Sunday, October 11, 2015

Physics 9702 Doubts | Help Page 209

  • Physics 9702 Doubts | Help Page 209

Question 1004: [Current of Electricity > Resistance]
(a) On Fig. 5.1, sketch the temperature characteristic of a thermistor.

(b) A potential divider circuit is shown in Fig. 5.2.

The battery of electromotive force (e.m.f.) 12 V and negligible internal resistance is connected in series with resistors X and Y and thermistor Z. The resistance of Y is 15 kΩ and the resistance of Z at a particular temperature is 3.0 kΩ. The potential difference (p.d.) across Y is 8.0 V.
(i) Explain why the power transformed in the battery equals the total power transformed in X, Y and Z.
(ii) Calculate the current in the circuit.
(iii) Calculate the resistance of X.
(iv) The temperature of Z is increased.
State and explain the effect on the potential difference across Z.

Reference: Past Exam Paper – June 2015 Paper 22 Q5

Solution 1004:
(a) It is a curved line showing decreasing gradient with temperature rise. It be a smooth line not touching temperature axis, not horizontal or vertical anywhere.

(i) There is (no energy lost in battery because) no/negligible internal resistance

Current I = V / R
{The correct potential difference should be selected across the relevant resistor to obtain the correct circuit current. The current in the circuit is the same everywhere as it is a series connection.
Current through Y = p.d. across Y / resistance of Y = 8 / 15×103 (this is the easiest)
Current in complete circuit = e.m.f / total resistance in circuit. Here the resistance of Z should be found from the potential divider equation. The same goes for the other equivalent calculations – where we need to calculate additional quantities to find the current. So, you are recommended to use the above one.}
Current I = 8 / 15×103             or 1.6 / 3.0×103           or 2.4 / 4.5×103           or 12 / 22.5×103
Current I = 0.53 × 10–3 A

{p.d. across X = 12 – p.d. across Y – p.d. across Z  
p.d. across Y = 8.0 V and p.d. across Z = IR = 3.0×103 × 0.53×10–3}
p.d. across X = 12 – 8.0 – (3.0×103 × 0.53×10–3)       (= 2.4 V)
RX {= V / I} = 2.4 / (0.53 × 10–3)

{Here, we find the total resistance first and from there, calculate the resistance of X.}
Rtot = 12 / 0.53 × 10–3 (= 22.5 × 103 Ω)
RX = (22.5 – 15.0 – 3.0) × 103

RX = 4.5(2) × 103 Ω

The resistance decreases and hence, the current (in the circuit) is greater. This causes the p.d. across X and Y to be greater, hence the p.d across Z decreases.

OR explanation in terms of potential divider:
RZ decreases, so RZ / (RX + RY + RZ) is less. Therefore the p.d. across Z decreases.

Question 1005: [Electromagnetism]
Negatively-charged particles are moving through a vacuum in a parallel beam. The particles have speed v.
The particles enter a region of uniform magnetic field of flux density 930 μT. Initially, the particles are travelling at right-angles to the magnetic field. The path of a single particle is shown in Fig. 7.1.

The negatively-charged particles follow a curved path of radius 7.9 cm in the magnetic field.
A uniform electric field is then applied in the same region as the magnetic field. For an electric field strength of 12 kV m–1, the particles are undeviated as they pass through the region of the fields.
(a) On Fig. 7.1, mark with an arrow the direction of the electric field.

(b) Calculate, for the negatively-charged particles,
(i) the speed v,
(ii) the ratio     charge / mass  

Reference: Past Exam Paper – June 2010 Paper 41 Q7

Solution 1005:
(a) The arrow should be pointing up the page
{The direction of an electric field is from positive to negative – that is, it shows the direction of the electric force on a positive charge. For the particle to pass undeviated, the electric force should act towards on it. Since this is a negatively-charged particle, the ‘positive’ should be down (so that the particle is attracted to it). Thus, the arrow points vertically upwards.}

{The magnitude of the electric force (= Eq) is equal to the magnitude of the magnetic force (= Bqv).}
Eq = Bqv
Speed v = (12 × 103) / (930 × 10–6)
Speed v = 1.3 × 107 m s–1

{The centripetal force is provided by the force on the charged particles due to the magnetic field.}
Bqv = mv2 / r  
q/m = (1.3 × 107) / (7.9 × 10–2 × 930 × 10–6)  
q/m = 1.8 × 1011 C kg–1

Question 1006: [Waves]
Light reflected from the surface of smooth water may be described as a polarised transverse wave.
(a) By reference to the direction of propagation of energy, explain what is meant by
(i) a transverse wave,
(ii) polarisation.

(b) A glass tube, closed at one end, has fine dust sprinkled along its length. A sound source is placed near the open end of the tube, as shown in Fig. 5.1.

The frequency of the sound emitted by the source is varied and, at one frequency, the dust forms small heaps in the tube.
(i) Explain, by reference to the properties of stationary waves, why the heaps of dust are formed.

(ii) One frequency at which heaps are formed is 2.14 kHz.
The distance between six heaps, as shown in Fig. 5.1, is 39.0 cm.
Calculate the speed of sound in the tube.

(c) The wave in the tube is a stationary wave. Explain, by reference to the formation of a stationary wave, what is meant by the speed calculated in (b)(ii).

Reference: Past Exam Paper – June 2007 Paper 2 Q5

Solution 1006:
(i) The vibrations (in the plane) is normal to the direction of energy propagation

(ii) The vibrations are in one direction (normal to direction of propagation)

(i) At (displacement) antinodes / where there are no heaps, the wave has maximum amplitude (of vibration). At (displacement) nodes / where there are heaps, the amplitude of vibration is zero/minimum. The dust is pushed to / settles at (displacement) the nodes.

{From the diagram,} 2.5λ = 39 cm
Speed v = fλ
Speed v = 2.14 × 103 × 15.6 × 10-2
Speed v = 334 m s-1

(c) The stationary wave is formed by interference / superposition / overlap of
EITHER wave travelling down tube and its reflection          OR or two waves of same (type and) frequency travelling in opposite directions.
Speed is the speed of the incident / reflected waves.

Question 1007: [Current of Electricity]
A box with four terminals P, Q, R and S contains two identical resistors.

When a battery of electromotive force (e.m.f.) E and negligible internal resistance is connected across PS, a high-resistance voltmeter connected across QR reads E/2.
Which diagram shows the correct arrangement of the two resistors inside the box?

Reference: Past Exam Paper – June 2015 Paper 13 Q36

Solution 1007:
Answer: A.
One method to answer this question would be to draw the battery and voltmeter in each of the four circuits.

The reading across QR is E / 2. This means that half of the resistance is across QR. [B and C are incorrect]
In circuit B, the voltmeter reading would be E since both resistors are found between the terminals of the voltmeter (S and R are at the same potential in this case).
In this circuit C, a wire connects Q and R so these points must be at the same potential, and a voltmeter between these points will read zero.

Circuit D is not a complete circuit, so the reading would be zero. In circuit A, the voltmeter is indeed connected across one of the 2 resistors and the circuit is complete. The reading would then be E / 2.

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