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

Tuesday, January 21, 2014

AP Physics Kinematics Practice Questions (Multiple Choice)



“Being ignorant is not so much a shame, as being unwilling to learn.”
– Benjamin Franklin


Many questions (with solution) involving kinematics in one dimension and two dimensions have been posted on this site earlier. You may click on the label ‘kinematics’ below this post to access them. After obtaining the first result, you will have to click on the ‘older posts’ tab to access all the posts in this section. Alternatively you may try a search for ‘kinematics’ using the search box provided on this page.

Today we shall discuss a few more multiple choice practice questions on kinematics:

(1) A car travels from station A and to station B separated by a distance of d km. The average speeds of the car while covering the first and second halves of the distance are v1 and v2 respectively. What is the average speed of the car for the entire trip from station A to station B?

(a) √(v1v2)

(b) (v1v2)/(v1+v2)

(c) (v1+v2)/2

(d) (v1v2)/(v1v2)

(e) (2v1v2)/(v1+v2)

The times taken for covering the first and second halves of the trip are d/2v1 and d/2v2 respectively.

Therefore, the total time taken to cover the entire istance d is (d/2v1 + d/2v2) = d[(1/2v1) + (1/2v2)] = d[(v1+v2)/2v1v2]

The average speed v for he entire trip is is given by

             v = d/d[(v1+v2)/2v1v2] = (2v1v2)/(v1+v2)

(2) An object has acceleration. Then

(a) its speed must be decreasing

(b) its speed must be increasing

(c) its speed must be decreasing or increasing

(d) its direction must be changing

(e) its speed or direction must be changing

An object moving with varying speed has acceleration. But this does not mean that all accelerated objects must move with a varying speed. For instance, an object in uniform circular motion has constant speed, even though it has a centripetal acceleration. Its direction of motion changes continuously and it is the change in direction that makes it an accelerated object.

For an object to be in accelerated motion, it is enough that its speed or direction of motion changes. Therefore, the correct option is (e).

(3) A bullet is fired from a gun in a direction inclined at angle θ with respect to the horizontal ground. Which one among the following graphs represents the plot of the vertical velocity v of the bullet against time t between the instant of firing and the instant just before the bullet hits the ground? (Take the upward direction as positive).
 




At the instant of firing, the bullet has the highest vertical velocity. When the bullet rises up, its vertical velocity goes on decreasing linearly (because of gravity) and at the highest point of its trajectory the vertical velocity becomes zero. The bullet then starts falling down with linearly increasing speed. In other words, the vertical velocity of the bullet becomes negative and its magnitude goes on increasing until it hits the ground.

The vertical velocity of the bullet as a function of time is therefore correctly represented by  graph (b).

(4) The adjoining figure shows forces F1 and F2 with their lines of action in the XY plane and acting on a particle.P.
If F1 = a1 î + b1 ĵ
and F2 = a2 î + b2 ĵ where î and ĵ are unit vectors in the x-direction and y-direction respectively, which one among the following statements is correct?
(a) a1, b1, a2, and b2 are positive.
(b) a1 and  b1 are negative where as a2 and b2 are positive.
(c) a1 is negative where as b1, a2, and b2 are positive.
(d) a1, b1 and a2 are positive where as b2 is positive.
(e) a1, b1, a2, and b2 are negative.
Imagine the rectangular components of F1 and F2. You can easily see that the x-component of F1 is along the negative x-direction while the y-component is along the positive y-direction. The x-component of F2 is along the positive x-direction while the y-component is along the positive y-direction.
This means that a1 is negative where as b1, a2, and b2 are positive [Option (c)].

(5) A small object at the foot of a smooth inclined plane AB (Fig.) is projected up along the inclined plane with an initial speed v. The object returns before reaching the top of the inclined plane and after reaching the foot of the plane, it moves further along a smooth horizontal surface AC. Which one among the following graphs represents the variation of the speed v of the object against time t? 
of the object therefore gets decreased uniformly and becomes zero when the object reaches its highest position on the incline. Then the object retraces its path with uniformly increasing speed until it reaches the foot of the incline. Then it moves along the horizontal surface AC with uniform speed.
The above facts are correctly represented by the  graph (d).

Tuesday, November 12, 2013

AP Physics B - Multiple Choice Practice Questions on Physical Optics



“One of the deep secrets of life is that all that is really worth the doing is what we do for others.”
– Lewis Carroll

Today we shall discuss a few typical multiple choice questions on physical optics:

Questions  (1) and (2) are based on the following statement:

A beam of monochromatic light having wave length λ, frequency f and intensity I in air enters a glass slab of refractive index 1.5. After traveling through the slab, the beam of light emerges through the opposite face of the slab and passes through air.

(1) Inside the slab the wave length and frequency of the light are respectively 

(a) 1.5 λ and 1.5 f

(b) λ/1.5 and f

(c) 1.5 λ and f

(d) λ and f

(e)  λ and 1.5 f

The frequency of the light is unchanged where as the speed of the light is decreased within the slab. Note that the refractive index of a medium is the ratio of the speed of light in free space (or air at ordinary pressures) to the speed in the medium. Therefore, the speed of light (v) in the glass slab is given by

            v = c/1.5 where ‘c’ is the speed of light in free space (or air).

The wave length λ, frequency f and speed v are related as

            v = f λ  

Or,       λ = v/f

Since the speed in glass is (1/1.5) times the speed in air, the wave length in glass is (1/1.5) times the wave length in air [Option (b)].

(2) The energy of the photons in glass is

(a) 1.5 times the energy in air

(b) 1/1.5 times the energy in air

(c) the same as the energy in air

(d) 3 times the energy in air

(e) 1/3 times the energy in air

The energy E of a photon is given by

             E = hf where h is Planck’s constant and f is the frequency. Since the frequency of the light is the same in both air and glass, the energy of the photons in glass is the same as the energy in air [Option (c)].

(3) White light is passing through a transparent plastic slab. Inside the slab

(a) the green component  travels with maximum speed

(b) the green component  travels with minimum speed

(c) the violet component  travels with maximum speed

(d) the red component  travels with maximum speed

(e) all components travel with the same speed

The refractive index of any transparent medium is maximum for light of violet colour and minimum for light of red colour.

[This is why violet rays are deviated most and red rays are deviated least while traveling through a glass prism, producing dispersion of white light].

The speed of light (v) in the plastic slab is given by

            v = c/n where ‘c’ is the speed of light in free space (or air) and n is the refractive inex of the slab.

Since the refractive index is the least for light of red colour, it follows that red component  travels with maximum speed [Option (d)].

(4) In an experiment with Young’s double slit, a student measures the intensity of the central maximum of the interference pattern as I. If one of the slits is covered, what will be the intensity at the position of the central maximum?

(a) I/2

(b) I/4

(c) I/(2)

(d) I

(e) 2I

When both slits are open, suppose the resultant amplitude (of light wave) at the position of the central maximum is a. If one of the slits is covered, the amplitude at the position of the central maximum becomes a/2.

Intensity is directly proportional to the square of the amplitude. Therefore we have

            I α a2 and

            I1 α (a/2)2

 Therefore I1 = I/4, as given in option (b).

(5) Four laser sources produce light waves y1, y2, y3, and y4 given (with usual notations) by

             y1 = a sin ωt

             y2 = a sin 2ωt

             y3 = 2a sin (ωt + φ)

             y4 = 2a sin (3ωt + φ)                                                               

Superposition of which two waves can produce interference fringes?

(a) y1 and y2

(b) y2 and y3

(c) y3 and y4

(d) y1 and y4

(e) y1 and y3

Waves y1 and y3 have the same wave length and hence they can produce interference fringes.

[The angular frequencies of y1 and y3 are equal].

The correct option is (e).

You will find a few useful multiple choice questions in this section here.

Friday, September 13, 2013

AP Physics B & C - Electrostatics - Multiple Choice Practice Questions


“Iron rusts from disuse, stagnant water loses its purity and in cold weather becomes frozen; so does inaction sap the vigors of the mind.”
– Leonardo da Vinci
 

Questions on electrostatics were discussed on many occasions on this site. You can access them by clicking on the label ‘electrostatics’ below this post or by trying a search for ‘electrostatics’ using the search box provided on this page. Today we shall discuss a few more questions in this section.
(1) Right angled triangle ABC is located in a uniform electric field E. Sides AC and BC have lengths 0.5 m and 0.3 m respectively and BC is at right angles to the electric field lines (Fig.). If the electric potential difference between A and C is 80 V, what is the magnitude of the electric field E?
(a) 80 V/m
(b) 160 V/m
(c) 200 V/m
(d) 100 V/m
(e) 16 V/m
The electric potential at the point B is the same as that at the point C (since the straight line BC is at rjght angles to the direction of the uniform electric field E). Therefore the potential difference between points A and B is 80 V. The electric field E is directed along AB and hence the magnitude of E is (80 V)/(0.4 m) = 200 V/m.
[AB = √(AC2 – BC2) = √(0.52 – 0.32)  = 0.4 m]
(2) In the above question, what is the component of electric field along the direction AC?
(a) 80 V/m
(b) 160 V/m
(c) 200 V/m
(d) 100 V/m
(e) 40 V/m
The potential difference between the points A and C is 80 V and the distance between these points is 0.5 m. Therefore, the component of electric field along the direction AC is (80 V)/(0.5 m) = 160 V/m.

(3) A cube of side a has a charge Q at its centre (Fig.). What is the electric flux through one face of the cube?
(a) Q/ε0a
(b) Qa/ε0
(c) Q/6ε0
(d) Q/8ε0
(e)0/6
The electric flux over a closed surface, according to Gauss theorem, is Q/ε0 where Q is the net charge enclosed by the surface and ε0 is the permittivity of free space.
[You can understand the above even without knowing Gauss law. You know (from inverse square law) that the electric field at distance r from a point charge Q is Q/4πε0r2. The electric field at any point is the electric flux through unit area held with the plane of the area perpendicular to the electric field lines. Imagine a spherical surface of radius r such that the charge Q is at the centre. Since the area of the spherical surface is 4πr2 the electric field at any point on the surface must be Ф/4πr2 where Ф is the total electric flux produced by the charge Q. Therefore we have
            Q/4πε0r2 = Ф/4πr2
This gives Ф = Q/ε0 as stated in Gauss law.]
Since the charge Q is at the centre of the cube, all the six faces of the cube receive equal electric flux so that the flux through one face of the cube is Ф/6 = Q/6ε0

(4) The effective capacitance between terminals A and B in the network shown in the adjoining figure is
(a) 16 μF
(b) 8 μF
(c) 6 μF
(d) 16/3 μF
(e) 8/3 μF
Since the capacitors C1, C2, C3 and C4 are of values satisfying the condition C1/C2 = C3/C4, the junction of C1 and C2 is at the same potential as the junction of C3 and C4 (as in the case of a balanced Wheatstone brige). Therefore the capacitors C5 and C6 connected across the diagonal of the bridge have no effect and can be ignored.
The network therefore simplifies to the series combination of C1 and C2 connected in parallel with the series combination of C3 and C4.
Series combined value of C1 and C2 = (3×6)/(3+6) = 2 μF
Series combined value of C3 and C4 = (1×2)/(1+2) = 2/3 μF
Therefore the effective capacitance between terminals A and B = 2 μF + (2/3) μF = 8/3 μF

The capacitors C1 and C4 in the above question are short circuited and the circuit then gets modified as shown in the figure. What is the effective capacitance between the terminals A and B in this situation?   
(a) 11 μF
(b) 9 μF
(c) 8 μF
(d) 7.5 μF
(e) 6.5 μF
Have a careful look at the circuit. You will find that one plate of C2, C3 and C5 is connected to terminal A. The other plate of C2 as well as C3 is connected to terminal B while that of C5 is connected 6through C6 to terminal B. The series combination of C5 and C6 gives an effective capacitance of 1μF. Thus we have three capacitances 1μF, 1μF and 6μF connected in parallel across the terminals A and B.


Therefore the effective capacitance between terminals A and B in this case is 1μF + 1μF + 6μF = 8μF.