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.

Try to answer the question. I’ll be back soon with a model answer for your benefit.

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

A free response practice question on electromagnetic induction was posted on 10^th 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 (μ₀I/πa)[(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 = (μ₀I/ 2π) [1/x – 1/(a+x)]

Magnetic flux linked with the strip = B dA = Badx

[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 = _a∫^2a Badx = (μ₀Ia/2π) _a∫^2a [1/x – 1/(a+x)]dx

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

Therefore, Ф = (μ₀Ia/2π) [ln x – ln (a+x)] between limits x = a and x = 2a.

Or, Ф = = (μ₀Ia/2π) ln [x/(a+x)] between limits x = a and x = 2a.

= (μ₀Ia/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 V_average induced in the loop is given by

V_average = 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, V_average = = (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 /4πaρ) ln (4/3) = (5μ₀ /πρ) 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.

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.

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 13^th 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 (C_p) 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 MC_v∆T where M is the mass of the sample of the gas, C_v 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 C_v, the number of moles in the sample of the gas is to be used in the place of M.

Now, C_v = C_p – 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.

AP Physics B & C – Few More Multiple Choice Practice Questions on Friction

"What you are will show in what you do."

– Thomas A. Edison

Questions involving friction were discussed on this site earlier. You can access them by clicking on the label ‘friction’ below this post or by trying a search for ‘friction’, using the search box provided on this page.

Today we will discuss some more multiple choice practice questions on friction:

(1) A uniform chain is placed on a rough horizontal table so that one end of the chain hangs down over the edge of the table. When 20% of the length of the chain hangs over the edge, it starts sliding. What is the coefficient of static friction between the chain and the table?

(a) 0.1

(b) 0.15

(c) 0.25

(d) 0.35

(e) 0.5

When the chain just begins to slide, the frictional force between the chain and the table is maximum and is called limiting frictional force which is equal to μ_sN where μ_s is the coefficient of static friction between the chain and the table and N is the normal force.

Here N = 0.8mg and μ_sN = 0.2mg where m is the mass of the chain.

[Note that the weight of 20% of the length of the chain (= 0.2mg) balances the frictional force μ_sN].

Therefore, μ_s×0.8 mg = 0.2 mg from which μ_s = 0.25

(2) A wooden block is placed on a horizontal surface and a horizontal force equal to the limiting frictional force is applied on it. If the coefficient of static friction and the coefficient of kinetic friction are respectively 0.5 and 0.4 and the acceleration due to gravity is 10 ms^–2, the acceleration of the wooden block is

(a) 1 ms^–2

(b) 1.1 ms^–2

(c) 1.4 ms^–2

(d) 1.5 ms^–2

(e) zero

The limiting (or, maximum) frictional force is μ_s mg where μ_s is the coefficient of static friction, and mg is the weight of the wooden block. When the block moves along the horizontal surface, the frictional force acting is reduced to kinetic frictional force and is equal to μ_k mg where μ_k is the coefficient of kinetic friction.

[If the frictional force were not reduced, the block would have moved with uniform velocity since there would be no net force].

The net force acting on the block is μ_s mg – μ_k mg and the acceleration a of the block is given by,

a = (μ_s mg – μ_k mg)/m = (μ_s – μ_k) g = (0.5 – 0.4) × 10 = 1 ms^–2

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(3) A slab of mass 50 kg is placed on frictionless horizontal floor and a block of mass 10 kg is placed on the slab as shown. The coefficient of static friction and the coefficient of kinetic friction between the slab and the block are respectively 0.4 and 0.2 and the acceleration due to gravity is 10 ms^–2. A horizontal force equal to 30 N acts on the block. The acceleration of the slab will be

(a) 0.1 ms^–2

(b) 0.2 ms^–2

(c) 0.3 ms^–2

(d) 0.5 ms^–2

(e) 1.2 ms^–2

The limiting frictional force (μ_sN) between the block and the slab is given by

μ_sN = μ_sMg = 0.4×10×10 = 40 N

The block will not slide along the slab since the applied force (= 30 N) is less than the limiting frictional force. So the slab and the block will move together, with acceleration a given by

a = 30 N/(50+10)kg = 0.5 ms^–2.

[Since there is no relative motion between the slab and the block, the coefficient of kinetic friction does not play any role in the above problem. It just serves the purpose of a distraction].

The above questions may be ‘enjoyed’ by AP Physics B & C aspirants. The following questions are specifically for AP Physics C aspirants:

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(4) A slab A of mass M = 6 kg is placed on rough horizontal floor. The coefficient of kinetic friction between the floor and the slab is 0.1. A block B of mass m = 4 kg is placed on the slab (Fig.). The coefficient of static friction between the slab and the block is 0.2. When a horizontal force of magnitude F is applied on the slab, the block B just begins to slide along the slab A. What is the value of F? (Take g = 10 ms^–2).

(a) 40 N

(b) 30 N

(c) 15 N

(d) 10 N

(e) 8 N

The limiting frictional force between the slab and the block is μ_smg where μ_s is the coefficient of static friction between the slab and the block.

When the force F is applied on the slab, the entire system containing the slab and the block must move with an acceleration a so that the inertial force (= ma) on the block just balances the frictional force.

[The inertial force must be infinitesimally greater than the frictional force for the slipping to occur].

Therefore, we have μ_smg = ma so that a = μ_sg = 0.2×10 = 2 ms^–2

If the floor were smooth, the force to be applied on the slab to attain the slipping condition would have been a(M+m).

But since the floor is rough, the applied force has to overcome the force of kinetic friction [μ_k(M+m)g] between the floor and the slab.

Therefore, F = μ_k(M+m)g + a(M+m)

Thus F = (M+m)( μ_kg + a) = 10(1+2) = 30 N.

[The above question could be made a little more difficult if the coefficient of static friction (say, 0.25) between the floor and the slab also is given (to distract you). You should remember that once the slab slides along the floor, the friction called into play is kinetic friction and hence your answer will be unchanged].

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(5) A man of mass M = 80 kg tries to pull down a tree using a rope. He exerts the pulling force at an angle of 30º with the horizontal (Fig.). If the coefficient of static friction between the man’s feet and the ground is 0.6 and the gravitational acceleration is 10 ms^–2, the maximum pulling force he can exert (before he slips) is approximately

(a) 50 N

(b) 100 N

(c) 200 N

(d) 400N

(e) 600 N

If the man exerts too much pulling force his feet will slip and hence the maximum pulling force is determined by the coefficient of friction and the angle of pull. If T is the maximum tension produced in the rope, the maximum pulling force, F = T.

The vertical component of T which acts upwards reduces the normal force N and we have

N = Mg – T sin 30º = 80×10 – T/2

The maximum frictional force = μ_sN = 0.6×(800 – T/2).

The horizontal component of the maximum tension T = T cos 30º = T×(√3)/2

The frictional force should balance the horizontal component of tension so as to prevent the man from slipping.

Therefore we have

0.6×(800 – T/2) = T ×(√3)/2

Or, 960 – 0.6 T = 1.732 T so that T = 960/2.332 = 400 N, approximately