AP Physics B – Practice Questions (MCQ) on Wave Motion (including sound)

"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."

– Albert Einstein

Today we will discuss a few multiple choice practice questions involving wave motion. It will be useful to access earlier posts in this section by clicking on the label ‘waves’ or ‘wave motion (including sound)’. Here are the questions:

(1) When the atmospheric pressure is P the minimum resonating length of air column in a resonance column apparatus is L with a tuning fork of frequency f. When the atmospheric pressure increases by 0.5% the consequent change in the minimum resonating length will be

(a) 0.25%

(b) 0.5%

(c) 1%

(d) 2%

(e) zero

The speed v of sound in a gas is given by Newton-Laplace equation:

             v = √(γP/ρ) where γ is the ratio of specific heats of the gas, P is the pressure of the gas and ρ is the density of the gas.

When pressure P increases, the density ρ also increases by the same proportion so that the quantity P/ρ is a constant. Therefore, the speed of sound is unchanged with changes in pressure. This means that the wave length λ of sound and the resonating length are unchanged [Option (e)]

[Note that the minimum resonating length is λ/4]

(2) A tuning fork of frequency 512 Hz resonates with the air column in a pipe of length 15 cm closed at one end. This tuning fork does not resonate with the air column in any shorter pipes. For which one of the following closed pipe lengths will this tuning fork exhibit resonance with the air column?

(a) 20 cm

(b) 30 cm

(c) 45 cm

(d) 55 cm

(e) 60 cm

Since the tuning fork does not resonate with the air column in any shorter pipes, the resonating length is equal to λ/4 where λ is the wave length of sound in air. In this case there is a node at the closed end of the pipe and the next antinode is at the open end (Fig.). If the length of the pipe used is an odd multiple of λ/4, resonance will occur since in such cases the open end can be an antinode. Thus resonance is possible only if the closed pipe lengths are λ/4, 3λ/4, 5λ/4, 7λ/4 etc.

[Note that a standing wave can be obtained in a closed pipe with consequent resonance only if the the closed end is a noe an the open end is an antinode].

Since λ/4 = 15 cm we have 3λ/4 = 45 cm, 5λ/4 = 75 cm, 7λ/4 = 105 cm and so on. Option (c) gives the length of the pipe as 45 cm and it is the correct one.

(3) The frequencies of the 2^nd and 3^rd overtones of a vibrating string are 3f/4 and 2f respectively. The fundamental frequency of vibration of the string is

(a) f/8

(b) f/6

(c) f/4

(d) f/2

(e) f

The natural frequencies of vibration of a string are integral multiples of the fundamental (lowest) frequency. If the fundamental frequency is n, the frequency of the first overtone (or, the 2^nd harmonic) is 2n. The frequency of the 2^nd overtone (or, the 3^rd harmonic) is 3n and the frequency of 3^rd overtone (or, the 4^th harmonic) is 4n.

Therefore, considering the 2^nd overtone, we have

             3n = 3f/4 from which n = f/4.

(4) Sound waves producing interference have their amplitudes in the ratio 3 : 2. The intensity ratio of maximum and minimum of interference fringes is

(a) 27 :  8

(b) 25 : 1

(c) 3 : 2

(d) 9 : 4

(e) 6 : 4

The  resultant amplitudes at the interference maximum and the interference minimum are in the ratio (3+2) : (3 – 2) since the waves are in phase at the interference maximum and 180º out of phase at the interference minimum. Since the intensity is directly proportional to the amplitude, the intensity ratio of maximum and minimum of interference fringes is (3+2)² : (3 – 2)² = 25 : 1

Multiple Choice Practice Questions on AP Physics B Thermodynamics

“This time, like all times, is a very good one if we but know what to do with it”

– Ralph Waldo Emerson

You will find many practice questions (with solution) on thermodynamics on this site. You can access them by clicking on the label ‘thermodynamics’ below this post. Today we shall discuss a few more multiple choice practice questions in this section for the benefit of AP Physics B aspirants.

(1) Keeping the pressure constant, the temperature of m kg of a gas is raised through ΔT º C. If the specific heats of the gas at constant volume and constant pressure are C_v and C_p respectively, the increase in internal energy of the gas is

(a) m C_p ΔT

(b) m (C_p – C_v) ΔT

(c) m C_v ΔT

(d) m (C_p + C_v) ΔT

(e) m (C_p + C_v) ΔT/2

If the temperature increase occurs at constant volume, the entire energy supplied to the gas is used up in increasing the internal energy of the gas. The energy required for increasing the temperature of one kilogram of the gas through 1º C at constant volume is the specific heat of the gas at constant volume (C_v). When the temperature increase occurs at constant pressure, the gas has to expand and therefore has to do external work. The specific heat of the gas at constant pressure is equal to the energy supplied for increasing the temperature of one kilogram of the gas through 1º C at constant pressure. But the increase in the internal energy of 1 kg of the gas on heating through 1º C in this case also is C_v.

Therefore, the increase in internal energy (ΔU) of m kg of the gas when its temperature is raised through ΔT º C is given by

             ΔU = m C_v ΔT

(2) In the above question the work done by the gas is

(a) mC_pΔT

(b) m(C_p – C_v)ΔT

(c) mC_vΔT

(d) m(C_p + C_v)ΔT

(e) Zero

Keeping the pressure constant, when the temperature of m kg of the gas is raised through ΔT º C, the total energy required is mC_pΔT. Out of this the energy used in increasing the internal energy of the gas is mC_vΔT. The energy spent for doing work is therefore given by the difference between the two values and is equal to m(C_p – C_v)ΔT.

(3) A sample of gas contained in a cylinder undergoes a cyclic process shown by the adjoining PV diagram. Among the following statements which one is correct?

(a) Work is done on the gas during the process shown by AB

(b) Work done by the gas during the cycle ABCA is negative

(c) BC represents an isochoric  process

(d) CA represents an isobaric process

(e) Work done on the gas during the process CA is zero

During the process shown by AB the gas expands. Therefore work is done by the gas (and not on the gas). Statement (a) is therefore incorrect.

Since the cycle ABCA is clockwise, work done by the gas is positive. Statement (b) is therefore incorrect.

[Work done by the gas during its expansion represented by the area under the curve AB is greater than the work done on the gas during its contraction represented by the area under the curve BC]

During the process BC the volume of the gas changes and hence the process is not isochoric.

[Isochoric process is also known as isovolumetric process]

During the process CA the pressure of the gas changes and hence the process is not isobaric.

Work done on the gas uring the process CA is zero since CA represents an isochoric process. Therefore option (e) is correct.

[Note that work is done only if volume changes].

(4) A Carnot engine operates using a high temperature source at 400 K and a low temperature sink at 300 K. How much more efficient will this engine be if the temperature of the sink is reduced to 200 K?

(a) Twice as efficient

(b) Three times as efficient

(c) Four times as efficient

(d) Five times as efficient

(e) Six times as efficient

Efficiency (η) of Carnot engine is given by

            η = (Q₁ – Q₂)/Q₁ = (T₁ – T₂)/T₁ where Q₁ is the quantity of heat absorbed from the source at the source temperature T₁ and Q₂ is the quantity of heat liberated to the sink at the sink temperature T₂.

The efficiency η₁ when the source and sink are at 400 K and 300 K respectively is given by

            η₁ = (400 – 300)/400 = ¼

The efficiency η₂ when the source and sink are at 400 K and 200 K respectively is given by

             η₂ = (400 – 200)/400 = ½

The engine is therefore twice as efficient [Option (a)].

(5) A fixed mass of gas does 25 J of work on its surroundings and transfers 15 J of heat to the surroundings. The internal energy of the gas

(a) decreases by 15 J

(b) increases by 15 J

(c) remains unchanged

(d) decreases by 40 J

(e) increases by 40 J

The internal energy of the gas decreases when it does work on the surroundings and also when it transfers heat to the surroundings. The total decrease in the internal energy of the gas is therefore equal to 25 J + 15 J = 40 J [Option (d)].

You will find similar questions with solution here.