Measurements, Error & uncertainties Questions and Answers

 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

(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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