# Best Wishes to Graduating Students of 2016

Thank you graduating students of 2016!

It has been a very enjoyable journey tutoring all of you in preparation for your exams this year.

Examinations are always tough and you deserve a well-needed rest!

All the best for your results and I'll check in with you next year when they are released!

O-Level results will be released between 11 and 13 January 2017.

A-Level results will be released between 24 and 28 Feburary 2017.

Clive, iGCSE 2016

Ashley, iGCSE 2016

David, O-Level 2016

Yusong, A-Level 2016

Ehan, A-Level 2016

Gavin, A-Level 2016

Siew Mun, A-Level 2016

Jia Yee, A-Level 2016

Kimberly, A-Level 2016

# James 29 Sep 2016

## Question

Find the potential difference across the 2.0 Ω resistor.

Find the potential difference across the open gap in the circuit.

The easiest way to find the potential difference in this case is to assign values of potential at the various points.

We first start at the sides of the cell, A and B.

We can assign any value as long as the difference between A and B is 6.0 V and A is a higher value than B.

In this case, we choose A to be 7 V and B to be 1 V.

All connecting wires are assumed to have no resistance. This means that the value of potential at all points of the same wire are of the same value.

Hence, E has a value of 7 V and C has a value of 1 V.

Since the circuit is an open circuit and no current flows, from V = I R, the potential difference across the 2.0 Ω resistor is zero.

This means that the values of the potential at both sides of the resistor are the same.

From the diagram, we can see that the difference between the potential at C and D is 0 V.

Hence, the potential difference across the 2.0 Ω resistor is zero.

From the diagram, we can see that the difference between the potential at E and D is 6 V.

Hence, the potential difference across the open gap in the circuit is 6.0 V.

# Yusong 25 Aug 2016

## Question

The figure shows a long horizontal wire PQ carrying a steady current of 50A in the direction QP. A copper wire RS of diameter 0.40mm hangs horizontally at a distance of 0.15m below wire PQ using some threads.

A current flows through RS such that there is no tension in the threads.

Identify the direction is the current and draw a free body diagram of the forces acting on RS.

If the current in RS is flowing opposite to the current in PQ, the two wires will repel each other.

Since the wire is suspended in equilibrium, the resultant force must be zero.

Fup = Fdown
T1 + T2 = FB + W

We can see that it is not possible for the tension to be zero.

If the current in RS is flowing in the same direction as the current in PQ, the two wires will repel attract other.

Since the wire is suspended, the resultant force must be zero.

Fup = Fdown
T1 + T2 + FB = W

When the tensions in both threads are zero, we have

FB = W

Hence, current in RS must be flowing in the same direction as PQ, in the direction SR.

The free body diagram when there is no tension in the threads is as shown below:

# Jia Yee 25 Aug 2016

## Question

The figure below shows a region PQRS of a uniform magnetic field directed downwards into the plane of the paper.

Electrons, all having the same speed, enter the region of the magnetic field.

a) On the figure, show the path of the electrons as they pass through the magnetic field emerging from side QR.

b) A uniform electric field is also applied in the region PQRS so that the electrons now pass undeflected through this region. On the figure, mark with an arrow labelled E, the direction of the electric field.

c) The undeflected electrons in (b) each have charge -e, mass m and speed v. State and explain the effect if any, on the particles entering the region PQRS of the same magnetic and electric fields in (b) if the particles each have

i) charge -e, mass m and speed 2v.

ii) charge +e, mass m and speed v.

a) Since the electrons are moving to the right, it is the same as a current flowing to the left. By using Flemming’s left hand rule, the magnetic force on the electrons is downwards.

According to Flemming’s left hand rule, the magnetic force will always be perpendicular to the velocity of the electrons. This means that it provides a centripetal force. Hence, the electrons will move in a circular path as shown.

Once the electrons exit the magnetic field, they will move in a straight path since there is no resultant force acting on the electrons. We ignore the effects of the weight because the acceleration of due to the weight is too small to cause any significant change in the velocity of the electrons.

b) In order for the electrons to pass through undeflected, the resultant force on the electrons must be zero. Since the magnetic force on the electrons is downwards, the electric force acting on the electrons must be of the same magnitude and acting upwards.

In order for the electric force to be acting upwards, electric field lines must be directed downwards. (Since electrons are negatively-charged, they will experience an electric force in the opposite direction to the electric field lines.)

c)  When an electron is moving with a speed of v, it passes straight through without any deflection. The magnetic force and electric forces on the electrons are given by the following formulae:

FB= Bqv
FE = qE

Since the downwards magnetic force is now greater than the upwards electric force, the resultant force is downwards and the particle will curve downwards as it passes through the region.

ii) When a particle of the opposite charge (+e) and mass (m) and same speed (2v) moves through the field, the directions of the magnetic force and electric forces are now reversed.

The electric force acts downwards and the magnetic force acts upwards on the particle.

However, both forces are still of the same magnitude and the resultant force is still zero. Hence, the particle will still pass through the region undeflected.

# David 16 Aug 2016 - 2

## Question

The figure below shows the path of ray X as it passes through a thin converging lens AB.

Draw light rays to find the position of the image formed and determine the focal length of the lens AB.

Firstly, ray X is used to find the position of the base of the image. Since ray X comes from the base of the object, it will form the base of the image. Hence, the image is located along the blue line shown below:

Next, the ray from the tip of the object that passes through the optical centre is drawn.

The intersection point of the ray and the blue line is the location of the tip of the image.

The blue line is removed and a ray parallel to the principal axis from the tip of the object is added. This ray bends at the lens and reaches the tip of the image. The focal point can be found and the focal length can now be obtained.

The focal length is 10.0 cm.

# David 16 Aug 2016 - 1

## Question

A rectangular block of wood rests on the ground as shown in the diagram below.

Which of the following is the easiest way for a force F to topple the block?

For each block we identify the pivot point and the perpendicular distance of force F from the pivot point.

We can see that the perpendicular distance for D is the greatest. Force F will produce the greatest moment for D, hence it will topple the easiest.

# Farhan 05 Aug 2016

## Question

A piece of furniture was recovered from an archaeological site and carbon dating was used to estimate its age. It was assumed that the proportion of Carbon-14 to natural carbon (Carbon-12) in living wood was 1.25 x 10-12. If the number of particles emitted from 5 g of carbon prepared from the specimen was 21 per minute, how old is the specimen?
[The half-life of Carbon-14 is 5600 years and the mass number of natural carbon is 12.]

Since the proportion of Carbon-14 is very small compared to Carbon-12, we can assume for the sake of finding the number of carbon atoms that all of the carbon in the specimen is Carbon-12.

Given that the proportion of Carbon-14 to Carbon-12, we can find the original number of Carbon-14 atoms when the piece of furniture was living wood using the following approximation. (Refer to bottom for further discussion of this approximation.)

Using the half-life, we can calculate the decay constant of Carbon-14.

Hence, we can find the initial activity when the specimen was still living wood.

The final activity of the specimen is given by

Now, we can find the age of the specimen.

## Discussion on Approximation

Based on the phrasing of the question that

“the proportion of Carbon-14 to natural carbon is k”,

However, since the k is small compared to 1,

Hence, we arrive at

# Yusong 04 Aug 2016

## Question

A simple iron-core transformer is shown below:

Suggest why thermal energy is generated in the core when the transformer is in use.

Consider a section of the transformer core which is coloured blue as shown below:

As the magnetic field produced by the current in the primary coil increases, the magnetic flux linkage of the section of the coil will increase.

By Faraday’s Law, there will be an induced e.m.f. produced in the section of the core. By Lenz’s Law, there will be a current flowing in the section of the core which will produce a magnetic effect to oppose the change.

By the Right Hand Grip Rule, this means that a circular current must be flowing in the section of the core in order to produce the magnetic field. (This circular current is called the eddy current.)

Since an induced current is flowing in the core, by P = I 2 R, there will be power dissipated in the core and thermal energy is generated in the core of the transfomer.

# Ashley 02 Aug 2016 - 3

## Question

Which statement about an object moving in a straight line through air is correct?

A When it accelerates, the resultant force acting on it is zero.
B When it moves at a steady speed, the air resistance acting on it is zero.
C When it moves at a steady speed, the resultant force acting on it is zero.
D When it moves, there is a resultant force acting on it.

According to Newton’s First Law of Motion, when the resultant force is zero, a stationary object remains at rest and a moving object remains moving with constant velocity.

# Ashley 02 Aug 2016 - 2

## Question

The diagram shows a satellite that is moving at a uniform rate in a circular orbit around the Earth.

Which statement describes the motion of this satellite?

A It is accelerating because its speed is changing.
B It is accelerating because its velocity is changing.
C It is not accelerating but its speed is changing.
D It is not accelerating but its velocity is changing.

As the satellite orbits the Earth, the speed remains the same but the velocity changes.

Since the velocity is changing, there is acceleration.

# Ashley 02 Aug 2016 - 1

## Question

A parachutist is falling at terminal velocity, without her parachute open.

She now opens her parachute.

What is the direction of her motion, and what is the direction of her acceleration, immediately after she opens her parachute?

When the parachutist is falling at terminal velocity, the upwards air resistance is equal to the weight of the parachutist. The resultant force is zero and the velocity is downwards.

When the parachute opens, the air resistance increases and is now greater than the weight.

The resultant force is now upwards and the acceleration is upwards. The velocity is still downwards.

# Minqi 19 May 2016 - 4

## Question

A set-up to control the brightness of a bulb is shown below.

X and Y are to connected to a uniform circular ring which has a constant cross-sectional area and resistivity X is connected to the left side of the ring.

Where should Y be connected for the light to be the dimmest?

The circular ring and its connection point can be seen as a connection of 6 resistors of resistance R as shown below:

When Y is connected to A, the connection is as shown below:

The effective resistance is 0.833 R.

When Y is connected to B, the connection is as shown below:

The effective resistance is 1.33 R.

When Y is connected to C, the connection is as shown below:

The effective resistance is 1.5 R.

When Y is connected to D, the connection is as shown below:

The effective resistance is 1.33 R.

Hence, connecting Y to C will result in the greatest resistance and the light bulb will be the dimmest.

# Minqi 19 May 2016 - 3

## Question

The diagram below shows three parallel wires X, Y and Z placed in a horizontal plane. Wires X and Z carry current I in opposite directions. Wire Y carries a current of 3I in the direction shown and is equidistant from the other wires.

The magnitude of the force per unit length between two parallel wires placed a distance apart, each carrying a current of I is 2.0 x 10-6 N m-1. What is the direction and magnitude of the net force per unit length acting on the wire Z?

(The force per unit length is proportional to the product of the currents in the two wires and inversely proportional to the separation between them.)

Based on the information regarding the forces per unit length between two wires,

$${F \over m} = k {I_1 I_2 \over r}$$

Based on the value of the force given,

The force between Y and Z is attractive because the currents in both wires are flowing in the same direction.

The force between X and Z is repulsive because the currents in both wires are flowing in the opposite directions.

Hence, the resultant force per unit length is 5.0 x 10-6 N m-1 upwards towards Y.

# Minqi 19 May 2016 - 2

## Question

The figure shows a full scale diagram of equipotential lines of an electric field, with potential difference of 3.0 V between adjacent lines.

What is the closest estimate of the magnitude and direction of the field strength at P along the dotted line?

The electric field lines are from the positive charge to the negative charge, hence, the direction of the electric field strength is towards the left.

To calculate the electric field strength, we use the following relationship:

Since the diagram is drawn full-scale, we need to measure the distance d as shown in the diagram below using a ruler.

The potential difference between the 2 lines are 6.0 V.

Hence the electric field strength is $$\left | E \right | = \left | 6.0 \over d \right |$$ towards the left.

# Minqi 19 May 2016 - 1

## Question

A beam of initially unpolarised light passes through three polaroids P, Q and R. Polaroid P's axis of polarisation is vertical. Which orientation of polaroids Q and R with respect to the vertical axis will produce an emergent beam from polaroid R with maximum intensity.

Orientation with respect
to the vertical axis

Q         R
A     45o      45o
B     45o      90o
C     90o      180o
D     180o     60o

When a wave polarised in one axis passes through another polaroid with the axis at an angle of θ, only the component of the wave parallel to the new axis can pass through.

A1 = A0 cos θ

After passing through the second polaroid placed at an angle of ϕ, the final amplitude will be given by

A2 = A1 cos ϕ

Hence, the final intensity will be given by

A2 = A1 cos ϕ
= ( A0 cos θ ) cos ϕ

I2 = k ( A 2 )2
= k ( A0 cos θ cos ϕ )2
= k A02 cos2 θ cos2 ϕ
= I0 cos2 θ cos2 ϕ

For option A,
I2 = I0 cos2 θ cos2 ϕ
= I 0 cos2 45 o cos2 (45o - 45 o)
= 0.50 I0

For option B,
I2 = I 0 cos2 θ cos2 ϕ
= I0 cos2 45o cos2 (90o - 45o)
= 0.25 I0

For option C,
I2 = I0 cos2 θ cos2 ϕ
= I0 cos 2 90o cos2 (180o - 90o)
= 0

For option D,
I2 = I0 cos2 θ cos2ϕ
= I0 cos2180o cos2(180o - 60o)
= 0.25 I0

Hence, Option A will produce a beam with maximum intensity.

# David 08 May 2016 - 1

## Question

To avoid an accident the brakes of a car travelling at 30 ms-1 were applied suddenly. If the driver has a mass of 80 kg and it takes 3.0 s for the car to decelerate to rest what is the average force exerted by the safety belt on the driver during deceleration?

We first calculate the acceleration of the driver.

The force by the safety belt provides the force that causes the deceleration.

Hence, the average force exerted by the safety belt on the driver is 800 N.

# James 08 May 2016 - 2

## Question

What are the errors in the diagram below?

The Earth wire has to be connected to the metal casing.

The Live and Neutral wires must be connected to the heating coil.

The switch and fuse must be on the Live wire.

# James 08 May 2016 - 1

## Question

What does the following paragraph mean?

To improve the accuracy for measurement that is too small, measure a few together and then find the average. For example, if we are finding the density of a marble, we take the mass of eg 5 marbles and divide by volume of the 5 marbles – and not find the density of individual marbles. If we are finding the average diameter of a marble, we measure the length of eg 5 marbles in a straight line and then divide by 5 – and not find diameter of individual marble and take average.

When finding the measurement of a small quantity, it is more desirable to find the total quantity due to a few items and then divide to obtain the quantity of a single item.

For example, we have a stick which is 3.58854 cm long. When measuring the length using a ruler, we will obtain 3.6 cm because a ruler can only read to 0.1 cm.

However, if we take 10 identical sticks and join them end-to-end and measure, we can obtain a total length of 35.8 cm. By dividing, we can find the length of each stick.

We know that there are exactly 10 sticks, not 10.1 or 9.9, hence the value of 10 is infinitely accurate and precise. We can assume that it is correct to many significant figures. Hence, the limiting significant figures is due to the value 35.8.

Therefore, our final answer of 3.58 cm is to 3 significant figures.

This method allows you to obtain an answer to more decimal places than the limit of the measuring instrument.

We can proceed further and join 100 identical sticks together and get a value which is even closer to the true value of the length.

With 100 sticks, the total length measured would be 358.8 cm when using a measuring tape. Then, dividing to find the length of each stick,

However, once we join so many sticks together, there will be other factors contributing to an error in measurement. For example, we need to ensure that the sticks are identical, are joined in a straight line and there are no spaces between all the sticks.

# David 07 May 2016 - 5

## Question

38. The figure below shows a permanent magnet suspended from a spring placed right on top of an electromagnet.

What will happen to the permanent magnet once the switch S is turned on?

When the switch is closed, current will flow in the direction as shown by the arrows on the wires in the diagram below. Based on the right hand-grip rule, the magnetic field produced by the current will be as shown below.

This means that the top of the iron core will be a magnetic North pole and repel the North pole of the permanent magnet. Hence, the permanent magnet will be pushed upwards.

# David 07 May 2016 - 4

## Question

A stone is thrown vertically upwards with a velocity of 5.0 m/s. After time t, it reaches the original position. Neglecting air resistance, what is the time required for the ball to reach the highest position?