Saturday, May 23, 2015

Physics 9702 Doubts | Help Page 150

  • Physics 9702 Doubts | Help Page 150



Question 739: [Current of Electricity > Resistance]
A pencil is used to draw line of length 30 cm and width 1.2 mm. Resistivity of the material in the pencil is 2.0 × 10–5 Ω m and the resistance of the line is 40 kΩ.
What is the thickness of the line?
A 1.25 × 10–10 m
B 1.25 × 10–8 m
C 1.25 × 10–7 m
D 1.25 × 10–5 m

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q32



Solution 739:
Answer: C.
Resistance R = ρL / A
where ρ is the resistivity of the material, L is the length and A is the cross-sectional A

The line drawn can be considered to be a cuboid (in 3-dimension).
Let the thickness of the line be y.
Cross-sectional A = y × (1.2×10-3)

R = ρL / A
A = ρL / R
y × (1.2×10-3) = ρL / R
Thickness y = (2.0 × 10–5) (0.3) / (40000 × 1.2×10-3) = 1.25×10-7 m








Question 740: [Kinematics]
Diagram shows a velocity-time graph for a mass moving up and down on the end of a spring.
Which point represents velocity of the mass when at the lowest point of its motion?


Reference: Past Exam Paper – November 2014 Paper 13 Q7



Solution 740:
Answer: D.
At the lowest and highest point in the motion, the mass is momentarily at rest. So, at these points, the velocity should be zero. [A and C are incorrect]

Additionally at the extremes of the motion, the velocity changes direction.
If the mass was moving upwards towards the highest point, at the highest point, the velocity would changes from being previously positive to negative. That is, the mass will then be moving downwards afterwards. So, at the highest point, the velocity changes from positive to negative. This is point B in the graph.

If the mass was moving downwards towards the lowest point, at the lowest point, the velocity would changes from being previously negative to positive. That is, the mass is about to move upwards afterwards. So, at the lowest point, the velocity changes from negative to positive. This is point D in the graph.









Question 741: [Measurement > C.R.O]
Diagram shows a trace of a wave on a cathode-ray oscilloscope.
The vertical and horizontal gridlines have spacing of 1.0 cm. Voltage scaling is 4 V cm–1 and time scaling is 5 ms cm–1.

What are the amplitude and period of the wave?
amplitude / V              period / ms
A         1.5                               4
B         5.0                               10
C         6.0                               20
D         12.0                             20

Reference: Past Exam Paper – June 2011 Paper 11 Q4



Solution 741:
Answer: C.
The amplitude is given by the vertical spacing from the equilibrium position and the period is given by the horizontal spacing.

Note that the equilibrium position is found at the middle of the 3rd square along the vertical.
Amplitude = 1.5cm = 1.5 (4) = 6V
Period = 4cm = 4 (5) = 20ms







Question 742: [Pressure]
A child drinks liquid of density ρ through a vertical straw.
Atmospheric pressure is p0 and the child is capable of lowering the pressure at top of the straw by 10%. The acceleration of free fall is g.
What is maximum length of straw that would enable the child to drink the liquid?
A p0 / 10ρg                  B 9p0 / 10ρg                C p0 / ρg                      D 10p0 / ρg

Reference: Past Exam Paper – June 2003 Paper 1 Q20



Solution 742:
Answer: A.
Imagine a beaker containing the liquid and a vertical straw (not an inclined straw).

Atmospheric pressure p0 acts on the surface of the liquid.

Now, the child is sucking at top of the straw with a pressure of 0.9p0 (since the child is capable of lowering the pressure at the top of the straw by 10%). As the child sucks up the liquid rises up the straw of length h.

The difference in pressure at the surface of the liquid in the beaker (= p0) and the pressure at the top of the straw (= 0.9p0) is represented by the pressure due to the column of liquid in the straw (= hρg).

p0 – 0.9p0 = hρg
hρg = 0.1p0 = p0 / 10
Length of straw, h = p0 / 10ρg








Question 743: [Waves > Intensity]
Intensity I of a sound at a point P is inversely proportional to the square of the distance x of P from the source of the sound. That is

Air molecules at P, a distance r from S, oscillate with amplitude 8.0 μm.
Point Q is situated a distance 2r from S.
What is the amplitude of oscillation of air molecules at Q?
A 1.4 μm                     B 2.0 μm                     C 2.8 μm                     D 4.0 μm

Reference: Past Exam Paper – June 2008 Paper 1 Q26



Solution 743:
Answer: D.
Intensity I is proportional to the square of the amplitude (A2).

Now, the intensity I is also inversely proportional to the square of the distance x form the source (1/x2).
So, doubling the distance x causes the intensity to be (I / 4).

Recall: Intensity I is proportional to the square of the amplitude (A2).
Intensity I is proportional to 8.02
Intensity I/4 is proportional to (8.02 / 4)

So, at distance = 2r,
(Amplitude)2 = (8.02 / 4)
Amplitude = (8.02 / 4) = 8.0 / 2 = 4.0μm

OR
Since I α A2     and I α 1 / x2,
It can be concluded that A2 α 1 / x2
So, amplitude A α 1 / x
Thus, if distance x is doubled, amplitude A is halved.

OR
At distance = r, Intensity IP α 82
At distance = 2r, Intensity IQ = IP / 4 (as calculated above)
At distance = 2r, Intensity (IP / 4) α AQ2

OR
Since we are dealing with the same quantities in both cases, we can take the ratio of intensities and amplitudes at P and at Q.

IP / IQ = AP2 / AQ2
IP / (IP/4) = 82 / AQ2
4 = 82 / AQ2
Amplitude AQ = (8.02 / 4) = 8.0 / 2 = 4.0μm









Question 744: [Electric field]
Two vertical conducting plates X and Y are positioned so that they are separated by distance of 6.0 mm in air. A 60 V d.c. supply is connected as shown.

What is electric field strength at E, a point midway between the plates?
A 1.0 × 104 V m–1 towards X
B 1.0 × 104 V m–1 towards Y
C 2.0 × 104 V m–1 towards X
D 2.0 × 104 V m–1 towards Y

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q31



Solution 744:
Answer: B.
The direction of the electric field is from positive to negative. Since plate X is connected to the positive terminal of the supply, it is also positive. So, the direction of the electric field is from plate X to plate Y. [A and C are incorrect]

Electric field strength E = V / d = 60 / 0.006 = 1 × 104 Vm-1
The electric field strength between 2 parallel plates is constant anywhere between them.



11 comments:

  1. Replies
    1. For 11/M/J/11 Q.27, see solution 753 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-152.html

      Delete
  2. 9702/23/o/N/12 QUESTION 2 AND 3

    ReplyDelete
    Replies
    1. For Q3, check solution 1003 at
      http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-208.html

      For Q2, check solution 587 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-115.html

      Delete
  3. This is amazing!!! Whats your website??

    ReplyDelete
  4. I need help in maths. Can u give me any reference??

    ReplyDelete
  5. Can I also get reference for chemistry??

    ReplyDelete
    Replies
    1. For chemistry:
      http://chemistry-ref.blogspot.com/

      but it is not updated as the physics one.


      We do not have one for maths. But if anyone is willing to contribute to build up for these subjects, you are welcomed.

      Delete
    2. I would Love to contribute to help with producing these references for pure maths. how can i help??

      Delete
    3. You are welcomed.

      For more details, go to
      http://physics-ref.blogspot.com/2016/01/participating-to-physics-reference-blog.html

      Delete

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