Which is a possible displacement-time graph for particle Y?

Answer: Option D

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

Cells in parallel vs single cell

Answer: Option C

2013 SPP1Q2 Finding resultant force from 2 forces at 0 to 90 degree

Answer: Option A

Refer to the video tutorial for the explanation.

resultant force-time graph link to speed-time graph

Answer: Option A

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.

Answer to Q23: Option A

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.

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.

Option: A

 

Ratio of masses of same substance when heated

Two blocks X and Y, which are made of the same metal, are heated by heaters of the same power rating. The variation of temperature with time of both blocks is shown on the same graph below.

What is the ratio of the mass of X to that of Y?

A   1 : 3          B   1 : 4         C   3 : 1        D   4 : 1

Option: B

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

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.

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.

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.

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.