Help!!

[cs]

 CIE physics paper 1

June 2003 Q5
November 2003 Q34
June 2004 Q 26 and 29
November 2004 Q 37( If possible, please show me the value of I1 and I2 when e.m.f is 12V)
June 2005 Q7
Thank you!!

[Deadly_king]

June 03 No 5
Accuracy = 3 %

3% of 327.66 is almost equal to 10. Therefore the value should be nearest 10th which is 330m/s.

Answer is C

Nov 03 No 34
Overall resistance in circuit X is 3.5 Ohm while in circuit Y it is 5.0 Ohm.
p.d of battery V = 1.5V
Let p.d across 3.0 Ohm resistor be V_a in circuit X and that in circuit Y V_b
V_a= R_a/R_total x V
Therefore for circuit X....... V_a = 3/3.5 x 1.5 = 1.29V
for circuit Y ....... V_b = 3/5 x 1.5 = 0.9V
1.29 > 0.9 -----> p.d in X is greater than p.d in Y across 3.0 Ohm resistor.

This eliminates answers C and D.

P = V²/R
For circuit X.......P = 1.29²/3 = 0.55W
For circuit Y.......P = 0.9²/3 = 0.27W
Therefore power dissipated in X is greater than that dissipated in Y across 3.0 Ohm resistor.

Answer is B.

June 04 No 26 and 29
 
26. Energy per unit time (E) = Power(P) = Intensity x Area

Intensity(I) being proportional to the square of the amplitude (A²)

P = IS = A²S
By doubling the amplitude the power is multiplied by 4 (2²) but since area is halved power is halved too.
Net power developed hence turns out to be twice the previous power (2E)

Answer is B



29. The acceleration of an electron is always opposite to the direction of the electric field.

Answer is D

Nov 04 No 37

Ok........i'll take the E.M.F to be 12V.

First calculate the overall resistance across the circuit. It will come out to 3.Ohm. (If you cant figure it out, let me know).
Now we can find the current flowing through the circuit using R = 3.0 and E.M.F = 12
I = V/R = 12/3 = 4.0A

I'll take the second loop first. Since both resistance are identical (2.0 Ohm) the current will just be divided by two according to Kirchoff's current law. Hence I₂ = 2.0A

Now back to the first loop. Voltage across both resistances is constant. Therefore I is inversely proportional to R.
I₁R₁ = I₂R₂
I₁/I₂ = R₂/R₁ = 6/3 = 2
In other words I₁ = 2I₂
Since I₁ + I₂ = 4.0 ------> I₁ turns out to be 2.67A.

Hence I₁ > I₂
Total resistance in loop 1 being greater will require more voltage. therefore V₁ > V₂

Answer is A

Jun 08 No 7

You need to plot a Velocity against time graph to be able to understand that. You'll see that you will obtain a straight line graph with a constant negative gradient as air resistance is being neglected.

Gradient = g = -9.81 m/s²

Answer is B

Hope i've been of help. If you have any doubt, do ask. I'll try to elaborate more.

[cs]

@Deadly_king

Thank you so much, i really appreciate it. I got the calculation for Nov 03, No 34 correct! i am so poor at that.. +rep!!

I am preparing for my AS this OCT and its school holiday in my country here, couldn't believe that you solved the questions so fast.

[cs]

I have another question, Nov 2005 Q 37 (the "light level" part)

[astarmathsandphysics]

When I get home.

[Deadly_king]

@Deadly_king

Thank you so much, i really appreciate it. I got the calculation for Nov 03, No 34 correct! i am so poor at that.. +rep!!

I am preparing for my AS this OCT and its school holiday in my country here, couldn't believe that you solved the questions so fast.

Its ok dear..........am actually sitting for A2 this october. so am bound to be more prepared than you

Try to practice more papers......am sure you'll perform well then!

[Deadly_king]

I have another question, Nov 2005 Q 37 (the "light level" part)

As light intensity increases the resistance of the LED decreases.

This is the only thing you need to remember about LED. Just memorize that and you'll be able to do every question about it

Since resistance of LED decreases the voltage across it will increase as voltage is inversely proportional to its resistance.

I believe you understood the temperature part.

Answer is C

[cs]

But deadly_king, why is the answer low light level..? as you said the voltage increase, when resistance decrease- where the light intensity increase.. shouldn't it be high light level..? its there something wrong with my english that i misunderstood the answer?

[cs]

Sorry to bother anyone, but i cannot get the answer for June 2006 33,34 and 37. Any help will be greatly appreciated..

[Deadly_king]

But deadly_king, why is the answer low light level..? as you said the voltage increase, when resistance decrease- where the light intensity increase.. shouldn't it be high light level..? its there something wrong with my english that i misunderstood the answer?

Actually i made a small mistake. I apologise for that. seems you understood well

V = IR.........Voltage is directly proportional to resistance.

let me clear your mind.

The questions ask when both readings on V_T and V_L are high.

When V_L is high.........the resistance of the LED should be high. For the resistance to be high the light intensity should be low.

[cs]

May i ask why "When VL is high.........the resistance of the LED should be high." ?

[Deadly_king]

May i ask why "When VL is high.........the resistance of the LED should be high." ?

This is because the voltage V_L is directly proportional to the resistance of the LDR.
The equation V = IR proves that.
Voltage = Current x Resistance

*Another mistake.......its LDR......not LED!

[Deadly_king]

Sorry to bother anyone, but i cannot get the answer for June 2006 33,34 and 37. Any help will be greatly appreciated..

The answers are as follows :

33. B

34. A

37. A

If you need explanations ........feel free to ask

[cs]

Thank you,

I need explanation for June 2009, Q 10, 21 and 29

June 2008, Q 6,14, 22,26, 30, 31 (why downwards)

I feel really bad for asking so many questions.

[The Golden Girl =D]

I feel really bad for asking so many questions.

 u shldn't feel bad cuz we help everyone who needs smthn