AP Physics B – Simple and Interesting Fluid Mechanics

“There is no democracy in physics. We can't say that some second-rate guy has as much right to an opinion as Fermi.”

–Luis Walter Alvarez

The first free response question in the AP Physics B 2013 Examination was from fluid mechanics which provides ample scope for question setters to ask interesting questions. Today we shall discuss some interesting and useful points in this section.

An important point you need to note is that an object denser than a liquid, can displace liquid of  weight equal to that of the object, if the object forms part of a floating  body.  The same object submerged independently in the liquid will displace liquid of weight less than that of the object.

(1) To make the above point more clear, let us consider a large tank containing water of density 1000 kg/m³. A wooden block A of specific gravity 0.5 and volume 2 m³ is floating in the water (Fig. 1). An object B of specific gravity 5 and volume 0.1 m³ is placed on the wooden block. Will the wooden block still continue to float?

The wooden block A has mass 1000 kg as is clear from the relation,

             mass = volume×density

[Since the specific gravity of the wooden block is 0.5, its density is 0.5×1000 = 500 kg/m³. Therefore, the mass of the wooden block = 2×500 = 1000 kg.

Remember that the specific gravity or relative density of a body is the ratio of the density of the body to the density of water. Since water has density 1000 kg/m³ in the international system (SI) of units, the density of a body has numerical value 1000 times its specific gravity].

Mass of the object B placed on the wooden block is 0.1×5000 = 500 kg. Therefore the total mass of the wooden block and the object placed on it is (1000+500) kg = 1500 kg.

Since the volume of the wooden block is 2 m³ it can displace 2 m³ of water before getting submerged. The wooden block and the object placed on it will continue to float since the total volume of water to be displaced needs to be 1.5 m³ only.

[Mass of 1.5 m³ of water is 1500 kg].                                                                                    

In the above explanation we have used the law of floatation, which states that the weight of a floating body is equal to the weight of the displaced liquid.

(2) Now, consider a slightly different situation which is shown in the adjoining figure (Fig. 2). The same wooden block A and the denser object B as we considered above are shown here also; but the object is suspended from the wooden block using a rope of negligible mass. How much water will be displaced by the system (containing the wooden block and the object) in this case?

The object has specific gravity 5 and it is fully immersed in water. If it were not part of the floating system, it would have displaced water of volume equal to its own volume (which is 0.1 m³). Since the object is part of a floating system, it is capable of displacing water of mass equal to its own mass (which is 500 kg), by pulling the wooden block further down making use of the connecting rope. Thus the total mass of water displaced is 1500 kg and the total volume of water displaced is 1.5 m³, as in the case of the system shown in Fig. 1.

In Fig. 1 the water level in the tank is shown as L₁ where as in Fig. 2 the water level is shown as L₂. Obviously these two levels coincide.

(3) Now, suppose the rope connecting the object B to the wooden block A cut. The object B, being denser than water, will then sink and will rest at the bottom of the tank. How much water will be displaced by the wooden block and the object together in this case?

The wooden block is a floating body and hence it will displace water having mass equal to its own mass, which is 1000 kg. So the volume of water displaced by the wooden block is 1 m³. The object B will displace water of volume equal to its own volume, which is equal to 0.1 m³. Therefore, the total volume of water displaced by the wooden block and the object together in this case is 1.1 m³.

The water level in the tank will be lowered (compared to L₁) in this case.

(4) Let us once again consider the original system (Fig. 1) in which the denser object B (of specific gravity 5 and volume 0.1 m³) is placed on the wooden block. What must be the mass of the object B if the entire wooden block just begins to submerge?

Questions of the above type are often found in question papers of various examinations. When the wooden block is completely immersed in water the volume of water displaced is equal to the volume of the wooden block, which is 2 m³. The weight of the wooden block and the object together in this case is therefore equal to the weight of 2 m³ of water, which is 2000 kg. Since the mass of the wooden block is 1000 kg, the mass of the object must be 1000 kg.

We shall now consider another situation:

(5) Suppose the object placed on the wooden block (Fig. 1) has specific gravity 0.8 and the water level in the tank is noted. If the object is gently transferred to the water in the tank, will the water level in the tank change?

Since the specific gravity of the object is 0.8, it will float in water. Whether it is on the wooden block or independently outside, it will displace water of mass equal to its own mass. Therefore, there is no change in the water level in the tank.

(6) Now, suppose the object of specific gravity 0.8 (which we considered above) is tied to the wooden block using a rope of negligible mass so that it is fully immersed in the water in the tank, as shown in Fig. 3. Will the level of water in the tank change in this case?

Many among you might be confused about this situation. Understand that the wooden block and the object tied to it is still a floating system and the mass of water displaced must be equal to the mass of the floating system. Therefore there is no change in the water level in the tank.

We shall now consider the case of an ice block floating in water: 

(7) Consider a block of ice floating in a tank containing pure water (Fig. 4). If the ice melts without any change in the temperature, will the water level in the tank rise up, fall down or remain unchanged?

This is a very popular question and most of you know the answer: The water level will remain unchanged. But how do you justify your answer?

The block of ice is a floating body and the volume of water displaced by it has weight equal to its own weight. When the ice melts, the eaxtra water produced should fill exactly the volume originally occupied by the immersed portion of ice. Therefore, the water level in the tank is unchanged.

(8) Suppose a metallic bob is placed on the ice block so that it still floats in the water in the tank (Fig. 5). If the ice melts without any change in the temperature, will the water level in the tank change?

When the ice alone is there, the water level in the tank remains unchanged when the ice melts, as we have seen above. When the metallic bob also is present, it sinks when the ice melts and it displaces water of volume equal to its own volume. This volume is certainly less than the volume of water it displaces when it is part of the floating system formed along with the ice block. The water level in the tank is therefore lowered when the ice melts (without any change in the temperature).

(9) Now, suppose the metallic bob in the above question is replaced by a wooden bob (which is lighter than water). If the ice melts without any change in the temperature, will the water level in the tank change?

In this case there will be no change in the water level since the wooden bob floats in the water in the tank when the ice melts, displacing the same volume of water as before.

Multiple Choice Practice Questions on Atomic Physics and Quantum Effects for AP Physicsw B

“Being ignorant is not so much a shame as being unwilling to learn.”

– Benjamin Franklin

During March and April I have been unexpectedly busy and therefore could not post any thing here. Today I give you a few multiple choice practice questions on atomic physics and quantum effects along with their solution. This section will be interesting to most of you. Questions (with solution) in this section were posted on many occasions on this site. You may access them by clicking on the label ‘atomic physics and quantum effects’ below this post. Here are the additional questions:

(1) The energy of the electron in the hydrogen atom in the ground state is E₁. When the hydrogen atom is in its first excited state, the energy of its electron is E₂. Suppose we have a large number of hydrogen atoms in the ground state and radiation of frequency (E₂ – E₁) is incident on these atoms. What happens?

(a) All atoms will jump to the first excited state.

(b) All atoms will continue to remain in the ground state.

(c) Some of the atoms will jump to the first excited state.

(d) Most of the atoms will jump to the second excited state.

(e) Most of the atoms will get ionised.

Since the energy of each photon in the incident radiation is equal to the energy difference between the ground state and the first excited state of the hydrogen atom, the probability of absorption of energy is high so that some of the hydrogen atoms will jump to the first excited state. The correct option is (c).

(2) The simple Bohr model is applicable to which one of the following?

(a) Atoms with light nuclei

(b) Atoms with heavy nuclei

(c) Helium atoms

(d) Doubly ionised lithium

(e) Uranium atoms

Doubly ionized lithium has just one electron moving around the nucleus and hence it is hydrogen-like. Therefore the simple Bohr model is applicable to it [Option (d)].

(3) A proton is moving through a region of space where a constant electric field acts along its direction of motion. The de Broglie wave length of the proton

(a) decreases with time

(b) increases with time

(c) remains constant

(d) varies in a sinusoidal manner

(e) goes on increasing first and soon becomes constant

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and and p is the momentum.

Since the proton is accelerated by the electric field, its momentum goes on increasing with time. Therefore, the de Broglie wave length λ of the proton decreases with time [Option (a)].

(4) An electron is projected at an acute angle with respect to a uniform magnetic field so that it travels along a helical path. The de Broglie wave length of the electron

(a) varies periodically

(b) increases with time

(c) remains constant

(d) decreases with time

(e) goes on decreasing first and attains a constant value

A static magnetic field cannot change the kinetic energy of a charged particle. The momentum of the electron therefore remains unchanged. Therefore the de Broglie wave length of the electron remains constant.

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.