Firstly, I would like to caution students regarding the use of solutions when preparing for exams. They are certainly a double-edged sword and since you are planning to invest an enormous amount of time, effort and money into getting a degree, it would be a good idea to use solutions properly from now on. In all cases you should only consult a solution once you think your answer is correct or if you have spent a significant amount of time trying to reach an answer. Coming up with your own conclusion after several minutes of independent thought is far more worthwhile than learning directly from the solution itself. You should also make sure you are critical of all solutions presented to you. Not only will this help your understanding, but solutions may have errors. Therefore if you find that you can't reconcile a solution with yours then take the time to get in touch with the author to discuss the discrepancy. For example, please do contact me if you have spotted a mistake below or would like further guidance on any of the questions.

I will firstly set out a series of hints. Try to do the questions from these alone and if you are still having difficulty then scroll down to the full solutions.

Hints

Section 2 - Physics

1 a) (i) Work out how many degrees are turned through in one second. From this you can work out how long each degree takes.

(ii) Work out how long it takes the discs to turn through $30^{\circ}$ and then use this to evaluate the speed required to traverse $24cm$ in this time.

(iii) Think about the alternatives to the discs rotating $30^{\circ}$ - this is not the only option for molecules to travel through both slits.

(iv) Repeat the calculation for (ii) with the smallest possible angle of rotation apart from $30^{\circ}$.

(v) Repeat the calculation for (ii) being careful to select the correct values for the maximum and minimum angles turned through.

b) (i) Use the equations of constant acceleration to work out the time taken. Use this in conjunction with the number of degrees turned per second.

(ii) Think about how many degrees the disc must rotate before the hole is in the correct place for the first time after $t=0$. Substitute into your expression in b) (ii) and rearrange for $h$.

(iii) Firstly accept that there is no frequency of rotation that allows this to occur with the particle being released from above the holes. Turn your attention to the amount of time taken for the particle to fall from the upper disc to the lower disc and work from there.

2 a) (i) Use Ohm’s law remembering that the potential difference of the cells combine in series.

(ii) Cells in parallel don’t give a higher overall voltage but the current splits at the junction.

b) (i) Consider the internal resistance to be a resistor connected in series with the cell it belongs to.

(ii) Define an unknown current $I$ and use this to find the voltage dropped across each internal resistance (in terms of $I$). Equate the potential difference provided by the parallel combination to $IR$ for the $1.0\Omega$ resistor and solve for $I$. This yields the ammeter readings and applying Ohm’s law to the $1.0\Omega$ resistor gives the reading on the voltmeter.

c) If you can’t think of a worded explanation then evaluate $P=IV$ or $I^2R$ or $\frac{V^2}{R}$ for each resistor and select the circuit which draws the least power from the cells (this one will last longest). For the worded explanation, try to see which combination would draw the least current because the larger the current drawn, the quicker the cell will expire.

d) Given the constraints on $n$ and $N$, try a solution and see if it delivers the required power and potential difference.

e) (i) This is a straightforward calculation of the potential of the cell divided by the resistance of the whole circuit.

(ii) Place a resistor of $0.1\Omega$ in series with the cell. Either use a potential divider or work out the voltage dropped across the internal resistance to work out the voltage across the parallel combination then work from there.

(iii) Remembering that the internal resistance is still present, work out the resistance of the combination and apply $V=IR$.

(iv) This is very similar to (iii) but the reciprocals of the resistances form a convergent infinite geometric series.

Full solutions Section 2 - Physics

1 a) (i) $160\times360=57600^{\circ}$ turned in $1s$

One degree takes $\frac{1}{57600}=1.7\times10^{-5}s(2s.f.)$

(ii) The highest speed implies that the discs have only turned through $30^{\circ}$ and not some other angle. The time taken for the discs to rotate through the $30^{\circ}$ is

$\frac{30}{57600}=5.208…\times10^{-4}s$

$Speed=\frac{Distance}{Time}=\frac{0.24}{5.208\times10^{-4}}=460ms^{-1}(2s.f.)$

(iii) The speed calculated in (ii) is for molecules that make it through after the discs have rotated by $30^{\circ}$. However, they can also make it through if the discs rotate by $30^{\circ}$ plus an integer number of complete rotations on top of this (e.g. $390^{\circ}$, $750^{\circ}$ etc.). It takes more time for the discs to rotate through a larger angle and since $d$ is constant, this gives rise to more speeds.

(iv) This corresponds to $30+360=390^{\circ}$.

$390^{\circ}$ takes $\frac{390}{57600}=6.77…\times10^{-3}s$

$Speed=\frac{Distance}{Time}=\frac{0.24}{6.77\times10^{-3}}=35ms^{-1}(2s.f.)$

(v) Consider the fastest molecules that traverse the slits. They will just miss one side of the first slit then scrape past the opposite side of the second slit. In this event, the discs have rotated through $30^{\circ}$ minus the total width of the slit ($29.4^{\circ}$). For the slowest molecules the opposite is true and the rotation is $30.6^{\circ}$.

$V_{max}=\frac{0.24}{\frac{29.4}{57600}}$

$V_{min}=\frac{0.24}{\frac{30.6}{57600}}$

$V_{max}-V_{min}=57600\times0.24(\frac{1}{29.4}-\frac{1}{30.8})=18ms^{-1}(2s.f.)$

b) (i) $h=\frac{1}{2}gt^2$

$t=\sqrt{\frac{2h}{g}}$

$f$ revolutions per second

$360f$ degrees per second

$\theta=$number of degrees per second$\times$time$=360f\sqrt{\frac{2h}{g}}$

(ii) Let $\theta=360$

$360=360f\sqrt{\frac{2h}{g}}$

$\frac{2h}{g}=\frac{1}{f^2}$

$h=\frac{g}{2f^2}=1.2cm(2s.f.)$

(iii) If you attempt to do this question with the particle starting above the two holes, try to calculate that the ratio of number of rotations between the upper and lower discs would have to be $\sqrt{\frac{2}{5}}$, recognise that this isn’t possible and try a different approach. The minimum rotation will be for when the discs turn just $360^{\circ}$ during the time it takes for the particle to fall between the two discs.

Time taken to fall to the first disc$=t_1=\sqrt{\frac{2\times0.1}{g}}$

Time taken to fall to the second disc$=t_2=\sqrt{\frac{2\times0.25}{g}}$

$t_2-t_1=\sqrt{\frac{2}{g}}(\sqrt{0.25}-\sqrt{0.1})$

$f=\frac{1}{t_2-t_1}=12$ revolutions per second $(2s.f.)$

2 a) (i) $V=1.5+1.5=3V$

Using $V=IR$,

$A_1=A_2=\frac{3}{1}=3A$

(ii) The potential difference of identical cells in parallel is unaffected by the number of cells, so $V=3V$

$A_1=3A$ as before but since the current splits equally at the junction (by symmetry) we have $A_1=1.5A$.

b) (i) Taking the internal resistance to be in series with the cell, the total resistance of the circuit is now $1.2\Omega$. Using $V=IR$ we have $I=\frac{3}{1.2}=2.5A$ therefore $A_1=A_2=2.5A$.

For the voltmeter reading, $V=IR=2.5\times1=2.5V$

(ii) Define a current $I$ through the $1\Omega$ resistor. By symmetry we have a current of $\frac{I}{2}$ through each branch of the parallel part.

The voltage supplied by the cells (taking into account their internal resistances) is $3$ minus the voltage dropped across the internal resistance. This is equal to $V=IR$ for the $1.0\Omega$ resistor.

$V=3-\frac{0.1I}{2}=1.0\times I$

Therefore $A_1=2\frac{6}{7}$, $A_2=1\frac{3}{7}$, $V=2\frac{6}{7}$

c) My first thought was to work out the power dissipated in all resistors (sometimes using $P=IV$ and sometimes using $P=I^2R$) and select the lowest total power as follows:

a) (i) $P=IV=3\times3=9W$

(ii) $P=IV=3\times3=9W$

b) (i) $P_{Total}=2.5\times2.5+(2.5)^2\times0.1+(2.5)^2\times0.1=7.5W$

(ii) $P_{Total}=(2\frac{6}{7})^2+(1\frac{3}{7})^2\times0.1+(1\frac{3}{7})^2\times0.1=8\frac{4}{7}W$

The combination which draws the least total power will drain the cells slowest, so the answer is b) (i).

On second thoughts however, I believe the examiners were expecting a worded explanation:

Cells will run out least quickly for the smallest current drawn. This is the case for the circuit with the largest resistance. This will be for one of the circuits with the internal resistances, in particular the one which has series combinations rather than parallel because these have a greater resistance. Therefore the answer is b) (i).

d) There aren’t many possible combinations given the restrictions on $n$ and $N$. Since there are $10$ cells in total, we can only have $10$ branches of $1$ cell, $5$ branches of $2$ cells, $2$ branches of $5$ cells or a series combination of $10$ cells. Since $2.4V$ is required we need at least $5$ cells in series and the question isn’t about a series combination of $10$ cells, so try $2$ branches of $5$ cells.

The diagram shows the effect of having $2$ branches of $5$ cells. If this is to deliver $2.4V$ then $0.1V$ must be dropped across the $0.5\Omega$ resistance.

$I=\frac{V}{R}=\frac{0.1}{0.5}=0.2A$, therefore $I_{Total}=0.4A$

Work out the current required by the fan:

$I_{fan}=\frac{P}{V}=\frac{0.96}{2.4}=0.4A$

which agrees with the circuit in the diagram so $n=2$ and $N=5$.

e) (i) $R_{Total}=\frac{1}{\frac{1}{1}+\frac{1}{2}}=\frac{2}{2+1}=\frac{2}{3}$

$V=IR$

$I=\frac{V}{R}=\frac{1.5}{\frac{2}{3}}=2.25A$

(ii) $I_{cell}=\frac{1.5}{0.1+\frac{2}{3}}=1.957…A$

$V_{parallel}=1.5-0.1\times1.957$

$P_{R_2}=\frac{V_{parallel}^2}{R_2}=0.85W(2s.f.)$

(iii) $R_{Total}=\frac{1}{1+\frac{1}{2}+\frac{1}{4}}+0.1=\frac{4}{4+2+1}+0.1=\frac{4}{7}+0.1$

$I=\frac{1.5}{\frac{4}{7}+0.1}=2.23A(3s.f.)$

(iv) $R_{Total}=\frac{1}{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+…}$

$S_{\infty}=\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio ($|r|<1$)

$R_{Total}=\frac{1}{\frac{1}{1-\frac{1}{2}}}+0.1=\frac{1}{2}+0.1=0.6\Omega$

$I=\frac{V}{R_{Total}}=\frac{1.5}{0.6}=2.5A$

I will firstly set out a series of hints. Try to do the questions from these alone and if you are still having difficulty then scroll down to the full solutions.

Hints

Section 2 - Physics

1 a) (i) Work out how many degrees are turned through in one second. From this you can work out how long each degree takes.

(ii) Work out how long it takes the discs to turn through $30^{\circ}$ and then use this to evaluate the speed required to traverse $24cm$ in this time.

(iii) Think about the alternatives to the discs rotating $30^{\circ}$ - this is not the only option for molecules to travel through both slits.

(iv) Repeat the calculation for (ii) with the smallest possible angle of rotation apart from $30^{\circ}$.

(v) Repeat the calculation for (ii) being careful to select the correct values for the maximum and minimum angles turned through.

b) (i) Use the equations of constant acceleration to work out the time taken. Use this in conjunction with the number of degrees turned per second.

(ii) Think about how many degrees the disc must rotate before the hole is in the correct place for the first time after $t=0$. Substitute into your expression in b) (ii) and rearrange for $h$.

(iii) Firstly accept that there is no frequency of rotation that allows this to occur with the particle being released from above the holes. Turn your attention to the amount of time taken for the particle to fall from the upper disc to the lower disc and work from there.

2 a) (i) Use Ohm’s law remembering that the potential difference of the cells combine in series.

(ii) Cells in parallel don’t give a higher overall voltage but the current splits at the junction.

b) (i) Consider the internal resistance to be a resistor connected in series with the cell it belongs to.

(ii) Define an unknown current $I$ and use this to find the voltage dropped across each internal resistance (in terms of $I$). Equate the potential difference provided by the parallel combination to $IR$ for the $1.0\Omega$ resistor and solve for $I$. This yields the ammeter readings and applying Ohm’s law to the $1.0\Omega$ resistor gives the reading on the voltmeter.

c) If you can’t think of a worded explanation then evaluate $P=IV$ or $I^2R$ or $\frac{V^2}{R}$ for each resistor and select the circuit which draws the least power from the cells (this one will last longest). For the worded explanation, try to see which combination would draw the least current because the larger the current drawn, the quicker the cell will expire.

d) Given the constraints on $n$ and $N$, try a solution and see if it delivers the required power and potential difference.

e) (i) This is a straightforward calculation of the potential of the cell divided by the resistance of the whole circuit.

(ii) Place a resistor of $0.1\Omega$ in series with the cell. Either use a potential divider or work out the voltage dropped across the internal resistance to work out the voltage across the parallel combination then work from there.

(iii) Remembering that the internal resistance is still present, work out the resistance of the combination and apply $V=IR$.

(iv) This is very similar to (iii) but the reciprocals of the resistances form a convergent infinite geometric series.

Full solutions Section 2 - Physics

1 a) (i) $160\times360=57600^{\circ}$ turned in $1s$

One degree takes $\frac{1}{57600}=1.7\times10^{-5}s(2s.f.)$

(ii) The highest speed implies that the discs have only turned through $30^{\circ}$ and not some other angle. The time taken for the discs to rotate through the $30^{\circ}$ is

$\frac{30}{57600}=5.208…\times10^{-4}s$

$Speed=\frac{Distance}{Time}=\frac{0.24}{5.208\times10^{-4}}=460ms^{-1}(2s.f.)$

(iii) The speed calculated in (ii) is for molecules that make it through after the discs have rotated by $30^{\circ}$. However, they can also make it through if the discs rotate by $30^{\circ}$ plus an integer number of complete rotations on top of this (e.g. $390^{\circ}$, $750^{\circ}$ etc.). It takes more time for the discs to rotate through a larger angle and since $d$ is constant, this gives rise to more speeds.

(iv) This corresponds to $30+360=390^{\circ}$.

$390^{\circ}$ takes $\frac{390}{57600}=6.77…\times10^{-3}s$

$Speed=\frac{Distance}{Time}=\frac{0.24}{6.77\times10^{-3}}=35ms^{-1}(2s.f.)$

(v) Consider the fastest molecules that traverse the slits. They will just miss one side of the first slit then scrape past the opposite side of the second slit. In this event, the discs have rotated through $30^{\circ}$ minus the total width of the slit ($29.4^{\circ}$). For the slowest molecules the opposite is true and the rotation is $30.6^{\circ}$.

$V_{max}=\frac{0.24}{\frac{29.4}{57600}}$

$V_{min}=\frac{0.24}{\frac{30.6}{57600}}$

$V_{max}-V_{min}=57600\times0.24(\frac{1}{29.4}-\frac{1}{30.8})=18ms^{-1}(2s.f.)$

b) (i) $h=\frac{1}{2}gt^2$

$t=\sqrt{\frac{2h}{g}}$

$f$ revolutions per second

$360f$ degrees per second

$\theta=$number of degrees per second$\times$time$=360f\sqrt{\frac{2h}{g}}$

(ii) Let $\theta=360$

$360=360f\sqrt{\frac{2h}{g}}$

$\frac{2h}{g}=\frac{1}{f^2}$

$h=\frac{g}{2f^2}=1.2cm(2s.f.)$

(iii) If you attempt to do this question with the particle starting above the two holes, try to calculate that the ratio of number of rotations between the upper and lower discs would have to be $\sqrt{\frac{2}{5}}$, recognise that this isn’t possible and try a different approach. The minimum rotation will be for when the discs turn just $360^{\circ}$ during the time it takes for the particle to fall between the two discs.

Time taken to fall to the first disc$=t_1=\sqrt{\frac{2\times0.1}{g}}$

Time taken to fall to the second disc$=t_2=\sqrt{\frac{2\times0.25}{g}}$

$t_2-t_1=\sqrt{\frac{2}{g}}(\sqrt{0.25}-\sqrt{0.1})$

$f=\frac{1}{t_2-t_1}=12$ revolutions per second $(2s.f.)$

2 a) (i) $V=1.5+1.5=3V$

Using $V=IR$,

$A_1=A_2=\frac{3}{1}=3A$

(ii) The potential difference of identical cells in parallel is unaffected by the number of cells, so $V=3V$

$A_1=3A$ as before but since the current splits equally at the junction (by symmetry) we have $A_1=1.5A$.

b) (i) Taking the internal resistance to be in series with the cell, the total resistance of the circuit is now $1.2\Omega$. Using $V=IR$ we have $I=\frac{3}{1.2}=2.5A$ therefore $A_1=A_2=2.5A$.

For the voltmeter reading, $V=IR=2.5\times1=2.5V$

(ii) Define a current $I$ through the $1\Omega$ resistor. By symmetry we have a current of $\frac{I}{2}$ through each branch of the parallel part.

The voltage supplied by the cells (taking into account their internal resistances) is $3$ minus the voltage dropped across the internal resistance. This is equal to $V=IR$ for the $1.0\Omega$ resistor.

$V=3-\frac{0.1I}{2}=1.0\times I$

Therefore $A_1=2\frac{6}{7}$, $A_2=1\frac{3}{7}$, $V=2\frac{6}{7}$

c) My first thought was to work out the power dissipated in all resistors (sometimes using $P=IV$ and sometimes using $P=I^2R$) and select the lowest total power as follows:

a) (i) $P=IV=3\times3=9W$

(ii) $P=IV=3\times3=9W$

b) (i) $P_{Total}=2.5\times2.5+(2.5)^2\times0.1+(2.5)^2\times0.1=7.5W$

(ii) $P_{Total}=(2\frac{6}{7})^2+(1\frac{3}{7})^2\times0.1+(1\frac{3}{7})^2\times0.1=8\frac{4}{7}W$

The combination which draws the least total power will drain the cells slowest, so the answer is b) (i).

On second thoughts however, I believe the examiners were expecting a worded explanation:

Cells will run out least quickly for the smallest current drawn. This is the case for the circuit with the largest resistance. This will be for one of the circuits with the internal resistances, in particular the one which has series combinations rather than parallel because these have a greater resistance. Therefore the answer is b) (i).

d) There aren’t many possible combinations given the restrictions on $n$ and $N$. Since there are $10$ cells in total, we can only have $10$ branches of $1$ cell, $5$ branches of $2$ cells, $2$ branches of $5$ cells or a series combination of $10$ cells. Since $2.4V$ is required we need at least $5$ cells in series and the question isn’t about a series combination of $10$ cells, so try $2$ branches of $5$ cells.

The diagram shows the effect of having $2$ branches of $5$ cells. If this is to deliver $2.4V$ then $0.1V$ must be dropped across the $0.5\Omega$ resistance.

$I=\frac{V}{R}=\frac{0.1}{0.5}=0.2A$, therefore $I_{Total}=0.4A$

Work out the current required by the fan:

$I_{fan}=\frac{P}{V}=\frac{0.96}{2.4}=0.4A$

which agrees with the circuit in the diagram so $n=2$ and $N=5$.

e) (i) $R_{Total}=\frac{1}{\frac{1}{1}+\frac{1}{2}}=\frac{2}{2+1}=\frac{2}{3}$

$V=IR$

$I=\frac{V}{R}=\frac{1.5}{\frac{2}{3}}=2.25A$

(ii) $I_{cell}=\frac{1.5}{0.1+\frac{2}{3}}=1.957…A$

$V_{parallel}=1.5-0.1\times1.957$

$P_{R_2}=\frac{V_{parallel}^2}{R_2}=0.85W(2s.f.)$

(iii) $R_{Total}=\frac{1}{1+\frac{1}{2}+\frac{1}{4}}+0.1=\frac{4}{4+2+1}+0.1=\frac{4}{7}+0.1$

$I=\frac{1.5}{\frac{4}{7}+0.1}=2.23A(3s.f.)$

(iv) $R_{Total}=\frac{1}{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+…}$

$S_{\infty}=\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio ($|r|<1$)

$R_{Total}=\frac{1}{\frac{1}{1-\frac{1}{2}}}+0.1=\frac{1}{2}+0.1=0.6\Omega$

$I=\frac{V}{R_{Total}}=\frac{1.5}{0.6}=2.5A$