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

## Friday, December 24, 2010

### AP Physics C - Multiple Choice Practice Questions involving Electric Flux and Gauss’s Law

Gauss’s law (Gauss theorem) in electrostatics is a very useful tool for calculating electrostatic field produced by a system of charges when the system has some symmetry. You might have realized how simple the calculation is if you choose a suitable Gaussian surface. You should definitely remember the statement of Gauss’s law:

“Electric flux through any closed surface S = Q0 where Q is the total charge enclosed by the surface S and ε0 is the permittivity of free space.”

Today we will discuss a few multiple choice practice questions involving electric flux and Gauss’s law:

(1) A closed cubical surface of side a contains an electric dipole of moment p. The total electric flux through the cubical surface is

(a) zero

(b) 8a2p

(c) 6a2p

(d) 8p/ε0a

(e) 6p/ε0a

The electric dipole contains positive and negative charges of equal magnitude. Therefore the total charge inside the cubical surface is zero. By Gauss’s law the total electric flux through the cubical surface is zero [Option (a)].

(2) A uniform electric field E given by E = 200 ĵ NC–1 exists in a certain region of space. The electric flux through a square of area A = 0.04 î m2 located in this field is (î and ĵ are unit vectors along the x-direction and y-direction respectively)

(a) 5000 Nm2C–1

(b) 5000 Nm2C–1

(c) 8 Nm2C–1

(d) 8 Nm2C–1

(e) zero

The area vector by convention is perpendicular to the plane of the area. Since the area vector is directed along the x-axis, you know that the square lies in the yz plane. Therefore, there is no electric flux passing through the square and the correct option is (e).

[In vector notation electric flux is the scalar product (dot product) of E and A. Therefore we have flux Ф = E.A = 200 ĵ . 0.04 î = 0]. (3) A cubical closed surface has N inward electric flux lines passing through one face and N outward electric flux lines passing through the opposite face as shown in the adjoining figure. The flux lines exhibit convergence and there are no flux lines passing through any other face. If the magnitude of the electric field at the centre of the cube is E, the total charge inside the cube is

(a) ε0N

(b) – ε0N

(c) 2ε0N

(d) – 2ε0N

(e) zero

Since the inward flux and the outward flux have the same value, the net flux through the closed surface is zero. By Gauss’s law, we have

0 = q0 where q is the total charge inside the closed surface.

Therefore q = 0. (4) A point positive charge Q is placed at the centre of a cubical closed surface. The outward electric flux through one pair of opposite faces of the cube is

(a) zero

(b) Q0

(c) 2Q0

(d) Q/2ε0

(e) Q/3ε0

The total electric flux lines originating from the positive charge Q is Q0. Since these flux lines are distributed equally among the six faces (or three pairs of faces) of the cub e, the flux through one pair of opposite faces of the cube is Q/3ε0.

(5) A hemispherical bowl of radius R is kept inverted on a horizontal table (so that the mouth of the bowl is in contact with the surface of the table). Assume that the surface of the table coincides with the XZ plane of a right handed Cartesian coordinate system. If a uniform electric field E directed along the negative z-direction exists in the entire region, what is the electric flux through the mouth of the hemispherical bowl?

(a) πR2E

(b) πR2E/3

(c) πR2E/3

(d) πR2E

(e) Zero

As the mouth of the bowl is lying in the XZ plane the electric field E (which is along the negative z-direction) is parallel to the plane of the mouth. The electric flux through the mouth of the bowl is therefore zero. (6) A point charge Q is directly above the centre of a square of side a at distance a/2 from the centre (Fig.). The electric flux through the square is

(a) 2Q0

(b) Q0

(c) Q/2ε0

(d) Q/3ε0

(e) Q/6ε0

You may imagine the charge Q to be at the centre of a cube of side a. The total flux through all the six faces of the cube is Q0 by Gauss’s law. Now think of the square given in the question. Since it makes one face of the cube, the flux through it will be one sixth of the total flux. So the correct option is Q/6ε0.

Now, see some useful multiple choice questions (with solution) in this section here.

## Saturday, December 11, 2010

### AP Physics C - Multiple Choice Practice Questions on Transients in RC and RL circuits

“Nearly every man develops an idea, works it up to the point where it looks impossible, and then he gets discouraged. That’s not the place to become discouraged.”
– Thomas A. Edison
Electric transient phenomena associated with circuits containing capacitors and resistors (CR circuits) are important from theoretical as well as practical point of view and they find place in the AP Physics C course. Transient phenomena associated with circuits containing inductors and resistors (LR circuits) are not as important as those associated with CR circuits but you should not ignore them.
Today we will discuss some multiple choice practice questions in this section. It will be a good idea to get yourself equipped sufficiently for the exercise by referring to my on this site.
Here are the questions with their solution: (1) In the circuit shown the the battery has emf E and internal resistance r. The switch S is closed at time t = 0. If the voltage across the capacitor is 90% of the emf E of the battery after time t’, what is the value of t’?
(a) RC ln10
(b) e–0.9/RC
(c) e0.9/RC
(d) (R+r)C ln(0.9)
(e) (R+r)C ln10
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 volt, the charge Q coulomb on the capacitor after a time t seconds is given by
Q =Q0 (1 e–t/RC) where Q0 is the final (maximum) charge and ‘e’ is the base of natural logarithms.
Since the voltage V across the capacitor is Q/C, it follows that
V =V0 (1 e–t/RC) where V0 is the final (maximum) voltage which is the emf E of the battery.
Therefore we have
0.9 E = E[1 e–t’/(R+r)C] since the total resistance in the circuit is R+r.
This gives e–t/(R+r)C = 0.1
Or, et/(R+r)C = 10 from which t’ = (R+r)C ln10 (2) The adjoining figure shows a resistance R connected in series with an inductance L and a battery of emf E through a switch S. The internal resistance of the battery is negligible. Which one of following graphs represents the nature of variation of the potential drop V across the resistance with time? If an inductor of inductance L and a resistor of resistance R are connected in series with a battery, the current I in the circuit increases exponentially with time and is given by
I = I0 (1 e–Rt/L) where I0 is the final (maximum) current in the circuit.
This shows that the current at time t = 0 (immediately after closing the switch) is zero. As time t increases, the current increases exponentially and reaches the final maximum value I0 (= E/R) when t becomes infinite.
As V = IR the potential drop V across R also increases exponentially with time as indicated by graph (a). (3) A capacitor C charged to V0 volt is discharged through the series combination of two resistances R ohm and 2R ohm by closing the switch S (Fig.). The base of natural logarithms is e. What is the current in the circuit when 3RC seconds are elapsed after closing the switch S?
(a) V0/(RC)
(b) V0/(3RC)
(c) V0/(3R)
(d) V0/(RCe)
(e) V0/(3Re)
The time constant of the circuit is 3RC and hence the voltage across the capacitor after 3RC seconds must be V0/e. You can write this if you remember how the time constant of an ‘RC circuit’ is defined.
The discharge current at this instant is equal to (V0/e)/(3R) = V0/(3Re).
[The initial charge Q0 on the capacitor decays exponentially with time t as given by the equation, Q = Q0 e–t/3RC where Q is the charge at time t. The discharge current at time t is given by
I = dQ/dt = (Q0/3RC) e–t/3RC
The negative sign just shows that the discharge current is opposite to the charging current. Ignoring the negative sign, the discharge current when t = 3RC is Q0/3RCe.
But Q0/C = V0, the initial voltage across the capacitor.
Therefore I = V0/(3Re)].
(4) In question no.3 above, the voltage across the resistance R and the current in it immediately after closing the switch S are respectively
(a) 3V0, V0/(3R)
(b) 2V0/3, zero
(c) V0/3, V0/(3R)
(d) zero, zero
(e) Unpredictable
This is a very simple question. But some of you will have some doubts. Initially the voltage across the capacitor is V0 and the total resistance connected across this voltage is 3R. Therefore, voltage drop across the resistor of value R must be V0/3.
[Note that the initial discharge current I0 is given by
I0 = dQ/dt = (Q0/3RC) e0, appropriate to t = 0.
Ignoring the negative sign, I0 = (Q0/3RC) = V0/3R since Q0/C = V0
The initial voltage across R is I0R = V0/3].

## Monday, November 29, 2010

### 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].

## Sunday, November 14, 2010

### 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
(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)]. -->
(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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