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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Monday, March 31, 2008

### AP Physics B & C –Electric Circuits –Equations to be Remembered

You must remember the following equations to make you strong in answering multiple choice questions involving electric circuits:

(1) Electric current, I = nAvq where n is the number density (number per unit volume) of mobile charge carriers, A is the area of cross section of the conductor, v is the drift velocity of the charge carriers and q is the charge on each carrier.

Since the charge carriers in a conductor are electrons of charge e, the above expression becomes

I = nAve

(2) Resistivity ρ = RA/L where R, A and L are respectively the resistance, area of cross section and the length of the conductor.

(3) If a wire of resistance R is stretched so that its length becomes n times its original length, its area of cross section becomes (1/n) times its original area of cross section. The resistance of the stretched wire therefore becomes n2R.

[The stretch can be described also in terms of the radius of the wire. Thus if a wire of resistance R is stretched so that its radius becomes (1/n) times the original radius, its area of cross section becomes (1/n2) times the original area of cross section and its length becomes n2 times the original length. The resistance of the wire therefore becomes n4R].

(4) Conductivity s is the reciprocal of resistivity: s = 1/ρ

(5) Effective value (R) of resistances in series: R =R1 +R2 +R3 +R4 +….etc.

(6) Effective value (R) of resistances in parallel is given by the reciprocal relation,

1/R = 1/R1 + 1/R2 + 1/R3 +… etc.

(7) (a) The equivalent emf of a series combination of n cells is the sum of their individual emf’s:

Ɛ = Ɛ1 + Ɛ2 + Ɛ3 + Ɛ4 +……etc

(b) The equivalent internal resistance of a series combination of n cells is the sum of their internal resistances:

r = r1 + r2 + r3 + r4 +……etc

(8) If there an n cells of emf Ɛ1, Ɛ2, Ɛ3, …… Ɛn, and of internal resistances r1, r2 , r3, r4……. rn respectively, connected in parallel, the combination is equivalent to a single cell of emf Ɛeq and internal resistance req, such that

1/req = 1/r1 +1/r2 +1/r3 +……..+1/rn and

Ɛeq /req = Ɛ1/r1 + Ɛ2/r2 + Ɛ3/r3 +…….+Ɛn /rn

(9) Kirchhoff’s rules:

(i) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction.

(ii) Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells is zero.

(10) Power in a D.C. circuit = VI = I2R = V2/R where V is the voltage, I is the current and R is the resistance.

(11) If devices consuming powers P1, P2, P3, P4 ….etc. (at the same supply voltage V) are connected in parallel across the supply voltage V, the total power consumed by them is given by

P = P1+P2+ P3+ P4 +….etc.

(12) If devices consuming powers P1, P2, P3, P4 ….etc. (at the same supply voltage V) are connected in series and the series combination is connected across the supply voltage V, the total power consumed by them is given by the reciprocal relation,

1/P = 1/P1 +1/P2 +1/P3 +1/P4 +….etc.

(13) A cell will transfer maximum power to a load if the internal resistance of the cell is equal to the resistance of the load.

Transients in RC circuits are additionally included for the AP Physics C Examination. AP Physics C aspirants should therefore remember the following points also:

(i) When a capacitor of capacitance C farad is charged by connecting it in series with a resistance R ohm and a battery of emf V volts, the charge Q coulomb on the capacitor after a time t seconds is given by

Q =Q0 (1 e–t/RC)

where Q0 = CV which is the final maximum charge (at infinite time) on the capacitor and ‘e’ is the base of natural logarithm. [Q = Q0 exp(–t/RC)].

The product RC is the time constant of the RC circuit and has dimensions of time. If R is in ohm and C is in farad, RC is in seconds.

(ii) When a capacitor of capacitance C farad having initial charge Q0 coulomb is discharged through a resistance R ohm, the charge Q coulomb on the capacitor after a time t seconds is given by

Q =Q0 e–t/RC

[Or, Q = Q0 exp(–t/RC)].

The time constant of the RC circuit can be defined using the charging process as well as the discharge process. It is the time required for the charge to build up to [1– (1/e)] times (which is 63.2%) the final maximum charge. It can be defined also as the time required for the charge to decay to (1/e) times (which is 36.8%) the initial charge.

[The growth and decay of charge on a capacitor in a CR circuit is similar to the growth and decay of current through an inductor in an LR circuit. If an inductor of inductance L and a resistor of resistance R are connected in series with a battery, the nature of increase of current I through the circuit is exponential with time and is given by

I = I0 (1 e–Rt/L) where I0 is the final (maximum) current in the circuit.

On disconnecting the battery, the current in the circuit decays exponentially with time as given by the equation

I = I0 e–Rt/L

Here L/R is the time constant of the LR circuit].

Questions in this section will be discussed in due course. Meanwhile, find some useful multiple choice questions (with solution) in this section at physicsplus: Multiple Choice Questions on Direct Current Circuits

## Thursday, March 27, 2008

### AP Physics B – Answer to Free-Response Question (for practice) involving Compton Scattering

A free response question involving Compton Scattering was posted on 24th March 2008 for your practice. This was the question:

An X-ray photon of energy 31000 eV proceeds in the positive X-direction. It collides with a free electron at rest. After the collision the electron moves in the positive X-direction with a momentum greater than that of the X-ray photon. In the process, the wave length of the X-ray photon is changed by an amount dλ = 0.0486 Ǻ.

(a) What is the direction of motion of the photon after the collision? Justify your answer.

(b) Will the wave length of the photon decrease or increase? Justify your answer.

(c) Determine the wave length of the photon before collision.

(d) Calculate the kinetic energy transferred (by the photon) to the electron.

(a) The initial momentum of the photon electron system is the initial momentum of the photon which is in the positive X-direction. After the collision, the electron has a momentum in the X-direction. Since it is greater than the initial momentum of the photon, the final momentum of the photon must be in the negative X-direction so that the total final momentum of the system is equal to the initial momentum of the photon (in accordance with the law of conservation of momentum).

(b) The energy of the photon is given by

E = hc/λ where h is Planck’s constant, c is the speed of light in free space and λ is the wave length.

The wave length λ of the photon must be increased since the energy of the photon is decreased by transferring kinetic energy to the electron.

(c) The product of the wave length in Angstrom and the energy in electron volt in the case of any photon is 12400. Therefore, wave length of the X-ray photon before collision is

λ = 12400/31000 = 0.4 Ǻ

[The wave length λ can be found also by using the expression for the photon energy, E =hc/λ, where h is Planck’s constant and c is the speed of light in free space. The energy E should be in joule: E = 31000×1.6×10–19 joule. The value of λ will then be in metre].

(d) The initial energy of the photon = 31000 eV = 31000×1.6×10–19 joule = 4.96×10–15 J.

The final wave length of the photon is λ’ = (0.4 + 0.0486) Ǻ = 0.4486 Ǻ. Therefore, the final energy of the photon is hc/λ’ = (6.63×10–34×3×108)/(0.4486×10–10) = 4.4338×10–15 J.

The energy transferred by the photon to the electron = change in energy of the photon = (4.96×10–15 – 4.4338×10–15) J = 0.5262×10–15 J..

I give you a multiple choice question related to Compton scattering. This question will be quite simple once you understand the answer to the free response qestion we discussed above. Here is the question:

An X-ray photon of wave length λ proceeding in the positive X-direction suffers a head-on collision with a free electron at rest and retraces its path. If the wave length of the photon increases to λ’ in the process, the momentum transferred (by the photon) to the electron is (h = Planck’s constant, c = speed f light in free space)

(a) hc[(λ’ + λ)/ λλ]

(b) hc[(λλ)/ λλ]

(c) h[(λ’ + λ)/ λλ]

(d) h[(λλ)/ λλ]

(e) ) h[(λλ)

The initial momentum of the photon = h/λ

The final momentum of the photon = h/λ

The negative sign shows that the final momentum of the photon is in the negative X-direction.

The momentum transferred by the photon to the electron = change in momentum of the photon = [(h/λ – (–h/λ’) = h(1/λ + 1/λ’) = h[(λ’ + λ)/ λλ]

## Monday, March 24, 2008

### AP Physics B – A Free-Response Question (for practice) involving Compton Scattering

Here is a typical free-response question for your practice:

An X-ray photon of energy 31000 eV proceeds in the positive X-direction. It collides with a free electron at rest. After the collision the electron moves in the positive X-direction with a momentum greater than that of the X-ray photon. In the process, the wave length of the X-ray photon is changed by an amount dλ = 0.0486 Ǻ.

(a) What is the direction of motion of the photon after the collision? Justify your answer.

(b) Will the wave length of the photon decrease or increase? Justify your answer.

(c) Determine the wave length of the photon before collision.

(d) Calculate the kinetic energy transferred (by the photon) to the electron.

This question carries 10 points which may be divided among parts (a), (b), (c) and (d) as 3+2+2+3. Answer this question. I’ll be be back soon (with the answer, for your cofidence).

## Monday, March 17, 2008

### AP Physics B – Answers to Practice Questions (MCQ) on Kinetic Theory

Some typical multiple choice questions for practice were given to you in the post dated 15th March 2008. As promised, I give below the answers with explanation.

(1) If a diatomic gas molecule has an additional vibrational mode which contributes to both kinetic and potential energies, the sum of which is equal to kT where k is Boltzmann’s constant and T is the absolute temperature, what is the ratio of specific heats of the gas?

(a) 3/2

(b) 4/3

(c) 5/3

(d) 7/5

(e) 9/7

A diatomic molecule without vibrational mode has average energy equal to (5/2)kT since it has 5 degrees of freedom. Since the vibrational mode contributes an extra energy of kT, the total average energy per molecule is (5/2)kT+kT = (7/2)kT.

The molar specific heat at constant volume (Cv) is therefore equal to (7/2)kT×N where N is Avogadro’s number.

Therefore, Cv = (7/2)R and the molar specific heat at constant pressure (Cp) is [(7/2)R+R] = (9/2)R where R is universal gas constant.

Therefore, the ratio of specific heats, γ = Cp/Cv = 9/7

(2) Four moles of helium (a mono atomic gas) is contained in a barrel provided with a light, frictionless piston (fig.). The quantity of heat to be supplied to the gas for raising its temperature by 5 K is (universal gas constant, R = 8.31 J mol–1 K–1)

(a) 415.5 J

(b) 249.3 J

(c) 136.5 J

(d) 166.2 J

(e) 581.7 J

When heat is supplied to the gas, the internal energy of the gas is increased, raising the temperature of the gas. The pressure of the gas is unchanged since the piston moves outwards, doing work against the atmospheric pressure. It is the molar specific heat at constant pressure (Cp) that is involved here so that the quantity of heat (Q) to be supplied is given by

Q = no. of moles × molar specific heat at constant pressure × rise in temperature.

Helium being mono atomic, molar specific heat at constant pressure (Cp) is (5/2)R

Therefore, Q = 4×(5/2)R×5 = 4×(5/2)×8.31×5 = 415.5 J

(3) The intermolecular attraction is an important reason why real gases behave differently compared to ideal gas. But real gases too can behave like ideal gas at

(a) low temperature and low pressure

(b) low temperature and high pressure

(c) high temperature and high pressure

(d) high temperature and low pressure

(e) absolute zero (0 K) temperature

At high temperatures and low pressures the molecules are far apart and molecular interactions are negligible. Under these conditions even a real gas behaves like an ideal gas [Option (d)].

(4) A fixed mass of an ideal gas at pressure P is contained in a closed vessel of volume V. It is heated so that the root mean square velocity of the gas molecules is doubled. The thermal expansion of the vessel is negligible. Then the increase in pressure (P) of the gas is

(a) P

(b) 2P

(c) 3P

(d) 4P

(e) √2 P

The root mean square velocity of the gas molecules is given by c = √(3kT/m) where k is Boltzmann’s constant, T is the absolute temperature and m is the mass of the molecule. The r.m.s. velocity is therefore directly proportional to the square root of absolute temperature. Since the r.m.s. velocity is doubled on heating the gas, the temperature must be quadrupled.

By Charle’s law, we have PV/T = constant for a given mass of gas. When the temperature is quadrupled at constant volume, the pressure must be quadrupled. The final pressure is thus 4P and the increase in pressure is 3P.

(5) A cubical vessel of side contains N molecules of an ideal gas. If the mass of a molecule is m and the root mean square velocity is c, the pressure of the gas is

(a) mNc2 /33

(b) mNc2 /3

(c) mc2 /33

(d) 3mNc2 /3

(e) (1/3)mNc2

The pressure exerted by a gas is given by P = (1/3) ρc2 where ρ is the density of the gas and c is the root mean square velocity of the gas molecules. The density of the gas here is

ρ = Total mass/Total volume = mN/3

Therefore, pressure P = mNc2 /33

(6) Four moles of oxygen (diatomic) is mixed with eight moles of neon (mono atomic). If they are treated as ideal gases, the effective molar specific heat at constant volume of the mixture is (R = universal gas constant)

(a) (5/3) R

(b) (7/5) R

(c) (9/5) R

(d) (11/6) R

(e) (13/8) R

The molar specific heat at constant volume (Cv) of a diatomic gas is (5/2)R and that of a mono atomic gas is (3/2)R where R is universal gas constant.

Molar specific heat is the heat required to raise the temperature of one mole of the gas by 1 K. The quantity of heat required to raise the temperature of four moles of oxygen and eight moles of neon by 1 K is [4×(5/2)R + 8×(3/2)R] = 22 R and the total number of moles in the mixture is 12.

Therefore, molar specific heat of the mixture = 22 R/12 = (11/6) R

(7) The temperature at which the root mean square velocity of hydrogen gas molecules is twice that at 0º C is

(a) 20º C

(b) 40º C

(c) 819º C

(d) 859º C

(e) 1132º C

Since the root mean square velocity is directly proportional to the square root of absolute temperature, it will be doubled at 4 times the absolute temperature corresponding to 10º C. Therefore, the answer is 4×283 K = 1132 K = (1132 – 273) K = 859º C.

(8) Equal number of oxygen molecules (molar mass m1) and helium molecules (molar mass m2) are kept in two identical vessels. If they are at the same temperature, their pressures will be in the ratio

(a) m1/m2

(b) m2/m1

(c) √(m1/m2)

(d) √(m2/m1)

(e) 1

Most of you know that equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules. If equal numbers of molecules of different gases have the same temperature and volume their pressure must therefore be the same. The correct option is (e).

The expression for pressure in the form, P = nkT also is handy in working out similar problems. Since k is Boltzmann’s constant, the pressure depends only on the temperature and the number density (number per unit volume). You can easily obtain this expression by substituting for the r.m.s. velocity [c = √(3kT/m)] in the expression for pressure, P = (1/3) nmc2.

## Saturday, March 15, 2008

### AP Physics B – Practice Questions (MCQ) on Kinetic Theory

As promised in the post dated 13th March 2007, I give below some typical multiple choice questions on kinetic theory for practice. For the time being I leave these questions here without solution.

(1) If a diatomic gas molecule has an additional vibrational mode which contributes to both kinetic and potential energies, the sum of which is equal to kT where k is Boltzmann’s constant and T is the absolute temperature, what is the ratio of specific heats of the gas?

(a) 3/2

(b) 4/3

(c) 5/3

(d) 7/5

(e) 9/7

(2)
Four moles of helium (a mono atomic gas) is contained in a barrel provided with a light, frictionless piston (fig.). The quantity of heat to be supplied to the gas for raising its temperature by 5 K is (Universal gas constant, R = 8.31 J mol–1 K–1)

(a) 415.5 J

(b) 249.3 J

(c) 136.5 J

(d) 166.2 J

(e) 581.7 J

(3)
The intermolecular attraction is an important reason why real gases behave differently compared to ideal gas. But real gases too can behave like ideal gas at

(a) low temperature and low pressure

(b) low temperature and high pressure

(c) high temperature and high pressure

(d) high temperature and low pressure

(e) absolute zero (0 K) temperature

(4)
A fixed mass of an ideal gas at pressure P is contained in a closed vessel of volume V. It is heated so that the root mean square velocity of the gas molecules is doubled. The thermal expansion of the vessel is negligible. Then the increase in pressure (P) of the gas is

(a) P

(b) 2P

(c) 3P

(d) 4P

(e) √2 P

(5)
A cubical vessel of side contains N molecules of an ideal gas. If the mass of a molecule is m and the root mean square velocity is c, the pressure of the gas is

(a) mNc2 /3 3

(b) mNc2 /3

(c) mc2 /33

(d) 3mNc2 /3

(e) (1/3)mNc2

(6)
Four moles of oxygen (diatomic) is mixed with eight moles of neon (mono atomic). If they are treated as ideal gases, the effective molar specific heat at constant volume of the mixture is (R = universal gas constant)

(a) (5/3) R

(b) (7/5) R

(c) (9/5) R

(d) (11/6) R

(e) (13/8) R

(7) The temperature at which the root mean square velocity of hydrogen gas molecules is twice that at 10º C is

(a) 20º C

(b) 40º C

(c) 819º C

(d) 859º C

(e) 1132º C

(8) Equal number of oxygen molecules (molar mass m1) and helium molecules (molar mass m2) are kept in two identical vessels. If they are at the same temperature, their pressures will be in the ratio

(a) m1/m2

(b) m2/m1

(c) √(m1/m2)

(d) √(m2/m1)

(e) 1

Try to answer these questions. The essential points required for solving these questions were discussed in the post dated 13th March 2008, which you can access by clicking on the label ‘kinetic theory’ below this post.

I’ll be back with the solution in a day or two.