FOLLOW US ON TWITTER
SHARE THIS PAGE ON FACEBOOK, TWITTER, WHATSAPP ... USING THE BUTTONS ON THE LEFT


YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

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

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 133