• Difficult Questions From Physics Jamb 2004


    Admin Admin
    28-01-2016 12:03:00 +0000

    These questions presented a challenge for most of the students that attempted them. Lets all attempt to answer them.

    (Question 26) In a reverse biased junction diode, current flows in by.........?
    A. Electrons alone
    B. Majority carriers
    C. Minority carriers
    D. Positive holes alone

    (Question 33) The change in volume when 450kg of ice is completely melted is?
    [density of ice = 900 kgm-3, density of water = 1000 kgm-3]
    A. 0.5m3
    B. 0.45m3
    C. 0.05m3
    D. 4.50m3 UPDATE - ANSWER INCLUDED BELOW A big thank you to all that have participated in solving these questions. Though no one got the right answer it's encouraging to see you attempt to solve them. Here is a detailed breakdown of the solution for each of the questions above.

    (Question 33).
    The correct answer is C EXPLANATION
    When you here the word change it implies difference.
    So change in volume V = Volume of ice Vi - Volume of water Vw.
    Now our work is to find the volumes of ice and water.
    From formula, Density=mass/volume
    Then volume by subject formula = mass/density.
    For ice Vi = mass of ice/density of ice = 450/900 = 0.5m3
    For water Vw = mass of water(which is same with ice)/density of water = 450/1000 = 0.45m3
    Therefor change in volume V = 0.5m3 - 0.45m3 = 0.05m3

    (Question 26).
    The correct answer is C EXPLANATION
    For reverse bias junction diode, the negative terminal of the battery is connected to the P-type material, and the positive terminal of the battery to the N-type material.

    The negative potential attracts the holes (which are positively charged) away from the edge of the junction barrier(P-N junction) on the P side, while the positive potential attracts the electrons (negatively charged) away from the edge of the barrier on the N side. This action increases the barrier width because there are more negative ions on the P side of the junction, and more positive ions on the N side of the junction. This increase in the number of ions prevents current flow across the junction by majority carriers. However, the current flow across the barrier is not quite zero because of minority carriers crossing the junction.

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