Thursday, November 20, 2014

Physics 9702 Doubts | Help Page 18

  • Physics 9702 Doubts | Help Page 18

Question 95: [Force > Moment]
Diagram shows plan view of door which requires a moment of 12Nm to open it.

What is the minimum force that must be applied at door’s midpoint to ensure it opens?
A 4.8 N                       B 9.6 N                       C 15 N                        D 30 N

Reference: Past Exam Paper – November 2007 Paper 1 Q14



Solution 95:
Answer: D.
Moment = Force x perpendicular distance of force from pivot
For minimum force, the force should be applied perpendicularly to the door.
Midpoint = 0.80 /2 = 0.40m

Moment = F x d = 12Nm
F x 0.80 = 12
Minimum force, F = 12 / 0.40 = 30N









Question 96: [Current of Electricity > Potential difference and Resistance]
Power supply and solar cell are compared using potentiometer circuit shown.

e.m.f. produced by solar cell is measured on potentiometer.
Potentiometer wire PQ is 100.0 cm long and has resistance of 5.00 Ω. Power supply has e.m.f. of 2.000 V and solar cell has e.m.f. of 5.00 mV.
Which resistance R must be used so that galvanometer reads zero when PS = 40.0 cm?
A 395 Ω                      B 795 Ω                      C 995 Ω                      D 1055 Ω

Reference: Past Exam Paper – June 2014 Paper 12 Q35



Solution 96:
Answer: B.
When PS = 40.0cm, the potential difference across the length of wire PS is given by 2.000 V × (0.400 × 5.00 Ω / [R + 5.00 Ω]). (Potential divider equation)
{Since the total resistance of the 1m wire = 5.00Ω, the resistance of PS = 0.4 x 5.00Ω. [R + 5.00Ω] is the total resistance in the circuit.}
(A galvanometer is a type of sensitive ammeter: an instrument for detecting electric current.)
For the galvanometer to read zero, the e.m.f of the solar cell must be equal to that potential difference (given above) such that the potential difference across the galvanometer is zero.

So, by equating this p.d. to the e.m.f. of solar cell (= 5.00mV), resistance R can be obtained by rearranging the equation. 

2.000 V × (0.400 × 5.00 Ω / [R + 5.00 Ω] = 5 × 10-3 V
2.000 × (0.400 × 5.00) = (5 × 10-3) [R + 5.00]
4 = (5 × 10-3) [R + 5.00]
R + 5.00 = 4 / (5 × 10-3) = 800
R = 800 – 5 = 795 Ω
 

Resistance R is found to be 795 Ω.








Question 97: [Waves > Stationary waves]
Transmitter of electromagnetic waves is placed 45 cm from reflective surface.

Emitted waves have frequency of 1.00 GHz. Stationary wave is produced with node at the transmitter and node at the surface.
How many antinodes are in the space between transmitter and surface?
A 1                  B 2                  C 3                  D 4

Reference: Past Exam Paper – June 2013 Paper 12 Q27



Solution 97:
Answer: C.
Speed of wave, v = fλ

Electromagnetic waves travel at the speed of 3.0x108ms-1.
Frequency, f = 1.00GHz = 1x109Hz
Wavelength, λ = v / f = (3x108) / (1x109) = 0.3m

Distance between transmitter and reflective surface = 45cm = 0.45m
No. of periods present between transmitter and reflective surface = 0.45/0.3 = 1.5

An antinode is a point with maximum displacement.
1 period contains 2 antinodes + 1 antinode is present in the 0.5 (half) of a period.
Total number of antinodes = 2 + 1 = 3










Question 98: [Dynamics]
Supermarket trolley, total mass 30 kg, is moving at 3.0 m s–1. Retarding force of 60 N is applied to trolley for 0.50 s in opposite direction to trolley’s initial velocity.
What is trolley’s new velocity after application of the force?
A 1.0 m s–1                  B 1.5 m s–1                  C 2.0 m s–1                  D 2.8 m s–1

Reference: Past Exam Paper – November 2009 Paper 11 Q10 & Paper 12 Q9



Solution 98:
Answer: C.
Force F on trolley= Δp / Δt
where Δp is the change in momentum of the trolley in during a time Δt

Assuming v and u are in same direction,
Δp = m(v – u) = 30(v – 3)
Time of application of force, Δt = 0.5s
Force is in opposite direction and so, is taken as negative.

Δp / Δt = F
30(v – 3) / 0.5 = - 60.
New velocity, v = (0.5 x -60 / 30) + 3 = 2ms-1








Question 99: [Current of Electricity > Potentiometer]
In potentiometer circuit below, moveable contact is placed at N on the bare wire XY, such that galvanometer shows zero deflection.

Resistance of variable resistor is now increased.
What is the effect of this increase on potential difference across wire XY and on position of moveable contact for zero deflection?

Reference: Past Exam Paper – November 2002 Paper 1 Q35



Solution 99:
Answer: D.
The resistance of the wire being contributed to the total resistance in the circuit is from X to point N.

Let the resistance of the wire from point X to Y (with only the part within X and N contributing resistance to the circuit) = W and let the resistance of the variable resistor = R.

From potential divider equation,
Potential difference across length XY of the wire = V [W / (W+R)]
As the resistance of the variable resistor, R is increased, the p.d. across the wire decreases.  [A and B are incorrect]


For the galvanometer to show zero deflection, the potential difference across it should be zero (no current flows through it). That is, the potential at one side of the galvanometer should be equal to the potential at the other side (thus the potential difference is zero).

Initially (before the resistance of the variable resistor is increased), the galvanometer showed zero deflection. So, the potential difference across it was zero. In that case, the p.d. across XN was equal to the e.m.f. of the cell to which it is connected.

But an increase in the resistance of the variable resistor causes the p.d. across XY to decrease. Therefore, for the galvanometer to now show zero deflection, the resistance of wire should be increased (this causes the p.d. across it to increase – from the potential divider equation)

Resistance of the wire = ρL/A. So, the length of the wire, L (between X and N) should be increased. That is, the position of the moveable contact should be nearer to Y.





11 comments:

  1. in question 99, how is the galvanometer measuring the potential difference between XN and variable resistor?isn't the p.d, that it measures, equivalent to the p.d. JUST ACROSS XN?

    ReplyDelete
  2. First, do not forget that potential and potential difference are different things.

    XN is not a point, but a part of the wire. The galvanometer measures p.d. across 2 points.

    One end of the galvanometer is connected to the negative terminal of the battery and so, is at a potential of zero. The other end is connected at point N. So, the galvanometer shows the p.d. between these 2 points.

    The length XN of the wire is the part if the wire that is contributing some resistance to the circuit. Part XY of the wire is not contributing any resistance. So, point N and Y may be considered to be at the same potential.

    Now, the potential at N depends on both the resistance of the variable resistor and the length of wire being considered. Remember that N is moveable and any new position is still called N.

    ReplyDelete
    Replies
    1. what do you mean when you say "The length XN of the wire is the part if the wire that is contributing some resistance to the circuit. Part XY of the wire is not contributing any resistance. So, point N and Y may be considered to be at the same potential.Now, the potential at N depends on both the resistance of the variable resistor and the length of wire being considered. Remember that N is moveable and any new position is still called N."
      The galvanometer measures the p.d(in this case)between the negative terminal of the cell and N. So for the p.d. to be zero, the potential at N should also be 0, the same as that at the -ve terminal of cell, but this is not possible. p.d. at N can never be zero.
      Along with this, please do explain q37 and q29 in s13/paper 11.

      Delete
    2. First of all, there was an error in the explanation I provided at solution 99. For zero deflection, the p.d. across XN is equal to the E.M.F. of the cell to which the galvanometer is connected, NOT equal to the p.d. across the variable resistor. I already corrected it. The end connected to the terminal of the cell is actually connected to the cell itself, so the potential there would be equal to the e.m.f of the cell.

      By the way, there's another question on potentiometer above, you may read it if you want.

      As for the sentence you quoted, well this is the basis of potentiometer - that's how a potentiometer works.

      Delete
    3. both paper 11 questions are available at
      http://physics-ref.blogspot.com/2015/02/9702-june-2013-paper-11-worked.html

      Delete
  3. for question 96, I dont know why Im having trouble equating, can you show me how

    ReplyDelete
  4. in question 95 is this equation of any value F x 0.80 = 12?

    ReplyDelete
    Replies
    1. No, this is for the minimum value of F. If the force is not perpendicular or the distance has been changed, but it would have a different value.

      Delete
  5. Sir, how to do 9702/12/M/J/12 no 14 ? Thanks.

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2014/06/9702-june-2012-paper-12-worked.html

      Delete

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