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Physics: Post your doubts here!

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I am never able to complete the paper within an hour ... any tips anyone?
Phy P1: There are 7-8 'free' questions (Units, phases of matter,nuclear). Make sure you don't make a mistake in these. Some concepts are tested again and again with slight variations like Young Modulus, thermistors, inelastic collisions etc so look for patterns in the questions. Remember the formulas so you don't waste time recalling them.

Chem P1: I'm not taking Chem P1 this session but I found this as the hardest AS paper along with Phy P2 in my subjects (Maths, Phy & Chem). Again there are concepts which are repeated (Isomers,, type of organic reaction, ionization energy etc). Several concepts can be included in one question so learn how to tackle these questions. The last 10 are the hardest so make sure you leave enough time for these. Alternatively, you can do these at the start.

Always mark any question which you are not 100% sure of. When you are done with the paper review these and those which you got wrong. Look at the question that why did you get it wrong and how you need to avoid the mistake. Nearly all mistakes fall into four categories:
1. Content issue: You didn't know what to do as you didn't know the material.
Way to avoid these: Look up the material from where it was tested.
2. Comprehension issue: You weren't sure what the question was asking.
Way to avoid these: Re-read the question and underline the key points.
3. Time issue: You were running out of time.
Way to avoid these: If you're spending way more time on a question, skip it. Come back to it at the end and try to interpret the question what exactly is it asking.
4. Careless errors: These are the most frustrating ones.
Way to avoid these: Re-check all your answers at the end including the supposedly easy ones.

The first few questions on a topic are usually easy. Don't waste too much time on the difficult ones so you have to guess on the easier ones.
Usually all the information provided is important to answer a question. If you're doing a question without all of the information, you're probably wrong.
If you feel you don't remember many things to answer questions, do 1-2 papers with your notes/book open, but not more than this. This way you'll get the idea which things are usually essential to answer questions.
Don't waste too much time filling the answers but also make sure they are properly filled in. The best way I found to do this was to fill the answer in chunks. After doing all the questions from one page, I filled all the answers from that page and so on. Also, make sure you have filled the right answer for the right questions, especially when you have skipped a question. Otherwise, this can result in a great loss of marks.

Good luck :)
Additionally, remember the base units for some quantities like resistance, voltage etc.
 
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Helppp meeee!

Answer is C.

The answer is actually A and here's why.

As the object is accelerating, that mean's its speed is changing, and by changing I mean that it is increasing. When the acceleration peaks at the top and begins to turn, the velocity is not zero. So you can eliminate option C.

When the curve begins to descend, the object is STILL accelerating, it is not deceleration, so the speed increases but at a lower rate. So no, the velocity does not reach 0 at all. This eliminates B.

You're left with A and D, the key thing here is what happens when there is no acceleration. With D, the gradient is steeper, indicating the speed still increases, which is not possible since the acceleration is zero. So D is wrong.

The reason why A is the correct choice is because when the object stops accelerating, it moves at constant speed. So the gradient becomes flat/horizontal, there is no increase nor decrease.

Hope that helps!
 
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I don't understand this question, the answer is B but why?
The average height of the water molecules is h/2 so GPE before tap is opened is mgh/2. After the top is opened half of the water flows to Y so remaining height is h/4 and loss in GPE is mgh/4.
 
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For 16: 3= I1(6*10^3) I1= 0.5mA
3= Current flowing in 2nd loop(3*10^3) I= 1*10^-3
I2 is a combination of I1 and the current flowing in the 2nd loop, I2=1.5mA
Pd across 6k resistor= (0.5*10^-3)(6*10^3) = 3V
Pd across 3k resistor= (1*10^-3)(3*10^3) = 3V
Total pd = 6V
Whats the year? I might be able to find the answer.
Thanks. It's J1999
 
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Q10: How do we estimate this? any logic?:S
Screen Shot 2015-06-08 at 9.28.24 AM.png

Q4: Should I use 50N to find the power & hence the work done OR the horizontal component of 50? I think we shud use 50N because the question asks TOTAL WORK DONE( not the NET WORK DONE) ,right?

Screen Shot 2015-06-08 at 9.27.13 AM.png


Screen Shot 2015-06-08 at 9.28.02 AM.png

Help please
 
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