9702/May Jun/12/2015/Q6

A single sheet of aluminium foil is folded twice to produce a stack of four sheets. The total thickness of the stack of sheets is measured to be (0.80 ± 0.02) mm. This measurement is made using a digital caliper with a zero error of (−0.20 ± 0.02) mm.

What is the percentage uncertainty in the calculated thickness of a single sheet?

A 1.0%                        B 2.0%                        C 4.0%                        D 6.7%

Solution:

Answer: C

This challenging question involves both random and systematic error.

The zero error (−0.20) can be removed but its uncertainty (± 0.02) must be added to the measurement uncertainty.

True value of total thickness = 0.80 + 0.20 = 1.0 mm  

Total uncertainty in measurement = ± (0.02 + 0.02) = ± 0.04

So the four sheets have a true thickness of (1.00 ± 0.04) mm.

A single sheet would have a thickness of (0.25 ± 0.01) mm.

Percentage error = (0.01 / 0.25) × 100% = 4%

Reference: PYQ - Oct/Nov 2015 Paper 12 Q6

9702/May Jun/11/2018/Q34

In the circuit shown, the batteries have negligible internal resistance.

What are the values of the currents I1, I2 and I3?

Solution:

Answer: C

Reference: PYQ - May/Jun 2018 Paper 11 Q34

9702/Oct Nov/12/2016/Q24

A transverse progressive wave of wavelength λ is set up on a stretched string. The graph shows
the variation of displacement y with distance x at a particular instant of time. The displacement
where distance x = λ/8 is y₁.

What are the next two values of x where the displacement y is again equal to y₁?

Solution:

Answer: B

From the figure above, it is very obvious the option C and D are incorrect.

The answer is "B" is because the second value of x must more than λ.

Reference: PYQ - Oct/Nov 2016 Paper 12 Q24

9702/Oct Nov/11/2016/Q11

A car has mass m. A person needs to push the car with force F in order to give the car
acceleration a. The person needs to push the car with force 2F in order to give the car
acceleration 3a.

Which expression gives the constant resistive force opposing the motion of the car?

A ma
B 2ma
C 3ma
D 4ma

Solution:

Answer: A

Resultant force = applied force – resistive force

Ma = F- R

R = F- ma —-i

3ma = 2F – R

R = 2F – 3ma —–ii

Substitute for R in eq i

2F – 3ma = F – ma

F = 2ma

Therefore, R = 2ma – ma = ma

Resistive force = ma (A is the correct option)

Reference: PYQ - Oct/Nov 2016 Paper 11 Q11

9702/Oct Nov/13/2013/Q38

A 20 V d.c. supply is connected to a circuit consisting of five resistors L, M, N, P and Q.

There is a potential drop of 7 V across L and a further 4 V potential drop across N.

What are the potential drops across M, P and Q?

Solution:

Answer: C

This question can be easily tackled by considering the different loops present in the circuit and apply Kirchhoff’s laws to them.

Consider the loop: ‘+’ terminal supply – resistor L – resistor M – ‘-’ terminal supply

From Kirchhoff’s law, the sum of p.d. across any loop should be equal to the e.m.f. Analysis of the loop shows that a 7 V drop across resistor L must mean a 13V drop across M (to obtain a total 20V across L and M).

Notice that the direction of potential drop is also of significance.

Consider resistor L for example. The direction of potential drop is from left to right. Current flows from the ‘+’ terminal of the supply, so the junction on the left of resistor L should be at a higher potential (which is equal to 20V since there is no component between it and the ‘+’ terminal of the supply).

So, the direction of potential rise is from a greater value of potential to a small value of potential. Additionally, current flows from a greater potential to a smaller potential (as in the case of the ‘+’ terminal of the supply).

The junction between L and M is at a potential of 13V (since the right junction to which M is connected is at 0V as it is connected to the ‘-’ terminal of the supply).

There is a potential drop of 4V downwards across N, so current flows downwards. This means that the potential at the upper junction (between L and M) is greater than the lower junction (between P and Q) and the difference in potential is 4V.

Thus, lower junction (between P and Q) is at a potential of (13V – 4V =) 9V

So, the potential drop across resistor Q is 9V (since the right junction to which Q is connected is at 0V as it is connected to the ‘-’ terminal of the supply).

Finally, consider resistor P. Its terminal is at a potential of 20V and its right terminal is at a potential of 9V. Potential drop = 20 – 9 = 11V.

Important notice:

When considering a loop, it should start from one terminal of the supply and end at the other terminal. Only then will Kirchhoff’s law apply. For example, ‘+’ terminal supply – L – N – P – ‘+’ terminal supply is not a correct loop. You may notice that the sum of p.d. is not equal to the e.m.f. Additionally the flow of current is wrong.

Reference: PYQ - Oct/Nov 2013 Paper 13 Q38

9702/Oct Nov/13/2013/Q34

An electrical device of fixed resistance 20 Ω is connected in series with a variable resistor and a
battery of electromotive force (e.m.f.) 16 V and negligible internal resistance.

What is the resistance of the variable resistor when the power dissipated in the electrical device is 4.0W?

A 16 Ω 

B 36 Ω 

C 44 Ω 

D 60 Ω

Solution:

Answer: A

Power dissipated, P = I²R

For electrical device,

4.0 = I² (20)

Current I in circuit = 0.45A

Since this is a series circuit, the same current flows through the variable resistor.

Ohm’s law: V = IR

p.d. across electrical device = 0.45 (20) = 9.0V

Let the resistance of the variable resistor = R

From Kirchhoff’s second law, the sum of p.d. in a loop is equal to the e.m.f. in the circuit.

16 = 9.0 + 0.45R

Resistance R = 16Ω

Reference: PYQ - Oct/Nov 2013 Paper 13 Q34

9702/Oct Nov/13/2013/Q22

A lift is supported by two steel cables each of length 20 m.

Each of the cables consists of 100 parallel steel wires, each wire of cross-sectional area 3.2×10^–6 m². The Young modulus of steel is 2.1×10¹¹Nm^–2.

Which distance does the lift move downward when a man of mass 70 kg steps into it?
A 0.010 mm

B 0.020 mm

C 0.10 mm

D 0.20 mm

Solution:

Answer: C

Hooke’s law: F = ke

The lift is supported by 2 steel cables, each of which consists of 100 parallel steel wires. So, there is a total of 200 parallel steel wires in parallel.

For parallel spring,
Effective spring constant, k_eff = k₁ + k₂ + k₃ + … 

First, we need to find the spring constant, k for 1 steel wire.

For 1 wire,
Young modulus, E = stress / strain = (F/A) / (e/L) = FL / Ae
Hooke’s law: F = ke
Young modulus, E = (ke)L / Ae = kL / A
Spring constant, k = EA / L = (2.1 × 10¹¹) (3.2 × 10^–6) / 20 = 33600 Nm^-1 

Effective spring constant, k_eff = 200k
Hooke’s law: F = k_eff e
Mass of person = 70kg. Weight = mg = 700N           (take g = 10 ms^-2)
Extension, e = F / k_eff = 700 / (200 × 33600) = 0.00010m = 0.10 mm

Reference: PYQ - Oct/Nov 2013 Paper 13 Q22

9702/Oct Nov/13/2013/Q19

An electrical generator is started at time zero. The total electrical energy generated during the
first 5 seconds is shown in the graph.

What is the maximum electrical power generated at any instant during these first 5 seconds?

A 10 W
B 13 W
C 30 W
D 50 W

Solution:

Answer: C

Power = Energy / time

From the energy-time graph, the power generated is given by the gradient. Maximum electrical power generated at any instant is given by the gradient at that instant.

The steeper the graph, the greater the value of gradient and thus the greater the power generated. The graph is steepest between times t = 2s and t = 3s.

Consider the points: (2, 10) and (3, 40)

Maximum power = gradient = (40 – 10) / (3 – 2) = 30 W

Reference: PYQ - Oct/Nov 2013 Paper 13 Q19

9702/Oct Nov/13/2013/Q16

The graph shows how the total resistive force acting on a train varies with its speed.

Part of this force is due to wheel friction, which is constant. The rest is due to wind resistance.

What is the ratio wind resistance/wheel friction at a speed of 200 km h^–1?

A 4
B 5
C 8
D 10

Solution:

Answer: A

The wheel friction is constant and the wind resistance increases with speed.

When the speed is zero, the wind resistance is also zero. So, wheel friction = 8 kN.

At a speed of 200 km h^–1, the total resistive force is 40 kN.

Wheel friction = 8 kN

Wind resistance = 40 – 8 = 32 kN

Ratio = wind resistance / wheel friction = 32 / 8 = 4

Reference: PYQ - Oct/Nov 2013 Paper 13 Q16

9702/Oct Nov/13/2013/Q31

A small charge q is placed in the electric field of a large charge Q.

Both charges experience a force F.

What is the electric field strength of the charge Q at the position of the charge q?

Solution:

Answer: D

A small charge q is placed in the electric field of a large charge Q. Both charges experience force F.

Electric force F at position of charge q = Eq

Note that F also account for the magnitudes of the charges. The electric force can also be given by another formula (Coulomb’s law: F = Qq / 4πϵ₀r²) but we don’t need to calculate this here. The electric force is already given to be F.

Electric field is the force per unit charge. The charge in the expression is that on which the force acts (not the charge creating the field), so q is involved rather than Q.

Electric field strength of charge Q at position of charge q, E = F/q

Reference: PYQ - Oct/Nov 2013 Paper 13 Q31

9702/Oct Nov/13/2017/Q22

The graph shows how the displacement of a particle in a wave varies with time.

Which statement is correct?
A The wave has a period of 2 s and could be either transverse or longitudinal.
B The wave has a period of 2 s and must be transverse.
C The wave has a period of 4 s and could be either transverse or longitudinal.
D The wave has a period of 4 s and must be transverse.

Solution:

Answer: C

The period of the wave is 4s, so option A and B are incorrect.

The wave could be either transverse or longitudinal because the graph is not represent the actual shape of the longitudinal wave. Because the graph is only represent the displacement of the wave during specific time.

Reference: PYQ - Oct/Nov 2017 Paper 13 Q22

9702/Oct Nov/12/2017/Q7

The velocity-time graph for a train starting at one station and stopping at the next is shown.

Another train has double the acceleration but the same maximum speed and the same

deceleration.

Which velocity-time graph, on the same scale, shows the motion of this train between the same

stations?

Solution:

Answer: B

From option A to D, we know that all of the acceleration become double from beginning, after that same maximum speed and at the end same deceleration.

However, the question had mentions the same station for the train (which mean same displacement).

The displacement of the train = velocity x time
= the area of trapezium of the graph.
= 1/2 (11+5)(8)
= 64m

For option A, the displacement = 1/2 (9+5)(8) = 56m

For option B, the displacement = 1/2 (10+6)(8) = 64m

For option C, the displacement = 1/2 (8+4)(8) = 48m

For option D, the displacement = 1/2 (11+7)(8) = 72m

Reference: PYQ - Oct/Nov 2017 Paper 12 Q7

9702/Oct Nov/12/2017/Q22

When sound travels through air, the air particles vibrate. A graph of displacement against time for
a single air particle is shown.

Which graph best shows how the kinetic energy of the air particle varies with time?

Solution:

Answer: D

Kinetic energy = ½ mv²

A graph of displacement against time for a single air particle is shown.  The gradient of the displacement-time graph gives the velocity of the air particle at that point in time. This is done by calculating the gradient of the tangent at that point.

The gradient (and hence, velocity) is found to be zero at the maximum displacement (the tangent is horizontal) and maximum when the displacement is zero (the tangent is steepest).

Thus, at time = 0, T and 2T the velocity is zero and hence kinetic energy is zero. [A and C incorrect]

But, between time = 0 and T or between time = T and 2T the displacement is zero (in case) twice. So, the velocity (and kinetic energy) reaches its maximum value 2 times in each of the 2 intervals.[B is incorrect]

Reference: PYQ - Oct/Nov 2017 Paper 12 Q22

9702/May Jun/12/2015/Q24

Two light waves of the same frequency are represented by the diagram.

What could be the phase difference between the two waves?

A 150°
B 220°
C 260°
D 330°

Solution:

Answer: C

The 2 light waves are said to have the same frequency. The waves have a sinusoidal form.

Let the wave with the larger amplitude be wave A and the one with smaller amplitude be wave B.

Since both waves have a sinusoidal form, we can assume that any of the 2 waves will start at displacement = 0 (we take this point as the reference point), and then move up – just like the wave A at (0, 0).

{In fact, any point (at any displacement) could be taken as the reference point, but in this question, it is easier to consider that point.}

On the graph, the x-axis gives the phase angle.

At a phase of 0°, wave B has not yet reaching the starting (reference) point while wave A is already at that point. It is only at a phase of 100° that wave B reached the reference point (or we could say that wave B reaches this point AGAIN at 100°).

The solution of 100° is not available in the 4 choices, so we can say that wave B had actually already reached this point before – we need to find at which phase this was, according to the x-axis in the diagram.

The phase angle is actually between 0° and 359°. A phase difference of 360° is the same as a phase difference of 0° and a phase difference of 1° is the same as a phase difference of 361°, ….

The wavelength of a wave corresponds to a phase difference of 360°. Since the 2 light waves have the same frequency, it means that they are the same wavelength.

So, if wave B reached the reference point again at a phase of 100°, according to the x-axis, going back by a wavelength (by a phase difference of 360°), we can say that previously, wave B had reached the reference point at the phase of

Phase = 100 – 360 = – 260°

As for wave A, it reaches the reference point at a phase of 0° (according to the x-axis).

Thus, phase difference between wave A and B = 0 – (–260) = 260°

Reference: PYQ - May/Jun 2015 Paper 12 Q24

9702/Oct Nov/11/2009/Q9

The diagram shows two spherical masses approaching each other head-on at an equal speed u.
One has mass 2m and the other has mass m.

Which diagram, showing the situation after the collision, shows the result of an elastic collision?

Solution:
Answer: A

For an elastic collision,

Velocity of approach (before collision) = Velocity of separation (after collision)

{The above result can be obtained by considering that for elastic collision, both momentum and kinetic energy is conserved. Momentum, p = mv. Kinetic energy = ½mv². By equating the sum of momentum before collision to that after collision and by equating the sum of KE before collision to that after collision, 2 equations are obtained which can be simplified into the above stated result: Velocity of approach (before collision) = Velocity of separation (after collision). The proof will not be shown here}

{Approach means that the 2 sphere are coming towards each other and separation means that they are moving away from each other}

Before collision, velocity of approach = u + u = 2u

Consider A:

Velocity of separation = (u/3) + (5u/3) = 6u/3 = 2u

Consider B:

Velocity of separation = (u/6) + (2u/3) = 5u/6

Consider C:

Both spheres are moving in the same direction. So, speed of separation is the difference in the 2 speed here/

Velocity of separation = (2u/3) – (u/6) = 3u/6 = 0.5u

Consider D:

Since the spheres stick together, they are not separating. They move together.

Velocity of separation = 0

Only answer A gives velocity of separation = 2u.

Reference: PYQ - Oct/Nov 2009 Paper 11 Q9

9702/May Jun/12/2013/Q23

The diagram shows a large crane on a construction site lifting a cube-shaped load.

A model is made of the crane, its load and the cable supporting the load.

The material used for each part of the model is the same as that in the full-size crane, cable and
load. The model is one tenth full-size in all linear dimensions.

A             10⁰

B             10¹

C             10²

D             10³

Solution:
Answer: B

Stress = force per unit area = F / A

Let the force (weight) and the (cross-sectional) area in the full-size crane be F and A respectively.

Stress in cable of full-size crane = F / A

As mentioned in the question, the model is one-tenth full-size in all linear dimensions.

For the model,

The load, which has a cube-shaped load, is in 3-dimension. So, for each of the 3 dimensions, the length of the model is reduced by a factor of 1/10. Therefore, the volume of the load is reduced by a factor of (1/10)³ = 1 / 1000.

Weight = mg and Mass = density x volume. Since the same material is used, the density is the same. So, the mass is proportional to the volume. A reduction in the volume causes the mass to be reduced by the same factor. The weight, which depends on the mass (g is constant), is also reduced by the same factor.

Force (weight) in model = F / 1000

Similarly, the cross-sectional area (which depends on (diameter)²) will be reduced by a factor of (1/10)² = 1 / 100.

(Cross-sectional) Area in model = A / 100

Stress in cable of model crane = (F/1000) / (A/100) = 0.1 (F/A)

Ratio = (F/A) / 0.1(F/A) = 10  

Reference: PYQ - May/Jun 2013 Paper 12 Q23