Wednesday, May 6, 2015

Physics 9702 Doubts | Help Page 133

  • Physics 9702 Doubts | Help Page 133



Question 664: [Operational Amplifier]
(a) State three properties of ideal operational amplifier (op-amp).

(b) An amplifier circuit is shown in Fig.1.

(i) Calculate gain of the amplifier circuit.
(ii) Variation with time t of the input potential VIN is shown in Fig.2.

On axes of Fig.2, show variation with time t of the output potential VOUT.

Reference: Past Exam Paper – November 2013 Paper 43 Q9



Solution 664:
(a) Example:
zero output impedance / resistance
infinite input impedance / resistance
infinite (open loop) gain
infinite bandwidth
infinite slew rate

(b)
(i)
{The circuit is that of a non-inverting amplifier as the input VIN is applied directly to the non-inverting input. Voltage gain = VOUT / VIN = 1 + (Rf / R1)}
Gain = 1 + (10.8 / 1.2) = 10

(ii)
{VOUT = Gain × VIN = 10 × VIN
When VIN = 1.0 V, VOUT = 10 V. However, since the supply line is 9 V, the output potential gets saturated and 9.0 V.

Here, in order to get the correct shape of the graph, we use the equation VOUT = 10 × VIN, which gives us a straight line with gradient = 10. That is, it should be a straight line joining the points (VIN = 0, VOUT = 0) and (VIN = 1.0, VOUT = 10). However, this line should be drawn dotted since we know that the output gets saturated at 9 V. So, we draw a straight line joining the points (VIN = 0, VOUT = 0) and (VIN = 1.0, VOUT = 10) but ON this dotted straight line, we draw a straight line (not dotted) from VOUT = 0 up to VOUT = 9V. VOUT = 9V may be reached before VIN = 1.0V.
Basically, drawing the dotted line is just a side work to allow us to obtain the correct gradient of the graph. The extra dotted line (above VOUT = 9.0V) may be erased after drawing the correct line. Actually, it would be at VIN = 0.9 V}
Graph: straight line from (0,0) towards VIN = 1.0 V, VOUT = 10 V
{Since the output potential gets saturated at 9.0 V, the graph is a horizontal line at VOUT = 9.0 V up to the final value of positive VIN (that is up to VIN = 2.0 V)}
Horizontal line at VOUT = 9.0 V to VIN = 2.0 V
{As VIN changes from being positive to negative, VOUT also changes from positive to negative since this is a non-inverting amplifier. This is represented at a straight vertical line.}
correct +9.0 V → –9.0 V (and correct shape to VIN = 0)
{Now, the shape of the graph for negative values will be quite similar to that for the positive values. We need to repeat the process of drawing a straight line now joining (VIN = 0, VOUT = 0) [at the later time t] and (VIN = -1.0, VOUT = -10), but again, the complete straight line (not dotted) should stop at -9.0V since the amplifier gets saturated.}











Question 665: [Nuclear Physics]
Two horizontal metal plates are separated by distance d in vacuum. A potential difference V is applied across the plates, as shown in Fig.1.

A horizontal beam of α-particles from radioactive source is made to pass between the plates.
(a) State and explain effect on the deflection of the α-particles for each of the following changes:
(i) Magnitude of V is increased.
(ii) Separation d of the plates is decreased.

(b) Source of α-particles is replaced with a source of β-particles.
Compare, with a reason in each case, effect of each of the following properties on the deflections of α- and β-particles in a uniform electric field:
(i) charge
(ii) mass
(iii) speed

(c) Electric field gives rise to an acceleration of the α-particles and the β-particles.
Determine ratio
acceleration of the α-particles / acceleration of the β-particles

Reference: Past Exam Paper – November 2011 Paper 23 Q6



Solution 665:
(a)
(i) There is a greater deflection since there is a greater electric field / force on the α-particle
{Electric force = Eq = (V/d)q}

(ii) There is a greater deflection because there is a greater electric field / force on the α-particle

(b)
(i)
EITHER
The deflections are in the opposite directions because the 2 particles are oppositely charged
OR There is less deflection with the β-particles because β-particle has smaller charge

(ii) There is a smaller deflection with the α-particles because of its larger mass

(iii) There is less deflection with the β-particles because of its higher speed.

(c)
EITHER F = ma and F = Eq OR Acceleration a = Eq / m
{The electric field E is the same for both. So, the acceleration is directly proportional to the charge q and inversely proportional to the mass m.

An α-particle is a helium nucleus and has a charge of +2e. It contains 2 protons + 2 neutrons. That is, it contains 4 nucleons.
A charge of 1e = 1.6 × 10–19 C
Mass of 1 nucleon (unified atomic mass), u = 1.66 × 10–27 kg
[From the list of data given, u = 1.66 × 10–27 kg, but in the mark scheme, they used = 1.67 × 10–27 kg. I assume this is a mistake that the examiners did not penalize this time.]

A β-particle is the same as an electron.
Mass of an electron = 9.11 × 10–31 kg
In terms of u, mass of an electron = (1/2000) u

Acceleration of α = E (2e) / 4u = 2Ee / 4u
Acceleration of β = E (e) / (u/2000)
Ratio = 2e (u/2000) / e (4u)

EITHER
{Here, we are substituting the values.}
Ratio = [(2 × 1.6 × 10–19) × (9.11 × 10–31)] / [(1.6 × 10–19) × 4 × (1.66 × 10–27)]

OR
{Here, we are writing in terms of e and u.}
Ratio = [2e × (1/2000) u] / [e × 4u]

Ratio = 1 /4000           or 2.5 × 10–4    or 2.7 × 10–4










Question 666: [Measurement]
Speedometer in a car consists of a pointer which rotates. The pointer is situated several millimetres from a calibrated scale.
What could cause random error in the driver’s measurement of the car’s speed?
A The car’s speed is affected by the wind direction.
B The driver’s eye is not always in the same position in relation to the pointer.
C The speedometer does not read zero when the car is at rest.
D The speedometer reads 10 % higher than the car’s actual speed.

Reference: Past Exam Paper – November 2011 Paper 12 Q5



Solution 666:
Answer: B.
The wind direction affects the speed of the car, but this has no effect on the measurement of the speed by the driver. [A is incorrect]

If the speedometer do not read zero when the car is at rest, this would be a systematic error, not a random error. Similarly, if the speedometer reads 10% higher than the car’s actual speed, this would also be a systematic error. [C and D are incorrect]




9 comments:

  1. Here are the remaining questions for now:

    23/M/J/12 Q.2(c)

    21/O/N/12 Q.6(c)

    22/O/N/12 Q.3(b)

    21/M/J/13 Q.6(c)(d)

    22/O/N/13 Q.3(c)(ii)

    ReplyDelete
    Replies
    1. For 23/M/J/12 Q.2(c), see solution 670 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-134.html

      Delete
  2. Could you provide the solution of question 23 of October November paper 11?

    ReplyDelete
  3. Replies
    1. see solution 297 at
      http://physics-ref.blogspot.com/2015/01/physics-9702-doubts-help-page-50.html

      Delete

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