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]
Here are the remaining questions for now:
ReplyDelete23/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)
For 23/M/J/12 Q.2(c), see solution 670 at
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DeleteCould you provide the solution of question 23 of October November paper 11?
ReplyDeletewhich year?
DeleteOctober November 2011
ReplyDeleteOctober November 2011
ReplyDelete2011
ReplyDeletesee solution 297 at
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