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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 Va in circuit X and that in circuit Y Vb
Va= Ra/Rtotal x V
Therefore for circuit X....... Va = 3/3.5 x 1.5 = 1.29V
for circuit Y ....... Vb = 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.

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