Saturday, April 25, 2015

Physics 9702 Doubts | Help Page 122

  • Physics 9702 Doubts | Help Page 122

Question 617: [Operational Amplifier Circuits]
Amplifier circuit incorporating an operational amplifier (op-amp) shown in Fig.1.

(a) State
(i) name of this type of amplifier circuit
(ii) gain G in terms of resistances R1 and R2

(b) Value of R1 is 820 Ω. The resistor of resistance R2 is replaced with light-dependent resistor (LDR).
Input potential difference VIN is 15mV.
Calculate output potential difference VOUT for LDR having a resistance of
(i) 100Ω (LDR is in sunlight)
(ii) 1.0MΩ (LDR is in darkness)

Reference: Past Exam Paper – November 2010 Paper 41 & 42 Q9

Solution 617:
(i) Non-inverting (amplifier)

(ii) Gain G = 1 + R2/R1

Gain = 1 + 100/820 (= 1.12)  
(Gain = VOUT/VIN)
Output = (Gain x VIN = 1.12x15 =) 17mV

{Gain = 1 + [(1x106)/820] = 1220
Output = Gain x VIN = 1220 x 15 = 18397 mV = 18V
But the input supply line is only 9V. So, the op-amp saturates. The output is thus 9V (equal to the input supply line)}
Output = 9V  

Question 618: [Modulation]
(a) Radio communication may be either frequency modulated or amplitude modulated. Explain what is meant by modulation.

(b) Amplitude-modulated radio wave is to be used for broadcast of music having frequencies between 30 Hz and 4500 Hz. Radio station is broadcasting in the long wave waveband (wavelengths between 1 × 103 m and 1 × 104 m).
(i) bandwidth of the broadcast,
(ii) maximum number of radio stations that could operate simultaneously in the same area within the long wave waveband.

(c) On axes of Fig.1, sketch graph of a typical power spectrum for the radio station in (b).
Label the x-axis with values of frequency.

Reference: Past Exam Paper – November 2006 Paper 6 Q15

Solution 618:
(a) Modulations are variations in either the amplitude or the frequency of a wave
EITHER in synchrony with the displacement of information signal
OR in order to carry information on the wave

{The bandwidth is twice the (maximum) audio frequency.}
Bandwidth of the broadcast = 9 kHz

The LW (long wave) frequency range is 30 kHz → 300 kHz
{The bandwidth of 1 broadcast (used by 1 station) is 9 kHz as stated above. Now, the frequencies of the long wave waveband varies from 30 kHz → 300 kHz. So, 1 station could use frequencies 30 kHz to 38 kHz (there’s actually 9 here: 30, 31, 32, 33, 34, 35, 36, 37 and 38), another 39 kHz to 47 kHz, etc …
To find the maximum number of stations, we take (300 – 30) / 9}
Number of stations = 270 / 9 = 30


For the sketch, the carrier frequency should be shown as a vertical line and two sidebands. There should be a reasonable symmetry with the sideband indicating approx. 4500 Hz range.
{Note that the carrier frequency has be chosen between the range of 30 kHz and 300 kHz. (Of course, 30 kHz would not work because one of the sidebands would be outside the range.) The sidebands should each be shown 4500 Hz away from the carrier frequency.}

Question 619: [Electric Potential]
(a) Define electric potential at a point.

(b) Isolated solid metal sphere is positively charged.
Variation of the potential V with distance x from the centre of the sphere is shown in Fig.1.

Use Fig.1 to suggest
(i) why radius of the sphere cannot be greater than 1.0 cm,
(ii) that charge on the sphere behaves as if it were a point charge.

(c) Assuming that charge on the sphere does behave as a point charge, use data from Fig.1 to determine the charge on the sphere.

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q5

Solution 619:
(a) The electric potential at a point is defined as the work done in moving unit positive charge from infinity (to the point).

(i) In the sphere, the electric potential is constant.

For a point charge, the product Vx is constant.
{We need to use points from the graph to show that the products for these points are constant.}
The co-ordinates should be clear and we need to determine two values of Vx at least 4 cm apart.
{Consider the points (6, 30) and (2, 90) from the graph. The product of both of them give a value of 180. So, the product obtained is constant at different points.}
Conclusion made clear: since the product is constant, the charge on the sphere behaves as if it were a point charge.

{Vx = 180 Vcm = 180 × (1.0 × 10–2) Vm}
Charge q = 4πε0Vx = 4π × (8.85 × 10–12) × 180 × (1.0 × 10–2) = 2.0 × 10–10 C

Question 620: [Forces]
Two co-planar forces act on rim of a wheel. The forces are equal in magnitude.
Which arrangement of forces provides only a couple?

Reference: Past Exam Paper – November 2011 Paper 12 Q13

Solution 620:
Answer: D.
The system of forces can be provide both a resultant force and a resultant torque. We need only a couple, so the resultant force should be zero.

For the resultant force to be zero, the direction of the 2 forces should oppose each other. [A, B and C are incorrect]

A couple on an object consists of 2 equal forces acting in opposite directions to each other, they act through 2 different points, in such a way that they produce a turning effect.
E.g. steering wheel in a car


  1. Assalam alikum sir rather than doing questions that are already done can u do new questions
    For question 618 u gave wrong reference there is no paper 6. Could u please do more paper 1.

    1. I'm only soling questions that viewers ask through the comments. For those already solved, sometimes I add more details so that it is clear.

      Just let me know which questions you are having doubts and I'll try to help

    2. In the past, there was a paper 6 for Options, now I think it's no longer there.

  2. May ALLAH SUBHANAH tallah give u ajar for your this noble deed u have helped lot of students.

  3. oct/nov 2014 v42
    Q 6 (b) y do we multiply the force with 2 to find the variation??

    1. Check solution 700 at

  4. Q 619 b (i) part
    can you please explain a bit more.

    1. Even though this not directly shown in the graph, it is known that the electric potential starts to decrease from the surface of the sphere from further away and that the electric potential is constant inside the sphere (since the electric field intensity is zero inside the sphere). - This should be taken as a fact. It is beyond the syllabus to prove it.

      Since the graph starts at x = 1cm (and the part before x = 1cm is not shown [it is not be zero]), we can infer that before x = 1cm, we are inside the sphere.

      If the radius was greater than 1.0cm, the electric potential should be constant for some values of x after x = 1cm.

  5. can u solve question 4 from w13 paper 12?

    1. Go to

  6. For Q.618 How did the frequencies labelled on the graph come up? The sidebands should each be shown 4500 Hz away from the carrier frequency while the values on the graph are 95.5, 100, 104.5

    1. The values are given in kHz

    2. how to know the carrier frequency is 100kHz? Thanks

    3. the value 100 kHz is taken at random. what is important is that the sidebands are equal frequencies away.

  7. Can you please solve q9 w06 p1

    1. Go to solution 1104 at


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