Multiple Choice Practice Questions on Heat Transfer and Thermal Expansion for AP Physics B

“Genius is one percent inspiration and ninety nine percent perspiration”
– Thomas A. Edison

Today we will discuss some practice questions (MCQ) on heat transfer and thermal expansion. You may click here to obtain the essential points you need to note in this section.
Here are the questions:
(1) When water is heated from 0º C to 20º C its volume
(a) goes on increasing
(b) goes on decreasing
(c) remains constant up to 15º C and then increases
(d) first decreases and then increases
(e) remains constant up to 4º C and then increases
This question is meant just for checking your knowledge of the behaviour of water. Water has maximum density at nearly 4º C and hence the correct option is (d).
(2) 5 g of ice at 0º C is mixed with 10 g of water at 10º C. The temperature of the mixture is
(a) 0º C
(b) 2º C
(c) 2.5º C
(d) 5º C
(e) 7.5º C
In order to melt 5 g of ice (into water) without change of temperature 400 calories of heat are required since the latent heat of fusion for ice-water change is nearly 80 calories per gram. The heat that is released by 10 g of warm water at 10º C on cooling to 0º C is 100 calories only since the specific heat of water is 1 calorie per gram per Kelvin.
So the warm water can melt just a quarter of the amount of ice and the mixture will remain at 0º C [Option (a)].

(3) Equal masses of three liquids of specific heats C₁, C₂ and C₃ at temperatures t₁, t₂ and t₃ respectively are mixed. If there is no change of state, the temperature of the mixture is

(a) (t₁+ t₂ + t₃)/3

(b) (C₁t₁+ C₂t₂ + C₃t₃)/[3(C₁+ C₂ + C₃)]

(c) (C₁t₁+ C₂t₂ + C₃t₃)/ (C₁+ C₂ + C₃)

(d) 3(C₁t₁+ C₂t₂ + C₃t₃)/ (C₁+ C₂ + C₃)

(e) 3(t₁+ t₂ + t₃)

If the mass of each liquid is m, the total amount of heat (H) initially is given by

H = m(C₁t₁+ C₂t₂ + C₃t₃)

After mixing the same amount of heat is available. If the common temperature is t, we have

H = mt(C₁+ C₂ + C₃)

From the above equations, t = (C₁t₁+ C₂t₂ + C₃t₃)/ (C₁+ C₂ + C₃)

(4) The amount of heat required to raise the temperature of one mole of an ideal mono atomic gas through 2º C at constant pressure is (universal gas constant = R)
(a) 2 R
(b) 3 R
(c) 5 R
(d) 5R/2
(e) 7R/2
The molar specific heat of a mono atomic ideal gas at constant pressure (cp) is 5R/2 where R is the universal gas constant.
[The molar specific heat of a mono atomic ideal gas at constant volume (cp) is 3R/2 and in accordance with Meyer’s relation, cp = cv + R].
Therefore, the amount of heat required to raise the temperature of one mole of an ideal mono atomic gas through 2º C at constant pressure is 1×(5R/2) ×2 = 5R.

(5) Two identical rectangular strips, one of copper and the other of steel, are riveted as shown to form a bi-metal strip. On heating, the bi-metal strip will

(a) get twisted

(b) remain straight

(c) bend with steel on the convex side

(d) bend with steel on the concave side

(e) contract

On heating, the copper strip will suffer greater elongation and hence the bimetal strip will bend with the steel strip on the concave side.

[Bimetal strips are widely used in thermal switching applications such as automatic electric iron].

(6) Four cylindrical rods of different radii and lengths are used to connect two heat reservoirs at fixed temperatures t₁ and t₂ respectively. From the following pick out the rod which will conduct the maximum quantity of heat:

(a) Radius 1 cm, length 1 m

(b) Radius 1 cm, length 2 m

(c) Radius 2 cm, length 4 m

(d) Radius 3 cm, length 8 m

(e) Radius 0.5 cm, length 0.5 m

The quantity of heat conducted is directly proportional to the area of cross section and inversely proportional to the length of the rod (when the same temperature difference exists between the ends).

[Remember that the quantity of heat Q = KAdθ/dx where K is the thermal conductivity, A is the area of cross section and dθ/dx is the temperature gradient].

When you compare rod (a) with rod (b) you find that rod (a) can conduct better since its length is less than that of rod (b).

Rod (a) and rod (c) conduct equally since the cross section area as well as the length of rod (c) is 4 times that of rod (a).

Rod (d) is better than rod (a) since its cross section area is 9 times that of rod (a) while its length is only 8 times that of rod (a).

Rod (e) is worse than rod (a) since its cross section area is a quarter of that of rod (a) while its length is half that of rod (a).

Therefore, the rod which will conduct the maximum quantity of heat is (d).

[It is enough to compare the ratio of the area to length. The area is directly proportional to the square of the radius. Since the unit of radius is centimetre and that of length is metre in all cases, you can blindly compare the values of 1²/1, 1²/2, 2²/4, 3²/8 and (0.5)²/(0.5). The highest value is 3²/8].

AP Physics B - Multiple Choice Practice Questions on Standing Waves in Stretched Strings and Air Columns

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“The reason a lot of people do not recognize opportunity is because it usually goes around wearing overalls looking like hard work”

– Thomas A Edison

AP Physics B aspirants are expected to have a clear understanding of the standing wave modes for stretched strings fixed at both ends. They should also have a clear understanding of standing sound waves in pipes with either closed or open ends. A pipe closed at both ends is of no use and therefore of no interest to us. You should note that a closed pipe means a pipe closed at one end. An open pipe means a pipe open at both ends.

The following multiple choice practice questions are meant for checking your understanding basic points in respect of waves and the physics of standing waves (stationary waves) in stretched strings and air columns (in pipes).

(1) Sound does not pass through

(a) steel

(b) diamond

(c) nitrogen

(d) water

(e) vacuum

Sound requires a material medium for its propagation. So sound does not pass through vacuum.

(2) When the amplitude of a wave is increased by 50%, its intensity will be increased by

(a) 50%

(b) 100%

(c) 125%

(d) 150%

(e) 200%

Intensity of any wave is directly proportional to the square of the amplitude. Therefore, when the amplitude becomes 1.5 times (increment by 50%) the original value, the intensity becomes 2.25 times (1.5² times) the original intensity. The increment in intensity is 125% [Option (c)].

(3) Ultrasonic waves from a sonar undergoes refraction at the interface between water and air. Which one of the following characteristics of the wave remains unchanged?

(a) Wave length

(b) Speed

(c) Period

(d) Energy

(e) None of the above

The correct option is (c). The period (and of course frequency) of the wave remains unchanged.

(4) A stationary sound wave is produced in a resonance column apparatus using an electrically excited tuning fork. If P and Q are consecutive nodes, which one of the following statements is correct?

(a) If P is a position of condensation, Q is a position of rarefaction

(b) If P is a position of condensation, Q also is a position of condensation

(c) If P is a position of condensation, Q is a position of normal density (of air)

(d) Both P and Q are positions of normal density (of air)

(e) Both P and Q are positions of rarefaction

In a stationary wave the particles of the medium at the nodes will be always at rest. The phase of vibration of particles (of the medium) lying on one side of a node is opposite to the phase of vibration of particles lying on the opposite side. Therefore, if one node is a position of condensation, the next node is a position of rarefaction [Option (a)].

[Note that the particles at the antinodes will vibrate with maximum amplitude; but the air at the anti-node will have normal density (neither condensed nor rarefied]

(5) A cylindrical pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of its length is in water. The fundamental frequency of air column in this condition is

(a) 4 f

(b) 3 f

(c) 2 f

(d) f

(e) f/2

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In the fundamental mode there is a node at the middle of the open pipe and the anti-nodes are at the ends. When half of the pipe is dipped in water, there is a node at the water surface and in the fundamental mode the neighbouring anti node is at the open end, out side water (fig.). The distance from node to the neighbouring anti-node is λ/4 where λ is the wave length of sound. Evidently λ/4 = half the length of the pipe so that the wave length in the fundamental mode is the same in both cases. Therefore, the fundamental frequency is unchanged on dipping half the length of the pipe in water [Option (d)].

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(6) A stationary wave of frequency 30 Hz is set up in a string of length 1.5 m fixed at both ends. The string vibrates with 3 segments as shown in the adjoining figure. The speed of the wave along the string is

(a) 10 ms^–1

(b) 20 ms^–1

(c) 30 ms^–1

(d) 60 ms^–1

(e) 90 ms^–1

The distance between consecutive nodes (or anti-nodes) in a stationary wave is λ/2 where λ is the wave length. Therefore we have (from the figure) λ/2 = 0.5 m so that λ = 1m.

Since speed v = n λ where n is the frequency we have

v = 30×1 = 30 ms^–1

(7) What is the fundamental frequency of vibration of the string in the above question?

(a) 5 Hz

(b) 10 Hz

(c) 15 Hz

(d) 30 Hz

(e) 60 Hz

The speed of waves in the string is unchanged since the tension is unchanged. Since speed v = n₁λ₁ where n₁ is the fundamental frequency and λ₁ is the wave length in the fundamental mode of vibration, we have

n₁ = v/λ₁

In the fundamental mode of vibration, the entire length of the string forms a single segment (with anti-node at the middle and nodes at the ends). Therefore we have

λ₁/2 = length of string = 1.5 m so that λ₁ = 3 m.

Substituting, n₁ = v/λ₁ = 30/3 = 10 Hz.

[You can work out this problem in no time remembering that the fundamental frequency is one third of the frequency with which the string vibrates with three segments. If the string were originally vibrating with four segments, the fundamental frequency would be one fourth].

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Now see these multiple choice questions.

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 24^th 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.

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.