Circular Motion - Target Publications

1

Chapter 01: Circular Motion

Hints to Problems for Practice

1. For the second hand of a clock, T = 60 s, r = 4 102 m.

= 2 3.14

60

= 1.047 101 rad/s and

v = 4 102 1.047 101 = 4.188 103 m/s.

2. n = 600 rpm = 600

60= 10 rps, r = 0.5 m.

= 2 3.142 10 = 62.84 rad/s v = 0.5 62.84 = 31.42 m/s.

3. r = 20 102 m, n = 240

60= 4 rps.

= 2 3.14 4 = 25.12 rad/s and v = 20 102 25.12 = 5.024 m/s. 4. T = 60 min. = 3600 s, t = 20 min. = 1200 s,

= ? = 2

T

=

2

3600

=

1800

rad/s

= t = 1800

1200 = 2.093 rad.

5. r = 5 102 m, s = 7.25 102 m s = r

= 2

2

7.25 10

5 10

= 1.45 rad

6. v = r = 2

10= 0.2 rad/s.

Now, = t = 0.2 15 = 3 rad. 7. T = 24h = 24 3600s,

= 2

T

=

2 3.14

24 3600

= 7.27 105 rad/s.

8. v = r = 300

1500 = 0.2 rad/s.

v = 2 r / 5

t

or t =

2 r

5v

t = 2 3.14 1500

5 300

= 6.28 s

9. = 5 rad/s, r = 50 102 m, v = r = 50 10–2 5 = 2.5 m/s.

10. n1 = 500 rpm = 500

60rps, n2 = 0, t = 10 s, = ?

= 2 12 (n n )

t

=

2 3.14

10

5000

60

= 5.23 rad/s2

11. n1 = 600 rpm = 600

60= 10 rps and

n2 = 1200 rpm =1200

60= 20 rps, t = 20 s.

= 2 12 (n n )

t

=

2 (20 10)

20

= rad/s2 Now, 1 = 2 10 = 20 rad/s and 2 = 2 20 = 40 rad/s Using, 2 2

2 1 2 we get,

= 2 22 1

2

= 2(2 ) (400 100)

2

= 2 300 rad

No. of revolutions = 2 300

2 = 300

12. n1 = 100 rpm = 100 10

60 6 rps,

1 = 2n1 = 2 10

6=

20

6

rad/s

2 = 0, t = 15 s, = 2 rad

Using, = 2 1

t

=

0 20 / 6

15

= 20

90

=

2

9

rad/s2

Now using,

1 = 1t + 1

2 t2 we get,

1 = 220 1 215 15

6 2 9

= 300 15 15

6 9

= 50 25

1 = 25 rad Now, 1 rev. 2 rad. number of revolutions to cover 25 rad

= 25

2

= 12.5

Circular Motion 01

2

Std. XII Sci.: Physics Numericals

13. v1 = 72 kmh1 = 72 5

18 = 20 ms1.

d = 0.5 m, r = 0.5

2= 0.25 m, v2 = 0

no. of rotations = 20, = 20 2 = 40 rad = ?

1 = 1v

r =

20

0.25= 80 rad/s, 2 = 0

22 = 2

1 + 2

0 = (80)2 + 2 40 we get, = – 25.5 rad/s2

14. a = 2 2

r ta a = 2 2(123.62) (91.41)

= 153.74 m/s2.

15. = 2 12 (n n )

t

=

2 100

5

= 40 rad/s2

Now, aT = r = 5 102 40 = m/s2

16. ar = 2v

r=

20 20

10

= 40 m/s2.

a = 2 2r ta a = 2 240 30 = 50 m/s2.

17. a = 2v

r=

2 2r

r

= r2

a = 0.2 (0.8)2 = 0.128 m/s2 18. r = 9.8 m, ar = 8 9.8 m/s2, ar = r2 8 9.8 = 9.8 2 = 2.828 rad/s. 19. v = 20 m/s, r = 100 m, at = 5 m/s2,

ar = 2v

r =

20 20

100

= 4 m/s2.

a = 2 2r ta a = 2 24 5 = 6.403 m/s2.

20. T = 2 3.14

2

= 3.14 s

Now, F = mr2 = 10 1 (2)2 = 40 N 21. TBr = 24 9.8 N, r = 2m, m = 2 kg. 2

mqmr = TBrk

mq = BrkT

mr=

24 9.8

2 2

mq = 7.66 rad/s

n = 7.66

2 3.14= 1.22 rev/s.

Also, v = r = 2 7.66 = 15.32 m/s.

22. TBr = 2mv

r

50 = 21 v

0.5

v = 5 m/s

23. n = 300 rpm = 300

60rps = 5 Hz.

F = mr2 = mr. 42n2 = 0.2 2 4 3.142 52 F = 394.38 N.

24. = g

r

=

0.2 9.8

0.25

= 2.8 rad/s.

25. TBr = mr2 = mr 42n2.

n = 2

9.86

1 1 4 3.14

= 0.5 rps = 0.5 60 = 30 rpm.

26. i. vmax = (r Tmax/m)1/2 =

1

21.5 250

0.5

= 27.39 m/s ii. T = mr2 = 0.5 1.5 42 = 12 N

27. 1 = 2n = 2 80

60rad/s

21 1mr = 2

2 2mr

22 =

21 1

2

r

r

= 0.25 42

80 80

60 60

1

0.4

22 = 43.82

2 = 6.62 rad/s.

n2 = 6.62

2 3.14 60 = 63.25 r.p.m.

28. v = rg = 0.3 40 9.8 = 10.84 m/s

29. v = 18 km/hr = 18 5

18= 5 m/s,

v = rg

r = 2v

g=

25

0.25 9.8= 10.2 m.

3


30. v = rg = 2.2 62 9.8 = 36.56 m/s

31. v = 60 kmh1 = 60 5

18 m/s =

50

3m/s,

= 2v

rg=

250

340 9.8

= 0.71.

32. = tan1 2v

rg

= tan1 26

18 9.8

= 1132.

Now, = 2v

rg=

26

18 9.8= 0.2041

33. tan = 2v

rg= 244 / 3

36 9.8 = 0.6097

= tan–1(0.6097) = 31.37 = 90 – 31.37 = 58.62 = 5838

34. v = 75.6 km/h = 75.6 5

18= 21 m/s

= tan1 2v

rg

= tan1 221

90 9.8

= 26 34

35. C = 1.256 km = 1.256 103 m = 1256 m

r = 1256

2= 200 m

= tan1 2v

rg

= tan1 225

200 9.8

= tan1 (0.3189) = 1741

36. d = 320 m, r = 320

2= 160 m,

v = 144 km/h = 144 5

18= 40 m/s.

tan = 240

160 10

= tan–1 (1.00) = 45 37. v = 48 km/h

= 48 5

18=

40

3m/s,

= tan12v

rg

= tan1

240

3400 9.8

= 2.6

h = l sin = 1 sin 2.6 = 0.0454 m

38. v = rgd

2h=

100 9.8 2

2 1.5

= 25.56 m/s

= 2v

rg=

225.56

100 9.8

= 0.67

39. vmax =

1/2tan

rg1 tan

=

1

20.28 tan1555 9

1 0.28tan15

=

1

2495 0.5479

0.92498

= 17.12 m/s 40. v = rg tan or v2 = rg tan

r = v2/(g tan ) =2(250)

9.8 tan 22 = 15,785 m

41. = tan–12v

rg

= tan1 225

1800 9.8

= 22

h = l sin or l = h/sin

= 0.05

sin(2 2 ') = 1.412 m

42. vmax =

1/2tan

rg1 tan

=

1

20.4 tan1025 9.8

1 0.4 tan10

=

1

2245 0.5763

0.9295

= 12.325 m/s

43. r = 50 m, vmax = 60 5

18=

300

18=

50

3m/s,

= 0.5,

2maxv

rg =

tan

1 tan

250

350 9.8

= 0.5 tan

1 0.5tan

4


2500

9 5 98 =

0.5 tan

1 0.5tan

0.5669 = 0.5 tan

1 0.5tan

0.5669 – 0.5 0.5669 tan = 0.5 + tan 0.5669 0.5 = 0.2835 tan + tan 0.0669 = 1.2835 tan

tan = 0.0669

1.2835= 0.05212

= tan1 (0.05212) = 2.98 = 2 58 44. h = 2 2rl = 2 20.5 0.25 = 0.43 m

T = 2h

g= 2 3.14

0.43

9.8 1.32 s

Now, cos = 2

2

T g

4 l =

2

2

(1.32) 9.8

4 (3.14) 0.5

= 0.8659 = cos1 (0.8659) = 30 v = rg tan = 0.25 9.8 tan (30 )

v = 119 cm/s

45. n = 20

45= 0.44 Hz, n =

1

2g

h

h = 2 2

g

4 n=

2 2

9.8

4 3.14 0.44

h = 1.284 m

r = 2 2hl = 2 21.5 1.28 = 0.78 m

Now, T cos = mg or T = mg

cos =

mg

h

l

T = 0.05 9.8 1.5

1.28

= 0.57 N.

46. T = 2 2 1/2

mg

( r )l

l =

1

2 2 2

0.05 9.8 0.75

0.75 0.24

= 0.517 N

47. T = mg/cos = 0.3 9.8

cos12

= 3 N

T = 2 cos

g

l

= 2 3.14 1.15 cos12

9.8

T = 2.13 s 48. vT = rg = 3 9.8 = 5.422 m/s

vL = 5 rg = 5 3 9.8 = 12.124 m/s

vm = 3 rg = 3 3 9.8 = 9.391 m/s.

49. m = 4000 kg, r = 30 + 1.2 = 31.2 m,

v = 50.4 km/h = 50.4 5

18 = 14 m/s

Thrust t, N = 2

Topvm g

r

= – 4000 214

9.831.2

= 14072 N

vmax = rg = 31.2 9.8 = 17.49 m/s 50. v = rg = 15 9.8 = 12.124 m/s 51. R = 19.5 + 0.5 = 20 m

v = Rg = 20 9.8 = 14 m/s 52. v = rg = 5 9.8 =7 m/s

= v

r=

7

5= 1.4 rad/s

53. when TH = 0

TH = 2HmV

rmg = 0

vH = rg = 0.8 9.8 = 2.8 m/s

TL = 2HmV

r+ 5 mg

= 20.3 (2.8)

0.8

+ 5 0.3 9.8

= 2.94 + 14.7 = 17.64 N 54. r = 8 m,

v = rg = 8 9.8 = 8.854 m/s

= v

r=

8.854

8= 1.068 rad/s

n = 2

= 1.1068

2 3.14 60 = 10.57 rev./min

55. vL = 5rg = 5 25 9.8 = 35 m/s

56. TH = 2Hmv

r mg =

20.1 7

1

0.1 9.8

TH = 3.92 N

57. TL = 2Lmv

r + mg

45 = 2L0.5 v

0.5

+ 0.5 9.8

2Lv = 40.1

vL = 6.332 m/s

5


58. i. m = 0.4 kg v = 15 m/s, r = 1.2 m

= v

r =

15

1.2 = 12.5 rad/s

Ttop = m 2v

gr

= 0.4 215

9.81.2

= 71.08 N

ii. Ttop = m 2v

gr

= 0.4 215

9.81.2

= 78.92 N

59. At any point,

T.E. = 5

2mgr =

5

28 9.8 1 = 196 J

60. K.E.(bottom) K.E.(top) = 1

2m [5rg rg]

= 1

2m 4rg

= 2 mgr = 2 4 9.8 0.6 = 47.04 J. 61. Ttop = m (2r – g) = 0.6 ((15)2 1.8 – 9.8) = 237.12 N Tbottom = m(2r + g) = 0.6((15)2 1.8 + 9.8) = 248.88 N

62. = 2 22 1

2

= 2 2 24 (15 5 )

2 50 2

= 12.56 rad/s2

Now, t = 2 1

= 2 (15 5)

=

2 3.14(10)

12.56

= 5 s. 63. i. 2 = 1+t = 5+ 2 10 = 25 rad/s

ii. = 1 + 1

2 t2

= 5 10 +1

2 2 102 = 150 rad.

iii. aT = r2 = 2 0.1 = 0.2 m/s2.

64. = t + 1

2 t2

60 = 9t + 1

2 3t2

20 = 3t + 1

2 t2

t2 + 6t 40 = 0 t = 4 s

65. vmax = 50 km/h = 50 5

18 m/s 13.89 m/s,

= 0.26,

d = 0.04 km r = 0.04

2 = 0.02 km = 20 m

2maxv

rg =

tan

1 tan

213 89

196

=

0.26 tan

1 0.26 tan

0.9843 = 0.26 tan

1 0.26 tan

0.9843 – 0.9843 0.26 tan = 0.26 + tan 0.9843 – 0.26 tan = 0.26 + tan 0.26 tan + tan = 0.9843 – 0.26 1.26 tan = 0.7243

tan = 0.7243

1.26 = 0.5748

= 29.89 = 29 53

Hints to Multiple Choice Questions

1. v = 25 cm/s = 0.25 m/s, = 0.5 rad/s

Use formula v = r, r = 0.25

0.5= 0.5 m

2. = 2n

n = 8

2

= 4 60 = 240 rpm.

3. = /t = 2 /3600 rad/s

4. = t = 2

T

t =

2 3.1430

60

= 3.14 rad

5. = 2/t and v = r

= r 2

T

=

0.15 2 3.14

3600

= 2.62 104 m/s

6. = (2 – 1)/t

= 5 2.5

20

= 0.125 rad/s2

7. [Note: 1 = 0]

= 1 t + 1

2t2

6


100 = 1

2 (15)2

= 0.89 rad/s2

8. 2

2 – 21 = 2,

= 2 22 1

2

= 2 26.4 4.2

2 50

= 0.23 rad/s2

9. n1 = 50 rpm =50

60rps,

n2 = 100 r.p.m = 100

60rps, t = 20 s

= (2 – 1)/t,

= 2 12 n n

t

=

100 502

60 6020

= 0.26 rad/s2

10. ar = r2 = 0.12 (1.1)2 0.15 m/s2 11. = 2n = 2 14 rad/s, acp = 2 r = 42 n2r = 4 3.142 142 2 10–2 = 154.59 m/s2

12. a = 2 2

r ta a we get,

a = 2 229 12 = 841 144 = 985

31.38 m/s2

13. ar = 2r = 302 0.14 = 126 m/s2 ar : at = 126:0.24 = 525:1

14. 2r

r

= 600

= 2

600

=

2

3 rad/s2

Now, r = 0.2

r = 0.2

=

0.223

= 0.6

2 = 0.3 m = 30 cm

15. mr1 21 = mr2

22

r2 = 2

1 122

r

= 2

2

2

/ 4

= 8 cm

16. = 2n = 2 3.14 3 = 6 3.14 rad/s Now using, T = mr2 = 250 103 115 10–2 62 3.142 = 102.05 N

17. Tmax = 2maxmv / r

r = 2max

max

mv

T

= 20.2 (4.2)

300

=

3.528

300

= 0.01176 m = 1.176 cm r 1.18 cm 18. T = mv2/r, we get,

200 = 20.2 v

0.6

v = 24.49 m/s

Now using T = 2 r

v

= 2 3.14 0.6 / 24.49

= 0.154 s 19. v = r = r 2n

2rn = rg

42r2n2 = rg

n = 2

g

4 r

= 2

0.4 9.8

4 3.14 0.12

= 0.91 r.p.s. = 0.91 60 r.p.m = 54.6 r.p.m. ….[ Given that n = ]

20. T = mv2/r = 20.3 10

0.6

= 50 N

21. T = mv2/r v = (rT/m)1/2 = (0.75 15 9.8/0.5)1/2 14.85 m/s 22. v = rg = 0.6 42 9.8 = 15.71 m/s

For values of v greater that the 15.71 m/s, the car must slip-off.

23. v = 60 km/hr = 60 5

18 = 16.67 m/s,

r = 50 m

v = rg or

= v2/rg = 216.67

50 9.8

= 0.56. For values of

greater than 0.56, the car will not skid.

24. v = 80 km/hr, = 80 5

18 = 22.22 m/s,

= 0.2

v = rg

r = v2/g = 222.22

0.2 9.8 = 251.90 m

7


25. vmax = tan

rg1 tan

= 0.32 tan17

25 9.81 0.32 tan17

= 13.04 m/s

26. 2max

grdv

2h

r = 2maxv h 2

gd

=

2(22.02) 0.62 2

9.8 1.24

r = 49.48 m

27. v = 65 km/h = 65 5

18 = 18.06 m/s,

= 12 tan = v2/(rg)

r = v2/(g tan) = 218.06

9.8 tan12 = 156.6 m

28. r = 22 m,

v = 45 km/h = 45 5

18 = 12.5 m/s,

h = 1.1 m = v2/rg = tan–1 (v2/rg)

= tan–1 212.5

22 9.8

= 3556 Using h = l sin or l = h/sin we get l = 1.1/ sin 35 56 = 1.875 m 29. v = rg tan = 75 9.8 tan30

= 735 0.5774

= 424.389 = 20.6 m/s

30. vmax =tan

rg1 tan

= 0.34 tan 32

72 9.81 0.34 tan 32

= 29.4 m/s 31. v = rg tan = 0.22 9.8tan15

= 0.76 m/s 32. T cos = mg,

cos = 2 2rl

l=

2 20.9 0.25

0.9

cos = 0.961

Now using, T = mg/cos =0.15 9.8

0.961

= 1.53 N

33. T = 2 ( cos / g)l

= 2 3.14 1.2 cos16

9.8

= 2.155 s 34. T cos = mg T = mg/cos = (0.2 9.8) /cos 8 = 1.979 N

35. T = (mv2/r) + mg = m2v

gr

= 4026

9.82.5

= 968 N

36. Total Energy = E = 5mrg

2

= 5 0.1 0.4 9.8

2

= 0.98 J 37. v = 3rg = 3 0.8 9.8

= 4.85 m/s

38. v = 400 km/hr = 400 5

18= 111.11 m/s,

r = 100 m, m = 60 kg T = (mv2/r) – mg

= m 2v

gr

= 60

2111.1119.8

100

= 6819.40 N

39. E = 5mrg

2

m = 2E

5rg

m = 2 250

5 0.6 9.8

= 17 kg

40. l = 1.8 m (Note: l = r here)

vmin = rg = 1.8 9.8 = 4.2 m/s 41. mg = mr2

r = 2

g

=

2

9.8

1.5= 4.355

4.36 m 42. m = 200 g = 0.2 kg,

n = 60 rpm = 60

60 = 1 rps, r = 0.8 m

T = m 2v

gr

8


= m 2 2 2r 4 n

gr

= m [42n2r – g] = 0.2 [4 3.142 12 0.8 – 9.8] = 4.35 N 43. r = 18 m, h = 0.4 m, R = 18 + 0.4 = 18.4 m

v = Rg = 18.4 9.8

= 13.428 m/s 44. 2

2 = 21 + 2 we get,

22 = (0)2 + 2 12 30 = 720

2 = 720 26.83 rad/s

45. n1 = 600 r.p.m.=600

60r.p.s = 10 r.p.s.

n2 = 0, = 2 rad 2

2 = 21 + 2

= 2 22 1

2

= 2

10 2 n

2

= 2 2

14 n

2

= 2 24 10

2 2

= – 100 = – 3.14 100 = – 314 rad/s2

46. 22 – 2

1 = 2

= ( 22 – 2

1 )/2 = 2 212 10

2 40

= 0.55 rad/s2

47. = 2 1

t

= 2 12 n 2 n

t

= 40 20

4

= 5 rad/s2

= 1t + 1

2 t2

= 20 4 + 1

2 5 42

= 80 + 40 = 120

n = 2

= 120

2

= 60

1

Chapter 02: Gravitation


1. F = E S2

GM M

R

3.5 1022 =

11 24s

211

6.67 10 M 6 10

1.5 10

MS = 1.96 1030 kg

2. F = 2

GMm

R =

11

5 2

16.67 10 M M

81(3.85 10 )

= 1.999 1026 N 3. (R + h) = 4.23 107 m,

2h

2

g R

g (R h)

= 0.2243 m s–2

vc = 72 (R h) 2 3.142 4.23 10

T 24 60 60

= 3.076 103 m s–1

4. i. 2

hg R

g R h

h = 2 1

h = 2649 km

ii. 2

h'g R

g R h '

h = 711 km

5. vc = 2gR

(R h)

=6 2

6

9.8 (6.4 10 )

(6.4 0.9)10

= 7.4154 km s–1

T = c

2 (R h)

v

= 62 3.142 (6.4 0.9)10

7415.35

= 6185.45 s = 1 hr 43.2 min

6. i. vc = 2gR

R h = 7519.07 m s–1

ii. T = c

2 (R h)

v

= 1 hr 39 min

7. h = 1/32

2

T GMR

4

=1/32 11 24

2

(0.2 24 3600) 6.67 10 5.98 10

4 (3.142)

– (6.4 106) = 8050 km

8. vc = 2gR

(R h) = 2.38 103 m s–1

9. R + h = 1/32 2

M2

T gR

4

=

12 6 2 3

2

(27.5 24 3600) 9.8 (6.4 10 )

4

= 3.8575 105 km 10. vc = 2gR / (R h) = 7545.66 m s–1

11. i. vc = 2 6 2

6

gR 9.8(6.38 10 )

(R h) (6.38 0.25)10

= 7756.7 m s–1 7.76 km s–1

ii. T = 6

c

2 (6.38 0.25)102 (R h)

v 7756.7

= 5371 s 1 hr. and 29.4 min.

12. T = c

2 (R h)

v

= 62 3.142 (6.4 1)10

7364

= 6313.90 s 6314 s

Gravitation 02

2


13. 2

12

T

T =

3

1R

R

=

33

R4R

T1 = 0.650 yr.

14. 2 3

E E

M M

T r

T r

=

311

11

1.5 10

2.5 10

TM = 2.15 yrs. 15. T 3/2R

T 3/2(R h) 3/23/2 6

6

T R 6.4 10

T R h (6.4 0.6) 10

'

= 0.8742

T = 1.4

0.8742 = 1.60 hrs.

16. h1(K.E) =

1

GMm GMm

2(R h ) 6R

h2(K.E) =

2

GMm GMm

2(R h ) 10R

K.E. = h2(K.E) – h1

(K.E)

= GMm 1 1

R 10 6

= GMm

15R

= 11 24

6

6.67 10 6 10 400

15 6.4 10

K.E. = – 1.6675 109 J

P.E. = – 2(K.E.) = –2(–1.6675 109)

P.E. = 3.335 109 J

17. i. vc 1

R

A

B

c B

c A

v R

v R =

5

3

ii. T c

R

v

B

A

cA A

B B c

vT R 3 3 3 3

T R v 5 5 5 5

= 0.4648

iii. B.E. M

R

A A B

B B A

(B.E.) M R 2 5

(B.E.) M R 1 3

= 10 : 3

18. E = 2 2 1

1 2

h hgR m

(R h )(R h )

= gRm

12

= 6 39.8 6.37 10 2 10

12

= 1.0404 1010 J

19. K.E. = GMm

2(R h)

= 11 24 2

6

6.67 10 6 10 10

2(6.4 1.6)10

= 2.5013 109 J B.E. = K.E. = 2.5013 109 J T.E. = –K.E. = – 2.5013 109 J P.E. = –(2K.E.) = – 5.0026 109 J

20. K.E. = 6gRm 9.8 6.4 10 500

2 2

= 1.568 1010 J P.E. = – 2(K.E.) = – 3.136 1010 J B.E. = K.E. = 1.568 1010 J T.E. = – (K.E.) = – 1.568 1010 J

21. B.E. = GMm

2(R h)

B.E. = 11 24 3

6

6.67 10 6 10 10

2(6.4 4)10

= 1.92 1010 J

22. B.E. = GMm

2(R h)

= 11 24

6

6.67 10 6 10 50

2(6.4 0.6)10

B.E. = 1.4293 109 J (K.E.) = (B.E) = 1.4293 109 J (P.E.) = 2(K.E.) = 2.8586 109 J (T.E.) = (K.E.) = 1.4293 109 J

23. ve = M M2g R = 62 1.63 1.7 10

= 2.354 103 m s–1 = 2.354 km s–1

3


24. i. ve M

e

e

v M2

v M

' '

ve = 11.2 2 = 15.8392 km s–1

ii. ve 1

R

e

e

v R 1

v R 2

'

'

ve = 11.2

2 = 7.9196 km s–1

25. ve M

R

M

e

e

v

(v ) = m

m

RM

R M =

9

2


1. 2

2

mv Gmm

r (2r) v =

Gm

4r

The two particles will move on a circular path if they always remain oppositely directed and force of gravitation will act radially.

2. Let resultant force on sphere C be F.

F = 2 21 2 1 2F F 2F F cos60 =

2

2

3 Gm

4 R

Where, F1 = 2

2

Gm

4R along CA

F2 = 2

2

Gm

4R along CB

3. Let x be the distance between rocket and

moon, when the net gravitational force on rocket is zero, i.e., gravitational force on rocket due to earth = gravitational force on rocket due to moon.

Then, e m2 2

GM m GM m

(r )

x x

e

m

Mr

M

x

x

Solving, x 3.8 107 m 4. The force on mass m at P due to solid sphere is

F1 = 2 2

GM m GM m

(3R) 9R …(i)

Mass of cavity created,

M = 3

3

M 4 M(R / 2)

4 3 8R3

The force on mass m due to mass of cavity is

F = 2 2

G(M / 8) m GM m

(3R R / 2) 50R

The force on mass m due to remaining part of sphere

F2 = F1 – F = 41

450 2

GM m

R ….(ii)

From (i) and (ii) F2 : F1 = 41 : 50 5. Mass of element of rod of length dx at

distance x from point mass M = (M/L) dx Force experienced by point mass due to this

element is

dF = 2

2 2

GM (M / L)d GM d

L

x x

x x

Total force experienced by point mass due to whole rod is

F = 2L 2 2

2 2L

GM d GM

L 2L

x

x

6. F = 2

GMM

(2R) = G

23

2

4R

3(2R)

= G 2 44R

9

7. W = mg = m 2

GM

R

W = mg = m 2

GM

R

2

2

M R

M R

=

2

2

M / 7 R

M R / 2

4

W 0.77

= 0.4 kg wt

2R F2

F1

A

B C

2R 2R

60

4


8. e e m

m m e

R M g 80

R M g 6 = 3.65

9. g = 2

GM

R, g =

2

G(2M)

(3R)

g = 2

9g

g = 2.18 m/s2

10. hg

g=

2

2

R

(R h) i.e.,

64

100=

2

2

R

(R h)

Solving, h = 1600 km. 11. At h = R

2h

2

g R

g R h

=

2

2

R

2R=

1

4

gh = 10

4= 2.5 m/s2

12. g = 3

2 2

GM G 4 R 4G R

R 3R 3

1 1 1

2 2 2

g R

g R

13. s = ut + 1

2at2 As se = sm = h

2 2e e m m

1 1g t g t

2 2

2e m2m e

t g

t g i.e.

2m

2e

t

t= 6

tm2 = 2

e6t tm = e6t = 6 t 14. gE = g – R2

For given condition, gE = 3

g4

R2 = g – 3

g4

i.e. 2 = 1 g

4 R

= 3

10

4 6400 10

= 1

1600 rad/s

15. mg he = mg

6

hmoon.

hmoon = 1.2 6 = 7.2 m

16. vc = GM GM

R h r

;

c1 2

c 12

v r

v r =

9R

16R=

3

4

17. Orbital velocity is given by

vc = v0 = 1 2

GM GM

r r

where G and M

are constant. If radius r changes then, percentage change in

orbital velocity is given by

0

0

v

v

100 =

1 r

2 r

100 .…(1)

Given, radius decreases by 1 %

0

0

v

v

100 =

1

2 (–1) = 0.5 %

18. Critical velocity of satellite revloving close to

earth

vc = gR , = cv

R 2 =

2

gR

R

= 1.24 103 rad/s

19. T = 2 1

3 2

3

R 34 GG R3

20. T = 3 3R R

2 2GM G V

T = 23

3

(4R)

4G (4R)

3

= T

21. h = R

T = 3 3

2

2R 8R2

GM gR

No. of revolutions per day = 24 60 60

T

= 38.18

22. T2 r3

3 32 22 2

1 1

T r 4

T r 1

T2 = 8T1 = 8 years = 8 365 day = 2920 day

23. T2 d3 2

1

n d3

n2 d3 = constant

2 3 2 31 1 2 2n d n d

5


24. T2 r3

2 3

1 1

2 2

T r

T r

=

31

4

1

2

T

T =

1

8

25. 1

2

T

T=

3/21

3

T2 = 24

0.192 = 72 3 hrs

26. vc = GM

2R

K.E. = 2c

1 GMm mgRmv

2 4R 4

27. (P.E.)final = GMm

R nR

,

(P.E.)initial = GMm

R

(P.E.) = (P.E.)final (P.E.)initial

= GMm 1

1R (1 n)

= gmRn

n 1

28. (P.E.)initial = GMm

R

,

(P.E.)final =

GMm GMm

R h 2R

(h = R)

(P.E.) = (P.E.)final – (P.E.)initial

= GMm GMm

2R R

= GMm

2R

=1

mgR2

2

GMg

R

29. P.E. = –2GMm gR m

R h R h

In magnitude, 2gR m 1

R h 2

mgR

R + h = 2R

h = R

30. (P.E.)earth = GMm

R

(P.E.)planet =

MG m GMm2

R R2

Ratio is 1 : 1

31. P.E. = GMm GMm 1 GMm

RR 6 RR5

= 2

GM mR gmR

R 6 6

2

GMg

R

= gm5h 5

(mgh)6 6

32. (P.E.) = U1 = – 1

GMm

r

GM = 1 1U r

m

Weight W1 = mg1 = m 2

1

GM

r

and W2 = mg2 = m22

GM

r

= m 1 1 1 12 22 2

U r U r1

m r r

W2 = 9 7

9 9

4 10 10

10 10

N = 0.04 N

33. Total energy of satellite in circular orbit of

radius 2R E1 = P.E. + K.E.

= –2

GM m 1 GMm

2R 2 2R

= – GM m

4R

Total energy of satellite in circular orbit of radius 4 R

E2 = –2

GM m 1 GM GM mm

4R 2 2R 8R

Energy spent = E2 – E1 = mgR

8

=

6400 9.8 6.4 10

8

= 3.14 × 109 J

6


34. (K.E.)1 = 1

2mve

2 , ve = 2GM

R

(K.E.)2 = 1

2mvc

2 , vc = GM

R

1

2

K.E. 2=

K.E. 1

35. E = 22e

1 1mv m 2gR

2 2 = mgR

= 500 9.8 6.4 106 3.1 1010 J 36. Given h = Re

(P.E.)top = (K.E.)bottom

20

e

GMm 1mv

2R 2

0e

GMv

R

ee

2GMv

R

0

e

v 1

v 2 .

37. Oribital velocity of satellite,

0

GMv

r

Kinetic energy, E = 20

1 1 GMmmv

2 2 r

Kinetic energy for escaping body,

E = 2e

1 1 2GM GM mmv m 2E

2 2 r r

Additional energy needed = E – E = 2 E – E = E 38. ve = 2gR and ve = 2 g / 6 R / 4

ve = e2gR v 11km / s

24 24 24

39. R2 = 2R1, 1 = 2

1 23 31 2

M M

4 / 3 R 4 / 3 R

31 1

32 2

M R

M R

2

1

3e 2 1 2 1 2

3e 1 2 1 2 1

v M R R R R

v M R R R R

e2v = 11.2 2 = 22.4 km/s

40. 1

2

Rk

R ; 1

2

gr

g ; ve = 2gR

1

2

e 1 1

e 2 2

v g Rrk

v g R

41. ev 2 9 g 4R '

= 6 2gR = 6 11.2 = 67.2 km/s 42. M2 = 3M1 and R2 = 3R1

ve = 2GM

R

1

2

e 1 2 1 1

e 2 1 1 1

v M R M 3R1

v M R 3M R

43. ve = 2GM

R

Given e2v = e1

10v

e2

e1

v

v= 1

2

R

R

R2 = 64 km

1

Chapter 03: Rotational Motion

Hints to Practice for Problems 1. I = MK2 = 0.2 (0.4)2 = 0.032 kg m2

2. M.I. = 2ML

12 =

21 1

12

= 0.083 kg m2

3. R = Interatomic distance

2

I = 2 (MR2) = 2 [1.7 1027 (2 1010)2] = 1.36 1046 kg m2

4. R = Interatomic distance

2

I = 2 (MR2) = 2 [2.6 1026 (1.5 1010)2] = 1.17 1045 kg m2 5. For tangent in plane

I = 3

2(MR2) =

3

2 6 = 9 kg m2

For tangent perpendicular to plane I = 2 MR2 = 2 6 = 12 kg m2 6. M.I. for disc through centre.

Perpendicular to plane I = 1

2MR2

K = I

M

R = 2K = 2 6 = 8.4853 cm. i. M.I. for tangent perpendicular to plane

of disc,

I = 3

2MR2

K = 3 3

R 8.48532 2

= 10.3923 cm.

ii. M.I. along diameter,

I = 1

4MR2

K = 1

4R = 4.2427 cm.

iii. M.I. for tangent in plane of disc

I = 5

4MR2

K =5

4R

=5

4 8.4853

= 9.4869 cm

7. I = 1

2MR2

= 1

2(R2 t ) R2

= 1.584 107 g cm2

8. M.I. =25MR

4=

25 0.6 0.15

4

= 0.0169 kg m2

9. I = 2MR

2

MK2 = 2MR

2

K = R

2 =

0.12

2 = 0.085 m

10. ΔER = 1

2I 2 2

2 1

I = MK2 and = 2n

ΔER = 1

2 1.8 0.22 42 (82 62)

= 36.79 J 11. L = I

= 2

5MR2

2

T

= 2

5 1029 (5 1014)2

9

2

5 10

= 1.257 1049 g cm2 s1

12. n = 120

60= 2 r.p.s = 2 Hz

T = 1

n= 0.5 s

= 2 n = 4 rad/s R = D/2 = 3/2 m v = R = 6 m/s.

Rotational Motion 03

2


13. = I = (MK2) = 50 (0.5)2 25 = 312.5 N m 14. = 2n = 2(10) 2 2

2 1 = 2

= 400 10

40

rad/s2

= I

= 6 103 10

= 1.91 102 N m 15. n2 = 1 r.p.s. 2 = 1 + t

= 2 (1)

60

=

30

rad/s2

I =

20

/ 30

=

600

kg m2

16. 2 = 1 + t

= (20 4)

8

2 rad/s2

I = /

I = 5000

2= 2500 kg m2

17. L = I = 21MR

2

(2n)

= 1

2 2 (0.25)2 2 8

= 3.142 kg m2 s1

18. I = 1

2MR2 = 1.25 kg m2

L = I = (1.25) 20 = 25 kg m2 s1

ER = 1

2I2 =

1

2(1.25) (20)2 = 250 J.

19. P = 2 = I 2

= 2 1

t

=

2 (2) 0

10

P = 50 0.4 4 = 789.57 W.

20. = v 4

r 0.12

E = 1

2Mv2 +

1

2I2

E = 1

2 2 42 +

1

2 0.2

24

0.12

= 127.11 J

21. Neglecting M.I. of rod, M.I. of system = M (L/2)2 + M (L/2)2

= 2

22ML 1ML

4 2

= 1

2(0.2) (1.2)2

= 0.144 kg m2

ER = 1

2I2 =

1

2(0.144) (2 1)2 = 2.8425 J

22. (P.E)top = (K.E)bottom

Mgh = 1

2Mv2 +

1

2I2

For rolling cylinder,

I = 2MR

2, v = R

Mgh = 1

2M(R)2 +

1

2

2MR

2

2

2 = 2

4gh

3R

(ER)bottom = 1

2I2 =

2

2

1 MR 4gh

2 2 3R

= Mgh 6 9.8 8

3 3

= 156.8 J

23. E = ET + ER

= 1

2Mv2 +

1

2I2

E = 1

2Mv2 +

1

221

MR2

2 [ v = R]

= 1

2Mv2 +

1

4Mv2 =

3

4Mv2

K.E.of rotation

totalenergy =

2

2

1I

23

Mv4

=

2

2

1Mv

43

Mv4

= 1

3

1

3 100% = 33.33%

24. ET + ER = E

1

2Mv2 +

1

2I2 = E

1

2I2 =

1

221

MR2

2 = 1

4Mv2

3


1

2Mv2 +

1

4Mv2 = E or

3

4Mv2 = E

ER = 1

2I2 =

1

4Mv2 =

E

3

ER = Mgh

3=

20 9.8 4

3

= 261.33 J

25. (K.E.)bottom = (P.E.)top

1

2Mv2 +

1

2I2 = Mgh

1

2Mv2 +

1

221

MR2

2 = Mgh

3

4Mv2 = Mgh

h = 2 23v 3 7

4g 4 9.8

= 3.75 m

If s is the distance up the inclined plane, then

as sin = h

s,

s = h

sin =

3.75

sin 45 = 5.303 m.

26. I = 2 22MR Mh

5

= 2 223.1 0.25 3.1 0.25

5

= 0.27 kg m2 27. M.I.= (M.I.)A + (M.I.)B + (M.I.)C + (M.I.)D = 2

1Md + 22Md + 2

3Md + 24Md

= 4 (0.5)2 + 4 (0.5)2 + 4 (0.5)2 + 4 (0.5)2

= 4 kg m2

28. IC = 2ML

12

I0 = IC + Mh2 = 2ML

12+ M

2L

2

= 2ML

3

= 22.1 1

3

= 0.7 kg m2. 29.

I = 2ML

12+ Mh2

= 2

2M LM 0.1

12

= 22.62.6 0.1

12 = 0.243 kg m2.

30. IC = 0.25 Mh2

= 0.25 3 (0.2)2 = 0.13 kg m2 31. I = MR2 + Mh2 = 4.2 (0.4)2 + 4.2 (10.1)2 = 0.714 kg m2. 32. I0 = IC + Mh2

= 2MR

2 + Mh2 =

24.6 2

2

+ 4 (0.25)2

= 9.2 + 0.25 = 9.45 kg m2.

33. 2

2 2

1 1

T R

T R

T2 = 24 2

1

1

R / 2

R

= 6 hrs.

34. I1 = M.I. of disc about vertical axis I2 = M.I. of disc about same axis with wax Using parallel axes theorem, I1 (2n1) = (I1 + Mh2) (2n2) 125 I1 = 75 I1 + 75 (50 103 0.12) I1 = 7.5 104 kg m2 35. I1 1 + I2 2 = (I1 + I2)

= 1 1 2 2

1 2

I I (0.5 20 ) 0

(I I ) 2.5

n = 2

= 2 r.p.s. = 120 r.p.m.


1. Option (A) M.I. = 2MR

2

Option (B) M.I.= 2MR

4

Option (C) M.I. = 25MR

4

Option (D) M.I.= 23MR

2

h

0.2 m 1 m

new original

4


2. M.I. of solid sphere rotating about its axis,

I = 22MR

5

Radius of gyration, K = I

M

K = 22R

5 =

2R

5

3. I1 = 2ML

12,

I2 =2ML

3

1

2

I

I =

2ML

12

2

3

ML =

1

4

4. M.I. = M2 2R L

4 12

MK2 = M2 2R L

4 12

K =

1/22 20.6 1.2

4 12

= 0.458 m

5. M.I. of ring about an axis passing through its

centre and perpendicular to its plane, I = MR2 = 0.25 (0.5)2 = 0.0625 kg m2

6. M.I. in 1st case = MR2 = I = 2.62 kg m2

M.I.diameter =

2MR

2=

I

2=

2.62

2kg m2 = 1.31 kg m2

7. M.I. = 2ML

12

MK2 = 2ML

12 or K =

1/22L

12

=

1/221.25

12

= 0.36 m 8. M.I. of a ring about tangent perpendicular to

its plane, I = 2MR2

Radius of gyration, K = I

M

K = 2 R 9. M.I. of ring about transverse axis passing

through its centre I = MR2.

Radius of gyration K = 2IR

M

= R = D/2 = 0.5 D

10. M.I. = 1

4 MR2 =

1

4 2.5 (0.2)2

= 2n = 2 40 = 80 rad/s.

ER = 1

2I2

= 1

2

22.5 0.2

4

(80)2

= 789.57 790 J

11. ER = 1

2I 2 2

2 1

2 22 1 = R2 E

I

=

2 1500

1.5

22 = 2000 + 2

1

= 2000 + (2 2)2 = 2000 + 157.91 2 = (2157.75)1/2 = 2n

n = 1/ 22157.75

2 = 7.4 r.p.s.

12. ER = 1

2 I2

I = 2MR2 = 2 0.25 (0.1)2 = 5 103 kg m2

ER = 1

2 5 103 (2)2 = 0.01 J.

13. I = 2ML

12=

20.5 1

12

= 0.042 kg m2

ER = 1

2I2 =

1

2 0.042 (2 2)2 = 3.32 J

15. 1 = 2n1 = 2 7.5 = 47.1 rad/s, 2 = 0 = 2 no. of rotations = 2 9 = 18

= 2 2

2 1

2

=

21

2

=

247.1

36

= – 19.625 rad/s2 Negative sign indicates retardation, hence

neglecting,

I =

= 22 10

19.625

= 10.19 kg m2.

16. I = 2MR

4 =

20.5 0.15

4

= 2.8125 103 kg m2

= I = 2.8125 103 5 = 0.014 N m

5


17. 1 = 2 200

60 =

400

60 rad/s

= 2 1

t

=

400 10

60 12

= 400 1

60 12

rad/s2

Negative sign indicates retardation, hence neglecting,

= I = 0.8 400

60 12

= 1.4 N m.

18. a = 2

gsin3

= 1 3asin

2g

a = g

3

= 1 3 gsin

2 3 g

= 1 1sin

2

= 30

19. E = 2

22

1 KMv 1

2 R

For a sphere, K= 2

5R

E = 21 2Mv 1

2 5

= 7

10 1.5 12 = 1.05 J

20. v = 1/2

4gh

3

= 4.43 m/s

v2 = 0 + 2as

a = 2v

2s= 24.43

2 5.8= 1.69 m/s2.

21. K.E1 = K.E2

2 21 1

1 1I mv

2 2

25 36 20v

2 2

90 = 10 v2

v = 3 m/s 22. For hollow cylinder, v1 = (gh)1/2

h = 2v

g= 24.2

9.8= 1.8 m

For solid cylinder,

v2 = 1/2

4gh

3

= 1/2

4 9.8 1.8

3

= 4.85 m/s

23. a = gsin

2

= sin1 2a

g

= sin1 2 2

9.8

= 24 5 20

24. v1 = gh = g v1 = g = 3.13 m/s

v2 = 4gh

3=

4g

3 v2 = 3.62 m/s

v3 = 10gh

7=

10g

7 v3 = 3.74 m/s

v3 > v2 > v1 25. I = 2MR2

R = 1/2

I

2M

= 1/2

0.05

2

= 0.158 m

v = R = 3.6 0.025 = 0.5688 m/s

E = 21I

2 + 21

Mv2

= 1

2 0.05 (3.6)2 +

1

2 1 (0.5688)2

= 0.49 J. Nearest answer is (B)

26. ER = 2I

2

=

20.02 4

2= 0.16 J

E = ET + ER = 2.25 + 0.16 = 2.41 J.

27. E = 2

22

1 KMv 1

2 R

For a hollow cylinder, K = R

E = 21Mv 2

2 = Mv2

= 1.8 (2.8)2 = 14.11 J.

28. I = 2 22MR Mh

5

= 2

56 (0.9)2 + 6 (0.5)2

= 1.944 + 1.5 = 3.44 kg m2

29. I = 2ML

12 + Mh2

L2 = 2 12I Mh

M

L = 1/2

2 12I Mh

M

= 1/2

2 120.5 2.6 0.4

2.6

= 0.62 m.

6


30. Iz = Ix + Iy = 2Ix

Ix = zI

2=

2MR

4 =

23.8 0.25

4

= 0.059 kg m2.

31. Ix = 2

zC

I MRI

2 4

I0 = IC + Mh2

= 2MR

4+ MR2 = 25

MR4

= 256 0.5

4

= 1.875 kg m2.

32. I = 2

2MLMh

12

I = 21 L

12

+1 (0.4)2

L2 = [2 (0.4)2] 12 = 22.08 L = 4.69 4.7 m. 33. I0 = IC + Mh2 2

0MK = 2 2CMK Mh

2 2 2C 0K K h

= (0.28)2 (0.2)2 = 0.0384 KC = 0.196 m. 34. I0 = IC + Mh2 2

0MK = 2 2CMK Mh

K0 = 1/22 2CK h

= (0.52 + 0.22)1/2 = 0.54 m. 35. I0 = IC+ Mh2

h2 = (I0 IC).1

M = 2 2

0 C

1MK MK .

M

= 2 20 CK K

= (0.752 0.652) = 0.14 h = 0.37 m = d

36. I0 = IC + Mh2 = 2

5MR2 + Mh2

= 2

2 DM

5 2

+M2

D

2

= 2 22 D MD

M5 4 4

= 2

15

2MD

4

= 27 MD

5 4 = 27

MD20

37. From parallel axis theorem, I0 = IC + Mh2

= 22ML L

M12 4

= 2 2ML ML

12 16 =

27ML

48

38. Before coupling total angular momentum of

system L1 = I1 + 0 = I1 After coupling L2 = 2I According to conservation of angular

momentum, L1 = L2

I1 = 2I

= 1

2

= 1.6 kg m2/s

39. E = 21I

2

2 = 2E

I

= 1/2

2E

I

L = I = I 1/2

2E

I

= (2EI)1/2

= (2 25 0.42)1/2 = 4.58 kg m2/s

40. I1 = 2MR

2= 1

20.25

2= 0.03125 kg m2

L1 = I11 = 0.03125 2 L1 = 0.0625 kg m2/s

I2 = 2 20.5

2= 0.25 kg m2

L2 = 0.25 2 = 0.5 kg m2/s

2

1

L

L= 8

41. Since angular momentum must be conserved,

I11 = I22

I2 = 1 1

2

I

I0 IC

L/4 L/4 L/2

1

Target Publications Pvt. Ltd. Chapter 04: Oscillations

Hints to Practice Problems 1. Original length = 60 cm, x1 = 2 cm, f1 = 20 gm-wt, f2 = 500 gm-wt f = kx(numerically),

1

2

f

f= 1

2

kx

kx = 1

2

x

x

or x2 = 21

1

fx

f =

500

20 2

x2 = 50 cm New length = 60 cm + 50 = 110 cm 2. Comparing with,

2

2

d x

dt + 2x = 0 we get,

2 = 144 or = 12 rad/s

T = 2

12

= 0.523 s

n = 1

0.522 = 1.912 Hz

3. A = 0.06

2 = 0.03 m

x = A sint = Asin(2nt)

= 0.03 sin (2 20 1

120)

x 0.026 m 4. v = 2 2A x

3 = 2 2A 4 ….(i)

4 = 2 2A 3 ….(ii)

On squaring and then dividing equation (i) by equation (ii) we get,

2

2

9 A 16

16 A 9

On solving we get, A = 5 cm. From equation (i),

3 = 2 25 4 = 1 rad/s

T = 2

1

= 6.28 s

Path length = 2 A = 2 5 = 10 cm

5. 2 2A x = A

2

which on solving yields,

x = 3

2A = 0.866

6. K = f 1

2N / m.x 0.5

T = 0.5

22

= 3.14 s

7. A = 0.2

2 = 0.1m

a = 2x = 2 rad/s

v = 2 2A x

= 2 2 2(0.1) (0.06)

v = 0.16 m/s 8. For a second’s pendulum, T = 2s and L = original length

T L ….(i) If T1 and L1 are the values for faulty pendulum

then,

T1 1L where,

L1 = L + 4% of L = 1.04 L

T1 1.04L ….(ii) Dividing equation (ii) by (i),

1T

T=

1.04L

L = 1.04

T1 = 1.04 T = 1.0198 2 T1 2.04 s 9. Here, 2A = 4 or A = 2 cm

= 2

2 3

= 1

3rad/s

Now, v = a

2 2A x = 2x

On solving we get,

x = 3

2 A =

3

2 2 = 3 cm

= 1.732 cm

Oscillations 04

2


10. x = 7 sin 3 t6

A = 7, = 3 and

phase angle = 3t + 6

i. For t = 2s,

x = 7 sin(6 + 6

) = 7

1

2

= 3.5 cm

ii. v = 2 2A x

= 3 2 2(7) (3.5)

= 57.1057 cm/s iii. a = 2x, a = (3)2 3.5 a = 310.58 cm/s2

iv. Phase angle = 3(2) +6

Phase = 37π

6

rad

11. amax = 2A

= 216

4

= 2 rad/s

T = 2

, 2

2

= 1s

n = 1 1

T 1 = 1 Hz

12. max

max

a

v =

2A 8

A 4

on simplification gives, = 2 rad/s

T = 2

2

= 3.14 s

13. a = 4 cm/s2 when x = 4 cm, Using, a = 2x we get,

2 = a

x =

4

4 = 1 or = 1 rad/s

T = 2

= 2

1

= 2 3.14 = 6.28 s.

14. n = 1 k

2 m we get,

n = 31 4.9 10

2 100

n = 1.114 Hz

15. T = 2m

k

= 22

200

200

T = 2 s 16. x = 5 cm = 5 102 m,

A = 3 cm = 3 102 m

T = 2x

g = 2 3.14

25 10

9.8

T = 0.4486 s

n = 1

T =

1

0.4486 = 2.229 Hz

T.E. = 1

2m2A2

= 1

2 m

2

2

4

7

A2

= 1

20.5

2

2

4 (3.14)

(0.4486)

(3 102)2

= 0.04409

T.E. 4.41 102 J Now, v = vmax

( the mass is at mean position)

= A

= 3 102 2 3.14

0.4486

v = 0.42 m/s 17. K.E. = 3 P.E.

1

2m2(A2 x2) = 3

1

2m2x2

On simplifying we get,

x = A

2=

10

2= 5 cm

18. T.E. = 1

2m2A which on substitution and

solving gives,

A = 10

Now, v = 2 2A x

vx = 0 = 2 2A 0

= A

vx = 10 m/s

3

Chapter 04: Oscillations

19. x = 0.1 sin

t4

On comparing with x = A sint we get,

= 4

rad/s and A = 0.1 m

Given, x = 0.05 m and m = 0.05 kg

P.E. =1

2m2x2

=1

20.05

2

16

(0.05)2

= 3.85 105 J

K.E. = 1

2m2 (A2 x2)

= 1

2 0.05

2

16

[(0.1)2 (0.05)2]

= 1.155 104 J

T.E. = 1

2m2A2

= 1

20.05

2

16

(0.1)2

T.E. = 1.541 104 J

20. P.E.

K.E. =

2 2

2 2 2

1m x

21

m (A x )2

= 2

2 2

x

A x=

2

2 2

3

5 3

P.E.

K.E. = 9 : 16

21. K.E. = 2 P.E. ….(i)

1

2m2(A2 x2) = 2

1

2m2x2

on solving gives,

x = A

3= 5.774 cm

22. T = 1s

= 2

T

= 2 rad/s

P.E. = 1

2m2x2

= 1

2 20 42 (1)2

….[ x = 4 cm and A = 4 cm]

P.E. = 394.4 erg

K.E. = 1

2m2(A2 x2),

= 1

2 20 42 [(4)2 (1)2]

K.E. 5915.8 erg Now, k = m2 = 20 42 k = 788.77 dyne/cm 23. A = 0.2 m, x = 0.04 m, T = 6.28 s, m = 300 103 kg,

K.E. = 1

2m2(A2 x2)

= 1

2m

2

2

4

T

(A2 x2)

= 1

2 300 103

2

2

4 (3.14)

(6.28)

2 2[(0.2) (0.04) ]

K.E. = 5.76 103 J

P.E. = 1

2m2x2

= 1

2 300 10–3

2

2

(3.14)

(6.28) (0.04)2

P.E. = 2.4 104 J 24. Given that,

x1 = 12 sin 4 t6

x2 = 5 sin 4 t4

A1 = 12, 1 = 6

and A2 = 5, 2 =

4

R = 2 21 2 1 2 1 2A A 2A A cos( )

= 2 2(12) (5) 2(12)(5)cos6 4

17 cm 25. For the two given S.H.M.s, A1 = 3cm, A2 = 4cm Now,

R = 2 21 2 1 2 1 2A A 2A A cos(a )

i. For 1 2 = 0, cos 0 = 1

R = 2 2(3) (4) 2(3)(4)(1)

= 49 R = 7 cm

4


ii. For 1 2 = 60,

cos 60 = 1

2

R = 2 2 1(3) (4) 2(3)(4)

2

R 6.1 cm iii. For 1 2 = 90, cos 90 = 0

R = 2 2(3) (4) 2(3)(4)(0)

R = 5 cm 26. Two S.H.M.s are:

x1 = 5 sin 6 t3

x2 = 12 sin 6 t6

A1 = 5, 1 = 3

and

A2 = 12, 2 = 6

i. R = 2 21 2 1 2 1 2A A 2A A cos( )

= 2 2(5) (12) 2(5)(12)cos3 6

R = 16.52 cm

ii. = tan1 1 1 2 2

1 1 2 2

A sin A sin

A cos A cos

= tan1 5sin 12sin

3 6

5cos 12cos3 6

= tan–1 10.33

12.892

= 38 42

27. x = A

2

= 2

T

=

2

2

= rad/s

Now, x = A sint

A

2 = A sin t

1

2 = sint

t = 6

t = 1

6s

28. Given that, T1 = 1 s, L1 = 0.36 m L2 = 0.36 + 0.13 = 0.49 m

At a given place, T L

T1 = K 1L

T2 = K 2L

2

1

T

T = 2

1

L

L

T2 = 1 49

36or T2 = 1.166 s

T2 T1 = 1.166 1 = 0.1666 s

Loss in 24 hours

= 0.1666

1 24 = 3.998 h 4 h

29. If frequencies, time periods and acceleration

due to gravity for the 1st and 2nd case are n1, T1, g1 and n2, T2, g2 respectively then,

n1 = 50

3 60Hz and T1 = 2n1 =

2 50

180

T1 = 10

18

s

Similarly,

n2 = 49

3 60Hz and

T2 = 2 49

180

s

Using, T = 2L

g, we get,

1

2

T

T = 1 2

2 1

L g

L g

= 10

182 49

180

2

1

g

g=

50

49 ….( L1 = L2)

2

1

g

g =

2500

2401

2

1

g

g = 1.041

5


30. In a day, we have 24 60 60 = 86400 s. A seconds pendulum with T = 2 s

makes86400

2= 43200 = oscillations in a day

Time period of faulty pendulum,

T = 24 60 60 10

43200

T = 2.0023 s If length of faulty pendulum is L then,

T = 2L

g

L = 2

2

gT

4

L = 0.9962 m, on substitution. If L1 is length of seconds pendulum (T1 = 2 s) then,

L1 = 2

12

gT

4 = 0.9939 m, on substitution.

The required decrease = L L1 = 2.3 103 m Required decrease = 0.23 cm

31. Using A = bt

2m0A e

and applying it for case (1) and case (2) we get,

A1 = A0 e–50bT and

A2 = A0 e–150bT = A0

3(–50bT)e

A2 = A0

31

4

= 10 1

64 = 0.156 cm


1. x = 4 10–2 m

k = F

x (numerically)

= 2

2N

4 10 m = 50 Nm–1

Now, T = 2m

k =

0.125

50= 0.05 s

2. m = 10 103 kg

Using, Fmax = m2

2

4

T

A

2 = 10 103 2

2

4

( / 4)

A

A 3.1 m

3. Comparing given D.E. with

2

2

d x

dt = 2x we get,

= 10 and A = 2

cm

Now, vmax = A = 2

10 = 20 cm/s

4. P.E., u = 1

2kx2

2u = kx2 ( f = kx)

2u

f + x = 0

5. General equation of S.H.M. is y = A sin(t + )

= 0.4 sin2

t0.2 2

= 0.4 sin(10t + 2

)

y = 0.4 cos (10t) 6. Path length = distance between two extreme

positions of the particle in S.H.M. = AB. 8. At t = 0, x = r cos (0) at t = 2, x = r

Here, = 2

2

T

=

2

= 4 s

This means that the particle moves from one extreme position to other extreme position.

Distance covered = r + r = 2r 9. At t = 0,

y = sin3

y = sin60 = 3

2m

10. Equation of S.H.M. is,

y = A sint = A sin2 t

T

Here, y = A

2

A

2 = A sin

2 t

4

or t =

1

3s

6


11. For maximum velocity, = 0.

vmax = A 12. Equation of S.H.M. is,

x = 10 2 (sin 2t + cos 2t)

x = 10 2 sin 2t + 10 2 cos2t

Comparing this equation with the standard equation of S.H.M. which is,

x = A1 sint + A2 cost

we get,

A1 = 10 2 and A2 = 10 2

R = 2 21 2A A

= 2 2(10 2) (10 2)

A = 20 cm 13. Equation of S.H.M. is,

x = 8 sint + 6 sin(t + 3

)

Comparing with the standard equation,

x = A1sint + A2sin(t + ),

we get, A1 = 8 cm,

A2 = 6 cm and = 3

R = 2 21 2 1 2A A 2A A cos

= 2 2(8) (6) 2(8)(6)cos3

R 5 6 cm 14. amax = A2

A = max2

a

=

12

16

A = 0.75 cm

Path length = 2A = 2 0.75 = 1.5 cm

15. = 2

T

=

2

6

=

3

v = A cos t

2

=

3

A cos

3

3

A = 1.5 or 2A = 3 cm

16. v = 1

2 vmax

2 2A x = 1

2A

2 2A x = A

2

A2 x2 = 2A

4

x = 3A

2

17. Average speed = Distance covered

time

= 4A

T

Average speed = 4A2

= 2A

18. v = 2 2A x

v = 2

2 AA

2

= 3

2A

vmax = A v = 3

2vmax

19. a = 2x a = 0 when x = 0 20. Equation of S.H.M., x = 2 sin(2t).

dx

dt = 2 cos(2t).2

= 4 cos(2t)

a = 2

2

d x

dt = 4 [ sin(2t).2]

= 82 sin(2t) = 82 sin(2 0.25) |a| = 82 cm/s2 21. Equation of S.H.M. is,

y = 2 sin4

(t + 1)

= 2 sint

4 4

Comparing with y = A sin (t + )e we get,

A = 2 and = 4

amax = 2A = 2

16

2 =

2

8

cm/s2

7


22. A = 3

10

cm, n = 10 Hz

amax = 2A = (2n)2A

= 42 100 10

amax = 4000

cm/s2

23. Equation of S.H.M. is, x = 4 sin 3.14 (t + 2) cm

dx

dt = 4 cos 3.14 (t + 2) (3.14)

a = 2

2

d x

dt

= 4 sin [3.14(t + 2)] (3.14)2 at = 1.5 = 4 sin [3.14(1.5 + 2)] (3.14)2

= 4(3.14)2 sin3

2

= 4(3.14)2 ( 1) at = 1.5 = 42 cm s2

24. a = 2x or = a

x

= 2n

n = 1 a

2 x =

1 0.5

2 0.005

n 1.6 Hz 27. y1 = a sin(t kx) and y2 = b cos(t kx)

y2 = b sin(t kx + 2

)

Phase difference

= (t kx +2

) (t kx) =

2

29. y1 = 20 sin 2t and y2 = 25 cos 2t

y2 = 25 sin(2t + 2

)

Phase difference = (2t + 2

) 2t =

2

rad

30. Equation of S.H.M., x = A sin(t + ) = A sint cos + A cost sin Comparing with given equation, x = 8cost + 6sint

we get,

A sin = 8 and A cos = 6

Asin

Acos

= 6

tan = 4

3

31. As can be seen from the graph, one of the

S.H.M.s start from the extreme position

(X = A) corresponding to initial phase of 2

and the second S.H.M. starts from a point

midway between O and A,A

x2

corresponding to initial phase of 4

.

Thus, 1 = 2

and 2 =

4

Phase difference,

= 1 2 = 2

4

=

4

rad


2

2

d x

dt + x = 0

2 = or =

T = 2

= 2


x = 0.25 sin1

20t2

Comparing with

x = A sint we get,

= 20 T = 2

= 2

20

=

10

s

34. v = 2 2A x

v2 = 2(A2 x2) On substituting the values of v and x we get,

100 = 2A2 162 ….(i)

and 64 = 2A2 252 ….(ii)

Solve equations (i) and (ii) to obtain T = s

8


35. 2A = 4 cm

A = 4

2 = 2 cm = 0.02 m

v = A = 2 A

T

2

=

2 3.14 0.02

T

v = 0.08 s

36. v = dy

dt

= A cost Substitute v = 0.4 and t = 2 to obtain A = 1.44 m 37. Equation of S.H.M. is,

x = 10 cos 2π t3

Comparing given equation with x = A cos(t + ),

Using = 2n we get, n = 1 Hz 38. m = 100 103 kg,

n =1 k

2 m=

3

1 40

2 3.14 100 10

n 3 s1 39. n = 1.5 Hz

n 1

m

2

1

n

n = 1

2

m

m

Substitute m2 = 1m

4

n2 = 2n1 = 2 1.5 = 3 Hz

40. Using, =k

mand substitution we get,

f = 2.22 rad/s 41. a = 6 m/s2 x = 0.015 m a = 2x or

= a

x

On substitution we get, = 20 rad/s

42. P.E./A

2x =

22

2 2

1 Am

12 21 4m A2

43. K.E. = P.E.

1

2m2(A2 x2) =

1

2m2x2

on solving x = A

2

44. x = 1

2A

T.E. = 1

2m2A2

K.E. = 1

2m2(A2 x2)

= 1

2m2(A2

2A

4)

= 3

4(

1

2m2A2)

K.E. = 3

4 T.E.

45. K.E. = 1

4T.E.

1

2m2 (A2 x2) =

1

8m2 A2

On simplifying we get,

x = 3

2A

46. P.E.max = ETotal = 2 J 47. x = 5 cm = 5 102 m

1

2 m2A2 = 200

1

2m2x2 = 50

On dividing,

2

2

A

x = 4

A2 = 4 52 104 A2 = 102 or A = 0.1 m

48. K.E. = 2 2 21m (A x )

2

= 1

2m2

22 A

A4

K.E. = 3

4 16 = 12 J

9


49. k = 3 103 N/m x1 = 4 102m, x2 = 6 102 m For a spring,

P.E. = 1

2kx2

W = (P.E.)final (P.E.)initial

= 1

2k (x 2

2 x 21 )

= 1

2 3 103 (62 42) 10 4

= 3 N 50. R1 = 10 cm for the first particle For the second particle,

x2 = 5 sin 3t + 5 3 cos 3t

R2 = 2 21 2 1 2 1 2A A 2A A cos( )

= 2 2(5) (5 3) 2(5) (5 3)cos2

22R = 25 75 = 100 = 10 cm

1

2

R

R=

10

10 =

1

1

52. y1 = 10 sin 2 t3

y2 = 4[sin 2t + 3

4cos 2t]

A1 = 15

A2 = 2

2 3(4) 4

4

A2 = 16 9 = 5

1

2

A

A=

15

5 = 3 : 1

53. vmax = A vmax A velocity will be doubled. 54. The time period is independent of amplitude. 55. Linear momentum will be maximum for

v = vmax = A

T = 2

M2A2

2T

M = 2A2 = 2

maxv

vmax = 2T

M

Maximum linear momentum = M vmax

= M2T

M

pmax = 2MT 56. No. of oscillations for two pendulums are 9

and 6 respectively,

T = 2L

g

T

9 = 2 1L

g and

T

6 = 2 2L

g

1

2

L

L =

26

9

1

2

L

L =

4

9

57. gM = Eg

6

T = 2s for a seconds pendulum TE = TM

2 E

E

L

g = 2 M

M

L

g

On squaring and rearranging,

LM = E

E

L

g gM

LM = 1

6 m

58. A = 0A

8, t = 6 min

For a damped oscillator, amplitude,

A = t0A e

0A

8 = 6

0A e

e6 = 1

8

Amplitude after 2 min;

A1 = 20A e =

16 3

0A e

A = A0

1

31

8

= 0A

2

10


59. Initial mechanical energy of oscillator

E1 = 1

2kA2

bt

me

i.e. at t = 0, E1 = 1

2kA 2

because of damping, energy at time t, where b is damping constant.

E2 = 1

2 kA2

bt

me

,

Given that, E2 = 1E

2

1

2

E

E= 1

1

EE

2

= bt

m

1

e

= bt

me

bt

m= loge2

t = emlog 2

b = e0.4log 2

0.08= 5 loge2

60. Number of oscillations for two cases are 10

and 20 respectively Using A = A0e

bt Time for 10 oscillations, t1 = 10 T For the 1st case,

A1 = 0A

3 and t1 = 10 T

0A

3 = A0e

10bT

e10bT = 1

3

For the 2nd case, t2 = 20T A2 = bt2

0A e = A0e20bT

A2 = A0(e10bT)2 = A0

21

3

= 0A

9

1

Chapter 05: Elasticity


1. d = 8 103 m

r = d

2= 4 103 m

= 0.3

L100

= 3 103 L

F = 2r Y

L

= 3 2 3 113.14 (4 10 ) 3 10 L 20 10

L

= 3014.4 102 F = 3.014 105 N

2. F = 2r Y

L

= 24 113.14 5 10 0.012 1.23 10

1.2

= 0.9655 103

= 9.655 102

F = 965.5 N 3. Given that, for the two wires,

L1 = 2L2; r1 = 2r

2;

F1 = F2; Y1 = Y2

Using, = 2

FL

r Y, we get

1 = 1 12

1 1

F L

r Y and 2 = 2 2

22 2

F L

r Y

On dividing,

1

2

= 1 12

1 1

F L

r Y 2 2

22 2

F L

r Y= 1 1

21

F L

r Y

22 2

2 2

r Y

F L

= 2

1 22

2 1

L r

L r =

2

1 2

2 1

L r

L r

= 2 (2)2 = 8

1

2

= 8 : 1

4. Given that, for the two wires, Y1 = Y2; L1 = L2

1

2

F

F= 2; r1 = 2r2

1

2

= 1.5 mm = 1.5 103 m

Using,

= 2

FL

r Y we get,

1 = 1 12

1 1

F L

r Y and 2 = 2 2

22 2

F L

r Y

1

2

= 1 12

1 1

F L

r Y 2 2

22 2

F L

r Y= 1 1

21

F L

r Y

22 2

2 2

r Y

F L

= 2

1 2

2 1

F r

F r

= 2

12

2

= 1

2

1 = 2

2

=

31.5 10

2

= 0.75 103 m = 0.75 mm 5. d = 3 mm = 3 103 m r = 1.5 103 m

= 2

FL

r Y

= 23 11

30 2

3.14 1.5 10 1.1 10

= 7.7205 105 m

= 7.720 102 mm

6. = 2

mgL

r Y

= 3 2 11

3 9.8 5.06

3.14 (2.3 10 ) 2.131 10

= 4.2027 105 m 4.203 105 m

Elasticity 05

2


7. Side = 5 cm = 5 102 m x = 0.65 cm = 0.65 102 m A = (5 102) m2 = 25 104 m2 F = 0.25 N

Shearing stress = F

A

= 4

4

0.25 10

25 10

= 100 N/m2

Shearing strain = x

L=

2

2

0.65 10

5 10

= 0.13

8. d = 2 mm = 2 103 m r = 1 103 m,

Y = 2

mgL

r

Y = 3 2 3

5 9.8 1

3.14 (1 10 ) 1 10

= 15.605 109 Y = 1.56 1010 N/m2 9. LSt = 5 m, ASt = 3 105 m2 LCo = 3 m, ACo = 4 105 m2

St = Co ; FSt = FCo

Y = FL

A

St St St Co Co

Co St St Co Co

Y F L A

Y A F L

= St Co

Co St

L A

L A

= 5

5

5 4 10

3 3 10

= 20 : 9 10. F = 2 105 dyne,

L = , A = 2 cm2

Y = FL

A

= 52 10

2

= 1 105 dyne/cm2 11. L = 3 m, d = 2 mm = 2 103 m r = 1 103 m Y = 1.8 1011 Nm2 M = 50 kg, F = 50 9.8 = 490 N As the load is shared by four wires, Load on

each wire = F = mg/4.

Using,

= F / 4 L

A Y we get,

= 2

F / 4 L

d Y

= 2 113

50 9.8 / 4 3

1.8 103.14 10

= 65.021 105

= 6.502 104 m

= 0.65 mm 12. K = 3 106 N/m2,

dv

%v

= 0.2%,

K = dp

vdv

dv

v=

1dp

K

0.2

100=

6

1dp

3 10

dp = 60.2 3 10

100

= 0.6 104 = 6 103 N/m2 (numerically) 13. K = 1.25 1010 N/m2 dp = 20 atm = 20 105 N/m2 1 atm = 105 N/m2

K = 1

dpv

dv or

dv dp

v k we get,

dv

v=

5

10

20 10

1.25 10

= 16 105

= 0.016

dv

%v

0.016% (numerically)

14. Compressibility = 1

K

= 6

1

3 10

1

K = 3.33 107 m2/N

15. v1 = 100 litre, v2 = 100.5 litre, dv = 100.5 100 = 0.5 lit dp = 100 atm 1 atm = 1.013 105 Pa

3


K = 1

dpv

dv

K = 100 100atm

0.5

= 100 5100 1.013 10

0.5

K = 2.026 109 Pa (numerically) 16. dv = 0.0098%, dp = 100 atm

K = dp

vdv

we get,

dv P

v K

dv dp

v K

0.0098

100

=

100

K

K = 100 100

0.0098

K = 1.02 106 N/m2

17. v = 5 lit = 5 103 m3 dp = 10 atm = 10 1.013 105 N/m2

1

K= 5 1010 m2/N

1

K=

dv 1

v dp or

dv = 1

v dpK we get,

dv = 5 1010 5 103 10 1.013 105 dv = 25.325 107 m3

18. dp = 2.94 106 N/m2, v = 1000 cm3 = 1000 106 m3 = 103 m3 K = 6.9 1010 N/m2

K = dp

v.dv

dv = vdp

K=

3 6

10

10 2.94 10

6.9 10

= 2 1029.410 10

6.9

= 4.26 108 m3 4.3 108 m3

19. F = 2100 N, A = 3 106 m2 h = 0.1 m, x = 0.7 mm = 0.7 103 m

= Fh

Ax

= 6 3

2100 0.1

3 10 0.7 10

= 100 109

= 1 1011 N/m2 20. Side = 50 cm = 50 102 m A = (50 102)2 h = 5 cm = 5 102 m F = 9 104 N = 5.6 109 Pa Using,

= Fh

Ax we get,

x = Fh

A

x = 4 2

2 2 9

9 10 5 10

(50 10 ) 5.6 10

x = 3.214 106 m 21. Dimensions of plate are, 10 cm 10 cm 1mm A = 10 cm 10 cm = 10 102 m 10 102 m A = 102 m2 h = 1 mm = 1 103 m x = 1.2 10 mm x = 1.2 106 m = 5 1010 N/m2

= Fh

Ax

F = Ax

h

F = 10 2 6

3

5 10 10 1.2 10

1 10

F = 6 105 N

Shear strain = x

h

= 6

3

1.2 10

10

Shear strain = 1.2 103 22. L = 5 m, r = 1 mm = 1 103 m D = 2 103 m r = 1.5 108 m D = 3 108 m = 0.30

4


Using,

= DL

D

or

= DL

D

we get,

= 8

3

3 10 5

2 10 0.30

= 25 105 m

= 0.25 103 m 23. L = 3 m, D = 1 mm = 1 103 m

= 2 mm = 2 103 m

= 0.36

=

DDLDD

L

D = D

L

= 3 30.36 1 10 2 10

3

= 0.24 106 D = 24 108 m 24. r = 1 mm = 1 103 m D = 2 103 m M = 31.4 kg Y = 9 1010 N/m2, = 0.36 Using,

= 2

MgL

r Y and substituting in

= D.L

D.

we get,

= 2D Lr Y

D mgL

= 2D r Y

Dmg

D = 2

Dmg

r Y

= 3

3 2 10

0.36 2 10 31.4 9.8

3.14 (1 10 ) 9 10

D = 7.84 107

r = 77.8410

2

r = 3.92 107 m

25. = 0.42 1011 N/m2, K = 0.21 1011 N/m2,

9

Y=

1 3

K

9

Y=

11 11

1 3

0.21 10 0.42 10

119 10

Y

=

1 1

0.21 0.14

119 10

Y

=

0.35

0.21 0.14

Y = 119 0.21 0.14 10

0.35

Y = 0.756 1011 Y = 7.56 1010 N/m2 26. Y = 7.1 1010 N/m2 K = 7.7 1010 N/m2

9

Y=

1 3

K

10

9

7.1 10

10

1

7.7 10=

3

9 1

7.1 7.7 =

103 10

1.2676 0.1298 = 103 10

= 2.6368 1010 N/m2

Using,

= 3K 2

2 9K

we get,

= 10 10

10 10

3 7.7 10 2 2.6368 10

2 2.6368 10 9 7.7 10

= 10

10

(23.1 5.2736) 10

(5.2736 69.3) 10

= 17.8264

74.5736

= 0.239 0.24 27. Y = 3K (1 2) Y = 3 2.5 1011 (1 2 0.4) Y = 1.5 1011 dyne/cm2

28. Y = 2

MgL

r

= 3 2 3

10 9.8 3

3.14 (0.5 10 ) 3 10

= 124.84 109 Y = 1.248 1011 N/m2

5


Using, Y = 2 (1 + ) or

= Y

2(1 ) we get,

= 111.248 10

2(1 0.26)

= 0.4952 1011

= 4.952 1010 N/m2 5 1010 N/m2

29. L

= 0.1% =

0.1

100,

u = 1

L Y (strain)2

u = 1

2 1.1 1011

20.1

100

u = 5.5 104 J 30. L = 1.6 m, = 1.61 1.6 = 0.01 m

Stress = 2 102 N/m2

u = 1

2 stress strain

= 21 0.012 10

2 1.5

u = 0.67 J/m3

31. For the two cases, M1 = 2 kg, M2 = 7 kg, l1 = 0.5 mm = 0.5 103 m, l2 = 1 mm = 1 103 m, g = 9.8 m/s2 Using,

Work = 1

2Mgl

W = W2 W1

= 1

2(M2 l2 M1 l2)g

=1

2(7 1 103 2 0.5 103) 9.8

= 29.4 103 W = 2.94 102 J

32. u =1

2 Y (Strain)2

(Strain)2 = 2u

Y=

5

9

2 2.5 10

20 10

= 0.25 1014 Strain = 0.5 107 = 5 108 = 0.5 107

Stress = Y Strain = 20 109 5 108 = 10 102 N/m2 33. = 7800 kg/m3, m = 16 g = 16 103 kg, L = 250 cm = 2.5 m

= 1.2 mm = 1.2 103 m,

F = 80 N

Y = FL

A and A =

Volume

L

Y = FL L

m

= 2F L

m

= 2

3 3

80 7800 (2.5)

16 10 1.2 10

= 2.031 1011 N/m2

Using,

U = 1

2 F we get,

= 1

2 80 1.2 103

U = 4.8 102 J

Hints to Multiple Choice Questions 1. Load bearing capacity of a wire is independent

of its length.

2. Longitudinal strain = L

=

3

300= 0.01

3. Shear strain = L

L

=

3

2

0.6 10

60 10

= 4

1

6 10

6 10

= 103 = 0.001 rad.

4. Stress = F

A=

4

12

15 8 10 = 1000 N/m2

5. From, Y = FL

Al; l =

FL

AY

Increase in length on heating, l = l When the bar is permitted neither to expand

nor to bend, then

l = FL

AY= l

or F = A Y = 1 1012 105 100 = 109 dyne

6


6. Y = 2

T L

r or

2

T

r = constant

12

1

T

r= 2

22

T

ror 2

1

T

T=

2

2

1

r

r

= 2

2

1

= T2 = 4T1 = 4 500 = 2000 N 7. Breaking force area of cross section

1

2

F

F= 1

2

A

Aor

2

200

F= 1

1

A

2A

F2 = 400 kg-wt

8. Y = FL

Al or F = YA

l

l or F

A

l

1

2

F

F= 1 2

2 1

A

A

l

l=

21 222 1

r

r

l

l=

2

2

(2) 1

1 4 = 1 : 1

9. Y = stress required to double the length

of a wire

= 2

F

r=

7

4

2.4 10

2.4 10

= 1.0 1011 N/m2

10. Y = 2

MgL

r e

= 4 2 3

6.28 10 1

3.14 (10 ) 10

= 2 1012 N/m2

11. Y = Stress

Strain=

F / A

Strain=

F

A Strain

A = F

Y Strain=

4

9

100.3

7 10100

= 4

7

10

2.1 10 4.8 104 m2

12. Here, l = 4m = 400 cm, 2r = 5 mm or r = 2.5 mm = 0.25 cm F = 5 kg.wt = 5000 g.wt = 5000 980 dyne, Y = 2.4 1012 dyne/cm2

Y = 2

F

r

l

l

l = 2

Fl

r Y

= 2 12

(5000 980) 400

(22 / 7) (0.25) 2.4 10

= 0.0041 cm


K = 5 1010

= dv

dp v

dv = 5 1010 8 105 (10 103 cc) = 4 cc

14. K = dp

dv / v or dp = K

dv

v

hg = 8

3

14.7 10 0.3

100 9.8 10

= 7

5

4.5 10

10

= 450 m 15. Here, P = (1.230 105 1.02 105) = 0.21 105 Pa v/v = 15/100 = 0.15

Bulk modulus = p

v / v

= 50.21 10

0.15

= 1.4 105 Pa

16. Bulk modulus K = Fv

A v

P = F

A=

K v

v

= (2100) (0.002 /100 400)

400

= 0.042 kPa


K=

dv

vdp

dv = vdp C = 200 106 50 105 4 105 = 0.04 m3

18. = x

h =

36 10

0.2

= 3 102

= F

A=

6 2

1400

2 10 3 10

= 233.33 108 N/m2 233 108 N/m2 19. = 0.65 cm = 0.65 102 M,

= 5 cm = 5 102 m, F = 0.26 N,

= F

=

2 2

0.26

0.65 10 5 10

= 0.08 104 N/m2 = 800 N/m2

7


20. Modulus of rigidity = Shearingstress

Shearingstrain

Hence, shearing stress = F

A=

2

F

l

= 2 2

4

(6 10 )=

10000

9 Nm2

Shearing strain = l

l=

0.1

6= 0.016

= 10000

0.016 9 = 6.94 104 Nm2

7 104 Nm2.

21. = Y

12

= 2.2

12

= 1.1 1.0 = 0.1

22. Lateral Strain = L

L

= 20.25 0.2 10

2

= 2.5 104

23. L

L

=

6

100 100= 6 104

D

D

=

L

L

=

1

2 6 104 = 3 104

D

D

=

0.03

100= 0.03 %

24. 1 3

K

= 9

Y

1 3

K 3K =

9

Y

2

K=

9

Y

Y = 9K

2 = 4.5 K

25. = 1 Y

12 3K

= 11

11

1 1.5 101

2 3 10

= 1

4 = 0.25

26. = F L

A L

= 4

3

8 10 0.25

(0.25 0.05)(0.12 10 )

= 13.33109 N m2 For most materials, Y = 3 = 3 13.33 109= 4 1010 N m2.

27. Y = 2 (1 + )

Y

= 2(1 + ) = 2(1 + 2/3) = 2

5 10

3 3

28. Y = 3K (1 26) = 3 2.0 1011 (1 2 0.3) = 6 1011 0.4 = 2.4 1011 dyne/cm2

29. Y = 2 1010 N/m2

Strain = 0.04% = 0.06

100

Energy per unit volume

= 21Y Strain

2

= 2

101 0.042 10

2 100

= 1010 1.6 107 = 1600 joule/metre3

30. W = 21 YAx

2 L

= 11 6 61 2 10 1 10 3 3 10

2 3

= 0.3 J

31. W = 1

2YA

2

L

1

2

W

W=

2 21 1 22 22 2 1

r L

r L

= 2 2

r 0.1 L

2r 0.1 2L

= 1

8

W2 = 8 W1 = 16 J.

32. As Y = FL

A =

2

FL

r ,

r2L = FL

Y= constant

r12 1 = r2

2 2

2 = 2

r

2r

1 = 1

4

E = Energy/Volume = 1

2Y (strain)2

E (L)2 or 2

1

E

E=

2

2

1

=2

1

4

= 1

16

E1 : E2 :: 16 : 1. 33. Elastic energy per unit volume,

U = 1

F2

l = 2F

2AY

l

U 2

1

r(F and Y are constants)

A

B

U

U=

2

A B

B A

r

r

l

l= (3)

21

2

= 3

4

1

Chapter 06: Surface Tension

Hints to Problem for Practice

1. m = 1 310 kg, T = 7 102N/m, g = 9.8 m/s2

T = F

2l or l =

mg

2T we get,

l = 3

2

10 9.8

2 7 10

l = 7cm 2. Internal diameter = 5 102m

Thickness = 2 103 m r1 = 2.5 102 m T = 0.072N/m External radius = r2 = 2.8 102 m F = (2πr1+2πr2)T F = 2 3.14 (2.5 + 2.8) 102 0.072 = 6.28 5.3 0.072 102

= 2.396102 N 3. A = (100 50) cm2 = 50 104 m2

,

W = 3 10-4 J

W = TA or T = W

A,

T = 4

4

3 10

2 50 10

T = 0.03 Nm1

4. l = 20 102 m, b = 10 102 m h = 5 103 m, T = 70 103 N/m, = 0 Force acting on the plate, F = 2 ( l + h) T = 2 (20.5) 102 70 103

= 287 104

F = 0.0287 N 5. r1 = 0.02 m, r2 = 0.025 m, T = 0.075 Nm1

F = (2πr1 + 2πr2) T = 2 3.14 (0.02 + 0.025) 0.075 = 6.28 0.045 0.075 F = 0.02119 N

6. F = T 2 2πR,

F = 2 2 3.14 0.04 F = 0.045 N/m. 7. A1 = 2 103 m2,

A2 = 4 103 m2,

T = 3 102 Nm1 Using,

W = 2TA

= 2 3 102 (4 2) 103 = 12 105

W = 1.2 104 J 8. d = 0.2 cm R = 0.1cm = 0.1 102 m Number of droplets = 27,000 Using.

27,000 3 34 4r R

3 3

R3 = 27,000 r3 or R = 30r

r = 20.110 m

30

Now using,

W = T A we get,

W = T (n4πr2 4πR2)

= 7 102 4 3.14

2 2 2 20.127000 ( 10 ) (0.1 10 )

30

= 87.92 102 (30 106 1 106)

W = 2.55105 J 9. d1 = 7cm r1 = 3.5 cm, W = 36,960 erg,

T = 40 dyne cm1

W = 2T 2 22 14 r 4 r

36,960 = 2 40 4 3.14 2 22r 3.5

22

36960r 3

1004.8

22r 3 = 36.7834

22r = 39.7834

r2 6 cm on simplification

Surface Tension 06

2


10. R = 0.5 cm, r = 1 mm = 0.1 cm

3 34 4n r R

3 3

33

3

R 0.5n

0.1r

n = 125 11. d = 1 102 m r = 0.5 102 m T = 2.5 102 Nm1 W = 2TA W = 2 2.5 102 4 3.14 (0.5 102)2 = 5 12.56 102 0.25 104

W = 1.57 105 J. 12. W = 2T 2 2

2 14 r 4 r

= 2 60 4 3.14 (25 4) = 480 3.14 21 W = 31651erg 13. d = 0.5 mm, r = 0.25 mm = 0.025 cm = 13.6 g/cm3, T = 545 dyne cm1, = 130

h =2T cos 2 545 cos130

r g 0.025 13.6 980

= 700.638

333.2

h = 2.1 cm 14. h = 4 cm, r = 0.04 cm, = 1 g/cm3, g = 980cm/s2, = 0

T = hr g

2cos

T = 4 0.04 1 980

2 cos

= 156.8

2 1

T = 78.4 dyne cm1

15. hw = 5 cm, hHg = 1.54 cm Tw : THg = 2/13, rw = rHg w : QHg = 1/13.6, w = 0

cos = hr g

2T

Hgw w w w

Hg Hg Hg Hg w

Tcos h r

cos h r T

= 5 1 13

11.54 13.6 2

cos Hg = 0.6444 Hg = cos1(0.6444) Hg = 130

16. r = 1 mm = 1 103 m, h = 0.536 102 m T = 0.405 N/m, = 13600 kg/m3, g = 9.8m/s2

cos = hr g

2T

,

=2 3

1 0.536 10 10 13600 9.8cos

2 0.405

= 1 0.7143cos

0.81

= 15152 17. hw = 10 cm = 10 102 m, hHg = 3.42 102 m, w = 103 kg/m3, Hg = 13600 kg/m3, w = 0, Hg = 135

T = hr g

2cos

we get,

HgW w w w

Hg Hg Hg Hg

cosT h r

T h r cos

= 2 3

2 3

10 10 10 0.7071

3.42 10 13.6 10 1

W

Hg

T

T=

3535

23256= 0.152 : 1

18. = 0.8 g/cm3, T = 50 dyne/cm, = 0, r1 = 0.3 mm = 0.3 101 cm r2 = 0.6 mm = 0.6 101 cm

h2 h1 = 1 2

2T cos 1 1

g r r

= 1 1

2 50 cos0 1 1

0.8 980 0.3 10 0.6 10

= 1

100 10 10

8 98 10 3 6

h2 h1 = 21.258 cm 21.26 cm

Hints to Multiple Choice Questions 1. l = 3 102 m; b = 2 102 m; d = 1 103 m; s = 7.0 102 Nm1

Force on the plate due to surface tension is F = 2 (3 102 + 2 102) 7.0 102

= 7 103 N

3


2. Weight W = 2rT or 1

2

W

W= 1

2

T

T=

30

60=

1

2 3. The force on disc = T circumference = 7 10–2 2 R

= 7 10–2 22

7 (15 10–2) 2

= 6.6 102 N

4. F = 2AT

t

= 1

2 4 75

0.8 10

dyne

= 7.5 103 dyne 5. 1 dyne/cm = 103 N/m 108 dyne / cm = 108 103 N/m = 105 N m1

6. 2 2r T = 5500 105 105

T = 55500 7 10

4 22 6.25

dyne cm1

= 70 dyne cm1

7. F = (T1 T2) = (45 28) 4

= 68 dyne 8. 2T 2 r g

or 2 2Tr

g

or 2T

rg

9. 343 3

4r3 =

3

4 R3 or R = 7r

Increase in surface area = 343 4 r2 4R2

= 2343 4 R

49

4R2 = 24 R2

= 24 2

d

2

= 6d2

Change in surface energy = 6 d2 S 10. F = T (2r1 + 2r2) = T 2 (1.4 + 1.6) 102 = 0.07 2 3.14 3 10–2 = 1.32 10–2 N 11. T0 = 70 dyne/cm, Tx = 65.3 dyne/cm

T = 0 0 2 1T (1 ) T 1 t t

65.3 = 3270 1 2.7 10 (t 0)

32

65.31 2.7 10 t

70

2 3

65.3 1t 1

70 2.7 10

= 34.7 10

70 2.7

t2 25C 12. Increase in area,

A = 2(8 6 2) = 2 28cm2 [Film has 2 free surfaces]

Film = 56 10-4 m2

As, work done,

W = Surface tension increase in area

6.0 104 = S 56 104

S 0.1 Nm1

13. Area of film = 2 (8 102 5 102)

= (80 104 m2) S.T. = Energy /area

Energy = S.T. Area

Energy = 0.03 (80 104)

= 2.4 104 J

14.

20 R

4P

10 R

4P =

R

4

R

1 =

2R

1

1R

1

or R = 21

21

RR

RR

=

4

2

2.5 4 10

1.5 10

m

= 210 10

1.5

cm =

400

3 1000m

= 6.6 102m 15. Here, r = 1 cm = 1 102m; Density of oil,

= 0.8 103 kgm3;

H = 3 mm = 3 103m Presure due to 3mm column of oil,

P = hg = (3 103) (0.8 103) 9.8

= 3 0.8 9.8 Pa In case of a soap bubble,

P = 4S

r

Or S = 2Pr 3 0.8 9.8 1 10

4 4

6 102 Nm1

4


16. Here, S = 0.06 Nm1; r1 = 3 cm = 0.03 m; r2 = 6 cm = 0.06 m Since bubble has two surfaces, initial surface

area of the bubble

= 2 4π 21r = 2 4π (0.03)2

= 72π 104m2

Final surface area of the bubble

= 2 4π 22r = 2 4π (0.06)2

= 288π 104m2

Increase in surface area = 288π 104 72π 104

= 216π 104m2

work done = S increase in surface area = 0.08 216 π 104

= 0.0054J 17. Here, r = 30.0 mm = 3 cm; F = 3.03 gf =

3.03 980 dyne. Since, the liquid is touching the ring, both inside as well as outside therefore, force acting on the ring due to surface tension is given by

F = 2 (S circumference ring) = 2 (S 2πr) = 4Sπr

= 4 S 22

7 4 dyne

As F = F,

22

4 S 47

= 4 980

or S = 4 980 7

4 22 4

= 78 dyne cm1

18. If r is the radius of small droplet and R is the radius of big drop, then according to question,

3 3 34 4R 10 r

3 3 or r =

R0.1R

10

= 0.1 102m = 103m Work done = surface tension increase in area = 35 102 [103 4π (103)2

4π (102)2] = 4 103 J

19. cos 30 = 9

= 9

cos30 =

9 2

3

=

18 3

3

= 6 3 cm

20. R3 = 4 r3

4

3 R3 = 4

4

3 r3

R = 4 3

1

r 21. W = 2 4R2 T ; R is increased by a factor

of 3, so W is increased by a factor of 9. 22. If n droplets coalesce to form a single big drop

then,

3 31 2

4 4n r r

3 3 or

323

1

rn

r

Gain in thermal energy = work done against surface tension

JQ = W J.m.c. = T A

J.m.c. = T. 4r23

1 2

1 1

r r

J. 32

4r

3

.c. = T. 324 r

1 2

1 1

r r

J.1

13

1 = T.1 2

1 1

r r

….[For water, = 1 C.G.S. units and c = 1 C.G.S. units]

= 3T

J 1 2

1 1

r r

23. Here, h = 3.75 cm = 3.7 5 102 m ; S = 7.5 102 Nm1; = = 0,

As h = 2Scos

r g

; so 2r = 4Scos

h g

= 2

2 3

4 7.5 10 cos0

3.75 10 10 10

= 8 104 m = 0.8 mm

24. h2 – h1 = g

cosT2

12 r

1

r

1

= g

cosT4

12 D

1

D

1

= 33

o2

101010

0cos1074

1 1

4 6

= 228 10 1 1

10 4 6

m

= 2.3 10–3 m = 2.3 mm

5


25. hg = 2S

r or h =

2S

r g

= 2 75

0.01 1 1000

= 15 cm

26. h = gr

cosT2

m

w

T

T

=

m

w

h

h

w

m

cos

cos

m

w

= 7.5

2.45

0cos

135cos

6.13

1

1

6.5 27. = 0.0027 /C T = T0 (1 t) T = 70 [1 (0.0027 40)] = 70 [1 0.108] = 70 0.892 T = 62.44 dyne/cm

28. 2

1

r

r =

1

2

h

h or

5.1

3.4 =

3

2

29. height gravity.sp

1

2

1

h

h =

1

2

)g.s(

)g.s(

2h

5 =

1

25.1

h2 = 4.5

1.5 = 3 cm

30. h = rdg

cosT2 or h

r

1 or

1

2

h

h =

2

1

r

r

A2 = 1

9A1 or r2

2 = 1

9 r1

2 or 2

2

21

r

r = 9

2

1

r

r = 3

1

2

h

h = 3 or h2 = 3 3 = 9 cm.

1

Chapter 07: Wave Motion

Hints to Problems for Practice 1. v = n

= v

n

= 350

50

= 7 m. Now using,

= 2 x

x = 2

= 7

32

x = 1.17 m

= 2

T

t

= 2πn t = 2π 50 0.01 = π rad

= 180 2. Equation of Simple harmonic wave is,

y = 0.01 sin 100π x

t40

= 0.01 sin 2π 5x

50t4

Comparing given equation with

y = A sin 2π x

nt

we get,

A = 0.01 m, n = 50 Hz,

x = 4

0.8m5

Using, v = n we get, v = 50 0.8 = 40 m/s. 3. v = 100 m/s = 10000 cm/s, A = 2 cm, n = 100 Hz, t = 5s, x = 200 cm

v = n or =v

n

= 10000

100 = 100 cm,

y = A sin 2

(vt x)

y = 2 sin 2

100

(10000t x)m ....(1)

= 2 sin 2π (100 5 2)

= 2 sin 2π (498)

y = 0 m [ sin (n2π) = 0 where n = 0,1,2]

4. y = A sin 2

(vt x)

y = 0.05 sin 2

0.03

(50t x) m

y = 0.05 sin 2 3

t x

0.030.5998 10

5. A = 5 cm = 5 102m, n = 5Hz,

v = 40 cm/s = 40 102 m/s,

x = 38 cm = 38 102m, t = 1s

= v

n

= 240 10

5

= 8 102 m

y = A sin 2

(vt x)

y = 0.05 sin 2

0.08

(0.4t x)

= 0.05 sin 25 π (0.4 1 0.38)

[ sin n

2

=

2

, n = 1,3,5..]

= 0.05 sin 2

y = 0.05 m

= 2 x

= 22 0.8 10

0.08

= 5

rad.

Wave Motion 07

2


6. Equation of S.H. progressive wave is, y = 0.001sin 4π (100t 10 x)m = 0.001sin 2π (200t 20 x)

y = 0.001sin 2π x

200t0.05

Comparing this equation with

y = A sin 2π x

nt

, we get,

i. amplitude = 0.001m

ii. T = 31 15 10 s

n 200 = 0.005 s

iii. n = 200 Hz. iv. = 0.05 m

7. v = n or n = v

n = 83 10

287.3

n = 1.044 MHz 8. Equation of a S. H. Progressive wave is,

y = 0.04 sin 2320t x

0.08

y = 0.04 sin x

2 4000t0.08

Comparing this equation with

y = A sin 2π x

nt

we get,

A = 0.04 m, n = 4000 Hz and x = 0.08 m Using, v = n we get, v = 4000 0.08 v = 320 m/s

9. For A, n1 = 400

125Hz3.2

For B, the frequency may be 130 Hz or 120 Hz.

The wavelength of B may be 400

130 or

400

120

i.e., 3.10 m or 3.33 m 10. For the first and third wave, the phase

difference = π Their resultant amplitude = 8 5 = 3m This has a phase difference of π/2 with the

second wave

R = 2 23 4 = 5 m

11. Two sound waves are,

y1 = 0.5 sin 2πx

104t0.5

and

y2 = 0.5 sin 2π x

100t

Comparing with,

y = A sin 2π x

nt

we get,

A = 0.5m, 1 = 0.52m, n1 = 104 Hz, n2 = 100 Hz. n = n1 n2 = 104 100 n = 4 beats/s

1 2

v vn

For the first wave, v = n1 1

= 104 1.5 v = 52 m/s v = n2 2

2 = 2

v

n

= 52

100

2 = 0.52 m 12. No. of beats / sec. = 6 Time interval between a waxing and nearest

waning

= 1 2

1 1 1 1 1s

n n 2 8 2 16

13. Original beat frequency = 5 beats/s Frequency of fork A = nA = 330 Hz Frequency of fork B= nB When the fork B is loaded with wax, number

of beats heard = 6 beats/s. For the 1st case: nB nA = Beat frequency nA nB = Beat frequency nB = 330 + 5 = 335 Hz nB = 330 5 = 325 Hz

5 Hz 5 Hz nA nBnB

325 Hz 330 Hz 335 Hz

3


For the 2nd case: After loading,

nA nB = 6

nB nA = 6

nB = 336 Hz

nB = 324 Hz As the frequency of the fork reduces after

being loaded with wax, hence the original frequency of fork B should be 325 Hz

14. 1 = 0.6 m, 2 = 0.61 m

n = n1 n2 = 10 Hz

n = n1 n2

n = 1 2

v v

10 = v1 1

0.6 0.61

10 = v (0.61 0.6)

0.6 0.61

v = 366 m/s.

Substituting for v in 1 2

v vand

we get,

n1 = 610 Hz and n2 = 600 Hz 15. np = 256 Hz, Beat frequency = 4 beats/s

nQ = 256 + 4 = 260 Hz or

nQ = 256 4 = 252 Hz After filing, beat frequency decreases. Hence, the frequency of fork Q must increase.

Original frequency of fork Q should be 260 Hz.

16. When the train moves away from the

stationary observer then,

n= n

svv

v

From formula,

1852 = n335.3

335.3 26.822

1852 = n335.3

362.122

n = 1852 362.122

335.3

n = 2000 Hz.

17. i. When siren is moving towards the listener:

n = s

v

v v

n = 340

340 10 1000

n = 1030.3 Hz ii. When siren is moving away from the

listener:

n = s

v

v v

n = 340

340 10 1000

n = 971.4 Hz 18. Here, vs = v0

= 72 km h1 = 20 ms1; v = 332 ms1; n = 540 Hz Apparent frequency before the engines cross

each other:

n = 0

s

v vn

v v

=

332 20540

332 20

n = 609.2 Hz Apparent frequency, after the engines cross

each other:

Here, n = 0

s

v v

v v

= 332 20

540332 20

n = 478.64 Hz

Hints to Multiple Choice Questions 1. Comparing with y = acos (t+kx) we get,

k = 2

= 0.01π

= 200 cm Also, it is given that phase difference between particles

= 3

.

Hence, Path difference between them

x = 2

= 2 3

= 100

3

= 33.3 cm

4


2. Displacement equation is

y = 0.03 sin π (2t 0.02x)

Comparing with the standard equation

y = Asin2πt x

T

,

2 2

0.02 or0.02

= 100 m

1

= 0.01 per m or 1 per cm.

3. Maximum particle velocity (vp)max = a

And wave velocity, v = k

Ratio, p maxv a k

kav

4. Equation is

y = 10 sin 2

(50t x) ….(1)

Comparing eq. (1) with standard wave

equation, given by

y = A sin (t kx) ....(2)

we have,

= 50

252

2

25T

T = 2

0.08s25

5. y = 2sinx

3 t12

where = 3π and k = 12

3

v 36cm / sk

12

7. Phase diff. = 2 x

3

x = 3 2 6

8. y = a sin t x

3 6

Comparing it with standard equation

y = a sin 2t x

T

2 1

T 3 or T = 6s and

2 1

6

112 12v 2ms

T 6

Distance travelled by wave in 10 s = v t = 2 10 = 20 m 9. The longitudinal wave given by x = x0 sin [2π(nt x / )]

= x0 sin 2

2 nt x

The maximum particle velocity = A = x0 (2πn) The wave velocity = n Here we compare the given equation with the

equation x = x0 sin (t kx) From the equation, 2πnx0 = 4n

Or = 0x2

10. For interference phenomenon, we know that

when two waves of equal frequency propagate in same medium and same direction, then interference phenomenon takes place.

Here waves, y1 = a sin (t + 1) and y3 = asin (t + 2) have equal frequency, hence they will produce

interference 11. Equations of waves y1 = cos(6t 3x)

= sin 6t 3x2

and y2 = sin 6t 3x4

Therefore, phase difference between the two waves is

= 6t 3x 6t 3x4 2

4 2 4

5


12. Let 1 and 2 represent angles of the first and second waves, then

2 = 2

[(vt x) + x0]

and 1 = 2

(vt x)

But x0 = 2

2 1 = π

Hence, phase difference, = π. So, amplitude of the resultant wave

R = 2 2A B 2ABcos

= 2 2A B 2ABcos

= 2A B = A B

Or R = A B 13. The given waves are

y1 = 104 sin [25t+(x/30)+0.5] m

and y2 = 104 cos [25t+(x/30)] m

or y2 = 104sin 25t x / 302

m

sin cos2

Hence, the phase difference between the waves is

= 0.52

rad

= 3.14

0.52

rad

= (1.57 0.5) rad

1 rad 14.

R = 2 2P Q 2PQcos

= 2 2 2A A 2A cos120

R = A

15. y1 = A sin (t)

y2 = A sin t2

Resultant amplitude,

R2 = A2 + A2 + 2A2cos2

R2 = 2A2 + 2A2 0 R2 = 2A2

Or R = 2A However, both will have the same frequency

on superimposing. 16. We have the progressive wave given by y = 2 cos 1.571 (100t x) = 2 cos 2π (100t x) = 2 cos (2π 100t 2πx) = 2π 100,

And time period T = 2

= 1

s100

17. 1

2

I 25

I 16

Using, I a2

So,

2

1 1

2 2

I a

I a

2

1

2

a25

16 a

1

2

a25

16 a 1

2

a 5

a 4

a1 : a2 = 5 : 4

18. 2

1 12

2 2

I a 4

I 1a

1

2

a 4

a 1

1

2

a 2

a 1

2

max 1 22

min 1 2

I (a a )

I (a a )

=

22

22

(3a )

(a )

=

2

2

3

1=

9

1

Thus, Imax : Imin = 9 : 1

A A

120

60

A

6


19. Intensity (amplitude)2

I 2 2

2

1

a 8

a 4

= 4 : 1

20. The equations of motion are y1 = 2a sin (t kx) y2 = 2a sin (t kx ) Now, the equation of resultant wave is given

by, y = y1 + y2 = 2a sin (t kx) + 2a sin(t kx )

y = 2a t kx t kx

2sin2

cos t kx t kx

2

y = 4a cos2

sin t kx

2

...(1)

Now, comparing eq. (1) with y = A sin (t kx), we have resultant

amplitude A = 4acos2

.

21. v = n

n = v

The frequency corresponding to wavelength 1,

n1 = 1

v 330110Hz

3.0

The frequency corresponding to wavelength 2,

n2 = 2

v 330100Hz

3.3

Hence, number of beats per second = n1 n2 = 110 100 = 10

22. The frequency of beats is given by n = n1 n2

= 1 2

v v

= 1 1

v0.6 0.61

8 = v (1.6667 1.6393) = v 0.02736

v = 18292 ms

0.02736

23. Difference between frequencies of two consecutive forks is 3.

f = f1 + (n 1) (d)

(arithmetic progression)

f1 = 2f, n = 26, d = 3

f = 2f + (26 1) (3)

f = 75 Hz.

Frequency of 18th tuning fork is

f18 = f1 + (18 1)(3)

= 2 75 + 17 (3)

= 150 51

= 99 Hz

24. Frequency f = 6

2Hz3

f = v 1 2

1 1

2 = 300 2

1 1

3

2

1 1 1

150 3

2

1 1 1

3 150

2

1 50 1 49

150 150

2 = 150

3m49

25. Let the frequency of tuning fork C be

x Hz. The frequency of tuning fork A is

x + 2

100x = 1.02 x

The frequency of tuning fork B is

x 2.5

100x 0.98 x

Beat frequency of tuning fork A and B is

1.02x 0.98x = 0.04x

Thus, 0.04x = 8

x = 8

200 Hz0.04

7


26. n1 = v u

v

n and n2 = v u

v

n

Number of beats per second

x = n1 n2 = v u

v

n v u

v

n

= v u v u

nv

= 2u n 2u

v

27. When the source is coming to the stationary

observer,

n = s

vn

v v

or 700 = 350

n350 40

or n = 700 310

Hz350

= 620 Hz

when the source is moving away from the stationary observer,

n = s

vn

v v

= 350 700 310

350 40 350

= 0.897 620

556 Hz 28. vs = 20 ms1

v0 = 40 ms1

Apparent frequency n = n 0

s

v v

v v

n = n 340 40

340 20

n = n 380

320

n = 38

30n

Fractional change in frequency

38n n 6 332n 32 16

29. Wavelength = sv v

v

= 40

330330

6330

= 40 330 55

330

= 40385

47cm330

30. As motion of the source is always

perpendicular to the direction of propagation of sound, so there will be no Doppler’s effect.

31. From the relation

n = sound observer

sound source

v vn

v v

v0 = 0, because observer is stationary,

n = n 0

v 1f

v 7v

8 8

n 8

n 7

32. Frequency of reflected sound heard by the bat

n = n 0

s

v v

v v

= n 0

s

v v

v v

= n 0

b

v v

v v

= n 6330 510

330 5

= 1030 103 Hz 33. n1 n2 = 3% n Source moving towards observer,

n1 = n

svv

v

Source receding away from observer,

n2 = n s

v

v v

n

svv

v n

svv

v = 3 % n =

3n

100

8


v

ss vv

1

vv

1 =

3

100

v

2

s2

ss

vv

vvvv =

3

100

s3v

v =

3

100

Or vs = 100

v =

340

100 = 1.1 m/s

34. v0 = 1

v8

n1 = n

v

vv 0

n

n1 = v v / 8

v

=

9

8

n

nn1 100 = 1

8 100 = 12.5 %

35. f1 = f

1vv

v, f2 = f

2vv

v

2

1

f

f =

1

2

vv

vv

= 350 25

350 50

= 325

300 =

13

12 =

1112

1

Chapter 08: Stationary Waves

Hints to Problems for Practice 1. y = 4 cos (x/3) sin (40t) Comparing with

y = 2A cos 2 x

sin 2nt we get,

2A = 4 or A = 2 cm

2 x

= x

3

or = 6 cm

2nt = 40 t or n = 20 Hz v = n = 20 6 = 120 cm/s 2. Distance between two consecutive nodes (d) =

25 cm v = 300 m/s = 300 102 cm/s.

d = 2

25 = 2

or = 50 cm

Now using, v = n we get,

n = v

=

2300 10

50

= 600 Hz

3. i. v = n

= v 350

n 1750 = 0.2 m

ii. Distance between two adjacent nodes =

2

=

0.2

2 = 0.1 m

iii. vmax = Amax = 2 A (2n) = 2 1 102 2 3.14 1750 = 219.8 m/s

4. y = 0.05 cos 2 x

0.5

sin 2 t

0.002

Comparing with,

y = 2A cos 2 x

sin

2 t

T

we get,

i. 2A = 0.05 or A = 0.025 m

ii. 2 t

T

=

2 t

0.002

or T = 0.002

n = 1

T =

1

0.002= 500 Hz

iii. = 0.5 m

5. n1 = 325 Hz, 1 = 67 cm = 67 102 m,

2 = 65 102 m

n11 = n22

325 67 102 = n2 65 102

n2 = 325 67

65

= 335 Hz

number of beats = n2 n1 = 335 325 = 10 6. L = 1.5 m

For second overtone, L = 3

2

= 1.5

2

=

1.5

3= 0.5 m

Distance between a node and the antinode

close to it = 4

=

0.5

2= 0.25 m

7. = 75 cm = 75 102 m,

M = 6 g = 6 103 kg, n = 100 Hz.

m = M

=

3

2

6 10

75 10

= 0.08 101

= 8 103 kg/m

n = 1 T

2 m

T = 4 n22m

= 4 1002 (75 102)2 8 103

= 1.8 5 104 104 103 = 180 N 8. M = 0.5 g = 0.5 103 kg, = 0.5 m,

T = 19.6 N,

m = M

=

30.5 10

0.5

= 103 kg/m

n = 1 T

2 m

i. n = 3

1 19.6

2 0.5 10

= 2196 10

= 140 Hz ii. Frequency of 3rd overtone = 4 n = 4 140 = 560 Hz

Stationary Waves 08

2


9. m = 103 kg/m, T = 25.6 N, n = 480 Hz, n1 = 640 Hz

v = T

m=

3

25.6

10 = 25600 = 160 m/s

= v 160 1

n 48 3 m

480 = n T

2 m

640 = (n 1)

2

=T

m,

640 n +1

480 n

4n = 3n + 3 n = 3

480 = 3

3 25.6

2 10=

3160

2

3 160

2 480

=

1

2= 0.5 m

10. 1 = 110 cm, n1 : n2 = 3 : 2

Length of 1st segment = 110 2

5= 44 cm

Length of 2nd segment = 110 3

5= 66 cm

11. T1 = 156 N, T2 = 39 N, L1 = L2 = 1 m d1 = d2 or r1 = r2 A1 = A2 1 = 8 103 kg/m3, 2 = 2 103 kg/m3

Now, m = M

L=

v

L

=

AL

L

= A = r2

m1 = 21 1r and m2 = 2

1 1r

Using, n = T

m we get,

1

2

n

n= 1 2

2 1

T m

T m =

21 2 2

22 1 1

T r

T r

1

2

n

n= 1 2

2 1

T

T

=3

3

156 2 10

39 8 10

=

39

39= 1 : 1

12. 1 : 2 = 1 : 2, T1 = T2, m1 = m2

Using, n = 1 T

2 m we get,

1

2

n

n= 1 2

2 1

T

T

= 2 1 = 2

1 = 2 : 1

13. n1 = 75 Hz, 2 = 1

3

, T2 = 4T1

Using, n = 1 T

2 m we get, T = 4n22m

2

1

n

n= 1 2

2 1

T

T

n2 = 75 3 4 = 225 2 = 450 Hz 14. 1 = 60 cm = 60 102 m,

2 = 55 102 m, n1 n2 = 10

Using, n11 = n22 or

1

2

n

n = 2

1

= 55

60=

11

12

n2 = 12

11 n1 = 1.09 n1

n2 > n1 n2 n1 = 10 1.09 n1 n1 = 10

n1 = 10

0.09= 111.11 Hz

15. = 1 m, 1 2 = 4 mm = 4 103 m,

n1 n2 = 2,

1 2 = 1 and 1 2 = 0.004

Adding these equations we get,

21 = 1.004 or 1 = 0.502 m

2 = 1 0.502 = 0.498 m

As 1 > 2,

n1 < n2

n2 n1 = 2 or n2 = n1 + 2 Now using,

n11 = n22 we get,

n1 0.502 = (n1 + 2) 0.498 502 n1 498 n1 = 2 498 4 n1 = 2 498

n1 = 996

4= 249 Hz

n2 = 249 + 2 = 251 Hz 16. 2 = 0.99 1, n1 n2 = 6

As 2 < 1, n2 > n1 n2 n1 = 6

Using, n11 = n22 we get, n2 = n1 + 6

n11 = (n1 + 6) 0.99 1

3


n1 = 0.99 n1 + 6 0.99 0.01 n1 = 5.94

n1 = 5.94

0.01 = 594 Hz

17. n1 = 320 Hz, n2 = 480 Hz,

2 = (1 12) cm

Using, n11 = n22 we get,

320 1 = 480 (1 12)

480 12 = 480 1 320 1

160 1 = 5760

1 = 5760

160= 36 cm

18. Let 1, 2 and 3 be lengths of three segments

of the wire having respective frequencies n1, n2 and n3 respecively

1 + 2 + 3 = 110 ….(1)

And n1 : n2 : n3 = 1 : 2 : 3

1

2

n

n =

1

2 ….(2)

1

3

n

n=

1

3 ….(3)

As T and m are constant for three segments, n11 = n22 = n33.

1

2

n

n= 1

2

….(4)

1

3

n

n= 3

1

….(5)

From eq. (2) and (4) we get,

2

1

1

2

or 2 = 1

2

From eq. (3) and (5) we have,

3

1

1

3

or 3 = 1

3

From eq. (1) we get,

1 = 60 cm, 2 = 60

2= 30 cm and

3 = 60

3= 20 cm

19. p1 = 2, m2 = 4 m1, T1 = T2, L1 = L2, N1 = N2

N = p T

L m

1

1

p

m = 2

2

p

m or

p2 = p1 2

1

m

m= 2 1

1

4m

m= 4

20. p1 = 2, T1 = 64 gm-wt, p2 = 8

T121p = T2

22p

T2 = 2

1 122

T p

p=

2

2

64 2

8

=

64 4

64

= 4 gm-wt

21. p = 8, m = 0.3 g/m = 0.3 103 kg/m, L = 2.8 m , T = 1.47 N

N = p T

2L m

= 3

8 1.47

2 2.8 0.3 10

= 1.428 70 = 99.96 100 Hz 22. T = 4 103 9.8 N, p = 5, L = 80 cm = 0.8 m,

N = p T

L m

N =3

5

5 4 10 9.8

0.8 9.8 10

= 6.25 20 = 125 Hz 23. p1 = 4, T1 = 8 980 dyne p2 = 2 4 = 8, m1 = m2, L1 = L2, N1 = N2

21 1T p = 2

2 2T p

T2 = T1 2

1

2

p

p

= 7840 105 2

4

8

= 7840

4 = 1960 dyne

= 1960

g980

= 2 g 24. N = 520 Hz, L = 0.5 m, m = 1 104 kg/m, T = 1.69 N.

i. N = p T

2L m

p = 2 NL m

T

4


= 2 520 0.5 410

1.69

= 520 2110

1.69

= 0.769 520 102 = 399.88 102 = 3.99 4

ii. For parallel position, p = 4

2= 2

25. T1 = 120 gm-wt = (p0 +120) 980 dyne,

p1 = 2, T2 = (p0 + 26.25) 980 dyne P2 = 4, T3 = p0 980 dyne, p3 = ? when p0 = 0 Let p0 = mass of pan

Using, 21 1T p = 2

2 2T p we get,

(p0 + 120) 980 22 = (p0 + 26.25) 980 42 p0 + 120 = 4 (p0 + 26.25) 3p0 = (120 105)

p0 = 15

3 = 5 g

Now using,

21 1T p = 2

3 3T p we get,

125 22 = 5 23p

23p =

125 4

5

23p = 100

p3 = 10 26. p1 = 8 for perpendicular position

L1 = 1L

2, T1 = T2

Now, m = m

L

m1 = 1

m

Land m2 =

2

m

L

m2 = 1

m

L / 2 1

2

m

m = 1

1

m / L

m / L / 2

Using,

N = 1 1

1 1

p T

2L m and N11 = 1 2

2 2

p T

L m

We get,

1 2

1 2

p p

2L L or

p2 = 1 2

1

p L

2 L =

8 1

2 2 = 2

27. n2 n1 = 240 Hz

5v 3v

4L 4L = 240 Hz

2v

4L= 240 or

v

4L= 120

Frequency of 3rd overtone,

n3 = 7 v

4L= 7 120 = 840 Hz

28. n = 420 Hz, u = 19.25 cm = 19.25 102 m,

v = 336 m/s

v = 4n ( + 0.3d)

336 = 4 420 (19.25 102 + 0.3 d)

336

0.19254 420

1

0.3= d

d = (0.2000 0.1925) 1

0.3 = 2.5 cm

29. 1 = 20 cm, 2 = 63 cm

e = 0.3 d = 2 13

2

d = 63 3 20

2 0.3

= 3

0.6 =

1

0.2= 5 cm

30. v = n,

= v

n =

340

1000 = 34 102 m

2

= 17 102 m = 17 cm.

31. e = 2 13

2

e = 48 3 15

2

=

3

2 = 1.5 cm

32. n11 = n22

n2 = 1 1

2

n

= 320 24

16

= 480 Hz

33. v = 350 m/s,

= 68.8 cm = 68.8 102 m,

d = 2 cm = 2 102 m

n = v

2( 2e)

n2 = 4n

= 4 v

2( 0.6d)

5


= 2 2

4 350

2(68.8 10 0.6 2 10 )

= 2

1400

2 70 10 = 1000 Hz

34. = 48 cm = 48 102 m,

d = 3 cm = 3 102 m, v = 350 m/s.

n = v

2L=

v

2( 0.6d)

n = 2 2

350

2(48 10 0.6 3 10 )

= 2175 10

49.8

= 3.514 102 351.4 Hz

35. n1 n2 = 26

10= 2.6/s

1 = 80 cm = 80 102 m, 2 = 81 cm = 81 102 cm

n1 1 = n2 2

1 2

2 1

n 801.0125

n 81

i.e. n1 > n2. n1 n2 = 1.0125 n2 n2 = 2.6 or n2 = 208 Hz

Using, v = 2n2 2 we get,

v = 2 208 81

100= 336.96 337 m/s

36. 1 = 18 cm = 18 102 m,

2 = 55 cm = 55 102 m, n = 512 Hz

Using, v = 2 n (2 1) we get,

v = 2 512 (55 18) 102 = 378.88 379 m/s 37. 1 = 15.4 cm = 15.4 102 m,

2 = 0.486 m, v = 340 m/s.

e = 2 13

2

e = 0.486 0.154 3

2

=

0.024

2

= 0.012 m = 1.2 cm

v = 4 n ( + e)

n = 340

4(0.154 0.012)=

340

0.664 = 512 Hz

38. 1 = 16 cm = 16 102 m,

2 = 52 cm = 52 102 m, v = 350 m/s

Using, e = 2 13

2

we get,

i. e = 2 252 10 3 16 10

2

= 2 102 = 2 cm ii. e = 0.3 d

d = 2

0.3= 6.67 cm

iii. n =v

4L

=2

350

4(16 0.3 6.67) 10

= 486.1 Hz 39. 1 = 16.2 cm = 16.2 102 m,

2 = 50.2 cm = 50.2 102 m,

n = 512 Hz.

e = 2 13

2

e = 2(50.2 3 16.2) 10

2

= 21.6

102

= 0.8 102 m = 0.8 cm

v = 4 n ( + e)

v = 4 512 (16.2 + 0.8) 102 = 348.16 m/s

348.2 m/s


1. y = 1

20cos (x) sin (300 t)

Comparing with,

y = 2A cos 2 x

sin

2 t

T

we get,

2 t

300 tT

T = 1

150

n = 150 Hz

6


2. y1 = 0.01 sin (2t + 3x) y2 = 0.01 sin (2t + 3x) According to superposition principle, the

resultant displacement is y = y1 + y2 = 0.01 [sin (2t 3x) + sin (2t + 3x)] y = 0.01 2 sin 2t cos 3x y = (0.02 cos 3x) sin 2t = R sin 2t where R = 0.02 cos 3x = amplitude of the

resultant standing wave. At x = 0.2 m, R = 0.02 cos 3x = 0.02 cos 3 0.2 = 0.02 cos 0.6 (radian)

= 0.02 cos180

0.6

= 0.02 cos 34.4 = 0.02 0.83 m R 1.7 cm 3. Equation of stationary wave is y1 = a sin kx cos t, and equation of progresssive wave is y2 = a sin (t kx) = a(sin t cost kx cos t sin kx)

At x1 = 1

2k

and x2 =

5

3k

,

sin kx1 or sin kx2 is zero. neither x1 nor x2 is node.

x = x2 x1 = 5

3k

1

2k

=

7

6k

As x = 7

6k

, therefore,

2

k

> x >

k

But 2

k

= . So > x >

2

.

In case of a stationary wave, phase difference between any two points is either zero or .

1 = and 2 = k x = k7

6k

=

7

6

1

2

= 76

=

6

7

4. y = 14 cos x

8

sin (60 t)

Comparing with

y = 2A cos 2 x

sin

2 t

T

we get, Amptitude = 14 cm, 2nt = 60 t n = 30 Hz

2 x

= x

8

= 16 cm

and v = n = 30 16 = 4.8 m/s

[v = 10 cm/s is wrong] 5. When the stone is suspended in air

n = a

1

W1

2L m

When the stone is suspended in water,

n = w

2

W1

2L m

a

1

W

L = w

2

W

L or a

w

W

W=

2122

L

L

Specific gravity of stone

= a

a w

W

W W=

w

a

1W

1W

=

2221

1

L1

L

= 21

2 21 2

L

L L =

2

2 2

(50)

(50) (28) 1.5 g/cc

6. y = 6 sinx

18

cos (80t)

Comparing with standard stationary wave

equation, we get x2

= x

18

or = 36

Distance between node and antinodes

= 4

= 9 cm

7. = v

n=

3401.3m

256

Shortest distance for maximum amplitude

[antinode] is 0.3m4

8. 4 N an 3 A, spacing = 1.5 Å

From the figure, 3

2 = 1.5

= 1 Å

A A A

N N N N

7


9. y = 0.1 sin (x + 20t) = 0.1 sin 2 x 20t

2 2

Comparing with

y = A sin 2 t x

T

, we get

T = 10

n = 10

and = 2

v = n = 20 m/s

v = T

m we get,

T = v2m = 400 1.5 104 T = 6.0 102 N 10. 2 = 21, and d2 = 2d1

r2 = 2r1

f 1

ror

2

1

f

f = 1 1

2 2

r

r

= r

(2 )(2r)

= 4

1

f2 = f/4 11. M = 0.015 kg, L = 2.7 m, T = 80 N

m =L

M =

0.015

2.7

v = m

T =

80 2.7

0.015

= 120 m/s

12. n = 1 T

2L m

n2 = 2

T

4L m

m = 2 2

T

4L n=

2 4

4 9.8

4 7 7 10 4 10

= 5 104 kg/m

13. Beat frequency = 6

4Hz,

dT

T% = 1%

n T

dn

n 100 =

1

2

dT100

T

= 1

2 1% = 0.5%

dn = 0.5

100n =

6

4

n = 600

2

n = 300 Hz.

14. n = 200 Hz, dT

T 100 = 2%,

n T

dn

n 100 =

1

2

dT100

T

=

1

2 2% = 1%

dn

n=

1

100

dn = 1

100 200 = 2

15. For two loops, the frequency will be 2n

frequency of tuning fork will be 256 2 = 512 Hz

16. n 1

r or 1

2

n

n = 2

1

r

r =

2

1

17. n = P T

2L mfor same n, T and m,

P

L= constant

when L is tripled, no. of loops should be 3.

18. n = 1 T

2L m

n2 = 2

1 T

m4L

T = 4 L2n2 m

= 4 3

4

3

4 8 103 104

= 180 N 19. n1 = 300 Hz, n2 = 150 Hz,

1 = 1.5 cm

Using,

n11 = n22 we get,

300 1.5 = 150 2

2 = 3 m

8


20. T1 = 25 kg-wt, n2 = 1

2n1

Using,

n T ,

2

1

n

n = 2

1

T

T = 1

1

n / 2

n=

1

2

2

1

T

T =

1

4

1 2

1

T T

T

= 1

1

4 =

3

4

T1 – T2 = 3

4 25 19 kg-wt.

21. Fundamental frequency first harmonic = n = 125 Hz. 3rd overtone 4th harmonic = 4n = 4 125 Hz = 500 Hz. 22. n = 300 Hz, beat frequency = 5/s

n 1

l

n

n 5= 2

1

l

l

300

295=

60

59 = 2

1

l

l

23. T2 = 2T1, 2 = 1

1

2 , d2 = 2d1

r2 = 2r1 Using,

n = 2

1 T

2L r we get,

2

1

n

n=

2

2 1 1

1 2 2

T r

T r

= 1 2

24 1

= 1

n2 = n1 = n 24. 1 = 20 cm, 2 = 18 cm,

Beat frequency = 4/s Using,

1

2

n

n= 2

1

= 18

20=

9

10

n1 < n2 and n2 = n1 + 8

1

1

n

n 8=

9

10 or 10 n1 = 9 n1 + 72

n1 = 72 Hz

25. 21p T1 = 2

2p T2

25 4.5 = 16 T2 T2 7 gm-wt. 26. Frequency of the fork = 2n = 180 Hz. 27. 2

1p T1 = 22p T2

52 (61 + 14) = 72 T2 25 75 = 49 T2

T2 = 25 75

49

= 38 gm-wt

The weight that should be removed (75 38) = 37 gm-wt.

28. n1 = n2

1

2

L

L= 1

2

T

T

L2 = L1 2

1

T

T = 70

2.5

10= 35 cm

29. When is wavelength and l the length of pipe

and n the frequency of note emitted and v the velocity of sound in air, then

n = v

(fundamental note)

= 350

175= 2 m

But, = 4l

l = 4

=

2

4= 0.5 m

30. = 20 cm, d = 1 cm, r = 0.5 cm, v = 340 m/s

Using,

n =v

2l=

2

340

2 20 10 ….

2

= 8.5 102 m = 850 Hz n1 = 2 850 = 1700 Hz n2 = 2 1700 = 3400 Hz 31. Fundamental frequency of closed pipe = 320 Hz

Using n = v

4l we get,

v = 320 4l

If rd

1

3

of the pipe is filled with water, then

remaining length of air column is 2

3

l.

9


Now, fundamental frequency = v 3v2 8

43

l l

and first overtone = 3 fundmental frequency.

= 3 3v

8l=

3 320 4 3

8

l

l

= 1440 Hz

32. In first overtone mode,

l = 3

4

4

=

3

l

4

=

1.8

3

= 0.6 m Pressure variation will be maximum

displacement nodes, i.e., at 0.6 m from the open end.

33. 4

= l1 + e and

3

4

= l2 + e

or 2

1

e

e

l

l = 3

or l2 = 3l1 + 2e = 3 18 + 2 0.8 56 cm

34. Using, e = 2 13

2

we get,

e = 61.3 3 19.8

2

= 61.3 59.4

2

=

1.9

2= 0.95 cm

35. Using relation v = n we get,

= v

n=

340

340 = 1m

If length of resonance columns are l1, l2 and l3 then,

l1 = 4

=

1

4m = 25 cm (for first resonance)

l2 = 34

=

3

4= 75 cm (for second resonance)

l3 = 5

4

=

5

4m = 125 cm (for third resonance)

This case of third resonance is impossible because total length of the tube is 120 cm. So, minimum height of water

= 120 75 = 45 cm

36. For open pipe, f1 = v

2l

and for closed pipe,

f2 = v

44

l

= v

l= 2 f1

or 1

2

f

f =

1

2 37. Let l be the length of pipe, v the speed of

sound. The fundamental tone or first harmonic of

closed tube,

n1 = v

4l

For open tube,

n2 = v

2l

n2 = 2n1 Given that, n1 = 480 Hz

n2 = 2 480 = 960 Hz 38. Given that, fo fc = 3 ….(1) Frequency of fundamental mode for a closed

organ pipe,

fc = c

v

4L

Similarly frequency of fundamental mode for an open organ pipe,

fo = o

v

2L

Given Lc = Lo

fo = 2fc ….(2) From Eqs. (1) and (2), we get, fo = 6 Hz and fc = 3 Hz

Closed tube

l

A

/4

N

A

N

A

/2

A

N

A

N

l = 1.8 m

/4

2 /4

10


When the length of the open pipe is halved, its frequency of fundamental mode is

of = o

vL

22

= 2fo = 2 6 Hz = 12 Hz When the length of the closed pipe is doubled,

its frequency of fundamental mode is

of = c

v 1

4(2L ) 2

c

1f

2 3 = 1.5 Hz

Hence, number of beats produced per second is

of cf = 12 1.5 = 10.5 11

39. Here, 4

= 100

300 = 3 4

and 500 = 5

4

Hence, the pipe is closed at one end. 40. Length of first pipe l1 = 20 cm Length of second pipe l2 = 20.5 cm Frequency of first pipe f1 is given by,

1

v

2l=

v

2 20=

v

40Hz

Frequency of second pipe f2 is given by,

2

v

2l =

v

2 20.5 =

v

41Hz

(where v is velocity of sound) Hence, the beat frequency f = f1 f2

0.2 = v v v

40 41 40 41

v = 0.2 40 41 v = 328 m/s

1

Chapter 09: Kinetic Theory of Gases and Radiation


1. i. c = 1 2 2 2 3 4 5 5 6 610

= 3.6 m/s

ii. 2c = 2 2 21 2 ..... 6

10

=

164

10

= 16 m/s iii. crms = 16 = 4 m/s 2. c1 = 4 km/s, c2 = 6 km/s, c3 = 9 km/s, c4 = 10 km/s, c5 = 12 km/s.

c = ?; 2c = ?; crms = ?

c = 4 6 9 10 12

5

=

41

5= 8.2 km/s

2c =2 2 2 2 24 6 9 10 12

5

= 75.4 km/s

crms = 2c = 8.683 km/s 3. CO = 460.9 m/s at T = 0C + 273 = 273 K, MO = 32, MH = 2, CH = ?

H

O

C

C= O

H

M

M

CH = 460.9 32

2= 1843.6 m/s

4. c =3 8400 400

32

= 561.24 m/s

5. CH at TH = 400 K, MH = 2 CO at TO = 800 K, MO = 32,

H

O

C

C = OH

O H

MT

T M =

400 32

800 2

H

O

C

C = 2.828 : 1

6. T = 0C + 273 = 273 K, kB = 1.38 1023 J/mol K NA = 6.02 1023/g mol = 6.02 1026 k mol M = 64 g = 64 103 kg.

c = B3k T

m = B A3k TN

M

c = 23 23

3

3 1.38 10 273 6.02 10

64 10

= 3

3 1.38 273 6.02

64 10

c = 3106.31 10

c 326 m/s 7. T1 = 0C + 273 = 273 K,

P = 1.0 105 Nm2

= 1.98 kg m3

T2 = 30C + 273 = 303 K

c = 3P

c1 = 53 1 10

1.98

= 389 ms1

Now,

c2 = c1 2

1

T

T = 389

303

273 410 ms1

8. T1 = 0 + 273 = 273 K, c1 = 460 ms1,

T2 = 40 + 273 = 313 K, M1 = 32, M2 = 4

2

1

c

c= 2 1

1 2

T M

T M =

313 32

273 4

c2 = 460 3.0286 1394 m/s. 9. T1 = 273 K,

c2 = 2c1

2

1

c

c= 2

1

T

T

T2 =

2

2

1

c

c

T1

= (2)2 273

T2 = 1092 K

Kinetic Theory of Gases and Radiation 09

2


10. d = 2 1010 m, r = 1010 m,

2

1

2n

or n =

2

2

2

= 8 20

1

2 2.4 10 4 10

n = 263.3157 10

2

= 2.3443 1026 molecules/m3

11. nv = 2.79 1025 molecules/m3,

= 2.2 108 m,

= 2v

1

n

=v

1

n =

25 8

1

3.14 2.79 10 2.2 10

= 201000 10

3.14 2.79 2.2

= 7.203 1010 m

12. = 2v

1

n

= 25 10 2

1

3.14 3 10 (2 10 )

= 2.7 107 m 13. m = 4.55 107 kg, V = 1 litre = 103 m3

P = 105 N/m2, c = 350 m/s

N = 5 3

2 25 2

3PV 3 10 10

mc 4.55 10 (350)

= 5.382 1021

14. N

V = APN

RT=

5 2610 6.02 10

8310 300

= 2.415 1025

15. N

V =

2

3P

mc

N

V=

5

26 2

3 1.013 10

5.313 10 (461.2)

= 2.687 103 1031

= 2.687 1028 / m3

16. m = 64 g = 64 103 kg, V = 22.4 lit = 22.4 103 m3

T = 300 K, M = 32

PV = nRT = m

RTM

P = 3

3

64 10 8310 300

32 22.4 10

= 2.226 105 N/m2

17. N

V= 6.8 1015 /cm3 = 6.8 1021 m3,

c = 1.9 103 m/s, NA = 6.02 1026/k mol, M = 2

N

V= A

2

3PN

Mc

P =

2

A

NMc

V3N

= 21 3 2

26

6.8 10 2 (1.9 10 )

3 6.02 10

= 2.7185 10

13600 9.8

= 2.039 104

= 0.2039 mm of Hg 18. P = 1.1 105 N/m2, T = 300 K, M = 28, R = 8.314 J/mol k

c = 3RT

M=

3 8.314 300

28

= 16.35 m/s

P = 1

3c2 =

2

3P

c=

5

2

3 1.1 10

(16.35)

= 1.234 103 kg/m3

19. 24 3N10 / m

V , c = 500 m/s,

M = 2.568 1026 kg

P = 21 MNc

3 V = 21 N

Mc3 V

= 24 26110 2.568 10 500 500

3

= 21.4 102 = 2.14 103 N/m2

20. m = 15 g = 15 103 kg, M = 17.03 g/mol, R = 8.31 J/kmol T = 37C + 273 = 310 K

K.E. = 3

2 PV =

3

2nRT =

3

2

mRT

m

K.E. = 33 15 10 8310 310

2 17.03

= 3403 J

3


21. T = 127C + 273 = 400 K, R = 8310 J/kmol K,

NA = 6.03 1026 / kmol

K.E. per molecule = A

3 RT

2 N

= 26

3 8310 400

2 6.03 10

8.3 1021 J 22. M = 32,

P = 76 cm of Hg = 76 13.6 980,

NA = 6.022 1023 / mol,

R = 8.314 107 erg mol K T = 273 K

i. K.E. per cm3 =3

2P =

3

276 13.6 980

= 1.519 106 erg/cm3

ii. K.E. per mol = 3

2RT

= 3

2 8.314 107 273

= 3.405 1010 erg/mol

iii. K.E. per g = 3 RT

2 M

= 103.405 10

32

= 1.064 109 erg/g

iv. K.E. per molecule = A

3 RT

2 N

= 7

23

3 8.314 10 273

2 6.022 10

= 5.654 1014 erg/molecule 23. (K.E.)He = 9.353 105 J/kg MH = 2, MHe = 4

K.E. 1

M

HeH

He H

M(K.E.)

(K.E.) M

(K.E.)H = 9.353 105 4

2

= 1.8706 106 J/kg

24. Cv = 5 kcal/kmolC, = 1.4, R = 8.3 J/g mol K = 8300 J/k mol K.

= p

V

C

C

Cp = Cv = 1.4 5 = 7 kcal/ kmolC

Cp Cv = R

J or J =

p v

R

C C=

8300

7 5

= 4150 J/kcal 25. cP = 0.21 kcal/kg K, = 1.41, J = 4200 J/kcal

= p

v

c

c

cv = pc

=

0.21

1.41 = 0.1489 kcal/ kgK

Now, cp cv = R

J

R = J(cp cv) = 4200 (0.21 0.1489) = 256.62 J/kg K 26. Cp = 6.78 103 kcal/ mol K, Cv = 4.79 103 kcal/mol K R = 8.34 J/m = 8340 J/kmol K Cp Cv = R

J = p v

R

C C =

3

8340

(6.78 4.79) 10

4191 J/kcal 27. = 1.775 kg/m3, T = 27C +273 = 300 K P = 105 N/m2, cp = 846 J/kg K

R = PV

T=

5 110

T

=

510

300 1.775

= 187.79 J/kg K. cv = cp R = 846 187.79 = 658.21 J/kg K. 28. J = 4200 J/kcal, At NTP : V = 22400 lit = 22.4 m3/kmol P = 1.013 105 N/m2, T = 273 K Cp Cv = ? Using, PV = RT, Cp Cv = R

R = PV

T=

51.013 10 22.4

273

= 8311.79 J/kmol K

= 8311.79

4.2cal/kmol K

= 1978.99 = 1.979 kcal/ kmol K

4


29. dQ = 40 k cal, dU = 84000 J, J = 4200 J/cal

dQ = dU dW

J

dW = J.dQ dU = 4200 40 84000 J = 84000 J

30. dQ = dU dW

J

J = dU dW

dQ

=

12000 9000

5

= 4200 J/kcal

31. dQ = dU dW

J

dU = J.dQ dW = 4200 15 51000 = 12000 J 32. dQ = 40 kcal, dU = 8372 J, J = 4186 J/kal

dQ = dU dW

J

dW = J. dQ dU = 4186 40 8372 = 1.59068 105 J

= 51.59068 10

4186

= 38 kcal. 33. Qi = 270 J, Qr = 35.5 J,

Using, r = r

i

Q

Q =

35.5

270= 0.13

34. In 5 min, Qi = 5 1500 = 7500 J

Using, r = r

i

Q

Q=

450

7500= 0.06

Now using, a + r + t = 1, we get, t = 1 (a + r) = 1 (0.6 + 0.06) = 0.34

35. idQ

dt= 200 J/s, a = 0.8, r = 0.12, Qa = ?, Qr = ?,

Qt = ?, t = 15s. Using, a + r + t = 1, we get, t = 1 (0.8 + 0.12) = 0.08 Qi = 200 15 = 3000 J

Now, a = a

i

Q

Q or Qa = a,

Qa = 0.8 3000 = 2400 J Qr = r Qi = 0.12 3000 = 360 J Qt = t Qi = 0.08 3000 = 240 J.

36. Qa = 30% of Qi = 0.3 Qi

a

i

Q

Q= a = 0.3

a + r = 1 (for athermanous body) r = 1 a = 1 0.3 = 0.7 37. r = 0.01, a = 0.7 Using, a + r + t = 1, we get t = 1 (0.01 + 0.7), t = 1 0.71 = 0.29 38. side = 10 cm = 10 102 m,

dQ

dt= 60 w, e = 0.5, E = ?, Eb = ?

Using, E = 1 dQ

A dt we get,

E = 1 2

160

6 (10 )

= 10 102 = 1000 J/m2s

Using, e = b

E

Ewe get,

Eb = E

e=

1000

0.5= 2000 J/m2s.

39. P = 10 W, e = 0.8, t = 1 min = 60 s.

a = e, a = a

i

Q

Q for t = 0.

Energy incident in 1 min = 10 60 = 600 J a + r = 1 r = 0.2, Qr = i r = 600 0.2 = 120 J. 40. T = 127C + 273 = 400 K Eb = T4 Eb = 5.67 108 (400)4 = 1451.5 J/m2 s

41. a = e = b

E

E= a

i

Q

Q

Qa = b

E

E Qi =

701800

112 = 1125 J

42. m = 4.2 kg, A = 0.05 m2,

dQ

dt= 0.05 k/min =

0.5

60 k/s, c = 420 J/kg K,

Eb = 400 W/m2.

dQ

dt= m.s.

d

dt

dQ

dt= 4.2 420

0.5

60= 14.7 J/s

5


Now, E = dQ

Adt=

14.7

0.05= 294 W/m2

e = b

E 294

E 400 = 0.735

As e = a a = 0.735 43. Eb1 = 10 W/cm2, T1 = 1000C + 273 = 1273 K

Eb2 = 10 103 W/cm2, T2 = ?

Eb = T4

1

2

Eb

Eb=

424

1

T

T

T2 = 421

1

EbT

Eb

= 3

410 10(12 3)

10

= 4 4

410 (1273)

10

= 4

10 1273

10

= 7158 K 44. Eb = 1 kcal/m2 s

1 cal = 4.2 J

Eb = 4.2 103 J/m2s

Eb = T4

T4 = bE

=

3

8

4.2 10

5.73 10

= 124.210

57.3

T = 344.2

1057.3

= 0.520 103

T = 520 K 45. mT = b

m = 32.88 10

6150

= 4.683 104 103

= 4683 1010 m = 4683 Å 46. m = 4753A = 4753 108 cm,

b = 0.2893 cmK.

mT = b

T = 8

0.2893

4753 10 = 6.0867 105 108

= 6087 K.

47. TA = 2TB.

mT = b, A

B

m A

m B

T

T

= 1

Bm =

Am A

B

T

T

= 4.8 107 2

= 9.6 107 m

= 9600 1010 m

= 9600 Å 48. A = 100 cm2 = 100 104 m2,

T = 227C + 273 = 500 K,

e = 0.2, T0 = 27C + 273 = 300K,

= 5.7 108 W/m2 k4

dQ

dt= e A (T4 4

0T )

= 0.25.7108 100 104 (5004 3004)

= 620.16 102 = 6.202 J/s 49. V = 125 cm3 = 125 106 m3,

A = 6 25 104 m2 ( cube has 6 faces)

T 127C +127 = 400 K,

= 5.67 108 S.I unit,

J = 4200 J/kcal, dt = 10 min = 10 60 = 600s

dQ = A T4dt

dQ = (5.67108) (625104) (400)4 600

= 13.06 105 108 104 108 102

= 13060 J

= 13060

4200 = 3.109 kcal 3110 cal

50. Q

t= 60W, l = 0.5 m, r = 3 105 m,

= 5.67 108 J/m2s K4,

Q

t= A T4

60 = 42.7291 1013 T4

T4 = 60

42.7291 1013 =

12600 10

42.7291

T = 34600

1042.7291

= 1.935 103

= 1935 K

6


51. dQ

dt= 87.55W, T =127C+273 = 400 K.

4dQAT

dt

A = 4

dQ 1

dt T

= 87.55 8 4

1

5.67 10 (400) =

87.55

1451.52

= 6.032 102 m2

Side2 = 26.032 10

6

= 1.0053 102

Side = 1.0026 101 = 0.10026 m Volume = (0.10026)3 1 103 m3

52. 1

2

d

d= 2 : 1 1

2

r

r = 2 : 1, 1

2

T

T= 1 : 1

1 2

dQ dQ:

dt dt

= ?

dQ

dt = A T4

1 2

dQ dQ:

dt dt

= 1

2

A

A=

2122

4 r

4 r

=

2

1

2

r

r

= 2

2

1

= 4 = 4 : 1

53. T1 = 527 C + 273 = 800 K, T2 = 227 C + 273 = 500 K

1

2

dQ

dtdQdt

= ?

Using, 4dQT

dt

for a given body we get,

1

2

dQ

dt

dQ

dt

= 4

142

T

T =

4800

500

= 6.5536 : 1

54. 1

d

dt

= 1.6 C/min, 1 = 70C,

2 = 40 C, 0 = 30C, 2

dQ

dt

= ?

d

dt

= K ( 0)

2

1

d

dtd

dt

= 2 0

1 0

2

d

dt

= 2 0

l 01

d

dt

2

d

dt

= 1.6 10

40= 0.4 C/min.

55. For the metal sphere,

1 = 60C, 2 = 52C, t = 5 min.

1 = 52C, 2 = 44C, t = 7.5 min.

0 = ?, 3 = ?

d

dt

= K ( 0)

60 53

5

= K 0

60 52

2

8

5 = K (56 0) ….(1)

Now, 52 44

7.5

= K 0

52 44

2

8

7.5= K (48 0) ….(2)

Dividing eq. (1) by eq. (2) we get,

8

5

7.5

8 = 0

0

56

48

1.5 = 0

0

56

48

72 1.50 = 56 0 0r 0 = 32 C

Using equation for 3,

344

10

= K 344

322

….(3)

Dividing eq.(3) by eq. (2) we get,

344 7.5

10 8

=

1

16

4432

2

(44 3) 7.5 16 = 80 44

322

120 (44 3) = 40 (44 + 3) 2560

5280 1203 = 1760 + 403 2560

5280 1760 + 2560 = 403 + 1203

1603 = 6080 3 = 6080

160= 38C

7


56. 1

d

dt

= 3C/min at 1 = 70C,

2

d

dt

= 1.5 C/min at 2 = 50C,

3

d

dt

at 3 = 40 C.

Using,

d

dt

= K ( 0) we get,

1

d

dt

= K (70 30) ….(1)

2

d

dt

= K (50 30) ….(2)

1

2

d

dtddt

= 0

0

70

50

= 1.5

70 0 = 100 20 or 0 = 30C

3

d

dt

= K (40 30) ….(3)


3

d

dt

3

=

10

40

3

d

dt

= 0.75C/min

57. 1

d

dt

= 0.5 C/s, 1 0 = 50C,

2

d

dt

= ? when 1 0 = 30C.

Using, 2

d

dt

= K ( 0) we get,

2

d

dt

= 2

d

dt

2 0

1 0

( )

( )

= 0.5 30

50

= 0.3C/s 58. r1 = 2 cm = 2 102 m, r2 = 3 cm = 3 102 m, 1 = 80C, 2 = 90C, 0 = 30C, T1 = 353 K, T2 = 363 K, T0 = 303 K.

1 2

dQ dQ:

dt dt

= ?; 1 2

d d:

dt dt

= ?

1 2

dQ dQ:

dt dt

= 4 42

1 012 4 42 2 0

(T T )4 r

4 r (T T )

=

22 4 4

2 4 4

2 10 353 303

3 10 363 303

= 4

9

9

9

7.0985 10

8.9342 10

= 0.3531

d

dt

= K ( 0)

1 2

d d:

dt dt

= 1 0

2 0

= 80 30

90 30

= 50

60 = 5 : 6

59. 2

d

dt

: 1

d

dt

= 1

3

d

dt

= K ( 0)

1

2

d

dtddt

= 2K ( 30)

K (50 30)

= 1

3

3 (2 30) = 20 32 = 110 2 = 36.67 C

60. T =

12 4r 5

R 6

=

12 48

5 8

1.52 10 1400

7.5 10 5.67 10

= 1

2 6 8 40.20266 10 246.91 10

= 1

14 40.04107 246.91 × 10

= 1

12 44.107 246.91 × 10

= 5.643 103 = 5643 K

61. S = T42

R

r

= 25

4 88

7 10(6000) 5.7 10

1.48 10

= 1296 1012 22.37 106 5.7 108 = 165251.66 102 = 1652.51 J/m2s.

8



1. c = 2 2 2 22 4 6 14

4

ms1

= 63 ms1 = 7.94 ms1 8 ms1

2. R.M.S. velocity = rmsc = P3

=0.09

10 1.0133 5 1838 m/s

3. 2Ov =

1

3 2Hv

32

RT3 =

1

3

2

)273(R3

32

T =

1

9

2

273

T 485 K = 212 C 4. T1 = 27 + 273 = 300 K T2 = 327 + 273 = 600 K

Crms T Temperature is increased by a factor of 2. So,

r.m.s. speed is increased by a factor of 2 . 5. crms of Nitrogen at 0 C

i.e. 273 K is crms =O

3RT

M=

28

273R

MO = molecular weight The r.m.s. velocity of Oxygen at temperature

T is crms =OM

RT3=

32

RT3

32

RT3 =

28

R2733

T = 28

273 32 = 7

8 273 = 8 39 = 312

T = 312 – 273 = 39 C

6. 2

1

c

c= 2 1

1 2

T M

T M

2c

200=

1 1

2 2

2c

200=

1

4 =

1

2

c2 = 200

2 = 100 m/s

7. t = v

=

23 10 m

8s

= 3.75 s

8. The gas equation PV = nRT

or V

n=

RT

P ....(1)

Molecules per unit volume

N1 = 1

1

V

n =

1

1

RT

P=

RT

P

and N2 = 2

2

V

n =

2

2

RT

P =

)3/T(R

P2

= RT

P6 = 6

P

RT

= 6N1 ....from (1)

N1 : N2 = 1 : 6

9. P = 1

3c2

1

2

c

c = 1 2

2 1

P

P

= 3

2

10. crms=P3P =

3

c2=

2 2(400) 4.5 10

3

= 16 104 1.5 10–2 = 24 100 = 2.4 103 N/m2. 11. From Boyles law, P1 V1 = P2 V2 Here P2 = 1.25 P1

2 1

1 2

V P 1 4

V P 1.25 5

V2 = 14

V5

V = V1 V2 = V1 14V

5 = 1V

5

% change = 1

1

V / 5

V 100 =

100

5 = 20%

Volume decreases by 20%.

12. P ' 188 273 461

P 38 273 311

= 1.48

P = 1.48 1 = 1.48 atmosphere 1.5 atm 13. m2 = 2 m1 P1V = n1RT or P2V = n2RT

2

1

P

P= 1 1

2 2

n m

n m =

2

1 or P2 = 2 P1

= 2 105 Pa

9


14. PV = P 90V

100 or

P

P =

10

9

P

PP 100 =

10 9

9

100 = 1

1009

= 11.11% 15. Pressure = P = 1011 N/m2 Temperature T = 27C = 27 + 273 = 300 K k = 1.38 1026 J/molecule Kelvin.

N

V

= P

kT

N

V

= 11

26

10

1.38 10 300

= 1310

3 1.38 = 2.42 1012/m3

16. 4 g of He gives NA molecules 16 g of He gives 4 NA molecules. 17. Let x = depth of the lake. From Boyle’s law P1V1 = P2V2 V2 = 6 V1 P1 = (74 13.6 g + x 1 g) P2 = 74 13.6 g V(74 13.6 + x)g = 74 13.6 g 6V 74 13.6 + x = 6 74 13.6 x = (6 – 1) 74 13.6 = 5 74 13.6 = 5032 cm. x 50 m.

18. Kinetic energy = 2

3 PV

V

E.K=

2

3 P =

2

3 105

= 1.5 105 J.

19. K.E. T or 1

2

K.E.

K.E. = 1

2

T

T

20

2

3.3 10

K.E.

=

27 273

127 273

= 300

400

K.E.2 = 203.3 10 400

300

= 4.4 10–20 J.

20. 2

3kT = 4.5 1021

kT = 3

2 4.5 1021 = 3 1021

Now, PV = nkT

n = PV

kT =

5 6

21

1.5 10 300 10

3 10

= 1.5 1022

21. K.E. T 2

1

.E.K

.E.K =

2

1

T

T =

27 273

t 273

1

2

K.E.

3K.E. =

300

t 273

t + 273 = 900 T = 627 C.

22. Ek = 2

3kT =

2

3

AN

RT

= 23

3 8.3 400

2 6 10

J = 830 10–23 J

= 8.3 1021 J 23. The average kinetic energy of all the gases at

any given temperature is same.

24. Average K.E. of molecules = 2

3 kT

Energy gained by electron = eV = 1.6 10–19 2

2

3 1.38 10–23 T = 1.6 10–19 2

T = 4

36.4 10 6410

1.38 3 4.14

1.5 104 K

25. The mass of the molecules remain constant.

Hence when momentum increases, only their speed increases.

2 1

1

E E

E

100 =

2 22 1

21

U U

U

100

=2 2

2

(110) (100)

(100)

100

= 21%

26. vc = dT

dU

dU = vc dT =

2

3 R dT

= 2

3

4200

8300 2 = 5.93 6

27. p

v

c

c = 1.4 or cv = pc

1.4

cp cv = R or cp pc

1.4 = R

cp 1

11.4

= R

10


cp = 1.4R

0.4=

1.4 1.8

0.4

6.3 cal/g mol C 28. t = 60 40 = 20 C, n = 2, Q = 300 J

cP = Q

n t =

300

2 20 = 7.5

29. p

v

c

c= 1.4 or cp = 1.4 cv

cp cv = R or 1.4 cv cv = 8400 0.4 cv = 8400 or cv = 21000 J/kmol K 30. pc T = vc T + W

W = (cp – cv)T Fraction of heat converted into work

= Q

W

= Tc

T)cc(

P

VP

= 1 – 1

= 1 – 5

3 =

5

2

31. cp cv = R

MJ

J = p v

R

M(c c ) =

8310

2(4000 3000)

= 8310

2 1000= 4.155 J/cal

32. cp – cv = J

R or cv = cp –

J

R

7 –8.30

4.2= 7 – 1.976 = 5.024 cal/molC

cv = Tn

Q

Q = cv nT = 5.024 4 10 201 cal 33. P T3 ….(1) For an adiabatic change, P1T = constant

P1 1

T P 1T ….(2)

From (1) and (2), we find that 1

= 3

= 3 3 2 = 3 = 3

2

34. It is given that, P Tx ….(1) For an adiabatic change, P1T = constant

P1 1

T or P 1T ….(2)

and for the monoatomic gas, = 5

3

Comparing (1) and (2), we get

x = 1

but = 5

3

x =

5

35

13

=

5

32

3

= 5

3

3

2 =

5

2

35. Heat required to increase the temperature at

constant volume is Qv = nCv dT = 5 Cv (80 60) = 100 Cv But dQ = dU + dW. When the volume is kept

constant W = PdV = 0 dQ = dU Qv = 100 CV = dU = 40 J

Cv = 40

100= 0.4 J/mol K

Heat capacity = nCV = 5 0.4 = 2.0 J/K 36. From the first law of thermodynamics, dQ = dU + dW dU = dQ dW The change in internal energy = 55 25 = 30 J. The change in the internal energy of a system

depends only upon its initial and final states and is independent of the actual path followed. Hence we get the same change i.e. 30 J in U, whether the system goes from P Q R or directly from P R.

37. When we compress the gas, the work done on

the gas is negative. Work done = dW = PdV = 30 (4) = 120 J Now, dQ = dU + dW dU = dQ dW = dQ PdV dU = 80 (120) = 200 J 38. From first law thermodynamics, dQ = dU + dW

dU

dQ is the part of heat supplied to the gas that

is used for increasing the internal energy

and dU

dQ= v

p

nC dT

nC dT= v

p

C

C=

1

But for a monoatomic gas, = p

v

C

C=

5

3

11


dU

dQ =

1

5 / 3 =

3

5

dU = 3

5

dQ

Percentage of heat used for increasing the

internal energy = 3

5 100 = 60%

39. dQ = dU + dW but dW = 0 dQ = dU = n Cv dT In this case, n = 4 mol, dT = 100

Cv for an ideal monoatomic gas = 3

2R

dQ = 4 3

2R 100 = 600 R

40. = 20%

= 1 2

1

T

T= 1 2

1

T T

T

20

100 =

1

5= 1 2

1

T T

T

5T1 5T2 = T1 4T1 = 5T2 = 5 (273 + 30) = 1515 T1 = 378.75 K 106C

41. = 1 2

1

T

T= 1

300

450=

1

3

and = 1

W

Q

Q1 = W

=

200

1/ 3 = 600 J

42. As, 2

1

Q

Q= 2

1

T

T

T2 = 2

1

Q

Q

T1 = 300

400400

= 300 K = (300 273) = 27C 43. Efficiency of the engine = 40% = 0.4 Rate of energy supplied i.e. Power = 1000 W

For any engine, = Power output

Power input

0.4 = Power output

1000

Output power = 1000 0.4 = 400 W 44. T1 = 300 K, T2 = 273 K T1 T2 = 27 K

To freeze 5 g of water at 0C, the quantity of heat that must be transferred (removed) from water to the surrounding is

Q2 = mass latent heat = 5 80 = 400 cal

and = 2Q

W= 2

1 2

T

T T

W = 2 1 2

2

Q (T T )

T

=

400 27

273

40 cal

45. T1 = 273 + 27 = 300 K and T2 = 273 + 6 = 279 K

= 2

1 2

T

T T

= 279 279

300 279 21

13

46. = 2Q

W= 2

1 2

T

T T

= 275

300 275=

275

25 = 11

Q2 = W = 1 11 = 11 J 47. For a refrigerator, the coefficient of

performance,

= 2

1 2

T

T T =

230

250 230=

230

20 = 11.5 and

= 2Q

W Q2 = W = 11.5 1 = 11.5 J

where Q2 = Heat from the cold reservoir W = work done on it Q1 = Heat transferred to the room By the law of conservation of energy, Q1 + Q2 + W = 11.5 + 1 = 12.5 J 48. For athermanous body, t = 0 a = 0.55 a + r + t = 1 or 0.55 + r + 0 = 1 r = 0.45 50. Qa + Qr + Qt = Q 4 + 2 + Qt = 8 or Qt = 2

Now, t = tQ

Q=

2

8 = 0.25

51. a = aQ

Q =

336

486 = 0.691 0.7

e = a = 0.7 52. For athermanous body, t = 0 a = 38 % = 0.38 a + r + t = 0 0.38 + r + 0 = 1 r = 0.62

12


53. Q = 200 J/min, A = 4 cm2 = 4 104 m2

T = 427 + 273 = 700 K Q = eAtT4

e = 4TAt

Q

e = 4 8 4

200

4 10 60 5.67 10 (700)

= 12 8

5

6 5.67 2401 10 10

= 6.12 105 104 e 0.6 54. E = 2.835 watt/cm2 = 2.835 104 watt/m2 Using, E = T4

T = 1/41/4 4

8

E 2.835 10

5.67 10

= 1

12 40.5 10

= 0.84 103 = 840 K 55. E T4

4 4

2 2

1 1

E T 650

E T 273

(2.38)4 32

E2 = 32 E 56. P = AeT4 = 0.3 10–40.3 5.67 10–8(28 102)4 = 9 5.67 6.146 105 108 1014 31 W

57. E = Q

tA

0.25 = Q

20 0.04

Q = 5 0.04 = 0.2 kcal

58. E =Q

At= 2

Q

6 t=

2

0.6

6 (0.04) 100

= 0.625 0.63

59. 3

4r3 = a3 or

3/1

4

3

a

r

2/3 1/32

1 12

2 2

Q A 4 r 4 3:1

Q A 6a 6 4 6

60. T = 127 +273 = 400 K,

Q

t = 1459.2 J/s

Q

t= AT4

A = 4

(Q / t)

6T =

8 4

1459.2

5.7 10 (400) = 1 m2

61. As T

1m ,

Temp. of other star must be T/3. 62. According to Wien’s displacement law,

2

mb 0.298 10

T 4000

= 7450 Å

63. = 470 nm = 470 109 m

T =m

b

=

3

9

2.898 10

470 10

6166 K

64.

mstar sun

sun m star

T 720 4

T 540 3

65. 2

1

m 1

m 2

T

T

=

3000

2000 =

3

2

2m =

3

2

1m

= 3

2 3 106 =

9

2 106

= 4.5 106 m 66. T = 273 + 227 = 500 K T0 = 273 + 27 = 300 K

dt

dQ = Asur (T

4 –T04) = (6r2) (T4 –T0

4)

85.5 = 5.7 10–8 6r2 (625 – 81) 108 15 = r2 6 544 or 15 = 3264 r2 r2 = 4.596 103 m r = 6.78 102m Volume = r3 = 3.1 104 m3 = 0.31 103 m3

67. dt

dQ = A e (T4 –T0

4)

= 5.67 10–8 100 10–4 0.3 (6004 – 3004) = 5.67 100 0.3 10–12 (129.6 109

– 8.1 109) = 1.701 10–10 121.5 109 20.7 J/s 68. T = 227 + 273 = 500 K

Heat radiated per sec. = dQ

dt = AT4

13


Heat radiated in 20 sec. = 20 AT4

= 20 5.7 108 120 104 (500)4

= 8.55 106 108 104 108

= 855 J 69. From the Stefan’s law, heat radiated

= A t T4

= 200 10–4 60 5.710–8 (500)4

= 4.275 107 104 = 4275 J 70. Energy radiated from sun per unit time

= 4r2 Energy reached to earth where r = radius of earth orbit

ES = 4 3.14 (1.5 1011)2 1400

= 4 3.14 2.25 14 1024

= 3.14 126 1024 J

E = A eT4

T4 = 24

16 8

E 3.14 126 10

A e 4 3.14 49 10 1 5.67 10

= 0.113 1016

T = 0.5798 104 = 5798 K. 71. T1 = 427 + 273 = 700 K T2 = 227 + 273 = 500 K T0 = 27 + 273 = 300 K Using,

E = (T4 T04) we get,

EA = (T14 T0

4) and EB = (T24 T0

4)

A

B

E

E=

4 4

4 4

(700) (300)

(500) (300)

=

4 4

4 4

7 3

5 3

= 2401 81

625 81

= 2320

544 = 4.26 4

72. Rate of loss of heat per second

= A (T4 – T04) = (4 R2) (T4 – T0

4)

1dt

dQ

= 4R1

2 (T4 – T04) and

2dt

dQ

= 4R2

2 (T4 – T04)

2

1

)dt/dQ(

)dt/dQ(=

22

21

R

R

73. Rate of loss of heat Q A

1 1

2 2

Q A

Q A =

21

22

4 r

4 (r )

= 2

2

6

12=

1

4

74. Mass of ice melted/minute = 10

10 = 1 g

Quantity of heat used = 1 80 cal = 80 cal Area of lens = r2 = 3.14(2)2 = 12.56 cm2

Amount of heat received/min/cm2 from sun

= 80

12.56 = 6.37 6 cal/cm2 min

75. P T4 and P 1/d2

So, 2

1

P

P =

42

1

T

T

21

2

d

d

2

1

P

P = (2)4

21

2

= 22 = 4

P2 = 4P1 = 4P.

76. 4

4

E E [T T]

E T

=

4

4

[T (6 /100)T]

T

1 + E

E

= (1 + 0.06)4 = (1.06)4

E

E

= 1.2624 – 1 = 0.2624

E

E

100 = 0.2624 100

= 26.24% 26% 77. Applying Newton’s law of cooling

56 54

5

= K

56 5430

2

....(1)

41 39

t

= K

41 3930

2

....(2)

Dividing eq. (1) by eq. (2),

t

5=

25

10 or t = 12.5 min = 750 s

78. Applying Newton’s law of cooling,

70 50

10

= K(60 – 25) = K(35) and

K = 2

35

Now, 50

10

= K

5025

2

50

10

=

2

35

5025

2

350 – 7 = 100 + 2 – 100 = 38.88 39 C

14


79. Using Newton’s law of cooling,

dt

d = K(t –t0)

2 = K (60 – 25) = 35 K ....(1)

dt

d= K (40 – 25) = 15 K. ....(2)

d / dt

2

=

15

35 or

dt

d=

30 6

35 7

0.9 C/min

80. d

dt

= K (T T0),

0.3 = K(50) ....(1)

dt

d 2 = K(20) ....(2)

Divide eq. (1) by eq. (2) 2

0.3d

dt

= 50

20= 2.5

2d

dt

=

0.3

2.5= 0.12 C/s

81. dt

d 1 = K(1–0); d1 = 70 – 60 = 10 C

o10 C

5 = K 0

70 60

2

2 = K (65 – 0) ....(1)

Similarly, dt

d 2 = K (2– 0)

d2 = 60 – 50 = 10 C

o

2

10 C

t = K 0

60 50

2

2

10

t = K (55 – 0) ....(2)

From (1) and (2), 5

t 2 = 0

0

65

55

As 65 – 0 > 55 – 0, t2 > 5 min. 82. R ( 0).

Hence R1 (55 25) and R2 (45 25)

Therefore, R1 / R2 = 3/2.

Since same amount of heat is lost in both the cases, therefore R1t1 = R2t2.

This gives t2 = 3/2 8 = 12 minutes

83. d = 61 59 = 2 C dt = 4 min

d

dt

= K

0

21

2

4

2 = K (60 31) = 29 K .…(1)

Also, t

2 = K (50 31) = 19 K .…(2)

Dividing equation (1) by equation (2),

4

t = 29K

19K

t = 6 min

84. R1 = d

dt

, we get, R1 =

1

5

t

t1 = 1

5

R =

0

560 65

K2

t1 = 0

5

K 62.5 ....(1)

t2 = 2

5

R =

0

5

K 57.5 ....(2)

t3 = 3

5

R =

0

5

K 52.5 ....(3)

Comparing eq. (1), (2) and (3) we get, t1 < t2 < t3

85.

4

1

2

1

2

T

T

E

E

or

4/1

2

1

2

1

E

E

T

T

T1 = 6 103 (16 104)1/4 = 6 103 2 10 = 12 104 K 86. A = 20 m2, m = 200 g = 200 103 kg If solar constant is S, then SAt = mst

SA

tmst

= 7

7 4

200 1 (100 0) 60 4.2 10

8.4 10 20 10

= 7

7 5

100 100 60 4.2 10

8.4 10 10

= 6 4.2

8.4

= 3s

1

Chapter 10: Wave Theory of Light


1. = c

v

v = 8c 3 10

43

= 2.25 108 m s1

2. = sin i

sin r

r = sin1 sin i

= sin1 sin30

1.5

= 192817 3. i = 60, angle of relection r = 60 From figure angle of refraction r = 30

= sin i sin 60

sin r sin 30

= 1.7321

4. = c

v=

8

8

3 10

1.4 10

2.14

5. = sin i

sin r

r = sin1 sin i

= sin1 sin 45

4 / 3

= 321

r

i

d

d=

cos r

cosi= 1.19

6. vg = 8

air

g

v 3 10

1.667

= 1.7996 108 1.8 108 m s1

vw = 8

8 1air

w

v 3 102.25 10 ms

1.333

7. 6 17

1 12 10 m

5 10

8. 8

8

c 3 101.5

v 2 10

sin i

sin r

1 sin 60r sin

1.5

= 3516

9. i

r

d

d=

1,

2

i

r

d cosi

d cos r

1 i

r

dr cos cos70

d

= 4650

10. m = 7

m

4 10

1.6

= 2.5 107m

vm = 8c 3 10

1.6

= 1.875 108 m s1

m = 14m

m

v7.5 10 Hz.

11. vg = 8

8

g

c 3 101.875 10

1.6

m s1

Decrease in velocity = c vg

= 3108 1.875 108

= 1.125 108 m s1

12. 8

7

c 3 10

4 10

= 7.5 1014 Hz

1

= 2.5 106 m1

8

8 1liq

c 3 10v 2.4 10 ms

1.25

liq as frequency remains same.

8

liqliq 14

v 2.4 10

7.5 10

= 3.2 107m = 3200 Å

60

N

60

30

ri

30 60

r

Wave Theory of Light 10

2


13. = tan ip ip = tan1(1.33) = 53 4 14. = tan ip ip = tan1 (2.4142) = 67.5 r = 90 ip = 22.5 15. ip = tan1() = tan1(1.536) = 5656 r = 90 ip = 334

16. = C

1 1

sin i sin 45

= 1.4142

ip = tan1() = 54.74 5444 17. = tan ip = tan60 = 1.7321 r = 90 ip = 30

18. = 1 v / c

1 v / c

2 2

'

29

10

440 10

6560 10

= 1 v / c

1 v / c

0.45 = 1 v / c

1 v / c

Solving, v 1.138 108 m s1

19. = 1 v / c

1 v / c

= 60000.25

1.75

= 6000 (0.38) 2268 Å. Doppler shift = 6000 2268 = 3732 Å

20. = c

v

= 8

3

3 10

30 10

0.64 = 6400 Å.

21. v =

c

v = 4

5400

3 108

= 2.22 105 m s1. Negative sign indicates the star is receding

away.


1. In figure OA = 4 cm, AB = 3 cm.

OB = 2 24 3 = 5 cm

= 1 1 OB

sin AB / OB AB

=

5

3

From = 8c 3 10

5 / 3

m/s

= 1.8 108 m/s

2. = sin i

sin r

r = sin1 sin50

4 / 3

= 354

3. i = 90 40 = 50 r = 90 55 = 35

= sin i

sin r= 1.3356

4. = 8

7

c 3 10

3.9 10

= 7.692 1014 Hz

5. 14

8

1 4.5 10

c 3 10

= 1.5 106 m1

6. vg = 8c 3 10

3 / 2

2 108 m/s

T = 3

8g

t 4 10

v 2 10

= 2 1011 s

7. = sin i

sin r

r = sin1 sin 40

1.5

= 2522

i

r

d cosi

d cos r = 0.8478

8. 3

airair 7

air

d 18 10

6 10

= 3 104

9. T = 2

g g

8g a

dt 4 10 1.5

v c 3 10

= 2 1010 s 10. i = 90 48 = 42, r = 42 10 = 32

= sin i sin 42

sin r sin32

= 1.2627

3

Chapter 10: Wave Theory of Light

11. a a

m m

v 4560

v 3648

1.25 = 5

4 12. ag = 1.5, ad = 2.4

gd = a d

a g

= 2.4

1.5= 1.6

13. ao = sin i

sin r, sin r =

0.6428

1.45 = 0.4433

r = 1r , ow = w1

2 o

sin r

sin r

sin r2 = o 1

w

sin r

= 0.4833,

r2 = 28.9

14. Tg Tw = g w8 8

g w

t t 2.4 2.4

c c 2 10 2.25 10

= 1.333 109 s 1.3 ns

15. = sin i sin 2r 2sin r cos r

sin r sin r sin r = 2cos r

r = cos1 2

i = 2r = 2 cos1

2

16. ip = 60 = tan ip = 3 17. ip = 58 = tan ip = 1.6 18. Here, ip = 60,

=c

v= tan ip = tan 60 = 3

v = c

3 =

83 10

3

= 3 108

19. = tan ip = a

w

= 4800 4

3600 3 = 1.333

ip = tan1 (1.333) = 537 This gives the Brewster angle for the light of

given wavelength incident on water surface.

20. iC = sin1 5

8

iC = sin–1 5

8

sin iC =5

8

= C

1 8

sin i 5 = 1.6

Also, = tan ip ip = tan1 1.6 = 5759 Hence closest option is (C) 21. tan ip = = 1.57 ip = tan1 (1.57) = 5730 r = 90 ip = 3230 = 32.5 22. Deviation = ip r = 22 Also, ip + r = 90 Solving, r = 34

23. = v

c =

3

8

100 10

3 10

5700 = 1.9 Å

24. = v

c =

0.4c

c 4 107

= 1.6 107 Hz = = 4 107 1.6 107 = 2.4 107 Hz

25. v =

c

v = 0.05

100 3 108 = 1.5 105 m/s

As wavelength is decreasing, the star is moving close to earth.

26. vplane = 1

c2

= 3

9

1 32 10

2 8 10

3 108

= 600 m s1 = 2160 km/hr 27. Here, the source is moving away from the

stationary observer,

v = 10 8

5

c 3 10 10

4 10

= 7.5 106 cm/s 28. Velocity of light (in vacuum) is an absolute

constant and does not depend upon the motion of the observer.

1

Chapter 11: Interference And Diffraction


1. x = 100.5

N2

= 211

2

N = 211

As N is odd multiple of 2

, given point is dark.

60.3 106 = 100.5

= 660.3 10

100.5

= 6 107 m = 6000 Å 2. Wavelength decreases by 2000 Å new wavelength 2 = 4000 Å n11 = n22

n2 = 12 6000

4000

= 18

3. x = N2

N = 3

10

0.116 10 2

5800 10

= 400

As N is even multiple of 2

, given point is

bright.

4. 3x = (2 3 1) D

2d

= 7

3

5 1.5 6 10

2 0.12 10

= 0.01875 m = 1.875 cm

5. d = 3 2

12

d u 4 10 20 10

v 80 10

= 103 m

= 3 3Xd 0.4 10 10

D 1

= 4 107 m = 4000Å

6. x7 x1 = (7 1)D

d

= 7

3

6 5 10 1

0.5 10

= 6 103 m

7. X = 1

3cm =

210

3

m

d = 7

2

5 10 1.5 3

10

= 2.25 104 m

= 0.225 mm.

8. i. D = 3 3

9

Xd 1.3 10 0.3 10

620 10

= 0.629 m

ii. 1 1

2 2

X D

X D

X2 = 2D

DX1 = 2 1.3 103

= 2.6 103 m = 2.6 mm 9. d = 1 2d d = 4 1 = 2 mm 10. d = 10 + 65 = 75 cm = 0.75 m,

d = D

X

=

7

2

6 10 0.75

0.03 10

= 1.5 103 m

11. Y R(n 1) D (n) D

d d

R

Y

n 1

n

= 6400

5600

8n = 7(n + 1) n = 7

12. d = 1d u 0.7 30

v 70

= 0.3 cm = 0.3 102 m

X = 7

2

D 5.892 10 1

d 0.3 10

= 1.964 104 m = 0.1964 mm.

13. = Xd

D

= 2 20.058 10 0.1 10

1

= 5.8 107 m = 5800 Å

14. X = 7

3

D 6 10 1

d 0.6 10

= 10 104 m = 103 m = 1 mm

Interference And Diffraction 11

2


15. d = 1d u

v=

2 2

2

0.7 10 30 10

70 10

d = 0.3 102 m

X = D

d

x100 = 100 X = 100 D

d

= 2 10

2

10 5892 10 1

0.3 10

= 1964 105 m

16. d = 3 2

12

d u 1.5 10 40 10

v 60 10

= 103 m

= 3 3Xd 0.58 10 10

D 1

= 5.8 107 m = 5800 Å 17. Distance between first minimum and central

maximum

1

Dx

d

d = 6

3

0.5 10 1

5 10

= 104 m

18. = 3 3Xd 5 10 0.2 10

D 2

= 5 107 m = 5000 Å

19. d = 1.22

D

d = 71.22 5 10

1

= 6.1 107 rad

20. d = 75.8 10

2sin 2(sin 45 )

= 4.1 107 m

R.P = 1

d= 2.44 106 m1

21. R.P. = 6

1 1

d 4.8 10

= 2.083 105 m

D = 1.22

d

=9

6

1.22 650 10

4.8 10

= 0.1652 m

Radius R = D

2= 0.0826 m

22. d = 71.22 1.22 5.6 10

D 2

= 3.416 107 rad

23. d = 1.22

D

R.P. = 7

D 5.08

1.22 1.22 6 10

= 6.94 106 m1

24. D = 1.22

d

D = 7

6

1.22 5.8 10

6 10

= 0.1179 m = 11.79 cm


1. Here, x

=

6

10

3.75 10

5000 10

=37.5

5

= 7.5 =15

2

path diff. is odd integral multiple of /2 Hence the point must be dark.

2. 2

1 12

2 2

I a 4I 1

I a 36I 9

1

2

a 1

a 3 a2 = 3a1

2 2

max 1 2 12 2

min 1 2 1

I (a a ) 16a

I (a a ) 4a

=

4

1

Imax = 4I and Imin = I

3. 2

1 12

2 2

I a 25

I a 4

1

2

a 5

a 2

2 2

max 1 22 2

min 1 2

I (a a ) (5 2) 49

I (a a ) (5 2) 9

4. Shift in number of fringes

N = 6

10

( 1)t (1.5 1) 3 10

3000 10

= 5 fringes 5. For a dark point, path difference is odd

multiple of 2

17 = 342

hence point will be bright

similarly for 15 .

3

Chapter 11: Interference And Diffraction

1

17 does not result into natural number

multiple of 2

hence discarded.

17

2 = 17

2

= odd multiple of

2

hence dark.

6. Path diff. = 154

,

phase diff = 15 2

4

=

15

2

7. I = 2 2

1 2 1 2a a 2a a cos

= (2A)2 + (5A)2 + 2(2A) (5A) cos 60 = 39 A2

8. For dark illumination, path diff = 2

= 2 P.D. = 2 2945 = 5890 Å = 0.589 m 9. Let I1 = I2 = I, then resultant intensity for path

difference and phase difference = 0 is,

IR = I + I + 2 I I cos 0 = 4I = K

For path difference /3,

Phase diff. = 2

3

rad.

IR = I + I + 2 I I cos 2

3

= 2I + 1

2I2

= I =

K

4 10. For D and d constant,

2 2

1 1 1

1 3

4

2 = 3

41 = 0.15

11. d = 74 10

0.1180

= 2.29 104 m

12. When a bright fringe is formed opposite to one

of the slits x = d/2

path diff. 2xd d d d

D 2 D 2D

If it is nth order bright fringe, then path diff. = 2d

2D= n or n =

2d

2D

13. For fifth order bright fringe of 1, x5 = 51 For sixth order dark fringe of 2,

x6 = (2 6 1) 2

2

= 211

2

As x5 = x6 10 1 = 11 2 14. (n + 1) b = nr

n 1

n

=

5r

5b

7.5 10

5 10

=

3

2

n = 2

15. X = D

d

=

7

3

6.4 10 1.8

6 10

= 1.92 104 m = 0.192 mm

16. 2x = (2n 1) D

d

= 2x d

1.5D

=

3 310 0.9 10

1.5 1

= 6 105 cm

17. For = 8000 Å = 8 107 m, d = 0.2 mm = 2 104 m, D = 200 cm = 2m, n = 3.

x = n7

4

D 3 8 10 2

d 2 10

= 2.4 102 m = 2.4 cm 18. For and d being unchanged, X D 19. 5400 p = 6000 (p 1) p = 10 20. 20% Fringe width increases by increasing

distance by 15 cm

Original distance D = 100

20 15 = 75 cm

21. 6 2x x 11X 3X 8X

2 2 2 = 4.2 mm

X = 1.05 mm 22. d = 1 2d d 9 4 = 6 mm

23. D = 3 3

6

xd 0.54 10 10

0.54 10

= 1m

D = u + v 1 = 0.1 + v v = 0.9 m = 90 cm

24. 1 1

2 2

X

X

X2 = 0.32 4200

5600

= 0.24 mm

X % = 0.32 0.24

0.32

100 = 25%

4


25. For minima, d = n D

x

=

7

3

1 6 10 1.5

2.5 10

= 3.6 104 m = 0.36 mm 26. For first minimum, d sin = 1

d = sin

= 106000 10

sin 30

= 1.2 106 m 27. For minima, d sin = n

sin = 7

6

5 10

d 10

= 0.5

= sin1 (0.5) = 30 28. Condition for 3rd minima = 31 Condition for 2nd maxima

= (2 2 + 1) 2

2

= 25

2

31 = 25

2

61 = 52

29. x1 + x2 =7

4

2 D 2 6 10 1.5

d 6 10

= 3 mm

30. distance between central and first maximum

be x.

x 1 1

2 2

x

x

x2 = 3.5 5400

4200= 4.5 mm

31. n = d sin For 1st minimum, n = 1.

32. sin = d

= sin1

2.8

4

= 4426

As spread is on either side = 4426 33. For microscope limit of resolution in terms of

distance d

1 1

2 2

d

d

d2 = 0.3 3200

4800

= 0.2 mm

34. R.P. = 7

D 1.44

1.22 1.22 7.2 10

= 1.639 106

35. R.P. = D

1.22 =

7

1

1.22 5.5 10

= 1.49 106 = 0.149 107

36. R.P. 1

1

2

(R.P.) 8000 4

(R.P.) 6000 3

37. Limit of resolution d = x 1.22

d D

x = 7 3

2

1.22 d 1.22 6 10 1.6 10

D 12 10

= 9.76 103 m = 9.76 mm

38. Limit of resolution of eye = = 1.22

D

= 7

3

1.22 5 10

3 10

= 2.03 104 rad

If d is the maximum distance at which dots are just resolved, then

= 31mm 10

d d

= 2.03 104

d = 3

4

10

2.03 10

= 5 m

39.

Resolution limit = d = x

1.22d D

d = xD

1.22

= 3 3

7

4 10 3 10

1.22 5 10

20 m

40. R.P. = 2N.A.

N.A. = 7 67.8 10 10

2

= 0.39

41. = 2 d sin = 2 1 4.8 107 (sin 30) = 4.8 107 m = 4800 Å 42. For a microscope,

limit of resolution (d) = 2 N.A.

= 106000 10

2 0.12

= 2.5 106 m = 25 107 m

D

y

1

Chapter 12: Electrostatics


1. N = 0

q

or q = N 0

q = 18 3.14 105 8.85 1012 = 50 C

2. N = 0

q

k or k =

q

NK

k = 8

12

17.7 10

2500 8.85 10

= 8

3. q = 8.85 C = 8.85 106 C, l = 10 cm = 10 102 m, 0 = 8.85 1012 C2/Nm2

= 0

q

=

6

12

8.85 10

8.85 10

= 106 Nm2/C

4. q = 20 C = 20 106 C, r = 10 cm = 10 102 m

= 6

120

q 20 10

8.85 10

= 2.26 106 Nm2/C

5. = E ds cos =

cos = 86.6

200 0.5 = 0.866

= cos1 (0.886) = 30 6. ds = 100 cm2 = 100 104 m2, E = 100 N/C = E ds cos | | = 100 100 104 cos 90

= 0 ( cos 90 = 0)

= 100 100 104 cos 90

= 1 Wb ( cos 0 = 1)

30 = 100 100 104 cos (90 30)

= 1 1

2= 0.5 Wb

7. A = 100 m2, q = 44.25 C = 44.25 106 C, k = 1 (for air), 0 = 8.85 1012 C2/Nm2

E = 0

q

kA

E = 6

12

44.25 10

1 100 8.85 10

= 5 104 N/C. 8. = 120 C/m2 = 120 106 C/m2, k = 3

E = 0k

= 6

12

120 10

3 8.85 10

= 4.52 106 N/C 9. R = 0.1 m, r = 1 m, = 1.77 C/m = 1.77 106 C/m, k = 2

Using, E = 0

1

kA

2

r

= 9 69 10 2 1.77 10

2 (1)

= 1.593 104 V/m. 10. q = 109 C, l = 20 cm = 20 102 m, r = 20 cm = 20 102 m. k = 1 (for air)

E = 0

2

4 k r

= 0

1 2q 1

4 kr

l

= 9 109 9

2

2 10

20 10

2

1

1 20 10

= 450 V/m. 11. R = 2 mm = 2 103 m, = 0.5 C/m2 = 0.5 106 C/m2, k = 3.14, r = 4m,

E = 0

R

k r

= 0

4 R

4 kr

= 0

1 4 R

4 kr

= 9 3 69 10 4 3.14 2 10 0.5 10

3.14 4

= 9 V/m

Electrostatics 12

2


12. q = 100 C = 100 106 C, r = 1 m

N = 6

120

q 100 10

8.85 10

= 1.13 107 Nm2/C

E = 2

0

1 q.

4 r

= 9 109 6

2

100 10

(1)

= 9 105 V/m 13. = 0.885 C/m = 0.885 106 C/m,

k = 1 (for air), 0 = 8.85 1012 C2/Nm2

f = 2

02

f = 6 2

12

(0.885 10 )

2 8.85 10

= 0.7832

17.7 = 0.04425

= 4.425 102 N/m2

14. ds = 5m2,

q = 8.85 C = 8.85 106 C, k = 10

= 6q 8.85 10

ds 5

= 5.9498 107 C/m2

F = 2

0

ds

2 k

= 27

12

5.949 10 5

2 8.85 10 10

= 14

11

10 10

10

= 102 N. 15. ds = 40 cm2 = 40 104 m2, q = 0.2 C = 0.2 106 C

Using, E = 0

= 0

q

ds

= 6

4 12

0.2 10

40 10 8.85 10

= 5.6 106 V/m

Now using, f = 2

02

= 2

0

q / ds

2

= 26 4

12

0.2 10 / 40 10

2 8.85 10

= 23 2

12

5 10 10

2 8.85 10

= 1.4124 102 = 141.24 N/m2

16. ds = 0.5 m2, q = 10 C = 10 106 C

= 6q 10 10

ds 0.5

= 2 105 C/m2

F1 =2

0

ds

2

= 25

12

2 10 0.5

2 8.85 10

= 2.26 N

F2 =2

0

ds

2 k

= 2.26

5= 0.452 N

17. V = 0

1 q

4 k R

=

99 10 q

5 R

We get, q = 9

V R 5

9 10

= 6

9

2 10 1 5

9 10

= 1.11 103 = 1110 C

Now using, = 2

q

4 R=

6

2

1110 10

4 3.14 (1)

= 8.838 105 C/m2

Using, f = 2

02 k

= 25

12

8.838 10

2 8.85 10 5

= 10

12

78.11 10

88.5 10

= 88.26 N/m2

18. u = 1

2 E2

= 1

2 8.85 1012 (104)2

= 4.425 104 J/m3 19. u = 2 108 J/m3, k = 1 (for air)

Using, u = 1

20 E

2 or E = 0

2u

=8

12

2 2 10

8.85 10

= 67.23 V/m

3


20. q = 20 C = 20 106 C, r = 20 cm = 20 102 m

E = 2

0

1 q

4 r

E = 9 109

6

22

20 10

20 10

= 45 105 V/m Now using,

u = 1

2 0E

2

= 1

2 8.85 1012 (45 105)2

= 89.60 J/m3

21. d = 2 mm = 2 103 m, V = 200 V

E = V

d=

3

200

2 10= 105 V/m

u = 1

20 E

2 = 1

2 8.85 1012 (105)2

= 4.425 102 J/m3 = 0.04425 J/m3

22. u = 1

2 0KE2 or E =

0

2u

k

=12

2 8.85

8.85 10 2

= 106 V/m

23. A = 1.21 m2, V = 400 V, k = 6 d = 0.05 cm = 0.05 102 m = 5 104 m

E = V

d

E = 4

400

5 10

= 8 105 V/m

u = 1

2 0kE2

u = 1

2 8.85 1012 6 (8 105)2

= 8.85 3 64 1012 1010 = 1699 102 = 16.99 17 J/m3

24. V = 200 V, A = 20 cm2 = 20 104 m2, d = 0.1 mm = 0.1 103 m = 104 m,

E = V

d

E = 4

200

10 = 2 106 V/m

u = 1

20E

2

u = 1

2 8.85 1012 (2 106)2

= 1

2 8.85 4

= 17.70 J/m3. 25. d = 1 mm = 103 m, C = 1 F = 106 F

C = 0A

d

or A =

0

Cd

= 6 3

12

10 10

8.85 10

= 1000

8.85

113 m2

26. dc = 8 cm = 8 102 m, rc = 4 102 m, ds = 20 cm = 20 102 m rs = 10 102 m C = Cs

0A

d

= 40rs

2cr

d

= 4

d = 2cr 1

r 4

d = 2 2

2

(4 10 )

4 10 10

= 4 11610 10

4

= 4 103 m = 4 mm 27. D = 6 cm = 6 102 m

R1 = 3 102 m,

d = 0.05 cm = 0.05 102 m

C = 0A

d

=

21 0R

d

C = 22 12

2

3.14 3 10 8.85 10

0.05 10

= 5 103 104 1012 102

= 50 1012 F = 50 pF

4


Now using,

C = 40R2 = 50 1012

R2 = 12

12

50 10

4 3.14 8.85 10

R2 = 0.45 m. 28. C1 = 12 F = 12 106 F, d2 = 2 d,

A2 = 1

2A1

C = 0A

d

2 0 12

1 1 0 2

A dC

C A d

= 2 1

1 2

A d

A d

2

1

C 1 1

C 2 2

C2 = 1

1C

4 =

112

4

C2 = 3 F. 29. C = 14 F = 14 106 F, V = 250 V, work done = energy stored

U = 1

2CV2 =

1

2 14 106 (250)2

= 7 106 625 102

= 4375 104 = 0.4375 J 30. C = 8 F = 8 106 F, V = 2000 V,

U = 1

2 CV2

U = 1

2 8 106 (2000)2

= 4 106 4 106 = 16 J 31. C = 2 F = 2 106 F, E = 4 J

E = 21 Q

2 C

Q2 = 2 EC = 2 4 2 106 = 16 106

Q = 4 103 C 32. C = 20 F = 20 106 F, Q = 10 mC = 10 103 C = 102 C

U = 21 Q

2 C

U = 22

6

101

2 20 10

=4

5

1 10

2 2 10

= 0.25 10 = 2.5 J

33. C0 = 10 F = 10 106 F, V0 = 1000 V, k = 4

E0 = 1

2 C0

20V

E0 = 1

2 10 106 (103)2

= 5 J. Now, C = k C0 = 4 10 106 F

1

2CV2 = 5

1

2 4 10 106 V2 = 5

V2 = 66

5 2 110

4 10 10 4

V = 1

2 103 = 500 V.

34. C = 0.2 f = 0.2 106 F, V1 = 1.1 kV, k = 7.5

U = 21CV

2

U = 1

20.2 106 (1.1 103)2

= 0.1 106 1.21 106 = 0.121 J Now using,

U = 1

2kCV2

V2 = 2U

kC=

6

2 0.121

7.5 0.2 10

= 61.2110

7.5 = 0.1613 106 V

V = 60.1613 10

= 0.4016 103 = 401.6 V 35. CAB = 2 F + 4 F = 6 F CBC = 3 F + 1 F = 4 F

AD

1 1 1 1

C 6 4 12 =

6 1

12 2

CAD = 2 F

5


36. 3 F, 4 F and 5 F are in parallel, C = 3 + 4 + 5 = 12 F Now, 12F 4 F

1 1 1 4

C 12 4 12

C = 3 F ….(i) Also, 4 F 2 F C = 4 + 2 = 6 F Now, C and 3 F are in series

1

C =

1 1 3 1

6 3 6 2

C = 2 F ….(ii) Finally, C and C are in parallel, From equation (i) and (ii), CAB = 3 + 2 = 5 F 37. 6 F 10 F Ceq = 6 + 10 = 16 F, V = 100 V Now, 16 F and 4 F are in series.

1

1 1 1 5

C 16 4 16

C1 = 61610

5 F = 3.2 F

The charge acquired by the entire combination

= C1 V = 16

1005

= 320 C As Ceq and 4 F are in series, they have the

same charge = 32 105 C = 320 C

P.D. across 4 F = 5

6

32 10

4 10

= 80 V

As 6 F and 10 F are parallel,

V6 = V10 = 5

6

32 10

16 10

= 20 V.

Charge across 6F = 20 6F = 120C and charge across 10 F = 20 10 F = 200 C 38. C1 = 2 F = 2 106 F, V = 150 V, C2 = 3 F = 3 106 F,

CEq. = 1 2

1 2

C C

C C

CEq. = 6 6

6

2 10 3 10 6

(2 3) 10 5

106

= 1.2 106 C Now, charge across each capacitor remains the

same in series.

Q1 = Q2 = CEq. V = 1.2 106 150 = 180 106

= 1.8 104 C

V1 = 6

16

1

Q 180 10

C 2 10

= 90 V

V2 = 6

26

2

Q 180 10

C 3 10

= 60 V

39. Let C1 and C2 be the individual capacitances.

CS = 1 2

1 2

C C

C C and CP = C1 + C2

4 = 1 2

1 2

C C

C C .…(1)

and C1 + C2 = 18 .…(2)

4 = 1 2C C

18

or C1C2 = 72

C2 = 1

72

C

Substituting in equation (2) we get,

C1 + 1

72

C = 18

21C 18 C1 + 72 = 0

(C1 12) (C1 6) = 0 C1 = 12 or C1 = 6 C2 = 18 12 = 6 or C2 = 18 6 = 12 The capacitances are 6 F and 12 F. 40. C1 = 4 F, C2 = 5 F, C1 C2 Ceq = 4 + 5 = 5 = 9 F Q = CV Q1 = C1V = 4 106 100 = 4 104 C Q2 = C2V = 5 106 100 = 5 104 C

U = 1

2CV2

U1 = 1

2 C1V

2

= 1

2 4 106 (100)2

= 2 106 104

= 2 102 = 0.02 J

6


U2 = 1

2 C1V

2

= 1

2 5 106 (100)2

= 2.5 106 104

= 2.5 102

= 0.025 J 41. C1 | | C2

Ceq = C1 + C2 = 20 F + 10 F

= 30 F

Using, Q = CV we get,

Q1 = C1V1 = 20 106 500

= 10 103 C

Q2 = C2V2 = 10 106 200

= 2 103 C

Net charge, Q = Q1 + Q2 = 12 103 C

Now using, V = eq

Q

C we get,

Common potential, V = 3

6

12 10

30 10

= 4 3

5

10

10

= 400 V 42. C1 = 5 F, C2 = 10 F, C3 = 20 F, V = 210 V

eq

1 1 1 1 7

C 5 10 20 20

Ceq = 20

7 F = 2.857 F

Net charge across the combination

= 20

2107 = 600 C

As C1, C2 and C3 are in series, they have the

same charge = 600 C

P.D. across 5 F = 600 C

5 F

= 6

6

600 10

5 10

= 120 V

P.D. across 10 F = 6

6

600 10

10 10

= 60 V

P.D. across 20 F =6

6

600 10

20 10

= 30 V

43. C1 = 20 F, C2 = 30 F V1 = 1000 V, V2 = 0 C1 C2 Ceq = C1 + C2 = 20 + 30 = 50 F Total charge = Q1 + Q2 = C1V1 + 0 = 20 106 1000 = 2 102 C Common potential,

V = 2

6

2 10

50 10

= 0.04 104 = 400 V

44. i. s

1

C=

1 2 3

1 1 1

C C C

= 1 1 1 7

2 4 8 8

Cs = 8

7 F = 1.14 F

q = Cs V = 1.14 10 = 11.4 C ii. CP = C1 + C2 + C3 = 2 + 4 + 8 = 14 F q1 = C1V = 2 10 = 20 C q2 = C2V = 4 10 = 40 C q3 = C3V = 8 10 = 80 C 45. Side of cube = 5 cm = 0.05 m, k1 = 8,

E = 200 V/m, k2 = 6

u = 1

2k 0 E

2

u1 = 1

2 8 0 200 200

u2 = 1

2 6 0 200 200

= 12 104 J 0 = 12 104 8.85 1012 = 1.063 106 J Total energy = u volume E2 = u2 (0.05)3 = 1.063 106 (0.05)3 = 1.3288 1010 J 1.33 1010 J Since, u k,

E k or 1

2

E

E= 1

2

k

k=

8

6=

4

3

E1 = 4

3E2 =

4

3 1.33 1010

E1 = 1.77 1010 J 1.77 1010 J

7


46. A = 1 m2, V = 300 V, d = 0.01 cm = 104 m, k = 7, C = 0Ak

C = 12

4

8.85 10 1 7

10

Now using,

U = 1

2CV2

U = 1

2

122

4

8.85 10 7(300)

10

= 8.85 3.5 108 9 104 U = 278.78 104 J = 2.79 102 J 47. Initial capacitance,

s

1

C=

1 2

1 1

C C =

1 1

4 6 =

5

12F

Cs = 12

5= 2.4 F

Final capacitance, C = Cs + 0.64 = 2.4 + 0.64 = 3.04 F Now, (4 + x) is in series with 6.

1

3.04=

s

1

C=

1 1

(4 x) 6

= 6 4 x

6(4 x)

3.04 = 24 6x

10 x

30.4 24 = 6x 3.04x 6.4 = 2.94x

x = 6.4

2.96 = 2.162 F 2 F

48. E = 2

20

R

kr

From formula,

R2 = 2

0E kr

= 4 12 2

7

4 10 8.85 10 (2)

64 10

R2 = 6.9472 101 R = 0.83 m


1. n = Q

e =

6

19

3.2 10

1.6 10

= 2 1013

2. F = 0

1

4 1 2

2

q q

r =

9 12

2

9 10 10 10

10

= 9 N

3. E = 3

F 4.5

q 2 10

= 2.25 103 N/C

= 2250 N/C

4. E = 04

1

2r

q =

9 6

2 2

9 10 4 10

(10 10 )

= 3.6 106 N/C.

5. Volume of drop V = 4

3r3

Mass of oil drop = volume density m = V For stationary drop, mg = q E

E = 3mg V g 4 r g

q q 3 4e

= 7 3 3

19

4 (2 10 ) 2 10 9.8

3 4 1.6 10

= 8 3.14 9.8 8

4.8 4

1021 103 1019

= 102.6 10 V/m 1.03 103 V/m

6. F 1

k

F2 = F1 1

2

k

k = 20

1

5 = 4 N

7. N = 6

120

q 4.425 10

8.85 10

= 5 105 V m 8. The cube has six surfaces and as the charge is

at its centre, it will produce equal number of lines of forces through each surface.

The charge Q will produce in all 0

Q

lines

of force.

Each surface will allow 0

Q

6

= 8

12

2.124 10

6 8.85 10

= 400

C1 C2

X

Q

4 F

P

6 F

8


9. ds = 2 mm2 = 2 106 m2, = 60, d = Eds cos 60

= 12 106 2 106 1

2

= 12 Nm2/C 10. T.N.E.I. = q = 10 + 6 2 + 4 + 2 = 0

11. E = 0

R

k r

= 0

R 4

4 kr

E = 6 3 91.5 10 2 10 4 3.14 9 10

3.39 1

= 339.12

3.39

100 V/m

12. E = 2

0

1 q

4 r

4 104 = 9 109 q

9

q = 4

9

4 10

10

= 4 105 C = 40 C

13. A = 0.2 m2, Q = 20 C = 20 106 C

= 5

1

Q 2 10

A 2 10

= 104 C/m2

14. Q = 800 C = 800 106 C,

R =3

2m = 1.5 m

K = 4.8 109 N/m2

Bulk Modulus (K) = Volume stress

Volume strain

K= 2F / ds

volume strain 2k (Volumestrain)

Volume strain = 2

0

1

2k K

= 2

20

Q 1 1

4 R 2k K

….2

Q

4 R

= 2

2 4 90

Q

16 R 2 4.8 10

= 26

2 4 12 9

800 10

16 (3.14) (1.5) 8.85 2 10 4.8 10

= 12 4

4 3

64 10 10

6.785 10 10

= 9.43 109

15. F = 2 2

0 0

ds Q

2 2 ds

= 6 2

12

(17.7 10 )

2 8.85 10 2

= 12

12

313.29 10

35.4 10 1

= 8.85 N

16. = 26.5 C/m2 = 26.5 106 C/m2

dF

ds=

2

02k

dF

ds=

12

12

26.5 26.5 10

2 1 8.85 10

.... [ k = 1 for air]

= 39.67 N/m2 39.7 N/m2

17. l = 250 cm = 250 102 m, q = 1C, k = 1

F = 2 2 2

20 0 0

q q

2 k 2 kds 2 k2

l ( ds = l2 + l2)

= 2

12 2 2

(1)

4 8.85 10 (250 10 )

= 6 4 12

1

2.21 10 10 10

= 4.52 109 N

18. f = 1

20 kE2

f = 1

2 8.85 1012 5 (400)2

= 22.125 1012 16 104

= 3.54 106 N/m2

19. f =2

02 k

=2

20

q

2 kds=

2

2 20

q

2 k(4 R )

= 2

2 40

q

32 k R ....( ds = 4R2)

f = 6 2

12 2 4

(8 10 )

32 8.85 10 1 9.86 (4 10 )

= 12

5 20

64 10

7.148 10 10

= 8.953 103 N/m2 20. E = 2 106 V/m , k = 1

u = 1

20kE2

= 1

2 8.85 1012 1 (2 106)2

= 17.7 J/m3

9


21. u = 2

1 0 E

2 = 2

1 0 2

2

V

R

u = 2

1

212 3

22

8.85 10 10

2 10

= 1.1 102 J/m3

22. E = 400 V/m Volume = 600 ml = 600 103 m3,

u = 21E

2

= 1

2 8.85 1012 (400)2

= 70.8 108 Energy stored in 600 103 m3 of air = 70.8 108 600 103 = 42.48 108 J

23. u = 1

20 kE2

E2 = 5

120

2u 2 3.54 10

k 8.85 10 2

= 0.4 107 = 4 106

E = 64 10 = 2 103 N/C

24. W = 1

20kE2 dV

W = 1

2 8.85 1012 4 (104)2 (2 104)

= 17.7 2 108 J = 35.4 108 J = 3.54 107 J

25. x = 80 4

x x100 5

Using, u E2 oru

'u=

2x

x

= 2

4

5

= 16

25 or u =

16u

25.

26. Q = 20 mC = 20 103 C, C = 10 F = 10 106 F

V = Q/C = 3

6

20 10

10 10

= 2000 V

27. V = 12 V, d = 0.1 cm = 0.1 102 m,

Q = 6 F = 6 106 F

E = V/d = 3

12

10 = 12 103 V/m

F = QE = 6 106 12 103 = 72 103 = 0.072 N 28. Q1 = 50 C, Q2 = 150 C, k1 = 1 V1 = V2

2 2 2

1 1 1

C Q k

C Q k

2

1

k

k =

150 C

50 C

= 3

2k

1 = 3 or k2 = 3

29. Total charge Q = Q1 + Q2 = C1V1 + C2V2

= 20 10–6 500 + 10 10–6 200 = (10000 + 2000) 10–6

= 12000 10–6 Effective capacitance Cp = C1 + C2 = 20 F + 10 F = 30 F

Common potential difference

= Q

C =

6

6

1030

1012000

= 400 V

30. C1 = 6 F, d2 = 1d

2

Using, C 1

d we get,

2 1 1

1 2 1

C d d

C d d / 2 = 2

C2 = 2C1 = 2 6 = 12 F 31. r = 10 cm = 10 102 m, d = 0.6 cm = 0.6 102 m = 6 103 m, k = 4, t = 2 103 m

C = 0A

td t

k

and A = r2

C = 12 2 2

33 3

8.85 10 3.14 (10 10 )

2 106 10 2 10

4

= 14

3

27.789 10

4.5 10

= 6.17 1011 = 61.7 1012 F 62 pF 32. Q = 30 C, V1 = +6 V, V2 = 6 V

C = V

Q =

30

6 ( 6) =

30

12 = 2.5 F

10


33. R = 6400 km = 6400 103 m C = 4 0R = 4 3.14 8.85 1012 6400 103 = 7113.98 107

711 F 34. t = 0.2 cm, Capacity of capacitor = C

Q = CV = 0AV

d

.…(1)

After inserting a slab, capacitance becomes C1 and charge remains same. Q = C1V1

By increasing a distance, we get same potential difference as in first case.

Q = C2V .…(2)

2

1

C =

0

d 0.2 0.24

A

+ 0

0.2

k A

= 0

d 0.04

A

+ 0

0.2

k A

From (1) and (2), C = C2 or C

1 =

2C

1

0

d

A =

0

d 0.04

A

+ 0

0.2

k A

d = d – 0.04 + 0.2

k

0.2

k= 0.04

k = 0.2

0.04 = 5.

35. d = 4 103 m, V = 400 V

u = 2

0 2

1 V

2 d

u = 1

2 8.85 1012

4

6

16 10

16 10

= 4.425 102 J/m3 4.4 102 J/m3

36. C = 40 F = 40 106 F, Q = 4000 106 C = 4 103 C

W = 2 6

6

Q 16 10

2C 2 40 10

=16

80 = 0.2 J

37. E1 = 2

1CV 2

1 , E2 = 1

2 2

2CV

2

1E

E =

2122

V

V

E2 = 222

1

V

V E1 =

2

2

24

6 1E = 16 E1

38. E = 80 J, C = 6.4 F = 6.4 106 F

E = 2

1 CV2

V = 2E

C =

6

2 80

6.4 10

= 625 10

= 5 103 V = 5000 V 39. C = 4 F = 4 106 F, Q = 20 C = 20 106 C

C = Q/V or V = Q

C

Energy = 2

1 2Q

C=

2

1 6 2

6

(20 10 )

4 10

E = 50 106 J = 5 10–5 J 40. C1 = 3 F = 3 106 F, V1 = 20 V, C2 = 5F = 5 106 F, V2 = 80 V

U = 1

2CV2

U = 1

2 C1

21V +

1

2 C2

22V

= 1

2[3 106 20 20 + 5 106 80 80]

= 1

2 [12 104 + 320 104] = 4332

102

= 166 104 J 1.7 102 J 41. C = 8 F = 8 106 F, V1 = 15 V, V2 = 25 V

Using, U = 1

2CV2 we get,

Increase in energy

= 2

1 2

2CV – 2

1 CV1

2 = 2

1C( 2

2V – 21V )

= 2

1 8 10–6 (625 – 225)

= 4 10–6 400 = 16 10–4 J. 42. C = 6 F = 6 106 F, (V) = 150 V Heat produced = Electrical energy stored

U = 2

1CV2

= 2

1 (6 106) (150)2

= 6.75 104 106 0.07 J

11


43.

CXY = C C

2 2 = C

44. Join M and N together. Similarly join P and Q. Then the given circuit gets modified as shown

in the above diagram.

Ceq = 4 F + 4 F + 4 F = 12 F

45. PQ QR

PS RS

C C

C C

This is balanced Wheatstone’s bridge circuit. Therefore, there is no charge on capacitor

which is connected in the middle branch (PQ). The equivalent circuit becomes

s1

1

C =

1

3 +

1

3 =

2

3

1s

C = 3

2F

Now,s2

1

C=

1

3 +

1

3 =

2

3

2SC =

3

2F

CP = 1s

C + 2sC

= 3

2F +

3

2F

= 3 F.

46. From the circuit (I), equivalent capacitance of parallel capacitors

1PC = 4 + 4 = 8 F

Equivalent capacitance of series connection of

4 F and 1PC is given by,

1s

4 8C

4 8

=

32

12F =

8

3F

Similarly, the equivalent capacitance for circuit (II) is,

2s

8 4C

8 4

=

8

3F

Now 1s

C ,2sC and C are in parallel.

From given condition, C + 8 8

3 3 = 6

C + 16

3 = 6

C = 6 16

3=

2

3F

47. On connecting X to P, 6 F capacitor is

charged to a constant potential (E). As connection of X is switched over to Q, the

total charge on 6 F capacitor is shared between 6F and 3F capacitors, which is

6 6

6 3 9

=

2

3 of original charge.

48. Capacitance of first capacitor (C1) = 25 F = 25 106 F and its voltage (V1) = 400 V; Capacitance of the second capacitor (C2) = 15 F = 15 106 F and its voltage (V2) = 200 V. We know that for parallel combination, the common potential (V),

= 1 1 2 2

1 2

C V C V

C C

=6 6

6 6

(25 10 400) (15 10 200)

(25 10 ) (15 10 )

X Y X Y C

C C

C C C

CC

C

4 F

C A B4 F

4 F

P R

3 F 3 F

3 F 3 F

A

B

C

I

4F

4F

4F

4F

4F 4F

II

12


= 6

6

13000 10

40 10

= 325 V. 49. Charge on capacitor = Q = CV = 5 106 10 = 50 C Total capacity = C = C1 + C2

Final P.D., V= C

Q =

1 2

50 C

C C

= 2

50 C

2

= C1 + C2 or C2 = 25 C1

= 25 5 = 20 F 50. Q = C1V = 40 6 = 240 C

Cs = 6 12

18

= 4 F

Now, V = s

Q

C =

240

4 = 60 V

V2 = 2

Q

C =

240

12 = 20 V

1

Chapter 13: Current Electricity

Hints to Problems for Practice 1. Let P, Q and R be the junction points as

shown in the figure. Let currents through PQ and QR be I2 and I3 respectively.

Applying Kirchhoff’s 1st law to i. junction P we get, 6 2 I2 = 0 I2 = 4 A ii. to junction Q we get, I2 + 6 2 I3 = 0 I3 = 4 + 6 2 = 8 A iii. to junction R we get, I3 + 6 I 9 = 0 I = 8 + 6 9 = 5A 2. V2 = V1 + 10,

R = V

I

1 2

1 2

V V

I I or 1 1V V 10

0.5 1.5

1.5 V1 = 0.5 V1 + 5 or V1 = 5 V

R = 1

1

V

I =

5

0.5= 10

3. Applying Kirchhoff’s 2nd law to loop

ABCDA, 1 I1 2 I2 = 10 20 I1 2 I2 = 10 ....(1) Applying Kirchhoff’s 2nd law to loop AEFDA, 1 I1 + 30 I = 10 I1 + 30 I = 10 ....(2)

Applying Kirchhoff’s 2nd law to loop BEFCB, 2 I2 + 30 I = 20 I2 + 15 I = 10 ....(3) Applying Kirchhoff’s 1st law to junction P, I = I1 + I2 From (2) we get, I1 + 30 I1 + 30 I2 = 10 31 I1 + 30 I2 = 10 ....(4) Solving (1) and (4) we get, I1 = 3.04 A Substituting eq.(1) we get, I2 = 3.48 A 4. Let A be the junction as shown in the figure. Applying Kirchoff’s first law to junction A we

get, I1 + I2 = I3 + I4 I4 = (I1 + I2 I3) = (20 + 10 16) = 14 A 5. The circuit can be redrawn as: Applying Kirchoff’s voltage law to loops

AEFDA and EFCBE we get, I1 2 + (I1 + I2) 8 + 3 I1 + 2 I1 3 = 0

….(1) and I2 2 + (I1 + I2) 8 + 3 I2 + (1 I2) 4 = 0

….(2)

6 A6 A

9 A2 A2 A

6 A

I

P Q I3

R

I2

I4

I1 = 20 A I3 = 16 A

I2 = 10 A

A

8 4V

2

D I2 3 I1F3

A B

C

I11

3V

2 I2

R3

2 E

8

6

4

8

E1 A r1

r2E2

D

B C

I1

10 V

20 V

1

2

P I2

I

E F R = 30

Q

Current Electricity 13

2


From eq.(1) we get, 2 I1 + 8 I1 + 8 I2 + 3 I1 + 2 I1 3 = 0

15 I1 + 8 I2 = 3 or I2 = 13 15I

8

….(3)

From eq.(2) we get, 2 I2 + 8 I1 + 8 I2 + 3 I2 + I2 4 = 0 14 I2 + 8 I1 = 4 or 8 I1+ 14 I2 = 4 …(4) Substituting value of I2 in eq.(4) we get,

8 I1 + 14 13 15I

8

= 4

8 I1 + 1

7(3 15I )

4 = 4

32 I1 + 21 105 I1 = 16

73 I1 = 5 or I1 = 0.068 = 5

A73

From eq.(3) we get,

I2 =

53 15

738

=

3 1.0274

8

I2 = 0.2466 A. P.d. across R3 = (I1 + I2) 8 = (0.068 + 0.2466) 8 = 0.3146 8 = 2.517 2.52 V

6. I = V

Ror R =

V

I=

4.8

3= 1.6

I = E

R r

3 = 5

1.6 r

4.8 + 3r = 5 or 3r = 0.2 r = 0.0667 7. 10 , 10 , 10 and 20 form a

Wheastone’s network. Let S be the shunt needed across 20 resistor.

For balance condition,

10 10

20S1020 S

20S

20 S= 10

20 S = 200 + 10 S 10 S = 200 S = 20

8. For balance condition,

P S

Q R

20 30

50 R

R = 1500

20= 75

Now, X = 50 is connected across Q,


P S

R'QXQ X

20 30

50 50 R'50 50

20 30

25 R' or R =

750

20

R = 37.5 9. L1 = 30 cm, L2 = 40 cm, r1 = 0.5 m,

r2 = 0.6 mm, l1 = ?

Let R1 and R2 be introduced in the left and right gaps respectively.

Now, R L

Aor R

2

L

r

22

1 1 1 1 22

2 2 2 2 1

R L / r L r

R L / r L r

= 2

30 0.6

40 0.5

= 27

25

R1 = 27 x and R2 = 25x

….[x is the common multiple]


1

2

27x

25x

l

l

2

1

25

27

l

l

2 1

1

52

27

l l

l

1

100 52

27

lor l1 = 51.9 cm

3


10. Let S be the shunt across R4.

Reff = 4

4

R S

R S

= 30S

30 S


31

2 4

RR

R R

3 9

30S630 S

or 1 9(30 S)

2 30 S

1 = 3(30 S)

5S

5 S = 90 + 3 S S = 45 11. P and Q are connected in series across one gap

and a known resistance of 27 is connected in the other gap to obtain the null point at 40 cm from the end corresponding to series combination of P and Q

As per the balance condition of metre bridge,

P Q

27

=

40

100 40=

40

60=

2

3

3P + 3Q = 54 P + Q = 18 ….(1) Now, P and Q are in parallel and known

resistance becomes 27 – 21 = 6 .

PQP Q

6

=

2

3

PQ

(P Q)=

12

3 = 4

PQ = 4(P + Q) PQ = 4(18) = 72 ….From (1) P(18 – P) = 72 ….From (1) 18P P2 = 72 P2 18P + 72 = 0 (P 6) (P 12) = 0 P = 6 or P = 12 Substituting this in eq. (1), Q = 18 6 = 12 if P = 6 OR Q = 18 12 = 6 if P = 12 The values of P and Q are 6 and 12 . 12. Let X be the unknown resistance R1 = X, R2 = 50 , l1 = 40 cm, l2 = 10 40 = 60 cm


X 40 2

50 60 3

X = 100

3 = 33.33

13. The two halves of the ring formed by its two

diametrically opposite points form a parallel combination of the resistances of each half.

Let R be the resistance of each half.

R1 = R||R = 2R R

2R 2 , R2 = 15

l1 = 40 cm, l2 = 100 40 = 60 cm For balance condition,

R40 22

15 60 3

R

2=

2

3 15

R

2= 10 or R = 20

Resistance of ring = 2 20 = 40 14. Let R1 and R2 be the two unkown resistances.

When R1 and R2 are connected in series, for the balance condition,

1 2R R 50

9 50

or R1 + R2 = 9 ....(1)

When R1 and R2 are connected in parallel, for the balance condition,

1 2

1 2

R R

R R 50

2 50

or 1 2

1 2

R R

R R= 2 ....(2)

from (1) and (2),

1 2R R

9 = 2 or R1R2 = 18 ....(3)

from (1), R1 = 9 R2. Substituting in (3), (9 R2) R2 = 18 9R2 2

2R = 18 or 22R 9R2 + 18 = 0

On solving, R1 = 3 , R2 = 6 15. For the balance condition,

X 2

Y 3 or X =

2

3Y ....(1)

In the second case,

X 20 1

Y 4

....(2)

4X 80 = Y

4


42

Y3

80 = Y ....From (1)

8Y 240 = 3Y or 5Y = 240 Which on solving gives, Y = 48

From (1), X = 2

3 48 = 32

16. X : Y = 2 : 3 or X 2

Y 3 ....(1)

Also, for balance condition,

X 30 5

Y 30 6

6 X + 180 = 5Y + 150 6 X 5Y = 30 ....(2) From (1),

X = 2

3Y ....(3)

Subtituting in (2), we get,

6 2

3Y 5Y = 30

Y = 30 Y = 30 From (3),

X = 2

3 30 = 20

17. Two equal resistances are introduced in the

two gaps of a metre bridge. Let R be the value of each resistance and let 1 be the balancing

length from left end of the wire. As per the balance condition for a metre

bridge we get,

R

R= 1

1100

1 = 100 1 or 2 1 = 100

1 = 50 cm and 2 = 50 cm

Now, R in the left gap is shunted with value R.

R R

R RR

= 1

1100

2R

2R R= 1

1100

1

2 = 1

1100

2 1 = 100 1 3 1 = 100

1 = 100

3= 33.33 cm and

2 = 100 33.33 = 66.67 cm

Shift in the null-point

= 1 1 = 50 33.33

Shift = 16.67 cm to the left

If R in the left gap is connected with another resistance R in series then we have,

R R

R

= 1

1100

2 = 1

1100

or 200 2 1 = 1

3 1 = 200

1 = 200

3

1 = 66.67 cm

Shift in null point = 66.67 50

Shift in null point = 16.67 cm to the right. 18. Two resistances P and Q are introduced in the

two gaps of a metre bridge and ratio of two parts of the wire is 1 : 3.

As per the balancing condition for a metre bridge we get,

P 1

Q 3 or Q = 3 P .…(1)

Now, P and Q are increased by 25 each and the new ratio is 3 : 7.

P 25

Q 25

= 3

7

7P + 175 = 3Q + 75

7P + 175 = 3(3P) + 75 ….[From (1)]

175 75 = 9P 7P

P = 50

Q = 3 50 = 150

19. E = 1.02 V, 1 = 150 cm = 1.5 m,

S = 4 , 2 = 1.2 m

R

1

2

r 1.5 5

X 1.2 4

….(1)

5


Now, X = r s 4r

r s 4 r

r r 4

X 4

….(2)


r 4 5

4 4

or r = 5 1

r = 1 20. L = 4 m, R = 8 , E = 2 V, r = 0, l1 = 217 cm = 2.17 m, l2 = 200 cm = 2, R = 15

r = R 1 2

2

l l

l

r = 152.17 2

2

= 1.275

1.3

21. R

L= 1 m, E1 = 1.4 V,

l1 = 280 cm = 2.8 cm

K = V IR

L L

K = I 1 = I E1 = Kl1

1

1

E

l= K = I

I = 1.4

2.8 = 0.5 A

22. E = Kl

1 1

2 2

E

E

l

lor

2

1.5 1.8

1

l

l2 = 1.2 m 23. K = 103 V/cm = 103 102 V/m = 101 V/m

K = V

Lor V = KL = 10–1 4 = 0.4 V

I = V 0.4

R 4 = 0.1 A

Now, I = S

E

R r R

0.1 = s

2

4 2 R or 0.6 + 0.1 Rs = 2

Rs = 14

24. K = 5 103 V/cm = 5 103 102 = 0.5 V/m, l = 216 cm = 2.16 m.

E = Kl E = 0.5 2.16 = 1.08 V 25. L1 = 10 m, 1 = 250 cm = 2.5 m

L2 = 10 + 1 = 11 m R L

1 1

2 2

R L 10

R L 11 ….(1)

But R1 1 and R2 2

1 1

2 2 2

R 2.5

R

….(2)


2

10 2.5

11

or 2 = 11 2.5

10

= 2.75 m

2 = 275 cm 26. L = 2 m, R = 5 , RS = 998 , E1 = 4 mV = 4 103 V, E = 2V, r = 2 ,

I = S

E

R r R =

2

5 2 998 = 1.99 103 A

V = IR = 1.99 103 5 = 9.95 103 V

K =3V 9.95 10

L 2

= 4.975 103 V/m

Now, E1 = K1

1 = 1E

K=

3

3

4 10

4.975 10

= 0.804 m

27. I = E

R r

I = 2

25 0= 0.08 A

Now using, V = IR = 0.08 25 = 2 V

K = V 2

L 4 = 0.5 V/m

28. I = E 2.1 2.1

R r 9 1.5 10.5

= 0.2 A

Now using,

K = V IR 0.2 9

L L 10

= 0.18 V/m

Using, E1 = Kl1 or l1 = 1E

Kwe get,

l1 = 1.08

0.18= 6 m

6


29. R

L= 0.1 /cm = 100 /cm,

E1 = 1.5 V, l1 = 300 cm = 3 m, E2 = 1.4 V

1 1

2 2

E

E

l

l

l2 = l1 2

1

E

E= 3

1.4

1.5= 2.8 m = 280 cm


1. Using E = I (R + r) we get, E = 2(8 + r) .…(1)

E = 5 (2 + r) .…(2) [ 4 || 4 = 2 ]

Comparing equation (1) and (2), 8 + r = 2.5 (2 + r) 8 + r = 5 + 2.5 r

1.5 r = 3 or r = 3

1.5= 2

2. I = E

R r or 1 =

4

R 1

R + 1 = 4 or R = 3 Let E = E + E = 2 E = 2 4 = 8 V R = R + r + r = 5 For two batteries of 4 V connected in series,

I = 8

5 = 1.6 A

3. Equivalent resistance in the series circuits are, R1 = 3X, R2 = 3X, R3 = 3X Effective resistance of the parallel circuit is,

1

R=

1

3X+

1

3X +

1

3X

R = 3X

3= X = 1

Now, E = IR

I = E

R =

3

1 = 3 A

4. Initial current I = V

R =

12

R and new current

when resistance is increased by 4 ,

I =V

R 4=

12

R 4

From given condition I I = 0.5 A

12 12

R R 4

= 0.5

1 1

12R R 4

= 0.5

R 4 R

R R 4

= 0.5

12

4

R R 4 =

0.5

12

1

R R 4 =

0.5

4 12

R2 + 4R = 96 R2 + 4R 96 = 0 (R 8) (R + 12) = 0 Either R = 8 or 12 but, ve resistance is

not possible. R = 8 5. Here, R1 = 3 , R2 = 3 , R3 = 6 , Total e.m.f. of circuit = E2 – E1 = 6 – 1.5 = 4.5 V Total resistance = r1 + r2 + R1 + (R2|| R3)

= 0.4 + 0.8 + 3 + (3 || 6) = 4.2 +

9

63

= 4.2 + 2 = 6.2

Total current i = R

E =

4.5

6.2 0.73 A

For E1, current is negative. V1 = E1 – ir1 = 1.5 – (–0.4 0.73) = 1.5 + 0.29 = 1.79 V 1.8 V For E2, V2 = E2 – ir2 = 6 – (0.8) (0.73) = 6–0.58 = 5.42 V 5.4 V 6. In the circuit, resistances 10 , 10 , 20

are in parallel. Voltage drop across them is same. Voltage drop across path 1 = V1 = I1 R1 = 0.5 10 = 5 V (C) is incorrect. Voltage drop across path 2 = V2 = I2 R2

5 V = I2 10 or I2 = 10

5 = 0.5 A

Voltage drop across path 3 = V3 = I3 R3

5 = I3 20 or I3 = 20

5 = 0.25 A.

(B) is incorrect. Total current I = I1 + I2 + I3

= 0.5 + 0.5 + 0.25 = 1.25 A VX = 25 5 = 20 V (D) is incorrect.

X = 25.1

20=16 (A) is correct.

7


7. A = 10 , B = 5 , D = 4 , C = 4 For the bridge to balance,

A C

B D

Now, A

B=

10

5= 2 : 1 and

= 4

4= 1 : 1

i. If 10 be connected in series with A then we get,

A 10

B

=

10 10

5

=

20

5= 4 : 1

This is not equal to 1:1. ii. If 10 be connected in series with B,

then we get,

A

B 10=

10

5 10=

10

15= 2 : 3

This is not equal to 1 : 1. Hence option (C) is not correct. iii. If 5 be connected in parallel with B

then we get,

A

B || 5=

10

5||5=

1025

10

= 100

25 = 4 : 1

This is not equal to 1 : 1. Hence option (D) is not correct. iv. If 10 be connected in parallel with A

we get,

A||10

B=

10||10

5=

100

205

= 5

5= 1 : 1

Hence, A C

B D = 1 : 1

Hence option (B) is correct. 8. As current through G is zero, then it is

balanced Wheatstone’s bridge.

4 6

8 X or X =

48

4= 12 .

9. PQ and QR are in series Equivalent resistance of PQ and QR,

1SR = 2 + 3 = 5

PS and RS are in series Equivalent resistance of PS and RS,

2SR = 2 + 3 = 5 .

Now 1SR , 5 and

2SR are in parallel.

Equivalent resistance is given by

1 2P S S

1 1 1 1

R R R 5

P

1 1 1 1 3

R 5 5 5 5

RP = 5

3

10. LP

NP

R

R= LM

MN

R

R

15 6

X || 8 3

=

15 6 || 6

4 4 || 4

21

8X3

8 X

= 18

4 2 = 3

21 8 X

11X 24

= 3

56 + 7X = 11X + 24 4X = 32 X = 8 11. To balance the bridge, we use shunt resistance

D with X such that, D = D || X

Let balance condition be A

B =

C

D '

3

3 =

3

D ' or D = 3

D = D || X = DX

D X

Substituting the values we get,

3 = 4X

4 X

4X = 12 + 3 X or X = 12 12. The circuit becomes balanced Wheatstone’s

network. Potential difference between B and D

is 0.15 20

30 40

13. AD

AB

R

R =

CD

BC

R

R

4 4

4 X

= 6 || 3

5 || 5

8

4 X =

18 / 9

25 / 10

8


8

4 X = 2

25

10 =

4

5

2

4 X =

1

5

4 + X = 10 X = 6 14. Current through QX and RY is zero due to

symmetry of network. Reff = (R1 +R4+ R6) || (R2 +R8 + R7) = 6 || 6 = 3

I = V

Reff. =

3

3 = 1 A

15. A

B

= A

B =

1

2

2A = B A : B = 1 : 2 When A and B are increased by 10 each we

get, A 10 3

B 10 4

4A + 40 = 3B + 30 4A + 40 = 6A + 30 ....[ B = 2 A]

2A = 10 or A = 5

5 10 3

B 10 4

or B = 10 A : B = 5 : 10 = 1 : 2

16. In 1st case, QP RR =40

60=

2

3

Rp = 3

2RQ ...(1)

In 2nd case, resistance in the right gap is, RQ||10 = 'RR10R10 QQ

Now, RR P =1 or RP = R

Rp = Q

Q

10R

10 R ....(2)

From (1) and (2) or 3

2RQ =

Q

Q

R10

R10

3

1 =

QR10

5

or 10 + RQ = 15

RQ = 5 and RP = 10/3 17. Let X be the smaller resistance in the metre

bridge, X = 40 cm

R = 100 40 = 60 cm

Now, X

R

= X

R

or 40

60 =

X

R

or X

R =

2

3

R = 1.5 X ....(1) From second condition,

X 10

R

=

80

100 80

or X 10

R

= 4

R = X 10

4

....(2)

Equating (1) and (2) we get,

X 10

4

= 1.5 X

6X = X + 10 5X = 10 X = 2

18. X

30 =

100

.…(1)

p

1

X =

X

1+

4

X =

5

X

Xp = X

5

X / 5

R = x

x100

.…(2)

Equating equations (1) and (2), we get

X

30 = X

5R

5R = 30 or R = 6 19. Eeff = 4 – 4 = 0 or potential gradient = 0 20. Current through the cell,

I = E

R r

=

5

40 5=

5

45A =

1

9A

Reading of voltmeter V = IR = 1

9 40 = 4.4 V

21. E =

L

V or =

E

p.g.

=

3

1.08

3 10 100 =

10.8

3 = 3.6 m.

9


22. E

E

1.08 =

60

40

E = 2

3 1.08 = 3 0.54 = 1.62 V

23. V

E =

200

150 =

4

3

V = 3

4E =

3

4 1.1 = 0.825 V

Using, E = V + ir

i = r

)VE( =

1.1 0.825

0.5

= 0.275

0.5= 0.55 A

24. r = R

1

1

= 3600

1200

= 6

25. P.D. across potentiometer wire = 2 V Potential gradient = LV

= 1002 V/cm

Now, E = LV .

E = 1002 .60 = 2 6

10= 1.2 V

26. 1

2

E

E = 1 2

1 2

where, 1 = 240 cm

1.5

1.1 = 2

2

240

240

2640 + 112 = 3600 152

262 = 960

2 = 960

26 37 cm

27. 1

2

E

E =

6

8 6 =

63

2

or 1

2

E

E = 3 ....(1)

When length of potentiometer wire is increased by 2 m, then new length

= 8 + 2 = 10 cm Let the balance point is at a distance of x m

from one end.

In this condition 1

2

E

E =

x

10 x

x

10 x= 3 ....[From (1)]

x = 30 3x 4x = 30

x = 30

4= 7.5 m

28. RAB = 4 10 = 40

i = 4

40 20 =

4

60 =

1

15

V = i RAB = 1

15 40 2.67 V

L

V =

3.67

10 = 0.3 V/m.

29. E = 2 volt, r = 5 , = 1000 cm,

R = 20 , Current through the circuit,

I = E

R r=

2

20 5 =

2

25 = 0.08 A

V = IR = 0.08 A 20 = 1.6 V

Potential gradient = V

= 1.6 / 1000

= 1.6 103 V/cm.

30. 1

21

E

EE =

200

150 =

4

3

1 + 1

2

E

E =

4

3

1

2

E

E =

1

3

2

1

E

E =

3

1 = 3: 1

1

Chapter 14: Magnetic Effect of Electric Current


1. B = 0nI

2R

B = 74 10 100 1

2 0.1

= 6.28 104 T

2. B = 0NI = 0n

I

= 4 107 500

50.5

= 6.28 103 T

3. B = 0 2I

4 r

= 74 10 2 8

4 0.03

= 5.33 105 T

4. B = 0NI

= 4 107 280 20

= 7.04 103 T

5. B = 0 2I

4 r

= 7

2

4 10 2 5

4 4 10

= 2.5 105 T

6. i

v

S

S =

/ I

/ IR

= R

Sv = iS

R =

3

1 div

10 A 200

= 1 div

0.2V =

5div

1V

= 3

5div

10 mV

= 5 103 div/mV

7. I = k

nAB

=

9

4

1.5 10 5

50 12 10 0.025

= 5 106 A = 5 A

8. Sv = V

=

3

50

25 10 = 2 103 div/V

Si = 1

20div/A

= 6

0.05 div

10 A = 5 104 div/A

Rg = i

v

S

S =

4

3

5 10

2 10

= 25

9. I = k

n AB

= 8

2 2 2

1.5 10 0.5

60 4 10 2.5 10 2.5 10

= 5 106 A = 5 A 10. Si R

10

2 =

50

R R = 10

where R is total resistance of combination.

1

R =

1 1

G S

where S is required shunt resistance connected in parallel with galvanometer.

1 1

10 50 =

1

S

S = 500

50 10 = 12.5

11. Si = 0.105 rad/A = 0.105 106 rad/A

B = iS k

nA

= 6 9

3

0.105 10 25 10

10

= 2.625 Wb/m2

12. Si = n AB

k

= 3 3 2

8

100 40 10 25 10 10

10

= 105 deg A1 = 0.1 deg A1

Magnetic Effect of Electric Current 14

2


13. I

1

2

= 1

2

I

I

1 = 30 = 30

180

=

6

rad

I2 = 1 2

1

I

= 20 0.314 6

3.142

12 A

14. R = g

VG

I

= 3

2510

0.25 10

= 99990 in series 15. Ig = 6 mA, Vg = 15V

G = g

g

V

I= 2500

S = g

g

IG

I I

= 3

3

6 102500

3 (6 10 )

S = 2500

499

16. Vg = 50 div 1 mv/div = 50 mV.

Ig = 50

5 = 10 mA

G = g

g

V

I= 5

i. S = g

g

IG

I I

= 3

3

10 10 5

5 (10 10 )

= 350 10

4.99

=

5

499

ii. R = g

VG

I

= 3

1005

10 10

= 9995 in series.

17. S = g

g

IG

I I

= 50

3964950

= 4

18. Ig = I

n I = n Ig

Since, S = g

g

IG

I I

= g

g g

IG

nI I

= g

g

IG

(n 1) I

= G

n 1

19. G = g

g

V

I=

500

20= 25

i. S = g

g

IG

I I

=0.02

2510 0.02

= 25

499

= 0.05

ii. R = g

VG

I =

3

10025

20 10

= 4975 in series. 20. I = Is + Ig Is = I Ig From, Is S = Ig G (I Ig) 294 = 6 Ig 49 (I Ig) = Ig I = 50 Ig

Ig = I 1

50

21. i. S = g

g

IG

I I

= 3

3

2 10100

(2 2 10 )

= 0.002

1001.998

= 100

999

ii. g

VG

I =

3

5100

2 10

= 2400 in series

3


22. Intially current through galvanometer be I which is maximum current.

Upon shunting, deflection reduces to 1/5th

original Ig = I

5

S = g

g

IG

I I

= g

g g

IG

5I I

4 = G

4

G = 16 .

23. R = g

VG

I

= 3

10200

5 10

= 1800 in series

24. Ig = iS

=

20

10= 2 mA.

Vg = vS

=

20

2= 10 mV

G = g

g

V

I= 5

i. S = g

g

IG

(I I )

S = 0.002

52 0.002

= 0.01

1.998 =

10

1998 =

5

999

ii. R = g

VG

I

= 20

50.002

= 9995 in series.

25. Current for full scale deflection.

Ig = 10 div 1 mA/div = 10 mA

R = g

VG

I =

3

5015

10 10

= 4985 in series.

26. S = g

g

IG

I I

Ig = 0.02 I

1 = 0.02I

GI 0.02I

G = 0.98

0.02= 49

27. V = nVg 250 = n (25) n = 10 R = (n 1) = 500 (9) = 4500 in series. 28. Reading = 10 V per 5 div. Total div = 20

Required voltage = 10 20

5

V = 40 V

R = g

VG

I =

4012

0.02

= 1988 in series. 29. Ig = 10 mA = 102 A, Vg = 5 V

G = g

g

V

I = 500

V = n Vg n = 100

5= 20

R = G (n 1) = 500 19 = 9500 in series.

30. Emax = 2(qBR)

2m

= 19 2

27

(1.6 10 0.35 0.5)

2 1.67 10

= 1.4671 106 eV 1.47 MeV

31. f = Bq

2 m

= 19

27

1 1.6 10

2 17 1.67 10

= 15.25 106 Hz = 15.25 MHz

v = q BR

m=

19

27

1.6 10 1 0.4

1.67 10

= 3.83 107 m/s. Emax = e V Potential difference

4


V = 2(q BR)

e2m

V = 19 2

19 27

(1.6 10 1 0.4)

1.6 10 2 1.67 10

= 7.67 106 eV = 7.67 MeV


1. Bcentre = 0nI

2R

. When R is doubled, the B is

halved. 2. B = 0NI

For N same, I = I

2

B = B

2

3. Bcentre = 0I

2R

If the wire is bent into n turns, then

R = R

n

New magnetic inductor, B = 0n I

2R

B =

0n I

2 R / n

= n2B

For, n = 2, so B = 4 B. 4. Magnetic field due to a long current carrying

conductor,

B = 0 2I

4 r

BL = 107 2 30

1002

BL = 3 104 tesla B = 4 104 tesla

BResultant = L2 2

||B B = 5 104 T

5. B = 0nI

2R

=

7 24 10 10 0.50.4

22

= 1.57 104 N/A m. 6. B = 0NI

N = 20 turns/cm

= 20 102 turns/m B = 4 107 20 102 4 = 32 104 T = 3.2 103 T 8. Force on PQ = BIL

Length QR = L/ cos

Force on QR = BI (L/ cos ) sin

F(PQ) : F(QR) = 1 : 1

9. B = 0 2I

4 r

= 7 2 60

102

= 60 107 = 0.6 105 T 10. F = BIl sin = 3 2.5 1 sin 90 = 7.5 N 11. N = 10 turns/cm

= 10 102 turns/m = 103 turns/m

I = 0

B

N=

3

7 3

12 10

4 10 10

= 30

A

12. Magnetic field at the centre of a long solenoid

is given by,

B = 0NI Where N = number of turns per unit length. B NI

1

2

B

B = 1

2

N

N1

2

I

I

= 200 I

100 I / 3

1

2

B

B= 6

B2 = 2

1B 6.28 10

6 6

= 1.05 102 Wb m2

13. At distance a/2 current is,

I = I

2

2

a

2a

= 2

2

i a

4 a

= i

4

B1 = 0 2I

4 a / 2

= 0 i

4 4

BR

BL

2 cm

B

5


At distance 2a current is I = i

B2 = 0 2I

4 2a

= 0 i

4 a

Hence, 1

2

B

B=

1

1

14. Si = nAB

k

2

1

S

S = 2

1

n AB k

k n AB = 2

1

n

n

5

4 = 2n

28 n2 = 35

No. of turns should be increased by 7. 15. For same current n

16. For same, Si 1

I

17. = nIBA

k=

6

9

1 200 10 0.06 0.01

5 10

= 24

18. I = k

nAB

I 1

B(when other quanties are constant)

Hence, if field is doubled, current should be halved.

19. nIBA = k 60 I 500 6 = 18

I = 18

60 50 6 =

18

18000=

1

1000= 103A

20. Sv = nBA

kR

For remaining parameters constant, Sv 1

R

21. Given, sI = Is + s

200I

100= s

120I

100; R = 2R

Then, initial voltage sensitivity,

Vs = sI

R

New voltage sensitivity,

sV = sI

R

= s

120I

100

1

2R= s

3V

5

% decrease in voltage sensitivity

= s s

s

V V100

V

=

s s

s

3V V

5 100V

= 40%

22. R = k = 4.2 109 30 = 1.26 107 N m.

23. As, I = k

nBA

k = nBAI

= 4 4 360 90 10 5.0 10 20 10

18

= 3.0 107 N m per degree

24. Si = nAB

k=

4 2

8

10 20 10 6 10

0.8 10

= 1.5 105 rad/A

25. Sv = iS

R =

1 div / mA

50 = 2 102 div/mV

= 2

3

2 10 div

10 V

= 20 div/V

26. Let R be voltmeter resistance and n be total

no. of divisions in voltmeter. When current of value ig flows through it, voltage recorded by each division is,

gi R

n= V ….(i)

After connecting resistance,

gi (R 1470)

n

= 50 V ….(ii)

Dividing (ii) by (i)

R 1470

R

= 50

Solving, R = 30 . 27. I = Ig + Is and Ig G = Is S 7.5 G = (92.5 ) S

S = 7.5 G

92.5

=

0.3G

3.7

28. gI

I =

S

S G

5

100 =

S

S 95

20 S = S + 95 S = 5 29. In ammeter, galvanometer and shunt is parallel

(G || S). Hence resistance of ammeter is

1

R=

1 1

G S

R = G S

G +S

6


30. In ammeter, glavanometer and shunt is parallel (G || S). Hence resistance of ammeter is

1

R =

1 1

60 5 R =

60

13 4.62

31. I = g

S GI

S

= 2 60

202

= 620 mA = 0.62 A 33. Shunt required to increase the range to n times

is,

S = G

n 1

= 0.018

(10 1)= 0.002

34. Potential difference across ammeter and shunt

resistance is same.

ig R = (i ig) S

S = g

g

i R

i i =

100 13

750 100

= 1300

650 = 2

35. S = g

g

IG

I I

For I = 50 A, Ig = 10 A

12 = 10

G50 10

G = 48 . 36. For V = nVg , R = (n 1) G

n = 12

1.5 = 8

R = 8 1500 = 12,000 .

37. R = g

VG

I

Ig = V

R G =

10

512 = 0.019 A 0.02 A

38. Without additional resistance, range it will

cover will be given by

Vg = Ig = 2 103 200 = 0.4 V = 400 mV

39.

Ig = V 10V

G R (5 0.05)k

=

10V

10.1k

= 1.98 103A = 1.98 mA To obtain I = 1 A

S = 3

g

3g

I G 1.98 10 100

I I 1.5 (1.98 10 )

0.13

40. R = g

VG

I =

3

100

10 10 25

= 10000 25 = 9975

41. R = V

I=

50

10 = 5 .

actual value measured by voltmeter is less than real value.

42. To increase range of ammeter, resistance wire

should be connected in parallel.

S = g

g

IG

I I

= 3

3

20 1025

[5 (20 10 )]

S = 3500 10

4.98

=

0.5

4.98=

50

498

R = S = A

= SA

=

2 4

7

50 2.49 10 10

498 5 10

= 0.5 m

44. Emax = 22R2f2m = 2 2 (0.8)2 (107)2 1.6 1027 = 2.0213 1012 J = 1.263 107 eV = 12.63 MeV

45. f = qB

2 m=

19

27

1.6 10 2

2 3.14 1.67 10

30.5 106 Hz

(i ig)

S

ig i

G

Ammeter

S

IgI

I

GR

Voltmeter

7


46. Kinetic energy of a charged particle is given by E = qV

If V remains same, then E q Substituting qp = e; qe = e ; q = 2e, we get Ep : Ee : E = 1 : 1 : 2

47. E = 2 2 2B q r

2m

For an particle, q = 2e and m = 4 mp

p

E

E =

2 2 2p

2 2 2p

2(m )B (2e) r

2(4m ) B (e) r

= 4

4 = 1

E = Ep

48. r = mv

Bq=

m

q

v

B =

v

Specific charge B

= 5

7 2

2 10

5 10 4 10

= 0.1 m

1

Chapter 15: Magnetism

Hints to Problems for Practice 1. 2l = 6 cm = 6 102 m, m = 500 Am M = m.2l M = 500 6 102 = 30 Am2

2. M = m.2l

2l = 6

60 = 0.1 m

L = 6

25

l

L = 6

0.15 = 0.12 m.

3. M = m.2l

2 l = 20

200 = 0.1 m

l = 0.05 m

L = 6

25 l = 0.12 m

4. m = 50 Am, 2 l = 20 cm = 0.2 m i. Magnetic moment = M = m 2l = 50 0.2 = 10 Am2 ii. = 30, B = 0.10 T Now, = MB sin = 10 0.1 sin 30 = 0.5 Nm 5. M = 5 Am2, r = 50 cm = 0.5 m

Baxis = o3

2M

4 r

Baxis = 7

3

10 2 5

(0.5)

= 8 106 Wb/m2

6. Baxis = 4 104 tesla, r = 20 cm = 0.2 m

Baxis = o3

2M

r

M = 3

axis

o

B r

24

= 4 3

7

4 10 (0.2)

2 10

M = 16 Am2

7. r = 40 cm = 0.4 m, B = 104 tesla

Baxis = o3

2M

4 r

m = 3

axis

o

B r

24

m = 4 3

7

10 (0.4)

2 10

= 3.2 10 = 32 Am2

8. M1 = 1 Am2, M = 2 Am2,

r = 20

2 cm = 10 cm = 0.1 m

B at the midpoint joining the centres of magnets

Baxis = o3

2M

r

1axisB = 107

3

2

(0.1)

= 2 104 Wb/m2 and

2axisB = 107

3

2 2

(0.1)

= 4 104 Wb/m2 For the case, when similar poles of the

magnets face each other the resultant magnetic induction, =

2 1axis axisB B

= 4 104 2 104 = 2 104 Wb/m2 For case, when their dissimilar poles face each

other The resultant magnetic induction, =

2 1axis axisB B

= 4 104 + 2 102 = 6 104 Wb/m2

9. M = 0.1 Am2, r = 10 cm = 0.1 m

B = o3

M

4 r

= 107

3

0.1

(0.1) = 105 T

Magnetism15

2


10. raxis = 2 requator

o o3 3axis equator

2M M2

4 r 4 r

2

3equator

axis

r

r

= axis

eqator

B

B

axis

equator

B

B= 2

31

2

= 1 : 4 11. raxis : requator

o3

axis axis

oequator3eq

2M

4B rMB

4 r

,

axis

eq.

B2 :1

B

12. Baxis = Bequator , r = 0.2 m

o3axis

2M

4 r

= o

3equator

2M

4 r

3 3 3equator axis

1 1r r (0.2)

2 2

= 0.008

2 = 0.004 m

requator = 3 0.004 = 0.1587 m 13.

(axis) (equater )P QB 54 B

o3P

2M

r

= 54 o

3Q

M

r

3p

3Q

r 2

54r =

1

27

p

Q

r

r = 1 : 3

14. Paxis = Qequator

3

Qo o3 3

pp p

r2M 2M2 1

4 4 rr r

p 3

Q

r2

r or rp : rQ = 1.26 : 1

15. B = 2o3

M3cos 1

4 r

B = 107 2

2

2.4 13 1

2(0.1)

= 105 2.4 7

4

= 3.1752 104 Wb/m2

16. B = 2o3

M3cos 1

4 r

B = 107 2

3

10 13 1

2(0.2)

= 6

3

10 7

48 10

= 1.654 104 Wb/m2

= tan1 1

tan 602

= tan1 1

1.7322

= tan1 (0.8660)

= 4053 17. r = 20 cm = 0.2 m, M = 8 Am2, = (90 30) = 60

B = 2o3

M3cos 1

4 r

B = 107 2

3

8 13 1

2(0.2)

= 104 7

4 = 1.3229 104

= 1.323 104 tesla

18. B = 2o3

M3cos 1

4 r

B = 107 2

3

2.4 13 1

2(0.2)

= 0.3 104 7

4

= 4 105 Wb/m2

= tan1 1

tan2

= tan1 1

32

= tan1 (0.866) = 4054

3


19. M1 = 5 Am2, M2 = 40 Am2 Here, B1 + B2 = 0

0 01 23 3

2M 2. 0

4 d 4 (30 d)

1 23 3

M M0

d (30 d)

3 3

5 40

d (30 d)

= 0

3 3

1 8

d (30 d)

= 0

3 3

1 8

d (30 d)

(30 d)3 = 8d3

30 d = 2d

3d = 30

d = 10 cm. 20. M = 6.25 Am2, d = 25 cm = 25 102 m B1 = BH

…[B1 and BH have same directions

and same magnitude]

0H3

M. B

4 d

BH = 7

2 3

10 6.25

(25 10 )

= 13

12.5010

25

= 4 105 Wb/m2. 21. M = 20 Am2. M = 2l m For case (i),

The pole strength becomes 1

2 its original

value or M

2

Maxis = 1

M2

= 1

202

Maxis = 10 Am2 For case (ii),

Magnetic length = 2

2

l = l

Meq = 1

2 20 = 10 Am2.

22. i. M = NIA The current in the coil,

I =M

NA=

5.6

200 0.04= 0.7 A

ii. The magnitude of the torque, = MB sin

= 5.6 0.3 sin 30 = 5.6 0.3 1

2

= 0.84 Nm 23. A = 2 cm 5 cm = 10 10–4 m2 = 10–3 m2, I = 2 A, B = 1.5 T, = 60 i. M = IA = 2 10–3 = 0.002 Am2

ii. = MB sin = 2 10–3 1.5 3

2

= 1.5 1.732 10–3 = 2.598 10–3 Nm

24. M0 = 1.728 10–23 A.m2 , e

m 1.759 1011 kg

M0 = e

e

2mL0

L0 = 0

e

2M

(e / m )

=23

11

2 1.728 10

1.759 10

= 1.965 10–34 kg.m2/s. 25. M = I A = 10 6.8 104 = 6.8 103 Am2 The direction of magnetic moment is

perpendicular to the plane of the loop and away from the observer.

26. r = 0.5 Å = 0.5 1010 m, f = 109 MHz = 1015 Hz Equivalent magnetic moment (M)

I = q

T =

e

T = ef

M = IA = I.r2 As a revolving electron is equivalent to

electric current, using I = ef, I = 1.6 1019 1015 = 1.6 104 A M = IA = I r2 M = 1.6 104 3.14 (0.5 1010)2 = 1.6 3.14 0.25 104 1020 = 1.256 1024 Am2

4


27. The coercivity of 2 103 Am1 of the bar magnet implies that magnetic intensity H = 2 103 Am1 (in opposite direction) is required to demagnetise it.

Here, number of turns per unit length of the solenoid,

n = 50

10= 5 turns cm1 = 500 turns m1

B = o n I

I = 3

o

B H 2 10

n n 500

= 4 A

28. Mean radius of toroid,

r = 10 12

2

= 11 cm = 11 102 m

Number of turns per unit length =

n = 2

2200

2 3.14 11 10

= 3.18 103 m1 B = o r nI

r = 0

B

nI=

7 3

2.5

4 3.14 10 3.18 10 0.50

1250 29. = 0.10 T A1 m, 0 = 4 107 T A1 m

r = 7

o

0.1

4 10

= 7.96 104 8 104 m = 8 104 1 8 104

30. H = 1600 Am1; = 2.4 105 Wb, A = 0.1 cm2 = 0.1 104 m2 Now, magnetic induction,

B = 5

4

2.4 10

A 0.1 10

= 2.4 T

= B 2.4

H 1600 = 15 104 T mA1

Therefore, relative permeability,

r = 4

7o

15 10

4 10

= 1194

Again, B = o (H + I)

I = 7

o

B 2.4H 1600

4 10

= 19.1 105 1600 = 19.08 105 Am1 m = r 1 = 1194 1 = 1193.

31. H = 2 103Am1, I = 4.8 102 Am1 and T = 280 K

m = 2

3

I 4.8 10

H 2 10

= 2.4 105

Aluminium is a paramagnetic substance and hence obeys Curies’ law. If m is susceptibility of aluminium at temperature T (= 320 K), then

m

m

T

T

m = m T

T

m = 2.4 105 280

320

= 2.1 105

If I is intensity of magnetisation of aluminium at 320 K, then

I = m H = 2.1 105 2 103 = 4.2 102 Am1

Hints to Multiple Choice Questions 1. M = m 2l

m = M

2l =

4

0.08 = 50 Am

2. Pole strength does not deped on length. 3. When it is divided into two equal parts by

cutting it along the axis, the pole strength of

each part becomes m

2 with no change in its

magnetic length.

M = m 2l and M = m

2 2l =

M

2

Magnetic moment of each part,

= M

2 =

16

2 = 8 Am2

4. As magnetic moment pole strength area

of cross-section, i.e., M m A M1 : M2 : M3 = 1 : 3 : 5. 5. = M B sin

M = 0.04

Bsin 0.2 sin 45

0.3 Am2

6. = 30, W = BM cos 30 = 3 MB

2;

MB = 2W

3

5


= MB sin = 2W

3 sin 30

= 2W 1

23 =

W

3

7. As the magnet is short,

B 3

1

d

33

1 23

2 1

B d 24cm

B 12cmd

= 8

8. Transverse position Equatorial position

longitudinal position Axial position

Tr

long

B

B=

3

M

d

3d

2M =

1

2

9. As magnetic moments are directed along SN,

angle between M

and M

is = 120. Resultant magnetic moment

= 2 2M M 2.M.M.cos120

= 2 2 2M M 2M ( 1/ 2) = M 10. Baxis = 4 Bequator ….(Given)

03

A

2M

4 (d )

= 4 03

B

M.

4 (d )

3 3A 8

1 2

d d

Taking the cube roots of both sides, we get

A

1

d=

3

B

2

d

dB = 3A2d = 1.259 8 = 10.07 cm 10 cm

11. M = m 2l = 15 0.2 = 3 Am2 The point is at a distance of 30 cm from either

pole. Hence it must be on the equator.

Beq = 02 2 3/2

M

4 (d )

l

But (d2 + l2)1/2 = 20 cm = 0.2 m

Beq = 107 3

3

(0.2) =

7

3

3 10

8 10

= 0.375 104

B = 3.75 105 Wb/m2

12. Baxis = 0

4

2 2 2

2Md

(d ) l

M = m 2l = 25 0.15 = 3.75 Am2

and l 0.08 m, d = 0.2 m

Baxis = 107 2 2 2

2 3.75 0.2

[(0.2) (0.08) ]

= 7

2

1.5 10

3.36100

= 35.71 105

1.30 104 Wb/m2

13. In figure (a) at neutral point P,

BH = 03

M

4 d

In Fig. (b) Net magnetic induction at P = resultant of

03

2M

4 d

= 2 BH along horizontal and BH along

vertical

= 2 2H H H(2B ) (B ) 5 B

14. Magnetic diopole moment, M = nIA = nI (r2)

= 5 10 22 7 7

7 100 100

= 0.77 Am2 along Z axis

15. B = oI

2r

=

7

2

4 10 2.5

2 6.28 10

= 2.5 105 Wb m2

16. 2r = L

r = L

2

M = IA = I r2 = I2

2

L

4

M = 2IL

4

N

S

E W

n

s

(a)

P

BH

Beq

N

S

E W

nsBH

BaxisP

(b)

6


17. L = length of wire and n = number of turns

L = n(2R) = n(2.R

2)

n = 2n M = nIA = nI(R2) ….(i) New magnetic moment,

M = nIA = (2n).I.2

R

2

M = nI2R

2

….(ii)

Dividing equation (ii) by (i),

M

M

=

1

2

18. a = 10 cm = 10 102 m, B = 4 105 Wb/m2

B = o 2 i

4 a

4 105 = 106 i 3.14 2

i = 5

6

4 10

2 3.14 10

= 0.637 10 = 6.37 A 19. The magnetic moment of the solenoid, M = (IA) N = 2.0 2.0 104 1000 = 0.4 Am2 Using, = MB sin = 0.4 0.16 sin 30 0.03 Nm

20. M = IA = 2er

T but T =

2 r

v

M = 2ev evrr

2 r 2

M = 19 6 101.6 10 3.2 10 0.5 10

2

= 1.6 3.2 0.25 1023 = 1.28 1023 Am2 1.3 1023 Am2 21. The Bohr magneton is the magnetic moment

associated with an electron due to its orbital motion in the ground state of the hydrogen atom and its value is given by

M0 = e

eh

4 m= 9.28 1024 Am2

In this case, the mass of the muon = 200 me

But its charge = e Its orbital magnetic moment

= 24

e

eh 9.28 10

4 (200m ) 200

= 4.64 1026 Am2

22. time (t) = Distance travelled

Velocity

= 2r r r( 2)

Velocity(v) v

I = Q

t =

Qv

r ( 2)

M = I A = 2Qv r

r( 2) 2

=

rQv

2( 2)

23. r = 0.5 Å = 0.5 1010 m, = 1010 MHz = 1016 Hz The revolving electron is equivalent to a

current – carrying loop,

I = q

T =

e

T = e

1

T

Using, M = IA = (e) r2 M = 1.6 1019 1016 3.14 (0.5 1010)2 = 1.6 3.14 0.25 10320 = 1.256 1023 Am2

24. m = 90 g = 90 103 kg, M = 3 Am2, d = 6 g/cm3 = 6000 kg/m3 Volume of the magnet,

V = 3

2

m 90 10

d 60 10

= 1.5 105 m3

I = 5

M 3

V 1.5 10

= 2 105 A/m

25. A = 0.4 cm2 = 0.4 104 m2, = 3.2 105 Wb, H = 500 A/m

B = A

= 5

4

3.2 10

0.4 10

= 0.8 Wb/m2

= B 0.8

H 500 2 103

= 2 103 N/m2

7


26. When space inside the toroid is filled with air, Using, Bo = o H When filled with tungsten, B = H = o r H = o (1 + m) H Percentage increase in magnetic field

induction.

= o

o

(B B ) 100

B

= o m

o

H 100

H

= m 100 = 6.5 105 100 = 0.0065%

27. r = 2

1

L 45

L 15 = 3 and

m = r 1 = 3 1 = 2. 28. o = 4 107 (SI units) r = 2 102 (SI units) = 0 r = 4 107 2 102 = 8 105

29. T1 = 127 + 273 = 400 K

From Curie’s law, 1

T

2 1

1 2

T

T

but it is given that 2

1

1

3

2

1 400

3 T

T2 = 1200 K = (1200 273) = 927C

30. = 0.2, 1

T= 4 103/K

T = 3

1 1000

44 10

= 250 K

According to Curie’s law, = C

T

or C = T = 0.2 250 = 50 K

1

Chapter 16: Electromagnetic Induction


1. dt = 25 s, e = 2 10–3

e = d

dt

or d = edt

= 2 103 25 = 0.05 Wb 2. A1 = 15 cm 5 cm = 75 10–4 m2, A2 = 10 cm 10 cm = 100 10–4 m2 B1 = 0.8 Wb/m2, B2 = 1.4 Wb/m2, dt = 1s,

e = d

dt

=

d

dt

(B.A) = 1 1 2 2B A B A

dt

e = 4 40.8 75 10 1.4 100 10

1

= 80 104 = 8 10–3 e = –0.008 V 3. n = 200, A = 0.16 m2, dB = 0.4 Wb/m2, dt = 0.02s

e = n.d

dt

= n.

A dB

dt

= 200 0.16 0.4

0.02

= 640 V 4. nA = 2m2 , B1 = 0.08 Wb/m2, B2 = 10% of B1 = 0.1 B1, dt = 0.6 s

e = d

dt

=

d(n A B)

dt

= 2 1n A(B B )

dt

e = 2 1 2 1 1(B B ) 2(B 0.1B )

dt dt

= 2 0.9 0.08

0.6

= 0.24 V

5. A = 4m2, e = 0.32 V,

B2 = 20% of B1 = 1

5B1. dt = 0.5 s,

e = 1 2A(B B )

dt

0.32 = 1 1

14(B B )

50.5

1 10.32 0.5 4 0.32 25

B or B4 5 16

= 0.05 Wb/m2

6. A = 500 cm2 = 500 104 m2, n = 1000, B = 4 105 T, = 180, dt = 0.1s

e =d

dt

1 2( )

dt

= n A B(1 cos )

dt

= 4 51000 500 10 (1 cos180 ) 4 10

0.1

= 104 500 104 (1+1) 4 105 = 103 4 105 = 4 102 = 40 mV 7. d = 12 103 6 103 = 6 103 Wb, dt = 0.1s

e = d

dt

=

36 10

0.1

= 60 mV

8. A = 5 cm 10 cm = 50 cm2 = 50 10–4 m2,

B = 1 10–3 Wb/m2, 2 = 0, dt = 3 s

e = –d

dt

e = 2 1

dt

e = – 1(0 A B )

dt

=

4 350 10 10

3

= 65 10

3

= 1.67 10–6 V e = 1.67 V and d = 5 10–6 Wb 9. A = 4m2, B1 = 0.05 Wb/m2,

B2 = 20% of B1 = 1

5 B1, dt = 10 s.

e = –A dB

dt = 1 2A(B B )

dt

e = 4 (0.05 – 1

5 0.05)

1

10

= 4 4

5 0.05

1

10

e = 0.016 V = 16 mV

Electromagnetic Induction 16

2


10. r = 7.5 cm = 7.5 10–2m, n = 1500,

dB = (67 – 14) 10–3 = 53 10–3T,

dt = 2.3 s,

e = nA. dB

dt

e = 1500 3.14 (7.5 10–2)2 353 10

2.3

e = 2

515 3.14 7.5 5310

2.3

= 0.61V 11. l 0.3 m, v = 10 m/s, B = 5 10–5 T,

e = Blv

e = 5 10–5 0.3 10 = 1.5 10–4 V

= 0.15 mV 12. e = Blv = 0.2 2 5 = 2 V

13. l = 1.5 m, v = 60 km/h = 60 5

18 = 16.67 m/s,

BH = 3.6 10–5 Wb/m2, = 46, tan 46 = 1.035

e = Blv = BH lv tan

e = 3.6 105 1.5 16.67 1.035

= 93.16 105 V

= 931.6 V

14. l = 2

m, B = 5 10–3 Wb/m2,

e = 20 mV = 20 10–3 V.

e = 1

2Bl2 =

1

2Bl2 2f

f = 2 2

2e e

B 2 B

l l

= 3

3

20 104

5 10

= 1 r.p.s. = 60 r.p.m.

15. e = 1

2Bl2

e =1

2 5 105 (0.5)2 2 3.14 10

= 5 0.25 3.14 104

= 3.925 104 V

16. r = 20 cm = 20 10–2 m, n = 1200 r.p.m.

= 1200

20r.p.s.60

B = 100 G = 100 104 T

e = Bf2

e = 100 104 20 (20 102)2

….( = r)

= 102 20 400 104

= 80 104 V

= 8 mV 17. d I = 10 – 5 = 5A, dt = 0.15, e = 5 V

e = L dI

dt L =

edt

dI

L = 5 0.1

5

= 0.1 H

18. n = 2000, I = 5A, B = 0.4 106 Wb

n = LI

L = 42000 0.4 10

5

= 1.6 104 = 0.16 mH. 19. e = 10 V, L = 500 mH = 500 103 H,

dI2 = 500 mA = 500 103 A

e = L.dI

dt

10 = 500 103 1

1

dI

dt

1

1

dI

dt =

3

10

500 10= 20 A/s

1

1

dI

dt = 2

2

dI

dt ….[Given]

20 = 3

2

500 10

dt

dt2 = 2.5 102 s 20. L = 10 mH = 10 10–3 H, I = 5A, t = 20 s

e = L. dI

dt

e = 10 103 0

20 ...[ I is constant.]

e = 0.

3


21. dI

dt = 25 A/s, e = 10 V

e = dI

M.dt

M edIdt

M = 10

25 = 0.4 H

22. M = 5 mH = 5 103 H, dI

dt = 250 A/s

e = M.dI

dt

= 5 103 250 = 1.25 V 23. m = 10 H, dI = 20 15 = 5A, dt = 0.01 s

e = M.dI

dt

= 10 5

0.01= 5 103 V

24. e = 5 mV = 5 10–3 V, dI

dt = 20 A/s

e = M.dI

dt

= 35 10

20

= 0.25 mH

25. M = 1.5 H, dI = 20 0 = 20A, dt = 0.05 s, Ns = 800

es = M.dI

dt

es = 1.5 20

0.05

es = 600 V d = M.dI = 1.5 20 = 30 Wb 26. n1 = 2000, n2 = 300, = 0.3 m,

A = 1.2 103 m2, dI = 2(2) = 4A, dt = 0.25 s

For a solenoid, M = o 1 2n n A

= 7 34 10 2000 300 1.2 10

0.3

= 590.432 10

0.3

= 301.44 105

e = dI

mdt =

5301.44 10 4

0.25

= 4823.04 105 = 4.8 102 V

27. np = 50, ns = 100, ep = 220 V Transformer action is only pertaining to A.C.

supply. es = 0

28. s

p

n

n = 200 : 1, ep = 220 V

s s

p p

e n

e n

es = 220 200

1= 44 103

= 44000 V 29. Ns : Np = 20 : 1, Ip = 4 A

ps

p s

IN

N I

Is = Ip p

s

N

N

= 4 1

20= 0.2 A

30. s

p

N

N= 50 : 1, Ip = 20 A, ep = 220 V

ps s

p p s

Ie N

e N I

se

220= 50 =

s

20

I

es = 220 50 = 11000 V

and Is = 20

50= 0.4 A

31. e = 75% = 0.75, Pout = 36 W,

e = out out

in in in

P P

P V I

Iin = out

in

P

V e=

36

220 0.75= 0.218 A

32. P0 = 4 104 W, e = 75% = 0.75

e = 0

in

P

P

Pin = 4

0P 4 10

e 0.75

Pin = 5.33 104 W 33. e = 100 sin 100 t Comparing with e = eo sin t epeak = 100 V and = 100 2f = 100 or f = 50 Hz

4


34. n = 500, A = 500 cm2 = 500 104 m2, B = 4 105 Wb/m2,

f = 600/ min = 600

60= 10 /s

eo = n AB(2f) eo = 500 500 104 2 3.14 10

4 105 = 628 104 = 62.8 mV 35. = 60 cm = 60 102 m,

b = 4 cm = 4 102 m, n = 1000, f = 50 rps., eo = 60 V e0 = n A B = n A B 2f

B = oe

2 nAf

= 2 2

60

2 3.14 1000 60 10 4 10 50

= 1

1

2 3.14 6 4 5 10

= 7.962 103 Wb/m2

36. f = 60 Hz, e = oe

2, = 90

e = eo sin (t + )

oe

2= eo sin (2ft + 90) = eo cos 2ft

1

2 = cos 2 60

1

2 = cos 60 (2)t

cos 60 = cos 120 t

120 t = 60 or t = 60

120=

1

2=

1s

360

37. Io = 10 A, f = 60 Hz, t = 1

s360

,

I = Io sin t = Io sin 2ft

I = 10 sin (2 60 1

360)

I = 10 sin (60) = 10 3

2

= 8.66 A 38. n = 100, A= 0.01 m2, f = 50 r.p.s., B = 0.05 T, R = 30 eo = nAB = nAB 2f eo = 100 0.01 0.05 2 3.14 50 = 15.7 V

Irms = oe

R

Irms = 5

A36 6

Pmax = erms Irms

Pmax = 15.7 6

=

15.7 3.14

6

= 8.2268 W 8.23 W 39. erms = 110 V

erms = oe

2

eo = rms2 e

= 1.414 110 = 155.54 V 40. I = 60 sin 200 t Comparing with I = Io sin 2 ft we get, Io = 60 A and f = 100 Hz

Irms = oI 6030 2 A

2 2

41. Io = 10 A

Irms = oI 10

2 2 = 7.07 A

42. R = 500 , e = 10 sin (120 t),

t = 1

s360

Comparing with e = eo sin t eo = 10 V

Io = oe

R

Io = 10

500= 2 102 A

Irms = oI

2

Irms = 0.02

2 = 0.01414 A

= 1.414 102 A I = Io sin t

I = 0.02 sin (120 1

360)

= 0.02 sin 60 = 0.02 3

2

= 1.732 102 A

5


43. R = 99 + 9901 = 10,000 = 104 ,

Eo = 100 2 V

Io = oe

R Io =

4

100 2

10

Io = 22 10 A

As ammeter gives r.m.s. value of current,

Irms = 2

oI 2 10

2 2

= 0.01 A

44. R = 160 , e = 230 V, I = 0.25 A

Z = e

I Z =

230

0.25= 920

Z = 2 2LR X

XL = 2 22 2R Z 920 160

XL = 905.98 906

XL = 2fL

L = LX

2 f

= 906

2 3.14 50 2.88 H

45. eo = 220 V, f = 50 Hz,

L = 253 mH = 253 103 H, R = 9

Z = 2 2LR X

= 2 2R (2 fL)

= 2 2 2 2R 4 f L

= 2 2 2 3 29 4 (3.14) (50) (253 10 )

= 81 6311

= 6392 = 79.9

Io = oe

Z

Io = 220

79.9= 2.75 A

Irms = oI

2

Irms = 2.75

2 = 1.945 1.95 A

46. L = 1mH = 1 103H, eo = 0.5 V, f = 50 Hz

rms

rms

e2 fL

e

o o

oo

e / 2 e2 fL

II / 2

Io = o3

e 0.5

2 fL 2 3.14 50 1 10

= 1.592 103 103 = 1.592 A 47. R = 50 , XL = 120 , erms = 260 V

Z = 2 2LR X

Z = 2 250 120 = 2500 14400

Z = 130

Irms = rmse

Z =

260

130 = 2 A

48. eo = 100 V, f = 50 Hz,

L = 0.2 mH = 0.2 103 H

XL = 2fL

= 2 3.14 50 0.2 103

= 62.8 103

= 0.0628

Io = o

L

e

X

= 100

0.0628 = 1.59236 103

= 1592.36 A 49. ID.C. = 5A, VD.C. = 50 V, VA.C. = 100 V, F = 50 Hz, IA.C. = 2 A When connected to D.C., L = 0

R = D.C.

D.C.

V 50

I 5 = 10

XL = r.m.s.

r.m.s.

e

I

XL = 100

2 = 50

XL = 2fL

L = LX 50

2 f 2 3.14 50

L = 1

6.28 = 0.159 H

6


50. L = 0.5 H, erms = 230 V, f = 50 Hz XL = 2fL XL = 2 3.14 50 0.5 XL = 157

Ipeak = rms

L

e2

X =

230 1.4142.071A

157

51. C = 100 F = 100 106 f, f = 50 Hz.

Xc = 1

2 fC

= 6

1

2 3.14 50 100 10

= 2

1

3.14 10 =

100

3.14

= 31.847 31.85 52. C = 10 F = 10 10–6 F, e = 15 V,

f = 5 kHz = 5 103 Hz

Xc = 3 6

1 1

2 fC 2 3.14 5 10 10

= 3.18

I = c

e

X = 15

3.18 = 4.72 A

53. L = 2 mH = 2 10–3H, e = 2 V,

f = 1 kHz = 103 Hz

f = 1

2 LC or C =

2 2

1

4 f L

C = 22 3 3

1

4 3.14 10 2 10

= 3

61012.6 10 F

78.877

= 12.6 F.

IL = 3 3

L

e 2

X 2 3.14 10 2 10

= 0.159 A.

IC = c

e

X

= 2 2 3.14 103 12.6 10–6

IC = 158.26 103 A = 0.158 A 54. R = 10 , L = 5H, f = 50 Hz, C = 0.4 F = 0.4 10–6 F, erms = 226 V. XL = 2 fL XL = 2 3.14 50 5 XL = 1570

Xc = 1

2 fC

Xc = 6

1

2 3.14 50 0.4 10

Xc = 7.96 103

Z = 2 2L CR (X X )

= 2 210 (1570 7960)

= 7100 4.083 10

Z = 6390

I = rmse 220

Z 6390 = 3.443 102 A

Vl = I. XL = 3.443 102 1570 = 54.05 V

Vc = I.XC = 3.443 102 7960 = 274.06 V

Vr = I.R = 3.443 102 10 = 0.3443 V 55. E = 310 sin 314 t, L = 0.1 H,

C = 25 F = 25 106 F, R = 24 Comparing with

E = eo sin t,

eo = 310 V, = 314 rad/s

= 2f

2f = 314 or f = 314

2

f = 50 Hz

erms = oe

2

= 310

2= 219.2 V

XL = 2fL

XL = 2 3.14 50 0.1

= 31.4

Xc = 1

2 fC

Xc = 6

1

2 3.14 50 25 10

= 1.274 104 106

= 127.4

Z = 2 2L CR (X X )

Z = 2 224 (31.4 127.4)

= 576 9216 = 9792 = 98.95

7


cos = 1

R

Z

= cos1 R

Z

= cos1 24

98.95

= cos1 (0.2425)

= 75.96 76 56. L = 3 H = 3 10–3H, C = 10 F = 10 10–6 F

= 10–5 F, R = 10 , e = 1 V, f = 1kHz = 103 Hz

Z = 2 2L cR (X X )

Z =

2 5 3

23 5

10 (2 3.14 10 3 10

1)

2 3.14 10 10

= 2100 (18.84 15.924)

= 100 8.503

Z = 10.42

I = e 1

=Z 10.42

= 9.59 10–2 A.

el = I XL

el = 9.59 10–2 18.84 = 1.807 V

ec = I XC

ec = 9.59 10–2 15.924 = 1.527 V

er = I R

er = 9.59 10–2 10 = 0.959 V = 0.96 V

57. L = 2

H

, f = 60 Hz, = 45

XL = 2 fL = 2 60 2

= 240

tan = LX

R R = LX

tan

R = 240

tan 45 =

240

1 = 240

58. R = 50 , P = 5000 W P = R 2

rmsI

Irms =P

R =

5000

50= 100 = 10 A

Io = rms2 I 2 10

= 10 2 = 14.1 A

eo = IoR = 14.1 50 = 705 V er.m.s. = 705 1.414 ≈ 500 V

I = 5000

500 = 10A

59. erms = 110 V, R = 11

Irms = oI

2

Io = Irms 2

Io = 10 2 A

Irms = oI

2 =

10 2

2 = 10 A

P = R 2rmsI = 11 102 = 1100 W

60. erms = 100 V, f = 50 Hz, R = 50

erns = 0e

2

0e = 100 2 = 141.4 V

I0 = 0e

R

= 141.4

50 = 2.83 A.

Pav = 2

rmse

R

Pav = 2(100)

50 = 200 W

Energy spent = 200 5 3600 s = 3.6 106 J 61. P = 25 W, e0 = 100 V, P = rms rmse I

Irms = rms

P

e

250.25 2

100

2

0.354 A 62. L = 0.11 mH = 0.11 103 H, R = 15 , C = 60 F = 60 106 F, eo = 240 V

fr = 1

2 LC

fr = 3 6

1

2 3.14 0.11 10 60 10

= 4

1

6.28 0.8124 10 = 0.196 104

= 1960 kHz

8


At resonance, Z = R

I = 0e 240

R 15 = 16 A

63. erms = 0.1 V, L = 100 H = 100 106 H,

C = 200p F = 200 1012 F, R = 2 ,

fr = 1

2 LC

fr = 6 12

1

2 3.14 100 10 200 10

= 7

1

6.28 1.414 10

fr = 0.1126 107 = 1126 kHz

At resonance, Z = R

Irms = rmse

R=

0.1

2 = 0.05 A

XL = 2 fr L

= 2 3.14 1126 103 100 106

= 7.071 105 103

XL = 707.1

VL = Irms XL = 0.05 707.1

= 35.35 V

As, XL = XC at resonance,

VC = 35.35 V 64. R = 10 , L = 0.1 H,

C = 50 F = 50 106 F, eo = 50 V,

f = 50 Hz

XL = 2fL

= 2 3.14 50 0.1 = 31.4

Xc = 1

2 fC

Xc = 6

1

2 3.14 50 50 10

= 6.369 105 106

= 63.69

cos = R

Z

cos = 10

63.69= 0.1570

= cos1 (0.1570) = 80.96

65. L = 100 mH = 100 103 H,

C = 0.1 F = 0.1 106 F, R = 200

At resonance, Z = R = 200

fr = 1

2 LC

fr = 3 6

1

2 3.14 100 10 0.1 10

= 34

1 1010

6.28 10 6.28

= 1.592 103 = 1592 Hz 66. L = 100 mH = 100 103 H, fr = 50 Hz

fr = 1

2 LC

C = 2 2

r

1

4 f L=

2 2

1

4 3.14 50 0.1

= 1

985.96 = 1.014 104 F

67. erms = 100 V, L = 2 mH = 2 103 H,

fr = 1000 kHz = 106 Hz

For resonance, C = 1

L

C = 2 2 2

r

1 1

L 4 f L

= 2 12 3

1

4 (3.14) 10 2 10

= 1.27 1011 F

C = 12.7 pF

IL = rms

L

e

X = rmse

2 fL

= 6 3

100

2 3.14 10 2 10

= 0.1

4 = 7.96 103 A

IL 8 mA

Ic = r.m.s. r.m.s.

c L

e e

x x

Ic = 8 mA

9


68. R = 100 , L = 0.5 H,

C = 100 F = 100 10–6 = 10–4 F,

e0 = 240 V, f = 50 Hz

Z = 2 2L CR (X X )

Z =

2

4

100 (2 3.14 50 0.5

1)

2 3.14 50 10

= 21000 157 31.8471

= 21000 125.153

= 25663.2734

Z = 160.197 = 160.2

Imax = 240

160.2 = 1.498 A

For resonance, fr = 1

2 LC

= 4

1

2 3.14 0.5 10

= 3

1

6.28 7.071 10

fr = 22.519 22.52 Hz

At resonance, Z = R

Irms = oI

2= oe

2 R=

240

2 100

= 1.697 1.7 A 69. L = 60 mH = 60 103 H,

C = 25 F = 25 106 F

fr = 1

2 LC

fr = 3 6

1

2 3.14 60 10 25 10

= 10

1

2 3.14 60 250 10

= 510

2 3.15 60 250

fr = 130 Hz

Hints to Multiple Choice Questions 1. N = 500, A = 103 m3, B = 103 T, = NBA = 500 103 103

= 5 104 weber 2. d = 4 104 20% of 4 104 = 80% of 4 104

= 484 10

10

e = d

dt

= 48 4 10

10 0.4

= 0.8 mV

3. R = 20 , 1 = 0.75 Wb, 2 = 0.25 Wb,

dt = 1

s10

e = d

dt

=

0.25 0.75

1

10

= 0.5 10 = 5 V

I = e

R =

5

20 = 0.25 A

q = It = 0.251

20= 0.0125 C = 12.5 mC

4. R = 50 , t = 2 s

i = R

e =

R

dt/d =

R

1

dt

d(3 t2 – 4t +4)

i = 6 t 4

R

=

6 2 4

50

=

8

50= 0.16 A

5. B = 4 103 T, A = 200 cm2 = 200 104 m2 n = 100, e = 0.2 V

e = t

=

t

nAB

t = e

nAB

t = 4 3100 (200 10 ) 4 10

0.2

= 40 103 = 0.04 s 6. e = 4 105 V, v = 8 m/s, B = 0.5 104 Wb/m2 e = Blv

l = e

Bv=

5

4

4 10

0.5 10 8

= 0.1 m

10


7. Given:

v = 54 km/hr = 54 5

18m/s

l = 1 m, B = 3 104 T e = vlB

= 54 5

18 1 3 104

= 45 10–4 V = 4.5 mV 8. t = 0.4 s, A1 = 62 104 m2, A2 = 100 104 m2, B1 = 2T, B2 = 2.5 T

|e| = 2 1 2 2 1 1B A B A

t t

|e| =4 42.5 (100 10 ) 2 (3.14 36 10 )

0.4

= 4 4250 10 226.08 10

0.4

= 23.92

0.4 104 = 59.8 104

60 104 V 6 mV 9. dI = 4 2 = 2 A, e = 4.4 V, dt = 0.02 s

L =dt

edI

= 4.422 10

2

= 44 mH

10. For steady current, dI

dt= 0

e = M dI

dt = 0

11. M = 3 103H, = 50 rad/s Ip = I0 sin t

dIp

dt = I0 cos t

eS = M dIp

dt = MI cos t

Max. eS = 3 103 8 50 1

( cos t = 1)

= 1.2 = 3.768 V 3.8 V 12. s = 1.2 Wb, dIp = 6 A s = MdIP

M = s

PdI

=

1.2

6 = 0.2 H

13. Join them in parallel for min. inductance

P

1

L=

1 1 1 11

2 4 6 12

Lmin = 12

H11

1.1 H

14. The inductors are joined in parallel

p

1

L=

1 1 1

5 5 5 =

3

3

LP = 5

3= 1.666 1.67 H

15. C = 2 F = 2 106 F, dV/dt = 6 V/s

Displacement current Id = 00

d

dt

but = EA = VA

d

where A = area of the plates d = distance between the plates.

ID = 0

d VA

dt d

= 0A dV

d dt

but C = 0A

d

= capacity of the capacitor

ID = dV

Cdt

= 2 106 6

= 12 106A = 12 A 16. According to property of continuity, Charging current = Displacement current =

0.8 A 17. L = 80 mH = 80 103 H, I = 0.7 A

E = 1

2LI2 =

1

2 80 103 (0.7)2

= 0.0196 J 0.02 J

18. p p

s s

e N

e N

pN240

60 75

Np = 300. 19. There is no change in frequency.

s

p

N

N= 10 : 1, ep = 220 V

p p

s s

N e

N e

10

1 p

s s

V 220

V V

Vs = 220

10 = 22 V

11


20. Vp = 100 volt, Vs = 15 volt, Is = 1 A

s

p

V

V = p

s

I

I

IP = ss

p

VI

V =

151

100

= 0.15A = 150 mA

21. W = 80 W, V = 20 V

Using, Is = W

V=

80

20= 4 A

Efficiency = s s

p p

V I

V I=

20 4

220 0.4

= 0.909 91% 22. N = 400, B = 5 103 T, A = 0.02 m2, t = 0.02 s

e = 2NAB

t

= 32 400 0.02 5 10

0.02

= 4000 103 = 4 V 23. NA = 2000 cm2 = 2000 104 m2 = 0.2 m2,

B = 2

T, n = 500 r.p.m =

500r.p.s.

60 E0 = NAB = 2nNAB

= 2 500

60 0.2

2

= 4.7 V 24. emax = 40 mV = 40 103 V, n = 20, A = 102 m2, B = 0.2 tesla emax = nAB 40 103 = 20 102 20 102 40 103 = 4 102 = 1 rad/s 25. t = 1.6 ms = 1.6 103 s e = 8 cos(2ft)

= 8 cos 32 50 1.6 10

= 8 cos 6

= 8

3

2= 4 3 volt

26. B = 2 102 T, r = 20 cm = 0.2 m, R = 10 ,

= t = 90, f = 120

60 = 2 r.p.s.

Alternating current induced in the coil is given by,

I = Io sin t = 2 fnBA

sin90R

= 2 22 2 1 2 10 0.2

110

3 103 A = 3 mA

27. r = 8 cm = 8 102, B = 0.04 T,

= 40 rad/s

e = 2

1Bl2

e = 2

1 4 103 (8 102)2 40

= 512 106 = 512 V 28. erms = 120 V, f = 50 Hz, R = 20

Irms = rmse

R=

120

20 = 6 A

Imax = 2 Irms = 6 2 A 29. Comparing with

V = V0 sin (t + )

V = 40 sin (80 t + 0.4)

V0 = 40 V 30. V = 220 V, VR = 120 V

220 = 2 2L120 V or 48400 = 14400 + 2

LV

VL = 34000 184 V 31. Eo

= 100 V, C = 2 F = 2 106 F, = 20 r.m.s.

Irms = 2

CE

X

E 0

C

rms .... c

1X

C

= 6100 20 2 10

2

= 2 103 = 2 mA

32. L = LX 157

2 f 2 3.14 100

= 0.25 H

33. XL f

L 2

L 1

2

2

X f 2

X f 1

2LX = 2 20 = 40

12


34. C = 20 F = 20 106 F, f = 50 Hz Impedence of circuit Z = Xc

Z = fC2

1

=

6

1

2 50 20 10 =

500

35. The reactance of a capacitor, Xc = 1

C

= 1

2 fC

If the frequency (f) is increased, the reactance decreases and hence the current (I) increases.

As displacement current (Id) = conduction current, Id will be tripled if the frequency is tripled.

36. Xc = fC2

1

or XC

f

1

c

c

X

X

=

f

f =

50

150 =

1

3

Xc = CX

3 =

24

3 = 8

37. L = 5 H, C = 200 F = 2 104 F The impedance of combination

Z =

fC2

1fL2

= 2 50 5 4

1

2 50 2 10 1570 16 1554 38. R = 40 , XL = 20 , = 200 rad/s,

I = e

Z

Z = e

I =

60

2 = 30

Z2 = R2 + (Xc XL)2 302 = 202 + (Xc –XL)2 (Xc –XL)2 = 900 – 400 = 500

Xc –XL = 10 5 = 22.36 Xc = XL + 22.36 = 20 + 22.36 = 42.36 42

39. Phase difference = 3 2

= 3 2

=

5

6

= 150

40. R = 50 , XL = 200 , Xc = 150

tan = L cX X

R

=

200 150

50

tan = 1 = 45

41. R = 200 , f = 800 Hz Figure below shows the graph for the given

case,

tan = R

L

X

X=

R

L

V

V =

R

L

L = R ....[ tan = tan 45 = 1]

L = R

= f2

R

= 200

2 3.14 800 = 0.0398

= 39.8 10–3 H 40 mH 42. L = 2.1 H, R = 120

= tan12 fL 2 50 2.1

R 120

tan1(5.5)

43. e = 5 sin(t + 90) or i = 2 sint

There is phase difference of 2

between E and

I P = 0 44. Vo = 40 V, Io = 40 103 A

Pavg = 2

cosIV 00

=340 40 10 cos( / 3)

2

= 0.4 W 45. eo = 40 V, Io = 400 mA = 400 103 A Power = erms Irms cos

Power = 40

2

3400 10

2

cos 30

= 8 3

2

= 4 1.73

= 6.92 7 W 46. N = 50, A = 1 m2, = 20 rad/s, B = 4 102 T, R = 400 Maximum power dissipated in the circuit,

Pmax = Erms Irms = 0 0E I

2 2 = 0 0E I

2

VL

VR

V

= 45

13


But I0 = 0E

R

Pmax = 0 0E E

R 2

= 20E

2R but E0 = NAB

Pmax = 2(NAB )

2R

Pmax = 2 2(50 1 4 10 20)

2 400

Pmax = 40 40

2 400

= 2 W

47. V = 3 cos(t) = 3 sin t2

Phase difference between V and I is 2

Power factor = cos 2

= 0

Power dissipated = 0 48. OQ = Irms cos . This contributes for power.

PQ = Irms sin . This does not contribute for power. Hence it is wattless.

Iwattless = Irms sin

3 3 = 6 sin

sin = 3

2

= 60

Power factor = cos = cos 60 = 1

2

49. R = 4 , XL = 3 For an L – R circuit

Power factor (cos ) = 2 2

L

R

R X

= 4

16 9=

4

5 = 0.8

50. L = 0.4 H, C = 4 F = 40 106 F

fr = LC2

1

fr = 5

1

2 3.14 0.4 4 10

= 6

1

6.28 16 10 =

1000

25.12

= 39.8 40 Hz 51. VR = 40 V, R = 800 , = 250 rad/s, C = 5 F = 5 106 F

At resonance, L = 1

Cand Z = R

Current through the circuit,

I = RV

R=

40

800 =

1

20A

Voltage across L is given by VL = IXL = I L

But L = 1

C

VL = I

C =

6

1 / 20

250 5 10 =

5

6

4 10

10

= 40 V

52. VT = 150 V For resonance, VL = VC VT = VR = 150 V 53. V = 12V, I = 3A, f = 50 Hz, e = 30 V

R = V

I and Z =

e

I

R = 12

3 = 4 and Z =

30

3= 10

Power factor = cos = R

Z=

4

10= 0.4

54. L = 0.8 H, C = 5 F = 5 106 F At resonance,

= 1

LC=

6

1

0.8 5 10 =

310

4

= 500 rad/s 55. In series resonance, Z = R = 80 56. In series resonance, P.D. across

L = P.D. across C = 180 V 57. C = 25 F = 25 106 F, = 500 m The oscillating frequency,

f = 1

2 LC 2f =

2

1

4 LC

L = 2 2

1

4 f C ....(1)

but f = velocityof light(c)

wavelength( )

2

f = 2

2

c

=

8 83 10 3 10

500 500

= 0.36 1012

= 36 1010 From eq. (1),

L = 10 6

1

4 10 36 10 25 10

2.8 104 105 2.8 109 H

O

Q

PIrms

Erms

1

Chapter 17: Electrons and Photon


1. Energy of the incident photon, E = hc

i. For violet light,

E1 = hc

= 34 8

9 19

(6.63 10 ) (3 10 )

(390 10 ) 1.6 10

= 3.19 eV. ii. For yellow-green light,

E2 =

34 8

9 19

6.63 10 3 10

550 10 1.6 10

= 2.26 eV. iii. For red light,

E3 =

34 8

9 19

6.63 10 3 10

760 10 1.6 10

= 1.64 eV.

2. = 29 8

34

pc 4.95 10 3 10

h 6.6 10

= 2.25 1013 Hz.

= 8

13

c 3 10

2.25 10

= 1.33 10–5 m.

3. p = 6 19

8

E 2 10 1.6 10

c 3 10

= 1.067 10–21 kg m/s

4. n = P

E

E = hc

n = P

hc

= 3

34 8

20 10 400

6.63 10 3 10

4.02 1031 photons/s.

5. max max1 2(K.E.) (K.E.) = h2 h1

(8.92 2.31) 1019 = h(2 5 1014)

2 = 19

34

6.61 10

6.63 10

+ 5 1014

= 14.97 1014 Hz.

2 = 8

142

c 3 10

14.97 10

= 2.004 107 m = 2004 Å

6. 0 = 0

hc

W

= 34 8

19

6.63 10 3 10

5 1.6 10

2.4863 107 m = 2486.3 Å

7. n = P

hc

= 7

34 8

10 5.986 10

6.63 10 3 10

= 3.0096 1019 photons/s.

8. W0 = 0

hc

0 = 34 8

19

6.63 10 3 10

2 1.6 10

= 6.216 10–7 m = 6216 Å.

9. i. 0 = 0

c

= 8

7

3 10

6 10

= 5 1014 Hz.

ii. W0 = 34 8

7 190

hc 6.62 10 3 10

6 10 1.6 10

= 2.07 eV.

10. = 8

7

c 3 10

6.8 10

= 4.41 1014 Hz

0 = 19

034

W 2.4 1.6 10

h 6.63 10

= 5.79 1014 As 0 , photoelectric emission is not

possible.

Electrons and Photon17

2


11. W0 = 0

hc

h = 0 0W

c

=

19 106.7 1.6 10 1972 10

3 10

= 7.047 1034 J s. 12. W0 = h0

0 = 19

34

2 1.6 10

6.63 10

= 4.83 1014 Hz.

13. W0 = 0

hc

0 = 34 8

19

6.63 10 3 10

1.9 1.6 10

= 6.543 107 m = 6543 Å

(K.E.)max = hc

W0

=34 8

19

6.63 10 3 10

1.9 1.6 10

(1.9 1.6 1019)

= 1.9325 1019 J (K.E.)max = 1.208 eV (K.E.)max = eV0 = 1.208 eV V0 = 1.208 V 14. For 2 1

0 01 2eV eV = hc

1 2

1 1

.

02eV

=

7 7

hc 1 1e 6

e 4 10 6 10

= 6e 34 8 7

19 14

6.63 10 3 10 10

1.6 10 12 10

= 6e 1.04e 02

V = 4.96 V

15. (K.E.)max = hc 0

1 1

= 34 8

19

6.6 10 3 10

1.6 10

7 7

1 1

6 10 8 10

=34 8 7

19 7 7

6.6 10 3 10 1 10

1.6 10 24 10 10

= 0.5156 eV

16. i. eV0 = hc

W0

eV0 = 34 8

19 7

6.63 10 3 10

1.6 10 3.6 10

2

eV0 = 3.45 2 = 1.45 eV V0 = 1.45 V ii. (K.E.)max = eV0 = 1.45 eV

17. hc

= W0 + K.E.max

hc

=

0

hc

+ K.E.max.

= 34 8

197

6.63 10 3 101.56 1.6 10

2.5 10

hc

= 1.0452 1018

= 34 8

18

6.63 10 3 10

1.0452 10

= 1.903 107 m

18. E = hc 0

1 1

E = 6.63 1034 3 108

8 8

1 1

12 10 24 10

= 34 8

19 8

6.63 10 3 10

1.6 10 24 10

= 5.1797 eV

19. h0 = h 1

2 mv2

h0 = 6.63 1034 8.2 1014 1

2 9.1

1031 (6.5 105)2 = 54.366 1020 19.22 1020

0 = 20

34

35.146 10

6.63 10

= 5.3 1014 Hz.

20. For 2 1

max max1 21 2

1 1(K.E.) (K.E.) hc

h = 19 19

78

14

3.62 10 0.972 10

2 103 10

15 10

= 19 14

8 7

2.648 10 15 10

3 10 2 10

= 6.62 1034 J s

3


1 0

hc hc

+ max1

(K.E.)

34 8

70

hc 6.62 10 3 10

3 10

3.62 1019

= 3 1019 J

0 = 34 8

19

6.62 10 3 10

3 10

= 6.62 107 m. = 6620 Å

21. W0 = hc

eV0.

W0 = 34 8

7

6.63 10 3 10

3 10

(1.6 1019 2.1)

= (6.63 1019 3.36 1019) J

= 19

19

3.27 10

1.6 10

eV

= 2.04 eV.

22. v = c

10 = 3 107 m/s.

hc

= W0 +

1

2 mv2

= 4.53 1.6 1019 +

1

2 9.1 1031 9 1014

= 7.248 1019 + 4.095 1016

= 34 8

3 19

6.63 10 3 10

4.102 10 10

= 4.849 1010 m

4.85 Å.

23. (K.E.)max = hc 0

1 1

(K.E.)max = 6.63 1034 3 108

7 7

1 1

15 10 4 10

= 34 8 7

14

6.63 10 3 10 10

4 10

J

= 19

19

4.9725 10

1.6 10

eV

= 3.11 eV

24. (K.E)max = hc 0

1 1

(K.E)max = 6.63 1034 3 108

8 8

1 1

15 10 27.5 10

= 25 8

16

1.989 10 (27.5 15) 10

15 27.5 10

= 6.027 1019 J = 3.767 eV. Also, eV0 = (K.E.)max = 3.767 eV. V0 = 3.767 V 25. For 1 2

max max2 1(K.E.) (K.E.) = hc

2 1

1 1

.

(2.04 0.6) 1.6 1019 = hc 1 2

1 2

h = 19 14

8 7

1.44 1.6 10 3.333 2.4 10

3 10 (3.333 2.4) 10

= 6.58 1034 J s.

26. K.E. = hc

W0

=34 8

7

6.6 10 3 10

5 10

1.9 1.6 1019

= 0.58 eV


1 E = hc

=

34 8

7

6.63 10 3 10

5 10

= 3.978 1019 J

2. n = 3

3134 7

P 66.3 1010

h 6.63 10 10

3. = pc

h=

29 8

34

13.2 10 3 10

6.6 10

= 6 1013 Hz.

4. E = nh

n = E

hc

n

t =

E

t hc

= 7 7

34 8

9 10 6.6 10

6.6 10 3 10

= 3 1012

6. p = E

c =

hc /

c

=

h

=

34

11

6.63 10

10

= 6.63 1023 kg m/s

4


8. K.E. = 1

2 mv2

= 31 12

19

9.1 10 64 10

2 1.6 10

eV

= 182 eV. 9. E = h and p = h/λ

E

p =

h

(h / )

= = c.

10. V0 = 21 mv

2 e =

31 12

19

9 10 16 10

2 1.6 10

= 45 V

11. W0 = 0

hc

=

34 8

10

6.63 10 3 10

3315 10

= 6 1019 J

12. 0AW =

h

e

eV

= 34 14

19

(6.6 10 ) (1.8 10 )

1.6 10

= 0.74 eV

0BW =

34 14

19

6.6 10 2.2 10

1.6 10

= 0.91 eV

Since incident energy 0.825 eV is greater than 0.74 eV and less than 0.91 eV, so photoelectrons are emitted from metal A only.

13. W0 = 0

hc

hc = 0 01 1W = 0 02 2

W .

02 = 0 01 1

02

W

W

=

2.4 5480

4.6

= 2859 Å

14. 0 = 0

hc

W =

34 8

19

6.63 10 3 10

4.5 10

= 4420 Å

15. W0 = 0h

e

=

34 14

19

6.63 10 2.8 10

1.6 10

= 1.16 eV

16. W0 0

00 1 1

0 02 2

W

W

01

02

=

4.8

6.4 =

3

4 =

75

100.

17. W0 0 and W0 0

1

00 B B

0 0A A

hc

W

W h

8

70 B

140 A

3 10W 4.5 10

W 2 10

= 14

14

3 10

4.5 2 10

=

1

3

18. W0 = 0

hc

, i.e., W0

0

1

or 0

0

1

W

0 0 0

0 0 0

W W

W W / 2

= 2

0 = 2 0.

19. (K.E.)max = 0

hc 1 1

e

= 34 8 9

19 18

6.63 10 3 10 180 10

1.6 10 660 480 10

= 0.71 eV 20. For 1 2

(K.E.)1 (K.E.)2 = hc 1 2

1 1

= 34 8 9

19 18

6.63 10 3 10 150 10

1.6 10 450 600 10

= 0.69 eV 0.7 eV. 21. eV0 = E W0 = 11 eV V0 = 11 V.

22. 0 011

hceV W

and 0 02

2

hceV W

0 02 1e V V =

2 1

1 1hc

= 1 2

1 2

hc

02V = 3.11 + 4.8 = 7.91 V

23. Since negative sign of potential only indicates

that it is retarding potential, neglecting it, W0 = eV0 = 11 9 = 2 eV. 24. Max. K.E. of electron = K.E.max = eV0 = 3 eV If E is the energy of the incident radiation and

W0 is the work function then E = K.E.max + W0 = 3 + 6.2 = 9.2 eV Wavelength of incident radiation,

=

hc 12400 eV A 12400

E E eV 9.2

= 1348 Å

The radiation of this wavelength belongs to ultraviolet region.

5


25. 1

2max

1mv

2 = 4 W0 W0 = 3 W0 ….(I)

and

2max2

1mv

2 = 3 W0 W0 = 2 W0 ….(II)

Dividing (I) by (II)

1

2

2

max

max

v

v

= 3

2

1

2

max

max

v 3

v 2

26.

max A 01

max B 02

K.E. h

K.E. h

A 0

B 0

x

y

On solving,

0 = B Ax y

x y

27. E = hc

W0 and 2E =

hc W0

2E E = hc hc

E + hc

=

hc

= hc

E hc

.

28. eV = hc 0

1 1

….(I)

e(4 V) = hc 0

2 1

….(II)

Dividing (II) by (I)

4 = 0

0

2 1

1 1

On solving,

0 = 3

2

29. 2max1

1mv

2 = h (30) h0.

1

2m(8 106)2 = 2h 0. ….(I)

And new frequency = 30 0 = 20.

2max2

1mv

2 = h(20) h0 = h0 ….(II)

Dividing (I) by (II)

26

2max2

8 10

V

= 2

6

6max2

8 10V 4 2 10 m / s

2

.

30. As, 0

hc hc

= eV0

0

hc hc

= e (3V)

0

1 1 3eV

hc

….(i)

and 0

hc hc

= e (1 V)

0

1 1 1eV

2 hc

.…(ii)

solving (i) and (ii), 0 = 4 . 31. eV0 = hv W0

= 34 15

19

6.6 10 10

1.6 10

2.5 eV

= 4.13 2.5 = 1.63 eV V0 = 1.63 V 1.6 V.

1

Chapter 18: Atoms, Molecules and Nuclei

Hints to Problems for Practice 1. rn = r1 n2 r3 = 0.53 1010 32 = 4.77 1010 m

mv = nh

2 r

mv = 34

10

3 6.63 10

2 3.14 4.77 10

6.64 1025 kg m/s

v = 25

31

6.64 10

9.1 10

= 0.7297 106 m/s

a = v2/r

a = 6 2

10

(0.7297 10 )

4.77 10

= 1.116 1021 m/s2

2. p = nh

2 r

p3 = 34

10

3 6.63 10

2 3.14 4.77 10

6.64 1025 kg m/s 3. rn = r1n

2 r2 = 0.53 1010 22 = 2.12 1010 m = 2.12 Å

En = E1/n2

E2 = 2

13.6

2

= 3.4 eV

4. rn n2

1

2

r

r =

2

2

7

2=

49

4

r7 = 2.12 49

4= 25.97 Å

5. rn n2

2

22

3

r 2 4

r 3 9 2

3

d

d= 4 : 9

6. rn = 26.5 107 mm = 26.5 1010 m = 26.5 Å

rn = 26.5

2= 13.25 Å

rn n2

2

12

n

r 1

r n n2 = n

1

r 13.25

r 0.53 = 25

n = 5 7. d1 = 1.06 Å, dn = 16.96 1010 m rn n2

n

1

r

r= n2

n2 = n

1

r

r =

8.48

0.53 = 16

n = 4

8. L = nh

2

L = 341 6.63 10

2 3.14

= 1.056 1034 kg m2/s

9. rn = r1n

2 r2 = 0.53 22 = 2.12 Å

L = nh

2

L = 342 6.63 10

2 3.14

L = 2.11 1034 kg m2/s.

p = L

r

p = 34

10

2.11 10

2.12 10

= 9.952 1025 kg m/s.

10. rn = Kn2 r1 = K (1)2 and rn = K (n)2

n

1

r

r = n2

n = n

1

r

r =

13.25

0.53 = 25

n = 5 11. rn = kn2 r3 = k (3)2 = 9k and r5 = k(5)2 = 25k.

3

5

r 9

r 25

Atoms, Molecules and Nuclei18

2


r3 = 132.5 1011 9

25

= 47.7 1011 m = 4.77 Å

12. T = 2 3

304

4 hn

me

T = 12 2 34 3

31 19 4

4 (8.86 10 ) (6.6 10 )

9.1 10 (1.6 10 )

= 1.51 1019 s 13. T3 = 4.132 1015 s, T n3

4

3

T

T=

3

3

4

3 T4 = 4.132 1015

64

27

= 9.794 1015 s

14. f = v

2 r

v2 = 1v

n=

62.245 10

2

….

1v

n

= 1.1225 106 m/s

f2 = 2

2

v

2 r =

6

10

1.123 10

2 3.14 2.125 10

f2 = 8.415 1014 Hz

15. v = nh

2 mr

v = 34

31 10

1 6.63 10

2 3.14 9.1 10 0.53 10

= 6.63

2 3.14 9.1 0.53 107

v = 2.188 106 m/s

f = v

2 r

f = 6

10

2.188 10

2 3.14 0.53 10

= 162.188 10

2 3.142 0.53

= 6.573 1015 Hz 16. E3 = 1.51 eV,

En 2

e

n

2

42

3

E 3

E 4 or E4 = 1.51

9

16= 0.85 eV

17. E1 = 13.6 eV,

En = l2

E

n

E3 = 13.6

9

= 1.51 e/v

Eexcitation = Ehigher Elower Eexcitation = 1.51 + 13.6 = 12.09 eV 18. P.E. = 27 eV. P.E. = 2 K.E.

K.E. = 1

2 27 = +13.5 eV

T.E. = K.E. T.E. = 13.5 eV. 19. 1 L

= 12.15 nm = 12.15 109 m

2

1 R

n

1 L

1

=

2

R

(1)= R ….(1)

2

1 B

1 R R

42

1 B =

4

R= 4 1 L

= 4 1.215 107 = 4.86 107 = 486 nm Similarly,

For 1 P ,

21 P

1 R R

93

1 P =

9

R= 9 1 L

= 9 1.215 107 = 10.935 107 = 1093.5 nm 20. En n2

n

1

E

E=

2

1

n

E3 = 13.6

9

= 1.51 eV

h = E3 E1

= 34

1.51 ( 13.6)

6.63 10

= 2.92 1015 Hz

3


21. R = 1.094 107 m1 For Balmer series shortest line p = 2, n = wavelength,

i. = R 2 2

1 1

p n

= 1.094 107 2 2

1 1

2

= 71.094 10

4

= 0.2735 107

Bs =

1

= 3.656 107 = 3656 Å

For largest wavelength, p = 2, n = 3

= 72 2

1 11.094 10

2 3

= 1.094 107 5

36

= 0.1519 107

Bs =

7

1

0.1519 10= 6.582 107

= 6583 Å ii. For Paschen series, For shortest wavelength, P = 3, n =

= 2 2

1 1 RR

3 9

1

=

R

9 =

9

R= 8.226 107

= 8226 Å For largest wavelength, P = 3, n = 4

= R 2 2

1 1

3 4

= 7R

144

= 144

7R= 18.804 107

= 18804 Å 22. E1 = 13.6 eV.

En = 12

E

n

For n = 2, E2 = 13.6

4

= 3.4 eV

E3 = 13.6

9

= 1.51 eV

E4 = 13.6

6

= 0.85 eV

23. 2 2

1 1 1R

p n

For 1st member of Balmer series, p = 2, n = 3.

2 2

B

1 1 1R

2 3

5R

36 ….(1)

For 1st member of Lyman series, p = 1, n = 2

L

1

=

2 2

1 1R

1 2

= 3R

4 ….(2)

Dividing eq. (1) eq. (2),

L

B

= 5R / 36

3R / 4=

5R 4 5

36 3R 27

L = B 5

27 = 6563 1010

5

27

L = 1215.14 1010 = 1215.14 Å. 24. For 2nd L2

= 5400 Å member of Lyman’s

series, p = 1, n = 3

Using, 2 2

1 1 1R

p n

we get,

2 2

L2

1 1 1 3RR

1 3 9

….(1)

For member by Lyman series, p = 1, n = 2

L1

1

= R

2 2

1 1

1 2

= 3R

4 ….(2)

Dividing eq. (1) by eq. (2),

L1

L2

=

8R 4

9 3R =

32

27

L1 = L2

32

27 = 5400

32

27

L1 = 6400 Å

25. = 6563 Å, p = 2, n = 3

1

= R

2 2

1 1

p n

10

1

6563 10 =

2 2

1 1R

2 3

or

10

1

6563 10 = R

5

36

R = 1.097 107 m1

4


26. Eexcitation = Ehigher Elower 12.09 = Ehigher (13.6) or Ehigher = 1.51 eV n = 3

27. = R2 2

1 1

p n

i. Shortest wavelength of Lyman Series, p = 1, n =

= 1.094 107 2 2

1 1

1

1

= 1.094 107 1

= 9.141 108 = 914 Å. For longest wavelength, p = 1, n = 2.

72 2

1 1 11.094 10

1 2

= 1.094 107 3

4

= 7

4

3 1.094 10 = 1.219 107

1219 Å. ii. For shortest line of Balmer series, p = 2, n = .

72 2

1 1 11.09 10

2

= 7

4

1.094 10

= 3656 Å For longest line of Balmer series, p = 2, n = 3.

72 2

1 1 11.094 10

2 3

= 1.094 107 5

36

= 7

36

1.094 5 10 = 6581 Å.

iii. For shortest line of paschen series,

p = 3, n = .

72 2

1 1 11.094 10

3

= 1.094 107 1

9

= 7

9

1.094 10 = 8.227 107

= 8227 Å For longest line of paschen series, p = 3,

n = 4.

72 2

1 1 11.094 10

3 4

= 1.094 107 7

144

= 77

14418.804 10

1.094 10 7

= 18804 Å 28. E2 E1 = h = E

E2 (13.6) 12.09

E2 = (12.09 13.6) = 1.51 eV

Electron can be raised to the third Bohr orbit. The possible wavelengths are:

i. for n3 n2

hc

= E3 – E2

= 34 8

3 2

hc 6.63 10 3 10

E E 1.51 ( 3.4)eV

=

26

19

19.89 10

3.4 1.51 1.6 10 J

= 6577 Å ii. for n3 n1

= 3 1

hc

E E =

34 86.63 10 3 10

1.51 13.6 eV

= 8 34

19

6.63 3 10 10

12.09 1.6 10 J

= 1.028 107 m 1028 Å iii. for n2 n1

= 2 1

hc

E E =

34 86.63 10 3 10

3.4 13.6 eV

= 26

19

19.89 10

10.2 1.6 10 J

= 1.2188 107 m = 1219 Å

29. For maximum speed, n = 1 … n

1v

n

En = 4

2 2 20

e m

8 n h

5


E =

419 31

2 2212 34

1.6 10 9.1 10

8 8.85 10 1 6.63 10

= 2.17 1018 J.

30. As En 2

1

n, Emax = E1 for n = 1

For most energetic photon, p = 1, n = .

1

= R

2 2

1 1

p n

1

= 1.094 107

2 2

1 1

1 Co

= 1.094 107

= c

= 3 108 1.094 107

= 3.282 1015 Hz E = h E = 6.63 1034 3.828 1015 = 21.76 1019 J

= 19

19

21.76 10

1.6 10

= 13.6 eV 31. H

= 6563 Å

For H line of Balmer series, p = 2, n = 3.

Using, H

1

= R 2 2

1 1 5R

362 3

….(1)

For H line of Balmer series, p = 2, n = 4.

H

1

= R 2 2

1 1 3R

2 4 16

….(2)


H

H

5R 16 20

36 3R 27

H = 6563

20

27 = 4861 Å

32. For ground state, n = 1.

En = 4

2 2 20

e m

8 n h

E1 =

419 31

2 2212 34

1.6 10 9.1 10

8 8.85 10 1 6.63 10

= 21.65 J = 19

21.65

1.6 10

= 13.53 eV

For second line of Balmer series, p = 2, n = 4.

1

=

2 2

1 1R

p n

1

= 7

2 2

1 11.97 10

2 4

= 1.97 107 3

16

= 16

1.97 3 107 = 2.707 107

= 2707 Å 33. (P.E.)n = 1 = 27.2 eV

P.E. = 2

K

n

(P.E.)1 = 2

K

1 and

(P.E.)n = 4 = 2

K

4

2n 4

2

n 1

(P.E.) 1 1

P.E. 164

(P.E.)n = 4 = 27.2

16

= 1.7 eV

34. E = 12.1 eV 13.6 eV = 1.5 eV E1 = 13.6 eV.

n2 = 1

n

E

E

n2 = 13.6

1.5

9

n = 3 35. Einitial = 0.85 eV = 0.85 1.6 1019 J

Efinal = 10.2 eV = 10.2 1.6 1019

h = Einitial Efinal

hc

= 0.85 1.6 1019 + 10.2 1.6 1019

34 86.63 10 3 10

= 9.35 1.6 1019

= 26

19

6.63 3 10

9.35 1.6 10

= 1.330 107

= 1330 Å

36. En = 12

E

n

For longest wavelength in Lyman series, p = 1, n = 2

6


2 2

1 1 1R

1 2

1 3

R4

= 7

4 4

3R 3 1.097 10

= 1214.66 Å 1215 Å For shortest wavelength in Lyman series

p = 1, n =

2 2

1 1 1R

p n

For shortest wavelength,

1

= R

2 2

1 1

1

or 1

R

or = 7

1 1

R 1.097 10

0.911 107

= 911 Å 37. E1 = 13.6 eV

K.E. = T.E.

= ( 13.6) = 13.6 eV

Also, P.E. = 2 K.E.

= 2 13.6

= 27.2 eV

38. E = hc

E = 34 8

10

6.63 10 3 10

4800 10

= 4.144 1019 J

E = 19

19

4.144 10

1.6 10

= 2.59 eV.

39. E = h E = 6.6 1034 8 1014 = 52.8 1020 J E = mc2

m = 2

E

c

=

20

28

52.8 10

3 10

=

52.8

9 1020 1016

= 5.87 1036 kg.

40. T = 400 years, N = 25% of N0 N = 0N

4

N = 0n

N

2 0N

4= 0

n

N

2 n = 2

Time of disintegration = 400 2 = 800 years 41. T = 1580 years = 1580 365 86400 s

= 0.693

T=

0.693

1580 365 86400

= 1.39 1011 s1

42. N = 0N

16, T = 20 days

N = N0 n

1

2

0N

4= N0

t101

2

t102 = 4 = 22

t

10= 2

t = 20 days

43. N0 = 106, T = 1

2 min = 30 s, t = 15 s

n = t

T

n = 15

30=

1

2

N = N0 n

1

2

N = 106

1

21

2

= 610

2

= 610 2

2

= 61.414 10

2

= 0.707 106

N 7 105

44. = h

p =

h

mv

v = 34

31 10

h 6.63 10

m 9.1 10 5200 10

= 1401 m/s

7


45. = h

mv

= 34

27 6

6.63 10

6.62 10 8 10

= 0.1252 1013 = 1.252 1014 m 46. V = 100 kV = 100 103 V = 105 V

= h

2meV

= 34

31 19 5

6.63 10

2 9.1 10 1.6 10 10

= 3.885 1012 m

47. = h h

p mV

= 346.63 10

0.5 10

= 13.26 1035

= 1.326 1034 m 48. K.E. = 144 eV = 144 1.6 1019 J

K.E. = 21mv

2

v = 2K.E.

m=

19

31

2 144 1.6 10

9.1 10

= 1250.6374 10

v = 7.11 106 m/s

= h

mv

= 34

31 6

6.63 10

9.1 10 7.11 10

= 1.02 1010 m = 1.02 Å

49. = h

2meV

= 11 2

h

2 1.72 10 m V

( e = 1.72 1011 m)

= 34

31 11

6.63 10

9 10 2 1.72 1000 10

= 0.3972 1010 m

= 0.3972 Å

50. rn n2 r1 = K (1)2, r5 (5)2, r10 – K(10)2 and r25 = K (25)2

5

1

r

r= 25 r5 = 25 r1 = 25 0.53

= 13.25 Å

Similarly, r10 = 0.53 100 = 53 Å and

r25 = 0.53 625 = 331.25 Å.

Hints to Multiple Choice Questions 1. rn = n2r1 r2 = 4 0.53 1010 m

f2 = 2

2

v

2 r

f2 = 6

11

2.5 10

2 3.14 4 5.3 10

= 1877 1012 Hz

2. v 1

n and R n2

R 2

1

v

R2 2

1

(v / 3) i.e.,

2

9

v

2

22

R 9 v

R v 1 = 9

R2 = 9 R

3. En = 12

E

n,

n2 = 13.6

3.4

= 4 = n2 n = 2

Angular momentum of the second orbit

= 34h h 6.63 10

22 3.14

= 2.11 1034 J s 4. rn n2, r3 9, r3 = 4.77 Å, r5 25

5

3

r 25

r 9 or r5 =

254.77

9 = 13.25 Å

5. rn n2 0.53 12 and 212 n2

2n

1=

212

0.53

n = 212

0.53= 20

8


6. r n2 and orbital area r2 i.e. A n4

4

14

2

A 1 1

A 2 16

n = 2 gives the first excited state.

7. p 1

n 3

1

p

p=

1

3

p3 = 1p

3=

246 10

3

= 2 1024 kg m/s

8. vn 1

n

10

4

v 4 1

v 10 2.5

v10 = 45 10

2.5

= 2 104 m/s

9. L n L5 5 and L12 12

L 12 = L5 12

5 =

345.525 10 12

5

= 13.26 1034 J s

10. I = q q qv

2 rT 2 rv

I = 19 6

10

1.6 10 2.5 10

2 3.14 0.5 10

= 1.27 mA

1.3 mA

11. L5 L3 = 5h

2

3h

2=

h

= 2.11 1034 J s

12. T = 12 r

v

=

10

6

2 3.14 0.53 10

2.18 10

= 1.527 1016 s 13. rn n2 r4 = 16 r1 = 16 0.53 1010 m

T = 10

46

2 r 2 3.14 0.53 10 16

v 4.28 10

1.24 1015 s 14. T1 : T2 = 8 : 27.

T n3 3132

n8

27 n

1

2

n

n = 3

8 2

27 3

1

2

r

r =

2122

n

n =

4

9

15. n1 = 2, n2 = 3

T = 2 r

v

But r n2 and v

1

n

T1 23 and T2 33

T n2 n or T n3

3

1

2

T 2 8

T 3 27

16. 3

1

n 1

1

1, 3

1

27

3 = 16

1 5.4 10

27 27

= 0.2 1016

= 2 1015 rad/s

17. En = 2

13.6

n

and En = 1.51 eV.

1.51 = 2

13.6

n

n2 = 9 n = 3

18. En 2

1

n

E3 1

9 and E2

1

4

2

3

E 9

E 4

E2 = 3

9E

4 =

9

4 1.51 3.4 eV

19. P.E. = 2 (T.E.) = 2 0.86 = 1.72 eV 20. K.E. = (T.E.) = (3.4) = 3.4 eV 21. E = 13.6 Z2

For He, Z = 2, E = 13.6 4 = 54.4 eV

Ionisation energy = 54.4 eV

22. P.E. 2

1

n

(P.E.)2 1

4 and

(P.E.)4 1

16

(P.E.)2 = 4 P.E.4 = 6.8 eV and

(K.E.)2 = 2(P.E.) 6.8

2 2

= 3.4 eV.

23. For series limit, ni =

For Balmer series, nf = 2, ni =

9


B = 4

R

For Paschen series, nf = 3, ni =

P = 9

R

P = B

9 96400

4 4 = 14400 Å

24. The number of photons emitted (emission lines) when the electron jumps from n = 4 to n = 1 is given by

N = n(n 1) 4 3

2 2

= 6

25. For H, p = 2, n = 3 and H, p = 2, n = 4

H

1 1 1 5RR

4 9 36

H

36

5R

and

H

1 1 1 3RR

4 16 16

H

16

3R

H

H

27

20

26. 2 2

1 1 1R

p n

For transition, p = 1, n = 2

1 1

R 14

=

3R

4

= 4

3R

For transition, p = 1, n = 3

1

= 1

R 19

= 8R

9

= 9

8R

= 9 / 8R

4 / 3R

= 27

32 =

27

32

27. For Lyman series limit, p = 1, n =

2

L

1 1 1R

1

L = 1

R

For Balmer series P = 2, n =

B = 4

R B = 4 L = 3644 Å

28. Longest wavelength for Lyman series = 4

3R

and for Balmer series = 36

5R

B

L

36 3R 27

5R 4 5

As 1

,

B

L

5

27

L

B

27

5

29. For H line p = 2, n = 4 For H line p = 2, n = 5

Using, 2 2

1 1 1R

p n

H = 16

3R, H =

100

21R

H H 3R 100

16 21R =

4690 100

112

H = 4187.5 Å 4200 Å 30. For shortest wavelength in Lyman series,

P = 2, n = 2

B

1 1R 0

2

B = 7

4 4

R 1.097 10

= 3.646 107 m = 3646 Å 31. n = 3, p = 1

N = (n p)(n p 1) 2 3

2 2

= 3

32. E1 = 13.6 eV, n = 2

En = 12

E

n

E2 = 13.6

4

= 3.4 eV

33. R (A)1/3

1/3 1/3

O O

H H

R A 16

R A 4

= 2

34. A = 343 R = R0 A

1/3 = 1.2 1015 (343)1/3

= 1.2 1015 7 = 8.4 1015 m = 8.4 1013 cm.

10


35. Since, E c2,

energy released becomes 1

16

It is reduced by 1 1

16 =

15

16th part

36. Deuteron (1H

3) has two nucleons Binding energy of 2 dueteron atoms = 2 [2 1.3] = 5.2 MeV Helium atom (2He4) has four nucleons. Binding energy of a single helium atom = 4 7.2 = 28.8 MeV The energy released = B.E. of 2He4 B.E. of two deuteron atoms = 28.8 5.2 = 23.6 MeV

37. 4 n

1 1 1

16 2 2 n = 4

t = 1/2 hour = 30 minutes

T =t

n=

30minutes

4= 7.5 minutes

38. If r1 and r2 are the atomic radii and A1 and A2

are the mass numbers of the target metal and 2He4 respectively then,

1/3

1 1

2 2

r A

r A

3

1 1

2 2

A r

A r

= [(15)1/3]3 = 15

But A2 = 4 1A

4= 15

A1 = 60 Z = A No. of neutrons = 60 25 = 35 Atomic number of target atom = 35

39. 0

N 1

N 20

t = 0N2.303log

N=

2.303 4.2log 20

0.693

= 2.303 4.2 1.301

0.693

=

12.584

0.693

= 18.15 18 days. 40. In all other nuclear reactions, both charge

(atomic number) and mass number (A) are conserved.

This condition is satisfied in (A), (B) and (C). But in (D) 11 + 1 9 + 4 (mass numbers are

not equal.) Hence this reaction is not possible.

41. n 12/4 3

0

N 1 1 1 1

N 2 2 2 8

Fraction decayed = N0 N = N0 0N

8= 0

7N

8.

42. By using the conservation of mass number for

the reaction 24 4 x 1

12 2 14 0Mg He Si n

we get, 24 + 4 = x + 1 x = 27 43. For the element X, Binding energy = 200 7.4 = 1480 MeV For A, Binding energy = 110 8.2 = 902 MeV For B, Binding energy = 80 8.1 = 648 MeV Energy released in the reaction = (902 + 648) 1480 = 1550 1480 = 70 MeV 44. Suppose the missing term is ZXA. 13 + A = 10 + 4, A = 1 7 + Z = 5+ 2 Z = 0 It must be neutron. 45. t1 = 10 h, t2 = 20 h.

1t

0

Ne

N

75

100 = e or e =

3

4

and N

100= e2 =

22 3 9

(e )4 16

= 56%

46. vn 1

n v1 = 2v2 = 2 2.1 106 m/s

= 4.2 106 m/s

= 34

31 6

h 6.63 10

mv 9.1 10 4.2 10

= 0.1735 109 m = 1.73 Å

47. = h

p for the electron.

p = h

(from the Broglie’s equation)

for photon, E = h = hc

Both of them have the same wavelength.

p h 1

E hc c

= 8

1

3 10= 3.33 109

11


48. = 12.27

V=

12.27

268 0.75 Å

49. e

v

v= 3 and e

=

4

1

1.624 10=

410

1.624

For the particle, = h

mvor m =

h

v

For the electron, e = e e

h

m v

me = e e

h

v

e e e e

e

v vm h

m v h v

m = 410

1.624311

9.1 103

= 1.867 1027

1.9 1027 kg 50. E = 250 eV = 250 1.6 1019 J

= h h

mv 2mE

= 34

27 19

6.63 10

2 1.876 10 250 1.6 10

= 0.171 1011 m 17 1013 m

51. v = h

m

=34

31 10

6.63 10

9.1 10 2 10

= 3.64 106 m/s 3.6 106 m/s

52. K.E. = eV and K.E. 21mv

2

v = 19

31

2eV 2 1.6 10 84

m 9.1 10

= 12172.810

9.1

= 5.43 106 m/s 53. De-Broglie wavelength of photon is,

= h

p

p = 34

10

6.63 10

5000 10

= 1.326 1027 kg-m/s

54. = h

2mE

2 = 2h

2mE

E = 2

2

h

2m=

34 2

31 10 2

(6.6 10 )

2 9 10 (2 10 )

E = 68

51 19

6.6 6.6 10eV

18 2 2 10 1.6 10

= 0.378 102

E = 37.8 eV 55. For the grain of sand,

= h

mv =

34

6

6.63 10

10 20

( 1 mg = 106 kg)

= 3.315 1029 m = 33.15 1030 m

1

Chapter 19: Semiconductors

Hints to Problems for Practice 1. Resistance in forward bias

Rf = V

I

= 3

0.01

2 10 = 5

2. Rr = V

I

= 6

6.8 4

(13 6) 10

= 2.8

7 106 = 4 105

3. Rr = V

I

= 62

13 8

I 40 10

I2 – (40 10–6) = 3

5

250 10 = 0.02103

I2 = (20 + 40) 106 = 60 A 4. Rs = 150 + 50 = 200 Rp = 200 // 200 = 100

I = p

E 0.7

R

=

6 0.7

100

= 5.3 10–2 A = 53 mA

5. Rd = V

I

= 3

0.01

10 10 = 1

6. I = I1 + I2

I1 = Z

L

V

R =

5

100 =

1

20 A = 50 mA

I = ZE V

R

=

10 5

80

=

1

16A = 62.5 mA

I2 = I – I1 = 12.5 mA 7. For half wave rectifier, finput = foutput = 50 Hz For full wave rectifier, 2 finput = foutput = 120 Hz 8. Voltage drop across resistor VR = E VZ = 6 – 2 = 4 V

Current through series resistor

IR = VR / R = 4

200 = 20 mA.

The smallest value of load resistance for which stabilisation occurs is that which allows a negligible ( 0) zener current.

Current thought the load resistance, IL = IR = 20 mA.

Smallest Value of RL = Z

Z

V

I=

3

6

20 10

= 300 9. 0 c

(I ) = 1.412 mA

IC = 0 cI

2 1 mA [ Irms = 0I

2]

= C

B

I

I =

3

6

10

20 10

= 50

10. Ic = 90% IE

IE = C100I

90 =

10 10

9

11 mA

11. = C

B

I

I =

3

6

3 10

100 10

= 30

IE = IC + IB = (3 + 0.1) mA = 3.1 mA

= 1

= 30

31 = 0.97

12. For C.E. transistor

a.c. = C

B

I

I

= 3

6

10

50 10

= 20

ΔIE = ΔIC + ΔIB = (1000 + 50)106 = 1050 106 A = 1050 A

a.c. = a.c.

a.c.1

= 20

21

13. a.c. = C

B

I

I

= 3

6

3 10

20 10

= 150

14. For C.E transistor:

i. a.c. = C

B

I

I

= 3

6

30 10

30 10

= 100 E

R VZ

Semiconductors19

2


ii. Ri = BE

B

V

I

= 3

6

30 10

30 10

= 1000

iii. gm = a.c.

iR

=

100

1000 = 0.1

iv. VA = o

i

R

R a.c.

= 3

3

5 10

10

100 = 500

15. For C. B. transistor,

Current gain = 1

= 75

76

= C

E

I

I IC = IE =

75

76 5 = 4.93 mA

16. = 1

= 100

101 = 0.99

17. VA = o

i

R

R =

35000

2050 50 = 853.66

18. a.c. =

o a.c.

i a.c.

V

V

o a.c.V = 80 0.4 = 32 mV

19. = C

E

I

I

IE = 2

0.98 = 2.0408 mA

IB = IE – IC = 0.0408 mA = 40.8 A 20. For a C. E. amplifier,

VA = o

i

R

R =

5

2 20 = 50


1. D2 in reverse bias will have infinite resistance

and will not conduct R = Resistance of D1 + 70 + 200 = 300.

I = V

R =

12

300= 0.04 A

2. R = V

I =

D

3

e.m.f . V

6 10

= 3

4 0.4

6 10

= 600 3. Diode being in reverse biased mode will not

conduct so A2 will read 0.

I1 = 5V

20 = 0.25 A so A1 will read 0.25 A

4. VB > VA hence, p type of diode should be connected towards B and n-type towards A.

I = B A

0

V V 1 ( 6)

R R 50 200

= 0.02 A = 20 mA

5. E = h

= 19

34

2 1.6 10

6.62 10

= 4.8 1014 Hz

5 1014 Hz. 6. Potential difference across resistor is V = 1 (3) = 2 V

I = V 2V

R 100

= 20 mA.

7. Reverse resistance = 6

V 1

I 0.5 10

=2106 .

8. Forward resistance =3

V 0.7 0.6

I (15 5) 10

= 3

0.1

10 10 = 10

9. R = effective3

V 12 0.6=

I 2×10

=

3

11.4

2 10 = 5.7103

10. Band gap = 2.5 eV

Corresponding wavelength = g

hc

E

34 8

19

6.63 10 3 10

2.5 1.6 10

= 4960 Å

Nearest option is (C). 11. As nenh = 2

in

ne = 216 32

i22 3

h

10 mn

n 5 10 m

= 2 109 m3

12. The intrinsic concentration of electron-hole

pairs is given by,

2in = nenh

ne = 2i

h

n

n =

219

21

10

10 = 1017/m3

13. Voltage across RL = 5 V

I = 3

L

V 5V

R 10

= 5 mA

14. For full wave rectifier f0 = 2 fi = 120 Hz. 15. For half wave rectifier f0 = fi = 80 Hz. 16. d. c. output Vo = Id RL = 30 100 = 3000 mV = 3 V

3

Chapter 19: Semiconductors

17. RL = o3

d

V 6

I 50 10

= 120

18. VA = o

i

V

V V0 = 60 30 mV

= 1800 mV = 1.8 V

19. VA = o

i

R

R = 40

800

200

= 160

20. = 99

1 100

= 0.99

21. = 50 / 51 50 51

501 51 1151

= 50

22. IC = 90% IE = 0.9 IE and IB = IE IC = 0.1 IE = 1.2 mA. 24. ΔIB = ΔIE ΔIC = 6.86 6.20 = 0.66 mA = 660 A

25. = 1

1 1

=

1 1

=

1 1

= 1

26. VA = o

i

R

R

= 50 100

600

= 50/6

PA = o

i

R

R 2 =

2600 50

100 6

= 416.67

Nearest answer is 420.

27. f = 1

2 LC

1

2

f

f= 2 2

1 1

1L C

L C

= 1/2 1/2

2 2

1 1

L C 2L 4C

L C L C

= (8)1/2 = 2 2

f2 = 1f

2 2

f2 = f

2 2 [ f1 = f]

[Note: Using shortcut (4)]

28. Voltage gain = Resistance gain Current gain

o o

i i

V R

V R 50

Vo = 4000 10

50500 1000

= 4 V

29. Af = A

1 A

10 = A

1 0.09A

10(1 + 0.09 A) = A 10 = 0.1 A A = 100 30. As per Berkhausen criterion, A = 1

1

Chapter 20: Communication System


1. d = 62Rh 2 6.4 10 150

= 4.382 104 m

=

6

22 4

50,00,000 5 10

d 4.382 10

= 8.288510–4/m2

= 828.85 / km2.

2. 3

t r 6

d 40 10h h

2R 2 6.4 10

= 11.18

rh 11.18 70.3 2.795

hr = 7.815 m

3. h = 232

6

25 10d

2R 2 6.4 10

= 48.83 m

4. i. d = 62Rh 2 6.4 10 144

= 42.933 km ii. population covered = d2 = 103 (42.933)2 = 57.91 lakhs iii. For d = 2d.

h = 2 22d 2d

2R R

=3 2

6

2 (42.933 10 )

6.4 10

= 576.01 m. Increase in height = h – h = 432.01 m.

5. d1 = 12h R

On increasing the height

h2 = 121

100 h1

d2 = 2 1

1212h R 2 h R

100

= 1 1

11 112h R d

10 10

d2 – d1 = 1 1 1

11 1d d d

10 10

Hence, the percentage increase in the transmission range of the TV tower,

2 1 1

1 1

d d d /10100 100 10%

d d

6. i. For Am = Ac Amax = Ac + Am = 2 Ac

Amin = Ac – Am = 0.

m = max min c

max min c

A A 2A 01

A A 2A 0

ii. For Am = 1.5 Ac Amax = 2.5 Ac, Amin = –0.5 Ac

m = c c

c c

2.5A 0.5A

2.5A ( 0.5A )

= 1.5

7. l = 8

6

c 3 10

4 f 4 30 10 4

= 2.5 m.

8. d = 62Rh 2 6.4 10 300

62 km.

As 80 km > 62 km, given both frequencies cannot be propagated via space wave propagation.

Fc = 12max9 N 9 10 9MHz.

So frequency signal lesser than Fc i.e. 7.5 MHz can be propagated via sky wave propagation.

75 MHz being greater than Fc, satellite communication is used.

Hints to Multiple Choice Questions 1. Area covered = d2 = 2

2Rh = 2 Rh

= 22

7 2 (6.4 106) 80 m2 = 3217 km2.

2. Fc = 9 (Nmax)

1/2

Nmax =

2 2c c

F F

9 81

= 2612 10

81

= 1.78 1012 m3

Communication system20

2


3. l = 8

6

c 3 10

4 f 4 20 10 4

= 3.75 m

4. fLSB = fc – fm = 2510 – 12 = 2498 kHz. fUSB = fc + fm = 2510 + 12 = 2522 kHz. 5. Fc = 9 (Nmax)

1/2

Nmax = 2c2

F

9 =

2710

81 = 1.2 1012 m3

6. BW = f2 – f1 = 3100 – 300 = 2800 Hz.

7. m = m

c

A

A Am = m Ac

= 0.75 20 = 15 V

8. m = max min

max min

A A

A A

= 12 4 8

0.512 4 16

9. fm = 400 Hz to 1 kHz = 0.4 kHz to 1 kHz. fUSB = 50 + 0.4 to 50 + 1 = 50.4 kHz to 51 kHz fLSB = 50 – 1 to 50 – 0.4 = 49 kHz to 49.6 kHz 10. d = 2Rh

= 2

Population covered

πd 6

30,00,000

2 6.4 10 125

= 5.9683 10–4 / m2 = 596.83 / km2

597 / km2

11. d = 2Rh

h = 2 3

6

d (55 10 )

2R 2 6.4 10

= 236.33 m

Minimum height of tower is 236.33 m. so from given options nearest value is 238 m.

SAMPLE C

ONTENT

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