Here is the Final UPCAT Sample Quiz in the Science Subtest. If you find any errors in the quiz, please dont hesitate to let us know about it through the comments section. Moreover, if you have a better solution set or explanation that is provided in the solution sets, please dont hesitate to share it too.

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## Physics Quiz

*Physics Quiz*. You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%%

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**and please dont forget to click the share button to share the quiz with your classmates!^^**ps. you still have to adjust your scores for the UPCAT Penalty of correct - 1/4 wrong.

Question 1 |

A | Doubled |

B | Tripled |

C | Quartered |

D | Halved |

Question 2 |

A | Car A went on reverse at t = 6 secs and stopped moving at t = 3 secs and resumed motion at t = 4 secs. |

B | Car A stopped moving at t = 6 secs and is moving at constant velocity at t = 3 secs until it accelerated at t = 4 secs. |

C | Car A went on reverse at t = 6 secs and is moving at constant velocity at t = 3 secs until it accelerated at t = 4 secs. |

D | Car A decelerated at t = 6 secs and is moving at constant velocity at t = 3 secs until it accelerated at t = 4 secs. |

Question 3 |

A | The block is exerting force on Carrie. |

B | Both a and b. |

C | Carrie is exerting force on the block. |

D | Neither a and b. |

Question 4 |

A | A cart moving downhill on a mountain slope |

B | A ball thrown upwards |

C | A car running along a curve |

D | A book on the table |

Question 5 |

A | t = 2 secs to t = 4 secs. |

B | t = 6 secs to t = 8 secs. |

C | t = 4 secs to t = 6 secs. |

D | t = 0 secs to t = 2 secs. |

Question 6 |

A | It is doubled. |

B | It is halved. |

C | It is quadrupled. |

D | It is quartered. |

Question 7 |

A | ordinary speed |

B | average speed |

C | instantaneous speed |

D | uniform speed |

Question 8 |

A | Applied |

B | Gravity |

C | Tension |

D | Normal |

Question 9 |

A | A scalar has direction while a vector doesn’t. |

B | A vector is larger than a scalar. |

C | A vector has direction while a scalar doesn’t. |

D | A scalar is a typed of vector. |

Question 10 |

^{2})

A | 2341.12 W |

B | 5500 W |

C | 1567.13 W |

D | 1718.75 W |

Question 11 |

A | 3 m/s ^{2} |

B | 0.25 m/s ^{2} |

C | 4 m/s ^{2} |

D | 1 m/s ^{2} |

^{2}

Question 12 |

A | bounces off, changes direction |

B | passes through, bounces off |

C | bounces off, changes speed |

D | stops, bounces off |

Question 13 |

A | \(\frac{Kg\cdot m}{s}\) |

B | m/s2 |

C | Joules |

D | Newton |

Question 14 |

A | 0 N |

B | 800 N |

C | 4800 N |

D | 48000 N |

Question 15 |

A | The bigger the object the more inertia it has. |

B | Inertia is the force that stops motion. |

C | Inertia is a force that keeps objects at rest or relative motion. |

D | Inertia is directly proportional to speed. |

Question 16 |

A | None of the above |

B | a and b |

C | Potential energy only exists when an object is at rest. |

D | Potential energy is dependent with altitude. |

Question 17 |

A | Increasing |

B | Decreasing |

C | Constant |

D | Massive |

Question 18 |

A | Ohm |

B | Coulomb |

C | Watt |

D | Volt |

Question 19 |

A | The voltage rating of each battery. |

B | The sum of the voltage rating divided by the resistance of the load. |

C | The sum of their voltage ratings. |

D | The voltage rating divided by the resistance of the load. |

Question 20 |

A | The initial velocity of projectile dictates the peak height of the projectile. |

B | The acceleration of the projectile at its peak is 0 m/s2. |

C | The speed of a projectile is always constant. |

D | The acceleration has both upward and downward acceleration. |

Question 21 |

A | Expandable |

B | Elastic |

C | Inelastic |

D | Conserved |

Question 22 |

A | Speed |

B | Velocity |

C | Momentum |

D | Acceleration |

Question 23 |

A | 5 m/s2 |

B | 8.25 m/s2 |

C | 9 m/s2 |

D | 7 m/s2 |

Question 24 |

A | The force of the ball against the bat. |

B | Muscular effort in the player’s arm. |

C | The air resistance on the ball. |

D | The grip of the player’s hand on the bat. |

Question 25 |

A | ultrasonic |

B | supersonic |

C | mega sonic |

D | infrasonic |

Question 26 |

A | Kinetic energy is always positive. |

B | Free falling objects have increasing kinetic energy. |

C | Kinetic energy depends upon the position of the object. |

D | If two objects have the same mass then the one that moves faster has the greater kinetic energy. |

Question 27 |

A | Watt |

B | \(\frac{kg\centerdot {{m}^{2}}}{{{s}^{2}}}\) |

C | \(N\centerdot m\) |

D | Joule |

Question 28 |

A | 2.5 m/s |

B | 2.14 m/s |

C | 15 m/s |

D | 25 m/s |

Question 29 |

A | Joules |

B | Watts |

C | Hertz |

D | Torr |

Question 30 |

A | Acceleration due to gravity is constant |

B | Acceleration due to gravity is imaginary |

C | Gravity changes the mass of an object |

D | Gravity is affected by altitude |

Related posts:

- UPCAT Sample Questions: General Science
- UPCAT Sample Test: Basic Math
- UPCAT Sample Questions: Biology
- UPCAT Sample Exam: Chemistry
- UPCAT Review Questions: Algebra
- UPCAT Review Questions: Sample Trigonometry / Geometry Quiz
- UPCAT Review Questions: Sample Parts of Speech/General Grammar Rules Quiz
- UPCAT Sample Questions: Syntax and Mechanics
- UPCAT Reviewer / Sample Quiz: Diction and Vocabulary
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- UPCAT Mock Exam: Reading Comprehension

Number 13

Mali yung computation,

Vf=Vi+at

120=0+a(5)

a=120/5

a=24m/s

so kung force hinahanap F=ma

F=(800kg)(24m/s)

F=19200 N

wala yung sagot sa choices

22/30 🙁 practice pa moreeeee!bring it on!

Correction: L1 = 80cm not L2. L2 = 80cm x 500N / 1500N, L2 = 26.67cm (means that 1500N is 26.67cm from the fulcrum).

Haha tnx d2 mas nging broad yung mind ko sa lalabas sa UPCAT <3

Help me po,Im taking UPCAT this August,give me somd tips,hints sa mga pweding lumabas this year’s UPCAT,09273336362,number ko,text or call me nalang f u can help me. Thanks in advance

Help me po,Im taking UPCAT this August,give me somd tips,hints sa mga pweding lumabas this year’s UPCAT

kailan po yung UPCAT for batch 2012-2013?

The UPCAT for AY 2012-2013 freshmen was held in August 2011. The UPCAT for AY 2013-2014 freshmen (those who want to become UP Freshies in June 2013) will be administered in August 2012.

i need to practice more :)) i only got 12 :))

At least you still have time to practice. Incoming 4th year HS student? 🙂

@Arvi, Is that how it is? But I thought the heavier person should be much closed to the center so that the seesaw would balance. Since the distance of the lighter person is 20cm from the center, shouldn’t the distance of the heavier person be smaller?

Pls re-read. It’s 20cm from the left side. So it’s 80cm from the center given the length of seesaw is 2m (200cm), half of it being 1m (100cm). I agree with Arvi.

Hope I helped! 🙂

kelan po deadline ng application form ng ustet

Hi!

I’m Arvi from UP Diliman, Eng’g Batch ’03.

I’d like to correct the solution to question no. 25. Following your line of explanation, L1/ F1 = L2/ F2, we can’t simply express the moment arms (L1 & L2) proportional to the forces (F1 & F2) in this manner. In Physics, this is governed by statics, wherein the summation of forces and moments must equal to zero at equilibrium (to prevent the see-saw from moving or changing its horizontal position.) Hence, the correct proportion should have been L1/F2 = L2/F1 or L1xF1 (moment of F1) = L2 x F2 (moment of F2) using the fulcrum (center of see-saw) as our reference for the moments. Wherein L1 & L2 are the moment arms, F1 & F2 are the forces or weights of the bodies. Once you get L2, subtract it from 1m or 100cm to get the answer.

Moreover, using the center of the see-saw as our point to get the summation of moments, L1 would be 80cm from the fulcrum. F1 is 500N. The moment of force would be F1 x L1 (500N x 80cm = 40000 N.cm, counterclockwise) This moment must be equal to the moment of force, F2 = 1500N in order to maintain equilibrium (F2 will cause a moment in the clockwise direction). The position, x, from the fulcrum multiplied by F2 must be equal to 40,000 N.cm. Solving for x; 40000 = 1500x, x = 26.67cm. Hence, the answer must be 100 – x = 100 – 26.67 = 73.33 from the right side of the see-saw.

Therefore, the answer in not in the given choices. Just my two cents.

ang galing galing nman..

i agree on your equation ARVI, which is L1/F2 = L2/F1, but with respect to the given length of the see-saw which is 2meters (200cm), the distance from the fulcrum on the left side is 180cm (L1=180cm). F1 and F2 are the weight of the boy and the man (F1=500N, F2=1,500N). Thus. L2= (L1*F1)/F2,

L2=(180cm*500N) / 1,500N

L2=90,000cmN / 1,500N

L2=60cm

Hence, the answer remains to be 60cm (which is included on the choices) however the explanation should be edited.

Fulcrum is in the middle of the seesaw, therefore L2=80cm not 180cm.