Oxford Physics Admission Interview 2: Questions and Answers

It’s difficult to concisely say how my first interview left me feeling, but if I were forced to choose a word, I’d go with “bemused”. I’d gotten through one question fairly cleanly (like, I thought?), and one only with a significant amount of assistance from the tutors. Was that good? Was that expected? Having arrived 10 minutes early to the second interview on account of being extremely over-cautious in my estimates of how long it would take me to walk from my room to Tom Quad, there was nothing for it but to agonise over the dilemma whilst listening to the muffled voices in the room. Again I speculated how big the effect of having a good/bad candidate precede me would be.

After an awkward brushing past the previous candidate – it was a different one to before my first interview, so at least they were mixing it up that way – a face poked through the door and informed me that the interviewers would be with me in a minute. That was good. That was another unknown removed.

The setup was much the same as in the first interview: two tutors on one side of a desk upon which was some paper and biros, and me on the other. Once again, there was absolutely zero chit-chat about what I’d written in my personal statement, and question one came out…

Question 1

Imagine that you have two clocks. You can’t tell anything about their inner workings, just watch them tick. You take them both to the moon, and one goes crazy.

This was made no less funny by his saying it in a thick German accent. (I mean, he was German, he wasn’t putting it on just for the question.) I wouldn’t rise to that kind of cheap shot though, and merely nodded my understanding: sure, crazy clock. What of it? The other tutor took over…

Intuitively, what might you put this difference down to?

“GRAVITY?” I exploded, before quickly qualifying it with the more sober “I mean, a less-strong gravitational field?”

Dramatic shifts in seating positions.

What sort of clock might be affected by the gravitational field strength?

At this point, being aware that they were supposed to ask you about AS-Level stuff, I had an idea of where they might be going. I answered that a pendulum clock would.

If I were to tell you that the field strength on the surface of the moon is approximately a sixth of that on the surface of the earth, what would be the difference between the two clocks?

Ha! Rumbled. Fortunately, the formula for the time period of a pendulum clock was fresh in my mind from AS Level physics, and, thanks to that interview, still is to this day:

T=2\pi\sqrt\frac{l}{g}

… where l is the pendulum length and g is the field strength. In trying to compare the behaviour of the two clocks, I wrote two versions for earth-clock and moon-clock:

T_E=2\pi\sqrt\frac{l}{g_E}

T_M=2\pi\sqrt\frac{l}{g_M}

And the interviewer had just told me that g_M=\frac{1}{6}g_E, so I could write

T_M=2\pi\sqrt\frac{l}{g_E/6}

… or, to form a more direct comparison I suppose,

T_M=\sqrt6T_E

There was, again, that awkward period where I felt I’d gotten to an answer and the tutors just stared at me, so I vocalised the result I’d just derived with “the time period of the clock on the moon is root six times the time period of the one on the earth.”

And root six is about, what?

WHAT THE HELL KIND OF QUESTION IS THAT? Looking back, I suppose they were trying to see whether I thought that was a realistic answer or that alarm bells should be ringing, but at the time all I could think was “I’m being asked what the square root of 6 is in my Oxford interview.” I reasoned that it must be slightly above 2, so went with “about 2.3”. No feedback other than another awkward pause. The next question came out.

Question 2

Intuitively, what angle to the ground would you fire a cannon at in order to get the maximum range on the projectile?

I knew full-well that the answer was 45 degrees, having proved it at some stage during AS Level mechanics, and also having read recently that some guy had been the first to prove it. I was up front with this, but the interviewer was very keen that I was answering out of “intuition” rather than knowing the answer. Ah. Yes. Do let’s play that game. Talking about a reasonable compromise between vertical motion (for time of flight) and horizontal motion (for a greater rate of range progression), I went with 45 degrees with the air of one speculating that next summer might be warm.

Predictably, the extension was to prove this result, so I cracked out a cheeky diagram (framed prints are available – contact me):

Screen Shot 2014-08-20 at 12.50.14

… and wrote down the expressions for the vertical and horizontal components of the starting velocity:

v_V=Vsin\theta

v_H=Vcos\theta

The time of flight is decided solely by the vertical motion: the projectile undergoes constant acceleration due to gravity, until it reaches a vertical velocity of the same magnitude but in the opposite direction, by which point it will hit the ground again. (Neglecting effects due to air resistance and other important real-world factors obviously.) Using one of the constant acceleration equations (difference between initial and final velocities is equal to the product of the acceleration and time of flight), and denoting the time of flight as t and the acceleration as g, you get:

2Vsin\theta=gt

ie, the time of flight is

t=\frac{2Vsin\theta}{g}

In the horizontal direction, no forces act, so the range of the projectile is just speed multiplied by time:

r=v_Ht

r=(Vcos\theta)(\frac{2Vsin\theta}{g})

r=\frac{2V^2sin\theta cos\theta}{g}

This is all we need to answer the question – an expression for r as a function of θ, involving some constants V and g.

At this stage in the proceedings, the deities of trig identities threw me down a bone in the form of a memory. To find the maximum of this with respect to theta, you’d normally have to go through the rigmarole of differentiating, but by a lucky quirk of fate the relation sin(2\theta)=2sin\theta cos\theta was fresh in my mind, so I rewrote the above as

r=\frac{V^2}{g}sin(2\theta)

At this stage I took a chance that the interviewers would back me to know what a sine curve looks like, and that the (first) maximum occurs when the argument of the sine function is equal to 90 degrees, ie r_{max} occured when

2\theta_{max}=90^{\circ}

\theta_{max}=45^{\circ}

The tutors looked at each other. One was clearly satisfied with this, and the other clearly wanted me to go through the differentiation rigmarole. Entertainingly, for some reason neither of them wanted to say this, so there was a good deal of grunting and widened eyes before one of them caved in and came out with the next question.

Question 3

You have a blob of metal that you can deform in any way you like. You start off with it in the form of a cuboid, and measure the resistance between each “end”. Now you want to deform it into another cuboid, where the resistance between the ends is double that of the original cuboid. How do you change its shape?

Ie, I wanted to go from something like this…

Screen Shot 2014-08-20 at 13.33.36

… to something like this…

Screen Shot 2014-08-20 at 13.34.28

… so that the resistance across the block in the latter case is double that of the block in the first place.

Given that we were talking about changing the length and cross-sectional area of a piece of metal, it seemed that the resistivity formula would be relevant, so I wrote that down:

R=\frac{\rho l}{A}

… where ρ is the resistivity of the material, l is the length of material, and A is the cross-sectional area.

One of the tutors encouraged me to write down “before” and “after” equations, and impose the relation between them that the “after” resistance be twice the “before” resistance.

R_1=\frac{\rho l_1}{A_1}

R_2=\frac{\rho l_2}{A_2}

R_2=2R_1

At this stage I supposed I needed some relationship between the various lengths and areas in order to eliminate some stuff and solve, so given that I was only changing the shape of the same bit of metal, I argued that the volumes before and after must be the same. If I multiplied the numerators and denominators on each side by the appropriate length, I could express the two in terms of volumes before and after thus:

\frac{\rho l_2^2}{A_2l_2} = 2\frac{\rho l_1^2}{A_1l_1}

\frac{\rho l_2^2}{V_2} = 2\frac{\rho l_1^2}{V_1}

But, same piece of metal, so V_1=V_2, so…

l_2^2=2l_1^2

… ie,

l_2=\sqrt{2}l_1

So there you go, stretch it into a cuboid about 1.4 times as long and you’d have double the resistance. The interviewers nodded. There were a few very short questions about some areas of physics I’d mentioned in my personal statement, which amounted to little more than “so, you like that, do you?” And with that, my grand total of 40 minutes under the Oxford admissions microscope were over. In a haze of surreality I went and got my train.