Physics Paper 2, May/June. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
Weakness/Remedies
Strength

Question 14

(a)

When a positively charged conductor is placed near a candle flame, the flame spreads out as shown in the diagram above.
Explain this observation.

(b)A proton moving with a speed of 5.0 x 105 ms-1; enters a magnetic field of flux density 0.2 T at an angle of 30° to the field. Calculate the magnitude of the magnetic force exerted on the proton.[proton charge = 1.6X 10-19C]

(c)

)

The diagram above illustrates 9.0 V battery of internal resistarIce 0.5Ωconnected to two resistors of values 2.0Ωand R Ω A1,A2 and A3, are ammeters of negligible internal resistances. If A1 reads 4.0 A, calculate the:
(i) equivalent resistance of the combined resistors 2.0Ω and RΩ
(ii) currents through A2 and A3;
(iii) value of R.

Observation

Part (a). Poor responses were given to the expected explanation of the observed spreading out of the candle flame.
Part (b). Correct responses were quite high
Part (c). Generally, candidates did not show enough knowledge of what was required and the attendant mathematical computation due to poor interpretations of the given circuit diagram.
14. (a)The candle flame ionizes the air around it. .
The positively charged conductor attracts the negative charges in the air and repels the positive charges.

(b)F = qvB Sin θ   .
= 1.6 X 10-19 X 5.0 X 105 X 0.2 X Sin 30°
= 0.8 X 10-1

(c)(i)E = I (Rc + r)
9 = 4(Rc + 0.5)
R: = 9/4-0.5
= 1.75Ω
(ii) Lost volt = Ir = 4.0 X 0.5 =2.0 V
V oltage across 2Ωand RΩ
V = 9.0-2.0 =7.0V
Current in A2 = I2 = V/R2= 7/2 = 3.5A
Current in A3 =I3 = V/R = 7/14 = 0.5A
OR
A3 = A1 - A2
= 4.0 - 3.5 = 0.5A
(iii) 1/RC = 1/R + 1/R2
1/1.75 = 1/R + 1/2
1/R = 1/1.75- 1/2
R = 14Ω