Q.1
The diameter d of a uniform wire was measured using a micrometer, reading to ±0.01 mm, and the following results were obtained:
1.02 mm 1.02 mm 1.01 mm 1.02 mm 1.02 mm
The micrometer gives a reading of -0.02 mm when the wire is removed and the jaws are closed.
At the appropriate precision, what is the value of d?
A 1.0 mm B 1.00 mm C 1.038 mm D 1.04 mm
Answer: D
Average reading = (1.02 + 1.02 + 1.01 + 1.02 + 1.2)/5 = 1.018
≡ 1.02 ± 0.01 mm (with appropriate precision)
d = 1.02 – (–0.02) = 1.02 mm
Q.2
In an experiment, a student calculated the speed of sound to be 327.66 m s-1 and estimated the accuracy of his result to be ±3%.
Which of the following shows his result with the appropriate number of significant figures?
A 300 m.s-1 B 327.7 m.s-1 C 328 m s-1 D 330 m s-1
Answer: D
±3%. Of 327.66 ≈ 10 (1sf)
So, it is only meaningful to express the result to the nearest 10 m s-1.
Q.3
Which one of the following will involve random errors in the measurement?
A not allowing for zero error on a moving-coil voltmeter
B not subtracting background count rate when determining the count rate from a radio active source
C stopping a stopwatch at the end of a race
D using the value of g as 10 N kg-1 when calculating weight from mass
Answer: C
The (human) error in stopping a stopwatch will cause a scatter of readings around the true value.
Q.4
The diameter of a wire was measured by a student and the following readings were obtined:
1.52 mm, 1.48 mm, 1.49 mm, 1.51 mm, 1.49 mm.
What is the most appropriate way for the student to report the diameter of the wire?
A (1.50 ± 0.01) mm B 1.5 mm C (1.498 ± 0.012) mm D 1.498 mm
Answer: A
Deduce that the appropriate precision of the readings is ± 0.01 mm.
Average reading = (1.52 + 1.48 + 1.49 + 1.51 + 1.49)/5 = 1.498 mm
≡ (1.50 ± 0.01) mm
Q.5
Several pairs of readings are plotted on a graph and the gradient of the best-fit line is used to determine the measurand. Which of the following statements about this method is incorrect?
A It reduces the effects of random errors
B It may be possible to identify ‘poor’ readings.
C It may be possible to identify and avoid random errors.
D It may be possible to identify and avoid systematic errors.
Answer: C
Random errors may be identified and reduced but cannot be avoided.
Q.6
The diameter of a steel ball was determined by using a metre rule to measure four similar balls in a row.
The positions on the scale were estimated to be as follows:
X (1.0 ± 0.2) cm
Y (5.0 ± 0.2) cm
Which of the following gives the of the diameter with its associated uncertainty?
A (1.0 ± 0.05) cm
B (1.0 ± 0.1) cm
C (1.0 ± 0.2) cm
D (1.0 ± 0.24) cm
Answer: B
Diameter of a steel ball
= [(5.0 ± 0.2) – (1.0 ± 0.2)]/4
= (1.0± 0.1) cm
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