Wednesday, November 26, 2014

Physics 9702 Doubts | Help Page 26

  • Physics 9702 Doubts | Help Page 26

Question 146: [Waves > Graph]
Graph shows how height of a water surface at a point in a harbour varies with time t as waves pass the point.

What are p and q?

For June 2008 Paper 1 Q25:



For November 2013 Paper 13 Q25:
p                      q
A         displacement   period
B         displacement   wavelength
C         amplitude        period
D         amplitude        wavelength


Reference: Past Exam Paper – June 2008 Paper 1 Q25 & November 2013 Paper 13 Q25



Solution 146:


June 2008 Paper 1 Q25 - Answer: B.
November 2013 Paper 13 Q25 – Answer: A.

The x-axis indicates the time t. So, q gives the period (which is in units of time) and not the wavelength (which is in units of length).

The y-axis indicates the height of the water surface. The term ‘amplitude’ is used for the maximum displacement. As p does not show the maximum displacement, it is not the amplitude. p only shows the displacement.









Question 147: [Electric Field > Equipotential surface]
Diagram shows two points P and Q which lie 90° apart on circle of radius r.
Positive point charge at centre of the circle creates an electric field of magnitude E at both P and Q.

Which expression gives work done in moving a unit positive charge from P to Q?
A 0                  B E × r                        C E × (πr / 2)               D E × (πr)

Reference: Past Exam Paper – June 2014 Paper 11 Q29 & June 2010 Paper 12 Q28 & June 2010 Paper 13 Q29



Solution 147:
Answer: A.
Work done = Force x Distance moved in direction of force

The force on the charge moved from P to Q does not change if it is moved along the circle.
The force is always at right-angles to the motion, so no work is done.

The problem can also be considered in terms of the electrical potential. The charges P and Q are at the same potential (from the +ve charge at the centre) and so zero work done is the correct answer.

It is incorrect to use W = Ex which would result in answer C.










Question 148: [Waves > Diffraction]
Light passes through diffraction grating ruled at 1000 lines per cm and same wavelength of light also passes through two narrow slits 0.5 mm apart. Both situations produce intensity maxima and minima on screen.
Which statement about separation of the maxima on the screen and the sharpness of the maxima is correct?
A Diffraction grating maxima are less widely spaced and are less sharp than the two-slit maxima.
B Diffraction grating maxima are less widely spaced and are sharper than the two-slit maxima.
C Diffraction grating maxima are more widely spaced and are less sharp than the two-slit maxima.
D Diffraction grating maxima are more widely spaced and are sharper than the two-slit maxima.

Reference: Past Exam Paper – June 2014 Paper 12 Q26



Solution 148:
Answer: D.
The diffraction grating maxima are more widely spaced (since the length of a line in the diffraction grating {= 1/1000 = 0.001cm = 0.01mm} is less than the separation of the 2 narrow slits {= 0.05mm}).

{Consider the spots formed to be black and white. A higher sharpness would cause the spots formed to be ‘whiter’ or ‘blacker’. That is, they can be more distinguished from each other.}
Additionally, the diffraction grating maxima are sharper than the two-slit maxima (since more interference occurs from the waves that get diffracted from more slits in the diffraction grating, - the amplitudes of more diffracted waves contribute to every single maxima).










Question 149: [Current of Electricity > Internal Resistance]
Cell has electromotive force (e.m.f.) of 6 V and internal resistance R. External resistor, also of resistance R, is connected across cell, as shown.

Power P is dissipated by external resistor.
Cell is replaced by a different cell that has e.m.f. of 6 V and negligible internal resistance.
What is the new power that is dissipated in external resistor?
A 0.5P                         B P                  C 2P                D 4P

Reference: Past Exam Paper – June 2014 Paper 13 Q34



Solution 149:
Answer: D.
For the initial cell, total resistance on circuit = R + R = 2R
Current in circuit, I (= V / 2R) = 6 / 2R = 3/R
Power dissipated by the external resistor, P = I2R.

When a different cell with negligible internal resistance is used, the total resistance in the circuit is halved (total resistance in circuit = R) and thus, the current is doubled (from Ohm’s law: V = IR giving current, I = V / R = 6/R).

When the current is doubled, power dissipated (P = I2R) increases by a factor of 4 (since it depends on I2).










Question 150: [Dynamics > Laws of Motion]
Which statement about Newton’s laws of motion is correct?
A First law follows from the second law.
B Third law follows from the second law.
C Conservation of energy is a consequence of the third law.
D Conservation of linear momentum is a consequence of the first law.

Reference: Past Exam Paper – June 2009 Paper 1 Q7



Solution 150:
Answer: A.
It is important to note that the conservation laws (of energy and linear momentum) are NOT consequences of the third or first (Newton’s) law of motion but the conservation laws are fundamental laws to which the other laws are subject to. [C and D incorrect]

Newton’s first law of motion states that a body continues at rest or at constant velocity unless a resultant (external) force acts on it.
Newton’s second law of motion states that the resultant force is equal to the rate of change of momentum.

Newton’s first law does follow from the second law as follows. By using the second law for a body on which the resultant force is zero, the acceleration will also be zero, and so the first law is thus proved from the second law.








Question 151: [Measurements > Estimates]
Which statement is incorrect by factor of 100 or more?
A Atmospheric pressure is about 1 × 105Pa.
B Light takes 5 × 102s to reach us from Sun.
C Frequency of ultra-violet light is 3 × 1012Hz.
D Life-span of a man is about 2 × 109s.

Reference: Past Exam Paper – June 2013 Paper 13 Q3



Solution 151:
Answer: C.
“The frequency of ultra-violet light is 3 × 1012Hz” is incorrect. The correct frequency is from about 1014Hz to 1017Hz.









Question 152: [Forces > Equilibrium]
Diagrams show two ways of hanging same picture.

In both cases, string is attached to same points on the picture and looped symmetrically over a nail in a wall. Forces shown are those that act on the nail.
In diagram 1, string loop is shorter than in diagram 2.
Which information about magnitude of the forces is correct?
A R1 = R2        T1 = T2
B R1 = R2        T1 > T2
C R1 > R2        T1 < T2
D R1 < R2        T1 = T2

Reference: Past Exam Paper – June 2014 Paper 11 Q11 & November 2009 Paper 11 Q12 & November 2009 Paper 12 Q11



Solution 152:
Answer: B.
The weight of the picture is the same, so, the upward tension is the same. i.e. R1 = R2. [C and D incorrect]
{From this point itself, it is clear that T1 cannot be equal to T2. The solution may be chosen as B from here itself. Explanations are available below to show that B is actually the correct answer}

Consider the acute angles (angles between 0o and 90o) between the T’s and the vertical. Call these θ. The acute angle formed in diagram 1 (θ1) is clearly larger than in diagram 2 (θ2).

For each diagram, the sum of the vertical downward components of the tensions should be equal to the weight, W which is equal to R (R1 = R2).
2T1cosθ1 = Weight, W            and Weight, W= 2T2cosθ2.
Therefore, T1cosθ1 = T2cosθ2

From the graph of cosine, it can be seen that as acute angle θ increases (angle between 0o and 90o), the value of cosθ decreases. Since angle θ1 > angle θ2, the value of cosθ1 < value of cosθ2. So, for the equation (T1cosθ1 = T2cosθ2) to hold, tension T1 > tension T2.









Question 153: [Current of Electricity > Electrons]
Electric current is passed from thick copper wire through a section of thinner copper wire before entering second thick copper wire as shown.

Which statement about current and speed of electrons in wires is correct?
A Current and speed of the electrons in the thinner wire are both less than in the thicker copper wires.
B Current and speed of the electrons is the same in all the wires.
C Current is the same in all the wires but speed of the electrons in the thinner wire is greater than in the thicker wires.
D Current is the same in all the wires but speed of the electrons in the thinner wire is less than in the thicker wire.

Reference: Past Exam Paper – November 2013 Paper 13 Q33



Solution 153:
Answer: C.
In the situation described, the thinner copper wire acts as a ‘resistor’ since a smaller cross-sectional area is now available for the electrons to flow.

It is a common misconception among many learners of physics to think that electrons slow down in a resistor. But they do not. The electrons speed up to release energy (as it is known that thermal energy is released).
 (A similar situation to this is that water speeds up when it moves from a river to a waterfall.)

Since the wires are connected in series, the current is the same in each of them.










Question 154: [Current of Electricity > Potentiometer]
Unknown e.m.f. E of cell is to be determined using a potentiometer circuit. Balance length is to be measured when the galvanometer records a null reading.
What is the correct circuit to use?

Reference: Past Exam Paper – June 2008 Paper 1 Q38



Solution 154:
Answer: B.
For the galvanometer to give a null reading, the potential difference across it should be zero. This can only occur when the position at which it is connected on the wire gives a potential difference (on the wire) equal to the cell of e.m.f. E. The cell should be connected such that its potential opposes the potential on the wire.  When the potential difference across the galvanometer is zero, no current flows through it and the galvanometer records a null reading. Only circuit B provides such a situation.

Current flows from the positive terminal to the negative terminal. So, the positive terminal of the cell cannot be connected directly to the galvanometer because this will always cause a current to flow through it. [A is incorrect] {Circuit A may be used to calibrate the potentiometer}

Similarly, the galvanometer cannot be connected directly to the negative terminal of the main supply because all the current would return to the negative terminal of the main supply, causing the galvanometer to always give a non-zero reading. [C and D are incorrect]









Question 155: [Electric Field > Moving charge]
Diagram shows path of a charged particle through a uniform electric field, having vertical field lines.

What could give a path of this shape?
A positive charge travelling left to right in a field directed downwards
B positive charge travelling right to left in a field directed downwards
C negative charge travelling right to left in a field directed upwards
D negative charge travelling left to right in a field directed downwards

Reference: Past Exam Paper – June 2013 Paper 13 Q30



Solution 155:
Answer: D.
The direction of the electric field lines is (away) from positive towards negative.
So, if the field is directed downwards, a positive change would experience a force downwards while a negative charge would experience a force upwards. Similarly, if the field is directed upwards, a positive charge would experience a force upwards and a negative charge would experience a force downwards.

One important deduction that should be made from the diagram is that the charge is actually entering the field lines at an angle downward. So, for any of the possible choices for answers, it should be accounted that initially, the path would be along the path of motion of the charge (that is, initially at an angle downward).

Consider A:
If it’s entering from the left, the right part of its path should be downwards since the field is directed downwards. [A is incorrect]

Consider B:
If it’s entering from the right, the left part of its path should be downwards since the field is directed downwards. [B is incorrect]

Consider C:
If it’s entering from the right, the left part of its path should be downwards since the field is directed upwards. [C is incorrect]

Consider D:
By entering from the left (at an angle downwards), the right part of the path should be upwards since the field is directed downwards. [D is correct]




29 comments:

  1. For question 11 from November 2012 paper11 why is the momentum is not conserved?

    ReplyDelete
    Replies
    1. The question is explained as solution 177 at
      http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-29.html

      Delete
  2. plz solve 2008 june paper2 question 6

    ReplyDelete
    Replies
    1. Check Question 45 at
      http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-8.html

      Delete
  3. Salam!
    in q152, shouldn't R1 be equal to the sum of weight and the downward vertical components of the tension in the 2 strings(i.e. R1=2T1sinθ1+W) and shouldn't R2=2T2sinθ2+W as the only vertically upward forces present which can balance the vertical components of the tensions(T1 and T2) and weight are R1 and R2?
    so that way the answer should be D

    ReplyDelete
    Replies
    1. Wslm.
      R1 is equal to the weight of the picture. THe weight of the picture CAUSES a tension T1 in the string at each hole in the picture. The tension in the strings are DUE TO the weight of the picture. SO, by considering the tensions, we are already accounting for the weight of the pictures. The weight and the tensions are not separate forces. THe tensions in the strings arise due to the weight,

      SO, the sum of vertical components of the tensions T1 should be equal to R1.

      Similarly, R2 is also equal to the weight of the pictures. Since it is the same picture, R1 = R2.

      Re-read explanations above to further details

      Delete
  4. plz solve for question 1,2,3 from November 2012 paper 23

    ReplyDelete
    Replies
    1. Question 2 is explained as solution 587 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-115.html

      Delete
  5. In question 152, the force R is equal to the weight plus the sin components of the tension. Since the angle changes, the sin components would also change, hence affecting the value of R. But why is that not considered?

    ReplyDelete
    Replies
    1. No, you are mistaken.
      R = T1 + T2 = W
      The sum of tensions is equal to the weight. The tensions are downwards because of the way to picture is hung.

      R is the reaction (from Newton's 3rd law) of the 2 tensions.

      Delete
  6. 19. A conveyor belt is driven at velocity v by a motor. Sand drops vertically on to the belt at a rate of
    mkgs–1.
    What is the additional power needed to keep the conveyor belt moving at a steady speed when
    the sand starts to fall on it?
    A. 1/2 mv B. mv C. 1/2mv2 D. mv 2

    Can you solve this please??

    ReplyDelete
    Replies
    1. See solution 1017 at
      http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-212.html

      Delete
  7. why is that the unknown cell in potentiometer has to have the positive terminal and the batteries positive terminal on the same side of the circuit if u understand what I am trying to say?

    ReplyDelete
    Replies
    1. Try to read again why choice A is eliminated. I think it's already explained above.

      Delete
  8. Q155 Why the direction of electric field is form bottom(positive) to the top(negative), why not the other way around?

    ReplyDelete
    Replies
    1. this is how the electric field is DEFINED. so, we use it this way

      Delete
  9. lets say a ball is thrown vertically upwards from a cliff top.So when calculating the time the ball takes to return back on cliff top, why do we take acceleration as -g ?
    when the ball goes up the acceleration IS -g ;but on returning, isn't it accelerating downwards at 'g'?

    ReplyDelete
    Replies
    1. Acceleration due to gravity is a vector and is always downwards. When the ball is thrown up, the displacement is upward, while g is downward (so, it is opposite). Thus we take it as -ve.

      When the ball moves down, the direction of motion and acceleration are the same, so g is +ve.

      Delete
  10. when an object floats on liquid, is the magnitude of weight equal to the upthrust?
    or is it that the upthrust has to be greater than the weight for the object to be floating over a liquid?

    ReplyDelete
    Replies
    1. Yes, they are equal. But this is not limited to floating objects. If an objects remains at the same depth the forces are still equal.

      Delete
  11. please tell me where i am going wrong. :)
    in order to lift up an object from ground, my book says that the force needed to lift it up should be equal in magnitude to the weight of the object.and it also says that since the resultant force would then be zero,the objects is lifted up at a steady speed.but how is it then obeying the 1st law of newton;while it says that AN OBJECT WILL REMAIN STATIONARY OR IN IT'S STATE OF CONSTANT MOTION UNLESS A RESULTANT FORCE ACTS UPON IT.

    ReplyDelete
    Replies
    1. initially, there is also the normal force acting upwards. as soon as it is lifted, it force vanishes and it is the force exerted by the person that is equal to the weight.

      Delete
  12. If we connect an LED and a resistor to a microcontroller in series,then does it matter if we connect the LED before the resistor?Will the current remain the same if we connect the resistor first?If yes, then what is the use of connecting the resistor, if the current that flows before and after it, is the same?

    ReplyDelete
    Replies
    1. the current depends on the overall resistance in the circuit. so wherever the resistor is placed, the overall resistance is still the same. current does not decrease as it flows in a series circuit.

      Delete
  13. Q149 can we use P=VI instead? But when I used P=VI i got the answer as 2P instead of 4P

    ReplyDelete
  14. Replies
    1. no. this is only a derivation. it does not represent the 2nd law.

      2nd law: rate of change of momentum

      Delete

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