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    does anyone have them? lol

    cause im working through them as prep for my interview (not at trinity, but it does have a similar style "quiz" at interview).

    thanks
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    If you post the problems up, I'm sure you will get a few people chipping in with solutions when they have time.
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    http://www.trin.cam.ac.uk/show.php?dowid=4

    :P

    i spent an hour last night on it, and did 1 which is a "show that". i got "((pi)^3)/6 - (pi)/4" for 3. 1600 for 4. "65/129" for 5.

    ill stick up more of my answers once i get round to doign some more
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    right, ive spent 2h10m on them now (according to watch). and got the following answers. if anyone could confirm them or not that would be great thanks (2 and 9 blank cause i havnt done them yet)

    1. show that...
    2.
    3.\frac{\pi^3}{6} - \frac{\pi}{4}
    4. 1600
    5. 65/129
    6. a) mu > tan(theta)
    b) \sqrt{2dg(\mu(cos(theta))-(sin(theta))}
    7. show that...
    8. root(2gl)
    9.
    10.i) 35/76
    ii) 25/57
    iii) 77/228


    and i havnt rigorously checked my answers to be correct so dont hold me to the answers lol.
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    For 5, I get 5/516.

    Edit: I now 65/129.
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    I'm fairly sure the answer to 5 reduces to something 'nice'. (Both from memory of seeing a solution and doing it myself).
    I don't agree with the answer to 8.

    I agree with 3,4 and 10. I haven't looked at 6 yet.

    Edit: 6 looks OK too.
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    (Original post by Glutamic Acid)
    For 5, I get 5/516.

    Edit: I now 65/129
    (Original post by DFranklin)
    I'm fairly sure the answer to 5 reduces to something 'nice'.
    I get 5/6.
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    I get 2/3. (Roughly, you are 128 times more likely to pick an unbiased coin, but 256 times more likely to get 8 heads using a biased coin. So biased coin | 8 heads is twice as likely as unbiased coin | 8 heads).
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    (Original post by DFranklin)
    I get 2/3. (Roughly, you are 128 times more likely to pick an unbiased coin, but 256 times more likely to get 8 heads using a biased coin. So biased coin | 8 heads is twice as likely as unbiased coin | 8 heads).
    I get 5/6.

    P ( biased | 8 heads ) = \displaystyle \frac{\frac{1}{129}}{\frac{1}{12  9} + \frac{128}{129} \left ( \frac{1}{2} \right )^8} = \frac{2}{3}

    P ( 'biased | 8 heads) = 1/3

    P (heads) = 1/3 * 1/2 + 2/3 = 5/6 ?

    EDIT: I think you just forgot that the unbiased coin could produce a head again
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    if i remember correctly.

    for 5, i said that there is a 1/129 chance the bias coin is chosen and that there is a 128/129 chance that an ubias coin was chosen. if the bias was chosen, the 9th head would be a certainty, so thats 1/129 x 1. the other possibly is that the unbias is chosen, and that the 9th head has a probability of 1/2. so that would be 128/129 x 1/2

    add up the 2 possbilities and i get 65/129.
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    Just look at what Simon has said in his post above, his answer is correct.

    You have not used the information that there has already been 8 heads (which makes it more likely that the coin chosen was the biased one)
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    aaaa i see where i went wrong thanks
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    any help with 8? if im wrong i think i misunderstoof the question lol.

    thanks
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    (Original post by SouthernFreerider)
    any help with 8? if im wrong i think i misunderstoof the question lol.

    thanks
    I get a smiliar answer, but I may well have missed something too.

    I've treated the rod as a particle located at its centre of mass, and used energy conservation laws:

     \frac{1}{2}mv^2 = mgh \implies v = \sqrt{2gh} ,

    and as the change in vertical displacement of the particle is l/2, the speed of the particle when the rod is first vertical is  \sqrt{gl} . However, the question asks for the speed of the bottom of the rod. This has to travel double the distance in the same time, and so must move at double the speed, which gives an answer of  2\sqrt{gl} .

    I may be wrong though.
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    For those on here that have had interviews at Trinity, how does this quiz compare to yours in terms of difficulty?
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    (Original post by SimonM)
    EDIT: I think you just forgot that the unbiased coin could produce a head again
    Actually, I was just careless about reading the question and just gave the prob that the coin was biased as the final answer.

    Sorry.
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    (Original post by Daniel Freedman)
    I get a smiliar answer, but I may well have missed something too.

    I've treated the rod as a particle located at its centre of mass,
    This is an oversimplification. The rod has rotational energy too, so you need to consider the moment of inertia.
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    I can't seem to get an analytical solution to 8. :blushing: I have a D.E. of the form \ddot{y} = -\frac{2g}{l} \sin (y) but, unless I've forgotten a DE method, this equation is not solvable. (My coefficient might be wrong, but I think the form is correct.)

    Have you solved 8, Dave, or do you also think there is no analytical solution? (which doesn't seem likely, given the context of the question)
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    You can solve it by energy conditions. Let I be the MI of the rod about the pivot point (which you need to calculate or look up). Then when rotating with angular velocity w, the K.E. = w^2 I / 2. But the gravitational energy gained is mgl/2. Equate, solve for w, and then the final speed is lw.
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    Oh right, thanks. I was doing sort-of the right thing, but just over-complicated it.
 
 
 
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