PAPER 1 MAY/JUNE 2013 PHYSICS DOUBTS HERE

[hino]

I've got quite a lot of problems. Would be extremely grateful if anyone could explain as detailed as they could.

O/N '10: P12 -- 22, 25, 27, 28
O/N '10: P11 -- 11, 34
O/N '09: P11 -- 10, 14, 15, 22, 28
M/J '09: P1 -- 9, 10, 11, 13, 15, 18

Sorry for any inconvenience in advance.
Requesting to be answered as soon as possible?

--Thank you, Aqua x

0/n 09 p11 Q 28
F=EQ F=mg
Q=F/E m=F/g
Q/m=F/E÷F/g
=F/E x g/F
=g/E
as for the polarity,its negative(it can be seen quite clearly from the diagram)

 

[h4rriet]

Thank you! However, what does 'I' stand for in 9?

Impulse. Which is also delta momentum.

 

[Oishee Asif]

Aaah, purely awesome.
Thanks again!
MashaAllah, bless you!

 

[Champ101]

hey guys can anyone explain me why a voltmeter should have an infinite resistance?

 

[hino]

In the voltmeter case, you are providing the current an alternative pathway to flow once you have introduced the voltmeter in the circuit. Hence, in order to minimize the amount of current using this path (created by the inclusion of the voltmeter) you want the resistance of the voltmeter to be very large compared to the circuit component it is measuring. Higher resistance means that most of the current will follow through the original path and hence leads to minimal change to the circuit conditions.

 

[Champ101]

In the voltmeter case, you are providing the current an alternative pathway to flow once you have introduced the voltmeter in the circuit. Hence, in order to minimize the amount of current using this path (created by the inclusion of the voltmeter) you want the resistance of the voltmeter to be very large compared to the circuit component it is measuring. Higher resistance means that most of the current will follow through the original path and hence leads to minimal change to the circuit conditions.

thankyou! so basically it has a high resistance so that it cannot take any current, hence our calculation is accurate (no loss of current). is it?

 

[Champ101]

The filament of a 240 V, 100W electric lamp heats up from room temperature to its operating temperature. As it heats up, its resistance increases by a factor of 16. What is the resistance of this lamp at room temperature?

 

[hino]

thankyou! so basically it has a high resistance so that it cannot take any current, hence our calculation is accurate (no loss of current). is it?

your welcome! and yes you got it right

 

[hino]

The filament of a 240 V, 100W electric lamp heats up from room temperature to its operating temperature. As it heats up, its resistance increases by a factor of 16. What is the resistance of this lamp at room temperature?

I=P/V
I=100/240
=0.416A
R=V/I
=240/0.416
=576Ω
since this resistance is 16 times the resistance that was at room temp
the resistance at RT=576/16
=36Ω

 

[Bluejeans#]

Impulse. Which is also delta momentum.

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf
kindly please explain Q: 35,34,32,29,27,26,25,22,20,26,25,13

 

[yousef]

hino

hello bros!
i have doubt in electricity questions ...(attached files) + this two questions
help will be appreciated


1__Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.
The springs are stretched, separately, by a force that is gradually increased from zero up to a
certain maximum value, the same for each spring. The work done in stretching spring P is WP,
and the work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP =
1/4 WQ B WP =1/2 [1] WQ C WP = 2WQ D WP = 4WQ
i get C ... but the answer is B


2___A wave of amplitude a has an intensity of 3.0Wm–2.
What is the intensity of a wave of the same frequency that has an amplitude 2a
A_12 .. i used I>A^2 and still not coming

 

[hino]

hino
h4rriet

hello bros!
i have doubt in electricity questions ...(attached files) + this two questions
help will be appreciated


1__Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.
The springs are stretched, separately, by a force that is gradually increased from zero up to a
certain maximum value, the same for each spring. The work done in stretching spring P is WP,
and the work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP =
1/4 WQ B WP =1/2 [1] WQ C WP = 2WQ D WP = 4WQ
i get C ... but the answer is B


2___A wave of amplitude a has an intensity of 3.0Wm–2.
What is the intensity of a wave of the same frequency that has an amplitude 2a
A_12 .. i used I>A^2 and still not coming

Q1 first we need to find out the extension for both the springs
force is the same for both the springs that is equal to 'F'
force for P F=2k*x
x=F/2k
force for Q F=kx
x=F/k
Wp=1/2*k*x^2=1/2*2k*(F/2k)^2
Wq=1/2*k*x^2=1/2*k*(F/k)^2
Wp/Wq=1/2*2k*(F/2k)^2÷1/2*k*(F/k)^2
Wp=1/2Wq Ans

 

[hino]

hino
h4rriet

hello bros!
i have doubt in electricity questions ...(attached files) + this two questions
help will be appreciated


1__Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.
The springs are stretched, separately, by a force that is gradually increased from zero up to a
certain maximum value, the same for each spring. The work done in stretching spring P is WP,
and the work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP =
1/4 WQ B WP =1/2 [1] WQ C WP = 2WQ D WP = 4WQ
i get C ... but the answer is B


2___A wave of amplitude a has an intensity of 3.0Wm–2.
What is the intensity of a wave of the same frequency that has an amplitude 2a
A_12 .. i used I>A^2 and still not coming

Q2 I=A^2
we ll be using the unitary method
if a^2=3
then 2a^2=I
I=12Wm-2 Ans

 

[hino]

hino

hello bros!
i have doubt in electricity questions ...(attached files) + this two questions
help will be appreciated


1__Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.
The springs are stretched, separately, by a force that is gradually increased from zero up to a
certain maximum value, the same for each spring. The work done in stretching spring P is WP,
and the work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP =
1/4 WQ B WP =1/2 [1] WQ C WP = 2WQ D WP = 4WQ
i get C ... but the answer is B


2___A wave of amplitude a has an intensity of 3.0Wm–2.
What is the intensity of a wave of the same frequency that has an amplitude 2a
A_12 .. i used I>A^2 and still not coming

Q36 for you to find the p.d you first need to calculate the voltage across both the points X and Y
current at X Ix=V/R
Ix=2/15=0.13
since current is same at all points in a series combination
voltage at Vx=0.13*5=0.66
current at Y Iy=2/15=0.13
Vy=0.13*10=1.33
now subtract the two voltages to get the potential difference
P.d=Vy-Vx
=1.33-0.66
=0.67=2/3 Ans

 

[hino]

hino

hello bros!
i have doubt in electricity questions ...(attached files) + this two questions
help will be appreciated


1__Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.
The springs are stretched, separately, by a force that is gradually increased from zero up to a
certain maximum value, the same for each spring. The work done in stretching spring P is WP,
and the work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP =
1/4 WQ B WP =1/2 [1] WQ C WP = 2WQ D WP = 4WQ
i get C ... but the answer is B


2___A wave of amplitude a has an intensity of 3.0Wm–2.
What is the intensity of a wave of the same frequency that has an amplitude 2a
A_12 .. i used I>A^2 and still not coming

Q35 increased resistance across variable resistor means more p.d across it leaving decreased p.d across the XY
as for the position of the movable contact i myself am not sure but i guess to maintain the galvanometer at zero we ll have to make use of the max. length of the wire so that the resistance would increase hence we ll position it near to Y

 

[yousef]

Q35 increased resistance across variable resistor means more p.d across it leaving decreased p.d across the XY
as for the position of the movable contact i myself am not sure but i guess to maintain the galvanometer at zero we ll have to make use of the max. length of the wire so that the resistance would increase hence we ll position it near to Y

Thanks ... ; thanks alot bro ...!

You cleared a Big doubt which i had for Weeks ....
q31 not dicussed .. i think its not clear??

 

[hino]

hino

hello bros!
i have doubt in electricity questions ...(attached files) + this two questions
help will be appreciated


1__Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.
The springs are stretched, separately, by a force that is gradually increased from zero up to a
certain maximum value, the same for each spring. The work done in stretching spring P is WP,
and the work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP =
1/4 WQ B WP =1/2 [1] WQ C WP = 2WQ D WP = 4WQ
i get C ... but the answer is B


2___A wave of amplitude a has an intensity of 3.0Wm–2.
What is the intensity of a wave of the same frequency that has an amplitude 2a
A_12 .. i used I>A^2 and still not coming

Q31 total resistance across the circuit =1/6+1/3 +2=4 ohm
current across the whole circuit=12/4=3A
following kirchoffs first law the current at the junction just before the 2Ω is going to be 3A
so the voltage across the 2Ω resistor =2*3=6V
the voltage across the parallel combination of resistors (6Ω and 3Ω) is going to be the rest of 6V
since the voltage is the same across the parallel combination,voltage across 6Ω resistor is equal to 6V
so the current across the 6Ω resistor is
I=6/6=1A Ans

 

[hino]

Thanks ... ; thanks alot bro ...!

You cleared a Big doubt which i had for Weeks ....
q31 not dicussed .. i think its not clear??

glad i was able to help

 

[hino]

i have posted the last one as well,was a bit hard to explain,hope you get it

 

[abrar]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s08_qp_1.pdf
can any one help me with Q6, Q13, Q14