Wednesday, July 12, 2017

A cathode-ray oscilloscope (c.r.o.) is connected to an alternating voltage. The following trace is produced on the screen. ...





Question 1
A cathode-ray oscilloscope (c.r.o.) is connected to an alternating voltage. The following trace is produced on the screen.


The oscilloscope time-base setting is 0.5 ms cm-1 and the Y-plate sensitivity is 2 V cm-1.
Which statement about the alternating voltage is correct?
A The amplitude is 3.5 cm.
B The frequency is 0.5 kHz.
C The period is 1 ms.
D The wavelength is 4 cm.





Reference: Past Exam Paper – June 2014 Paper 12 Q3





Solution 1:
Answer: B.

Amplitude is the maximum voltage. It can be obtained by considering the vertical position of the trace from the central line on the screen.
Y-plate sensitivity = 2V cm-1
Amplitude = 3.5 cm = 3.5 × 2 = 7 V
[A is incorrect]

Frequency = 1 / period = 1 / T
A value of time (not ‘wavelength’) can be obtained horizontally. [D is incorrect]
Time-base setting = 0.5 ms cm-1
Period T = 4 cm = 4 × 0.5 = 2 ms       [C is incorrect]
Frequency = 1/T = 1 / (2×10-3) = 0.5 kHz

Saturday, June 24, 2017

What, to two significant figures, are the period, the frequency and the amplitude of the wave represented by the graph?






Question 1
What, to two significant figures, are the period, the frequency and the amplitude of the wave represented by the graph?


period / s        frequency / Hz           amplitude / m
A          0.0027             370                              0.0067
B          0.0031             320                              0.013
C         0.0035             290                              0.0067
D         0.0042             240                              0.013





Reference: Past Exam Paper – June 2014 Paper 12 Q22





Solution:
Answer: C.
A graph of displacement against time is given. The y-axis gives displacement in ‘m’ and the x-axis gives time in ‘ms’.

Period can be directly obtained from the time axis.
3.5 waves correspond to 12.1 ms.
3.5 T = 12.1×10-3 s
Period T = 12.1×10-3 / 3.5 = 0.0035 s

Frequency = 1 / T = 1 / 0.0035 = 290 Hz

Amplitude is the maximum displacement. This can be obtained from the y-axis.
Amplitude = 6.7 mm = 0.0067 m

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

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