AP Physics B- Additional Multiple Choice Practice Questions on Kinetic Theory and Thermodynamics

Often you will get simple questions to answer. Don’t be too impulsive while answering seemingly simple questions The intention of the question setter in many cases will be to check whether you have mastered the fundamental principles. She may make a question a little bit indirect and you may tend to commit mistakes. It will be a good practice to find time to check your answers, especially in the case of multiple choice questions. Here are some simple questions on kinetic theory and thermodynamics.

(1) In the gas equation, 3PV = RT where R is universal gas constant, V stands for

(a) volume of 3 kg of gas

(b) volume of 3 g of gas

(c) volume of 3 mole of gas

(d) volume of 1/3 kg of gas

(e) volume of 1/3 mole of gas

The gas equation for n moles is

PV = nRT

The gas equation given in the question is

PV = (1/3) RT

The equation is therefore for 1/3 mole and V stands for the volume of 1/3 mole of gas [Option (e)].

(2) Identical balls each of mass m and moving with the same velocity v hit a fixed surface normally and elastically. If there are n hits per second, what is the average force exerted (by the balls) on the surface?

(a) mnv/2

(b) mnv

(c) 2mnv

(d) mnv²/2

(e) 2mnv²

Since the collision is elastic, the balls will rebound with the same speed. The initial momentum of each ball is mv and the momentum after collision with the surface is –mv, the negative sign indicating the reversal of direction. The change of momentum due to the collision is –mv – mv = – 2mv. Each collision therefore transfers a momentum 2mv to the surface. Since there are n collisions per second, the momentum transferred per second to the surface (which is the rate of change of momentum) is 2mnv.

Rate of change of momentum is force according to Newton’s second law and hence the average force exerted (by the balls) on the surface is 2mnv [Option (c)].

(3) The root mean square speed of molecules of a gas in a container maintained at temperature T is 600 ms^–1. If 25% of the gas molecules leak out at this temperature, the root mean square speed of the remaining molecules is

(a) 300 ms^–1

(b) 600 ms^–1

(c) 600√2 ms^–1

(d) 1200 ms^–1

(e) 2400 ms^–1

The root mean square speed of gas molecules is directly proportional to the square root of absolute temperature. [This follows from the postulate of kinetic theory which states that the average kinetic energy of gas molecules is directly proportional to the absolute temperature].

Since the temperature remains constant, the root mean square speed is unchanged [Option (b)].

(4) A cylinder containing a gas is divided into two parts A and B by a tight fitting fixed piston (Fig.). The pressure and volume of gas in the two parts A and B respectively are (P, 2V) and (4P, V). If the piston is released so that it can slide freely under isothermal condition, then the volume of the gas in the two parts A and B are respectively

(a) V, 2V

(b) 1.5V, 1.5V

(c) V, 2V

(d) 2V, V

(e) 0.5V, 2.5V

When the piston is left free, it moves towards the left and the pressures in the two parts get equalized. If this common pressure is P’, we have (since the temperature is unchanged)

P×2V + 4P×V = P’×3V, from which P’ = 2P

If V₁ and V₂ are the final volumes of gas in the parts A and B respectively, we have

P×2V = P’V₁ and 4P×V = P’V₂

Substituting for P’ we obtain V₁ = V and V₂ = 2V [Option (a)].

[You can solve the above problem in no time if you note that the product of pressure and volume of gas in part B is twice the product of pressure and volume in part A. Since the final pressure is the same, the volume of gas in part B has to be twice the volume in part A. Let us consider another example:

What would be the final volumes in parts A and B if the initial pressure- volume states were (2P, 3V) and (12P, 2V) respectively?

The pressure volume products are 6PV and 24PV. So the total volume of 5V is to be divided in the ratio 6:24, that is 1:4. We obtain V₁ = 5V×1/5 = V and V₂ = 5V×4/5 = 4V].

(5) A Carnot engine working between 300 K and 800 K has a work output of 1000 joule per cycle. The amount of heat absorbed by the engine from the source is

(a) 400 J

(b) 500 J

(c) 800 J

(d) 1200 J

(e) 1600 J

In the case of a Carnot engine we have the expression for efficiency h given by

h = (Q₁ – Q₂)/Q₁ = (T₁– T₂)/T₁ where Q₁ is the quantity of heat absorbed from the source at higher temperature T₁ and Q₂ is the quantity of heat liberated to the sink at lower temperature T₂.

Since Q₁ – Q₂ is the work output, we have

1000/Q₁ = 500/800 from which Q₁ = 1600 J

(6) The height of a water fall is h metre. Assume that the entire energy due to the fall is converted into heat. If the specific heat capacity of water is c Jkg^–1K^–1 and the acceleration due to gravity is g ms^–2, the temperature difference between the top and bottom of the fall in degree Kelvin is

(a) c/gh

(b) gh/c + 273

(c) gh/c

(d) cgh

(e) cgh + 273

Water loses gravitational potential energy while falling down and gains an equivalent thermal energy, resulting in a rise of temperature. Considering a mass m of water, if ∆T is the rise in temperature, we have

mgh = mc∆T

Therefore, ∆T = gh/c

You will find some useful multiple choice questions (with solution) in this section here.

AP Physics B & C- Electrostatics- Practice Multiple Choice Questions on Electric Potential

“Your worth consists in what you are and not in what you have”

–Thomas A. Edison

Multiple choice as well as free response practice questions on electrostatics (with solution) were posted on this blog on many occasions. You can access them either by clicking on the label ‘electrostatics’ below this post or by trying a search for ‘electrostatics’, making use of the search box on this page. In both cases you would need to click on the ‘older posts’ and ‘newer posts’ buttons to access all posts. Today we will discuss a few more multiple choice practice questions in this section.

The following questions are meant for AP Physics B as well as AP Physics C aspirants:

(1) In a region of the atmosphere near the earth’s surface there exists an electric field of intensity 100 Vm^–1directed vertically downwards. Consider a line segment of length 4 m inclined at 60º with the vertical (fig.). What is the potential difference between the points A and B?

(a) 200 V

(b) 100 V

(c) 60 V

(d) 50 V

(e) 10 V

Since the electric field is 100 Vm^–1and is directed vertically, a difference in height of 1 m corresponds to a potential difference of 100 V. The difference in height between the points A and B is AB cos60º = 4×½ = 2 m.

Therefore, the potential difference between points A and B is 200 V.

(2) A positive point charge Q located at point A produces an electric field of 50 NC^–1 at point B. If an equal and opposite charge (– Q) is placed additionally at the point B, the electric field and electric potential at the mid point of the line AB are respectively

(a) 400 NC^–1, 50 V

(b) 400 NC^–1, 0 V

(c) 200 NC^–1, 50 V

(d) 200 NC^–1, 0 V

(e) 100 NC^–1, 0 V

The electric fields due to the positive charge at A and the negative charge at B are directed from A to B and they will get added at the mid point of the line AB. The magnitudes of these fields are equal and is 200 NC^–1 each, since the field is quadrupled when the distance is halved (in accordance with the inverse square law). The resultant electric field at the mid point of the line AB is 400 NC^–1.

The electric potential at the mid point of the line AB is zero since the positive potential produced by the positive charge is canceled by the equal negative potential produced by the negative charge [Option (b)].

(3) Two conducting spheres A and B have radii R and 3R respectively. They are given charges Q and 3Q respectively. If they are connected by a copper wire,

(a) a charge Q/2 will flow from B to A

(b) a charge Q/2 will flow from B to A

(c) a charge Q will flow from B to A

(d) a charge Q will flow from A to B

(e) no charge will flow through the wire

The potentials of A and B are the same. Hence no charge will flow through the wire.

[Note that the potentials of A and B are (1/4πε₀)(Q/R) and (1/4πε₀)(3Q/3R), which are equal].

(4) Point charges 1μC, 2μC, –3μC and 4μC are arranged in free space at the corners A, B, C and D of a square ABCD of side √2 m. If 1/4πε₀ is approximately equal to 9×10⁹ Nm²C^–2, the electric potential at the centre of the square is

(a) 1.8×10⁴ V

(b) 3.6×10⁴ V

(c) 2.7×10⁴ V

(d) –2.7×10⁴ V

(e) –6.3×10⁴ V

The diagonal of the square has length 2 m and the distance of the centre from the corners is 1 m. The positive charges of total value 7μC will produce (at the centre) a total positive potential of (1/4πε₀) (7×10^–6 /1) volt = (9×10⁹) (7×10^–6) volt = 6.3×10⁴ volt.

[Note that the potential due to a point charge Q at distance r is (1/4πε₀)(Q/r)].

The negative charge of value –3μC will produce (at the centre) a negative potential of (1/4πε₀) (–3×10^–6 /1) volt = (9×10⁹) (–3×10^–6) volt = –2.7×10⁴ volt.

Therefore, the electric potential at the centre of the square is (6.3×10⁴ –2.7×10⁴) volt = 3.6×10⁴ volt.

(5) The potential on an equipotential surface is V volt. The work done in moving a charge q through a distance d along the equipotential surface is

(a) directly proportional to V, q and d

(b) directly proportional to V and q but inversely proportional to d

(c) directly proportional to V and d but inversely proportional to q

(d) directly proportional to q and d but inversely proportional to V

(e) independent of V, q and d

The correct option is (e) since the work done in moving a charge along the equipotential surface is zero.

[Since the surface is equipotential, you don’t require a force to move the charge along the surface].

The following questions are meant for AP Physics C aspirants; but AP Physics B aspirants also can ‘enjoy’ them:

(6) Two identical thin rings A and B of radius R are arranged coaxially with a separation R between their centres. Ring A carries positive charge 2Q where as ring B carries positive charge Q. What is the external work requied to move a positive charge q from the centre of A to the centre of B?

(a) (1/4πε₀)[Q/(√2 R)]

(b) (1/4πε₀)(2Q/R)

(c) (1/4πε₀)[Q/(√2 R) – Q/R]

(d) (1/4πε₀)[Q/R – Q/(√2 R)]

(e) (1/4πε₀)[Q/(√2 R) + Q/R]

With reference to the adjoining figure, the electric potential V₁ at the centre of ring A is given by

V₁ = (1/4πε₀)[2Q/R + Q/(√2 R)]

[The first term is due to the charge on A while the second term is due to the charge on B].

The electric potential V₂ at the centre of ring B is given by

V₂ = (1/4πε₀)[Q/R + 2Q/(√2 R)]

[The first term is due to the charge on B while the second term is due to the charge on A].

The potential difference V between the centres of A and B is given by

V = V₁ the centre of A V₂ = (1/4πε₀)[2Q/R – Q/R + Q/(√2 R) – 2Q/(√2 R)]

Or, V = (1/4πε₀)[Q/R – Q/(√2 R)]

Since the centre of A is at a higher positive potential, a positive charge q placed at the centre of A will move by itself towards the centre of B. So the work to be done by the external agency is negative and the answer to the question is – V and we have

– V = (1/4πε₀)[Q/(√2 R) – Q/R], as given in option (c).

(7) A total charge (positive) of 1.11×10^–9 C is sprayed non uniformly on a non conducting ring of radius 1 m. If E represents the electric field at any point in space due to this charge distribution and r = 0 represents the centre of the ring, what is the value (in joule per coulomb) of the line integral ∫– E.dr from r = ∞ to r = 0?

(a) zero

(b) ∞

(c) 1

(d) 10

(e) 100

The quantity given by the line integral from infinity to the centre of the ring is the work done in bringing unit positive charge from infinity to the centre of the ring. This is the electric potential at the centre of the ring and is equal to (1/4πε₀)(Q/R) where Q is the total charge on the ring and R is the radius of the ring.

Since 1/4πε₀ = 9×10⁹ Nm²C^–2 (very nearly), we obtain the answer as 9×10⁹×1.11×10^–9/1 = 10, very nearly.

Well, you will find a very useful post in this section here.

Geometric Optics for AP Physics B – Additional Multiple Choice Questions

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

Equations to be remembered in geometric optics were discussed in the post dated 29^th December 2007, followed by some questions with solution in subsequent posts. You can access all those posts either by clicking on the label ‘geometric optics’ (or ‘optics’) below this post or by performing a search within this blog. Today we will discuss a few more multiple choice questions (for practice) in this section:

(1) A converging lens is made of 5 layers of transparent material (fig.). Layers 1, 3 and 5 are of refractive index n₁ where as layers 2 and 4 are of refractive index n₂. A beam of monochromatic light from a distant object, proceeding parallel to the principal axis, is incident on this lens. How many images will be formed?

(a) 1

(b) 2

(c) 3

(d) 5

(e) Nil

Each layer can function as a converging lens of its own focal length f determined by the refractive index n and the radii of curvature R₁ and R₂, as given by lens maker’s equation,

1/f = (n – 1) (1/R₁ – 1/R₂)

Since the lens is made of layers of two different materials, n has two values (n₁ and n₂) and so the focal length f has two values. Therefore there will be two images [Option (b)].

(2) A converging lens is made of 3 portions (1, 2 and 3 in fig.) of three different transparent materials of refractive indices n₁, n₂ and n₃ respectively. A beam of monochromatic light from a distant object, proceeding parallel to the principal axis, is incident on this lens. How many images will be formed?

(a) 1

(b) 2

(c) 3

(d) 6

(e) 9

This lens can be imagined to be made of three different lenses in contact. The combination can have a single focal length and hence there will be just one image [Option (a)].

(3) light is incident normally on a crown glass plate of thickness 3 mm. If the refractive index of crown glass is 1.5 and the speed of light in free space is 3×10⁸ ms^–1 , the time taken by light to travel this thickness of glass is

(a) 1.5×10^–5 s

(b) 3×10^–8 s

(c) 2×10^–11 s

(d) 1.5×10^–11 s

(e) 1.5×10^–12 s

The speed v of light in glass is given by

v= c/n

where c is the speed in free space and n is the refractive index of glass.

Therefore v = 3×10⁸/1.5 = 2×10⁸ ms^–1

The time t taken by light to travel 3 mm (= 3×10^–3 m) thickness of glass is given by

t = 3×10^–3/ 2×10⁸ = 1.5×10^–11 s.

(4) A point source of light (S) is arranged at a depth d in a large water tank. If the critical angle of water is C, what is the area of the water surface that transmits the light from the source?

(a) π d² sin²C

(b) π d² cot²C

(c) π d² tan²C

(d) π d² cos²C

(e) π d² cosec²C

All the light rays within a cone of semi vertical angle C will be transmitted (fig.). The area of the water surface that transmits the light from the source is the area of the circle of radius r given by r = d tanC, as shown in the figure.

[Note that the rays incident at the water surface at an angle C will graze the surface where as the rays incident at at an angle greater than C will be totally reflected back into the water].

Therefore, the area of the water surface that transmits the light from the source is given by

π r² = π (d tanC)² = π d² tan²C

(5) Pick out the incorrect statement:

(a) The refractive index of a material depends on the frequency of the light used for the measurement

(b) The speed of light in a medium depends on the refractive index of the medium

(c) Critical angle of water is smaller for violet light compared to that for red light

(d) In all material media yellow light travels with the same speed

(e) In free space the speed of light is a constant irrespective of the wave length of light.

The speed v of light in a medium is related to the refractive index of the medium as

v = c/n where c is the speed of light in free space and n is the refractive index of the medium.

Different media have different refractive indices for light of a given colour (as determined by the frequency) and for any given medium, the refractive index is maximum for violet light and minimum for red light. There is nothing special about the yellow light and as with light of any other colour, the speed will be different in different material media.

Therefore, the only incorrect statement in the above question is (d).