Sunday, October 4, 2015

Physics 9702 Doubts | Help Page 206

  • Physics 9702 Doubts | Help Page 206

Question 992: [Alternating Current > Transformer]
A student is asked to design a circuit by which a direct voltage of peak value 9.0 V is obtained from a 240 V alternating supply.
The student uses a transformer that may be considered to be ideal and a bridge rectifier incorporating four ideal diodes.
The partially completed circuit diagram is shown in Fig. 6.1.

(a) On Fig. 6.1, draw symbols for the four diodes so as to produce the polarity across the load as shown on the diagram.

(b) Calculate the ratio
number of turns on the secondary coil / number of turns on the primary coil

Reference: Past Exam Paper – June 2010 Paper 41 Q6

Solution 992:

NS / NP = VS / VP
Peak voltage V0 = √2 × Vrms
{VS = 9.0 V. VP = √2 × 240 V since 240V is the Vrms of the primary alternating voltage}
ratio = 9.0 / (√2 × 240) = 1/38            or 1/37             or 0.027

Question 993: [Current of Electricity > Resistance]
A wire RST is connected to another wire XY as shown.

Each wire is 100 cm long with a resistance per unit length of 10 Ω m–1.
What is the total resistance between X and Y?
A 3.3 Ω                       B 5.0 Ω                       C 8.3 Ω                       D 13.3 Ω

Reference: Past Exam Paper – June 2015 Paper 11 Q38

Solution 993:
Answer: C.
Resistance per unit length of wire = 10 Ω m–1

Resistance of wire RST = 1m × 10 Ω m–1 = 10 Ω

Since the wire XY is 100cm, the sum of XR + TY = 100 – 50 = 50cm
Sum of resistance of (XR+TY) = 0.5m × 10 Ω m–1 = 5 Ω
Resistance of wire RT = 0.5m × 10 Ω m–1 = 5 Ω

Hence the arrangement is effectively a 10 Ω resistor RST in parallel with a 5 Ω resistor RT, and in series with another 5 Ω resistor (XR+TY).

Effective resistance of parallel combination = [1/10 + 1/5]-1 = 3.3 Ω
The total resistance is therefore 5 Ω + 3.3 Ω = 8.3 Ω.

Question 994: [Electric field]
A charged point mass is situated in a vacuum. A proton travels directly towards the mass, as illustrated in Fig. 4.1.

When the separation of the mass and the proton is r, the electric potential energy of the system is UP .
The variation with r of the potential energy UP is shown in Fig. 4.2.

(a) (i) Use Fig. 4.2 to state and explain whether the mass is charged positively or negatively.

(ii) The gradient at a point on the graph of Fig. 4.2 is G.
Show that the electric field strength E at this point due to the charged point mass is given by the expression
Eq = G
where q is the charge at this point.

(b) Use the expression in (a)(ii) and Fig. 4.2 to determine the electric field strength at a distance of 4.0 cm from the charged point mass.

Reference: Past Exam Paper – June 2012 Paper 42 Q4

Solution 994:
(a) (i) As the separation r decreases, the electric potential energy Up decreases / work got out due to attraction (attractive force) between the mass and the proton. So, the point mass is negatively charged.

Electric potential energy is the product of charge and electric potential. {Up = qV}
The electric field strength is the (electric) potential gradient (rate of change of potential with respect to displacement). {E = dV / dr – may remove the ‘d’s}
So, the electric field strength is equal to the gradient of the potential energy graph divided by charge.
{G = gradient = Up / r
Replace V = Up / q in E = dV / dr.
So, E = Up / qdr = G/q. Hence, Eq = G}

(b) Draw a tangent at the point (4.0, 14.5) obtaining a gradient of 3.6×10-24.                      
Field strength = (3.6×10-24) / (1.6×10-19) = 2.3×10-5 Vm-1

Question 995: [Dynamics > Resultant force]
A wooden block rests on a rough board. The end of the board is then raised until the block slides down the plane of the board at constant velocity v.

Which row describes the forces acting on the block when sliding with constant velocity?

frictional force on block          resultant force on block
A         down the plane                       down the plane
B         down the plane                       zero
C         up the plane                             down the plane
D         up the plane                             zero

Reference: Past Exam Paper – November 2008 Paper 1 Q12 & June 2013 Paper 13 Q13

Solution 995:
Answer: D.
Since the block slides down with constant velocity, the acceleration is zero. Thus, the resultant force on the block is zero. [A and C are incorrect]

Frictional force opposes motion. Since the block is moving downwards along the plane, the friction force would be up the plane.


  1. Please consider answering ALL of the following questions:
    4/O/N/02 Q.5(b),Q.6(c)(i)
    6/O/N/02 Q.11(a)(b)
    6/O/N/03 Q.9
    04/M/J/04 Q.8(a),(b)(i),(ii)1.
    06/M/J/04 Q.9(b)(iii),Q.11(b)
    06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)
    04/M/J/05 Q.7(a)
    06/O/N/05 Q.8(b),Q.10(a)
    04/M/J/06 Q.6(a),(c),Q.7(b)
    06/M/J/06 Q.14(b)
    04/O/N/06 Q.3(c)
    06/O/N/06 Q.3(b)
    05/M/J/07 Q.2(d)
    04/O/N/07 Q.7(b)(i),(c),Q.10(c)
    04/M/J/08 Q.5(b),Q.9(b),Q.11(a)(ii)
    41/O/N/09 Q.6(a),(b)(i),Q.10
    41/M/J/10 Q.7(a)
    51/M/J/10 Q.2(d)

    1. For 04/M/J/08 Q.11(a)(ii), see solution 997 at

  2. can u please answer june 2015 paper 11 question 15,18,23,25,26,32,33

    1. For Q15, see solution 1054 at

  3. This is so helpful!! Thanks a lot.

  4. Thank you for doing this bro really appreciate it


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