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University of the Philippines College of Science PHYSICS 72 SET A First Semester 2011-2012 Final Examination

Physics 72 Final Exam Set a Draft 3 (With Feeling Answer Key)

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University of the Philippines College of Science

PHYSICS 72

SET A

First Semester 2011-2012 Final Examination

National Institute of Physics 1st Sem AY 2011-2012 Physics 72

1 | A

INSTRUCTIONS: Choose the best answer and shade the corresponding circle on your answer sheet. To change your answer, cross-out and sign your original answer and then shade your new answer. No computational devices allowed (e.g. calculators, mobile phones). Following instructions is part of the exam.

Useful formulas:

Area Volume

Sphere (radius = r)

4 r2

4

3 r

3

Cylinder (radius =r, height = h)

2 rh

r2h

Useful constants: e -1.60 x 10-19 C me 9.1 x 10-31 kg o 8.854 x 10-12 C2/Nm2

k 8.988 x 109 Nm2/C2

θ 0 π/6 π/3 π/4 π/2

sin θ 0 1

2 3

2

2

2

1

cos θ 1 3

2

1

2 2

2

0

tan θ 0 1

3 3

1 ∞

1. Charges at equilibrium. Consider three identical heart-shaped conductors (labeled B, Y and E) with net charges +q, -4q and +3q as shown in the figure. Determine the final charges on each conductor at electrostatic equilibrium if B is put in contact with Y, while Y is placed near, but not in contact with E.

A. B: 0, Y: 0, E: 0 B. B: -4q, Y: +q, E: +3q C. B: +q, Y: -4q, E: +3q D. B: -3q, Y: -3q, E: +3q E. B: -3q/2, Y: -3q/2, E: +3q

2. Force versus field. Consider a charge +Q

placed at the origin which causes an electric field E = Eo j on a point charge –q at point S. Determine the electric field and the electric force that it will cause if the charge –q is moved at point T, which is equidistant from the origin as point S.

A. E = Eo i; F = 0 B. E = Eo i; F = qEo i

C. E = Eo i; F = -qEo i D. E = -Eo i; F = qEo i E. E = -Eo i; F = -qEo i

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B and Y are in contact, adding both charges results to -3q, dapat hati sila, so -3q/2 each. Walang mangyayari kay E, since not in contact.
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Pumunta lang naman ung -q charge sa x-axis. Para mahanap ung force, formula lang.

National Institute of Physics 1st Sem AY 2011-2012 Physics 72

2 | A

3. Electric dipole. Consider two identical, but oppositely oriented, dipoles (labeled C and K) that are placed in a region of uniform electric field directed to the right. Compare the net

force (F) and the net torque (T) acting on each dipole.

A. FC = FK; TC = TK B. FC = FK; TC = -TK C. FC > FK; TC = TK D. FC > FK; TC = -TK E. FC < FK; TC = TK

4. Flux you. Consider the Gaussian surfaces SI,

SII and SIII enclosing charges as shown at the right. If Q > 0, how are the electric fluxes through each surface related?

A. ΦI > ΦII >ΦIII B. ΦIII > ΦII >ΦI C. ΦI = ΦII >ΦIII D. ΦI = ΦII =ΦIII E. ΦI > ΦII =ΦIII

For the next two numbers, consider a uniformly charged sphere with radius RS and volume charge density ρ with a concentric cavity of radius RC.

5. nEar. What is the electric field magnitude at a

point a distance r < RC? A. Zero B. ρ(RS

3-RC3)/(3ε0r

2) C. ρ(r3-RC

3)/(3ε0r2)

D. ρ(RS3-r3)/(3ε0r

2)

E. ρRS3/(3ε0r

2)

6. therE. What is the electric field magnitude at a point a distance r > RS?

A. Zero B. ρ(RS

3-RC3)/(3ε0r

2) C. ρ(r3-RC

3)/(3ε0r2)

D. ρ(RS3-r3)/(3ε0r

2) E. ρRS

3/(3ε0r2)

7. Potential. Two charged metal spheres are connected by a wire. Sphere A is larger

than sphere B. Which of the following is true about the magnitudes of the electric

potential on each surface?

A. VA > VB D. VA = VB B. VA ≥ VB E. It depends on the charges on the spheres. C. VA ≤ VB

RC

RS

ρ

-Q

Q 2Q

Q SI

SII

SIII

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Net force is always zero for a dipole immersed UNIFORM in electric field. Use right hand rule to know the net torque of each dipole.
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Gauss' Law lang pfouwz.
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There are no charges located in the cavity of the given sphere.
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A gaussian sphere r > Rs encloses the whole given sphere. Tatanggalin mo lang ung bungal.
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Electric potential is not dependent on the size of the figure. It is only dependent on the charges it contains, since the two spheres are CONDUCTING and are IN CONTACT or connected, we can say that its potentials are equal.
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National Institute of Physics 1st Sem AY 2011-2012 Physics 72

3 | A

8. Electric Field. The electric potential in a region of space is given by V(x,y,z) = (10 V/m)x3 + (20 V/m)xy + (15 V/m)z2. What is the z-component of the electric field in this region?

A. -(30 V/m)z k D. -(7.5 V/m)z j

B. -(15 V/m)z i E. (30 V/m)z k C. (15 V/m)z j

9. P.E. Two positive point charges are separated by a distance R. If the distance

between the charges is reduced to R/2, what happens to the total electric potential energy of the system?

A. It is doubled. B. It remains the same. C. It increases by a factor of 4. D. It is reduced to one-half of its original value. E. It is reduced to one-fourth of its original value.

For the next two numbers, consider the circuit diagram shown below. The potential difference Vab is 12 V.

10. Equivalent Capacitance. What is the equivalent capacitance of the circuit? A. 1 μF D. 4 μF B. 1/3 μF E. 4/3 μF C. 3 μF

11. Potential Energy. What is the total potential energy stored in the network of

capacitors? A. 24 μJ D. 216 μJ B. 72 μJ E. 288 μJ C. 96 μJ

12. Dielectric. Consider an isolated parallel-plate capacitor with surface charge

density |σ| on each plate. Initially the two plates are separated by Teflon with dielectric constant K = 2. If we replace the Teflon with Mica of K = 4, which of the following statement/s will be TRUE?

I. The electric field between the plates will double. II. The electric field energy density will double.

III. The capacitance will double.

A. I only. D. I and II. B. II only. E. II and III. C. III only.

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Math 55 principles, kunin lang ung NEGATIVE gradient nung function, since z-component lang ung kelangan, kunin lang ung NEGATIVE derivative ng polynomial na may z.
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Use formula for two point charges: U = (kq1q2)/r
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For parallel connections, simply add their capacitance, for series, reciprocalize(LOL) first then add, and then reciprocalize (LOL X2) again.
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Use formula for potential energy: U = 1/2(C)(V^2)
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-nope, dapat halved
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-nope, decrease by 1/4 its original
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this is based on the formula: u = 1/2(epsilon)(E^2)
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and E = Eo/K

National Institute of Physics 1st Sem AY 2011-2012 Physics 72

4 | A

13. silindrho. Consider an ohmic conducting cylinder with ends connected to a voltage source V. What will happen to the resistivity of the cylinder through its ends if its length is halved?

A. Quartered D. Doubled

B. Halved E. Quadrupled C. Remains the same

For the next two numbers, consider the network shown in the right figure. Assume that the battery is ideal and is equal to 5 V. The following are the values of each resistor: R1 = 0.5 Ω, R2 = 1 Ω, R3 = 2 Ω, R4 = 2 Ω.

14. Resistance. What is the equivalent resistance of the network?

A. 1 Ω D. 19/6 Ω B. 2 Ω E. 11/2 Ω C. 5/2 Ω

15. Current. What is the current passing through R2 if the current passing through

R1 is 2 A? A. 2 A D. 8 A B. 4 A E. 10 A C. 6 A

16. Selectah. A velocity selector is shown in the figure. The

electric field is -50 N/C i and the magnetic field is -25 T k. The charge of the particle is 1 C. What velocity will allow the charge to move out at the lower end of the selector?

A. 2.5 m/s towards + y-axis B. 2.5 m/s towards - y-axis C. 2 m/s towards + y-axis D. 2 m/s towards - y-axis E. 0.5 m/s towards the - y-axis

For the next two numbers, consider a wire loop parallel to an external uniform magnetic field as shown. The loop has radius r0 and is carrying a current I directed clockwise.

17. Force. What is the direction of the net magnetic force on the current loop?

A. Into the page B. Out of the page C. Up the page D. Down the page E. Zero

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all materials have their own resistivity, it never changes if you change the material's dimensions, it only changes when temperature changes.
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DI KO SURE 'TO. Ginawa kong parallel ang R2 at R4, then R1, R3, pati ung result ng R2 and R4 ay connected via series connection.
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Apply Kirchoff's rules here: 5V - (2A)(.5ohms) - I(1ohms) = 0 I = 4A
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Di ko sure, pero meron sa Arciaga, formula ng selector: Vselect = E/B. As for direction, dapat pababa sya, as asked by the question.
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Net magnetic force is always zero on a closed loop immersed in a UNIFORM magnetic field.
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National Institute of Physics 1st Sem AY 2011-2012 Physics 72

5 | A

18. Torque. What is the direction of the net torque on the current loop? A. Into the page B. Out of the page C. Up the page

D. Down the page E. Zero

19. 4 Protons. At an instant of time, four

protons (I, II, III and IV) are located at the four corners of a square with side length s and are moving with equal speeds in different directions pointed to by the arrows as shown in the right figure. Which of the protons will produce the largest magnitude of magnetic field at point P located exactly at the center of the square?

A. I B. II C. III D. IV E. All protons have equal magnitudes of magnetic field at point P.

20. Alternating. Consider six infinitely long wires carrying equal currents and

equidistant with each other. The direction of the current for each wire is shown in the figure below. What is the direction of the net magnetic field at point P?

A. Pointing to the right B. Pointing to the left C. Pointing downward D. Pointing upward E. No direction because the net magnetic field at point P is zero.

21. Bubble. Consider two single loops of wire having

different radii (r1 = ½ r2) as shown in the figure. If the two wires carry equal currents but opposite in direction as shown by the arrows, what is the direction of the net magnetic field at point P located at the center of both wires?

A. Out of the page B. Into the page C. Downward D. Upward E. No direction because the net magnetic

field at point P is zero.

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Use torque = (mu) x(cross) B mu is the direction of the current of the loop, in the case of the problem, dapat into the page. Use right-hand rule.
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Use the Inverse Square Law for magnetic fields: something something... (qv x r(hat)) / r^2 , impt lang dito ung cross product, nagiging max lang ang cross product pag 90 deg ung angle.
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look at the current carrying wire nearest point P. use right hand rule and biot-savart's law.
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Ampere's Law, since the directions of the currents are equal and opposite, they will cancel each other.
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National Institute of Physics 1st Sem AY 2011-2012 Physics 72

6 | A

22. Experiment. The north pole of a permanent bar magnet is pushed along the axis of a coil as shown below. The pointer of the sensitive voltmeter connected to the coil moves to the right and gives a maximum reading of 8 units. The experiment is repeated but on this occasion, the south pole of the magnet enters

the coil at twice the previous speed. Which of the following gives the maximum deflection of the pointer of the voltmeter?

V

N Sa x is o f c o i l

A. 8 units to the right B. 8 units to the left C. 16 units to the right D. 16 units to the left E. There is no deflection.

23. Induced E. Which of the following statement/s is/are TRUE about induced electric field?

I. It is not conservative. II. Its field lines do not form closed loops.

III. It is parallel to the changing magnetic field that induces it.

A. I only B. II only D. I and III only C. III only E. I and II only

24. French TGV. Trains move through the earth’s

magnetic field at rather high speeds. The typical value of the earth’s magnetic field is 0.5 x 10-4

T. The French TGV train reach speeds of up to 100 m/s moving on tracks about 2 m. At top speed moving perpendicular to the earth’s magnetic field, what potential difference is induced across the tracks as the wheels roll?

A. 0.02 V D. 0.1 V B. 0.01 V E. 0 V C. 0.2 V

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Motional EMF. Use formula: dE = (v x B) dot dl or E = vbl E here is EMF not E. Field just to be clear. since doubled ang speed, doubled din ang reading, and since south ang pinasok, ibang direction naman.
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Yes
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Nope, dapat anti-parallel.
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Yes, since sinabi na non-conservative sya.
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Motional EMF: dE = vblsin90 = 100(0.5x10^-4)(2m) = .01
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again, E here is EMF, not E. Field.

National Institute of Physics 1st Sem AY 2011-2012 Physics 72

7 | A

25. Inductance. At a given instant, the current and self-induced emf in an inductor are directed as indicated in the right figure. If the magnitude of the induced emf is 3 V and the current

decreases at a rate of 2 A/s, which of the following correctly describes the potential and inductance across points a and b?

A. potential increases from a to b and the inductance is equal to -1.5 H. B. potential drops from a to b and the inductance is equal to -1.5 H. C. potential increases from a to b and the inductance is equal to 1.5 H. D. potential drops from a to b and the inductance is equal to 1.5 H. E. potential is unchanged from a to b and the inductance is -1.5 H.

26. R-L Circuits. The figure below shows three circuits with identical batteries,

inductors, and resistors. Which of the following relationships correctly describes the time (t) for the current to reach 50% of its equilibrium value after the switches are closed?

A. tB > tA < tC B. tB > tA > tC C. tB < tA > tC D. tB < tA < tC E. tB = tA = tC

27. R-L-C Circuit. Which among the following plots correctly describes the charge as

a function of time of an R-L-C circuit with R = 4Ω; L = 6μH and C = 8μF?

28. All low shall pass. In electronics, a current passing through a circuit is recorded as signals. A signal is filtered out if the current is not allowed to pass. Which circuit element is the most suitable series filter so that high frequency signals cannot pass through?

A. Resistor D. Transformer B. Capacitor E. Antenna C. Inductor

For the next two numbers: A series RLC circuit with resistance R, capacitance C, and

inductance L is powered by an emf ℰ = ℰ 0 sin t where the frequency is below resonance.

A dielectric slab is then inserted into the capacitor.

29. RLC. How does the impedance change? A. It increases. D. It changes from time to time. B. It decreases. E. It will continuously decrease. C. It is unchanged.

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use V = L(dI/dt) L = V/ (dl/dt)
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negative ang dI/dt mo, so decreasing V.
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Time constant = L/R, bigger R means lesser time.
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Charge exponentially decays. so A ang sagot.
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other circuit elements do not function as 'current stoppers'
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based on the formula of impedance: Z = sqrt(R^2 + ((XL - XC)^2)) and since XC = 1/(omega)C, dielectric application means higher C, therefore lower XC and lower Z.

National Institute of Physics 1st Sem AY 2011-2012 Physics 72

8 | A

30. Resonance. If the dielectric has a dielectric constant κ. How does the resonance frequency change after the slab is inserted?

A. It increases by a factor of κ

B. It decreases by a factor of κ

C. It increases by a factor of κ1/2. D. It decreases by a factor of κ1/2. E. It is unchanged.

For the next two numbers, consider a linearly-polarized wave with E = Eo cos[(6 m-1) z + (2c rad/s)t] i propagating in a medium with unknown refractive index.

31. SB. Determine the appropriate directions of the Poynting vector (S) and the oscillating magnetic field (B)?

A. S: +i; B: +j D. S: +k; B: -j B. S: -i; B: -j E. S: -k; B: -j C. S: +k; B: +j

32. Index. What is the value of the medium’s refractive index?

A. ½ D. 3 B. 1 E. 4 C. 2

33. Pressured. Light is normally incident on the cross sectional

area of a heart-shaped surface whose dark-painted left half is 100% absorbing while the other half is 100% reflecting. Compare the total force (F) acting upon each surface.

A. Fleft = Fright D. Fleft = 4 Fright B. Fleft = 2 Fright E. Fleft = ¼ Fright C. Fleft = ½ Fright

34. Refraction. A parallel beam of light in air makes an angle of 30 with the surface

of a glass plate. The angle between the refracted part of the beam and the surface of the glass is 60. What is the refractive index of the glass?

A. 13

D. 3

2

B. 12

E. 3

3

C. 3

35. TIR. Which of the following statement/s is/are ALWAYS TRUE about the

conditions for total internal reflection to occur? I. Only if the material in which the ray would travel is air.

II. Only if the angle of incidence is larger than the critical angle. III. Only when a ray is incident on the interface with a second material whose

index of refraction is smaller than that of the material in which the ray is traveling.

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resonance frequency is independent of any dielectric.
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E = Eocos(kz + wt), + sign means, -z direction or -k ang Poynting, apply right hand rule para mahanap ang direction ng B.
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given ka ng wave number, k = 6 = 2pi/lambda and w = 2c = 2pi(f) manipulate para magamit sa formula ng v = lambda(f), v = c/3, refractive index, n = c/v = c/ (c/3) = 3
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absorbing means only half radiation pressure, reflecting means twice rading pressure, so F(right), which is the reflective has 4 times rad. pressure than F(left)
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snell's law: n1sin(theta1) = n2sin(theta2), n1 = 1.00 sin (60) = n2sin (30) n2 = sqrt(3) *remember that angles used in snell's law should be measured with respect to the NORMAL of the surface.
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Nope, dapat higher, RI Yes
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Nope, again, dapat higher RI ang incident.
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National Institute of Physics 1st Sem AY 2011-2012 Physics 72

9 | A

A. I only D. I and II only B. II only E. I and III only C. III only

36. Two Topics. A parallel beam of unpolarized light in air is incident at an angle of 60

from the normal on a plane glass surface. The reflected beam is completely linearly polarized. What is the refractive index of the glass?

A. 1 D. 3

2

B. 12

E. 3

3

C. 3

37. Plane refracting surface. A dolphin named Flipper is held captive in one of the

pools in Ocean Park. He is 8.0 m beneath the pool’s surface when he saw a small bird flying directly above him. If the bird’s actual height as measured from the pool’s surface is 6.0 m. How far from the pool’s surface does the bird appear to the dolphin? (nair = 1, npool = 4/3)

A. 4.5 m D. 10.6 m B. 6.0 m E. 14.0 m C. 8.0 m

38. Spherical Mirror. Kobe, who stands 6’6’’ (or

6½ ft), looks at himself on a convex mirror

with |R| = 2 ft. If Kobe is 3 ft away from the vertex of the mirror, what is the height of Kobe’s image formed by the convex mirror?

A. 13/4 ft. B. 13/7 ft. D. 13/10 ft. C. 13/8 ft. E. Kobe’s image is infinitely tall.

39. Thin Lens. Consider two

identical converging thin lenses 20 cm apart with focal length magnitude of 4 cm. If an object is placed 6 m to the left of the first lens, where is the final image located as measured from the first lens? (Note: figure is to scale)

A. 5 cm D. 25 cm B. 8 cm E. 28 cm C. 12 cm

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Want to see my Brewster's Angle? tan(theta) = nglass/ nair tan 60 = nglass nglass = sqrt(3)
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use object-image rel. with refractive indices, and R = infinity for a plane surface.
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again use object-image rel. minus refractive indices, and f = - 1ft (definition) also use formula for magnification
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you can use ray diagrams since naka-scale naman or solve it using object image rel., note that the first image formed acts as second object for the final image.

National Institute of Physics 1st Sem AY 2011-2012 Physics 72

10 | A

40. Path diff. Two coherent sources S1 and S2 with

wavelength of 50 μm were made to interfere at point P as shown in the figure. If the path traversed

by S1 and S2 were 400 μm and 225 μm respectively, what can be said about the intensity of the interference pattern at point P?

A. The intensity is at maximum. B. The intensity is at minimum. C. The intensity is 1/2 of its maximum value. D. The intensity is 1/4 of its maximum value. E. The intensity is 1/8 of its maximum value.

41. Young’s double slit experiment. In a Young’s double

slit experiment, the observed interference pattern is shown in the right figure. If the distance between the slit and the screen is decreased, which of the following will most likely be the observed new interference pattern in the screen? (Note: The broken line corresponds to the center of the interference pattern.)

42. Anti-UV. If you want to coat a glass (n = 3/2) with a material with index of refraction of 6/5 so that UV light with wavelength of 300 nm will all be reflected, what should be the thickness of the coating?

A. 62.5 nm D. 150 nm B. 75 nm E. 300 nm C. 125 nm

43. Fringe width. Light of wavelength λ = 400 nm is used in a single slit diffraction

setup with three slit width settings (a1 = 0.1 μm, a2 = 10 μm, a3 =100 μm). Arrange the three in order of increasing width of the central bright fringe. A. a1 < a2 < a3 D. a3 < a2 < a1 B. a1 < a3 < a2 E. a3 < a1 < a2 C. a2 < a1 < a3

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adding wavelengths results to 625 (mu)m, which is not an integral multiple of lambda, which is 50, actually, 625 divided by 50 results to multiple m + 1/2, therefore destructive interference.
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y = mR(lambda) / d, R is decreased so y also decreases, B.
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apply Phase shift principles here.
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w = mR(lambda)/ a. higher a means lower width of central bright fringe, or CBM.
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11 | A

44. Snake eyes. In a two–slit diffraction experiment in which each slit has width a and separated by a distance d = 4a, how many bright fringes can be counted inside the diffraction central maximum?

A. 3 D. 6

B. 4 E. 7 C. 5

45. Green lantern. Green light of wavelength λ = 540 nm is used in three diffraction

experiments with varying number of slits (N1 = 2, N2 = 4, N3 = 6). Each slit is very narrow, and any two adjacent slits are separated by the same distance d. Arrange the slit numbers in order of increasing brightness of the principal maxima. A. N1 < N2 < N3 D. N3 < N2 < N1 B. N1 = N2 < N3 E. N3 = N1 < N2 C. N1 = N2 = N3

REMOVAL EXAM 27 OCTOBER 2011

10:00 a.m. - 12:00 p.m. Room: NIP R201

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remember this formula: 2d/a - 1
Clemente
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Clemente
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Clemente
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Clemente
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Clemente
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Intensity of light increases as number of slit increases, as dictated by the formula: Imax = N^2*Islit