Evan's Space

Wonders of Physics

Leave a comment

2016 Pure Physics P1 Q40 – Voltage and Power of Bulb Y when X is switched off

2016 PPp1q40

Solutions: Option C
[There is another similar question in 2013 Nov Pure Physics P1Q40. The answer is Option C too. Refer to the last section of this post]

Refer to the 2 videos below. The 1st one is faster if you know that the bigger the resistance, by proportion, the bigger the potential difference of the component as it will take a larger portion of the emf wrt to the other components in series. This method is especially useful for MCQ.

If you are still unsure, you may put in values to find the I, V and P across the components. This will be more tedious and time-consuming.

Another similar question is 2013 Nov Pure Physics P1 Q40.

2013 PP p1q40

Solutions: Option C

Leave a comment

Identifying what lens, focal length and image from 2 rays – PP2010P1Q23 and SP2014P1Q11

These 2 questions are actually the same. Q23 is from 2010 Pure Physics P1 while Q11 is from 2014 Sci Physics P1. Take a look at these 2 questions. If you are not sure, view the video below for the explanation.

Capture1 PP 2010

Answer to Q23: Option A

Capture2 2014SP

Answer to Q11: Option D

If you do not know how to answer these 2 questions, view this video and also refer to the lens summary below.



N2010 P2Q12 Fuel gauge using variable resistor



View the video on how the gauge works.

As the fuel level drops, the float which stays on the fuel surface will descend. The rod which is attached to the float will turn clockwise about the pivot X. As the rod turns, the resistance on the variable resistor increases. This increases the resistance of the circuit. Hence the current flowing through the circuit will decreases, causing the needle to deflect more to the left, indicating towards E (empty). Thus the reading on the fuel gauge decreases.



(i) Both the fixed resistor and sensor are in series.

Total effective resistance Re = 5000 + 1000 = 6000 ohms

V = IR
12 = I x 6000
I = 0.0020 A

Hence potential across Y, V = IR (where I is constant in a series circuit)
= 0.0020 x 1000
= 2.0 V

(ii) When the temperature increases and the resistance of sensor Y decreases, the total effective resistance of the circuit decreases. Current flowing through the circuit increases.

Since V = IR, where the R of the fixed resistor is a constant 5000 ohms, as current I increases, the potential difference across the 5000 ohms resistor will increase.

Leave a comment

N2007P1Q9 Pure Physics – Energy needed to pump water up a building

An apartment block receives water from a nearby reservoir. A pump is necessary to lift the water into a storage tank at the top of the building.

The gravitational field strength is 10 N / kg. How much energy does the pump supply to lift each kilogram of water into the tank?

GPE up 10 m

Solutions: 100J

Water will always find its own level. Hence with or without the pump, the water level in the pipe in the building will be of the same level as the reservoir. Hence, the pump is only needed to pump water up a height of 10m only (instead of the 50m).

Energy, E = mgh = 1 kg x 10 N/kg x 10m = 100 J

1 Comment

N2010 P1 MCQ Q12 Sci Physics 2010 O-Level

Which distance is equal to the focal length of a lens?
A. The distance between a distant object and its image.
B. The distance between the image of a close object and the centre of the lens.
C. The distance between the image of a distant object and the centre of the lens.
D. The distance between two principal foci.
Solution: Option C.

Recall that only reflected rays from distant object (object placed far away) are considered parallel. When these parallel light rays pass through the converging lens, they converge to a point, Focal Point (F).
Only then, the distance between the (sharp) image of distant object to the centre of the lens, optical centre (C), is known as focal length (f).

Refer to the diagrams below.


Also refer to another question on lens: http://evantoh.posterous.com/2010/11/09/converging-lens-important-concepts

1 Comment

N2004 P1 Q15 – Pressure P1V1 = P2V2

Water if depth 10 m exerts a pressure equal to atmospheric pressure. An air bubble rises to the surface of a lake which is 20 m deep. When the bubble reaches the surface, its volume is 6 cm3.



 What was the volume of the air bubble at the bottom of the lake?


A) 2 cm3          B) 3 cm3         C) 12 cm3      D) 18 cm3



Solutions: Option A
Since the air bubble is enclosed, PV at A is equal to PV at B.

                  PAVA = PBVB

(Patm + Pwater) VA = (Patm) VB

        (10 + 20) VA = 10 x 6

                      VA = 60 / 30

                           = 2 cm3


Leave a comment

N2005/P2/A1 – Tension in Spring

The figure below shows a 0.40 kg mass hanging at rest from a spring.


(a) State what is meant by the mass of an object.

Mass is the amount of matter in the body. SI unit is kg.

(b) (i) On the figure above, draw an arrow showing the ling of action and the direction for each of the two forces that act on the mass. Write the name of the force next to each arrow.


(ii) The gravitational field strength is 10 N/kg. Calculate the values of the two forces you have drawn in (i).

W = mg = 0.40 x 10 = 4 N
Weight = Tension = 4 N (Newton’s 1st law, net force = 0N)

(c) The mass is pulled downwards and then released. Explain, in terms of any changes in the forces acting on the mass, why the mass accelerates upwards.

As the spring is pulled downwards, there is a gain in elastic potential energy. The tension in the spring increases compared to previously.
Since Tension > Weight, there is an resultant force acting upwards, hence there is an acceleration upwards.

<p><a href=”http://vimeo.com/61785429″>closed loop triangle mpeg4</a> from <a href=”http://vimeo.com/user10931667″>evantoh</a> on <a href=”http://vimeo.com”>Vimeo</a>.</p>

1 Comment

N2004P1Q4 – Dynamics + Kinematics

Trolley X and trolley Y are joined by a stretched spring. Trolley X has twice the mass of trolley Y.


When the trolleys are released, the acceleration of X is   to the right.  What is the initial acceleration of trolley Y to the left?Solutions:Key concept here is Newton’s 3rd Law. Action equals reaction. Both trolleys are connected by the spring, hence the tension force experienced by each trolley is equal and opposite.

Fx    =    Fy
ma       =    ma
2m x 2    =    m x a
a        =   4 m/s2

Another important concept to know. In Fig A, a box of m is pulled with a force of 40 N. Given that the frictional force is 10 N. The resultant force is 30N. Box is accelerating.


In Fig B, if another identical box m is added on top of the existing box (no change to the floor), the total mass is 2m, the frictional force now will double = 20 N. Hence resultant force now will be 20N. The acceleration of the box will be lower than previously.

Leave a comment

N2007P1Q17 – Thermal Properties

Four bars, all exactly the same size, are each placed with one end in boiling water.The times taken for the temperature of the other end to increase by 2 oC are measured.

Material of bar Time for 2oC rise / s
Aluminium 10
Copper 5
Cork 800
Styrofoam 1200

To make a large metal tank with the least heat loss, which materials should be used for the tank and its insulation?

Tank Insulation
A Aluminium Cork
B Aluminium Styrofoam
C Copper Cork
D Copper Styrofoam

Solutions: Option B

This question can be a bit tricky. It depends on which approach you view the question. Simply using conduction will be a much easier way to get the answer. The hint to use conduction is from the first paragraph. Time taken for temp of the other end to increase by 2 oC is about conduction.

To build such a tank, metal (aluminium or copper) has to be used and insulation on the external wall (cork or styrofoam).

To contain boiling water with least heat loss, both tank and insulation have to be good insulator (poor conductor) to reduce heat lost to surrounding. Hence tank should be aluminium and insulation should be styrofoam (option B)

On the other hand, if you approach the question in term of heat capacity, it will be a bit tedious and you do not have the values.Firstly, if aluminium takes a longer time to rise by 2 oC, it has a higher specific heat capacity. But you cannot merely look at specific heat capacity. You have to look at heat capacity as the mass of the container is important.

Copper: Density = 8940 kg/m3 and specific heat capacity = 400 J/kg.K

Aluminium: Density = 2700 kg/m3 specific heat capacity = 900J/kg.K

Assumptions: Volume of copper and aluminium are the same (same shape of container) say 0.02 m3, Initial temperature of metal is 30oC

Considering the amount of thermal energy gained from water as metal temperature reaches 80oC

Heat lost by water = heat gain by metal


Mass of copper = 8940 x 0.02 = 178.8 kg

Heat lost be water = heat gained by copper = mcθ = 178.8 x 400 x (80-30) = 3576000 J


Mass of Aluminium = 2700 x 0.02 = 54 kg

Heat lost be water = heat gained by aluminium = mcθ = 54 x 900 x (80-30) = 2430000J

From the calculation, it is obvious that the aluminium gains lesser thermal energy from the water, hence water will remain warmer compared to when using copper. Hence using this approach, it is still Option B. But important to know that you have to consider heat capacity [C], not specific heat capacity (c). [ C = mc]

Leave a comment

N2007P1Q27 – Current Electricity – Brightness of bulbs

X and Y are lamps with filaments made of the same material. The filament of lamp X is thicker and shorter than that of lamp Y. Each lamp is connected to the mains and switched on. Which is the brighter lamp and which lamp has the larger resistance?

Brighter lamp Larger resistance

Solutions: Option B

Using the formula, R = ρl/A     (ρ: resistivity (Ωm), l: length (m) and A: cross-sectional area (m2) Since both are of the same material, resistivity ρ is the same. X is thicker (cross-sectional area) and shorter, X will have lower resistance than Y.

Hence Y has larger resistance. Both are then connected to mains.

You can assume they are connected individually as shown below. Potential difference across each bulb is the same as the mains.


Since brightness depends on power, using P = IV, bulb X is brighter.


N2005P1Q24 Sound – Reflection of sound from 2 walls

A man who is standing at a point X between two parallel walls (as shown in the diagram) fires a starting pistol.


He hears the first echo after 0.6s and another one after 0.8s. How long after firing the pistol will he hear the next echo?


View video tutorial to visualise the distance travelled.

Time taken to hear next echo = 0.6 + 0.8 = 1.4s

The distance travelled by sound (left and right) from X is the same during 3rd and 4th echos.

Hence 3rd and 4th echos actually coincide. And time taken is 0.6 + 0.8 = 1.4s.

1 Comment

N2008P1Q15 – Kinetic Model of Matter

Gas inside a cylinder is heated slowly to a higher temperature. The pressure inside the cylinder remains constant as the piston moves outwards.


How do the speed of the gas molecules and their rate of collision with the piston compare with their initial values at the lower temperature?


In short:  Temperature increases, Kinetic Energy increases, Rate of Collision decreases, Average Force on wall increases, Pressure constant.

As temperature increases, the speed of molecules increases, the kinetic energy of the air molecules increases.

As piston is free to move, it will move to the right such that the pressure remains constant (equal to atmospheric pressure outside). As the piston moves to the right, the volume inside the piston increases

Surface area in which the air molecules collide increases.

The rate of collision decreases as the number of molecules remains constant. With higher KE of molecules, the molecules will collide the wall with greater force. Though rate of collision decreases, with each collision having greater impact force, the average force acting on the wall of piston increases.

Since P = F / A, with greater force F, over a bigger area A, the pressure P remains constant. (Compared with previously,  smaller F over smaller A, but P constant)  

Misconception: Many think that the rate of collision remains the same, which is wrong. Apparently the effect of volume increases is more significant, hence rate of collision decreases, even though they collide with greater impact force. Hence overall force on wall still increases. If the speed of the molecules increases but the pressure remains constant, then the molecules must collide less frequently. If the rate of collision stayed the same, the pressure would increase.  If the rate of collision increased, the pressure would increase even more.