Thursday, May 14, 2015

Physics 9702 Doubts | Help Page 141

  • Physics 9702 Doubts | Help Page 141



Question 695: [Sensing devices]
(a) State name of an electrical sensing device that will respond to changes in
(i) length
(ii) pressure

(b) Relay is sometimes used as the output of a sensing circuit.
Output of particular sensing circuit is either +2V or -2V.
On Fig.1, draw symbols for relay and any other necessary component so that the external circuit is switched on only when the output from the sensing circuit is +2V.






Reference: Past Exam Paper – June 2010 Paper 42 & 43 Q10

Solution 695:
(a)
(i) Strain gauge
(ii) Piezo-electric / quartz crystal / transducer

(b) For the circuit, a coil of relay is connected between the sensing circuit output and earth and the switch is across the terminals of the external circuit.
A diode is connected in series with the coil with a correct polarity for the diode (downwards). A second diode with correct polarity is connected (upwards and parallel to coil of relay).
{Note that the following explanations have been taken directly from the application booklet.
The diode D1 conducts only when the output is positive with respect to earth and thus the relay coil is energized only when the output is positive. When the current in the relay coil is switched off, a back e.m.f. is generated in the coil that could damage the sensing circuit. A diode D2 is connected across the coil to protect the sensing circuit from this back e.m.f.}









Question 696: [Current of Electricity]
(a) A wire has length 100 cm and diameter 0.38 mm. Metal of the wire has resistivity 4.5 × 10–7 Ω m.
Show that resistance of the wire is 4.0 Ω.

(b) The ends B and D of the wire in (a) are connected to a cell X, as shown in Fig.1.

Cell X has electromotive force (e.m.f.) 2.0 V and internal resistance 1.0 Ω.
A cell Y of e.m.f. 1.5 V and internal resistance 0.50 Ω is connected to the wire at points B and C, as shown in Fig.1.
Point C is distance l from point B. The current in cell Y is zero. Calculate
(i) current in cell X,
(ii) potential difference (p.d.) across the wire BD,
(iii) distance l.

(c) Connection at C is moved so that l is increased. Explain why the e.m.f. of cell Y is less than its terminal p.d.

Reference: Past Exam Paper – November 2014 Paper 23 Q6



Solution 696:
(a)
Resistance R = ρl / A
Cross-sectional area, A = [π × (0.38 × 10–3)2] / 4       (= 0.113 × 10–6 m2)
Resistance R = (4.5 × 10–7 × 1.00) / ([π × (0.38 × 10–3)2] / 4) = 4.0 (3.97) Ω

(b)
(i) Current І = V / R = 2.0 / 5.0 = 0.4(0) A

(ii) p.d. across BD {= IR} = 4 × 0.4 = 1.6 V

(iii)
{Since the current in cell Y is zero, the p.d. across BC should be equal to the e.m.f. of cell Y, that is 1.5 V. BC is a distance l.}
p.d. across BC (l) = 1.5 (V)
{For a wire, the resistance is proportional to the length. R = ρl / A. The p.d. across the wire is V = IR, so the p.d. V is proportional to the resistance of the wire, which is itself proportional to the length.
100cm of wire has a p.d. of 1.6V across it.
l cm of wire has a p.d. of (l / 100) × 1.6 V across it. But, as deduced above, it is equal to 1.5V. So, the distance l can be obtained.}
BC (l) = (1.5 / 1.6) × 100 = 94 (93.75) cm

(c) {This question has been DISCOUNTED from the paper. I believe the correct question should have been ‘Explain why the e.m.f. of cell Y is greater than its terminal p.d.’}
{As the distance l is increased, the p.d. across BC is no longer equal to (it is greater than) the e.m.f.of the cell Y. So, a current flows in the cell Y. Since the cell Y has some internal resistance, there will be some lost volts inside the cell. So, the p.d. across its terminal would be less than the e.m.f.}
The p.d. across wire not balancing e.m.f. of cell OR the cell Y has current. So, energy is lost or lost volts due to internal resistance.










Question 697: [Measurement > Prefixes]
A signal has frequency of 2.0 MHz.
What is period of the signal?
A 2 μs                         B 5 μs                         C 200 ns                                  D 500 ns

Reference: Past Exam Paper – November 2010 Paper 11 Q1 & Paper 13 Q4



Solution 697:
Answer: D.
Frequency f = 2.0 MHz = 2.0×106 Hz

Period = 1 / f = 1 / (2.0×106) = 0.5×10-6 = 500×10-9 = 500ns









Question 698: [Kinematics + Dynamics]
A helicopter has a cable hanging from it towards the sea below, as shown in Fig.1.

A man of mass 80 kg rescues child of mass 50.5 kg. The two are attached to the cable and are lifted from sea to the helicopter. Lifting process consists of an initial uniform acceleration followed by a period of constant velocity and then completed by a final uniform deceleration.
(a) Calculate combined weight of the man and child.

(b) Calculate tension in the cable during
(i) initial acceleration of 0.570 m s–2,
(ii) period of constant velocity of 2.00 m s–1.
                                                      
(c) During final deceleration the tension in the cable is 1240 N. Calculate this deceleration.

(d) (i) Calculate time over which the man and child are
1. moving with uniform acceleration,
2. moving with uniform deceleration.

(ii) Time over which the man and child are moving with constant velocity is 20 s. On Fig.2, sketch a graph to show variation with time of the velocity of the man and child for the complete lifting process.

Reference: Past Exam Paper – June 2011 Paper 23 Q3



Solution 698:
(a) {Combined mass = 80 + 50.5 = 130.5kg}
Combined weight = m × g = 130.5 × 9.81 = 1280 N

(b)
(i)
Resultant force F = ma
{Tension is upwards while the weight acts downwards. The resultant force / acceleration is upwards.}
T – 1280 = 130.5 × 0.57
Tension T = 1280 + 74.4 = 1350 N

(ii) Tension T = 1280 N
{For constant velocity, the resultant force is zero. Thus, tension = weight.}
                                                      
(c)
{Resultant force F = ma
For a deceleration, the direction of the resultant force should be opposite to the direction of motion. Motion is upwards, so the resultant force is downwards.
Tension – Weight = Resultant force = ma      (a is negative here)}
1240 – 1280 = 130.5 × a
a = (–) 0.31 m s–2

(d) (i)
1.
{Initial acceleration a = 0.570 m s–2. Initial velocity, u = 0. Final velocity, v = 2.00 m s–1.
a = (v – u) / t}
Time t {= (v – u) / a = 2.00 / 0.570} = 3.5s

2.
{Deceleration a = - 0.31 m s–2. Initial velocity, u = 2.00ms-1. Final velocity, v= 0.
a = (v – u) / t}
Time t {= (v – u) / a = -2.00 / -0.31} = 6.5s

(ii) The graph has a basic shape with the correct points.










Question 699: [Alternating Current]
An ideal transformer has 5000 turns on its primary coil. It is to be used to convert mains supply of 230 V r.m.s. to an alternating voltage having a peak value of 9.0 V.
(a) Calculate number of turns on the secondary coil.

(b) Output from the transformer is to be full-wave rectified. Fig.1 shows part of the rectifier circuit.




On Fig.1, draw
(i) diode symbols to complete diagram of the rectifier such that terminal A of the resistor R is positive with respect to terminal B,
(ii) symbol for a capacitor connected to provide smoothing of potential difference across the resistor R.

(c) Fig.2 shows variation with time t of the smoothed potential difference V across the resistor R.



(i) State interval of time during which the capacitor is being charged from the transformer.
(ii) Resistance of the resistor R is doubled. On Fig.2, sketch variation with time t of the potential difference V across the resistor.

Reference: Past Exam Paper – June 2007 Paper 4 Q4



Solution 699:
(a)
r.m.s. output = 9/√2 or peak input = 230√2
NS / NP = VS / VP
{NS / 5000 = (9/√2) / 230        or NS / 5000 = 9 / 230√2}
NS = 138 → 140 turns

(b)
(i) The four diodes should be correctly positioned (regardless of output polarity) giving correct output polarity (all ‘point to left’)
{The diodes should be drawn such that current always passes point A first, before passing point B.}

(ii) The capacitor should be shown in parallel with R


 




(c)
(i) time t1 to time t2

(ii) The sketch has the same peak values with the ripple reduced and a reasonable shape



8 comments:

  1. 11/O/N/10 Q.31
    12/O/N/10 Q.26,33

    ReplyDelete
    Replies
    1. For 11/O/N/10 Q.31, see solution 704 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-142.html

      Delete
  2. In solution 699, the answer comes to be different if I use the peak input rather than the rms output, please help.

    ReplyDelete
    Replies
    1. If you use the peak input, you should use 9V for the output. Both methods are given in red above.
      Look at the division signs correctly.

      Delete
  3. Also, can you please explain the last part of the same question, why has the ripple reduced? Your help would be much appreciated

    ReplyDelete
    Replies
    1. Introducing a resistor causes the smoothing effect on the output potential (see your notes for further details). This causes the ripples to be reduced - that is, the potential TENDS towards a straight line (though it is not a straight line).

      So, increasing the resistance causes the smoothing to be more effective. The potential approaches a straight line even more.

      Delete
  4. Is there a difference between increasing capacitance and increasing resistance on smoothing? Because I remember doing a past paper question where the only acceptable answer for better smoothing was increasing C. THe mark scheme rejected increasing R.

    ReplyDelete
    Replies
    1. The effect of resistance is shown above.

      The bigger the C, the less the ripple will be.

      Try to view the question again. Maybe it was a bit different

      Delete

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