5054/11

Physics 5054/11October/November 2012

Cambridge O-Level · Multiple Choice · answer key with instant marking and worked solutions

40
questions
40
marks
60
minutes

Topics Forces · Energy, Work and Power · Thermal Properties of Matter · Electric Circuits · Magnetic Effect of a Current and the d.c. Motor · Physical Quantities and Measurement · +17 more

Tap an option under each question to check it — your score builds as you go.

Q11MPhysical Quantities and MeasurementFree sample

Vernier calipers read to one tenth of a millimetre.

Which reading is given to this precision?

Options

A   3.3 cm3.3\ \text{cm}
B   3.31 cm3.31\ \text{cm}
C   3.310 cm3.310\ \text{cm}
D   3.312 cm3.312\ \text{cm}

DifficultyMedium-Easy
Worked solution

Working

A vernier caliper reading to one tenth of a millimetre has a least count of 0.1 mm0.1\ \text{mm}.

In centimetres,

0.1 mm=0.01 cm0.1\ \text{mm} = 0.01\ \text{cm}

So a reading written in cm must be given to two decimal places.

  • A 3.3 cm3.3\ \text{cm} is to 0.1 cm0.1\ \text{cm} (one millimetre) — too coarse.
  • B 3.31 cm3.31\ \text{cm} is to 0.01 cm0.01\ \text{cm} — correct.
  • C 3.310 cm3.310\ \text{cm} and D 3.312 cm3.312\ \text{cm} imply a precision of 0.001 cm0.001\ \text{cm}, which is finer than the instrument allows.

Answer

B

Final answer

B

Detailed explanation

Walkthrough

The reading of a vernier caliper is only reliable to one tenth of a millimetre, which is its least count. This means the instrument can distinguish lengths that differ by 0.1 mm0.1\ \text{mm}.

Since 1 cm=10 mm1\ \text{cm} = 10\ \text{mm}, one tenth of a millimetre is

0.1 mm=0.01 cm0.1\ \text{mm} = 0.01\ \text{cm}

So if the measurement is written in centimetres, it should have two decimal places. The last digit is the hundredths place, and that is the digit the instrument can actually read.

Now look at the options:

  • 3.3 cm3.3\ \text{cm} has only one decimal place, so it suggests a precision of 0.1 cm0.1\ \text{cm}, which is one whole millimetre. That is too coarse for a vernier caliper.
  • 3.31 cm3.31\ \text{cm} has two decimal places, so it is given to 0.01 cm0.01\ \text{cm}, exactly the least count of the instrument.
  • 3.310 cm3.310\ \text{cm} and 3.312 cm3.312\ \text{cm} have three decimal places, which would suggest a precision of 0.001 cm0.001\ \text{cm}, or 0.01 mm0.01\ \text{mm}. That is finer than a vernier caliper that reads to one tenth of a millimetre can measure.

Therefore the correct answer is B.

Key Takeaways

  • The number of decimal places in a measurement tells you the precision of the instrument.
  • A vernier caliper reading to one tenth of a millimetre measures to 0.01 cm0.01\ \text{cm}.
  • Always convert the least count into the same unit as the reading before comparing options.
  • A trailing zero can change the apparent precision, so 3.310 cm3.310\ \text{cm} is not the same as 3.31 cm3.31\ \text{cm} in terms of precision.

Common Mistakes

  • Choosing A because it is the shortest reading. The reading must match the instrument's precision, not just look simple.
  • Choosing C or D because they have more decimal places. More decimal places do not mean more accurate if the instrument cannot measure that finely.
  • Confusing 0.1 mm0.1\ \text{mm} with 0.1 cm0.1\ \text{cm}. They are different: 0.1 mm=0.01 cm0.1\ \text{mm} = 0.01\ \text{cm}.

Things to Be Careful About

  • Always check the unit of the reading. In this question the options are in cm, but the least count is given in mm.
  • Remember that 1 cm=10 mm1\ \text{cm} = 10\ \text{mm}, so the decimal place shifts by one when converting.
  • A measurement such as 3.31 cm3.31\ \text{cm} means the value is known to the nearest 0.01 cm0.01\ \text{cm}, which is exactly the precision of the vernier caliper given.
Techniques used
convert the least count of the instrument into the unit used in the readingcompare the number of decimal places in each optionidentify the option whose precision matches the instrument

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