1

Chapter 01 : Electrostatics

1. q = ne

n = 18

19

5.8 10

1.6 10

−

−

××

= 36.25

As n is not an integer, such a charge cannot

exist.

2. Fnet = 1 2

2

0

1 q q

4 rπε(110/100) (90/100) times

i.e., 99

100 times

∴ Net force = 99

100× 100 = 99 N

3. Force on – q1 due to q2 is

1 212 2

Cq qF

b= along X-axis

Force on – q1 due to − q3 is

1 313 2

Cq qF

a= at ∠θ with negative direction of

Y-axis.

∴ x component of force on – q1 is

Fx = F12 + F13 sinθ = Cq12 3

2 2

q qsin

b a

+ θ

i.e., Fx ∝ 2 3

2 2

q qsin

b a

+ θ

4. Asq

Aσ = , to increase the charge density, area

‘A’ should be made very small. 5.

Since the centre of the square lies at origin,

hence each quadrant will have the charge that

cancels the charge of diagonally opposite

quadrant. This results in the zero net charge on

square. 6. At neutral point,

2 2 2 2

0 0

1 20 1 Q

4 (20 10 ) 4 (40 10 )− −× = ×

πε × πε ×

∴ Q = 80 C 7. τ = pE sin θ

∴ p = 26

4

9 10

Esin 10 sin30

−τ ×=

θ °

p = 1.8 × 10–29

Cm

8. According to Gauss’ law,

φ = 0

q

ε

If the radius of the gaussian surface is

increased the outward flux remains constant as

it depends only on the charge enclosed by the

surface.

9. V = 0

1

4πε

q

R

V1 = 0

1

4πε 1

1

q

R and V2 =

0

1

4πε 2

2

q

R

Both the sphere having same potential

V1 = V2

1

1

q

R = 2

2

q

R

∴ 1

2

q

q = 1

2

R

R

10. V = 0

1 q

4 rπε i.e., V ∝ q …Q r is constant

∴ He

H

V

V= He

H

q

q=

2

1

∴ VHe = 2(VH) = 54.4 V 11. φE = E × 4πr

2

Also φE = 0

q

ε

∴ 0

q

ε= E × 4πr

2

q = 4πε0Er2 = 4πε0(Ar)r

2 = 4πε0Ar

3

Substituting values,

q = 3

9

100 (0.20)

9 10

××

C = 8.89 × 10–11

C

12. Work done = change in potential energy

= 0

1

4πε 1 2

1

q q

r –

0

1

4πε1 2

2

q q

r

= 0

1

4πε q1q2

1 2

1 1

r r

−

= 9 × 109 × 12 × 10

–6 × 8 × 10

–6

× 2 2

1 1

4 10 10 10− −

− × ×

= 9 × 96 × 10–3

× 100 100

4 10

−

= 39 96 10 600

40

−× × ×

= 13 J 13. C = 4πε0R

C = 4 × 3.142 × 8.85 × 10−12

× 0.9

= 100.10 × 10−12

∴ C = 100 pF

P

σ0(xy)

(0, 0) O

Electrostatics01

1

Chapter 02 : Current Electricity

1. I = q

t =

n e

t

n = 1 million = 106

∴ I = 6 19

3

10 1.6 10

10

−

−

× ×

= 1.6 × 10–10

A. 2. Resistivity depends only on the material of the

conductor.

3. 1 2R R 9+ = and 1 2

1 2

R R2

R R=

+ ⇒ 1 2R R 18=

2

1 2 1 2 1 2R R (R R ) 4R R− = + −

∴ R1 – R2 = 81 72 3− =

on solving,

R1 = 6 Ω, R2 = 3 Ω 6. Let I1 and I2 be the current through AB and

AD respectively.

To find current through galvanometer Ig,

applying Kirchhoff’s 2nd

law to loop ABDA,

− 5I1 − 10 Ig + 15 I2 = 0

∴ − I1 − 2Ig + 3I2 = 0 ….(i)

Applying Kirchhoff’s 2nd

law to loop BCDB,

−10 (I1 − Ig) + 20 (I2 + Ig) + 10 Ig = 0

∴ −10I1 + 10Ig + 20I2 + 20 Ig + 10Ig = 0

∴ −10I1 + 20I2 + 40Ig = 0

∴ −I1 + 2I2 + 4Ig = 0 ….(ii)

Subtracting equation (ii) from equation (i),

−I1 + 3I2 − 2Ig = 0

−I1 + 2I2 + 4Ig = 0

+ − −

I2 − 6Ig = 0 ∴ I2 = 6Ig ….(iii)

From equations (ii) and (iii),

−I1 + 2(6Ig) + 4Ig = 0

∴ −I1 + 12 Ig + 4Ig = 0

∴ I1 = 16Ig ….(iv)

Adding equations (iii) and (iv),

I1 + I2 = 6Ig + 16Ig

∴ I1 + I2 = 22Ig

∴ 2 = 22 Ig (Q I1 + I2 = 2)

∴ Ig = 2

22=

1

11 A

7. I = 4E E 3E 3 12

5r R 5r R 5(0.2) 20

− ×= =

+ + += 1.7 V

8. In balance condition, no current will flow

through the branch containing S.

9. R1000 = V2/750 and R200 = V

2/P;

Now, R1000 = R200 (1 + α × 800)

So, 2

V

750 =

2V

P(1 + 4 × 10

–4 × 800)

∴ P = 750 (1 + 0.32) = 990 W.

10. Drift velocity vd = I

nAe =

2

I 4

n D e

×π

i.e., vd ∝ 2

1

D

∴ d1

d2

v

v =

2

2

2

1

D

D =

21

2

= 1

4 12. When connected in series,

total resistance of 5 resistors = 5R

∴ Power dissipated = 2V

5R = 5

∴ 2V

R = 25 ….(i)

When connected in parallel,

total power = 5 2V

R

= 5 × 25 = 125 W

13. Drift velocity, vd = µE

µ = dv

E = d

v

V / l

VE

= Q

l

µ = 0.2 2

100

×

= 4.0 × 10

−3 m

2V−1

s−1

14. Let current through 6 volt battery be I1 and

5 volt battery be I2. The circuit can be drawn

as:

Applying Kirchhoff’s second law to loop

ABCDEFA,

8 (I1 + I2) + 2 I1 = 6

∴ 10 I1 + 8 I2 = 6

∴ 5 I1 + 4 I2 = 3 ….(i)

Applying Kirchhoff’s second law to loop

BCDEB,

8 (I1 + I2 ) + 2 I2 = 5

8 I1 + 10 I2 = 5 ….(ii)

I2

I1

8 ΩDC

B E

A Fr = 2 Ω

E = 6 volt

r = 2 Ω

E = 5 volt

Current Electricity02

2

2

Physics Vol-II (Med. and Engg.)

Multiplying equation (i) by 5 and equation (ii)

by 2 then subtracting,

1 2

1 2

1

25I 20I 1516I 20I 10

9I 5

+ =+ =

− − −=

∴ I1 = 5

A9

From equation (i),

I2 =

53 5

9

4

− × =

2

36 =

1

18A

∴ I1 = 5

9A and I2 =

1

18A

Current through external resistance,

I = I1 + I2 = 5

9+

1

18=

11

18A

18. P = I

2R. Current is same so P ∝ R.

In the first case, it is 3R, in second case it is

2R

3, in third case it is

R

3and in fourth case

the net resistance is 3R

2.

i.e., III II IV I

R R R R< < <

∴ III II IV I

P P P P< < <

19. I

A = nevd

∴ vd = I

Ane =

6 26 19

5

4 10 5 10 1.6 10− −× × × × ×

∴ vd = 1

40 1.6× =

1

64m/s

20. P

Q =

1

3 ⇒ Q = 3P

and P 40

Q 40

+

+ =

3

5

∴ P 40

3P 40

++

= 3

5

∴ P = 20 Ω and Q = 3P = 60 Ω 21. S = X || 10 = 8 Ω

∴ 10X

10 X+ = 8

∴ 10X = 80 + 8X

∴ 2X = 80

∴ X = 40 Ω 23. 0.9 (2 + r) = 0.3 (7 + r)

∴ 6 + 3r = 7 + r

∴ r = 0.5 Ω

24. R = 2

8

2 2

50 1050 10

A (50 10 )

−−

−

ρ ×= × ×

×

l= 10

–6 Ω

25. 3

1 1

3

2 2

R (1 t ) 50 (1 3.92 10 20)

R (1 t ) 76.8 (1 3.92 10 t)

−

−

+ α + × ×= ⇒ =

+α + ×

∴ t ≈ 167 °C

26. V

E =l

; E is constant (voltage gradient)

∴ 1 2

1 2

V V=

l l ⇒

1.1 V

140 180=

∴ 180 1.1

V 1.41V140

×= =

1

Chapter 03 : Magnetic Effect of Electric Current

2. The magnetic field B = µ0nI

= 4π × 10−7

× 4000

1/ 2 × 2

Nn

= Q

l

= 64 π × 10−4

T

= 2.01 × 10−2

T ≈ 2 × 10−2

T 3. The field is non-zero only inside the core

surrounded by the windings of the toroid. The

field outside the toroid is zero. 5. F = IlB sin θ = 1.2 × 0.5 × 2 × sin (90°) = 1.2 N 6. The force per unit length is,

2

0 2 IF

4 R

µ= ×

π

If R is increased to 2R and I is reduced to I / 2,

the force per unit length becomes,

2

0 2(I / 2)F

4 2 R

µ′ = ×

π=

2

0 2I 1 F.

4 R 8 8

µ× =

π

7. τ = N I A B sin θ

τ = 1 × 10 × 0.01 × 0.1 × sin 90° = 0.01 Nm 8. Area of coil, A = 8 cm × 6 cm = 48 cm

2

= 48 × 10−4

m2

Maximum torque τ = NBIA

= 2000 × 0.2 × 200 × 10−3

× 48 × 10−4

Nm

= 0.384 Nm ≈ 0.4 Nm.

9. S = G

n 1−, n =

50

10 = 5

12 = G

5 1− =

G

4

∴ G = 48 Ω 10. τ = NIABcos (60°) ⇒ τ = NIAB sin (90°− 60°)

= 500 × 0.2 × 4 × 10–4

× 10–3

×1

2

τ = 2 × 10–5

N-m

11. S = NAB

k

To increase the sensitivity of MCG, N, A and

B should be large but k should be small.

12. Current sensitivity = I

θ=

NAB

k

= 4

8

80 5 10 5

10

−

−

× × ×

= 20 × 106 rad/A

= 20 rad/µA

13. gI 10 S

I 100 S G= =

+

∴ 10 1

10 G 10=

+

∴ G = 100 − 10

∴ G = 90Ω

14. R = g

V

I – G

∴ 0 = 3

V

3 10−× – 100

∴ V = 100 × 3 × 10–3

= 0.3 V 15. The magnetic induction at the centre of a

circular coil of N turns, radius R and carrying

a current I is given by,

B = 0 NI

2R

µ

Here B1 = 0 NI

2R

µand B2 =

( )0N 3I

2R

µ

Now B = 2 2

1 2(B ) (B )+

= 0 NI

2R

µ(1 3)+

∴ B = 02 NI

2R

µ=

0 NI

R

µ

16. r = mv

qB =

v

(q / m)B

= 5

7

2 10

2.5 10 0.05

×

× ×

= 0.16 m = 16 cm

18. S = I

θ =

1

K =

NAB

k

S ∝ N

S

S

′ =

N

N

′

N′ = 125

100 × 48 =

5

4 × 48 = 60

19. RS = g

V

I − G =

15

0.5 − 2

= 150

5− 2

= 30 − 2

= 28 Ω

Magnetic Effect of Electric Current 03

2

2

Physics Vol-II (Med. and Engg.)

20. B1 = 0 I

.4 a

µ θπ

, B2 = 0 I

.4 b

µ θπ

,

as b < a

B2 > B1

∴ Field due to ABCD = B2 − B1 = 0I 1 1

4 b a

µ θ − π

21. The field induction at O due to straight part of

conductor is, B1 = 0

4

µπ

2I

R

The field induction at O due to circular coil is,

B2 = 0

4

µπ

2 I

R

π

Both the fields will act in opposite direction,

hence the total field of induction at O will be,

B = B2 – B1 = 0

4

µπ

2I

R (π – 1)

∴ B = 0I

2 R

µπ

(π – 1)

1

Chapter 04 : Magnetism

1. M ∝ iA 2. M = iA = 12 × 7.5 × 10−4 = 9 × 10−3 Am2 As current in loop is clockwise, direction of

magnetic dipole moment vector is downward. 3. The magnetic dipole moment of the earth, M = iA ∴ M = i(πR2)

∴ i = 2

M

Rπ =

21

12

6.4 10

3.14 6.4 6.4 10

×× × ×

∴ i = 4.97 × 107A ∴ ≈ 5 × 107 A 4. Magnetic dipole moment, M = NiA = Ni.(πr2)

= 5 × 10 × 22

7 ×

7

100 ×

7

100

∴ M = 0.77 Am2 The direction of M is perpendicular to the

plane of the coil. Hence it is along the Z-axis.

5. Mo = e v r

2, Lo = m v r

∴ e

2 m= gyromagnetic ratio = o

o

M

L

6. ν = 6.8 × 109 MHz = 6.8 × 1015 Hz r = 1.06/2 = 0.53 Å = 0.53 × 10–10 m M = i A = e ν π r2

= (1.6 × 10–19) × (6.8 × 1015) ×22

7

× (0.53 × 10–10)2 ∴ M = 9.7 × 10–24 A-m2

7. Magnetic field intensity,

Ba = 0

4

µπ

3

2M

rand Be = 0

4

µπ

3

M

r

∴ B ∝ Mr−3 n = – 3

8. Baxis = 0

4

µπ×

3

2M

d= B

Bequator = 0

4

µπ×

3

M

d =

1

20

3

2M

4 d

µ × π =

1

2Baxis

But, Baxis = B

∴ Bequator = B

2

9. Ba = 0

4

µπ

× 3

2M

r = 10–7 ×

2 3

2 4

(10 10 )−

×

×

∴ Ba = 8 × 10–4 T.

10. Be = 0

4

µπ

.3

M

r= 10–7 ×

3

0.2

(0.08) = 3.9 × 10–5 T

11. Torque on a bar magnet in earth’s magnetic

field (BH) is HMB sin ,τ = θ τ will be maximum

if sin θ = maximum i.e., θ = 90o. 12. τ = MB sin θ ∴ τ = (mL) B sin θ ∴ 25 × 10−6 = (m × 5 × 10−2) × 5 × 10−2 × sin 30 ∴ m = 2 × 10−2 A-m

13. From τ = M B sin θ, M = B sin

τ

θ

∴ M = 24.5 10

0.25 sin30

−×× °

= 0.36 JT–1

15. Angle between geographic meridian and

magnetic meridian = Declination i.e., θ = 58′ Also, angle of dip (δ) is given by,

tan δ = V

H

B 0.3

B 3= ×

4

4

10

0.3 10

−

−×=

1

3

∴ δ = 30°

16. tan δ = V

H

B

B

tan δ′ = V

H

B

B′ = V

H

B

B cos45°

tan δ′ = tan tan30

cos45 cos45

δ °=° °

= 2

3= 0.8164

∴ δ′ = 39°14′. 17. When the magnet is placed with its N pole

towards north of earth, neutral points are obtained on the equatorial line and B2 = BH i.e.,

02 2 3/ 2

M

4 (d )

µ

π + l = BH

∴ BH = 7

2 2 3/ 2

10 0.4

(0.1 0.05 )

− ×+

= 2.86 × 10−5 T

∴ BH = 0.286 G 19. According to Curie law, for a paramagnetic

material,

χ ∝ T

1 ∴ 2

1

χχ

= 1

2

T

T

∴ 3

χ

χ =

2

273 27

T

+ =

2

300

T

∴ T2 = 300 × 3 = 900 K ∴ T2 = 900 − 273 = 627 °C

Magnetism 04

2

Physics Vol-II (Med. and Engg.)

2

20. For a temporary magnet, the hysteresis loop should be long and narrow.

21. I = M

V =

6

3

4 2 1.25 10−× × ×

= 3 × 105 A-m2/m3

∴ I = 3 × 105 A/m 22. M = Niπr2 = 300 × 15 × π × (7 × 10−2)2

∴ M = 69.27 Am2

23. The point P lies on axial line of magnet N1S1.

∴ B1 = 0 13

1

2M

4 (O P)

µ⋅

π =

7

3

10 2 12.5

(0.05)

− × ×

∴ B1 = 0.02 T (along PX) The point P lies on the equatorial line of the

magnet N2S2.

∴ B2 = 0 23

2

M

4 (O P)

µ⋅

π =

7

3

10 12.5

(0.05)

− ×

∴ B2 = 0.01 T (along PY) The resultant magnetic field at point P,

B = 2 21 2B B+ = 2 2(0.02) (0.01)+

∴ B = 2.236 × 10−2 T

24. →

τ = M→

× B→

= MB sin θ At θ = 90°, τmax = MB = 10−4 Nm At θ = 30°, τ = MB sin 30°

∴ τ = 10−4 × 1

2=

1

2× 10−4 Nm

25. µ = B 1.6

H 1000= = 1.6 × 10−3 Tm A−1

µr = 3

70

1.6 10

4 10

−

−

µ ×=

µ π× ≈ 1.3 × 103

χ = µr − 1 = 1.3 × 103 − 1 = 1.299 × 103 ≈ 1.3 × 103

5 cm 5 cm

Y1

B2 B

θ

S1 O1 N1B1P O2

X

S2

N2

1

Chapter 05 : Electromagnetic Induction and

Alternating Current

2. The resistance of copper loop is less than that

of aluminium loop, hence induced current will

be more in the copper loop as compared to that

of the aluminium loop.

3. Charge induced, q = d

R

− φ

= (0 NBA) NBA

R R

− −=

= 450 0.02 100 10

2

−× × ×

∴ q = 5 × 10−3

C.

5. e = MdI

dt

For e = M, dI

dt= 1

6. N1 = 50 turns/cm = 5000 turns/metre = 5000 l

turns in l metre

N2 = 200

A = 4 cm2 = 4 × 10

−4 m

2

M = 0 1 2N N Aµ

l

= 7 44 10 5000 200 4 10− −π× × × × ×l

l

∴ M = 5.024 × 10−4

H

7. XC = 1

2 Cπν

XC ∝ 1

ν

C 2

C 1

(X )

(X ) = 1

2

ν

ν =

50

200 =

1

4

(XC)2 = 4

4 = 1 Ω

8. XL = ωL = 2πνL

L = LX

2πν =

50 7

2 22 50

×

× × = 0.16 H

9. I = RV 36

R 90= = 0.4 A

As V2 = 2 2

R LV V+

∴ VL = 2 2

RV V− = 2 2120 36− = 114.5 V

As VL = I(XL)

∴ XL = LV 114.5286.2

I 0.4= = Ω

As XL = 2πνL

∴ L = LX 286.2

2 2(3.14)60=

πν= 0.76 H

10. The resonant frequency is given by,

ν0 = 1

2 LCπ=

6 8

1

2 100 10 4 10− −×π × × ×

= 6

1

2 2 10−π× ×

∴ ν0 = 610

4π=

25

π× 10

4 Hz

11. Z = ( )22

L CR X X R+ − =

Hence I = 100

10 = 10 A.

12. E = 2 2

R C LV (V V )+ −

= 2 2(40) (80 40)+ −

∴ E = 1600 1600+ = 2(1600) = 40 2 V.

13. ν0 = 1

2 LCπ =

( )6

1

2 5 80 10−

π × ×

= 50

2π =

25

π Hz

14. E = E0 sin (ωt + φ)

Erms = 0E

2=

200

2

Power, P = Erms Irms cosφ

∴ Irms = rms

P

E cos× φ =

1000 2

200 cos60× °= 10 2 A

15. E0 = NABω and I0 = 0E

R

∴ I0 = NAB

R

ω =

1000 2 0.2 60

6000

× × × = 4 A

18. As M = 0 1 2N N Aµ

l

∴ M becomes 4 times.

19. tan φ = LX

R =

2 fL

R

π=

400 (15/16 )

300

π× π =

5

4

∴ φ = tan–1

(5/4)

20. P

S

N

N = P

S

E

E ⇒

S

75

N =

120

2400 =

1

20

∴ NS = 75 × 20 = 1500.

Electromagnetic Induction and Alternating Current 05

1

Chapter 06 : Electromagnetic Waves

1. uE = 2

0

1E

2ε

= 121

8.85 102

−× × × (170)2

∴ uE = 1.28 × 10−7

J/m3

3. Here, By = 10

–6 sin[7 × 10

11 t + 400 πx]

The Y-component of the magnetic field is

given by,

By = B0 sin 2πt

T

+ λ

x

Comparing the given equation with the above

equation,

2π

λ = 400 π

∴ λ = 2

400 = 5 × 10

–3 m

4. λ = c

ν =

8

6

3 10

3.4 10

××

= 88.24 m

8. The pressure exerted is given as,

Pressure (P) = I

c =

8

1.2

3 10× =

8

2

N0.4 10

m

−×

10. Suppose the charge on the capacitor at time t

is Q, the electric field between the plates of

the capacitor is 0

QE .

A=ε

The flux through the area considered is

φE = 0

Q.4A

Aε =

0

4Q

ε

∴ The displacement current,

ID = E0

d

dt

φε = 0

0

4 dQ

dt

ε

ε = 4I.

11. B0 = 0E

c =

8

90

3 10× = 3 × 10

–7 T

12. Charge oscillating sinusoidally is given by

Q = q0 sin ωt

Displacement current,

ID = dq

dt = q0 ω cos ωt

(ID)max = q0 ω = q0 × 2πν

= 2 × 10–6

× 2 × 3.14 × 8 × 105

≈ 10 A

14. E = Bc = 40 × 10−6

× 3 × 108 = 12,000 V/m.

15. v = r r

c

µ ε

= 83 10

1.42 2.31

×

×= 1.65 × 10

8 m/s

Electromagnetic Waves 06

1

Chapter 07 : Ray Optics

1. All others give the images which are either

diminished or of same size. 4. Brightness depends upon aperture or size

while image formation is independent of size

of the mirror.

5. µ = c

v =

100

70 = 1.43

6. m = f

f u− ∴ – 4 =

f

f ( 12)− − ∴ f = –9.6 cm

R = 2 × f = – 19.2 cm

7. 1

v +

1

u =

1

f

∴ 1

v =

1

f –

1

u =

1

18−–

1

27− =

1

54

−

∴ v = – 54 cm

Negative sign of v indicates the screen should

be placed in front of the concave mirror as it is

a real image.

8. Q m = v

u− also

1 1 1

f v u= + ⇒

u u1

f v= +

⇒ u u

1v f

− = −v f

u f u

−⇒ =

− so

fm

f u=

−

9. µg = c

v ∴ v =

g

c

µ =

83 10

1.5

× = 2 × 10

8 ms

–1

v = Thicknessof glassplate(d)

Timerequired to travel through it (t)

∴ t = d

v=

3

8

2 10

2 10

−×

× = 10

–11 s

10. µ = c

v =

8

8

3 10

1.5 10

×

× = 2

ic = sin–1

1

µ = sin

–1 1

2

ic = 30°

11. Lens formula 1

v –

1

u =

1

f

u is always negative, v is positive.

12. m = f

f u+ ∴ – 2 =

1

31

u3+

∴ u = – 0.5 m

−ve sign follows from the sign conventions.

13. f = 1

P =

1

25 = 0.04 m = 4 cm

M.P. = D

1f

+

= 25

14

+

= 7.25

14. M = – o

e

f

f ∴ – 10 = – o

e

f

f

∴ fo = 10 fe and fo + fe = 44

fe = 4 cm ∴ fo = 40 cm.

15. 2

v

µ – 1

u

µ = 2 1

R

µ −µ

∴ 1

v –

1.5

3− =

1 1.5

5

−

− ∴ v = – 2.5 cm

16. 1

f = (µ – 1)

1 2

1 1

R R

−

∴ 1

20+ = (1.55 – 1)

1 1

R R

− −

∴ R = 40 × 0.55 = 22.0 cm

17. wµg = 1.54

1.33, ic = sin

–1

1.33

1.54

= 59° 42′ 18. P = 15.5 – 5.5 = 10 D

∴ f = 1

P = 0.10 m

19. δ = A (µ – 1), δb = A (µb – 1), δr = A (µr – 1)

∴ D2 = A (1.525 – 1)

∴ D1 = A (1.520 – 1)

⇒ D2 > D1

20. M = – o

e

f

f Negative sign for inverted image.

∴ – 5 = – o

e

f

f

∴ fo = 5 fe and fo + fe = 24

∴ fo = 20 cm, fe = 4 cm 21. For the person to see nearby objects clearly

(at 25 cm), the image should be formed at

75 cm by the corrective lens. The image will

then act as an object and the final image will

be formed on retina. The corrective lens is

convex lens.

u = – 25 cm, v = – 75 cm

1

v –

1

u =

1

f ∴

1

75− –

1

25− =

1

f

∴ f = 37.5 cm = 0.375 m

P = 1

f (m) =

1

0.375 = + 2.67 D

22. 1

f = P = (µ – 1)

1 2

1 1

R R

−

∴ P = (1.5 – 1) 1 1

R R

− − +

∴ P = – 0.5 × 2

R = −0.5 ×

2

0.3 = –

10

3 D

Ray Optics07

1

Chapter 8 : Wave Optics

1. As medium changes there is no change in

frequency whereas wavelength in medium

become 1

µ times in air.

3. The ray near the base has to travel larger

distance through prism hence it is delayed than

the ray near the apex which has to travel very

less distance through prism, hence emerges

first.

7. cg = a

g

c

µ=

83 10

1.5

× = 2 × 10

8 m/s

t = g

d

c =

8

8

4 10

2 10

×

× = 2 s

8. w

g

c

c = wµg =

g

w

µ

µ

∴ cw = g

w

µ

µ × cg =

( )

( )

3 / 2

4 / 3× 2 × 10

8

∴ cw = 8910 m/s

4×

9. Angle made with surface = 30°

∴ i = 90° − 30° = 60°

1.5 = sini

sin r

∴ sin r = sin i

1.5 =

sin60

1.5

° = 0.5773

∴ r = 35°17′

Ratio of the width

2

1

W

W=

cos r

cos i =

cos 35 17

cos 60

′°

° =

0.8163

0.5 = 1.633

10. ∆λ = 706 − 656 = 50 nm = 50 × 10

−9 m

As v

c

∆λ=

λ

∴ v = c∆λ

×λ

= 9

8

9

50 103 10

656 10

−

−

×× ×

×

= 2.3 × 107 ms

−1

Wave Optics08

1

Chapter 09 : Interference of light

1. The phase difference of 2π corresponds to

path difference of λ. Hence the phase difference of δ corresponding to path difference x will be

2π

λ =

x

δ or

2 xπ

λ = δ

2. 93 λ = 0.0465 mm = 465 × 10–7 m

∴ λ = 746510

93−× = 5 × 10–7 m = 5000 Å

3. 1

2

I

I =

2122

a

a =

25

4

∴ 1

2

a

a =

5

2

max

min

I

I=

21 2

21 2

(a a )

(a a )

+

− =

( )

( )

2

2

5 2

5 2

+

− =

49

9 4. Fringes overlap so that it looks like uniform

illumination. 6. Different points in filament emit light

independently and have no fixed phase relationship. Hence, interference is not possible.

8. β = β′

∴ D

d

λ =

D

2d

′λ

∴ D′ = 2D 9. yb = yr ∴ (n + 1)λb = nλr ∴ (n +1) × 5 × 10–5 = n × 7.5 × 10–5 ∴ 5n + 5 = 7.5 n ∴ 2.5 n = 5 ∴ n = 2

10. β = D

d

λ

β ∝ D ∴ Decrease in D by 25% will bring 25%

decrease in β.

11. yn = n D

d

λ and ny′ =

n D

d

′ ′λ

As they coincide yn = ny′

∴ n

n′ =

′λ

λ =

900

750 =

6

5

5y′ = y6 = n D

d

λ =

9

3

6(750 10 )2

(2 10 )

−

−

×

×= 4.5 mm

12. For incoherent waves, Imax = nI0

∴ n = max

0

I

I =

32

2 = 16

Interference of Light 09

1

Chapter 10 : Diffraction and Polarisation

of light .11.3

1. 2

1

x

x= 2

1

λ

λ =

3600

5400

∴ x2 = x12

1

λ λ

= (0.3) × 3600

5400 = 0.2 mm

6. dθ = 1′ =

o1

60

= 180 60

π ×

radian

If d is the actual distance between the pillars

and D is distance between the pillars and the

person then,

d

D = dθ

∴ d = D dθ = 15000 × 180 60

π×

≈ 4.36 m

7. According to Brewster’s law,

µ = tan ip

∴ ip = tan–1

(µ)

8. dmin = 10

3

1.22 x 1.22 6600 10 5

D 3 10

−

−

λ × × ×=

×

= 2.01 × 10–3

m

9. R.P. = 7

D 4.2

1.22 1.22 6 10−=

λ × ×

= 6710

1.22× = 5.74 × 10

6

10. a sin θ = λ ∴ a = sin

λθ

= 75 10

sin1

−×°

= 75 10

0.0175

−×

∴ a = 0.028 mm 11. Depending upon amount of diffraction

resolution of an optical instrument is observed. 12. Intensity of light transmitted by the first

polariser is half of the intensity of unpolarised

light = 18 Wm–2

13. The width of central maxima = 2 D

d

λ

= 7

3

2

2 2.1 5 101.4 10 m 1.4 mm

0.15 10

−−

−

× × ×= × =

×

14. R.P. of telescope ∝ (diameter of aperture) 16. According to Brewster’s law,

µ = 1.57 = tan ip

∴ ip = 57.5°, r = 90° – 57.5° = 32.5° 17. Sound waves cannot be polarised because they

are longitudinal. Light waves can be polarised

because they are transverse.

18. Using law of Malus,

Intensity of light transmitted from 1st polaroid,

I1 = A2 cos

2 θ = A

2 cos

2 40°

∴ Intensity of emergent light,

I2 = I1 cos2 40°

= A2 cos

2 40 cos

2 40

= 0.34 A2

Diffraction and Polarization of Light 10

1

Chapter 11 : Dual Nature of Matter and

Radiation

1. Light is packets of energy called as quantas

which is photon. 4. Stopping potential is independent of intensity.

6. E = φ + K.Emax or hc

λ = φ + eV0

∴ λ = 0

hc

eVφ+

7. Einstein’s photoelectric equation is

21mv

2= hv − φ

φ = hν – 21mv

2

9. Given

max

1

hcK.E. = −φ

λ ….(i)

and max

2

hc2K.E. = − φ

λ ….(ii)

Dividing equation (ii) by equation (i), we get

2

1

h c

2h c

− φ λ =

− φ λ

which gives 2 1

1 2

hc(2 )φ = λ −λ

λ λ

10. λ = h

p=

h

mv.

11.

12. λ0 = hc

φ =

( ) ( )34 8

19

6.6 10 3 10

4.125 1.6 10

−

−

× × ×

× ×

= 3000 × 10−10 m = 3000 Å

13. λ ∝ 1

v ⇒ If v = 0, λ = ∞

14. K.E. = eV = 1.602 × 10–19 × 100 = 1.602 × 10–17 J 15. E = 180 eV = 180 × 1.6 × 10−19 = 2.88 × 10−17 J

Now λ = h

2mE

∴ λ = 34

131 17 2

6.6 10

2 9 10 2.88 10

−

− −

×

× × × ×

= 34

24

6.6 10

7.2 10

−

−

××

= 0.916 × 10−10 m = 0.9 Å

I

50°

V = 54V

φ 0

Dual Nature of Matter and

Radiation 11

1

Chapter 12 : Atoms and Nuclei

1. Energy released = E4 − E1

= −2

13.6

4−

2

13.6

1

−

= 12.75 eV 3. In each α-emission, the mass number

decreases by 4 and atomic number decreases

by 2. In each beta emission, the mass number

remains unchanged, but atomic number

increases by 1. 4. T= 50.8 days

∴ λ = 0.693

T=

0.693

50.8 = 1.36 ×10

−2 per day

N = 40% of N0 = 0.4 N0

As, 0N

N= e

λt

∴ ln 0N

N

= λt

∴ t =

0Nn

N

λ

l

= ( )

2

n 2.5

1.36 10−×

l=

2

0.916

1.36×10−

= 67.37 days

6. Energy required 2 2

13.6 13.6

n 10= = = 0.136 eV

7. Shortest wavelength comes from n1 = ∞ to

n2 = 1 and longest wavelength comes from

n1 = 6 to n2 = 5 in the given case. Hence,

2 2

min

1 1 1R

1

= − λ ∞ = R

∴ 2 2

max

1 1 1 36 25 11R R R

5 6 25 36 900

− = − = = λ ×

∴ max

min

λλ

=

900

11R

R

× =

900

11

8. E3 = − 13.6

9= − 1.51 eV

E4 = −13.6

16= − 0.85 eV

∴ E4 − E3 = 0.66 eV 10. Reaction (C) is incorrect because the mass

number is not conserved.

12. R = Ro 1

3A = 1.1 × 10−15

× 1

3(16)

= 2.77 × 10−15

m 14. For the ionization of second He electron, He

+

will act as hydrogen like atom.

Hence ionization potential = Z2 × 13.6

= (2)2 × 13.6

= 54.4 eV

16. 0

21115 115

48 50Cd Sn

β − →

17. nα = A A 200 168 32

84 4 4

′− −= = =

n β = (2nα − Z + Z′) = (16 − 90 + 80)

= 16 − 10 = 6

18. Wave number 1

ν =λ

= 8

1

5896 10−× = 16961 per cm 19. Two months = 2 half-lives. The activity of the

sample will become 2

1

2, i.e., one-fourth in

2 months. 20. As charge number is fixed (= 92), therefore,

number of protons and electrons is same. As

atomic weight is greater by 3, therefore, 238

92 U

contains 3 more neutrons. 21. Penetration power of γ is 100 times of β, while

that of β is 100 times of α. 22. For Lyman series,

Lyman 2 2

max

c 1 1 3RcRc

(1) (2) 4

ν = = − = λ

For Balmer series,

Balmer 2 2

max

c 1 1 5RcRc

(2) (3) 36

ν = = − = λ

∴ Lyman

Balmer

27

5

ν=

ν

23. For ground state, n = 1

For first excited state, n = 2

As rn ∝ n2

∴ radius becomes 4 times. 24. E = mc

2 = (1 × 10

–3)(3 × 10

8)

2 = 9 × 10

13 J

25. nα = A A

4

′−=

232 2086

4

−=

nβ = (2nα – Z + Z′) = (12 − 90 + 82) = 4 27. Nuclear fusion takes place in stars which

results in joining of nuclei accompanied by

release of tremendous amount of energy.

28. As is known, P.E. = −2 K.E. or P.E.

K.E. = −2

Atoms and Nuclei12

1

Chapter 13 : Electronic Devices

1. Donor atoms occupy the energy levels between conduction and valence band placed towards conduction band.

4. For intrinsic semiconductor , ne = nh 5. Both the impurities possess 3 valence

electrons. 6. In case of half-wave rectifier, we get output

current only for positive input as diode is forward biased.

10.

15. Id.c. = 61

π= 19.4 mA

∴ d.c. power output = Id.c.2 × RL

= (19.4 × 10−3)2 × 800 = 0.3 watt 16. P = VI = 9.1 × 40 = 364 mW 17. Vo = Vz = 8 volt 18. The breakdown field = 106 V/m, width of

depletion region = 2.5 × 10–6 m ∴ Vbreakdown = E × d = 106 × 2.5 × 10−6 ∴ Vbreakdown = 2.5 V 19. VBB = IBRB + VBE

∴ RB = BB BE

B

V V

I

− = 6

5.5 0.5

10 10−

−

× = 500 kΩ

20. Av = L

i

R

r

β and Ap =

2L

i

R

r

β = βAv

∴ p

v

A

A = β = 62

21. IE = IC + IB = 25 + 1 = 26 mA

∴ α = C

E

I

I =

25

26

22. I = 3

150 50

5 10

−

× = 20 mA

IL= 5 mA ….[Given] ∴ Zener current IZ = 15 mA

23. IB = i

i

V

R =

0.01

1000 = 0.01 × 10−3 A = 0.01 mA

∴ IC = βIB = 50 × 0.01 = 0.5 mA = 500 µA

24. α = 0.95

∴ β = 1

α

−α=

0.95

1 0.95− = 19

Also, β = C

B

I

I

∆

∆

∴ ∆IC = β(∆IB) = 19 × 0.4 = 7.6 mA 25. The energy of emission,

E = hν = hc

λ =

34 8

10

6.62 10 3 10

5890 10

−

−

× × ×

×

= 3.37 × 10−19 J

= 19

19

3.37 10

1.6 10

−

−

×

× = 2.11 eV

∴ For, λ = 5890 Å, E = 2.11 eV ∴ The condition for emission of electrons is, hν > Eg. But here, hν < Eg [Eg = 3eV] ∴ For emission of electrons, λ < 5890 Å is a must.

B

E C

+

−

−

+Forward Biased

Reversed Biased

Electronic Devices13

1

Chapter 12 : Atoms and Nuclei

1. ν = 1

2 LCπ =

9 6

1

2 2 10 45 10− −π × × ×

= 710

6π = 0.53 MHz

2. Maximum distance covered by space wave

communication

= 6 42Rh 2 6.4 10 100 12.8 10 m= × × × = ×

< 4 × 104 m

= 40 km

Since the distance between receiver and

transmitter is 80 km, the wave can be only

propagated by a sky wave.

∴ νc = ( )1/2

max9 N

= ( )1/ 212

9 10

∴ νc = 9 MHz 6. fSB = fc ± fm = 1800 ± 0.8 = 1800.8 kHz and

1799.2 kHz

7. c c

2Lν= =

λ L

2

λ = Q

83 10

2 0.3

×=

× = 500 MHz.

8. Time = 3

2

8

4000 101.33 10 s

3 10

−×= ×

× ≈ 13 ms

10. Number of stations

= B.W.

2 Highest modulating frequency×

= 240,000

2 12000×

= 10 11. d = 2hR

= 62 250 6.4 10× × ×

= 56.568 × 103m ≈ 56.6 km

12. d = 2Rh

Population covered

= πd2 × population density

= π(2Rh) × ρ

= 22

7× 2 × 6400 × 0.1 × 4000

= 1.6 × 107

13. Since, total power

2

a

PtPcm

12

= +

Here, ma = 0.6

∴ Pc = 2

8 8

(1 0.18)0.61

2

= ++

= 8

6.8 kW1.18

=

Communication Systems14