download neet ug/ jee(main) physics book (absolute vol 2

1

Chapter 01 : Electrostatics

1. q = ne

n = 18

19

5.8 101.6 10

= 36.25

As n is not an integer, such a charge cannot exist.

2. Fnet = 1 2

20

1 q q4 r

(110/100) (90/100) times

i.e., 99100

times

Net force = 99100

100 = 99 N 3. Force on – q1 due to q2 is

1 212 2

Cq qFb

along X-axis

Force on – q1 due to q3 is

1 313 2

Cq qFa

at with negative direction of

Y-axis. x component of force on – q1 is

Fx = F12 + F13 sin = Cq12 32 2

q q sinb a

i.e., Fx 2 32 2

q q sinb a

4. 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.

5. At neutral point,

2 2 2 20 0

1 20 1 Q4 (20 10 ) 4 (40 10 )

Q = 80 C 6. = pE sin

p = 26

4

9 10Esin 10 sin30

p = 1.8 10–29 Cm

7. According to Gauss’ law,

= 0

q

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.

8. V =

0

14

qR

V1 = 0

14

1

1

qR

and V2 = 0

14

2

2

qR

Both the sphere having same potential V1 = V2

1

1

qR

= 2

2

qR

1

2

qq

= 1

2

RR

9. V =

0

1 q4 r

i.e., V q …{ r is constant }

He

H

VV

= He

H

qq

= 21

VHe = 2(VH) = 54.4 V 10. E = E 4r2

Also E = 0

q

0

q

= E 4r2

q = 40Er2 = 40(Ar)r2 = 40Ar3

Substituting values,

q = 3

9

100 (0.20)9 10

C = 8.89 10–11 C 11. Work done = change in potential energy

= 0

14

1 2

1

q qr

–0

14

1 2

2

q qr

= 0

14

q1q2 1 2

1 1r r

= 9 109 12 10–6 8 10–6

× 2 21 1

4 10 10 10

= 9 96 10–3 100 100

4 10

= 39 96 10 600

40

= 13 J

P 0(xy)

(0, 0) O

Electrostatics01

2

Absolute Physics Vol - II (Med. and Engg.)

F2 F1

q1

x

q q d2

d2

12. At surface,

E = 0

= 6

12

1.77 108.85 10

= – 2 105 NC–1

13. The two forces experienced

by q1 are

F1 = F2 = 12

0 2

1 qq4 d

2

x

The horizontal components of F1 and

F2 will cancel each other.

Hence net vertical force on q1 will be

F = F1 cos + F2 cos = 1

1/ 22 20 2 2

2 qq4 d d

2 2

x

x x

=

32 2

21

0

2qq d4 2

x x

For F to be maximum, dFdx

= 0.

This gives x = d/2 2

3

Chapter 02 : Capacitors

1. As qA

, to increase the charge density, area

‘A’ should be made very small. 2. C = 4 k0R = 4 3.14 1 8.85 1012 6408 103 712.7 F 3. When two capacitors are connected in series

combination

S

1C

= 1

1C

+ 2

1C

= 1 2

1 2

C CC C

CS = 1 2

1 2

C CC C

= 154

F

15 (C1 + C2) = 4 C1 C2 ….(i) when two capacitors are connected in parallel

combination C1 + C2 = 16 ….(ii) Putting equation (ii) in equation (i) 15 16 = 4 C1 C2 15 4 = C1 C2 60 = C1 C2 C1 (16 –C1) = 60 ….From (ii) C1

2 – 16C1 + 60 = 0 C1

2 –10C1 – 6C1 + 60 = 0 (C1 – 10) (C1 – 6) = 0 C1 = 10 F or C1 = 6 F That means value of capacitors are 6 F and

10 F 4. Initial energy Ui = 1

2 C1V1

2 + 12

C2V22

If both the capacitors are connected in parallel their combination capacitance is

C = C1 + C2 Let q = q1 + q2 CV = C1V1 + C2V2 (C1 + C2) V = C1V1 + C2V2

V = 1 1 2 2

1 2

C V C VC C

Final energy, Uf = 12

CV2

Loss in energy = Ui – Uf

= 12

C1V12 + 1

2 C2V2

2 – 12

CV2

= 12

C1V12 + 1

2 C2V2

2

– 12

(C1 + C2) 2

1 1 2 2

1 2

C V C VC C

= 12

C1V12 + 1

2 C2V2

2

– 12

2 2 2 21 1 1 2 1 2 2 2

1 2

C V 2C C V V C VC C

= 12

2 2

1 2 1 1 2 2

1 2

C C (V 2V V V )C C

= 2

1 2 1 2

1 2

C C (V V )2(C C )

5. With temperature rise, dielectric constant of liquid decreases.

6. Capacitance C = . Since potential V

decreases, C will increase. Reason is also correct but these are two independent statements.

7. C = QV

= 4010 ( 10)

= 4020

= 2 F 8. U = 1

2CV2 = 1

20Ad V2

At any instant, let the separation between plates be x

U = 12

0Ax V2

dUdt

= 120 AV2(–1) 2

1x

dxdt

= – 12

20

2

AVx

(v)

i.e., potential energy decreases as (1/x2). 9. U = 1

2 CV2

V = 2UC

= 6

2 50100 10

= 610

= 103 V = 1000 V 10. C = 0Ak

d

qV

Capacitors02

4

4


1. I = q

t = n e

t

n = 1 million = 106

I = 6 19

3

10 1.6 1010

= 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 R 2R R

1 2R R 18

21 2 1 2 1 2R R (R R ) 4R R

R1 – R2 = 81 72 3 on solving, R1 = 6 , R2 = 3 6. I = 4E E 3E 3 12

5r R 5r R 5(0.2) 20

= 1.7 V 7. In balance condition, no current will flow

through the branch containing S. 8. R1000 = V2/750 and R200 = V2/P; Now, R1000 = R200 (1 + 800)

So, 2V

750 =

2VP

(1 + 4 10–4 800)

P = 750 (1 + 0.32) = 990 W. 9. Drift velocity vd = I

nAe = 2

I 4n D e

i.e., vd 2

1D

d1

d2

vv

= 2221

DD

= 21

2

= 14

11. 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

12. Drift velocity, vd = E

= dvE

= dvV / l

…. VE

l

= 0.2 2100

= 4.0 103 m2V1s1

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

by 2 then subtracting,

1 21 2

1

25I 20I 1516I 20I 10

9I 5

I1 = 5 A9

From equation (i),

I2 = 53 59

4

= 2

36 = 1

18A

I1 = 59

A and I2 = 118

A

Current through external resistance,

I = I1 + I2 = 59

+ 118

= 1118

A 17. P = I2R. Current is same so P R. In the first case, it is 3R, in second case it is

2 R3

, in third case it is R3

and in fourth case

the net resistance is 3R2

.

i.e., III II IV IR R R R III II IV IP P P P

I2

I1

8 DC

B E

A Fr = 2

E = 6 volt

r = 2

E = 5 volt

Current Electricity03

5

Chapter 03 : Current Electricity

18. IA

= nevd

vd = IAne

= 6 26 19

54 10 5 10 1.6 10

vd = 140 1.6

= 164

m/s 19. P

Q = 1

3 Q = 3P

and P 40Q 40

= 35

P 403P 40

= 35

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

10X10 X

= 8

10X = 80 + 8X 2X = 80 X = 40 22. 0.9 (2 + r) = 0.3 (7 + r) 6 + 3r = 7 + r r = 0.5 23. R =

28

2 2

50 1050 10A (50 10 )

l = 10–6

24.

31 1

32 2

R (1 t ) 50 (1 3.92 10 20)R (1 t ) 76.8 (1 3.92 10 t)

t 167 C 25. VE

l; E is constant (voltage gradient)

1 2

1 2

V V

l l 1.1 V

140 180

180 1.1V 1.41V140

6

6


2. The magnetic field B = 0nI

= 4 107 40001/ 2

2 .... Nn

l

= 64 104 T = 2.01 102 T 2 102

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 = 1.2 0.5 2 sin (90) = 1.2 N 6. The force per unit length is,

2

0 2 IF4 R

If R is increased to 2R and I is reduced to I / 2, the force per unit length becomes,

2

0 2(I / 2)F4 2 R

= 2

0 2I 1 F.4 R 8 8

7. = N I A B sin = 1 10 0.01 0.1 sin 90 = 0.01 Nm 8. Area of coil, A = 8 cm 6 cm = 48 cm2 = 48 104 m2

Maximum torque = NBIA = 2000 0.2 200 103 48 104 Nm = 0.384 Nm 0.4 Nm. 9. S =

Gn 1

, n = 5010

= 5

12 = G

5 1 =

G4

G = 48 10. = NIABcos (60) = NIAB sin (90 60)

= 500 0.2 4 10–4 10–3 12

= 2 10–5 N-m 11. S =

NABk

To increase the sensitivity of MCG, N, A and B should be large but k should be small.

12. Current sensitivity = I

= NAB

k

= 4

8

80 5 10 510

= 20 106 rad/A = 20 rad/A

13. gI 10 SI 100 S G

10 110 G 10

G = 100 10 G = 90 14. R =

g

VI

– G

0 = 3

V3 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 NI2R

Here B1 = 0 NI2R and B2 =

0N 3I

2R

Now B = 2 21 2(B ) (B )

= 0 NI (1 3)2R

B = 02 NI2R

= 0 NIR

16. r = mv

qB = v

(q / m)B

= 5

7

2 102.5 10 0.05

= 0.16 m = 16 cm

18. S = I

= 1K

= NAB

k

S N

SS

= NN

N = 125100

48 = 54

48 = 60

19. RS = g

VI

G = 150.5

2

= 150

5 2

= 30 2 = 28

Magnetic Effect of Electric Current 04

7

Chapter 04 : Magnetic Effect of Electric Current

20. B1 = 0 I.4 a

, B2 = 0 I.4 b

,

as b < a B2 > B1

Field due to ABCD = B2 B1 = 0I 1 14 b a

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

conductor is, B1 = 0

4

2IR

The field induction at O due to circular coil is,

B2 = 0

4

2 IR

Both the fields will act in opposite direction, hence the total field of induction at O will be,

B = B2 – B1 = 0

4

2IR

( – 1)

B = 0I2 R

( – 1)

8


8

1. M iA 2. M = iA = 12 7.5 104 = 9 103 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(R2)

i = 2

MR

= 21

12

6.4 103.14 6.4 6.4 10

i = 4.97 107A 5 107 A 4. Magnetic dipole moment, M = NiA = Ni.(r2)

= 5 10 227

7100

7100

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

e2 m

= gyromagnetic ratio = o

o

ML

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

= (1.6 10–19) (6.8 1015) 227

(0.53 10–10)2 M = 9.7 10–24 A-m2

7. Magnetic field intensity,

Ba = 0

4

3

2Mr

and Be = 0

4

3

Mr

B Mr3 n = – 3 8. Baxis = 0

4 3

2Md

= B

Bequator = 0

4 3

Md

= 12

03

2M4 d

= 12

Baxis

But, Baxis = B

Bequator = B2

9. Ba = 0

4

3

2Mr

= 10–7 2 3

2 4(10 10 )

Ba = 8 10–4 T.

10. Be = 0

4

. 3

Mr

= 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 , will be maximum if sin = maximum i.e., = 90o.

12. = MB sin = (mL) B sin 25 106 = (m 5 102) 5 102 sin 30 m = 2 102 A-m 13. From = M B sin , M =

B sin

M = 24.5 10

0.25 sin30

= 0.36 JT–1

15. Angle between geographic meridian and

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

tan = V

H

B 0.3B 3

4

4

100.3 10

= 1

3

= 30 16. tan = V

H

BB

tan = V

H

BB

= V

H

BB cos45

tan = tan tan30cos45 cos45

= 2

3= 0.8164

= 3914. 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

M4 (d ) l

= BH

BH = 7

2 2 3/2

10 0.4(0.1 0.05 )

= 2.86 105 T

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

material,

T 2

1

= 1

2

TT

3

= 2

273 27T =

2

300T

T2 = 300 3 = 900 K T2 = 900 273 = 627 C

Magnetism 05

9

Chapter 05 : Magnetism

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

21. I = MV

= 6

34 2 1.25 10

= 3 105 A-m2/m3

I = 3 105 A/m 22. M = Nir2 = 300 15 (7 102)2

M = 69.27 Am2

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

B1 = 0 13

1

2M4 (O P)

= 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

M4 (O P)

= 7

3

10 12.5(0.05)

B2 = 0.01 T (along PY) The resultant magnetic field at point P, B = 2 2

1 2B B = 2 2(0.02) (0.01) B = 2.236 102 T 24.

= M

B

= MB sin At = 90, max = MB = 104 Nm At = 30, = MB sin 30

= 104 12

= 12 104 Nm

25. = B 1.6

H 1000 = 1.6 103 Tm A1

r = 3

70

1.6 104 10

1.3 103

= r 1 = 1.3 103 1 = 1.299 103 1.3 103

5 cm 5 cm

Y1

B2 B

S1 O1 N1

B1P O2X

S2

N2

10

10

Physics Vol-II (Med. and Engg.)

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 = dR

= (0 NBA) NBAR R

= 450 0.02 100 10

2

q = 5 103 C. 5. e = MdI

dt

For e = M, dIdt

= 1 A/s 6. N1 = 50 turns/cm = 5000 turns/metre = 5000 l

turns in l metre N2 = 200 A = 4 cm2 = 4 104 m2

M = 0 1 2N N Al

= 7 44 10 5000 200 4 10 l

l

M = 5.024 104 H 7. XC = 1

2 C

XC 1

C 2

C 1

(X )(X )

= 1

2

= 50200

= 14

(XC)2 = 44

= 1 8. XL = L = 2L

L = LX2

= 50 72 22 50

= 0.16 H 9. I = RV 36

R 90 = 0.4 A

As V2 = 2 2R LV V

VL = 2 2RV V = 2 2120 36 = 114.5 V

As VL = I(XL)

XL = LV 114.5 286.2I 0.4

As XL = 2L

L = LX 286.22 2(3.14)60

= 0.76 H

10. The resonant frequency is given by,

0 = 12 LC

= 6 8

12 100 10 4 10

= 6

12 2 10

0 = 610

4= 25

104 Hz

11. Z = 22

L CR X X R

Hence I = 10010

= 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. 0 = 1

2 LC =

6

1

2 5 80 10

= 502

= 25

Hz 14. E = E0 sin (t + )

Erms = 0E2

= 2002

Power, P = Erms Irms cos

Irms = rms

PE cos

= 1000 2200 cos60

= 10 2 A 15. E0 = NAB and I0 = 0E

R

I0 = NABR = 1000 2 0.2 60

6000 = 4 A

18. As M = 0 1 2N N A

l

M becomes 4 times. 19. tan = LX

R = 2 fL

R = 400 (15/16 )

300 = 5

4

= tan–1 (5/4) 20. P

S

NN

= P

S

EE

S

75N

= 1202400

= 120

NS = 75 20 = 1500.

Electromagnetic Induction and Alternating Current 06

11

Chapter 06 : Electromagnetic Waves

1. uE = 20

1 E2

= 121 8.85 102

(170)2

uE = 1.28 107 J/m3

3. Here, By = 10–6 sin[7 1011 t + 400 x] The Y-component of the magnetic field is

given by,

By = B0 sin 2 tT

x

Comparing the given equation with the above equation,

2

= 400

= 2400

= 5 10–3 m

4. = c

= 8

6

3 103.4 10

= 88.24 m 8. The pressure exerted is given as,

Pressure (P) = Ic

= 8

1.23 10

= 82

N0.4 10m

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

The flux through the area considered is

E = 0

Q .4AA

= 0

4Q

The displacement current,

ID = E0

ddt

= 00

4 dQdt

= 4I.

11. B0 = 0Ec

= 8

903 10

= 3 10–7 T 12. Charge oscillating sinusoidally is given by Q = q0 sin t Displacement current,

ID = dqdt

= q0 cos t

(ID)max = q0 = q0 2 = 2 10–6 2 3.14 8 105 10 A

14. E = Bc = 40 106 3 108 = 12,000 V/m. 15. v =

r r

c

= 83 10

1.42 2.31

= 1.65 108 m/s

Electromagnetic Waves 07

12


12

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. = c

v = 100

70 = 1.43

6. m = f

f u

– 4 = ff ( 12)

f = –9.6 cm R = 2 f = – 19.2 cm 7. 1

v + 1

u = 1

f

1v

= 1f

– 1u

= 118

– 127

= 154

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. m = v

u also 1 1 1

f v u u u 1

f v

u u1v f

v f

u f u

so fmf u

9. g = cv

v = g

c

= 83 10

1.5 = 2 108 ms–1

v = Thicknessof glassplate(d)Timerequired to travel throughit (t)

t = dv

= 3

8

2 102 10

= 10–11 s

10. = cv

= 8

8

3 101.5 10

= 2

ic = sin–1 1

= sin–1 12

ic = 30 11. Lens formula 1

v – 1

u = 1

f

u is always negative, v is positive.

12. m = ff u

– 2 = 13

1 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. = D1f

= 2514

= 7.25

14. M = – o

e

ff

– 10 = – o

e

ff

fo = 10 fe and fo + fe = 44 fe = 4 cm fo = 40 cm.

15. 2

v – 1

u

= 2 1

R

1v

– 1.53

= 1 1.55

v = – 2.5 cm

16. 1f

= ( – 1) 1 2

1 1R R

120

= (1.55 – 1) 1 1R R

R = 40 0.55 = 22.0 cm 17. wg = 1.54

1.33, ic = sin–1 1.33

1.54

= 59 42

18. P = 15.5 – 5.5 = 10 D

f = 1P

= 0.10 m

19. = A ( – 1), b = A (b – 1), r = A (r – 1) D2 = A (1.525 – 1) D1 = A (1.520 – 1) D2 > D1 20. M = – o

e

ff

Negative sign for inverted image.

– 5 = – o

e

ff

fo = 5 fe and fo + fe = 24 fo = 20 cm, fe = 4 cm

Ray Optics 08

13

Chapter 08 : Ray Optics

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

1v

– 1u

= 1f

175

– 125

= 1f

f = 37.5 cm = 0.375 m

P = 1f (m)

= 10.375

= + 2.67 D 22. 1

f = P = ( – 1)

1 2

1 1R R

P = (1.5 – 1) 1 1R R

P = – 0.5 2R

= 0.5 20.3

= – 103

D

14


14

1. As medium changes there is no change in

frequency whereas wavelength in medium

become 1

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

= 83 10

1.5 = 2 108 m/s

t = g

dc

= 8

8

4 102 10

= 2 s

8. w

g

cc

= wg = g

w

cw = g

w

cg =

3 / 24 / 3

2 108

cw = 89 10 m/s4

9. Angle made with surface = 30 i = 90 30 = 60

1.5 = sinisin r

sin r = sin i1.5

= sin601.5

= 0.5773

r = 3517 Ratio of the width

2

1

WW

= cos rcos i

= cos 35 17cos 60

= 0.81630.5

= 1.633 10. = 706 656 = 50 nm = 50 109 m

As vc

v = c

=

98

9

50 10 3 10656 10

= 2.3 107 ms1

11. The phase difference of 2 corresponds to

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

2

= x or 2 x

=

12. 93 = 0.0465 mm = 465 10–7 m

= 7465 1093

= 5 10–7 m = 5000 Å

13. 1

2

II

= 2122

aa

= 254

1

2

aa

= 52

max

min

II

= 2

1 22

1 2

(a a )(a a )

=

2

2

5 25 2

= 49

9 14. Fringes overlap so that it looks like uniform

illumination. 16. Different points in filament emit light

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

18. =

Dd = D

2d

D = 2D 19. yb = yr (n + 1)b = nr (n +1) 5 10–5 = n 7.5 10–5 5n + 5 = 7.5 n 2.5 n = 5 n = 2 20. = D

d

D Decrease in D by 25% will bring 25%

decrease in . 21. yn = n D

d and ny = n D

d

As they coincide yn = ny

nn

=

= 900750

= 65

5y = y6 = n Dd =

9

3

6(750 10 )2(2 10 )

= 4.5 mm 22. For incoherent waves, Imax = nI0

n = max

0

II

= 322

= 16

Wave Optics and Interference of Light 09

15

Chapter 10 : Diffraction and Polarisation of light

.11.3

1. 2

1

xx

= 2

1

= 36005400

x2 = x12

1

= (0.3) 36005400

= 0.2 mm

6. d = 1 = o1

60

= 180 60

radian

If d is the actual distance between the pillars and D is distance between the pillars and the person then,

dD

= d

d = D d = 15000 180 60

4.36 m

= 436 cm 7. According to Brewster’s law, = tan ip ip = tan–1 () 8. dmin =

10

3

1.22 x 1.22 6600 10 5D 3 10

= 2.01 10–3 m 9. R.P. = 7

D 4.21.22 1.22 6 10

= 67 101.22

= 5.74 106 10. a sin =

a = sin

= 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

= 7

32

2 2.1 5 10 1.4 10 m 1.4 mm0.15 10

14. R.P. of telescope (diameter of aperture) 16. According to Brewster’s law, = 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 cos2 = A2 cos2 40 Intensity of emergent light, I2 = I1 cos2 40 = A2 cos2 40 cos2 40 = 0.34 A2

Diffraction and Polarisation of Light 10

16


16

1. Light is packets of energy called as quantas

which is photon. 4. Stopping potential is independent of intensity. 6. E = + K.Emax or hc

= + eV0

= 0

hceV

7. Einstein’s photoelectric equation is

21 mv2

= hv

= h – 21 mv2

9. Given

max1

hcK.E.

….(i)

and max2

hc2K.E.

….(ii)

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

2

1

h c

2h c

which gives 2 11 2

hc (2 )

10. = h

p= h

mv.

11.

12. 0 = hc

= 34 8

19

6.6 10 3 104.125 1.6 10

= 3000 1010 m = 3000 Å 13. 1

v If v = 0, =

14. K.E. = eV = 1.602 10–19 100 = 1.602 10–17 J

15. E = 180 eV = 180 1.6 1019 = 2.88 1017 J

Now = h2mE

= 34

131 17 2

6.6 10

2 9 10 2.88 10

= 34

24

6.6 107.2 10

= 0.916 1010 m = 0.9 Å

I

50

V = 54V

0

Dual Nature of Matter and Radiation 11

17

Chapter 12 : Atoms and Nuclei

1. Energy released = E4 E1

= 2

13.64

213.61

= 12.75 eV 3. In each -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

= 0.693T

= 0.69350.8

= 1.36 102 per day

N = 40% of N0 = 0.4 N0

As, 0NN

= et

ln 0NN

= t

t = 0Nn

N

l=

2

n 2.51.36 10

l = 2

0.9161.36×10

= 67.37 days

6. Energy required 2 2

13.6 13.6n 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 2min

1 1 1R1

= R

2 2max

1 1 1 36 25 11R R R5 6 25 36 900

max

min

=

90011 RR

= 900

11

8. E3 = 13.6

9= 1.51 eV

E4 = 13.616

= 0.85 eV

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

number is not conserved. 12. R = Ro

13A = 1.1 1015

13(16)

= 2.77 1015 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. 02 1115 115

48 50Cd Sn

17. n = A A 200 168 32 84 4 4

n = (2n Z + Z) = (16 90 + 80) = 16 10 = 6 18. Wave number 1

= 8

15896 10

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

sample will become 2

12

, 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 is 100 times of , while that of is 100 times of .

22. For Lyman series,

Lyman 2 2max

c 1 1 3RcRc(1) (2) 4

For Balmer series,

Balmer 2 2max

c 1 1 5RcRc(2) (3) 36

Lyman

Balmer

275

23. For ground state, n = 1 For first excited state, n = 2 As rn n2 radius becomes 4 times. 24. E = mc2 = (1 10–3)(3 108)2 = 9 1013 J 25. n = A A

4 = 232 208 6

4

n = (2n – 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

18


18

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

= 19.4 mA

d.c. power output = Id.c.2 RL

= (19.4 103)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 2.5 106 Vbreakdown = 2.5 V 19. VBB = IBRB + VBE

RB = BB BE

B

V VI = 6

5.5 0.510 10

= 500 k 20. Av = L

i

Rr

and Ap = 2

L

i

Rr

= Av

p

v

AA

= = 62 21. IE = IC + IB = 25 + 1 = 26 mA

= C

E

II

= 2526

22. I = 3

150 505 10

= 20 mA

IL= 5 mA ….[Given] Zener current IZ = 15 mA 23. IB = i

i

VR

= 0.011000

= 0.01 103 A = 0.01 mA

IC = IB = 50 0.01 = 0.5 mA = 500 A

24. = 0.95

= 1

= 0.951 0.95

= 19

Also, = C

B

II

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

E = h = hc

= 34 8

10

6.62 10 3 105890 10

= 3.37 1019 J

= 19

19

3.37 101.6 10

= 2.11 eV

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

B

E C

+

+Forward

Biased Reversed Biased

Electronic Devices 13

19

Chapter 12 : Atoms and Nuclei

1. = 12 LC

= 9 6

12 2 10 45 10

= 710

6 = 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.

6. fSB = fc fm = 1800 0.8 = 1800.8 kHz and

1799.2 kHz 7. c c

2L

L

2

83 10

2 0.3

= 500 MHz. 8. Time =

32

8

4000 10 1.33 10 s3 10

13 ms

10. Number of stations

= B.W.2 Highest modulating frequency

= 240,0002 12000

= 10 11. d = 2hR = 62 250 6.4 10 = 56.568 103m 56.6 km 12. d = 2Rh Population covered = d2 population density = (2Rh)

= 227 2 6400 0.1 4000

= 1.6 107 13. Since, total power

2

a

PtPc m12

Here, ma = 0.6

Pc = 2

8 8(1 0.18)0.61

2

= 8 6.8 kW1.18

Communication Systems 14

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