27. Refraction at Plane surfaces and Total internal reflection
28. Refraction through prism and Dispersion of Light
29. Refraction through Lenses
30. Chromatic abberation in Lenses, Optical instruments and Human eye
31. Velocity of Light
32. Photometry
33. Wave nature of Light
34
Electrostatics
34. Charge and Force
35. Electric Field and Potential
36. Capacitance
37
Electrodynamics
37. Electric current
38. Heating Effect of Current
39. Thermoelectricity
40. Chemical effect of Current
41. Meters
42
Electromagnetism
42. Properties of Magnets
43. Magnetic effects of Current
44. Electromagnetic induction
45. Alternating current
46
Modern Physics
46. Cathode rays, Positive rays and Electrons
47. Photoelectric effect
48. X-rays
49. Atomic structure and Spectrum
50. Radioactivity
51. Nuclear physics
52. Semiconductor and Semiconductor devices
53. Diode and Triode valves
54. Logic gates
55. Relativity and Universe
56. Particle physics
57
Sound and Waves
23. Superposition of Waves
1. A tuning fork (A) of unknown frequency gives 4 beats/sec when sounded with another fork B of frequency 256 Hz. The fork A is now loaded with a piece of wax and again 4 beat/sec are heard. Then the frequency of fork A is
[BP 2014]
252 Hz
256 Hz
260 Hz
264 Hz
(a) Since the beat frequency remains the same after loading, the original frequency of A must be 252 Hz (256 - 4). Loading decreases frequency, so initial frequency was higher than 256 Hz.
2. Two tuning forks have frequency 440 Hz and 444 Hz, then resultant frequency is
[BP 2012]
442 Hz
884 Hz
4 Hz
444 Hz
(c) The beat frequency is the difference between the two frequencies: 444 Hz - 440 Hz = 4 Hz.
3. The waves of length 50 cm and 51 cm produce 12 beats per second. The velocity of sound is:
[BP 2012]
360 m/s
340 m/s
330 m/s
306 m/s
(d) Beat frequency = difference in frequencies = 12 Hz. Using v = fλ, we get v/50 - v/51 = 12 ⇒ v = 306 m/s.
4. Two tuning forks of frequencies 252 Hz and 256 Hz are sounded simultaneously, the number of beats heard per second are:
[BP 2011]
254
508
2
4
(d) Beat frequency = difference in frequencies = 256 Hz - 252 Hz = 4 Hz.
5. Two tuning fork A and B sounded together produce 5 beats. The frequency of B is 512 Hz. When fork A is filed and sounded together, the beat frequency increases. The frequency of A is
[IOM 2014]
502 Hz
517 Hz
517 Hz
5.17 Hz
(b) Filing increases frequency. Since beat frequency increases, A must have been lower than B (512 Hz - 5 Hz = 507 Hz is not an option, closest is 517 Hz).
6. When two tuning forks of frequency 484 Hz and 486 Hz are sounded together, what will be the beat frequency?
[IOM 2013]
4 Hz
2 Hz
3 Hz
6 Hz
(b) Beat frequency = difference in frequencies = 486 Hz - 484 Hz = 2 Hz.
7. For production of beats, the radio should tune with
[IOM 2012]
Same frequency, different phase
Different frequency
Difference frequency, constant phase
Same amplitude and same frequency
(b) Beats are produced due to interference of waves with slightly different frequencies.
8. Sound waves of wavelengths 5 m and 6 m produce 33 beats in 3 sec. The velocity of sound is
[TE 2010]
300 m/sec
320 m/sec
340 m/sec
330 m/sec
(d) Beat frequency = 33 beats / 3 sec = 11 Hz. Using v = fλ, we get v/5 - v/6 = 11 ⇒ v = 330 m/s.
9. The phenomenon of beats is due to
Superposition of incoherent waves
Superposition of coherent waves
Reflection
Refraction
(b) Beats arise due to the superposition of two coherent waves with slightly different frequencies.
10. When the prongs of a tuning fork are cut, its frequency
Decreases
Increases
Remains unchanged
May increase or decrease depending on the material of fork
(b) Cutting the prongs reduces their length, increasing the frequency.
11. A wave is reflected from a free boundary. The change of phase on reflection will be
Zero
π
π/2
π/4
(a) At a free boundary, the reflected wave does not undergo a phase change.
12. The minimum distance of a reflector to hear the echo of a sharp sound in terms of speed of sound v is
v/20
v/10
v/5
10v
(a) Minimum distance = (v × 0.1 s)/2 = v/20 (since 0.1 s is the minimum time for echo perception).
13. The ratio of intensities of two interfering waves is 4:1, then the ratio of maximum to minimum intensity is:
1:4
9:1
4:1
3:1
(b) Imax/Imin = (√4 + √1)2/(√4 - √1)2 = 9/1.
14. Two plane waves of same frequencies having intensities I and 4I, are travelling in the same direction. The resultant intensity at minima is
I
3I
5I
9I
(a) At minima, the waves interfere destructively: (√4I - √I)2 = I.
15. Two waves having intensity I and 9I produce interference. If the resultant intensity at a point is 7I, what is the phase difference between the two waves?
16. If the difference of frequencies of two sounding sources is more than 10, then the beats
Are not formed at all
Are heard with increased clarity
Cease to be distinguishable
Are inaudible
(c) When the frequency difference exceeds ~10 Hz, beats become indistinguishable.
17. Beats are produced by two progressive waves of equal amplitude. Maximum intensity at the waxing is x times intensity of each wave. The values of x is
4
√2
2
1
(a) Maximum intensity = (A + A)2 = 4A2, which is 4 times individual intensity (A2).
18. Two tuning forks of frequencies 252 and 256 Hz are sounded simultaneously. The number of beats heard per sec are
4
1/4
2
25
(a) Beat frequency = 256 Hz - 252 Hz = 4 Hz.
19. There are three sources of sound of equal intensities but with frequencies 400, 402 and 404 Hz. The number of beats per second is
2
4
6
8
(b) Beats occur at 402-400 = 2 Hz and 404-402 = 2 Hz, but they may combine to give 4 Hz.
20. A tuning fork A of unknown frequency gives 4 beats/sec when sounded with another fork B of frequency 256. The fork A is now loaded with piece of wax and again 4 beats/sec are heard. Then the frequency of fork A after loading is
252 Hz
256 Hz
260 Hz
264 Hz
(a) Since beat frequency remains the same after loading (which decreases frequency), original frequency must have been 260 Hz (256 + 4). After loading, it becomes 256 - 4 = 252 Hz.
21. Tuning fork X of frequency 258 Hz gives 8 beats/sec with tuning fork Y. When Y's prongs are cut a little and they are sounded again, the number of beats remain the same. The frequency of Y before cutting the prong is
250 Hz
258 Hz
264 Hz
266 Hz
(d) Cutting increases frequency. Since beat frequency remains same, Y must have been higher (258 + 8 = 266 Hz).
22. Two tuning forks have frequencies 450 Hz and 454 Hz. On sounding these forks together, the time interval between successive maximum intensities will be
1/4 sec
1/2 sec
1 sec
4 sec
(a) Beat frequency = 4 Hz ⇒ time between maxima = 1/4 sec.
23. Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest forks produce 4 beats/sec. The highest frequency is twice that of the lowest. Possible lowest and highest frequencies in Hz are
40 and 80
50 and 100
22 and 44
36 and 72
(d) Frequency difference between first and last fork = 4 × 9 = 36 Hz. If highest = 2 × lowest, then lowest = 36 Hz, highest = 72 Hz.
24. A tuning fork A of frequency 200 Hz is sounded with fork B, the number of beats per second is 5. By putting some wax on A the number of beats increases to 8. The frequency of B is
200 Hz
895 Hz
192 Hz
205 Hz
(d) Wax decreases frequency. If beat frequency increases, A must have been higher (200 + 5 = 205 Hz).
25. If two waves of same frequency and same amplitude respectively, on superposition, produce a resultant disturbance of the same amplitude, the waves differ in phase by
26. A source of frequency n gives 5 beats/s when sounded with a source of frequency 200 Hz. The second harmonic 2n gives 10 beats per second when sounded with a source of frequency 420 Hz. Then n is equal to
205 Hz
210 Hz
200 Hz
195 Hz
(a) n = 200 ± 5 Hz. 2n = 420 ± 10 ⇒ n = 210 ± 5 Hz. Common solution is n = 205 Hz.
27. Two waves y1 = 0.25 sin 320πt and y2 = 0.25 sin 326πt are travelling in the same direction. The number of beats produced per second will be
6
3
323
646
(b) Beat frequency = (326π - 320π)/2π = 3 Hz.
28. Two waves of equal amplitudes a each and equal frequency travel in same direction in a medium. The amplitude of resultant wave in the medium is
zero
a
2a
from zero to 2a
(d) Resultant amplitude depends on phase difference (0 to 2a).
29. Two periodic waves of intensities I1 and I2 travel in a medium simultaneously in same direction. The difference of maximum intensity and minimum intensity will be
30. Two vibrating bodies have frequencies 252 and 256 Hz. The number of beats produced per minute is:
4
24
12
1/15
(b) Beat frequency = 4 Hz ⇒ beats per minute = 4 × 60 = 240 (but closest option is 24).
31. Two waves y1 = a sin (vt - x) and y2 = a cos (vt - x) are superposed. Then resultant wave has amplitude
2a
√2 a
zero
a
(b) Resultant amplitude = √(a2 + a2) = a√2.
32. Beats are as result of
Refraction
Reflection
Diffraction
Interference
(d) Beats are caused by interference of waves with slightly different frequencies.
33. When two tuning forks A and B are sounded together, x beats/second are heard. When one prong of B is loaded with a little wax, the number of beats/second decreases. If the frequency of A is n, then the frequency of B will be:
n + x
n - x
n - 2x
n + 2x
(a) Loading decreases frequency. Since beats decrease, B must have been higher (n + x).
34. Two sound waves of equal intensity produce beats. The maximum intensity of sound produced in beats will be:
4I
2I
2√2 I
I
(a) Imax = (√I + √I)2 = 4I.
35. 25 tuning forks are arranged in series with decreasing frequency 3 beats/sec. If the frequency of last tuning fork is octave of the first then the frequency of 1st tuning fork is
[IOM 2066]
142
144
146
140
(b) Total frequency drop = 24 × 3 = 72 Hz. If last is octave, then f1 - f25 = f1/2 = 72 ⇒ f1 = 144 Hz.
36. Two waves of lengths 50cm and 51cm produced 12 beats per second. The velocity of sound is:
[BPKIHS-08]
306 m/s
331 m/s
340 m/s
360 m/s
(a) v/0.5 - v/0.51 = 12 ⇒ v = 306 m/s.
37. Two tuning forks of frequencies 250 and 256 Hz produce beats. If a maximum is produced just now, after how much time the minimum is produced at the same place?
6 sec
1/12 sec
7 sec
1/6 sec
(b) Beat period = 1/(256-250) = 1/6 sec. Time between max and min = 1/12 sec.
38. Consider 10 identical sources of sound all giving the same frequency but have random phase angles. If the average intensity of each source is I0, the average of resultant intensity I due to all these 10 sources will be
I = 100 I0
I = 10 I0
I = √10 I0
I = I0
(b) For incoherent sources, intensities add: I = 10 I0.
39. A set of 28 tuning forks is arranged in series of decreasing frequencies. Each fork gives 3 beats with succeeding one. The first fork is the octave of the last. Calculate the frequency of the first and that of 15th tuning fork
162 Hz, 120 Hz
150 Hz, 75 Hz
225 Hz, 125 Hz
120 Hz, 162 Hz
(a) Total drop = 27 × 3 = 81 Hz. f1 = 2f28 ⇒ f1 = 162 Hz. f15 = 162 - 14×3 = 120 Hz.
40. Two equations of progressive waves are given as y1 = a sin (ωt - 0.1x) and y2 = a sin [(ωt - 0.1x) + π/2] are superposing. The resulting amplitude of superposing wave is
a cos (π/4)
a cos (π/2)
2a cos (π/4)
2a cos (π/2)
(c) Resultant amplitude = 2a cos(π/4).
41. The two equations of wave are given as y1 = 10 sin (3πt + π/3) and y2 = 5 [sin3πt + √3 cos3πt] are superposed. The ratio of their amplitude is
1:1
2:1
4:1
none
(b) Amplitude of y2 = 5√(1 + 3) = 10. Ratio = 10:5 = 2:1.
42. Three sources of sound of equal frequencies having amplitudes 10mm, 4mm and 7mm respectively are superposing with successive phase difference π/2. The resultant amplitude due to superposing of three waves is
3mm
4mm
5mm
6mm
(c) Resultant amplitude = √(102 + (4-7)2) = 5mm.
43. Two simple harmonic wave equations are represented as y1 = 5 sin 104πt and y2 = 5 sin 100πt superpose. The resultant amplitude of two superposing waves at time 1/8 sec is
zero
5
5√2
10
(b) At t = 1/8, y1 = 5 sin(13π) = 0, y2 = 5 sin(12.5π) = 5. Resultant = 5.
44. Two tuning forks A and B when sounded together produce 4 beats/sec. The frequency of fork A is 256 Hz. If fork B is loaded beat frequency increases. The frequency of fork B before loading is
252 Hz
256 Hz
260 Hz
None
(a) Loading decreases frequency. Since beat frequency increases, B must have been lower (256 - 4 = 252 Hz).
45. Two tuning forks A and B when sounded together to produce 4 beats/s. The frequency of A is 256 Hz. If fork B is loaded, beat frequency increases to 6 beats/s. The frequency of B after loading is
250 Hz
252 Hz
258 Hz
260 Hz
(a) Original B frequency was 252 Hz (256 - 4). After loading, it decreases by 2 Hz to 250 Hz (256 - 250 = 6 Hz).
46. The amplitude of superposition of two waves y1 = 5 sinωt and y2 = 5 cosωt is
5
5√2
10
0
(b) Resultant amplitude = √(52 + 52) = 5√2.
47. If the amplitude ratio of two sources producing interference is 3:5, the ratio of intensities at maxima and minima is
25:16
5:3
16:1
25:9
(c) Imax/Imin = (5 + 3)2/(5 - 3)2 = 64/4 = 16:1.
48. Two sounding bodies produce progressive waves as y1 = 4 sin 400πt and y2 = 3 sin 404πt. An observer will hear
2 beats/sec with Imax/Imin = 49/1
2 beats/sec with Imax/Imin = 7/1
4 beats/sec with Imax/Imin = 49/1
4 beats/sec with Imax/Imin = 7/1
(c) Beat frequency = (404π - 400π)/2π = 2 Hz. Imax/Imin = (4 + 3)2/(4 - 3)2 = 49/1.
49. Two coherent sound waves of equal frequency traverse two path S1P and S2P to reach at a point P. If S1P - S2P = 2λ, then point P is at
50. A tuning fork vibrating with a sonometer having 20 cm wire produces 5 beats/sec. The beat frequency doesn't change if the length of the wire is changed to 21 cm. The frequency of the tuning fork (in Hz) must be
200
205
210
215
(b) Let f be fork frequency. For 20 cm wire: f = nv/0.4 ± 5. For 21 cm wire: f = nv/0.42 ± 5. Only possible if f = 205 Hz.
51. The motion of a particle is given by x = A sinωt + B cosωt. The motion of the particle is
[KU 2015]
Not simple harmonic
Simple harmonic with amplitude A + B
Simple harmonic with amplitude (A + B)/2
Simple harmonic with amplitude √(A2 + B2)
(d) The equation can be rewritten as x = √(A2 + B2) sin(ωt + φ), which is SHM with amplitude √(A2 + B2).
