Physics Paper 2 help required

[no.mercy]

The new topic included in the syllabus is the transistor and bistable circuits..can someone provide me its notes pls

The syllabus states:

25. Electronic Systems
Note: There is no compulsory question set on Section 25 of the syllabus. Questions set on topics within
Section 25 are always set as an alternative within a question.
Content
25.1 Switching and logic circuits
25.2 Bistable and astable circuits
Learning Outcomes
Candidates should be able to:
(a) describe the action of a bipolar npn transistor as an electrically operated switch and explain its use in
switching circuits.
(b) state in words and in truth table form, the action of the following logic gates, AND, OR, NAND, NOR and
NOT (inverter).
(c) state the symbols for the logic gates listed above (American ANSI Y 32.14 symbols will be used).
(d) describe the use of a bistable circuit.
(e) discuss the fact that bistable circuits exhibit the property of memory.
(f) describe the use of an astable circuit (pulse generator).
(g) describe how the frequency of an astable circuit is related to the values of the resistive and capacitative
components.

 

[no.mercy]

Can someone help me out that how to find the answer of the following question..its really confusing:



The answer given in the marking scheme is D

click on the frog to see the image

 

[VelaneDeBeaute]

isnt opening

 

[no.mercy]

Here it is in the attachments

 

[Student-OR-Slave]

this is how a Bistable system works. there are two inputs, suppose R and S...and two outputs Q and Q-dash. Now those who have studied logic gates in computers will n
know truth tables.

Initially all inputs are zero.
When S=1, R=0:
We will suppose inputs at S to be 1(S itself) and 0(the one coming from above)(See the image)
Since it is a NOR gate, the output will be 0.
Now this output goes to the R logic gate which also is a NOR gate. Here both R and the other input are 0 so the output will be 1. Thus when R=0 and S=1, Q output will be activated while Q-dash will be off.
.... End of the first part.

Now when S=0 R=0continued from previous mechanism)
At S, S itself is zero and the input from above is 1. Since R is 0, Q output will remain activated and Q-dash 0.
....End of Second Part

Now when we make R=1 and S=0:
At R, R itself will be 1 while input from below will be 0. So Q output will be 0 while at S, both inputs are 0 so output will be one. Result will be Q goes off while Q-dash is activated.
....End

This is how save and delete command in computers basically work.
If you notice, we can substitute S with the Save command and R with Remove or Delete command while Q with memory and Q-dash with Erase memory. When we click on save, storage is done. After that, the memory doesnt get erased. it retains the storage until we press on delete.
Srry the truth table aint gud enough...i made it in Pant lol! ( i noe!)
btw...anybody notice the computer paper seemed as if made from the booklets here!

 

[Usman17]

@Student: Yeahhh dude it was definitely made from the booklet rofl

 

[no.mercy]

student can you tell me how to find the anwer of the MCQ...the answer is D but i am totally confused on how to find it

 

[BaKa]

sorry i am getting confused on how to find the density its really hard try to look up the year and i will tell u how to find the answer from past papers

also i donno how to state all the logic gates hense it becomes difficult so just look at this if u want some help
http://answers.yahoo.com/question/index ... 943AAgOGCI

 

[no.mercy]

bascially this question was given in the mock paper of Apex grammar school karachi..they had given so much difficult MCQ`s i was not able to solve most of them

Here is the attachment if you want

 

[maestro007]

(b) state in words and in truth table form, the action of the following logic gates, AND, OR, NAND, NOR and
NOT (inverter).

here you go:
AND: out put is 1 if each and every input is 1
OR: out put is 1 if one or more inputs are 1
NAND: out put is not 1 if each and every input is 1
NOR: output is not 1 if one or more inputs are 1
NOT: output is always the opposit of the input.

hope this helps!

 

[Jinkglex]

to the mcq above, i have a possible explanation..

mass of water = 500
mass of sand = 892
together they make 1392
but that is 200 in excess, as the mass of beaker should be included, for which we need the initial two masses to equal 1192 (1392-1192=200)
892/200= 4.46.

I have no idea how this works into a plausible explanation though :s

 

[VelaneDeBeaute]

Mercy i think i ve worked it out..!
To find the density...v need mass and volume..!
Mass of sand...--> Beaker 3(1200) - Beaker 1(308) = 892 g.
For water's density, 1g occupies 1 cm3....
so water's volume in Beaker 2= volume of whole beaker= same.
Water's volume in Beaker 2 =Beaker 2(808)-Beaker 1(308) = 500 cm3.
Similarly, water's volume in last beaker= its mass and its mass in last beaker is 1500-1200=300 g...so d=300 cm3
So if total beaker's vol is 500 and water's vol in last beaker is 300, sand's volume is 200 cm3.
Now v use the formula.
d=m/v
=892/200=4.46 g/cm3. :wink:

 

[ISCLHRTSR]

Ghalya is right. Here's the numeric version of his answer:

Mass of empty beaker = 308 g
Mass of beaker + water = 808 g
=> Mass of water = 500 g

=> Volume of beaker = 500 cm^3

Mass of beaker + sand = 1200
=> Mass of sand = 892

Mass of beaker + Water(II) + sand = 1500
=> Mass of water (II) = 300
=> Volume of water (II) = 300
=> Remaining volume = 200

Therefore, density of sand = mass/ volume occupied = 892/200 = 4.46

 

[VelaneDeBeaute]

thanx ISCLHRTSR...will u plz edit ur HE to a SHE...really pisses me off..!

 

[workaholic]

Plz explain this mcq.
Which statement describes how the steel bar can be demagnetised?
A Reverse the d.c. supply and gradually decrease the current in the circuit.
B Reverse the d.c. supply and gradually increase the current in the circuit.
C Use an a.c. supply and gradually decrease the current in the circuit.
D Use an a.c. supply and gradually increase the current in the circuit.

i dont understand y one shud decrease current and not increase
Answer is C....

 

[Maleeha]

heehee.. ghalya.. funny u being called a HE..lol.. n bani was always so tensed..haha

 

[ravaneous]

no.mercy the answer to your mcq

first find the mass of water = mass of water + container - mass of container = 500
volume of water = density x volume found in first step = 500
mass of sand = mass of sand + container - mass of container = 892
mass of water in fourth container = 1500 - 892 - 308 = 300

displaced water = 500 - 300 = 200 which is volume of sand

now find the density of sand by dividing 892 by 200 which is 4.46
hopes this helps

 

[CaptainDanger]

Oct Nov 2010 Paper 21 Last part... I manage to get one mark but how to find out the answer??

 

[Maleeha]

Oct Nov 2010 Paper 21 Last part... I manage to get one mark but how to find out the answer??

(present activity/actual activity) = (1/2)^n
where n is no. of half lives
350/1400 when simplified means 1/4 n that becomes (1/2)^2
this means 2 half lives.. 2(14.3) =28.6
hope u get it..

 

[CaptainDanger]

Oh thanks... 14.3 is what I found again now.... Should have read the part from start....