Friday, March 6, 2015

Physics 9702 Doubts | Help Page 78

  • Physics 9702 Doubts | Help Page 78

Question 414: [Waves]
A wave motion has the equation
a = a0 sin (ωt – kx).
The graph shows how the displacement a at a fixed point varies with time t.

Which one of the labelled points shows a displacement equal to that at the position x = π/2k at time t = 0?
A P                  B Q                 C R                  D S                  E The point cannot be determined from the information given.

Reference: Past Exam Paper – N78 / II / 9

Solution 414:
Answer: B.
a = a0 sin (ωt – kx)

When x = π/2k and t = 0,
Displacement a = a0 sin (ω(0) – k(π/2k)) = a0 sin (– π/2) = –a0

Note that a0 is the maximum displacement (amplitude). So, point Q is at a displacement equal to that calculated.

Question 415: [Current of Electricity]
The diagram shows the relation between the direct current I in a certain conductor and the potential V across it. When V < 1.8V, the current is negligible.

Which statement about the conductor is correct?
A It does not obey Ohm’s law but when V > 1.8V its resistance is 4Ω.
B It does not obey Ohm’s law but when V = 3V its resistance is 10Ω.
C It obeys Ohm’s law when V > 1.8V and when V = 3V its resistance is 10Ω.
D It obeys Ohm’s law when V > 1.8V and but its resistance is not constant.

Reference: Past Exam Paper – J98 / I / 14

Solution 415:
Answer: B.
Ohm’s law: V = IR

When Ohm’s law is obeyed, the above equation holds for all values of V, I and R. When V < 1.8V, the current is negligible. So, Ohm’s law is not obeyed. [C and D are incorrect]

The resistance of the conductor is obtained by taking the value V / I at any point (not gradient or inverse gradient – resistance is NOT given by R = ΔV/ΔI).

When V = 3V, I = 300mA
Resistance R = V / I = 3 / (300x10-3) = 10Ω

Question 416: [Electric field strength]
2 parallel plates P and Q are separated by distance of 7.6 mm in vacuum. There is potential difference of 250 V between plates, as illustrated. 

Electrons are produced at X on plate P. Electrons accelerate from rest and travel to plate Q. Electric field between plates may be assumed to be uniform.
(i) Determine force on an electron due to electric field.
(ii) Show that change in kinetic energy of electron as it moves from plate P to plate Q is 4.0x10-17J.
(iii) Determine speed of electron as it reaches plate Q.

(b) Positions of plates are adjusted so that electric field between them is not uniform. Potential difference remains unchanged.
State and explain effect, if any, of this adjustment on speed of an electron as it reaches plate Q.

Reference: Past Exam Paper – November 2008 Paper 2 Q4

Solution 416:
EITHER Force = e (V / d)      OR E = V / d
Force = (1.6x10-19) (250 /{7.6x10-3}) = 5.3x10-15N

EITHER ΔEk = eV                             OR ΔEk = Fd
ΔEk = (1.6x10-19)(250)                   ΔEk = (5.3x10-15)(7.6x10-3)
ΔEk = 4.0x10-17J

ΔEk = ½ mv2
4.0x10-17 = ½ (9.1x10-31) v2
v = 9.4x106ms-1

v2 = 2as           and      a = F/m
v2 = 2(5.3x10-15)(7.6x10-3) / (9.11x10-31)
v = 9.4x106ms-1

(b) {Energy = charge x potential difference. This energy is converted to kinetic energy of the electron. Since the p.d. is unchanged, the energy also remains the same. So, the speed also remains the same.}
The speed depends on the (electric) potential difference, so speed is always the same.

Question 417: [Electromagnetism]
(a) Two similar coils A and B of insulated wire are wound on to a soft-iron core, as illustrated.

When current I in coil A is switched on and then off, variation with time t of the current is shown in Fig.

On Fig, draw a graph to show variation with time t of the e.m.f. E induced in coil B.

(b) Fig is the circuit of a bridge rectifier.

Alternating supply connected across PR has an output of 6.0V r.m.s.
(i) On Fig, circle diodes that are conducting when R is positive with respect to P.
(ii) Calculate maximum potential difference between points Q and S, assuming that diodes are ideal.
(iii) State and explain how capacitor may be used to smooth output from the rectifier. You may draw on Fig if you wish.

Reference: Past Exam Paper – June 2002 Paper 4 Q6

Solution 417:
(a) The sketch should have peaks in opposite directions in the correct regions. No e.m.f. is induced when the current is constant. There should be a correct shape for one of the pulses.

{Faraday’s aw: e.m.f. induced is proportional to the rate of change of magnetic flux (this is proportional to the current). So, when current is not changing (even if the current is not zero), no e.m.f. is induced – as in section 1 and 3. Compared to section 2, the rate of change of current in the last section is more significant – so, the e.m.f. induced is greater.
Lenz’s law: e.m.f. induced acts in such a way as to produce effects to oppose the change producing it. When the ‘change in current’ is positive (that is, current is increasing), the e.m.f is negative, and when the ‘change in current is negative’ (that is, current is decreasing), the e.m.f. induced is positive.}

(i) The 2 correct diodes should be circled.

(ii) Maximum p.d. Vmax = √2 x Vrms = 8.48V

(iii) A capacitor could be connected across SQ {in parallel with load}. The capacitor discharges through the load when the potential difference (p.d.) / current in the load reduces. Thus, it maintains the p.d. across the load. (or other relevant comment)

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