# 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 V

_{IN}is shown in Fig.2.
On axes of Fig.2, show variation
with time t of the output potential V

_{OUT}.**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 V

_{IN}is applied directly to the non-inverting input. Voltage gain = V_{OUT}/ V_{IN}= 1 + (R_{f}/ R_{1})}
Gain = 1 + (10.8 / 1.2) = 10

(ii)

{V

_{OUT}= Gain × V_{IN }= 10 × V_{IN}
When V

_{IN}= 1.0 V, V_{OUT}= 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 V

_{OUT}= 10 × V_{IN}, which gives us a straight line with gradient = 10. That is, it should be a straight line joining the points (V_{IN}= 0, V_{OUT}= 0) and (V_{IN}= 1.0, V_{OUT}= 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 (V_{IN}= 0, V_{OUT}= 0) and (V_{IN}= 1.0, V_{OUT}= 10) but ON this dotted straight line, we draw a straight line (not dotted) from V_{OUT}= 0 up to V_{OUT}= 9V. V_{OUT}= 9V may be reached before V_{IN}= 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 V

_{OUT}= 9.0V) may be erased after drawing the correct line. Actually, it would be at V_{IN}= 0.9 V}
Graph: straight line from (0,0)
towards V

_{IN}= 1.0 V, V_{OUT}= 10 V
{Since the output
potential gets saturated at 9.0 V, the graph is a horizontal line at V

_{OUT}= 9.0 V up to the final value of positive V_{IN}(that is up to V_{IN}= 2.0 V)}
Horizontal line at V

_{OUT}= 9.0 V to V_{IN}= 2.0 V
{As V

_{IN}changes from being positive to negative, V_{OUT}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 V

_{IN}= 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 (V

_{IN}= 0, V_{OUT}= 0) [at the later time t] and (V_{IN}= -1.0, V_{OUT}= -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 Î±-particles
from radioactive source is made to pass between the plates.

**(a)**State and explain effect on the deflection of the Î±-particles for each of the following changes:

(i) Magnitude of V is increased.

(ii) Separation d of the plates is decreased.

**(b)**Source of Î±-particles is replaced with a source of Î²-particles.

Compare, with a reason in each case,
effect of each of the following properties on the deflections of Î±- and
Î²-particles in a uniform electric field:

(i) charge

(ii) mass

(iii) speed

**(c)**Electric field gives rise to an acceleration of the Î±-particles and the Î²-particles.

Determine ratio

acceleration of the Î±-particles / acceleration of the Î²-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 Î±-particle

{Electric force = Eq =
(V/d)q}

(ii) There is a greater deflection
because there is a greater electric field / force on the Î±-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
Î²-particles because Î²-particle has smaller charge

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

(iii) There is less deflection with
the Î²-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 Î±-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 Î²-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 Î± = E (2e) / 4u = 2Ee / 4u

Acceleration of Î² = 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

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-134.html

Could 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

Deletehttp://physics-ref.blogspot.com/2015/01/physics-9702-doubts-help-page-50.html

Thank you

ReplyDelete