# Physics 9702 Doubts | Help Page 217

__Question 1036: [Applications > Sensing devices > Op-amp]__
An electronic sensor may be
represented by the block diagram of Fig. 10.1.

**(a)**State suitable sensing devices, one in each case, for the detection of

(i) change of temperature,

(ii) pressure changes in a sound
wave.

**(b)**The ideal operational amplifier (op-amp) shown in Fig. 10.2 is to be used as a processing unit.

(i) State the value of the output
potential V

_{OUT}for an input potential V_{IN}of +0.5 V. Explain your answer.
(ii) A sensing device produces a
variable potential V

_{IN}.
The variation with time t of V

_{IN}is shown in Fig. 10.3.
On the axes of Fig. 10.3, sketch the
variation with time t of the output potential V

_{OUT}.**Reference:**

*Past Exam Paper – November 2014 Paper 43 Q10*

__Solution 1036:__**(a)**

(i) thermistor / thermocouple

(ii) quartz crystal / piezoelectric
crystal

**OR**transducer / microphone**(b)**

(i)

V

_{OUT}= –5 V
{This is an inverting
amplifier.}

The inverting input is positive OR V

^{–}is positive OR V_{–}> V^{+}, so V_{OUT}is negative
The op-amp has very large / infinite
gain and so it saturates.

(ii)

For the sketch:

V

_{OUT}switches from (+) to (–) when V_{IN}is zero
V

_{OUT}is +5 V or –5 V
V

_{OUT}is negative when V_{IN}is positive (or v.v.)

__Question 1037: [Waves > Interference]__
Wave generators at points X and Y
produce water waves of the same wavelength. At point Z, the waves from X have
the same amplitude as the waves from Y. Distances XZ and YZ are as shown.

When the wave generators operate in
phase, the amplitude of oscillation at Z is zero.

What could be the wavelength of the
waves?

A 2 cm B 3 cm C
4 cm D 6 cm

**Reference:**

*Past Exam Paper – June 2015 Paper 13 Q29*

__Solution 1037:__**Answer: C.**

The amplitude of oscillation at Z is zero, so destructive interference
occurs at Z and the difference between lengths XZ and YZ must be an odd number
of half-wavelengths.

Path difference between XY and YZ = 32 – 24 = 10cm

This path difference must be an odd number of half-wavelengths.

(n + ½) Î» = 10cm

Wavelength Î» = 10 / (n + 0.5)

Put n = 0, Wavelength Î» = 10 / (0 + 0.5) = 20cm

Put n = 1, Wavelength Î» = 10 / (1 + 0.5) = 6.67cm

Put n = 2, Wavelength Î» = 10 / (2 + 0.5) = 4cm [C is correct]

Put n = 3, Wavelength Î» = 10 / (3 + 0.5) = 2.86cm

…

__Question 1038:__

__[Applications > Operational Amplifier]__
The circuit of Fig. 10.1 may be used
to indicate temperature change.

The resistance of the thermistor T
at 16 °C is 2100 Î© and at 18 °C, the resistance is 1900 Î©. Each resistor P has
a resistance of 2000 Î©.

Determine the change in the states
of the light-emitting diodes R and G as the temperature of the thermistor
changes from 16 °C to 18 °C.

**Reference:**

*Past Exam Paper – November 2009 Paper 41 Q10*

__Solution 1038:__
{V

^{+}is always 1.00V since it is connected between the 2 resistors P which are of equal resistances. So, the potential of 2V is divided equally.
V

^{–}depends on the potential divider formed by thermistor T and resistor P.
At 16 °C, the resistance
of T = 2100 Î©. V

^{–}= 2 × [2000 / (2000+2100)] = 0.98V}
at 16 °C, V

^{+}= 1.00 V and V^{–}= 0.98 V OR V^{+}> V^{–}
{Since V

^{+}> V^{–}. the output is positive.}
at 16 °C, output is positive

{Both diodes are connected
to earth line (0V). Current flows from a relatively positive potential to a
relatively negative one. Since the output is positive, the current flows from
the op-amp to the earth line. Thus, only diode R is ‘on’.}

diode R is ‘on’

__and__diode G is ‘off’
{As the temperature rises,
the resistance of thermistor T decreases.

At 18 °C, the resistance
of T = 1900 Î©. V

^{–}= 2 × [2000 / (2000+1900)] = 1.03V
Since V

^{–}> V^{+}, the output is negative and thus current flows from the earth line towards the op-amp. Only diode G is ‘on’.}
as temperature rises, diode R goes
‘off’

__and__diode G goes ‘on’

__Question 1039: [Kinematics > Air resistance]__
A ball is thrown horizontally in
still air from the top of a very tall building. The ball is affected by air resistance.

What happens to the horizontal and
to the vertical components of the ball’s velocity?

**horizontal component of velocity vertical component of velocity**

A decreases
to zero increases
at a constant rate

B decreases
to zero increases
to a constant value

C remains
constant increases
at a constant rate

D remains
constant increases
to a constant value

**Reference:**

*Past Exam Paper – November 2010 Paper 11 Q7 & Paper 13 Q11*

__Solution 1039:__**Answer: B.**

This was the question on the whole
paper with the lowest percentage of correct answers.

The vertical velocity of the ball
will be affected by both the force of gravity and air resistance and the
horizontal velocity of the ball will also be affected by air resistance.

The vertical velocity will rise to a
constant value (terminal velocity) where the air resistance equals to the weight
of the ball.

The horizontal velocity cannot stay constant
over a considerable time but will fall to zero due to air resistance opposing
its motion. The question states that the ball is affected by air resistance;
this cannot be just its vertical motion.

Unlike the vertical velocity where
the force of gravity causes it to increases and the air resistance opposes this
increase, the horizontal velocity is affected only by air resistance, which causes
it to decreases and there is no force that increases the horizontal velocity.

Hello, In solution 1036 how do we know its an Inverting Amp? Like for an Inverting dont we need resistors in the circuit like the whole Rf ordeal? :/ I have my cie tomorrow, i'd appreciate if you reply ASAP

ReplyDeletethe input (V_IN) is connected to V-.

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