AP Physics B & C - Multiple Choice Practice Questions on One Dimensional Kinematics

"Whenever you are confronted with an opponent, conquer him with love."

– Mahatma Gandhi

Let us discuss a few interesting multiple choice practice questions on one dimensional motion. Here are some questions beneficial for AP Physics B as well as AP Physics C aspirants:

(1) The adjoining figure shows the velocity time graph of an object. Total displacement suffered by the object during the interval when it has non-zero acceleration and retardation is

(a) 80 m

(b) 70 m

(c) 60 m

(d) 40 m

(e) 30 m

The object has non-zero acceleration and retardation during the time intervals from 5 sec to 15 sec and from 20 sec to 30 sec. The total area under the velocity time graph during these intervals gives the required displacement.

Displacement from 5 sec to 15 sec = 30 m

Displacement from 20 sec to 30 sec = 40 m

Therefore, total displacement = 70 m

(2) Successive positions (x) of an object (moving from left to right) at equal time intervals are shown in the following figure:

Which one among the following position-time graphs best represents the motion of the object? 

 

Change of position is slowest in the beginning and in the end. The motion is best represented by graph (c).

[Note that graph (d) is not the answer since the change of position in the beginning and in the end is shown as fastest in it]

(3) Which one among the following velocity-time graphs best represents the motion of the object mentioned in question no.(2)? 

The graph (c) is the answer.

The following questions are meant for AP Physics C aspirants:

(4) A particle projected vertically upwards attains the maximum height h in time t. While returning from the highest point it takes an additional time t₁ to fall to the height h.2. If air resistance is negligible, how is t₁ related to t?

(a) t₁ = t/2

(b) t₁ = √(t/2)

(c) t₁ = t/√2

(d) t₁ = t/√3

(e) t₁ = t/3

For the upward motion we have

             0 – u² = – 2gh …………(i)

[We have used the equation of motion, v² – u² = 2as. The sign of the gravitational acceleration g is negative since it is opposite to the direction of the velocity of projection u]

Using the equation of motion, v = u + at we have

             0 = u – gt from which u = gt

Substituting this value of u in Eq (i), we have

             g²t² = 2gh

Therefore, t =√(2h/g) ……..(ii)

For the fall through h/2 from the highest point we have

             h/2 = 0×t₁ + ½ gt₁²

[We have used the equation of motion, s = ut + ½ gt²]

Therefore, t₁ =√(h/g)

Comparing this with the value of t given in Eq (ii) we obtain t₁ = t/√2.

(5) A ball projected vertically up has the same vertical displacement h at times t₁ second and t₂ second. If air resistance is negligible, the maximum height reached by the ball is

(a) gt₁t₂/2

(b) g(t₁² + t₂²)/2

(c) g(t₁ + t₂)²/4

(d) g(t₁ + t₂)²/8

(e) 2g(t₁ + t₂)²

The time taken by the ball to move from height h to the top of its trajectory and back to the height h is t₂ – t₁. Therefore, the time taken to move from height h to the top of the trajectory is (t₂ – t₁)/2.

The total time taken for the upward journey (from ground to the top most point) is evidently t₁ + t₂ – t₁)/2 = (t₁ + t₂)/2.

Time taken for the return journey (from the top most point to the ground) also is equal to (t₁ + t₂)/2. Therefore, the maximum height H (using the equation, s = ut + ½ gt²) is given by

             H = 0 + ½ g [(t₁ + t₂)/2]²

Or, H = g(t₁ + t₂)²/8

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

AP Physics B - Multiple Choice Practice Questions on Atomic Physics and Quantum Effects

"Men often become what they believe themselves to be. If I believe I cannot do something, it makes me incapable of doing it. But when I believe I can, then I acquire the ability to do it even if I didn't have it in the beginning

– Mahatma Gandhi

 

AP Physics 2012 exams are just a few days away. Your final preparations for the exam must be in full swing and there is no time to waste. Today I give you a few multiple choice practice questions on atomic physics and quantum effects. Questions in this section posted earlier on this site (with solution) can be accessed by clicking on the label ‘atomic physics and quantum effects’ given below this post. Or, you may try a search for ‘atomic physics and quantum effects’ using the search box provided on this page.

(1) The adjoining figure shows the graphical relation between the frequency of incident radiation and the magnitude of stopping potential in the case of metals A and B. Note that the straight line graphs have the same slope. Which one among the following statements is correct in this case?

(a) A and B have the same work function

(b) A and B have the same threshold wave length

(c) Maximum kinetic energy of photoelectron in the case of metals A and B is directly proportional to the increment in frequency of incident radiation over the threshold frequency.

(d) Metal B is a better photosensitive material than metal A.

(e) For a given change in the frequency of incident radiation, the changes in stopping potentials are different for metals A and B

Since we have straight line graphs for both metals, maximum kinetic energy of photoelectron is directly proportional to the increment in frequency of incident radiation over the threshold frequency. Therefore option (c) is correct.

(2) The de Broglie wave length of a particle can be reduced to half its initial value by changing its kinetic energy to

(a) half the initial value

(b) twice the initial value

(c) three times the initial value

(d) four times the initial value

(e) a quarter of the initial value

Kinetic energy E of a body of mass m is given by

             E = p²/2m where p is the momentum

[This follows from E = ½ mv² = m²v²/2m. Here v is the velocity and mv = p]

The above relation shows that the kinetic energy becomes four times when the momentum is doubled.

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant.

Therefore, the de Broglie wave length of a particle can be reduced to half its initial value by changing its momentum to twice the initial value. Evidently the kinetic energy of the particle the becomes four times the initial value [Option (d)].

[Suppose the above question is modifie as follows:

A particle has de Broglie wave length λ when its kinetic energy is E. What additional kinetic energy is to be aded to it in order to reduce the de Broglie wave length to λ/2?

(a) E

(b) 2E

(c) 3E

(d) 4E

(e) E/4

The answer is 3E since you are asked to find the additional kinetic energy].

(3) Uranium (atomic number 92) has an isotope of mass number 235. It can undergo successive disintegrations to get transformed into lead (₈₂Pb²⁰⁷). How many α-particles and β-particles are emitted during this transformation?

(a) α = 7, β = 4

(b) α = 4, β = 3

(c) α = 7, β = 0

(d) α = 7, β = 7

(e) α = 4, β = 7

Beta particle emission does not affect the mass number. In order to reduce the mass number by 28 (from 235 to 207), the number of α-particles to be emitte must be 7. Since each α-particles carries two fundamental units of positive charge, the atomic number of the end product gets reduced by 14. But the final product (₈₂Pb²⁰⁷) has its atomic number reduced by 10 only. The extra 4 units must be obtained by the emission of four β-particles. The correct option therefore ia (a).   

[Note that when a β-particles (electron) is emitted from the nucleus, the nuclear charge increases by one unit. This happens as a result of the transformation of a neutron in the nucleus into a proton].

(4) Fundamental forces in nature are gravitational force, electromagnetic force, nuclear force and weak force. If these forces act over very short distances of the order of nuclear dimensions, how do you arrange them in decreasing order (starting with the strongest?

(a) Gravitational force, electromagnetic force, nuclear force, weak force

(b) Electromagnetic force, gravitational force, nuclear force, weak force

(c) Electromagnetic force, nuclear force, weak force, gravitational force

(d) Gravitational force, nuclear force, electromagnetic force, weak force

(e) Nuclear force, electromagnetic force, weak force, gravitational force

The correct option is (d).

[Don’t get carried away by the term ‘weak force’. The weakest force is gravitational force where as the strongest is nuclear force].

(5) Two protons are separated by a distance of 50 Ǻ. If the electromagnetic force between them is F₁ and the nuclear force between them is F₂, which one among the following is the most reliable statement?

(a) F₁ >> F₂

(b) F₂ >> F₁

(c) F₁ > F₂

(d) F₂ > F₁

(e) F₂ = F₁

This question is similar to question No. (4) in the sense that it is meant for checking your knowledge of nuclear physics. The correct option is (a). Nuclear  force is a very short range force. At a separation of 50 Ǻ which is very large compared to the size of a nucleus, nuclear force (strong interaction) between two protons is negligible compared to the electrostatic force. Therefore the correct option is (a).

Physics B & C - Multiple Choice Practice Questions on Electric Circuits

"Nearly every man who 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

Today we will discuss a few multiple choice practice questions on  electric circuits. Questions in this section were discussed on earlier occasions. You may click on the label ‘electric circuits’ below this post to access all posts in this section. Here are the questions with their solution:

(1) The voltmeter in the shown shown has a resistance of. The internal resistance of the 10 V battery is negligible. What is the reading of the voltmeter?

(a) 2 V

(b) 3.3 V

(c) 4 V

(d) 5V

(e) 6 V

The voltmeter of resistance 100 KΩ is connected across a 100 KΩ resistor and these two give a parallel combined resistance value of 50 KΩ. The total resistance in series with the battery is 100 KΩ + 100 KΩ + 50 KΩ = 250 KΩ.

The current in the series circuit is 10 V/250 KΩ and hence the potential difference across the parallel combination of the voltmeter and the 100 KΩ resistor is (10 V/250 KΩ) × 50 KΩ = 2 V [Option (a)].

[We retained the resistances in KΩ itself in the above expressions so that we could obtain the potential difference in volts in a convenient manner].

 

(2) The adjoining figure shows junctions J₁ and J₂ of an electric circuit. Currents at these junctions sufficient for arriving at the answer are indicated in the figure. What is the value of the current I?

(a) 1 A

(b) 3 A

(c) 4 A

(d) 8 A

(e) 10 A

In accordance with Kirchoff’s current law (KCL), the current I₁ flowing outwards from junction J₁ towards junction J₂ is 5 A – 2 A = 3 A.

[Algebraic sum of currents at a junction is zero according to KCL:

             5 A – 2 A + I₁ = 0

Therefore, I₁ = – 3 A, the negative sign showing that the current flows outwards from the junction].

Total current flowing towards junction J₂ is 3 A + 5 A = 8 A.

Since a current of 4 A is shown as flowing out from junction J₂, the remaining current I₂ that has to flow out must be 4 A [Option (c)].

(3) A 2 μF capacitor is connected in series with a 1 μF capacitor and a 60 V power supply. The potential difference across the 2 μF capacitor is

(a) 60 V

(b) 40 V

(c) 30 V

(d) 20 V

(e) zero

The effective capacitance connected across the power supply is the series combined value of the 2 μF and 1 μF capacitors which is (²/₃) μF

[If capacitors C₁ and C₂ are  connected in series, their effective capacitance C is given by the reciprocal relation, 1/C = 1/ C₁ + 1/C₂ so that C = C₁C₂/( C₁ + C₂)]

The charge Q supplied by the power supply is given by

             Q = CV = (²/₃) × 60 = 40 μC

[We obtain the charge in micro coulomb since the capacitance is in μF].

The capacitors hold the same charge in series connection. Therefore, the voltage across the 2 μF

capacitor is Q/C₁ = 40 μC/2 μF = 20 volt [Option (d)].

[You will be able to work out this problem in no time if you remember that the voltages are distributed among series connected capacitors in inverse proportion to their capacitances].

The following questions are specifically meant for AP Physics C aspirants:

(4) The resistance R of any uniform wire of length L and cross section area A is related to the resistivity (or, specific resistance) ρ of its material as

             R = ρL/A

Suppose you have n wires of the same length L and the same area of cross section A made of materials of resistivities ρ, 2ρ, 3ρ, 4ρ, ……nρ respectively. If they are connected in series, what will be the resistivity of the material of the combined wire?

(a) n ρ

(b) n(n+1)ρ

(c) n ρ/2

(d) (n+1)ρ/2

(e) n(n+1)ρ/2

The resistances of the 1^st, 2^nd, 3^rd, 4^th, ……n^th wires are respectively given by R₁ = ρL/A, R₂ = 2ρL/A, R₃ = 3ρL/A, R₄ = 4ρL/A, ……. R_n = nρL/A.

The total resistance R of the series combination is given by

             R = R₁ + R₂ + R₃ + R₄ + ……+ R_n

Or, R = (ρL/A)(1 + 2 + 3 + 4 + ……n), on substituting for R₁, R₂, R₃ etc.

This gives R = (ρL/A)[n(n + 1)/2] = n(n + 1) ρL/2A

The length of the compound wire is nL and its area of cross section is A. If the resistivity is ρ its resistance can be written as

             R = ρnL/A

Therefore ρ = RA/nL = [n(n + 1) ρL/2A](A/nL) = (n+1)ρ/2

[Even without writing all the above steps you could have written the answer, arguing that the resistivity of the combined wire must be the average value of the resitivities of the individual wires, since the increments in resistivity occur in a regular manner. Therefore ρ = (ρ+2ρ+ 3+4ρ+ ……+nρ)/n = (n+1)ρ/2]

(5) In the circuit shown in the adjoining figure, the ammeter reading is zero when the switch S is closed. If the batteries are of zero internal resistance, the value of resistance X is

(a) 100 Ω

(b) 200 Ω

(c) 300 Ω

(d) 400 Ω

(e) 600 Ω

Since the current through the ammeter is zero, the voltage drop across the series combination of X and 600 Ω must be 4 V.

[The battery emf of 4 V and the p.d. across the series combination of X and 600 Ω have tu be equal and and in opposition to attain the condition of zero current through the ammeter].

The 6 V battery drives a current through the three resistors and 4 V is dropped across the series combination of X and 600 Ω. The remaining 2 V is dropped across the 400 Ω resistor.

Since the potential drop across the series combination of X and 600 Ω is 4V, we have

              X + 600 Ω = 400 Ω

Therefore X = 200 Ω

[If you want to use Kirchoff’s laws and write down mathematical steps for solving the above problem, here is how you will proceed:

             For the loop containing the 4 V battery, ammeter, X and 600 Ω, we have

             (I₁ + I₂)(X+600) = 4 where I₁ and I₂ are the currents supplied by the 4 volt and 6 volt batteries respectively.

Since I₁ = 0 we have

             I₂(X+600) = 4 …………..(i)

             For the loop containing the 6 V battery, 400 Ω, X and 600 Ω, we have

             I₂(400+X+600) = 6 ……..(ii)

Dividing Eq. (i) by Eq. (ii), we have

             (X+600) /(400+X+600) = 2/3

This gives X = 200 Ω]

AP Physics B & C - Multiple Choice Practice Questions on Motion in Two Dimensions including Projectile Motion

"If man is not rising upward to be an angel, he is sinking downward to be a devil. He cannot stop at the beast."

– Samuel Taylor Coleridge

Let us discuss a few questions (MCQ) involving motion in two dimensions. Questions in common for AP Physics B as well as AP Physics C aspirants are discussed first. Questions specifically meant for AP Physics C aspirants are discussed next.

(1) The adjoining figure represents the path of an ant on a horizontal floor between the instants t₁ and t₂. The velocities of the ant at positions A an B at the instants t₁ and t₂ are represented by the vectors V₁ and V₂ respectively. Which one among the following vectors best represents the net acceleration of the ant during the interval t₂ – t₁?

The acceleration a is given by

a = (V₂ – V₁)/(t₂ – t₁)

The direction of the vector V₂ – V₁ is given correctly in option (b).

[To find V₂ – V₁ you have to add the vector V₂ to the vector –V₁ using the parallelogram law (for convenience). On reversing the direction of V₁ and using it along with V₂ to form the parallelogram, you get the diagonal which points along the vector shown in option (b)].

The correct option is indeed (b).

(2) The path of a particle projected from point A with velocity v at an angle θ with respect to the horizontal is shown in the adjoining figure. Which one among the following graphs best represents the vertical component (v_y) of the velocity of the particle as a function of time t? (Assume that air resistance is negligible)

Since the motion of the particle is under gravity, the vertical component of velocity decreases uniformly as the particle moves up. At maximum height the vertical component of velocity momentarily becomes zero. Then the particle moves down and the vertical component of velocity goes on increasing uniformly until it hits the ground. The correct graph is (d).

[Note that the upward direction of the vertical component of velocity is taken as positive and the downward direction is taken as negative].

(3) In the above question, if g represents the gravitational acceleration, the values of acceleration of the particle at points A, B and C of the trajectory are

(a) greater than g, zero and less than g respectively

(b) greater than g, equal to g and less than g respectively

(c) less than g, zero and greater than g respectively

(d) less than g, equal to g and greater than g respectively

(e) same and equal to g

The gravitational force is the same everywhere along the trajectory and hence the acceleration of the particle is the same (equal to g).

The correct option is (e).

The following questions are specifically meant for AP Physics C aspirants:

(4) A particle starts from the origin at time t = 0 with velocity 2 ĵ and moves in the x-y plane with a constant acceleration of 2 î + 4 ĵ where î and ĵ are unit vectors along the x-direction and y irection respectively. What will be the x-coordinate of the particle when its y-coordinate becomes 12 m?

(a) 4 m

(b) 6 m

(c) 6.8 m

(d) 8 m

(e) 12 m

The position vector r_t of the particle at the instant t is given by

r_t = v₀t + ½ at² where v₀ is the initial velocity and a is the constant acceleration.

We have v₀ = 2 ĵ and a = 2 î + 4 ĵ

Therefore, r_t = 2 ĵ t + (½) (2 î + 4 ĵ)t² = t² î + (2t + 2t²) ĵ

The above equation shows that the x-coordinate of the particle at time t is t² and the y-coordinate is (2t + 2t²)

The time t at which the y-coordinate becomes 12 metre is given by

2t + 2t² = 12

Or, 2t² + 2t – 12 = 0

This gives t = [–2 ±√(4 + 96)]/4 = 2 seconds, ignoring the negative time.

Since the x-coordinate of the particle is t², its value when the y-coordinate becomes 12 m (at time 2 seconds) is 4 m [Option (a)].

(5) What is the speed of the particle in the above question at the time t = 1 s?

(a) 2 ms^–1

(b) √(10) ms^–1

(c) √(20) ms^–1

(d) √(40) ms^–1

(e) 8 ms^–1

The position vector r_t of the particle at time t, as shown above is given by

r_t = t² î + (2t + 2t²) ĵ

The velocity v_t of the particle at time t is given by

v_t = dr_t/dt = 2t î + (2 + 4t) ĵ

When the time t = 1 s, the velocity of the particle is 2 î + 6 ĵ.

The speed of the particle at time 1 s is the magnitude of the above velocity and is equal to √(2² + 6²) = √(40) m

You will find a useful post in this section here.