Saturday, September 19, 2015

Physics 9702 Doubts | Help Page 200

  • Physics 9702 Doubts | Help Page 200

Question 968: [Current of Electricity > Power]
A 100 Ω resistor conducts a current with changing direction and magnitude, as shown.

What is the mean power dissipated in the resistor?
A 100 W                     B 150 W                      C 250 W                      D 400 W

Reference: Past Exam Paper – November 2012 Paper 12 Q36

Solution 968:
Answer: C.
First, consider the period of the alternating current.
Look at the period just next to the y-axis. The period of the signal consists of a first section where the current I = 2A, followed by another section where I = -1A.

Power dissipated in a resistor = I2R
When current I = 2A (1st half of the cycle), Power dissipated = (2)2 × 100 = 400W
When current I = -1A (2nd half of the cycle), Power dissipated = (-1)2 × 100 = 100W

So, mean power = (400+100) / 2 = 250W

Question 969: [Kinematics > Graph]
The diagram shows a velocity-time graph.
What is the displacement during the last 2 seconds of the motion?
A 6 m              B 12 m                        C 18 m                        D 24 m

Reference: Past Exam Paper – November 2009 Paper 11 Q7 & Paper 12 Q6

Solution 969:
Answer: C.
In a velocity-time graph, the displacement is given by the area under the graph.

Displacement = area under graph for last 2s = area of trapezium
Displacement = ½ × sum of parallel sides × width = ½ × (6+12) × (2) = 18m

Question 970: [Radioactivity]
Uranium-236 (23692U) and Uranium-237 (23972U) are both radioactive.
Uranium-236 is an α-emitter and Uranium-237 is a β-emitter.
(a) Distinguish between an α-particle and a β-particle.

(b) The grid of Fig. 7.1 shows some proton numbers Z on the x-axis and the number N of neutrons in the nucleus on the y-axis.

The α-decay of Uranium-236 (23692U) is represented on the grid. This decay produces a nucleus of thorium (Th).
(i) Write down the nuclear equation for this α-decay.
(ii) On Fig. 7.1, mark the position for a nucleus of
1. Uranium-237 (mark this position with the letter U),
2. Neptunium, the nucleus produced by the β-decay of Uranium-237 (mark this position with the letters Np).

Reference: Past Exam Paper – June 2008 Paper 2 Q7

Solution 970:
α-particle: either helium nucleus         OR contains 2 protons + 2 neutrons   OR 42He
β-particle: either electron        OR 0-1e

Choose any 2 of the following:
The speed of α speed is less than that of β.
α-particles have discrete values of speed/energy while β-particles have a continuous spectrum.
EITHER The ionising power of α-particles is much greater than that of β-particles              OR The range of or α-particles is much less than that of β-particles.

(i) 23692U                 23290Th              +          42He
{Since it is an alpha decay, a helium nucleus must be produced. We can then calculate the corresponding proton and nucleon number of the other product.}

1. correct position for U at Z = 92, N = 145
{Since it is still uranium, the proton number would still be 92. Thus, the neutron number = 237 – 92 = 145}

2. correct position for Np relative to U i.e. Z + 1 and N – 1
{23792U                    23793Np             +          0-1e

Proton number = 93. Neutron number = 237 – 93 = 144}

Question 971: [Nuclear Physics]
Each of the nuclei below is accelerated from rest through the same potential difference.
Which one completes the acceleration with the lowest speed?
A 11H                           B 42He                                     C 73Li                         D 94Be

Reference: Past Exam Paper – June 2007 Paper 1 Q40

Solution 971:
Answer: C.
A force is needed to cause an acceleration.
Electric force F = Eq = (V/d) q = Vq / d
This should be equal to F = ma (the resultant force on the nuclei.).
So, ma = Vq/d giving a = Vq / md
Since V and d are constant for all particles, the acceleration a depends on q/m.

Hence, the acceleration would depend on the charge and mass of the particle. So, the q/m ratio is important. This ratio is also called the specific charge.

The particle that completes the acceleration with lowest speed is that one having the lowest value of this ratio.

The mass is the nucleon number while the charge is taken to be equal to the proton number (since we are dealing with nuclei, and not the atoms).

For 11H, ratio = 1/1 = 1
For 42He, ratio = 2/4 = 0.5
For 73Li, ratio = 3/7 ≈ 0.429
For 94Be, ratio = 4/9 ≈ 0.44


  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)
    04/O/N/08 Q.7(c)
    04/M/J/09 Q.11(b)(iii)
    41/O/N/09 Q.6(a),(b)(i),Q.10
    41/M/J/10 Q.6(a),Q.7(a)
    51/M/J/10 Q.2(d)
    41/M/J/11 Q.8(a)

    1. For 04/M/J/09 Q.11(b)(iii), see solution 973 at


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