• Common Questions Ask in Post UTME Physics Exam Day7

    Physics

    Chima Sunday (Tutor)
    23-06-2016 17:30:00 +0000

    Try your best to solve these questions. Show working on paper when necessary and include a picture of your hand written Solution in your response. Wish you the best!

    1. A body of density 0.8kgm-3 weighs 120N in air. It is suspended with a string with half of its volume immersed in a liquid of density 0.6kg/m-3. Find the tension in the string
    A. 75N
    B. 65N
    C. 45N
    D. 35N
    (uniben)

    Updated Solution
    Note all quantity is in its SI unit
    density of body = 0.8; weight in air = 120; recall that weight = mg
    So m = weight/g = 120/10 = 12
    recall that density = m/V; V = m/density = 12/0.8 = 15
    If the half of the volume is immersed in water, it therefore follows immediately from Archimedes that the volume of water displaced is half of the volume, so the mass of liquid displace is given thus, m = V × density = 15/2 × 0.6 = 4.5
    thus weight of liquid displace = 4.5 × 10 = 45 = Upthrust U
    from principle of flotation, U = Weight in air - Weight in liquid
    45 = 120 - Weight in liquid
    Weight in liquid = 120 - 45 = 75
    So Answer is Option A

    2. Some quantity of hot water at a temperature T is added to warm water at a temperature of 25oc in the ratio 1:4. Determine T if the final temperature of the mixture is 30oc
    A. 35oc
    B. 30oc
    C 50oc D. 55oc
    E. 10oc
    (futo)

    Updated Solution
    Heat lost quantity of water m = mc(T - 30)
    Heat gain by quantity of water M = Mc(30 - 25)
    neglecting heat loss to the environment, so mc(T - 30) = Mc(30 - 25)
    m(T - 30) = M(5)
    from question ratio of quantity is 1:4, so m/M = 1/4 and M/m = 4
    thus T - 30 = M/m(5) = 4 × 5 = 20
    T = 20 + 30 = 50
    Answer is Option C

    3. A uniform rod 10m long is balanced on a pivot placed at its midpoint, a girl of mass 30kg sits on arm of the rod at a point 3m away from the pivot. What mass can be place 2m away from the other end of the rod to keep the rod horizontal?
    A. 30kg
    B. 45kg
    C. 20kg
    D. 40kg
    (abu)

    Updated Solution
    From principle of moment, ACM = CWM
    So taking moment at the pivot
    30 × 3 = m(5 - 2)
    90 = 3m
    m = 30
    In fact we can even guess the answer rightly, without actually solving it, if you really understand the physical interpretation of principle of moment.
    So the correct answer is A



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