ =

76.

Quality factor ω ωL Q= r = r ; R ∆ω 1 1 1 As ωr = , hence Qα√L, Qα and Q α ; R C LC 1 Q= ; ω r CR

(a) (b) (c)

(X L )res

(X C )res

or

(d)

Q=

(e)

ƒ  Q =  r   ∆ƒ 

77. (a)

At resonance, peak voltages are (VL)res = e0Q; (b) (VC)res = e0Q ; (c) (VR)res = e0

78. (a) (b) (c) (d) (e)

Conductance, susceptance and admittance Conductance, G= (1/R); Susceptance, S = (1/X); SL = (1/XL) and SC=(1/XC) =ωC; admittance Y = (1/Z); Impedance add in series while admittance add in parallel

79.

Power in AC circuits 1 Pav = E0I0 cos φ = Erms Irms cos φ; 2 Pav Re al power = Power factor, cos φ = Virtual power E rms I rms

(a) (b)

R

or

or ∆f =

R

;

ƒr Q

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

21

Formula Booklet – Physics XII

(c)

Cos φ = (R/Z)

(d)

(i) (ii) (iii)

e 2rms R 0 C only : φ = – 90 = –π/2, cos φ = 0, Pav = 0 0 L only : φ = 90 = π/2, cos φ = 0, Pav = 0

(iv)

Series RL or RC: φ = tan–1 

R only : φ = 0, cos φ = 1, Pav = I2rms R =

 XL  –1  X   or φ = tan  C   R   R 

Pav = Erms Irms cos φ = (iv)

80. (a) (b) (c) (d) (e) (f)

1 = Z

2

 1 1   ; +  − 2  R  X L XC  1

Y = G 2 + (S L − SC ) ; 2

I0 = E0Y; S − SC ; tan φ = L G  1 R 2   −  LC L2  ; LC   in parallel resonance circuit, impedance is maximum, admittance is minimum and current is minimum.

ωr =

1

or ωr =

(a)

Cp = Np 

(b)

 ep  e  s

(f) 82. 83.

= I 2rms R

Parallel LCR circuit

Transformer:

(d) (e)

Z2

E2 R –1 –1  X − X C  2 Series LCR: φ = tan φ = tan  L  , PAV = rms2 = I rms R, R Z   At resonance, φ = 0, cos φ = 1 and Pav = I2rms R = E2rms/R

81.

(c)

E 2rms R

 dφ   dφ   and es = Ns    dt   dt 

  Np =  N   s

   

 I   ep   Np   Q ep Ip = es Is, so  s  =   =    Ip     es   Ns  Step down: es < ep, Ns < Np and Is > Ip Step up : es > ep, Ns > Np and Is < Ip e I  Efficiency, η =  s s   ep Ip    AC generator: e = e0 sin (2πƒt), (where e0 = NBAω)

DC motor: E−e I= (a)   R  (b) IE = Ie = I2R (c)

Back emf e efficiency, η   =  E  Applied emf

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

22

Formula Booklet – Physics XII

LIGHT 1.

Intensity of light

(a)

Spherical wave front: (i) I =

(b) (c) 2. 3. 4. (a)

P 2

, (ii) amplitude ∝

1 r 1

4 πr 1 Cylindrical wave front: (i) I ∝ , (ii) amplitude ∝ r r Plane wave front: (i) I ∝ r0, (i) A ∝ r0 (i.e. I and A are both constants)

Law of reflection: Angle of incidence (i) = Angle of reflection (r) sin i Law of reflection: Snell’s law: η = sin r Other relations v c 2η1 = 1 and η = v2 v

(c)

λ air or η η1 sin i = η2 sin r

5.

Electromagnetic nature of light

(a)

The magnitude of E and B are related in vacuum by: B=

(b)

(b) (c)

λmedium =

→

→

vmedium =

(Q νmedium = νair)

→

→

→

v air η

E C

→

E and B are such that E x B is always in the direction of propagation of wave 1 1 and v= c= µ0ε0 µε

(d)

Refractive index, η = √(µr εr) (µr = µ/µ0 and εr = ε/ε0) For non–magnetic material, µr ≈ 1 and η = √(εr)

(e)

The EM wave propagating in the positive x–direction may be represented by: Ey = E0 sin (kx – ωt) and Bz = B0 sin (kx – ωt)

6.

Energy transmitted by an electromagnetic wave

(a)

Energy density of electromagnetic wave is: u = ue + um =

(b)

As for EM wave, B = u=

1 1 B2 ε0 E2 + 2 2 µ0

E 1 and = √ (µ0 ε0), hence C c

1 1 E2 1 1 ε 0E 2 + = ε 0 E 2 + ε0 E 2 = ε 0E 2 2 2µ0 c 2 2 2

(c)

Time averaged value of energy density is: u =

7.

Intensity of an electromagnetic wave

(a)

In a medium: I =  ε0 E02  v

(b)

1  In free space: I =  ε 0 E02  c 2  

8.

Pointing vector

1 2

1 ε0 E 0 2 2

 

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

23

Formula Booklet – Physics XII

(a)

→

→

→

S = ExH= 2

→ → 1 → → ( E x B ) = c 2ε 0 ( E x B ) µ0 2

(b) (c) (d)

S = cε0E = √(ε0/µ0)E I = S and S = cu Impedance of free space, Z = √ (µ0/ε0) ≅ 377 ohm

9. (a)

Pressure of EM Radiation Change in momentum (normal incidence)

(b)

(c)

∆p =

U S A∆t = c c

∆p =

2 U S A∆t (reflector) = c c

(absorber)

Pressure (normal incidence) P=

S =u c

(absorber)

P=

2S = 2u c

(reflector)

Pressure for diffused radiation P=

1S 1 = u 3 c 3

(absorber)

2 S 2 = u (reflector) 3 c 3 Quantum theory of light: Energy of photon, E = hν = hc/λ E h = Momentum, p = c λ Rest mass of photon = 0 2 Mass equivalent of energy, m = (E/c ) P=

10. (a) (b) (c) (d) 11. (a)

Inclined mirrors: number of images When 3600 is exactly divisible by θ0 and 3600/θ0 is an even integer then the number of images formed is n=

(b)

360 − 1 (whatever may be location of the object) θ

0 0 When 360 is exactly divisible by θ and 360/θ) is an odd integer, then the number of images formed is

360 − 1 (for symmetrical placement) θ 360 (for unsymmetrical placement) = θ

n=

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

24

Formula Booklet – Physics XII

(c)

0 When 360 is not exactly divisible by θ, then the number of images formed is = integer value of n (where n = 360/θ)

12. (a)

Reflection amplitude and intensity When a ray of light is incident (with angle of incidence i ≈ 0) from a medium 1 of refractive index η1 to the plane surface of medium 2 of refractive index η2, then reflection amplitude is η − η2 R= 1 η1 + η 2

(b)

The ratio of the reflected intensity and the incident intensity is:

13.

Refraction of light sin i η= ; (b) sin r 1 1 η2 = ; (d) 2η1

(a) (c)

1

η2 =

I r  η1 − η 2 = I i  η1 + η 2

2

  . 

sin θ1 ; sin θ 2

Cauchy’s relation: η = A +

B λ2

14. (a) (b)

Parallel slab Angle of incidence, i = Angle of emergence, e Lateral shift = [t sin (i – r)/cos r]

15.

Composite block: η1 sin θ1 = η2 sin θ2 = η3 sin θ3 = constant

16.

Apparent depth R t a= (where R = Real depth) = η η If there is an ink spot at the bottom of a glass slab, if appears to be raised by a distance

(a) (b)

x=t–a=t– (c)

 1 t = t 1 −  , where t is the thickness of the glass slab η  η

If a beaker is filled with immissible transparent liquids of refractive indices η1, η2, η3 and individual depths t1, t2, t3 respectively, then the apparent depth of the beaker is: t t t a= 1 + 2 + 3 η1 η 2 η3

18.

1 η For a luminous body at a depth d inside a liquid: Radius of bright circular patch at the surface d r = d tan iC = η2 − 1

19. 20. (a) (b)

For optical fibre: sin i ≤ Prism: i+e=A+δ r1 + r2 = A;

17.

(c) (d)

Total internal reflection: Critical angle iC is given by: sin iC =

[(n

2

]

/ n1 ) − 1 2

 A + δm  sin    2  At minimum deviation: i = e and r1 = r2. Hence, η = A sin   2 For small angle prism: δ = (η–1) A

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

25

Formula Booklet – Physics XII

21. (a) (b)

Dispersion: δred < δviolet because ηred < ηviolet Angular dispersion: θ = δV – δR = (ηV–ηR)A

(c)

Dispersive power: ω =

(d)

Dispersion without deviation: (i) δC + δF = 0

δ V − δ R ηV − ηR ηB − ηR (In practice) = = δY ηY − 1 ηY − 1

or

(η − 1) AF =− C AC (η F − 1)

(ii) Also, angular dispersion, θ=AC (ηC–1) (ωC–ωF) (e)

Deviation without dispersion: (i) θC + θF = 0 (ii) Also,

or,

AF η − ηCR = − CV AC η FV − ηFR

δ ωF = − CY δ FY ωC

22.

Principle of superposition: y = y1 + y2

23. (a) (b) (c)

Superposition of waves of equal frequency and constant phase difference Resultant wave amplitude, a = √(a12 + a22 + 2a1a2 cos φ) Resultant wave intensity, I = I1 + I2 + 2√(I1I2) cos φ 2 If a1 = a2 = a0, and I1 = I2 = I0, then a = 2a0 cos (φ/2) and I = 4I0 cos (φ/2)

24. (a)

Constructive interference conditions: φ = 2nπ ≡ 0, 2π, 4π, 6π,….. or, ∆ = nλ ≡ 0, λ, 2λ, 3λ, ……

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

amax = a1 + a2 2 Imax ∝ (a1 + a2) 2 Imax = I1 + I2 + 2 √ (I1I2) = (√I1 + √I2) Imax = 4I0 ; If I1 = I2 = I0

25.

Destructive interference

(a)

conditions: φ = (2n –1) π ≡ π, 3π, 5π,…… or, ∆ = (2n–1)

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

amin = a1 – a2 2 Imin ∝ (a1 – a2) 2 Imin = I1 + I2 – 2√(I1I2) = (√I1 – √I2) Imin = 0 if I1 – I2 = I0

26.

Young’s double slit experiment 2π 2π Phase difference, φ = (S2P – S1P) = x path difference λ λ A = 2a0 cos (φ/2) and I 4I0 cos2 (φ/2) Position of nth fringe on the screen: nDλ (i) for bright fringe, x n = d ( 2 n − 1) Dλ (ii) for dark fringe, x n = 2d

(a) (b) (c)

27.

Fringe width:

(a)

Linear fringe width, β =

λ λ 3λ 5λ = , , ,......... 2 2 2 2

Dλ d

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

26

Formula Booklet – Physics XII

λ d

(b)

Angular fringe width, α =

(c)

βliquid =

(d)

βwater =

28. (a)

When a thin shit is introduced in the path of one of the interfering waves: (η–1) t = nλ (η − 1) t β Shift of the central fringe = λ

(b)

β air air

ηliquid

or λliquid =

Fringe visibility: V =

30. (a) (c)

Frensel’s biprism: d = 2a (η–1) α ; (b) β = (Dλ/d);

I max − I min I max − I min

d = √(d1d2) (d) dliquid < dair, for example, dwater = dair/4  ηg − 1   βliquid > βair ; βliquid = βair   ηg − ηt   

(e)

(b) 32.

Newton’s rings: Diameter of nth dark fringe, Dn = √(4nλR) D 2n +p − D 2n D 2 − D 2n and η = n2+ p λ= 4 pR D' n + p − D' 2n Thin films: For reflected light 2ηt cos r = nλ

(Dark fringe)

1 2ηt cos r = (n– )λ 2

33. (a) (b)

ηliquid

3 βair 4

29.

31. (a)

λ air air

(Bright fringe)

Diffraction: a sin θ = nλ (a = width of slit) Half angular width of central maxima, θ = sin–1 (λ/a) 2

(c)

(d) (e)

 sin φ   Intensity distribution of the screen I = I0   φ  πa y where, φ = and I0 = Intensity at central point of screen λD 1.22 λ Limit of resolution of telescope: θ = a 1 a Resolving power of telescope = = θ 1.22λ

34. (a)

Spherical mirrors: Focal length: ƒ = (R/2)

(b)

Mirror formula:

(c)

Newton’s formula: ƒ2 = xy

1 1 1 = + ƒ v u

(x and y are the distances of the object and image from the principal focus respectively)

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

27

Formula Booklet – Physics XII

I v ƒ ƒ−v =− = = o u ƒ−u ƒ

(d)

Linear magnification: m =

(e)

Longitudinal magnification: m = −

35. (c)

Spherical lenses: A single spherical surface: η 2 η1 (η 2 − η1 ) − = [For an object placed in a medium of refractive index η1] (i) v u R η1 η 2 (η1 − η 2 ) − = [For an object placed in a medium of refractive index η2] (ii) v R u R (iii) First principal focus: ƒ1 = where η = η2/η1 (η − 1) ηR (iv) Second principal focus: ƒ2 = (η − 1) v / η2 (v) Magnification: m = u/η1

(d)

Lens Maker’s formula: (i)

 1 1  η2 1   or, = − 1  −  ƒ  η1 R R 2  1

v2 u2

 1 η1 η1 1   − = (η2 − η1 )  −  v u R R 2  1

[When medium is same on both sides of the lens] (ii)

η3 η1  η 2 − η1   η3 − η 2 + − = v u  R 1   R 2

  

[When different medium exist on two sides of the lens] (e)

Biconvex or biconcave lens of the same radii for two surfaces:

(f)

Linear magnification: m =

(g)

Power of lens: P=

(h)

Lenses in contact: (i) (iii)

I v ƒ−v ƒ = = = O u ƒ ƒ+u

1 ƒ

1 1 1 ; = + ƒ ƒ1 ƒ 2

(ii)

P = P1 + P2

For lenses separated by a distance d =

1 1 1 d = + − ƒ ƒ1 ƒ 2 ƒ1ƒ 2

(i)

Achromatic lens combinations: Condition of achromatism,

36.

Silvering at one surface:

(a)

1 2 (η − 1) = ƒ R

1 1 1 1 2 2 (η - 1) = + + = = F ƒl ƒ m ƒl ƒl R

ω ω' =− ƒy ƒ 'y

R η Fig. 1

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

28

Formula Booklet – Physics XII

(b)

(c)

η

1 2 1  (η − 1)  2 2n = + =2 + = R F ƒl ƒ m  R  R

R

Fig. 2

η

R1

 1 1 2 1 1  2 + = + = 2(η − 1)  +  F ƒl ƒ m  R1 R 2  R 2

R2

Fig. 3 37. (a)

(b)

(c)

Optical Instruments Astronomical Telescope: ƒ0 ƒe

(i)

For normal adjustment: m =

(ii)

For near–point adjustment: m =

ƒ0  ƒ  1 + e  ƒe  D

Simple Microscope: D ƒ

(i)

For normal adjustment: m =

(ii)

For near–point adjustment = m = 1+

D ƒ

Compound Microscope: (i) (ii)

D    ƒe  v  D For near–point adjustment: m = 0 1 +  u0  ƒe 

For normal adjustment: m =

v0 u0

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

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

29

Formula Booklet – Physics XII

MODERN PHYSICS CATHODE RAYS AND POSITIVE RAYS 1. (a) (b)

Cathode rays Thomson identified cathode rays as an electron beam. 11 Specific charge q/m as measured by Thomson is: (q/m) = 1.759 x 10 Coulomb/Kg

2. (a) (b)

Positive rays Positive rays were discovered by Goldstein. (q/m) for positive rays is much less than that of electrons.

3. (a) (b) (c) (d) (e) (f)

Motion of charge particle through electric field (Field ⊥ to initial velocity) 2 2 The path is parabolic: y = (qE/2mu )x The time spent in the electric field: t = (L/u) The y–component of velocity acquired: vy = (qEL/mu) 2 The angle at which particle emerges out tan θ = qEL/mu The displacement in y-direction, when the particle emerges out of the field: y1=(qEL2/2mu2) 2 The displacement on the screen = Y = (qELD/mu )

4. (a) (b) (c)

Motion of charged particle through magnetic field (Field ⊥ to initial velocity) The path is circular with radius: r = (mu/qB) Momentum of the particle: p = qBr The deflection on the screen: X = (qBLD/mu)

5. (a)

Mass spectrographs Thomson’s mass spectrograph (i) Traces on the screen are parabolic in nature (ii) Inner parabola corresponding to heavy M white outer parabola to light M. (iii) The upper portion of parabola is due to small v ions, while lower portion is due to high v ions. (iv) Only v = ∞ ions can reach vertex of parabola. 2 2 (v) Equation of parabola: X = (B LD/E) (q/M) Y = K (q/M) Y

(b)

Brain bridge mass spectrograph (i) Velocity selector: v = (E/B) (ii) Other relations: r = (Mv/qB’) = (ME/qBB’) (whre B’ is the magnetic field in dome); d =2r; (d2 – d1) ∝ (M2–M1) ; M1 : M2 = d1 : d2 [where d1 and d2 are the distances of traces 1 and 2 from the slit S2 of velocity selector].

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

30

Formula Booklet – Physics XII

PHOTOELECTRIC EFFECT 6. 7.

Work function W = h h c hc hc Threshold wavelength: λ0 = = = v 0 hv 0 W

Threshold frequency: v0 =

(To calculate λ0, use hc = 1240 (eV) (nm) = 1.24 x 10–6 eV) (m) 8. (a) (b) 9. 10. (a) (c) 11. (a) (b) (c)

12.

Maximum kinetic energy of emitted photoelectrons 1 Kmax = mv2max = eV0 2 1 1   Kmax = hv – W = h (v – v0) = hc  −   λ λ0  h Slope of (V0 v) graph = e Energy, momentum and mass of a photon hc Rest mass of photon = 0 (b) E = hv = λ E h E h p= = (d) m = 2 = c λ cλ c Number of photons:

(

)

Intensity Watt/m 2 hv Power (Watt ) number of photons incident per second, np = hv number of electrons emitted per second = (efficiency of surface) x number of photons incident per second. 2 number of photons per sec per m , np =

(a)

Compton wavelength: h =2.426 pm λc = m0C

(b)

Change in wavelenth, (λ’ – λ) = λc (1– cos φ)

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

31

Formula Booklet – Physics XII

ATOMIC STRUCTURE 13. (a)

Rutherford’s α–particle scattering 4 N(θ) ∝ cosec (θ/2)

(b)

Impact parameter, b =

14.

Distance

15.

Bohr’s atomic model nh L = mvr = 2π hc hv = Ei = Ef = λ Radius of nth orbit:

(a) (b) (c)

of

(Ze )cot (θ/ 2) , 2

(4 π ε 0 ) E

closest

approach:

(where E =

r0 =

1 mu2 = KE of the α–particle) 2

2 Ze2 (where (4π ε0 )E

(iii) (iv)

Ratio of radii: r1:r2 : r3 = 1 : 4 : 9 ; rN : rHe+ : rLi++ = 1 :

rn ∝

n2 , Z

=

1 mu2 2

=

KE

(ii) rn =

1 1 : =6:3:1 2 3

Velcotiy of electron in nth orbit: 2π Ke 2 1 Z c  Z (where α = = = fine structures constant) (i) vn =  = αc n  137  n ch 137 1 1 (ii) v1 : v 2 ; v3 = 1 : : = 6 : 3 : 2 2 3 st (iii) v1 = velocity of electron is 1 orbit of H–atom = (c/137)

(e)

Total energy of electron: 2 (i) Potential energy, U = – (kZe /r) 1 (ii) K = mv2 = (kZe2/2r) 2 2 (iii) E = K + U = – (kZe /2r) = (U/2) = –K (iv) K = – (U/2) or U = 2K = 2E

(f)

En = –

13.6 Z 2

the

n2 Z

(d)

(v)

of

α–particle)

  h2    4 π 2 m ke 2    2 2 2 Bohr’s radius: a0 = (h /4π mke ) = 0.529 Å

(i)

E

eV = –

Z 2  2π 2 m k 2e 4 

h2 n2 n 2  2 Ionization energy = – E1 = + (13.6Z )eV (i) For H–atom, I.E. = 13.6 eV + (ii) For He – ion, I.E = 54.4 eV ++ (iii) For Li –ion, I.E. = 122.4 eV

(g)

Ionization potential: (i) For H–atom, I.P. = 13.6 V + (ii) For He ion, I.P. = 54.42

(h)

Series formula (wave number v = 1/λ

 

=−

2.18 x 10-18 Z2 n2

J

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

32

Formula Booklet – Physics XII

 1 1 1  = RZ 2  2 − 2  n  λ  1 n2 

(i)

2π 2 m k 2 e 4 ch

3

= 1.097 x 10 7 m -1

Series formula for H–atom (i)

Lyman series:

1 1   = R 1 − 2 , n = 2, 3, 4, ......∞ λ  n 

1 1   1 = R  2 − 2 , n = 3, 4, 5....∞ λ 2 n   1 1   1 (iii) Paschen series: = R  2 − 2  , n = 4, 5, 6....∞ λ n  3 1 1   1 (iv) Brackett series: = R  2 − 2 , n = 5, 6, 7….∞ λ n  4 1 1 1   (v) P–fund series: = R  2 − 2 , n = 6, 7, 8 ….∞ λ n  5 Series limits (λmin) (i) Lyman: λmin = 912 Å (ii) Balmer: λmin = 3645 Å (iii) Paschen: λmin = 8201 Å

(ii)

(j)

where R =

Balmer series:

16.

Number of emission lines from excited state n = n(n–1)/2

17. (a)

Time period of revolution Tn ∝ (n3/Z2) ; (b) T1 = 1.5 x 10–16 sec ; (c) T1 : T2 : T3 = 1 : 8 : 27

18.

Frequency of revolution

(a)

vn ∝ (Z2/n3);

19. (a)

Current due to orbital motion 2 3 In ∝ (Z /n ) ; (b) I1 = 1 mA

20. (a)

Magnetic field at nucleus due to orbital motion of electron Bn ∝ (Z3/n5) ; (b) B1 = 12.5 Tesla

21. (a) (b)

Magnetic moment: Mn = (eL/2m) = (nhe/4πm); –24 2 M1 = (eh/4πm) = µB = Bohr Magneton = 9.27 x 10 Am

22.

Magnitude of angular momentum: L = √[l(l+1)] (h/2π)

23. (a)

Angle of angular momentum vector from z–axis cos θ = [ml√{l(l+1)}]; (b) the least angle is for ml = l i.e. cos θmin = [l/√{l(l+1})]

24.

Magnitude of spin angular momentum

(b) v1 = 6.6 x 1015 Hz ; (c) v1 : v2 : v3 = 1 :

S = √[s (s+1)] (h/2π) =

25. (a)

3 (h/2π) 2

1 1 : 8 27

X–RAYS

Continuous X–rays: vmax = (eV/h) ; (b) λmin = (hc/eV) = (12400/V) Å

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

33

Formula Booklet – Physics XII

26. (a)

28.

Characterictic X–rays: λKα < λLα < λMα ; (b) vKα > vLα > vMα 3cR Frequency of Kα line : v (Kα) = (Z–1)2 = 2.47 x 1015 (Z–1)2 4 2 2 Wavelength of Kα line: λ(Kα) = [4/3R(Z–1) ] = [1216/(Z–1) ]Å

29.

2 Energy of Kα X–ray photon: E(Kα) = 10.2 (Z–1) eV

30. (a) (b)

Mosley’s law: 2 5 v = a (Z–b) , where a = (3cR/4) = 2.47 x 10 Hz For Kα line, b = 1; (c) √v α Z

31.

Bragg’s law: 2d sin θ = nλ

32.

Absorption formula: I = I0 e

33.

Half–value thickness: x1/2 = (0.693/µ)

34. (a) (b) (c) (d)

For photons: E = hv = (hc/λ) ; p = (hv/c) = (E/c) = (h/λ) ; 2 2 m = (E/c ) = (hv/c ) = h/cλ rest mass = 0, charge = 0, spin = 1 (h/2π)

35.

Matter waves:

(a)

de Broglie wavelength, λ = h = h =

(b)

(i)

For electron λe =

(ii)

For proton, λp =

(iii)

For alpha particle λα =

27.

–µx

MATTER WAVES

(c)

p

(g)

V 0.286 V

h 2m E

=

h 2mqV

[Q E =

1 2

mv 2 = qV ]

Å

Å 0.101 V

Å h 3 m KT

3    E = kT  2  

For neutron or proton: λ = (25.2/√T) Å [if E = (3/2) kT, average energy] 30.8 Å [if E = kT, most probable energy] but λ= T

The wavelength of electron accelerated by potential difference of V volts is: λe =

12.27

Å v Hence, accelerating potential required for obtaining de Broglie wavelength for as electron is: V=

(e) (f)

12.27

For particle at temperature T : λ = (i)

(d)

mv

150.6 λ2e

volt

Condition for obtaining stable orbit: 2πrn = nλ The phase velocity of a de Broglie wave of wavelength λ and frequency v is Ε h mc 2 h c2 = = i.e. v p > c. v p = vλ = x x h mv h mv v Group velocity, vg = (dω/dk). It is found that group velocity is equal to particle velocity i.e., v g = v

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

34

Formula Booklet – Physics XII

RADIOACTIVITY 36.

Decay law: (a) (dN/dt) = – λN ; (b) N = N0 e λ ; (c) (N/N0) = (1/2)

37.

Half life and decay constant: (dN/dt ) ; (b) λT = log 2 or T = (0.693/λ) λ=− e N

(a)

– t

t/T

or λ = (0.693/T)

38. (a)

Mean life: τ = (1/λ) or λ=(1/τ);

39. (a) (e) (g)

Activity: – t t/T R = |dN/dt| ; (b) R = λN ; (c) R = R0e λ ; (d) (R/R0) = (1/2) ; 10 1 Becquerel = 1 dps ; (f) 1 curie = 1 ci = 3.7 x 10 dps; 6 6 1 Rutherford = 1Rd = 10 Rd = 10 dps

40.

Decay of active mass:

(a)

m = m0 e

41.

Radioactive equilibrium: NAλA = NBλB

42.

Decay constant for two channels: (a) λ = λ1 + λ2 ; (b) T =

43.

Gamma intensity absorption: (a) I = I0e

–λt

(b) T = 0.693τ or τ = 1.443 T

; (b) (m/m0) = (1/2)t/T ; (c) N =

6.023 x 10 23 x m A

–µx

T1 T2 T1 + T2

; (b) Half value thickness, x1/2 = (0.693/µ)

NUCLEAR PHYSICS 44.

Atomic mass unit: (a) 1 amu = 1.66 x 10–27 kg ; (b) 1 amu ≡ 1u ≡ 931.5 MeV

45. (a)

(c)

Properties of nucleus 1/3 where R0 = 1.2 fermi Radius: R = R0A 4 4   3 3 Volume: V αA Q V = 3 πR = 3 π R 0 A    17 3 Density: ρ = 2.4 x 10 Kg/m (ρ is independent of A)

46.

Mass defect: ∆M = Zmp + (A–Z) mn – M

47. 48.

Packing fraction: ƒ = ∆/A = mass excess per nucleon [∆ = –∆M = mass excess] 2 Binding energy: ∆E = BE = (∆M)c

49. (a) (b) (c)

Binding energy per ncuelon: BEN = (BE/A); BEN for Helium = 7.1 MeV/nucleon BEN for Deuterium = 1.1 MeV/nucleon

(b)

ELECTRONICS 50. (a) (b) (c)

Richarson equation 2 –W/KT where A = 60 x 104 A/K2m2 J = AT e 2 11600 W/T –23 –5 J = AT e [∴ K = Boltzmann’s constant = 1.38 x 10 J/K = 8.62 x 10 eV/K Hence, 1/K = 11600 kelvin/eV] I = AT2Se–W/KT

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

35

Formula Booklet – Physics XII

3/2

51.

Child’s law: Ip = KVp

52. (a)

Diode resistance –1/2 (iii) Rp ∝ I −p 1 / 3 Static plate resistance: (i) Rp = (Vp/ Ip); (ii) Rp ∝ Vp

(b)

Dynamic plate resistance: (i) rp = (∆Vp/∆ Ip); (ii) rp ∝ v −p 1 / 2 ; (iii) rp ∝ I −p 1 / 3 .

53. (a)

Triode Constants:  ∆Vp   rp =  ;  ∆I p    Vg = constant

(b)

 ∆I p gm =   ∆Vg 

(c)

 ∆Vp µ=   ∆Vg 

(d)

µ = rp x gm ; (e) rp ∝ Ip

  ;   I p =consant

[K = constant of proportionality]

–1/3

55.

Vp    Plate current equation: Ip = K  Vg + µ   Cut off voltage: Vg = – (Vp/µ)

56.

Triode as an amplifier:

(a)

Ip = (µVg/RL + rp); (b)

54.

(c) 57.

  ;   Vp =cons tan t

3/ 2

A = (µRL/RL + rp) rp   ; (d) µ = A 1 + (e) A = µ/2 if RL = rp Amax = µ;   RL  Conductivity of semi conductors − E / 2 KT

(b)

Intrinsic: (i) σ = e (neµe + nh µh) ; (ii) σ = σ0e g Extrinsic: (i) n–type : σ = eneµe; (ii) p–type : σ = enhµh

58. (a)

Transistor: IE = IC + IB

(a)

(b)

(IB << IE, IB << IC) (i ) α = ΙC , α ac = ∆IC Current gains: ΙE ∆ΙΕ

(ii )

(c)

1/3

; (f) gm ∝ Ip

β=

ΙC ∆ IC , βac = IB ∆ IB

Relation between α and β: α =

β 1+ β

or

β=

α 1− α

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

36

Formula Booklet – Physics XII

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.

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

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

37

Formula Booklet – Physics XII

Never say Quit! When things go wrong, as they sometimes will, When the road you are trudging seems all uphill. When the funds are low and debts are high, And you want to smile, but you have to sigh. When care is pressing, you’re down a bit. Rest if you must, but never quit. Life is queer, with its twists and turns, As every one of us, sometimes learns. And many a fellow turns about, When he might have won, if he had stuck it out. Stick to your task, though the pace seems slow, You may succeed with just another blow. Often the goal is nearer than, It seems to a faint and faltering man. Often the struggler has given up. When he might have captured the victor’s cup. And he learned too late, when the night slipped down, How close he was to the golden crown. Success is failure turned inside out, The silver tints of the clouds of doubt. And you never can tell how close you are, It may be near when it seems afar. So stick to the fight when you are hardest hit, It’s when things seem worst, that you must never quit! Edwin Markham

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

38

Formula Booklet – Physics XII

Put Forth Your Best… And You’ve Already Won! The contest last for just moments, though the training’s taken years. It was’nt the winning alone that was worth the work and tears. The applause will be forgotten, the prize will be misplaced. But the long hours of practise will never be a waste. For in trying to win, you build a skill. You learn that winning depends on will. You never grow by how much you win, you only grow by how much you put in. So any new challenge you’ve just begun, Put forth your best and you’ve already won!

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

39

Formula Booklet – Physics XII

You can… if you think you can! If you think you are beaten, you are; If you think that you dare not, you don’t; If you’d like to win, but think you can’t, It’s almost certain you won’t. If you think you’ll lose, you’ve lost, For out in the world you find, Success begins with a fellow’s will, It’s all in the state of mind. Often many a race is lost, Before even a step is run, And many a coward fails, Before even his work’s begun. Think big, and your deeds will grow Think small, and you’ll fall behind. Think that you can, and you will, It’s all in the state of mind. If you think you’re outclassed, you are; You’ve got to think high to rise; You’ve got to be sure of yourself, Befoe you can ever win a prize. Life’s battles don’t always go, To the stronger or faster man, BUT SOONER OR LATER THE MAN WHO WINS, IS THE ONE WHO THINKS HE CAN. –Edwin Markham

QUEST - Power Coaching for IITJEE 1, Vigyan Vihar, Near Anand Vihar, Delhi – 92. Ph: 55270275, 55278916 E-16/289, Sector 8, Rohini, Delhi – 85, Ph: 55395439, 30911585

40