# Evan's Space

## 2017PurePhyP2Q5 Two points on the rope wave with displacement-time graphs given

In this this question, the displacement-time graphs are given, which are different from displacement-distance graphs.

In the displacement-time graphs of A and B, they show the displacement of that particular point at different timing. E,g, at t = 0s, the A is at the rest position (0 displacement) and at time 0.2 s it is at the maximum displacement. This means A is going up from t = 0 s to 0.2 s.

Solutions:
(a) Amplitude: 1.5 cm
(b)(i) Frequency is the number of complete waves produced in 1 second.
(ii) period T = 0.8s, f = 1/T = 1/0.8 = 1.25 Hz
(c) Closest possible positions of A and B, (refer to the video), is when the
time taken for the wave to move from A to B is T/4 = 0.8/4 = 0.2 s.
speed = distance/time = 38/0.2 = 190 cm/s approx. 200 cm/s
(ii) There are various possibilities in which B can be 38 cm to the right of A. Besides T/4, it can be 1.25T or 2.25 T etc. Hence the speed can be other values.

Refer to the video explanation below

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

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.

Solutions: Option C

## resultant force-time graph link to speed-time graph

Refer to the video tutorial below for explanation.

## 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.

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

## Which is better to cool the food?

Other examples in our daily lives:

In some supermarket, the seafood are placed outside of air-conditioned place. The seafood is kept cold by putting crushed ice covering the seafood to keep the them cold and fresh.

Refer to this Sci Physics question N2008P2Q6(b)

Solutions:
For the solid that does not melt, when thermal energy is absorbed from the surrounding  food, its temperature starts to rise. So it is not so effective at keeping the food cool.
For ice-pack, when thermal energy is absorbed from the surrounding food, it starts to melt. During melting process, a much larger quantity of thermal energy is absorbed from the food to melt per unit mass of ice, the temperature remains constant at 1oC, and the melting process is long. Hence ice-pack is more effective at keeping the food cool.

Density of ice – Why ice floats on water?

Will whole lake be frozen during winter?

## 2013SPp1q17 What is the new P when V across resistor is doubled?

Solutions: C

Alternatively, you may put values to work out using the 2 basic formulae of P = IV and V = IR. Refer the video below.

## N2010 P2Q12 Fuel gauge using variable resistor

Solutions:

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.

Solutions:

(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.

## OLevel SP P1 Q12 2010 focal length of lens using distant object

Which distance is equal to the focal length of a lens?

(A) the distance between a distant object and the 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

Solutions: Option C

Focal length f is the distance between the focal point and the centre of the lens (optical centre).

Note that only when parallel rays of light enter a converging lens, the rays will converge to a point. That point is considered to be focal point F (principal focus). The distance between focal point F and the optical centre is the focal length f. Refer to below.

As none of the options is similar to the above definition, you have to consider that the rays from a distant (far away) object are considered parallel. Hence the sharp image formed on the screen is considered the forcal point F of the lens and the distance between the image and the optical centre is the focal length f.

If it is a close object, the rays entering are not considered parallel. Hence even if the rays converged to a point, that point is NOT focal point F and the distance between this converged point and the optical centre is NOT focal length f.

Click the following posts for other lens concepts:

https://evantoh23.wordpress.com/2017/08/19/different-lens-ray-diagram-questions/

https://evantoh23.wordpress.com/2010/11/09/20101109converging-lens-important-concepts/

## 2013 Nov Sci Phy P2 Q9 – Pressure and Moment

Solutions:

(a) (i) A bigger force than F can be obtained due to the level system and hydraulic system.
Level system: Applying principle of moments, the anticlockwise moment by the F is equal to the clockwise moment by the force on piston A (note that the handle is pushing the piston down, but the piston A is pushing on the handle upwards – action = reaction). As the perpendicular distance from F to the pivot is greater than the perpendicular distance of the force by piston to the pivot, the force on the piston A is greater than F at handle.
Hydraulic system: As the pressure transmitted in the liquid is the same, pressure at piston A = pressure at piston B. As P = F/A and area of piston A is smaller than area of piston B, a larger force is obtained in piston B.
Hence these two systems allow the force on piston B to be greater than F at the handle.

(a) (ii) Both liquid and gas molecules are in a continuously random motion. But in liquid, the molecules are closely packed together and able to slide around one another. There is very little empty space between the molecules hence liquid is not compressible. Gas molecules are far apart from one another, hence gas can be easily compressed.

(b) P = F/A = 12000 / 0.060 = 200 000 Pa

(c) Velocity is the vector quantity while speed is a scalar. As the car goes round the bend, the direction of the car changes. Hence velocity is changing even though speed is constant.

Likewise, as the velocity is changing, the car is considered to have an acceleration (not in the sense of increasing speed though).

## N2007P1Q27 brightness of bulb

Solutions: Option B

As lamp X is thicker and shorter, these 2 factors make the resistance of X lower than Y.
Recall: Length increases, R increases & cross-sectional area increases, R decreases.
[R = pL/A]

As both bulbs are connected to the same mains, (assume 240V), the potential difference across the bulbs are 240 V.

Brightness of the bulb depends on power, P = IV or V2/R, since V is constant, I across X is higher due to lower R,
hence power of X is greater than power of Y, hence X is brighter.

## N98 P1 Q20 Light Reflection – Technique to remember

The diagram shows four lamps in front of a plane mirror. The card prevents the observer at X from seeing the lamps directly, although the image of one lamp can be seen in the mirror.

Which lamp’s image can be seen?

Solution: B

## 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?

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

## N2010 P1 MCQ Q12 Sci Physics 2010 O-Level

###### 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

## 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

## N2007P1Q33 – Electromagnetism

A copper wire is taped to two wooden blocks which stand on a sensitive balance. When there is a current in the wire and a magnet is held in the position shown, the balance reading increases.

Which other arrangement of magnet and current will give the same increased reading?

Solutions: Option D

This question is rather direct. Applying FLHR, the force acting on the copper wire is downwards. Hence using FLHR again on the 4 options, only D creates the same downward force.

But what I wish to highlight in this question is for you to compare this with the other question which I posted earlier. They seems similar but they are different. View the other related question.

## 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

Copper

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

Aluminium

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]

## N2007P1Q28 – Current Electricity – R of bulb increases as I increases

Voltage-current readings were obtained for different electrical components. Which readings are for a 3 V, 0.06 A torch bulb?

 A Voltage / V 0 1 2 3 Current / mA 0 6 12 18 B Voltage / V 0 1 2 3 Current / mA 0 25 45 60 C Voltage / V 0 1 2 3 Current / mA 0 20 40 60 D Voltage / V 0 1 2 3 Current / mA 0 10 20 30

Solutions: Option B

The rating of the bulb is 3V, 0.06A.

Hence only if a potential difference of 3V is provided across the bulb, the ideal current of 0.06A will flow through and the bulb will operate at its optimal brightness.

To solve this question, you have to find the resistance R of the bulb,  R = V/I = 3 /0.06 = 50Ω

As this is a filament bulb, the resistance increases as current increases. Filament bulb is a non-ohmic conductor.

Hence apply R = V/I for each component, the one which gives a increasing R will be the bulb.

A and D is out as when voltage is 3V, the current flowing through is not 60mA. C is out as the resistance is fixed at 50Ω throughout, not a characteristic of a bulb.

Hence only Option B is correct.

## 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 A X X B X Y C Y X D Y Y

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?

Solutions:

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.

## 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?

Solutions:

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.