Sunday, October 25, 2015

Physics 9702 Doubts | Help Page 217

  • 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 VOUT for an input potential VIN of +0.5 V. Explain your answer.

(ii) A sensing device produces a variable potential VIN.
The variation with time t of VIN is shown in Fig. 10.3.

On the axes of Fig. 10.3, sketch the variation with time t of the output potential VOUT.

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)
VOUT = –5 V
{This is an inverting amplifier.}
The inverting input is positive             OR V is positive        OR V > V+, so VOUT is negative
The op-amp has very large / infinite gain and so it saturates.

(ii)
For the sketch:
VOUT switches from (+) to (–) when VIN is zero
VOUT is +5 V or –5 V
VOUT is negative when VIN 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) λ = 10cm
Wavelength λ = 10 / (n + 0.5)
Put n = 0, Wavelength λ = 10 / (0 + 0.5) = 20cm 
Put n = 1, Wavelength λ = 10 / (1 + 0.5) = 6.67cm
Put n = 2, Wavelength λ = 10 / (2 + 0.5) = 4cm [C is correct]
Put n = 3, Wavelength λ = 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 Ω and at 18 °C, the resistance is 1900 Ω. Each resistor P has a resistance of 2000 Ω.
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 Ω. 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 Ω. 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.



2 comments:

  1. 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

    ReplyDelete
    Replies
    1. the input (V_IN) is connected to V-.

      Delete

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