Physics: Post your doubts here!

[JD REBORN]

6B.

 

[leosco1995]

q)35 of w02 p1,Q)35 of s04 p1,Q)37 of s07) p1,Q)32 and Q)37 of w08 p1,q)34 and q)36 of s11 p1.

I'll help you out with the S11 P1 ones (assuming you mean P12).

Q34: As soon as I finished reading the question, I thought of the formula V = Ir + IR (r being internal resistance, R being external resistance, I being the current and V being the EMF). Now, if you increase R, you are increasing the EMF according to the above equation. This means the answer is without a doubt C.

Q36:
A: If P is closed and Q is closed, then the current is going to reach the green lamp. You don't even need to look at R/S/T.
B: This time Q is open, so the current won't go through the PQ route as in A. However, the current also flows through point R. Since R is closed, if either S or T is also closed, the current will go through the lamp. Since S is closed, the current will take the RS route and reach the green lamp.
C: Q is open, so current won't flow through the PQ route.. R is also open, so basically the current won't even reach the green lamp at all. Hence, C is correct.
D: PQ route is open, but R is closed and T is also closed, so the current will flow through the RT route and reach the green lamp.

 

[JD REBORN]

I'll help you out with the S11 P1 ones (assuming you mean P12).

Q34: As soon as I finished reading the question, I thought of the formula V = Ir + IR (r being internal resistance, R being external resistance, I being the current and V being the EMF). Now, if you increase R, you are increasing the EMF according to the above equation. This means the answer is without a doubt C.

Q36:
A: If P is closed and Q is closed, then the current is going to reach the green lamp. You don't even need to look at R/S/T.
B: This time Q is open, so the current won't go through the PQ route as in A. However, the current also flows through point R. Since R is closed, if either S or T is also closed, the current will go through the lamp. Since S is closed, the current will take the RS route and reach the green lamp.
C: Q is open, so current won't flow through the PQ route.. R is also open, so basically the current won't even reach the green lamp at all. Hence, C is correct.
D: PQ route is open, but R is closed and T is also closed, so the current will flow through the RT route and reach the green lamp.

sorry but i meant 11

 

[leosco1995]

No problem.

36. The power to X is going to increase for sure, so rule out A & B. Why? Because when W is removed, more current will flow to X and hence the power there increases.

Now, when W is removed, the resistance of the first block (W + X) increases. This means the entire circuit will now have more resistance, and hence less current. Less current means there will be less power in Y and Z. So the answer should be D.

34. You can calculate the resistance using (ρL/A). The length will be the cube root of the cube (V^1/3) and the area will be square of the length (so V^2/3) . Plug in the values and you will get the answer C.

 

[unique840]

6B.

first find out the resistance of 1 resistor. (230^2) / 1000 = 52.9 ohms

i) in parallel, the total resistance is 1/R = 1/52.9 + 1/52.9 = 26.45 ohms
power lost = V^2 / R
= (230^2) / 26.45 ohms
= 2000 W
= 2kW

ii) in series, the total resistance is R1 + R2 = 52.9 + 52.9 = 105.8 ohms
power = (230^2) / 105.8 = 5oo W
= 0.5 kW

iii) 1 in series and 2 in parallel so total resistance will be 26.45 + 52.9 = 79.35 ohms
power = (230^2) / 79.35 = 666.67 W
= 0.667 kW

 

[JD REBORN]

first find out the resistance of 1 resistor. (230^2) / 1000 = 52.9 ohms

i) in parallel, the total resistance is 1/R = 1/52.9 + 1/52.9 = 26.45 ohms
power lost = V^2 / R
= (230^2) / 26.45 ohms
= 2000 W
= 2kW

ii) in series, the total resistance is R1 + R2 = 52.9 + 52.9 = 105.8 ohms
power = (230^2) / 105.8 = 5oo W
= 0.5 kW

iii) 1 in series and 2 in parallel so total resistance will be 26.45 + 52.9 = 79.35 ohms
power = (230^2) / 79.35 = 666.67 W
= 0.667 kW

love u thnx

 

[unique840]

love u thnx

ur welcum

 

[Ahmed Ali Akbar]

Please help me out in Question 35,,17 and 22...of oct/nov 10 Paper13.answer for 17 is D , D for 22,,,

 

[omg]

anybody having papers of 2000 and 2001?????? even older 1999 or 1998?????? pls if u do give me the linkkk !!!! p4!!

 

[JD REBORN]

Please help me out in Question 35,,17 and 22...of oct/nov 10 Paper13.answer for 17 is D , D for 22,,,

For 35 when slider is at X voltmeter will read 4V.When it is at Y,if u see the diagram closely u will see that the slider is connected to battery.So we will have to take potential at Y as 4V.It would have been B if the voltmeter was in place of battery.I havent studied syllabus relating to 22.

 

[abcde]

28 of http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s02_qp_1.pdf
Can't figure out how to express the distance between the screen and the slits. :S

 

[regmimanish]

help needed ? notes for opamp ?? plzz

 

[Nibz]

28 of http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s02_qp_1.pdf
Can't figure out how to express the distance between the screen and the slits. :S

Aoa!

As we move from O, the signal detector decreases, i.e, at O there is constructive interference (Maxima)
At X there is minima (destructive interference)
path difference of X from S2 & S1 is given by ( n + 1/2 ) lambda
At X, n = 0
So ( 0 + 1/2 ) lambda = lambda/2
S2X - S1X = lambda / 2

Ans C.

Hope this makes sense.

 

[abcde]

Aoa!

As we move from O, the signal detector decreases, i.e, at O there is constructive interference (Maxima)
At X there is minima (destructive interference)
path difference of X from S2 & S1 is given by ( n + 1/2 ) lambda
At X, n = 0
So ( 0 + 1/2 ) lambda = lambda/2
S2X - S1X = lambda / 2

Ans C.

Hope this makes sense.

W.S!
Thank you! :]

 

[Ahmed Ali Akbar]

For 35 when slider is at X voltmeter will read 4V.When it is at Y,if u see the diagram closely u will see that the slider is connected to battery.So we will have to take potential at Y as 4V.It would have been B if the voltmeter was in place of battery.I havent studied syllabus relating to 22.

Thanks,.but could you please explain me question 17 of this paper...

 

[unique840]

Thanks,.but could you please explain me question 17 of this paper...

total power at constant speed = f(total) * v
f1v1 - f2v2 = total power
m1 g v - m2 g v = power
gv(m1-m2) = total power

 

[Ahmed Ali Akbar]

thanks..a lot for clearing my concept about power at constant speed/......))

 

[unique840]

ur welcum

 

[Ahmed Ali Akbar]

need help in mcq no 20 of may june 07 paper 1...

 

[CaptainDanger]

need help in mcq no 20 of may june 07 paper 1...

Inner edge is always in compression and outer edge in tension in such cases... Maybe someone else has a better explanation to it...