Sunday, September 7, 2014

9702 November 2005 Paper 2 Worked Solutions | A-Level Physics

  • 9702 November 2005 Paper 2 Worked Solutions | A-Level Physics

Paper 2

Question 1
Pressure is defined as the force per unit area

Base units of pressure:

Pressure p at depth h in an incompressible fluid of density ρ is given by
p = ρgh            where g is acceleration of free fall
Use base units to check homogeneity of equation:
Base unit of ρ: kgm-3
Base unit of g: ms-2
So, base unit of hρg: (m)(kgm-3)(ms-2) = (kgm-1s-2)
This is the same unit as pressure

Question 2

{Detailed explanations for this question is available as Solution 669 at Physics 9702 Doubts | Help Page 134 -}


Question 3

Stone on a string is made to travel along horizontal circular path, as shown. Stone has constant speed.


Acceleration is defined as the change in velocity per time (taken)


Explain whether stone is accelerating:

Velocity is a vector / Velocity has both a magnitude and a direction. Since the direction is changing, the stone must be acceleration.


Stone has weight of 5.0N. When string makes an angle of 35o to vertical, tension in string = 6.1N, as illustrated.

Resultant force acting on stone in position shown:
Vertical component: 6.1cos(35) = 4.99N
So, it has no resultant vertical force (since weight = 5.0N)
Other component: 6.1sin(35) = 3.5N horizontally

A scale is shown. A triangle of correct shape is drawn. Resultant = 3.5 ± 0.2N Horizontal ± 3o.

Question 4

{Detailed explanations for this question is available as Solution 1024 at Physics 9702 Doubts | Help Page 214 -}

 Question 5
{Detailed explanations for this question is available as Solution 975 at Physics 9702 Doubts | Help Page 201 -}

Question 6

2 horizontal metal plates X and Y are at distance 0.75cm apart. Positively charged particle of mass 9.6x10-15kg situated in vacuum between plates, as illustrated. Potential  difference between plates adjusted until particle remains stationary.


Which plate, X or Y, is positively charged:

The electric force must be upwards (on the positive charge). So, plate Y is positive.


Potential difference required for particle to be stationary between plates is found to be 630V.


Electric field strength between plates:

E = V / d = 630 / (0.75x10-2) = 8.4x104NC-1

Charge on particle:
(Electric force = Gravitational force)
qE = mg
(g used to 1 dp since other values also given to 1dp)
q = mg / E = [(9.6x10-15) x 9.8] / (8.4x104) = 1.12x10-18C


Question 7

{Detailed explanations for this question is available as Solution 911 at Physics 9702 Doubts | Help Page 186 -}

Question 8

{Detailed explanations for this question is available as Solution 1087 at Physics 9702 Doubts | Help Page 230 -}


  1. hi sir i was wondering what the voltmeter in question 7 measures ? is it the voltage across cp ?

    1. It measures the potential difference between points B and M. Note that points C and P may be considered to be the same since there is no component across them.

      I should remind you though that wire PQ is not the same as the other wires since it has some resistance, unlike the others, whose resistances are assumed to be negligible.

    2. Thanks but im still a bit confused
      how does the voltmeter show 0 reading and how is the potential difference across M and Q 2.7 V?

    3. Firstly, note the difference between 'potential at a point' and 'potential difference between 2 points'. A voltmeter measures the potential difference between 2 points (This is easy to understand: if only one end of the voltmeter is connected, no reading is obtained).

      The voltmeter shows zero when the potentials at B and M are the same. That is, the potential DIFFERENCE between them is zero.

      Wire PQ has some resistance. The circuit is a potential divider circuit, that is, the wire is connected in such a way that only the length of wire from one end to point M is considered. Since the resistance of the wire depends on the length, the resistance of this part of the wire being considered is NOT equal to the total resistance of the wire.

      Since the voltmeter is showing a reading of zero, it means that the potential difference is zero. That is, the difference in potentials at B and M is zero. So, the potentials at B and M is not difference. They are the same. So, potential at M is 2.7V.

      Note that since points A and Q are connected to the negative terminal of the supply, they are both at zero potentials. So, point Q is at a potential of 0V and point M is at a potential of 2.7V. Therefore, potential difference between Q and M is 2.7V.

      Be sure to understand the terms 'potentials' and 'potential difference'.

    4. okay sir does that mean that the potential difference between P and M is 4.5 - 2.7 ?
      and can you please explain the next question, that is, The thermistor is warmed slightly. State and explain the effect on the length of wire
      between M and Q for the voltmeter to remain at zero deflection.

    5. Yes, the p.d. between P and M is (4.5 - 2.7).

      As for the next part, as stated above, the resistance of the thermistor decreases as the temperature increases. From the potential divider equation (V1 = [R1 / (R1+R2)] Vin), the p.d. across the thermistor decreases. This causes the p.d. across the 1200 ohm resistor to greater.

      A greater p.d. across the 1200 ohm resistor means that the potential at B is smaller than previously.

      So, for zero deflection, the potential at M should also be smaller. From the same logic as for the resistor, the p.d. across P and M should thus be greater. From Ohm's law: potential difference V = IR and the resistance of a wire is proportional to the length. So, the length PM should be greater for a greater p.d. across P and M. Therefore, the length of wire between Q and M is smaller,

  2. Can a I get a broader explanation of 2b)?

  3. can you please add may june 5 question numer 6 b i

  4. add question number 6 b i may june 05 paper2

    1. Check solution 28 at

  5. In Question # 7, b (iii), why is it that the distance between M and Q gets shorter, why not longer? How can we differentiate which end of the wire M should get closer to, why does it come closer to Q and not P?

    1. The solution has been updated. Check again.

      You are also welcomed to go through the first set of comments above. Someone asked about this before.

  6. could u pls add june 2005 q25 and q29

    1. For Q25, check solution 921 at

  7. how did the intensity be (1/9)I?
    where did 4 go? and
    isn't the resultant intensity 1*10^-8?
    very confused!! :(

  8. how is resultant intensity (1/9)I?
    i am confused because resultant intensity is 1*10^-8

  9. I don't understand question 5
    how to calculate proportion? please add full solution.

  10. can q4 part a be explained in a little more detail?
    how can we take a tangent at zero m/s? how can that give initial acceleration?

  11. can u explain q8 (ii)

  12. can u please explain q8 b (ii)the graph part

    1. The solution has been updated. See again.

    2. thanks I understood that.Can u explain like why is kinetic energy graph a curve in this question.

    3. As the ball falls through the air, its initial acceleration is equal to the acceleration of free fall. But due to air resistance, the resultant acceleration decreases until a terminal velocity is reached.

      EK depends on the velocity. Let discuss the motion in terms of velocity. Initially, the velocity increases, so EK increases. The rate of increase of velocity (acceleration) decreases, so the increase in EK becomes less as the ball falls. Terminal velocity is a constant velocity, so EK is also constant.

  13. can u do may june 2005 paper 2 q5 (b)

    1. See solution 1099 at


If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | 9702 November 2005 Paper 2 Worked Solutions | A-Level Physics