Q#1
Consider the following statements about sound passing through a gas.(A) The pressure of the gas at a point oscillates in time.
(B) The position of a small layer of the gas oscillates in time.
(a) Both (A) and (B) are correct.
(b) (A) is correct but (B) is wrong.
(c) (B) is correct but (A) is wrong.
(d) Both (A) and (B) are wrong.
Answer: (a)
Sound waves in a gas propagate due to the oscillation of the pressure at a point. The compressions and the rarefactions move through a point in the gas thus oscillating the pressure at that point, the layer of the gas does not oscillate. Hence (a).
Q#2
When we clap our hands, the sound produced is best described by
(a) p = p₀ sin(kx-⍵t)
(b) p = p₀ sinkx cos⍵t
(c) p = p₀ coskx sin⍵t
(d) p = Σp₀ₙ sin(kₙx-⍵ₙt).
Here p denotes the change in pressure from the equilibrium value.
Answer: (d)
When we clap, different regions of both the palms do not strike each other with the same pressure. Hence different regions of the palms produce different sounds having different pressure amplitudes, wave numbers and frequencies. So the pressure at a point is given as summation of pressures at that point at that instant. Hence the option (d).
Q#3
(a) p = p₀ sin(kx-⍵t)
(b) p = p₀ sinkx cos⍵t
(c) p = p₀ coskx sin⍵t
(d) p = Σp₀ₙ sin(kₙx-⍵ₙt).
Here p denotes the change in pressure from the equilibrium value.
Answer: (d)
When we clap, different regions of both the palms do not strike each other with the same pressure. Hence different regions of the palms produce different sounds having different pressure amplitudes, wave numbers and frequencies. So the pressure at a point is given as summation of pressures at that point at that instant. Hence the option (d).
Q#3
The bulk modulus and the density of water are greater than those of air. With this much of information, we can say that velocity of sound in air
(a) is larger than its value in water
(b) is smaller than its value in water
(c) is equal to its value in water
(d) cannot be compared with its value in water.
Answer: (d)
The velocity of sound in a fluid is given as V = √(B/ρ)
where B is the bulk modulus and ρ is the density of the fluid.
It is clear from the relation that if B increases V also increases but when ρ increases V decreases. Since B and ρ are both greater than air for water, B increases V but ρ decreases V for water. So until the ratio, B/ρ is available for both the water and air the velocity of sound in these mediums cannot be compared. Hence the option (d)
Q#4
(a) is larger than its value in water
(b) is smaller than its value in water
(c) is equal to its value in water
(d) cannot be compared with its value in water.
Answer: (d)
The velocity of sound in a fluid is given as V = √(B/ρ)
where B is the bulk modulus and ρ is the density of the fluid.
It is clear from the relation that if B increases V also increases but when ρ increases V decreases. Since B and ρ are both greater than air for water, B increases V but ρ decreases V for water. So until the ratio, B/ρ is available for both the water and air the velocity of sound in these mediums cannot be compared. Hence the option (d)
Q#4
A tuning fork sends sound waves in air. If the temperature of the air increases, which of the following parameters will change?
(a) Displacement amplitude.
(b) Frequency.
(c) Wavelength.
(d) Time period.
Answer: (c)
When the sound wave is being sent by a tuning fork, its displacement amplitude, frequency and the time period will remain the same even if the temperature changes. Since the velocity of the sound V ∝ √T where T is the temperature in °K. Thus with the increase in temperature V increases, but 𝛌 = V/ν. Here ν is constant hence with the increase in V the wavelength 𝛌 also increases. Hence the option (c).
Q#5
(a) Displacement amplitude.
(b) Frequency.
(c) Wavelength.
(d) Time period.
Answer: (c)
When the sound wave is being sent by a tuning fork, its displacement amplitude, frequency and the time period will remain the same even if the temperature changes. Since the velocity of the sound V ∝ √T where T is the temperature in °K. Thus with the increase in temperature V increases, but 𝛌 = V/ν. Here ν is constant hence with the increase in V the wavelength 𝛌 also increases. Hence the option (c).
Q#5
When sound wave is refracted from air to water, which of the following will remain unchanged?
(a) Wave number.
(b) Wavelength.
(c) Wave velocity.
(d) Frequency.
Answer: (d)
When the sound wave is refracted from air to water, due to the change in B and ρ the velocity of the sound V changes. Hence the options (a) and (c) are not correct.
Since the vibrating molecules of the air at the air-water interface will transfer its vibrations to the water molecules at the same frequency, the frequency of the sound wave in water will remain the same and wavelength will change. hence the option (d).
Q#6
(a) Wave number.
(b) Wavelength.
(c) Wave velocity.
(d) Frequency.
Answer: (d)
When the sound wave is refracted from air to water, due to the change in B and ρ the velocity of the sound V changes. Hence the options (a) and (c) are not correct.
Since the vibrating molecules of the air at the air-water interface will transfer its vibrations to the water molecules at the same frequency, the frequency of the sound wave in water will remain the same and wavelength will change. hence the option (d).
Q#6
The speed of sound in a medium depends on
(a) the elastic property but not on the inertia property
(b) the inertia property but not on the elastic property
(c) the elastic property as well as the inertia property
(d) neither the elastic property nor the inertia property.
Answer: (c)
The speed of the sound in a fluid is
(a) the elastic property but not on the inertia property
(b) the inertia property but not on the elastic property
(c) the elastic property as well as the inertia property
(d) neither the elastic property nor the inertia property.
Answer: (c)
The speed of the sound in a fluid is
V = √(B/ρ)
and in a longitudinal solid rod
V = √(Y/ρ)
For extended solids, V is a complicated function of B as well as G (Shear Modulus). B, Y and G are measures of elastic properties and ρ is the measure of inertia property. Hence the speed of sound in a medium depends on the elastic as well as the inertia property, thus option (c).
Q#7
Two sound waves move in the same direction in the same medium. The pressure amplitude of the waves are equal but the wavelength of the first wave is double the second. Let the average power transmitted across a cross-section by the first wave be P₁ and that by the second wave be P₂. Then
(a) P₁ = P₂
(b) P₁ = 4P₂
(c) P₂ = 2P₁
(d) P₂ = 4P₁.
Answer: (a)
Let the cross-section area = A, if the intensity of the first wave at this cross-section = I₁ and of the second wave = I₂, then
(a) P₁ = P₂
(b) P₁ = 4P₂
(c) P₂ = 2P₁
(d) P₂ = 4P₁.
Answer: (a)
Let the cross-section area = A, if the intensity of the first wave at this cross-section = I₁ and of the second wave = I₂, then
P₁ =AI₁ and P₂ = AI₂.
But the intensity I = p₀²V/2B
Since for a given medium velocity of sound V and the bulk modulus B are constant, I depends only on the pressure amplitude p₀. Since in the given problem pressure amplitude of both waves are same hence I₁ = I₂
P₁ = P₂
So the option (a).
Q#8
When two waves with same frequency and constant phase difference interfere,
(a) there is a gain of energy
(b) there is a loss of energy
(c) the energy is redistributed and the distribution changes with time
(d) the energy is redistributed and the distribution and the distribution remains constant in time.
Answer: (d)
The energy at a point due to the sound wave is proportional to the square of the pressure amplitude. When two waves with the same frequency and constant phase difference interfere, the energy is redistributed because the resultant pressure amplitude p₀ is given by
p₀² = p₀₁² + p₀₂² + 2p₀₁ p₀₂ cosδ
here p₀₁ and p₀₂ are pressure amplitudes of the two waves and δ is the phase difference. Since δ is constant here, so p₀ remains constant in time. Thus the energy redistribution remains constant in time. So, the option (d).
(a) there is a gain of energy
(b) there is a loss of energy
(c) the energy is redistributed and the distribution changes with time
(d) the energy is redistributed and the distribution and the distribution remains constant in time.
Answer: (d)
The energy at a point due to the sound wave is proportional to the square of the pressure amplitude. When two waves with the same frequency and constant phase difference interfere, the energy is redistributed because the resultant pressure amplitude p₀ is given by
p₀² = p₀₁² + p₀₂² + 2p₀₁ p₀₂ cosδ
here p₀₁ and p₀₂ are pressure amplitudes of the two waves and δ is the phase difference. Since δ is constant here, so p₀ remains constant in time. Thus the energy redistribution remains constant in time. So, the option (d).
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