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UPCAT Mock Test: Physics

UPCAT Sample QuizHere 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.

You might want to read more about the UPCAT after taking the quiz.   Here is a compilation of all our UPCAT tips and strategies organized into one post:  UPCAT: Your Ultimate Guide. You might also want to check out our 101 UPCAT Tips on the fanpage.  Finally, you might want to join the discussion and be updated on college entrance exam news and tips on our Fanpage or be a part of the 2011 UPCAT Review group where students like you help each other review for the UPCAT.

Physics Quiz

Directions: For each statement or question, choose the letter of the word or expression that, of those given, best completes or answers the question. ► Calculators of any kind are not permitted. All numbers used are real numbers.
Start
Congratulations - you have completed Physics Quiz. You scored %%SCORE%% out of %%TOTAL%%. Your performance has been rated as %%RATING%% Post your Results on our Fanpage to view how well you did compared to others [UPCAT is a ranking system afterall.:D] 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.
Your answers are highlighted below.

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17 thoughts on “UPCAT Mock Test: Physics

  1. 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

  2. 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

    • 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.

  3. @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! 🙂

  4. 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.

    • 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.

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