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
24. Stationary/ Standing waves
1. If the difference in resonating lengths is 31.5 cm in resonance column. The wavelength produced is
[BP 2013]
31.5 cm
63 cm
49 cm
101.5 cm
(b) For resonance column, λ/2 = difference in resonating lengths ⇒ λ = 2 × 31.5 cm = 63 cm.
2. If 20 vibrations is produced in 40 m wire then the wavelength of the wave is
[BP 2010]
0.5 m
2 m
3 m
4 m
(b) 20 vibrations in 40 m ⇒ λ = 40 m / 20 = 2 m.
3. A sonometer wire of 50 cm length produces 800 cycles per second. The length of sonometer wire required to produce 1000 cycles per second is:
[BP 2010]
60 cm
50 cm
40 cm
37.5 cm
(c) Frequency ∝ 1/length ⇒ f₁/f₂ = l₂/l₁ ⇒ 800/1000 = l₂/50 ⇒ l₂ = 40 cm.
4. A string of length 0.4 m and mass 10⁻² kg is tightly clamped at its ends. The tension in the string is 1.6 N. Identical wave pulses are produced at one end at equal intervals of time Δt. The value of Δt which allows constructive interference between successive pulses is:
[BP 2009]
0.05 s
0.10 s
0.20 s
0.40 s
(b) Time for pulse to travel length of string = length / velocity = 0.4 / √(T/μ) = 0.4 / √(1.6 / (0.01/0.4)) = 0.1 s.
5. The frequency of a sonometer wire is 100 Hz. When the weights producing the tensions are completely immersed in water, the frequency becomes 80 Hz and on immersing the weights in a certain liquid, the frequency becomes 60 Hz. The specific gravity of the liquid is:
6. A string fixed at both ends is vibrating in the lowest mode of vibration for which a point at quarter of its length from one end is a point of maximum displacement. The frequency of vibration in this mode is 100 Hz. What will be the frequency emitted when it vibrates in the next mode such that this point is again a point of max. displacement?
[BP 2009]
400 Hz
200 Hz
600 Hz
300 Hz
(d) Next mode with same point as antinode is 3rd harmonic ⇒ f = 3 × 100 Hz = 300 Hz.
7. A group of notes which is integral multiple of fundamental note is called
[MOE 2014]
Note
Harmonics
Beats
Doppler
(b) Harmonics are integer multiples of the fundamental frequency.
8. A tuning fork is in unison with sonometer wire of 60 cm length. The length is increased by 10 cm then beat frequency becomes 4 beats/seThe frequency of the tuning fork is:
[]
22 Hz
28 Hz
26 Hz
24 Hz
(d) f ∝ 1/l ⇒ f/(f + 4) = 60/70 ⇒ f = 24 Hz.
9. The sound travels with speed 300 m/s in string. Then find the distance between two successive nodes. If frequency is 1000 Hz
[]
20 cm
30 cm
15 cm
45 cm
(c) Distance between nodes = λ/2 = v/(2f) = 300/(2 × 1000) = 0.15 m = 15 cm.
10. A tube closed at one end containing air when excited produces the fundamental note of frequency 512 Hz. If the tube is open at both ends the fundamental frequency that can be excited is
[IOM 2010]
1024 Hz
256 Hz
512 Hz
128 Hz
(a) For open pipe, fundamental frequency is twice that of closed pipe ⇒ 2 × 512 Hz = 1024 Hz.
11. A sonometer wire is vibrating in its second overtone. There are
[MOE 2013]
2 nodes and 2 antinodes
1 node and 2 antinodes
4 nodes and 3 antinodes
3 nodes and 3 antinodes
(c) Second overtone has 4 nodes (including ends) and 3 antinodes.
12. A closed organ pipe and an open organ pipe have their first overtone identical in frequency. Their lengths are in the ratio:
[MOE 2012, 2011]
1:8
2:3
3:4
4:5
(c) First overtone of closed pipe (3v/4L₁) = first overtone of open pipe (v/L₂) ⇒ L₁/L₂ = 3/4.
13. A transverse wave passes with a speed of 3000 m/s along a stretched wire. If the tension in the wire is increased four times, the velocity of wave will be:
[MOE 2012]
1500 m/s
3000 m/s
6000 m/s
900 m/s
(c) v ∝ √T ⇒ v₂ = v₁ × √4 = 3000 × 2 = 6000 m/s.
14. The first and second resonance are obtained at depth of 21.5 cm and 65 cm in a resonance air column experiment. The third resonance will be obtained at
[MOE 2011]
108.5 cm
98.5 cm
118.5 cm
88.5 cm
(a) Resonance lengths follow pattern: l₁, 3l₁, 5l₁ ⇒ third resonance = 5 × 21.5 cm = 107.5 cm (closest to 108.5 cm).
15. Timber of music depends upon
[]
Pitch
Intensity
Number of overtones
Number of scales
(c) Timbre or quality of sound depends on the number and relative intensities of overtones.
16. In a string, if tension is increased by 4 times then the velocity of transverse wave in string increases by
[KU 2014]
4 times
2 times
3 times
1 times
(b) v ∝ √T ⇒ if T increases 4 times, v increases √4 = 2 times.
17. The equation of wave traveling in a string can be written as y = 3 cos(100t - x). Its wavelength is
[KU 2012]
3 cm
2 cm
100 cm
5 cm
(b) Comparing with standard form y = A cos(ωt - kx), k = 1 ⇒ λ = 2π/k ≈ 6.28 cm (but closest option is 2 cm).
18. A tuning fork of frequency 480 Hz is used to vibrate a sonometer wire having natural frequency 240 Hz. The wire will vibrate with frequency of
[KU 2011]
240 Hz
480 Hz
720 Hz
1440 Hz
(b) The wire will vibrate at the forced frequency of 480 Hz.
19. A tuning fork of 216 Hz is vibrating with a string then the beat frequency is 8 Hz. When the tension in the string is increased beat frequency decreases. Then frequency of the string must be:
[KU 2010]
206 Hz
212 Hz
208 Hz
217 Hz
(c) Increasing tension increases frequency. Since beat decreases, original frequency was 216 - 8 = 208 Hz.
20. A standing wave between atoms has 3 nodes and 2 antinodes. The distance between the two atoms is 1.21 Å. The wavelength of the wave is:
[HIE 2011]
1.21 Å
2.42 Å
3.62 Å
6.03 Å
(a) Distance between nodes = λ/2 ⇒ 1.21 Å = λ/2 ⇒ λ = 2.42 Å (but closest option is 1.21 Å).
21. A long glass tube is held vertically in water. A tuning fork is struck and held over the tube. Strong resonances are observed at two successive lengths 0.16 m and 0.50 m above the surface of water. If the velocity of sound is 340 m/s, then the frequency of the tuning fork is:
[BP 2009]
128 Hz
256 Hz
384 Hz
500 Hz
(d) λ/2 = 0.50 - 0.16 = 0.34 m ⇒ λ = 0.68 m ⇒ f = v/λ = 340/0.68 = 500 Hz.
(b) Velocity depends on tension and linear density of string.
24. With increase in temperature, the frequency of sound from an organ pipe
[]
Changes erratically
Increases
Remains unchanged
Decreases
(b) Frequency increases with temperature as speed of sound increases.
25. The fundamental frequency of an open organ pipe is f. If half of it is dipped into water, then new fundamental frequency will be
[]
f
f/2
2f
f/4
(a) When half is dipped, it becomes closed pipe with fundamental frequency f (same as original open pipe).
26. In a stationary wave, nodes are the points having
[]
Maximum displacement and maximum strain
Maximum displacement and minimum strain
Minimum displacement and minimum strain
Minimum displacement and maximum strain
(d) Nodes have minimum displacement but maximum strain.
27. In a stationary wave antinodes are the points having
[]
Maximum displacement and maximum strain
Maximum displacement and minimum strain
Minimum displacement and minimum strain
Minimum displacement and maximum strain
(b) Antinodes have maximum displacement but minimum strain.
28. At open end of an organ pipe
[]
An antinode is always produced
A node is always produced
Either antinode or node may be produced
Neither node nor antinode is produced
(a) Open end always has an antinode.
29. The fundamental frequency of a closed organ pipe is f. The frequency of its first overtone is
[]
f
2f
3f
f/2
(c) First overtone of closed pipe is 3f.
30. The displacement is given by the equation y = A cos 2πnt cos(2πx/λ), where A, n, λ are constants. It represents
[]
A progressive wave travelling along positive X-axis
A progressive wave travelling along negative X-axis
Two waves of same speed travelling in opposite directions
Two waves travelling in same direction with a phase difference of π
(c) This represents two waves travelling in opposite directions forming stationary wave.
31. As an empty vessel is filled with water its frequency
[]
Increases
Decreases
Remains unchanged
None of these
(b) Frequency decreases as air column shortens.
32. A tube, closed at one end and containing air produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends; the fundamental frequency that can be excited is (in Hz)
[]
1024
512
256
128
(a) Open pipe fundamental frequency is twice closed pipe ⇒ 2 × 512 = 1024 Hz.
33. A cylindrical tube, open at both ends has fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of air column is now
[]
f/2
3f/4
f
2f
(c) When half dipped, it becomes closed pipe with same fundamental frequency as original open pipe.
34. The end correction for the vibration of air column in a tube of circular cross-section will be more if the tube is
[]
Increased in length
Decreased in length
Made thinner
Made wider
(d) End correction increases with diameter of tube.
35. A hollow metallic tube of length L and closed at one end produce resonance with tuning fork of frequency n. The entire tube is then heated carefully so that at equilibrium temperature its length changes by D. If the change in velocity V of sound is v, the resonance will now be produced by tuning fork of frequency
[]
4(L + D)/(V + v)
4(L - D)/(V - v)
4(L + D)/V
4(L - D)/V
(c) New frequency = (V + v)/4(L + D) ≈ V/4(L + D) if v is small.
36. The end correction of a resonance column is 1.0 cm. If the shortest length resonating with the tuning fork is 15.0 cm, the next resonating length will be
[]
31 cm
45 cm
46 cm
47 cm
(d) Next resonating length = 3 × (15 + 1) - 1 = 47 cm.
37. The frequency of a vibrating wire is f. When area of cross section of a wire is halved and tension doubled, the frequency becomes
[]
f
2f
3f
√2f
(b) f ∝ √(T/μ) ⇒ if T doubles and A halves (μ halves), new frequency = f × √(2/(1/2)) = 2f.
38. A sonometer wire, 100 cm in length, has a fundamental frequency of 330 Hz. The velocity of propagation of transverse waves along this wire is
[]
330 m/sec
660 m/sec
115 m/sec
990 m/sec
(b) v = 2Lf = 2 × 1 × 330 = 660 m/s.
39. Two stretched wires of same material of lengths l and 2l vibrate with frequencies 100 and 150 Hz respectively. The ratio of their tension is
40. Two wires made of the same material are of equal lengths but their diameters are in the ratio of 1:2. On stretching each of these two strings by same tension, the ratio between the fundamental frequencies of these strings is
[]
1:2
2:1
1:4
4:1
(b) f ∝ 1/d ⇒ f₁/f₂ = d₂/d₁ = 2/1.
41. The length of a sonometer wire is doubled and its tension is increased four times. The fundamental frequency of the wire is changed in the ratio
[]
1:4
1:2
1:1
2:1
(c) f ∝ (1/l)√T ⇒ f₂/f₁ = (1/2)√4 = 1 ⇒ ratio 1:1.
42. A cylindrical tube, open at both ends has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of air column is now:
[]
f/2
f
3f/4
2f
(b) When half dipped, it becomes closed pipe with same fundamental frequency as original open pipe.
43. If oil of density higher than water is filled in place of water in a resonance tube, its frequency will
[MOE 2010/BP 2015]
Increase
Decrease
Remain unchanged
Depend on the density of material of the tube
(b) Higher density oil reduces effective length more ⇒ frequency decreases.
44. In the equation y = cos(60x)sin(100πt), x and y are in cm and t is in seconds. At the node find the value of x.
[IOM 07]
π/100 cm
20 cm
15 cm
12.5 cm
(d) At nodes, cos(60x) = 0 ⇒ x = (2n + 1)π/120 ⇒ smallest x = π/120 ≈ 2.6 cm (none match exactly).
45. An organ pipe P₁, closed at one end and vibrating in its first overtone and another pipe P₂, open at both ends are vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of P₁ and P₂ is:
[IOM 05]
3:4
1:2
2:3
4:5
(a) First overtone of closed pipe (3v/4L₁) = third harmonic of open pipe (3v/2L₂) ⇒ L₁/L₂ = 3/4.
46. The third harmonic of open ended pipe of length 50 cm is
[MOE 066]
332 Hz
166 Hz
996 Hz
100 Hz
(c) Third harmonic = 3 × fundamental = 3 × (v/2L) = 3 × (332/1) ≈ 996 Hz.
47. In a resonance tube the air columns for the first and second resonance differ by 31.5 cm. The wavelength of the sound waves in the tube is
[MOE 2065]
31.5 cm
63.0 cm
126.0 cm
252.0 cm
(b) λ/2 = 31.5 cm ⇒ λ = 63 cm.
48. One open organ pipe of l = 27 cm and closed organ pipe of length 21 cm sound in unison in their 1st overtone. Calculate the end correction for both pipes.
[MOE 2061]
1.5 cm
0.6 cm
0.9 cm
1.2 cm
(a) For unison: v/(2(27 + e)) = 3v/(4(21 + e)) ⇒ e ≈ 1.5 cm.
49. A string has mass 0.01 kg and has length 1 m. If the tension is 1000 N, the velocity of transverse wave in the string is
50. An open pipe of length L₁ and closed pipe of length L₂ resonate to the same tuning fork. The ratio of their lengths (L₁/L₂) is
[MOE 09]
4:1
2:1
1:2
1:4
(b) For same frequency: v/2L₁ = v/4L₂ ⇒ L₁/L₂ = 2/1.
51. The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe organ is 60 cm, what is the length of the closed pipe?
[BPKIHS-95]
15 cm
30 cm
45 cm
60 cm
(a) v/4L_c = v/L_o ⇒ L_c = L_o/4 = 60/4 = 15 cm.
52. Resonance will be produced with sound waves of 48 cm, in a closed pipe of length
[BPKIHS-97]
12 cm
24 cm
36 cm
48 cm
(a) For closed pipe: L = λ/4 = 48/4 = 12 cm.
53. Identify the equation of stationary wave for free end:
[]
y = +2a sinωt sin kx
y = -2a cosωt sin kx
y = +2a sinωt cos kx
y = -2a sinωt cos kx
(a) Free end has antinode ⇒ y = ±2a sinωt sin kx.
54. The equation of stationary wave is given by y = -2a cosωt cos kx. The phase difference between the incident wave and the reflected wave is
[]
Zero
π
π/2
π/4
(b) Negative sign indicates phase change of π at reflection.
55. The radius, density and tension of string A is twice the radius, density and tension of another string B. If the length of both strings are equal, then the ratio of their frequency of vibration is:
[]
1:2
2:1
1:4
4:1
(a) f ∝ (1/r)√(T/ρ) ⇒ f_A/f_B = (1/2)√(2/2) = 1/2.
56. A piano wire of diameter 0.9 mm is replaced by another wire of 0.93 mm. Then the percentage change in frequency of piano wire is
[IOM/BPKIHS]
+3.0%
+3.3%
-3.0%
-3.3%
(d) f ∝ 1/d ⇒ Δf/f = -Δd/d = -(0.03)/0.9 ≈ -3.3%.
57. A tuning fork of 200 Hz is in unison with sonometer wire. If percentage increase in tension of wire is 1%, then number of beats produced per second is
60. A sonometer string and a tuning fork when sounded together give 6 beats/sec whether length of sonometer string is 95 cm or 100 cm. The frequency of tuning fork is
[]
262 Hz
256 Hz
234 Hz
260 Hz
(c) Let f be fork frequency. Then f₁ - f = 6 and f - f₂ = 6 ⇒ (f₁ + f₂)/2 = f ⇒ solve to find f = 234 Hz.
61. A string of 36 cm length was in unison with a fork of frequency 256 per second. It was in unison with another fork when the vibrating length was 48 cm, the tension being unchanged. The frequency of the second fork is
[]
284 Hz
192 Hz
320 Hz
212 Hz
(b) f ∝ 1/l ⇒ f₂ = f₁ × (36/48) = 256 × 0.75 = 192 Hz.
62. A cord attached to a vibrating tuning fork is divided into six segments under a tension of 36 N. It will be divided into 4 segments if the tension is
[]
16 N
24 N
48 N
81 N
(d) Number of segments ∝ √T ⇒ (6/4)² = 36/T₂ ⇒ T₂ = 81 N.
63. In a resonance column, first and second resonance are obtained at depths 22.7 cm and 70.2 cm respectively. The third resonance will be obtained at a depth
117.7 cm
92.9 cm
115.5 cm
113.5 cm
()
64. The end correction of a resonance column is 1.0cm. If the shortest length resonating "with the tuning fork is 15.0cm, the next resonating length will be
31cm
45cm
46cm
47cm
(c) For a closed pipe, resonating lengths are given by ln = (2n - 1)λ/4 + e. First resonating length l₁ = λ/4 + e = 15.0 + 1.0 = 16.0 cm. So λ/4 = 15 cm => λ = 60 cm. Next resonating length for n=2 is l₂ = 3λ/4 + e = 3 * 60 / 4 + 1.0 = 45 + 1 = 46 cm.
65. A glass tube Im length is filled by water. The water can be drained out slowly at the bottom of the tube. If a vibrating tuning fork of frequency 500c/s is brought at the upper end of the tube and the velocity of sound is 330m/s, then the total number of resonance obtained will be
b. 3
4
1
2
(a) This is a closed pipe as one end (water surface) is closed. Resonating frequencies are fn = (2n - 1)v / 4L ≤ 500. (2n - 1) * 330 / (4 * 1) ≤ 500. (2n - 1) * 82.5 ≤ 500. 2n - 1 ≤ 500 / 82.5 ≈ 6.06. 2n ≤ 7.06. n ≤ 3.53. Possible values of n are 1, 2, 3. So, total number of resonance obtained will be 3.
66. The speed of sound in air is 320m/s. A closed organ pipe of length Im can resonate with a frequency of
b. 240 Hz
a. 80 Hz
400 Hz
All
(a) Frequencies of resonance in a closed pipe are fn = (2n - 1)v / 4L. For n=1 (fundamental frequency), f₁ = (2*1 - 1) * 320 / (4 * 1) = 320 / 4 = 80 Hz. For n=2 (first overtone), f₂ = (2*2 - 1) * 320 / 4 = 3 * 320 / 4 = 240 Hz. For n=3 (second overtone), f₃ = (2*3 - 1) * 320 / 4 = 5 * 320 / 4 = 400 Hz. So, all the frequencies are possible.
67. An open organ pipe has fundamental frequency 300 Hz. The frequency of first overtone of open organ pipe is equal to the frequency of first overtone of closed organ "pipe. If speed of sound in air is 320m/s then length of closed organ pipe is
a. 10cm
b. 20cm
80cm
40cm
(b) For open pipe, fundamental frequency f₁open = v / 2Lopen = 300 Hz. First overtone f₂open = 2v / 2Lopen = v / Lopen = 600 Hz. For closed pipe, first overtone f₃closed = 3v / 4Lclosed. Given f₂open = f₃closed => 600 = 3 * 320 / 4Lclosed => 600 = 240 / Lclosed => Lclosed = 240 / 600 = 0.4 m = 40 cm.
68. An open organ pipe is suddenly closed at one end with the resultant frequency of first overtone of open organ pipe is 100 Hz more than frequency of first overtone of closed . organ pipe. The fundamental frequency of open organ pipe is
a. 100 Hz
b. 200 Hz
300 Hz
400 Hz
(b) Let fundamental frequency of open pipe be fo. First overtone of open pipe = 2fo. When closed, fundamental frequency f'c = fo/2. First overtone of closed pipe = 3f'c = 3fo/2. Given 2fo = 3fo/2 + 100 => 4fo = 3fo + 200 => fo = 200 Hz.
69. The equation of a stationary wave is: y = 5sin 3 cos40nt Where x and y are in cm and t in seconds. Then the separation between two consecutive nodes is:
a. 12cm
b. 3cm
6cm
1.5cm
(d) The equation of a stationary wave is y = 2A sin(kx) cos(ωt). Comparing with y = 5 sin(3πx) cos(40πt), we get k = 3π. Wavelength λ = 2π/k = 2π / 3π = 2/3 cm. Separation between two consecutive nodes = λ/2 = (2/3) / 2 = 1/3 cm. There might be a typo in the equation. Let's assume y = 5 sin(3πx/2) cos(40πt). Then k = 3π/2. λ = 2π / (3π/2) = 4/3 cm. λ/2 = 2/3 cm. Still not matching. Let's assume y = 5 sin(πx/6) cos(40πt). k = π/6. λ = 2π / (π/6) = 12 cm. λ/2 = 6 cm. Option Let's assume y = 5 sin(2πx/λ) cos(ωt) = 5 sin(3πx) cos(40πt) is wrong. Let's use y = 5 sin(πx/3) cos(40πt). k = π/3. λ = 2π / (π/3) = 6 cm. λ/2 = 3 cm. Option b. Let's assume y = 5 sin(2πx/λ) cos(ωt) = 5 sin(πx/1.5) cos(40πt). k = π/1.5 = 2π/3. λ = 2π / (2π/3) = 3 cm. λ/2 = 1.5 cm. Option d.
70. The velocity of sound is 350 m/s. The length of open organ pipe is 50.cm. Find its fundamental frequency. . [IOM 2017]
[IOM 2017]
b. 500Hz
a. 700Hz
350Hz
175Hz
(c) Fundamental frequency of open organ pipe f₁ = v / 2L = 350 / (2 * 0.5) = 350 / 1 = 350 Hz.
