Friday, May 22, 2015

Physics 9702 Doubts | Help Page 149

  • Physics 9702 Doubts | Help Page 149



Question 733: [Current of Electricity]
Six resistors, each of resistance R, are connected as shown.

Combined resistance is 66 kΩ.
What is the value of R ?
A 11 kΩ                      B 18 kΩ                      C 22 kΩ                      D 36 kΩ

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q37



Solution 733:
Answer: D.
For the parallel combination of 2 resistors,
Combined resistance = [1/R + 1/R]-1 = R/2

For the parallel combination of 3 resistors,
Combined resistance = [1/R + 1/R + 1/R]-1 = R/3

Total combined resistance = R + R/2 + R/3 = 11R / 6
11R / 6 = 66
Resistance R = 6 × 66 / 11 = 36 kΩ










Question 734: [Waves > Diffraction]
Diffraction can be observed when a wave passes an obstruction. Diffraction effect is greatest when the wavelength and the obstruction are similar in size.
For waves travelling through air, what is the combination of wave and obstruction that could best demonstrate diffraction?
A microwaves passing a steel post
B radio waves passing a copper wire
C sound waves passing a human hair
D visible light waves passing a gate post

Reference: Past Exam Paper – November 2012 Paper 13 Q28



Solution 734:
Answer: A.
In this question, we need to estimate the sizes involved.

Below is the approximate range of wavelengths for the waves:
Microwave: 1mm – 1m
Radio wave: 1m – 100 000km
Visible light: 390nm – 750nm
Sound waves: 17mm – 17m (in audible frequencies)

A steel post is quite similar in size as the wavelength of microwaves. [A is correct]
As for the others, there is a huge mismatch in size between the wavelength and the obstacle.










Question 735: [Waves]
Which statement about sound waves in air at constant temperature is correct?
A Amplitude is inversely proportional to velocity.
B Frequency is inversely proportional to wavelength.
C Velocity is proportional to wavelength.
D Wavelength is proportional to amplitude.

Reference: Past Exam Paper – June 2011 Paper 11 Q23



Solution 735:
Answer: B.
Speed v = fλ
Frequency f = v / λ      (The frequency is inversely proportional to wavelength.)

At constant temperature, the velocity of the sound waves is constant, even if the wavelength changes. In such a case, the frequency also changes and the speed is still constant.  Thus, as the wavelength increases, the speed does not increase or decrease – it remains constant. So, the velocity is NOT proportional to wavelength at constant temperature. [C is incorrect]

Choices A and C are also incorrect / irrelevant.










Question 736: [Waves > Intensity]
Sound wave X has intensity 1012 times greater than that of sound wave Y.
By how much is amplitude of X greater than the amplitude of Y?
A 106 times
B 3.16 × 106 times
C 5 × 1011 times
D 1012 times

Reference: Past Exam Paper – June 2007 Paper 1 Q22



Solution 736:
Answer: A.
Intensity is proportional to (amplitude A)2.

The intensity of sound wave X is 1012 times greater than that of sound wave Y.
Let the amplitude of sound wave X by AX and that of sound wave Y be AY.
AX2 is 1012 times greater than AY2 
AX is √(1012) = 106 times greater than AY










Question 737: [Matter > Deformation]
Suspended copper wire is gradually loaded until it is stretched just beyond the elastic limit, and it is then gradually unloaded.
Which graph (with arrows indicating the sequence) best illustrates variation of the tensile stress with longitudinal strain?


Reference: Past Exam Paper – June 2003 Paper 1 Q19



Solution 737:
Answer: B.
Stress = Force / Area
Strain = Extension / Original Length

As the load is increased (stress increases), the extension of the wire increases (strain increases).

From Hooke’s law, the extension is proportional to the force (strain is proportional to stress) up to the limit of proportionality, which is before the elastic limit. So, up to this point, the graph is a straight line. [A is incorrect]

The loading part (when the load is being increased) is shown by the upward arrow. From Hooke’s law, when the load is zero, the extension is also zero. So, the loading part should pass through the origin. [C is incorrect]

Beyond the limit of proportionality, the strain is no longer proportional to the stress. Since the wire has been stretched just beyond the elastic limit, the loading part (upwards arrow) should be curved at some point.

So, the deformation is no longer elastic – that is when the load is removed, the wire does not regain its original length. However, as the load is decreased, some of the extension also decreases. However, this part of the graph is a straight line, not curved. [D is incorrect]









Question 738: [Dynamics > Newton’s 3rd law]
A ball falls vertically and bounces on the ground.
The following statements are about forces acting while the ball is in contact with the ground.
Which statement is correct?
A The force that the ball exerts on the ground is always equal to the weight of the ball.
B The force that the ball exerts on the ground is always equal in magnitude and opposite in direction to the force the ground exerts on the ball.
C The force that the ball exerts on the ground is always less than the weight of the ball.
D The weight of the ball is always equal in magnitude and opposite in direction to the force that the ground exerts on the ball.

Reference: Past Exam Paper – June 2004 Paper 1 Q10



Solution 738:
Answer: B.
Force is a vector – it has both a magnitude and a direction.

Newton’s 3rd law states that when 2 bodies are in contact, the force they exert on which other is equal and opposite to each other. [B is correct]

The time of contact of the ball with the ground can be described as follows. The ball just hits the ground, then it is in contact with the ground and finally, the ball is just about to leave the ground.
During these processes, the forces change accordingly.

As the ball just hits the ground, the force it exerts on the ground at that instant is equal to the weight. [C is incorrect] After this instant, the ball (which was falling down) comes to a stop momentarily. Thus, the force on the ball is reduced to zero. (If the force on the ball was ALWAYS equal to the weight, the ball would just keep on accelerating downwards) [A is incorrect]

The weight of the ball always acts towards the ground (downward). As stated above, the force on the ball changes during the time of contact. As the ball is just about to leave the ground to move up, the resultant force on the ball has increased and is now upwards. But the weight is always downwards. [D is incorrect]



9 comments:

  1. Replies
    1. For 11/M/J/11 Q.25, see solution 129 at
      http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-24.html

      Delete
  2. M/J/08 Q.26 paper12

    ReplyDelete
    Replies
    1. Check solution 743 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-150.html

      Delete
  3. Oct/Nov 2014 P41 Q5 b i & ii part

    ReplyDelete
  4. Oct/Nov 2014 P41 Q5 b (i) & (ii) part

    ReplyDelete
    Replies
    1. Check solution 619 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-122.html

      Delete
  5. Salam! in q 738, when the ball is about to move upwards, the force exerted by the ground on it is upward n greater than its weight which is why it accelerates upwards ryt?
    for option A, this is what i think, let me know if this is wrong.
    1. In absence of air resistance if the ball fallas and hits the ground, the force it exerts on the ground will be equal to its weight. (and if not, how can the ball not exert W because that is the force with which it fell, right?)
    2. but if air resistance is appreciable, then the ball would exert force on the ground equal to the resultant of its W and resistive force.
    Is this why A and D are both wrong?

    ReplyDelete
    Replies
    1. Actually in this question, we should not be thinking of the forces in terms of weight and air resistance.

      This question is mostly on Newton's laws of motion. From the third law itself, choice B is correct.

      Now, from Newton's 2nd law of motion, the force is equal to the rate of change of momentum. Remember, the ball is falling - so it has some velocity. When it hits the ground, its speed is momentarily zero. So, there is a change in velocity.
      Momentum = mv. Since velocity changes, momentum also changes, thus giving rise to a force (Newton's 2nd law).

      Since the way the velocity changes is different at different stages, the FORCE we are talking about differs from the weight.

      Delete

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