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

## Thursday, October 28, 2010

### AP Physics B – Answer to Free Response Practice Question on Thermodynamics

“Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time”

– Thomas A. Edison

A free response practice question on thermodynamics was posted on 24th October 2010 for AP Physics B aspirants. As promised, I give below a model answer (along with the question) for your benefit. The adjoining figure shows a fixed cylindrical vessel of inner diameter 10 cm fitted with a smooth, light piston connected to a light spring of spring constant 4000 Nm–1. The other end of the spring is attached to an immovable support. The cylinder contains an ideal gas at 27º C. The spring is initially in its released condition and the initial volume of the gas in the cylinder is 0.8×10–3 m3. When heat is supplied to the gas, it expands and pushes the piston through 10 cm. The atmospheric pressure is 105 pascal. Now answer the following questions:

(a) What are the initial pressure and the initial absolute (Kelvin) temperature of the gas?

(b) Calculate the potential energy acquired by the spring because of the expansion of the gas.

(c) Calculate the final pressure of the gas in the cylinder.

(d) Calculate the final temperature of the gas in the cylinder.

(a) The initial pressure P1 of the gas is the same as the atmospheric pressure 105 Nm–1. The initial temperature T1 of the gas is (27 + 273) K = 300 K.

(b) The potential energy of the spring is U = ½ kx2 where k is the spring constant and x is the compression of the spring.

Therefore, U = ½ ×4000×(0.1)2 = 20 J.

(c) The final pressure of the gas is the sum of the atmospheric pressure and the pressure due to the elastic force developed in the spring.

Elastic force = kx.

Since this force acts over the area A of the piston, pressure due to the elastic force is kx/A = kx/πR2 where R is the radius of the piston.

Therefore, final pressure P2 of the gas = 105 + (4000×0.1)/(π×0.052) = 105 + 0.51×105 = 1.51×105 pascal.

(d) The final temperature T2 of the gas is given by

P1V1/T1 = P2V2/T2

Therefore, T2 = (P2V2 T1)/ (P1V1)

The final volume V2 of gas = Initial volume + πR2x = 0.8×10–3 + (π×0.052×0.1) = 0.8×10–3 + 0.79×10–3 = 1.59×10–3 m3

Therefore, final temperature T2 = (1.51×105×1.59×10–3 ×300)/(105×0.8×10–3) = 900 K.

## Saturday, October 23, 2010

### AP Physics B – Free Response Practice Question on Thermodynamics

Today I give you a free response practice question (for AP Physics B) on thermodynamics. You can access all posts on thermodynamics on this site by clicking on the label ‘thermodynamics’ below this post or by trying a search using the search box provided on this page.

Here is the question: The adjoining figure shows a fixed cylindrical vessel of inner diameter 10 cm fitted with a smooth, light piston connected to a light spring of spring constant 4000 Nm–1. The other end of the spring is attached to an immovable support. The cylinder contains an ideal gas at 27º C. The spring is initially in its released condition and the initial volume of the gas in the cylinder is 0.8×10–3 m3. When heat is supplied to the gas, it expands and pushes the piston through 10 cm. The atmospheric pressure is 105 pascal. Now answer the following questions:

(a) What are the initial pressure and the initial absolute (Kelvin) temperature of the gas?

(b) Calculate the potential energy acquired by the spring because of the expansion of the gas.

(c) Calculate the final pressure of the gas in the cylinder.

(d) Calculate the final temperature of the gas in the cylinder.

The above question carries 10 points. You have about 11 minutes for answering it.

## Wednesday, October 13, 2010

### AP Physics C – Answer to Free Response Practice Question on Electromagnetic Induction

A free response practice question on electromagnetic induction was posted on 10th October 2010. As promised, I give below a model answer (along with the question) for your benefit.

[You can access earlier posts in this section either by clicking on the label 'electromagnetic induction' below this post or by trying a search for 'electromagnetic induction' using the search box provided on this page]. Two infinitely long straight parallel wires W1 and W2, separated by a distance ‘a’ in free space, carry equal currents I flowing in opposite directions as shown in the adjoining figure. A square loop PQRS of side ‘a’, made of nichrome wire of resistance ρ Ω per metre is arranged with its plane lying in the plane of the wires W1 and W2 so that the sides PQ and RS of the loop are parallel to the wires W1 and W2. The side PQ of the loop is at a distance ‘a’ from the wire W2. Now, answer the following questions in terms of the given quantities and fundamental constants:

(a) Determine the magnetic flux density at a point midway between the wires W1 and W2.

(b) Determine the magnetic flux density at the mid point of the square loop PQRS.

(c) Calculate the magnetic flux through the loop PQRS.

(d) What is the average emf induced in the loop when the current through the wires is switched off in a time of 50 ms?

(e) When the current through the wires is switched off, it is found that at a certain instant t, the current decays at the rate of 40 As–1. Calculate the current induced in the loop PQRS at the instant t.

Indicate the direction of the current in the loop and justify your answer.

The magnitude of the magnetic flux density B at a point distant a from an infinitely long straight conductor carrying current I is given by

B = μ0I/2πa where μ0 is the magnetic permeability of free space

The magnetic flux density at a point midway between the wires W1 and W2 is the resultant of the magnetic flux densities produced by these wires. Each wire produces magnetic flux density of magnitude μ0I/2πa.

These fields are directed perpendicular to the plane containing the wires and outwards (towards the reader) and hence they add up to produce a resultant flux density of magnitude μ0I/2πa + μ0I/2πa= μ0I/πa.

[Since the magnetic flux density is a vector, you should not forget to mention its direction].

(b) At the mid point of the square loop PQRS the magnetic fields due to the wires W1 and W2 are directed perpendicular to the plane containing the wires. But the field due to the wire W2 is directed into the plane of the figure (away from the reader) where as the field due to the wire W1 is directed outwards (towards the reader). The resultant field is directed into the plane of the figure (away from the reader) since the wire W2 produces stronger field.

The magnitude of the resultant magnetic flux density at the mid point of the square loop PQRS is [(μ0I)/(2π×3a/2)] – [(μ0I)/(2π×5a/2)]

This is equal to (μ0Ia)[(1/3) (1/5)] = 2μ0I/15πa (c) To find the magnetic flux through the loop PQRS, consider a strip of very small width dx at distance x from the wire W2 as shown in the figure. The resultant magnetic flux density B at the strip is given by

B = 0I/ 2π) [1/x – 1/(a+x)]

[dA is the area of the strip of length a and width dx]

Magnetic flux Ф linked with the entire loop PQRS is given by

Ф = a2a Badx = 0Ia/2π) a2a [1/x – 1/(a+x)]dx

[The limits of integration are x = a and x = 2a].

Therefore, Ф = 0Ia/2π) [ln x – ln (a+x)] between limits x = a and x = 2a.

Or, Ф = = 0Ia/2π) ln [x/(a+x)] between limits x = a and x = 2a.

= 0Ia/2π) [ln(2/3) – ln(1/2)]

= (μ0Ia /2π) ln (4/3) ...............................(i)

[The unit of magnetic flux density is tesla (or, weber per mrtre2) and the unit of magnetic flux is weber].

(d) The average emf Vaverage induced in the loop is given by

Vaverage = Rate of change of magnetic flux

= Change of magnetic flux/ Time

= [(μ0Ia /2π) ln(4/3) – 0]/(50×10–3) since the current through the wires is switched off in a time of 50 ms

Therefore, Vaverage = = (10μ0Ia /π) ln (4/3)

(e) The emf V induced in the loop PQRS at the instant t is given by

V = dФ/dt, the negative sign appearing because of Lenz’s law.

Ignoring the negative sign, we have from equation (i)

V = dФ/dt = [(μ0a /2π) ln (4/3)](dI/dt)

Here dI/dt = 40 As–1, as given in the question.

Substituting for dI/dt, we have V = (20μ0a /π) ln (4/3)

The induced current in the loop = V/R where R is the resistance of the loop, which is equal to 4aρ ohm.

Therefore, induced current = (20μ0a /4π) ln (4/3) = (5μ0 ρ) ln (4/3) ampere.

The direction of the induced current in the loop is clockwise, as indicated in the figure.

The resultant magnetic field is directed normally into the plane of the loop and is decreasing when the current is switched off. The induced current should oppose this change and should therefore produce a magnetic field acting in the same direction (normally into the plane of the loop). This is made possible by the clockwise flow of the induced current.

## Sunday, October 10, 2010

### AP Physics C - Free Response Practice Question on Electromagnetic Induction

“Non-violence leads to the highest ethics, which is the goal of all evolution. Until we stop harming all other living beings, we are still savages”

– Thomas A. Edison

Today I will give you a free response practice question on electromagnetic induction. You may try to answer this question within 15 minutes. Here is the question:

Two infinitely long straight parallel wires W1 and W2, separated by a distance ‘a’ in free space, carry equal currents I flowing in opposite directions as shown in the adjoining figure. A square loop PQRS of side ‘a’, made of nichrome wire of resistance ρ Ω per metre is arranged with its plane lying in the plane of the wires W1 and W2 so that the sides PQ and RS of the loop are parallel to the wires W1 and W2. The side PQ of the loop is at a distance ‘a’ from the wire W2. Now, answer the following questions in terms of the given quantities and fundamental constants:

(a) Determine the magnetic flux density at a point midway between the wires W1 and W2.

(b) Determine the magnetic flux density at the mid point of the square loop PQRS.

(c) Calculate the magnetic flux through the loop PQRS.

(d) What is the average emf induced in the loop when the current through the wires is switched off in a time of 50 ms?

(e) When the current through the wires is switched off, it is found that at a certain instant t, the current decays at the rate of 40 As–1. Calculate the current induced in the loop PQRS at the instant t.

Indicate the direction of the current in the loop and justify your answer.

This question carries 15 points. Try to answer it. I’ll be back soon with a model answer for your benefit.

## Wednesday, October 6, 2010

### AP Physics B - Multiple Choice Practice Questions on Kinetic Theory of Gases

Essential points to be remembered in kinetic theory of gases were discussed in the post dated 13th March 2008. Questions on kinetic theory of gases were discussed subsequently. You can access all posts related to kinetic theory of gases by clicking on the label, ‘kinetic theory’ below this post. To access older posts you need to click on the ‘older posts’ button.

Today we will discuss a few more typical multiple choice questions on kinetic theory of gases:

(1) The root mean square (R.M.S.) speed v of the molecules of an ideal gas is given by the expressions,

v = √(3RT/M ) and

v = √(3kT/m ) where R is universal gas constant, T is the absolute (Kelvin) temperature, M is the molar mass, k is Boltzman’s constant and m is the molecular mass. The R.M.S. speed of oxygen molecules (O2) at temperature T1 is v1. When the temperature is doubled, if the oxygen molecules are dissociated into atomic oxygen, what will be R.M.S. speed of oxygen atoms? (Treat the gas as ideal).

(a) v1/2

(b) v1

(c) √2 v1

(d) 2v1

(e) 4v1

We have v1 = √(3RT1/M ) or

v1 = √(3kT1/m )

On dissociation the molar mass as well as the molecular mass gets halved. Using the second equation, the R.M.S. speed v after dissociation is given by

v = √[3k×2T1/ (m/2 )] = 2√(3kT1/m ) = 2v1

(2) Four moles of an ideal diatomic gas is heated at constant volume from 20º C to 30º C. The molar specific heat of the gas at constant pressure (Cp) is 30.3 Jmol–1K–1 and the universal gas constant (R) is 8.3 Jmol–1K–1. The increase in internal energy of the gas is

(a) 80.3 J

(b) 303 J

(c) 332 J

(d) 880 J

(e) 1212 J

The increase in internal energy is MCvT where M is the mass of the sample of the gas, Cv is the specific heat at constant volume and T is the rise in temperature of the gas. If we use the molar specific heat of the gas at constant volume for Cv, the number of moles in the sample of the gas is to be used in the place of M.

Now, Cv = Cp R = 30.3 – 8.3 = 22 Jmol–1K–1.

Therefore, the increase in internal energy of the gas is 4×22×10 = 880 J.

(3) In the case of real gases, the equation of state, PV = RT (where P, V and T are respectively the pressure, volume and absolute temperature), is strictly satisfied only if corrections are applied to the measured pressure P and the measured volume V. The corrections for P and V arise respectively due to

(a) intermolecular attraction and the size of molecules

(b) size of molecules and expansion of the container

(c) expansion of the container and intermolecular attraction

(d) kinetic energy of molecules and collision of molecules

(e) intermolecular attraction and collision of molecules

In kinetic theory of gases it is assumed that there is no force between molecules But there is actually intermolecular attraction which reduces the pressure. So the correction for P arises due to intermolecular attraction.

The entire volume V of the container is not available for the molecules since the molecules have a finite size. The assumption (in kinetic theory) that the molecules are point masses without appreciable volume is incorrect. So the correction for V arises due to the size of molecules.

The correct option is (a).

(4) Gases exert pressure on the walls of the container because the gas molecules

(a) collide one another

(b) exert intermolecular attraction

(c) possess momentum

(d) expand on absorbing heat

(e) exert repulsive force

Because of the momentum of the gas molecules, they collide with the walls of the containing vessel and momentum transfer takes place, resulting in a force on the walls. Pressure is force per unit area. The basic reason for the pressure is the momentum of the gas molecules [Option (c)].

Now, see similar questions with solution here.