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Measurement - Reading Measurement Scales

Grade 6IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Understanding Units: Measurement scales use various units such as millimeters (mmmm), centimeters (cmcm), grams (gg), kilograms (kgkg), and milliliters (mlml). Always identify the unit on the scale before reading.

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Calculating Scale Intervals: To find the value of each small graduation (mark) on a scale, use the formula: Value per interval=Difference between two labeled marksNumber of gaps between them\text{Value per interval} = \frac{Difference\ between\ two\ labeled\ marks}{{Number\ of\ gaps\ between\ them}}.

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Precision and Accuracy: The smallest division on a scale determines its precision. For example, a ruler with mmmm marks is more precise than one with only cmcm marks.

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Reading Technique: Always look at the scale perpendicular to the mark (eye level) to avoid Parallax Error, which is a perceived shift in the position of an object when viewed from an angle.

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Zero Error: Ensure the measurement starts at the 00 mark. If a scale does not start at 00, subtract the starting value from the final reading.

📐Formulae

Value per Division=Major Mark2−Major Mark1Total Number of IntervalsValue\ per\ Division = \frac{{Major\ Mark_{2} - Major\ Mark_{1}}}{{Total\ Number\ of\ Intervals}}

10 mm=1 cm10\ mm = 1\ cm

100 cm=1 m100\ cm = 1\ m

1000 g=1 kg1000\ g = 1\ kg

1000 ml=1 l1000\ ml = 1\ l

💡Examples

Problem 1:

A measuring cylinder has major markings at 20 ml20\ ml and 30 ml30\ ml. There are 55 equal intervals between these two marks. If the liquid level is 22 marks above the 20 ml20\ ml line, what is the volume?

Solution:

Value per interval=30 ml−20 ml5=10 ml5=2 mlValue\ per\ interval = \frac{{30\ ml - 20\ ml}}{5} = \frac{{10\ ml}}{5} = 2\ ml Reading=20 ml+(2×2 ml)=20 ml+4 ml=24 mlReading = 20\ ml + (2 \times 2\ ml) = 20\ ml + 4\ ml = 24\ ml

Explanation:

First, find the value of each small mark by dividing the difference between the labeled values (10 ml10\ ml) by the number of gaps (55). Each mark represents 2 ml2\ ml. Since the level is 22 marks above 20 ml20\ ml, we add 4 ml4\ ml to 20 ml20\ ml.

Problem 2:

A weighing scale shows 0 kg0\ kg and 1 kg1\ kg with 1010 subdivisions in between. The needle points to the 7th7^{th} mark after 00. What is the mass in grams?

Solution:

Value per division=1 kg−0 kg10=0.1 kgValue\ per\ division = \frac{{1\ kg - 0\ kg}}{10} = 0.1\ kg Reading=7×0.1 kg=0.7 kgReading = 7 \times 0.1\ kg = 0.7\ kg 0.7 kg×1000=700 g0.7\ kg \times 1000 = 700\ g

Explanation:

Identify the value of one division by dividing the range (1 kg1\ kg) by the number of intervals (1010), resulting in 0.1 kg0.1\ kg per mark. The 7th7^{th} mark is 0.7 kg0.7\ kg. To convert to grams, multiply by 10001000.

Problem 3:

On a 15 cm15\ cm ruler, there are 1010 small divisions between the 4 cm4\ cm and 5 cm5\ cm marks. If an object ends at the 8th8^{th} small mark after 4 cm4\ cm, what is its length in mmmm?

Solution:

Value per division=5 cm−4 cm10=0.1 cm=1 mmValue\ per\ division = \frac{{5\ cm - 4\ cm}}{10} = 0.1\ cm = 1\ mm Length=4 cm+(8×0.1 cm)=4.8 cmLength = 4\ cm + (8 \times 0.1\ cm) = 4.8\ cm 4.8 cm×10=48 mm4.8\ cm \times 10 = 48\ mm

Explanation:

Each small division on a standard metric ruler represents 0.1 cm0.1\ cm or 1 mm1\ mm. The reading 4.8 cm4.8\ cm is converted to mmmm by multiplying by 1010.