Sunday, April 10, 2016

Physics 9702 Doubts | Help Page 237

  • Physics 9702 Doubts | Help Page 237

Question 1106: [Current of Electricity > Internal resistance]
A 12 V battery with internal resistance 0.50 Ω is connected to two identical filament lamps L1 and L2 as shown in Fig. 6.1.

Fig. 6.1
The lamps are connected to the battery via switches S1 and S2. The power rating of each lamp is 48 W for a potential difference of 12 V.
(a) S1 is closed and S2 open.
State and explain whether the power transformed in L1 is 48 W.

(b) S2 is now also closed.
(i) State and explain the effect on the current in L1.
(ii) State and explain the effect on the resistance of L1.

Reference: Past Exam Paper – November 2015 Paper 22 Q6

Solution 1106:
Internal resistance causes lost volts. The p.d. across lamp is less than 12 V, so, the power is less than 48 W
{There will be ‘lost volts’ due to internal resistance of the battery and thus the terminal p.d. will be below 12 V. Power (= VI) is also less than 48 W.}

(i) There is greater lost volts OR the p.d. across the cell/lamp is reduced, there is less current in the lamp.
{Lost volts = Ir           where I is the current flowing in the circuit and r is the internal resistance of the battery. By closing S2, the overall resistance in the circuit decreases (since the lamps are in parallel) and so, the current flowing in the circuit increases (I = V/R). Since current I has increased, there is greater lost volts (= Ir) as more current is supplied by the battery. This would result in the terminal p.d. being reduced and the power output of each lamp to be less.}

(ii) The p.d. across lamp/current in lamp decreases, hence resistance decreases
{We need to apply the I-V characteristics of a lamp to this circuit. For a filament lamp, the resistance is small when the current is small.}

Question 1107: [Nuclear Physics > Fusion]
One likely means by which nuclear fusion may be achieved on a practical scale is the D-T reaction.
(a) State what is meant by nuclear fusion.

(b) In the D-T reaction, a deuterium (21H) nucleus fuses with a tritium (31H) nucleus to form a helium-4 (42He) nucleus. The nuclear equation for the reaction is
21H       +          31H      - - - >               42He     +          10n        +          energy
Some data for this reaction are given in Fig. 9.1.
mass / u
deuterium (21H)                       2.01356
tritium (31H)                            3.01551
helium-4 (42He)                       4.00151
neutron (10n)                            1.00867
Fig. 9.1

(i) Calculate the energy, in MeV, equivalent to 1.00 u. Explain your working.

(ii) Use data from Fig. 9.1 and your answer in (i) to determine the energy released in this D-T reaction.

(iii) Suggest why, for the D-T reaction to take place, the temperature of the deuterium and the tritium must be high.

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q9

Solution 1107:
(a) In nuclear fusion, ‘light’ nuclei combine to form a ‘heavier’ nuclei

{u = 1.66×10–27 kg}
EITHER energy = c2Δm                     OR energy = (3.00 × 108)2 × 1.66×10–27
energy = 1.494 × 10–10 J
{To convert J into eV, we divide by the charge of an electron (1.60 × 10–19 C). To convert eV to MeV, we divide by 106. Thus, to convert J to MeV, we divide by (1.60 × 10–19 × 106 = 1.60 × 10–13).}
Energy = (1.494 × 10–10) / (1.60 × 10–13)
Energy = 934 MeV (3 s.f.)     

{Δm = (mass of deuterium + mass of tritium) – (mass of helium-4 + mass of neutron)}
Δm = (2.01356 + 3.01551) – (4.00151 + 1.00867)
Δm = 5.02907 – 5.01018
Δm = 0.01889 u
{1 u = 934 MeV}
Energy = 0.01889 × 934
Energy = 17.6 MeV

(iii) A high temperature means high speeds / kinetic energy of the nuclei. Deuterium and Tritium nuclei could collide despite repelling one another.
{There would be repulsion between the deuterium and the tritium nuclei. So, the kinetic energy of these nuclei would need to be very high so that the nuclei could approach one another and thus fusion could take place.}

Question 1108: [Current of Electricity > Charge]
A 12 V battery is charged for 20 minutes by connecting it to a source of electromotive force (e.m.f.). The battery is supplied with 7.2 × 104 J of energy in this time.
How much charge flows into the battery?
A 5.0 C                       B 60 C                         C 100 C                       D 6000 C

Reference: Past Exam Paper – June 2009 Paper 1 Q31

Solution 1108:
Answer: D.
Power P = VI              and I = charge / time = Q / t
Power = Energy / time

Energy / t = VQ / t
12 × Q = 7.2 × 104
Amount of charge Q = 7.2 × 104 / 12 = 6000 C

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