×

# GCE ‘O’ Level 2010 Oct/Nov Physics 5058 (MCQ) Paper 1 Suggested Answers & Solutions

(36)
Tuition given in the topic of Little Miss Loi the Science Tutor (archived) from the desk of at 11:16 pm (Singapore time)

Hello again everyone.

Once again, my 师姐 and I have drafted list of suggested answers for this year’s Physics 5058 MCQ Paper 1, which appears to contain a fair bit of bizarre controversies pertaining to skydiving, weird diodes, push-ups etc.

2010 O-Level October/November Physics 5058 Paper 1 Suggested Solutions

With all the “Confirm I’m right and you’re wrong!!!!!” flying around, please leave a comment if you spot any mistake, typo and/or wish to violently disagree with clarify any of the explanations given.

Good luck for those of you with your remaining papers! WE’RE ALMOST THERE!!!

### Revision Exercise

To show that you have understood what Miss Loi just taught you from the centre, you must:

1. Aaron commented in tuition class

2010
Nov
11
Thu
11:26pm

1

I have some doubts about question 29 on the diode one. Will put my explanation tmr see if it sounds logical to you because my answer is C instead. Hope there will be answer for the chem paper tmr too 😀

2. qwerty commented in tuition class

2010
Nov
11
Thu
11:36pm

2

wah sian la. so many careless mistakes. 33/40 pray hard i can get my a2 or even better scrape out an a1. time to burn some extra joss sticks

3. uthopia commented in tuition class

2010
Nov
12
Fri
12:16am

3

Why can't question 1 be B? I thought it is possible to use the micrometer to measure thickness of the copper pipe?

• 2010
Nov
12
Fri
5:59pm

3.1

@uthopia: This is unfortunately one of those ambiguous ones ...

To find the volume of the copper in the pipe, we subtract the volume obtained from its outer diameter do by the volume obtained from its inner diameter di i.e.

Volume of copper =

As the outer diameter is already known to be 2 cm, we'll need to measure its inner diameter which may be directly measured by calipers (via its inside jaws) or indirectly by subtracting 2 × its thickness measured by a micrometer (its thickness should be well within the range of a micrometer since outer diameter is only 2cm).

However to measure its "several metres long" length l it's more appropriate to use a tape rather than a rule, which typically can only measure up 1-2 metres (including those your cher used to smack your palms in days of yore ...) so the best answer in this context is D.

Of course there are the counter-arguments that tape measures are also called "retractable flexible rule", or how long exactly is "several meters" etc. so I do hope B can be accepted as well.

I'll have to amend and update the explanation for Q1 in the solutions later ...

4. Michelle commented in tuition class

2010
Nov
12
Fri
5:48am

4

In my opinion, answer for question 3 is D.

I dare say this because at C, I think the parachute is not completely open yet. That's why it's still deccelerating for such a long period of time (which is aprox 5 sec).

At D, the parachute is already at the constant speed which indicates that the parachut is fully open. And it's 1-2s before landing so even better, we can see that the parachute is already open at D.

I'm not like 100% sure if my suggestion is reasonable but I guess it worth suggesting.. Actually I saw some guy doing skydiving and I noticed that he immediately reached a very low constant speed when the parachute was open completely. It took a while to decelerate when he opened at first because the parachute was not open fully yet.

Lol this question is so controversial... I hate doing this :/

5. jingles commented in tuition class

2010
Nov
12
Fri
5:57am

5

The rock and planet question should be A, not B. The rock changes direction and slows down. the rock has inertia so it should resist the change in motion caused by the planet's gravitational pull. it should slow down. much like how asteroids slow down as they reach a planets orbit, unless they are directly impacting the planet.

6. tormented commented in tuition class

2010
Nov
12
Fri
10:03am

6

Hi Little Miss Loi. I like this website a lot as it is a good source of info. For Q17, shouldn't it be C? For option B, it says 'air is a poor conductor of heat and STOPS the heat...'. Even though air is a poor conductor of heat, it will still conduct heat, but more slowly, and the rate of conduction of heat by air is mayb so slow that it will not produce the specific latent heat of fusion or specific heat capacity of water; thus ice remains cold. The word 'stops' in B is very suspicious. As for C, it looks more like the answer to me. Opinions?

• Charis replied to tormented’s comment

2010
Nov
12
Fri
4:37pm

6.1

Q17) It is baked in a very hot oven. The metal dish conducted heat from the oven surroundings to the ice cream (hotter to lower).

B may be vague in the sense that it 'stops' heat but its the most suitable answer.

• 2010
Nov
12
Fri
5:12pm

6.2

The good heat conductivity of the metal dish in an oven is more to facilitate the transfer of heat energy into whoever whatever's being cooked inside. If it can conduct heat away from the ice cream to keep it cold the moment it's taken out of the oven, then our sponge cake would have to become ice-cold as well - not very tasty isn't it?

In this question, the pudding is placed in a very hot oven until the top of the egg whites turns almost chao tar brown. In such a horrifying scenario, the poor ice-cream would've melted in an instant if not for

• the sponge cake protecting it from the heat conducted via the metal dish from the bottom
• the air bubbles within the egg whites effectively 'stopping' the heat from traveling downwards to the ice-cream.

And yup I fully agree with you that being a poor conductor will never completely stop all heat from reaching the ice-cream. But in this context, the air bubbles within the very hot egg whites can almost be considered to be 'stopping' the heat from melting the ice cream that it is in contact with.

As Charis has mentioned, in the absence of a better answer (maybe something about the sponge cake being a poor conductor of heat as well?), B is the best option we have here.

I'll update the solutions with this added explanation later ...

7. Yip Bing You commented in tuition class

2010
Nov
12
Fri
11:36am

7

yikes, overlooked the cm for qns 13 ><
5072/01 over, looking forward to the suggested answers. But still gt Bio...

8. charis commented in tuition class

2010
Nov
12
Fri
12:13pm

8

The answer to Question 4 is apparently D - same initial acceleration and same terminal velocity because terminal velocity is affected by surface area, and since they both have the same size and shape, it should be the same. Clarification?

• 2010
Nov
12
Fri
7:33pm

8.1

@charis: To save a bit of time, I'll quote this wall of text from wikipedia:

A free-falling object achieves its terminal velocity when the downward force of gravity equals the upward force of drag. This causes the net force on the object to be zero, resulting in an acceleration of zero.

As the object accelerates (usually downwards due to gravity), the drag force acting on the object increases, causing the acceleration to decrease. At a particular speed, the drag force produced will equal the object's weight (mg). At this point the object ceases to accelerate altogether and continues falling at a constant speed called terminal velocity (also called settling velocity). Terminal velocity varies directly with the ratio of weight to drag. More drag means a lower terminal velocity, while increased weight means a higher terminal velocity.

Basically in Q4, both objects start with the same acceleration i.e. g (~10 m/s2) when dropped.

As they fall, air resistance increases and their velocities will continue to increase as long the air resistance is less than their weight (there is a net resultant force downwards ⇒ acceleration) until the point where their respective terminal velocities are reached (i.e. air resistance = their weight ⇒ net resultant force is zero ⇒ no acceleration)

As illustrated in the diagram above, it will take a shorter time (t1) for the air resistance to be equal to the weight of the lighter object.

At this point, the lighter object stops accelerating and maintains its terminal velocity v1 but at the same time the air resistance has not yet matched the weight of the heavier object and hence it continues to accelerate ... until it has reached its own higher terminal velocity v2 at the later time t2.

I'll update the solutions with the diagram later ...

9. maggie commented in tuition class

2010
Nov
12
Fri
12:50pm

9

according to your explanation, why is option B not chosen for qn 29?

• 2010
Nov
12
Fri
9:07pm

9.1

@maggie: Since there is only one diode in the circuit, half-wave rectification is performed.

From the diagram for option A below, the diode is forward biased in intervals 0 to 1 and 2 to 3, and hence a current flows through the circuit (when p.d. > 6) and there is a p.d. of 0.2 V across the resistor.

For intervals 1 to 2 and 3 to 4 in the diagram, the diode is reverse biased. Current flow is prohibited and hence there is no p.d. across the resistor during these intervals, so option B (and D) is ruled out.

• 2010
Nov
13
Sat
6:18pm

Question 29 the one on diode why is the answer not C?
"The diode only starts to conduct when there is a potential difference (p.d.) across it greater than 0.6V in one direction" means that it will only conduct 0.2V because it didnt allow the 0.6V to pass?

But the p.d. supplied is always 0.8V alternating and the 0.8V exceeds the 0.6V, so does that mean that the whole 0.8V should pass? Besides, they are talking about potential DIFFERENCE. Should it be than even if the current cannot flow to the resistor via the other end, the p.d. should still be 0.8V? On the bottom wire, potential is at 0.8V. On top, it can be at 0V if the diode is reverse biased and 0V also if the diode is forward biased since the 0.8V will get 'consumed' by the resistor?

After talking so much I now think that the answer is D leh. I think there may be a trick in the word "potential difference"? Don't know ...

Anyway I'm looking forward to the pure chem paper 1 answer.

10. Rahdom commented in tuition class

2010
Nov
12
Fri
3:19pm
11. Jack commented in tuition class

2010
Nov
12
Fri
5:11pm

11

What a paper! What's your lowest score here??

12. hie commented in tuition class

2010
Nov
12
Fri
10:42pm
13. hie commented in tuition class

2010
Nov
12
Fri
10:49pm

13

for qstn 36, if i am not wrong..the max induced emf occurs at the higher part of the coil not in the center. and is that is the case,this is a poor qstn as its based on assumption..so if the magnet is still able to pass thru the top of the coil at least half of it,i think it will have higher emf as it has higher speed- Vsq = Usq + 2AS

14. hie commented in tuition class

2010
Nov
13
Sat
12:42am

14

oh and so my conclusion is raising the spring will inc max induced emf

• 2010
Nov
13
Sat
6:44pm

14.1

@hie: Typically in such a scenario, the magnitude of the induced e.m.f. depends on:

• the no. of turns in the coil
• the speed at which the magnet is moving towards and away from the coil
• the strength of the magnetic field

With the first two factors being constant in our scenario (assuming we don't anyhow go and play with the spring), since the strength of the magnet's magnetic field is weaker the farther from the magnet, raising the spring higher and effectively locating the magnet farther away will decrease the maximum emf that can be induced.

I suppose you are referring to the textbook example of inducing emf via moving a magnet in and out of a solenoid. In that experiment, the induced emf is higher "at the higher part of the coil" because the magnet is still moving (high rate of change of magnetic flux). Textbook diagrams typically show no induced emf when the magnet is inside the coil with the sole reason being that the magnet is momentarily stationary (before it changes direction).

However, if you were to move the magnet at the same speed both inside and outside the coil, you'll actually get a higher induced emf inside the coil since more turns of wire are 'cutting' through the magnetic field as the magnet moves through the coil.

15. charrrr commented in tuition class

2010
Nov
13
Sat
1:17am

15

hello! any chances you'll upload the suggested answers for o'levels 2010 combined science(phy,chem) 5116/01 soon? hope to check the answers asap 😀 thank you!

16. joanne commented in tuition class

2010
Nov
13
Sat
2:44pm

16

will u be posting the mcq ans for pure chem ?

17. Moriarty678 commented in tuition class

2010
Nov
13
Sat
7:01pm

17

I disagree with 3 and 7. As you reach terminal the drop in speed is greater at b than at C. Therefore, C cannot possibly be the answer, as the drop in speed should surely be more when the chute is opened than when it has not yet.

I also disagree with 7. The planet's gravitational pull is TOWARDS ITS CENTRE. This means, if the rock is not going towards the centre, it would pull the rock towards its centre, thereby decreasing its speed towards its original trajectory. The rock also resists this new force acting upon it by its inertia, causing it to have a force on the opposite direction against the gravitational pull. Therefore, it changes direction, and decreases in speed. unless it was headed towards the CENTRE of the planet, but from the diagram, we can see that it was not.

18. Person commented in tuition class

2010
Nov
14
Sun
4:20pm
19. Ball commented in tuition class

2010
Nov
14
Sun
10:24pm

19

Hello. I think the answer for question 4 is D. Coz terminal velocity is affected by the surface area only?

20. James commented in tuition class

2010
Nov
17
Wed
1:25am
21. Champs05 commented in tuition class

2015
Jul
15
Wed
1:28am
• ### Latest News

A New Year. A New Hope.

Hybrid joss sticks math tuition sessions are continuing to be conducted both online and onsite at Novena in 2024.

Please check our latest 2024 Jφss Sticks weekly secondary / O level maths group tuition schedule for updates.

Subscribe now to Miss Loi's Maths Exam Papers are now available for immediate online purchase.

* Please refer to the relevant Ten-Year Series for the questions of these suggested solutions.