# HSC Physics MC Thread (1 Viewer)

#### Mr_Kap

##### Well-Known Member
I'm goin to be having a lot of questions on Multiple Choice for Past HSC Physics so i thought i would post them in one thread.

First question:

#### crankymidget

##### New Member
The disc is moving in same direction relative to the magnetic field. This creates creates a DC current. According to Lenz's law, the current will oppose the change in flux that created it. The disc is moving toward brush Y, therefore the current will flow from Y to X.

#### Mr_Kap

##### Well-Known Member
The disc is moving in same direction relative to the magnetic field. This creates creates a DC current. According to Lenz's law, the current will oppose the change in flux that created it. The disc is moving toward brush Y, therefore the current will flow from Y to X.
is that even legit reasonin??

#### Mr_Kap

##### Well-Known Member
And another fucked question. WHY IS THE ANSWER A? not B??:

Last edited:

#### crankymidget

##### New Member
is that even legit reasonin??
Haha. I couldn't illustrate my point very eloquently. MY reasoning was that the, the x brush is on the axel and the disc is rotating toward the y brush. Thats how I reasoned it in my head. I can't say for sure if this is correct, but it's the conclusion I came to.

I've had a look at some worked answers since and this is what i've found. "We know that because there is relative motion between the conductor and the magnetic field, a current will be induced. As the conductor is always moving in the same direction relative to the magnetic field, a direct current will be induced; hence the answer is either B or C.
We know that either the current will flow from X to Y (current flowing down), or Y to X (current flowing up). Also we know from Lenz’s Law that the induced current will be in such a direction that the magnetic field created by the induced current will oppose the original changing magnetic field. Hence the current must flow from Y to X for this to be true."

My excel success book says something similar, but does not give an answer as to working out the relative motion.

##### Member
And another fucked question. WHY IS THE ANSWER A? not B??:

When the current is turned on in X, at the start a magnetic field is induced to oppose the increase in voltage, and therefore generate a back EMF, however it will eventually reach a constant voltage as it is a DC circuit and there will eventually be no magnetic field. A similar field will be created to oppose the voltage change when the switch is turned off but it will be in the opposite direction. Because there is only a changing magnetic field when the switch is turned off and on and not while it is operating at full voltage, a current will only be induced in coil Y when this field is changing, so therefore only when the switch at X is turned on and off. The induced voltage in Y when the switch is turned on will be in the opposite direction as that induced when the switch is turned off, so it would be A.

#### Mr_Kap

##### Well-Known Member

What are the approimate work functions of both these metals??? is it where the lines start???
So is metal y at 9 and metal x at 4??

I though the work function is where the line meets the "KINETIC ENERGY" axis (y-intercept), and the absolute value of that is the work function??

#### InteGrand

##### Well-Known Member

What are the approimate work functions of both these metals??? is it where the lines start???
So is metal y at 9 and metal x at 4??

I though the work function is where the line meets the "KINETIC ENERGY" axis (y-intercept), and the absolute value of that is the work function??
$\bg_white Recall that the lines are E=hf-W_M, where W_M is the work function of the metal M. Note that all metals thus have the same slope for the line, and the only difference is the vertical intercept, which is determined by the metal's work function W_M.$

$\bg_white For a metal M, let f_{0, M} be the frequency where E is 0 (this is the \textit{threshold frequency} for that metal). Then 0=hf_{0, M}-W_M\Rightarrow W_M = hf_{0,M}. In other words, the work function for the metal is h (Planck's constant) times the horizontal intercept (threshold frequency of that metal).$

#### kawaiipotato

##### Well-Known Member

What are the approimate work functions of both these metals??? is it where the lines start???
So is metal y at 9 and metal x at 4??

I though the work function is where the line meets the "KINETIC ENERGY" axis (y-intercept), and the absolute value of that is the work function??
Yes you're correct.
Note that
$\bg_white KE = hf - \phi where \phi is the work function$

$\bg_white let KE = 0,$

$\bg_white \phi = hf$
$\bg_white where h is just Planck's constant and f is the x-intercept (so 9x10^14 Hz for metal Y)$

#### Zlatman

##### Member

What are the approimate work functions of both these metals??? is it where the lines start???
So is metal y at 9 and metal x at 4??

I though the work function is where the line meets the "KINETIC ENERGY" axis (y-intercept), and the absolute value of that is the work function??
The line has the formula: EK = hf - W

So yeah, when the value of the work function occurs when the frequency is zero, i.e. the y-intercept (as you said).

But, we can also find it out using the x-intercepts. When EK = 0, W = hf.

The THRESHOLD FREQUENCY for Metal X is at 4 x 10^-14 and for Metal Y is at 9 x 10^-14. Therefore, the WORK FUNCTION for Metal X is h * (4 x 10^-14) and the work function for Metal Y is h * (9 x 10^-14).

(LOL, NEVER MIND AGAIN - done above twice)

#### InteGrand

##### Well-Known Member

What are the approimate work functions of both these metals??? is it where the lines start???
So is metal y at 9 and metal x at 4??

I though the work function is where the line meets the "KINETIC ENERGY" axis (y-intercept), and the absolute value of that is the work function??
$\bg_white Reading off the graph then, we have$

$\bg_white W_X = hf_{0,X}\approx 6.626 \times 10^{-34}\text{ J s}\times 4\times 10^{14}\text{ Hz} (since the horizontal intercept for metal X is about 4 (based on a glance, use a ruler in the actual exam for more accurate value))$

$\bg_white \approx 2.65 \times 10^{-19}\text{ J}.$

$\bg_white Similarly, W_Y =hf_{0,Y}\approx 6.626 \times 10^{-34}\text{ J s}\times 8\times 10^{14}\text{ Hz}\approx 5.30 \times 10^{-19}\text{ J}.$

#### Crisium

##### Pew Pew
The line has the formula: EK = hf - W

So yeah, when the value of the work function occurs when the frequency is zero, i.e. the y-intercept (as you said).

But, we can also find it out using the x-intercepts. When EK = 0, W = hf.

The THRESHOLD FREQUENCY for Metal X is at 4 x 10^-14 and for Metal Y is at 9 x 10^-14. Therefore, the WORK FUNCTION for Metal X is h * (4 x 10^-14) and the work function for Metal Y is h * (9 x 10^-14).

(LOL, NEVER MIND AGAIN - done above twice)
This is like the 5th time this week LOL

#### InteGrand

##### Well-Known Member
$\bg_white The key thing to remember therefore for calculating work functions is the equation W_M=hf_{0,M}, i.e. multiply the frequency intercept for that metal's E-f graph by h.$

#### Zlatman

##### Member
This is like the 5th time this week LOL
ikr, ffs, need to get faster at typing up my answers, lol

##### Member

What are the approimate work functions of both these metals??? is it where the lines start???
So is metal y at 9 and metal x at 4??

I though the work function is where the line meets the "KINETIC ENERGY" axis (y-intercept), and the absolute value of that is the work function??
Its where it meets the y-axis, so if you have the equation hf=hf0+KE and you re arrange it to give your variables as KE and f, then you get KE=hf-hf0 then in the form y=mx+b, you get the y intercept to equal -hf0, which is also equal to the work function, and the gradient is equal to h from mx=hf

#### Mr_Kap

##### Well-Known Member

i thought it would be an INCREASING straight line.

Because it is a radial magnetic field and THETA is always 0 degrees it must be some form of straiht line, but i thought if it speeds up then the torqu needs to be larger

EDIT: I READ THE ANSWER WRONG. IT WAS D which was a straight line with negative gradient

Last edited:

#### vitamin D

##### Active Member
2014 question 15 (sorry dont know how to post pics on here yet)

#### Mr_Kap

##### Well-Known Member

Why is it B?

i thought it would be an INCREASING straight line.

Because it is a radial magnetic field and THETA is always 0 degrees it must be some form of straiht line, but i thought if it speeds up then the torqu needs to be larger

Last edited:

#### vitamin D

##### Active Member

Why is it B?

i thought it would be an INCREASING straight line.

Because it is a radial magnetic field and THETA is always 0 degrees it must be some form of straiht line, but i thought if it speeds up then the torqu needs to be larger
"radial" means constant torque, hence the straight line for torque. Not good enough at phys to explain in depth, sorry!

edit: just saw the real answer, ignore everything i said.

Last edited: