Machine-translated by Google. Questions are authored in English.
Top

Measurement Errors and Standards Questions and Answers

Answer: C

Correct answer: shortcomings in the instrument's construction or calibration

Instrumental errors belong to the instrument: friction in the bearings, a scale divided unevenly, a spring that has aged, a calibration that has drifted. They are systematic and reproducible, which is what makes them correctable - a calibration reveals them and a correction removes them.

Please log in or register to report this question.

Answer: D

Temperature, humidity, stray fields and mechanical vibration all shift a reading, and the correction for each has to be known before the measurement is trusted. Shielding and temperature control are the usual remedies.

Please log in or register to report this question.

Answer: A

Correct answer: a mirror behind the scale or a knife-edge pointer

Parallax arises when the observer’s eye is not directly above the pointer, so the pointer appears displaced against the scale behind it. A mirror strip lets the observer line the pointer up with its own reflection, which guarantees a perpendicular view; a knife-edge pointer serves the same purpose by minimising the gap.

Please log in or register to report this question.

Answer: B

Shunting the circuit with a finite resistance changes the very voltage being measured, and the effect grows as the circuit's own resistance approaches the meter's. A high input impedance is what makes the error negligible.

Please log in or register to report this question.

15.

A voltmeter of 10 kΩ resistance measuring across a 10 kΩ resistor in a divider of two equal 10 kΩ resistors will read

Answer: C

Correct answer: one third of the supply rather than one half

The meter’s 10 kΩ parallels the lower resistor, giving \( 10 \parallel 10 = 5 \) kΩ against the upper 10 kΩ. The reading is therefore \[ \frac{5}{5+10} = \frac{1}{3} \] of the supply, against a true value of one half - a 33 per cent error caused entirely by the act of measuring. A voltmeter must have a resistance high compared with the circuit it measures.

Please log in or register to report this question.

Answer: D

Correct answer: adds resistance in series with the circuit

An ammeter is inserted in the path of the current, so its own resistance is added to the circuit and reduces the very current it is measuring. The requirement is therefore the opposite of a voltmeter’s: an ammeter should have a resistance small compared with the circuit it is placed in.

Please log in or register to report this question.

Answer: A

It says how much the readings disagree with one another, which is a statement about precision rather than accuracy. A small standard deviation with a large systematic error is a perfectly consistent set of wrong answers.

Please log in or register to report this question.

18.

The standard deviation of the mean of \( n \) readings is smaller than that of a single reading by a factor of

Answer: B

Correct answer: \( \sqrt{n} \)

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}. \] The square root is what makes averaging a case of diminishing returns: four readings halve the uncertainty, but a hundred are needed to reduce it tenfold. Beyond a point, effort is better spent on removing systematic error, which averaging cannot touch.

Please log in or register to report this question.

19.

Probable error of a single reading is related to its standard deviation by a factor of about

Answer: C

Correct answer: 0.6745

The probable error is the deviation within which half the readings fall, and for a normal distribution that is \( 0.6745\sigma \). It is an older statistic, largely displaced by the standard deviation and by stated confidence intervals, but it appears in the classical measurement literature and is worth recognising.

Please log in or register to report this question.

20.

Ten readings of a quantity give a mean of 100.0 and a standard deviation of 0.5. The standard deviation of the mean is nearest

Answer: D

Correct answer: 0.16

\[ \sigma_{\bar{x}} = \frac{0.5}{\sqrt{10}} = \frac{0.5}{3.162} = 0.158. \] The mean is therefore far better determined than any single reading - but only against random error. If the instrument reads systematically high, all ten readings are high and this figure says nothing about it.

Please log in or register to report this question.

Loading…
Tags: Measurement Errors and Standards Questions and Answers || Measurement Errors and Standards MCQ Questions and Answers || Measurement Errors and Standards GK Questions and Answers || Measurement Errors and Standards GK MCQ Questions || Measurement Errors and Standards Multiple Choice Questions and Answers || Measurement Errors and Standards GK || GK on Measurement Errors and Standards || Electrical Measurements Questions and Answers || Electrical Measurements MCQ Questions and Answers || Electrical Measurements GK Questions and Answers || GK on Electrical Measurements