Saturday, May 13, 2017

The diagram shows a circuit containing three resistors in parallel. The battery has e.m.f. 12 V and negligible internal resistance.




Question 2
The diagram shows a circuit containing three resistors in parallel.

The battery has e.m.f. 12 V and negligible internal resistance. The ammeter reading is 3.2 A.
What is the resistance of X?

A 2.1 Ω           B 4.6 Ω           C 6.0 Ω           D 15 Ω





Reference: Past Exam Paper – November 2008 Paper 1 Q35





Solution:
Answer: D.
For a parallel combination, the p.d. across each branch is equal and in this case, the p.d. across each branch is equal to the e.m.f. of 12 V.

To know the resistance of X, we need to know the p.d. across it (= 12V) and the current flowing through it (unknown).
The current of 3.2A coming from the terminal of the battery splits at the junction.
Current across the 10Ω resistor = V/R = 12/10 = 1.2 A
Current through X = 3.2 – 1.2 – 1.2 = 0.8 A

From Ohm’s law, Resistance of X = V/I = 12 / 0.8 = 15 Ω

Monday, May 1, 2017

A uniform string is held between a fixed point P and a variable-frequency oscillator, as shown in Fig. 5.1.




Question 1
A uniform string is held between a fixed point P and a variable-frequency oscillator, as shown in Fig. 5.1.

Fig. 5.1
The distance between point P and the oscillator is L.
The frequency of the oscillator is adjusted so that the stationary wave shown in Fig. 5.1 is formed.
Points X and Y are two points on the string.
Point X is a distance 1/8 L from the end of the string attached to the oscillator. It vibrates with frequency f and amplitude A.
Point Y is a distance 1/8 L from the end P of the string.

(a) For the vibrations of point Y, state
(i) the frequency (in terms of f ), [1]
(ii) the amplitude (in terms of A). [1]

(b) State the phase difference between the vibrations of point X and point Y. [1]

(c) (i) State, in terms of f and L, the speed of the wave on the string. [1]

(ii) The wave on the string is a stationary wave.
Explain, by reference to the formation of a stationary wave, what is meant by the
speed stated in (i). [3]





Reference: Past Exam Paper – November 2009 Paper 22 Q5





Solution 1:
(a)
(i) f                                          [B1]

(ii) A                                        [B1]
{Amplitude means maximum displacement.}


(b) 180o           or π rad            (unit necessary)                       [B1]
{When talking about the phase of a point, we need to consider its displacement and the direction of motion/vibration.
Phase difference between 2 points means the difference in phase about the 2 points.
Point X and Y are at the same displacement (one is +ve and the other -ve). However, they moving in opposite direction. So, they are out of phase (phase difference = 180°).}

[For more explanation on Phase difference, see solution 318 at http://physics-ref.blogspot.com/2015/01/physics-9702-doubts-help-page-55.html]


(c) (i) Speed of wave = f x L                                                  [B1]

(ii)
The wave is reflected at the end / at P.                       [B1]
EITHER The incident and reflected waves interfere OR the two waves travelling in opposite directions interfere.                                              [M1]
The speed is the speed of incident or reflected wave / one of these waves.  [A1]

Monday, April 10, 2017

A student attempts to show the interference of light using two identical green LEDs.




Question 1
A student attempts to show the interference of light using two identical green LEDs.
Which statement explains why the experiment will not succeed?
A The light waves from the sources are not coherent.
B The light waves from the sources do not have the same amplitude.
C The light waves from the sources have a range of wavelengths.
D The light waves from the sources are not monochromatic.





Reference: Past Exam Paper – June 2014 Paper 13 Q28





Solution 1:
Answer: A.
Interference occurs when the light waves are coherent (that is, they have constant phase difference and the same frequency). So, if they light waves from the sources are not coherent, the experiment will not succeed.

Amplitude is not a condition for interference.

Since the LEDs are identical, they are monochromatic and have the same wavelength.
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